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https://courses.lumenlearning.com/beginalgebra/chapter/7-3-1-solving-radical-equations/
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Module 8: Roots and Rational Exponents
Solving Radical Equations
Learning Outcomes
Solve Radical Equations
Isolate square roots in equations and solve for a variable
Identify extraneous solutions to radical equations
Square Roots and Completing the Square for Solving Radical Equations
Use square roots to solve quadratic equations
Complete the square to solve a quadratic equation
Using the Quadratic Formula to Solve Quadratic Equations
Write a quadratic equation in standard form and identify the values of a, b, and c in a standard form quadratic equation.
Use the Quadratic Formula to find all real solutions.
Solve application problems requiring the use of the Quadratic Formula.
A basic strategy for solving radical equations is to isolate the radical term first, and then raise both sides of the equation to a power to remove the radical. (The reason for using powers will become clear in a moment.) This is the same type of strategy you used to solve other, non-radical equations—rearrange the expression to isolate the variable you want to know, and then solve the resulting equation.
Solutions to Radical Equations
The solutions of [latex]x^2=a[/latex] are called the square roots of a.
When a is positive, a > 0, [latex]x^2=a[/latex] has two solutions, [latex]+\sqrt{a},-\sqrt{a}[/latex]. [latex]+\sqrt{a}[/latex] is the nonnegative square root of a, and [latex]-\sqrt{a}[/latex] is the negative square root of a.
When a is negative, a < 0, [latex]x^2=a[/latex] has no solutions.
When a is zero, a = 0, [latex]x^2=a[/latex] has one solution: a = 0
Just to drive home the importance of the concept that when a is negative, a < 0, [latex]x^2=a[/latex] has no solutions, we will restate it in words. If you have a negative number under a square root sign as in this example,
There will be no real number solutions.
There are two key ideas that you will be using to solve radical equations. The first is that if [latex]a=b[/latex], then [latex]{{a}^{2}}={{b}^{2}}[/latex]. (This property allows you to square both sides of an equation and remain certain that the two sides are still equal.) The second is that if the square root of any nonnegative number x is squared, then you get x: [latex]{{\left( \sqrt{x} \right)}^{2}}=x[/latex]. (This property allows you to “remove” the radicals from your equations.)
Let’s start with a radical equation that you can solve in a few steps:[latex]\sqrt{x}-3=5[/latex].
Example
Solve. [latex]\sqrt{x}-3=5[/latex]
Show Solution
Add 3 to both sides to isolate the variable term on the left side of the equation.
Collect like terms.
Square both sides to remove the radical, since [latex]{{(\sqrt{x})}^{2}}=x[/latex]. Make sure to square the 8 also! Then simplify.
Answer
[latex]x=64[/latex] is the solution to [latex]\sqrt{x}-3=5[/latex]
To check your solution, you can substitute 64 in for x in the original equation. Does [latex]\sqrt{64}-3=5[/latex]? Yes—the square root of 64 is 8, and [latex]8−3=5[/latex].
Notice how you combined like terms and then squared both sides of the equation in this problem. This is a standard method for removing a radical from an equation. It is important to isolate a radical on one side of the equation and simplify as much as possible before squaring. The fewer terms there are before squaring, the fewer additional terms will be generated by the process of squaring.
In the example above, only the variable x was underneath the radical. Sometimes you will need to solve an equation that contains multiple terms underneath a radical. Follow the same steps to solve these, but pay attention to a critical point—square both sides of an equation, not individual terms. Watch how the next two problems are solved.
Example
Solve. [latex]\sqrt{x+8}=3[/latex]
Show Solution
Notice how the radical contains a binomial: [latex]x+8[/latex]. Square both sides to remove the radical.
[latex]{{\left( \sqrt{x+8} \right)}^{2}}={{\left( 3 \right)}^{2}}[/latex]
[latex]{{\left( \sqrt{x+8} \right)}^{2}}=x+8[/latex]. Now simplify the equation and solve for x.
Check your answer. Substituting 1 for x in the original equation yields a true statement, so the solution is correct.
Answer
[latex]x=1[/latex] is the solution to [latex]\sqrt{x+8}=3[/latex].
In the following video we show how to solve simple radical equations.
Example
Solve. [latex]1+\sqrt{2x+3}=6[/latex]
Show Solution
Begin by subtracting 1 from both sides in order to isolate the radical term. Then square both sides to remove the binomial from the radical.
[latex]\begin{array}{r}1+\sqrt{2x+3}-1=6-1\\sqrt{2x+3}=5\,\,\,\,\,\,\,\,\,\,\{{\left( \sqrt{2x+3} \right)}^{2}}={{\left( 5 \right)}^{2}}\,\,\,\end{array}[/latex]
Simplify the equation and solve for x.
Check your answer. Substituting 11 for x in the original equation yields a true statement, so the solution is correct.
Answer
[latex]x=11[/latex] is the solution for [latex]1+\sqrt{2x+3}=6[/latex].
Solving Radical Equations
Follow the following four steps to solve radical equations.
Isolate the radical expression.
Square both sides of the equation: If [latex]x=y[/latex] then [latex]x^{2}=y^{2}[/latex].
Once the radical is removed, solve for the unknown.
Check all answers.
Identify Extraneous Solutions
Following rules is important, but so is paying attention to the math in front of you—especially when solving radical equations. Take a look at this next problem that demonstrates a potential pitfall of squaring both sides to remove the radical.
Example
Solve. [latex]\sqrt{a-5}=-2[/latex]
Show Solution
Square both sides to remove the term [latex]a–5[/latex] from the radical.
[latex]{{\left( \sqrt{a-5} \right)}^{2}}={{(-2)}^{2}}[/latex]
Write the simplified equation, and solve for a.
[latex]\begin{array}{r}a-5=4\a=9\end{array}[/latex]
Now check the solution by substituting [latex]a=9[/latex] into the original equation.
It does not check!
[latex]\begin{array}{r}\sqrt{9-5}=-2\\sqrt{4}=-2\2\ne -2\end{array}[/latex]
Answer
No solution.
Look at that—the answer [latex]a=9[/latex] does not produce a true statement when substituted back into the original equation. What happened?
Check the original problem: [latex]\sqrt{a-5}=-2[/latex]. Notice that the radical is set equal to [latex]−2[/latex], and recall that the principal square root of a number can only be positive. This means that no value for a will result in a radical expression whose positive square root is [latex]−2[/latex]! You might have noticed that right away and concluded that there were no solutions for a.
Incorrect values of the variable, such as those that are introduced as a result of the squaring process are called extraneous solutions. Extraneous solutions may look like the real solution, but you can identify them because they will not create a true statement when substituted back into the original equation. This is one of the reasons why checking your work is so important—if you do not check your answers by substituting them back into the original equation, you may be introducing extraneous solutions into the problem. In the next video example, we solve more radical equations that may have extraneous solutions.
Have a look at the following problem. Notice how the original problem is [latex]x+4=\sqrt{x+10}[/latex], but after both sides are squared, it becomes [latex]{{x}^{2}}+8x+16=x+10[/latex]. Squaring both sides may have introduced an extraneous solution.
Example
Solve. [latex]x+4=\sqrt{x+10}[/latex]
Show Solution
Square both sides to remove the term [latex]x+10[/latex] from the radical.
[latex]{{\left( x+4 \right)}^{2}}={{\left( \sqrt{x+10} \right)}^{2}}[/latex]
Now simplify and solve the equation. Combine like terms, and then factor.
[latex]\begin{array}{r}\left( x+4 \right)\left( x+4 \right)=x+10\{{x}^{2}}+8x+16=x+10\{{x}^{2}}+8x-x+16-10=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\{{x}^{2}}+7x+6=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\left( x+6 \right)\left( x+1 \right)=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]
Set each factor equal to zero and solve for x.
[latex]\begin{array}{c}\left( x+6 \right)=0\,\,\text{or}\,\,\left( x+1 \right)=0\x=-6\text{ or }x=-1\end{array}[/latex]
Now check both solutions by substituting them into the original equation.
Since [latex]x=−6[/latex] produces a false statement, it is an extraneous solution.
[latex]\begin{array}{l}-6+4=\sqrt{-6+10}\\,\,\,\,\,\,\,\,-2=\sqrt{4}\\,\,\,\,\,\,\,\,-2=2\\text{FALSE!}\\\-1+4=\sqrt{-1+10}\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3=\sqrt{9}\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,3=3\\text{TRUE!}\end{array}[/latex]
Answer
[latex]x=−1[/latex] is the only solution
Example
Solve. [latex]4+\sqrt{x+2}=x[/latex]
Show Solution
Isolate the radical term.
[latex]\sqrt{x+2}=x-4[/latex]
Square both sides to remove the term [latex]x+2[/latex] from the radical.
[latex]{{\left( \sqrt{x+2} \right)}^{2}}={{\left( x-4 \right)}^{2}}[/latex]
Now simplify and solve the equation. Combine like terms, and then factor.
[latex]\begin{array}{l}x+2={{x}^{2}}-8x+16\\,\,\,\,\,\,\,\,\,\,\,0={{x}^{2}}-8x-x+16-2\\,\,\,\,\,\,\,\,\,\,\,0={{x}^{2}}-9x+14\\,\,\,\,\,\,\,\,\,\,\,0=\left( x-7 \right)\left( x-2 \right)\end{array}[/latex]
Set each factor equal to zero and solve for x.
[latex]\begin{array}{c}\left( x-7 \right)=0\text{ or }\left( x-2 \right)=0\x=7\text{ or }x=2\end{array}[/latex][latex][/latex]
Now check both solutions by substituting them into the original equation.
Since [latex]x=2[/latex] produces a false statement, it is an extraneous solution.
[latex]\begin{array}{r}4+\sqrt{7+2}=7\4+\sqrt{9}=7\4+3=7\7=7\\text{TRUE!}\\4+\sqrt{2+2}=2\4+\sqrt{4}=2\4+2=2\6=2\\text{FALSE!}\end{array}[/latex]
Answer
[latex]x=7[/latex] is the only solution.
In the last video example we solve a radical equation with a binomial term on one side.
Square Roots and Completing the Square
Quadratic equations can be solved in many ways. In the previous section we introduced the idea that solutions to radical equations in general can be found using these facts:
Solutions to Quadratic Equations
The solutions of [latex]x^2=a[/latex] are called the square roots of a.
When a is positive, a > 0, [latex]x^2=a[/latex] has two solutions, [latex]+\sqrt{a},-\sqrt{a}[/latex]. [latex]+\sqrt{a}[/latex] is the nonnegative square root of a, and [latex]-\sqrt{a}[/latex] is the negative square root of a.
When a is negative, a < 0, [latex]x^2=a[/latex] has no solutions.
When a is zero, a = 0, [latex]x^2=a[/latex] has one solution: a = 0
A shortcut way to write [latex]\sqrt{a}[/latex] or [latex]-\sqrt{a}[/latex] is [latex]\pm \sqrt{a}[/latex]. The symbol [latex]\pm[/latex] is often read “positive or negative.” If it is used as an operation (addition or subtraction), it is read “plus or minus.”
Example
Solve using the Square Root Property. [latex]x^{2}=9[/latex]
Show Solution
Since one side is simply [latex]x^{2}[/latex], you can take the square root of both sides to get x on one side. Don’t forget to use both positive and negative square roots!
[latex]\begin{array}{l}x^{2}=9\\,\,\,x=\pm\sqrt{9}\\,\,\,x=\pm3\end{array}[/latex]
Answer
[latex]x=\pm3[/latex] (that is, [latex]x=3[/latex] or [latex]-3[/latex])
Notice that there is a difference here in solving [latex]x^{2}=9[/latex] and finding [latex]\sqrt{9}[/latex]. For [latex]x^{2}=9[/latex], you are looking for all numbers whose square is 9. For [latex]\sqrt{9}[/latex], you only want the principal (nonnegative) square root. The negative of the principal square root is [latex]-\sqrt{9}[/latex]; both would be [latex]\pm \sqrt{9}[/latex]. Unless there is a symbol in front of the radical sign, only the nonnegative value is wanted!
In the example above, you can take the square root of both sides easily because there is only one term on each side. In some equations, you may need to do some work to get the equation in this form. You will find that this involves isolating [latex]x^{2}[/latex].
Example
Solve. [latex]10x^{2}+5=85[/latex]
Show Solution
If you try taking the square root of both sides of the original equation, you will have [latex]\sqrt{10{{x}^{2}}+5}[/latex] on the left, and you can’t simplify that. Subtract 5 from both sides to get the [latex]x^{2}[/latex] term by itself.
[latex]10x^{2}+5=85[/latex]
You could now take the square root of both sides, but you would have [latex]\sqrt{10}[/latex] as a coefficient, and you would need to divide by that coefficient. Dividing by 10 before you take the square root will be a little easier.
[latex]10x^{2}=80[/latex]
Now you have only [latex]x^{2}[/latex] on the left, so you can use the Square Root Property easily.
Be sure to simplify the radical if possible.
[latex]\begin{array}{l}{{x}^{2}}=8\\,\,\,x=\pm \sqrt{8}\\,\,\,\,\,\,=\pm \sqrt{(4)(2)}\\,\,\,\,\,\,=\pm \sqrt{4}\sqrt{2}\\,\,\,\,\,\,=\pm 2\sqrt{2}\end{array}[/latex]
Answer
[latex]x=\pm 2\sqrt{2}[/latex]
In the following video we show more examples of solving simple quadratic equations using square roots.
Sometimes more than just the x is being squared:
Example
Solve. [latex]\left(x–2\right)^{2}–50=0[/latex]
Show Solution
Again, taking the square root of both sides at this stage will leave something you can’t work with on the left. Start by adding 50 to both sides.
[latex]\left(x-2\right)^{2}-50=0[/latex]
Because [latex]\left(x–2\right)^{2}[/latex] is a squared quantity, you can take the square root of both sides.
[latex]\begin{array}{r}\left(x-2\right)^{2}=50\,\,\,\,\,\,\,\,\,\,\x-2=\pm\sqrt{50}\end{array}[/latex]
To isolate x on the left, you need to add 2 to both sides.
Be sure to simplify the radical if possible.
[latex]\begin{array}{l}x=2\pm \sqrt{50}\\,\,\,\,=2\pm \sqrt{(25)(2)}\\,\,\,\,=2\pm \sqrt{25}\sqrt{2}\\,\,\,\,=2\pm 5\sqrt{2}\end{array}[/latex]
Answer
[latex]x=2\pm 5\sqrt{2}[/latex]
In this video example, you will see more examples of solving quadratic equations using square roots.
Completing the Square to Solve a Quadratic Equation
Of course, quadratic equations often will not come in the format of the examples above. Most of them will have x terms. However, you may be able to factor the expression into a squared binomial—and if not, you can still use squared binomials to help you.
Some of the above examples have squared binomials: [latex]\left(1+r\right)^{2}[/latex] and [latex]\left(x–2\right)^{2}[/latex] are squared binomials. If you expand these, you get a perfect square trinomial.
Perfect square trinomials have the form [latex]x^{2}+2xs+s^{2}[/latex] and can be factored as [latex]\left(x+s\right)^{2}[/latex], or they have the form [latex]x^{2}–2xs+s^{2}[/latex] and can be factored as [latex]\left(x–s\right)^{2}[/latex]. Let’s factor a perfect square trinomial into a squared binomial.
Example
Factor [latex]9x^{2}–24x+16[/latex].
Show Solution
First notice that the [latex]x^{2}[/latex] term and the constant term are both perfect squares.
[latex]\begin{array}{l}9x^{2}=\left(3x\right)^{2}\\,\,\,16=4^{2}\end{array}[/latex]
Then notice that the middle term (ignoring the sign) is twice the product of the square roots of the other terms.
[latex]24x=2\left(3x\right)\left(4\right)[/latex]
A trinomial in the form [latex]r^{2}-2rs+s^{2}[/latex] can be factored as latex^{2}[/latex].
In this case, the middle term is subtracted, so subtract r and s and square it to get latex^{2}[/latex].
[latex]\begin{array}{c}\,\,\,r=3x\s=4\9x^{2}-24x+16=\left(3x-4\right)^{2}\end{array}[/latex]
Answer
[latex]\left(3x–4\right)^{2}[/latex]
You can use the procedure in this next example to help you solve equations where you identify perfect square trinomials, even if the equation is not set equal to 0.
Example
Solve. [latex]4x^{2}+20x+25=8[/latex]
Show Solution
Since there’s an x term, you can’t use the Square Root Property immediately (or even after adding or dividing by a constant).
Notice, however, that the [latex]x^{2}[/latex] and constant terms on the left are both perfect squares: [latex]\left(2x\right)^{2}[/latex] and [latex]5^{2}[/latex]. Check the middle term: is it [latex]2\left(2x\right)\left(5\right)[/latex]? Yes!
[latex]4x^{2}+20x+25=8[/latex]
A trinomial in the form [latex]r^{2}+2rs+s^{2}[/latex] can be factored as [latex]\left(r+s\right)^{2}[/latex], so rewrite the left side as a squared binomial.
latex^{2}=8[/latex]
Now you can use the Square Root Property. Some additional steps are needed to isolate x.
[latex]\begin{array}{r}2x+5=\pm \sqrt{8}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\2x=-5\pm \sqrt{8}\,\,\,\,\,\\x=-\frac{5}{2}\pm \frac{1}{2}\sqrt{8}\end{array}[/latex]
Simplify the radical when possible.
[latex]\begin{array}{l}x=-\frac{5}{2}\pm \frac{1}{2}\sqrt{4}\sqrt{2}\\x=-\frac{5}{2}\pm \frac{1}{2}(2)\sqrt{2}\\x=-\frac{5}{2}\pm \sqrt{2}\end{array}[/latex]
Answer
[latex]x=-\frac{5}{2}\pm \sqrt{2}[/latex]
One way to solve quadratic equations is by completing the square. When you don’t have a perfect square trinomial, you can create one by adding a constant term that is a perfect square to both sides of the equation. Let’s see how to find that constant term.
“Completing the square” does exactly what it says—it takes something that is not a square and makes it one. This idea can be illustrated using an area model of the binomial [latex]x^{2}+bx[/latex].
In this example, the area of the overall rectangle is given by [latex]x\left(x+b\right)[/latex].
Now let’s make this rectangle into a square. First, divide the red rectangle with area bx into two equal rectangles each with area [latex]\frac{b}{2}x[/latex]. Then rotate and reposition one of them. You haven’t changed the size of the red area—it still adds up to [latex]bx[/latex].
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The red rectangles now make up two sides of a square, shown in white. The area of that square is the length of the red rectangles squared, or [latex]{{\left( \frac{b}{2} \right)}^{2}}[/latex].
Here comes the cool part—do you see that when the white square is added to the blue and red regions, the whole shape is also now a square? In other words, you’ve “completed the square!” By adding the quantity [latex]\left(\frac{b}{2}\right)^{2}[/latex] to the original binomial, you’ve made a square, a square with sides of [latex]x+\frac{b}{2}[/latex].
Notice that the area of this square can be written as a squared binomial: [latex]\left(x+\frac{b}{2}\right)^{2}[/latex].
Finding a Value that will Complete the Square in an Expression
To complete the square for an expression of the form [latex]x^{2}+bx[/latex]:
Identify the value of b;
Calculate and add [latex]\left(\frac{b}{2}\right)^{2}[/latex].
The expression becomes [latex]x^{2}+bx+\left(\frac{b}{2}\right)^{2}=\left(x+\frac{b}{2}\right)^{2}[/latex].
Example
Find the number to add to [latex]x^{2}+8x[/latex] to make it a perfect square trinomial.
Show Solution
First identify b if this has the form [latex]x^{2}+bx[/latex].
[latex]\begin{array}{c}x^{2}+8x\b=8\end{array}[/latex]
To complete the square, add [latex]\left(\frac{b}{2}\right)^{2}[/latex].
[latex]b=8[/latex], so [latex]\left(\frac{b}{2}\right)^{2}=\left(\frac{8}{2}\right)^{2}[/latex]
Simplify.
[latex]\begin{array}{c}x^{2}+8x+\left(4\right)^{2}\x^{2}+8x+16\end{array}[/latex]
Check that the result is a perfect square trinomial. [latex]\left(x+4\right)^{2}=x^{2}+4x+4x+16=x^{2}+8x+16[/latex], so it is.
Answer
Adding [latex]+16[/latex] will make [latex]x^{2}+8x[/latex] a perfect square trinomial.
Notice that [latex]{{\left( \frac{b}{2} \right)}^{2}}[/latex] is always positive, since it is the square of a number. When you complete the square, you are always adding a positive value.
In the following video, we show more examples of how to find a constant terms that will make a trinomial a perfect square.
You can use completing the square to help you solve a quadratic equation that cannot be solved by factoring.
Let’s start by seeing what happens when you complete the square in an equation. In the example below, notice that completing the square will result in adding a number to both sides of the equation—you have to do this in order to keep both sides equal!
Example
Rewrite [latex]x^{2}+6x=8[/latex] so that the left side is a perfect square trinomial.
Show Solution
This equation has a constant of 8. Ignore it for now and focus on the [latex]x^{2}[/latex] and x terms on the left side of the equation. The left side has the form [latex]x^{2}+bx[/latex], so you can identify b.
[latex]\begin{array}{r}x^{2}+6x=8\b=6\end{array}[/latex]
To complete the square, add [latex]{{\left( \frac{b}{2} \right)}^{2}}[/latex] to the left side.
[latex]b=6[/latex], so [latex]{{\left( \frac{b}{2} \right)}^{2}}={{\left( \frac{6}{2} \right)}^{2}}={{3}^{2}}=9.[/latex]
This is an equation, though, so you must add the same number to the right side as well.
[latex]x^{2}+6x+9=8+9[/latex]
Simplify. Check that the left side is a perfect square trinomial. [latex]\begin{array}{r}\left(x+3\right)^{2}=x^{2}+3x+3x+9=x^{2}+6x+9\end{array}{r}[/latex], so it is.
[latex]\begin{array}{r}x^{2}+6x+9=17\x^{2}+6x+9=17\(x+3)^{2}=17\end{array}[/latex]
Answer
[latex]x^{2}+6x+9=17[/latex]
Can you see that completing the square in an equation is very similar to completing the square in an expression? The main difference is that you have to add the new number ([latex]+9[/latex] in this case) to both sides of the equation to maintain equality.
Now let’s look at an example where you are using completing the square to actually solve an equation, finding a value for the variable.
Example
Solve. [latex]x^{2}–12x–4=0[/latex]
Show Solution
Since you cannot factor the trinomial on the left side, you will use completing the square to solve the equation.
Rewrite the equation with the left side in the form [latex]x^{2}+bx[/latex], to prepare to complete the square. Identify b.
[latex]\begin{array}{r}x^{2}-12x=4\,\,\,\,\,\,\,\,\b=-12\end{array}[/latex]
Figure out what value to add to complete the square. Add [latex]{{\left( \frac{b}{2}\right)}^{2}}[/latex] to complete the square, so [latex]{{\left( \frac{b}{2} \right)}^{2}}={{\left( \frac{-12}{2} \right)}^{2}}={{\left( -6 \right)}^{2}}=36[/latex].
Add the value to both sides of the equation and simplify.
[latex]\begin{array}{l}x^{2}-12x+36=4+36\x^{2}-12x+36=40\end{array}[/latex]
Rewrite the left side as a squared binomial.
[latex]\left(x-6\right)^{2}=40[/latex]
Use the Square Root Property. Remember to include both the positive and negative square root, or you’ll miss one of the solutions.
[latex]x-6=\pm\sqrt{40}[/latex]
Solve for x by adding 6 to both sides. Simplify as needed.
[latex]\begin{array}{l}x=6\pm \sqrt{40}\\,\,\,\,=6\pm \sqrt{4}\sqrt{10}\\,\,\,\,=6\pm 2\sqrt{10}\end{array}[/latex]
Answer
[latex]x=6\pm 2\sqrt{10}[/latex]
In this last video, we solve more quadratic equations by completing the square.
You may have noticed that because you have to use both square roots, all the examples have two solutions. Here is another example that’s slightly different.
Example
Solve [latex]x^{2}+16x+17=-47[/latex].
Show Solution
Rewrite the equation so the left side has the form [latex]x^{2}+bx[/latex]. Identify b.
[latex]\begin{array}{c}x^{2}+16x=-64\b=16\end{array}[/latex]
Add [latex]{{\left( \frac{b}{2} \right)}^{2}}[/latex], which is [latex]{{\left( \frac{16}{2} \right)}^{2}}={{8}^{2}}=64[/latex], to both sides.
[latex]\begin{array}{l}x^{2}+16x+64=-64+64\x^{2}+16x+64=0\end{array}[/latex]
Write the left side as a squared binomial.
[latex]\left(x+8\right)^{2}=0[/latex]
Take the square roots of both sides. Normally both positive and negative square roots are needed, but 0 is neither positive nor negative. 0 has only one root.
[latex]x+8=0[/latex]
Solve for x.
[latex]x=-8[/latex]
Answer
[latex]x=-8[/latex]
Take a closer look at this problem and you may see something familiar. Instead of completing the square, try adding 47 to both sides in the equation. The equation [latex]x^{2}+16x+17=−47[/latex] becomes [latex]x^{2}+16x+64=0[/latex]. Can you factor this equation using grouping? (Think of two numbers whose product is 64 and whose sum is 16).
It can be factored as latex(x+8)=0[/latex], of course! Knowing how to complete the square is very helpful, but it is not always the only way to solve an equation.
Use the Quadratic Formula to Solve Quadratic Equations
Quadratic formula
You can solve any quadratic equation by completing the square—rewriting part of the equation as a perfect square trinomial. If you complete the square on the generic equation [latex]ax^{2}+bx+c=0[/latex] and then solve for x, you find that [latex]x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}[/latex]. This equation is known as the Quadratic Formula.
This formula is very helpful for solving quadratic equations that are difficult or impossible to factor, and using it can be faster than completing the square. The Quadratic Formula can be used to solve any quadratic equation of the form [latex]ax^{2}+bx+c=0[/latex].
The form [latex]ax^{2}+bx+c=0[/latex] is called standard form of a quadratic equation. Before solving a quadratic equation using the Quadratic Formula, it’s vital that you be sure the equation is in this form. If you don’t, you might use the wrong values for a, b, or c, and then the formula will give incorrect solutions.
Example
Rewrite the equation [latex]3x+2x^{2}+4=5[/latex] in standard form and identify a, b, and c.
Show Solution
First be sure that the right side of the equation is 0. In this case, all you need to do is subtract 5 from both sides.
[latex]\begin{array}{c}3x+2x^{2}+4=5\3x+2x^{2}+4–5=5–5\end{array}[/latex]
Simplify, and write the terms with the exponent on the variable in descending order.
[latex]\begin{array}{r}3x+2x^{2}-1=0\2x^{2}+3x-1=0\end{array}[/latex]
Now that the equation is in standard form, you can read the values of a, b, and c from the coefficients and constant. Note that since the constant 1 is subtracted, c must be negative.
[latex]\begin{array}{l}2x^{2}\,\,\,+\,\,\,3x\,\,\,-\,\,\,1\,\,\,=\,\,\,0\\,\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\\,ax^{2}\,\,\,\,\,\,\,\,\,\,bx\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c\\\,\,a=2,\,\,b=3,\,\,c=-1\end{array}[/latex]
Answer
[latex]2x^{2}+3x–1=0;a=2,b=3,c=−1[/latex]
Example
Rewrite the equation [latex]2(x+3)^{2}–5x=6[/latex] in standard form and identify a, b, and c.
Show Solution
First be sure that the right side of the equation is 0.
[latex]\begin{array}{c}2\left(x+3\right)^{2}–5x=6\2(x+3)^{2}–5x–6=6–6\end{array}[/latex]
Expand the squared binomial, then simplify by combining like terms.
Be sure to write the terms with the exponent on the variable in descending order.
[latex]\begin{array}{r}2\left(x^{2}+6x+9\right)-5x-6=0\2x^{2}+12x+18–5x–6=0\2x^{2}+12x–5x+18–6=0\2x^{2}+7x+12=0\end{array}[/latex]
Now that the equation is in standard form, you can read the values of a, b, and c from the coefficients and constant.
[latex]\begin{array}{l}2x^{2}\,\,\,+\,\,\,7x\,\,\,+\,\,\,12\,\,\,=\,\,\,0\\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c\\\,\,\,\,\,\,a=2,\,\,b=7,\,\,c=7\end{array}[/latex]
Answer
[latex]2x^{2}+7x+12=0;\,\,a=2,b=7,c=12[/latex]
Solving a Quadratic Equation using the Quadratic Formula
The Quadratic Formula will work with any quadratic equation, but only if the equation is in standard form, [latex]ax^{2}+bx+c=0[/latex]. To use it, follow these steps.
Put the equation in standard form first.
Identify the coefficients, a, b, and c. Be careful to include negative signs if the bx or c terms are subtracted.
Substitute the values for the coefficients into the Quadratic Formula.
Simplify as much as possible.
Use the [latex]\pm[/latex] in front of the radical to separate the solution into two values: one in which the square root is added, and one in which it is subtracted.
Simplify both values to get the possible solutions.
That’s a lot of steps. Let’s try using the Quadratic Formula to solve a relatively simple equation first; then you’ll go back and solve it again using another factoring method.
Example
Use the Quadratic Formula to solve the equation [latex]x^{2}+4x=5[/latex].
Show Solution
First write the equation in standard form.
[latex]\begin{array}{r}x^{2}+4x=5\,\,\,\x^{2}+4x-5=0\,\,\,\\a=1,b=4,c=-5\end{array}[/latex]
Note that the subtraction sign means the constant c is negative.
[latex]\begin{array}{r}{{x}^{2}}\,\,\,+\,\,\,4x\,\,\,-\,\,\,5\,\,\,=\,\,\,0\\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\a{{x}^{2}}\,\,\,+\,\,\,bx\,\,\,+\,\,\,c\,\,\,=\,\,\,0\end{array}[/latex]
Substitute the values into the Quadratic Formula. [latex]x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}[/latex]
[latex]\begin{array}{l}\x=\frac{-4\pm \sqrt{{{(4)}^{2}}-4(1)(-5)}}{2(1)}\end{array}[/latex]
Simplify, being careful to get the signs correct.
[latex]x=\frac{-4\pm\sqrt{16+20}}{2}[/latex]
Simplify some more.
[latex]x=\frac{-4\pm \sqrt{36}}{2}[/latex]
Simplify the radical: [latex]\sqrt{36}=6[/latex].
[latex]x=\frac{-4\pm 6}{2}[/latex]
Separate and simplify to find the solutions to the quadratic equation. Note that in one, 6 is added and in the other, 6 is subtracted.
[latex]\begin{array}{c}x=\frac{-4+6}{2}=\frac{2}{2}=1\\\text{or}\\x=\frac{-4-6}{2}=\frac{-10}{2}=-5\end{array}[/latex]
Answer
[latex]x=1\,\,\,\text{or}\,\,\,-5[/latex]
You can check these solutions by substituting [latex]1[/latex] and [latex]−5[/latex] into the original equation.
| | |
--- |
| [latex]\begin{array}{r}x=1\x^{2}+4x=5\\left(1\right)^{2}+4\left(1\right)=5\1+4=5\5=5\end{array}[/latex] | [latex]\begin{array}{r}x=-5\x^{2}+4x=5\,\,\,\,\,\\left(-5\right)^{2}+4\left(-5\right)=5\,\,\,\,\,\25-20=5\,\,\,\,\,\5=5\,\,\,\,\,\end{array}[/latex] |
You get two true statements, so you know that both solutions work: [latex]x=1[/latex] or [latex]-5[/latex]. You’ve solved the equation successfully using the Quadratic Formula!
The power of the Quadratic Formula is that it can be used to solve any quadratic equation, even those where finding number combinations will not work.
In teh following video, we show an example of using the quadratic formula to solve an equation with two real solutions.
Most of the quadratic equations you’ve looked at have two solutions, like the one above. The following example is a little different.
Example
Use the Quadratic Formula to solve the equation [latex]x^{2}-2x=6x-16[/latex].
Show Solution
Subtract 6x from each side and add 16 to both sides to put the equation in standard form.
[latex]\begin{array}{l}x^{2}-2x=6x-16\x^{2}-2x-6x+16=0\x^{2}-8x+16=0\end{array}[/latex]
Identify the coefficients a, b, and c. [latex]x^{2}=1x^{2}[/latex], so [latex]a=1[/latex].Since [latex]8x[/latex] is subtracted, b is negative. [latex]a=1,b=-8,c=16[/latex]
[latex]\begin{array}{r}{{x}^{2}}\,\,\,-\,\,\,8x\,\,\,+\,\,\,16\,\,\,=\,\,\,0\\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\downarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\a{{x}^{2}}\,\,\,+\,\,\,bx\,\,\,+\,\,\,\,c\,\,\,\,=\,\,\,0\end{array}[/latex]
Substitute the values into the Quadratic Formula.
[latex]\begin{array}{l}x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\x=\frac{-(-8)\pm \sqrt{{{(-8)}^{2}}-4(1)(16)}}{2(1)}\end{array}[/latex]
Simplify.
[latex]x=\frac{8\pm \sqrt{64-64}}{2}[/latex]
Since the square root of 0 is 0, and both adding and subtracting 0 give the same result, there is only one possible value.
[latex]x=\frac{8\pm \sqrt{0}}{2}=\frac{8}{2}=4[/latex]
Answer
[latex]x=4[/latex]
Again, check using the original equation.
[latex]\begin{array}{r}x^{2}-2x=6x-16\,\,\,\,\,\\left(4\right)^{2}-2\left(4\right)=6\left(4\right)-16\16-8=24-16\,\,\,\,\,\,\8=8\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\end{array}[/latex]
In the following video we show an example of using the quadratic formula to solve a quadratic equation that has one repeated solution.
In this video example we show that solutions to quadratic equations can have rational answers.
Applying the Quadratic Formula
Quadratic equations are widely used in science, business, and engineering. Quadratic equations are commonly used in situations where two things are multiplied together and they both depend on the same variable. For example, when working with area, if both dimensions are written in terms of the same variable, you use a quadratic equation. Because the quantity of a product sold often depends on the price, you sometimes use a quadratic equation to represent revenue as a product of the price and the quantity sold. Quadratic equations are also used when gravity is involved, such as the path of a ball or the shape of cables in a suspension bridge.
A very common and easy-to-understand application is the height of a ball thrown at the ground off a building. Because gravity will make the ball speed up as it falls, a quadratic equation can be used to estimate its height any time before it hits the ground. Note: The equation isn’t completely accurate, because friction from the air will slow the ball down a little. For our purposes, this is close enough.
Example
A ball is thrown off a building from 200 feet above the ground. Its starting velocity (also called initial velocity) is [latex]−10[/latex] feet per second. (The negative value means it’s heading toward the ground.)
The equation [latex]h=-16t^{2}-10t+200[/latex] can be used to model the height of the ball after t seconds. About how long does it take for the ball to hit the ground?
Show Solution
When the ball hits the ground, the height is 0. Substitute 0 for h.
This equation is difficult to solve by factoring or by completing the square, so solve it by applying the Quadratic Formula, [latex]x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}[/latex]. In this case, the variable is t rather than x. [latex]a=−16,b=−10[/latex], and [latex]c=200[/latex].
[latex]t=\frac{-(-10)\pm \sqrt{{{(-10)}^{2}}-4(-16)(200)}}{2(-16)}[/latex]
Simplify. Be very careful with the signs.
[latex]\begin{array}{l}t=\frac{10\pm \sqrt{100+12800}}{-32}\\,\,=\frac{10\pm \sqrt{12900}}{-32}\end{array}[/latex]
Use a calculator to find both roots.
t is approximately [latex]−3.86[/latex] or [latex]3.24[/latex].
Consider the roots logically. One solution, [latex]−3.86[/latex], cannot be the time because it is a negative number. The other solution, [latex]3.24[/latex] seconds, must be when the ball hits the ground.
Answer
The ball hits the ground approximately [latex]3.24[/latex] seconds after being thrown.
The area problem below does not look like it includes a Quadratic Formula of any type, and the problem seems to be something you have solved many times before by simply multiplying. But in order to solve it, you will need to use a quadratic equation.
Example
Bob made a quilt that is 4 ft [latex]\times[/latex] 5 ft. He has 10 sq. ft. of fabric he can use to add a border around the quilt. How wide should he make the border to use all the fabric? (The border must be the same width on all four sides.)
Show Solution
Sketch the problem. Since you don’t know the width of the border, you will let the variable x represent the width.
In the diagram, the original quilt is indicated by the red rectangle. The border is the area between the red and blue lines.
Since each side of the original 4 by 5 quilt has the border of width x added, the length of the quilt with the border will be [latex]5+2x[/latex], and the width will be [latex]4+2x[/latex].
(Both dimensions are written in terms of the same variable, and you will multiply them to get an area! This is where you might start to think that a quadratic equation might be used to solve this problem.)
You are only interested in the area of the border strips. Write an expression for the area of the border.
Area of border = Area of the blue rectangle minus the area of the red rectangle
Area of border[latex]=\left(4+2x\right)\left(5+2x\right)–\left(4\right)\left(5\right)[/latex]
There are 10 sq ft of fabric for the border, so set the area of border to be 10.
Multiply [latex]\left(4+2x\right)\left(5+2x\right)[/latex].
Simplify.
Subtract 10 from both sides so that you have a quadratic equation in standard form and can apply the Quadratic Formula to find the roots of the equation.
Factor out the greatest common factor, 2, so that you can work with the simpler equivalent equation, [latex]2x^{2}+9x–5=0[/latex].
Use the Quadratic Formula. In this case, [latex]a=2,b=9[/latex], and [latex]c=−5[/latex].
[latex]\begin{array}{l}x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\x=\frac{-9\pm \sqrt{{{9}^{2}}-4(2)(-5)}}{2(2)}\end{array}[/latex]
Simplify.
[latex]x=\frac{-9\pm \sqrt{121}}{4}=\frac{-9\pm 11}{4}[/latex]
Find the solutions, making sure that the [latex]\pm[/latex] is evaluated for both values.
Ignore the solution [latex]x=−5[/latex], since the width could not be negative.
Answer
The width of the border should be 0.5 ft.
In this last video, we show how to use the quadratic formula to solve an application involving a picture frame.
Summary
A common method for solving radical equations is to raise both sides of an equation to whatever power will eliminate the radical sign from the equation. But be careful—when both sides of an equation are raised to an even power, the possibility exists that extraneous solutions will be introduced. When solving a radical equation, it is important to always check your answer by substituting the value back into the original equation. If you get a true statement, then that value is a solution; if you get a false statement, then that value is not a solution.
Completing the square is used to change a binomial of the form [latex]x^{2}+bx[/latex] into a perfect square trinomial [latex]{{x}^{2}}+bx+{{\left( \frac{b}{2} \right)}^{2}}[/latex], which can be factored to [latex]{{\left( x+\frac{b}{2} \right)}^{2}}[/latex]. When solving quadratic equations by completing the square, be careful to add [latex]{{\left( \frac{b}{2} \right)}^{2}}[/latex] to both sides of the equation to maintain equality. The Square Root Property can then be used to solve for x. With the Square Root Property, be careful to include both the principal square root and its opposite. Be sure to simplify as needed.
Quadratic equations can appear in different applications. The Quadratic Formula is a useful way to solve these equations, or any other quadratic equation! The Quadratic Formula, [latex]x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}[/latex], is found by completing the square of the quadratic equation [latex][/latex].
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Unit 16: Radical Expressions and Quadratic Equations, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology and Education. Located at: License: CC BY: Attribution
Ex 2: Solve Radical Equations - Square Roots. Authored by: James Sousa (Mathispower4u.com) . Located at: License: CC BY: Attribution
Ex 1: Solve a Basic Radical Equation - Square Roots. Authored by: James Sousa (Mathispower4u.com) . Located at: License: CC BY: Attribution
Ex 4: Solve Radical Equations - Square Roots. Authored by: James Sousa (Mathispower4u.com) . Located at: License: CC BY: Attribution
Ex 2: Completing the Square - Real Irrational Solutions. Authored by: James Sousa (Mathispower4u.com) . Located at: License: CC BY: Attribution
Ex: Creating a Perfect Square Quadratic Trinomial Expression. Authored by: James Sousa (Mathispower4u.com). Located at: License: CC BY: Attribution
Ex 2: Solving Quadratic Equations Using Square Roots. Authored by: James Sousa (Mathispower4u.com). Located at: License: CC BY: Attribution
Ex 1: Solving Quadratic Equations Using Square Roots. Authored by: James Sousa (Mathispower4u.com). Located at: License: CC BY: Attribution
Ex: Quadratic Formula - One Real Rational Repeated Solution. Authored by: James Sousa (Mathispower4u.com) . Located at: License: CC BY: Attribution
Ex: Quadratic Equation App: Find the Width of a Frame (Factoring). Authored by: James Sousa (Mathispower4u.com). Located at: License: CC BY: Attribution
Ex2: Quadratic Formula - Two Real Irrational Solutions. Authored by: James Sousa (Mathispower4u.com). Located at: License: CC BY: Attribution
Ex: Quadratic Formula - Two Real Rational Solutions. Authored by: James Sousa (Mathispower4u.com) . Located at: License: CC BY: Attribution
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CC licensed content, Original
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Screenshot: Quadratic Equation. Provided by: Lumen Learning. License: CC BY: Attribution
CC licensed content, Shared previously
Unit 16: Radical Expressions and Quadratic Equations, from Developmental Math: An Open Program. Provided by: Monterey Institute of Technology and Education. Located at: License: CC BY: Attribution
Ex 2: Solve Radical Equations - Square Roots. Authored by: James Sousa (Mathispower4u.com) . Located at: License: CC BY: Attribution
Ex 1: Solve a Basic Radical Equation - Square Roots. Authored by: James Sousa (Mathispower4u.com) . Located at: License: CC BY: Attribution
Ex 4: Solve Radical Equations - Square Roots. Authored by: James Sousa (Mathispower4u.com) . Located at: License: CC BY: Attribution
Ex 2: Completing the Square - Real Irrational Solutions. Authored by: James Sousa (Mathispower4u.com) . Located at: License: CC BY: Attribution
Ex: Creating a Perfect Square Quadratic Trinomial Expression. Authored by: James Sousa (Mathispower4u.com). Located at: License: CC BY: Attribution
Ex 2: Solving Quadratic Equations Using Square Roots. Authored by: James Sousa (Mathispower4u.com). Located at: License: CC BY: Attribution
Ex 1: Solving Quadratic Equations Using Square Roots. Authored by: James Sousa (Mathispower4u.com). Located at: License: CC BY: Attribution
Ex: Quadratic Formula - One Real Rational Repeated Solution. Authored by: James Sousa (Mathispower4u.com) . Located at: License: CC BY: Attribution
Ex: Quadratic Equation App: Find the Width of a Frame (Factoring). Authored by: James Sousa (Mathispower4u.com). Located at: License: CC BY: Attribution
Ex2: Quadratic Formula - Two Real Irrational Solutions. Authored by: James Sousa (Mathispower4u.com). Located at: License: CC BY: Attribution
Ex: Quadratic Formula - Two Real Rational Solutions. Authored by: James Sousa (Mathispower4u.com) . Located at: License: CC BY: Attribution
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find the distance of the line 4x+3y+5=0 from the origin - Brainly.in
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Find the distance of the line 4x+3y+5=0 from the origin
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Given :
Equation of line : 4x+3y+5=0
Point = (0,0)
To find :
The distance of the line 4x+3y+5=0 from the origin(0,0).
Solution :
The distance between the point (x,y) and line Ax+By+C=0 is given by:
Therefore the distance of the line 4x+3y+5=0 from the origin (0,0) is 1unit.
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July 2013 - Volume 117 - Issue 1
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Letters to the Editor: Letter to the Editor
Context Sensitive Decrement Times of Remimazolam
Schnider, Thomas Prof. Dr. Med.; Minto, Charles FANZCA, PhD
Author Information
Institut für Anästhesiologie, Kantonsspital St. Gallen, Switzerland, University of Berne, Berne, Switzerland, thomas.schnider@kssg.ch
Anaesthetist Sydney, New South Wales, Australia
Conflicts of Interest: Thomas Schnider was a paid consultant to PAION AG Germany.
Drs. Borkett and Johnson were invited to respond but declined.
Anesthesia & Analgesia 117(1):p 285, July 2013. | DOI: 10.1213/ANE.0b013e3182942954
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Context Sensitive Decrement Times of Remimazolam
Anesthesia & Analgesia117(1):285, July 2013.
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103
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https://www.mathworksheetsland.com/6/tapediagrams.html
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Reading and Making Tape Diagrams Worksheets
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Home>Grade Levels>Grade 6>
Reading and Making Tape Diagrams Worksheets
Tape diagrams are a tool that is used to represent a mathematical situation. This visual diagram gets its name because it looks like it represents a series of tape pieces. It is a rectangular visual model that resembles a piece of tape that is divided as per the word problem. It is also known as a divided bar model, strip diagram, fraction strip, and length model. The key here is to create identical rectangles. Each division in the rectangle will denote a new section of tape or part of the whole. As you will see below, these visuals can be used to help you solve complex problems by representing word problems in a foundational form. The worksheets and lessons in this section will help you learn to read and create your own tape diagrams for all types of different situations.
Aligned Standard: Grade 6 Proportional Relationships - 6.RP.A.3
You may also see this work referred to as a Bar Model. They tend to go hand and hand. The basic premise is that we are using visual diagrams (pieces of tape or bars) to define parts of a whole.
School Composition Step-by-step Lesson- What is the ratio of boys to girls?
Guided Lesson - How much money did Peter go to the store with? How many runs did Rich account for?
Guided Lesson Explanation - We test both skills here.
Making Tape Diagrams Practice Worksheet - All these problems ask kids to make their own diagrams.
Making and Reading Diagrams Practice 2- Remember these are used vehicles.
Answer Keys - These are for all the unlocked materials above.
Answers are included in these pages. Check out page 2 of each file.
Quiz - We fire 6 problems at you from all directions. Example: It took 4 buses to take 120 students on a field trip. How many students would be on each school bus?
Homework 1 - This sheet is the best mix of problems in the entire set.
Homework 2 - Painters use a ratio of 5 parts blue paint to 3 parts red paint; when mixing the purple paint for Vikings Stadium. How much blue paint would they need to make 16 gallons of purple paint?
Practice Sheet 3 - Proportions on the length of ribbons and wooden boards.
Practice Sheet 4 - Students vote for what they want to eat during a class party.
Practice Sheet 5 - These questions are more focused on reading premade diagrams. You will work on problems like this: Lisa raised $45 for the animal shelter, which was 5 times as much money as Helen raised. How much money did Helen raise?
What Are Tape Diagrams in Math?
Tape diagrams are mathematical graphic organizers that students can create to model many different types of real-world math situations. They use various rectangular arrays in drawing formats to recreate these situations. Each rectangle usually represents a section of a ratio. This serves as a great visual reminder of all the aspects found in a word problem. There are commonly three main uses of tape diagrams:
A) Part Whole Model - This is a quick way to display multiple items that contribute towards a whole. It almost acts like a pie or circle chart in a rectangular format. These can be used to model all of the four basic math operations (addition, subtraction, multiplication, and division). Here is a look at a tape diagram that models the types of drinks in your cooler if the values were 6 waters, 4 juices, 2 sodas:
As you can see from the diagram the length of the rectangle is in sync will the size of the value. The rectangle for 6 is bigger than the rectangle for 4 or 2. We can use this to learn that there are a total of 12 drinks in the cooler. This model can also be used to make visuals of basic algebraic equation. For example, building off of what we have explored so far. If we were not told the number of waters but knew there were 12 drinks in the cooler. We could use that information to model: x + 4 + 2 = 12.
B) Addition Comparison - The model is used to help in situations of addition or subtraction. With this tape diagram we are comparing two different quantities by using those math operations. In most cases we are using two tapes with one section missing. That section represents the value we are attempting to determine. For example, let’s compare the number of waters to the number of sodas, from the previous scenario that we described.
This allows us to setup a comparison using either addition or subtraction. This is a great way to begin to use algebra to model situation because this model allows us to approach these equations from both angles. We could recreate this situation by using either:
Addition: 2 + x = 6
Subtraction: 6 - x = 2
You could also use this to learn to solve those one-step problems.
C) Ratio Models -This is used to compare multiplication or division scenarios. We use this to compare equal groups. These are helpful for any problem that uses the phrases "as many as". The ratio model below displays a 1 to 6 ratio. This indicates that for every 1 box type, you have 6 box of the other type.
Tape diagrams pivot into fraction modeling as well. We will cover those in our fraction units and continue on from here. We wanted to focus you on the use of tape diagrams with math operations, at this point.
How Tape Diagrams Help Students Progress with Higher Levels of Math
This technique of using graphic diagrams to model math problems is usually introduced to students later in the elementary grade levels. It will often make an appearance when students are exploring proportions because they lend themselves to make concrete comparisons. Some students will have a very strong affinity to this technique, and it will lead them to learn how to make adjustments to model a great deal more than simply proportions. As they begin to learn operations with fractions, they will turn back to skill and it is very helpful. When used properly this method can lead students towards thinking about problems algebraically. I have seen students consistently use this skill up through Algebra 1 and have great success staying organized when tackling tough word problems. On an occasion with my accountant, she explained my taxes using this technique. We encourage teachers through the middle level and even high school level to remind their students of the tape diagram method for solving troublesome exercises.
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104
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https://arc.aiaa.org/doi/10.2514/1.J059734
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Volume 59, Issue 3
Open AccessRegular Articles
New Development of Classical Actuator Disk Model for Propellers at Incidence
Corrections for this article
Correction: New Development of Classical Actuator Disk Model for Propellers at Incidence
Rafael L. Rubin and
Dan Zhao
Rafael L. Rubin
University of Canterbury, Christchurch 8041, New Zealand
Ph.D. Candidate, Department of Mechanical Engineering; Private Bag 4800. .
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and
Dan Zhao
University of Canterbury, Christchurch 8041, New Zealand
†Associate Professor, Department of Mechanical Engineering, Private Bag 4800; . Associate Fellow AIAA.
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Published Online:14 Dec 2020
Abstract
The actuator disk model thrust formula was mathematically expanded in series and divided into two parts to show that propellers at incidence comprise an axial and a wing lift equivalent component. Both components share a common induced speed 𝑤. This is done by considering an enhanced disk area for momentum balance, to match Glauert’s hypothesis mass flowrate. To shed light on the theoretical developments, wind tunnel tests were conducted on a two-bladed propeller at angles of incidence ranging from 0 to 90 deg. The wing component is shown to increase with airflow velocity and angle of incidence. The axial component decreases with 𝑉, for all angles. The generally observed thrust increase with angle of incidence is explained, by the theory, to be mostly due to the wing component contribution. The theory also explains why at angles of incidence above 60 deg propellers inherently behave differently than at lower angles. While thrust decreases with 𝑉 at lower angles, it grows with airspeed at an angle of incidence of approximately 60 deg or higher. This behavioral inversion happens as the wing component positive sensitivity to 𝑉 overcomes the negative sensitivity of the axial component. A simplified formula is presented for predicting thrust at a given angle, based only on propellers data at an angle of incidence of 0, regardless of blade geometry.
| Nomenclature | |
--- |
| AOA, 𝛼𝑝 | propeller’s angle of incidence, deg |
| 𝛼slp | slip-stream angle of incidence at rotor disk, deg |
| 𝛼ult | slip-stream ultimate angle of incidence, deg |
| 𝐶𝑇 | thrust coefficient, equal to 𝑇/𝜌𝑛2𝐷4 |
| 𝐷 | propeller’s diameter, m |
| 𝑒 | theoretical entrainment factor |
| 𝒆𝑵 | direction versor of propeller normal force due to angle of incidence |
| 𝒆𝑻 | thrust direction versor or propeller axial direction |
| 𝐹𝑥 | wind tunnel net force measured on 𝑥 axis, N |
| 𝐹𝑧 | wind tunnel net force measured on 𝑧 axis, N |
| 𝐽 | advance ratio, equal to 𝑉/𝑛𝐷 |
| 𝑀𝑝 | propeller’s pitching moment due to angle of incidence, Nm |
| 𝑀𝑦 | wind tunnel net moment measured on 𝑦 axis, Nm |
| ˙𝑚 | mass flowrate, kg/s |
| 𝑁𝑝 | propeller’s normal force due to angle of incidence, N |
| 𝑛 | propeller’s frequency, Hz |
| 𝑝 | static pressure, Pa |
| 𝑝atm | atmospheric pressure, Pa |
| 𝑞 | freestream dynamic pressure, equal to 0.5𝜌𝑉2, Pa |
| RPM | propeller’s revolution per minute, rpm |
| 𝑆disk | propeller’s swept area or rotor disk area, m2 |
| 𝑆eff | propeller’s momentum balance effective area, m2 |
| 𝑆wing | equivalent wing area for 𝑇wing calculation, m2 |
| 𝑇 | propeller’s thrust, N |
| 𝑇axial | axial component of thrust, N |
| 𝑇wing | wing lift equivalent component of thrust, N |
| 𝑉 | freestream or wind velocity, m/s |
| 𝑽 | freestream velocity vector, m/s |
| 𝑉disk | slip-stream velocity at rotor disk, m/s |
| 𝑽disk | slip-stream velocity vector at rotor disk, m/s |
| 𝑉ult | slip-stream ultimate velocity, m/s |
| 𝑽ult | slip-stream ultimate velocity vector, m/s |
| 𝑊𝐹 | wing factor |
| 𝑤 | propeller’s average induced velocity at rotor disk, m/s |
| 𝒘 | propeller’s average induced velocity vector, m/s |
| 𝑤𝑜 | propeller’s average induced velocity at disk in static condition, m/s |
| 𝑤ult | ultimate induced velocity, m/s |
| ∂/∂𝛼𝑝 | derivative or sensitivity with respect to AOA, 1/deg |
| ∂/∂𝑉 | derivative or sensitivity with respect to 𝑉, s/m |
| 𝜀, 𝜖 | angle between 𝑻 and 𝑽disk, deg |
| 𝜖ult, 𝜖ult | angle between propeller axial direction and 𝑽ult, deg |
| 𝜌 | atmospheric density, kg/m3 |
| 𝜎 | standard deviation |
I. Introduction
Large interest in vertical takeoff and landing (VTOL) tilt-mode aircraft development appeared in the late 1940s, requiring research on propellers performance for a wide range of angles of incidence. Recently, VTOL unmanned aerial vehicles (UAVs) projects are causing the resurgence of the subject for small scaled propellers. In most cases, propellers are designed either for near-axial flight for airplanes or to perform as helicopter rotors. The transition between these flight modes in tilt-mode vehicles is mostly brief. Nonetheless, understanding and modeling propellers behavior at all angles is crucial to determine those vehicles performance.
In the past century, several models with different accuracies and complexities were developed. The simplest model to assess propeller thrust performance is the classical momentum theory or actuator disk model. The model was first introduced for marine propellers by Rankine in 1865 and Froude in 1889. Later, in the beginning of the 20th century, it was adopted for airscrews with the advent of the airplane. The theory applies the basics of fluid dynamics conservation laws [mass (continuity), momentum, and energy] to provide a general understanding of the propeller performance. Its assumptions disregard the propeller blades’ geometry and details of the flow about them. The propeller is considered as an infinitely thin actuator disk that impels a sudden increase in pressure on the fluid as it flows across its surface. This causes an acceleration of the flow, so an induced velocity increment is modeled at the disk. The flow is assumed to be incompressible and inviscid and rotation is neglected. Also, the velocity and pressure over the disk are considered to be uniform, and so thrust is distributed evenly over its surface. The major limitations of the theory are that it does not take profile drag losses of the propeller blades nor blades’ tip vortices and rotation effects into account. Despite the simplifications assumed, the patterns of velocities and pressures in the actuator disk model have been verified experimentally , and although it is not very accurate for power estimates , it provides a very good approximation for thrust. The theory is, however, not expected to yield a good basis for rotor in plane analysis, and hence it is not suitable for propeller design by itself. The derivation of the equations involved in the classical axial momentum theory for propellers can be found in several sources [5–8].
Subsequent models with higher complexity arose in the wake of the actuator disk theory. General or extended momentum theory, where the incorporation of rotation is added to the model, was developed by Betz . The blade element theory, which considers the geometry of the blades, was first devised by Drzewiecki in 1920. It ignored the effect of the induced flow inside the stream tube , as defined in the momentum theory, taking into account only the freestream velocity 𝑉 and the propeller rotation for every blade element analysis. The more precise blade element-momentum theory (BEMT) incorporated the induced velocity. Other variations of BEMT calculate the induced velocity in different ways. Goldstein’s classical vortex theory related the induced velocity to the bound circulation around every blade element. Vortex theory studies were later enhanced by Theodorsen . Extensive compilations of rotor aerodynamics analysis by Joukowsky in light of momentum theories and under vortex theories are presented in Refs. and , respectively.
Previous relevant studies to incorporate nonaxial flow conditions were made by Ribner [16,17], who devised formulas, using BEMT, to calculate side forces and moments that appear when propellers operate at incidence. The development assumed low angles of incidence. De Young later modified the Ribner formulas and expanded the use for high angles and also analyzed the effects of incidence on thrust. Experimental investigations have also been made in order to understand the behavior of full-scale propellers at a wide range of incidence in Refs. [19–21], in which it has been shown that thrust increases with increasing AOA and that thrust grows with the advance ratio 𝐽, at high AOA, as opposed to the effect at low angles. Recently, experiments on several small scaled propellers for UAV applications [22,23] also showed that for AOA<60 deg, 𝐶𝑇 diminishes with the advance ratio 𝐽, and vice versa for higher angles of incidence. Previous studies on wind turbines under yaw conditions are reported in Refs. [24–31].
Glauert [7,32] conducted an analysis on helicopter rotors in forward flight using momentum theory. Although it is mentioned that no solid proof has been given, he introduced a thrust formula for angles of incidence that reverts to the classical static thrust equation in the case of zero forward speed, and at the other extreme at fast forward flight, the equation takes the form of the elliptic wing lift formula, implying that the rotor behaves as a wing in that case. Bramwell investigated the validity of Glauert’s formula by solving the linearized Euler’s equations with small disturbances in the velocity field, showing that, for lightly loaded propellers, the model is valid for the axial case and for high speeds at 90 deg, regardless of blade geometry. Moreover, in the latter case, the linear theory appears to hold also for nonuniform load distributions. Glauert’s hypothesis for propellers at incidence has been widely used also for a wide angle of attack (AOA) range [3,4,8,11].
The appeal of the actuator disk momentum theory simplicity attracted further investigation to determine its limitations and possible improvements of its accuracy. Van Kuik [34,35] mentions that the average induced velocity calculated with the assumptions of the classical theory is underestimated. For a uniform load, a singularity at the disk edge must be included through a correction represented by a discrete vortex carrying an edge force. Goorjian also states an inconsistency in the general momentum theory, as mutual interferences between different annular elements are not considered. The existence of this inconsistency was also known to Glauert, but only in recent years, the errors associated with it have been quantified. This has been done by Sorensen and Mikkelsen , Van Kuik and Lignarolo , and Bontempo and Manna for the axial momentum theory, and in the general case, to Sorensen , and Bontempo and Manna [41,42]. Conway developed an analytical closed solution for the linearized actuator disk with arbitrary radial load distributions. As an extension of the linearized solution, a semi-analytical method was then developed for a nonuniform heavily loaded disk in . Pitt and Peters presented a linear, unsteady actuator model considering the dynamic inflow behavior of helicopter rotors. A model for axial and for skewed flows by Morillo and Peters presents the solution of the complete velocity field above the disk by converting the potential flow equations into ordinary differential ones. Recently, Rosen and Gur developed an axisymmetric axial actuator disk model including radial and tangential induced speed components. The model defines a pressure ratio factor that depends on the blades geometry to conclude the momentum balance, for which calculations are performed iteratively. Later, Kominer and Rosen adapted the model for asymmetric skewed inflows.
Although the previously cited actuator disk models improve accuracy, many of them are cumbersome to implement and end up missing the advantage of the momentum theory simplicity. Another approach for high accuracy is to use computational fluid dynamics (CFD) to solve the three-dimensional (3D) flow equations. These models may include viscosity, nonuniform propeller inflows, rotation, and more complicating assumptions such as the addition of nacelles and wing interferences in the analysis [22,49,50]. Often, high-complexity CFD models are coupled with experimental tests for validation. In general, they are also computational expensive and time consuming to implement.
In this Paper, we conduct theoretical studies by following the classical momentum theory and Glauert’s hypothesis for aerodynamic propellers at incidence. By considering an enhanced mass flow rate through the disk, this new development allows for a mathematical decomposition of thrust into two parts: 𝑇axial, which is dependent on the axial component of the oncoming wind speed, and 𝑇wing which is sensitive to the wind component parallel to the rotor plane. This is done in Sec. II. The development quantifies the contribution of each component to total thrust and helps clarify, under the scope of momentum theory alone, why rotors in forward flight behave as wings and why thrust increases with AOA. To validate the theoretical findings, experimental wind tunnel tests are conducted on a two-blade propeller at angles of incidence ranging from 0 to 90 deg. This is described in Sec. III. The effects of revolutions per minute, AOA, and oncoming flow velocity on thrust measurements are experimentally evaluated, and the propeller‘s behavior is explained, through the influence of the two components, in Sec. IV. Furthermore, as a consequence of the analysis, a simplified formula is derived to allow for the estimation of propellers thrust when at incidence, based on the performance data measured at AOA=0. A discussion on slip-stream parameters in response to AOA is also provided. Key findings are summarized in Sec. V.
II. Momentum Theory for Propellers at Incidence
A. Theoretical Entrainment Factor 𝑒
The classical momentum theory models an inviscid flow, with a uniform pressure jump and an average uniform induced speed at the actuator disk. Here, rotation is neglected. To apply the principles of momentum theory at incidence, it is first necessary to define the mass flow rate through the disk and the boundaries of the stream tube. As the airflow velocity at the disk 𝑉disk is not normal to the disk plane (see Figs. 1a and 1b), initially one would assume the boundaries of the stream tube touching the disk rim and the normal component of 𝑉disk, in other words, 𝑉cos𝛼𝑝+𝑤, to estimate the mass flux through the disk. However, it has been determined experimentally that the rotor entrains air from outside the rim stream tube . The phenomenon can be explained to be caused by radial pressure gradients effects not modeled in the classical momentum theory. The radial pressure gradients cause an increase in mass flow that grows with AOA, by means of tip vortices. A rigorous description of the phenomenon can be found in Refs. [34–37]. As the radial momentum balance is disregarded, the uniform pressure load at the disk would cause an infinite radial pressure gradient at the edge and a velocity singularity, which are inconsistent. This inconsistency in the model can only be resolved through the addition of edge vortices, which act as natural concentrators (see Refs. [34,51]), which increase the mass flow through the disk. This effect grows with AOA and with the speed component in the rotor plane (see Ref. ).
To incorporate an increase in mass flow rate to the model, an entrainment factor 𝑒 is applied to the disk area, and a new effective area 𝑆eff=𝑒𝑆disk is defined. Then, ⋅𝑚 can be expressed as
˙𝑚=𝜌(𝑉cos𝛼𝑝+𝑤)𝑆eff(1)
Shapiro also postulates the hypothesis of a much larger region of induced flow affected by the rotor at incidence and assumes a wider stream tube, where the mass of air that takes part in the exchange of momentum is the mass flux through a projected area normal to 𝑉disk and equivalent to 𝑆disk, regardless of the angle of incidence. In other words, this wider effective stream tube can be defined by flipping the rotor disk area to a normal position relative to 𝑉disk (see Fig. 1). The mass flow rate in the effective stream tube would then be given as
˙𝑚=𝜌𝑉disk𝑆disk(2)
Now, stretching the disk area at the original position to reach the new stream tube boundaries will define a new enhanced or effective disk area 𝑆eff that is adopted for the mass flux calculation in Eq. (1). The following development is based on the assumption that Eqs. (1) and (2) are equivalent for the definition of the mass flowrate ˙𝑚. This concept that determines the mass flowrate through a circle, with an area defined by the propeller diameter and normal to 𝑉disk, comes from rotor analysis in forward flight and uses the analogy to the wing theory [3,5,7,52]. It was presented in Glauert’s hypothesis .
The freestream velocity projected on the propeller’s reference frame can be shown as
𝑉=−𝑉 cos𝛼𝑝 𝒆𝑻+𝑉sin𝛼𝑝 𝒆𝑵(3)
By following the momentum theory, where the velocity at the disk is the sum of the axial induced velocity by the propeller and the freestream speed, 𝑽disk=𝑽+𝒘, then it is quite obvious that the following two relationships hold:
𝑽disk=−(𝑉cos𝛼𝑝+𝑤)𝒆𝑻+𝑉sin𝛼𝑝 𝒆𝑵(4a)
|𝑉disk|=√(𝑉cos𝛼𝑝+𝑤)2+(𝑉sin𝛼𝑝)2(4b)
As illustrated in Fig. 1b, it can be determined from vector geometric relationship that
𝑉disk cos𝜀=(𝑉cos𝛼𝑝+𝑤)(5)
Substituting Eqs. (2) and (5) into Eq. (1) leads to the theoretical entrainment factor 𝑒 being determined as
𝑒=𝑉disk(𝑉cos𝛼𝑝+𝑤)=√√⎷1+(sin𝛼𝑝)2(cos𝛼𝑝+𝑤/𝑉)2=1cos𝜀(6)
Hence, the effective area and the disk area are related to each other as
𝑆eff=𝑆diskcos𝜀(7)
With 𝑒 and the mass flow rate predicted, now we consider the momentum balance in the propeller axis direction 𝒆𝑻, between the ultimate wake section far downstream in the stream tube and section 1 far upstream of the propeller. As the classical momentum theory disregards the wake rotation effects and viscosity, then thrust can be written as
𝑇=˙𝑚(𝑉ultcos𝜀ult−𝑉cos𝛼𝑝)(8)
by applying geometry relationship from Fig. 1b it is possible to see that 𝑉ultcos𝜀ult=𝑉cos𝛼𝑝+𝑤ult, which leads to
𝑇=˙𝑚𝑤ult(9)
Substituting Eq. (1) in Eq. (9) allows thrust to be rewritten as
𝑇=𝜌(𝑉cos𝛼𝑝+𝑤)𝑆eff𝑤ult(10)
Considering the streamlines of the stream tube between section 1 and immediately before the disk (−) and from immediately after the disk (+) to the ultimate wake, the following two Bernoulli equations are obtained:
𝑝1+12𝜌𝑉2=𝑝−disk+12𝜌𝑉−2disk(11a)
𝑝ult+12𝜌𝑉2ult=𝑝+disk+12𝜌𝑉+2disk(11b)
Here, 𝑉−disk=𝑉+disk≜𝑉disk for continuity through the disk, and 𝑝1=𝑝ult=𝑝atm. This far wake pressure recovery can only be completed in the inviscid actuator disk model, as the wake rotation is neglected [7,37,39]. At the disk, there is a jump in pressure Δ𝑝 impelled by the propeller. Thus, 𝑝+disk=𝑝−disk+Δ𝑝. Manipulating Eqs. (11a) and (11b) and applying the pressure jump definition leads to
12𝜌(𝑉2ult−𝑉2)=Δ𝑝(12)
where 𝑉ult as illustrated in Fig. 1b is given as
𝑉2ult=(V+𝑤ult cos𝛼𝑝)2+(𝑤ultsin𝛼𝑝)2(13)
By substituting Eq. (13) into Eq. (12) and considering that the force associated with the jump in pressure is the thrust 𝑇, then the pressure jump can be rewritten as
Δ𝑝=𝑇𝑆eff=12𝜌((𝑉+𝑤ultcos𝛼𝑝)2+(𝑤ultsin𝛼𝑝)2−𝑉2)(14)
Substituting Eq. (10) into Eq. (14) and expanding terms in the rhs leads to
𝜌(𝑉cos𝛼𝑝+𝑤)𝑤ult=12 𝜌(𝑤2ult+2𝑤ult𝑉cos𝛼𝑝)(15a)
𝑤ult=2𝑤(15b)
The thrust 𝑇 can be expressed by using Eqs. (15b) and (10) as
𝑇=2𝜌(𝑉cos𝛼𝑝+𝑤)𝑒 𝑆disk𝑤(16)
On the other hand, 𝑇 can also be rewritten from Eqs. (2), (9), and (15b) as
𝑇=2𝜌𝑉disk𝑆disk𝑤(17)
Equation (17) is the well-known equation of thrust from Glauert’s hypothesis . Substituting Eq. (4b) into Eq. (17) leads to a quartic relationship in terms of 𝑤 as
(𝑇2𝜌𝑆disk)2=𝑉2𝑤2+2𝑉𝑤3cos𝛼𝑝+𝑤4(18)
As thrust can be expressed as 𝑇=𝜌𝑛2𝐷4𝐶𝑇 and the advance ratio of the propeller is defined as 𝐽=𝑉/𝑛𝐷, then Eq. (18) can be rewritten as
(2𝑛2𝐷2𝐶𝑇𝜋)2=𝑉2𝑤2+2𝑉𝑤3cos𝛼𝑝+𝑤4(19a)
(2𝐶𝑇𝜋𝐽2)2=(𝑤𝑉)2+2(𝑤𝑉)3cos𝛼𝑝+(𝑤𝑉)4(19b)
According to Eq. (19a), an increase in the propeller frequency 𝑛, or in revolutions per minute, is associated with an increase in 𝑤. The same effect can be observed in terms of 𝐽 from Eq. (19b), where a raise in 𝐽, or an increase in 𝑉 relative to 𝑛 or revolutions per minute, will cause a decrease in 𝑤/𝑉 and, vice versa, a rise in the ratio of the propeller rotation to wind speed 𝑉 will incur in an increase in 𝑤/𝑉. For the particular case of static thrust, where 𝑉=0, one obtains the classical induced velocity for hover condition 𝑤|𝑉=0,
𝑤|𝑉=0=√𝑇2𝜌𝑆disk(20)
For AOA=0 deg (i.e., no incidence) and if 𝑇>0, the only physically valid solution of the four roots for Eq. (18) is 𝑤|AOA=0 deg as (see Ref. )
𝑤|AOA=0 deg=12√(2𝑇𝜌𝑆disk)+𝑉2−𝑉2(21)
Equation (21) can be rewritten as
(𝑤𝑉)AOA=0 deg=12√(8𝐶𝑇|AOA=0 deg𝜋𝐽2)+1−12(22)
The derived equations from the classical momentum theory for propellers at a given incidence angle allows one to estimate the theoretical slip-stream angle at the disk 𝛼slp and at the ultimate wake 𝛼slpult. As illustrated in Fig. 1b, it can be shown from Eqs. (13), and (15b) that
𝛼slpult=arcsin⎛⎜⎝2(𝑤/𝑉)sin𝛼𝑝√1+4(𝑤/𝑉)cos𝛼𝑝+4(𝑤/𝑉)2⎞⎟⎠(23)
𝑉ult=𝑉√1+4(𝑤𝑉)cos𝛼𝑝+4(𝑤𝑉)2(24)
From Fig. 1b, (𝛼slp) at the disk is related to 𝛼𝑝 and 𝜀 as
𝛼slp=𝛼𝑝−𝜀=𝛼𝑝−arctan(sin𝛼𝑝cos𝛼𝑝+(𝑤/𝑉))(25)
Further analysis shows
𝛼slp=arcsin⎛⎜⎝(𝑤/𝑉)sin𝛼𝑝√1+2(𝑤/𝑉)cos𝛼𝑝+(𝑤/𝑉)2⎞⎟⎠(26)
It has been shown that the angle 𝜀 influences 𝑇 through the entrainment factor 𝑒. For no incidence flight, 𝜀 equals 𝛼𝑝 at 0 deg. For 𝛼𝑝=90 deg,
𝜀=arctan(𝑉𝑤)(27)
Figure 2a illustrates the variation of the entrainment factor 𝑒 determined by using Eq. (6) with 𝜀. The effects of AOA and 𝑤/𝑉 on 𝑒 is shown in Fig. 2b. It can be seen that at low angles of incidence 𝑒 tends to be unity. The effective area 𝑆eff is similar to the rotor disk area, as the propeller is operating at wind speeds near the axial condition. 𝑆eff is increased with 𝑒 and therefore with AOA. As AOA approaches 90 deg and 𝑉cos𝛼𝑝 tends to zero, 𝑒 tends to infinity, in theory at very low 𝑤/𝑉 so as to increase the area 𝑆eff in order to maintain the same finite value of mass flow rate ˙𝑚 through the rotor, according to Eqs. (1), and (2). In these extreme cases, the angle 𝜀 is approaching 90 deg, as 𝑉disk is almost parallel to the rotor disk. The increase in 𝑒 becomes more relevant at high angles of incidence and at very low 𝑤/𝑉 values.
B. Entrainment Factor Expansion and Introducing Axial and Wing Equivalent Components of Thrust
The entrainment factor 𝑒 can be expanded as a Maclaurin series as
𝑒=1cos𝜀=1+𝜀22+5𝜀424+61𝜀6720+277𝜀88064+50521𝜀103628800+…(28)
By naming the higher-order terms of the rhs of Eq. (28) as ∑terms(𝜀),
𝑒=1cos𝜀=1+∑terms(𝜀)(29)
Applying Eq. (16) and using Eq. (29) to replace the entrainment factor leads to the thrust 𝑇 being determined as
𝑇=2𝜌(𝑉cos𝛼𝑝+𝑤)𝑆disk 𝑤+2𝜌(𝑉cos𝛼𝑝+𝑤)𝑆disk 𝑤∑terms(𝜀)(30)
It can be seen that the first term on the rhs of Eq. (30) has the form of an axial thrust, as the disk area is constant and independent of the entrainment factor 𝑒 for any angle of attack. There is only the component of the incoming wind velocity in the axial direction, with its absolute value changing with cos 𝛼𝑝. Thus, that term is named the 𝑇axial component of thrust, and it is defined as
𝑇axial=2𝜌(𝑉cos𝛼𝑝+𝑤)𝑆disk𝑤(31)
The second term on the rhs of Eq. (30) can be rewritten by expressing ∑terms(𝜀)=1/cos𝜀−1 from Eq. (29) and by using the following relationship that can be obtained from the geometric graph in Fig. 1b:
(𝑉cos𝛼𝑝+𝑤)=𝑉sin𝛼𝑝√(1cos𝜀−1)(1cos𝜀+1)(32)
After further simplification, we notice that the resulting expression has the form of the wing lift formula from wing theory [3,5,7,52]. Therefore, we define it as the wing equivalent component of thrust 𝑇wing as
𝑇wing=2𝜌𝑉sin𝛼𝑝𝑤𝑆disk√1−cos𝜀1+cos𝜀(33)
Because it is assumed in the disk actuator theory that the induced speed 𝑤 is constant across the rotor disk, 𝑇wing can be considered to be equivalent to the lift produced by an elliptic wing subjected to a wind speed of magnitude 𝑉sin𝛼𝑝 and having the wing surface area equal to
𝑆wing=𝑆disk√1−cos𝜀1+cos𝜀(34)
The factor multiplying 𝑆disk and defining the equivalent wing area will be regarded as the wing factor 𝑊𝐹. It can be expressed also as function of 𝑒 as
𝑊𝐹=√1−cos𝜀1+cos𝜀=√𝑒−1𝑒+1(35)
Substituting Eq. (6) on Eq. (35) and simplifying it leads to 𝑊𝐹 being given as
𝑊𝐹=sin𝛼𝑝√1+2(𝑤/𝑉)cos𝛼𝑝+(𝑤/𝑉)2+cos𝛼𝑝+(𝑤/𝑉)(36)
Figure 3a shows the variation of 𝑊𝐹 with the angle 𝜀. For low 𝜀, 𝑊𝐹 implies a very small equivalent wing area, vanishing at 𝜀=0 deg (where also 𝛼𝑝=0 deg and the wing component vanishes). For 𝜀→90 deg, 𝑊𝐹 tends to unity. It yields then a full wing equivalent area of 𝑆disk. It can be seen from Fig. 3b that 𝑊𝐹 can only reach unity for 𝑤/𝑉→0, or at high speeds, and at AOA=90 deg, which corresponds also to 𝜀→90 deg. For AOA≠0 deg, 𝑊𝐹 only vanishes asymptotically as 𝑤/𝑉→∞, when hovering.
In the extreme case, at AOA→90 deg and 𝑤/𝑉→0, where 𝑊𝐹→1, thrust can be shown as
𝑇=𝑇axial+𝑇wing→2𝜌𝑆disk𝑤2+2𝜌𝑉𝑤𝑆disk(37)
Here, it can be seen that 𝑇axial takes the form of the static thrust formula and it contributes to total thrust even at forward speeds. 𝑇wing converges to the wing lift formula with full area 𝑆disk in the case of a rotor at fast forward flight (𝑤≪𝑉), when 𝑇axial is negligible. As 𝑇wing becomes dominant, the propeller behaves as a wing. These two expressions are mentioned by Glauert in the development of his thrust hypothesis in the analysis of helicopters in forward flight. Ariza presents results from 3D numerical simulations showing induced vortices produced by the rotor when at high angles of attack in the same manner as finite wing vortices. It is suggested that a possible explanation for the phenomenon in those conditions is that the propeller could behave as an elliptic wing. Johnson also mentions that the helicopter rotor behaves as a wing at forward speeds and that these two expressions are the limits of thrust when 𝑉≫𝑤 and when in hover, where 𝑉=0. Although he claims that there is no theoretical justification for the approach at intermediate forward speeds, good agreement has been found with measured rotor performance and with vortex theory results, suggesting therefore that it should be accepted for the entire range of speeds. McCormick also mentions the analogy of the propeller at forward flight to an elliptic wing when 𝑤/𝑉→0 and when 𝑉=0 to the hover case. The derivations of Eqs. (31) and (33) in this Paper allow us to assume that indeed the two components 𝑇axial and 𝑇wing are always present and comprise the total thrust for the whole operational envelope, at any angle of incidence and velocity.
From Eqs. (31) and (33), it is possible to calculate the ratio 𝑇wing/𝑇axial to show the contribution of the wing component to the axial component on thrust, according to
𝑇wing𝑇axial=2𝜌 𝑉sin𝛼𝑝𝑤 𝑆disk √1−cos𝜀1+cos𝜀2𝜌(𝑉cos𝛼𝑝+𝑤)𝑆disk𝑤=tan𝜀 √1−cos𝜀1+cos𝜀(38)
Figure 4 shows the variation of 𝑇wing/𝑇axial with AOA and 𝑤/𝑉, determined by Eqs. (6) and (38). The region where 𝑇wing≥𝑇axial is highlighted. It can be seen that the contribution of 𝑇wing to the total thrust 𝑇wing+𝑇axial increases with AOA and with 𝑉. As 𝑤/𝑉 is reduced to less than 0.6 and AOA grows higher than 60 deg, eventually 𝑇wing overcomes 𝑇axial and becomes dominant, in other words, 𝑇wing/𝑇axial>1.
At high AOA and at high speeds (𝑤/𝑉→0), then 𝑇wing/𝑇axial→∞, so 𝑇 is composed mainly by 𝑇wing. In those cases, 𝑇→2𝜌𝑉𝑤𝑆disk. At 𝑉=0 in hover, it becomes 𝑇=2𝜌𝑆disk𝑤2. This agrees well with the findings reported by Glauert . By manipulating Eqs. (35), (36), and (38), it is possible to rewrite 𝑇=𝑇axial(1+𝑇wing/𝑇axial) as
𝑇=𝑇axial⎧{⎨{⎩1+(sin𝛼𝑝)2[cos𝛼𝑝+(𝑤𝑉)]⎡⎢⎣√1+2(𝑤𝑉)cos𝛼𝑝+(𝑤𝑉)2+cos𝛼𝑝+(𝑤𝑉)⎤⎥⎦⎫{⎬{⎭(39)
III. Experimental Wind Tunnel Tests
A. Experimental Setup and Tests Procedures
The experimental tests were conducted at the University of Canterbury’s closed circuit subsonic wind tunnel that has a rectangular cross section of 0.9 m in height by 1.20 m in width, providing maximum test speeds of 60 m/s. The force balance is a six-axis JR3 45E15A4 sensor capable of measuring the forces and moments in the three-axis 𝑥𝑦𝑧. It can stand loads of up to 400 N with nominal accuracy of ±0.25%. A fourth-order Butterworth analog filter with eight different cutoff frequencies ranging from 6.3 to 926 Hz can be selected via jumper plugs. The default frequency at the wind tunnel is set to 6.3 Hz for maximum noise attenuation. An aluminum rig with variable AOA settings was manufactured to hold a 6 mm carbon fiber square rod and the 3D-printed motor/propeller assembly, shown in Fig. 5.
The propeller used is a two-blade, 6.0 in. diameter, 4.5 in. pitch propeller, manufactured by HQProp, powered by an Ethix TeamBlacksheep Silk V2 2345 V electric motor. An 80A T-motor electronic speed controller (ESC) is used to control the motor. A power supply continuously provides 15.7 V and electric current set according to power requirement to maintain a desired revolutions per minute, limited at around 20 A for motor constraints. The input electric current for the motor is measured through a current meter that communicates with data acquisition hardware (DAQs) from National Instruments to a desktop PC prepared with a LabVIEW program that records electric power, current, voltage, and motor revolutions per minute. A Monarch remote optical light-emitting diode (LED) sensor able to measure up to 250,000 rpm from up to 0.9 m distance, at a maximum of 45 deg angle, is installed at the wind tunnel test section facing the motor that is prepared with a reflective tape on top of half of its circumference. An exponential smoothing filter for revolutions per minute readings was applied in the LabVIEW program. Another LabVIEW program is used for the force measurement and includes a low-pass filter with adjustable cutoff frequencies. Oscillatory nature of thrust, moments, and lateral forces from propellers at incidence are mitigated by the use of the filter. A 1 Hz cutoff frequency has been chosen throughout the tests in order to achieve more stable average readings. Experimental test data are compiled and analyzed with MATLAB®.
Propeller thrust and normal forces reading were acquired at wind tunnel speeds ranging from zero to 25 m/s and motor speed rotations from 9000 to 18,000 rpm. Angles of attack ranged from 0 to 90 deg. Electric input motor power is also measured and analyzed for each test condition. For a wind speed of 25 m/s and AOA higher than 60 deg, it was not possible to reach revolutions per minute higher than 15,000 as power requirement exceeded motor limitations.
B. Experimental Data Postprocessing
Before the experimental tests, the holding set without the propeller was exposed to the same wind speeds and AOA values as of the propeller tests in order to evaluate the resistance of the rig. The forces readings acquired were fitted to spline surface functions. These are netted from the final propeller tests readings. The extra effect of the propeller slip stream on the holding sting was not evaluated, and the total net thrust could be expected to be somewhat higher. However, the slip stream is expected to impinge only on the sting behind the propeller and not on the entire holding set.
Equations (40a) and (40b) enable the calculation of 𝑇 and 𝑁𝑝 from forces 𝐹𝑥 and 𝐹𝑧 measured at the wind tunnel tests, according to the scheme depicted in Fig. 6. The analysis of experimental test data is performed by using the thrust values acquired for every test condition and subsequently inputting those into Eq. (18) to calculate the associated induced speeds 𝑤. Once 𝑤 is obtained, then all the remaining calculations can be performed.
𝐹𝑥=(cos𝛼𝑝)𝑇−(sin𝛼𝑝)𝑁𝑝(40a)
𝐹𝑧=(sin𝛼𝑝)𝑇+(cos𝛼𝑝)𝑁𝑝(40b)
The experimental data were acquired for around 20 s per experimental test. Batches of three tests were performed for every tested condition, in other words, RPM, AOA, and 𝑉, comprising around 80 points per condition. The experimental measurements were averaged, and the precision error evaluated as one standard deviation and presented in percentage terms relative to the average for 𝑉 and RPM. The errors for the forces 𝐹𝑥, and 𝐹𝑧 are presented in N, rather then in percentage values, as in many experimentally tested cases the very small average values obtained would result in meaningless very high percentage error values. A total of 182 tested conditions were measured. The figure shown in Appendix C, presents the measurements errors obtained for all tested conditions. The maximum error for 𝑉 was around 5%, although the vast majority of the tests showed errors below 2.0%. The maximum errors observed for RPM, 𝐹𝑥, and 𝐹𝑧 were 0.8%, 0.14 N and 0.65 N, respectively. The errors in 𝑇 were calculated based on the 𝐹𝑥, and 𝐹𝑧 errors, according to error propagation rules, 𝜎𝑇=√(cos𝛼𝑝𝜎𝐹𝑥)2+(sin𝛼𝑝𝜎𝐹𝑧)2. The cumulative distribution function for all 𝜎𝑇 points is presented in the subgraph 𝑓 in Appendix C, where it is shown that for 85% of the cases 𝜎𝑇<0.1 N and that the measurement errors tend to a log-normal distribution behavior as verified by a corresponding data fit.
IV. Results and Discussions
A. Thrust Measurements Analysis
Figure 7a shows thrust performance of the propeller as a function of RPM at different wind speeds, at no incidence (AOA=0 deg). As expected from the momentum theory, the freestream velocity increase at constant RPM (or a raise in 𝐽) will cause a reduction in the available thrust . It is also noticeable from propeller tests results presented in previous works [19–23,53,54], and the thrust exhibits a parabolic variation with RPM as in Refs. [19–23,53–56]. Figure 7b illustrates the influence of AOA on thrust for a constant wind speed 𝑉=20 m/s. The thrust 𝑇 is increased, as the angle of incidence rises. This finding is consistent with the previous concluding remarks in Refs. [19–23].
Figure 8 presents the thrust measured from the propeller as function of AOA and RPM, for a) 𝑉=10 m/s and b) 𝑉=20 m/s. Again, from both graphs at AOA=0 deg, it can be seen that the thrust is reduced with the increase of 𝑉. For instance, at 18,000 rpm, in Fig. 8a, 𝑇=6 N, whereas, in Fig. 8b, 𝑇=4 N, approximately. As AOA rises, 𝑇 increases in both cases. However, the slope of thrust increase with AOA ∂𝑇/∂𝛼𝑝 is higher in the case of the higher speed. Another interesting observation from Fig. 8 is that at low RPM and low AOA, no thrust is available, which is also seen in Fig. 7. In that region, the propeller will perform in a windmill state until a sufficient value of RPM is achieved. For example, in Fig. 8a in which 𝑉=10 m/s, no thrust is generated before the propeller reaches around 5000 rpm at low angles of attack. For the case of 𝑉=20 m/s in Fig. 8b, the value required is 9500 rpm at no incidence, and the region of no thrust achievable (𝑇≤0) is wider, reaching up to AOA≈60 deg. However, as AOA is increased, less RPM is required to achieve some thrust in both cases.
Figure 9 shows how thrust 𝑇 is varied with AOA, for 𝑉=10,𝑉=15, and 𝑉=20 m/s and rotation speeds of 12,000, 15,000, and 18,000 rpm. In all tested cases, the thrust 𝑇 is found to always increase with AOA, more noticeably for AOA>20 deg. A higher slope ∂𝑇/∂𝛼𝑝 is associated with a higher 𝑉, for a given RPM. Therefore, the analysis here means to be extended also to the advance ratio 𝐽, as an increase in velocity, at a given constant RPM, represents also an increase in 𝐽, and vice versa for a decrease. This difference in slopes will cause a change in thrust sensitivity to 𝑉 at high angles of incidence; while at low angles 𝑇 decreases with 𝑉 (∂𝑇/∂𝑉<0), for AOA around 60 deg and over, 𝑇 eventually starts to increase with increased 𝑉 (∂𝑇/∂𝑉>0). Similar results for 𝐶𝑇 growing with AOA and the inversion of behavior at a higher AOA are found in the experimental tests in Refs. [21–23].
Figure 10a shows the thrust 𝑇 contour surface at 15,000 rpm. The analysis here, at constant RPM, again means that an increase or decrease in 𝑉 is also associated with a corresponding increase or decrease in 𝐽. Again, at a low AOA value, 𝑇 is decreased as 𝑉 is growing. However, at a high AOA value, beyond 60 deg, the thrust variation behavior inverts to increase with 𝑉. Also, 𝑇 is always increasing with AOA at constant speeds. 𝑇axial is presented in Fig. 10b, and 𝑇wing is depicted in Fig. 10c, calculated according to Eqs. (17), (31), and (33). 𝑇axial behaves in the same way for all angles of incidence (AOA), decreasing as 𝑉 is increased. 𝑇wing is rising with 𝑉 and AOA, peaking at AOA=90 deg and at higher speeds. The slope of 𝑇wing increase with AOA, (∂𝑇wing/∂𝛼𝑝) also grows with 𝑉. Note in Figs. 10a and 10b the slight increase in 𝑇 with AOA for 𝑉=0 (static thrust tests performed at different AOA) is probably associated with ground effects from the wind tunnel. The expected behavior would be the same static thrust at all incidence angles.
Figure 11 shows the measured 𝑇 and its components 𝑇axial and 𝑇wing calculated for the cases of fixed velocities 𝑉=10 m/s, 𝑉=15 m/s, 𝑉=20 m/s and RPM values of 9000 and 15,000. The increase of thrust with AOA (∂𝑇/∂𝛼𝑝>0) follows to a great extent the behavior of 𝑇wing (∂𝑇wing/∂𝛼𝑝>0) from which it can be inferred the wing component to be mostly the reason for the phenomenon. Also, it can be observed from Figs. 11b, 11d, 11f and 11a, 11c, 11e that as 𝑉 (and 𝐽) is increased the sensitivity slopes of 𝑇wing to AOA are also increased. 𝑇wing contribution below AOA≈30 deg is negligible even at high advance ratios 𝐽 as the wing factor 𝑊𝐹 and sin𝛼𝑝 are low at small angles. At AOA=0 deg, 𝑇wing vanishes, and so 𝑇=𝑇axial, which formula reverts to the classical momentum theory thrust equation at no incidence. 𝑇wing relevance starts at AOA>30 deg, and a high 𝐽 value causes the contribution of 𝑇wing relative to 𝑇axial to become very important in the region around AOA≈60 deg, where 𝑇wing eventually surpasses 𝑇axial as illustrated in Figs. 11a–11f. This can only be observed for 𝑤/𝑉 lower than around 0.6 (see Fig. 4), which is associated with a higher 𝐽 [see Eq. (19b)]. At very high angles and as 𝐽 is increased, 𝑇wing composes an ever-larger part of 𝑇. Note also that as 𝐽 is increased (decreased 𝑤/𝑉) the crossing of the two components is possible at a lower angle than 90 deg, toward 60 deg (Figs. 11a, 11c, 11e and 11b, 11d, 11f). This is in accordance with the theoretical prediction behaviors as shown in Fig. 4.
As 𝑇axial is predominant at angles AOA<60 deg, 𝑇 is decreased with increasing speed (and 𝐽), and accordingly (∂𝑇/∂𝑉<0). For higher angles, 𝑇 is growing with 𝑉 (∂𝑇/∂𝑉>0) (see Fig. 11a, 11c, 11e at 9000 rpm and Figs. 11b, 11d, 11f at 15,000 rpm). The inversion from negative to positive sensitivity (see the figure in Appendix B) could be explained as the influence of 𝑇wing on 𝑇 (∂𝑇wing/∂𝑉>0.) being predominant over 𝑇axial (∂𝑇axial/∂𝑉<0) in that region. In Ref. , several experimental tests were performed for 𝐶𝑇 vs 𝐽 and AOA, ranging from 0 to 85 deg on three different full-scale propellers at many different blade pitch angles, where the results obtained showed inversions of slope (∂𝐶𝑇/∂𝐽) ocurring at angles of incidence higher than 60 deg, for all blade angle configurations. It is interesting to note that this inversion is in the region of high AOA, where the theory predicts that 𝑇wing and its effects begin to overcome 𝑇axial if under a large enough velocity 𝑉 (high 𝐽) (see Fig. 4). A possible explanation for this would be that different blade configurations would have different slope sensitivities of 𝑇 to 𝑉(∂𝑇/∂𝑉) as different propellers should have different 𝑇 surfaces vs 𝑤/𝑉 and AOA, and so (∂𝑇wing/∂𝑉>∂𝑇axial/∂𝑉) would happen at different angles for different propellers. However, we assume that for any propeller the relation between 𝑇wing and 𝑇axial must follow Eq. (38), as illustrated in Fig. 4, to fulfil the momentum theory assumptions.
𝑇axial is relatively insensitive to AOA at low advance ratios (up to 𝐽=0.53 for the propeller tested) and up to angles around 60 deg (see Figs. 11a, 11b, 11d, and 11f). A decrease in 𝑇axial at higher AOA is noticed in all cases, being more accentuated as 𝐽 is growing (𝑤/𝑉 diminishing), where also 𝑇wing growth is more intense. At higher advance ratios, 𝑇axial is increased with AOA to peak at around 60 deg for the propeller studied (see Figs. 11c and 11f). As 𝑇wing tends to be the main component of 𝑇 at high 𝐽 and AOA, the thrust formula is indeed reduced to the second term of Eq. (37) or 𝑇→2𝜌𝑉𝑤𝑆disk as predicted by Glauert and described in Ref. for helicopterrotors at high translational speeds (high 𝐽), when the propeller behaves as a wing.
B. Simplified Formula for Estimating 𝑇 at Incidence from Data at AOA=0 Deg
The thrust measurements obtained previously from the wind tunnel experimental tests at any AOA allow us to calculate 𝑤/𝑉 from Eq. (18) and estimate 𝑇axial and 𝑇wing at any point. However, it is not possible to measure these two components separately except in the extreme cases of hover and at high 𝑉 (high 𝐽, low 𝑤/𝑉), at AOA→90 deg, when 𝑇axial is negligible and 𝑇wing becomes the main component of 𝑇, i.e. 𝑇→𝑇wing as per Eq. (37), which is in accordance with Glauert . To prove the validity of the theory, we assume the values of thrust measured at AOA=0 deg and calculate the value of 𝑤/𝑉 from either Eq. (18) or Eq. (19), extrapolating it to any AOA, but using 𝑇 obtained at AOA=0 deg. This assumption comes from the fact that 𝑇axial is relatively insensitive to AOA as seen in Fig. 11 unless at high speeds. Figure 12 illustrates 𝑤/𝑉 calculated from measured 𝑇 data for all angles against the approximation model of 𝑤/𝑉 in solid lines. It is seen that there is a good match of the simplified model with the measurement data down to 𝑤/𝑉=0.2 or 𝐽 up to 0.53. For 𝐽>0.53, there is a detachment past AOA=30 deg that grows with 𝐽. At 𝐽=0.87 and AOA=0 deg, the propeller is windmilling and 𝑇|AOA=0 deg =0 as in Fig. 11e, so the model is not suitable as it predicts 𝑤=0 for all angles.
Figure 13 presents the results obtained from actual thrust measurements against predicted 𝑇 through the simplified model according to Eq. (39), using the values of 𝑇 at AOA=0 deg. Also, 𝑤/𝑉 projected to be used in Eq. (39) is obtained using 𝑇 at AOA=0 deg in either Eq. (18) or Eq. (19) as illustrated in Fig. 12. As the simplified model relies on a projection of thrust at AOA=0 deg and as 𝑇axial is relatively constant up to AOA around 60 deg, at low advance ratios (see Fig. 11 for 𝐽<0.53), good agreement is observed between the model and the actual thrust (see Figs. 13a, 13b, and 13d). In those tested cases, 𝑤/𝑉 projected from the simplified model is very close to 𝑤/𝑉 obtained from experimental data (see Fig. 12). At around AOA>70 deg, the model overestimates the practical test data. This should be expected as the model assumes a 𝑇axial constant for all incidence angles, but a decrease in that component at a higher AOA can be observed from Figs. 11a, 11b, and 11d.
A small detachment of 𝑤/𝑉 projected by the model, from measured 𝑤/𝑉 is seen at 𝐽=0.53 in Fig. 12, which will cause the start of the detachment of 𝑇 estimated by the model from the real 𝑇 measured, as in Fig. 13f. As 𝐽 grows, the detachment of 𝑇 projected by the model increases as in Fig. 13c until the model eventually loses validity as in Fig. 13e. This happens when 𝑇|AOA=0 deg measured at no incidence is vanished (windmill/brake state), alongside 𝑤/𝑉, which is used by the model to project 𝑇 at incidence. The growing detachments of the model 𝑤/𝑉, from 𝑤/𝑉 calculated based on the real experimental data, occur in consonance with 𝑇axial no longer being relatively constant and similar to 𝑇|AOA=0 deg at angles up to around 60 deg, for increasing 𝐽 values. In these cases, the 𝑇axial growth with AOA is not captured by the simplified model that underestimates 𝑇 up to those angles.
C. Slip-Stream Parameters
Table 1 summarizes the calculation of the relevant angles and velocities, as depicted in Fig. 1, from the experimental data. At given AOA, RPM, and 𝑉 conditions and from the thrust 𝑇 measured, calculations are performed according to Eqs. (4b), (6), (15b), (18), (23), (24), (26), (31), and (33). 𝑇wing and 𝑇axial are also presented at different flow conditions.
For AOA=30 deg, the theoretical entrainment factor 𝑒 is close to unity, as expected from the momentum theory. For increasing 𝑉 values, 𝛼slp gets smaller as it is harder for the propeller to turn the flow, especially at low RPM (high 𝐽, low 𝑤/𝑉), which is in agreement with Ref. , and if AOA is high, so is 𝑒 whenever 𝛼slp is small. The angle 𝛼slp-ult is always greater than 𝛼slp at the disk as 𝑤ult is higher than 𝑤 [see Fig. 1 and Eq. (15b), 𝑤ult=2𝑤]. At high AOA, the contribution of 𝑇wing becomes more relevant, and for 𝑤/𝑉 values lower than 0.6 and angles higher than 60 deg, 𝑇wing eventually surpasses 𝑇axial as in the last two cases where AOA=90 deg and 𝑤/𝑉=0.518 and 0.16. Note the entrainment factor 𝑒 is also high (𝑒=2.2 and 𝑒=6.3) accompanied by a high angle 𝜀 in these instances.
Figure 14a depicts the behavior of 𝛼slp as a function of 𝐽 and AOA for the propeller being tested in this Paper. It is seen that for static tests (𝐽=0)𝛼slp=AOA. At no incidence (AOA=0 deg), 𝛼slp also vanishes as the flow is axial. As 𝐽 is increased, 𝛼slp tends to diminish, and for a constant 𝐽, 𝛼slp grows with increased AOA. Figure 14b shows that the angle 𝜀 is varied with AOA and with 𝐽. For static test (𝐽=0),𝜀=0 deg as 𝒘 is aligned with 𝑻. At AOA=90 deg, 𝜀 tends asymptotically to 90 deg with the growth of 𝐽 (which implies growth in 𝑉/𝑤). The analysis of 𝜀 is of interest as it defines the entrainment factor 𝑒 and the wing factor 𝑊𝐹 and therefore is directly related to 𝑇wing, as indicated in Eqs. (33) and (35) and shown in Fig. 3. The maximum 𝜀 is associated with the highest 𝑇wing, which occurs at high 𝐽 (low 𝑤/𝑉 values) and high angles of incidence (see Fig. 10c).
V. Conclusions
In this Paper, a series of experimental tests was conducted to investigate the aerodynamic performance of a two-blade propeller operating at angles of incidence (AOA) ranging from 0 to 90 deg. This is done in in a closed-loop wind tunnel at University of Canterbury. An alternative approach on analyzing aerodynamic thrust performance is proposed. It is based on the classical momentum theory by defining a theoretical entrainment factor 𝑒 accounting for the mass flow rate through an enhanced area of the rotor disk to equalize Glauert’s hypothesis mass flow definition. The factor 𝑒 is found to depend on the angle 𝜀, defined as the angle between 𝑽disk and the thrust vector 𝑻. It is shown mathematically that the thrust 𝑇 consists of two components: one is the axial component 𝑇axial, and the other is the wing lift equivalent component 𝑇wing.
𝑇axial behaves similarly to a propeller operating at no incidence under an axial stream of magnitude 𝑉cos𝛼𝑝. Therefore, for a given RPM, it is decreased with increasing airspeed (∂𝑇axial/∂𝑉<0), for all angles of incidence. 𝑇axial is shown to be not so sensitive to AOA at low airspeeds (low 𝐽), especially at AOA<60 deg, while at high airspeeds (high 𝐽), it grows with the angle of incidence up to AOA≈60 deg. At higher angles, 𝑇axial is decreased with increasing AOA. This decrease is more intense at higher speeds.
𝑇wing provides the equivalent lift of an elliptic wing under velocity 𝑉 sin 𝛼𝑝 of magnitude with a variable area equal to 𝑆disk multiplied by a factor 𝑊𝐹. This factor ranges from zero at AOA=0 deg to 𝑊𝐹→1 as AOA approaches 90 deg at high airspeed 𝑉. 𝑇wing was found to rise with AOA and 𝑉. Also, the slope (∂𝑇wing/∂𝛼𝑝) is increased with 𝑉 and (∂𝑇wing/∂𝑉) is increased with AOA, for a given RPM.
The theory shows that 𝑇wing surpasses 𝑇axial at AOA≈60 deg or above and at high speeds (𝑤/𝑉<0.6). Also, as the theoretical ratio 𝑇wing/𝑇axial stems from the momentum theory development, which disregards blade geometry and propeller design, the authors assume that it would be valid for any propeller. The influence of 𝑇wing on 𝑇 overcomes that of 𝑇axial at high angles and speed as in those conditions the positive sensitivity (∂𝑇wing/∂𝑉) is larger than the negative sensitivity value of (∂𝑇axial/∂𝑉).
Thrust is found to be decreased with 𝑉 (and 𝐽) (∂𝑇/∂𝑉<0) at low angles, as 𝑇axial is dominant, whereas at around AOA≈60 deg or higher and at high airspeeds (and 𝐽), 𝑇 changes its behavior to increase with increasing 𝑉, in other words, (∂𝑇/∂𝑉>0). This behavior is interpreted as being the consequence of 𝑇wing impact overcoming the contribution of 𝑇axial effects. Different propellers and blade configurations should present different surfaces of 𝑇 vs 𝑤/𝑉 and AOA. Therefore, different slope sensitivities (∂𝑇wing/∂𝑉), (∂𝑇axial/∂𝑉) and [∂/∂𝑉(∂𝑇wing/∂𝛼𝑝)] should be expected, for a given RPM. It is believed that this could explain the inversion of ∂𝑇/∂𝑉 at different angles of incidence for different propellers seen in other studies. However, the ratio 𝑇wing/𝑇axial must follow the theory, and the inversion of thrust behavior should happen around the region where 𝑇wing becomes relevant and (∂𝑇wing/∂𝑉) overcomes (∂𝑇axial/∂𝑉), which happens at high angles and at high speeds (𝑤/𝑉<0.6). This verification could be done in further work.
At AOA→90 deg, 𝑇axial converges to the static thrust formula, being always present in the thrust composition even in forward flight. At hovering condition, when 𝑇wing vanishes, then 𝑇axial becomes the sole contributor of 𝑇, in which case the theory is proven to be consistent with the traditional formula 𝑇=2𝜌𝑆disk𝑤2. The proposed theory is also shown to agree completely with the classic momentum theory at AOA=0 deg. To a great extent, the increase of 𝑇 with AOA is due to the wing component. 𝑇wing peaks to compose most of the thrust at very high speed and high AOA. In this case, the proposed theory is shown to be consistent with Glauert’s hypothesis, reverting to the wing lift formula used for helicopter rotors in fast forward flight, 𝑇→2𝜌 𝑉𝑤 𝑆disk.
The development presents an alternative simplified formula for estimating 𝑇 at any AOA up to 90 deg for any propeller, based on data acquired at no incidence. (𝑇 or 𝐶𝑇 and 𝑉, RPM, or 𝐽). The formula showed good agreement with the experimental results for the propeller tested up to intermediate advance ratios and AOA≈80 deg.
A concluding remark is that one could assume the thrust of a propeller at incidence to be interpreted as the thrust of a propeller in axial flow condition under incoming speed of 𝑉cos𝛼𝑝 with a thrust addition equivalent to the lift produced by an elliptic wing of area 𝑆disk𝑊𝐹, under incoming speed 𝑉 sin 𝛼𝑝, where both components share a common induced speed 𝑤.
F. N. CotonAssociate Editor
Appendix A: General Thrust Formula for Propellers at Incidence Based on Thrust at AOA=0 Deg
A final equation for thrust prediction based on propellers data at AOA=0 deg is presented based on the assumption of 𝑇axial relatively constant with AOA up to around 60 deg for intermediate 𝐽 values. The simplified equation is an approximation based on Eq. (39),
𝑇𝑇|𝐴=0 deg=⎧{⎨{⎩1+(sin𝛼𝑝)2[cos𝛼𝑝+(𝑤𝑉)]⎡⎢⎣√1+2(𝑤𝑉)cos𝛼𝑝+(𝑤𝑉)2+cos𝛼𝑝+(𝑤𝑉)⎤⎥⎦⎫{⎬{⎭(A1)
where the value of 𝑤/𝑉 is obtained from solving Eq. (18) or Eq. (19), while using thrust data acquired at AOA=0 deg.
Appendix B: Measured Different Thrust Behaviors for Tested AOA Range
Constant AOA surfaces are depicted in Fig. B1, which shows 𝑇 as function of 𝑉 and RPM. It can be seen that below AOA≈60 deg, 𝑇 decreases with 𝑉, or (∂𝑇/∂𝑉<0), for any given RPM, while for AOA>60 deg, 𝑇 grows with 𝑉, or (∂𝑇/∂𝑉>0). This can be interpreted as the contribution of 𝑇wing influence overcoming that of 𝑇axial, in other words, (∂𝑇wing/∂𝑉>∂𝑇axial/∂𝑉) at high angles, and vice versa at low angles.
Appendix C: Experimental Measurement Error Analysis
Acknowledgments
The authors would like to express appreciation to the technical staff who helped make the experimental tests achievable, Julian Murphy, Julian Philips, Kenneth Brown, David Fanner, Natalia Kabaliuk, and Tony Doyle, and to Emeric Jago, an undergraduate engineering student. The first author would like to thank Dan Zhao for supervising his Ph.D. research and encouraging this work. The authors would also like to thank the anonymous reviewers for their insightful and detailed comments.
Tables
Table 1 𝑇 measured, 𝑇wing, and 𝑇axial calculated from the theory and slip-stream parameters
| AOA, deg | RPM | 𝑉, m/s | 𝐽 | 𝑤/𝑉 | 𝑇, N | 𝑇axial,𝑁 | 𝑇wing, N | 𝑒 | 𝜀, deg | 𝛼slp, deg | 𝛼slp_ult, deg | 𝑉disk, m/s | 𝑉ult, m/s |
| 30 | 17,864 | 10.4 | 0.23 | 0.832 | 7.001 | 6.715 | 0.285 | 1.0 | 16.4 | 13.6 | 18.8 | 18.4 | 26.8 |
| 30 | 14,989 | 15.5 | 0.41 | 0.301 | 4.050 | 3.723 | 0.327 | 1.1 | 23.2 | 6.8 | 11.2 | 19.7 | 24.0 |
| 30 | 9024 | 20.1 | 0.87 | 0.006 | 0.103 | 0.089 | 0.014 | 1.2 | 29.8 | 0.2 | 0.3 | 20.2 | 20.3 |
| 60 | 17,802 | 10.5 | 0.23 | 0.938 | 7.668 | 6.569 | 1.100 | 1.2 | 31.1 | 28.9 | 40.0 | 17.6 | 26.6 |
| 60 | 15,002 | 15.7 | 0.41 | 0.386 | 5.204 | 3.721 | 1.483 | 1.4 | 44.4 | 15.6 | 25.7 | 19.5 | 24.2 |
| 60 | 9074 | 19.9 | 0.86 | 0.081 | 1.480 | 0.824 | 0.655 | 1.8 | 56.1 | 3.9 | 7.4 | 20.8 | 21.7 |
| 90 | 17,865 | 10.3 | 0.23 | 1.155 | 8.289 | 6.267 | 2.022 | 1.3 | 40.9 | 49.1 | 66.6 | 15.8 | 26.0 |
| 90 | 14,861 | 15.6 | 0.41 | 0.518 | 6.296 | 2.896 | 3.400 | 2.2 | 62.6 | 27.4 | 46.0 | 17.6 | 22.5 |
| 90 | 8979 | 19.9 | 0.87 | 0.160 | 2.828 | 0.446 | 2.382 | 6.3 | 80.9 | 9.1 | 17.7 | 20.2 | 20.9 |
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Copyright © 2020 by Rafael Levy Rubin and Dan Zhao. Published by the American Institute of Aeronautics and Astronautics, Inc., with permission. All requests for copying and permission to reprint should be submitted to CCC at www.copyright.com; employ the eISSN 1533-385X to initiate your request. See also AIAA Rights and Permissions www.aiaa.org/randp.
Topics
Aerodynamic Performance
Aerodynamics
Aerospace Sciences
Airspeed
Computational Fluid Dynamics
Equations of Fluid Dynamics
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Wind Tunnels
Keywords
Disk Actuator Theory
Propeller Performance
Wind Tunnel Tests
Airspeed
Angle of Attack
Blade Element Theory
Mass Flow Rate
Vortices
Helicopter Rotor
Numerical Simulation
Acknowledgments
The authors would like to express appreciation to the technical staff who helped make the experimental tests achievable, Julian Murphy, Julian Philips, Kenneth Brown, David Fanner, Natalia Kabaliuk, and Tony Doyle, and to Emeric Jago, an undergraduate engineering student. The first author would like to thank Dan Zhao for supervising his Ph.D. research and encouraging this work. The authors would also like to thank the anonymous reviewers for their insightful and detailed comments.
Received7 May 2020
Accepted2 September 2020
Published online14 December 2020
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https://commons.case.edu/cgi/viewcontent.cgi?article=1206&context=wsom-ops-reports
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Published Time: Thu, 03 Apr 2025 19:01:26 GMT
Case Western Reserve University Case Western Reserve University
Scholarly Commons @ Case Western Reserve Scholarly Commons @ Case Western Reserve
University University
Research Reports from the Department of
Operations Operations Department
9-1-1977
Finding the General Solution of a Linear Diophantine Equation Finding the General Solution of a Linear Diophantine Equation
Susumu Morito
Harvey M. Salkin
Follow this and additional works at:
Part of the Business Administration, Management, and Operations Commons
Recommended Citation Recommended Citation
Morito, Susumu and Salkin, Harvey M., "Finding the General Solution of a Linear Diophantine Equation"
(1977).
Research Reports from the Department of Operations . 207.
This Report is brought to you for free and open access by the Operations Department at Scholarly Commons @
Case Western Reserve University. It has been accepted for inclusion in Research Reports from the Department of
Operations by an authorized administrator of Scholarly Commons @ Case Western Reserve University. For more
information, please contact digitalcommons@case.edu .
CWRU authors have made this work freely available. Please tell us how this access has benefited or impacted you! i FLE COPY
TECHNICAL MEMORANDUM NO. 429
FINDING THE GENERAL SOLUTION
OF A LINEAR DIOPHANTINE EQUATION
by
Susumu Morito
Harvey M. Salkin
Department of Operations Research
Case Western Reserve University
Cleveland, Ohio 44106
September 1977
iThe authors would like to express their appreciation to Professor
Dong Hoon Lee (Department of Mathematics, Case Western Reserve
University) for his time and helpful discussions.
Part of this work was supported by the Office of Naval Research under
contract number NO0014-67-A-0404-0010.
§TABLE OF CONTENTS
Page
ABSTRACT . . . . . . . . . . . . . ii
INTRODUCTION . . . . . . .. 1
THE ALGORITHM . . . . . 4
EXAMPLE AND DISCUSSION . . . . . . . 9
REFERENCES . . . . . . . . . 13
i
ABSTRACT
A new procedure for finding the general solution of a linear
diophantine equation is given. As a byproduct, the algorithm finds
the greatest common divisor (gcd) of a set of integers. Related re-
sults and discussion concerning existing procedures, are also given.
ii
INTRODUCTION
This note presents an alternative procedure for computing the
denoted by ged (al,a2,•••,an)' and for finding the general solution
greatest common divisor (gcd) of a set of n integers al,a2,··•, an'
of a linear diophantine equation in which these n integers appear as
coefficients. A classical procedure to find the gcd of n integers is
based on the repeated application of the standard Euclidean Algorithm
for finding the ged of two integers. More specifically, it repeatedly
uses the argument: gcd (al'a 2' ..., an) = gcd (gcd (al'82), a),0,0'an)'
A more efficient algorithm, which ia related to the procedure pre
sented here, for computing the gcd was given by Blankinship .
Weinstock developed a precedure to find a solution for a
linear diophantine equation, and Bond later showed that the Wein-
stock Algorithm can be applied repeatedly to find the general solution
for a linear diophantine equation.
In this note, we present an alternative approach to find the gen-
eral solution, and show that the algorithm produces (n - 1) n-dimentional
v-ectors wit_11_integer- components w ose inte-g-er linear combination generates
all solutions which satisfy the linear diophantine equation with the
' right-hand side 0. We call a set of these (n - 1) "generating" vectors as
a generator. It is easy to show that the generator is not unique for
n k 3. In fact, for n k 3 there exist infinitely many generators. The pro-
posed algorithm has certain desirable characteristics for computer imple-
mentation compared to the Bond Algorithm. Specifically, the Bond Algorithm
1
generally produces generating vectors whose (integer) components are
mostly huge numbers (in absolute value) . This often makes computer im-
plementation unwieldy . The approach, presented here, was initially
suggested by Walter Chase of the Naval Ocean Systems Center (San Diego,
California), in a slightly different form for solving the radio frequency
intermodulation problem .
For illustrative purposes, we will continuously use the following
example with n = 3: (al'a2'a3) = (8913, 5677, 4378). Or, we are inter-
ested in the generator of:
8913 x+ 5677 x+ 4378 x =0 1 2 3
It turns out that the Bond Algorithm produces the two generating
vectors (5677, -8913, 0) and (2219646, 3484888, -1), whereas the procedure
we propose gives (cf,Section 3) (-57, 17, 94) and (61, -95, -1).
Three obvious results are now given without proof. Throughout this
paper,
we
assume that
the right-hand side of a linear
diophantine equation
ao, if it is nonzero, is an integer multiple of d = gcd (al'a2'...'an).
This is because of the well-known result which says that a linear dio-
phantine equation has a solution if and only if al is divisible by d, and,
if d divides al there are an infinite number of solutions.
Lemma 1. Consider the following two equations:
(1) alxl + a2x2 •••+ anxn = 0
(2) alxl + a2x2 o,o+ anxn = ao.
2
Assume that (xp '...'xF ) is the generator of (1). Then all solutions n-1
x = (xi) of (2) can be expressed in the form
(3) x = x - kl xF1 + k2]CF2 o+ kn-1 ]CF . ' n-1
where X is any solution satisfying (2) and kl'k2 '..., kn-1 are any integers.
: Lemma 2. If al = al + 22a2 + 23a 3 +... £nan for some integers 22' 3""'2n' .
then the gcd (al'a2'...'an) = gcd (al'a2' an
Lemma 3. If al + £2a2 + £3a 3 +•••+ f a =0 for some integers 22'£3'...'En' nn
then the gcd (al,a2,...,an) = gcd (a2'.oo'all)'
Notice, for example, Lemma 3 is true because if d = gcd (a2'...'an) then
n
al = ( I Zi ).d for some integers Zi (2 <i <n). Thus, gcd (al 'a2'... 'an) =
i=2
gcd (al,d) = d.
Finding the general solution of a linear diophantine equation having a
right-hand side different from zero (say a 4 0) is straightforward, because
of Lemma 1, if the generator and one solution for (2) is known. The algorithm
we propose first finds a solution, say xd, for the linear diophantine equation
with right-hand side d = gcd (al'a2'...'an) as well as the generator. Then a
solution for (2) can be found as (a /d) xd.
3
THE ALGORITHM
We now present the algorithm to find the general solution of the
linear diophantine equation (2). The method is based on Lemma 1, namely,
it finds the generator (xF 'xF ,...,xF ) of (1) as well as any one
12
n-1 0
solution x of (2), so that any solution of (2) can be expressed as in (3).
0
A solution x of (2) is found as a by-product of finding the generator.
We list the steps.
Step 0. Setk=1. b(1)= ,
(1)- 1- (1) -
a , and N = n. '
1
al, 02 - a2'"0' un n
(1) (1)
Also let x(bl ) = (1,0,...,0), x(b2 ) = (0,1,0,...,0) '...,
x(b )= (0,···,0,1), where x(b) denotes the solution of (2) with
(1)
n
right-hand side al = b.
Stepl. Find integers 22,2 '..., ZN so that they satisfy
3
b (k)=£ b (k)+r 0 S r<b (k)
122 2, 22
r =g b(k)+r. OEr<b(k)
2 3 3 3' 3 3
r = £ b rN, O 5 rN = b; ,<bNCk)
(k)
N-1 N N
and thus
b (k)--= 2 b (k) + E b(k) --(k) + ...+2 b + b' 1 22 3 3 NN 1
Step 2. Find a solution x(b ) for alxl + a2x2 ··· + anxn = bl as follows:
x(bl) = x(bl(k) ) - 22 'x(b2(k) ) - 23 ' x(b Ck) ) - ... - £N ' x(bN(k) )
Step 3. If bl = 0, x(bl) is one of the generating vectors. Eliminate one
variable, i.e., N=N-1, and set bl - U2 ' 2 3 '••'' (k+1) - 1. (k) b (k+1) = b (k)
bil(k+1) = bN ] . If N=l, go to Step 4 (termination). If N>1,
increment the iteration count (i.e., k = k+1) and return to Step 1.
4
I
If bl 4 0, set bl(k+1) = b (k) , b (k+1) = b (k) b (k+1)
""' N 1'
= b
k = k+1 and return to Step 1.
(k+1)
Step 4. We now have (n - 1) generating vectors for (1), and b l is
the gcd (al'a2'...'a ). A solution for (2) can be found as
0 (k+1)
b (k+1) x (b 1 ) . Stop.
1
We now give three results which show the validity of the algorithm.
Theorem 1. There is a one-to-one correspondence between the solutions of
(4) and (5):
(4) a x1 + a x
... +a x =0 1 22 n n
(k) (k) +b (k) y = 0 (5) bl Yl + b2 Y2 + n n
(k) (k) (k)
Here bl 0 0'b 2 4 0' o00 ' n i
b 0 0 correspond to the values for b
obtained in kth iteration of Step 1, as far as N = n.
Proof. Consider the following two equations corresponding to any two con-
secutive iterations of the algorithm:
Ckth iteration) bl Yl + b2 y2 +"'+bn Yn = 0
(k) (k) (k) (k) (k) (k)
-Ck) -
-C-k)
-(k) .(k+1)
(k) _(k+1)
(k + lst iteration) (b 1 - £2b2 -' ' '- 2.nbn )l 22
+b Y
(k)
(k+1)
+...+ b Y =0
n n
The second equation can be written as
b (k)y (k+1)+ b (k) (y (k+1) _ 2 y (k+1))+. . + b (k) (yn k+l) - 2.n Yl(k+1) )= 0..
1 1 2 2 21
IThis means
(k)
(k+1) (k) = (ktl) _ £ (k+1) (k) (k+1) (k+1)
Yi = Yi ' 2 2 2 1 ""'Yn = Yn - 2n l
Using vector-matrix notation, we have
i 1 y,CH i 1-1 1 0 0 1 1 l
(k+1) 1
Ck) 2 (k+1) y -£ 0 1 0
232
(k) . . (k+1)
Y==T Y
(k) -£ 0. . .1 (k+1)
lyn 1
1 0 ... . 0 1 ;.
Notice that |det T| (i.e., the absolute value of the determinant of T) = 1.
We now show inductively on k that there exists a matrix M such that
X1
X2
(k)
X=
=M Y
Ixn
which satisfies |det M| =1. Clearly for the first iteration, T=M and
|det T| = |det M| = 1. Assume that there exists a matrix M' with |det M'| = 1
(k) (k)
(k+1) (k+1)
such that x = M' y Substituting y =T Y , we get x = M'T y
(k+IF
As |det(M'T)| = |detM'| x|det T| = 1. Thusx -My where M = M'T and
|det M| = 1.
It is well known (e.g., see ) that if there exists a matrix M such that
x=M y with |det M| =1, there is a one-to-one correspondence between the
solutions x and y. Thus the theorem is proved. Q.E.D.
6
Theorem 2. If (yF ' YF ' . . . 'Y F ) is the generator of (5 ), the corres-
12
n-1
ponding (xp ' xp ' . . . ' xF ) is the generator of (4 ) .
12 n-1
Proof. Assume that (x F ' 2"", XF ) is not the generator of (4 ).
Then there exists a solution vector x satisfying (4) such that it cannot
1 n-1
be expressed as integer linear combination of xF , xF2 '..., XF However,
n-1
because of the one-to-one correspondence (Theorem 1), there exists a
unique y which corresponds to x (i. e. ,M y=x) , and there are integers
Bl, 82'"'' Bn-1 such that y = Blyfl + 82YF2 + + Bn-1 yF as n-1
(yfl' YF2"", YF ) is the generator. However, n-1
x =My = M(BlyF + 82 F2 '' '+ Bn-1 F 1 n-1
= B l M Y F + 82 M YF +"'+ Bn-1 M YF
1 2 n-1
= Bl XF + 82 xf "'+ Bn-1 XF '1 2 n-1
and thus a contradiction. Q.E.D.
Theorem 3. Assume that d = gcd (al'a2' ..., an) = gcd (a2'...'an). Then
0
the general solution of (6) can be expressed as x = kx + x', where k is
§ an integer, x any solution of (7), and x' the general solution of (8).
(6 ) alxl + a2x2 + an xn = 0
(7) al + a2x2 , + an xn = 0
(8) a x + ... +a x =0
2 2
n n
Proof. Since d divides al, we have for £ integer, al = ld, and thus there
are solutions x2,x3,•••,xn to (7). This means, (6) has solutions when xl is
7
"
fixed to any integer. Clearly, x is any such solution to (6) in which
0
Xl = 1. Observe that all solutions for (6) can be characterized by fixing
Xl to any integer k and solving (6) in the remaining variables, x2'x3'...xn.
More specifically, for xl fixed to k, we want all solutions which satisfy (6)':
.
(6) a x +...+ a x = -a k2 2 n n 1
.
From Lemma 1, however, solutions for (6) can be expressed as a sum of a
solution for (6) and the general solution for (8) . T h u s,x=k x+
0
(kl][Fl + k2]cp. +...+ kn-2 xfn-2) is the general solution for (6) for integer,
2
kl'k2'...,kn-2' where kx is a solution satisfying_(6)-, and (X.Fl,XF2""'X?n-2 .
is the generator of ( 8) with xl = 0• Setting x'= I k.x means that X.
i=l 1 Fi'
is any solution to (8), and hence the result. O.E.D.
I3. EXAMPLE AND DISCUSSION
Table 1 lists the computational process to find the generator (xF ,xF )
for a 3-variable diophantine equation with the right-hand side equal to zero.
57 /61\
The two
vectors <-94 < -1
17
and 1 -95 1 form the generator.
From Theorem 1, there is a one-to-one relationship between (9) and (10):
(9) 8913 x I + 5677 x2 + 4378 x =0
: (10) 10 Yl 5 Y2 3 73=0.
The relationship is x = M y , where
-3 27 4
M=
-3 -10 13 |det M| = 1.
10 -42 -25
From Theorem 2, the generator (yF , yF2) of (10), if found, will be translated
to
the generator (xF , xF ) = (M yF , M yF ) of (9 ) 12
1j
Iteration 10 of the algorithm (cf Table 1) finds a solution y = 1-2
\O)
for (10) , and from Theorem 3, the general solution for (10) can be found as
k -2 + y', where y' =
Y2 is the general solution for (10) with Yl = 0.
C O ' -- -Y3 ----
Iterations 11 through 13 are performed to find the general solution for (11) :
(11) 5 y2 + 3 Y 3 = 0.
It can easily be checked that the general solution for (11) is y' = E < 3
0
1 0 -5
for £ integer. Thus -2 , _3 form a generator for (10) .
0
In general, whenever the final remainder (i.e., bl') of Step 1 in each
iteration becomes 0, we obtain a vector which is one of the n-1 generating
9
TABLE 1
ALGORITHM COMPUTATIONS
I -
Iteration b (k)=£ b (k) +E b (k) + b' x x xk 1
22 33 1 1 2 3
8913 1 0 0
5677 0 1 0
4378 0 0 1
1 8913 1(5677) + 0(4378) + 3236 1 -1 0
5677 1(4378) + 0(3236) + 1299 0 1 -1
3
4378 1(3236) + 0(1299) + 1142 -1 1 1
3236 2(1299) + 0(1142) + 638 1 -3 2
5
1299 1(1142) + 0 (638) + 157 1 0 -2
6
1142 1 (638) + 3 (157) + 33 -5 4 5
7
638 4 (157) + 0 (33) + 10 -3 -3 10
8
157 4 (33) + 2 (10) + 5 27 -10 -42
9 33 3 (10) + 0 (5) + 3 4 13 -25
10 10 2 (5) + 0 (3) + 0 -57 17 94
11 5 1 (3) + 2 23 -23 -17
12 3 1 (2) + 1 -19 36 -8
13 2 2 (1) + 0 61 -95 -1
10 vectors, and the size of problem (i.e., the number of variables) is re-
duced by 1.
Theorem 3 shows that this elimination of one variable at a time
guarantees the generating characteristic. After the problem is reduced,
the same arguments (i.e., Theorems 1 through 3) will be applied to the
reduced problem, sequentially. Eventually a two-variable problem will
be solved which yields the (n - 1) st or last generating vector, and the
process terminates.
From Lemmas 2 and 3, the last nonzero remainder in the algorithm
gives the greatest common divisor of al'a2'...'a . In the example, de-n
5677, and 4378. To see this note that, the gcd (8913, 5677, 4378) is equal
tailed in Table 1, the last nonzero remainder is 1 and is the gcd of 8913,
to gcd (10, 5, 3) by Lemma 2, which, in turn, is equal to gcd (5,3) by
Lemma 3. Repeating the same argument gives gcd (5,3) =
gcd (3,2) =
gcd (2,1) = gcd (1) = 1, or gcd (8913, 5677, 4378) = 1.
Finally,
Table 1 displays a
solution for
the equation with
the
right-
hand side equal to 1 = gcd (8913, 5677, 4378). The general solution for
the equation with the right-hand side a can then be expressed as:
< -19 11
/ -57 j 4.- -r- 1 -95 1-7-- - -.---
< 61 j
ao < 36 1---+ 'l -R: 1
2 /\ -1 /
where kl' k2 are integers.
Remarks.
An examination of the algorithm indicates that the divisions in
Step 1 can be made computationally more efficient by using the least
11
absolute remainder rather than the positive remainder. Specifically, we
(k)
find Zi (i = 2,...,N) such that |ri is minimized (0 S ri S bi ) in
Step 1, rather than using ri' where 0 5 ri S bi. This change allows the
proofs of the theorems to go through essentially unchanged.
The preceding discussion can be used to show that the Blankinship
Algorithm for finding the ged of n integers will also find the general
solution of a linear diophantine equation. Specifically, the algorithm
presented here can be regarded as a modified Blankinship Algorithm,
where the modification is in selecting the operators (according to Blankin-
ship' s terminology) . The Blankinship Algorithm, on the other hand, can be
regarded as a special case of our method, where £2 E £3 E E = O in
"' n-1 -
Step 2 of the algorithm presented here.
12 .
REFERENCES
W. A. Blankinship, "A New Version of the Euclidean Algorithm", American
Mathematical Monthly 70(7), 742-745 (1963).
R. Weinstock, "Greatest Common Divisor of Several Integers and an
Associated Linear Diophantine Equation," American Mathematical Monthly
67(7), 664-667 (1960).
J. Bond, "Calculating the General Solution of a Linear Diophantine
Equation", American Mathematical Monthly 74 (8), 955-957 (1967) .
W. Chase, "The Indirect Threat Algorithm", Technical Memorandum, Naval
Electronics Laboratory.Center, San Diego, California, November 1975.
S. Morito and H. M. Salkin, "A Comparison of Two Heuristic Algorithms
for a Radio Frequency Intermodulation Problem", Technical Memorandum,
Department of Operations Research, Case Western Reserve University,
Cleveland, Ohio, October 1977.
T. Saaty, Optimization in Integers and Related Extremal Problems,
McGraw-Hill, New York, 1970.
H. M. Salkin, Integer Programming, Addison-Wesley, Reading, Mass., 1975.
|
106
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https://www.youtube.com/watch?v=BAa0N9vvD0s
|
Volume of a Rectangular Prism | 5th Grade Math
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welcome to fifth grade math with mr. J so in this video we are going to be covering how to find the volume of rectangular prisms now rectangular prisms are three-dimensional shapes so what volume is is finding out the amount of space that object takes up three dimensional space so there's a couple formulas we use in there up top here to find the volume of a rectangular prism now we're going to happen the number one and just go through a few examples here so you can get this down now the V stands for volume in our formula and then this L right here stands for length times width times height or another formula you may see is this capital B which stands for the area of the base times the height now these formulas are basically the same thing because length times width will give you the area of the base the area of the base which you just multiplied by height so either way you're going to get the same answer and they're basically the same thing I'm going to use this one right here length times width times height so I always like to write my formula out this is always a good habit to get into and then we're going to plug in our values for length width and height so volume equals length times width times height now we go over to our model here and we see the length is 1 foot times our width right here is 3 feet times our height which is 2 feet then we multiply those together and that will give you the volume of that rectangular prison how many cubic feet would fill that prison how much space it takes up so 1 times 3 is 3 times 2 is 6 now we always want to label our answer and we look back at our model and we see feet and what we do for volume is we put a little three up here and this means six cubic feet okay six cubic feet that's how many cubic feet it would take to fill that rectangular prism or you could say six feet cubed either one now let's go to number two here write out our formula volume equals length times width times height plug in length going back is five our width is four and our height is 15 so what we would do now is multiply these out when we would get to 300 inches cubed or 300 cubic inches five times four is 20 times 15 is 300 300 inches cubed all right let's go to the number three volume equals length times width times height plug in our length going back is for our width is 10 and our height is two centimeters so we multiply out four times 10 is 40 times two 80 centimeters cubed formula plug-in answer and then let's go to number four now there's something special about number four and I put this in very specifically because you have to pay attention before you start some of these problems one thing should pop out about the height on this rectangular prism it doesn't make sense in comparison to the others as you notice the width is for that side is not as long as that one foot it's because we have inches here and this is feet for our height if that happens you need to convert everything to the same unit of measurement it's going to be easier to convert feet into inches and then plug in so one foot is 12 inches so we know our height is going to be 12 inches now everything is in inches and we're good to go so formula volume equals length times width times height plug in our length is 2 inches with this 4 and our height is 12 inches remember don't put 1 we need inches 2 times 4 is 8 and 8 times 12 is 96 so 96 cubic inches so there you have it there is how you find the volume of rectangular prisms your formula length times width times height or again you may see the area of the base base times height sometimes both are basically the same thing so it's now now time for you to try some on your own over at the mastery check I dropped the link below the Oregon you are going to try 4 examples to see if you have it down thanks a lot for watching and until next time peace
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Solved 5. In a certain college, 25 percent of the students | Chegg.com
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5. In a certain college, 25 percent of the students failed mathematics, 15 percent failed chemistry, and 10 percent failed both mathematics and chemistry. A student is selected at random. a) If the student failed chemistry, what is the probability that he or she failed mathematics? b) If the student failed mathematics, what is the probability that he or she
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Question: 5. In a certain college, 25 percent of the students failed mathematics, 15 percent failed chemistry, and 10 percent failed both mathematics and chemistry. A student is selected at random. a) If the student failed chemistry, what is the probability that he or she failed mathematics? b) If the student failed mathematics, what is the probability that he or she
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In a certain college, 25 percent of the students failed mathematics, 15 percent failed chemistry, and 10 percent failed both mathematics and chemistry. A student is selected at random. a) If the student failed chemistry, what is the probability that he or she failed mathematics? b) If the student failed mathematics, what is the probability that he or she failed chemistry? c) What is the probability that the student failed mathematics or chemistry? d) What is the probability that the student failed neither mathematics nor chemistry?
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Solving Polynomial Inequalities Note: This material is to supplement Section 3.6.
Unlike linear inequalities, polynomial inequalities cannot always be solved with just algebra, and other techniques will need to be used. We will learn to solve inequalities once they are of the form where one side of the inequality is a polynomial and the other is 0. If one side of the inequality is not 0, we need to do some algebra to change it into that form.
Example. Solve x2 < x + 2 First we will need to use algebra to change the inequality so that one side is a polynomial and the other is 0. We can do this by subtracting x and adding 2 to both sides yielding, x2 −x −2 < 0.
In this example we are asking for what x-values is the function f(x) = x2 −x −2 negative, or on what interval(s) is the graph below the x-axis In Section 3.4, we have learned how to graph polynomial functions using end-behaviors, x-intercepts, multiplicities, and sign charts. Our goal is to create a sign chart, a chart that shows where the function is positive or negative.
In order to create the sign chart for the function we need to identify the x-intercepts, and thus we must factor the function.
f(x) = x2 −x −2 = (x −2)(x + 1) and thus the x-intercepts are at x = 2 and x = −1.
The intermediate value theorem tells us that the function will be always positive or always negative on the intervals between and around the x-intercepts. There are then 3 intervals that we are concerned with, (−∞, −1), (−1, 2), and (2, ∞).
We will look at three different ways to finish this problem by creating a sign chart, all of which can be used for any problem. It is your choice on which technique works best for you.
Technique 1: Test Point Method One way to determine the sign of the function on each interval is to plug in a test point on the interval and determine the sign, we will call this the Test Point Method. It doesn’t matter what test point you use as long as it is not the x-intercepts themselves (which will yield an answer of 0). Lets choose easy values x = −2 for the first interval, x = 0 for the second, and x = 3 for the third. Also note that we are not concerned about the value of the function at these points but only the sign of the function. It is much easier to determine the sign by plugging in the point into the factored form of the function.
x −4 −3 −2 −1 0 1 2 3 4 f(−2) = (−2 −2)(−2 + 1) =( – )( – ) (+) f(0) = (0 −2)(0 + 1) =( – )(+) ( – ) f(3) = (3 −2)(3 + 1) =(+)(+) (+) Technique 2: Stacking Factors The Stacking Factors method is very similar to the Test Point Method, and is merely just a different way of visualizing and organizing the information. It can be especially useful if there are a large number of factors (if the degree of the polynomial is large). For each factor (in this example (x + 1) and (x −2)) we will determine whether it contributes a positive or negative for each x-value. This is easy for linear factors as it will always be negative to the left of the associated x-intercept and positive to the right. Then we will count the number of negatives to determine if the function is positive or negative on that interval.
x −4 −3 −2 −1 0 1 2 3 4 (x + 1) (x −2) f(x) ( – ) (+) (+) ( – ) ( – ) (+) (+) ( – ) (+) Technique 3: Using Properties of Polynomial Functions Another way to determine the sign chart is to look at properties of polynomials discussed in Section 3.4. For instance since the degree of f(x) is even and the leading coefficient is positive, then both the right and left end behaviors tend to infinity and thus the first and the last intervals must both be positive.
Furthermore since the multiplicities of both zeros are 1, the function will cross the x-axis at these points, giving that the sign on the middle interval is negative. Note that only some of this information is needed in order to obtain the sign chart.
Then given the sign chart we now have that f(x) is negative on the interval (−1, 2) and is thus our answer.
ANSWER: (-1,2) .
Example. Solve x2 −x −2 ≥0.
To complete this problem we will use the same sign chart as in Example 1 since we are dealing with the same function. This time however we are looking for when the function is positive OR zero. This means that we must include the x-intercepts in our interval(s). The function is positive or zero on the intervals (−∞, −1) and (2, ∞), and we will use a union (∪) to combine these intervals.
ANSWER: (-∞, −1) ∪(2, ∞) .
Example. Solve x3 + x2 −5x + 9 ≥6 To begin this problem we must first put it in the correct form, that is a polynomial on one side and 0 on the other. After subtracting 6 from both sides we obtain f(x) = x3 + x2 −5x + 3 ≥0.
Then we must factor the function using techniques learned in Section 3.5. to obtain f(x) = (x −1)2(x + 3) ≥0.
There are then three intervals we are concerned with, (−∞, −3), (−3, 1), and (1, ∞).
Using Properties of Polynomial Functions In this example we will first create the sign chart using end-behaviors and multiplicities of zeros. Since the degree of f(x) is odd and the leading coefficient is positive, the right end behavior tends to ∞and the left end behavior tends to −∞, and thus the sign on (−∞, −3) is negative and the sign on (1, ∞) is positive.
x −4 −3 −2 −1 0 1 2 ( – ) (+) Then since the multiplicity of the zero x = 1 is even, the function will touch the x-axis, making the middle interval (−3, 1) positive, or alternatively since the multiplicity of the zero x = −3 is odd, the function will cross the x-axis at this point giving the same result.
x −4 −3 −2 −1 0 1 2 ( – ) (+) (+) Test Point Method We can also find the sign chart using the test point method.
x −4 −3 −2 −1 0 1 2 3 f(−4) = (−4 −1)2(−4 + 3) =( – )2( – ) ( – ) f(0) = (0 −1)2(0 + 3) =( – )2(+) (+) f(2) = (2 −1)2(2 + 3) =(+)2(+) (+) Stacking Factors Or we can use the Stacking Factors method to create the sign chart. Note that we will consider (x−1)2 as two factors, i.e. (x −1)(x −1) although they will contribute the same signs.
x −4 −3 −2 −1 0 1 2 3 (x + 3) (x −1) (x −1) f(x) ( – ) (+) (+) ( – ) ( – ) (+) ( – ) ( – ) (+) ( – ) (+) (+) You can use the Stacking Factors method by thinking of (x −1)2 as one object. This is especially useful since the multiplicity is even and thus it will always be positive (where an odd multiplicity wouldn’t change the sign).
x −4 −3 −2 −1 0 1 2 3 (x + 3) (x −1)2 f(x) ( – ) (+) (+) (+) (+) (+) ( – ) (+) (+) Regardless of technique we arrive at the same sign chart and can then find our answer. Finally, note that we must also include the x intercepts since the problem calls for ≥0, so our answer will be [−3, 1]∪[1, ∞), which simplifies.
ANSWER: [-3,∞) Things to Remember • We must first use algebra to modify the inequality so that there is a polynomial on one side and 0 on the other side. This is a common mistake in problems like these.
• The goal is to create a sign chart, each of the three techniques used generate the same sign chart.
We only need to use ONE of these techniques, the choice is up to you. We have a lot of information (much of which we do not need), all of this information works together and can be used to check an answer quickly.
• Be careful of < versus ≤. If we include zero, we must include the x-intercepts as well in our answer often leading to using closed brackets instead of open parentheses.
• The final answer must be in interval notation and simplified. As in the last example, directly reading the sign chart does not give the simplified answer.
• In the Test Point Method, we do NOT care what the function value is, only what the sign is (this will save a lot of time and anguish).
• Also in the Test Point Method, it is much easier to plug the test point into the factored form of the polynomial.
• In the Stacked Factors method, if a linear factor as even multiplicity, it will always contribute a positive value, and if odd multiplicity it will contribute the same sign as if the multiplicity were one.
Solving Rational Inequalities Note: This material is to supplement Section 3.7.
In this section we will solve inequalities where one side of the inequality is a rational function and the other is zero. As before with polynomial inequalities, we may first need to use algebra to manipulate an inequality into this form.
Example. Manipulate the inequality 6 −x (x + 2)(x −3) > −1 so that there is a zero on one side of the inequality and a rational function on the other.
First we will add 1 to both sides, and then rewrite the left hand side as a rational function using a common denominator of (x + 2)(x −3).
6 −x (x + 2)(x −3) + 1 > −1 + 1 ⇒ 6 −x (x + 2)(x −3) + (x + 2)(x −3) (x + 2)(x −3) > 0 ⇒6 −x + (x + 2)(x −3) (x + 2)(x −3) > 0 ⇒6 −x + (x2 −x −6) (x + 2)(x −3) > 0 ⇒ x2 −2x (x + 2)(x −3) > 0 ⇒ x(x −2) (x + 2)(x −3) > 0 ANSWER: x(x −2) (x + 2)(x −3) > 0 Example. Solve 6 −x (x + 2)(x −3) > −1.
The first step of this problem is to transform the inequality so that there is a rational function on the left and zero on the right (as done in the previous example). So we have the equivalent problem: Solve x(x −2) (x + 2)(x −3) > 0.
Just as we did for polynomial inequalities, we will solve this by finding a sign chart using either the Test Point method or the Stacking Factors method. One difference from polynomial inequalities, though, is that we must look at both x-intercepts and vertical asymptotes, as the function can change sign at both.
The function f(x) = x(x −2) (x + 2)(x −3) has x-intercepts at x = 0 and x = 2 (by setting the numerator equal to zero), and vertical asymptotes x = −2 and x = 3 (by setting the denominator equal to zero).
There are then 5 intervals that we are concerned with (−∞, −2), (−2, 0), (0, 2), (2, 3), and (3, ∞).
Test Point Method We will need to test a point in each of the 5 intervals above. Below solid lines represent x-intercepts and dotted lines represent vertical asymptotes. We will choose test points −3, −1, 1, 5 2, 4 (note that sometimes we cannot choose integer values).
x -3 -2 -1 0 1 2 3 4 5 2 f(−3) (−3)(−3−2) (−3+2)(−3−3) ( −)( −) ( −)( −) (+) (+) (+) f(−1) (−1)(−1−2) (−1+2)(−1−3) ( −)( −) (+)( −) (+) ( −) ( – ) f(1) (1)(1−2) (1+2)(1−3) (+)( −) (+)( −) ( −) ( −) (+) f( 5 2) ( 5 2 )( 5 2 −2) ( 5 2 +2)( 5 2 −3) (+)(+) (+)( −) (+) ( −) ( – ) f(4) (4)(4−2) (4+2)(4−3) (+)(+) (+)(+) (+) (+) (+) Stacked Factors Method The Stacked Factors method is the same as with polynomial inequalities with the addition of including the factors associated with vertical asymptotes (those in the denominator). Just as before linear factors are negative before the associated x-intercepts/vertical asymptotes and positive afterwards.
x -3 -2 -1 0 1 2 3 4 5 2 x (x −2) (x + 2) (x −3) f(x) ( – ) ( – ) (+) (+) (+) ( – ) ( – ) ( – ) (+) (+) ( – ) (+) (+) (+) (+) ( – ) ( – ) ( – ) ( – ) (+) (+) ( – ) (+) ( – ) (+) The question asks for when the function is greater than zero, so we want to identify the positive intervals which are (−∞, −2), (0, 2), and (3, ∞), which gives an answer: ANSWER: (-∞, −2) ∪(0, 2) ∪(3, ∞) .
Example. Solve 6 −x (x + 2)(x −3) ≤−1.
Note that this is the same problem as above with ≤instead of >. Thus we are looking at the same sign chart but identifying the negative intervals AND the x-intercepts x = 0 and x = 2. (Remember we are not including the vertical asymptotes because they are not included in the domain).
ANSWER: (-2,0 ] ∪[ 2, 3) .
Things to Remember • We use the same techniques as when solving polynomial inequalities but we need to add the factors associated with vertical asymptotes as well.
• We must have one side of the inequality be 0. If this is not the case, some algebra (often involving common denominators) to obatin this.
• If we have ≤or ≥we must include x-intercepts but NOT vertical asymptotes. This is why it is important to differentiate between them in the sign charts).
• Just like with polynomial inequalities, in order to determine the x-intercepts (and vertical asymp-totes) we need to factor both the top and the bottom.
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3 x 3 Magic Square SUPER PUZZLE... Can you solve it??
Math Puzzles
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Posted: 5 Sep 2024
In today's mind-bending puzzle, we’ve got a classic magic square! Your task: find the missing numbers from 1 to 9 and place them in the empty squares. But there's a twist! Every row, column, and diagonal must have the same sum—are you up for the challenge? 🤓
💥 Test your skills and see if you can crack this before the explanation! 💥
🕒 Timestamps:
0:00 - Introduction
0:05 - Math Puzzle
0:33 - Huzefa's Explanation
🔥 Don't forget to like, share, and subscribe if you enjoy the challenge! Let us know in the comments if you solved it before the explanation and what you think the magic sum is! ✨
MathPuzzles #MagicSquare #PuzzleChallenge #BrainTeasers #MathGenius #LogicalThinking #FunWithNumbers
Transcript:
Introduction [Music] Math Puzzle what's up everybody and welcome back to another math puzzle in this one the question is as follows find the missing numbers 1 through n for each Square so that every row column and diagonal have the same sum so this is a good old magic square question so you got to fill in digits 1 through n in these remaining squares hit that pause button and see if you can figure it out when you're ready hit play and I'll give you the Huzefa's Explanation explanation so the first part of this puzzle is figuring out what is that magic sum that every row column and big diagonal can add up to that can be the same for each and every one and that is actually going to be 15 now the way that we get there is you're going to add up all the digits 1 through 9 and take the average of those so you're going to add them up and divide by 9 that's how you get 15 now you can see that right here we have a filled out column 2 + 9 is 11 + 4 is indeed 15 now we have numbers 1- 9 written out here and we see that 2 9 4 and 1 are already gone now we got to figure out how to place the rest of these values so I'm going to start here I see that I only have one empty space and I have nine and one that adds up to 10 so to get to 15 we know that this middle spot must be five next I'm going to go here since we already have two out of three for that diagonal 2 + 5 is 7 we need an 8 to get us to 15 now I've got 4 and 8 that gives me 12 what's going to get us to 15 that's going to be three right here next I'm going to look at this one and I know in this column I have 8 + 1 which is 9 to get to 15 I need a six so that goes right there and last but not least we're going to place the seven in this spot and we can now see that every row column and big diagonal going those ways all add up to 15 that's how you do it done I hope you guys enjoyed this math puzzle and if you did please click that like button if you want to see more math puzzles on the regular make sure to click subscribe thank you guys so much for joining see see you in the next video take it easy
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Cylinder volume & surface area (video) | Khan Academy
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Cylinder volume & surface area
SCCCR.Math: 78CM.8.MGSR.1.1, 8.MGSR.1.1, 8GCM.GS.PAFR.2.1, GS.MGSR.1.1, GS.PAFR.2.1
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A cylinder's volume is π r² h, and its surface area is 2π r h + 2π r². Learn how to use these formulas to solve an example problem.Created by Sal Khan.
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Richard Fan 8 years ago Posted 8 years ago. Direct link to Richard Fan's post “what is the value of pi” more what is the value of pi
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Anshul Ojha 5 years ago Posted 5 years ago. Direct link to Anshul Ojha's post “Pi is irrational, which m...” more Pi is irrational, which means the number doesnt have any regular patterns and goes on, one could say, forever. Pi starts of like this, 3.1415926535897932384626433832, but You can estimate 3.14 or 3.141. Pi can also be expressed as a fraction, 22/7, which is helpful when your radius/diameter is a multiple of 7 because that gives you a whole number as your area
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talinkumar 2 years ago Posted 2 years ago. Direct link to talinkumar's post “What is the cylinder volu...” more What is the cylinder volume formula? I didn't see him put it in the video.
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Kim Seidel 2 years ago Posted 2 years ago. Direct link to Kim Seidel's post “The volume of a cylinder ...” more The volume of a cylinder = π r² h
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macy hudgins 6 years ago Posted 6 years ago. Direct link to macy hudgins's post “I'm confused. I get how y...” more I'm confused. I get how you do surface area of a circle and why you have to do it twice, but I don't get how or what Sal did with the part in the middle. Can anybody help?
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Polina Vitić 6 years ago Posted 6 years ago. Direct link to Polina Vitić's post “To calculate the surface ...” more To calculate the surface area of the middle part of the cylinder, I like to think of it as a rectangle. The area of a rectangle is length x width.
In this case, length = the height of the cylinder and width = the circumference of the end of the cylinder (the circle).
The length is given, and the width can be calculated using the formula for circumference of a circle. Once you have both values, just multiply l x w to find the area, then add it to the area of the two circles to find the total surface area.
It might help to picture a can of soup - imagine you use a can opener to remove the top & bottom. Then imagine you take the can and cut it from top to bottom. If you flattened it out, you would have a rectangle. The cut side would be the height of the can, and the other side would be the part that wrapped around the top or bottom of the can.
Hopefully this helps make sense why that side is the same length as the circumference of the circle at the end of the cylinder!
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mianaw 2 years ago Posted 2 years ago. Direct link to mianaw's post “Think im more confused no...” more Think im more confused now than when I started
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☺ DarronA ☺ 2 years ago Posted 2 years ago. Direct link to ☺ DarronA ☺'s post “Probably just pressure or...” more Probably just pressure or something. I was a little intimidated when I saw that this video was like 8 minutes, but I turned out fine.
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Ryan Kwan 10 years ago Posted 10 years ago. Direct link to Ryan Kwan's post “In a question on surface ...” more In a question on surface area, it asks me to answer in a multiple of pi. How do I do that?
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Scott Moore 6 years ago Posted 6 years ago. Direct link to Scott Moore's post “Once you have simplified ...” more Once you have simplified your expression in terms of pi, go no further. Just say the answer is 24π or whatever it is.
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30280 6 months ago Posted 6 months ago. Direct link to 30280's post “Will I most likely need t...” more Will I most likely need this at some point in my life, or is more likely I won’t. If you could give examples that would be great. 🤠👍
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iAD63! 6 months ago Posted 6 months ago. Direct link to iAD63!'s post “As an adult, you will rea...” more As an adult, you will realize that you cannot live without a trash can. Thus, you will go to buy a trash can. You realize that there are no more shopping centers, and commerce is purely online. You look at a few trash cans, but you don't know which one is the right size. Some of them are really wide, but others are really tall. You want to buy the one that can contain the most trash. So, you solve for the volume of the cylinder. After, you are able to find out which trash can can contain the most trash. You buy that trash can. You live happily forever after.
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WoLf 2 years ago Posted 2 years ago. Direct link to WoLf's post “how do you find the awnse...” more how do you find the awnser in 3-5 easy steps?
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lilbona 2 years ago Posted 2 years ago. Direct link to lilbona's post “Sorry if I'm late, but he...” more Sorry if I'm late, but here you go:
Volume: Use the formula v=πr^2 h, where r=radius and h=height.
Surface area: a=ch, where c=circumference and h=height.
If you want the whole formula, it is a=π(2r)(h), where r=radius and h=height. (2r is the same thing as the diameter.)
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James Blagg✝ 5 years ago Posted 5 years ago. Direct link to James Blagg✝'s post “How do you find the surfa...” more How do you find the surface area for a sphere?
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Ian Pulizzotto 5 years ago Posted 5 years ago. Direct link to Ian Pulizzotto's post “The surface area of a sph...” more The surface area of a sphere with radius r is given by the formula
S = 4pi r^2.
Have a blessed, wonderful day!
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pandainspace128 5 months ago Posted 5 months ago. Direct link to pandainspace128's post “alright. so what we're tr...” more alright. so what we're trying to do in this video is just to get the overall surface area and volume of the cylinder, for volume we want to get the area of the top which is just Pi x R squared, for example if the radius was 5 inches we would get 25 cm sqaured and then get it multiplied by pi, making it 25pi (cm squared), then we multiply it with the height of the cylinder for the volume, volume is just 3d area, For surface area, it's essentially 3d Perimeter so we do the same thing to find the top area of the circle, but we don't multiply it with length, we actually multiply it by 2 because we want to count the bottom circle aswell, Then we get the circumference which is just diameter x pi, which for example the diameter being 10 would give us 10 pi cm, then we multiply the circumfrence with the length of the cylinder and get something like 35 Pi cm Squared, Cm changes to squared when multiplying it with other cm values btw, then we add the values and we get the surface area, this might be confusing so just put this text in chatgpt for a detailed version but honestly u got this bro. just remember that you're trying to multiply the circles area and height for the volume and for the surface area all we want to do is get the area of the top circle, multiply it by 2, then get the circumference of the circle and multiply it with the height, then we add the area of the 2 circles and the circumference x height, after adding these we have our value!. and if you divide this formula by 1/3 you get the formula for the volume of a cone.
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I HAVE PLAYED THESE GAMES BEFORE 8 months ago Posted 8 months ago. Direct link to I HAVE PLAYED THESE GAMES BEFORE's post “why are we here again” more why are we here again
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Video transcript
Let's find the volume of a few more solid figures and then if we have time, we might be able to do some surface area problems. So let me draw a cylinder over here. So that is the top of my cylinder. And then this is the height of my cylinder. This is the bottom right over here. If this was transparent, maybe you could see the back side of the cylinder. So you could imagine this kind of looks like a soda can. And let's say that the height of my cylinder, h, is equal to 8. I'll give it some units. 8 centimeters, that is my height. And then let's say that the radius of one of the top of my cylinder of my soda can, let's say that this radius over here is equal to 4 centimeters. So what is the volume here? What is the volume going to be? And the idea here is really the exact same thing that we saw in some of the previous problems. If you can find the surface area of one side and then figure out how deep it goes, you'll be able to figure out the volume. So what we're going to do here is figure out the surface area of the top of this cylinder, or the top of the soda can. And then we're going to multiply it by its height. And that'll give us a volume. This will tell us essentially, how many square centimeters fit in this top. And then if we multiply that by how many centimeters we go down, then that'll give us the number of cubic centimeters in this cylinder or soda can. So how do we figure out this area up here? Well the area at the top, this is just finding the area of a circle. You could imagine drawing it like this. If we were to just look at it straight on, that's a circle with a radius of 4 centimeters. The area of a circle with a radius 4 centimeters, area is equal to pi r squared. So it's going to be pi times the radius squared, times 4 centimeters squared, which is equal to 4 squared, is 16, times pi. And our units now are going to be centimeters squared. Or another way to think of these are square centimeters. So that's the area. The volume is going to be this area times the height. So the volume is going to be equal to 16 pi centimeters squared, times the height, times 8 centimeters. And so, when you do multiplication, the associative property, you can kind of rearrange these things, and the commutative property. It doesn't matter what order you do it if it's all multiplication. So this is the same thing as 16 times 8. Let's see. 8 times 8 is 64. 16 times 8 is twice that. So it's going to be 128 pi. Now you have centimeter squared times centimeters. So that gives us centimeters cubed. Or 128 pi cubic centimeters. Remember, pi is just a number. We write it as pi, because it's kind of a crazy, irrational number, that if you were to write it, you could never completely write pi. 3.14159 keeps going on, never a repeat. So we just leave it as pi. But if you wanted to figure it out, you can get a calculator. And this would be 3.14 roughly, times 128. So it would be close to 400 cubic centimeters. Now, how would we find the surface area? How would we find the surface area of this figure over here? Well, part of the surface area of the two surfaces, the top and the bottom. So that would be part of the surface area. And then the bottom over here would also be part of the surface area. So if we're trying to find the surface area, it's definitely going to have both of these areas here. So it's going to have the 16 pi centimeters squared twice. This is 16 pi. This is 16 pi square centimeters. So it's going to have 2 times 16 pi centimeters squared. I'll keep the units still. So that covers the top and the bottom of our soda can. And now we have to figure out the surface area of this thing that goes around. And the way I imagine it is, imagine if you're trying to wrap this thing with wrapping paper. So let me just draw a little dotted line here. So imagine if you were to cut it just like that. Cut the side of the soda can. And if you were to unwind this thing that goes around it, what would you have. Well, you would have something. You would end up with a sheet of paper where this length right over here is the same thing as this length over here. And then it would be completely unwound. And then these two ends-- let me do it in magenta-- these two ends used to touch each other. And-- I'm going to do it in a color that I haven't used yet, I'll do it in pink-- these two ends used to touch each other when it was all rolled together. And they used to touch each other right over there. So the length of this side and that side is going to be the same thing as the height of my cylinder. So this is going to be 8 centimeters. And then this over here is also going to be 8 centimeters. And so the question we need to ask ourselves is, what is going to be this dimension right over here. And remember, that dimension is essentially, how far did we go around the cylinder. Well, if you think about it, that's going to be the exact same thing as the circumference of either the top or the bottom of the cylinder. So what is the circumference? The circumference of this circle right over here, which is the same thing as the circumference of that circle over there, it is 2 times the radius times pi. Or 2 pi times the radius. 2 pi times 4 centimeters, which is equal to 8 pi centimeters. So this distance right over here is the circumference of either the top or the bottom of the cylinder. It's going to be 8 pi centimeters. So if you want to find the surface area of just the wrapping, just the part that goes around the cylinder, not the top or the bottom, when you unwind it, it's going to look like this rectangle. And so its area, the area of just that part, is going to be equal to 8 centimeters times 8 pi centimeters. So let me do it this way. It's going to be 8 centimeters times 8 pi centimeters. And that's equal to 64 pi. 8 times 8 is 64. You have your pi centimeters squared. So when you want the surface area of the whole thing, you have the top, you have the bottom, we already threw those there. And then you want to find the area of the thing around. We just figured that out. So it's going to be plus 64 pi centimeters squared. And now we just have to calculate it. So this gives us 2 times 16 pi, is going to be equal to 32. That is 32 pi centimeters squared, plus 64 pi. Let me scroll over to the right a little bit. Plus 64 pi centimeters squared. And then 32 plus 64 is 96 pi centimeters squared. So it's equal to 96 pi square centimeters, which is going to be a little bit over 300 square centimeters. And notice, when we did surface area, we got our answer in terms of square centimeters. That makes sense, because surface area, it's a two-dimensional measurement. Think about how many square centimeters can we fit on the surface of the cylinder. When we did the volume, we got centimeters cubed, or cubic centimeters. And that's because we're trying to calculate how many one by one by one centimeter cubes can we fit inside of this structure. And so that's why it's cubic centimeters. Anyway, hopefully that clarifies things up a little bit.
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1969 IMO Problems
Problems of the 11th IMO 1969 in Romania.
Contents
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1 Problem 1
2 Problem 2
3 Problem 3
4 Problem 4
5 Problem 5
6 Problem 6
Problem 1
Prove that there are infinitely many natural numbers with the following property: the number is not prime for any natural number .
Solution
Problem 2
Let be real constants, a real variable, and Given that , prove that for some integer .
Solution
Problem 3
For each value of , find necessary and sufficient conditions on the number so that there exists a tetrahedron with k edges of length , and the remaining edges of length 1.
Solution
Problem 4
A semicircular arc is drawn on as diameter. is a point on other than and , and is the foot of the perpendicular from to . We consider three circles, , all tangent to the line . Of these, is inscribed in , while and are both tangent to and to , one on each side of . Prove that , and have a second tangent in common.
Solution
Problem 5
Given points in the plane such that no three are collinear. Prove that there are at least convex quadrilaterals whose vertices are four of the given points.
Solution
Problem 6
Prove that for all real numbers , with , the inequality is satisfied. Give necessary and sufficient conditions for equality.
Solution
1969 IMO
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Cystic Hygroma in an Adult Male: A Diagnostic Challenge
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Cystic Hygroma in an Adult Male: A Diagnostic Challenge
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Cystic Hygroma in an Adult Male: A Diagnostic Challenge
Document Type : Case Reports
Authors
Varsha Bhatia
Rahul Taksande
Divyaja Sondankar
Dept. of Pathology, Grant Government Medical College and Sir JJ Group of Hospitals,Byculla, Mumbai, Maharashtra, India
10.7508/ijp.2015.04.010
Abstract
Cystic Hygroma is a well-recognized lesion in paediatric patients but it is very rare in adults. We report here a case of Cervical Cystic Hygroma in a 45 yr old male. The reasons to present this case are 1) Its rarity in adults 2) Unusual age of presentation 3) Simulates various benign and malignant lesions at this site 4) Is a diagnostic challenge and 5) Role of histology for definitive postoperative diagnosis.
Keywords
Cystic Hygroma
Case Report
##### Full Text
Introduction
Cystic Hygromas (lymphangiograms) are soft tissue tumor of disputed pathogensis originally reported by Redenbacker in 1828 and the name "Cystic Hygroma" was coined by Werner in 1834 (1). There are two histogenic theories: Firstly, cystic hygroma is due to congenital blockage of lymphatic system or secondly it is a true neoplasm (1). It mainly occurs in the head and neck region but can be present at other anatomical locations like thorax, shoulder, abdomen, pharynx and mediastinum. These cysts vary in size from a few millimeter to more than several centimeters in diameter (2, 3). They are classified into septate (multiloculated) or nonseptate (single cavity) cystic lesions.
Although common in childhood, occurrence in adults is quite uncommon and fewer than 150 reports of adult cervicofacial cystic hygromas have been reported in the literature (4, 5). It presents a diagnostic challenge in adults as compared to children and definitive diagnosis is usually based on post-operative histology (4).
Case Report
A 45-year-old male came to Surgery Department with complaint of swelling on right side of neck since 2 years. Patient was asymptomatic 2 years back after which he noticed a swelling in the neck; which was initially single and small and had gradually increased. Patient also complained of pain in right ear and difficulty in deglutition. Patient did not give any history of trauma or upper respiratory tract infection. On examination, there was a large swelling in the cervical region on the right side of neck measuring 15x9 cm. The swelling was non-tender, nonfluctuant, brilliantly translucent extending into both anterior and posterior triangle of the neck. The patient underwent a Computed Tomography (CT) scan of neck and thorax (plain + contrast) and ultrasonography (USG) which showed a large non enhancing hypodense multiloculated thin walled cystic lesion on the right side of the neck extending from submandibular to supraclavicular region involving anterior as well as posterior triangle with mass effect. A preoperative diagnosis of lymphatic malformation suspicious of cystic hygroma was made.
The patient underwent surgical exploration of the right side of neck and excision of the mass was done. The mass was sent for histopathological examination. The surgeons also performed total thyroidectomy as they were suspecting the lesion to be thyroid malignancy or metastasis. We received five ill-defined tissue masses measuring 8x4x3 cm, 7x5x4 cm, 5x4x2.5cm, 4x3x1.5 cm, 3x1.5x1cm. Externally they were irregular, soft in consistency, gray white and congested. After cutting open, they were large cystic masses with multiple small cysts within i.e. multiloculated and containing serous colourless fluid with smooth inner lining (Fig.1). The wall of the cyst was thin as well as thick without any solid component. We also received total thyroidectomy specimen, which showed normal lobes of thyroid on external and cut surface. Histopathology revealed large, irregular, dilated lymphatic channels lined by a single layer of benign flattened endothelium separated by collagenous stroma (Fig. 2). The stroma showed a dense lymphocytic infiltration with formation of lymphoid follicles (Fig. 3). Few cysts contain eosinophilic coagulum with occasional lymphocytes (Fig. 4). The diagnosis was given as Cystic Hygroma. Histopathology of both lobes of thyroid revealed normal thyroid follicles filled with colloid. There was no evidence of malignancy.
Fig.1: Gross photograph showing multiple variable size cystic cavities with smooth inner lining
Fig.2: The microphotograph shows large,irregular dilated lymphatic channels lined by attenuated bland endothelial cells with variable thick collagenous wall and lymphoid aggregates. (H&E, ×200)
Fig.3: The microphotograph shows collagenous wall showing lymphoid aggregates. (H&E, ×400)
Fig.4: The microphotograph shows lymphatic channel with eosinophilic coagulum containing lymphocytes. (H&E, ×400)
Discussion
Cystic hygroma is a benign congenital malformation of the lymphatic system. Almost 80% to 90% of cystic hygromas present itself before the end of second year of life. It is considered as an uncommon differential diagnosis of a progressively enlarging neck mass in adulthood (6). It can be misdiagnosed with other cystic neck swellings in adults like branchial cleft cyst, dermoid cyst, lipomas, hemangiomas, thyroid mass, tumours and metastatic disease (4, 6). Although after clinical and radiological examination, the above-mentioned diagnosis may look similar to cystic hygroma but histopathology plays a critical role in differentiating it from others.
Gow et al. (4) described in his case report the most common adult presentation as asymptomatic painless swelling with trauma and upper respiratory tract infection as possible precipitating factors for onset of cystic hygroma.In our case, the patient had asymptomatic painless swelling with no identifiable cause. Kalsotra et al. (7) in their case report described that massive enlargement of hygroma very rarely compresses major structures of neck such as larynx, trachea, oesophagus, branchial plexus and great vessels (4,7) and give rise to clinical symptoms. In our case, it was a large swelling causing mass effect with patient complaining of difficulty in deglutition and pain in the right ear.
Morsley SE (8) proposed in his case report that it is essential to evaluate intrathoracic extension of cystic hygroma, which might be seen in 10% of cases, and this is done by USG/CT/Magnetic Resonant Imaging (MRI) (9).Our patient also underwent CT scan/USG but swelling was restricted to supraclavicular region involving anterior and posterior triangle causing mass effect.
The treatment of choice for cystic hygroma is surgical excision without which there is a danger of infection and fistula formation. Saxena et al. (6) described that cystic hygroma might infiltrate in to adjacent structures which can lead to recurrence in 10-15% of cases, if complete excision is not done.
In addition, total thyroidectomy was carried out on the patient as surgeons suspected this cystic swelling as thyroid mass/metastatic lesion. The gross and histological examination of thyroid was normal with no evidence of malignancy. Thus, this case report emphasizes the utility and the role of histopathogical diagnosis of cervical cystic lesions in suspected cases of malignancy on clinical examination.
There are three histological subtypes of cystic hygroma- capillary lymphangioma (composed of small lymphatics), cavernous lymphangioma (Composed of large lymphatics), cystic lymphangioma (cystic hygroma -composed of large lymphatics spaces with collagen and smooth muscle) (10). Our case was of cystic lymphangioma.
Cystic hygroma is not commonly seen in clinical practice in adult patients. The paucity of literature and atypical presentation pose a diagnostic challenge due to rarity of this lesion. This case was unusual because a large cervical cystic hygroma presented denovo in an adult with no history of trauma or upper respiratory tract infection. Complete excision was done and follow up of the patient was uneventful.
The authors would also like to emphasize the importance of clinical, pathological and radiological correlation for the management of such rare lesions.
Acknowledgements
The authors declare that there is no conflict of interests.
##### References
Thawley SE, Panje WR, Batsake JG. Lindeberg RD. Comprehensive management of head and neck tumours. Saunders:New York;1999.
Veeraraghavan G, Denny C, Lingappa A. Cystic Hygroma in an Adult; a case report. Libyan J Med 2009;4(4):160-1.
Hamoir M, Plouin-Gaudon I, Rombaux P, Francois G, Cornu AS, Desuter G et al. Lymphatic malformation of head and neck, a reterospective review and a support for staging. Head Neck 2001; 23(4):326-37.
Gow L, Gulati R, Khan A, Mihaimeed F. Adult onset cystic hygroma: a case report and review of management. Grand rounds 2011;11:5-11.
Kraus J, Plzák J, Bruschini R, Renne G, Andrle J, Ansarin M, et al. Cystic Lymphangioma of the neck in adults: a report of three cases. Wien Klin Wochenschr. 2008;120(7-8):242-5.
Saxena P, Chandra D. Cystic Hygroma of neck in an adult female: Case Report & literature review. Int J Surg 2010;22(2).
Kalsotra P, Prusty A, Gupta A, Manhas M. Cervical cystic hygroma in an Adult. JK Science 2001;3(4):191-3.
Morley SE, Ramesar KC, Macleod DA. Cystic Hygroma in an adult: a case report. J R Coll Surg Edinb 1999;44(1):57-8.
Karakas O1, Karakas E, Boyaci FN, Yildizhan M, Demir S, Saglam MA et al. Cervicomediastinal Giant Cystic Hygroma: A Case Report. J Clin Med Res 2013;5(1):61-3.
Guner Ali, Aydin A, Celik F. Cystic Hygromas in Adults: Reports of Two cases. Medical Journal of Bakirkoy 2006;2(3):101-3
)
Volume 10, Issue 4 - Serial Number 4
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Pages 310-313
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Bhatia, V. , Taksande, R. and Sondankar, D. (2015). Cystic Hygroma in an Adult Male: A Diagnostic Challenge. Iranian Journal of Pathology, 10(4), 310-313. doi: 10.7508/ijp.2015.04.010
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MLA
Bhatia, V. , , Taksande, R. , and Sondankar, D. . "Cystic Hygroma in an Adult Male: A Diagnostic Challenge", Iranian Journal of Pathology, 10, 4, 2015, 310-313. doi: 10.7508/ijp.2015.04.010
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HARVARD
Bhatia, V., Taksande, R., Sondankar, D. (2015). 'Cystic Hygroma in an Adult Male: A Diagnostic Challenge', Iranian Journal of Pathology, 10(4), pp. 310-313. doi: 10.7508/ijp.2015.04.010
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CHICAGO
V. Bhatia , R. Taksande and D. Sondankar, "Cystic Hygroma in an Adult Male: A Diagnostic Challenge," Iranian Journal of Pathology, 10 4 (2015): 310-313, doi: 10.7508/ijp.2015.04.010
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VANCOUVER
Bhatia, V., Taksande, R., Sondankar, D. Cystic Hygroma in an Adult Male: A Diagnostic Challenge. Iranian Journal of Pathology, 2015; 10(4): 310-313. doi: 10.7508/ijp.2015.04.010
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113
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https://brilliant.org/wiki/does-sqrtatimes-sqrtb-sqrtab/
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Do Square Roots Always Multiply?
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Nazmus sakib,
Rishik Jain,
Eli Ross,
and
Calvin Lin
Jimin Khim
contributed
This is part of a series on common misconceptions.
True or False?
For all real numbers a and b,
a×b=ab.
Why some people say it's true:
By the rules of exponents, a×b=a21×b21=(ab)21=ab. This seems to work out for positive real numbers; e.g., if a=4 and b=9, then 4×9=2×3=6, just as 4×9=36=6.
Why some people say it's false:
There are other cases to consider besides positive real numbers. In those other cases, this "identity" may fail.
This statement is false. In particular, a×b=ab is true except when a and b are both negative.
Proof:
When x≥0, square roots are fairly straightforward, but keep in mind the ⋅ operator is a function that gives exactly one value. For example, 4=2. However, when x<0, there is no real number whose square is x. Thus, when x<0, we define x=i−x, where i=−1 (the imaginary unit). For example, −4=2i.
First, let's check what happens when exactly one of a or b is negative. For example, if a<0 and b>0, then a×b=i−a×b. Since −a>0 and b>0, we can multiply these square roots to obtain
i−a×b=i−a×b.
Then, since a×b<0, we have a×b=i−a×b, so the two sides are equal!
However, we also need to check what happens when both a and b are negative. In this case,
a×b=i−a×i−b=i2ab=−ab.
On the other hand, since a and b are both negative, ab>0. Thus,
a×b=−ab =ab,
so the statement does not hold true. □
Rebuttal: Do we really need to take the square root before multiplying the radicands?
Reply: Yes! The square root sign is essentially a fractional exponent, and we have to take care of exponents first according to order of operations.
Rebuttal: Why doesn't the "squaring both sides" of the equation argument work?
Reply: Keep in mind that x2=y2 does not imply that x=y. While (a×b)2=(ab)2, this does not mean that a×b must be equal to ab.
Rebuttal: My teacher told me that a×b=ab.
Reply: Well, hopefully your teacher mentioned that this only holds true under certain conditions! In particular, it is true when a and b are both positive, which is where you usually encounter such equations. It is also true if exactly one of a or b is negative, but it is false when both are negative, as shown above.
Rebuttal: Why must we say that −x=ix for x<0? Couldn't we set −x=−ix as well? For example, (−2i)2=−4.
Reply: No! Remember that for x≥0, x is defined as the non-negative number whose square is x. Similarly, the convention is that, for x<0, x is defined as −1 times the non-negative number whose square is −x.
Note that ⋅ is a function, so it's important that it is defined to have a single output for each input. Many complicated issues (which are tackled in the subject of complex analysis) would arise if we did not use these conventions!
Want to make sure you've got this concept down? Try these problems:
−2
2
2
4
−2×−1=?
The correct answer is: −2
-6
5
6
36
−4×−9=?
The correct answer is: -6
Before we invoke ab=ab, we must first pull out the negatives from underneath each of the radicals. This means
−4−9=(i4)(i9)=(2i)(3i)=6i2=6(−1)=−6
Note: When we evaluate square roots of non-positive numbers, we are choosing the branch of the square root to take. When involved in an operation, we have to use the same branch of the square root. For example, we do not say that i+i=0 or that i2=1, where one of the i is on the negative branch of the other.
True
False
True or False?
−9×−16=−12
The correct answer is: True
Let i=−1
−9=3i and −16=4i.
Therefore, −9×−16=3i×4i=12i2=−12
See Also
Square Roots
Complex Numbers
List of Common Misconceptions
Cite as:
Do Square Roots Always Multiply?.
Brilliant.org.
Retrieved
10:38, September 28, 2025,
from
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114
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https://physicspages.com/pdf/Relativity/Four-momentum%20conservation%20in%20the%20electron-photon%20collision.pdf
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FOUR-MOMENTUM CONSERVATION IN THE ELECTRON-PHOTON COLLISION Link to: physicspages home page.
To leave a comment or report an error, please use the auxiliary blog and include the title or URL of this post in your comment.
Post date: 22 June 2021.
We consider a photon colliding with a stationary electron (in the lab frame) and producing an extra electron-positron pair (so we have 2 elec-trons and a positron after the collision; the photon gets absorbed so it dis-appears). What is the minimum energy of the incoming photon (in the lab frame) to achieve this?
First, we can look at the problem in the centre of momentum (COM) frame. There, the electron travels at some speed v (the photon, of course, travels at v = 1) and after the collision all 3 particles are at rest. From conservation of momentum we must have (pe +pγ)2 = −(3m)2 (1) where pe is the momentum of the electron and pγ is the momentum of the photon before the collision, and m is the mass of an electron or positron.
Multiplying this out, we get p2 e +p2 γ +2pe ·pγ = −m2 −0+2pe ·pγ (2) = −9m2 (3) pe ·pγ = −4m2 (4) Since the scalar product on the LHS is invariant, it must also be true in the lab frame. There, p′ e = [m,0] since the electron is at rest, and p′ γ = [E,−E].
Therefore p′ e ·p′ γ = −mE (5) = pe ·pγ = −4m2 (6) E = 4m (7) This gives us the minimum energy of the photon in the lab frame.
1 FOUR-MOMENTUM CONSERVATION IN THE ELECTRON-PHOTON COLLISION 2 We can work out the wavelength of this photon by using E = h/λ where Planck’s constant has the value h = 1240 × 10−9 eV m and the electron mass is m = 0.51×106 eV. We get λ = h E (8) = 1240×10−9 4×0.51×106 (9) = 6.08×10−13m (10) This is well into the gamma ray region of the electromagnetic spectrum.
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https://support.microsoft.com/en-us/office/keeping-leading-zeros-and-large-numbers-1bf7b935-36e1-4985-842f-5dfa51f85fe7
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Excel
Enter and format data
Format data
Format data
Keeping leading zeros and large numbers
Keeping leading zeros and large numbers
Applies To
Excel for Microsoft 365 Excel for Microsoft 365 for Mac Excel 2024 Excel 2024 for Mac Excel 2021 Excel 2019 Excel 2016
Do you ever import or enter data in Excel that contains leading zeros, like 00123, or large numbers like 1234 5678 9087 6543? Examples of these are social security numbers, phone numbers, credit card numbers, product codes, account numbers, or postal codes. Excel automatically removes leading zeros, and converts large numbers to scientific notation, like 1.23E+15, in order to allow formulas and math operations to work on them. This article deals with how to keep your data in its original format, which Excel treats as text.
Set automatic data conversions
Important:This feature is available in Excel for Microsoft 365, Excel for Microsoft 365 for Mac, Excel 2024, Excel 2024 for Mac.
Use Excel's Automatic Data Conversions feature to change Excel's default behavior for the following automatic data conversions:
Removing leading zeros from numerical text and converting to a number.
Truncating numerical data to 15 digits of precision and converting to a number displayed in scientific notation.
Converting numerical data surrounding the letter "E" to scientific notation.
Converting a continuous string of letters and numbers to a date.
For more information, see Set automatic data conversions.
Convert numbers to text when you import text data
Use Excel's Get & Transform (Power Query) experience to format individual columns as text when you import data. In this case, we're importing a text file, but the data transformation steps are the same for data imported from other sources, such as XML, Web, JSON, etc.
Select the Data tab, then From Text/CSV next to the Get Data button. If you don’t see the Get Data button, go to New Query > From File > From Text and browse to your text file, then press Import.
Excel will load your data into a preview pane. Press Edit in the preview pane to load the Query Editor.
If any columns need to be converted to text, select the column to convert by clicking on the column header, then go to Home > Transform > Data Type > select Text.
Tip:You can select multiple columns with Ctrl+Left-Click.
4. Next, select Replace Current in the Change Column Type dialog, and Excel will convert the selected columns to text.
5. When you're done, select Close & Load, and Excel will return the query data to your worksheet.
If your data changes in the future, you can go to Data > Refresh, and Excel will automatically update your data, and apply your transformations for you.
Use a custom format to keep the leading zeros
If you want to resolve the issue just within the workbook because it's not used by other programs as a data source, you can use a custom or a special format to keep the leading zeros. This works for number codes that contain fewer than 16 digits. In addition, you can format your number codes with dashes or other punctuation marks. For example, to make a phone number more readable, you can add a dash between the international code, the country/region code, the area code, the prefix, and the last few numbers.
| Number code | Example | Custom number format |
---
| Socialsecurity | 012345678 | 000-00-0000 012-34-5678 |
| Phone | 0012345556789 | 00-0-000-000-0000 00-1-234-555-6789 |
| Postalcode | 00123 | 00000 00123 |
Steps
Select the cell or range of cells that you want to format.
Press Ctrl+1 to load the Format Cells dialog.
Select the Number tab, then in the Category list, select Custom and then, in the Type box, type the number format, such as 000-00-0000 for a social security number code, or 00000 for a five-digit postal code.
Tip:You can also select Special, and then select Zip Code, Zip Code + 4, Phone number, or Social Security Number.
Find more information about custom codes, see Create or delete a custom number format.
Note:This will not restore leading zeros that were removed prior to formatting. It will only affect numbers that are entered after the format is applied.
Use the TEXT function to apply a format
You can use an empty column next to your data and use the TEXT function to convert it to the format you want.
| Number code | Example (In cell A1) | TEXT function and new format |
---
| Socialsecurity | 012345678 | =TEXT(A1,"000-00-0000")012-34-5678 |
| Phone | 0012345556789 | =TEXT(A1,"00-0-000-000-0000")00-1-234-555-6789 |
| Postalcode | 00123 | =TEXT(A1,"00000")00123 |
Credit card numbers are rounded down
Excel has a maximum precision of 15 significant digits, which means that for any number containing 16 or more digits, such as a credit card number, any numbers past the 15th digit are rounded down to zero. In the case of number codes that are 16 digits or larger, you must use a text format. To do this, you can do one of two things:
Format the column as Text
Select your data range and press Ctrl+1 to launch the Format Cells dialog. On the Number tab, select Text.
Note:This will not change numbers that have already been entered. It will only affect numbers that are entered after the format is applied.
Use the apostrophe character
You can type an apostrophe (') in front of the number, and Excel will treat it as text.
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https://www.cliffsnotes.com/study-notes/19581195
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Exploring Osmosis: Observing Plasmolysis in Elodea Cells - CliffsNotes
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Exploring Osmosis: Observing Plasmolysis in Elodea Cells
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Griffith UniversityWe aren't endorsed by this school
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SET 1041SCG
Subject
Biology
Date
Sep 10, 2024
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Biology
1041SCG Biological Systems Trimester 1, 2020 Laboratory 2 Osmosis
1041SCG - Biological Systems (Trimester 1, 2020) Laboratory Manual Laboratory #2 - Osmosis Page 1 Assessment:Before you leave the laboratory, you will need to: q Answer the questions about plasmolysis in Elodea cells (page 3) and have this checked by you demonstrators(worth 2% of your final mark). q You will have to submit a lab report via Turnitin based on the results you will generate in this lab. See the section "Lab report" below for more detailed instructions on what to include. (worth 10% of your final mark). Objectives 1)Observe plasmolysis in Elodea. 2)Observe the effects of different concentrations of salt solutions on potato cores. 3)Infer the relationship between weight loss and rate of osmosis. 4)Use a graph to display data and draw conclusions. Background Osmosis is the process whereby water moves across a cell membrane by diffusion. Diffusion takes place when the molecules of a substance tend to move from areas of higher concentration to areas of lower concentration. Cells must tightly control the process of osmosis - otherwise they die. For example, if you place a red blood cell in pure (distilled) water, it will quickly take up water until it bursts. That is why plasma, the liquid portion of our blood, is made of water with proteins and dissolved salts, preventing the unnecessary gain of water by our blood cells. In plants, osmosis is just as important. Plants with too little water wilt. This happens when water moves out of the cells by osmosis. Without this water there is little pressure inside the cells and the plant can no longer support itself against the pull of gravity. However, after watering the plant, the cells become "re-inflated" with water and the plant again stands upright. The effect of water loss on plant cells is shown in the diagram below. Figure 1.A.Plant cell placed in pure water. This cell will become inflated because the water outside the cell is at a higher concentration than the water inside the cell. As water moves in by osmosis the vacuole fills up and presses out against the cell wall. B.The same cell in water equal to the concentration inside the cell. This cell has no overall gain or loss of water because whatever moves out will be replaced by water moving in. C.A cell placed in a salt solution. This cell will lose water as the water moves by diffusion from higher to lower concentration. The cytoplasm of this cell has shrunken in a process called plasmolysis.
1041SCG - Biological Systems (Trimester 1, 2020) Laboratory Manual Laboratory #2 - Osmosis Page 2 In this lab activity you will observe the effects of osmosis on plant cells. In the first part, you will observe the process of plasmolysis in a freshwater aquatic plant, Elodea. In the second part, you will conduct an experiment, measuring the weight of pieces of potato immersed in solutions of varying salt concentration to see how much water moves in and out of cells at different salt solutions. Materials Part 1 •Elodea leaves •microscope •microscope slide and coverslip •dropper •salt solution (5%) •distilled water Part 2 •potato •cork borer •knife or razor blade •beakers, baby food jars, or plastic cups •salt solutions (1%, 3%, 5%) •distilled water •balance •ruler Part A - Observing plasmolysis in Elodea leaves 1.Remove a single leaf from an Elodea plant and place it on a microscope slide. Use a dropper to place a drop of aquarium water on the leaf (note that this is freshwater). Cover the leaf with a coverslip and place the slide under a microscope. 2.Focus on the leaf under high power. Move the slide around to find an area where you can clearly see single cells. They will appear rectangular and look like green bricks in a wall. The illustration below shows how they look. The green ovals inside the cells are chloroplasts. After a while you may see them start to move around the cell. This is called "cytoplasmic streaming". 3.Leave the slide on the microscope. Using the dropper, place a drop of 5% salt solution on the right edge of the coverslip. Place a small piece of paper towel on the left edge of the coverslip. As the fresh water soaks into the paper towel it will draw the salt water under the coverslip. This is illustrated below.
1041SCG - Biological Systems (Trimester 1, 2020) Laboratory Manual Laboratory #2 - Osmosis Page 3 4.Observe the same cells you were looking at before adding the salt solution. You will notice that the cell contents have shrunken down, leaving a space between the cytoplasm and the cell wall. This is called plasmolysis. Make a sketch of the appearance of a few of the cells in the space 5.Repeat Step 3 using distilled water. It may take a number of tries to remove all of the salt water. When successful, the cells will appear as normal. 6.Clean up the slide and coverslip and return the microscope to its proper place. Questions 1.In which direction was the water moving during plasmolysis? Into or out of the cell? 2.What prevents the Elodea cells from completely collapsing? (use this section to take notes during your lab)
1041SCG - Biological Systems (Trimester 1, 2020) Laboratory Manual Laboratory #2 - Osmosis Page 4 Part B - An experiment to measure osmosis in potato cells You will be conducting an experiment to measure osmosis in potato cells. By immersing potatoes in solutions of varying salt concentration and comparing their weight before and after 24h in solution, you will be able to determine if there are any differences in weight between the different experimental groups. Note that while the methods for this experiment are described below, you should think about these details before and consider a) why you are doing this experiment (in other words, what are you investigating, really?), and b) whether this is the best way to do this (there are some prompts in italics below to help you think about the protocol). Your lab demonstrators will be happy to discuss the experimental plan with you in the lab. Methods - Day 1 1.Come up with a unique group name that you and your partner(s) will remember. 2.Label 4 containers with your group name and the following: "distilled water", "1% salt", "3% salt", and "5% salt". 3.Using the cork borer, make 12 cylinders from your potato. Trim them with a knife until they are 3 cm long (do you think it is important for them to be the same size? Why or why not?) Caution: be very careful with the cork borer and knife. Always cut away from yourself. Make sure there is no peel left on the core (why is this important?). If a cork borer is not available, you may also cut strips of potato using a knife. 4.Place three cores in each of the containers and cover them temporarily (why three cores? Why not more, or less?). When you are ready, remove the cores, blot them briefly on a piece of paper towel (think about how you will blot them ... what is especially important about this step?), and find the mass of each group of three using the balance. Mass all three of them together, not separately. You should mass them to the nearest 0.01 g. Record your data (think about where you should record this data). 5.Immediately return the cores to the correct container and cover them with the correct solution. Place a lid on each container and set them aside for 24h (does it matter that you do exactly 24h? What would happen if you did less, or more? What is especially important about this step?). Day 2 6.After 24 h, remove each set of three cores from their containers. Briefly blot them with a paper towel to remove excess water (think about the question on blotting earlier ... what was important about it?). Quickly find the mass of each group of 3 and record the mass (what should you be recording these new data on?). 7.Make observations of the texture, colour and flexibility of the cores. Record these observations. 8.Calculate and record the change in mass (think about how you should express this change in mass ... imagine you had a 1g potato that lost 0.1g and a 10g potato that lost 1g - which one "lost the most"?). 9.Make a graph of your data (remember "DRY-MIX": you should plot the Dependent variable, the Response ("the variable being measured") on the Y-axis and the Manipulated variable ("the variable that is changed or controlled"), the Independent variable on the X-axis ... so which one was the dependent variable in this experiment? Also, what % of salt is distilled water?.
1041SCG - Biological Systems (Trimester 1, 2020) Laboratory Manual Laboratory #2 - Osmosis Page 5 Your lab report IMPORTANT:Note that while you will perform the experiments in the lab as a group, this lab report is an individual assessment. You can (and are encouraged to) discuss your assignment with your colleagues, but your assignment must be your own work, and not copied from each other. The latter would be plagiarism. There are significant penalties for plagiarism. Your lab report will be formatted as a typical scientific paper, but we will provide some of the sections for you. See below to see what sections we will provide and what section you will need to complete: •Title(you) •Author(you) •Abstract(you) •Introduction (we will write this) •Materials and methods (we will write this) •Results(you): your results sections must include: - a brief written description of the most important data trends in the tables and figures, and a "call out" to those tables and figures in your text - ONE table that presents both your group's data and a meaningful summary of the whole class data (the latter will be provided on blackboard) [a meaningful summary means that you should use descriptive statistics on the class data, as we discussed in the week 2 workshop]; - ONE carefully selected figure(ie a graph) that visually summarises the data; and - a one sentence explanation for why your chose that particular style of graph as a note in the title of the graph. Note that you will lose marks for having more than ONE table or ONE figure (although the latter can be a multi-panel figure) - you need to be able to meaningfully summarise data! •Discussion(you): for the discussion, you only need a brief paragraph that explains what the data actually tells you, AND you need to work into your discussion an answer to the following 5 questions: 1. Which solution caused the potato cores to gain the most weight? 2. Was water moving into or out of the potato cells in this case? How can you tell? 3. Which solution caused the potato cores to lose the most weight? Which way was the water moving in this case? 4. What is the natural % salt concentration inside of a potato cell, as determined from your experiment? How can you tell? 5. Why do you think the workers at the grocery store spray fresh water on the vegetables on display? Explain this in terms of what happens in "osmosis". •Conclusions(you) •References(you, although we will provide some too from the intro and the M&M sections that you will need to include) Your lab report will be worth 10% of your final mark. More information about how to write your lab report will be covered in workshops 2 and 3, and in this lab of course. Laboratory practical adapted from the University of Texas Southwestern osmosis demonstration lab,
Page 1 of 6
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Finding the gradient with directional derivatives
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The problem I am working on is this:
Find ∇f(a,b)∇f(a,b) for the differential function f(x,y)f(x,y) given the directional derivatives:
D(i+j)/2 f(a,b)=3 2–√D(i+j)/2 f(a,b)=3 2 , D(3 i−4 j)/5 f(a,b)=5 D(3 i−4 j)/5 f(a,b)=5
I've done some reading on Directional Derivatives to see if I could then solve them, but I can't make head or tails of this problem. Could someone guide me through this or point me in the right direction?
Thanks in advance.
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asked Mar 23, 2015 at 21:33
Kommander KittenKommander Kitten
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Perhaps the formula D a f=(∇f)⋅a D a f=(∇f)⋅a might help.user137731 –user137731 2015-03-23 21:34:54 +00:00 Commented Mar 23, 2015 at 21:34
The thing that is confusing me is the (i+j)/2(i+j)/2 part. Where do I use that? Oh sorry, didn't see the a a. Okay... so I dot product the function with what exactly?Kommander Kitten –Kommander Kitten 2015-03-23 21:36:32 +00:00 Commented Mar 23, 2015 at 21:36
That's your a a in my formula.user137731 –user137731 2015-03-23 21:36:58 +00:00 Commented Mar 23, 2015 at 21:36
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Hints:∇f(a.b)=(D i f(a,b),D j f(a,b))∇f(a.b)=(D i f(a,b),D j f(a,b)), write i i and j j in terms of (i+j)/2(i+j)/2 and (3 i−4 j)/5(3 i−4 j)/5, and:
i=α⋅i+j 2+β⋅3 i−4 j 5⟹D i f(a,b)=α⋅(3 2–√)+β⋅(5),i=α⋅i+j 2+β⋅3 i−4 j 5⟹D i f(a,b)=α⋅(3 2)+β⋅(5),
and similarly for the other one.
I'll do the first one for you. Write:
(1,0)=α(1 2,1 2)+β(3 5,−4 5)⟹{α 2+3 β 5=1 α 2−4 β 5=0,(1,0)=α(1 2,1 2)+β(3 5,−4 5)⟹{α 2+3 β 5=1 α 2−4 β 5=0,
so that β=5/7 β=5/7 and α=4/7 α=4/7. Hence:
D i f(a,b)=D 4 7(i+j 2)+5 7(3 i−4 j 5)f(a,b)=4 7 D i+j 2 f(a,b)+5 7 D 3 i−4 j 5 f(a,b).D i f(a,b)=D 4 7(i+j 2)+5 7(3 i−4 j 5)f(a,b)=4 7 D i+j 2 f(a,b)+5 7 D 3 i−4 j 5 f(a,b).
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edited Mar 23, 2015 at 22:00
answered Mar 23, 2015 at 21:37
Ivo TerekIvo Terek
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I don't understand what you mean by "write i i and j j in terms of (i+j)/2(i+j)/2 and (3 i−4 j)/5(3 i−4 j)/5 Kommander Kitten –Kommander Kitten 2015-03-23 21:48:18 +00:00 Commented Mar 23, 2015 at 21:48
If I told you to write (1,0)(1,0) as a combination of (1/2,1/2)(1/2,1/2) and (3/5,−4/5)(3/5,−4/5) would you understand? There is more than one possible combination, but the final outcome will be the same.Ivo Terek –Ivo Terek 2015-03-23 21:49:36 +00:00 Commented Mar 23, 2015 at 21:49
A bit clearer, but still unsure of myself. Could you explain further? I feel like I'm missing something fairly obvious.Kommander Kitten –Kommander Kitten 2015-03-23 21:54:20 +00:00 Commented Mar 23, 2015 at 21:54
I'll add an edit. By the way, I just noticed that we will have a unique combination, I looked too fast.Ivo Terek –Ivo Terek 2015-03-23 21:56:06 +00:00 Commented Mar 23, 2015 at 21:56
@Kommander see if you can do the second one Ivo Terek –Ivo Terek 2015-03-23 22:01:17 +00:00 Commented Mar 23, 2015 at 22:01
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https://engineering.purdue.edu/~wassgren/teaching/ME30800/NotesAndReading/CompressibleFlow_Introduction_LectureNotes.pdf
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Compressible Flow – Introduction Compressible Flow – Introduction Thermodynamics Review Some Definitions u := specific internal energy h := specific enthalpy = u + p/r cv := specific heat at constant volume cp := specific heat at constant pressure k := specific heat ratio = cp/cv (k = 1.4 for air) R := gas constant (Rair = 287 J/(kg.K) = 53.3 (ft.lbf)/(lbm.degR)) adiabatic := no heat transfer reversible := system and surroundings can be restored exactly to their initial states isentropic := no change in entropy, i.e., Ds = 0 adiabatic and reversible => isentropic T-s diagrams are frequently used to illustrate processes Ideal Gas Relations (use absolute pressure and temperature; R is the gas constant, not the universal gas constant)) and Perfect Gas Relations perfect gas = ideal gas with constant specific heats (cv,air = 718 J/(kg.K) = 133 (ft.lbf)/(lbm.degR)) (cp,air = 1005 J/(kg.K) = 187 (ft.lbf)/(lbm.degR)) isentropic process involving a perfect gas: , , !p = ρRT !
u2 −u1 = cv T ( )dT T 1 T2 ∫ !
h 2 −h 1 = cp T ( )dT T 1 T2 ∫ !cp = cv + R !
cp = kR k −1 !
cv = R k −1 !
ds = cv T ( )dT T −Rdρ ρ = cp T ( )dT T −Rdp p !u2 −u1 = cv T 2 −T 1 ( ) !h 2 −h 1 = cp T 2 −T 1 ( ) !
s2 −s1 = cv lnT 2 T 1 −R ρ2 ρ1 = cplnT 2 T 1 −Rln p2 p1 !
T 2 T 1 = ρ2 ρ1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ k−1 !
p2 p1 = T 2 T 1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ k k−1 !
p2 p1 = ρ2 ρ1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ k 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 0 200 400 600 800 1000 1200 1400 1600 specific heats (cp, cv) [kJ/(kg.K)] temperature [K] cp cv air Compressible Flow – Introduction 1st Law for steady flow of a gas with one inlet and one outlet and uniform inlet and outlet properties From COM: . Combining with 1st Law: For an adiabatic (qinto = 0) flow with no work other than pressure work (won = 0): or For a perfect gas (Dh = cpDT): ! d dt eρdV CV ∫ =0! steady ( ) ! " # $ # + h+ 1 2V 2 + gz negligible!compared to!other!terms!since its!a!gas % ⎛ ⎝ ⎜ ⎜ ⎜ ⎜ ⎞ ⎠ ⎟ ⎟ ⎟ ⎟ ρurel ⋅dA ( ) =d & m ! " # $ # CS ∫ = & Qinto!CV + & Won!CV ! !
m= !
min = !
mout !
Δ h+ 1 2V 2 ( )= qinto!CV + won!CV !
Δ h+ 1 2V 2 ( )= 0 !h+ 1 2V 2 = constant !cpT + 1 2V 2 = constant
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Inequalities and equations - creating sets from quadratic equations.
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My question is just making sure that my working is correct and that I understand properly (self teaching, can get confused...)
So question :
Find the set of values for which
x2−4x−12<0
I can factor this to
(x−2)(x+6)<0
So this means that x=2 or x=−6
This then get expressed as
−2<x<6
I can't really explain why though...
Is this because x−2=0 so x>−2 ...?
If I create a graph of y=(x−2)(x+6) the range of values between the two x intercepts are −6 to 2....
So from that I would expect the equalities to be
−6<x<2
not
−2<x<6
I'm not sure what I'm missing here?
Thanks
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asked Oct 16, 2014 at 11:29
baxxbaxx
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Your factorisation should be (x+2)(x−6)
– Mark Bennet
Commented
Oct 16, 2014 at 11:31
dear me how stupid... so then it all makes sense, there's no confusion after that really... should I delete this post? I'm not too sure on the convention there... thanks though Mark.
– baxx
Commented
Oct 16, 2014 at 11:34
Let x1 and x2 be the two roots that satisfy x2−4x−12=0. Assume x1<x2. What you should do is then divide the real number line in three intervals, namely (−∞,x1), (x1,x2) and (x2,+∞) and find out for each interval if the expression x2−4x−12 is positive or negative.
– Ritz
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x2−4x−12=(x+2)(x−6)
is a parabola that opens up (a "smiling" one) or a convex upwards one, and it vanishes at x=−2,6 . If you make even a raw drawing of this, you will clearly see that
x2−4x−12<0⟺−2<x<6
The above's already a complete, mathematical formal proof if you've already studied quadratic functions and parabolas from a geometric/algebraic point of view.
One can also prove the inequality in a purely algebraic way:
(x+2)(x−6)<0⟺⎧⎩⎨x+2>0andx−6<0orx+2<0andx−6>0
The second possibility above is impossible as this would imply x>−2∧x>6 , and there's no such a thing, so it must be the first one, which is the same we got in the first part above.
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answered Oct 16, 2014 at 11:36
TimbucTimbuc
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Thanks for that Timbuc! I messed up the factoring and then spent ages trying to reinvent maths for some stupid reason! :/
– baxx
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Oct 16, 2014 at 11:40
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Circular Motion on Banked Surfaces With Friction
This topic is part of the HSC Physics course under the section Circular Motion.
HSC Physics Syllabus
analyse the forces acting on an object executing uniform circular motion in a variety of situations, for example:
- cars moving around horizontal circular bends
- a mass on a string
- objects on banked tracks
Circular Motion on Banked Surfaces WITH Friction
This video covers the forces involved in a car undergoing uniform circular motion on a banked surface with friction. The video walks through free body diagrams and relevant derivation of equations for calculating the centripetal force and velocity.
Mastering the concepts of circular motion, especially on banked surfaces with friction, is crucial for HSC Physics students. This concise guide dives into the fundamental principles of this topic, ensuring clarity, retention, and academic success.
The Basics: Circular Motion
When an object moves in a circular path, it is said to be in circular motion. A crucial concept here is centripetal force - the net force causing the object to move in a circular path, acting towards the centre of the circle.
Banked Surfaces: Why the Incline?
Banked surfaces, often seen in racetracks or curved roads, are inclined planes designed to assist objects (like cars) in maintaining circular motion without skidding. The banking angle helps balance the necessary forces, reducing the reliance on friction.
Role of Friction in Circular Motion on Banked Surfaces
In a real-world scenario, no surface is entirely frictionless. Friction plays a critical role:
Static Friction: Prevents the car from skidding and helps in providing the necessary centripetal force.
Kinetic Friction: Acts if the car starts skidding.
In cases where an objects were to remain at the same position on the banked surface while undergoing uniform circular motion, only static friction applies.
The frictional force, especially static friction, assists the car in navigating the curve without losing grip. Too little friction, and the car would skid outwards; too much, and it would skid inwards.
When a mass is moving at its ideal velocity in uniform circular motion on a bank surface, no assistance from static friction is required. If the mass' speed increases from the ideal velocity, static friction will start to act down the banked surface to increase the centripetal force, in order to keep the mass in the same position on the banked surface. Conversely, if the mass' speed decreases from the ideal velocity, static friction will start to act up the banked surface to decrease the centripetal force and keep the mass in the same position on the banked surface.
Balancing the Forces
For a car moving on a banked surface with friction, several forces come into play:
Gravitational Force: Acts vertically downward.
Normal Force: Perpendicular to the surface.
Frictional Force: Can act towards or away from the centre, depending on the circumstances.
Resultant Centripetal Force: This is the net force required for circular motion.
For a car to move safely around a banked curve, the horizontal component of the normal force combined with the frictional force (if needed) should equal the necessary centripetal force.
When the car is moving faster than the ideal speed (when no friction is needed), static friction acts down the banked surface to prevent the car from skidding up the surface.
In this case, the horizontal components of static friction and normal force contribute to the centripetal force.
When the car is movingslowerthan the ideal speed, static friction acts up the banked surface to prevent the car from skidding down the surface.
In this case, the horizontal component of static friction acts away from the direction of centripetal force, resulting in a smaller magnitude of centripetal force. A smaller centripetal force permits the car to move on the banked surface in uniform circular motion with a slower speed.
BACK TO MODULE 5: ADVANCED MECHANICS
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Computing discrete residues of rational functions
Carlos E. Arreche, Department of Mathematical Sciences, The University of Texas at Dallas, USA, arreche@utdallas.edu
Hari Sitaula, Department of Mathematical Sciences, Montana Technological University, USA, Hsitaula@mtech.edu
DOI:
ISSAC '24: International Symposium on Symbolic and Algebraic Computation, Raleigh, NC, USA, July 2024
In 2012 Chen and Singer introduced the notion of discrete residues for rational functions as a complete obstruction to rational summability. More explicitly, for a given rational function f(x), there exists a rational function g(x) such that f(x) = g(x + 1) − g(x) if and only if every discrete residue of f(x) is zero. Discrete residues have many important further applications beyond summability: to creative telescoping problems, thence to the determination of (differential-)algebraic relations among hypergeometric sequences, and subsequently to the computation of (differential) Galois groups of difference equations. However, the discrete residues of a rational function are defined in terms of its complete partial fraction decomposition, which makes their direct computation impractical due to the high complexity of completely factoring arbitrary denominator polynomials into linear factors. We develop a factorization-free algorithm to compute discrete residues of rational functions, relying only on gcd computations and linear algebra.
CCS Concepts: • Computing methodologies → Algebraic algorithms; Discrete calculus algorithms;
Keywords: discrete residues, creative telescoping, rational summability, difference Galois theory, factorization-free algorithm
ACM Reference Format:
Carlos E. Arreche and Hari Sitaula. 2024. Computing discrete residues of rational functions. In International Symposium on Symbolic and Algebraic Computation (ISSAC '24), July 16--19, 2024, Raleigh, NC, USA. ACM, New York, NY, USA 9 Pages.
1 INTRODUCTION
Let KK be a field of characteristic zero, and consider the field K(x)K(x) of rational functions in an indeterminate x with coefficients in KK. First formulated in [Abr71], the rational summation problem asks, for a given f(x)∈K(x)f(x)∈K(x), to construct g(x),h(x)∈K(x)g(x),h(x)∈K(x) such that
f(x)=g(x+1)−g(x)+h(x)f(x)=g(x+1)−g(x)+h(x)
(1.1)
and the degree of the denominator of h(x) is as small as possible. Such an h(x) is called a reduced form of f(x). The rational summation problem has a long and illustrious history [Abr71, Abr75, Abr95, Kar81, Moe77, MS95, Pau95, Pir95, Pol08, PS95]. It is clear that the problem admits a solution (by the well-ordering principle), and that such a solution is not unique, because for any solution (g(x), h(x)) we obtain another solution (g(x) − h(x), h(x + 1)) since the degree of the denominator of h(x + 1) is the same as that of h(x). In comparing the approaches in op. cit., one can then ask for the denominator of g(x) to be also as small as possible, and/or to compute some (any) solution (g(x), h(x)) to (1.1) as efficiently as possible. We refer to the introduction of [PS95] for a concise summary and comparison between most of these different approaches.
Every algorithm for solving the rational summation problem also addresses, as a byproduct, the rational summability problem of deciding, for a given f(x)∈K(x)f(x)∈K(x), whether (just yes/no) there exists g(x)∈K(x)g(x)∈K(x) such that f(x) = g(x + 1) − g(x), in which case we say f(x) is rationally summable. There are various algorithms for addressing this simpler question, designed to forego the usually expensive and often irrelevant computation of the certificate g(x), which are presented and discussed for example in [BCCL10, CHKL15, CS12, GGSZ03, GHLZ22, HW15, Mat00] and the references therein.
We center our attention on the approach to rational summability proposed in [CS12]. The discrete residues of f(x)∈K(x)f(x)∈K(x) are constants defined in terms of the complete partial fraction decomposition of f(x), and have the obstruction-theoretic property that they are all zero if and only if f(x) is rationally summable. Computing these discrete residues directly from their definition is impractical due to the high computational cost, or possible infeasibility, of factoring the denominator of f(x) into linear factors. We propose here algorithms for computing these discrete residues relying only on gcd computations and solving systems of linear equations in KK. To be clear, the discrete residue data of an arbitrary f(x) are in general algebraic over KK. We submit that it would be perverse to avoid expensive factorizations throughout the algorithm, only to demand them at the very end! Inspired by [BS93, Thm. 1] our output consists of pairs of polynomials with coefficients in KK: one whose roots are the places where f(x) has non-zero discrete residues, the other whose evaluation at each such root gives the corresponding discrete residue (see §3 for a more detailed description). Of course, any user who wishes to actually see the discrete residue data of f may use the KK-polynomials produced by our algorithms to compute them explicitly to their heart's content and at their own risk.
Let us now describe our general strategy for computing discrete residues (cf. Algorithm 4). We apply iteratively Hermite reduction to f(x) in order to reduce to the special case where the denominator of f(x) is squarefree. Then we compute a reduced form ˉf(x)¯f(x) of f(x) whose denominator is both squarefree and shift-free, so that the discrete residues of f(x) are the classical first-order residues of ˉf(x)¯f(x). The factorization-free computation of the latter is finally achieved by [Tra76, Lem. 5.1].
Our proposed algorithms to compute discrete residues are obtained by combining in novel ways many old ingredients. Indeed, Hermite reduction is very old [Her72, Ost45], and its iteration in Algorithm 1 is already suggested in [Hor71, §5] for computing iterated integrals of rational functions. And yet, we have not seen this approach being more widely used in the literature, and it seems to us a good trick to have to hand. Indeed, we wonder whether it could provide a reasonable alternative, at least in some cases and for some purposes, to the algorithm in [BS93] for symbolically computing complete partial fraction decompositions over the field of definition. Having thus reduced via Algorithm 1 to the case where f(x) has squarefree denominator, many of the varied earlier approaches to the summation and summability problems seem to accidentally collide into essentially the same procedure when restricted to this simpler situation. In this sense, our own reduction procedure described in §5 strikes us as eerily similar to the one presented in [GGSZ03, §5] over 20 years ago — that ours may look simpler is a direct consequence of its being restricted to a simpler class of inputs. The simplicity of our approach allows us to exercise a great deal of control over the form of the outputs, in ways that are particularly useful for developing some extensions of our basic procedures, elaborated in §7. It is not obvious to us (but it would be interesting to see) how the same goals might be better accomplished differently, say by combining the reduction of [GGSZ03] with the symbolic complete partial fraction decomposition algorithm of [BS93].
Our interest in computing discrete residues is motivated by the following variant of the summability problem, which often arises as a subproblem in algorithms for computing (differential) Galois groups associated with (shift) difference equations [Arr17, Hen98, HS08, vdPS97]. Given several f1(x),…,fn(x)∈K(x)f1(x),…,fn(x)∈K(x), compute (or decide non-existence) of 0≠v=(v1,…,vn)∈Kn0≠v=(v1,…,vn)∈Kn such that
v1f1(x)+⋯+vnfn(x)=gv(x+1)−gv(x)v1f1(x)+⋯+vnfn(x)=gv(x+1)−gv(x)
(1.2)
for some gv(x)∈K(x)gv(x)∈K(x). Even if one wishes to compute the certificate gv(x) explicitly, it seems wasteful to perform a rational summation algorithm n times for each fi(x) separately to produce (gi(x), hi(x)) as in (1.1) as an intermediate step, because there is no guarantee that hv(x) = ∑ivihi(x) has smallest possible denominator, so one may need to perform the algorithm an (n+1)st(n+1)st time to hv(x) in order to decide summability (this last step may be avoidable by making a more careful sequence of interrelated reduction choices; see e.g. [CHKL15, §5]). These inefficiencies are exacerbated in the more general context of creative telescoping problems [WZ92, Zei90, Zei91], where the unknown vi∈Kvi∈K are replaced with unknown linear differential operators Li∈K[ddx]Li∈K[ddx] (cf. [HS08, Prop. 3.1]). We refer to [Che19] for a succinct and illuminating discussion of the history and computational aspects of creative telescoping problems, and in particular how the “fourth generation” reduction-based approaches bypass the computation of certificates, as our motivating problem (1.2) illustrates.
Our approach based on discrete residues makes it very straightforward how to accommodate several fi(x) simultaneously as in (1.2), which adaptation is less obvious (to us) how to carry out efficiently using other reduction methods. On the other hand, we share in the reader's disappointment that we offer hardly any theoretical or experimental evidence supporting the efficiency of our approach in contrast with other possible alternatives. In fact, we expect that our approach will not be universally more efficient than some future adaptation of [GGSZ03, §5] or [CHKL15, §5] (for example) to the situation of (1.2), but rather that it can be a useful complement to them. We also expect that the conceptual simplicity of our approach will be useful in developing analogues to other related (but more technically challenging) contexts beyond the shift case, such as q-difference equations, Mahler difference equations, and elliptic difference equations, for which the corresponding notions of discrete residues have also been developed respectively in [CS12], [AZ22, AZ23], and [HS21].
2 PRELIMINARIES
2.1 Basic notation and conventions
We denote by NN the set of strictly positive integers, and by KK a computable field of characteristic zero in which it is feasible to compute integer solutions to arbitrary polynomial equations with coefficients in KK. Such a field is termed canonical in [Abr71]. We denote by ¯K¯¯¯¯K a fixed algebraic closure of KK. We do not assume KK is algebraically closed, and we will only refer to ¯K¯¯¯¯K in proofs or for defining theoretical notions, never for computations.
We work in the field K(x)K(x) of rational functions in a formal (transcendental) indeterminate x. For f(x)∈K(x)f(x)∈K(x), we define
Note that σ is a KK-linear field automorphism of K(x)K(x) and Δ = σ − id is only a KK-linear map with ker(Δ)=Kker(Δ)=K. We often suppress the functional notation and write f instead of f(x), σ(f) instead of f(x + 1), etc., when no confusion is likely to arise.
A proper rational function is either 0 or else has numerator of strictly smaller degree than that of the denominator. We assume implicitly throughout that rational functions are normalized to have monic denominator. Even when our rational functions are obtained as (intermediate) outputs of some procedures, we will take care to arrange things so that this normalization always holds. In particular, we also assume that the outputs of gcd and lcm procedures are also always normalized to be monic. An unadorned gcd or lcm or deg means that it is with respect to x. On the few occasions where we need a gcd with respect to a different variable z, we will write gcdz. We write ddxddx (resp., ddzddz) for the usual derivation operator with respect to x (resp., with respect to z).
2.2 Partial fraction decompositions
A polynomial b∈K(x)b∈K(x) is squarefree if b ≠ 0 and gcd(b,ddxb)=1gcd(b,ddxb)=1. Consider a proper rational function f=ab∈K(x)f=ab∈K(x), with deg(b) ≥ 1 and gcd(a, b) = 1. Suppose further that b is squarefree, and that we are given a set b1,…,bn∈K[x]b1,…,bn∈K[x] of monic non-constant polynomials such that ∏ni=1bi=b∏ni=1bi=b. Then necessarily gcd(bi, bj) = 1 whenever i ≠ j, and there exist unique non-zero polynomials a1,…,an∈K[x]a1,…,an∈K[x] with deg(ai) < deg(bi) for each i = 1, …, n such that f=∑ni=1aibif=∑ni=1aibi. In this situation, we denote
ParFrac(f;b1,…,bn):=(a1,…,an).ParFrac(f;b1,…,bn):=(a1,…,an).
(2.1)
We emphasize that the computation of partial fraction decompositions (2.1) can be done very efficiently [KT77], provided that the denominator b of f has already been factored into pairwise relatively prime factors bi, which need not be irreducible in K[x]K[x]. One can similarly carry out such partial fraction decompositions more generally for pre-factored denominators b that are not necessarily squarefree. But here we only need to compute partial fraction decompositions for pre-factored squarefree denominators, in which case the notation (2.1) is conveniently light.
2.3 Summability and dispersion
We say f∈K(x)f∈K(x) is (rationally) summable if there exists g∈K(x)g∈K(x) such that f = Δ(g). For a non-constant polynomial b∈K[x]b∈K[x], we follow the original [Abr71] in defining the dispersion of b
For a reduced rational function f=ab∈K(x)f=ab∈K(x) with gcd(a, b) = 1 and b∉Kb∉K, the polar dispersion pdisp(f) ≔ disp(b).
We denote by ¯K/Z¯¯¯¯K/Z the set of orbits for the action of the additive group ZZ on ¯K¯¯¯¯K. For α∈¯Kα∈¯¯¯¯K, we denote
the unique orbit in ¯K/Z¯¯¯¯K/Z containing α. We will often simply write ω∈¯K/Zω∈¯¯¯¯K/Z whenever there is no need to reference a specific α ∈ ω.
3 DISCRETE RESIDUES AND SUMMABILITY
Definition 3.1 ([CS12, Def. 2.3]) Let f∈K(x)f∈K(x), and consider the complete partial fraction decomposition
f=p+∑k∈N∑α∈¯Kck(α)(x−α)k,f=p+∑k∈N∑α∈¯¯¯¯Kck(α)(x−α)k,
(3.1)
where p∈K[x]p∈K[x] and all but finitely many of the ck(α)∈¯Kck(α)∈¯¯¯¯K are zero for k∈Nk∈N and α∈¯Kα∈¯¯¯¯K. We define the discrete residue of f of order k∈Nk∈N at the orbit ω∈¯K/Zω∈¯¯¯¯K/Z to be
dres(f,ω,k):=∑α∈ωck(α).dres(f,ω,k):=∑α∈ωck(α).
(3.2)
The relevance of discrete residues to the study of rational summability is captured by the following result.
Proposition 3.2 ([CS12, Prop. 2.5])f∈K(x)f∈K(x) is rationally summable if and only if dres(f, ω, k) = 0 for every ω∈¯Kω∈¯¯¯¯K and k∈Nk∈N.
As pointed out in [CS12, Rem. 2.6], the above Proposition 3.2 recasts in terms of discrete residues a well-known rational summability criterion that reverberates throughout the literature, for example in [Abr95, p. 305], [Mat00, Thm. 10], [GGSZ03, Thm. 11], [AC05, Cor. 1], and [HW15, Prop. 3.4]. All of these rely in some form or another on the following fundamental result of Abramov, that already gives an important obstruction to summability.
Proposition 3.3 ([Abr71, Prop. 3]) If a proper rational function 0≠f∈K(x)0≠f∈K(x) is rationally summable then pdisp(f) > 0.
Discrete residues are also intimately related to the computation of reduced forms for f, in the sense that, as discussed in [CS12, §2.4], every reduced form h of f as in (1.1) has the form
h=∑k∈N∑ω∈¯K/Zdres(f,ω,k)(x−αω)kh=∑k∈N∑ω∈¯¯¯¯K/Zdres(f,ω,k)(x−αω)k
(3.3)
for some arbitrary choice of representatives αω ∈ ω. Conversely, for any h of the form (3.3), Proposition 3.2 immediately yields that f − h is rationally summable. An equivalent characterization for h∈K(x)h∈K(x) to be a reduced form is for it to have polar dispersion 0 (see [Abr75, Props. 4 & 6]). By (3.3), knowing the dres(f, ω, k) is thus “the same” as knowing some/all reduced forms h of f. But discrete residues still serve as a very useful organizing principle and technical tool, for both theoretical and practical computations.
For a given f∈K(x)f∈K(x), our goal is to compute polynomials Bk(x),Dk(x)∈K[x]Bk(x),Dk(x)∈K[x] for each k∈Nk∈N such that (Bk, Dk) = (1, 0) if and only if dres(f, ω, k) = 0 for every ω∈¯Kω∈¯¯¯¯K (which holds for all but finitely many k∈Nk∈N) and, for the remaining k∈Nk∈N, we have 0 ≤ deg(Dk) < deg(Bk) and Bk is squarefree with disp(Bk) = 0. These polynomials will have the property that the set of roots α∈¯Kα∈¯¯¯¯K of Bk is a complete and irredundant set of representatives for all the orbits ω∈¯K/Zω∈¯¯¯¯K/Z such that dres(f, ω, k) ≠ 0, and for each such root α∈¯Kα∈¯¯¯¯K such that Bk(α) = 0, we have dres(f, ω(α), k) = Dk(α).
4 ITERATED HERMITE REDUCTION
It is immediate that the polynomial part of f∈K(x)f∈K(x) in (3.1) is irrelevant, both for the study of summability as well as for the computation of discrete residues. So in this section we restrict our attention to proper rational functions f∈K(x)f∈K(x).
Our first task is to reduce to the case where f has squarefree denominator. In this section we describe how to compute fk∈K(x)fk∈K(x) for k∈Nk∈N such that, relative to the theoretical partial fraction decomposition (3.1) of f, we have
fk=∑α∈¯Kck(α)x−α.fk=∑α∈¯¯¯¯Kck(α)x−α.
(4.1)
Of course we will then have by Definition 3.1
dres(f,ω,k)=dres(fk,ω,1)dres(f,ω,k)=dres(fk,ω,1)
(4.2)
for every ω∈¯K/Zω∈¯¯¯¯K/Z and k∈Nk∈N.
Our computation of the fk∈K(x)fk∈K(x) satisfying (4.1) is based on iterating classical so-called Hermite reduction algorithms, originally developed in [Her72, Ost45] and for which we refer to the fantastic modern reference [Bro05, §2.2,§2.3].
Definition 4.1 For proper f∈K(x)f∈K(x), the Hermite reduction of f is
where g,h∈K(x)g,h∈K(x) are proper rational functions such thatand h has squarefree denominator.
The following Algorithm 1 computes the fk∈K(x)fk∈K(x) satisfying (4.1) by applying Hermite reduction iteratively and scaling the intermediate outputs appropriately.
Lemma 4.2 Algorithm 1 is correct.
Letting ||f||=m∈N||f||=m∈N denote the highest order of any pole of f∈K(x)f∈K(x), note that f1,…,fm∈¯K(x)f1,…,fm∈¯¯¯¯K(x) defined by (4.1) are uniquely determined by having squarefree denominator and satisfying
f=m∑k=1(−1)k−1(k−1)!dk−1dxk−1(fk).f=m∑k=1(−1)k−1(k−1)!dk−1dxk−1(fk).
(4.3)
Defining inductively g0 ≔ f and
(gk,ˆfk):=HermiteReduction(gk−1)⟺gk−1=ddx(gk)+ˆfk(gk,^fk):=HermiteReduction(gk−1)⟺gk−1=ddx(gk)+^fk
(4.4)
for k∈Nk∈N as in Algorithm 1, we obtain by construction that all gk,ˆfk∈K(x)gk,^fk∈K(x) and every ˆfk^fk has squarefree denominator. Moreover, ||gk|| = ||gk − 1|| − 1 = m − k, and therefore the algorithm terminates with gm = 0. Moreover, it follows from (4.4) thatTherefore the elements (−1)k−1(k−1)!ˆfk(−1)k−1(k−1)!^fk are squarefree and satisfy (4.3), so they agree with the fk∈¯K(x)fk∈¯¯¯¯K(x) satisfying (4.1). □
Remark 4.3 As we mentioned in the introduction, we do not expect Algorithm 1 to be surprising to the experts. What is surprising to us is that this trick is not used more widely since being originally suggested in [Hor71, §5]. We expect the theoretical cost of computing HermiteList(f)HermiteList(f) iteratively as in Algorithm 1 is essentially the same as that of computing HermiteReduction(f)HermiteReduction(f) only once. This might seem counterintuitive, since the former is defined by applying the latter as many times as the highest order m of any pole of f. But the size of the successive inputs in the loop decreases so quickly that the cost of the first step essentially dominates the added cost of the remaining steps put together. This conclusion is already drawn in [Hor71, §5] regarding the computational cost of computing iterated integrals of rational functions.
5 SIMPLE REDUCTION
The results of the previous section allow us to further restrict our attention to proper rational functions f∈K(x)f∈K(x) with simple poles, which we write uniquely as f=abf=ab with a,b∈K[x]a,b∈K[x] such that b is monic and squarefree, and either a = 0 or else 0 ≤ deg(a) < deg(b).
Our next task is to compute a reduced form ˉf∈K(x)¯f∈K(x) such that f−ˉff−¯f is rationally summable and ˉf¯f has squarefree denominator as well as polar dispersion 0, which we accomplish in Algorithm 3 . As we mentioned already in the introduction, many algorithms have been developed beginning with [Abr71] that can compute such a reduced form, even without assuming f has only simple poles.
Algorithm 3 requires the computation of the following set of integers, originally defined in [Abr71].
Definition 5.1 For 0≠b∈K[x]0≠b∈K[x], the (forward) shift set of b is
The following Algorithm 2 for computing ShiftSet(b)ShiftSet(b) is based on the observation already made in [Abr71, p. 326], but with minor modifications to optimize the computations.
Lemma 5.2 Algorithm 2 is correct.
As pointed out in [Abr71, p. 326], ShiftSet(b)ShiftSet(b) is the set of positive integer roots of the resultant R(z)∈K[z]R(z)∈K[z] defined in Algorithm 2, which is the same as the set of positive integer roots of the square-free part R(z)/gcdz(R(z),dRdz(z))R(z)/gcdz(R(z),dRdz(z)). It is clear that R(0) = 0, and since we do not care for this root, we are now looking for positive integer roots of the polynomial ˜R(z)~R(z) defined in Algorithm 2 . It follows from the definition of R(z) that R(ℓ) = 0 if and only if R(− ℓ) = 0 for every ℓ∈¯Kℓ∈¯¯¯¯K (not just for ℓ∈Zℓ∈Z), and we see that this property is inherited by ˜R(z)~R(z). Since z∤˜R(z)z∤~R(z), the even polynomial ˜R(z)=T(z2)~R(z)=T(z2) for a unique T(z)∈K[z]T(z)∈K[z], whose degree is evidently half of that of ˜R(z)~R(z). □
Remark 5.3 The role of the assumption that KK be canonical (cf. §2.1) is made only so that one can compute ShiftSet(b)ShiftSet(b). We note that in [GGSZ03, §6] a much more efficient (and general) algorithm than Algorithm 2 is described, which works for b∈Z[x]b∈Z[x]. We remark that in order to compute ShiftSet(b)ShiftSet(b) in general, it is sufficient to be able to compute a basis {w1, …, ws} of the QQ-vector subspace of KK spanned by the coefficients of the auxiliary polynomial T(z)∈K[x]T(z)∈K[x] defined in Algorithm 2 . Indeed, we could then write T(z)=∑sj=1wjTj(z)T(z)=∑sj=1wjTj(z) with each Tj(z)∈Q[z]Tj(z)∈Q[z] and simply compute the set of (square) integer roots of ˜T(z)=gcd(T1,…,Ts),~T(z)=gcd(T1,…,Ts), which is the same as the set of (square) integer roots of T(z).
The previous Algorithm 2 to compute ShiftSetShiftSet is called by the following Algorithm 3 to compute reduced forms of rational functions with squarefree denominators.
Proposition 5.4 Algorithm 3 is correct.
As in Algorithm 3, let b denote the denominator of f and let S=ShiftSet(b)S=ShiftSet(b). Then indeed if S = ∅, pdisp(f) = 0, so f is already reduced and there is nothing to do. Assume from now on that S ≠ ∅, and let us consider roots of polynomials in ¯K¯¯¯¯K. For each ℓ ∈ S, the roots of gℓ ≔ gcd(b, σ− ℓ(b)) are those roots α of b such that α − ℓ is also a root of b. Therefore the roots of G = lcm(gℓ: ℓ ∈ S) are those roots α of b such that α − ℓ is also a root of b for some ℓ∈Nℓ∈N (because all possible such ℓ belong to S, by the definition of S). It follows that the roots of b0 ≔ b/G are those roots α of b such that α − ℓ is not a root of b for any ℓ∈Nℓ∈N. In particular, disp(b0) = 0. We call b0 the divisor of initial roots.
Now the roots of bℓ ≔ gcd(σ− ℓ(b0), b) are those roots α of b such that α − ℓ is a root of b0, i.e., the roots of b which are precisely ℓ shifts away from the initial root in their respective ZZ-orbits. It may happen that bℓ = 1 for some ℓ ∈ S, because even though each ℓ ∈ S is the difference between two roots of b, it might be that no such pair of roots of b involves any initial roots of b0. Writing N ≔ {0} ∪ {ℓ ∈ S | deg(bℓ) ≥ 1}, it is clear that
∏ℓ∈Nbℓ=bandgcd(bℓ,bj)=1forℓ≠j.∏ℓ∈Nbℓ=bandgcd(bℓ,bj)=1forℓ≠j.
(5.1)
Therefore we may uniquely decompose f into partial fractions as in (2.1) with respect to the factorization (5.1) as called by Algorithm 3 ,
Now this ˉf¯f is a sum of proper rational functions with squarefree denominators, whence ˉf¯f also is proper with squarefree denominator. Since σℓ(bℓ) = gcd(b0, σℓ(b)) is a factor of b0 for each ℓ ∈ N and disp(b0) = 0, we conclude that pdisp(ˉf)=0pdisp(¯f)=0. Finally, for each ℓ ∈ N − {0} we see that
whence f−ˉff−¯f is a sum of rationally summable elements, and is therefore itself rationally summable. □
Remark 5.5 As we stated in the introduction, Algorithm 3 strikes us as being conceptually similar to the one already developed in [GGSZ03, §5], but its description is made simpler by our restriction to rational functions with simple poles only. Having a procedure that is easier for humans to read is not necessarily a computational virtue. But it is so in this case, because the relative simplicity of Algorithm 3 makes it also nimble and adaptable, enabling us in §7 to easily modify it to pursue other related applications beyond plain rational summability.
6 COMPUTATION OF DISCRETE RESIDUES
Now we wish to put together the algorithms presented in the earlier sections to compute symbolically all the discrete residues of an arbitrary proper f∈K(x)f∈K(x), in the sense described in §3. In order to do this, we first recall the following result describing the sense in which we compute classical residues symbolically by means of an auxiliary polynomial, and its short proof which explains how to actually compute this polynomial in practice.
Lemma 6.1 ([Tra76, Lem. 5.1]) Let f=ab∈K(x)f=ab∈K(x) such that a,b∈K[x]a,b∈K[x] satisfy a ≠ 0, deg(a) < deg(b), gcd(a, b) = 1, and b is squarefree. Then there exists a unique polynomial 0≠r∈K[x]0≠r∈K[x] such that deg(r) < deg(b) and
Since the set of poles of f (all simple poles) is the set of roots of b, we know the classical first-order residue c1(α) of f at each α∈¯K such that b(α) = 0 satisfies 0≠c1(α)=a(α)/dbdx(α). Using the extended Euclidean algorithm we find the unique 0 ≠ r in K[x] such that deg(r) < deg(b) and r⋅ddx(b)≡a(modb). □
For f∈K(x) satisfying the hypotheses of Lemma 6.1, we denote
FirstResidues(f):=(b,r),
(6.1)
where b,r∈K[x] are also as in the notation of Lemma 6.1. We also define FirstResidues(0):=(1,0), for convenience. With this, we can now describe the following simple Algorithm 4 to compute a symbolic representation of the discrete residues of f.
Theorem 6.2 Algorithm 4 is correct.
It follows from the correctness of Algorithm 1 proved in Lemma 4.2 that f has no poles of order greater than m, whence by Definition 3.1 every non-zero discrete residue of f has order at most m. Consider now ˉfk:=SimpleReduction(fk), which by the correctness of Algorithm 3 is such that fk−ˉfk is summable. We prove the correctness of Algorithm 4 for each k = 1, …, m depending on whether ˉfk=0 or not.
In case ˉfk=0, Algorithm 4 produces (Bk, Dk) = (1, 0). In this case we also know that fk is summable, and therefore by (4.2) dres(fk, ω, 1) = dres(f, ω, k) = 0 for every ω∈¯K/Z. Thus the output of Algorithm 4 is (vacuously) correct in this case because the constant polynomial Bk = 1 has no roots.
Suppose now that ˉfk≠0. It follows from the definition of (Bk,Dk):=FirstResidues(ˉfk) as in (6.1) that Bk is the denominator of the proper rational function ˉfk, and therefore Bk is non-constant, squarefree, and has disp(Bk) = 0, by the correctness of Algorithm 3 proved in Lemma 5.4. Let us denote by ˉck(α) the classical first order residue of ˉfk at each α∈¯K (note that ˉfk has only simple poles, so there are no other residues). We obtain from Lemma 6.1 that Dk ≠ 0, deg(Dk) < deg(Bk), and Dk(α)=ˉck(α) for each root α of Bk. Since ˉfk has at most one pole in each orbit ω∈¯K/Z (this is the very meaning of pdisp(ˉfk)=0), it follows that ˉck(α)=dres(ˉfk,ω(α),1) for every α∈¯K. To conclude, we observe that
for each ω∈¯K/Z; the first equality follows from the summability of fk−ˉfk established in the proof of Proposition 5.4, and the second equality is (4.2). □
Remark 6.3 As we mentioned in §3, the knowledge of a reduced form h for f is morally “the same” as knowledge of the discrete residues of f. And yet, the output ((B1,D1),…,(Bm,Dm)) of Algorithm 4 has the following deficiency: it may happen that for some j ≠ k, we have dres(f, ω, k) ≠ 0 ≠ dres(f, ω, j), and yet the representatives αj, αk ∈ ω such that Bj(αj) = 0 = Bk(αk) may be distinct, with αj ≠ αk. In many applications, this is not an issue because summability problems decompose into parallel summability problems in each degree component, as we see from Proposition 3.2. Actually, the systematic exploitation of this particularity was the original motivation of Algorithm 1 and remains its raison d’être. But it is still unsatisfying that the different Bk associated to the same f are not better coordinated, and this does become a more serious (no longer merely aesthetic) issue in further applications to creative telescoping, where the discrete residues of different orders begin to interact. We explain how to address this problem in Remark 7.2, when we have developed the requisite technology.
7 EXTENSIONS AND APPLICATIONS
In this section we collect some modifications to the procedures described in the previous sections to produce outputs that allow for more immediate comparison of discrete residues across several rational functions and across different orders.
We begin with the parameterized summability problem (1.2) described in the introduction. Let f=(f1,…,fn)∈K(x)n be given, and suppose we wish to compute a K-basis for
V(f):={v∈Kn | ⟨v,f⟩ is summable},
(7.1)
where ⟨•, •⟩ denotes the usual inner product. By Proposition 3.2,
v=(v1,…,vn)∈V(f)⟺n∑i=1vi⋅dres(fi,ω,k)=0
(7.2)
for every ω∈¯K/Z and every k∈N. If only we knew how to write down this linear system explicitly, we would be able to solve for the unknown v. We can apply Algorithm 4 to each fi, and obtainBut it may happen that many different αi,k∈¯K all belong to the same orbit ω∈K/Z and satisfy Bi, k(αi, k) = 0, which leads to very undesirable bookkeping problems.
To address this kind of problem, we introduce in Algorithm 5 a generalization of Algorithm 3 that computes reduced forms for several f1, …, fn compatibly, so that whenever fi and fj have non-zero residue of order a given k at a given orbit ω if and only if Bi, k and Bj, k have a common root α∈¯K such that ω = ω(α). For this purpose, we may assume as in §5 that the fi are proper and, thanks to Algorithm 1, that they all have squarefree denominators.
Corollary 7.1 Algorithm 5 is correct.
The proof is very similar to that of Proposition 5.4, so we only sketch the main points. The key difference is that now b0 has been defined so that for each root α of b0 and each root αi of bi belonging to ω(α) we have that αi−α∈Z≥0. The roots of bi, ℓ are precisely those roots of bi which are ℓ steps away from the unique root of b0 that belongs to the same orbit. By construction, the denominator of each ˉfi is a factor of b0, which has disp(b0) = 0 as before. □
Remark 7.2 Algorithm 5 also allows us to fix the deficiency discussed in Remark 6.3. For a non-zero proper f∈K(x), let us define (f1,…,fm):=HermiteList(f) as in Algorithm 4 . If we now set (ˉf1,…,ˉfm):=SimpleReduction+(f1,…,fm), instead of ˉfk:=SimpleReduction(fk) separately for k = 1, …, m, we will no longer have the problem of the Bk being incompatible.
More generally, we can combine Algorithm 5 with the modification proposed in the above Remark 7.2 to compute symbolic representations of the discrete residues dres(fi, ω, k) of several f1,…,fn∈K(x), which are compatible simultaneously across the different fi as well as across the different k∈N. This will be done in Algorithm 6, after explaining the following small necessary modification to the FirstResidues procedure defined in (6.1). For an n-tuple of proper rational functions f = (f1, …, fn) with squarefree denominators, suppose FirstResidues(fi)=:(bi,ri) as in (6.1), and let b ≔ lcm(b1, …, bn). Letting ai ≔ numer(fi) and di:=bbi, we see that gcd(bi, di) = 1 because b is squarefree, and therefore by the Chinese Remainder Theorem we can find a unique pi∈K[x] with deg(pi) < b such that
Then we see that pi(α) is the first-order residue of fi at each root α of b. We define
Corollary 7.3 Algorithm 6 is correct.
This is an immediate consequence of the correctness of Algorithm 5 proved in Corollary 7.1, coupled with the same proof, mutatis mutandis, given for Theorem 6.2. □
Algorithm 6 leads immediately to a simple algorithmic solution of the problem of computing V(f) in (7.1).
Proposition 7.4 Let f = (f1, …, fn) with each 0≠fi∈K(x) proper. Let DiscreteResidues+(f)=(B,D) with
be as in Algorithm 6 and let V(f) be as in (7.1). Then
V(f)={v∈Kn | n∑i=1viDi,k=0 for each 1≤k≤m}.
(7.3)
For each v∈Kn and α∈¯K such that B(α) = 0,
dres(⟨v⋅f⟩,ω(α),k)=n∑i=1viDi,k(α)
(7.4)
by the correctness of Algorithm 6 proved in Corollary 7.3. Moreover, since deg(Di, k) < deg(B) (whenever Di, k ≠ 0) and B is squarefree, we see that for each given 1 ≤ k ≤ m the expression (7.4) is zero for every root α of B if and only if the polynomial ∑iviDi, k ≡ 0 identically. We conclude by Proposition 3.2. □
Remark 7.5 One can produce without too much additional effort a variant of Proposition 7.4 that computes more generally the K-vector space of solutions to the creative telescoping problem obtained by replacing the unknown coefficients vi∈K in (1.2) with unknown linear differential operators Li∈K[ddx].
8 CONNECTIONS WITH GALOIS THEORY
Our interest in computing the vector space V(f) in (7.1) is due to the fact that it is isomorphic to the difference Galois group of the block-diagonal difference system σ(Y) = (A1⊕…⊕An)Y with diagonal blocks Ai=(1fi01). Indeed, this difference Galois group consists of block-diagonal matrices G(v) = G(v1)⊕…⊕G(vn) with diagonal blocks G(vi):=(1vi01) for v = (v1, …, vn) ∈ V(f) (cf. [Har08, Prop. 2.1]).
Another application of the procedures developed here to Galois theory of difference equations arises from the consideration of diagonal systems
σ(Y)=diag(r1,…,rn)Y,where r1,…,rn∈K(x)×
(8.1)
As shown in [vdPS97, §2.2], the difference Galois group of (8.1) iswhere E⊆Zn is the subgroup of e = (e1, …, en) such that
re11⋯renn=σ(pe)/pe
(8.2)
for some pe∈K(x)×. Now suppose (8.2) holds, and let us write fi:=ddx(ri)/ri for each i = 1, …, n and ge:=ddx(pe)/pe, so that we have
e1f1+⋯+enfn=σ(ge)−ge.
(8.3)
At first glance, this looks like a version of problem (1.2), but it is even more special because the fi have only first-order residues, all in Z. So we can compute the Q-vector space V of solutions to (8.3) using SimpleReduction+(f1,…,fn), just as in Proposition 7.4, as a preliminary step. Then we can compute a Z-basis e1, …, es of the free abelian group ˜E:=V∩Zn. Since each gej has only simple poles with integer residues, one can compute explicitly pej∈K(x) such that ddxpej=gejpej, and thence constants γj∈K× such that rej=γj. This reduces the computation of E from the defining multiplicative condition (8.2) in K(x)× modulo the subgroup {σ(p)/p | p∈K(x)×} to the equivalent defining condition in K× :
9 EXAMPLE
Let us conclude by illustrating some of our procedures on the following example considered in both [MS95, PS95]. In order to make the computations easier for the human reader to follow, we have allowed ourselves to write down explicitly in this small example the irreducible factorizations of denominators. We emphasize and insist upon the fact that none of our procedures performs any such factorization.
Consider the rational function
f:=1x3(x+2)3(x+3)(x2+1)(x2+4x+5)2.
(9.1)
We first compute HermiteList(f)=(f1,f2,f3) with
f1=787x5+4803x4+9659x3+9721x2+9502x+500818000(x2+1)(x+3)(x2+4x+5)(x+2)x;f2=−787x3+3372x2+4696x+103018000(x2+4x+5)x(x+2);andf3:=−7x−1300(x+2)x;
using Algorithm 1 . We apply the remaining procedures to f1 only – the remaining f2 and f3 are similar, and easier. Denoting
b:=(x2+1)(x+3)(x2+4x+5)(x+2)x
(9)
the monic denominator of f1, we compute with Algorithm 2 (or see by inspection) that ShiftSet(b)={1,2,3}. The factorization b = b0b1b2b3 computed within Algorithm 3 is given by
b0=(x+3)(x2+4x+5);b1=gcd(σ−1(b0),b)=x+2;b2=gcd(σ−2(b0),b)=x2+1;XXb3=gcd(σ−3(b0),b)=x.
The individual summands in the partial fraction decomposition of f1 with respect to this factorization are given by
a0b0=−13391x2+37742x−92931080000(x+3)(x2+4x+5);a1b1=1250(x+2);a2b2=−7x−18000(x2+1);a3b3=31333750x.
The reduced form ˉf1=a0b0+σ(a1b1)+σ2(a2b2)+σ3(a3b3) is given byThe first pair of polynomials (B1,D1)∈K[x] computed by Algorithm 4 is given by
Let us compare this output (B1, D1) with the discrete residues dres(f, ω, 1) of f according to the Definition 3.1 in terms of classical residues. We see from the factorization of the denominator of f in (9.1) that its set of poles is
and that each of these poles belongs to one of the three orbitsTherefore, f has no discrete residues outside of these orbits, and we verify that the set of roots {−3,√−1−2,−√−1−2} of the polynomial B1 correctly contains precisely one representative from each of these orbits (subject to our verification below that the first-order discrete residues of f at these orbits are actually non-zero!). We can compute directly in this small example that the classical first-order residues of f at the poles in ω(0) are given byand at the poles in ω(±√−1) are given by
Res1(f,±√−1)=±√−1−716000;andRes1(f,±√−1−2)=∓1119√−1−53380000.
Finally we can verify directly that the polynomial D1 correctly computes the first-order discrete residues of f at all three orbits ω(0) and ω(±√−1) according to Definition 3.1:and
dres(f,ω(±√−1),1)=±√−1−716000+∓1119√−1−53380000==∓557√−1−28440000=D(±√−1−2).
Both authors gratefully acknowledge the support of NSF grant CCF-1815108 and the Arreche's UT Dallas Startup grant. We are also thankful to Michael Singer, Shaoshi Chen, and the anonymous referees for all their helpful comments and suggestions. An earlier version of this work formed part of Sitaula's PhD thesis [Sit23].
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Lymphangioma: Practice Essentials, Background, Pathophysiology
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Tools & References>Dermatology
Lymphangioma
Updated: Apr 18, 2023
Author: Robert A Schwartz, MD, MPH; Chief Editor: Dirk M Elston, MDmore...;)
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Sections
Lymphangioma
Sections Lymphangioma
Overview
Practice Essentials
Background
Pathophysiology
Etiology
Epidemiology
Prognosis
Patient Education
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Overview
Practice Essentials
Lymphangiomas are uncommon, hamartomatous, congenital malformations of the lymphatic system that involve the skin and subcutaneous tissues. The classification of lymphangiomas lacks a standard clear definition and universal application, in part because of the nature of lymphangiomas, which represent a clinicopathologic continuum. The classification most frequently used divides these lesions into 2 major groups based on the depth and the size of these abnormal lymph vessels. The superficial vesicles are called lymphangioma circumscriptum. The more deep-seated group includes cavernous lymphangioma and cystic hygroma. Many categorize cystic hygroma as a variant of cavernous lymphangioma. Note the image below.
A 16-year-old obese boy with large unilateral verrucous lymphangioma.
View Media Gallery)
Also see Oral Lymphangiomas and Cervicofacial Lymphangiomas.
Signs and symptoms
Lymphangiomas may affect a number of sites, including the scrotum. Multilocular lymphangiomatosis of the scrotum is unusual. The site of a lymphangioma may determine its clinical signs and symptoms. For example, a lacrimal gland lymphangioma may result in acute proptosis and inferonasal displacement of an eye.
Note the images below.
A 44-year-old woman with plaque on her forearm since birth that clinically appeared to be typical lymphangioma circumscriptum. Histologically, however, it had features of both hemangioma and lymphangioma.
View Media Gallery)
Close-up demonstrating the clinical morphology to better advantage.
View Media Gallery)
Lymphangioma circumscriptum
Lymphangioma circumscriptum involves small clusters of vesicles measuring about 2-4 mm. These clear vesicles can vary from pink to red to black secondary to hemorrhage. Although it may appear localized to the dermis, this neoplasm frequently extends deeply and laterally. The lesions can have a warty appearance on their surface; as a result, they are often confused with warts. Shah et al report a lymphangioma presenting on the penis.
Cavernous lymphangioma
Typically, cavernous lymphangiomas appear as subcutaneous nodules with a rubbery consistency. They may have large dimensions. The overlying skin has no lesions or changes. The area of involvement varies, ranging from lesions smaller than 1 cm in diameter to larger lesions that involve an entire limb. Rarely, nasolacrimal lymphomas may be seen, with one first evident with unilateral blood tearing.
Cystic hygroma
Cystic hygromas are usually larger than cavernous lymphangiomas, and they more commonly occur in the neck and parotid area. Often, deep cavernous lymphangiomas are not evident on superficial examination, but cystic hygromas are detected with ease because of their size and location. These large cystic lesions are soft and translucent. Congenital and acquired lymphangiomas of the vulva are rare. [6, 7] Diffuse lymphangiomas may go unnoticed, evident as asymptomatic, erythematous flat, indurated, or atrophic plaques with or without swelling with or without any surface changes.
Fetal lymphangioma
Fetal lymphangioma may involve fetal skin and occasionally mucosa and subcutaneous tissue, typically on neck or axilla. The fetal chest wall cystic lymphangioma is rare.
Diagnostics
The diagnosis of lymphangiomas is based mainly on the clinical history and findings from physical examination and conventional light microscopy.
Also see Histologic Findings.
MRI can help define the degree of involvement and the entire anatomy of the lymphangioma lesion. MRI can help prevent unnecessary extensive, incomplete surgical resection, because of the association with a high recurrence rate.
The scrotum is not a usual site, but a painless scrotal swelling was documented as a lymphangioma by typical sonography and MRI findings, followed by excision and pathologic confirmation.
Immunohistochemical study is useful in differentiating lymphangiomas from hemangiomas in difficult cases. Test results with factor VIII – related antigen are positive for hemangiomas but negative or weakly positive in the endothelium of lymphangiomas. Immunohistochemical studies for laminin show the typical multilayered basal lamina of normal blood vessels and the discontinuous basal lamina in lymphangiomas.
Dermoscopic findingsmay aid in the diagnosis of cutaneous lymphangioma circumscriptum. [11, 12, 13] Nodules filled with clear fluid show light brown lacunas surrounded by paler septa. Those tinged with blood may have focal reddish areas inside the lagoons, pink diffuse coloration, and/or reddish to violaceous lacunar structures. In those lacunae that contained blood, it accumulated in the lowest part. Thus, they are characterized by a lacunar pattern and with a marked hematic content may be indistinguishable from a hemangioma.
Management
The preferred treatment for lymphangiomas is complete surgical excision. On the basis of the Whimster hypothesis, the large subcutaneous cisterns should be removed to prevent the lesion from resurfacing.
Also see Treatment.
Next: Background
Background
Lymphangiomas can occur anywhere in the skin and the mucous membranes. The most common sites are the head and the neck, followed by the proximal extremities, the buttocks, and the trunk. However, they sometimes can be found in the intestines, the pancreas, and the mesentery. Deeper cystic lesions usually occur in areas of loose and areolar tissue, typically the neck, the axilla, and the groin. Their skin involvement ranges from small, well-demarcated areas to large, diffuse regions with unclear borders.
Lymphangioma circumscriptum, the common form of cutaneous lymphangioma, is characterized by persistent, multiple clusters of translucent vesicles that usually contain clear lymph fluid (often compared with frog spawn). These vesicles represent superficial saccular dilations from underlying lymphatic vessels that occupy the papilla and push upward against the overlying epidermis. Each skin lesion may range from a minute vesicle to a small bulla-sized lesion. These vesicles can be clear or vary from pink to dark red because of serosanguineous fluid and hemorrhage. These vesicles often are associated with verrucous changes, which give them a warty appearance.
In the case of lymphangioma circumscriptum, the underlying lesions constitute abnormal dilated lymph vessels involving the upper part of the dermis. The sites of predilection are the proximal extremities, trunk, axilla, and oral cavity, especially the tongue. Involvement in other areas, such as the scrotum, is not uncommon. Lymphangioma circumscriptum has a high recurrence rate after excision because of its deep component (see Pathophysiology).
Cavernous lymphangioma are also uncommon and usually arise during infancy. The most common sites are the head and neck areas and, less frequently, the extremities. These lesions are seated deep in the dermis, forming a painless swelling or thickening of the skin, mucous membranes, and subcutaneous tissue. Unlike lymphangioma circumscriptum, the overlying skin usually is uninvolved. Occasionally, patients report pain when the involved area is pressed. The affected area may be 1 cm, it may be as large as several centimeters in diameter, or it may involve an entire extremity. Upon examination and palpation, lipomas or cysts can be mistaken for these lesions. Lymphangioma circumscriptum can occur in conjunction with cavernous lymphangioma and cystic hygroma.
Some authors categorize cystic hygroma or cystic lymphangioma as an independent entity. Many authors agree that cystic hygroma is a form of cavernous lymphangioma in which the degree of involvement and character is determined by its location. These congenital lesions are deeply seated in areas of areolar or loose connective tissue. They appear early in life as large soft-tissue masses, usually on the axilla, neck, or groin. These lesions are soft, vary in size and shape, and tend to grow extensively if not surgically excised. Typical lesions are multilocular cysts filled with clear or yellow lymph fluid. Usually, cystic hygroma is diagnosed clinically with its large size, location, and translucence.
Terminology for these lesions can be confusing. Some lymphangiectasias have sometimes been called acquired lymphangiomas, secondary lymphangiomas and acquired lymphangioma circumscriptum.
Previous
Next: Background
Pathophysiology
In 1976, Whimster studied the pathogenesis of lymphangioma circumscriptum. According to Whimster, the basic pathologic process is the collection of lymphatic cisterns in the deep subcutaneous plane. These cisterns are separated from the normal network of lymph vessels, but they communicate with the superficial lymph vesicles through vertical, dilated lymph channels.
Whimster postulated that these cisterns might arise from a primitive lymph sac that fails to connect with the rest of the lymphatic system during its embryonic development. A thick coat of muscle fibers that cause rhythmic contractions line these sequestered primitive sacs. Rhythmic contractions increase the intramural pressure, causing dilated channels to protrude from the walls of the cisterns toward the skin. He suggested that the vesicles seen in lymphangioma circumscriptum are outpouchings of these dilated projecting vessels.
Whimster's observations are supported by those of lymphangiographic and radiographic studies. These studies revealed that large multilobulated cisterns extend deep in the dermis and laterally beyond the obvious clinical lesions. These deep lymphangiomas show no evidence of communication with the adjacent normal lymphatics. The cause for the failure of these primitive lymph sacs to connect to the rest of the lymphatic system is not known.
Some lymphangiomas may represent vascular malformations during embryonic development rather than as true neoplasms. Vascular endothelial growth factor (VEGF)–C and VEGF receptor-3 are active in the formation of lymphangiomas. Based upon their expression, superficial lymphangiomas more likely result from peripheral lymphatic dilatation than from a growth factor.
Fetal lymphangioma is presumably a result of failure in lymphatic drainage.
Previous
Next: Background
Etiology
The reason that these embryonic lymph sacs remain disconnected from the rest of the lymphatic system is not known. Fetal lymphangioma is presumably a result of failure in lymphatic drainage, documented on routine prenatal ultrasonographic studies. Vulva lymphangioma circumscriptum may be secondary to pelvic lymphatic obstruction, linked in some patients to rectal adenocarcinoma, cervical carcinoma, or endometrial carcinoma. [7, 21]
Previous
Next: Background
Epidemiology
Frequency
Lymphangiomas are rare. They account for 4% of all vascular tumors and approximately 25% of all benign vascular tumors in children.
Its incidence in India has been estimated at 1.2-2.8 cases per 1000 newborns.
Race, sex, and age
No racial predominance is reported for lymphangiomas.
Equal sex incidences are reported for lymphangiomas in most studies. Some groups have reported that lymphangioma circumscriptum is more common in females than in males, while others report a 3:1 male-to-female ratio.
Lymphangioma can become evident at any age, but the greatest incidence occurs at birth or early in life. About 50% of lymphangiomas are seen at birth, and most lymphangiomas are evident by the time the patient is aged 5 years. It may be documented in fetuses too.
Previous
Next: Background
Prognosis
Lymphangiomas are benign hamartomatous malformations instead of true neoplasms. The prognosis for lymphangiomas is excellent.
Rarely do cutaneous lymphangiomas interfere with the well-being of patients. Patients are expected to live a full healthy life, and they usually seek medical intervention because of cosmetic reason.
Lymphangiomas represent hamartomatous malformations with no risk of malignant transformation. In the case of cystic hygroma, total surgical excision is appropriate to prevent complications such as respiratory compromise, aspiration, and infections in critical areas, such as the neck. Lymphangiomas have a strong tendency for local recurrence unless they are completely excised. Recurrent episodes of cellulitis and minor bleeding are not uncommon.
Previous
Next: Background
Patient Education
Patients should receive reassurance. Lymphangiomas represent benign lymphatic malformations and not premalignant lesions. Patients should be aware of the risk of recurrence.
Previous
Clinical Presentation
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References
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Media Gallery
A 44-year-old woman with plaque on her forearm since birth that clinically appeared to be typical lymphangioma circumscriptum. Histologically, however, it had features of both hemangioma and lymphangioma.
Close-up demonstrating the clinical morphology to better advantage.
A 16-year-old obese boy with large unilateral verrucous lymphangioma.
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Contributor Information and Disclosures
Author
Robert A Schwartz, MD, MPH Professor and Head of Dermatology, Professor of Pathology, Professor of Pediatrics, Professor of Medicine, Rutgers New Jersey Medical School
Robert A Schwartz, MD, MPH is a member of the following medical societies: Alpha Omega Alpha, American Academy of Dermatology, New York Academy of Medicine, Royal College of Physicians of Edinburgh, Sigma Xi, The Scientific Research Honor Society
Disclosure: Nothing to disclose.
Coauthor(s)
Geover Fernandez, MD, FAAD Staff Physician, Department of Dermatology, Rutgers New Jersey Medical School
Geover Fernandez, MD, FAAD is a member of the following medical societies: Alpha Omega Alpha, American Academy of Dermatology, American Society for MOHS Surgery
Disclosure: Nothing to disclose.
Specialty Editor Board
David F Butler, MD Former Section Chief of Dermatology, Central Texas Veterans Healthcare System; Professor of Dermatology, Texas A&M University College of Medicine; Founding Chair, Department of Dermatology, Scott and White Clinic
David F Butler, MD is a member of the following medical societies: Alpha Omega Alpha, American Academy of Dermatology, Association of Military Dermatologists, Phi Beta Kappa, Texas Dermatological Society
Disclosure: Nothing to disclose.
Jeffrey J Miller, MD Associate Professor of Dermatology, Pennsylvania State University College of Medicine; Staff Dermatologist, Pennsylvania State Milton S Hershey Medical Center
Jeffrey J Miller, MD is a member of the following medical societies: Alpha Omega Alpha, American Academy of Dermatology, Society for Investigative Dermatology, Association of Professors of Dermatology, North American Hair Research Society
Disclosure: Nothing to disclose.
Chief Editor
Dirk M Elston, MD Professor and Chairman, Department of Dermatology and Dermatologic Surgery, Medical University of South Carolina College of Medicine
Dirk M Elston, MD is a member of the following medical societies: American Academy of Dermatology
Disclosure: Nothing to disclose.
Additional Contributors
Carrie L Kovarik, MD Assistant Professor of Dermatology, Dermatopathology, and Infectious Diseases, University of Pennsylvania School of Medicine
Carrie L Kovarik, MD is a member of the following medical societies: Alpha Omega Alpha
Disclosure: Nothing to disclose.
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Lec 16 | MIT 6.042J Mathematics for Computer Science, Fall 2010
MIT OpenCourseWare
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Posted: 31 Dec 2012
Lecture 16: Counting Rules I
Instructor: Marten van Dijk
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Intro the following content is provided under a Creative Commons license your support will help MIT open courseware continue to offer high quality educational resources for free to make a donation or view additional materials from hundreds of MIT courses visit MIT opencourseware at ocw.mit.edu so this week we are going to talk about uh counting uh tonight is uh a problem set 8 uh due um for this week uh we will post a new problem set tonight as well um Counting is uh very important the rest of the semester after this week we we'll actually explain a probability Theory and that's all based on Counting so um we going to teach you this week a whole toolkit of all kinds of ways how to count and as you can see we're going to talk about a lot of different kinds of rules a mapping rule we talk about a principle pital principle another rule a product rule and a sum Rule and these are all ways to count uh one thing by counting another thing actually what we are going to talk about is how to uh count say a difficult uh set of objects and then we will map those objects to something that we can count in a much easier way okay so let's uh start with a lot of definitions actually so uh we have to to talk about sets sequences and permutations to start off with so the Sets and sequences definition of a set is uh as follows a set is actually an unordered an unordered collection of distinct elements so as an example we may have uh say a set that contains the items the elements a b and c well if you reorder these elements it doesn't really matter it's still the same set we can also write it as say the set c a b on the other hand if you have a collection in which say the element a appears twice well these two are not distinct so this is not a set so this is not a set but it is a collection and in and and you may call this a multi set but we not go into that so we will be talking about sets and we are interested usually in the cardinality of a set so what's that the cardinality or size um is defined as follows it's just cardinality it's just a number of elements that the set s really has so it's the number um of elements in s and and how do we denote this this is denoted by uh two vertical bars around uh the letter that represents the set so we denote this as follows now uh when we talk about sets we may also be interested in uh ordered collections of elements and that's what we call a sequence so a sequence is defined as follows a sequence is an ordered collection of uh ele elements and we also call these elements uh components or terms and these elements do not necessarily have to be distinct so not necessarily distinct now as an example so how do we not this for sets we use these this type of notation uh like this symbol uh for sequences we just have a very simple type of brackets just the round bracket so an example could be um well the elements a b and c uh the first entry or the first term of a sequence uh depending on whether you look at it from the left or the right that may differ uh is here a b and then C another way another one could be a b and a and as you can see the element a occurs twice in this sequence so we're going to relate uh sets and sequences and uh let's talk about the Permutations permutation so a permutation of a set is defined as follows a permutation of a set s is actually a sequence that contains every element in s exactly once so every element in s occurs exactly once and uh as an example we may look at the set that we have described over here that set a b and c and well how many uh permutations are there so first one I may just order the elements in a b and c so I have say the sequence a and then B and then C uh there are many more I can for example start with b and then C and then may cycle through to a that's another permutation of these three elements and I can do that once more I can for example uh start with uh C and then a and then B and it turns out that we can do a few more um we can also start with c and then B and then a we sort of revered we have refers the order that we had over here into its opposite so first C then B then a and we can look at the cycle there there shift um we start with ba a and then we rotate through to C which is what we did over here when we went from this permutation to this one so we have B A and then C and then we may start with a and we have a c and b it turns out that this is actually uh all the possible permutations for of of of this set so there are six of these and uh in general uh how many permutations are there of a set of n elements so let's have a look here what we did so if I want to create a permutation of a set I may start off by selecting for the first term uh a b or c so I have actually three choices so the first term has uh three choices for the second term over here well for example suppose my first term was B then the second term well must be an element that is different from B is yet part of s so I have only two choices for for the second term either C or a so the second term has two choices and once I've chosen the second one well if I've already chosen B and C there's only one element left in the set over here only a so I have only one choice left so in this way we may count the total number of permutations of this set as the number of choices I have for the first term times the number of choices for the second term 3 2 times the number of choices for the third term which is only only one so the third um term third term has only one choice now in general we can do this for any permutation and uh if you want to count the number of permutations of a set with n elements well it turns out that that is equal to n n minus one just like here we have three elements 3 3 - one and so on n n -1 nus 2 Etc and all the way up to one and this is n factorial and you've seen this already when we talked about Sterling's formula and how to approximate this now this type of reasoning we will generalize later on when we come to the generalized uh product rule and uh but this is already a first example of how we go about this so permutations relates uh sets and sequences so now we go on to Define more special functions so permutation is one kind of mapping so let's now Define uh functions and then we will talk about a few different uh flavors of those uh uh we talk about subjective functions injective functions and bjective functions and the whole idea is is that if I can use a a mapping from one set to another set with that satisfies some of those properties can say something about how their cardinalities are related and that's what we want we want to count okay so the Functions definition of a function is as follows so a function uh F from X to Y is actually a relation uh between the sets X and Y and we say that um oh with a property that every single element every element of X is actually related to exactly one element of Y and uh we will call uh X to be the uh domain of the function X F and Y we will call the range or image of the function f so let's give an example and see uh a couple of examples of a of first of all of a function and then uh relations that are actually not functions so suppose we have a mapping from X which just contains the elements a b and c just in line with this example over here and we have a mapping F that maps to um uh the set Y and the Y is just the numbers one two and three well I could map for example A to 1 uh B to 3 and C to 3 now this is a function because um every element of X is mapped to exactly one element of y a is just mapped to one B is also mapped to an element only three C is mapped to three as well and we usually write this as uh F evaluated in a is equal to 1 and FB is equal to three and F of c isal equal to three as well now what is not a function well I could for example add another Edge over here if I wanted to well this is not a function because B is now mapped to two elements two and three and that's not what's covered by this definition so this is not a function I could also remove say an edge well in this case b is not met to anything at all and that's not a function either so we really have the property for functions that there is exactly one outgoing Arrow if you want to think about it as being a graph from uh uh from each element in X to exactly one element in in y so now we can talk about a few definitions um so we will talk about these few Surjective Injective properties subjective injective and bjective and then we can start to do a few interesting uh examples so a function f that goes from X to Y is called surjective if um every single element oops if every single element of Y is mapped to at least one is mapped to at least once so what does that mean uh to at least once so every element of Y so uh say uh one for example here here is mapped to at least once well to one the element one we have mapped a so that's great but for example the element two is not mapped to at all so this particular example is not subjective but we will come to a few examples that are so here we have the distinction that every element of Y so every single element of Y is mapped to exact at least once then injective is defined in a similar fashion but now every element of Y is not mapped to at least once but at most once so let's have a look over here and that's also not true for this example actually because three is mapped to two times so it's not mapped to at most once so this example is also not injective because in if a function is injective then every element of of Y of the range is mapped to at most one on bjective is if we have that every element of Y is met to exactly once and we can see that uh by a function is bjective if and only if it is both subjective and injective so bactive if and only if we have both the properties surjective as well as injective so let's give a couple of uh examples so it's a first example we may have the set X and Y we have 1 two and three inated two elements A and B over here one is mapped to a two is mapped to a and three is mapped to B and now we can see that every element in y is mapped to at least once this one is mapped to two times this one is mapped to once so this one is actually subjective so that's great um another example of something that is injective is if we have say one two and three and a b c and d one is Ma say to a 2 to B 3 to D well in this case we have that it is injective because every element of Y is mapped to at most ones once once zero times and once so this one is injective and this one is not injective right because this one is mapped to two times this one is not subjective but because this one is not covered at all it's only me to Once okay so let us talk again about permutations um so when we talked about permutations we can define a mapping using a per liation um that is an example of a bje so let's do that and then we can come to the mapping rle and we can start to do some counting so for example um if we have a uh permutation A1 up to a n of uh so let this be a permutation uh of the set s that contains all the elements A1 up to a n so this is just one example of a permutation and now we may Define the following function we say that Pi um evaluated in AI gives us output I actually what we mean here is that um if we take an element in s then this one is mapped to under this uh function uh to I if and only if a is in the E position in this permutation so if and only if a is in the e um uh term in the permutation so in this case we know that Pi is bjective and why is this well uh we know from the definition of a permutation that in a permutation this just a sequence in which every element of s occurs exactly once so that means that every uh position is covered exactly once by an element of s and that is exactly the definition over here which says that every element in the range is covered exactly once uh is mapped to exactly once by an element in the domain so this is a an example of a Mapping Rule bje okay so now that we have defined functions in the special properties let's talk about uh the mapping rule um which we will do over here and uh now for the first time we start to uh talk about the cardinalities of sets and how they related to one another so the mapping rule is that uh first of all if F if f is a function from X to Y and if f is actually subjective well what do we know we actually know that the cardinality of the the number of elements in the domain is at least the number of elements in the range and why is that if you look at the definition of subjectivity well we know that every element of Y is covered by some element in X at least once and all uh the elements in in in X are mapped to exactly one element in y so we know that the cardinality of X is at least Y because every element in y is mapped by some unique distinct element in X and if a function f is injective well in that case we know that the reverse relation holds so uh inequality holes um the cardinality of X is at most the cardinality of Y so why is that well every element in in in an injective function right every element is mapped to at most one element uh by so so so every element in y is mapped to by at most one element in X so we know that all the elements in X are mapped to some element in y but uh every element in y cannot map two by more than two more than two times by something in the domain so we know that this inequality HS okay for a bjective function we have the that both these properties hold and we will have an equality over here so oops if this one is bjective we have that the cardinalities are equal to one another and this is also called the uh bje rule so let's give an example uh where we want to um find out how many ways there are to uh select uh 12 donuts from five varieties so let's see how that would work and the whole idea is that we're going to uh Define the set that we want to count which is all these possible configurations of donuts over five varieties of flavors and um and then we're going to map these to another structure that we can understand a little bit better so let's do this so it's an example um let X be uh all the ways to select say 12 donuts from five varieties so let's give an example um for example we may have two donuts and they are in the chocolate favorite basket so we have chocolate and suppose we have no um Donuts in the lemon filed uh version of a donut suppose we have a whole bunch of donuts uh say six of those that are uh with sugar we have some that are glazed say two and finally we have just a couple of plain Donuts so this would be a configuration that is in X because we have 1 two 3 four five varieties and we have uh 12 donuts uh two over here six here two and another two 12 donuts that are selected from these five varieties now if we are going to uh try to represent such a configuration that's usually how we think about counting then we may map this to a 01 sequence so how do we do this we can just map the donuts to zeros the divider between the two baskets uh is a one so this is a one then we have no zeros between these two ones because there are no Donuts in the lemon field basket so we have a one that Maps is is is the mapping from this uh divider over here we got six zeros we have another one that is this divider two zeros two Donuts in the Glazed version and so on so what do we see here we have a 01 sequence where we have 12 zeros and uh we have one two three four ones and we can see that this mapping is actually bjective because if I have two zeros I can map them back to Donuts the divider I can M the one I can map back to uh the divider between two baskets so let uh y be the set of uh 12 uh all kinds of configurations of 12 all kinds of sequen is oops maybe I will not take this one out uh let's do this one so if you're going to define y as the [Music] set of all uh 16 bit sequences with exactly four ones then we know that by the bje rule we have created this bje over here that the cardinalities of X and Y are exactly equal so now we know that by the bijection rule we have been able to count uh the number of elements in X by counting something else which is really how we proceed in these types of proofs so if now uh we are now able to just count these types of objects and later on next lecture we will actually figure out the formula for this one so this is an example of how we can use the braction rule now another example is one in which uh we're going to count the Counting Subsets subsets so I'll give a lot of examples through these two lectures and also the problem set as you will see will have a lot of uh small little Parts with all kinds of countings that you will need to do applying different rules so let's talk about how to count uh subsets of a set X so what we want is a bje from subsets of a set X containing of say say one up to n uh so the integer one up to n to n bit sequences we know that we can do this if we Define a projection as follows um so he map a set a subset s under a mapping F to a bit sequence B1 B2 all the way to BN uh via the following relation bi is computed as either a one or a zero it's computed as a one if I is in s and it's a zero if I is not in s now we know that this is a bje so if we have a bit sequence then we can construct from this mapping the corresponding subset if you have a subset we can use it mapping to construct the corresponding sequence the corresponding bit sequence so let the uh so so so how many uh nbit uh sequences are there well there are 2 to the power n n bit sequences why is that well we have two choices for B1 a zero or a one two choices for B B2 Z or one and so on so we have two 2 two choices over here so we have two to the power n choices for the number for the for for a bit sequence so there are two to the power n number of bit sequences and this is actually equal because of this bje rule that we have described over here this is equal to the total number of possible ways to select subsets of x so this is the number of subsets of an N element uh set so this is a very a nice way to uh to demonstrate how we can use the bje rule to count something that appears to be much more harder to think about to grasp at least for me it's harder to grasp so I have a subset that can be any size in uh in a set of n elements and now I can find this really easygoing mapping that I can show to be bjective and all of a sudden I know how to count it because I can just look at uh the image and count those types of objects in this case n bit sequences I get a really easygoing number that I can compute fairly easily and now I have computed something much counted something much more complex so this is how we generally will think about these things okay so let's uh talk now about the generalized pitchon Hall principle so we have covered quite a lot of uh definitions right now uh so we explained the functions mapping rules so now we come to the generalized pitchon Hall principle and a few other rules Okay so Pigeonhole Principle so what about a generalized pitchen Hall principle this is actually uh the following counting argument if I know that the cardinality of a X is more than K the cardinality of a set Y what do I know well I know that for all functions F that have domain X and range y I know that there must exist k + one different elements oh different elements uh of X that are mapped to the same element in y and if we take a specific case uh k equals 1 we will actually call this the pitch and hole principle and let me just demonstrate that by the famous example of the pigeons so well if I have um more pigeons than the number of holes than they can fly into I know for sure that there exists a hole that uh two pigeons will visit so that's where the name comes from so let me write it out so an example is um if I have more than n pigeons so the pigeons from the set X and say they fly into n holes and the holes is my set y well then I know that the cardinality of X is more than the cardinality of Y I have more pitons than there are holes so I know that at least two uh pigeons will fly into the same hole so for the generalized case well how can we uh prove something like that well we could use for example something like a contradiction for example uh suppose that um you know for all uh cess one as suppose that the negation is true so we assume so how do we prove this usually so assume we have this he want to prove this well suppose that's not true suppose there's a mapping F such that for all k + one different elements of x uh well they are not mapped to the same elements in y but what I really know then is that every element in y is mapped to at most by K distinct elements of x so that means that um the total number of elements of X must be at most K Y and that's not true it's larger by assumption so of a contradiction so this is very general principle though and it's worth writing it all out because this is a famous uh and a rule that we will use in uh counting and it leads to interesting results okay so let's uh give another example um let's think about Boston in Boston we have say half a million non-bal people it turns out that um there are at least three people uh that have the exact same number of hairs on their head that's kind of weird how how do we know that I cannot point out any three in Boston that have the same number of hairs I have no idea but somehow I can count and use this principle and tell you that it must be true that in Boston with 500,000 people there are three of them that are not bald so we exclude the bald people because that would be easy they all have zero hairs but say non- bald people that um they actually have the same number of hairs so how do we do that how can we make such types of Nonconstructive Proof conclusion so say Boston has uh about 500,000 uh non-bal people and let's call this set X because we're going to use the pitol principle so our claim is that there exists three people in Boston such that they have the same number of hairs on their head so how do we do this well let's uh well we know uh that we may generally assume that any person has at most 200,000 hairs on their head so the number of hairs um on AE is at most uh 200,000 so how should I Define my set Y in order to apply this pitching hole principle so what do I do so I want to have a mapping right from uh from all the people uh from all the people uh the set X to their number of hairs so the number of hairs on one's head is going to be the set Y and uh what do we know we know that the cardinality of Y is 200 at most 200,000 actually the way we defined it it's it's exactly 200 200,000 and uh the set X has a cardinality of about 500,000 so what do we know we can apply our generalized Pinal principle it's very surprising because we notice that X is more than 2 the cality of Y 2 200,000 is less than 500,000 so I knowed by this particular uh principle um this particular mapping must have the property because this holds for all mappings that there are at least k + 1 2 + 1 three different people three different people in Boston out of the set X that are mapped to the same element in white that means that they have the same number of hairs on their head so this is kind of really surprising we can make a statement Without Really inspecting every single person's head but we can still make a statement about uh um about the fact that there are three different people in Boston that have the exact same number of hairs so this is a an example of a non-constructive proof and I will give another one and it's a very important uh principle actually it's a new technique that you haven't seen before so far we have uh been uh constructively proving all kinds of properties using induction mainly um and this is what is called a non-constructive proof because I cannot give a specific example that demonstrates that this claim is is true but yet I've have shown that it is true but in a non-constructive way without an example okay so what about another one for example we may pick uh 10 arbitrary uh twod digigit numbers so pick 10 arbitrary uh double digit numbers and oh we can uh pick any uh uh sequence of numbers I'm just picking a few and you may add a few two I don't know two seven 14 I don't know 31 25 60 92 uh and so on so I have 1 two 3 four five six seven eight uh I don't know nine oh nine and another one say 91 or something like that um so I have 10 double digit numbers it turns out that I can show to you that there are two different two subsets that uh if if I look at the sum of their elements so I look at the sum of the elements of the first subset and look at the sum of the elements of the second subset said that I can find two subsets that have an equal sum now if you just look at those numbers and I've now picked like 10 arbitrary double digit numbers well usually it's pretty hard to figure out whether that's really true or not maybe I have been selecting the numbers in such a way that it's easy to see um I mean we can still try to wrap our minds around it and try to really solve this constructively by giving an example it turns out that we can prove this statement and we will use pitchon Hall principle and we do not even have to actually oh you have a question oh sure we can make a double digit so uh I can put a zero before too if I want to but uh but yeah just select something else it doesn't really matter um so uh yeah so what we going to show now is that through the pitchon hole principle we can prove that there are two subsets that have the same sum and just by inspection it's a very hard problem to solve um so I do not even give you an example but we can still show this so how can we go ahead about with this so let's think together about this problem so I want to choose two sets X and Y and somehow I want to have a mapping right I want to have a mapping from uh any uh double digit uh set of numbers uh somehow I want to map that to sums because that's what I'm interested in I'm interested in sums and I want to show something about how hard it is or or or or I want to show something about uh subsets of uh of these double digit numbers so what do I do I take X as The Collector of subsets of these numbers and uh I want to show that there are at least two subsets that map to the same sum so let's first count how many we have here we already did this we made a mapping from subsets to uh to uh binary sequences uh bit sequences and in this case we have uh uh 10 numbers so we have 2 to the^ 10 possible um subsets so this is equal 2024 now Y is going to be uh the sum of uh a subset so uh what do I know um I knowe that the uh possible sums range from zero all the way to well what's the maximum sum that I can have out of uh 10 double digit uh a subset of 10 double digit num so I can select all the 10 um elements in this set and they are double digits so at most they are 99 so I know that this set is really the set of all possible sums now we know that 1024 is more than 990 so the cardinality of X is more than cality of Y so by the pitch andhole principle we know that um there exist at least two different elements of X in our case there exist two different subsets that map to the same sum uh the same elements in why the same sum so now we have shown that uh even though we did not show we have not shown any particular example that demonstrates this claim that are two different subsets that that have the same sum we still got a prove using counting that this is true so this is called a non-constructive proof uh let me write it down and this is uh a great way of proving uh properties okay so now we can continue with um um with another uh definition where we look at uh another property over here we talks about surjectivity injectivity and bctiv um and Division Rule now we will talk about uh the following property we say that a k to one mapping K to one function is um f from X to Y actually Maps exactly K elements K elements to of x to every element of Y so what do we know well we can have the following counting rule that we call the division Rule and it says that if f is such a type of function so if f is uh uh K 21 well then we know that the cardinality of the domain is equal to um K times the cardinality of Y and why is that well exactly K elements of X mapped to each element of Y so I have the first element of Y we have K elements of X that map to it the second element of Y K elements mapped to that one so we know that the domain is exactly K times uh the range K times the size of the range now this division rule actually generalizes the bje rule uh which I've put over there the projection Rule and why is that well that's because um a function is a bje if and only if it is actually one uh 2 one so if you replace K by one we have that uh exactly one element of X is met to every element in y and that's the definition of a bje and the bje rule says that if you have a bje then the cardinality of the domain is equal to the is equal to the cardinality of the range so for k equals 1 here so let's give an example on how this works um I think we can take this out actually so let's give an example using a chessboard um where we have two identical Rooks and we want to count the number of ways we can put them on the chessboard in such a way that uh the two Rooks cannot uh uh see one another meaning that uh the Rooks are on different rows and on different columns so let's give an example so the Example example is like this so how many ways how many ways do we have uh to place uh two identical Rooks um on a chessboard in such a way that no row uh or column is shared so how can we do this um well uh for for example let's look at a uh chess board and suppose we have a rook over here and a rook over here and say the first Rook is on Row one and on column one and the second Rook is on row two uh R2 on row R2 and on the column that is indexed by C2 so how can we describe such configurations well I could uh describe this by using a sequence in which I look at the placement of the first rook and then describe the placement of the second Rook so I may have R1 C1 and then R2 and C2 so this could be uh a way to describe the positioning of these Rooks and I could create a mapping F that is doing this for me and uh so if I call this uh an example of uh a valid uh so let's y be the set of valid rot configurations and this is one example of it so this is part of this set and if I Define X is all the sequences all the sequences R1 C1 R2 and C2 such that well uh The Rook over here does not share a row with the Rook uh that is described by this position so R1 is not equal to R2 and they also do not share a column so the rot uh the first rot has column C1 the second one is on column C2 so also C1 and C2 should be different um so these sequences are really placements right so this describes Rook uh one this describes Rook two and the whole combination is really um a placement so now I have described a function f that Maps a sequence that describes the position of the first and the second R Rook Maps a SE such a sequence to an element in Y which is a valid Rook configuration so now let's uh have a look at how we can apply the division rule so um is this function by bjective is that true so so is it true that um every uh so I have a mapping that goes from here to here um but is it true that every valet uh root configuration is mapped to exactly once is that is that true is this the only representation that I can use to that will the only sequence that will map using this this function f into a fet configuration yep that's true so you can switch uh Rook one and Rook two um and it will still look the same the two Rooks are identical they look exactly the same so we have again the exact same configuration and uh we can see that this particular uh sequence also maps to the Sim it just switch swap uh the positions so we have R2 C2 R1 and C1 also Maps under F to that same configuration and it's exactly uh those are the exact two possibilities that can happen to that that map to this configuration every F configuration is mapped to to exactly two times so now we can use a division rule because f is 2 to one so f is two 2 one what does that mean if you apply the division rule it means that the cardinality of uh all those sequences is equal to two times the cardinality of all fell configurations or in other words uh the cality of all the valid configurations is the cality of all those possible sequences divided by two so now we can start counting X over here so how do we do that well we're going to use something similar as what we did when we were counting permutations and I will generalize that in a moment um so how do we go about this well let's have a look if I have um R1 uh C1 and r 2 and C2 so this is a sequence so how many choices do I have well a chessboard has eight rows so I can choose eight possible rows for the first Rook it also has eight columns so I have eight possible choices for the column but what about the second Rook well the second Rook uh can be on any row except the one that I've already selected for the first so that is 8 minus one we have seven possible choices to select the row for the second Rook it must be different from the one that was already selected and I have seven possible choices and similarly for this particular column as well the column has to be different so how many choices do I have well it's not eight it's one less because I've already selected the one for the first Rook so I have seven choices so the cardinality of X is actually equal to 8 8 7 7 so it's 8 7^ 2ar / 2 so now we have counted by using the division rule we saw that we have to divide this by two we have counted the total number of valid configurations so now we are going to generalize this uh principle that we have talked about here and um we will do that over here I think yeah and that's the generalized product rule so the generalized product rule is as follows uh it's essentially uh saying that if we have a set of of uh of of of of sequences of length k then how can we count those well if we know the following property so let me first write out the generalized generalized uh product rule is as follows uh let s be a set of length K sequences then I know that if there are um N1 possible first entries and if I know that once I've selected my first entry there are n two possible second entry or and two possible second entries for each first entry and if I continue like this so my choice for the third term in the sequence is uh always uh I've have always N3 possible choices given my selection for the first two entries so if I have that property that continues in that way so let me write it out so we have n three possible uh third entries for each combination in this case of uh first together with second entries and if I continue this all the way to NK so NK possible K entries for each combination of all the previous entries so then I know that the set s can be counted as well I have N1 possible choices for the first entry once I've chosen and fixed that one I have N2 possible choices for the second then n three possible choices for the third and then go all the way to NK now remember that well let's first talk about it from the perspective of the chess problem here I got eight possible choices for R1 given R1 I don't care I still have eight possible choices for the for the column here so if eight choices here but for the third one once I have uh selected R1 and C1 I only have seven choices left for R2 and seven choices left for C2 so that's an example where we use this particular U generalized product rule also uh when we were counting the number of permutations we were saying we can fix the first entry of a permutation in N ways if I have a permutation Over N elements and then I have the second uh uh um entry the second term of a permutation has only n minus one choices because I've already chosen one then the next one has n minus three choices because I've already selected two of them so have only n minus two choices left then I have minus three choices because you've already selected three and so on and I get n factorial so that's the same kind of uh principle that we have here so let me give an example uh where we can see uh how this works um so so what do we do um in this example we want to uh count the number of committees so it's the exact same kind of principle that we're going to talk about so we're going to count a number of committees uh described by sequence XYZ where X the first one is say the uh leader of the committee uh this the second one in indicates say the secretary uh the third one is some consultant so they're all different they have all different roles um such a committee is selected from n members and in how many ways can I do this well I have n ways to choose my first uh uh term in the sequence I have n ways to choose the leader so there's n ways to choose X how many ways I have to choose y well if I've chosen already a leader I need to choose someone else so I have n minus one members left n minus one ways to choose uh y um I'm just not allowed to choose X and then I will have n minus two ways to choose C except X and Y and so for each X I have only n minus one ways to choose y for each X and Y I have only n min-2 ways to choose Z so if I multiply all this together I get n N - 1 n minus 2 to choose uh all these Comm this is the total number of POS committees that I can select from an N member set of people so let's go to a little bit of a different uh example that uses the same principle um okay let's uh make some space uh in this second problem we going to count uh I will Define to you a defective dollar bill it's not really the effective but it's a property that we will assign to a dollar bill so and you can check for yourself whether you have one in your wallet so let's define an effective dollar bill to have uh the property that if you look at the uh 8bit um at the 8bit serial number um it has the property that uh some of the digits appear more than once so some digit appears more than once in the 8bit serial number so you can check your own wallet and check for your $1 bills and check whether you have a defective dollar it seems to be pretty specific and rare property right well check your dollar check your one uh che check your dollar bills you'll figure out that you have uh probably defec of dollar bill in your wallet so that's kind of weird so this seems this property if you look at that it seems to be something that is maybe a little bit more common than we thought it is it seems to be so special so let's do a counting argument and find out what's happening here so let's uh look at a fraction of uh the N defective so we count the opposite uh of the non-defective uh bills well that's the number of non uh defective serial numbers divided by uh the total number of serial numbers and let's call these uh small X and Y and count these so let's see um so first of all let's count why well that's easy I have eight digits in my serial number so I have 10 choices for the first digit 10 choices for the second one and so on in total I have 10 10 10 10 the^ 8 choices what about X well I'm using uh again our generalized um product rule over here well if I'm going to have a nondefective dollar bill then all the digits in the eight digit serial number have to be different so for the first digit I have 10 choices now that I've selected my first digit I have nine digits for my nine digits for my second choice for my second digit in the serial number then I have eight possible choices seven because I've already selected three and I cannot choose those anymore time six 5 4 3 and now I have chosen an eight digit serial number in which all the digits are different okay so how many are these well this is actually equal to 10 factorial / 2 factorial and it turns out to be something like 1, 84,4 possible choices so now let's look at the fraction it turns out if you divide this by this we get a really very small fraction this is actually equal to 1. 18144 per so a very small fraction is non defective so almost all the dollars are sort of defective it simply means that they have the spe special property that uh some digit occurs occurs more than once so it's kind of interesting so we can already see that by counting we can we it's sometimes a little bit counterintuitive because if I would see this particular property I would in first instance think this is a very special property but it's not true it's very common it turns out now a special uh case of the generalized product rule is the product rule um and this is defined as follows um we're going to first OFA to find a product over sets and definition is that the product of a set A1 with A2 up to a n is actually equal to the set of sequences where each element so the first entry is selected from the first set the second entry is selected from the second set and so on and now the product rule tells us by just applying that reasoning over here that um the cardinality of the product of all those sets is actually equal to uh the cardinality of the first set multiplied by the cality of the second set and so on up to the cardinality of the last set because we have this number of choices for the very first element over here this number of choices for the second one and so on and we apply that rule and we see that this is the result um now and we use this in in specific to count all the uh nbit sequences we have exactly uh two choices for the first bit uh we have the set 01 in our example times the set 01 over here and so on and that's how we derived that uh we have 2 2 two to the power uh n choices for an N bit sequence so now we come to the sum Rule and we will give uh an example for that so the sum rule states that if we look at sets then we may be able to count their Union and we will consider a very specific case in next lecture we will talk about the general uh case uh so the sum rule is the f is the following uh counting uh mechanism if the sets A1 up to a n are all disjoined so they are disjoint sets then we know that uh if I try to count the union of all those sets it's going to be the sum of the separate cardinalities so let me just write it out actually so it's the cardinality of A1 Plus A2 all the way to a m why is this well all the sets are disjoined so there are no intersections between sets uh that contain elements all the intersections are empty so counting the union is really counting each separate set and that's why we have the sum and in next lecture we will talk about the inclusion exclusion Rule and then we will take into account that we have intersections that are nonempty but let's give now an example um where we um talk about where we count the number of passwords with certain probability with certain properties and we will apply all these different rules TOS together and that's the type of problems that you would like to be able to solve so in our example in our last example here we have that passwords have the following property uh they are uh six to eight symbols so that's property one we have that the very first symbol must be special in the sense that uh it is a letter and this can be an upper or lower case and say that the other uh symbols are actually uh letters or digits so let's count the total number of possible passwords we're going to use the sum rule um so let's define um what kind of sets we are taking the symbols from so the first set is for the first symbol which we call F of first we have all the letters a b c in lower case and then all of them in uppercase and in total we have 52 elements in this set for the second symbol and the or or the other symbols we have all these uh letters but also all the digits 0 1 up to 9 and in this this set actually has cardinality 62 we added 10 digits so let's talk about um the uh actually we like to use this in the Su rule as well so how do we count what kind of possibilities do we really have so let's describe this set of uh password explicitly in a formula so let P be the set of passwords of possible passwords and this one is actually equal to well I need to choose a first symbol and then I need to choose a second symbol and a third and a fourth and a fifth and a six that's one possibility I use six symbols I can also use seven or eight symbols but this is one of them we also denote this as s to the^ five that's an equivalent notation um the other possibilities for passwords are that we first choose an entry from f a letter and then then we will need to choose another six symbols in total we have seven and we have another possibility where we choose a first symbol and then we choose seven other symbols so this has eight symbols this has in total seven this one has in total six symbols so this covers all the possible passwords so let's count them we know that these sets are all different they are distinct these are sequences that have six entries seven and eight entries so if you look at the cardinality of P it's actually equal to the sum by the sum rule that we just described over there um is equal to the very first one f s ^ 5 plus the cardinality of f um the product of f with s to the power six and we have f um cross product uh I mean times s the^ 7 so this is by application of the uh sum Rule and now we can apply the product rule very simply which is this one so that's uh equal to the cardinality of f times the cality of s to the^ five and then we have the same rule applied to this one it's the cardinality of ftimes the cardinality of s to the^ 6 and then we have the cardinality of ftimes the cality of s to the^ 7 um now it's simply plug in these numbers and then you will have the total number of passwords that you uh that you can select from it turns out to be about 1.8 10 to the^ 14 so here we have applied both the sum Rule and a product rule so in general you will see that you have to apply multiple rules together in order to um find an answer to your counting problem and you will see that on the problem set and we will give a few more examples in next lecture and we will start talking about the generalization of the sum rule called inclusion exclusion and we will give you another type of proof technique called com called combinatorial proofs all right good good luck with the problems
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Phys 325 Discussion 31 – Drag Force Intuition We learned last week that the resistance exerted by fluids on moving objects is well described by a linear drag force that’s proportional to the object’s speed plus a quadratic drag force that’s proportional to the object’s speed-squared: fair = −(bv + cv2) ˆ v = flin + fquad (Recall that this expression is only valid as long as the object moving through the air is travelling at speeds well below the speed of sound.) For spherical objects of diameter D the constants b and c have the forms b = βD and c = γ D2 , where β and γ properties of the medium through which the object is travelling. When that medium is air at STP2 we get these specific values: linear constant: b = β D where βair = 1.6 × 10–4 N·s/m2 quadratic constant: c = γ D2 where γair = 0.25 N·s2/m4 Let’s use these relations to develop some intuition about air resistance in everyday situations!
Q1. Baseballs and Beach Balls Taylor 2.1; Checkpoints 3 (a) Consider a ball of diameter D that flies through the air at STP. Show that the ratio of the quadratic to the linear drag force is fquad / flin = 1.6 ×103 s/m2 ( ) Dv .
(b) Taking the case of a baseball, which has regulation size D = 7 cm, find the approximate “threshold” speed vthresh of the baseball at which the two drag forces are equally important. Since baseball speeds are usually given in miles per hour (mph), you might want to use that unit to get a feel for your results → use the convenient approximate relation 1 m/s ≈ 2 mph to convert from metric. Use your finding to address these questions: • For what approximate range of speeds is it safe to treat the drag force as purely quadratic?
• Under normal conditions is it a good approximation to ignore the linear term? (c) Answer the same questions for a beach ball of diameter 70 cm. Q2. Origin of the Linear Term Taylor 2.2 The origin of the linear drag force on a sphere in a fluid is the viscosity of the fluid. According to Stokes’ Law, the viscous drag on a sphere is flin = 3π η Dv where v is the viscosity (see footnote4) of the fluid, D is the sphere’s diameter, and v is its speed (as before). (a) Use Stokes’ Law to derive the form given above: flin = bv with b = βD. (You’ll get a formula for β.) (b) Given that the viscosity of air at STP is η = 1.7 × 10–5 N·s/m2, verify the value of βair given above.
1 The assignments are numbered by week — there was no discussion #2 because of the holiday last week. 2 STP stands for “Standard Temperature and Pressure” and refers to T = 0° C and P = 100 kPa ≈ 1 atmosphere. 3 Q1 (b) vthresh ≈ 0.02 mph → quadratic totally dominates for typical baseball speeds of 80–100 mph (c) vthresh ≈ 0.001 m/s → quadratic totally dominates for typical beach ball speeds of a few m/s 4 Viscosity is not an easy concept to understand. For now, you can’t do better than to read Taylor’s textbook: he has a wonderful footnote to this question (problem 2.2) that provides an excellent physical definition of viscosity. Q3. Origin of the Quadratic Term Taylor 2.4 The origin of the quadratic drag force on any projectile in a fluid is the inertia of the fluid that the projectile has to push as it moves forward. (a) Assuming the projectile has a speed v and a cross-sectional area A (i.e. the area perpendicular to its velocity), and that the fluid through which it moves has a mass density ρ (units: mass/volume), show that the rate at which the projectile displaces fluid (units: mass/time) is ρAv . Hint: make a sketch!
(b) Making the simplifying assumption that all of this fluid is accelerated to the speed v of the projectile, show that the net drag force on the projectile is ρAv2 . Summary: We now have a decent understanding of the quadratic v2 dependence of “everyday” air resistance: one factor of v comes from the amount of fluid that must be pushed per unit time, and the other comes from the speed to which that fluid must be accelerated. FYI: It is not true that all the fluid the projectile encounters is accelerated to the full speed v, but one might reasonably guess that the actual force would have the form fquad =κ ρ Av2 , where κ is a number less than 1 that would depend on the shape of the projectile (e.g. κ would be small for a streamlined body and larger for a body with a flat front end). As it happens, this guess is correct, and for a sphere the factor κ is found to be 1/4. (c) Show that the expression above for fquad reproduces the form at the top of the previous page, with the constant c = γD2 as advertised. Your work will give you an expression for γ; use it to verify our value for γair given that air at STP has density ρ = 1.29 kg/m3 and that a sphere has κ = 0.25 .
Q4. A Brief Interlude: Magnetism and Math ≈ Taylor 1.45 Another velocity-dependent force is the Lorentz force, F = q v × B, that a magnetic field B exerts on a moving particle of charge q and mass m. Applying Newton, we get a = (q v × B) / m . Prove these useful results: • If a(t) is always orthogonal to v(t) — as in the magnetic-force case — then | v(t) | never changes; • and the converse: if | v(t) | is constant, then a(t) must be orthogonal to v(t) .
Stuck? …. Are you sure? ☺ … see Lecture 2A where we calculated the time-derivative of a vector magnitude. Q5. The Reynolds Number Taylor 2.3; Checkpoints 5 (a) Show that the ratio of the quadratic and linear drag forces on a moving sphere is fquad / flin = R / 48 , where the dimensionless Reynolds number is R ≡Dvρ /η . FYI: The Reynolds number is a common measure of the relative importance of linear and quadratic drag, so you should know about its existence. When R is very large (e.g. for big, fast objects) the quadratic force dominates; when R is very small (e.g. for small, slow objects) the linear force dominates … more or less. ☺ The factor of 48 between R and the quadratic-to-linear drag ratio is for spherical objects; objects of other shapes have different constants of proportionality.
(b) Find the Reynolds number for a baseball (diameter 7 cm) that is thrown through STP air (density 1.29 kg/m3 and viscosity 1.7 × 10–5 N·s/m2) at a world-class pitching speed of 100 mph. (c) Find the Reynolds number for a steel ball bearing (diameter 2 mm) that moves at 5 cm/s through glycerin (density 1.3 g/cm3 and viscosity 12 N·s/m2 at STP). 5 Q5 (b) R ≈ 3 × 105 >> 48 → quadratic term dominant (c) R ≈ 0.01 << 48 → linear term dominant
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Simple Probability Puzzles From Brilliant.Org — And a Word about ChatGPT and GROK -
Pascal Bercker
8 min readSep 23, 2024
These are fairly simple probability puzzles from the educational website Brilliant.org who often sponsor STEM-oriented YouTube videos. My only added-value objective here is to show how to model these with simple Bayesian Belief Networks (I use Netica but there are others) something they never seem to do (at least not yet). I will also provide some animated gifs to show how dynamic and interactive Bayesian networks can be.
Let’s begin.
At first, I was confused because my initial thinking was that you should be equally likely to get heads or tails. And that’s true enough, but that’s not quite the question being asked. Let’s look at the details.
If you forgot the binomial theorem you could just “brute force” your way through this, always keeping in mind that probability is mostly just counting. We could just create 8 nodes, one for each coin flip with two equally possible states {tails, heads} (the default mode). We need only count the number of heads which we do here by using the variable index (where tails = 0 and heads = 1). (I use the “plus” function simply because I can easily copy and paste the sequence c1 through c8).
This is part of the internal table (with 2⁸ = 256 entries) doing the counting for our boolean indicator node:
Of course it’s easier with the binomial distribution and a boolean indicator node. Let me show you how.
Notice that we get much more than just the distribution. For example, see what would happen if we were to set our boolean node to TRUE:
(Yes, I know it initially asked about the probability of more tails than heads, but just go with this instead! It’s the same logic.)
Imagine a well-shuffled deck of cards numbered 1–20. Each number is equally likely but note the dependence between M and N.
To ensure that M does not equal N we need a simple equation for node N:
P (N | M) = M == N ? 0 : 1
If M = N then false (0) otherwise true (1).
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The other equation compares M and N. Is “M > N?” true?
P (B | M, N) = M > N ? 1 : 0 P(B |M,N) = M> N ?1 : 0
If M > N then Molly wins otherwise Nia wins. Clearly, neither has a better chance of winning than the other. As a bonus feature, suppose that we set the nod to TRUE, what do you think will happen to the distribution to M and N?
Necessarily, whatever M is, it cannot be 1 since it’s greater than N, and N cannot be 20 for the same reason. As a bonus, we also see the mean (14 for M and 7 for N, which makes sense given that we are here assuming that M won this round).
BEFORE we compile an equation:
The empty internal table for who wins:
P (wins | A, Z) = A > Z ? wins == Anela : A < Z ? wins == Zeke : wins == draw P(wins |A,Z) = A> Z ? wins ==Anela : A< Z ? wins ==Zeke : wins == draw
Once compiled we get the updated network:
For the sake of illustration, let’s see what happens to the probability distributions when given who wins, or maybe draws. Watch (20 seconds):
This is why a Bayesian network is more fun and dynamic to play with than mere static calculations.
Presumably, the dog is just randomly drawing the balls and are equally likely to be drawn:
Colored socks in a drawer in a dark room:
First, we graph out the generic causal network.
We need some simple equations. First, we need to establish the initial prior distribution of the available socks:
P (draw1 | ) = draw1 == R ? 4/18 : draw1 == B ? 6/18 : 8/18 4 18 6 18 8 18
Then the conditional probability of the second draw, depending on the color of the color of the first drawn sock:
P (draw2 | draw1) = draw1 == R ? ( draw2 == R ? 3/17 : draw2 == B ? 6/17 : 8/17) : draw1 == B ? ( draw2 == R ? 4/17 : draw2 == B ? 5/17 : 8/17) : draw1 ==Y ? ( draw2 == R ? 4/17 : draw2 == B ? 6/17 : 7/17) : 0 3 17 6 17 8 17 4 17 5 17 8 17 4 17 6 17 7 17 0
And a simple equation for the boolean indicator node:
P (match2 | draw1, draw2) = draw1 == draw2 ? 1 : 0 P 1 0
Once compiled we get our answer:
We can reuse our initial model and add an extra node with a slightly more complicated equation. Note that the 3rd draw will have to take into account what happened with draw1 and draw2. And the indicator node will have three states, one for each of the possible answers above to see which is the most likely. First the generic model before computing anything:
We reuse the equations we had before for draw1 and draw2 but the third draw is a bit more complicated. We could just count the possible socks colors of the remaining 16 socks (we started with 18) in each of 9 cases:
The logic is simple (but tedious). For example, if the first two socks drawn are red, only 2 red socks remain, along with 6 blue ones and 8 yellow ones. So on and so forth. We can also write out an equation, but this being too tedious, I “cheated” and asked chatGPT to just give me the equivalent equation!
P(draw3 | draw1, draw2) = draw1 == R ? ( draw2 == R ? (draw3 == R ? 2/16 : draw3 == B ? 6/16 : 8/16) : draw2 == B ? (draw3 == R ? 3/16 : draw3 == B ? 5/16 : 8/16) : draw2 == Y ? (draw3 == R ? 3/16 : draw3 == B ? 6/16 : 7/16) : 0) :draw1 == B ? ( draw2 == R ? (draw3 == R ? 3/16 : draw3 == B ? 5/16 : 8/16) : draw2 == B ? (draw3 == R ? 4/16 : draw3 == B ? 4/16 : 8/16) : draw2 == Y ? (draw3 == R ? 4/16 : draw3 == B ? 5/16 : 7/16) : 0) :draw1 == Y ? ( draw2 == R ? (draw3 == R ? 3/16 : draw3 == B ? 6/16 : 7/16) : draw2 == B ? (draw3 == R ? 4/16 : draw3 == B ? 5/16 : 7/16) : draw2 == Y ? (draw3 == R ? 4/16 : draw3 == B ? 6/16 : 6/16) : 0) : 0 2 16 6 16 8 16 3 16 5 16 8 16 3 16 6 16 7 16 0 3 16 5 16 8 16 4 16 4 16 8 16 4 16 5 16 7 16 0 3 16 6 16 7 16 4 16 5 16 7 16 4 16 6 16 6 16 0 0
For the indicator node we have this table to fill out (which would be tedious by hand):
The easiest approach is to isolate those that are all the same or all different; the rest is guaranteed to have 2 of the same colors and 1 different.
match (draw1, draw2, draw3) = (draw1 == draw2 && draw2 == draw3) ? same :(draw1 != draw2 && draw2 != draw3 && draw1 != draw3) ? all_diff : same2_diff1
This is the resulting table:
The resulting network:
For the fun of it let’s see what happens when we activate the network and cycle through all 3 possible scenarios for the “match” node. We want to see what happens to the other nodes to see if that makes sense and matches our expectations. (What we do shows up in green).
Here is the dart board. She just throws the dart randomly at the board.
This is really just a simple calculation and there is nothing obvious to model with a Bayesian network here. So let’s just calculate the relevant blue and orange areas and take the ratios.
But I asked chatGPT and GROK to solve this problem as well. I was so confused that I almost lost confidence in my obvious solution! I asked a friend to comment on the “solution” it returned. He was in no doubt and took out his red pen!
ChatGPT also had the basic details right:
But then it gets it all wrong. It starts out on the right path but quickly deviates:
Finally I asked for an image of that square dart board as described. This is what it produced.
## Brilliant | Learn interactively
Brilliant - Build quantitative skills in math, science, and computer science with hands-on, interactive lessons.
brilliant.org
## Courses
Master key skills in probability and data analysis
brilliant.org
## Norsys - Netica Application
Netica is a powerful, easy-to-use, complete program for working with belief networks and influence diagrams. It has an…
www.norsys.com
Probability
Educational
Binomial Distribution
Bayesian
Netica
## Written by Pascal Bercker
125 followers
·150 following
Philosophy | philosophy of science | logic | Bayesian networks and logic puzzles | keen on the use of Bayesian networks to solve Smullyan-type logic puzzles
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2024 AMC 12A Problems/Problem 16 - AoPS Wiki
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ResourcesAops Wiki 2024 AMC 12A Problems/Problem 16
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2024 AMC 12A Problems/Problem 16
Contents
1 Problem
2 Solution 1A (Trivial/Easy solve)
3 Solution 1B (12fact bash)
4 Solution 2
5 Solution 3
6 Solution 4
7 Solution 5
8 Video Solution, Fast, Quick, Easy!
9 Video Solution 1 by SpreadTheMathLove
10 See also
Problem
A set of tokens — red, white, blue, and black — is to be distributed at random to game players, tokens per player. The probability that some player gets all the red tokens, another gets all the white tokens, and the remaining player gets the blue token can be written as , where and are relatively prime positive integers. What is ?
Solution 1A (Trivial/Easy solve)
We have ways to handle the red/white/blue balls distribution on the denominator. Now we simply for the numerator in order to handle the black balls and distinguishable persons. The solution is therefore or
Remarks - Notice we let balls and persons be distinguishable to increase ease of calculations
~polya_mouse
Solution 1B (12fact bash)
We have total possible arrangements of distinct tokens. If we imagine the first tokens of our arrangement go to the first player, the next go to the second, and the final go to the third, then we can view this problem as counting the number of valid arrangements.
Firstly, the tokens are not all distinct, so we multiply by , , , and to account for the fact that the red, white, blue, and black tokens, respectively can switch around from where they are.
Letting denote red, denote white, denote blue, and denote black, then our arrangement must be something like . The three players are arbitrary, so we multiply by ; then, the player who gets the reds has possible arrangements, the player who gets the whites has possibilities, and the player who gets the blacks has possibilities. Our total on top is thus , and the denominator is . Firstly, we have the in the numerator cancel out part of the denominator; we thus have the following:
Our answer is
~Technodoggo
Solution 2
Assume all of them are distinct even though some have the same color,
Total possibility = (choosing 4 random token for each person)
Next, assume that all the token are already in 3 different groups (Note: 3! Ways to do so since 3 people)
We then distribute the 6 distinct black token into these 3 different groups (So 1,2,3 token for each group)
There are a total of ways in doing so
Thus the answer is
So the answer is
~lptoggled
Solution 3
We first assume there are designated red, white, and blue token players that will receive all of their respective one.
Consider each non-black token: The probability of the red player getting the first red token is , because each player has 4 empty token "slots" for a total of 12. It follows that the probability of the player receiving all 3 red tokens is , the white token player is , and the blue token player is .
The combined probability is .
Finally, we multiply the probability by to remove our initial assumption to get .
The requested sum is .
~SilverRush
Solution 4
Process start, first player get tokens. The probability that he gets 3 red tokens and 1 black token is
There is possible arrangements (RRRB,RRBR,RBRR,BRRR) and 3 possibilities who is the first, so the probability that some player gets 3 red tokens and 1 black token is
After that second player get tokens. The probability that he gets 2 white tokens and 2 black tokens is
There is possible arrangements (WWBB,WBWB, WBBW, BBWW, BWBW, BWWB) and 2 possibilities who is the second, so the probability that some player gets 2 white tokens and 2 black tokens is
The third player gets last tokens - 1 blue and 3 black tokens.
The desired probability is .
To check the result suppose that first and some) player gets 1 blue and 3 black tokens. The probability is
The probability that second (and some) player gets 3 red tokens and 1 black token is
The desired probability is
vladimir.shelomovskii@gmail.com, vvsss
Solution 5
Arranging these 12 tokens in a row, there are total of arrangements.
Player one takes the first to the fourth tokens, player two takes the fifth to the eighth, and player three takes the rest. There are 6 cases of desired arrangements:
Case 1. Player one has three red tokens (and one black), player 2 has two white tokens and player 3 has 1 blue token. There are total of arrangements in this case.
Other cases (such as player 1 has two white tokens, player 2 has three red, and player has one blue token) are all similar.
The number of desired arrangements for all cases is .
The target probability is
-J.Z.
Video Solution, Fast, Quick, Easy!
Video Solution 1 by SpreadTheMathLove
See also
2024 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 15Followed by
Problem 17
1•2•3•4•5•6•7•8•9•10•11•12•13•14•15•16•17•18•19•20•21•22•23•24•25
All AMC 12 Problems and Solutions
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.
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Organism Size
Animal Biology
Land-living Mammals
Elephant Anatomy
Animal Comparison
Animal Characteristics
Animal Size
5
What is the size of elephants compared to other land animals?
Matthew Haynes
I read a lot about animals
·
Author has 9.5K answers and 16.1M answer views
·
1y
The biggest elephant currently alive is the ten ton (ish) Craig, from Kenya. Giants like that used to be more common, but decades of poaching and ivory hunting has made them exceptionally rare. Still, a standard (but still large) bull would be closer to seven tons.
The biggest non-elephant on land is the white rhinoceros, which is usually about two and a half tons, is big at three, and the record is a bit over four. Meaning a rhinoceros is about a third the size of an elephant. A hippo is about the same size; only marginally smaller than the rhino.
The biggest animal you are likely to have seen
The biggest elephant currently alive is the ten ton (ish) Craig, from Kenya. Giants like that used to be more common, but decades of poaching and ivory hunting has made them exceptionally rare. Still, a standard (but still large) bull would be closer to seven tons.
The biggest non-elephant on land is the white rhinoceros, which is usually about two and a half tons, is big at three, and the record is a bit over four. Meaning a rhinoceros is about a third the size of an elephant. A hippo is about the same size; only marginally smaller than the rhino.
The biggest animal you are likely to have seen up close is a horse. A riding horse is about 800 to 1,000 pounds; even a huge draft horse like a Clydesdale weighs on average just under a ton. So a large bull elephant is fourteen times the size of the biggest horse you are likely to have ever seen, and even if you’ve seen a Clydesdale, it’s less than a seventh the size of an elephant.
On the off-chance you haven’t seen a horse, I am confident you’ve seen a dog. So how does an African elephant compare to a dog? Well, the dog everyone seems to want is a Labrador. A Labrador retriever is up to about eighty pounds. That makes it one tenth the size of a small riding horse, and therefore about a one hundred fiftieth the size of a big elephant. In fact, each of Craig’s tusks are heavier than a lab.
Cats are also popular pets, so let’s compare elephant to them as well. Cats are usually about seven to ten pounds, with big ones getting up to twenty pounds or so. The math of cat vs elephant winds up very easy, then: a large elephant is 2,000 times the size of a cat.
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Related questions
What are the different sizes of elephants and how do they compare to other animal species?
Where is the biggest elephant found?
Is the elephant a vengeful animal?
What animal is stronger than it looks?
Which is the longest living land animal?
Matthew Haynes
I read a lot about animals
·
Author has 9.5K answers and 16.1M answer views
·
1y
Originally Answered: What are the different sizes of elephants and how do they compare to other animal species?
·
Present-day elephants range between three and eleven tons, between eight and thirteen feet tall, and between twenty and thirty-five feet long. The African forest elephant holds the low end of that spectrum, the Asian elephant sits in the middle, and the African bush elephant is the largest of the three.
A white rhinoceros or hippopotamus can on rare occasion break into the lower end of the forest elephant’s weight range, and the killer whale reaches more or less the exact maximum as the African bush elephant. Only certain rare beaked whales most people are unfamiliar with match the Asian elepha
Present-day elephants range between three and eleven tons, between eight and thirteen feet tall, and between twenty and thirty-five feet long. The African forest elephant holds the low end of that spectrum, the Asian elephant sits in the middle, and the African bush elephant is the largest of the three.
A white rhinoceros or hippopotamus can on rare occasion break into the lower end of the forest elephant’s weight range, and the killer whale reaches more or less the exact maximum as the African bush elephant. Only certain rare beaked whales most people are unfamiliar with match the Asian elephant well.
This is all very well and good, but most people have never and will never get particularly close to a hippo, rhino, or killer whale, let alone an obscure beaked whale. Indeed, I would advise against getting up close and familiar with any of them unless you’re a trained zoologist. So how does an elephant compare to some animals I can be confident we are all intimately familiar with?
The most popular pet on earth is the dog, and, per Country Living, the most popular dog is the French bulldog (which unfortunate as I know only one French bulldog owner who is at all pleasant and she acknowledges she’s in the minority there). Frenchies are between eleven and thirteen inches tall and at most twenty-eight pounds. This means a small forest elephant is eight times the height and almost two hundred fifteen times the weight of a Frenchie. A very large bush elephant would be thirteen times the height of the Frenchie and weighs as much as almost 786 of them. Suffice to say elephants are very big. Conveniently, the second-most popular dog, the Labrador, is about twice the size of the Frenchie, meaning a large bull elephant is merely three hundred and ninety-two times the weight and six and a half times the height.
The other most popular pet is the cat. An average house cat weighs about ten pounds, which makes for nice, easy math as an average Asian elephant weighs about ten thousand pounds. The elephant is thus about a thousand times the weight of the cat. In couldn’t find height data for house cats, but they are conveniently a foot long, making them one-twenty-fifth the length of the same Asian elephant cow.
Robert L. Marshall
Senior Software Developer (2010–present)
·
7y
Related
Is the elephant the largest land animal?
Yes, currently the male African Bush Elephant is the largest land animal. However, the largest land animal ever discovered was a dinosaur called Argentinosaurus.
Here is a picture of it next to an Elephant, and a Human.
Yes, currently the male African Bush Elephant is the largest land animal. However, the largest land animal ever discovered was a dinosaur called Argentinosaurus.
Here is a picture of it next to an Elephant, and a Human.
Claire Jordan
Worked at National Health Service (NHS)
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Author has 61.1K answers and 555M answer views
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9mo
Related
Why are there no small elephants?
There are some moderately small, rhino-sided elephants, and have been even smaller ones on islands in the past. But if they got too small they would no longer look like elephants. Their thick, pillar-like legs to support their weight and huge ears to radiate heat are functions of their large size: a really small elephant would just look like a tapir with a longer nose and tusks. It might even be furry: this little chap, who is about the size of a brown rat, is a close relative of the elephant.
There are some moderately small, rhino-sided elephants, and have been even smaller ones on islands in the past. But if they got too small they would no longer look like elephants. Their thick, pillar-like legs to support their weight and huge ears to radiate heat are functions of their large size: a really small elephant would just look like a tapir with a longer nose and tusks. It might even be furry: this little chap, who is about the size of a brown rat, is a close relative of the elephant.
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Virendra Kumar Jain
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The African Savanna (Bush) elephant is the world's largest land animal – with
adult males, or bull elephants, standing up to 3m high and weighing up to 6,000kg on average
. Males only reach their full size at 35-40 years - that's well over half their lifespan as wild elephants can live for up to 60-70 years.
Firdous Rani
BSc in Botany & Zoology, Quaid E Azam University Islamabad (Graduated 2021)
·
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They're the world's largest land animal.The African Savanna (Bush) elephant is the world's largest land animal – with adult males, or bull elephants, standing up to 3m high and weighing up to 6,000kg on average.
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Related
Where is the biggest elephant found?
African savanna elephant
The African savanna elephant is the largest elephant species, while the Asian forest elephant and the African forest elephant are of a comparable, smaller size. Asian elephants differ in several ways from their African relatives, with more than 10 distinct physical differences between them.
African savanna elephant
The African savanna elephant is the largest elephant species, while the Asian forest elephant and the African forest elephant are of a comparable, smaller size. Asian elephants differ in several ways from their African relatives, with more than 10 distinct physical differences between them.
Matthew Haynes
Works at The Church of Jesus Christ of Latter-day Saints (Mormons)
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Author has 9.5K answers and 16.1M answer views
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Related
Elephants are the largest land animals. What are the top ten largest land animals (no dinosaurs)?
Barring dinosaurs, the largest land animals ever to have lived were:
Palaeoloxodon
Indricotherium
Mammuthus
Deinotherium
Elephas
Purrusaurus
Sarcosuchus
Deinosuchus
Loxodonta
Megatherium
That’s as far as I know, anyway. My inclusion of
Elephas
may be erroneous as
Elephas recki
is sometimes considered a species of
Palaeoloxodon
(were I to remove it from the list, the rhinoceros
Elasmotherium
would be included in 10th place). In any case, half of them are elephants of one sort or another; aside from these there are three crocodiles, a rhinoceros, and a giant ground sloth. Were I to include dinosaurs, none
Barring dinosaurs, the largest land animals ever to have lived were:
Palaeoloxodon
Indricotherium
Mammuthus
Deinotherium
Elephas
Purrusaurus
Sarcosuchus
Deinosuchus
Loxodonta
Megatherium
That’s as far as I know, anyway. My inclusion of Elephas may be erroneous as Elephas recki is sometimes considered a species of Palaeoloxodon (were I to remove it from the list, the rhinoceros Elasmotherium would be included in 10th place). In any case, half of them are elephants of one sort or another; aside from these there are three crocodiles, a rhinoceros, and a giant ground sloth. Were I to include dinosaurs, none of these would crack even the top 50, as even the largest of them is about 1/4 the weight of the largest titanosaurs, all of which cluster in the 70–90 ton range.
Today, the top ten are:
African bush elephant
Asian elephant
African forest elephant
white rhinoceros
hippopotamus
Indian rhinoceros
Javan rhinoceros
black rhinoceros
giraffe
gaur
The fifth living rhino, the Sumatran rhino, falls in eleventh place. The positions of white rhino and hippo are interchangeable according to one’s source, as are the positions of black and Javan rhino. Without the Pleistocene extinctions, the list would be composed almost entirely of elephants, and would read thus (assuming the world as it was about 12,000 years ago):
Columbian mammoth
African bush elephant
woolly mammoth
Megatherium
Asian elephant
Stegodon
Cuvieronius
American mastodon
Elasmotherium
African forest elephant
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“This is the key to healthy aging,” says Norris. “I’m living pro
Just look at the legendary Chuck Norris’s advice since he is now a whopping 81 years old and yet has MORE energy than me. He found a key to healthy aging… and it was by doing the opposite of what most of people are told. Norris says he started learning about this revolutionary new method when he noticed most of the supplements he was taking did little or nothing to support his health. After extensive research, he discovered he could create dramatic changes to his health simply focusing on 3 things that sabotage our body as we age.
“This is the key to healthy aging,” says Norris. “I’m living proof.”
Now, Chuck Norris has put the entire method into a 15-minute video that explains the 3 “Internal Enemies” that can wreck our health as we age, and the simple ways to help combat them, using foods and herbs you may even have at home.
I’ve included the Chuck Norris video here so you can give it a shot.
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Related
What are some possible explanations for why no animals are larger than elephants (or close in size)?
Bones have a limited maximum strength, due to the way they’re built.
If an animal grows too large, its bones would not be able to support the weight, and they would break. That would cause the animal to starve and die and not reproduce. That would lead to eventual extinction.
Mr Darwin says.
And if the bones didn’t break, the connective tissue between the bones and the animal would not be able to support the weight, and they would snap.
That’s why land-dwelling whale ancestors returned to the ocean. They could grow no larger on land but, buoyed in the water, the stress on their limbs was largely e
Bones have a limited maximum strength, due to the way they’re built.
If an animal grows too large, its bones would not be able to support the weight, and they would break. That would cause the animal to starve and die and not reproduce. That would lead to eventual extinction.
Mr Darwin says.
And if the bones didn’t break, the connective tissue between the bones and the animal would not be able to support the weight, and they would snap.
That’s why land-dwelling whale ancestors returned to the ocean. They could grow no larger on land but, buoyed in the water, the stress on their limbs was largely eliminated.
Whales still have vestigial hips and legs, but they only use them to play soccer on Saturday nights.
And to head off arguments: the biggest dinosaurs were not as heavy as the modern blue whale.
Seraphina Aizen
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Author has 8.3K answers and 6.6M answer views
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Related
What are some possible explanations for why no animals are larger than elephants (or close in size)?
Lots of animals are larger than elephants.
Lots of animals are larger than elephants.
Ciro Pabon
Lives in Cali, Colombia (2008–present)
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Author has 8.1K answers and 21.7M answer views
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3y
Related
Are there non-mammals that are bigger than elephants?
Of course, even if you restrict yourself to animals, no plants.
A very large elephant weighs 10 tons and measures 6 metres.
Adult whale sharks weighs up to 45 tons following reports or 21 tons measured and up to 20 metres in length.
As for longer animals that weigh less there are plenty:
Basking sharks are longer at 12 meters measured, but weigh only 5 tons.
The rare oarfish has been measured up to 13 metres, but weighs less than one ton.
Lion Mane Jellyfish has been measured at 75 metres in tentacle spread and Portuguese Man O´war measures up to 53 metres (one stung me when I was 16 and left me sca
Of course, even if you restrict yourself to animals, no plants.
A very large elephant weighs 10 tons and measures 6 metres.
Adult whale sharks weighs up to 45 tons following reports or 21 tons measured and up to 20 metres in length.
As for longer animals that weigh less there are plenty:
Basking sharks are longer at 12 meters measured, but weigh only 5 tons.
The rare oarfish has been measured up to 13 metres, but weighs less than one ton.
Lion Mane Jellyfish has been measured at 75 metres in tentacle spread and Portuguese Man O´war measures up to 53 metres (one stung me when I was 16 and left me scars for life, hurt like hell, the thing was huge).
Tapeworms have been measured at 30 metres in length.
Parasite round worms of whales reach 9 metres.
Beluga sturgeon, white shark, reef manta ray, Greenland shark and tiger shark are bigger in length all measuring 7 metres but weigh even less at 3 tons.
Colossal squid has been measured at 14 metres.
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Lives in Kochi, Kerala, India (2022–present)
·
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Is the elephant the largest animal on earth?
No, Elephants are not the largest animal on the earth.
Blue whales are the largest animals ever known to have lived on Earth. These magnificent marine mammals rule the oceans at up to 100 feet long and upwards of 200 tons.
No, Elephants are not the largest animal on the earth.
Blue whales are the largest animals ever known to have lived on Earth. These magnificent marine mammals rule the oceans at up to 100 feet long and upwards of 200 tons.
D. Legrand
I'm obsessed with increasing my knowledge in the subject of Biology
·
8y
Related
How big is an elephant?
Like David said it depends on the location of where the elephants originated. An Asian elephant or African forest elephant could possibly weigh up to 6,000 lbs. while an African bush elephant could weigh up to 13,000 lbs. and stand 11–14 ft. tall.
Now compare that to the average man. Geez.
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Where is the biggest elephant found?
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Invariance of a tensor under coordinate transformation
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I know, that a tensor is a mathematically entity that is represented using a basis and tensor products, in the form of a matrix, and changing a representation doesn't change a tensor, is kind of obvious.
So does the invariance of a tensor under coordinate transformation mean what I stated above or does it mean that under a set of particular transformation the representation of a particular tensor also doesn't change.
Quoted from Wikipedia:
A vector is invariant under any change of basis, so if coordinates transform according to a transformation matrix L, the bases transform according to the matrix inverse L−1, and conversely if the coordinates transform according to inverse L−1, the bases transform according to the matrix L.
Can someone please shed some light on this?
tensor-calculus
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edited Jun 30, 2014 at 20:56
Qmechanic♦
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My answer from a few days ago might be what you're looking for. If not, I am sure we already have discussed what you are asking somewhere else on the site.
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Regarding what you quoted: a vector is represented by the sum of a set of basis vectors times the vector components. If the components transform according to L, then the bases will transform according to L−1, which means that when you multiply the bases with the components (to make the vector), you will get the same result every time (since L⋅L−1=I). This is what is meant by invariance.
Invariance of a tensor means basically what you stated above- the tensor itself doesn't change under a change of coordinates (like I explained). However, the tensor's components can very well change.
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answered Jun 30, 2014 at 20:58
Physics LlamaPhysics Llama
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I guess there is two different notions of invariance of tensors. First notion is that if you look at a tensor as a mapping then the first notion of invariance is what you mentioned above. The other notion of invarance is that you do transformation but the " component" of metric does not change. For instance, if we do Lorentz transformation then the Minkowski metric is invariant, meaning that the component will be +1 , -1 , -1 ,-1 .
I might be wrong!!
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edited Feb 21, 2016 at 6:15
user36790
answered Nov 18, 2015 at 21:15
Mohammad AbediMohammad Abedi
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Malinowski should be Minkowski?
– Kyle Kanos
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Nov 18, 2015 at 21:48
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In probability theory and statistics, the continuous uniform distribution or rectangular distribution is a family of symmetric probability distributions. The distribution describes an experiment where there is an arbitrary outcome that lies between certain bounds. The bounds are defined by the parameters, a and b, which are the minimum and maximum values. The interval can be either be closed (eg. [a, b]) or open (eg. (a, b)). Therefore, the distribution is often abbreviated U (a, b), where U stands for uniform distribution. The difference between the bounds defines the interval length; all intervals of the same length on the distribution's support are equally probable. It is the maximum entropy probability distribution for a random variable X under no constraint other than that it is contained in the distribution's support.
rectangular distribution probability theory probability distribution
1. Definitions
1.1. Probability Density Function
The probability density function of the continuous uniform distribution is:
: [math]\displaystyle{ f(x)=\begin{cases} \frac{1}{b - a} & \mathrm{for}\ a \le x \le b, \[8pt] 0 & \mathrm{for}\ x\lt a\ \mathrm{or}\ x\gt b \end{cases} }[/math]
The values of f(x) at the two boundaries a and b are usually unimportant because they do not alter the values of the integrals of f(x) dx over any interval, nor of x f(x) dx or any higher moment. Sometimes they are chosen to be zero, and sometimes chosen to be 1/(b − a). The latter is appropriate in the context of estimation by the method of maximum likelihood. In the context of Fourier analysis, one may take the value of f(a) or f(b) to be 1/(2(b − a)), since then the inverse transform of many integral transforms of this uniform function will yield back the function itself, rather than a function which is equal "almost everywhere", i.e. except on a set of points with zero measure. Also, it is consistent with the sign function which has no such ambiguity.
Graphically, the probability density function is portrayed as a rectangle where (b-a) is the base and (1/(b-a)) is the height. As the distance between a and b increases, the density at any particular value within the distribution boundaries decreases. Since the probability density function integrates to 1, the height of the probability density function decreases as the base length increases.
In terms of mean μ and variance σ2, the probability density may be written as:
: [math]\displaystyle{ f(x)=\begin{cases} \frac {1}{2 \sigma \sqrt{3}} & \mbox{for }-\sigma\sqrt{3} \le x-\mu \le \sigma\sqrt{3} \ 0 & \text{otherwise} \end{cases} }[/math]
Example 1. Using the Uniform Probability Density Function
For random variable X
X~U(0,23)
Find P(2 < X < 18):
P(2 < X < 18) = (18-2)(1/(23-0)) = 16/23.
In graphical representation of uniform distribution function [f(x) vs x], the area under the curve within the specified bounds displays the probability (shaded area is depicted as a rectangle). For this specific example above, the base would be (18-2) and the height would be (1/23).
Example 2. Using the Uniform Probability Density Function (Conditional)
For random variable X
X~U(0,23)
Find P(12 < X | X > 8):
P(X > 12 | X > 8) = (23-12)(1/(23-8))=11/15.
The example above is for a conditional probability case for the uniform distribution: given X > 8 is true, what is the probability that X > 12. Conditional probability changes the sample space so a new interval length (b-a) has to be calculated, where b is 23 and a is 8. The graphical representation would still follow Example 1, where the area under the curve within the specified bounds displays the probability and the base of the rectangle would be (23-12) and the height (1/15).
1.2. Cumulative Distribution Function
The cumulative distribution function is:
: [math]\displaystyle{ F(x)= \begin{cases} 0 & \text{for }x \lt a \[8pt] \frac{x-a}{b-a} & \text{for }a \le x \le b \[8pt] 1 & \text{for }x \gt b \end{cases} }[/math]
Its inverse is:
: [math]\displaystyle{ F^{-1}(p) = a + p (b - a) \,\,\text{ for } 0\lt p\lt 1 }[/math]
In mean and variance notation, the cumulative distribution function is:
: [math]\displaystyle{ F(x)= \begin{cases} 0 & \text{for }x-\mu \lt -\sigma\sqrt{3} \ \frac{1}{2} \left( \frac{x-\mu}{\sigma \sqrt{3}} +1 \right) & \text{for }-\sigma\sqrt{3} \le x-\mu \lt \sigma\sqrt{3} \ 1 & \text{for }x-\mu \ge \sigma\sqrt{3} \end{cases} }[/math]
and the inverse is:
: [math]\displaystyle{ F^{-1}(p) = \sigma\sqrt{3}(2p-1) +\mu\,\, \text{ for }0 \le p \le 1 }[/math]
1.3. Generating Functions
Moment-generating function
The moment-generating function is:
: [math]\displaystyle{ M_x = E(e^{tx}) = \frac{e^{tb}-e^{ta}}{t(b-a)} \,! }[/math]
from which we may calculate the raw moments m k
: [math]\displaystyle{ m_1=\frac{a+b}{2}, \,! }[/math]
: [math]\displaystyle{ m_2=\frac{a^2+ab+b^2}{3}, \,! }[/math]
: [math]\displaystyle{ m_k=\frac{1}{k+1}\sum_{i=0}^k a^ib^{k-i}. \,! }[/math]
For the special case a = –b, that is, for
: [math]\displaystyle{ f(x)=\begin{cases} \frac{1}{2b} & \text{for}\ -b \le x \le b, \[8pt] 0 & \text{otherwise}, \end{cases} }[/math]
the moment-generating functions reduces to the simple form
: [math]\displaystyle{ M_x=\frac{\sinh bt}{bt}. }[/math]
For a random variable following this distribution, the expected value is then m1 = (a + b)/2 and the variance is m2 − m12 = (b − a)2/12.
Cumulant-generating function
For n ≥ 2, the nth cumulant of the uniform distribution on the interval [-1/2, 1/2] is Bn/n, where Bn is the nth Bernoulli number.
1.4. Standard Uniform
Restricting [math]\displaystyle{ a=0 }[/math] and [math]\displaystyle{ b=1 }[/math], the resulting distribution U(0,1) is called a standard uniform distribution.
One interesting property of the standard uniform distribution is that if u1 has a standard uniform distribution, then so does 1-u1. This property can be used for generating antithetic variates, among other things. In other words, this property is known as the inversion method where the continuous standard uniform distribution can be used to generate random numbers for any other continuous distribution. If u is a uniform random number with standard uniform distribution (0,1), then x = Inverse of F(u) generates a random number x from any continuous distribution with the specified cumulative distribution function F.
1.5. Relationship to Other Functions
As long as the same conventions are followed at the transition points, the probability density function may also be expressed in terms of the Heaviside step function:
: [math]\displaystyle{ f(x)=\frac{\operatorname{H}(x-a)-\operatorname{H}(x-b)}{b-a}, \,! }[/math]
or in terms of the rectangle function
: [math]\displaystyle{ f(x)=\frac{1}{b-a}\,\operatorname{rect}\left(\frac{x-\left(\frac{a+b}{2}\right)}{b-a}\right) . }[/math]
There is no ambiguity at the transition point of the sign function. Using the half-maximum convention at the transition points, the uniform distribution may be expressed in terms of the sign function as:
: [math]\displaystyle{ f(x)=\frac{ \sgn{(x-a)}-\sgn{(x-b)}} {2(b-a)}. }[/math]
2. Properties
2.1. Moments
The mean (first moment) of the distribution is:
: [math]\displaystyle{ E(X)=\frac{1}{2}(a+b). }[/math]
The second moment of the distribution is:
: [math]\displaystyle{ E(X^2) = \frac{b^3 - a^3}{3b - 3a}. }[/math]
In general, the n-th moment of the uniform distribution is:
: [math]\displaystyle{ E(X^n) = \frac{b^{n+1} - a^{n+1}}{(n+1)(b-a)} = \frac{1}{n+1} \sum_{k=0}^n a^k b^{n-k}. }[/math]
The variance (second central moment) is:
: [math]\displaystyle{ V(X)=\frac{1}{12}(b-a)^2 }[/math]
2.2. Order Statistics
Let X1, ..., Xn be an i.i.d. sample from U(0,1). Let X(k) be the kth order statistic from this sample. Then the probability distribution of X(k) is a Beta distribution with parameters k and n − k + 1. The expected value is
: [math]\displaystyle{ \operatorname{E}(X_{(k)}) = {k \over n+1}. }[/math]
This fact is useful when making Q–Q plots.
The variances are
: [math]\displaystyle{ \operatorname{V}(X_{(k)}) = {k (n-k+1) \over (n+1)^2 (n+2)} . }[/math]
See also: Order statistic § Probability distributions of order statistics
2.3. Uniformity
The probability that a uniformly distributed random variable falls within any interval of fixed length is independent of the location of the interval itself (but it is dependent on the interval size), so long as the interval is contained in the distribution's support.
To see this, if X ~ U(a,b) and [x, x+d] is a subinterval of [a,b] with fixed d > 0, then
: [math]\displaystyle{ P\left(X\in\left [ x,x+d \right ]\right) = \int_{x}^{x+d} \frac{\mathrm{d}y}{b-a}\, = \frac{d}{b-a} \,! }[/math] which is independent of x. This fact motivates the distribution's name.
2.4. Generalization to Borel Sets
This distribution can be generalized to more complicated sets than intervals. If S is a Borel set of positive, finite measure, the uniform probability distribution on S can be specified by defining the pdf to be zero outside S and constantly equal to 1/K on S, where K is the Lebesgue measure of S.
3. Related Distributions
If X has a standard uniform distribution, then by the inverse transform sampling method, Y = − λ−1 ln(X) has an exponential distribution with (rate) parameter λ.
If X has a standard uniform distribution, then Y = Xn has a beta distribution with parameters (1/n,1). As such,
The standard uniform distribution is a special case of the beta distribution with parameters (1,1).
The Irwin–Hall distribution is the sum of n i.i.d. U(0,1) distributions.
The sum of two independent, equally distributed, uniform distributions yields a symmetric triangular distribution.
The distance between two i.i.d. uniform random variables also has a triangular distribution, although not symmetric.
4. Statistical Inference
4.1. Estimation of Parameters
Estimation of maximum
Minimum-variance unbiased estimator
: Main page: German tank problem
Given a uniform distribution on [0, b] with unknown b, the minimum-variance unbiased estimator (UMVUE) for the maximum is given by
: [math]\displaystyle{ \hat{b}_\text{UMVU}=\frac{k+1}{k} m = m + \frac{m}{k} }[/math]
where m is the sample maximum and k is the sample size, sampling without replacement (though this distinction almost surely makes no difference for a continuous distribution). This follows for the same reasons as estimation for the discrete distribution, and can be seen as a very simple case of maximum spacing estimation. This problem is commonly known as the German tank problem, due to application of maximum estimation to estimates of German tank production during World War II.
Maximum likelihood estimator
The maximum likelihood estimator is given by:
: [math]\displaystyle{ \hat{b}_{ML}= m }[/math]
where m is the sample maximum, also denoted as [math]\displaystyle{ m=X_{(n)} }[/math] the maximum order statistic of the sample.
Method of moment estimator
The method of moments estimator is given by:
: [math]\displaystyle{ \hat{b}_{MM}= 2\bar{X} }[/math]
where [math]\displaystyle{ \bar{X} }[/math] is the sample mean.
Estimation of midpoint
The midpoint of the distribution (a + b) / 2 is both the mean and the median of the uniform distribution. Although both the sample mean and the sample median are unbiased estimators of the midpoint, neither is as efficient as the sample mid-range, i.e. the arithmetic mean of the sample maximum and the sample minimum, which is the UMVU estimator of the midpoint (and also the maximum likelihood estimate).
4.2. Confidence Interval
For the maximum
Let X1, X2, X3, ..., Xn be a sample from U( 0, L ) where L is the population maximum. Then X(n) = max( X1, X2, X3, ..., Xn ) has the density
: [math]\displaystyle{ f_n( X_{(n)} ) = n \frac{1}{L} \left(\frac{ X_{(n)} }{ L }\right)^{ n - 1 } =n \frac{ X_{(n)}^{ n - 1 } }{ L^n }, 0 \lt X_{(n)} \lt L }[/math]
The confidence interval for the estimated population maximum is then ( X(n), X(n) / α1/n ) where 100(1 – α)% is the confidence level sought. In symbols
: [math]\displaystyle{ X_{(n)} \le L \le X_{(n)} / \alpha^{1/n} }[/math]
4.3. Hypothesis Testing
In statistics, when a p-value is used as a test statistic for a simple null hypothesis, and the distribution of the test statistic is continuous, then the p-value is uniformly distributed between 0 and 1 if the null hypothesis is true.
5. Occurrence and Applications
The probabilities for uniform distribution function are simple to calculate due to the simplicity of the function form. Therefore, there are various applications that this distribution can be used for as shown below: hypothesis testing situations, random sampling cases, finance, etc. Furthermore, generally, experiments of physical origin follow a uniform distribution (eg. emission of radioactive particles). However, it is important to note that in any application, there is the unchanging assumption that the probability of falling in an interval of fixed length is constant.
5.1. Economics Example for Uniform Distribution
In the field of economics, usually demand and replenishment may not follow the expected normal distribution. As a result, other distribution models are used to better predict probabilities and trends such as Bernoulli process. But according to Wanke (2008), in the particular case of investigating lead-time for inventory management at the beginning of the life cycle when a completely new product is being analyzed, the uniform distribution proves to be more useful. In this situation, other distribution may not be viable since there is no existing data on the new product or that the demand history is unavailable so there isn't really an appropriate or known distribution. The uniform distribution would be ideal in this situation since the random variable of lead-time (related to demand) is unknown for the new product but the results are likely to range between a plausible range of two values. The lead-time would thus represent the random variable. From the uniform distribution model, other factors related to lead-time were able to be calculated such as cycle service level and shortage per cycle. It was also noted that the uniform distribution was also used due to the simplicity of the calculations.
5.2. Sampling from an Arbitrary Distribution
The uniform distribution is useful for sampling from arbitrary distributions. A general method is the inverse transform sampling method, which uses the cumulative distribution function (CDF) of the target random variable. This method is very useful in theoretical work. Since simulations using this method require inverting the CDF of the target variable, alternative methods have been devised for the cases where the cdf is not known in closed form. One such method is rejection sampling.
The normal distribution is an important example where the inverse transform method is not efficient. However, there is an exact method, the Box–Muller transformation, which uses the inverse transform to convert two independent uniform random variables into two independent normally distributed random variables.
5.3. Quantization Error
In analog-to-digital conversion a quantization error occurs. This error is either due to rounding or truncation. When the original signal is much larger than one least significant bit (LSB), the quantization error is not significantly correlated with the signal, and has an approximately uniform distribution. The RMS error therefore follows from the variance of this distribution.
6. Computational Methods
6.1. Sampling from a Uniform Distribution
There are many applications in which it is useful to run simulation experiments. Many programming languages come with implementations to generate pseudo-random numbers which are effectively distributed according to the standard uniform distribution.
If u is a value sampled from the standard uniform distribution, then the value a + (b − a)u follows the uniform distribution parametrised by a and b, as described above.
7. History
While the historical origins in the conception of uniform distribution are inconclusive, it is speculated that the term 'uniform' arose from the concept of equiprobability in dice games (note that the dice games would have discrete and not continuous uniform sample space). Equiprobability was mentioned in Gerolamo Cardano's Liber de Ludo Aleae, a manual written in 16th century and detailed on advanced probability calculus in relation to dice.
References
"Uniform Distribution (Continuous)". 2019.
Illowsky, Barbara (2013). Introductory Statistics. Rice University, Houston, Texas, USA: OpenStax College. pp. 296–304. ISBN 978-1-938168-20-8.
Casella & Berger 2001, p. 626
Nechval KN, Nechval NA, Vasermanis EK, Makeev VY (2002) Constructing shortest-length confidence intervals. Transport and Telecommunication 3 (1) 95-103
Walpole, Ronald (2012). Probability & Statistics for Engineers and Scientists. Boston, USA: Prentice Hall. pp. 171–172. ISBN 978-0-321-62911-1.
Dekking, Michel (2005). A modern introduction to probability and statistics : understanding why and how. London, UK: Springer. pp. 60–61. ISBN 978-1-85233-896-1.
Wanke, Peter (2008). "The uniform distribution as a first practical approach to new product inventory management". International Journal of Production Economics 114 (2): 811–819. doi:10.1016/j.ijpe.2008.04.004.
Bellhouse, David (May 2005). "Decoding Cardano's Liber de Ludo". Historia Mathematica 32: 180–202. doi:10.1016/j.hm.2004.04.001.
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Euler line - Wikipedia
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1 Triangle centers on the Euler lineToggle Triangle centers on the Euler line subsection
1.1 Individual centers
2 ProofsToggle Proofs subsection
2.1 A vector proof
3 PropertiesToggle Properties subsection
3.1 Distances between centers
4 RepresentationToggle Representation subsection
4.1 Equation
4.2 Parametric representation
4.3 Slope
5 Relation to inscribed equilateral triangles
6 In special trianglesToggle In special triangles subsection
6.1 Right triangle
6.2 Isosceles triangle
6.3 Automedian triangle
6.4 Systems of triangles with concurrent Euler lines
7 GeneralizationsToggle Generalizations subsection
7.1 Quadrilateral
7.2 Tetrahedron
7.3 Simplicial polytope
8 Related constructions
9 References
10 External links
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From Wikipedia, the free encyclopedia
Line constructed from a triangle
Euler's line, with the center of the nine-point circle
Medians (intersect at the centroid)
Altitudes (intersect at the orthocenter)
Perpendicular lines from the side midpoints (intersect at the circumcenter)
In geometry, the Euler line, named after Leonhard Euler (/ˈ ɔɪ l ər/OY-lər), is a line determined from any triangle that is not equilateral. It is a central line of the triangle, and it passes through several important points determined from the triangle, including the orthocenter, the circumcenter, the centroid, the Exeter point and the center of the nine-point circle of the triangle.
The concept of a triangle's Euler line extends to the Euler line of other shapes, such as the quadrilateral and the tetrahedron.
Triangle centers on the Euler line
[edit]
Individual centers
[edit]
Euler showed in 1765 that in any triangle, the orthocenter, circumcenter and centroid are collinear. This property is also true for another triangle center, the nine-point center, although it had not been defined in Euler's time. In equilateral triangles, these four points coincide, but in any other triangle they are all distinct from each other, and the Euler line is determined by any two of them.
Other notable points that lie on the Euler line include the de Longchamps point, the Schiffler point, the Exeter point, and the Gossard perspector. However, the incenter generally does not lie on the Euler line; it is on the Euler line only for isosceles triangles, for which the Euler line coincides with the symmetry axis of the triangle and contains all triangle centers.
The tangential triangle of a reference triangle is tangent to the latter's circumcircle at the reference triangle's vertices. The circumcenter of the tangential triangle lies on the Euler line of the reference triangle.: p. 447 : p.104, #211, p.242, #346 The center of similitude of the orthic and tangential triangles is also on the Euler line.: p. 447 : p. 102
Proofs
[edit]
A vector proof
[edit]
Let A B C{\displaystyle ABC} be a triangle. A proof of the fact that the circumcenterO{\displaystyle O}, the centroidG{\displaystyle G} and the orthocenterH{\displaystyle H} are collinear relies on free vectors. We start by stating the prerequisites. First, G{\displaystyle G} satisfies the relation
G A→+G B→+G C→=0.{\displaystyle {\vec {GA}}+{\vec {GB}}+{\vec {GC}}=0.}
This follows from the fact that the absolute barycentric coordinates of G{\displaystyle G} are 1 3:1 3:1 3{\displaystyle {\frac {1}{3}}:{\frac {1}{3}}:{\frac {1}{3}}}. Further, the problem of Sylvester reads as
O H→=O A→+O B→+O C→.{\displaystyle {\vec {OH}}={\vec {OA}}+{\vec {OB}}+{\vec {OC}}.}
Now, using the vector addition, we deduce that
G O→=G A→+A O→(in triangle A G O),G O→=G B→+B O→(in triangle B G O),G O→=G C→+C O→(in triangle C G O).{\displaystyle {\vec {GO}}={\vec {GA}}+{\vec {AO}}\,{\mbox{(in triangle }}AGO{\mbox{)}},\,{\vec {GO}}={\vec {GB}}+{\vec {BO}}\,{\mbox{(in triangle }}BGO{\mbox{)}},\,{\vec {GO}}={\vec {GC}}+{\vec {CO}}\,{\mbox{(in triangle }}CGO{\mbox{)}}.}
By adding these three relations, term by term, we obtain that
3⋅G O→=(∑c y c G A→)+(∑c y c A O→)=0−(∑c y c O A→)=−O H→.{\displaystyle 3\cdot {\vec {GO}}=\left(\sum \limits {\scriptstyle {\rm {cyc}}}{\vec {GA}}\right)+\left(\sum \limits {\scriptstyle {\rm {cyc}}}{\vec {AO}}\right)=0-\left(\sum \limits _{\scriptstyle {\rm {cyc}}}{\vec {OA}}\right)=-{\vec {OH}}.}
In conclusion, 3⋅O G→=O H→{\displaystyle 3\cdot {\vec {OG}}={\vec {OH}}}, and so the three points O{\displaystyle O}, G{\displaystyle G} and H{\displaystyle H} (in this order) are collinear.
In Dörrie's book, the Euler line and the problem of Sylvester are put together into a single proof. However, most of the proofs of the problem of Sylvester rely on the fundamental properties of free vectors, independently of the Euler line.
Properties
[edit]
Distances between centers
[edit]
On the Euler line the centroid G is between the circumcenter O and the orthocenter H and is twice as far from the orthocenter as it is from the circumcenter:: p.102
G H=2 G O;{\displaystyle GH=2GO;}O H=3 G O.{\displaystyle OH=3GO.}
The segment GH is a diameter of the orthocentroidal circle.
The center N of the nine-point circle lies along the Euler line midway between the orthocenter and the circumcenter:
O N=N H,O G=2⋅G N,N H=3 G N.{\displaystyle ON=NH,\quad OG=2\cdot GN,\quad NH=3GN.}
Thus the Euler line could be repositioned on a number line with the circumcenter O at the location 0, the centroid G at 2 t, the nine-point center at 3 t, and the orthocenter H at 6 t for some scale factor t.
Furthermore, the squared distance between the centroid and the circumcenter along the Euler line is less than the squared circumradiusR 2 by an amount equal to one-ninth the sum of the squares of the side lengths a, b, and c:: p.71
G O 2=R 2−1 9(a 2+b 2+c 2).{\displaystyle GO^{2}=R^{2}-{\tfrac {1}{9}}(a^{2}+b^{2}+c^{2}).}
In addition,: p.102
O H 2=9 R 2−(a 2+b 2+c 2);{\displaystyle OH^{2}=9R^{2}-(a^{2}+b^{2}+c^{2});}G H 2=4 R 2−4 9(a 2+b 2+c 2).{\displaystyle GH^{2}=4R^{2}-{\tfrac {4}{9}}(a^{2}+b^{2}+c^{2}).}
Representation
[edit]
Equation
[edit]
Let A, B, C denote the vertex angles of the reference triangle, and let x: y: z be a variable point in trilinear coordinates; then an equation for the Euler line is
sin(2 A)sin(B−C)x+sin(2 B)sin(C−A)y+sin(2 C)sin(A−B)z=0.{\displaystyle \sin(2A)\sin(B-C)x+\sin(2B)\sin(C-A)y+\sin(2C)\sin(A-B)z=0.}
An equation for the Euler line in barycentric coordinatesα:β:γ{\displaystyle \alpha :\beta :\gamma } is
(tanC−tanB)α+(tanA−tanC)β+(tanB−tanA)γ=0.{\displaystyle (\tan C-\tan B)\alpha +(\tan A-\tan C)\beta +(\tan B-\tan A)\gamma =0.}
Parametric representation
[edit]
Another way to represent the Euler line is in terms of a parameter t. Starting with the circumcenter (with trilinear coordinates cosA:cosB:cosC{\displaystyle \cos A:\cos B:\cos C}) and the orthocenter (with trilinears secA:secB:secC=cosB cosC:cosC cosA:cosA cosB),{\displaystyle \sec A:\sec B:\sec C=\cos B\cos C:\cos C\cos A:\cos A\cos B),} every point on the Euler line, except the orthocenter, is given by the trilinear coordinates
cosA+t cosB cosC:cosB+t cosC cosA:cosC+t cosA cosB{\displaystyle \cos A+t\cos B\cos C:\cos B+t\cos C\cos A:\cos C+t\cos A\cos B}
formed as a linear combination of the trilinears of these two points, for some t.
For example:
The circumcenter has trilinears cosA:cosB:cosC,{\displaystyle \cos A:\cos B:\cos C,} corresponding to the parameter value t=0.{\displaystyle t=0.}
The centroid has trilinears cosA+cosB cosC:cosB+cosC cosA:cosC+cosA cosB,{\displaystyle \cos A+\cos B\cos C:\cos B+\cos C\cos A:\cos C+\cos A\cos B,} corresponding to the parameter value t=1.{\displaystyle t=1.}
The nine-point center has trilinears cosA+2 cosB cosC:cosB+2 cosC cosA:cosC+2 cosA cosB,{\displaystyle \cos A+2\cos B\cos C:\cos B+2\cos C\cos A:\cos C+2\cos A\cos B,} corresponding to the parameter value t=2.{\displaystyle t=2.}
The de Longchamps point has trilinears cosA−cosB cosC:cosB−cosC cosA:cosC−cosA cosB,{\displaystyle \cos A-\cos B\cos C:\cos B-\cos C\cos A:\cos C-\cos A\cos B,} corresponding to the parameter value t=−1.{\displaystyle t=-1.}
Slope
[edit]
In a Cartesian coordinate system, denote the slopes of the sides of a triangle as m 1,{\displaystyle m_{1},}m 2,{\displaystyle m_{2},} and m 3,{\displaystyle m_{3},} and denote the slope of its Euler line as m E{\displaystyle m_{E}}. Then these slopes are related according to: Lemma 1
m 1 m 2+m 1 m 3+m 1 m E+m 2 m 3+m 2 m E+m 3 m E{\displaystyle m_{1}m_{2}+m_{1}m_{3}+m_{1}m_{E}+m_{2}m_{3}+m_{2}m_{E}+m_{3}m_{E}}+3 m 1 m 2 m 3 m E+3=0.{\displaystyle +3m_{1}m_{2}m_{3}m_{E}+3=0.}
Thus the slope of the Euler line (if finite) is expressible in terms of the slopes of the sides as
m E=−m 1 m 2+m 1 m 3+m 2 m 3+3 m 1+m 2+m 3+3 m 1 m 2 m 3.{\displaystyle m_{E}=-{\frac {m_{1}m_{2}+m_{1}m_{3}+m_{2}m_{3}+3}{m_{1}+m_{2}+m_{3}+3m_{1}m_{2}m_{3}}}.}
Moreover, the Euler line is parallel to an acute triangle's side BC if and only if: p.173 tanB tanC=3.{\displaystyle \tan B\tan C=3.}
Relation to inscribed equilateral triangles
[edit]
The locus of the centroids of equilateral triangles inscribed in a given triangle is formed by two lines perpendicular to the given triangle's Euler line.: Coro. 4
In special triangles
[edit]
Right triangle
[edit]
In a right triangle, the Euler line coincides with the median to the hypotenuse—that is, it goes through both the right-angled vertex and the midpoint of the side opposite that vertex. This is because the right triangle's orthocenter, the intersection of its altitudes, falls on the right-angled vertex while its circumcenter, the intersection of its perpendicular bisectors of sides, falls on the midpoint of the hypotenuse.
Isosceles triangle
[edit]
The Euler line of an isosceles triangle coincides with the axis of symmetry. In an isosceles triangle the incenter falls on the Euler line.
Automedian triangle
[edit]
The Euler line of an automedian triangle (one whose medians are in the same proportions, though in the opposite order, as the sides) is perpendicular to one of the medians.
Systems of triangles with concurrent Euler lines
[edit]
Consider a triangle ABC with Fermat–Torricelli pointsF 1 and F 2. The Euler lines of the 10 triangles with vertices chosen from A, B, C, F 1 and F 2 are concurrent at the centroid of triangle ABC.
The Euler lines of the four triangles formed by an orthocentric system (a set of four points such that each is the orthocenter of the triangle with vertices at the other three points) are concurrent at the nine-point center common to all of the triangles.: p.111
Generalizations
[edit]
Quadrilateral
[edit]
In a convex quadrilateral, the quasiorthocenter H, the "area centroid" G, and the quasicircumcenterO are collinear in this order on the Euler line, and HG = 2 GO.
Tetrahedron
[edit]
Main article: Tetrahedron §Properties analogous to those of a triangle
A tetrahedron is a three-dimensional object bounded by four triangular faces. Seven lines associated with a tetrahedron are concurrent at its centroid; its six midplanes intersect at its Monge point; and there is a circumsphere passing through all of the vertices, whose center is the circumcenter. These points define the "Euler line" of a tetrahedron analogous to that of a triangle. The centroid is the midpoint between its Monge point and circumcenter along this line. The center of the twelve-point sphere also lies on the Euler line.
Simplicial polytope
[edit]
A simplicial polytope is a polytope whose facets are all simplices (plural of simplex). For example, every polygon is a simplicial polytope. The Euler line associated to such a polytope is the line determined by its centroid and circumcenter of mass. This definition of an Euler line generalizes the ones above.
Suppose that P{\displaystyle P} is a polygon. The Euler line E{\displaystyle E} is sensitive to the symmetries of P{\displaystyle P} in the following ways:
If P{\displaystyle P} has a line of reflection symmetry L{\displaystyle L}, then E{\displaystyle E} is either L{\displaystyle L} or a point on L{\displaystyle L}.
If P{\displaystyle P} has a center of rotational symmetry C{\displaystyle C}, then E=C{\displaystyle E=C}.
Related constructions
[edit]
A triangle's Kiepert parabola is the unique parabola that is tangent to the sides (two of them extended) of the triangle and has the Euler line as its directrix.: p. 63
References
[edit]
^ abcKimberling, Clark (1998). "Triangle centers and central triangles". Congressus Numerantium. 129: i–xxv, 1–295.
^Euler, Leonhard (1767). "Solutio facilis problematum quorundam geometricorum difficillimorum" [Easy solution of some difficult geometric problems]. Novi Commentarii Academiae Scientarum Imperialis Petropolitanae. 11: 103–123. E325. Reprinted in Opera Omnia, ser. I, vol. XXVI, pp.139–157, Societas Scientiarum Naturalium Helveticae, Lausanne, 1953, MR0061061. Summarized at: Dartmouth College.
^Schattschneider, Doris; King, James (1997). Geometry Turned On: Dynamic Software in Learning, Teaching, and Research. The Mathematical Association of America. pp.3–4. ISBN978-0883850992.
^Edmonds, Allan L.; Hajja, Mowaffaq; Martini, Horst (2008), "Orthocentric simplices and biregularity", Results in Mathematics, 52 (1–2): 41–50, doi:10.1007/s00025-008-0294-4, MR2430410, S2CID121434528, It is well known that the incenter of a Euclidean triangle lies on its Euler line connecting the centroid and the circumcenter if and only if the triangle is isosceles.
^ abLeversha, Gerry; Smith, G. C. (November 2007), "Euler and triangle geometry", Mathematical Gazette, 91 (522): 436–452, doi:10.1017/S0025557200182087, JSTOR40378417, S2CID125341434.
^ abcdefAltshiller-Court, Nathan, College Geometry, Dover Publications, 2007 (orig. Barnes & Noble 1952).
^ abDörrie, Heinrich, "100 Great Problems of Elementary Mathematics. Their History and Solution". Dover Publications, Inc., New York, 1965, ISBN0-486-61348-8, pages 141 (Euler's Straight Line) and 142 (Problem of Sylvester)
^Scott, J.A., "Some examples of the use of areal coordinates in triangle geometry", Mathematical Gazette 83, November 1999, 472-477.
^ abWladimir G. Boskoff, Laurent¸iu Homentcovschi, and Bogdan D. Suceava, "Gossard's Perspector and Projective Consequences", Forum Geometricorum, Volume 13 (2013), 169–184.
^Francisco Javier Garc ́ıa Capita ́n, "Locus of Centroids of Similar Inscribed Triangles", Forum Geometricorum 16, 2016, 257–267 .
^Parry, C. F. (1991), "Steiner–Lehmus and the automedian triangle", The Mathematical Gazette, 75 (472): 151–154, doi:10.2307/3620241, JSTOR3620241.
^Beluhov, Nikolai Ivanov. "Ten concurrent Euler lines", Forum Geometricorum 9, 2009, pp. 271–274.
^Myakishev, Alexei (2006), "On Two Remarkable Lines Related to a Quadrilateral"(PDF), Forum Geometricorum, 6: 289–295.
^Tabachnikov, Serge; Tsukerman, Emmanuel (May 2014), "Circumcenter of Mass and Generalized Euler Line", Discrete and Computational Geometry, 51 (4): 815–836, arXiv:1301.0496, doi:10.1007/s00454-014-9597-2, S2CID12307207.
^Scimemi, Benedetto, "Simple Relations Regarding the Steiner Inellipse of a Triangle", Forum Geometricorum 10, 2010: 55–77.
External links
[edit]
An interactive applet showing several triangle centers that lies on the Euler line.
"Euler Line" and "Non-Euclidean Triangle Continuum" at the Wolfram Demonstrations Project
Nine-point conic and Euler line generalization, A further Euler line generalization, and The quasi-Euler line of a quadrilateral and a hexagon at Dynamic Geometry Sketches
Bogomolny, Alexander, "Altitudes and the Euler Line" and "Euler Line and 9-Point Circle", Cut-the-Knot
Kimberling, Clark, "Triangle centers on the Euler line", Triangle Centers
Archived at Ghostarchive and the Wayback Machine: Stankova, Zvezdelina (February 1, 2016), "Triangles have a Magic Highway", Numberphile, YouTube
Weisstein, Eric W."Euler Line". MathWorld.
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https://www.youtube.com/watch?v=aV_TlYeZ0G4
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End Behavior of ln(x)
MathDoctorBob
66200 subscribers
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Description
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Posted: 31 May 2011
Calculus: Using the exponent rule for natural logarithm, we verify the end behavior of the graph of ln(x). That is, we show that the limit as x goes to infinity of ln(x) is positive infinity and that the limit as x goes to 0 from the right is negative infinity. We interpret the result as an area.
5 comments
Transcript:
now that we have the exponent rule for natural logarithm let's take another look at its graph recall natural log of x is defined as definite integral from 1 to X of 1/ t with respect to T the way we interpret this take the graph of 1 over T then we're going to take the area under the graph between 1 and X now natural log of x is only going to be defined when X is greater than zero if x is between 0 and one we're going to want to switch the limits of integration and that's going to introduce a minus sign so we're between 0 and one natural log of x is negative now let's recall what we have for the graph of natural log of x first we have some special values natural log of 1 is equal to zero natural log of e is equal to 1 e is roughly 2.718 so it's going to be a special number then we'll have that the domain of natural L of X is just x s than zero so we're only going to have graph in this region here then we have that the derivative of natural log of x is 1 /x okay on the region X greater than zero so it's going to be differentiable where it's defined that means it's going to be continuous so we could draw the graph without picking up our pencil then we'll have from the derivative that our functions increasing if we take the second derivative we'll see that's concave down so the only thing we have left is to check on the end behavior and we're going to get that using the exponent Rule now our exponent rule rule for logarithm States we take natural log of a some power R we're allowed to take the exponent R and just bring it down in front we're going use this to show The Following two limits first I'll have the limit as X goes to plus infinity natural log of x is equal to plus infinity so as we let X go off to the right that's going to say that the values of natural log of x are going to grow without bound they grow very slowly okay you'll not from the graph it looks like it could go off to a horizontal ASM toote if it wants but it's just going to keep growing then I want the limit as X Goes To Zero from the right of natural log of x is going to be equal to minus infinity so as we come into Zero from the right the values of natural log of x are just going to spike down to minus infinity so we're going to have a vertical ASM toote of zero now to get our limits the trick we're going to use is substitution so the idea is in natural L of X I want to replace x with another function it's going to be easier to work with for our first limit we let X be equal to 2 to the U so the idea here if I let U go off to Infinity 2 to the U is going to be equal to plus infinity to see that just pick your favorite values for you so you go 0 1 2 3 4 put that in two to the U you're going to get 1 2 4 8 16 that thing's definitely going to grow without bound now to do the limit what we're going to do is I start off with what I want limit as X goes to plus infinity natural log of x I'm going replace x with 2 to the U now as U goes to Infinity 2 to the U goes to Infinity so that means X is going to go to infinity and that way I can hand off this limiting part here so we're really looking at limit as U goes to plus infinity of natural log of 2 the U now I can use the exponent rule to bring the U down so we do that then you'll not natural log of two is just a number it's roughly 693 so I can just pull that out in front of the limit as a constant so we're going to be left with natural log of two which is a number times the limit as U goes to plus infinity of U so that limit is just going to be plus infinity our constant out in front is roughly 693 it's a positive number so our limit is going to go to plus infinity as promised next want to show the limit as X goes to Zer from the right of natural log of x is equal to minus infinity here we're going to do the substitution x = 1 U we take the limmit as U goes to Plus plus infinity of 1/ U we get zero so if I let U go off to the right the values of 1/ U are to get driven down to zero we also note as those values come down they're always going to be positive numbers so this will be the same as saying that X is going to go to Zero from the right or through positive numbers now we substitute in our limit we'll have natural log of 1/ U then our limit is going to be from U going to plus infinity instead of X going to Zer from the right now 1 U is equal to U minus1 so I can use the exponent rule for logarithm to bring the minus one down in front so we have this expression here if I replace U with X we have the limit that we worked out on the previous board so it's going to go to plus infinity now there's a minus sign in front so our answer here is going to be minus infinity as promised it's a final note we interpret our first limit so we have that the limit as X goes to plus infinity natural log of x equals plus infinity recall natural log of x is going to be the area under the graph of 1 overt between 1 and X so if we're going to let X go off to plus infinity what we're doing we're take these areas and just keep adding a little bit more and a little bit more as we go off to Infinity the result is you take all of this area it's going to be infinite so if you want to paint that area you need an infinite amount of paint now this will be important later on we do sequences in series we're going to be interested in taking infinite list of numbers and trying to add them up so one will be interested in be 1 plus a half plus a 3 plus a 4th and so on we're going to be able to use this result to show that the sum of this list is going to be equal to plus infinity so I'll leave it as an exercise for you to figure out what this picture here has to do with this result okay if you can't wait sequences in series
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https://learnaboutstructures.com/Identifying-Zero-Force-Members
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3.3 Identifying Zero Force Members | Learn About Structures
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3.3 Identifying Zero Force Members
>>When you're done reading this section, check your understanding with the interactive quiz at the bottom of the page.
Zero force members in a truss are members which do not have any force in them (obviously...). There are two rules that may be used to find zero-force members in a truss. These are described below and illustrated in Figure 3.3.
Case 1 At a TWO member joint: If those members are NOT parallel AND there are no other external loads (or reactions) at the joint THEN both of those members are zero force members.Case 2 In a THREE member joint: If TWO of those members ARE parallel AND there are no other external loads (or reactions) at the joint THEN the member that is not parallel is a zero force member.
Figure 3.3: Examples of Zero Force Members
Two examples, one for each case, are shown in Figure 3.3. By applying equilibrium to the appropriate joints, we can see why the members shown do not have any force.
For the case 1 example, members AB and AC are zero force members. This may be shown to be the case by solving the equilibrium equations (1)(1) at joint A.
∑i=1 n F x i=0;∑i=1 p F y i=0;(1)(1)∑i=1 n F x i=0;∑i=1 p F y i=0;
For vertical equilibrium (y y-direction), the vertical component of F A C F A C is the only vertical force:
F A C y∑i=1 p F y i∴F A C y F A C sin θ∴F A C=F A C sin θ=0=0=0=0 as long as θ≠0 F A C y=F A C sinθ∑i=1 p F y i=0∴F A C y=0 F A C sinθ=0∴F A C=0 as long as θ≠0
Therefore F A C F A C is a zero-force member. If we now apply horizontal equilibrium (x x-direction), we have two horizontal forces, F A B F A B and the horizontal component of F A C F A C:
F A C x∑i=1 n F x i∴−F A C x−F A B F A B=F A C cos θ=0=0=0=−F A C x=0 F A C x=F A C cosθ=0∑i=1 n F x i=0∴−F A C x−F A B=0 F A B=−F A C x=0
Therefore, F A B F A B must also be a zero force member. In this example F A C x F A C x and F A B F A B are both negative because the arrows both point to the left. This analysis works for any two loads at joint which are not parallel where there is no external load on the joint.
For case 2 in Figure 3.3, member BD is a zero force member. This may be shown to be the case by solving the equilibrium equations (1)(1) at joint B. For vertical equilibrium, the vertical component of F B D F B D is the only vertical force:
F B D y∑i=1 p F y i∴F B D y F B D sin θ∴F B D=F B D sin θ=0=0=0=0 as long as θ≠0 F B D y=F B D sinθ∑i=1 p F y i=0∴F B D y=0 F B D sinθ=0∴F B D=0 as long as θ≠0
Therefore F B D F B D is a zero-force member. This analysis was simplified because the members BC and AB were parallel to the x x-axis; however, the orientation of the x x-axis is arbitrary, an analysis will show that member BD is a zero force member as long as two of the members are parallel at the joint, even if they are not parallel to the x x- or y y-axis (try it out!).
Finding these zero force members up front can save analysis time and effort.
Book traversal links for 3.3 Identifying Zero Force Members
3.2 Calculating x and y Force Components in Truss Members
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3.4 Using Global Equilibrium to Calculate Reactions
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Chapter 1: Introduction and Review
Chapter 2: Stability and Determinacy
Chapter 3: Analysis of Determinate Trusses
3.1 Introduction
3.2 Calculating x and y Force Components in Truss Members
3.3 Identifying Zero Force Members
3.4 Using Global Equilibrium to Calculate Reactions
3.5 The Method of Joints
3.6 The Method of Sections
3.7 Practice Problems
Chapter 4: Analysis of Determinate Beams and Frames
Chapter 5: Deflections of Determinate Structures
Chapter 6: Influence Lines
Chapter 7: Approximate Indeterminate Frame Analysis
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https://www.quora.com/What-happens-if-I-mix-2-acids-of-the-same-concentration-and-volume-but-of-a-different-pH-for-example-pH-2-and-pH-6-What-is-the-resulting-pH
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5
What happens if I mix 2 acids of the same concentration and volume but of a different pH (for example, pH=2 and pH=6)? What is the resulting pH?
Michael Mombourquette
Retired Chemistry Prof, Church member, Knight of Columbus,
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Upvoted by
Łukasz Golon
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Chris Marlowe
, Bachelors degree in Chemistry from Rutgers University in 1976. Practice 40 years · Author has 6.8K answers and 17.7M answer views
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8y
Some answers missed a key point. if the two acids are the same concentration but have different pH values then they must have different Ka values. Let's pretend we know that the pH 2 acid is a strong acid. Then it's concentrations must be
1
×
10
−
2
.
And the other is a weak acid with a pKa that is much smaller. By adding two equal volumes, the concentrations are both halved. So now, they are both
5
×
10
−
3
. And since we can ignore the concentration of hydronium from the weak acid, the final pH will be
−
l
o
g
(
5
×
10
−
3
)
=
2.3
. If the stronger acid is not actually 100% dissociated,
Some answers missed a key point. if the two acids are the same concentration but have different pH values then they must have different Ka values. Let's pretend we know that the pH 2 acid is a strong acid. Then it's concentrations must be
1
×
10
−
2
.
And the other is a weak acid with a pKa that is much smaller. By adding two equal volumes, the concentrations are both halved. So now, they are both
5
×
10
−
3
. And since we can ignore the concentration of hydronium from the weak acid, the final pH will be
−
l
o
g
(
5
×
10
−
3
)
=
2.3
. If the stronger acid is not actually 100% dissociated,
i.e.
, it’s not a strong acid after all, then there will be some amount of additional dissociation and as a result, the pH calculation would be a bit more complicated than what we just did but the final result should be not too far from the same value 2.3.
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Related questions
Do acids with the same concentration have the same pH value?
What mixture is produced when a solution of 2 pH is mixed with 6 pH?
What is the pH of the resulting solution obtained by mixing equal volumes of two solutions with pH=3 and pH=6?
If two acids, HX and HY, have the same pH, would they have the same concentration?
Which solution is most acidic (that is, which one has the lowest pH)?
Assistant
Bot
·
1y
When mixing two acids of the same concentration and volume but with different pH values, the resulting pH will depend on the nature of the acids and their dissociation in water.
Understanding pH and Acidity
pH Scale: The pH scale ranges from 0 to 14, where lower values indicate higher acidity. A solution with pH 2 is much more acidic than one with pH 6.
Concentration: pH is a logarithmic measure of hydrogen ion concentration [H+]. A pH of 2 corresponds to a [H+] of 0.01M (10^-2 M), while a pH of 6 corresponds to a [H+] of 0.000001M (10^-6 M).
Mixing the Two Acids
Calculatin
When mixing two acids of the same concentration and volume but with different pH values, the resulting pH will depend on the nature of the acids and their dissociation in water.
Understanding pH and Acidity
pH Scale: The pH scale ranges from 0 to 14, where lower values indicate higher acidity. A solution with pH 2 is much more acidic than one with pH 6.
Concentration: pH is a logarithmic measure of hydrogen ion concentration [H+]. A pH of 2 corresponds to a [H+] of 0.01M (10^-2 M), while a pH of 6 corresponds to a [H+] of 0.000001M (10^-6 M).
Mixing the Two Acids
Calculating Hydrogen Ion Concentrations:
For the acid with pH 2: [H+]1=0.01M
For the acid with pH 6: [H+]2=0.000001M
Mixing Volumes: If you mix equal volumes of both solutions, the total volume doubles, and the total hydrogen ion concentration can be calculated as follows:
[H+]total=[H+]1+[H+]22
[H+]total=0.01M+0.000001M2≈0.010001M2=0.0050005M
Calculating the Resulting pH:
pH=−log([H+]total)≈−log(0.0050005)≈2.3
Conclusion
The resulting pH of the mixture will be approximately 2.3, which indicates the solution remains strongly acidic, primarily influenced by the more acidic solution (pH 2). The pH of the final solution is closer to that of the stronger acid due to its significantly higher hydrogen ion concentration compared to the weaker acid.
Aaron
BSC (Hons) in Chemistry & Mathematics, University of Leeds (Graduated 2007)
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Author has 51 answers and 84.1K answer views
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8y
The pH of the resulting mixture would be somewhere between the two, as the pH is a measure of the concentration of H+ ions in the solution:
pH = - log [H+]
(H+ ions come from the acids)
Since pH is a logarithmic scale, a solution of pH 2 actually has a concentration of H+ ions which is 10,000 times greater than that in a solution of pH 6. So, when adding the solutions together, the number of H+ ions in the pH 6 solution is practically negligible; you may as well be just be adding water to the pH 2 acid.
The effect is to double the volume of the pH 2 solution, and thus half the concentration of H+
The pH of the resulting mixture would be somewhere between the two, as the pH is a measure of the concentration of H+ ions in the solution:
pH = - log [H+]
(H+ ions come from the acids)
Since pH is a logarithmic scale, a solution of pH 2 actually has a concentration of H+ ions which is 10,000 times greater than that in a solution of pH 6. So, when adding the solutions together, the number of H+ ions in the pH 6 solution is practically negligible; you may as well be just be adding water to the pH 2 acid.
The effect is to double the volume of the pH 2 solution, and thus half the concentration of H+ ions. Using the equation above,
New pH = - log (0.5 [H+])
where [H+] is that of the original pH 2 acid.
Using theory of logs:
New pH = - log [H+] - log 0.5
= old pH + 0.30
= 2.3
By my reckoning the new pH will be around 2.3, which you may find quite surprising.
I think that's right, but I am currently quite tired so other chemists are welcome to check my assumptions and maths!
I hope the asker is familiar with logs, or doesn't need to follow the working out because I'm not really fancying the idea of explaining those!
Srirama Chandran
M.Sc from Sri Venkateswara University (Graduated 1989)
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Author has 146 answers and 291.1K answer views
·
8y
For entrance, just do like this:
If pH = 2 and oH = 6 (both < 7), then, pH of mixture = Lowe pH + 0.3 = 2.3
In case both values are higher than 7, pH of mixture = higher pH - 0.3
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Related questions
How much acid should be dosed in 20,000 litres of pH 6.85 tank to get 2 pH?
What is the resultant pH of the solution if equal volumes of two solutions with pH = 6 and pH = 8 are mixed? (Solve using Normality and Volume method)
Two solution the same concentration and volume are mixed if pH valua of one is 2 &the other is 6 . How calculated the pH of mixture?
Equal volumes of solution with pH=4 and pH=10 are mixed. What is the pH of the resulting solution?
How do I find find the pH of the resultant solution when 2 acids of equal volumes are mixed together and their pH is given (Ex:- 3 & 6)?
Muttineni Ajender Rao
Subject Matter Expert at Tata Consultancy Services (company) (2015–present)
·
7y
The resultant solution will have pH =2.301. The H+ concentration of first solution will be 10
2.The H+ Concentration of second solution will be 10-6. Assume the solutions mixing are of equal volume. Using the resultant molarity formula you get the resultant pH of the solution is 2.301.
Chris Specker
B. S. in Chemistry, University of North Alabama (Graduated 1995)
·
Author has 103 answers and 340.5K answer views
·
8y
Assuming they're both strong acids, you would have 0.01 moles of H+ per liter for one, and 10^-6 mol/L for the other. The resulting mixture would be be about 0.005 mol/L, or pH 2.3.
Of course, if one or both acids were weak, you would probably need to calculate acid-base equilibrium, which would require the pKa of any weak acids
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Antonio Daniels
Former Learner 2020- Chemist
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Author has 8K answers and 4.5M answer views
·
3y
strong acids???
If they have the same cocentration and volume the pH will be the same……and mixing them nothing will change
Devender Singh
MSc in Chemistry & Board of High School and Intermediate Education, Uttar Pradesh, Chaudhary Charan Singh University (Graduated 1990)
·
Author has 926 answers and 3.8M answer views
·
6y
Related
The equal volume of two HCL solutions of pH=3 and pH=5 were mixed. What is the pH of the resulting solution?
The pH of the resulting solution assume = x
The equal volume of two HCL solutions of pH=3 and pH=5 were mixed.
thanks
The pH of the resulting solution assume = x
The equal volume of two HCL solutions of pH=3 and pH=5 were mixed.
thanks
Ganta Naveen
Related
Equal volumes of two solutions, one of pH=6 and the other of pH=4, are mixed. What would be the pH of the resulting solution?
Related questions
Do acids with the same concentration have the same pH value?
What mixture is produced when a solution of 2 pH is mixed with 6 pH?
What is the pH of the resulting solution obtained by mixing equal volumes of two solutions with pH=3 and pH=6?
If two acids, HX and HY, have the same pH, would they have the same concentration?
Which solution is most acidic (that is, which one has the lowest pH)?
How much acid should be dosed in 20,000 litres of pH 6.85 tank to get 2 pH?
What is the resultant pH of the solution if equal volumes of two solutions with pH = 6 and pH = 8 are mixed? (Solve using Normality and Volume method)
Two solution the same concentration and volume are mixed if pH valua of one is 2 &the other is 6 . How calculated the pH of mixture?
Equal volumes of solution with pH=4 and pH=10 are mixed. What is the pH of the resulting solution?
How do I find find the pH of the resultant solution when 2 acids of equal volumes are mixed together and their pH is given (Ex:- 3 & 6)?
What happens when you mix sulphuric acid and water?
Which has a lower pH, dilute acids or concentrated acids?
What happens when you mix HCL and water?
What is the pH of a mixture solution of a 2-ml strong acid with a pH 3 and 3 ml strong base with pH 10?
Equal volumes of two solutions, one of pH=6 and the other of pH=4, are mixed. What would be the pH of the resulting solution?
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鸡兔同笼解法全集，总有一种适合你，重点推荐使用方法二，方法九恐怕你想不到
中国科学院大学 光学博士
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鸡兔同笼问题是小学阶段非常经典的数学问题，可以这么说，所有的小学生都学习过或见识过鸡兔同笼问题，而且有不少学生还为此吃尽了苦头。为什么鸡兔同笼问题在中国这么“流行”呢？因为这个问题本身就是我们中国人提出来的。
| | | | | | | | | |
--- --- --- ---
| 鸡 | 35 | 34 | 33 | 32 | 31 | 30 | …… | 23 |
| 兔 | 0 | 1 | 2 | 3 | 4 | 5 | …… | 12 |
| 总脚数 | 70 | 72 | 74 | 76 | 78 | 80 | | 94 |
很多学生和家长对这种方法不屑一顾，觉得这种方法既笨拙又麻烦，我并不这么认为，其实，对于小学低年级的学生而言，这种方法我倒是认为是最值得推荐的方法，因为在制作表格的过程中，学生需要自主地去探索鸡、兔在数量变化的时候，总脚数的变化特点，通过动手绘制以及用眼观察，分析比较得出，由于兔子的脚比鸡多两只，所以当鸡数少1只，兔数多1只的时候，总脚数会增加2只的规律性认识。而这正是培养学生探索精神，提升学生数学思维的重要途径。
经典解法二：灵机一动假设法
鸡兔同笼的解法中，我个人最喜欢的同时也是最推荐学生使用的就是假设法，因为假设法几乎能够解决所有类型的鸡兔同笼问题，即使题目进行了很大的改编和变形，假设法都能有效的对题目进行解析，当然，对于一些变形的鸡兔同笼问题，用假设法会比较烧脑。在实际应用中，假设法几乎是最经典，最有效率的一种方法，学生运用假设，将不同的情形（鸡和兔子的脚数不同）转化成相同的情形，有利于简化问题，理清思路。
鸡兔同笼问题的难点就在于每只鸡和每只兔子的脚数是不同的，这是问题的难点，但也是解决问题的关键点或者说是突破口，假设鸡和兔子的脚数相同，那么，题目就会大大简化，将复杂问题简单化，是解决数学问题的常见思路。
假设一：所有兔子都站起来，藏起2只脚。这样的话，每只鸡和每只兔子的脚数就相等了，都是2只，在这种情况下，一共有35个头，也就是说一共有35只动物，每个动物有2只脚，那么总的脚数=35×2=70只，这比题目给出的94只脚少了24只，想一想为什么少了？因为每只兔子都站了起来，收起了2只脚，一只兔子少2只脚，一共少了24只脚，所以一共有24÷2=12只兔子，再用35-12=23就是鸡的数量。
假设二：我们也可以把鸡假设成兔子，此时，所有鸡增加2只脚。这样的话，每只鸡和每只兔子的脚数就相等了，（都是4只），在这种情况下，一共有35个头，也就是说一共有35只动物，每个动物有4只脚，那么总的脚数=35×4=140只，这比题目给出的94只脚多了46只，想一想为什么这次脚又多了呢？因为每只鸡都多了2只脚，一只鸡多2只脚，一共多了46只脚，所以一共有46÷2=23只鸡，再用35-23=12就是兔子的数量。
经典解法三：公平交换代换法
实际上，我们还可以用一二年级时学到的变量代换的方法求解鸡兔同笼问题。用红圆圈代表鸡，用蓝圆圈代表兔子。根据题意，我们可以列出下面的算式：
这种方法也是我非常推荐的，因为这种方法虽然只是用到了一二年级的知识，但实质上却是方程思想的初步应用，是设未知数求解问题的雏形，在这道题中，我们分别用红圈和蓝圈代表鸡和兔子，本质上就是一种数学抽象，对提升学生的分析归纳问题能力有非常好的作用和效果。
当然，这种做法对四年级以下的学生来说，听是能听懂的，但让他们再做一次恐怕有些难度，因为这种做法的实质是方程解法，只不过用符号代替了x、y，对低年级的学生来说是有一定难度的。
我比较建议家长用这种方法尝试给孩子讲一下，看看孩子的反应，也看看孩子在今后遇到鸡兔同笼问题时，会采用哪一种方法，可以据此做一个对孩子理解力的初步判断，如果他仍然愿意并能够使用这种方法解决鸡兔同笼问题，那我觉得是可以提前给他讲讲方程的。
经典解法四：一目了然图形法
鸡兔同笼问题当然还可以采用图形的方法来解决，比如，下面我先用线段表示鸡和兔，蓝色的线段代表鸡的只数，红色线段代表兔的只数。
我们知道，一只鸡2只脚，一只兔子4只脚，我们在上图的基础上，向外拓展一下，形成下面的图形。
可见，蓝色区域的面积等于鸡×2，即鸡的脚数，红色区域的面积等于兔×4，即兔子的脚数。
这里边有一个条件我们不要忽略了，那就是鸡和兔子的数量一共是35只。这样我们可以构建出下面的图形。
从上图中可以计算出35×4=140，是整个图形的面积，从上面的分析中可以知道，多出来的阴影部分面积应该等于140-94=46，而这个长方形的宽=4-2=2，那么长就应该等于46÷2=23，也就是鸡的数量，进而我们可以得出兔子的数量是35-23=12只。
这种方法也是我比较推荐的，并不是说它计算的有多么快，多么便捷，而是说这种方法为我们拓宽了求解鸡兔同笼问题的视野，使我们从呆板单一的数字运算中，愉快地过渡到图形的世界之中，对于启发学生的数形结合思想，激发学生的创造力非常有帮助。
说到这里我要提醒一下家长，我用面积的方法来讲，家长们应该能看懂，但实际上，三年级的学生，如果按照课本的知识点讲解进度来说，这种方法可能是他听不懂的，当然，也有部分学生能听懂。
这种方法的表现是数形结合，实质是面积概念中的这个“积”的应用，所谓的积就是两个数相乘，从算式来看，就是35×4，从图形来看，就是长35，宽4的一个长方形的图形面积，显然，我们可以用图形的面积来代表两个数的乘积，这种思路会不会启发学生去解决行程问题，浓度问题，工程问题呢？
经典解法五：简单粗暴设x
设鸡有x只，因为鸡和兔子一共有35只，那么兔子就有35-x只，根据题意，一只鸡2只脚，所以鸡的脚数是2x只，一只兔子4只脚，所以兔子的脚数是4×(35-x)，我们知道总脚数是94只，所以可以列出下面的算式：
2x+(35-x)×4=94，解出x=23，即鸡有23只，所以兔子是12只。
对于高年级的学生，我是非常推荐用这种方法解题的，可以这么说，对于高年级的学生，设未知数的方法是首选的方法。因为设未知数列方程的方法既是最快的又是最简洁的，长期运用方程思想求解实际问题，对于提升学生的问题抽象能力有非常大的帮助。不过对于低年级的学生，我还是觉得应该慎重地向他们讲述此种方法，因为过早的学习方程解法，一方面对于低年级的学生来说，他们的认知水平有限，会造成他们的认知困扰，就像上面介绍的等量代换的方法一样，孩子能听懂，但自己动手做不出来。另一方面，如果他们能够理解并熟练掌握此种方法，一定会放弃其他的方法，这对于培养他们的探索能力、数形结合能力、分析归纳能力来说，简直就是个灾难！
经典接法六：二元一次方程组
好吧，我承认用二元一次方程组来解鸡兔同笼问题有点儿小题大做了，但这确实也是一种比较好的方法。
设鸡有x只，兔子有y只，那么根据题意，我们可以列出下面的方程组：
x+y=35 ①
2x+4y=94 ②
把第一个式子左右都乘以2，得到2x+2y=70 ③
再用②-③，得到2y=24，进而求得y=12，即兔子有12只，鸡有35-12=23只。
这种方法确实是简单粗暴，但问题是很多学生掌握不了，确实有他的局限性，对学有余力的学生，理解能力较强的学生，可以尝试讲这种方法。
经典解法七：图解法求解二元一次方程组
刚才提到了可以用二元一次方程组求解这类鸡兔同笼问题，很多同学可能会感到比较陌生，这里，我再介绍一种用作图的方法，解决二元一次方程组，看看同学们是否能够理解。
同样还是设鸡有x只，兔子有y只，那么根据题意，我们可以列出下面的方程组：
x+y=35 ①
2x+4y=94 ②
到这一步我们要开始变形了，用②÷①，得到 2x+4y/x+y=94/35=2\frac{24}{35}
也就是平均每只动物有2\frac{24}{35} 只脚。
从图中课件，矩形ABMP是鸡兔的总脚数94，与矩形ACDE（鸡的脚数）和矩形CBFG（兔的脚数）的和相等，那么，矩形PQDE的面积就应该等于矩形GFMQ的面积。
这样，就有 （4-2\frac{24}{35}）y=(2\frac{24}{35}-2)x
即 \frac{46}{35}y=\frac{24}{35}x ，所以 x:y=23:12
结合 x+y=35 得到 x=23，y=12
严格上说，这里的 \frac{2x+4y}{x+y}=\frac{94}{25}=2\frac{24}{35} 孕育着混合物加权平均的思想！
经典解法八：可爱乖萌的金鸡独立法
一只鸡2只脚，一只兔子4只脚，我们让它们的脚数都减少一半，也就是让鸡单脚着地，来一个金鸡独立，让兔子前肢收齐，两个后腿撑地，像下图那样。
这时，它们总的脚数应该是最初时的一半，即94÷2=47只。我们注意观察一下，此时，一只鸡有1只脚，头脚是一一对应的，一只兔子有2只脚，每只兔子的脚数比头数多1个。现在的情况是一共35个头，47只脚，鸡是头脚一一对应是不多的，那么多出来的脚都是兔子的，所以有兔子47-35=12只，知道了兔子的只数，很容易就算出鸡的只数是23只了。
这种方法从表面上看和假设法十分相似，但如果你仔细分析后就会发现，这种方法的妙处在于通过金鸡独立，鸡的头数和脚数一一对应了，一个头对应一只脚，那么多出来的脚就是兔子的了。因此，这种方法告诉我们一种解题思路：将其中一个动物的头脚数做到一比一，这样，总脚数与总头数之差就是另一个动物的脚数与头数之差，在这种情况下，问题得到了简化，直接可以算出另一个动物的只数。
经典解法九：滑稽搞笑的吹哨法
听口令：抬起一只脚！这时，鸡展示了它金鸡独立的功夫，兔子则蹑手蹑脚的抬起了一只脚。
此时，每个动物都少了一只脚，一共有35个动物，就是少了35只脚，现在的总脚数是94-35=59只。听口令：再抬起一只脚！这时，鸡整个蹲了下来，兔子则是两只后腿着地，如图所示：
此时，总脚数=59-35=24，注意观察我们发现，鸡已经没有脚了，也就是说剩下的24只脚都是兔子的，我们看图，现在的兔子有2只脚，一共24只脚，那么兔子就应该有24÷2=12只，那么，鸡就是35-12=23只。
对这种解法，我只能说I服了U。这种解法的妙处就在于通过两次吹哨，把鸡的腿变没了，彻底把题目简化成兔子的头脚问题，这种思路非常值得大家学习，如果孩子对鸡兔同笼问题感兴趣，我倒是建议家长可以尝试给孩子讲讲这种方法，激发他们继续探索解题方法的热情。
这种方法和假设法是有区别的，实际上，和上面提到的金鸡独立法也有区别，请同学们认真思考这三种方法的区别到底在什么地方。
经典解法十：插翅难飞法
一只鸡2只脚，一只兔子4只脚，但是鸡会飞啊，来一个大鹏展翅。
这个时候我们再来看看，一只鸡2脚2翅，也算是凑足了四肢了，这样，35个动物，每个动物都是四肢，一共有35×4=140，比题目中的94只脚多了46，这46就是展开的翅膀，我们知道一只鸡2只翅膀，所以46只翅膀就是46÷2=23只鸡，兔子就是35-23=12只。
这种解法本质上是假设法的一种变形，假设所有的动物都是兔子，都有四只脚（此方法认为是2脚2翅膀），然后运用假设法把题目做出来。不过，这种方法和假设法不同之处，在于运用了图形和想象，这样，有助于学生更好地理解。和上面介绍的吹哨法有异曲同工之妙。
经典解法十一：调转乾坤平均法
出于对求解题目计算的简化，我们把题目稍作修改，鸡兔共有20个头，共有50只脚。我们来看如何用平均法求解。
一只鸡2只脚，一只兔子4只脚，那么它们混合在一起，平均一个动物的脚数应该是大于2小于4的。从题目中我们可以看出，20个头，50只脚，平均下来一个动物2.5只脚（这是什么怪物？！）我们用线来表示如下:
我们把一只鸡2只脚，一只兔子4只脚也标记在线上。
从上图中可以看出，一只兔子的脚数比平均数多了1.5只，一只鸡的脚数比平均数少了0.5只，我们可以这么理解，一只兔子比平均数多出的1.5只脚，需要3只鸡来“拉平”，即一只兔子配3只鸡，可以配出2.5只脚的效果，这样，我们把动物一共分成4份，鸡占了3份，兔子占了1份。鸡就是20×¾=15只，兔子就是20×¼=5只。
对于学有余力的同学，极力推荐用这种方法思考鸡兔同笼问题。因为这种方法把鸡兔同笼问题和平均数问题联系在一起，对于提高对平均数的理解大有好处。不过这种方法由于涉及到各项占比的情况，所以对题目中数字的要求较高，计算的时候需要格外注意。
这种方法我觉得至少要等到孩子四年级下学期才能讲给孩子听，否则，无论是平均数的理解，还是所谓的占比（比和比例都是五年级下或者六年级的内容了），他听起来都会很吃力，当然，如果孩子很聪明，也有三年级的孩子能够听懂这种方法，甚至习惯于用这样的方法来解题的。
经典解法十二：
其实，这道题还可以这样考虑，既然鸡、兔的总头数是35，如果能求出鸡兔头数之差，把问题转化成和差问题，再利用和差公式就很容易算出两种动物的只数了。
如下图所示，设鸡有x只，有2x只脚（蓝色矩形），兔子有y只，有4y只脚（黄色矩形）。那么，两个蓝色矩形与两个黄色矩形一起，拼成了一个大的正方形ABCD，中间则形成一个中空的矩形PQMN。
矩形ABCD的面积是：
(4+2)(x+y)=6x+6y=6(x+y)=6×35=210
它等于两个蓝色矩形的面积加上两个黄色矩形面积，再加上中间中空的白色矩形PQMN的面积。
因此，有：
210=2×94+(4-2)(y-x)
则 y-x=11
再利用和差公式，很容易求出 x=12,y=23
我是优博数学，中科院理学博士，关注我带给你更多学习方法和解题思路方面的干货内容。
优博数学
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中国科学院大学 光学博士
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编辑于 2022-03-30 19:24
方法
方法学
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写下你的评论...
18 条评论
默认
最新
天晴
太感谢了
2023-02-06
优博数学
作者
希望对你有帮助
2023-02-08
zheng purple
受益匪浅，思路开阔！
06-24
硬贝壳
方法十一和方法七原理相同
2024-03-24
优博数学
作者
是的
2024-09-23
IMZHC
真棒，一个问题，这么多种解法，开了眼了
2022-11-13
哟哟闹了闹
嘚！去我收藏夹吃灰去吧！
2022-08-11
禁止焦虑
鸡兔同笼，共有260只脚，兔比鸡少19只，鸡和兔各有多少只？
2022-07-14
浑成
这种含有倍数关系的用分组法，这里把一只鸡和一只兔看做一组放一起，后面还剩下19 只鸡。260-19×2=222(条)，222÷（2+4）=37（组），37+19=56（只），故兔37 只，鸡56只。
2022-08-18
司徒六六
你出的这道比原题简单，只考虑脚就行了。
260-(19×2)＝222只脚
222/(2＋4)＝37只兔和鸡
37＋19＝56只鸡
∴ 37只兔，56只鸡
2022-12-21
清风2015
真棒，我把2学会告诉孩子
2022-04-26
优博数学
作者
对，会一种就行了。
2022-04-26
小头爸爸
我和博士原来就差在了鸡兔同笼上，让我这大专生感觉我来这个世界就是凑数的
2022-04-13
六六名被占用
专科是大学专科呢，
怎么不自信纳
中师生说这是算理塞
不是博士的内容嘛
2022-07-06
优博数学
作者
2022-04-26
粽子妈咪
真的是学习了，受益匪浅👍🏻
方法八中，是不是把47写成了46哇
2022-03-30
优博数学
作者
是的，谢谢您的提醒，我已经修改了。
2022-03-30
写下你的评论...
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优博数学
亲自讲授小初高数学物理（含竞赛）
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Lung cancer in never-smoker Asian females is driven by oncogenic mutations, most often involving EGFR | Oncotarget
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Published in Oncotarget V6I7, Mar 10, 2015
Clinical Research Papers:
Lung cancer in never-smoker Asian females is driven by oncogenic mutations, most often involving EGFR
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Oncotarget. 2015; 6:5465-5474.
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Sang Yun Ha_, So-Jung Choi, Jong Ho Cho, Hye Joo Choi, Jinseon Lee, Kyungsoo Jung, Darry Irwin, Xiao Liu, Maruja E. Lira, Mao Mao, Hong Kwan Kim, Yong Soo Choi, Young Mog Shim, Woong Yang Park, Yoon-La Choi and Jhingook Kim
Abstract
Sang Yun Ha 1,, So-Jung Choi 2,, Jong Ho Cho 3,, Hye Joo Choi 2, Jinseon Lee 2, Kyungsoo Jung 4, Darry Irwin 5, Xiao Liu 6,7, Maruja E. Lira 8, Mao Mao 9, Hong Kwan Kim 3, Yong Soo Choi 3, Young Mog Shim 3, Woong Yang Park 10, Yoon-La Choi 1,4,10, Jhingook Kim 2,3
1 Department of Pathology and Translational Genomics, Samsung Medical Center, Sungkyunkwan University School of Medicine, Seoul, Korea
2 Samsung Biomedical Research Institute, Samsung Medical Center, Sungkyunkwan University School of Medicine, Seoul, Korea
3 Department of Thoracic and Cardiovascular Surgery, Samsung Medical Center, Samsung Biomedical Research Institute, Sungkyunkwan University School of Medicine, Seoul, Korea
4 Samsung Advanced Institute for Health Sciences & Technology, Sungkyunkwan University School of Medicine, Seoul, Korea
5 Agena Bioscience, Sequenom, San Diego, CA, USA
6 BGI-Shenzhen, Shenzhen, China
7 Department of Biology, University of Copenhagen, Copenhagen, Denmark
8 Oncoloy Research Unit, Pfizer Worldwide Research and Development, San Diego, CA, USA
9 WuXi AppTec, Shanghai, China
10 Samsung Genome Institute, Samsung Medical Center, Sungkyunkwan University School of Medicine, Seoul, Korea
These authors contributed equally to this work
Correspondence to:
Jhingook Kim, e-mail: jhingookkim@gmail.com
Yoon-La Choi, e-mail: ylachoi@skku.edu
Keywords: non-small cell lung cancer, adenocarcinoma, never-smoker female, driver mutation, EGFR
Received: August 07, 2014 Accepted: December 16, 2014 Published: March 09, 2015
ABSTRACT
The aim of this study was to determine the distribution of known oncogenic driver mutations in female never-smoker Asian patients with lung adenocarcinoma. We analyzed 214 mutations across 26 lung cancer-associated genes and three fusion genes using the MassARRAY® LungCarta Panel and the ALK, ROS1, and RET fusion assays in 198 consecutively resected lung adenocarcinomas from never-smoker females at a single institution. EGFR mutation, which was the most frequent driver gene mutation, was detected in 124 (63%) cases. Mutation of ALK, KRAS, PIK3CA, ERBB2, BRAF, ROS1, and RET genesoccurred in 7%, 4%, 2.5%, 1.5%, 1%, 1%, and 1% of cases, respectively. Thus, 79% of lung adenocarcinomas from never-smoker females harbored well-known oncogenic mutations. Mucinous adenocarcinomas tended to have a lower frequency of known driver gene mutations than other histologic subtypes. EGFR mutation was associated with older age and a predominantly acinar pattern, while ALK rearrangement was associated with younger age and a predominantly solid pattern. Lung cancer in never-smoker Asian females is a distinct entity, with the majority of these cancers developing from oncogenic mutations.
INTRODUCTION
Lung cancer is the leading cause of cancer-related mortality, with 1.38 million annual deaths worldwide . Tobacco smoking is the main risk factor for lung cancer; however, approximately 25% of lung cancers worldwide occur in never-smokers [2, 3]. Moreover, the risk of lung cancer differs by race/ethnicity. In the United States, approximately 10% of patients with lung cancer are never-smokers , while in Asia, >30% of patients with lung cancer are never-smokers and ≥50% of lung cancers occur in women who are never-smokers . Never-smoker East Asian females have a tendency to develop adenocarcinoma, and these never-smokers exhibit higher treatment response rates to epidermal growth factor receptor tyrosine kinase inhibitors (EGFR-TKIs), such as gefitinib and erlotinib, than those with a history of tobacco smoking . In several phase III studies, significantly better response rates and longer progression-free survival were observed in advanced non-small cell lung cancer (NSCLC) patients harboring activating EGFR mutations who were treated with first-line EGFR-TKIs than those patients treated with doublet platinum-based chemotherapy [7, 8]. In addition, a fusion protein of the N-terminal portion of the echinoderm microtubule-associated protein-like 4 (EML4) gene and the intracellular signaling portion of the anaplastic lymphoma kinase (ALK) tyrosine kinase receptor has been identified in a small subset of NSCLC patients . Patients harboring the EML4-ALK fusion show unique clinicopathologic and physiological characteristics and respond positively to ALK inhibitors [10–12]. As for EGFR mutations, several reports have identified the EML4-ALK fusion protein predominantly in young female never-smokers with adenocarcinoma, although the presence of this fusion protein is mutually exclusive with EGFR mutation [9, 13]. More specifically, the EML4-ALK fusion gene was found in 23.7% of never-smoker female lung adenocarcinoma patients .
Lung adenocarcinoma in never-smoker females has been established as a distinct entity based on its particular epidemiologic, clinical, and biological characteristics. EGFR mutations and the EML4-ALK translocation are defined as driver mutations because these alterations are responsible for both initiation and maintenance of lung cancer. The discovery of driver oncogene genetic variants that are sensitive to molecular-targeted drugs is crucial for improvement of treatment strategies. Therefore, use of multi-mutational profiling in lung cancer studies is important for identification of driver gene alterations in order to validate the effectiveness of molecular-targeted therapies. In addition, the proportion of never-smokers with lung cancer is likely to increase as a result of smoking cessation and prevention programs. Thus, the aims of this study were to analyze the distribution of oncogenic driver mutations and to compare these mutations with clinicopathologic characteristics in female Asian never-smoker lung adenocarcinoma patients.
RESULTS
Patient population and histopathologic classification
The patient population is summarized in Table 1. All patients were women with no smoking history. Median age at diagnosis was 60 years (range, 29–81 years). Most patients (85.3%) had undergone lobectomy. Two patients had pleural metastasis at the time of surgery. Eleven (5.6%) patients had received neoadjuvant concurrent chemoradiation therapy. Among the 198 patients in this study, 104 (52.5%) patients had T1 stage tumors, 83 (41.9%) patients had T2 stage tumors, and 11 (5.6%) patients had T3 stage tumors. Nine patients did not undergo lymph node dissection and were excluded from the analysis of N stage. N0, N1, and N2 stage tumors were observed in 124 (65.6%), 25 (13.2%), and 40 (21.2%) patients, respectively. Two of the 198 cases were classified as minimally invasive adenocarcinomas with an invasive component of <5 mm. Invasive adenocarcinomas were classified as predominantly acinar (142 cases, 71.7%), predominantly papillary (18 cases, 9.1%), predominantly solid (17 cases, 8.6%), predominantly lepidic (9 cases, 4.5%), and predominantly micropapillary (1 case, 0.5%) patterns. Nine (4.5%) cases were classified as mucinous adenocarcinoma. Two minimally invasive adenocarcinomas were classified as predominantly lepidic for statistical analysis. In addition, all histologic patterns that were observed in >10% of the tumor area were recorded. An acinar pattern was observed in 169 (85.4%) cases. Papillary, lepidic, micropapillary, solid, and mucinous patterns were observed in 68 (34.3%), 36 (18.2%), 36 (18.2%), 33 (16.7%), and 17 (8.6%) cases, respectively.
Fusion gene and LungCarta analyses
The results of the fusion gene and LungCarta analyses are summarized in Figure 2 and Table 2, and a full list of mutations that were identified is shown in Supplemental Table 1. Among the 198 cases, driver mutations were detected in 157 (79%) cases. EGFR mutations were the most frequently found mutation in lung adenocarcinomas of female never-smokers (124 cases, 63%). EGFR mutations were detected in exon 19 in 47 (24%) cases, exon 20 in 16 (8%) cases, exon 21 in 58 (29%) cases, exons 18 and 21 in 2 (1%) cases, and exons 18 and 20 in 1 (1%) case. ALK rearrangement was observed in 14 (7%) of cases. The frequencies of KRAS, PIK3CA, TP53, ERBB2, and BRAF mutations were 4%, 2.5%, 2%, 1.5%, and 1%, respectively. ROS1 and RET gene rearrangements were each found in 2 (1%) cases. ALK, ROS1, and RET gene fusions were mutually exclusive with EGFR and KRAS mutations. The gene mutations were mutually exclusive with the exception of concurrent mutation of EGFR/PIK3CA (n = 3), EGFR/TP53 (n = 1), ALK/TP53 (n = 2), KRAS/PIK3CA (n = 1), and KRAS/ERBB2 (n = 1; Figure 3). The results of Sanger sequencing validation were consistent with the results of LungCarta analyses in all cases (Supplemental Table 2).
Table 1: Patient population
CharacteristicsNumber of patients (%)
Age, years median 60 (range, 29–81)
Histologic classification
Minimally invasive adenocarcinoma 2 (1.0)
Invasive adenocarcinoma
Lepidic predominant 9 (4.5)
Acinar predominant 142 (71.7)
Papillary predominant 18 (9.1)
Micropapillary predominant 1 (0.5)
Solid predominant 17 (8.6)
Mucinous adenocarcinoma 9 (4.5)
T stage
1 104 (52.5)
2 83 (41.9)
3 11 (5.6)
N stage1
0 124 (65.6)
1 25 (13.2)
2 40 (21.2)
M stage
0 196 (99.0)
1 2 (1.0)
Operation
Wedge resection 16 (8.0)
Lobectomy 169 (85.3)
Lobectomy + α213 (6.6)
Neoadjuvant CCRT3
Yes 11 (5.6)
No 187 (94.4)
1 Nine cases were excluded because lymph node dissection was not performed.
2 α includes lobectomy with wedge resection of the other lobe, bilobectomy, or pneumonectomy.
3 CCRT indicates concurrent chemoradiation therapy.
Relationship between driver mutation status and clinicopathologic characteristics
The relationship between the driver mutation status and histologic subclassification of adenocarcinoma is summarized in Table 2 and Figure 4A–4E. Tumors with EGFR mutation exhibited a high frequency of predominantly acinar patterns and an absence of a predominantly mucinous pattern. ALK rearrangements were frequently found in cases with a predominantly solid pattern. KRAS mutations were detected with high frequency in cases with predominantly mucinous tumors. BRAF mutations were detected only in predominantly papillary tumors. These four genes were significantly associated with the histologic subclassification. Mucinous adenocarcinoma tended to harbor a lower frequency of known driver gene mutations than other histologic subtypes (mutation rate, 44.4% vs. 80.4%; p = 0.090).
Figure 1: Process of patient (female never smoker with lung adenocarcinoma) selection in this study.(A) Distribution of lung cancer according to histology subtype. (B) Distribution of lung adenocarcinoma according to gender. (C) Distribution of lung adenocarcinoma from female according to smoking status. Patients with no medical record of smoking status were excluded. The record of year 2012 was not shown due to lack of smoking information.
Figure 2: Frequency of driver gene mutations in lung adenocarcinomas from East Asian never-smoker females.
EGFR mutations were associated frequently with lepidic and acinar patterns and infrequently with mucinous patterns in cases where the histologic pattern predominated in >10% of the tumor area. ALK rearrangement was more frequently detected in tumors with solid and mucinous patterns, while KRAS mutations were also commonly found in tumors with a mucinous pattern. Other mutations were not associated with histologic subtypes (Supplemental Table 3).
Patients with tumors harboring EGFR and BRAF mutations were older than those with wild-type tumors, whereas ALK gene fusions and ERBB2 mutations were associated with younger age at diagnosis (Supplemental Table 4). Driver gene mutation status was not associated with tumor stage.
Table 2: Relationship between driver mutation status and histologic subclassification of adenocarcinoma according to predominant type
TotalLepidic1Acinar2PapillaryMicropapillarySolidMucinous3p-value
n (%)198 (100)11 (5.6)142 (71.7)18 (9.1)1 (0.5)17 (8.6)9 (4.5)
Wild-type41 (20.7)3 (7.3)29 (70.7)2 (4.9)0 (0)3 (7.3)4 (9.8)0.446 Mucinous vs. others: 0.090
Mutant type157 (79.3)8 (5.1)113 (72.0)16 (10.2)1 (0.6)14 (8.9)5 (3.2)
EGFR 124 (62.6)7 (5.6)102 (82.3)9 (7.3)1 (0.8)5 (4.0)0 (0)< 0.001
ALK 14 (7.1)0 (0)6 (42.9)2 (14.3)0 (0)5 (35.7)1 (7.1)0.012
KRAS 8 (4.0)1 (12.5)1 (12.5)0 (0)0 (0)2 (25.0)4 (50.0)< 0.001
PIK3CA 5 (2.5)0 (0)3 (60.0)1 (20.0)0 (0)0 (0)1 (20.0)0.31
TP53 4 (2.0)0 (0)4 (2.8)0 (0)0 (0)0 (0)0 (0)1
ERBB2 3 (1.5)0 (0)2 (66.7)0 (0)0 (0)0 (0)1 (33.3)0.272
BRAF 2 (1.0)0 (0)0 (0)2 (100)0 (0)0 (0)0 (0)0.043
ROS1 2 (1.0)0 (0)0 (0)1 (50.0)0 (0)1 (50.0)0 (0)0.086
RET 2 (1.0)0 (0)1 (50.0)1 (50.0)0 (0)0 (0)0 (0)0.487
NRAS 1 (0.5)0 (0)0 (0)0 (0)0 (0)1 (100)0 (0)0.192
1 Two cases of minimally invasive adenocarcinoma were included in the lepidic predominant type.
2 Six cases with acinar pattern showed concurrent mutation of EGFR/PIK3CA (n = 3), EGFR/TP53 (n = 1), and ALK/TP53 (n = 2).
3 Two cases of mucinous adenocarcinoma showed concurrent mutation of KRAS/PIK3CA (n = 1) and KRAS/ERBB2 (n = 1).
Figure 3: Diagram demonstrating driver gene mutation status and clinicopathologic features in 198 adenocarcinomas from East Asian never-smoker females. Most mutations were mutually exclusive with the rare exception of concurrent mutation of EGFR/PIK3CA (n = 3), EGFR/TP53 (n = 1), ALK/TP53 (n = 2), KRAS/PIK3CA (n = 1), and KRAS/ERBB2 (n = 1). Patients who received concurrent chemoradiation therapy and those who did not undergo lymph node dissection were excluded in the analysis of TNM stage.
DISCUSSION
In this study, we evaluated the status of driver gene mutations in lung adenocarcinoma samples from 198 East Asian female never-smokers using the MassARRAY® LungCarta Panel and ALK, ROS1, and RET fusion assays. Approximately 79% of patients harbored driver gene mutations, and EGFR mutation (63%) was the most frequent driver mutation detected. Other genetic alterations occurred less frequently with ALK gene fusion, KRAS mutation, ERBB2 mutation, HER2 mutation, ROS1 rearrangement, and RET rearrangement occurring in 7%, 4%, 1.5%, 1%, 1%, and 1% of cases, respectively. Most mutations were mutually exclusive.
Figure 4: Frequency of driver gene mutations according to predominant histologic subtype.(A) Lepidic, (B) Acinar, (C) Papillary, (D) Solid, (E) Mucinous subtype. In acinar subtype, four cases with concurrent mutations of EGFR/PIK3CA (n = 3) and EGFR/TP53 (n = 1) are represented as an EGFR mutation, and 2 cases of ALK/TP53 mutation as an ALK mutation. In mucinous subtype, two cases with concurrent mutations of KRAS/PIK3CA and KRAS/ERBB2 are represented as a KRAS mutation.
Ethnic background is a well-established factor in NSCLC. Asian female non-smokers were the specific epidemiologic subgroup in this study, and a few previous large-scale studies have shown that Asian ethnicity is a prognostic factor of overall survival for NSCLC patients [15–17]. The link between ethnicity and cancer survival rates is likely due to differences in genetic background among ethnicities. Compared to Caucasians, East Asians with NSCLC have higher rates of EGFR mutations and lower rates of KRAS and LKB1 mutations . According to a recent meta-analysis of 94 studies, the rate of EGFR mutation in lung adenocarcinoma in Asians (47.9%) is higher than that in Westerners (19.2%), while KRAS (11.2%) and LKB1 (4.0%) mutation rates are lower in Asians than in Westerners (26.1% and 16.2%, respectively) . Smoking status is also an important factor in the development of lung adenocarcinoma. Lung cancers in never-smokers are more frequently associated with adenocarcinoma with EGFR mutations and less frequently with KRAS mutations . In a recent study of lung adenocarcinomas, mutation rates for EGFR (39%), KRAS (4%), ALK (15%), and HER2 (5%) in never-smoker groups differed from rates in current or former smoker groups (10%, 35%, 4%, and 1%, respectively) . Notably, the proportion of female lung cancers in never-smokers is much higher in East Asia than in Europe or the United States (60% vs 15–20%) .
Although many studies have investigated driver gene mutations in lung adenocarcinoma, only a few have focused specifically on Asian female never-smokers despite the demonstrated importance of this patient group for selection of targeted candidates for NSCLC therapy [22–24]. Zhang et al. analyzed EGFR, KRAS, ALK, HER2, and BRAF mutation in 349 Chinese never-smoker females with lung adenocarcinoma and discovered mutation rates of 76%, 5%, 4%, 2%, and 1% for EGFR, HER2, ALK, KRAS, and HER2, respectively, while only 12% of cases harbored no detected mutation . In an analysis of mutations of 10 driver genes (EGFR, KRAS, NRAS, HRAS, HER2, BRAF, ALK, PIK3CA, TP53, and LKB1) in 52 lung adenocarcinomas from East Asian never-smokers including 41 women , Sun et al. found similar rates of genetic alterations: EGFR mutation in 79% of cases, EML4-ALK fusion in 6% of cases, HER2 mutation in 4% of cases, and KRAS mutation in 2% of cases. In addition, only 10% of patients did not harbor any detected mutation in these genes. Ren et al. reported EGFR mutation in 70% and ALK rearrangement in 9.6% in adenocarcinomas from never-smoker Chinese women (see Supplemental Table 5 for summary of these studies) . The incidences of mutation detected in this study are consistent with those of previous studies.
Several important features distinguish our study from previous studies. In our study, we analyzed a total of 26 known oncogenes and 3 fusion genes. We discovered that 1% of lung adenocarcinomas from East Asian never-smoker female patients harbored gene rearrangements of ROS1 and RET, which have recently been recognized as driver genes in lung adenocarcinoma [28–30] but had not been evaluated in East Asian female never-smokers. We used a multiplex method that can be used in routine clinical practice and our results demonstrated that this method was comparable to traditional mutational analysis using polymerase chain reaction (PCR) for each gene. The use of a multiplexed PCR-based assay to genotype NSCLCs was recently demonstrated to be clinically feasible . Moreover, we confirmed well-known associations of driver gene mutations with histologic subtypes and clinical characteristics. EGFR mutation correlated positively with older age and a predominantly acinar pattern [25, 32, 33], while ALK rearrangement was associated with younger age and solid histology [34, 35]. KRAS mutation was detected more frequently in mucinous adenocarcinoma [25, 36]. Furthermore, most PIK3CA mutations coexisted with other mutations . The fact that our findings are consistent with those of previous studies supports the value of the multiplex method in mutation analysis and suggests that clinicopathologic associations are useful for determination of the priority of driver mutation tests.
The most outstanding result of our study was that driver gene mutations were detected in 79% of female never-smoker Asian patients. Targeted therapies for EGFR (erlotinib/gefitinib) and ALK (crizotinib) are currently available, and 70% of female never-smoker Asian patients with EGFR mutation (63%) or ALK rearrangement (7%) may benefit from such targeted therapies. An additional 10% of patients in our study may benefit from newly developed targeted drugs. A recent clinical trial described a partial response to the EGFR/HER2 inhibitor BIBW2922 in a HER2 mutant tumor , suggesting that this new drug is a promising treatment strategy. More comprehensive genomic analysis and deep sequencing may be necessary to identify genetic alterations in the remaining 20% of patients with no known mutations . Notably, known driver gene mutations were found less frequently in mucinous adenocarcinoma in our study, suggesting the possibility of mutation in another driver gene in this histologic subtype.
In conclusion, lung cancer in never-smoker Asian females is a distinct entity, with the majority of these lung cancers developing from oncogenic mutations.
METHODS
Study population
Specimens were obtained from Samsung Medical Center (SMC) in Seoul, Korea with prior informed patient consent and approval by the Institutional Review Board of Samsung Medical Center. Between January 2008 and January 2013, 3796 consecutive patients underwent pulmonary resection with curative intent for primary lung cancer at our institute.
After excluding 1307 patients with non-adenocarcinoma (Figure 1A), the remaining 2489 patients consisted of 1302 men and 1187 women (Figure 1B). Of the 1187 female patients with lung adenocarcinoma, 422 patients were excluded due to a lack of smoking information. Of the remaining 765 female patients, 94.9% (726 women) were never-smokers (Figure 1C). Among the 726 never-smoker female pulmonary adenocarcinoma cases, 198 had sufficient tissue for genomic analysis and were included in this study.
Data collection and histologic classification of adenocarcinoma
Study data were abstracted from in-hospital charts and electronic medical records by trained experienced nurses from the Departments of Medical Oncology, Surgical Oncology, Laboratory Medicine, Pathology, and Nursing at Samsung Medical Center. Baseline clinical characteristics included gender, age at diagnosis, smoking history, alcohol consumption status, date of diagnosis of advanced lung cancer, tumor histology, tumor stage, and ambulatory status at diagnostic work-up. Tumor stage was defined according to the seventh edition of the American Joint Committee on Cancer . Smoking history and alcohol consumption status were determined by self-reported answers on questionnaires. Never-smokers were defined as patients who smoked less than 100 cigarettes over their lifetime. Histologic classification was determined according to the International Association for the Study of Lung Cancer, American Thoracic Society, and European Respiratory Society classification of lung adenocarcinomas . All histologic patterns that covered at least 10% of the tumor area were recorded, and the predominant pattern was defined as the pattern that covered the largest portion of the tumor area.
DNA and RNA extraction
Genomic DNA or RNA was extracted from lung tumors or distant histologically normal lung tissue using standard protocols (RNeasy Mini Kit and QIAamp DNA Mini Kit, Qiagen, Valencia, CA, USA).
ALK, ROS1, and RET fusion assays
The nCounter™ gene expression assays were custom-designed and synthesized by NanoString Technologies (Seattle, WA, USA). Hybridization, sample cleanup, and digital reporter counts were performed according to the manufacturer’s protocol. RNA was obtained from fresh-frozen tissues using the Qiagen RNeasy Mini Kit (Qiagen, Valencia, CA, USA). RNA concentration was assessed by spectrophotometry using the Nanodrop 8000 (Thermo-Scientific, Wilmington, DE, USA).
Samples were processed according to the gene expression protocol of NanoString Technologies. Briefly, total RNA was hybridized to a multiplexed mixture of custom-designed nCounter™ capture and reporter probes complementary to ALK, ROS1, and RET target sequences (Supplemental Table 2) for at least 16 h at 65°C. The samples were cleaned up and processed using an automated nCounter™ Sample Prep Station (NanoString Technologies). Unhybridized probes were removed, and the hybridization complex was immobilized onto a cartridge and aligned. Fluorescently labeled, color-coded reporters were subsequently imaged on an nCounter™ Digital Analyzer (NanoString Technologies) set at 1155 fields of view. Raw reporter counts were collected using nSolver software v1.0 (NanoString Technologies).
LungCarta analysis
High-throughput multiplex mutation profiling was performed using the MassARRAY® LungCarta Panel Version 1.0 (Sequenom, San Diego, CA, USA). This panel permits screening of 214 mutations across 26 oncogenes and tumor suppressors with a limit of sensitivity of approximately 10% with the use of 480 ng DNA . DNA was amplified using the OncoCarta PCR primer mix, unincorporated nucleotides were inactivated by shrimp alkaline phosphatase, and a single base extension reaction was performed using extension primers that hybridize adjacent to the mutation. Multiplexed reactions were spotted onto the SpectroChipII (Sequenom) using the MassARRAY Nanodispenser. Peaks with different mass were resolved by matrix-assisted laser desorption/ionization time-of-flight on the MassARRAY Compact Analyzer. A predefined ratio of expected normal allele to mutant allele at a specific nucleotide position allows mutations to be detected using primer extensions at that specific position. Because of the multiplexing capabilities of this assay, multiple mutations are detected simultaneously using one panel. Further details of the multiplex methodology can be found in the protocol provided by the manufacturer.
Sanger sequencing
To validate the LungCarta analysis, we performed Sanger sequencing of EGFR in 11 selected cases (4 cases of EGFR E746_A750del, 4 cases of EGFR L858R, and 3 cases of KRAS G12D) according to the method described previously by our group .
Statistical analysis
Pearson’s chi-squared, Fisher’s exact, and independent t tests were used as indicated. All tests were two sided, and a p-value of < 0.05 was considered statistically significant. Statistical analyses were performed using SPSS software (SPSS Inc., Chicago, IL, USA).
ACKNOWLEDGMENTS
This study was supported by the Converging Research Center Program (No. 2013K000278) funded by the Korean government (The Ministry of Science, Information and Communications Technology, and Future Planning) and by the R&D Program for the Society of the National Research Foundation (NRF) funded by the Ministry of Science, ICT, and Future Planning (NRF-2013M3C8A1078501) and by a grant of the Korea Health Technology R&D Project through the Korea Health Industry Development Institute(KHIDI), funded by the Ministry of Health & Welfare, Republic of Korea (grant number:HI13C2096).
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Why does a maximum curve have a negative value of its second derivative I know it’s because that the gradinet becomes negative as you go from left to right but what about if you go from right to left shouldn’t it be positive? - Quora
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Why does a maximum curve have a negative value of its second derivative I know it’s because that the gradinet becomes negative as you go from left to right but what about if you go from right to left shouldn’t it be positive?
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Shubham Ghosh
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Why is a second order derivative of Maxima negative and positive for Minima? What is the concept?
I'll try to explain this in the simplest way possible for me. This might be a bit long, but it will be worth reading (I hope so). Let's get started!
You know that derivative can be thought of as a rate measure. First derivative tells us the slope of the tangent, that is, the change in y-coordinate of a function with respect to its x-coordinate.
You also know second derivative is nothing but the derivative of the first derivative, that is, derivative of slope of the tangent OR the rate of change of tangent, right?
Now, if the second derivative of a function is positive at a point, (say, a) it mean
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I'll try to explain this in the simplest way possible for me. This might be a bit long, but it will be worth reading (I hope so). Let's get started!
You know that derivative can be thought of as a rate measure. First derivative tells us the slope of the tangent, that is, the change in y-coordinate of a function with respect to its x-coordinate.
You also know second derivative is nothing but the derivative of the first derivative, that is, derivative of slope of the tangent OR the rate of change of tangent, right?
Now, if the second derivative of a function is positive at a point, (say, a) it means the slope is increasing in the neighborhood of a. Got it? For example, if acceleration of a body is positive, it means its velocity is increasing. Hope you got my point.
If the second derivative is positive, it means that the slope of the tangent is increasing from a negative (or small positive value) value to a positive value. In this case, it is called a concave up. It looks like a 'u' on the graph. The following picture illustrates this fact:
Note that in the left neighborhood of 0, the slope of tangent is negative, while in the right neighborhood it is positive. Hence, the slope of tangent is increasing i.e., the second derivative of the function is positive.
Also, note that this concave up condition at a point is possible iff the graph is attaining a low value (minima) at that point (0 in the above case).
Hence, second derivative is positive for minima.
Similarly, it can be shown that second derivative is negative for Maxima (concave down). Try to visualise it from the following graph. It's your homework ;)
If you have any doubt, feel free to ask.
Thanks.
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More answers below
When is second derivative positive or negative?
After you find the derivative of your curve, and find its value at a point of interest, the value may be positive, negative, or zero. What does it tell you about your curve if the derivative has a value of 0?
What does d e t∣∣(∂F∂x(x))∣∣d e t|(∂F∂x(x))| mean?
What does C[n 7]n≡n!(n−[n 5])![n 5]!C n[n 7]≡n!(n−[n 5])![n 5]! mean?
If the slope of a curve changes direction, then what will happen to its second derivative (if it exists)?
Badder Danbad
MBA-IS in Business Information Systems, University of Northampton (Graduated 2011)
·2y
The first derivative of the function f (x), which we write as f ′(x) or as df/dx , is the slope of the tangent line
to the function at the point x. To put this in non-graphical terms, the first derivative tells us how whether
a function is increasing or decreasing, and by how much it is increasing or decreasing.
This information is reflected in the graph of a function by the slope of the tangent li
Continue Reading
The first derivative of the function f (x), which we write as f ′(x) or as df/dx , is the slope of the tangent line
to the function at the point x. To put this in non-graphical terms, the first derivative tells us how whether
a function is increasing or decreasing, and by how much it is increasing or decreasing.
This information is reflected in the graph of a function by the slope of the tangent line to a point on the graph, which is sometimes describe as the slope of the function. Positive slope tells us that, as x increases, f (x) also increases. Negative slope tells us that, as x increases, f (x) decreases. Zero slope does not tell us anything in particular: the function may be increasing, decreasing, or at a local maximum or a local minimum at that point. Writing this information in terms of derivatives.
if df/dx (p) > 0, then f (x) is an increasing function at x = p.
if df/dx (p) < 0, then f (x) is a decreasing function at x = p.
if df/dx (p) = 0, then x = p is call...
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George Ivey
Former Math Professor at Gallaudet University · Author has 23.7K answers and 2.6M answer views
·2y
If a curve goes up to a maximum, then back down, the derivative is positive since the curve is going up, 0 at the maximum, then negative since the curve is going down.
Since the derivative goes from positive to negative it is decreasing so its derivative, the second derivative, is negative.
“If you go from right to left to right” everything is reverse but that is not relevant because that is not how derivatives are calculated! The definition of the derivative is lim h→0 f(x+h)−f(x)h lim h→0 f(x+h)−f(x)h, not, lim h→0 f(x)−f(x+h)h lim h→0 f(x)−f(x+h)h.
Upvote ·
Simon Tsai
Lives in Taiwan · Author has 4.9K answers and 2M answer views
·2y
Assume analyticity. The key observation is
f(x+h)=f(x)+h f′(x)+1 2 h 2 f′′(x)+O(h 3)f(x+h)=f(x)+h f′(x)+1 2 h 2 f″(x)+O(h 3)
If f′(x)=0 f′(x)=0, then as h→0 h→0, the only thing that matters is the second-order derivative: f(x)>f(x+h)f(x)>f(x+h) for all h≠0 h≠0 if and only if f′′(x)<0 f″(x)<0. I hope that this answer is of your help.
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More answers below
What happens to a curve when its second derivative changes sign?
What is the second derivative of 3x2?
Why is slope at instant (the derivative) called as instantaneous rate of change of "y" with respect to x, when the slope at an instant has nothing to do with 'y' values of the function?
If taking derivative means finding the slope of a curve, how would you explain taking a second derivative?
Why is it that when I plug an x value into the derivative, I get the slope of the x value on a curve? Does this have to do with the limit-based definition of derivative?
Dave Benson
trying to make maths easy. · Author has 6.1K answers and 2.1M answer views
·2y
If 2nd derivative is negative it means the curve is concave downwards irrespective of direction.
So the curve is a maximum at that point.
Upvote ·
Fred Oberlander
Graduate Degrees Compsci, Math, Stat, Finance, Law · Author has 1.9K answers and 483.6K answer views
·2y
approaching a local or global maximum of a convex region by incrementing x must cause monotonically reducing second derivative regardless of direction of approach, or what I think you mean by direction.
Upvote ·
Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Upvoted by
Richard Shearer
, MSc (Biochemistry) Biochemistry & Mathematics, University of the Witwatersrand · Author has 6.8K answers and 52.8M answer views
·1y
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Why is the rate of change of slope for a point of inflection 0? Why is the second derivative of an inflection point 0?
Please study this diagram…
If you understood the above related diagrams then I have answered your question but if you need a bit more then see below…
Let’s just consider the first diagram…
Slope at A = 1
Slope at B = 0
Slope at C = – 2
Slope at D = 0
Slope at E = 2
From A to B the slope is decreasing from 1 to zero at B
From B to C the slope becomes negative and continues to decrease until it reaches C where the slope = – 2
But then the slope starts increasing again until it reaches 0 at D
From D to E the slope increases to 2 at E
The point where the slope stops decreasing and starts increasing is a point
Continue Reading
Please study this diagram…
If you understood the above related diagrams then I have answered your question but if you need a bit more then see below…
Let’s just consider the first diagram…
Slope at A = 1
Slope at B = 0
Slope at C = – 2
Slope at D = 0
Slope at E = 2
From A to B the slope is decreasing from 1 to zero at B
From B to C the slope becomes negative and continues to decrease until it reaches C where the slope = – 2
But then the slope starts increasing again until it reaches 0 at D
From D to E the slope increases to 2 at E
The point where the slope stops decreasing and starts increasing is a point of inflection!
Q is an INFLECTION POINT
—————————————————————————————————
It is best learnt in person so I made these videos. Please watch…
POINTS OF INFLECTION 1
POINTS OF INFLECTION 2
POINTS OF INFLECTION 3
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Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views
·3y
Related
Why, when the second order derivative is less than zero, is it the maximum?
To understand the Second Derivative Test, consider these curves:
NOTICE THREE POINTS:
When the cubic has a MAXIMUM the 2nd derivative is a NEGATIVE number.
When the cubic has a MINIMUM the 2nd derivative is a POSITIVE number.
When the cubic has an INFLECTION point the 2nd derivative is ZERO.
Continue Reading
To understand the Second Derivative Test, consider these curves:
NOTICE THREE POINTS:
When the cubic has a MAXIMUM the 2nd derivative is a NEGATIVE number.
When the cubic has a MINIMUM the 2nd derivative is a POSITIVE number.
When the cubic has an INFLECTION point the 2nd derivative is ZERO.
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Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views
·2y
Related
What happens to the second derivative when we approach a local maximum or minimum?
The best way to see this is to have 3 separate graphs in line so that we can compare graphs of y = f(x), y = f’(x) and y = f’’(x)
Let’s use a nice cubic curve…
So as we approach a maximum point the second derivative is negative
and as we approach a minimum point the second derivative is positive.
Also as we approach a point of inflection the second derivative is zero.
Continue Reading
The best way to see this is to have 3 separate graphs in line so that we can compare graphs of y = f(x), y = f’(x) and y = f’’(x)
Let’s use a nice cubic curve…
So as we approach a maximum point the second derivative is negative
and as we approach a minimum point the second derivative is positive.
Also as we approach a point of inflection the second derivative is zero.
Upvote ·
99 18
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Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views
·Updated 1y
Related
Why does the second derivative d^2y/dx^2>0 give maxima? What is the graphical meaning of the second derivative test for maxima?
I made this poster for my classroom wall as a constant reminder to students about this concept.
I think you will find it self-explanatory.
Continue Reading
I made this poster for my classroom wall as a constant reminder to students about this concept.
I think you will find it self-explanatory.
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Carter McClung
B.S. in UPSC General Studies&Mathematics, The University of Texas at Dallas (Graduated 2006) · Author has 1.6K answers and 6.5M answer views
·7y
Related
Why, if the second derivative of a function is negative in optimization problems, is there a maximum in that point?
You’re leaving out one little bit:
If the first derivative is zero, it is a critical point.
It’s either a local maximum, local minimum, or a point of inflection. That’s because in the first two cases, you’re transitioning from positive to negative slope (or vice versa). In point of inflection case, you’re just taking a momentary break from a positive (or negative) slope to have a zero slope, then returning right to whatever slope sign you had before. No change of direction, so no max or min.
The second derivative deals with concavity. If your slope is zero and concave down, what does that mean?
Cl
Continue Reading
You’re leaving out one little bit:
If the first derivative is zero, it is a critical point.
It’s either a local maximum, local minimum, or a point of inflection. That’s because in the first two cases, you’re transitioning from positive to negative slope (or vice versa). In point of inflection case, you’re just taking a momentary break from a positive (or negative) slope to have a zero slope, then returning right to whatever slope sign you had before. No change of direction, so no max or min.
The second derivative deals with concavity. If your slope is zero and concave down, what does that mean?
Clearly this is all concave downward (negative 2nd derivative) and where the slope (first derivative) is zero we have a local maximum.
Similarly when the second derivative is positive and the slope is zero, we get a local minimum:
The last possibility is that the concavity is transitioning from concave up to concave down (or vice versa). If the slope is zero there it looks like this:
Here you should be able to see the concavity change at the point of inflection.
— — — -
Long story short:
First derivative zero? You have a critical point.
Then the second derivative tells you the concavity around that point and “tells its story.”
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Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views
·5y
Related
If taking derivative means finding the slope of a curve, how would you explain taking a second derivative?
I have two simple explanations for you.
Secondly, if the distance x metres of an object from a point P at t seconds is for t values from 0 to 3 seconds is given by…
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I have two simple explanations for you.
Secondly, if the distance x metres of an object from a point P at t seconds is for t values from 0 to 3 seconds is given by…
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9 2
9 1
B. S. Thomson
Lived in Vancouver, BC · Author has 1.2K answers and 2.9M answer views
·10mo
Related
Why is there a relative maximum where the second derivative of f is less than zero?
I find it curious that all the answers I have read here so far are at an informal level describing the intuition, not the mathematics. Some are misguided. If you ask this question over a beer these might be helpful. If this were a mathematics exam question in a rigorous course then it would be hard to search for partial credit in any of them.
So perhaps there is room for a different kind of answer. Turn on pedantic mode. This account simply makes precise the friendly advice in some of the other answers.
Q. Suppose f:R→R f:R→R is a differentiable function and that the second derivative f′′(x 0)f″(x 0) ex
Continue Reading
I find it curious that all the answers I have read here so far are at an informal level describing the intuition, not the mathematics. Some are misguided. If you ask this question over a beer these might be helpful. If this were a mathematics exam question in a rigorous course then it would be hard to search for partial credit in any of them.
So perhaps there is room for a different kind of answer. Turn on pedantic mode. This account simply makes precise the friendly advice in some of the other answers.
Q. Suppose f:R→R f:R→R is a differentiable function and that the second derivative f′′(x 0)f″(x 0) exists and is negative at a point x 0.x 0. Explain the conditions under which there is a relative maximum at x 0 x 0.
We need to appeal to this elementary theorem describing exactly what simple implications there are for a solitary nonzero derivative [positive or negative] at a point:
Theorem. Suppose g:R→R g:R→R and that the derivative g′(x 0)g′(x 0) exists and is negative [positive] at a point x 0 x 0. Then there is an interval (u,v)(u,v) containing x 0 x 0 so that
g(x)>g(x 0)>g(y)f o r a l l u<x<x 0g(x 0)>g(y)f o r a l l u<x<x 0<y<v.
[For positive g′(x 0)g′(x 0) reverse these.]
Argument I: From the inequality it is clear that a relative maximum or a relative minimum of g g cannot possibly occur at a point where g′g′ is positive or negative. It might occur at a point where g′g′is zero, but it might not. Hence we are encouraged to look at the second derivative.
From the theorem there is a simple corollary.
Corollary. Suppose f:R→R f:R→R and that the second derivative f′′(x 0)f″(x 0) exists and is negative at a point x 0 x 0. Then provided it is also true that f′(x 0)=0 f′(x 0)=0 the point x 0 x 0 is a relative maximum of f f. If, however, f′(x 0)f′(x 0) is not zero that point is neither a relative maximum or a relative minimum.
Proof: Under the assumptions that f′′(x 0)<0 f″(x 0)<0 and f′(x 0)=0 f′(x 0)=0 the theorem supplies an interval (u,v)(u,v) containing x 0 x 0 so that
f′(x)>0>f′(y)f o r a l l u<x<x 00>f′(y)f o r a l l u<x<x 0<y<v.
So f′f′ is positive on (u,x 0)(u,x 0) and is negative on (x 0,v)(x 0,v). A basic calculus theorem says such a function is increasing on u,x 0]u,x 0] and decreasing on [x 0,v).[x 0,v). So x 0 x 0 must be a strict local maximum.
Exercise 1. Prove the theorem.
Exercise 2. What would you say to a student who answers the first exercise this way: "Well if the derivative g′(x 0)g′(x 0) is negative then obviously the function must be decreasing!" [Don’t be gentle.]
Exercise 3. Give an example of a differentiable function ψ ψ such that ψ′(0)=1 ψ′(0)=1 but ψ ψ is not monotone in any interval (u,v)(u,v) that contains 0.0.
[Hint: See page 9 of
—————————-
Moral. The weaker calculus students tend to absorb the slogan “positive derivative implies increasing” uncritically. It is an important distinction to ask what a positive derivative at a single point signifies. And can you prove it?
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Related questions
When is second derivative positive or negative?
After you find the derivative of your curve, and find its value at a point of interest, the value may be positive, negative, or zero. What does it tell you about your curve if the derivative has a value of 0?
What does d e t∣∣(∂F∂x(x))∣∣d e t|(∂F∂x(x))| mean?
What does C[n 7]n≡n!(n−[n 5])![n 5]!C n[n 7]≡n!(n−[n 5])![n 5]! mean?
If the slope of a curve changes direction, then what will happen to its second derivative (if it exists)?
What happens to a curve when its second derivative changes sign?
What is the second derivative of 3x2?
Why is slope at instant (the derivative) called as instantaneous rate of change of "y" with respect to x, when the slope at an instant has nothing to do with 'y' values of the function?
If taking derivative means finding the slope of a curve, how would you explain taking a second derivative?
Why is it that when I plug an x value into the derivative, I get the slope of the x value on a curve? Does this have to do with the limit-based definition of derivative?
Can a function's derivative be zero if its original value was negative and its second one positive, or if its original value was positive and its second one negative?
Why does the value of the derivative change when the original function changes, even though they both represent the same slope at a specific point?
Is there a meaningful way of seeing the intuition of the power rule through graphical representation of the contributions of each derivative to the area under a curve of a function?
How do you find out if something has a positive, negative, or zero value for its second derivative function at a certain point?
Let f f be a twice differentiable function on [a,b][a,b], and f′(a)=f′(b)=0 f′(a)=f′(b)=0. How do I prove that there exist c∈(a,b)c∈(a,b) such that ∣∣f′′(c)∣∣≥4(b−a)2|f(b)−f(a)|?|f″(c)|≥4(b−a)2|f(b)−f(a)|?
Related questions
When is second derivative positive or negative?
After you find the derivative of your curve, and find its value at a point of interest, the value may be positive, negative, or zero. What does it tell you about your curve if the derivative has a value of 0?
What does d e t∣∣(∂F∂x(x))∣∣d e t|(∂F∂x(x))| mean?
What does C[n 7]n≡n!(n−[n 5])![n 5]!C n[n 7]≡n!(n−[n 5])![n 5]! mean?
If the slope of a curve changes direction, then what will happen to its second derivative (if it exists)?
What happens to a curve when its second derivative changes sign?
What is the second derivative of 3x2?
Why is slope at instant (the derivative) called as instantaneous rate of change of "y" with respect to x, when the slope at an instant has nothing to do with 'y' values of the function?
If taking derivative means finding the slope of a curve, how would you explain taking a second derivative?
Why is it that when I plug an x value into the derivative, I get the slope of the x value on a curve? Does this have to do with the limit-based definition of derivative?
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Pyramid of Biomass - Environment Notes
A graphical representation of biomass contained in a unit area at multiple trophic levels is called a biomass pyramid. It shows the link between biomass and trophic level, as well as the biomass available in each trophic level of an energy community at any given time. This article will explain to you about the Pyramid of Biomass which will be helpful in preparing the Environment syllabus for the UPSC Civil service exam.
| |
| Table of Contents 1. Pyramid of Biomass - Concept 2. Example of Biomass Pyramid 3. Pyramid of Biomass - Features 4. Types of Biomass Pyramid 4.1 Upward Pyramid 4.2 Inverted Pyramid 5. Limitations of Pyramid of Biomass 6. Conclusion 7. FAQs 8. MCQs |
Table of Contents
Pyramid of Biomass - Concept
| Other Relevant Links | |
--- |
| Pyramid of Numbers | Inverted pyramid of Biomass |
| Ecological Pyramids | Pyramid of Energy |
Example of Biomass Pyramid
Pyramid of Biomass - Features
| |
| Biomass Biomass refers to the total mass of all living or organic matter in an ecosystem at any particular time in ecological terms. Biomass can be divided into two categories: Species Biomass and Community Biomass. The entire mass of species in an ecosystem is referred to as species biomass. The entire mass of all the species that consider the defined community to be their habitat, on the other hand, is referred to as community biomass. When it comes to biomass measurement, species might range from humans to bacteria. |
Biomass
Types of Biomass Pyramid
Upward Pyramid
Inverted Pyramid
Limitations of Pyramid of Biomass
Conclusion
The pyramid of biomass was actually introduced to overcome the shortcomings of the pyramid of numbers as it does not take into account the biomass. Besides, the pyramid of biomass has more accuracy in comparison to the number pyramid. An upright biomass pyramid represents almost all of the world's ecosystems and biomes. However, the biomass pyramid can sometimes represent energy worth, but it doesn't always reveal how much energy is trapped at each energy level.
| Other Relevant Links | |
--- |
| Environment Notes | Functions of an Ecosystem |
| Ecology | Levels of Organisations in Ecology |
| Bio-Geo-Chemical Cycle | Ecological Succession |
| Pollutants And Trophic Level | Ecological Pyramids |
| Ecosystem | Environmental Ecology |
| Grassland Ecosystem | Indian Forest Types |
| Environmental Pollution | Mangroves |
FAQs
Q1: What is a Pyramid of Biomass?
Answer: A Pyramid of Biomass represents the total mass of living organisms present at each trophic level in an ecosystem. It is a graphical representation that shows the decrease in biomass as energy moves up through different trophic levels from producers to consumers.
Q2: How does the Pyramid of Biomass differ from the Pyramid of Numbers?
Answer: The Pyramid of Biomass represents the total mass of organisms at each trophic level, whereas the Pyramid of Numbers represents the number of organisms at each trophic level. While both indicate energy flow, the Pyramid of Biomass provides a more accurate reflection of the energy available at each level.
Q3: Why is the Pyramid of Biomass usually upright?
Answer: The Pyramid of Biomass is usually upright because the total biomass at the base (producers) is generally greater than at higher trophic levels (herbivores, carnivores). This is due to the loss of energy as it moves up the food chain.
Q4: Can the Pyramid of Biomass ever be inverted?
Answer: Yes, the Pyramid of Biomass can be inverted in certain ecosystems, such as aquatic environments, where the biomass of consumers (like fish) may exceed that of the producers (like phytoplankton) due to rapid turnover rates.
Q5: How does energy transfer in the Pyramid of Biomass?
Answer: Energy is transferred from one trophic level to the next in the Pyramid of Biomass, but there is a significant loss of energy (about 90%) at each level, primarily due to metabolic processes like respiration and heat loss. Only about 10% of the energy is passed on to the next level.
MCQs
a) Representation of energy flow in an ecosystem
b) Representation of total mass of organisms at each trophic level
c) Representation of the number of individuals at each trophic level
d) Representation of the size of organisms in an ecosystem
Answer: (B) See the Explanation
The Pyramid of Biomass represents the total mass of living organisms present at each trophic level in an ecosystem.
a) Grassland
b) Forest
c) Aquatic ecosystem
d) Desert
Answer: (C) See the Explanation
In aquatic ecosystems, the Pyramid of Biomass can be inverted because the biomass of consumers (like fish) may exceed that of the producers (like phytoplankton) due to their rapid reproduction and turnover rates.
a) Herbivores
b) Primary consumers
c) Producers
d) Carnivores
Answer: (C) See the Explanation
In a typical terrestrial ecosystem, producers (plants and other autotrophs) have the highest biomass as they form the base of the food chain.
a) Because energy is lost at each trophic level
b) Because producers consume more energy
c) Because the number of organisms increases
d) Because higher trophic levels have more organisms
Answer: (A) See the Explanation
The Pyramid of Biomass is narrower at the top because energy is lost due to respiration, metabolic processes, and heat as it moves up the food chain, leaving less energy for higher trophic levels.
a) It does not represent energy flow
b) It does not account for the number of organisms
c) It does not reflect the actual energy transfer
d) It can be inverted in some ecosystems
Answer: (D) See the Explanation
A limitation of the Pyramid of Biomass is that it can be inverted in certain ecosystems, like aquatic environments, where the biomass of consumers may be higher than that of producers.
GS Mains Questions and Model Answers
Q1: "The Pyramid of Biomass provides a clearer understanding of the energy flow in ecosystems compared to the Pyramid of Numbers." Critically analyze this statement.
Answer: The Pyramid of Biomass offers a clearer understanding of energy flow in ecosystems compared to the Pyramid of Numbers because it reflects the total mass of organisms at each trophic level, giving a more accurate representation of the energy available. While the Pyramid of Numbers only shows the number of organisms, it does not account for the size or mass of these organisms, which can vary significantly across trophic levels. For example, in a forest ecosystem, a few large trees (producers) may support a larger number of herbivores, but the biomass of the trees far exceeds that of the herbivores. The Pyramid of Biomass corrects this discrepancy by focusing on the mass and energy content, which provides a more accurate depiction of energy transfer through the ecosystem. However, in certain ecosystems like aquatic environments, the Pyramid of Biomass may be inverted, which limits its effectiveness in such cases.
Q2: Evaluate the importance of the Pyramid of Biomass in the study of ecosystem productivity and trophic dynamics.
Answer: The Pyramid of Biomass is crucial in studying ecosystem productivity and trophic dynamics as it helps illustrate how energy is distributed among different trophic levels. By representing the total biomass at each level, it provides insight into the efficiency of energy transfer within the food chain. Since energy is lost at each trophic level through metabolic processes, the Pyramid of Biomass typically shows a decrease in biomass from producers to higher trophic levels, which helps in understanding ecosystem productivity.
It also highlights the importance of primary producers in sustaining ecosystems, as they form the foundation of the energy pyramid. In ecosystems where the Pyramid of Biomass is inverted, such as aquatic systems, it draws attention to the rapid turnover of producers and the dynamic nature of energy flow. Overall, the Pyramid of Biomass is essential for understanding trophic interactions, ecosystem stability, and the flow of energy through different ecological levels.
Q3: Discuss how the Pyramid of Biomass can be used to assess the sustainability of an ecosystem. Provide examples to support your argument.
Answer: The Pyramid of Biomass can be an effective tool for assessing the sustainability of an ecosystem by illustrating how biomass and energy are distributed across different trophic levels. A healthy, sustainable ecosystem typically has a broad base of biomass at the producer level, which supports a smaller biomass of herbivores and even smaller biomass of carnivores. This structure ensures that there is enough energy at the lower levels to sustain higher trophic levels.
For example, in a terrestrial ecosystem like a grassland, the large biomass of grasses (producers) supports a smaller biomass of herbivores like deer, which in turn supports a smaller population of carnivores like lions. If this balance is disturbed, such as through overgrazing or habitat loss, the ecosystem may become unsustainable.
In contrast, in some aquatic ecosystems where the Pyramid of Biomass is inverted, the rapid turnover of primary producers like phytoplankton can still support a larger biomass of consumers. However, this may indicate that the system is more vulnerable to disturbances, such as overfishing or pollution, which can quickly disrupt the balance. The Pyramid of Biomass, therefore, helps in understanding how energy flow impacts ecosystem sustainability and resilience.
Previous Year Questions on Pyramid of Biomass
1. UPSC CSE 2020
Q1: Explain how the Pyramid of Biomass can be inverted in aquatic ecosystems.
Answer: In aquatic ecosystems, the Pyramid of Biomass can be inverted because the biomass of primary producers, such as phytoplankton, is relatively small compared to the biomass of primary consumers like zooplankton and fish. Phytoplankton have a rapid turnover rate and reproduce quickly, allowing them to sustain a higher biomass of consumers despite their own low mass. This results in an inverted pyramid, where the biomass of consumers exceeds that of producers.
2. UPSC CSE 2019
Q2: Discuss the significance of the Pyramid of Biomass in understanding energy flow in an ecosystem.
Answer: The Pyramid of Biomass is a critical tool for understanding energy flow in an ecosystem. It represents the total mass of living organisms at each trophic level, indicating how energy is distributed across the ecosystem. As energy moves up the pyramid from producers to herbivores and carnivores, there is a significant loss of energy, with only about 10% being transferred to the next trophic level. This helps explain why biomass decreases at higher trophic levels and highlights the inefficiency of energy transfer in ecosystems.
| Other Relevant Links | |
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Vulvar Cancer: 2021 Revised FIGO Staging System and the Role of Imaging
Mayur Virarkar
Mayur Virarkar
1 Department of Diagnostic Radiology, University of Florida College of Medicine, 655 West 8th Street, C90, 2nd Floor, Clinical Center, Jacksonville, FL 32209, USA; mayur.virarkar@jax.ufl.edu (M.V.); chandana.lall@jax.ufl.edu (C.L.)
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1, Sai Swarupa Vulasala
Sai Swarupa Vulasala
1 Department of Diagnostic Radiology, University of Florida College of Medicine, 655 West 8th Street, C90, 2nd Floor, Clinical Center, Jacksonville, FL 32209, USA; mayur.virarkar@jax.ufl.edu (M.V.); chandana.lall@jax.ufl.edu (C.L.)
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1,, Taher Daoud
Taher Daoud
2 Department of Diagnostic Radiology, The University of Texas MD Anderson Cancer Center, 1515 Holcombe Blvd., Houston, TX 77030, USA; tedaoud@mdanderson.org (T.D.); sanaz.javadi@mdanderson.org (S.J.); priya.bhosale@mdanderson.org (P.B.)
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2, Sanaz Javadi
Sanaz Javadi
2 Department of Diagnostic Radiology, The University of Texas MD Anderson Cancer Center, 1515 Holcombe Blvd., Houston, TX 77030, USA; tedaoud@mdanderson.org (T.D.); sanaz.javadi@mdanderson.org (S.J.); priya.bhosale@mdanderson.org (P.B.)
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2, Chandana Lall
Chandana Lall
1 Department of Diagnostic Radiology, University of Florida College of Medicine, 655 West 8th Street, C90, 2nd Floor, Clinical Center, Jacksonville, FL 32209, USA; mayur.virarkar@jax.ufl.edu (M.V.); chandana.lall@jax.ufl.edu (C.L.)
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1, Priya Bhosale
Priya Bhosale
2 Department of Diagnostic Radiology, The University of Texas MD Anderson Cancer Center, 1515 Holcombe Blvd., Houston, TX 77030, USA; tedaoud@mdanderson.org (T.D.); sanaz.javadi@mdanderson.org (S.J.); priya.bhosale@mdanderson.org (P.B.)
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1 Department of Diagnostic Radiology, University of Florida College of Medicine, 655 West 8th Street, C90, 2nd Floor, Clinical Center, Jacksonville, FL 32209, USA; mayur.virarkar@jax.ufl.edu (M.V.); chandana.lall@jax.ufl.edu (C.L.)
2 Department of Diagnostic Radiology, The University of Texas MD Anderson Cancer Center, 1515 Holcombe Blvd., Houston, TX 77030, USA; tedaoud@mdanderson.org (T.D.); sanaz.javadi@mdanderson.org (S.J.); priya.bhosale@mdanderson.org (P.B.)
Correspondence: vulasalaswarupa@gmail.com
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Gennaro Cormio: Academic Editor
Received 2022 Apr 13; Accepted 2022 Apr 29; Collection date 2022 May.
© 2022 by the authors.
Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (
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PMCID: PMC9102312 PMID: 35565394
Abstract
Simple Summary
Primary vulvar malignancy is a rare gynecological neoplasm constituting 5–8% of cases. Considering the prognostic capability of staging, FIGO has introduced the revised 2021 staging for vulvar cancer. We review the etiopathogenesis revised 2021 International Federation of Gynecology and Obstetrics (FIGO) classification and emphasize imaging in the staging of vulvar cancer.
Abstract
Vulvar cancer is a rare gynecological malignancy. It constitutes 5–8% of all gynecologic neoplasms, and squamous cell carcinoma is the most common variant. This article aims to review the etiopathogenesis revised 2021 International Federation of Gynecology and Obstetrics (FIGO) classification and emphasize imaging in the staging of vulvar cancer. The staging has been regulated by FIGO since 1969 and is subjected to multiple revisions. Previous 2009 FIGO classification is limited by the prognostic capability, which prompted the 2021 revisions and issue of a new FIGO classification. Although vulvar cancer can be visualized clinically, imaging plays a crucial role in the staging of the tumor, assessing the tumor extent, and planning the management. In addition, sentinel lymph node biopsy facilitates the histopathological staging of the draining lymph node, thus enabling early detection of tumor metastases and better survival rates.
Keywords: 2021 vulvar cancer staging, FIGO staging, imaging of vulvar cancer
1. Introduction
Primary vulvar malignancy is a rare gynecological neoplasm constituting 5–8% of cases . It is the fourth most common gynecological malignancy that usually affects post-menopausal women with a median age of 68 years [2,3]. The American Cancer Society estimates that around 6330 vulvar cancers will be diagnosed and approximately 1560 vulvar cancer deaths will occur in 2022. Squamous cell carcinoma (SCC) constitutes greater than 90% of vulvar cancer cases, followed by melanoma, adenocarcinoma, basal cell carcinoma, sarcoma, and undifferentiated type . Vulvar SCC (VSCC) is characterized by morphological variants, including keratinizing, basaloid, warty, and verrucous types. For the last ten years, there has been a 0.6% increase in the incidence of vulvar cancer per year and a 1.2% increase in mortality rates [4,5]. The increase in 5-yearly average incidence was higher in women < 60 years when compared to women of all ages (11.6% vs. 4.6%) . The age-standardized incidence is 1.4 per 100,000 women between the years 2003 and 2007.
Patients with vulvar cancer may be asymptomatic or present with pruritus, bloody discharge, palpable mass, or pain . Clinically, the lesion appears as thickened or discolored skin or a flat, raised, ulcerated, or plaque-like lesion . Around 59% of vulvar cancers demonstrate localized disease, whereas 30% and 6% metastasize to regional lymph nodes and distant sites, respectively [2,7]. The 5-year survival rate is 86%, 53%, and 19% for localized disease, regional spread, and distant spread, respectively .
2. Background
The vulva is a diamond-shaped structure comprising mons pubis, genitocrural folds, labia majora and minora, clitoris, vestibule, and Skene and Bartholin vestibular glands (Figure 1) . The skin of the labia is the most common vulvar area associated with cancer, while the clitoris and vestibular glands constitute the rare zones of the cancer [2,9]. Nearly 70% of vulvar cancers involve labia majora and minora, and 15–20% of cases involve the clitoris . It is critical to be familiar with the lymphatic drainage of the vulva for lymph node staging. The lymph nodes are drained primarily through three groups of superficial inguinal lymph nodes: (i) medial group: medial to the femoral and greater saphenous vein; (ii) intermediate group: near the femoral and saphenous vein; and (iii) lateral group: near lateral one-third of the groin . Then the lymphatics traverse the cribriform fascia to enter deep inguinal lymph nodes and later into paraaortic lymph nodes. Occasionally clitoris lymph drainage can bypass superficial inguinal lymph nodes and drain directly into deep inguinal lymph nodes or rarely into external iliac lymph nodes .
Figure 1.
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(A) Normal anatomy of vulvar region. (B) Normal description of location of lesions. (C) Regional and non-regional lymph nodes of the vulva. (D–F) Axial non-fat saturated T2-weighted images (WI) of a 36-year-old-female with normal anatomy. Image D demonstrates labia majora (long thick arrow) and labia minora (short, thin arrow). Image E shows the bulb of the vestibule (short, thin arrow), glans (asterisk), and crus (long thin arrow) of the clitoris. Image F demonstrates the ischiocavernosus muscle (dotted arrow).
The vulvar SCC can be divided into human papillomavirus (HPV)-dependent and HPV-independent subtypes based on etiopathogenesis. Around 43% of vulvar cancers are secondary to HPV . HPV 16 (77%) followed by HPV 33 (10.6%) and HPV 18 (3%) are the most common strains associated with VSCC . For either subtype of VSCC, vulvar intraepithelial neoplasia (VIN) is the precursor lesion defined by the histopathological atypia. VIN is classified into usual VIN (uVIN) (also termed as high-grade squamous intraepithelial lesion (HSIL)) and differentiated type VIN (dVIN) . The uVIN is associated with chronic HPV infection, whereas dVIN is HPV-independent and is associated with a chronic inflammatory condition, lichen sclerosis. Women with lichen sclerosis have a relative risk of 38.4 for the evolution of dVIN-associated VSCC . It is critical to distinguish between uVIN and dVIN as the latter is associated with a higher risk of malignant transformation (5.7% vs. 32.8%) . Genomic characterization of the lesion, as described in other sections, aids in differentiation among uVIN and dVIN. Furthermore, advancing age increases the risk of dVIN progression to the cancer . Usually, uVIN lesions transform into warty or basaloid variant SCC and affect younger women, constituting one-third of cases [13,14]. The dVIN lesions are the precursor types for keratinizing variant SCC and commonly affect older women.
3. Genomics of Vulvar Cancer
Although 80% of HSIL cases are associated with HPV DNA, only 1.5% and 28.6% of dVIN and VSCC cases are HPV-positive . HPV-positive vulvar cancer is characterized by the over-expression of E6 and E7 viral oncoproteins, which inactivate p53 and pRB (retinoblastoma transcriptional corepressor 1) protein. E6 oncoprotein degrades p53 by tagging it with E6-associated protein, a ubiquitin-protein ligase. E7 oncoprotein degrades pRB, releasing cellular E2F transcription factors responsible for cell-cycle activation. Consequently, cyclin-dependent kinase (CDK) inhibitors, including p16, p21, and p14, upregulate, leading to the downregulation of cyclinD1 and its complex formation with CDK4 . Of all the three CDK inhibitors, p16 is reliable in HSIL and can be used as a surrogate marker for HPV infection . It has a sensitivity and specificity of 100% and 98%, respectively, in the classification of VSCC into HPV-dependent and -independent types . The p16 is a tumor suppressor protein encoded by the CDKN2A gene. Normal cells arrest the cell cycle by complexing with CDK 4 and 6 and attenuating pRB phosphorylation, which is required for E2F release. HPV leads to over-expression of p16, resulting in continuous “block-positive” nuclear or nuclear and cytoplasmic staining extending from basal to superficial layer on the immunohistochemistry [6,15]. Lee et al. reported higher 5-year progression-free survival (65% vs. 16%) and overall survival (65% vs. 22%) in patients with p16-positive vulvar cancer when compared to the p16-negative disease .
In contrast to HPV-dependent vulvar cancers, somatic TP53 mutation is implicated in 30–80% of the dVIN . Nearly 58% of mutations are C: G to T: A transition types due to hydrolytic deamination of 5-methylcytosine residues . The dVIN shows solid nuclear staining on immunohistochemistry due to the accumulation of hyperstable and mutant p53 protein in the basal layer (7). The TP53 gene encodes p53 protein, a multifunctional transcription factor and a tumor suppressor. The p53 protein regulates cell-cycle progression, DNA integrity, and apoptosis and prevents genomic mutations. Hence, it is termed “the guardian of the genome.” Mutations in the TP53 gene lead to loss of tumor-suppressor activity of p53. The presence of TP53 mutation is associated with poor survival and short disease-free intervals compared to HPV-dependent SCC. Many chemotherapeutic agents act through the p53 pathway , and loss of the p53 tumor suppressor gene can result in chemoresistance and radio-resistance.
4. Role of Imaging in Vulvar Cancer
Although imaging is not an integral part of the vulvar cancer FIGO staging, it provides crucial information on the status of tumor extension to deeper tissues, lymph nodes, and distant organs. Due to its superior tissue resolution, MRI is the modality of choice to assess the vulvar anatomy. The normal vulva is hypo- to isointense on T1 weighted imaging (T1WI) and hyperintense on T2 weighted imaging (T2WI) sequences . The vestibular bulb and clitoral units are T2 hyperintense compared to muscle and exhibit avid enhancement on contrast administration. On imaging, vestibular bulbs appear as a U-shaped structure surrounding the lower vagina and urethra, whereas the clitoral appears as a curved structure. Compared to pelvic muscles, the bulbospongiosus and ischiocavernosus muscles are thin, and the hypointensity can be observed just underneath the labia majora and skin. The urethral complex has an outer hypointense and inner hyperintense layer, giving the target-like appearance to axial T2WI and contrast-enhanced T1WI sequences.
According to the European Society of Urogenital Radiology, a minimum field strength of 1.5 T is recommended in vulvar cancer evaluation. Based on MRI findings, around 83% of the vulvar lesions can be accurately classified . Patients are advised to fast for at least 4–6 h before the imaging, void the bladder, and are prescribed anti-peristaltic agents to limit the bowel movements. Vaginal distention with ultrasound gel aids in better identification of smaller vulvar tumors with or without vaginal infiltration. The patients should be supine and imaged with an eight-channel cardiac or a phased array pelvic coil and get the intravaginal gel. The MRI pelvic protocol and MRI-focused report for vulvar imaging are delineated in Table 1 and Table 2. They include axial T1WI, T2WI, fat-suppressed T1WI (FS-T1WI), and gadolinium-enhanced FS-T1WI. Axial diffusion-weighted imaging (DWI), with a similar angle to T2WI, adds value for a better tumor definition. Vulvar cancer is appreciated better on FS-T2WI than non-fat-suppressed sequences. The former can be explained by the high amount of fat in the perineal region which gets suppressed on fat-saturation images, improving the lesion conspicuity [23,24]. Axial T1WI and T2WI sequences with a large field of vision from aortic bifurcation to below the vulva determine the extent of the tumor alongside lymph node and pelvic bone metastases. Gadolinium-enhanced and axial 3D dynamic fat-suppressed T1WI sequences and a sagittal fat-suppressed T1 allow the assessment of tumor characterization and extent. The 3D dynamic images are acquired at a contrast resolution of 13–16 s for 3 to 5 min. Contrast-enhanced MRI assists in the visualization of smaller tumors and the involved adjacent organs such as the vagina, urethra, and anus. In a study by Kataoka et al., contrast administration improved tumor detection accuracy from 75% to 85% [22,23]. In addition to T1WI and T2WI sequences, axial FS-T2WI sequence and 3D dynamic gadolinium-enhanced axial T1WI sequence demonstrate and differentiate the recurrence of malignancy from radiation-induced fibrosis .
Table 1.
Protocol of pelvic MRI.
| Series Number | Series Description | FOV | Slice Thickness | Spacing | Frequency Encoding | Freq × Phase |
:---: :---: :---:
| 1 | Coronal T2 (to include the kidneys) | 420 | 5 | 0 | S/I | 288 × 192 |
| 2 | Sagittal T2 | 240 | 5 | 0 | A/P | 320 × 224 |
| 3 | rFOV Sagittal b = 50,600 | 240 | 5 | 0 | S/I | 96 × 80 |
| 4 | Axial T2 | 240 | 5 | 0 | L/R | 320 × 224 |
| 5 | AxialT1 | 240 | 5 | 0 | L/R | 320 × 224 |
| 6 | Axial DWI b = 50,400,800 | 380+ | 5 | 0 | L/R | 96 × 160 |
| 7 | Axial 3D Pre | 240 | 5 | −2.5 | L/R | 320 × 224 |
| 8 | Axial Dynamic 115 | 240 | 5 | −2.5 | S/I | 256 × 224 |
| 9 | +C Sagittal 3D Immediate delay | 240 | 5 | −2.5 | L/R | 256 × 224 |
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All patients receive vaginal gel. Ultrasound gel (60 cc; Aquasonic Clear Ultrasound Gel, Parker Laboratories) is used to distend the vagina. FOV: Field of Vision; DWI: Diffusion-Weighted Imaging.
Table 2.
MRI-focused report for locally advanced vulvar cancer staging.
| MRI Report Findings |
| Tumor dimension: maximum diameter |
| Tumor location: lateral/midline/multifocal |
| Clitoris involvement |
| Extension to adjacent organs/structures: urethra and/or vagina with caudo-cranial extension specification; lower one-third or upper two-third; urethral meatus; bladder; fourchette area, anus/rectum |
| Lymph node involvement: inguinofemoral and/or pelvic and/or abdominal |
| Additional findings: uterus; adnexa; kidneys; and pelvic bones |
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MRI: Magnetic Resonance Imaging.
5. International Federation of Gynecology and Obstetrics Staging of Vulvar Cancer
The International Federation of Gynecology and Obstetrics (FIGO) has determined the staging of vulvar cancer since 1969 . In 1988, FIGO classification surgical and pathological characteristics alone were considered in staging vulvar cancer. Later, in 2009, the revisions were made to include the tumor extension to adjacent structures and the number and extent of lymph node metastases. Considering the prognostic capability of staging, FIGO has introduced the revised 2021 staging for vulvar cancer (Table 3 and Figure 2). Imaging had no role in FIGO 2009 classification. However, the 2021 revision incorporated cross-sectional imaging findings into vulvar cancer staging, and this staging applies to all types of vulvar cancer except vulvar melanoma.
Table 3.
Revised 2021 FIGO staging of vulvar cancer.
| Stage | Characteristics |
:---: |
| I | Tumor confined to the vulva |
| | IA—Tumor size ≤ 2 cm and stromal invasion ≤ 1 mm |
| | IB—Tumor size > 2 cm or stromal invasion > 1 mm |
| II | Tumor of any size with extension to lower one-third of the urethra, lower one-third of the vagina, lower one-third of the anus with negative nodes |
| III | Tumor of any size with extension to the upper part of adjacent perineal structures or with any number of nonfixed, nonulcerated lymph nodes |
| | IIIA—Tumor of any size with disease extension to upper two-thirds of the urethra, upper two-thirds of the vagina, bladder mucosa, rectal mucosa, or regional lymph node metastases ≤ 5 mm |
| | IIIB—Regional lymph node metastases > 5 mm |
| | IIIC—Regional lymph node metastases with extracapsular spread |
| IV | Tumor of any size fixed to the bone or fixed, ulcerated lymph node metastases, or distant metastases |
| | IVA—Disease fixed to the pelvic bone or fixed or ulcerated regional lymph node metastases |
| | IVB—Distant metastases |
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Depth of invasion is measured from the basement membrane of the deepest, adjacent, dysplastic, tumor-free rete ridge (or nearest dysplastic rete peg) to the deepest point of invasion. Regional refers to the inguinal and femoral lymph nodes.
Figure 2.
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Revised 2021 FIGO staging of vulvar cancer. Stage I tumor confined to the vulva. (A) Stage IA tumor size less than equal to 2 cm and stromal invasion less than equal to 1 mm. (B) Stage IB tumor size more than 2 cm and stromal invasion more than 1 mm. (C) Stage II tumor of any size with extension to lower one-third of the urethra, lower one-third of the vagina, lower one-third of the anus with negative nodes. Stage III tumor of any size with extension to the upper part of adjacent perineal structures or with any number of nonfixed, nonulcerated lymph nodes. (D) Stage IIIA tumor of any size with disease extension to the upper two-thirds of the urethra, upper two-thirds of the vagina, bladder mucosa, rectal mucosa, or regional lymph node metastases less than equal to 5 mm. (E) Stage IIIB regional lymph node metastases more than 5 mm. (F) Stage IIIC regional lymph node metastases with extracapsular spread. Stage IV tumor of any size fixed to bone or fixed, ulcerated lymph node metastases, or distant metastases. (G) Stage IVA disease fixed to the pelvic bone or fixed or ulcerated regional lymph node metastases. (H) Stage IVB distant metastases.
5.1. Stage I
The stage I vulvar cancer classification remained unchanged from the 2009 FIGO staging except for the depth of invasion definition. In the 2009 classification, the depth of invasion is measured from the epithelial-stromal junction of the adjacent, most superficial dermal papilla to the deepest extent of tumor invasion. However, it is redefined in 2021 as a measurement from the basement membrane of the adjacent, deepest, dysplastic, tumor-free rete ridge to the deepest extent of the invasion . Stage I cancer signifies the tumor confined to the vulva. It is sub-classified into stage IA and IB disease based on tumor size and stromal invasion. In stage IA, the tumor measures ≤ 2 cm in size and invades stroma ≤ 1 mm from the basement membrane of the deepest tumor-free or dysplastic rete ridge. Stage IB is defined by the tumor > 2 cm size and >1 mm stromal invasion (Figure 3 and Figure 4). There is no adjacent structural, nodal, or distant organ involvement in stage I disease. The role of imaging in stage I disease is debatable as the tumor can be certainly evaluated by physical examination. Magnetic resonance imaging (MRI) is the imaging modality of choice in tumors > 2 cm, not confined to vulva and perineum, and >1 mm stromal invasion . There is limited literature on the efficacy of MRI in tumors without the former characteristics . Compared to muscles, the vulvar tumor appears as a solid hypo- to isointense on T1WI and intermediate- to hyperintense (“evil gray”) mass on T2WI sequences [23,26]. Hyperintensity on T2WI imaging sequences is due to internal necrosis in the case of larger tumors. Tumor demonstrates early arterial enhancement on dynamic contrast MRI and restricted diffusion on DWI sequences [23,26].
Figure 3.
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FIGO stage IA. A 43-year-old female with squamous cell carcinoma of the vulva. (A) Axial T2 weighted image and (B) post-contrast fat-saturated axial T1 weighted MRI image showing an enhancing 1.3 × 0.6 cm T2 intermediate signal biopsy-proven vulvar tumor (arrow).
Figure 4.
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FIGO stage IB. A 52-year-old female with squamous cell carcinoma of the vulva. (A) Axial T2 weighted image and (B) post-contrast fat-saturated axial T1 weighted MRI image showing an enhancing 3.4 × 2.3 cm T2 intermediate signal biopsy-proven vulvar tumor (arrow).
5.2. Stage II
Stage II represents any sized tumor extending to the lower one-third of the urethra, vagina, or anus without metastases to lymph nodes. There are no additional revisions in this stage from FIGO 2009 classification (Table 4 and Figure 5). In addition to the size of the tumor, MRI assesses the extension of locally advanced vulvar cancer to superficial (labia majora and minora and clitoris) and deep structures (urethra, vagina, and anus). Clinical examination alone is challenging in patients with urethral involvement and absent clinical findings. In such cases, disruption of the target appearance of the urethra on T2WI and contrast-enhanced T1WI is suggestive of a urethral infiltration .
Table 4.
International Federation of Gynecology and Obstetrics (FIGO) stage II definitions.
| FIGO Staging System | II |
:---: |
| 2009 | Tumor of any size with extension to the adjacent perineal structures (lower one-third of the urethra, lower one-third of vagina, anus) with no lymph node involvement |
| 2021 | Tumor of any size with extension to lower one-third of the urethra, lower one-third of the vagina, lower one-third of the anus with no lymph node involvement |
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Figure 5.
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FIGO stage II. A 57-year-old female with squamous cell carcinoma of the vulva. (A) Axial T2 weighted image and (B) coronal T2 weighted MRI image show a 3.4 × 1.4 cm vulvar tumor (asterisk) involving the right aspect of the lower one-third of the vagina (long thin arrow) and right ischio-anal fossa (short thick arrow). Anteriorly, the mass abuts the lower third of the urethra. No lymphadenopathy was noted. Note that the gel in the vagina helps better delineate the tumor.
In contrast, the absence of disruption on MRI reduces the likelihood of urethral infiltration. In patients with vaginal wall involvement, the disruption of low signal intensity of the vaginal wall on T2WI and contrast-enhanced T1WI may be observed . Similarly, the interruption of the low-signal intensity of the anal sphincter on T2WI and the muscles on post-contrast T1WI indicates anal wall involvement . In case of uncertainty, endo-anal ultrasound can be performed to delineate muscle layers better. DWI aids in assessing the tumor extension by demonstrating the continuation of restricted diffusion into the urethral wall, vaginal wall, or anal sphincter .
5.3. Stage III
Stage III of FIGO 2021 classification is divided into three substages A, B, and C (Figure 6). It includes vulvar tumors of any size extending to the upper part of adjacent structures or any number of nonulcerated and nonfixed lymph nodes. This contrasts with the FIGO 2009 classification, where tumor extension to upper parts of adjacent structures is now considered stage IVA (Table 5) [13,24]. Stage IIIA contains tumors of any size with extension to the upper two-thirds of the urethra, bladder mucosa, upper two-thirds of the vagina, rectal mucosa, or regional inguinofemoral lymph node metastases ≤ 5 mm. Stage IIIB is defined as tumor metastasis > 5 mm to regional lymph nodes. Stage IIIC contains tumors with regional lymph node metastases with extracapsular spread. The extracapsular spread is associated with a worse 5-year survival rate than patients with intranodal metastases alone (34% vs. 66%) .
Figure 6.
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FIGO stage III. A 49-year-old female with squamous cell carcinoma of the vulva. Axial T2 weighted MRI image shows infiltrative vulvar tumor (asterisk) involving, anteriorly, the lower one-third of the urethra (long thin arrow) and, posteriorly, the lower one-third of the anus (long thick arrow). Moreover, there is a metastatic right inguinal lymph node (short thick arrow). The findings correspond to FIGO stage IIIB.
Table 5.
International Federation of Gynecology and Obstetrics (FIGO) stage III definitions.
| FIGO Staging System | III | IIIA | IIIB | IIIC |
:---: :---:
| 2009 | Tumor of any size, with/without extension to adjacent perineal structures (lower third of urethra, the lower third of vagina, anus) with positive inguinofemoral lymph nodes | With one lymph node metastases (≥5 mm) or with 1–2 lymph node metastases (<5 mm) | With two or more lymph node metastases (≥5 mm) or with three or more lymph node metastases (<5 mm) | With positive nodes with extracapsular spread |
| 2021 | Tumor of any size with extension to upper parts of adjacent perineal structures or with any number of nonfixed and nonulcerated lymph nodes | Tumor of any size with disease extension to upper two-thirds of the urethra, upper two-thirds of the vagina, bladder mucosa, or regional lymph node metastases ≤ 5 mm | Regional lymph node (inguinofemoral) metastases >5 mm | Regional lymph node (inguinofemoral) metastases with extracapsular spread |
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Rectal mucosa, bladder mucosa, vagina, and urethral infiltration can be identified by visualizing the intermediate signal intensity of the vulvar tumor on T2WI. Multiplanar imaging is recommended for the accurate assessment of adjacent organ involvement. For instance, the sagittal plane is more beneficial in evaluating the caudo-cranial extension of the tumor through the fornices to the upper portion of the vagina, the anal sphincter to the rectum, and the urethra to the base of the bladder. Axial and coronal planes assess the lateral tumoral extension to the pelvic bones and muscles.
Lymph node metastases are the most important prognostic factor, if present, resulting in the reduction in the 5-year survival rate from 95% to 62% . The 5-year survival rate drops from 80% in node-negative patients to 50% in node-positive patients and 12% in patients with >4 positive lymph nodes . Around 88% of node-negative patients are free from disease in two years, whereas 60%, 43%, and 29% of patients with one, two, or ≥3 lymph nodes, respectively, are free from the disease within two years . Inguinofemoral nodes are considered regional lymph nodes and the superficial medial group of inguinal lymph nodes in the sentinel nodes in a large percentage of vulvar carcinoma . Lateral vulvar lesions usually drain to ipsilateral lymph nodes, whereas the lesions within 1 cm of the midline may drain bilaterally . Moreover, the tumor involving adjacent structures such as the vagina, urethra, or anus (above the dentate line) may spread directly to the pelvic lymph nodes.
The relevance of ultrasound in lymph node evaluation has been reported in several studies. It has a sensitivity and specificity ranging from 76 to 100% and 69 to 91%, respectively. This wide range can be explained by the several imaging features of lymph nodes such as short-axis diameter, long-axis to short-axis ratio, shape, fatty hilum, and vascularization. The increased peripheral vascularity and a spectral waveform with high resistance suggest metastatic lymph nodes. Ultrasound combined with fine-needle aspiration cytology can avoid unnecessary lymph node dissection. Hall et al. reported that the combination has 93% and 100% sensitivity and specificity, respectively . However, the limitations of ultrasound include operator dependency and sampling error in patients with nodal micro-metastases
Lymph node involvement can be suspected on MRI based on several criteria, including short-axis diameter > 1 cm, rounded morphology with long-axis to short-axis ratio < 1.3:1, cystic nodal change, loss of fatty hilum, and irregular contour [23,26]. Evaluation based on short-axis diameter > 10 mm has 89% and 91% sensitivity and specificity, respectively . The long-to-short-axis ratio is most accurate (85%), the presence of necrosis is most specific (93%), and readers’ confidence in lymph node metastases is most sensitive (87%) to estimate tumor staging . Ultrasmall superparamagnetic iron oxide (USPIO) contrast agents are the nanoparticles used to detect lymph node involvement on MRI. USPIO particles are incorporated into reticuloendothelial cells such as macrophages. On MRI, the accumulation of USPIO particles in the normal lymph nodes results in low signal intensity on T1WI and T2WI sequences. However, when the nodal reticuloendothelial cells are replaced by tumor cells, there is no uptake of USPIO particles and no signal change. In a recent meta-analysis of various tumors, including vulvar cancer, the incorporation of nanoparticles demonstrated a sensitivity and specificity of 88% and 96%, respectively .
CT is a commonly used preoperative assessment tool to evaluate nodal and extra-nodal spread of the tumor. CT aids in assessing the proximity of metastatic inguinal lymph nodes to the blood vessels, evaluating the pelvic lymph nodes, or planning postoperative management in patients with surgically proven inguinal metastases (8). The GROINSS-VII study recommended CT as a preferred imaging modality to exclude lymph nodal metastases before SLN biopsy. This has been questioned by Pounds et al., who reported the sensitivity and specificity of CT as 59% and 78%, respectively, in the evaluation of inguinal lymph nodes. In their study, over one-third of the patients with negative nodes on imaging had inguinal metastases (35).
5.4. Stage IV
Stage IV comprises tumors of any size fixed to the bone or with the presence of fixed ulcerated lymph node metastases or distant structural metastases (Table 6). Stage IV is subdivided into IVA and IVB. Stage IVA describes the characteristics of metastases manifesting as fixed or ulcerated lymph nodes or tumors fixed to the pelvic bone. Stage IVB signifies distant metastases (Figure 7). Pelvic lymph node involvement is considered distant metastases and is also categorized under FIGO stage IV [26,32]. The Cloquet node is an essential indicator of pelvic nodal spread as it is the lowest external iliac node located at the entrance of the femoral canal. Distant metastasis is rare and often preceded by local tumor recurrences. Lungs, liver, bone, lymph nodes, and skin are most involved. The prognosis is poor in patients with distant metastases, with a 2-year survival rate of 11.3% and a median survival rate from a diagnosis of 5.6 months .
Table 6.
International Federation of Gynecology and Obstetrics (FIGO) stage IV definitions.
| FIGO Staging System | IV | IVA | IVB |
:---: :---: |
| 2009 | Tumor invades adjacent structures or fixed and ulcerated lymph nodes or distant metastases | Upper urethral and/or vaginal mucosa, bladder mucosa, rectal mucosa, or is fixed to the pelvic bone Fixed or ulcerated inguinofemoral lymph nodes | Any distant metastases, including pelvic lymph nodes |
| 2021 | Tumor of any size fixed to the bone, or fixed and ulcerated lymph node metastases or distant metastases | Disease fixed to the pelvic bone or fixed or ulcerated regional (inguinofemoral) lymph node metastases | Distant metastases |
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Figure 7.
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FIGO stage IV. A 56-year-old female with squamous cell carcinoma of the vulva. (A) post-contrast fat-saturated axial T1 weighted MRI image shows an enhancing 5.2 × 2.8 cm vulvar tumor (asterisk). (B) Coronal lung window contrast-enhanced CT image shows pulmonary nodule (arrow). A biopsy of the nodule was positive for metastasis. The findings correspond to FIGO stage IVB.
According to National Comprehensive Cancer Network guidelines, the PET/CT with 18-fluorodeoxyglucose is recommended for patients with suspected metastases [33,34]. It is a non-invasive tool for pre-operative assessment and staging of vulvar cancer. The 18 F-FDG PET/CT detects primary vulvar lesions with 100% accuracy. However, its sensitivity (50–100%) and specificity (67–100%) ranges are vast, making its application less feasible in nodal staging. The overall sensitivity could be attributed to its limited ability to identify lymph node metastases < 5 mm [7,33]. PET/CT scans exhibit false-positive findings in the presence of inflammation and false-negative results in necrotic tissue. Triumbari et al. reported that PET/CT aids in excluding groin metastases in vulvar cancer patients with a sensitivity and specificity of 70% and 90%, respectively . FDG PET/CT can be supportive, in cases of suspected pelvic nodal or distant metastases, given its 100% negative predictive value . The mean SUV max for metastatic lymph nodes ranges from 6.1 to 11.0 . PET/CT also assists in evaluating the treatment response and detecting the disease recurrence and hence is considered superior to conventional imaging in this perspective. Albano et al. reported the sensitivity, specificity, and accuracy of PET/CT in detecting recurrent disease as 100%, 92%, and 98%, respectively . Early detection of recurrence aids to stratify the patients better and begin prompt therapies to improve outcomes and survival.
6. Treatment
As the tumor margin status has been accepted as a significant prognostic factor for tumor recurrence, radical local excision with at least a 1–2 cm margin is recommended in early-stage vulvar cancer (Figure 8, Figure 9 and Figure 10) . Margins can be reduced if the tumor is close to midline structures such as the clitoris, anus, or urethra to preserve their function. In case of positive margins (<8 mm from the tumor), re-excision, or, if unresectable or involving the urethra, anus, or vagina, adjuvant external beam radiotherapy is recommended . The total radiation dose shall be between 60 Gy and 70 Gy . Inguinofemoral lymphadenectomy (IFLD) is not recommended due to the low risk (<1%) of lymph node metastases in stage IA disease. Patients with stages IB or greater possess a 20–30% risk of lymphatic metastases; hence, unilateral lymphadenectomy or sentinel lymph node (SLN) biopsy should be performed if the tumor is <4 cm and located ≥ 2 cm from the midline . Compared to IFLD, SLN has a sensitivity of 92% and NPV of 97–98% in identifying metastases . For a tumor within 2 cm of the vulvar midline, bilateral IFLD or SLN biopsy is recommended [4,5,41]. Contralateral lymphadenectomy is advised for patients with positive lymph node metastases or undetectable SLN [4,5]. Positive SLN with <2 mm metastases can be treated with radiotherapy, sparing from groin lymph node dissection . The GROINSS-V trial reported that SLN metastases ≤ 2 mm have better disease-specific survival rates than SLN metastases > 2 mm (94% vs. 70%; p = 0.001) .
Figure 8.
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Primary treatment of vulvar cancer.
Figure 9.
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Post-surgical surveillance of vulvar cancer.
Figure 10.
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Lymph node evaluation and treatment in patients with vulvar cancer.
SLN mapping was introduced in 1977 by Cabanas and in 1994 for the first time in the vulvar cancer [23,43]. Previously, radical surgery with IFLD was the standard treatment for early-stage vulvar cancer. Although radical surgery improved the 5-year survival rates, the morbidity was significant due to the complications such as infections, pain, lymphocele, and chronic lymphedema of the lower extremities. These complications often prolong the duration of hospitalization and delay the adjuvant therapy introduction when needed. Lymphedema is reported in 30–70% of patients undergoing complete IFLD (48).
Moreover, only 10–26% of the dissected lymph nodes demonstrate metastases, while approximately 80% of patients were reported to be free from lymph node spread . Hence novel diagnostic tests with high sensitivity are required to exclude patients without lymph node metastases. Clinical examination alone is not reliable because 16–24% of patients have normal clinical findings, and 24–41% of clinically enlarged lymph nodes may have normal results in the histological examination . Imaging such as ultrasound, CT, MRI, and PET are reported to have a wide sensitivity range (45–86%), and negative likelihood ratio range (0.12–0.6) and hence are less reliable. The SLN biopsy reduced the postoperative morbidity due to lymphadenectomy without jeopardizing the LN metastases detection.
All the patients with stage III or IVA shall receive radiotherapy with concurrent chemotherapy (Figure 11). The same therapy covering the primary lesion is also recommended in patients with negative IFLD. Patients with residual tumors may be considered for resection, or if unresectable, external beam radiotherapy is recommended . Palliative care and better quality of life are the primary treatment goals in patients with distant metastases. Chemoradiotherapy can be contemplated at the primary tumor site for symptom relief. Incorporating systemic chemotherapy is still controversial in the case of metastatic vulvar cancer, and further studies are necessary to make any recommendations . However, platinum-based chemotherapy regimens alone or combined with another agent such as paclitaxel, vinorelbine, 5-Fluorouracil, or mitomycin C are the most frequently used . Gefitinib and trastuzumab combination regimen has shown efficacy by increasing the radiosensitivity of the vulvar lesion [45,46].
Figure 11.
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Treatment response assessment. A 55-year-old female with recurrent vulvar squamous cell carcinoma. (A) Axial and (B) sagittal T2 weighted MRI images show a large vulvar mass (asterisk) invading both the low rectum (thick arrow) and diffusely throughout the left semi circumferential anal canal (not shown), with additional invasion along the left posterolateral aspect of the upper two-thirds of the vagina (thin arrow). No evidence of enlarged lymph nodes. The findings correspond to FIGO stage IIIA. A Follow-up MRI after three months of chemoradiation was obtained. (C) Axial and (D) sagittal T2 weighted image MRI show an interval decrease in the size of the treated left vulvar mass invading the lower rectum, anal canal, left upper one-third of the vagina, and left lateral pelvic sidewall. (b, bladder; v, vagina; r, rectum).
7. Surveillance of Vulvar Cancer
Due to a recurrence rate between 12 to 40%, routine surveillance is recommended in all patients of VSCC . The GROningen International Study on SLNs in vulvar cancer (GROINSS-V) study, involving 403 patients, reported that the groin recurrence rate was 2.3% over 3.5 months in patients with negative SLN . Recurrence of vulvar cancer is common in the vulvar and perineal region and usually occurs within two years of clinical presentation and may benefit from tumor restaging with computerized tomography (CT) of the chest, abdomen, and pelvis and shall be managed through radical excision and adjuvant radiotherapy . Among women with positive nodes, the recurrence rate is higher within the first two years when compared to node-negative patients (33% vs. 5.1%) . After two years, the recurrence rates are similar (12%) irrespective of the node status . Patient surveillance should include history and careful clinical examination for vulvar lesions and cervical, perianal, and vaginal neoplasms, all associated with HPV infection . It is essential to emphasize the symptoms of tumor recurrence, including new lesion/mass, vulvar itching, leg or groin pain, lower extremity lymphedema, urinary symptoms, weight loss, and cough. Laboratory studies, imaging, and biopsies are recommended for all suspicious lesions. Surveillance of cancer patients often involves interdisciplinary coordination among specialists and multiple health care providers.
8. Conclusions
The revised FIGO classification for vulvar cancer has overcome the limitation of prognostic capability and guides the clinicians in the appropriate management of patients. Incorporating imaging into the classification is a paradigm shift from the older classification. MRI is superior to conventional imaging due to its excellent soft-tissue resolution and aids local vulvar cancer staging by assessing the involvement of adjacent tissues. CT and PET/CT imaging aid in determining the extent of the distant metastases. PET/CT is better in detecting lymph node metastases. As the staging and treatment of vulvar cancer evolve, radiologists must be familiar with the recent FIGO staging of vulvar cancer and help guide management.
Acknowledgments
We thank Kelly Kage for the beautiful illustrations.
Author Contributions
Conceptualization, M.V., P.B. and C.L.; validation, P.B. and C.L.; writing—original draft preparation, M.V. and S.S.V.; writing—review and editing, T.D., S.J., P.B., C.L. and M.V.; supervision, P.B. and C.L.; project administration, M.V. and S.S.V. All authors have read and agreed to the published version of the manuscript.
Conflicts of Interest
The authors declare no conflict of interest.
Funding Statement
This research received no external funding.
Footnotes
Publisher’s Note: MDPI stays neutral with regard to jurisdictional claims in published maps and institutional affiliations.
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1. Introduction
2. Background
3. Genomics of Vulvar Cancer
4. Role of Imaging in Vulvar Cancer
5. International Federation of Gynecology and Obstetrics Staging of Vulvar Cancer
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1970 AHSME Problems/Problem 32
Problem
and travel around a circular track at uniform speeds in opposite directions, starting from diametrically opposite points. If they start at the same time, meet first after has travelled yards, and meet a second time yards before completes one lap, then the circumference of the track in yards is
Solution
Let be half the circumference of the track. They first meet after has run yards, meaning that in the time has run yards, has run yards. The second time they meet is when is 60 yards before he completes the lap. This means that in the time that has run yards, has run yards. Because they run at uniform speeds, we can write the equation Cross multiplying, Adding to both sides and simplifying, we have Because is only half of the circumference of the track, the answer we are looking for is .
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1970 AHSC (Problems • Answer Key • Resources)
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Problem 31Followed by
Problem 33
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How to Determine If an Ordered Pair is a Solution to a System of Equations
GoTutor Math
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Description
23394 views
Posted: 22 Sep 2023
This algebra math tutorial explains how to determine whether an ordered pair is a solution to a system of linear equations. To determine if an ordered pair is a solution to a system of two equations, we substitute the values of the variables into each equation. If the ordered pair makes both equations true, it is a solution to the system. If the order pair does not satisfy at least one of the equations, it is not a solution to the system.
ordered pair solutions of equations, ordered pairs, ordered pair, system of equation, ged math, gcse math, sat math,
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Transcript:
In this lesson, we will discuss how to determine
whether an ordered pair is a solution to a system of equations. To determine if an ordered pair
is a solution to a system of two equations, we substitute the values of the variables into each
equation. If the ordered pair makes both equations true, it is a solution to the system. Let's take a
look at each exercise one by one. Is the ordered pair (2 , 0) a solution to this system of
equation? Let's find out. The first number of the ordered pair is the X value and the second number
is the Y value. Our first step is to substitute X with 2 and Y with 0 into both equations.
Let's start with the first equation. Replace X with 2 and Y with 0. Now, 2 times 2 is equal to 4. 4
minus 0 is 4, which indeed equals 4. So the order pair (2 , 0) satisfies the first equation. However,
for the ordered pair to to be a solution to the system, it must also work for the second equation.
As we did before, substitute X with 2 and Y with 0. 3 times 2 is 6, and 4 times 0 is 0. 6 plus 0 is 6, and 6 is indeed equal to 6. So it also satisfies the second equation. Therefore, since the ordered pair (2 , 0) makes both equations true it is a solution to this system of equations. Now let's move on to
the second example. Does the ordered pair (-8 , -4) satisfy this system of equations? -8 is the X
value and -4 is the Y value. Just like before, our initial step is to substitute these values into
both equations. Starting with the first equation, plug in -8 for X and -4 for Y. -5 times -4 equals 20. -8
plus 20 is 12, and indeed 12 is equal to 12. So the ordered pair satisfies the first equation. Let's
see now whether it works for the second equation. Replace X with -8 and Y with -4. -3 times 8 equals 24 and 24 plus 7 is 31. But -4 is not equal to 31 so it does not satisfy the second equation. Remember that for an
order pair to be a solution to the system it must satisfy both equations. Since the ordered pair (-8 , -4) does not satisfy the second equation, it is not a solution to this system of equations. Now
it's your turn to give it a try. Is the ordered pair (3 , 1) a solution to this system of
equations? Feel free to pause the video and work on it. The X value is 3 and the Y value is
1. Plug in 3 for X and 1 for Y. 7 times 3 equals 21, and 4 times 1 equals 4. 21 minus 4 equals 17. However, 17 is not equal to -1 so the ordered pair does not satisfy the first equation. At this point there is no need to
check for the second equation since it already fails to satisfy the first equation. Therefore, the
ordered pair (3 , 1) is not a solution to this system of equations. Thanks for watching please
consider to like this video and subscribing to the channel.
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The Q-value of the reactions in proton-proton cycle and the overall Q-value. | bartleby
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SciencePhysicsModern Physics
The Q-value of the reactions in proton-proton cycle and the overall Q-value.
The Q-value of the reactions in proton-proton cycle and the overall Q-value.
BUY
Modern Physics
3rd Edition
ISBN: 9781111794378
Author: Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Publisher: Cengage Learning
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Chapter 14 : Nuclear Physics Applications
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Section: Chapter Questions
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Problem 36P
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Nuclear Reactions Nuclear reactions are processes in which the fundamental characteristics of a nucleus changes. They change into a completely different nucleus. These processes occur naturally in an unstable nucleus. Bombarding an energetic particle can induce nuclear re…
Question
Chapter 14, Problem 36P
To determine
The Q-value of the reactions in proton-proton cycle and the overall Q-value.
Expert Solution & Answer
Answer to Problem 36P
The Q-value of the reactions in proton-proton cycle is 0.931 MeV,5.493 MeV,19.285 MeV and 12.861 MeV and the overall Q-value is 12.861 MeV.
Explanation of Solution
Write the first reaction in proton-proton cycle
H 1+H 1→H 2+e++ν(I)
Here, H 1 is the hydrogen, H 2 is the deuterium, e+ is the electron and ν is the neutrino.
Write the expression for Q-value of this reaction
Q 1=[2 m(1 H)−m(2 H)−m e+]c 2(II)
Here, Q 1 is the Q-value of the first reaction in the proton-proton cycle, m(1 H) the mass of the hydrogen nucleus, m(2 H) the mass of the deuterium, m e+ the mass of the positron and c is the speed of light.
Write the second reaction in proton-proton cycle
H 1+H 2→H 3 e+γ(III)
Here, H 3 e is the Helium-3 nuclei and γ is the gamma ray photon.
Write the expression for Q-value of this reaction
Q 2=[m(1 H)+m(2 H)−m(3 He)]c 2(IV)
Here, Q 2 is the Q-value of the second reaction in the proton-proton cycle and m(3 He) the mass of the tritium.
Write the third reaction in proton-proton cycle
H 1+H 3 e→H 4 e+e++ν(V)
Here, H 4 e is the helium-4 nucleus.
Write the expression for Q-value of this reaction
Q 3=[m(1 H)+m(3 He)−m(4 He)−m e+]c 2(VI)
Here, Q 3 is the Q-value of the third reaction in the proton-proton cycle and m(4 He) the mass of the helium-4 nuclei.
Write the fourth reaction in proton-proton cycle
H 3 e+H 3 e→H 4 e+H 1+H 1(VII)
Here, H 4 e is the helium-4 nucleus.
Write the expression for Q-value of this reaction
Q 4=[2 m(3 He)−m(4 He)−2 m 1 H]c 2(VIII)
Here, Q 4 is the Q-value of the fourth reaction in the proton-proton cycle.
Write the fourth reaction in proton-proton cycle
4(H 1)→H 4 e+2e++2 ν(IX)
Write the expression for Q-value of this reaction
Q=[4 m 1 H−m(4 He)−2 m e+]c 2(X)
Here, Q is the Q-value of the overall reaction of the proton-proton cycle.
Conclusion:
Substitute 1.007825 u for m(1 H), 2.014102 u for m(2 H), 0.000549 u for m e+ and 931.5 Mev/u for c 2 (II) to find Q 1
Q 1=2(1.007825 u)−2.014102 u−0.000549 u=0.931 MeV
Substitute 1.007825 u for m(1 H), 2.014102 u for m(2 H), 3.016029 u for m(3 He) and 931.5 Mev/u for c 2 (IV) to find Q 2
Q 2=(1.007825 u)−2.014102 u−3.016029 u=5.493 MeV
Substitute 1.007825 u for m(1 H), 3.016029 u for m(3 He), 4.002603 u for m(4 He)0.000549 u for m e− and 931.5 Mev/u for c 2 (VI) to find Q 3
Q 3=(1.007825 u)+(3.016029 u)−(4.002603 u)−(0.000549 u)=19.285 MeV
Substitute 1.007825 u for m(1 H), 3.016029 u for m(3 He), 4.002603 u for m(4 He) and 931.5 Mev/u for c 2 (VIII) to find Q 4
Q 4=2(3.016029 u)−(4.002603 u)−2(1.007825 u)=12.861 MeV
Substitute 1.007825 u for m(1 H), 4.002603 u for m(4 He), 0.000549 u for m e− and 931.5 Mev/u for c 2 (X) to find Q
Q=4(1.007825 u)−(4.002603 u)−2(0.000549 u)=25.709 MeV
Therefore, the Q-value of the reactions in proton-proton cycle is 0.931 MeV,5.493 MeV,19.285 MeV and 12.861 MeV and the overall Q-value is 12.861 MeV.
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Resonances, waves and fields
Their applications, physics, and math. -- Peter Ceperley
For a list of all topics discussed, scroll down to the very bottom of the blog, or click here.
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Strong damping - critical damping
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Simple Resonators - Shock excitation
2.3 Strong damping - critical damping
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For some applications, oscillations in resonant structures are not desirable. For example, shock absorbers in automobiles are designed to dampen out oscillations of the suspension, while still allowing the suspension to be responsive to bumps in the road. Damping is also required for engine and motor mounts, body panels and other parts of cars and aircraft, and read/write arms in computer hard drives. These parts all have the ingredients of a mass-and-spring resonator: a mass, a spring, and damping. Therefore the equations of the previous section are appropriate, if converted to general coefficients, such as were used in Table 6 on a previous posting. Table 7 below uses general coefficients and derives the frequency and decay constant for three damping strengths: underdamping, critical damping, and overdamping. Note that these terms are usually used in connection with applications designed to prevent oscillations and not with normal resonators in which oscillations are desirable as discussed in previous sections.
Note also that changing the damping in a mechanical resonator such as that in the animation involves changing something mechanical. For example, in the animation, adding damping might involve putting cotton filling or porous sponge material inside the rubber tire that supports the platform. The air flow through this material would absorb energy as the tire is distorted by the weight of the girl. A more effective method would be to use a separate air chamber, such as that inside the middle of an inflated inner tube and provide a hole in it to let air to pass from the outside atmosphere to inside this chamber and back the other way as the girl compresses the tire. By changing the hole size and thus the resistance to air flow, one could change the damping.
The animation above illustrates the effect of different values of the damping coefficient on the resulting "oscillations". Mouse over it (or click on it) to set it into motion. Select the type of damping you wish to observe. A little experimentation with the animation will convince you that critical damping leads to the fastest reaching of stationary equilibrium. Overdamping has a nice exponential decay with no oscillations, but it causes a slower decay than with critical damping. Underdamping causes over shooting, oscillations, and a slower reaching of stationary equilibrium. Of course, the exact values of the damping coefficients affect the speed at which equilibrium is reached, but coefficients adjusted for critical damping result in the fastest equilibrium.
Two more notes before we start the math:
Real shock absorbers as well as some other dampers in modern technology use more sophisticated methods than addressed in this posting, such as damping only in one direction of motion, or computer controlled damping.
Most of these posts in this series are directed towards applications that require oscillations to work and would be considered to be extremely "under damped" in the language of this particular posting. This particular post is odd in that it is directed towards applications requiring abatement of oscillations. We include it for completeness.
| Table 7. Equations for a heavily damped resonator | |
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| general second order differential equation (7a) | We repeat the general second order differential equation, last shown in Equation (6a) (of section 2.1) with general coefficients A, B, and C. In this case C is not the capacitance, but instead just the last coefficient in this general equation. |
| general solution to s (7b) | Here we use the complex equation, Equation (6h), from the last posting that gives exponential coefficient, s. We have substituted R = 2L/τ (5a) and LC = 1/ω02 (5c) to converted it to the general coefficients as used in Equation (7a). |
| breaking up s (7c) | Here we break up s into real and imaginary parts. a is the real part of s, i.e. Re(s), and b is the imaginary part of s, i.e. Im(s). |
| breaking up s (7d) | When we insert s in the form of (7c) into a typical damped oscillating parameter, such as y(t) = Aest, similar to the q(t) in Equation (6c), we get (7d). Looking at the extreme right of (7d), we see that eat is an exponential growth factor (for positive a's) or exponential decay factor (for negative a's). The eibt factor has a complex argument meaning that its real equivalent is a sinusoid. The A in this equation and also in the equations that follow is not the constant in Equation (7a) above. Instead the A as used here (and below) is the initial amplitude of oscillations. Also, as per the standard convention with complex phasors, we take only the real part of Equation (7d) for Equation (7e). |
| real part of (7d) (7e) | This is the real part of (7d). Comparing (7e) with (6b) of the previous posting, we see a is the damping coefficient (i.e. −1/τdecay), and b is the angular frequency (i.e. ωdecay). That is to say Re(s) = a = -1/τdecay and Im(s) = b = ωdecay . |
| τdecay = −1/a = −1/Re(s) (7f) ωdecay = b = Im(s) (7g) y(t) = Aet/τdecaycos ωdecayt (7h) | Here we summarize the above. |
| When we have strong damping, the equation under the square root sign of (7b) (the discriminant) can be positive, zero, or negative which greatly affects the resulting "oscillations". These three cases are called underdamped, critically damped, and overdamped, respectively. This is summarized as: 1. Underdamped, τ > 1/ω0 2. Critically damped, τ = 1/ω0 3. Overdamped, τ < 1/ω0 Often texts on this subject introduce a "damping ratio", ζ, defined as: ζ ≡ 1/(τω0) . In terms of the damping ratio, underdamping corresponds to ζ < 1, critical damping corresponds to ζ = 1, and overdamping corresponds to ζ > 1. We work out the solutions to the "oscillations" for these cases next: | |
| | Graphs of underdamping, critical damping, and overdamping | | | --- | graph of underdamping | graph of critical damping | graph of overdamping | | Fig. 1a. Underdamping | Fig. 1b. Critical damping | Fig. 1c. Overdamping | | |
| general solution to s (7b) decay constant for underdamped case (7i) angular decay frequency for the underdamped case (7j) equation for an underdamped oscillating paramer (7k) | The underdamped case where τ > 1/ω0 and the discriminant in (7b) is positive, i.e. ω02 − 1/τ2 > 0. This is basically the same case as we have been discussing in all the previous postings where we have nice oscillations. The oscillations of the parameter are as given by (7k). We see these decaying oscillations in Fig. 1a above. We have repeated Equation (7b) at the left (as well as for the next two cases below) because it is central to understanding the behavior of the resonator. | Oscillator type | τdecay= τ | ω0 | ωdecay | --- --- | | general oscillator with coefs A, B, and C | 2A/C from (6c) | √C/A from (6e) | use (7j) | | LRC circuit | 2L/R from (6i) | 1/√LC from (5c) | use (7j) | | mass-and-spring | 2m/R | √k/m | use (7j) | |
| general solution to s (7b) decay with critical damping (7l) | The critically damped case where τ = 1/ω0 and the discriminant in (7b) is zero , i.e. ω02 − 1/τ2 = 0. This means that ωdecay = Im(s) = 0 and oscillations do not occur. The real part of s is still present and from (7b) is given by τdecay = −1/Re(s) = τ . All "oscillating" parameters do not oscillate, but instead simply decay as given by (7l) and seen above in Fig. 1b. | Oscillator type | τdecay= τ | ω0 | ωdecay | --- --- | | general oscillator with coefs A, B, and C | 2A/C from (6c) | √C/A from (6e) | 0 | | LRC circuit | 2L/R from (6i) | 1/√LC from (5c) | 0 | | mass-and-spring | 2m/R | √k/m | 0 | |
| general solution to s (7b) decay with over damping (7m) inverse of decay time with over damping (7n) decay time with over damping (7o) | The overdamped case where τ < 1/ω0 and the discriminant in (7b) is negative , i.e. ω02 − 1/τ2 < 0. This means that the square root term in (7b) has another √−1 = i in front of it. Multiplying that i with the one already in (7b) we get an entirely real term and no imaginary term. Therefore by (7g) ωdecay = Im(s) = 0 and the parameter y(t) in (7h) does not oscillate as we see above in Fig 1c. We have a situation much like that in the critical damping case, only exponential decay as given by (7m). However, in this case, the now real square root term adds to the real 1/τ in (7b) making the decay time longer than in the critical damping case, shown in (7n) and (7o). Note: the real part of s determines the damping as per (7i). One confusing aspect of overdamping is that τ is no longer the operational decay time as it was in the underdamping and critical damping cases (where τ = τdecay). Instead, in the overdamped case we must consider τ to be just a collection of constants as shown in the table below, while the operational decay constant in the exponent in (7m) is the larger τdecay given by (7o). | Oscillator type | τdecay | τ | ω0 | ωdecay | --- --- | general oscillator with coefs A, B, and C | use (7o) | 2A/C from (6c) | √C/A from (6e) | 0 | | LRC circuit | use (7o) | 2L/R from (6i) | 1/√LC from (5c) | 0 | | mass-and-spring | use (7o) | 2m/R | √k/m | 0 | We can ask: why does a shorter τ lead to a longer τdecay in the overdamped case? What is going on here? The answer is: shortening τ involves adding more resistance to the damping term in the differential equation, i.e. making R larger in the LRC circuit or in the mass-and-spring resonator. A larger R, in addition to adding dissipation, also adds resistance to any current flow or motion for the two resonators. This added resistance actually slows down the settling to equilibrium once we have enough resistance and dissipation to kill the oscillations. Overdamping is just what its name implies: over doing the damping. ||
P. Ceperley, Jan. 2010.
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| All postings by author | previous: 2.2 Complex math derivation | up: Contents - simple resonators | next: 2.4 How a resonator works |
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| All postings by author | previous: 2.2 Complex math derivation | up: Simple resonators - Contents | next: 2.4 How a resonator works |
Simple Resonators - Shock excitation
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For some applications, oscillations in resonant structures are not desirable. For example, shock absorbers in automobiles are designed to dampen out oscillations of the suspension, while still allowing the suspension to be responsive to bumps in the road. Damping is also required for engine and motor mounts, body panels and other parts of cars and aircraft, and read/write arms in computer hard drives. These parts all have the ingredients of a mass-and-spring resonator: a mass, a spring, and damping. Therefore the equations of the previous section are appropriate, if converted to general coefficients, such as were used in Table 6 on a previous posting. Table 7 below uses general coefficients and derives the frequency and decay constant for three damping strengths: underdamping, critical damping, and overdamping. Note that these terms are usually used in connection with applications designed to prevent oscillations and not with normal resonators in which oscillations are desirable as discussed in previous sections.
Note also that changing the damping in a mechanical resonator such as that in the animation involves changing something mechanical. For example, in the animation, adding damping might involve putting cotton filling or porous sponge material inside the rubber tire that supports the platform. The air flow through this material would absorb energy as the tire is distorted by the weight of the girl. A more effective method would be to use a separate air chamber, such as that inside the middle of an inflated inner tube and provide a hole in it to let air to pass from the outside atmosphere to inside this chamber and back the other way as the girl compresses the tire. By changing the hole size and thus the resistance to air flow, one could change the damping.
The animation above illustrates the effect of different values of the damping coefficient on the resulting "oscillations". Mouse over it (or click on it) to set it into motion. Select the type of damping you wish to observe. A little experimentation with the animation will convince you that critical damping leads to the fastest reaching of stationary equilibrium. Overdamping has a nice exponential decay with no oscillations, but it causes a slower decay than with critical damping. Underdamping causes over shooting, oscillations, and a slower reaching of stationary equilibrium. Of course, the exact values of the damping coefficients affect the speed at which equilibrium is reached, but coefficients adjusted for critical damping result in the fastest equilibrium.
Two more notes before we start the math:
Real shock absorbers as well as some other dampers in modern technology use more sophisticated methods than addressed in this posting, such as damping only in one direction of motion, or computer controlled damping.
Most of these posts in this series are directed towards applications that require oscillations to work and would be considered to be extremely "under damped" in the language of this particular posting. This particular post is odd in that it is directed towards applications requiring abatement of oscillations. We include it for completeness.
| Table 7. Equations for a heavily damped resonator | |
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| general second order differential equation (7a) | We repeat the general second order differential equation, last shown in Equation (6a) (of section 2.1) with general coefficients A, B, and C. In this case C is not the capacitance, but instead just the last coefficient in this general equation. |
| general solution to s (7b) | Here we use the complex equation, Equation (6h), from the last posting that gives exponential coefficient, s. We have substituted R = 2L/τ (5a) and LC = 1/ω02 (5c) to converted it to the general coefficients as used in Equation (7a). |
| breaking up s (7c) | Here we break up s into real and imaginary parts. a is the real part of s, i.e. Re(s), and b is the imaginary part of s, i.e. Im(s). |
| breaking up s (7d) | When we insert s in the form of (7c) into a typical damped oscillating parameter, such as y(t) = Aest, similar to the q(t) in Equation (6c), we get (7d). Looking at the extreme right of (7d), we see that eat is an exponential growth factor (for positive a's) or exponential decay factor (for negative a's). The eibt factor has a complex argument meaning that its real equivalent is a sinusoid. The A in this equation and also in the equations that follow is not the constant in Equation (7a) above. Instead the A as used here (and below) is the initial amplitude of oscillations. Also, as per the standard convention with complex phasors, we take only the real part of Equation (7d) for Equation (7e). |
| real part of (7d) (7e) | This is the real part of (7d). Comparing (7e) with (6b) of the previous posting, we see a is the damping coefficient (i.e. −1/τdecay), and b is the angular frequency (i.e. ωdecay). That is to say Re(s) = a = -1/τdecay and Im(s) = b = ωdecay . |
| τdecay = −1/a = −1/Re(s) (7f) ωdecay = b = Im(s) (7g) y(t) = Aet/τdecaycos ωdecayt (7h) | Here we summarize the above. |
| When we have strong damping, the equation under the square root sign of (7b) (the discriminant) can be positive, zero, or negative which greatly affects the resulting "oscillations". These three cases are called underdamped, critically damped, and overdamped, respectively. This is summarized as: 1. Underdamped, τ > 1/ω0 2. Critically damped, τ = 1/ω0 3. Overdamped, τ < 1/ω0 Often texts on this subject introduce a "damping ratio", ζ, defined as: ζ ≡ 1/(τω0) . In terms of the damping ratio, underdamping corresponds to ζ < 1, critical damping corresponds to ζ = 1, and overdamping corresponds to ζ > 1. We work out the solutions to the "oscillations" for these cases next: | |
| | Graphs of underdamping, critical damping, and overdamping | | | --- | graph of underdamping | graph of critical damping | graph of overdamping | | Fig. 1a. Underdamping | Fig. 1b. Critical damping | Fig. 1c. Overdamping | | |
| general solution to s (7b) decay constant for underdamped case (7i) angular decay frequency for the underdamped case (7j) equation for an underdamped oscillating paramer (7k) | The underdamped case where τ > 1/ω0 and the discriminant in (7b) is positive, i.e. ω02 − 1/τ2 > 0. This is basically the same case as we have been discussing in all the previous postings where we have nice oscillations. The oscillations of the parameter are as given by (7k). We see these decaying oscillations in Fig. 1a above. We have repeated Equation (7b) at the left (as well as for the next two cases below) because it is central to understanding the behavior of the resonator. | Oscillator type | τdecay= τ | ω0 | ωdecay | --- --- | | general oscillator with coefs A, B, and C | 2A/C from (6c) | √C/A from (6e) | use (7j) | | LRC circuit | 2L/R from (6i) | 1/√LC from (5c) | use (7j) | | mass-and-spring | 2m/R | √k/m | use (7j) | |
| general solution to s (7b) decay with critical damping (7l) | The critically damped case where τ = 1/ω0 and the discriminant in (7b) is zero , i.e. ω02 − 1/τ2 = 0. This means that ωdecay = Im(s) = 0 and oscillations do not occur. The real part of s is still present and from (7b) is given by τdecay = −1/Re(s) = τ . All "oscillating" parameters do not oscillate, but instead simply decay as given by (7l) and seen above in Fig. 1b. | Oscillator type | τdecay= τ | ω0 | ωdecay | --- --- | | general oscillator with coefs A, B, and C | 2A/C from (6c) | √C/A from (6e) | 0 | | LRC circuit | 2L/R from (6i) | 1/√LC from (5c) | 0 | | mass-and-spring | 2m/R | √k/m | 0 | |
| general solution to s (7b) decay with over damping (7m) inverse of decay time with over damping (7n) decay time with over damping (7o) | The overdamped case where τ < 1/ω0 and the discriminant in (7b) is negative , i.e. ω02 − 1/τ2 < 0. This means that the square root term in (7b) has another √−1 = i in front of it. Multiplying that i with the one already in (7b) we get an entirely real term and no imaginary term. Therefore by (7g) ωdecay = Im(s) = 0 and the parameter y(t) in (7h) does not oscillate as we see above in Fig 1c. We have a situation much like that in the critical damping case, only exponential decay as given by (7m). However, in this case, the now real square root term adds to the real 1/τ in (7b) making the decay time longer than in the critical damping case, shown in (7n) and (7o). Note: the real part of s determines the damping as per (7i). One confusing aspect of overdamping is that τ is no longer the operational decay time as it was in the underdamping and critical damping cases (where τ = τdecay). Instead, in the overdamped case we must consider τ to be just a collection of constants as shown in the table below, while the operational decay constant in the exponent in (7m) is the larger τdecay given by (7o). | Oscillator type | τdecay | τ | ω0 | ωdecay | --- --- | general oscillator with coefs A, B, and C | use (7o) | 2A/C from (6c) | √C/A from (6e) | 0 | | LRC circuit | use (7o) | 2L/R from (6i) | 1/√LC from (5c) | 0 | | mass-and-spring | use (7o) | 2m/R | √k/m | 0 | We can ask: why does a shorter τ lead to a longer τdecay in the overdamped case? What is going on here? The answer is: shortening τ involves adding more resistance to the damping term in the differential equation, i.e. making R larger in the LRC circuit or in the mass-and-spring resonator. A larger R, in addition to adding dissipation, also adds resistance to any current flow or motion for the two resonators. This added resistance actually slows down the settling to equilibrium once we have enough resistance and dissipation to kill the oscillations. Overdamping is just what its name implies: over doing the damping. ||
The A in this equation and also in the equations that follow is not the constant in Equation (7a) above. Instead the A as used here (and below) is the initial amplitude of oscillations. Also, as per the standard convention with complex phasors, we take only the real part of Equation (7d) for Equation (7e).
We work out the solutions to the "oscillations" for these cases next:
| Graphs of underdamping, critical damping, and overdamping | | |
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| graph of underdamping | graph of critical damping | graph of overdamping |
| Fig. 1a. Underdamping | Fig. 1b. Critical damping | Fig. 1c. Overdamping |
| Oscillator type | τdecay= τ | ω0 | ωdecay |
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| general oscillator with coefs A, B, and C | 2A/C from (6c) | √C/A from (6e) | use (7j) |
| LRC circuit | 2L/R from (6i) | 1/√LC from (5c) | use (7j) |
| mass-and-spring | 2m/R | √k/m | use (7j) |
| Oscillator type | τdecay= τ | ω0 | ωdecay |
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| general oscillator with coefs A, B, and C | 2A/C from (6c) | √C/A from (6e) | 0 |
| LRC circuit | 2L/R from (6i) | 1/√LC from (5c) | 0 |
| mass-and-spring | 2m/R | √k/m | 0 |
We have a situation much like that in the critical damping case, only exponential decay as given by (7m). However, in this case, the now real square root term adds to the real 1/τ in (7b) making the decay time longer than in the critical damping case, shown in (7n) and (7o). Note: the real part of s determines the damping as per (7i).
One confusing aspect of overdamping is that τ is no longer the operational decay time as it was in the underdamping and critical damping cases (where τ = τdecay). Instead, in the overdamped case we must consider τ to be just a collection of constants as shown in the table below, while the operational decay constant in the exponent in (7m) is the larger τdecay given by (7o).
| Oscillator type | τdecay | τ | ω0 | ωdecay |
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| general oscillator with coefs A, B, and C | use (7o) | 2A/C from (6c) | √C/A from (6e) | 0 |
| LRC circuit | use (7o) | 2L/R from (6i) | 1/√LC from (5c) | 0 |
| mass-and-spring | use (7o) | 2m/R | √k/m | 0 |
P. Ceperley, Jan. 2010.
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My favorite reference books on waves:
Electromagnetic books that I use:
| Resonances, Waves, and Fields Contents |
| Wave-centric interpretation of quantum mechanics Three types of waves + Mathematical expressions for one dimensional traveling waves + Standing waves in one dimension + Waves in a ring resonator - standing and rotating waves + Traveling and standing waves in two and three dimensions - Cartesian coordinates + Standing waves and rotating waves in two dimensional circular resonators + Standing waves and rotating waves in three dimensional cylindrical resonators + Spherical harmonics + The Lagrangian approach to simple waves - several common waves that lack momentum + Why only J’s in a standard Bessel expansion? + Games with arrays of wave emitters + Octave programs used in emitter games Brunelleschis Dome - its structure and construction Origins of Newton's laws of motion Conformal mapping - solving Laplace's equation + Contents of conformal mapping postings + Introduction to conformal mapping + Conformal mapping II + Fluid flow patterns handled by other functions + Conformal mapping of electric and magnetic fields + Detailed conformal mappings Simple resonators, shock excitation + Introduction - various examples + Shock excitation + Complex math derivation + Strong damping - critical damping + How a resonator works + Energy in a resonator + Power loss in resonators + Energy and power by complex methods + Valid method of multiplying two complex phasors + Modulation, an application of phasor multiplication + Up/down conversion, an application of phasor multiplication + Q, quality factor Resonators - continuous excitation + Sinusoidal excitation of a resonator - contents + Damping + Driven oscillations + Various looks of resonance curves + Resonance peak properties + The resonance peak - qualitatively + More on the impact model + Mathematical response to a very short impulse + Summing impulses with calculus + Buildup and decay of envelope - Supplemental derivation + Q - a review and more + Coupling of waves to a resonator + Coupling strength + Waveguide excited resonator with circulator + Resonant scattering of waves - one dimensional cases + Circuit analysis of the acoustical circuit in Fig. 35 - Electrical model of the acoustical circuit in Fig. 35 of posting 3.15 - Looking at the resonator part of the circuit - Steady state response of the circuit: transmission line, coupling inductor and resonator - Transmission line resonators - Steady state response of a transmission line resonator via a Smith Chart - Differential equation solution for transient response and SPICE simulation - Summary of analysis of a transmission line excited resonator using a circulator + Circuit analysis of resonant scattering with two output channels + Separating out the radiated and simply reflected components - mathematical + Applying methods of 3.18 to the circuit of 3.17 + Derivation of wave properties for transmission lines + Reflection and absorption coefficients for transmission lines with loads + Comparison of series and parallel LRC resonant circuits Clocks + History of Mechanical Clocks with Animations: + Understanding a mechanical clock with animations: includes pendulum, balance wheel and quartz clocks Index for resonances, waves and fields Album of 24 flash animations showing the physics of oscillations, waves, and phasors The LRC circuit, the classical simple resonator: Marconi invents the wireless telegraph. Use of complex numbers in oscillations and waves: + Euler's formula and complex numbers: an introduction to and history of complex numbers. + Details on infinite series. + The technology of oscillations and waves demand new mathematical methods: a look at the mathematical difficulty of dealing with oscillations and waves. + Phasors + More on complex numbers: to prepare for using them for oscillations and waves. + Complex phasors + The algebra of complex phasors + True waves + Waves using complex phasors + Types of waves + Superposition and standing waves + Complex phasor representation of a standing wave + Reflection of waves as a process to make standing waves Water Waves: + Water waves + Water waves - mathematical derivation - part 1 + Water waves derivation 2 - dynamics of the free surface Fourier Analysis: + The spectrum of a waveform - Fourier Analysis + Mathematical definition of Fourier series + Sines, cosines, and phases + How good is a Fourier series of a function at reproducing the original function? + Symmetries in Fourier series Special relativity + Non-mathematical introduction to relativity - 9 part series + - NEW - Mathematics of special relativity - 19 part series Programming in swishMax and flash: + Matrix manipulation routines for swishMax + Ways to make Waves: programming and non-programming methods for animating waves in swishMax and flash. The Author Permitted use of material on this blog and how to download free animations from this site. Index for resonances, waves and fields |
Resonances, Waves, and Fields Contents
Contents
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http://cau.ac.kr/~mhhgtx/courses/LinearAlgebra/references/MadsenLightsOut.pdf
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Summation May 2010, pp. 36-40 Abstract—Microcontrollers are seen everywhere in everyday life. Items such as cell phones, remotes, and electronic games use them. This paper investigates the electronic game Lights Out that uses a microcontroller and explains how one could compute a winning strategy for the game. It demonstrates a couple of different methods and works through the proofs of them. Key Words—Lights Out, Linear Algebra, Microcontroller I. INTRODUCTION ICROCONTROLLERS are small computers on a single integrated circuit consisting internally of a relatively simple CPU, clock, timers, I/O ports, and memory. They are used in automatically controlled products and devices, such as automobile engine control systems, implantable medical devices, remote controls, power tools, and toys. The device that this paper is on is the electronic game Lights Out. In this paper, I will go over several different methods that can be used to find a solution for the game Lights Out. I will first go through a mathematical method that involves the uses of linear algebra to find a winning strategy using the fewest number of moves. Then I’ll talk about a method that doesn’t involve the use of an electronic device and one that was described to me by one of my colleagues. II. LIGHTS OUT RULES AND OBSERVATIONS A. Basic Rules Lights Out is an electronic game manufactured by Tiger Toys in 1995. It consists of 25 buttons, each of which can be lit up, which form a 5 by 5 array. Each button can be in one of two states (on or off). At the start of play, a pattern of lit buttons is chosen by the microcontroller. By pressing the buttons, the player can alter the state of some subset of the buttons according to the rule. The effect of pressing a button is to toggle the state of that button, and its immediate vertical and horizontal neighbors. The goal of the game is to turn all the buttons to the off state. The game has three modes. In the first mode, you are presented with 50 increasingly difficult puzzles. You have to Work presented to the college March 23, 2010 as the second talk in a series of three talks about microcontrollers Matthew A. Madsen currently pursuing his baccalaureate degree in mathematics at Ripon College in Ripon, WI 54971 USA (e-mail:madsenm@ripon.edu). solve one puzzle before you can move on to the next. The goal of this game is to solve each initial set-up in the minimum possible number of moves; however you are allowed to go over this minimum by ten. The second mode gives you another 1000 puzzles to solve. Mode three allows the player to enter your own puzzles, however not all the possible configurations of states are solvable. This will be demonstrated later in the paper. B. Observations The first initial observation about the game is that each button needs to be pressed no more than once. This observation comes from the fact that pressing a button twice is like not pressing it at all. Since pressing a button toggles the state of that button and of its immediate vertical and horizontal neighbors, pressing that same button again will reverse the states and toggle the buttons back to their original states. The key consequence of this observation is that all the calculations can be done in modulo 2. That is, pressing a button an even number of times is like not pressing it at all and pressing it an odd number of times is the same as pressing it only once. This consequence will be very important in calculating the solutions mathematically. The second initial observation is that the state of each button only depends on how many times it and its immediate vertical and horizontal neighbors are pressed. This observation indicates is that the order in which you press the buttons is irrelevant to the resulting configuration. A third observation comes directly from the first observation. If we start with the board completely off and press a sequence of buttons to get a configuration, then starting with that configuration and pressing the same sequence of buttons will result in all the lights turning off. Lights Out: Solutions Using Linear Algebra Matthew A. Madsen Department of Mathematics and Computer Science – Ripon College M Figure 1. An example of how to play the game. The green squares represent lights that are on. The red dots are the selection of the play. This starting configuration can be solved with only pressing four buttons 36 2010 Senior Seminar: Microcontrollers III. LINEAR ALGEBRA SOLUTION One way to guarantee a solution in the fewest number of buttons pressed is to use linear algebra. This can be done with basic matrix operations, some Gauss-Jordan elimination, and an understanding of the column and null space of a matrix. Since there are two states to the buttons, on and off, we can do all of our calculations in modulo 2 by letting one represent on and zero represent off. We can then think of the 5 by 5 array as a 25 by 1 vector, as in T b b b b b b ) , , , , , ( 25 24 3 2 1 " K = (1) where b1, b2, b3, … ,b24, b25 will represent the state of each button related to Fig. (2). This vector will be referred to as the configuration of the array. A configuration b K is obtained by pressing a sequence of buttons, which we will denote as T x x x x x x ) , , , , , ( 25 24 3 2 1 " K = (2) where x1, x2, x3, … ,x 24, x25 will represent whether that particular button is pressed. If we start with all the lights out and the configuration b K is obtained by strategy x K , then .
25 24 20 25 25 24 23 19 24 18 14 13 12 8 13 8 4 3 2 3 7 3 2 1 2 6 2 1 1 x x x b x x x x b x x x x x b x x x x b x x x x b x x x b + + = + + + = = + + + + = = + + + = + + + = + + = # # # # (3) It is now straightforward to rewrite this system of linear equations as the matrix product b x A K K = , where A is the 25 by 25 matrix ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = Z I O O O I Z I O O O I Z I O O O I Z I O O O I Z A . (4) In this matrix, I is the 5 by 5 identity matrix, O is the 5 by 5 matrix of all zeros, and Z is the matrix ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = 1 1 0 0 0 1 1 1 0 0 0 1 1 1 0 0 0 1 1 1 0 0 0 1 1 Z . (5) A key observation to make about Z is that it is symmetric, and therefore A is also symmetric. To find a solution for the A given configuration b K , we must find the strategy x K that satisfies b x A K K = . To do this, we will prove two theorems. A. Theorem 1 Given a starting configuration, we first need to check to see if the configuration is winnable. We say that configuration b K is winnable if there is a strategy x K that satisfies b x A K K = . Therefore, Theorem 1 states a configuration b K is winnable if and only if it is orthogonal to the two vectors 1 n K and 2 n K , where ( ) T n 0 , 0 , 1 , 1 , 0 , 1 , 0 , 1 , 0 , 1 , 1 , 1 , 0 , 1 , 1 , 1 , 0 , 1 , 0 , 1 , 0 , 1 , 1 , 1 , 0 1 = K (6) and ( ) T n 1 , 0 , 1 , 0 , 1 , 1 , 0 , 1 , 0 , 1 , 0 , 0 , 0 , 0 , 0 , 1 , 0 , 1 , 0 , 1 , 1 , 0 , 1 , 0 , 1 2 = K . (7) To prove this theorem, we first need to understand some definitions and theorems about the column and null space of a matrix. For configurationb K to be winnable, it must have a strategy x K that satisfies b x A K K = , meaning that configuration b K is winnable if and only if it belongs to the column space of A, denoted as Col (A). Since A is symmetrical, the row space of A, denoted as Row (A), is equal to Col (A). From the definitions of the row space and the null space, we know that Row (A) is equal to the orthogonal complements of the null space of A, denoted Null (A). Since the null space of a matrix is not affected by the Gauss-Jordan elementary row operations, let matrix E be the reduced row echelon form of matrix A. We can then see that Null (A) = Null (E). It then follows that b K is winnable if and only if it belongs to the orthogonal complements of Null (E). Since we are looking for the orthogonal complements of Null (E), we need to find an orthogonal basis for the Null (E). To find a basis for the Null (E), we will interpret matrix E as a system of equations and then solve for the dependent variables. We can see in the equations 25 25 24 24 25 24 23 25 24 3 24 2 25 1 25 24 23 25 24 3 24 2 25 1 0 0 0 0 e e e e e e e e e e e e e e e e e e e e e e e e = = + = = + = = = ⇒ + + = = + + = + = + = # # # # , (8) Figure 2. How the buttons on a 5 by 5 are viewed as a 25 by 1 vector 37 Madsen – Lights Out that by doing the calculations in modulo 2, that the negatives become positive. By writing the independent variables in terms of the dependent variables we get ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ + ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ = = = = = = = 1 0 1 1 0 1 0 1 1 1 1 0 25 24 25 24 23 3 2 1 # # # e e e e e e e e . (9) The resulting collection of vectors will be the basis for the Null (E). So a basis for the Null (E) is 1 n K and 2 n K , the same as (6) and (7). To check to see if these two vectors form an orthogonal basis, we must simply show the dot product of the two vectors is equal to zero. We find out that 8 2 1 = ⋅n n K K , but since we are doing our calculations in modulo 2, we see that 0 2 1 = ⋅n n K K , and that 1 n K and 2 n K are an orthogonal basis for the Null (E). Since b K is winnable if and only if it belongs to the orthogonal complements of Null (E), then b K is winnable if and only if it is orthogonal to the two vectors 1 n K and 2 n K , thus proving Theorem 1. B. Theorem 2 Theorem 2 states that if b K is winnable, then the four winning strategies are 2 1 2 1 n n b R n b R n b R b R x + + + + = K K K K K (10) where 1 n K and 2 n K are the same as (6) and (7) and R is the product of the elementary matrices which perform the reducing row operation, RA=E . Since E has two independent variables, 24 x and 25 x , then the four configurations we can choose for them are 1 1 1 0 0 1 0 0 25 24 x x (11) Let’s choose 0 24 = x and 0 25 = x . We can then see the product x x E K K = . We can then substitute in RA=E and b x A K K = to get b R x A R x E x K K K K = = = . (13) Therefore, one solution is b R x K K = . By looking at the other configurations for 24 x and 25 x , we can let 24 x and 25 x be as shown in (11). As a result, we can see that the four winning strategies are the four in (10). C. Graphical User Interface Using Theorems 1 and 2, we now have a way to mathematically check to see if a starting configuration is winnable. If it is winnable, we now have four equations that will find for a strategy that will turn all the lights out. To find the strategy with the fewest number of buttons pressed, you simply need to count the number of buttons that need to be pressed for each of the four strategies. Since we are dealing with large matrices, 25 by 25, doing this computation by hand would be long and tedious. With the help of a computer algebra system or programming language capable of handling matrices like Maple, Mathematica, or R, one could compute the winning strategies. Due to my current knowledge, I did my computation in R and created a graphical user interface where one could enter a starting configuration and the GUI would return “No Solution” if there was no winning strategy or it would return the winning strategy with the fewest number of moves. IV. SOLVING USING LIGHT CHASING A way that one could solve Lights Out without using a computer algebra system was first introduced to me by one of my colleagues . Upon listening to him explain this method to me, which he called the “Binary Method,” I did a little research and found that the method he was explaining was the Light Chasing Method. The idea of the Light Chasing Method is to separate the 5 by 5 array into a 1 by 5 and a 4 by 5 array and then work on turning all the lights out in the 1 by 5 array. Then you would continue to separate the remaining array until you got down to a 1 by 5 array. An easy way to think about this is by turning out all the lights on the top row. One achieves this by simply pressing the buttons on the second row that are directly underneath a lit buttons on the top row. The top row will then have all its lights off. By repeating this step for the second, third, and fourth rows (i.e. chase the lights all the way down to the bottom row), you may have solved the puzzle, but it is more likely that there will now be some lights left on in the bottom row. If so, there are only seven possible configurations. Depending on which configuration you are left with, you will need to press some buttons in the top row. You can determine which buttons you need to press from the Figure 3. The problem with the Light Chasing Method is that you will most likely end up pressing some buttons more that once, which means you won’t solve the game in the fewest number of buttons pressed and will end up replaying the board. A simple solution to this is once you get down to your bottom row, restart the board and start by pressing the corresponding buttons in the top row to clear the board. If you don’t have the Figure 3 with you, my colleague that 38 2010 Senior Seminar: Microcontrollers introduced me to this method explained that one should think of the five buttons in the top row as a five digit binary number staring at zero and increasing by one each time. If you press the buttons in the top row and it doesn’t clear the board then continue on to the next binary number. For example, first try 00000, 10000, 01000, and 00100 where a one represents pressing the corresponding button. If you continue this, you will find that there are only 32 different ways to press the buttons in the top row. We proved earlier in this paper that there are 4 winnable solutions, so an eighth of the possible ways to press the buttons in the top row will clear the board. Using a geometric distribution and the “Binary Method”, you have a 0.5512 probability of clearing the board within the first six attempts, with a mean of eight attempts. The other problem with solving lights out using Light Chasing is there is no way of checking to see if a board is winnable unless you go through all 32 different combinations. V. CHANGES TO THE GAME So far this paper has just talked about the original 5 by 5 Lights Out, but since its release in 1995, Tiger Toys have created several other variations to the game. The game Lights Out Cube consists of a 3 by 3 by 3 cube of lights and is played just like ordinary lights out. When you press a button at the edge of a 3x3 square, the affected lights wrap around onto the adjacent faces. The Lights Out Keychain, also referred to as Lights Out Mini, is a 4 by 4 array of lights that plays on a torus. Playing on a 4 by 4 torus there is a unique board because every possible starting configuration is solvable and each solution is unique. Lights Out Deluxe is a version of Lights Out played on a 6 by 6 array and offers some different game variations. These variations are slight changes in the rules. One variation is called Lit Only where you are only allowed to press buttons that are currently on. Tiger Toys’ latest Lights Out is called Lights Out 2000. This game plays 3-state Lights Out. The lights cycle through red, green, and off. This creates a significantly different and challenging game. There are also hundreds of different variations that could be made to the game. First of all, the board could take on almost any shape it wants. Whether it is an n by n board, a circle, or a randomly scattered board, as long as you know which buttons toggle when you press a given button, you can play Lights Out. VI. CONCLUSION This paper has shown a variety of different methods that can be used to solve the electronic game Lights Out. One could use a computer algebra system capable of handling matrices where you know you can get an answer right away or use a guessing method like the “Binary Method” where you keep trying different sequences until onefinds a solution. Having read this paper, I hope that I have inspired you to search the internet for an online version of Lights Out so you can practice the methods that I have talked about or a method of your own. If reading this paper has made you ambitious enough, go buy a microcontroller and try to recreate the game Lights Out. ACKNOWLEDGMENT I would like to acknowledge and thank the others members of my group, Jon Palecek and Prasoon Saurabh, for their input and help on my research. I would also like to thank our faculty advisor Professor Timothy Hess for his input and help with programming. I would also like to thank Ross Lewis for a great conversation about the game and introducing me to a different method for solving the game. Additionally, I would like to acknowledge Professor David Scott for helping me with various terms and calculations. REFERENCES Turning Lights Out with Linear Algebra, by Marlow Anderson and Todd Feil Mathematics Magazine, Vol. 71, No. 4 (Oct., 1998), pp. 300-303 Mathematical Association of America. An Easy Solution to Mini Lights Out, by Jennie Missigman and Richard Weida Mathematics Magazine © 2001 Mathematical Association of America. Two Reflected Analyses of Lights Out, by Óscar Martín-Sánchez and Cristóbal Pareja-Flores Mathematics Magazine © 2001 Mathematical Association of America. C. Haese, paper in the sci.nath newsgroup in 1998, available at math/98/lights-out Hohn, Franz E. Introduction to Linear Algebra. New York: The Macmillian Company, 1972. 161-188. Print. K. Banr, Lights Out Fan Club, at D.L. Stock, Merlin's magic square revisited, Amer Math. Monthly 96 (1989), 608-610. java version of the commercially marketed game Lights Out at Barile, Margherita. "Lights Out Puzzle." From MathWorld--A Wolfram Web Resource, created by Eric W. Weisstein. Lewis, Ross. Conversation with about alterative solution Figure 3. After chasing the lights down to the bottom row, you will have turned the lights out or have one of seven remaining configuration as shown in the left column. By pressing the buttons in the top row that are in the right column of the row of the remaining configuration and repeat the Chasing Lights Method you will turn all the lights out. 39 Madsen – Lights Out Matthew A. Madsen was born in Reedsburg, Wisconsin on November 7, 1988. He is currently pursuing his Baccalaureate degree in Mathematics and Physics from Ripon College in Ripon, Wisconsin and will be graduating in May 2011 after completing his final semester. He is currently employed by the Mathematics and Computer Science Department and the Physics Department as a Departmental Assistant while working part-time at Webster’s Pick ‘N Save in Ripon, Wisconsin as customer service and terminal operator. He has spent several years working as a lifeguard at the Wilderness Resort in Wisconsin Dells. Mr. Madsen is a member of the Society of Physics Students. In his free time he enjoys playing video games, hunting, watching movies and sports, and spending time with family and friends. 40
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Alternate functions of physiological anticoagulants - PubMed
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. 2025 Sep 2:S1538-7836(25)00540-9.
doi: 10.1016/j.jtha.2025.08.014. Online ahead of print.
Alternate functions of physiological anticoagulants
Teagan Prouse1,Alan E Mast2,Vera Ignjatovic3,Mirjana Kovac4,Zsuzsanna Bereczky5,Rinku Majumder6
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Affiliations
1 Department of Interdisciplinary Oncology, Louisiana State University Health Sciences Center, New Orleans, Louisiana, USA.
2 Thrombosis and Hemostasis Program, Versiti Blood Research Institute, Milwaukee, Wisconsin, USA.
3 Johns Hopkins All Children's Institute for Clinical and Translational Research, St. Petersburg, Florida, USA; Department of Pediatrics, School of Medicine, Johns Hopkins University, Baltimore, Maryland, USA.
4 Blood Transfusion Institute of Serbia, Hemostasis Department, Faculty of Medicine, University of Belgrade, Serbia.
5 Division of Clinical Laboratory Science, Department of Laboratory Medicine, Faculty of Medicine, University of Debrecen, Hungary.
6 Department of Interdisciplinary Oncology, Louisiana State University Health Sciences Center, New Orleans, Louisiana, USA. Electronic address: rmajum@lsuhsc.edu.
PMID: 40907701
DOI: 10.1016/j.jtha.2025.08.014
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Alternate functions of physiological anticoagulants
Teagan Prouse et al. J Thromb Haemost.2025.
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. 2025 Sep 2:S1538-7836(25)00540-9.
doi: 10.1016/j.jtha.2025.08.014. Online ahead of print.
Authors
Teagan Prouse1,Alan E Mast2,Vera Ignjatovic3,Mirjana Kovac4,Zsuzsanna Bereczky5,Rinku Majumder6
Affiliations
1 Department of Interdisciplinary Oncology, Louisiana State University Health Sciences Center, New Orleans, Louisiana, USA.
2 Thrombosis and Hemostasis Program, Versiti Blood Research Institute, Milwaukee, Wisconsin, USA.
3 Johns Hopkins All Children's Institute for Clinical and Translational Research, St. Petersburg, Florida, USA; Department of Pediatrics, School of Medicine, Johns Hopkins University, Baltimore, Maryland, USA.
4 Blood Transfusion Institute of Serbia, Hemostasis Department, Faculty of Medicine, University of Belgrade, Serbia.
5 Division of Clinical Laboratory Science, Department of Laboratory Medicine, Faculty of Medicine, University of Debrecen, Hungary.
6 Department of Interdisciplinary Oncology, Louisiana State University Health Sciences Center, New Orleans, Louisiana, USA. Electronic address: rmajum@lsuhsc.edu.
PMID: 40907701
DOI: 10.1016/j.jtha.2025.08.014
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The anticoagulant proteins antithrombin, protein C, protein S, and tissue factor pathway inhibitor are physiological regulators of the blood coagulation pathway and have complex roles in causing or modifying the severity of bleeding and clotting disorders. The anticoagulant activity, as well as alternative physiological functions, of these proteins also impact many other diseases beyond bleeding and clotting disorders. We searched PubMed for articles published between 2018 and early 2025 that pertained to alternative functions of these physiological anticoagulants in clinical disease and selected therapeutic applications in the following categories: cancer, cardiovascular disease, novel therapeutics, pregnancy health, inflammation and infection, blood-brain barrier protection, stroke, novel developments in thrombosis, and endothelial and platelet interactions. This review highlights recently identified physiological implications of antithrombin, protein C, protein S, and tissue factor pathway inhibitor functions that present an exciting therapeutic avenue for a diverse array of human diseases.
Keywords: antithrombin; protein C; protein S; tissue factor pathway inhibitor.
Published by Elsevier Inc.
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Conflict of interest statement
Declaration of competing interests A.E.M. receives research grant funding and honoraria from Novo Nordisk. The other authors declare no conflict of interest.
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Distance, speed and time formulae - Distance, Speed and Time - National 4 Maths Revision - BBC Bitesize
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Distance, Speed and Time Distance, speed and time formulae
Calculations involving distance, speed and time can be worked out using formulae. When doing these calculations, the units used should be consistent.
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Distance, speed and time formulae
Calculating distance given speed and time
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Calculating speed given distance and time
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Distance, speed and time formulae
All of the calculations in this section will be worked out using the distance, speed and time formulae.
An easy way to remember the formulae is to put distance, speed and time (or the letters D, S and T) into a triangle.
The triangles will help you remember these three rules:
D i s t a n c e=S p e e d×T i m e
T i m e=D i s t a n c e S p e e d
S p e e d=D i s t a n c e T i m e
On the next page there are some examples to work through.
Have paper and a pen handy, draw the distance, speed and time triangle on your paper, then try the examples.
Next page Calculating distance given speed and time
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Where Does Buoyancy Force Come From? [Physics of Fluid Mechanics #35]
Simmy Sigma
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I've gotten a lot of questions about a few examples I did where we had an object that was fully submerged in a body of water and why we didn't include the weight of the water above the object in our free-body diagram. In this video, I'll show you where this buoyant force comes from and how the weight of the water is in-fact incorporated into our calculations when calculating the pressures at the top and bottom of the object.
This lecture is part of the Physics of Fluid Mechanics course:
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Transcript:
okay in this video i want to talk about where buoyant force comes from so in a previous video i actually did an example where i had this container that was filled with water and inside of the container there was this block that was tethered down to the bottom of the container by a piece of string and when we drew the free body diagram of this block so i'm going to draw the free body diagram here we had a buoyant force acting up which i called f sub b we had the weight of the block coming down and then we had the tension in the string coming down and i got a few people reaching out and they had asked well why don't we consider the weight of the water above the block and even the weight of the water below the block because the block is submerged in water so there is weight from this body of water right here acting on top of the block and they are right there is weight acting on top of the weight or the block but we did include that weight or the weight difference in this buoyant force so the definition of buoyant force is the difference between the force at the bottom and the top caused by the pressures at the top and on the bottom so if we look at this example right here this block is submerged in a body of water so that there is pressure all over this block on every single side on the top the bottom the left the right and you'll notice intuitively that the pressures on the left and on the right side of the blocks they cancel out because they're symmetric but the pressure here at this line on top of the block is a little bit less than the pressure here on the bottom right the deeper you go down in water the higher the pressure and the higher the pressure the higher the force acting on a surface so you can see that the force acting on the bottom of the block due to the pressure at the bottom is a little bit bigger than the force acting on top of the block and so again this buoyant force that i'll call f sub b is equal to the force of the bottom minus the force at the top and those forces are due to the pressure at those surfaces times the area right so pressure times area gives us force so in this example or this video i want to derive this buoyant force and in a lot of other videos i've been saying that the buoyant force f sub b is equal to the weight of the liquid displaced in this case it's water so i'll just call it w sub w so the buoyant force is equal to the weight of the displaced water acting on the object so how do we go from here to there well that's what we're going to do in this video okay so i have two diagrams uh drawn here on the left and on the right and they are the same situation on the left we have this body of water and there is a blocked fully submerged in this liquid and on the right i just have this dotted outline here which represents the amount of water that got displaced due to this block so this volume and this volume are exactly the same and if we were to draw the free body diagram of this object at the very bottom we have pressure right and that pressure is going to cause some sort of a force which i'll call f sub bottom and that's equal to p bottom times a and at the top we have another force called f top and that's equal to p top times a and then we also have the weight of the block so this is mass of the block times gravity and due to this object being in static equilibrium we know that all these forces balance out so all the forces going up must equal all the forces going down in other words f sub bottom must equal mass times gravity of the block plus f sub top okay cool but how do we get buoyant force how do we get point force well remember buoyant force is the weight of the fluid displaced in other words all of this fluid has a mass and therefore a weight and that is the point force that's going to act on top of this block and push it upwards so on the displaced water diagram i'm going to have something similar i'm going to have f sub top which is the force due to the pressure at the top of that area times area so this is equal to pressure top times a and at the bottom we have f sub bottom and that's equal to the pressure at the bottom times the area now i'm going to label a couple things so this block has a height of d and i'm going to say the distance from the surface level of the water down to the bottom of the block i'll call that h so we'll need those in a little bit but i want to go back to the concept of pressure now if you remember pressure is equal to rho g h now this h is not this h that i've drawn here uh this is just a side note so just this is on its own rho g h the mass density of the fluid times gravity times the depth right the deeper you go the higher the pressure so based off that we can calculate p top now p top is going to be the mass density of the water times gravity times the distance from the surface down to the point that we're studying which is the top of the block or the top of the volume that we're looking at and if this is d and this is h then that must mean this distance is h minus d so this is going to be h minus d cool well what about the pressure at the bottom so the pressure at the bottom is more or less the same you have the mass density of the water times g and now the distance from the surface down to the bottom well that's h so i'm just going to call that h so cool we have pressure top and we have pressure bottom we can take those two terms and plug them into their respective equations here and we can use those forces then to figure out what the buoyant force is because remember the buoyant force is going to be equal to the pressure or the force at the bottom minus the force at the top cool so we can do that so i'm going to write that here the point force is f bottom minus f top and we know that f bottom is p bottom times a minus f top is p top times a so p bottom we calculated right there that was the mass density of the water times g times h and then we have times a minus p top which is this term right here that was the mass density of the water times gravity times h minus d and all that times a so if we keep going we have mass density of the water times gravity times h times a minus mass density of water times g times h times a plus mass density of water times g times d times a and you can see that this term and this term cancel out because they're exactly the same so this is going to go to zero and what we're left with is the mass density of the water times gravity times d times a that is going to be equal to the buoyant force so if i scroll down a little bit i want to bring up the concept of mass density so if you remember mass density in general is equal to mass volume and in this case we know that the volume of this space right here is going to be the surface area a times d so we have an area times a height or a depth so the volume is equal to a times d if you look at this equation right here there is our a times d term so i can rewrite this as the mass density of water times gravity times volume now if you look at this term right here the mass density definition that's mass over volume and if i rewrite that i can rewrite that as mass is equal to the mass density times volume right i just multiplied by v on both sides and i get this equation so we can substitute this in for the mass density times the volume in this equation so this is going to be mass of water times gravity and there you go that is why the buoyant force is equal to the weight of the water displaced and the weight of the water displaced is mass times gravity the mass of the water which is all of this mass right here times gravity
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https://ora.ox.ac.uk/objects/uuid:1628c9d4-388f-4144-b6b6-03d354fff71e/files/s5h73pw635
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ISSN 1471-0498
DEPARTMENT OF ECONOMICS DISCUSSION PAPER SERIES
SELECTION EFFECTS WITH HETEROGENEOUS FIRMS Monika Mrazova and J. Peter Neary
Number 588 December 2011
Manor Road Building, Oxford OX1 3UQ SELECTION EFFECTS WITH HETEROGENEOUS FIRMS ∗
Monika Mr´ azov´ a†
University of Surrey and CEP, LSE
J. Peter Neary ‡
University of Oxford and CEPR
December 24, 2011
Abstract
We provide a general characterization of which firms will select alternative ways of serving a market. If and only if firms’ maximum profits are supermodular in production and market-access costs, more efficient firms will select into the activity with lower market-access costs. Our result applies in a range of models and under a variety of assumptions about market structure. We show that supermodularity holds in many cases but not in all. Exceptions include FDI (both horizontal and vertical) when demands are “sub-convex” (i.e., less convex than CES), fixed costs that vary with access mode, and R&D with threshold effects.
Keywords: Foreign Direct Investment (FDI); Heterogeneous Firms; Proximity-Concentration Trade-Off; R&D with Threshold Effects; Super- and Sub-Convexity; Supermodularity.
JEL Classification : F23, F15, F12
∗
We are grateful to Arnaud Costinot for extensive discussions, and to Paola Conconi, Jonathan Dingle, Carsten Eckel, Peter Egger, Gene Grossman, Dermot Leahy, Marc Melitz, Giordano Mion, Lindsay Oldenski, Emanuel Ornelas, Gianmarco Ottaviano, Adrian Wood, Kresimir Zigic, and participants at various seminars and conferences for helpful comments. Monika Mr´ azov´ a wishes to thank the ESRC [grant number PTA-026-27-2479].
†
School of Economics, University of Surrey, Guildford, Surrey GU2 7XH, UK; e-mail: m.mrazova@surrey.ac.uk; URL:
‡
Department of Economics, University of Oxford, Manor Road, Oxford OX1 3UQ, UK; e-mail: pe-ter.neary@economics.ox.ac.uk; URL: 1 Introduction
Why do different firms choose to serve particular markets in different ways? Ten years ago, economists had little theory to guide them in thinking about such questions, though a growing body of empirical work had already documented systematic patterns in firm-level data that were unexplained by traditional theory. In the intervening decade, a new and exciting body of theoretical work has emerged which has placed these empirical findings in context and inspired further extensions and elaborations. The starting point of this recent literature is the explicit recognition that firms differ in one or more underlying attribute, typically identified with their productivity; and its central prediction is that more productive firms select into activities with higher fixed costs but lower variable costs. The locus classicus for this pattern of behavior is Melitz (2003), who showed that more efficient firms select into exporting, whereas less efficient ones serve the home market only. 1 Subsequent work in the same vein has shown that more efficient firms select into many different activities, such as producing in-house rather than outsourcing, as in Antr` as and Helpman (2004); serving foreign markets via foreign direct investment (FDI) rather than exports, as in Helpman, Melitz, and Yeaple (2004); paying higher wages as in Egger and Kreickemeier (2009) and Helpman, Itskhoki, and Redding (2010); and producing with more skill-intensive techniques as in Bustos (2011). Exploring the implications of firm heterogeneity has already had a profound effect on the study of international trade, and is increasingly being extended to other fields, for example by Ghironi and Melitz (2005) to international macroeconomics, by Davies and Eckel (2010) to international tax competition, and by Forslid, Okubo, and Ulltveit-Moe (2011) to environmental economics. This literature on heterogeneous firms prompts a number of observations. First, international trade is not the only field in economics where it has been noted that a firm’s superiority in one dimension may enhance its performance in others. The same
1A related result of this recent literature, also due to Melitz (2003), is a new source of gains from trade: trade liberalization encourages exit by less productive firms and entry by more productive ones, and so, even when the productivity of each individual firm is unchanged, aggregate productivity rises. However, recent work by Arkolakis, Costinot, and Rodr´ ıguez-Clare (2009) suggests that this effect operates in a similar fashion to the gains from trade in traditional models with homogeneous firms.
2idea, though expressed in very different ways, can be found in Milgrom and Roberts (1990), who argued that such a complementarity or “supermodularity” between different aspects of firm performance is typical of modern manufacturing. They also advocated using the mathematical tools of robust comparative statics to examine the responses of such firms to exogenous shocks, especially in contexts where variables may change by discrete amounts. This suggests that it may be worth exploring possible links between these two literatures, and possible payoffs to adapting the tools of robust comparative statics to better understand the behavior of heterogeneous firms. Second, the question immediately arises whether the results derived to date in the literature on heterogeneous firms and trade are robust. One dimension of robustness is that of functional form. All the papers cited above assume that consumers have Dixit-Stiglitz or constant-elasticity-of-substitution (CES) preferences, and all but Melitz (2003) assume that firm productivities follow a Pareto distribution. These assumptions have been relaxed in some papers; for example, Melitz and Ottaviano (2008) show that more efficient firms also select into exports when preferences are quadratic rather than CES. However, with existing techniques each small change in assumptions requires that the model be solved again in full, and, as a result, the question of robustness with respect to functional form has been relatively little explored. A different dimension of robustness is symmetry: existing studies typically assume that countries are identical, both in size and in the distribution of firm productivities. Does this matter for the results? Yet another dimension of robustness is market structure. All the literature on heterogeneous firms to date assumes that the industry is monopolistically competitive, so firms produce a unique product but are infinitesimal in their market. However, if successful firms are indeed large in every dimension, then monopolistic competition may not be an appropriate assumption about market structure. At least in some markets, it may be more plausible to allow for the emergence of a small number of large firms, competing strategically against each other, and possibly coexisting with a “monopolistically competitive fringe” as in Neary (2010). Clearly it would be desirable to know if the selection effects that have been derived assuming monopolistic competition are also likely to hold in oligopolistic 3markets. Finally, turning robustness on its head, we can ask whether the fact that more efficient firms engage in more activities is a universal tendency. Should we always expect more productive firms to engage in more and more complex activities? Or are there interesting counter-examples? In this paper we seek to illuminate these issues both substantively and technically. Our main substantive contribution is a general result on firm selection which fully characterizes the conditions under which what we call the “conventional sorting” pattern occurs: more efficient firms select into activities with lower marginal costs. We first prove this result in a simple though canonical context: that of a single monopoly firm choosing between serving a foreign market by either exports or horizontal FDI. We then show that, with appropriate qualifications, the result extends to a wide variety of market structures, including both monopolistic competition and oligopoly; and to a wide variety of firm decisions, including vertical FDI, in-house production versus outsourcing, and choice of technique. In all cases, the key consideration is how a firm’s own marginal cost of production interacts with the marginal cost of serving the market under different access modes. Our result reveals the unifying structure underlying a wide range of results in the literature, and also shows how they can easily be generalized in new and important ways. From a technical point of view, the paper contributes to the small but growing liter-ature which uses the techniques of monotone comparative statics, and in particular the concept of supermodularity, to illuminate issues in international trade. 2 Supermodularity arises very naturally in our context. Our interest is in comparing firms whose produc-tion costs differ by a finite amount, and in particular in comparing their behavior under different modes of serving a market, whose marginal costs also differ by a finite amount. Supermodularity imposes a natural restriction on the finite “difference-in-differences” of
2For more technical details on the application of monotone comparative statics to economics, see Milgrom and Roberts (1990), Milgrom and Shannon (1994), and Athey (2002). Other applications of supermodularity to international trade include Grossman and Maggi (2000), Costinot (2009), and Costinot and Vogel (2010), who use it to study problems of matching between different types of workers or between workers and sectors; Lim˜ ao (2005), who considers links between trade and environmental agreements; and Costinot (2007), who considers firm selection effects, though in a very different way to the present paper.
4the firm’s profit function which we need to sign in order to make this comparison. As we show, the profit function exhibits supermodularity under a wide range of assumptions, which allows us to generalize existing results and derive news ones with remarkably few restrictions on technology, tastes, or market structure. The plan of the paper is as follows. Sections 2 and 3 focus on a single monopoly firm which faces the decision on how to serve a foreign market, trading off the proximity benefits of foreign direct investment against the concentration advantage of producing at home and exporting. Section 2 introduces the setting and explains the restrictions implied by supermodularity. Section 3 formalizes the gains from tariff-jumping and derives our main result on how firms of different productivities will select into one or other mode of serving the foreign market. The remainder of the paper shows that our approach applies in a wide range of contexts, both old (including some of the most widely-used models in international trade), and new. Sections 4 and 5 look at alternative market structures, considering monopolistic competition and oligopoly respectively. Section 6 turns to ex-plore firm choices other than that between exports and FDI. It reviews a range of other applications, and considers the implications of heterogeneous and endogenous fixed costs of production. The overall message of the paper is that supermodularity holds in many cases but is not inevitable. Among the specific examples we give where supermodular-ity may be violated, and so the conventional assignment of firms to different modes of accessing foreign markets may be reversed, are FDI (both horizontal and vertical) with non-CES preferences, fixed costs that vary with access mode, and R&D with threshold effects.
2 Operating Profits and Supermodularity
In this section we consider a firm located in one country which contemplates serving consumers located in a foreign country. The maximum operating profits the firm can earn in the foreign country equal π (t, c ), where t is the access cost (tariffs and transport costs) it faces and c is an exogenous cost parameter, which we can think of as an inverse 5measure of productivity. The parameter c equals the firm’s marginal production cost in many applications, though not in all: we will see exceptions in Example 2 and Section 6 below. We assume that π is non-increasing (though not necessarily continuous) in both t
and c. Profits also depend on the firm’s choice variables and on other exogenous variables. However, the former have been chosen optimally and so are subsumed into the π function, while the latter are suppressed for convenience; we give some examples of each below. We define ∆ c as the finite difference between the values of a function evaluated at two different values of c, c1 and c2, with the convention that c1 is greater than or equal to c2.Applying this to the operating profit function π gives: ∆cπ (t, c ) ≡ π (t, c 1) − π (t, c 2) when c1 ≥ c2 (1) So, ∆ cπ (t, c ) is the profit loss of a higher-cost relative to a lower-cost firm and is always non-positive. Note that, when π (t, c ) is differentiable in c, ∆cπ(t,c )
c1−c2
reduces to the partial derivative πc as c1 approaches c2.3
We can now define what we mean by supermodularity in the context of our paper: 4
Definition 1. The function π (t, c ) is supermodular in t and c if and only if: ∆cπ (t1, c ) ≥ ∆cπ (t2, c ) when t1 ≥ t2. (2) Intuitively, supermodularity of π means that a higher tariff reduces in absolute value the cost disadvantage of a higher-cost firm. Putting this differently, the profit function exhibits the “Matthew Effect”: “to those who have, more shall be given”. Rewriting the definition we can see that supermodularity is equivalent to:
π (t2, c 2) − π (t1, c 2) ≥ π (t2, c 1) − π (t1, c 1) ≥ 0 when t2 ≤ t1 and c2 ≤ c1 (3)
3We use subscripts of functions to denote partial derivatives: e.g., πc≡∂π/∂c and πtc ≡∂2π/∂t∂c .
4More generally, following Milgrom and Roberts (1990) and Athey (2002), supermodularity can be defined in terms of vector-valued arguments: πis supermodular in a vector-valued argument when
π(x∨y)+ π(x∧y)≥π(x)+ π(y), where x∨y≡inf {z|z≥x, z ≥y}and x∧y≡sup {z|z≤x, z ≤y}.This is equivalent to the definition in the text when we set: x={c2, t 1}and y={c1, t 2}.
6Thus, when supermodularity holds, a lower tariff is of more benefit to a more productive firm. This might seem like the natural outcome, since a lower tariff contributes more to profits the more a firm sells, and we might expect a more productive firm to sell more. As we will see, this is often the case, but there are important counter-examples. When the first inequality in (2) is reversed, we say that the function is submodular. Note that, when π (t, c ) is differentiable in t and c, supermodularity of π implies that the second derivative πtc is positive as t1 approaches t2 and c1 approaches c2. In the dif-ferentiable case, supermodularity is analogous to Hicksian complementarity in consumer theory or strategic complementarity in game theory.
Example 1. A simple case which helps to fix ideas is that of a single-product monopoly firm with constant marginal cost and specific tariffs. Let p (x) denote the inverse demand function which the firm faces, where p and x denote its price and sales respectively. Its operating profits therefore equal:
π (t, c ) ≡ M ax
x
[{p(x) − c − t} x] (4) It is easy to check that the profit function is supermodular in t and c in this case. 5
Intuitively, a firm with higher production costs c has lower sales; hence its profits are reduced less by a rise in the tariff. Example 1 exhibits two key features: π is continuous in trade and production costs, and it depends only on their sum. If both these conditions hold, then supermodularity in
t and c is equivalent to convexity of π in both t and c: if π(t, c ) = π(t + c) and π is differ-entiable, then πtc = πcc . Our next example is a simple case where one of these conditions does not hold and as a result the profit function may not exhibit supermodularity.
Example 2. Consider next the same example as above except that marginal cost varies
5It follows from the envelope theorem that the first derivative of πwith respect to tis minus the initial level of sales: πt=−x(t, c ). Hence the second cross-partial derivative of profits is minus the partial derivative of sales with respect to c:πtc =−xc>0. To establish the sign of this term, differentiate the first-order condition p−c−t+xp ′= 0 to get: xc=−H−1. The expression H≡ − (2 p′+xp ′′ ) must be positive from the firm’s second-order condition. Hence we have that πtc =−xc=H−1>0, and so πin (4) is supermodular in tand c.
7with output. 6 Assume the firm’s problem is as follows:
π (t, c ) ≡ M ax
x
[{p(x) − C (c, x ) − t} x] (5) Here c is not equal to marginal cost, but rather it is a parameter representing the firm’s inverse productivity. The new expression C (c, x ) is the firm’s average variable cost: it depends positively on c and also varies with output x.7 (As we will see, the sign of the first derivative Cx, which indicates whether average variable cost rises or falls with output, is irrelevant to whether supermodularity obtains.) Now the second cross-derivative of the profit function equals the following: 8
πtc = H−1 (Cc + xC xc ) (6) The term H is positive from the second-order condition for profit maximization, and Cc
in the expression in parentheses is positive as already noted, both of which work in favor of supermodularity. However, these terms could be offset, and the profit function could be submodular, if C is submodular in its arguments {x, c } so the term Cxc is negative; that is, if the cost of production falls faster (or rises more slowly) with output for a firm with higher c (i.e., a less productive firm). Figure 1 illustrates this possibility. Firm 1 is less productive than firm 2 overall, but it is relatively more productive at higher levels of output. As a result its marginal cost curve M C 1 lies below that of firm 2 and so it has
lower marginal cost and (facing the same marginal revenue curve) higher output. The profit function in this case is therefore submodular rather than supermodular. The configuration shown in Figure 1, though not pathological, is somewhat contrived and of limited empirical relevance. In general, supermodularity will hold as long as static differences in efficiency between firms work in the same direction on average and at the margin, which seems the natural case. In later sections we will consider more plausible
6We are grateful to Dermot Leahy for suggesting this example.
7Total variable cost equals xC (c, x ), so marginal cost is C+xC x, which also varies with output.
8As in Example 1, the envelope theorem holds, πt=−x(t, c ), and so πis supermodular in tand
cif and only if xis decreasing in c:πtc =−xc.Direct calculation yields equation (6) where H≡−[2 p′+xp ′′ −(2 Cx+xC xx )] >0.
8pC(c1,x)
C(c2,x)MC 1
MC 2
MR
x
MR
x1x2
Figure 1: An Example of Submodularity examples of submodularity.
3 Selection into FDI versus Exporting
We return to the general case where π (t, c ) is unrestricted, and compare the relative profitability of different modes of serving the foreign market. We first restate in our notation the familiar proximity-concentration trade-off, and then derive a general result on which firms will select into exporting and FDI. 9
Exporting faces a higher access cost, so FDI has the advantage of proximity. However, it foregoes the benefits of concentration. In addition to operating profits, the firm must incur a fixed cost of serving the market, which differs depending on the mode of access. The fixed cost equals fX if the firm exports and fF if it engages in FDI and builds a plant in the market in question. We assume that fF is strictly greater than fX and that access costs conditional on FDI are zero. Total profits as a result of exporting are: ΠX = π (t, c ) − fX (7)
9Our formalization of the proximity-concentration trade-off follows Neary (2002).
9while total profits from locating a plant in the target market are: ΠF = π (0 , c ) − fF (8) We define the tariff-jumping gain γ as the difference between these two: 10
γ(t, c, f ) ≡ π(0 , c ) − π(t, c ) − f = Π F − ΠX (9) Here f ≡ fF − fX is the excess fixed cost of FDI relative to exporting, which by our earlier assumption is strictly positive. We can now apply the finite difference operator ∆c to the tariff-jumping gain: 11
∆cγ(t, c, f ) = ∆ cπ(0 , c ) − ∆cπ(t, c ) (10) Recalling the definition of supermodularity, we can sign this unambiguously, which gives our first result:
Lemma 1. If and only if the profit function π is supermodular in t and c, ∆cγ(t, c, f ) is negative.
The economic implications of this are immediate: if and only if π is supermodular in t and
c, the tariff-jumping gain is lower for higher-cost firms and higher for more productive ones. Since γ measures the incentive to engage in FDI relative to exporting, we can go further and state the central result of our paper:
Proposition 1. Assume that some firms serve the foreign market by exports and the remainder by FDI. Then, if and only if the profit function π is supermodular in t and c,higher-cost firms will select into exports, while lower-cost firms will select into FDI, for all admissible fixed costs f .
The sufficiency part of the proposition follows immediately from Lemma 1. The necessity
10 Strictly γis a trade-cost-jumping gain , but the shorter title is traditional and simpler.
11 To avoid confusion, we include famong the arguments of ∆ cγ(t, c, f ). However, this finite difference is independent of f, a point which will become crucial below.
10 part is more subtle and reflects the fact that we require the result to hold for all admissible fixed costs. A formal proof is in the Appendix. Here we give an intuitive account. F
XX
X
gFDI
fX
1
c
fF
1
c
f
f
Figure 2: The Conventional Sorting can Hold without Supermodularity The upper quadrant of Figure 2, based on Helpman, Melitz, and Yeaple (2004), gives a hypothetical illustration of total profits under the two modes of market access, as functions of inverse production costs. 12 The lower quadrant shows γ(t, c, f ), the difference between Π F and Π X . In the example shown, this curve is non-monotonic in c, and so the profit function is not supermodular in t and c. Despite this, the conventional sorting holds: there is a unique cost threshold, with all firms that have lower costs engaging in FDI and all those that have higher costs engaging in exports. This shows that supermodularity is not necessary for the conventional sorting to hold for arbitrary fixed costs: the necessary condition, that the two profit curves cross only once, is weaker. However, Figure 3 shows that supermodularity is necessary if the conventional sorting is to hold for all fixed costs. If γ is not monotonic in c, there always exists some level of f which leads to a violation of the conventional sorting. In the case shown in Figure 3, the level of fF , the fixed cost of FDI, is lower than in Figure 2, shifting the γ(t, c, f ) curve upwards such that the
12 Helpman, Melitz, and Yeaple (2004) assume CES preferences, in which case the total profit curves are linear in a decreasing transformation of costs; see Section 4 below for further discussion. More generally, the total profit curves must satisfy only two restrictions: (i) both must be upward-sloping, reflecting the assumption that operating profits are non-increasing in c; and (ii) Π Fmust lie everywhere above Π X
when fF=fX, reflecting the assumption that operating profits are non-increasing in t.
11 conventional sorting no longer holds: both the lowest- and the highest-cost active firms engage in FDI. Hence we can conclude that supermodularity of the profit function in t
and c is necessary as well as sufficient for higher-cost firms to select into exports, and lower-cost firms to select into FDI, for all admissible fixed costs f . F
XX
g
FDI XFDI
f
fX
1
c
fF
1
c
f
Figure 3: Absent Supermodularity, the Conventional Sorting Fails for Some Fixed Costs A striking feature of Proposition 1 is that it does not depend directly on fixed costs. While fixed costs affect the level of the tariff-jumping gain γ, they vanish when we compare across two firms using the finite difference operator ∆ c. Fixed costs are essential for a proximity-concentration trade-off, and hence they are necessary for the existence
of selection effects. However, they do not determine their direction . So statements like “Only the more productive firms select into the higher fixed-cost activity” are often true, but always misleading: they are true given supermodularity, but otherwise not. 13 What matters for the direction of selection effects is not a trade-off between fixed and variable costs, but whether there is a complementarity between variable costs of production and of trade. Putting this differently, for FDI to be the preferred mode of market access, a firm must be able to afford the additional fixed costs of FDI, but whether it can afford them or not depends on the cross-effect on profits of tariffs and production costs. When supermodularity prevails, a more efficient firm has relatively higher operating profits in
13 The quoted statement is from Oxford graduate trade lecture notes in late 2009.
12 the FDI case, but when submodularity holds, the opposite is true. Of course, all this assumes that fixed costs are truly fixed, both for a single firm as output varies, and for cross-section comparisons between firms. Matters are different if they depend on either t or c, as we shall see in Section 6. First, we turn to compare Proposition 1 with the result obtained by Helpman, Melitz, and Yeaple (2004). Our result is more general than theirs in that it places no restrictions on the functional form of the demand function: only mild restrictions on the maximized profit function π(t, c )are needed. However, at first sight our result seems more special since it holds only for the case of a single monopoly firm, whereas Helpman, Melitz, and Yeaple (2004) proved their result in a general equilibrium model with monopolistic competition. In the next section we show that this apparent limitation of our result is illusory. With suitable reinterpretation, our result holds in a large class of monopolistically competitive models, including that of Helpman, Melitz, and Yeaple (2004).
4 Monopolistic Competition
To see how our result extends to models of monopolistic competition, we need to address the issue of market structure per se, to explore demand systems other than the CES, and to examine the specification of transport costs. We consider these issues in turn.
4.1 Exports versus FDI with CES Preferences
As already noted, Helpman, Melitz, and Yeaple (2004) were the first to consider how firms of different costs will select into different modes of serving foreign markets. They considered this issue in a model of heterogeneous firms in monopolistic competition with CES or Dixit-Stiglitz preferences, iceberg transport costs, and a Pareto distribution of firm productivities. In our notation the variable-profit function for a typical firm in such a model is:
π (t, c ) = ( τ c )1−σ B (11) 13 where τ ≡ 1 + t ≥ 1 is an iceberg transport cost; 14 σ is the elasticity of substitution in demand, which must be greater than one; and B is a catch-all term which summarizes the dependence of the demand for one firm’s good on total expenditure and the prices of all other goods. 15 Consider first the partial equilibrium or firm-level case where B is taken as given. In that case, it is clear that the profit function is supermodular in t and
c:
πtc = ( σ − 1) 2 (τ c )−σ B > 0 (12) Hence, from Proposition 1, the ranking of firms by their mode of serving foreign markets established by Helpman, Melitz, and Yeaple (2004) follows immediately without any need to compare the levels of profits in different modes. In full industry equilibrium, the demand term B is endogenous. It depends directly on the level of total expenditure E and on the overall price index P in the market in question, while P in turn depends on all the variables that affect the global equilibrium, including at a minimum the number of active firms serving this market from every country i, the distribution of firm costs g (c), and the transport cost τ :
B = ˜B (E, P ) P = ˜P [{ni} , g (c) , τ ] (13) However, for the comparisons we wish to make, this endogeneity is not relevant. The price index and hence the demand term B would be affected by changes in transport costs which disturb the full equilibrium. But our concern is rather with characterizing the pattern of firm selection between different modes of serving foreign markets which face differences in transport costs. Since any pair of firms is infinitesimal relative to the mass of all firms, we can compare their choices while holding constant the actions of all other firms. Hence, partial equilibrium is the appropriate framework for the cross-section comparisons between different firms in the same equilibrium that we want to make. This key point can be made differently by considering Figure 4, which is based on
14 For continuity with previous sections we continue to write ex post profits as a function of t. This is not a restriction since ∂π/∂τ =πt.
15 In typical specifications, B= ( σ−1) σ−1A/σ σ, where Ais the constant term in the demand function
x=Ap −σ.Ain turn depends on nominal expenditure Eand the aggregate price index P:A=EP σ.
14 Π
ΠF
ΠX
F
F
X
X
fX
fF
c1−σ
1c1−σ
2
c1−σ
Firm 2 Firm 1
Figure 4: Inferring Selection Effects from Supermodularity Helpman, Melitz, and Yeaple (2004). Their approach, now standard in the literature, is to compute the general equilibrium of the world economy and then to investigate what pattern of selection effects it exhibits. Thus they calculate not only the profit functions ΠF and Π X , allowing for their dependence on expenditure and price indices in general equilibrium, but also their point of intersection, which is the threshold cost level at which a firm is indifferent between exports and FDI. By contrast, our approach is very different. We assume that an equilibrium exists, and that π is supermodular. We can then pick an arbitrary pair of firms, say those with the unit costs c1 and c2 in Figure 4. Rewriting the supermodularity condition ∆ cπ(t, c ) > ∆cπ(0 , c ), and adding −fF +fX to both sides gives a ranking of the two firms’ total profits when they engage in FDI rather than exporting:
π(t, c 1) − π(t, c 2) > π (0 , c 1) − π(0 , c 2) (14)
⇔ π(0 , c 2) − π(t, c 2) > π (0 , c 1) − π(t, c 1)
⇔ ΠF (c2) − ΠX (c2) > ΠF (c1) − ΠX (c1)Repeating this comparison for every pair of firms allows us to infer the qualitative prop-erties of the Π F and Π X loci without the need to calculate the full equilibrium. Our approach parallels that of Maskin and Roberts (2008), who show that all the central the-15 orems of normative general equilibrium theory can be proved using elementary methods once the existence of equilibrium has been established. Our approach cannot confirm that an equilibrium exists. However, by dispensing with computing one explicitly, it applies without specific restrictions on the functional forms of preferences, technology, or the distribution of costs; it avoids the need to assume that countries are symmetric; and it extends easily to considering firm choices other than that between exports and FDI, as we shall see in Section 6.
4.2 General Preferences
While the result in the last sub-section has already been derived by Helpman, Melitz, and Yeaple (2004), the strength of our approach is that it allows us to sign selection effects into FDI for a much wider class of preferences than the CES. Write the demand function perceived by the firm in inverse form, p = p(x), with no restrictions other than that consumers’ willingness to pay is decreasing in price, p′ < 0; and write the elasticity of demand as a function of sales: ε (x) ≡ − ∂x ∂p px = − pxp ′ . To determine which specifi-cations of demand favor the conventional sorting, we introduce the term “superconvex” demand: we define a superconvex demand function as one for which ln p is convex or more so in ln x.16 As we show in the Appendix, this is equivalent to the demand func-tion being equally or more convex than a constant-elasticity CES demand function (for which ε equals σ), and to one whose elasticity of demand is increasing in output, so εx
is non-negative. The case where demand is not superconvex, so ε is decreasing in x, we call subconvex, and, as noted by Krugman (1979), it may be considered the “normal” case: it implies plausibly that consumers are more responsive to price changes the greater their consumption; and it encompasses many of the most widely-used non-CES specifi-cations of preferences, including the quadratic (to be considered further below) and the
16 For a formal definition, and proofs of the statements that follow, see the Appendix, Section 8.2. The term “superconvexity” seems to be used, if at all, as a synonym for logconvexity, i.e., ln xconvex in p.(See Kingman (1961).) For related discussions, see Neary (2009), Zhelobodko, Kokovin, Parenti, and Thisse (2011), and Dhingra and Morrow (2011). All these papers assume that preferences are additively separable, though this is not necessary for our approach, since we only consider the demand function from the firm’s perspective.
16 translog. 17 Strict superconvexity is less widely encountered; an example is where the inverse demand function has a constant elasticity relative to a displaced or “translated” level of consumption: p = ( x − β)−1/σ with β strictly positive. 18
The importance of superconvexity in this context is shown by the following result:
Proposition 2. With general demands and iceberg transport costs, a sufficient condition for the profit function to be supermodular in t and c for all levels of output is that the demand function is weakly superconvex, i.e., the elasticity of demand is non-decreasing in output, εx ≥ 0.
The proof, given in the Appendix, follows by expressing the cross-partial derivative of the profit function in terms of the elasticity of demand and its responsiveness to output:
πtc = (ε − 1) 2 + xε x
ε − 1 − xε x
x (15) The denominator ε − 1 − xε x must be positive from the second-order condition. Hence the numerator shows that, whenever εx is strictly negative, submodularity may hold for sufficiently high x. We can be sure that supermodularity holds for all output levels only in the CES and strictly superconvex cases. 19
Intuitively, the result follows from another implication of superconvexity. A positive value of εx means that larger firms face a higher elasticity of demand. Since output is decreasing in c in this model ( xc < 0), this implies that, if and only if εx is positive, more productive firms face more elastic demand. Hence, they also have lower mark-ups, as measured by the Lerner Index, L ≡ p−τ c p , since L = 1
ε
.20 This implies that a
17
Feenstra (2003) shows how the translog can be adapted to allow for a variable number of varieties consumed, and so used in models of monopolistic competition with free entry. He also shows that it implies an elasticity of demand which is always increasing in price, and so, from Lemma 7 in the Appendix, is subconvex. (See Feenstra (2003), p. 85).
18
We are grateful to Rob Feenstra for suggesting this example.
19
In the CES case, when εx is zero, (15) reduces to πtc = ( σ − 1) x and is always positive. This is equivalent to equation (12), using the fact that output with CES preferences equals: x =
( σσ−1 τ c
)−σ
.
20
It can be checked that the Lerner index falls as costs fall if and only if εx is positive: dL dc = − εx
ε2
dx dc
.This result does not hold for the absolute mark-up, μ ≡ p − τ c = pε , which can be either increasing or decreasing in production cost. For example, it is increasing in c in the CES case: μ = τ c σ−1 , and so
dμ dc
= τσ−1 > 0; but decreasing in c in the quadratic case to be discussed further below: μ = A−τ c
2
, and so dμ dc = − τ
2
< 0.
17 more productive firm will have an incentive to expand output more in order to maximize profits. As a result, the Matthew Effect is stronger when εx is positive, sufficiently so that supermodularity is guaranteed. By contrast, when εx is negative, the Matthew Effect is weaker and so more productive firms may not benefit as much from avoiding the tariff by engaging in FDI. Proposition 2 is important for highlighting which classes of demand function are con-sistent with super- or submodularity, but it is only a sufficient condition. To determine whether a particular demand function exhibits supermodularity, we can use the necessary and sufficient condition given by the following:
Proposition 3. With general demands and iceberg transport costs, a necessary and suf-ficient condition for the profit function to be supermodular in t and c is that the sum of the elasticity and convexity of demand is greater than three.
The proof (given in the Appendix) proceeds by showing that equation (15) can be reex-pressed as follows:
πtc = ε + ρ − 32 − ρ , ρ ≡ − xp ′′
p′ (16) where ρ is our measure of convexity of demand. So, submodularity is more likely when demand is less elastic and more concave. In particular, it may arise for any linear or concave demand system, and even for demands that are “not too” convex. To illustrate how these results can be applied in practice, we consider two subconvex demand systems, one of which always implies that profits are supermodular and the other which implies submodularity for high values of output. The first is a version of the translated CES case already mentioned:
Lemma 2. The demand function p = ( x − β)−1/σ , with β strictly negative, is always subconvex, but the implied profit function is supermodular at all levels of output provided
σ ≥ 1.25 .
In this case, higher sales are associated with a lower demand elasticity and thus a higher markup, implying that more productive firms do not exhibit such a large difference in 18 output. Nevertheless, the elasticity of demand never falls sufficiently low to allow sub-modularity to emerge. 21
Our second example is the case of quadratic preferences, which have been studied in the context of heterogeneous firms by Melitz and Ottaviano (2008). Their model has been extended to the choice between exports and FDI by Nefussi (2006), but only by solving for the full general equilibrium. Using our approach it is easy to establish its properties and to show that its predictions for firm selection into FDI are ambiguous. Writing the inverse demand function as p = A − bx , we can express the elasticity of demand as a function of output: 22
ε(x) = A − bx bx (17) This is monotonically decreasing in x, so from Proposition 2 we know that for high values of output the profit function may be submodular. However, we need to check that this will happen for values of x that are admissible in equilibrium. To confirm this we specialize Proposition 2 to the quadratic case:
Lemma 3. With quadratic preferences and iceberg transport costs, the profit function π
is supermodular when defined on the interval c ∈ [ A
2τ
, ∞] and submodular when defined on the interval c ∈ [0, A
2τ
].
(The proof is in the Appendix.) Unlike the CES case, the profit function is therefore
submodular for low-cost exporters, although it continues to be supermodular for high-cost ones. Hence, provided both exporting and FDI are profitable in the relevant range, we can expect a threefold selection effect in this model: the very highest-cost firms will select into exporting, but so will the very lowest-cost ones, while intermediate-cost firms will select into FDI. 23 Figure 5 illustrates this configuration.
21 As shown in the Appendix, Section 8.5, ε=x−βxσand ρ=xx−βσ+1
σ. Though the demand function is subconvex for β < 0, it is always strictly convex. It follows that ε+ρ−3 = ε+σ+1
ε−3, which can be negative only for very low σ.
22 As always in monopolistic competition, the demand parameters Aand bare taken as given by firms, but are endogenous in general equilibrium. For example, in the Melitz-Ottaviano framework,
A≡γα +ηN ¯pγ+ηN and b≡γL, where α,γand ηare demand parameters, Lis market size, Nis the mass of firms, and ¯ pis the aggregate price index.
23 This case holds provided a number of boundary conditions are met: (i) exporting must be profitable, ΠX≡π(t, c )−fX>0, which requires: c < 1
τ
(A−2√bf X
); (ii) FDI must be profitable, Π F≡
19 X
FFigure 5: Selection Effects with Quadratic Preferences and Iceberg Transport Costs As already noted, the corollary given here generalizes the result of Nefussi (2006), dispensing with the assumptions of symmetric countries and a Pareto distribution of firm productivities which he makes. Our result also extends easily to explain the pattern of firm selection into exporting, export-platform FDI, and multi-market FDI when there is more than one foreign market. This is particularly convenient because, as Behrens, Mion, and Ottaviano (2010) show, it does not seem to be possible to compare two different FDI equilibria analytically when preferences are quadratic. The problem arises from the fact that all variables in any given equilibrium can be written as functions of the cost cutoff (the threshold level of marginal cost above which a firm finds it unprofitable to produce). 24
However, comparing two different cutoffs is extremely difficult. Our approach makes it unnecessary to do so: we assume that an equilibrium exists in which firms select into different modes of serving the market, and can then invoke our result on supermodularity
π(0 , c )−fF>0, which requires: c < A −2√bf F; and (iii) some selection must take place, i.e., the quadratic equation in cdefined by Π X= Π Fmust have two real roots, which requires: ( τ−1) A2>
4( τ+1) b(fF−fX). Note that we allow for a non-zero fixed cost of exporting, unlike Melitz and Ottaviano (2008). To solve their model in full, they have to assume that exports do not incur any fixed costs, in which case the demand parameter Aequals the marginal cost of the threshold firm in equilibrium. Our approach can accommodate fixed costs of exporting, so this property does not necessarily hold here.
24 Though this is only possible if there is no fixed cost of exporting, an assumption which our approach does not require, as discussed in the previous footnote.
20 to justify which mode is relatively more profitable for any pair of firms, and, by extension, for all firms.
4.3 General Transport Costs
The result in the previous sub-section that the largest firms select into exporting for a wide class of demand functions is not necessarily paradoxical. It may simply be viewed as yet another example of large firms’ “supermodular superiority”. 25 To the extent that the most efficient firms are more productive in all the activities in which they engage, then it is reasonable to assume that they also incur the lowest per unit transport costs. Perhaps they are able to avail of economies of scale in transportation, or to negotiate better terms with transport contractors. From that perspective, the assumption of iceberg transport costs can be seen as a convenient reduced-form way of modeling this superiority of more efficient firms. On the other hand, the suspicion remains that this result is an artifact of iceberg transport costs. It is stretching credulity to assume that the most efficient firms produce the cheapest icebergs, and, in particular, that highly efficient firms, with production costs close to zero, also incur negligible transport costs irrespective of distance. But this is what is implied by the iceberg assumption: to sell q units it is necessary to produce and ship τ q units, so the technology of transportation is identical to that of production: ( p − τ c )x = px − c(τ x ). To see how alternative specifications of transport costs affect the outcome, consider a general specification of the ex post profit function as the outcome of choosing output x to maximize ˜ π (x; τ, c ), the firm’s operating profits as a function of the exogenous variables
τ and c and an arbitrary level of output:
π (t, c ) ≡ M ax
x
˜π (x; τ, c ) (18) We can now express the desired cross-partial derivative of π in terms of second derivatives
25 We are grateful to Adrian Wood for suggesting this line of reasoning.
21 of ˜ π:26
πtc = ˜ πτ c + ˜ πτ x
dx dc = ˜ πτ c − ˜πτ x (˜ πxx )−1 ˜πxc (19) This shows that supermodularity of the profit function in t and c depends on the balance between two effects: a direct effect given by ˜ πτ c , which is the effect of a difference in production costs on the profit disadvantage of higher transport costs at a given level of output; and an indirect effect given by the second term on the right-hand side. The expression ˜ πxx is negative from the firm’s second-order condition, so the sign of the indirect effect depends on the product ˜ πτ x ˜πxc . This is presumptively positive; for example it must be so in the case of constant production costs and iceberg transport costs, when ˜πτ x = −c and ˜ πxc = −τ . This is the Matthew Effect from Section 2: it arises because a higher-cost firm is less vulnerable to a rise in transport costs since it has presumptively lower sales: both ˜ πτ x and dx dc are negative, so their product is positive. By contrast, the direct effect is less robust. In the case of iceberg transport costs it simply equals ˜πτ c = −x and is clearly the source of the potential for submodularity identified in the previous sub-section. It reflects the fact that a higher-cost firm loses more from a rise in transport costs (˜ πτ is more negative) since its cost of shipping one unit of exports is (τ − 1) c.It is immediate that the direct effect vanishes if transport costs and production costs are separable in the profit function ˜ π. This corresponds to the case where exports do not melt in transit, but trade costs are levied instead on the value of sales: ˜π (x; τ, c ) = R (x, τ ) − cx (20) Here net sales revenue accruing to the firm, R, depends in a very general way on the transport cost parameter. However, there is no interaction between transport costs and production costs. As a result, there is no direct effect in the supermodularity expression given by (19): total transport costs and hence ˜ πτ do not depend directly on c, implying that the direct effect ˜ πtc is zero. By contrast, the indirect effect is positive as before.
26 To derive this we use the envelope theorem to set πt= ˜ πτ, and totally differentiate the first-order condition ˜ πx= 0 to obtain dx dc =−(˜ πxx )−1˜πxc .
22 Hence, profits are supermodular in t and c for all levels of output and all specifications of demand when transport costs and production costs are separable in this way. Specific transport costs, already considered in Example 1 in Section 2 above, provide one example of (20). Another is where transport costs are ad valorem or proportional to price, so net sales revenue becomes: R(x, τ ) = xp (x)
τ
. Relative to the case of iceberg transport costs, the firm’s first-order condition is unchanged, but profits are deflated by τ :˜π (x; τ, c ) =
[p(x)
τ
− c
]
x. Similar derivations to those already given shows that equation (19) now becomes: πtc = −c(2 p′ + xp ′′ )−1 > 0. Thus the full effect is unambiguously positive for all demand systems, and so the profit function is always supermodular. Figure 6 illustrates the case of quadratic preferences and proportional transport costs. Clearly, the conventional sorting is now restored, and the model predicts that the most efficient firms will always engage in FDI rather than exporting. F
X
Figure 6: Selection Effects with Quadratic Preferences and Ad Valorem Transport Costs
5 Selection Effects in Oligopoly
The previous section, like almost all the recent literature on trade with heterogeneous firms, assumed that markets are monopolistically competitive. Rare exceptions to this generalization include Porter (2011), who shows that the more efficient firm in a duopoly 23 is more likely to engage in FDI than exporting, and Leahy and Montagna (2009) who show a similar result for outsourcing. It is desirable to establish whether similar results hold more generally when firms are large enough to exert market power over their rivals, so markets are oligopolistic. As already noted, this is of interest both as a check on the robustness of the results and also because, to the extent that more successful firms are likely to engage in a wider range of activities, the assumption that they remain atomistic relative to their smaller competitors becomes harder to sustain. If individual firms are no longer of measure zero then the arguments used in Section 4.1 no longer hold. If we wish to compare a firm’s profits under exporting and FDI, we can no longer assume that the industry equilibrium is unaffected by its choice. However, our earlier result still holds when we take behavior by rival firms as given. To illustrate with a simple example, consider the case where there are two rival U.S. firms, labeled “1” and “2”, both of which consider the choice between exporting to the EU and locating a foreign affiliate there. The payoffs to firm 1, conditional on different choices of firm 2, are given in Table 1. Thus, the first entry in the first row, π(t, c, X ) − fX gives the operating profits which it will earn if it exports to the foreign market, conditional on the rival firm 2 also exporting. We would expect this to be always less than the second entry,
π(t, c, F )−fX , which is conditional on firm 2 engaging in FDI: better market access by the rival presumably reduces firm 1’s profits, ceteris paribus . However, what matters for firm 1’s choice is the comparison between different entries in the same column, and it is clear that, conditional on a given mode of market access by firm 2, firm 1’s choice will reflect exactly the same considerations as in previous sections. Hence, provided supermodularity holds in each column, and in the columns of the corresponding table for firm 2, our earlier result goes through: when that is the case, more efficient firms will select into FDI and less efficient ones into exporting. While the central result derived earlier still holds, it has to be applied with care. One issue is that boundary cases have to be considered in detail. Depending on the configuration of the two firms’ costs, in the Nash equilibrium only one of them may serve the market at all, or do so via FDI. There may be no equilibria in pure strategies, in 24 Choice of Firm 2: Export FDI Export: π(t, c, X ) − fX π(t, c, F ) − fX
FDI: π(0 , c, X ) − fF π(0 , c, F ) − fF
Table 1: Payoffs to Firm 1 Given Choices of Firm 2 which case mixed strategy equilibria have to be considered. Finally, the necessity part of Proposition 1 does not survive. This is because, even when we allow for all values of fixed costs as in the proof of Proposition 1, supermodularity of the profit function conditional on rivals’ responses is necessary for the conventional sorting only at those points which are relevant to a particular Nash equilibrium. Thus it is conceivable that supermodularity might not hold over a range of the profit function; but if that range was never relevant for any value of fixed costs, then the conventional sorting would still apply.
6 Other Applications
6.1 Alternative Firm Choices
So far, our focus has been on the choice between exports and FDI. However, the same approach applies to a wide range of other firm choices. For example, we show in Mr´ azov´ aand Neary (2010) that similar considerations can be used to predict whether firms will select into export-platform or multi-plant FDI, when they are faced with the problem of how to serve a number of foreign markets. We also show below that our approach extends to the choice between in-house production and outsourcing, as in Antr` as and Helpman (2004), and to the choice between more and less skill-intensive technology, as in Bustos (2011). In these and many other cases, results analogous to those derived above apply: supermodularity between the firm’s own cost parameter and a parameter representing the marginal cost of the mode of accessing a market is necessary and sufficient for the standard selection effect, whereby more productive firms select into the access mode with lower marginal cost. 25 One key exception to our framework is the “zero-one” choice between serving a market or not, and a central example of this case is the decision on whether to engage in exporting or not, as originally considered by Melitz (2003). Selection effects in that case depend only on the level of total profits attainable by exporting, Π X ≡ π(t, c ) − fX , rather than on the difference between the profits attainable via exports, Π X , and those attainable via FDI, Π F . Selection effects between firms with different productivities thus depend in the Melitz case on the first derivative of the variable profit function rather than on the more subtle second cross-derivative which we have emphasized: π decreasing in c ensures the conventional sorting whereby there is a single threshold level of costs such that all firms with lower costs will export and all those with higher costs will serve the home market only. It is perhaps appropriate that, unlike many of its applications and extensions, the model of Melitz makes a much more robust prediction about selection effects, one which does not depend on the properties of the second derivative of the profit function. In the remainder of this section, we present some applications of our approach. We first look at the issues of both choice of location and choice of organizational form which arise when firms can vertically disintegrate. We then turn to show how our approach extends to the case where fixed costs differ between locations and between firms. Finally, we consider how selection effects can also be inferred in models where fixed costs are endogenous, determined by prior investments in variables such as technology, research and development (R&D), or marketing.
6.2 Vertical Disintegration and Choice of Organizational Form
Our discussion of FDI in previous sections concentrated on the horizontal kind, where the firm is considering how to serve a foreign market, and FDI involves effectively reproducing abroad the production facilities which are already located in the home country. This archetypal problem is also one in which differences between the two countries are not central: in particular, we assumed for convenience that the marginal cost of production was the same whether the firm engaged in exporting or in FDI. A different problem arises in the case of a firm whose goal is to serve its home market, but which faces two distinct 26 choices about its organizational form. On the one hand, it has the option of producing either at home or in a lower-cost location abroad. On the other hand, it can choose either to produce in-house or to outsource: the choice of whether or not to vertically integrate arises irrespective of where production is located. The classic treatment of this issue in a model with heterogeneous firms is by Antr` as and Helpman (2004), and we draw on their work in what follows. However, since our specification is a reduced-form one, it is also consistent with many other ways of modeling the choice of organizational form. Ignoring fixed costs for the present, the choice of production location and organiza-tional form will depend on the total operating profits that are realized in each case. We write this as:
π (w, ψ, c ) ≡ M ax
x
(1 − ψ) [ p(x) − wc ] x (21) Unlike in previous sections, we abstract from transport costs. In other respects the model is more complicated. First, the wage which the firm must pay differs between locations. If production takes place in the home country, which we will henceforth refer to as “North”, the wage equals wN , while if it takes place in “South” it is lower and equal to wS , with wS < w N . Second, the firm owner must use the services of the supplier of an intermediate input, the quality of which, though observable to both parties, is not contractible. 27 This leads to a profit loss due to incomplete contracting between the firm owner and the intermediate-input supplier, represented by the parameter ψ. Structural microfoundations for this parameter are provided in Antr` as and Helpman (2004). 28 Here we need only assume that it may differ both between location and organization form; in particular, we assume that, irrespective of location, it is lower when the firm vertically integrates than when it outsources and must contract with an outside supplier: ψVj < ψ Oj ,
j = N, S .29
27 Either the costs of writing a comprehensive contract are infinite, or the outcomes cannot be observed by a third-party arbitrator. In either case a complete contract cannot be enforced.
28 They assume that the input supplier’s outside option is zero, so it must be paid a fixed amount to persuade it to participate. The headquarters maximizes (1 −ψ)π−T−fwhere Tis the fixed amount that it must pay the supplier. This is equivalent to maximizing (1 −ψ)π−f.
29 Our reduced-form specification covers both the case where internalization eliminates all costs of incomplete contracting, as in Williamson (1975) and Grossman and Helpman (2002), and the case where even vertical integration in the North incurs some cost, as in Grossman and Hart (1986), Hart and Moore (1990), and Antr` as and Helpman (2004). Our approach can easily be extended to allow the efficiency
27 Considering only variable costs, the firm owner has an incentive to locate production in the South, and to produce in-house rather than outsource. Offsetting these differences in variable costs, the fixed costs of locating in the South are higher then in the North, irrespective of the choice of organizational form: f iS > f iN , i = O, V ; and the fixed costs of vertical integration or in-house production are higher then those of outsourcing, irrespective of the choice of location: f Vj > f Oj , j = N, S . This configuration of fixed and variable costs ensures that there is a trade-off between different modes of organization, and opens up the possibility that some selection by firms will take place. However, just as in previous sections, it does not predict the direction of selection effects, except in the special case of CES preferences considered by Antr` as and Helpman (2004).
Firm’s Decision Profits εx≥0εx<0HFDI with iceberg transport costs: π(t, c ) = ( p−τ c )xYes No HFDI with proportional transport costs: π(t, c ) = ( p/τ −c)xYes Yes Produce in North or South: π(w, c ) = ( p−wc )xYes No Produce in-house or outsource: π(ψ, c ) = (1 −ψ)( p−wc )xYes Yes
Table 2: Is the Profit Function Supermodular at all Levels of Output? To see this, we consider in turn the two choices which the firm must make. Consider first the simplest form of the choice of location, where we assume that there is no efficiency cost of incomplete contracting, so ψ in (21) equals zero. This corresponds to the choice between outsourcing to a Northern contractor and offshoring to a Southern contractor in Antr` as and Helpman (2004). Comparing the first and third row of Table 2, it is clear that the profit function in this case has exactly the same form as the profit function with iceberg transport costs in the horizontal FDI case, with the wage w playing the role of the iceberg cost parameter τ . Hence the results of previous sections apply immediately. From Proposition 1, if selection takes place, then more efficient firms will offshore and less efficient ones will outsource at home if and only if the profit function is supermodular in
w and c; while from Proposition 2 we can be sure that the profit function is supermodular
cost of incomplete contracting to depend on either firm productivity or wages or both: just replace ψin (21) by Ψ ( ψ, c, w ), where ψnow represents a structural parameter that determines the cost of incomplete contracting for given values of cand w.
28 for all output levels if and only if the elasticity of demand is constant or increasing in sales. Thus, without the need for any further analysis, we can conclude that more efficient firms select into outsourcing at home when preferences are CES, but not necessarily otherwise. For example, the sorting predicted by Antr` as and Helpman (2004) need not hold under quadratic or translog preferences, even with all of their other assumptions retained: very efficient firms are likely to choose to outsource at home rather than to offshore, since they are less affected by the higher wage in the North. On the other hand, with CES preferences, their results go through irrespective of how firm productivities are distributed or whether countries are symmetric or not, provided only that an equilibrium exists. 30
Consider next the simplest type of choice of organizational form, where we hold lo-cation constant and hence the wage is given. From the fourth row of Table 2, selection effects now depend on whether the profit function is supermodular in ψ and c. It turns out that this is always the case. The direct effect of a difference in costs between firms is straightforward: the higher-cost firm has lower profits, and as a result is less affected by a higher profit leakage due to incomplete contracts. In fact, the direct effect is simply ˜πψc = x. As for the indirect effect, it equals zero: although the higher-cost firm has lower sales, it is neither more nor less vulnerable to a rise in the profit leakage parameter
ψ, since this is analogous to a uniform profits tax on all firms: ˜ πψx = −˜πx = 0. Thus, adapting equation (19) to this case, we have:
πψc = ˜ πψc − ˜πψx (˜ πxx )−1 ˜πxc = x > 0 (22) This implies that the selection effect for outsourcing is very robust: more efficient firms select into the lower-ψ organizational form (i.e., producing in-house rather than outsourc-ing) for all preferences and productivity distributions. Finally, selection effects into vertical FDI are ambiguous in general. Relative to pro-
30 Note a further generalization which our approach makes possible. Antras and Helpman (2004) assume that fixed costs are always incurred in the North, so Π = π−wNf. Suppose instead that fixed costs are incurred in the country where production is located. In that case we can redefine variable profits as π(w, c ) = ( p−wc )x−wf ,w=wN, w S, and we can see that this makes no difference to whether πis supermodular or not in wand c. Hence the assumption made by Antr as and Helpman is not needed for the results. (The only complication is that, for the necessity part of the proof of Proposition 1, we need to assume that there is some component of fixed costs which is independent of w.)
29 ducing as an integrated firm at home, vertical FDI faces conflicting incentives. On the one hand, it incurs a higher cost of incomplete contracting, ψVS > ψ VN , since in the event of the relationship breaking down, the headquarters cannot expect to retain as large a share of profits when production is in the South. On the other hand, it incurs a lower wage rate, wS < w N . Thus the profit function is unlikely to be unambiguously supermodular or submodular in c and the cost vector [ ψ w ], and the predicted selection effects are likely to be highly sensitive to the specification of the model.
6.3 Heterogeneous Fixed Costs
Up to this point we have followed most of the literature on heterogeneous firms in assum-ing that fixed costs are the same for all firms and in all foreign locations. This is clearly unrealistic, and we need to examine whether our approach can be extended to the case where fixed costs differ between firms or locations. The previous analysis is unaffected if fixed costs vary with trade costs t only, so f
becomes f (t). For example, Kleinert and Toubal (2010) allow the fixed costs of a foreign plant to increase with its distance from the parent country, and show that this change in assumptions rationalizes a gravity equation for FDI, while Kleinert and Toubal (2006) show that it also avoids the counter-factual prediction that falling trade costs lower FDI. These are important insights, but the model’s predictions about selection effects are unchanged. The reason is simple: although the fixed cost varies with trade costs, the finite difference operator applied to the gain from FDI relative to exporting γ eliminates the fixed cost since ∆ cfF (t) = 0. While differences in fixed costs between locations clearly affect locational choice, they do so in the same way for all firms. Matters are more complicated if fixed costs vary with both production costs c and trade costs t. Technically, our approach can still be applied, but some care is needed. We now need to include any firm-specific fixed costs in the definition of operating profits. 31
Thus, let ˜ π(t, c ) denote operating profits net of firm-specific fixed costs, and define full
31 If the necessity part of Proposition 1 is to hold, we must also assume that there is a component of fixed costs which is common to all firms, as before.
30 operating profits π(t, c ) as follows:
π (t, c ) ≡ ˜π (t, c ) − I(t)fF (c) (23) Here, I(t) is an indicator function, which equals zero when t is positive (the exporting case), and equals one when t is zero (the FDI case). Thus the fixed cost of a foreign plant must be subtracted to get full operating profits, and this varies with the firm’s productivity. Now, there is an additional reason why supermodularity may not hold, depending on how fixed costs vary with productivity. Applying the finite difference operator to the full operating profits function (23) gives: ∆cπ(t, c ) − ∆cπ(0 , c ) = [∆ c ˜π(t, c ) − ∆c ˜π(0 , c )] + ∆ cfF (c) (24) The first term in parentheses on the right-hand side is the same as in previous sections. The second term is new, and shows that supermodularity is more likely to hold if fixed costs are higher for less efficient firms. Two examples from the literature illustrate how this effect can work in different direc-tions. The first is from Behrens, Mion, and Ottaviano (2010), who assume that a firm’s fixed costs are proportional to its variable costs, fF (c) = cf , so more efficient firms incur lower fixed costs of establishing a foreign plant. In this case, the final term in (24) be-comes: ∆ cfF (c) = ( c1 −c2)f which is strictly positive for c1 > c 2. Hence, supermodularity of π and so the conventional sorting pattern are reinforced in this case. A second example comes from Oldenski (2010), who develops a model of task-based trade in services. Because they use knowledge-intensive tasks disproportionately, higher-productivity firms in service sectors are more vulnerable to contract risk when located abroad. As a result, their fixed costs of FDI are decreasing in c: fF (c), f ′
F
< 0. As a result, the final term in (24) becomes ∆ cfF (c) = fF (c1) − fF (c2), which is strictly negative for c1 > c 2, so π may be submodular. In this case the conventional sorting may be reversed, as higher-productivity firms may find it more profitable to locate at home, and Oldenski presents evidence for this pattern in a number of U.S. service sectors. 31 6.4 Endogenous Fixed Costs
The previous sub-section considered fixed costs that differ between firms but in an exoge-nous way. By contrast, there are many ways in which a firm can influence the level of its fixed costs as well as its variable costs in each market. It is desirable to explore whether our approach extends to this case, where the firm faces more complex trade-offs. In fact, the earlier derivations go through with relatively little modification. All that is needed is to redefine the maximized profit function as the outcome of the firm’s choice of
both its output and its level of investment. To fix ideas, consider the case of investment in cost-reducing R&D. (Other forms of investment, such as in marketing or product innovation, can be considered with relatively minor modifications.) Let k denote the level of investment which the firm undertakes. For simplicity we concentrate on examples where k is a continuous variable. However, it is clear that our approach does not require this, and that supermodularity also applies to cases of a discrete choice between a finite number of techniques, as in Bustos (2011). Turning to specifics, we assume that investment in R&D incurs an endogenous fixed cost F (k) but reduces average production costs, now denoted C (c, k ). Here c is, just as in earlier sections, a parameter representing the firm’s exogenous level of costs (the inverse of its productivity), while k is chosen endogenously. C (c, k ) is increasing in c and decreasing in k, while fixed costs F (k) are increasing in k. The maximum profits which the firm can earn in a market, conditional on t and c, are:
π (t, c ) ≡ M ax
x,k
[{p (x) − C (c, k ) − t} x − F (k)] (25) Unlike earlier examples, the firm now has two choice variables rather than one. It can be checked that π is supermodular in t and c for many commonly used specifications of the cost functions F (k) and C (c, k ), so all our results apply in those cases too. However, there are also economically interesting examples where supermodularity is violated, and so the selection pattern of firms into different modes of serving foreign markets given by Proposition 1 is reversed. 32 To check whether the profit function (25) exhibits supermodularity in t and c, we proceed as in Example 1. The envelope theorem still applies, so the derivative of max-imum profits with respect to the tariff equals minus the level of output: πt = −x (t, c ). Hence it follows as before that: πtc = −xc. So, to check for supermodularity, we need only establish the sign of the derivative of output with respect to the cost parameter c.The first-order conditions for output x and investment k are:
p − C − t + xp ′ = 0 (26)
−xC k − F ′ = 0 (27) Totally differentiate these and write the results as a matrix equation:
2p′ + xp ′′ −Ck
−Ck −(xC kk + F ′′ )
dx dk
=
Ccdc + dt xC kc dc
(28) From the firm’s second-order conditions, the diagonal terms in the left-hand coefficient matrix must be negative, and the determinant of the matrix, which we denote by D, must be positive. 32 Solving for the effect of the cost parameter on output and substituting into
πtc gives:
πtc = −xc = D−1+
[
Cc (xC kk + F ′′ )
︸ ︷︷ ︸
+
− xC k
−
Ckc
]
(29) The first term inside the brackets is positive from the second-order conditions, which works in favor of supermodularity of π. The second could work either way. In particular, the term could be negative, and so supermodularity might not prevail, if Ckc is negative, so a lower-productivity firm benefits more from investment, in the sense that its costs fall by more; or, equivalently, if Cck is negative, so investment lowers the cost disadvantage of a lower-productivity firm. In some of the most widely-used models of R&D, the ambiguity of the sign of (29) can be resolved. We consider three examples in turn.
32 D=−(2 p′+xp ′′ ) ( xC kk +F′′ )−C2
k
33 6.4.1 Linear-Quadratic Costs
Following d’Aspremont and Jacquemin (1988), a widely-used specification of R&D costs, also applied to the study of FDI by Petit and Sanna-Randaccio (2000), assumes that the marginal cost function is linear while the investment cost function is quadratic in k:33
C (c, k ) = c0 − c−11 k F (k) = 12γk 2 (30) Firms may differ in either the c0 or c1 parameters, but it is clear that in either case output must be decreasing in c: Cck is zero if firms differ in c0 and positive if they differ in c1.34 Hence, the right-hand side of (29) is positive and supermodularity is assured for this specification of R&D costs.
6.4.2 Exponential Costs of R&D
An implausible feature of the d’Aspremont-Jacquemin specification is that the returns to investing in R&D are constant. 35 A more attractive and only slightly less tractable alternative due to Spence (1984) is also widely used: 36
C (c, k ) = c0 + c1e−θk F (k) = k (31) In this case investment lowers marginal production costs ( Ck = −θc 1e−θk < 0) but at a diminishing rate ( Ckk = θ2c1e−θk > 0), as illustrated in Figure 7(a) (drawn for c0 = θ =1); while the direct cost of investment increases linearly in k (F ′′ = 0). Once again, firms may differ in either the c0 or c1 parameters, and supermodularity is assured if they differ in c0. However, matters are different if firms differ in c1 (so we set c1 = c from now on). In this case, a lower-productivity firm benefits more from investment: Cck = −θe −θk < 0,
33 d’Aspremont and Jacquemin and Petit and Sanna-Randaccio also allowed for spillovers between firms.
34 When c=c1,Cck =c−2>0.
35 The linearity of Cin kalso suggests that the cost of production can become negative, though second-order conditions ensure that this never happens in equilibrium.
36 These specifications of C(c, k ) and F(k) come from Section 5 and from equation (2.3) on page 104 of Spence (1984), respectively.
34 01230 1 2 3 4 5
ckc1= 1
c1= 2
(a) Spence Model: a = 1
012300.5 11.5 2
ckc1= 1
c1= 2
(b) Threshold Effects in R&D: a = 2
Figure 7: Marginal Cost of Production as a Function of Investment and this effect is sufficiently strong that it exactly offsets the diminishing returns to investment. 37 As a result, equation (29) is zero under the Spence assumptions: other things equal, two firms with different cost parameters produce the same output. 38 Hence it follows that π (t, c ) is modular , i.e., both supermodular and submodular: the expression in Definition 1 holds with equality. To see the implications of this for the decision to engage in FDI, consider how two firms of different productivities will assess the relative advantages of exporting and FDI. For any given mode of accessing a market, both firms will produce the same output, the less productive firm compensating for its higher ex ante cost by investing more. 39 Hence their operating profits are also the same. It follows that both firms face exactly the same incentive to export or engage in FDI. We cannot say in general which mode of market access will be adopted, but we can be sure that both firms will always make the same choice. More generally, for any number of firms that differ in c1, all firms will adopt the same mode of serving the foreign market, so no differences in firm selection will be observed.
37
Formally, the semi-elasticities of both Cc and Ck with respect to k, Cck /C c and Ckk /C k, are equal to −θ.
38
Evaluating (29) gives: −Cc (xC kk + F ′′ ) + xC kCkc = − (e−θk ) ( xθ 2ce −θk ) + x (−θce −θk ) ( −θe −θk )
= −xθ 2ce −2θk + xθ 2ce −2θk = 0.
39
From (28), the effect of a difference in the cost parameter c on the level of investment is given by:
Dk c = (2 p′ + xp ′′ ) xC kc + CcCk. In general the first term on the right-hand side is ambiguous in sign while the second is negative. In the Spence case, the first term is positive and dominates the second, and the expression as a whole simplifies to: kc = θc .
35 6.4.3 R&D with Threshold Effects
The fact that the specification due to Spence is just on the threshold between super-and submodularity has implausible implications as we have seen. It also suggests that a less convex investment cost function would imply submodularity. Such a specification is found by adapting that of Spence as follows:
C (c, k ) = c0 + ce −θk a
, a > 0 F (k) = k (32) This is illustrated in Figure 7(b) (drawn for c0 = θ = 1 and a = 2). 40 For values of
a greater than the Spence case of a = 1, the cost function is initially concave and then becomes convex. 41 This justifies calling this specification one of threshold effects in R&D: low levels of investment have a relatively small effect on production costs whereas higher levels yield a larger payoff. In the FDI context this implies that firms will select into different modes of market access in exactly the opposite way to Proposition 2. Since profits are submodular in t and c, less efficient firms have a greater incentive to establish a foreign affiliate and carry out their R&D investment locally. By contrast, more efficient firms gain relatively little from further investment in R&D, and find it more profitable to concentrate production in their home plant and serve foreign markets by exporting. Hence the conventional sorting is reversed.
7 Conclusion
This paper has provided a novel approach to the central question of how different firms select into different modes of serving a market. In doing so, we give a unifying perspective on a large and growing literature, identify the critical assumptions which drive many ex-isting results, and develop an approach which can easily be applied to new ones. The key
40 This is the Gaussian distribution.
41 From (32), Ckk =−θack a−2e−θk a
(a−1−θak a). For 0 < a ≤1 this is always positive. However, for a > 1 it is negative for low kand then becomes positive. The point of inflection occurs where the expression in brackets is zero, which is independent of cand so (for given aand θ) is the same for all firms. In the case illustrated, with θ= 1 and a= 2, this occurs at k= 1 /√2. No firm will produce positive output below the inflection point, since Ckk must be positive from the second-order conditions. Note that, while the function is concave at some points and convex at others, it is log-concave everywhere.
36 criterion we highlight is extremely parsimonious: all that needs to be checked is whether the function giving the maximum profits a firm can earn in a market is supermodular in the firm’s own cost parameter, and in a second parameter measuring the marginal cost of accessing the market. We show that this property is necessary and sufficient for the con-ventional sorting pattern: more efficient firms select into the lower-variable-cost mode of serving the market, whereas less efficient firms select into the higher-variable-cost mode. We first proved this result for a special case where a single monopoly firm chooses be-tween exporting to a foreign market and engaging in foreign direct investment there. We then showed that our approach generalizes to other market structures, both oligopolistic and monopolistically competitive; and to a wide range of other firm choices, including between outsourcing and producing in-house, and between producing with more or less skill- or R&D-intensive techniques. Our criterion for selection effects is simple both in what it includes and in what it omits. Our approach does not require any special assumptions about the distribution of firm productivities nor about whether countries are symmetric. We are able to dis-pense with such assumptions because our approach sidesteps the key issue of existence of equilibrium. As Maskin and Roberts (2008) show in a different context, given that an equilibrium exists, its properties can often be established relatively easily. Since the impact effect of both tariffs and production costs is to lower profits, it is not so surprising that there are many cases where their cross effect is positive, so that supermodularity holds. Nevertheless, the restriction of supermodularity is a non-trivial one, and we have shown that there are many plausible examples where it does not hold. Perhaps most surprisingly, we have shown that, in the canonical case of horizontal FDI where exports incur iceberg transport costs, supermodularity fails to hold at every level of output for most widely-used demand systems, except the CES. Thus if preferences are quadratic or translog, and if selection is observed, it is likely that the most efficient firms will select into exporting rather than FDI. In the case of horizontal FDI, the source of this anomalous result can be traced to the assumption of iceberg transport costs: when higher productive efficiency translates into lower transport costs, the most efficient firms suffer 37 a lower transport penalty and so will select into exporting rather than FDI. However, in the case of choosing between outsourcing at home versus offshoring to a lower-wage location, our result continues to hold, and implies that for most non-CES preferences more productive firms will select into outsourcing at home, where their greater efficiency offsets the higher wage penalty they incur. We have also identified other plausible cases where supermodularity may fail, such as fixed costs which are higher for more efficient firms, and market-specific investment costs which are subject to threshold effects. Our results cast the role of fixed costs as determinants of selection effects in a new light. For example, in the choice between FDI and exports, a fixed cost of FDI is essential for a proximity-concentration trade-off to exist. For FDI to be the preferred mode of market access, a firm must be able to afford the additional fixed cost of FDI. However, fixed costs are irrelevant to the determination of which firms are in that category. What matters is the difference-in-differences effect on profits of the marginal costs of production and trade. When supermodularity prevails, a more efficient firm has relatively higher profits in the low-tariff case, but when submodularity holds, the opposite is true. In this paper we first highlighted the implications of this insight for selection into FDI, and then noted that the general point applies to other cases, including selection by more efficient firms into offshoring as in Antr` as and Helpman (2004), or into more skill-intensive techniques as in Bustos (2011). There are likely to be still other models which can be illuminated by our approach, and other contexts where the assumption of supermodularity helps to bound comparative statics responses.
8 Appendix
8.1 Proof of Proposition 1
Sufficiency is trivial. To prove necessity, let t > 0. If π is not supermodular in t and c,then there exist some c1 and c2 such that c1 > c 2 and π(t, c 1)−π(t, c 2) < π (0 , c 1)−π(0 , c 2). 38 Rearranging terms gives:
π(0 , c 1) − π(t, c 1) > π (0 , c 2) − π(t, c 2) (33) Thus γ(t, c 1, f ) > γ (t, c 0, f ) for all f ; i.e., the higher-cost firm has a greater incentive to engage in FDI than in exports. To prove that selection will actually occur for some fixed cost, choose f such that f = 12 [π(0 , c 1) − π(t, c 1) + π(0 , c 2) − π(t, c 2)]. Then for this f ,we have γ(t, c 1, f ) = π(0 , c 1) − π(t, c 1) − f ≥ 0 and γ(t, c 2, f ) = π(0 , c 2) − π(t, c 2) − f ≤ 0. Hence the higher-cost firm will serve the foreign market via FDI while the lower-cost firm will serve it by exports. Thus, if π is not supermodular in t and c we can always find a fixed cost such that the conventional sorting is reversed. It follows that supermodularity is necessary for the conventional sorting.
8.2 Superconvexity
Throughout, we measure the convexity of a function y = f (x) by the unit-free measure
ρf ≡ − xf ′′
f′
. In particular, the convexity of an inverse demand function p = p(x) is
ρ ≡ − xp ′′
p′
. Our formal definition of superconvexity is as follows:
Definition 2. Define P ≡ ln p and X ≡ ln x, and write P = ˆp(X). Then p(x) is superconvex if and only if ˆ p(x) is convex; i.e., if and only if ˆ ρ ≡ − X ˆp′′
ˆp′
≥ 0. Some implications of superconvexity are easily established:
Lemma 4. Superconvexity implies convexity but the converse does not hold. Proof. Differentiating ˆ p(X) yields: ˆ p′ = dP dX = xp ′
p
; and ˆ p′′ = dˆp′
dX
= xdˆp′
dx
= xp2 [p(xp ′′ + p′)
−x(p′)2] = xp ′
p
(xp ′′
p′
1 − xp ′
p
)
. Hence ˆ ρ equals: ˆρ = −X ˆp′′
ˆp′ = X
(
ρ − ε + 1
ε
)
(34) Clearly, ˆ ρ > 0 ⇒ ρ > 0.
Lemma 5. A demand function is superconvex if and only if it is more convex than a CES demand function with the same elasticity.
39 Proof. Differentiating the CES inverse demand function p = αx −1/σ gives: p′ = −α 1
σ
x−(1+ σ)/σ ;and p′′ = α σ+1
σ2
x−(1+2 σ)/σ . Hence we have εCES = σ and ρCES = σ+1
σ
, from which the result follows by inspection of equation (34).
Lemma 6. A demand function is superconvex if and only if its elasticity is increasing in sales. Proof. Differentiating the expression for the elasticity of demand, ε (x) = − p(x)
xp ′(x)
, yields:
εx = − 1
x + p (p′ + xp ′′ )(xp ′)2 = − 1
x (1 + ε − ερ ) (35) Comparison with (34) shows that εx and ˆ ρ always have the same sign: εx = εxX ˆρ.Finally, super-convexity can also be expressed in terms of the direct demand function
x = x(p), with elasticity e(p) ≡ − px ′
x
= ε [x (p)]:
Lemma 7. A demand function is superconvex if and only if its elasticity is decreasing in price. Proof. Differentiating the identity equating the two expressions for the elasticity of de-mand, e(p) = ε [x (p)], yields ep = εxp′, which from Lemma 6 implies that ep and ˆ ρ always have opposite signs: ep = p′ εxX ˆρ.
8.3 Proof of Proposition 2
Differentiating the profit function π (t, c ) = M ax
x
[p (x) − τ c ] x gives: πt = −cx ; and
πtc = −x − cdx dc = −x − τ c
2p′ + xp ′′ (36) We want to express the right-hand side in terms of ε and εx. First, solve (35) for p′ + xp ′′
in terms of εx, and add p′ to it. Next, use the definition of ε to eliminate p′, p′ = − pxε ,which gives the second-order condition in terms of ε and εx:2p′ + xp ′′ = − pxε 2 (ε − 1 − xε x) (37) 40 This confirms that the second-order condition 2 p′ +xp ′′ < 0 is equivalent to ε−1−xε x > 0. The last preliminary step is to use the first-order condition p − τ c + xp ′ = 0 to express
τ c in terms of p and ε: τ c = p + xp ′ = p − pε = ε−1
ε
p. (This is very familiar in the CES case.) Finally, substitute these results into (36):
πtc = −x + ε − 1
ε − 1 − xε x
εx (38) Collecting terms gives the desired expression in (15).
8.4 Proof of Proposition 3
The proof follows immediately by substituting for εx from (35) into the expression for
πtc in (15), and noting that, as in (15), the denominator must be positive from the second-order condition.
8.5 Proof of Lemma 2
For the demand function p = ( x − β)−1/σ , we have p′ = − 1
σ
(x − β)− σ+1
σ
and p′′ =
σ+1
σ2
(x − β)− 2σ+1
σ
. Hence ε = x−βx σ and ρ = xx−βσ+1
σ
. It follows immediately that εx = βx2 σ,and so the demand function is strictly subconvex ( εx < 0) if and only if β is negative. To establish for which values of σ the profit function is supermodular, rewrite the elasticity as ε = σ+1
ρ
. Hence we seek to minimize σ = ερ − 1 by choice of ε and ρ, subject to the supermodularity constraint from Proposition 3, ε + ρ ≥ 3. Solving gives the boundary values ε∗ = ρ∗ = 1 .5, which imply that the threshold value of σ is σ∗ = (32
)2−1 = 1 .25.
8.6 Proof of Lemma 3
Maximizing operating profits, π = ( p−τ c )x, yields the first-order condition, A−2bx = τ c ,which can be solved for optimal output: x = 12b (A − τ c ). Substituting back into the expression for profits gives the maximized operating profit function:
π(t, c ) = bx 2 = 14b (A − τ c )2 (39) 41 Hence the second cross-derivative is:
πtc = −x + τ c
2b = − 12b (A − 2τ c ) (40) This is clearly positive for c ≥ A
2τ
and negative for c ≤ A
2τ
, which proves the Lemma. 42 References
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Phi and Fibonacci in Kepler and Golden Triangles - The Golden Ratio: Phi, 1.618
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You are here: Home/Geometry/ Phi and Fibonacci in Kepler and Golden Triangles
Phi and Fibonacci in Kepler and Golden Triangles
May 13, 2012 by Gary Meisner21 Comments
Creating a Triangle based on Phi(or Pythagoras meets Fibonacci):
Pythagoras discovered that a right triangle with sides of length a and b and a hypotenuse of length c has the following relationship:
a² + b² = c²
A foundational equality of phi has a similar structure:
1 + Phi = Phi 2
( 1+ 1.618… = 2.618… )
By taking the square root of each term in this equality, we have the dimensions of a triangle, known as a Kepler Triangle, a right triangle based on this phi equality, where:
Side Length squared
per above Length,
or square root Length divided
by phi so c = 1
a 1 1 1 / Phi
b Phi√Phi 1 / √Phi
c Phi 2 Phi 1
This triangle is illustrated below. It has an angle of 51.83° (or 51°50′), which has a cosine of 0.618 or phi.
The Pythagorean 3-4-5 triangle is the only right-angle triangle whose sides are in an arithmetic progression. 3 + 1 = 4, and 4 plus 1 = 5. The Kepler triangle is the only right-angle triangle whose side are in a geometric progression: The square root of phi times Φ = 1 and 1 times Φ =Φ.
Although difficult to prove with certainty due to deterioration through the ages, this angle is believed by some to have been used by the ancient Egyptians in the construction of theGreat Pyramid of Cheops.
Other triangles with Golden Ratio proportions can be created with a Phi (1.618 0339 …) to 1 relationship of the base and sides of triangles:
The isosceles triangle above on the right with a base of 1 two equal sides of Phi is known as a Golden Triangle. These familiar triangles are found embodied inpentagramsandPenrose tiles.
Creating a Triangle based on Fibonacci numbers
No three successive numbers in the Fibonacci series can be used to create a right triangle. Marty Stange, however, contributed the following relationship in January 2007: Every successive series of four Fibonacci numbers can be used to create a right triangle, with the base and hypotenuse being determined by the second and third numbers, and the other side being the square root of the product of the first and fourth numbers. The table below shows how this relationship works:
Fibonacci Series The Fibonacci Triangle b’a c b”a² b’xb”a² + b’xb” = c² 0 1 1 2 1 0 1 1 1 2 3 1 3 4 1 2 3 5 4 5 9 2 3 5 8 9 16 25 3 5 8 13 25 39 64 5 8 13 21 64 105 169 8 13 21 34 169 272 441 13 21 34 55 441 715 1,156 21 34 55 89 1,156 1,869 3,025 34 55 89 144 3,025 4,896 7,921 55 89 144 233 7,921 12,815 20,736 89 144 233 377 20,736 33,553 54,289 144 233 377 610 54,289 87,840 142,129
Thus for the illustration highlighted in gold, Stange’s Treatise on Fibonacci Triangles reveals that a triangle with sides of 5 and the square root of 39 (e.g., 3 x 13) will produce a right triangle with a hypotenuse of 8.
As greater numbers in the series are used, the triangle approaches the proportions of the phi-based Kepler Triangle above, with a ratio of the hypotenuse to the base of Phi, or 1.618…
Filed Under: Geometry
Comments
Ken Landin says
March 14, 2013 at 5:22 pm
What a brilliant analysis by Stange. I loved your beautifully done illustrations, and how clear your descriptions were.
Anyone interested in phi should look up how using the Fibonacci numbers can create a series of rectangles that approach the phi ratio. Reply
2. N K Srinivasan says
May 19, 2013 at 3:14 pm
A very useful presentation and quite interesting! Reply
3.
June 23, 2014 at 3:44 pm
CORRECT REPETITION
————————
A very interesting topic.
My decoding Plato’s Timaeus “MOST BEAUTIFUL TRIANGLE” shows that Kepler / Magirus Triangle is a similar triangle, “not the same” and” not as beautiful” , but constituent to that of Plato’s:
This orthogonal scalene triangle has all its sides in ratio T and scalene angle ArcTan [ T ] , T=SQRT[Phi].
Its hypotenuse is T^3, its bigger side is T^2 and its smaller is T^1.
{
Shows the ruller and compass structure of T [ = sqrt(Phi) ] and Θ [ = ArcTan(T) ] ,
via known construction of Phi [ = T^2].
It is based on a fourth order equation : T^4 – T^2 -1 =0 }
For quadrature of circle, [ as proved imposibility by transcendentals ] is obtained , however, by the irrational value of 4/[SQRT(Phi)] , following the equation : [ Πσχ ]^4 + { [ Πσχ ]^2}4^2 – 4^4 =0
Ref:
And an AutoCad drawing:
© Copyright 1999 P. C. Stefanides.
Regards,
Panagiotis Stefanides Reply
June 14, 2021 at 5:13 am
REF:
GEOMETRIC CONCEPTS IN PLATO
Panagiotis Stefanides
Reply
Dennis ONeill says
September 6, 2021 at 1:57 pm
Wow, very cool, thanks for posting. Check out Ken Wheeler’s “Pythagoras, Plato and the Golden Ratio” available online free as a PDF. Reply
Iuliana says
April 15, 2016 at 10:06 am
Awesome explanation. Thank you!! Reply
5. pritam pagla says
July 25, 2016 at 11:12 pm
Really started loving golden ratio Reply
6.
February 19, 2019 at 12:14 pm
One more amazing Golden Triangle exists, although unknown long time.
It is described and represented here:
© 2003 The Watch Publisher Reply
April 25, 2019 at 12:14 pm
Completed further descriptions of the Great Golden Triangle,
as well as the Golden Section-s in the Vitruvian Man by Leonardo da Vinci updated here:
© 2019 The Watch Publisher Reply
Kathy says
October 6, 2019 at 1:03 pm
Because the powers of Phi, unlike other numbers, are in Phi ratio to one another AND are in an additive sequence where any member is the sum of the prior two, it naturally follows that
Right triangle sides a,b,c where c is hypontenuse. IF a = phi^n and c= phi^(n+1)
THEN b must equal phi^ ((n+2)/2) Reply
8. Giuseppe Fiorentino says
March 13, 2020 at 7:15 am
Beautiful work. A real joy. Thank You. Reply
9. Mr. M says
July 2, 2022 at 2:22 am
“No three successive numbers in the Fibonacci series can be used to create a right triangle.” ?
What about the sqrt of (3 successive) Fibonacci-numbers; not a Keppler-triangle, like 5 : 8 : sqrt(3×13) ?
The proportions between the sides of a sqrt3 : sqrt5 sqrt8 -triangle looks just as right ánd wrong to me, as I would simply use fractions to get a Keppler triangle; with c being sqrt(8/3), the base 1 and a hight of sqrt (5/3.) The more perfect with higher F-numbers Reply
July 9, 2022 at 4:41 am
Yes, but the sqrt3, sqrt5 and sqrt8 are NOT Fibonacci numbers. They are square roots of Fibonacci numbers, which is not the same thing. Good analysis though! Reply
July 12, 2022 at 2:41 am
A right triangle in the Euclidean plane is a Kepler triangle if and only if it is similar to a triangle with side lengths: 1 , sqrtPhi, Phi.
1 Aside, Fibonacci-numbers are (of course) not similar to those dimensions, nor are their roots. In that sense single Lucas nr. can be use to generate Phythagorian Kepler-triangles with similar dimensions The higher the number the closer.
123^0 : 123^1/20 : 123^1/10 Reply
July 22, 2022 at 9:17 am
I should have used the roots of Fibonacci-products: (2×13),(3×13),(5×13) for all sides, to make it an even better one. Reply
11.
July 26, 2022 at 4:25 am
I am sort of missing the link between the square sum of the 3 consecutive natural numbers i.e. 9 + 16 = 25 and the sum of the 3 consecutive Fibonacci numbers: 3 + 5 = 8 as 3 times those numbers gives: 9 + 15 = 24 which reveales the similarity a little bit more. as there is only 1 less on either side. Maybe it is all to obvious for math enthousiasts; it really was not to me. Reply
12.
August 27, 2022 at 9:44 am
ps I notice(d) that it is not typically Fibonacci:
as 7, 8, 15, 23 works too, or 12² + 11 x 35 = 23². Reply
August 29, 2022 at 9:49 pm
There are likely many combinations that will work. The article is just showing how Fibonacci sequence numbers consistently follow this pattern, which is not meant to imply that only those numbers work. Reply
August 30, 2022 at 6:56 am
Somewhat less mathemagical
(b-a)(a+b) = b²- a²
(5-3)(5+3) = 5²-3²=4²
5² = 4² + 3² Reply
November 8, 2022 at 5:23 am
144 is the only Fib.square nr. that can be related to 3x 2,3,5,8 (6,9,15,24) or (15-9)(15+9) and a 9,12,15 Pythagoras triangle (9²+12²=15²) but also to a Pythagoras 5²+12²=13² which gives a nice equation:
15²-9²=13²-5²
225-81=169-25
250=160+90 Reply
14.
December 9, 2022 at 8:47 am
I have found a different approach
(a+b)²-(b-a)²=4(ab)
(a²+b²)²-(b²-a²)²=4(ab)²
(1²+2²)²-(2²-1²)²=4(1×2)²
5²=4²+3²
13²=12²+5²
34²=30²+16²
(3²+5²)²=4(3×5)²-(5²-3²)²
Reply
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Published Time: 2023-05-13T19:51:54+00:00
18 Pseudoscience Examples (2025)
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18 Pseudoscience Examples
Written by Chris Drew (PhD)
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Dr. Chris Drew is the founder of the Helpful Professor. He holds a PhD in education and has published over 20 articles in scholarly journals. He is the former editor of the Journal of Learning Development in Higher Education. [Image Descriptor: Photo of Chris]
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| May 10, 2024
Pseudoscience is a term applied to any belief system that claims to have scientific validity despite lacking the rigorous scientific research required for it to be recognized by the scientific community.
Often, pseudoscience offer solutions (both physical and spiritual) to problems that science cannot solve. Many pseudosciences have long and storied histories and their origins cannot even be identified.
However, proponents of pseudosciences often argue that they have positive effects that may not be accessible through traditional scientific methods.
Note that I’m not applying judgment on any of the below examples of pseudoscience (I do some of them myself!). Rather, I’m presenting a range of concepts that lack scientific support due to lack of clear evidence based on the traditional scientific method.
Contentsshow
Pseudoscience Examples
Additional Examples of Pseudoscience
Conclusion
Pseudoscience Examples
1. Acupuncture
Acupuncture is an ancient healing practice from China that is believed by most scientists to be pseudoscientific. As a result, many countries do not cover in their national health insurance plans.
Acupuncture is said to promote physical balance and harmony by resolving imbalances in the flow of Qi (what might be considered life energy).
Qi is believed by practitioners to flow through the body along energy pathways called meridians. There are 14 major meridians. Each meridian is associated with a different organ or bodily system.
When Qi becomes blocked or imbalanced, it is believed to lead to pain or disease.
Acupuncture supposedly unblocks and re-balances Qi by stimulating various points along the meridians using thin needles.
However, there is debate over whether acupuncture works and science has found no evidence of Qi or meridians.
2. Astrology
Astrology is the pseudoscience of predicting the future based on the alignment of the stars.
Astrology is believed to be a false pseudoscience because it is based on the unproven premise that the positions of celestial bodies such as the planets, moon, sun, and constellations at the time of a person’s birth can influence their personality and future.
This claim has not been supported by scientific evidence. Nevertheless, proponents of Astrology believe that it can help you to reflect on your personality and life circumstances more effectively.
3. Chiropractic
Chiropractic is a field of alternative medicine that is based on the non-verified belief that misalignments of the spine, called “subluxations,” can be resolved through realignment and that this realignment has a wide range of flow-on effects for your health.
Chiropractors claim to be able to treat a wide variety of health problems by adjusting these subluxations; however, others (including many physiotherapists and western medical doctors) say that there is no evidence that this is effective.
Nevertheless, many people continue to report excellent benefits from chiropractic therapy and it is a very large industry. Furthermore, many extended health plans cover it.
4. Conversion Therapy
Conversion Therapy is a pseudoscientific practice that attempts to change an individual’s sexual orientation from homosexual or bisexual to heterosexual.
It was widespread among conservative religious groups in the the 1990s and early 2000s before being banned in many Western nations due to its harmful psychological effects and changing attitudes toward LGBT rights.
Conversion therapy has been condemned by major medical organizations as ineffective and potentially harmful.
5. Ear Candling
Ear candling, also known as ear coning or thermal auricular therapy, is an alternative medicine practice claimed to improve general health and well-being by lighting one end of a candle and inserting it into the ear canal.
Theoretically, the heat from the burning candle creates a vacuum that sucks out wax and other impurities from your ear.
However, there is no scientific evidence to support the idea that ear coning really works, and many professionals warn that ear candling can cause more harm than good.
In addition to causing burns and perforating the eardrum, ear candling can also push wax and debris further into the ear canal. As a result, many experts consider ear candling to be a pseudoscientific practice that should not be engaged with.
6. Faith Healing
This one is controversial because so many billions of people rely on prayer every day of their lives. While spiritually, there are many who say it works, there is no scientific way to prove it, andrespected impartial studiesthat have attempted to prove it have not turned up clear correlations.
Faith healing often relies on anecdotal evidence, which is not considered to be reliable evidence in a scientific setting.
Another reason why faith healing is considered pseudoscience is that it often relies on confirmation bias. In other words, many religious people really want faith healing to be proven to work because they have a lot of their identity and belief system invested in the idea.
Don’t get me wrong – prayer has many meaningful spiritual benefits – but the scientific community warns that relying on prayer alone and ignoring western medicine can be harmful.
7. Feng Shui
Feng shui involves organizing space to create positive energy. It relies on the belief that the arrangement of objects in a space can have an impact on the flow of energy.
Practitioners of feng shui claim that by rearranging their furniture and other objects in their rooms, you can improve your “health, wealth, and relationships.”
However, to date, there is no scientific evidence to support these claims.
In fact, many of the principles of feng shui can be traced back to ancient superstitions and highly non-scientific folklore. For example:
The placement of mirrors is said to be able to reflect negative energy away from a person.
The practice of hanging wind chimes is said to be able to attract positive energy.
I personally like some aspects of feng shui because it makes me think about my home’s layout and optimal placements of furniture to improve my life, but it’s got no true science behind it. And that’s okay!
8. Healing Crystals
Healing crystals are crystals that you place on your body or in your environment in order to promote physical, emotional, and spiritual well-being.
The idea is that each type of crystal has its own unique properties. For example:
Amethystcan supposedly help relieve stress
Rose quartzcan supposedly attract love.
There is no scientific evidence to support the ancient claims that certain crystals have certain powers. Nevertheless, many people find them to be helpful. Likely, this is due to the placebo effect.
9. Homeopathy
Homeopathy is another alternative medicine. It is based on the principle that “like cures like.” In other words, a substance that causes certain symptoms can also be used to treat those same symptoms.
A common example is onions causing crying. As a result, some people use onions to onset crying, which is believed to have the flow-on effect of clearing out allergies like hayfever that often also causes you to cry.
Although homeopathy is widely considered pseudoscientific because of its lack of evidence in clinical trials, it is embraced by millions of people around the world.
10. Hypnotherapy
Hypnotherapy is therapy that takes place under a state of hypnosis. Supposedly, when under hypnosis, the patient is more relaxed and can enter a state of focused attention.
Once the patient is in this state, the therapist can then work to address any subconscious issues that can only be treated when you’re in that state.
While hypnotherapy is widely used and many people say it has been extremely effective in achieving personal emotional breakthroughs, it is still considered somewhat controversial due to its lack of rigorous scientific basis.
Additional Examples of Pseudoscience
Magnet Therapy –Magnet therapy is the use of magnets with the idea that they can heal you. They thought to produce an electromagnetic field that can do things like improving blood flow and promoting healing on a molecular level.
Myers-Briggs Type Indicator –The Myers-Briggs Type Indicator is a debunked personality assessment instrument that is often used in the workplace and during professional retreats to help people reflect on their strengths and weaknesses in the workplace. It is based on the idea that people can be operationalized into 16 personality types based four categories: Extroversion/Introversion, Sensing/Intuition, Thinking/Feeling, and Judging/Perceiving. However, it is criticized for lacking content validity.
Numerology –The idea that numbers have spiritual meanings. If you keep seeing or dreaming of specific numbers, the idea is that they mean something spiritually. Similarly, your birth date may have some deeper meaning. It is not taken seriously by scientists.
Psychoanalysis –Psychoanalysis aims to help people understand their unconscious thoughts and emotions. It often relies on bizarre and debunked ideas like Freud’s oedipus complex. While it’s still widely used in psychology, the hard sciences tend to critique it as an unscientific idea (for more on this, see my article on pseudo-psychology).
Qigong –Qigong is an ancient Chinese practice that involves the use of breathing exercises and meditation to improve health and well-being. Some aspects of it, like meditation, are found to be very effective, while other ideas within Qigong like Qi (or life energy – discussed earlier) have no scientific basis.
Reflexology –Reflexology is a type of massage that focuses on the feet and hands. It relies on an as yet unproven idea that all organs of the body has a ‘touch point’ on the sole of your foot.
Reiki –Reiki is a natural healing art that uses the hands to help balance the body’s energy system. It believes in a life source or energy system that science cannot identify and does not come out of scientific research.
Aromatherapy –Aromatherapy is the use of essential oils to promote healing and wellness. Supposedly, these oils can help you improve your mood, reduce stress, and boost energy levels. Science tends to be skeptical of such claims.
Conclusion
This article isn’t meant to debunk or even criticize the above pseudosciences. In fact, some of them may end up being true (or at least having some tangible benefits to people’s lives). Nevertheless, they tend to be categorized as pseudosciences because they do not currently have sufficient scientific backing or clear evidence behind them that is based on rigorous scientific research in scholarly settings.
Chris Drew (PhD)
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Dr. Chris Drew is the founder of the Helpful Professor. He holds a PhD in education and has published over 20 articles in scholarly journals. He is the former editor of the Journal of Learning Development in Higher Education. [Image Descriptor: Photo of Chris]
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Vibrio vulnificus
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From Wikipedia, the free encyclopedia
Species of pathogenic bacterium found in water
| Vibrio vulnificus | |
--- |
| | |
| False-color SEM image of Vibrio vulnificus | |
| Scientific classification | |
| Domain: | Bacteria |
| Kingdom: | Pseudomonadati |
| Phylum: | Pseudomonadota |
| Class: | Gammaproteobacteria |
| Order: | Vibrionales |
| Family: | Vibrionaceae |
| Genus: | Vibrio |
| Species: | V. vulnificus |
| Binomial name | |
| Vibrio vulnificus (Reichelt et al. 1976) Farmer 1979 | |
| Synonyms | |
| Beneckea vulnifica | |
Vibrio vulnificus is a species of Gram-negative, motile, curved rod-shaped (vibrio), pathogenic bacteria of the genus Vibrio. Present in marine environments such as estuaries, brackish ponds, or coastal areas, V. vulnificus is related to V. cholerae, the causative agent of cholera. At least one strain of V. vulnificus is bioluminescent.
Increasing seasonal ocean temperatures and low-salt marine environments like estuaries favor a greater concentration of Vibrio within filter-feeding shellfish; V. vulnificus infections in the Eastern United States have increased eightfold from 1988–2018.
Infection with V. vulnificus leads to rapidly expanding skin infections by entering a wound causing cellulitis or even sepsis.: 279 V. vulnificus is also a source of foodborne illness. It was first isolated as a source of disease in 1976.
Strains
[edit]
Vibrio vulnificus is a species of gram-negative, motile, curved rod-shaped (bacillus), pathogenic bacteria of the genus Vibrio. Present in marine environments such as estuaries, brackish ponds, or coastal areas, V. vulnificus is related to V. cholerae, the causative agent of cholera.
The most harmful strains of V. vulnificus documented have been observed in three different forms. The first is in an anti-phagocytic polysaccharide capsule that protects the bacteria. By encapsulating the bacteria, phagocytosis, and opsonization can not occur, thus allowing the bacteria to continue throughout the organism it is in. The second way that V. vulnificus has been most harmful is with some of the toxins that it creates. These toxins are not part of the infection that V. vulnificus causes but instead they are part of a secondary infection in the gastrointestinal tract that most certainly will lead to systemic infection. Lastly, V. vulnificus has been seen to cause more harm in patients who have higher levels of iron.
Genome size
[edit]
The genome size of V. vulnificus is approximately 5.3 Mbp. The genome is organized into two circular replicons, similar to that of V. cholerae, which also has a large and a small chromosome. However, the genome of V. vulnificus is at least one megabase bigger.
Natural transformation
[edit]
Natural transformation is a bacterial adaptation for DNA transfer between individual cells. V. vulnificus was found to become naturally transformable during growth on chitin in the form of crab shells. The ability to now carry out transformation experiments in the laboratory should facilitate molecular genetic analysis of this opportunistic pathogen.[citation needed]
Signs and symptoms
[edit]
Vibrio vulnificus is an extremely virulent bacterium that can cause three types of infections:
Acute gastroenteritis from eating raw or undercooked shellfish: V. vulnificus causes an infection often incurred after eating seafood, especially raw or undercooked oysters. It does not alter the appearance, taste, or odor of oysters. Symptoms include vomiting, diarrhea, and abdominal pain.
Invasive sepsis can occur after eating raw or undercooked shellfish, especially oysters. V. vulnificus is 80 times more likely to spread into the bloodstream in people with compromised immune systems, especially those with chronic liver disease. When this happens, severe symptoms including blistering skin lesions and septic shock can sometimes lead to death. This severe infection may occur regardless of whether the infection began from contaminated food or an open wound.
Necrotizing wound infections can occur in injured skin exposed to contaminated marine water. V. vulnificus bacteria can enter the body through open wounds when swimming or wading in infected waters, or by puncture wounds from the spines of fishes such as stingrays. People may develop a blistering dermatitis sometimes mistaken for pemphigus or pemphigoid.
Among healthy people, ingestion of V. vulnificus can cause vomiting, diarrhea, and abdominal pain. In someone with a compromised immune system, particularly those with chronic liver disease, it can infect the bloodstream, causing a severe and life-threatening illness characterized by fever and chills, decreased blood pressure (septic shock), and blistering skin lesions. While men have been shown to be more at risk from this infection than women, co-morbidities such as alcoholic cirrhosis and diseases affecting the endocrine system (diabetes, rheumatoid arthritis, etc.) put a person far more at risk of developing an infection from V. vulnificus.
Pathogenesis
[edit]
Capsule: V. vulnificus has a capsule, made of polysaccharides, and is thought to protect against phagocytosis. The capsule also aids the bacteria in escaping opsonization. Different strains of the bacteria are capable of shifting through the unencapsulated and encapsulated forms. Mouse models have shown that the unencapsulated forms are avirulent. These same strains, however, are shown to have a higher predisposition to shift to the virulent encapsulated form when taken up by oysters.
Endotoxin: Like all gram-negative bacteria, V. vulnificus has LPS (lipopolysaccharide as the major component of its outer membrane). However, the LPS the bacteria produces isn't as efficient at triggering the immune system's release of tumor necrosis factor (TNF) alpha and other cytokines that produce shock syndromes. The capsular proteins the bacteria express, however, are capable of producing an immune response contributing to shock syndrome.
Exotoxin: V. vulnificus produces a number of extracellular toxins such as metalloprotease VvpE, cytolysin/hemolysin VvhA, and the multifunctional autoprocessing repeats-in-toxins (MARTX) toxin. While the VvhA and MARTX toxin are factors in the bacteria's virulence, in vivo studies in mice suggest that the MARTX toxin is more responsible for bacterial dissemination from the intestine to produce sepsis.
Iron: Growth of V. vulnificus is dependent on the amount of iron that is accessible to the bacteria. The observed association of the infection with liver disease (associated with increased serum iron) might be due to the capability of more virulent strains to capture iron bound to transferrin.
Type IV Pilus: V. vulnificus has genes that encode for the biosynthesis of a type IV pilus. These genes have been located on the large chromosome and may help V. vulnificus adhere to host cells and with motility.
Treatment
[edit]
Vibrio vulnificus wound infections have a mortality rate of around 25%. In people in whom the infection worsens into sepsis, typically following ingestion, the mortality rate rises to 50%. The majority of these people die within the first 48 hours of infection. The optimal treatment is not known, but in one retrospective study of 93 people in Taiwan, the use of a third-generation cephalosporin and a tetracycline (e.g., ceftriaxone and doxycycline, respectively) was associated with an improved outcome. Prospective clinical trials are needed to confirm this finding, but in vitro data support the suggestion that this combination is synergistic against V. vulnificus. Likewise, the American Medical Association and the Centers for Disease Control and Prevention (CDC) recommend treating the person with a quinolone or intravenous doxycycline with ceftazidime. The first successful documented treatment of fulminant V. vulnificus sepsis was in 1995. Treatment was ceftazidime and intravenous (IV) ciprofloxacin and IV doxycycline, which proved successful. Prevention of secondary infections from respiratory failure and acute renal failure is crucial. Key to the diagnosis and treatment were the early recognition of bullae in an immunocompromised person with liver cirrhosis and oyster ingestion within the previous 48 hours, and the request by the physician for STAT Gram staining and blood cultures for V. vulnificus. Amputation of limbs may be required. Vibrio vulnificus often causes large, disfiguring ulcers that require extensive debridement or even amputation.[citation needed]
Prognosis
[edit]
Vibrio vulnificus is the most common cause of death due to seafood in the United States, causing over 95% of deaths that are known to have occurred due to ingested seafood. If treatment with tetracycline or other cephalosporin antibiotics is initiated at the onset of symptoms and the full course followed, patients generally experience no long-term effects.
The worst prognosis is in those people arriving at the hospital in a state of shock. Total mortality in treated people (ingestion and wound) is around 33%.
People especially vulnerable are those with liver disease (especially cirrhosis and hepatitis) or immunocompromised states (some kinds of cancer, bone marrow suppression, HIV, diabetes, etc.). In these cases, V. vulnificus usually enters the bloodstream, where it may cause fever and chills, septic shock (with sharply decreased blood pressure), and blistering skin lesions. About half of those who contract blood infections die.
The consumption of contaminated raw oysters in those with chronic liver disease causes primary sepsis with a mortality rate of over 50%. Exposure of wounds to contaminated seawater or handling of contaminated seafood products causes infections with a mortality rate of about 25%.
Vibrio vulnificus infections also disproportionately affect males; 85% of those developing endotoxic shock from the bacteria are male. Females having had an oophorectomy experienced increased mortality rates, as estrogen has been shown experimentally to have a protective effect against V. vulnificus.
Epidemiology
[edit]
Further information: Effects of global warming on oceans
Vibrio vulnificus is commonly found in the Gulf of Mexico, where more than a dozen people have died from the infection since 1990. Most deaths at that time were occurring due to fulminant sepsis, either in the area of oyster harvest and ingestion, or in tourists returning home. Lack of disease recognition, and also of the risk factors, presentation, and cause, were and are major obstacles to good outcome and recovery.[citation needed]
Increasing seasonal temperatures and decreasing salinity levels seem to favor a greater concentration of Vibrio within filter-feeding shellfish of the U.S. Atlantic seaboard and the Gulf of Mexico, especially oysters (Crassostrea virginica). Scientists have frequently demonstrated the presence of V. vulnificus in the gut of oysters and other shellfish and in the intestines of fish that inhabit oyster reefs. The vast majority of people who develop sepsis from V. vulnificus became ill after they ate raw oysters; most of these cases have been men.
According to the CDC, V. vulnificus infections in the Eastern United States increased eightfold from 1988–2018 and the border of where these infections occur has migrated 48 km Northward each year. In July–August 2023, Connecticut, New York, and North Carolina, reported severe and fatal V. vulnificus infections. On August 8, 2025, a V. vulnificus infection was confirmed in a swimmer who had swum with an open wound at a Falmouth, Massachusetts beach.
Research
[edit]
While the treatment for V. vulnificus can be as straightforward as making the rapid choice of appropriate antibiotics, there have been cases in which the genes mutated, thus rendering antibiotics ineffective. While looking for an answer to this problem, researchers found that one way to stop the infection from spreading is to again mutate the bacteria. This mutation happens on the flagellum of the bacteria. When injected with flgC and flgE (two genes in the flagella that cause the mutation), the flagellum no longer function properly. When unable to move normally, the bacteria is no longer able to spread toxins through the body, thus decreasing the effect that V. vulnificus has on the body systemically.
History
[edit]
The pathogen was first isolated in 1976 from a series of blood culture samples submitted to the CDC in Atlanta. It was described as a "lactose-positive vibrio". It was initially given the name Beneckea vulnifica, and then finally the name Vibrio vulnificus by J. J. Farmer in 1979.
In 2005, health officials clearly identified strains of V. vulnificus infections among evacuees from New Orleans due to the flooding there caused by Hurricane Katrina.
In 2015, eight cases of V. vulnificus infection were reported in Florida, with two resulting in death.
In 2022, following Hurricane Ian, Lee County, Florida, saw a sharp rise in infections and deaths from V. vulnificus. By October 18, 2022, four deaths and 29 illnesses had been recorded since the landfall of the hurricane in late September.
In 2023, it was reported that a 40-year-old woman from California had been infected by Vibrio vulnificus after eating undercooked fish (tilapia). She had her four limbs removed to save her life. However, upon further investigation, local public health officials reported no evidence of V. vulnificus infection in her case.
See also
[edit]
Necrotizing fasciitis
References
[edit]
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^ Gold, Scott (September 6, 2005). "Newest Peril from Flooding Is Disease". Los Angeles Times.
^ Katy Galimberti (June 18, 2015). "Flesh-Eating Bacteria Kills Two in Florida as Water Temperatures Rise". msn.com. Archived from the original on August 5, 2016. Retrieved August 1, 2016.
^ Matza, Max (2022-10-19). "Florida flesh-eating illness cases spike after Hurricane Ian". BBC News. Archived from the original on 2022-10-18. Retrieved 2022-10-18.
^ Melissa Montoya; Amy Oshier (October 14, 2022). "Michigan man dies from bacteria while helping in Ian recovery". Wink News, Fort Meyers Broadcasting Company. Archived from the original on 21 October 2022. Retrieved 21 October 2022.
^ Alund, Natalie Neysa (2023-09-19). "California mother's limbs amputated after flesh-eating bacteria infection linked to fish: Report". USA Today. Archived from the original on 2023-09-21. Retrieved 2023-09-20.
^ O'Neill, Maggie (2023-09-21). "Health officials cast doubt on report of woman who had limbs amputated after bacterial infection from tilapia". Independent. Archived from the original on 2023-10-03. Retrieved 2023-10-19.
External links
[edit]
Large and detailed article on V. vulnificus at Todar's Online Textbook of Bacteriology
CNN video on vibrio vulnificus
Type strain of Vibrio vulnificus at BacDive - the Bacterial Diversity Metadatabase
| v t e Pseudomonadota-associated Gram-negative bacterial infections | |
--- |
| α | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- | | Rickettsiales | | | | | | | | | | | | | | --- --- --- --- --- --- | | Rickettsiaceae/ (Rickettsioses) | | | | --- | | Typhus | Rickettsia typhi + Murine typhus Rickettsia prowazekii + Epidemic typhus, Brill–Zinsser disease, Flying squirrel typhus | | Spotted fever | | | | --- | | Tick-borne | Rickettsia rickettsii + Rocky Mountain spotted fever Rickettsia conorii + Boutonneuse fever Rickettsia japonica + Japanese spotted fever Rickettsia sibirica + North Asian tick typhus Rickettsia australis + Queensland tick typhus Rickettsia honei + Flinders Island spotted fever Rickettsia africae + African tick bite fever Rickettsia parkeri + American tick bite fever Rickettsia aeschlimannii + Rickettsia aeschlimannii infection | | Mite-borne | Rickettsia akari + Rickettsialpox Orientia tsutsugamushi + Scrub typhus | | Flea-borne | Rickettsia felis + Flea-borne spotted fever | | | | Anaplasmataceae | Ehrlichiosis: Anaplasma phagocytophilum + Human granulocytic anaplasmosis, Anaplasmosis Ehrlichia chaffeensis + Human monocytotropic ehrlichiosis Ehrlichia ewingii + Ehrlichiosis ewingii infection | | | Hyphomicrobiales | | | | --- | | Brucellaceae | Brucella abortus + Brucellosis | | Bartonellaceae | Bartonellosis: Bartonella henselae + Cat-scratch disease Bartonella quintana + Trench fever Either B. henselae or B. quintana + Bacillary angiomatosis Bartonella bacilliformis + Carrion's disease, Verruga peruana | | |
| β | | | | | | | | | | --- --- --- --- | | Neisseriales | | | | --- | | M+ | Neisseria meningitidis/meningococcus + Meningococcal disease, Waterhouse–Friderichsen syndrome, Meningococcal septicaemia | | M− | Neisseria gonorrhoeae/gonococcus + Gonorrhea | | ungrouped: | Eikenella corrodens/Kingella kingae + HACEK Chromobacterium violaceum + Chromobacteriosis infection | | | Burkholderiales | Burkholderia pseudomallei + Melioidosis Burkholderia mallei + Glanders Burkholderia cepacia complex Bordetella pertussis/Bordetella parapertussis + Pertussis | |
| γ | | | | | | | | | | | | | | --- --- --- --- --- --- | | Enterobacteriales (OX−) | | | | --- | | Lac+ | Klebsiella pneumoniae + Rhinoscleroma, Pneumonia Klebsiella granulomatis + Granuloma inguinale Klebsiella oxytoca Escherichia coli: Enterotoxigenic Enteroinvasive Enterohemorrhagic O157:H7 O104:H4 + Hemolytic-uremic syndrome Enterobacter aerogenes/Enterobacter cloacae | | Slow/weak | Serratia marcescens + Serratia infection Citrobacter koseri/Citrobacter freundii | | Lac− | | | | --- | | H2S+ | Salmonella enterica + Typhoid fever, Paratyphoid fever, Salmonellosis | | H2S− | Shigella dysenteriae/sonnei/flexneri/boydii + Shigellosis, Bacillary dysentery Proteus mirabilis/Proteus vulgaris Yersinia pestis + Plague/Bubonic plague Yersinia enterocolitica + Yersiniosis Yersinia pseudotuberculosis + Far East scarlet-like fever | | | | Pasteurellales | | | | --- | | Haemophilus: | H. influenzae + Haemophilus meningitis + Brazilian purpuric fever H. ducreyi + Chancroid H. parainfluenzae + HACEK | | Pasteurella multocida | Pasteurellosis Actinobacillus + Actinobacillosis | | Aggregatibacter actinomycetemcomitans | HACEK | | | Legionellales | Legionella pneumophila/Legionella longbeachae + Legionnaires' disease Coxiella burnetii + Q fever | | Thiotrichales | Francisella tularensis + Tularemia | | Vibrionaceae | Vibrio cholerae + Cholera Vibrio vulnificus Vibrio parahaemolyticus Vibrio alginolyticus Plesiomonas shigelloides | | Pseudomonadales | Pseudomonas aeruginosa + Pseudomonas infection Moraxella catarrhalis Acinetobacter baumannii | | Xanthomonadaceae | Stenotrophomonas maltophilia | | Cardiobacteriaceae | Cardiobacterium hominis + HACEK | | Aeromonadales | Aeromonas hydrophila/Aeromonas veronii + Aeromonas infection | |
| ε | | | | --- | | Campylobacterales | Campylobacter jejuni + Campylobacteriosis, Guillain–Barré syndrome Helicobacter pylori + Peptic ulcer, MALT lymphoma, Gastric cancer Helicobacter cinaedi + Helicobacter cellulitis | |
| v t e Bacterial skin disease | |
--- |
| Gram +ve | | | | | | | | | | | | | | --- --- --- --- --- --- | | Bacillota | | | | --- | | Staphylococcus | Staphylococcal scalded skin syndrome Impetigo Toxic shock syndrome | | Streptococcus | Impetigo Cutaneous group B streptococcal infection Streptococcal intertrigo Cutaneous Streptococcus iniae infection Erysipelas / Chronic recurrent erysipelas Scarlet fever | | Corynebacterium | | | Clostridium | Gas gangrene Dermatitis gangrenosa | | Others | Listeriosis Erysipeloid of Rosenbach Mycoplasma | | | Actinomycetota | | | | --- | | Mycobacterium- related | Aquarium granuloma Borderline lepromatous leprosy Borderline leprosy Borderline tuberculoid leprosy Buruli ulcer Erythema induratum Histoid leprosy Lepromatous leprosy Leprosy Lichen scrofulosorum Lupus vulgaris Miliary tuberculosis Mycobacterium avium–intracellulare complex infection Mycobacterium haemophilum infection Mycobacterium kansasii infection Papulonecrotic tuberculid Primary inoculation tuberculosis Rapid growing mycobacterium infection Scrofuloderma Tuberculosis cutis orificialis Tuberculosis verrucosa cutis Tuberculous cellulitis Tuberculous gumma Tuberculoid leprosy | | Others | Cutaneous actinomycosis Nocardiosis Cutaneous diphtheria infection Arcanobacterium haemolyticum infection Group JK corynebacterium sepsis | | |
| Gram -ve | | | | | | | | | | --- --- --- --- | | Pseudomonadota | | | | --- | | Alpha | Endemic typhus Epidemic typhus Scrub typhus North Asian tick typhus Queensland tick typhus Flying squirrel typhus Trench fever Bacillary angiomatosis African tick bite fever American tick bite fever Rickettsia aeschlimannii infection Rickettsialpox Rocky Mountain spotted fever Human granulocytotropic anaplasmosis Human monocytotropic ehrlichiosis Flea-borne spotted fever Japanese spotted fever Mediterranean spotted fever Flinders Island spotted fever Verruga peruana Brill–Zinsser disease Brucellosis Cat-scratch disease Oroya fever Ehrlichiosis ewingii infection | | Beta | Gonococcemia/Gonorrhea/Primary gonococcal dermatitis Melioidosis Cutaneous Pasteurella hemolytica infection Meningococcemia Glanders Chromobacteriosis infection | | Gamma | Pasteurellosis Tularemia Vibrio vulnificus Rhinoscleroma Haemophilus influenzae cellulitis Pseudomonal pyoderma / Pseudomonas hot-foot syndrome / Hot tub folliculitis / Ecthyma gangrenosum / Green nail syndrome Q fever Salmonellosis Shigellosis Plague Granuloma inguinale Chancroid Aeromonas infection | | | Campylobacterota | Helicobacter cellulitis | | Other | Syphilid Syphilis Chancre Yaws Pinta Bejel Chlamydia infection Leptospirosis Rat-bite fever Lyme disease Lymphogranuloma venereum Gram-negative folliculitis Gram-negative toe web infection | |
| Unspecified pathogen | Abscess + Periapical abscess Boil/furuncle + Hospital furunculosis Carbuncle Cellulitis + Paronychia / Pyogenic paronychia + Perianal cellulitis Acute lymphadenitis Pilonidal cyst Pyoderma Folliculitis + Superficial pustular folliculitis + Sycosis vulgaris Pimple Ecthyma Pitted keratolysis Trichomycosis axillaris Necrotizing fasciitis Gangrene + Chronic undermining burrowing ulcer + Fournier gangrene Elephantiasis nostras Blistering distal dactylitis Botryomycosis Malakoplakia Pyomyositis Blastomycosis-like pyoderma Bullous impetigo Chronic lymphangitis Recurrent toxin-mediated perineal erythema Tick-borne lymphadenopathy Tropical ulcer |
| Taxon identifiers | |
--- |
| Vibrio vulnificus | Wikidata: Q135183 Wikispecies: Vibrio vulnificus BacDive: 17276 CoL: 5BBY8 EoL: 973197 EPPO: VIBRVU GBIF: 3222430 IRMNG: 11169103 ISC: 117833 ITIS: 967565 LPSN: vibrio-vulnificus NCBI: 672 NZOR: 13de6667-0332-4f3e-9190-f4ae17761b25 OBIS: 395086 Open Tree of Life: 755235 SeqCode Registry: 22345 WoRMS: 395086 |
| | |
--- |
| Authority control databases: National | |
Retrieved from "
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positional notation
a type of numeration in which the position of a digit affects its value.
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positional notation
noun
the method of denoting numbers by the use of a finite number of digits, each digit having its value multiplied by its place value, as in 936 = (9 × 100) + (3 × 10) + 6
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Origin of positional notation1
First recorded in 1940–45
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第十五章 分式方程
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第十五章 分式方程
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分式方程是继一元一次方程（整式方程）之后学习的另一类方程，它的特征主要体现在分母中含有未知数。求解这类方程的思路：要将分式方程转化为整式方程，即将新问题转化为旧问题，通过求解整式方程得到分式方程的解，而转化的方法是要在方程的两边同时乘以若干分式的最简公分母。去分母后整式方程有解，但分式方程并不一定有解，这是因为整式方程的解可能使得分母为0.
[x]
【归纳】：解分式方程的步骤：①去分母，化为整式方程；②解整式方程；③检验整式方程的解是否为分式方程的解，将整式方程的解代入最简公分母。若不为0，则整式方程的解就是分式方程的解；若为0，则分式方程无解，这个整式方程的解也称为分式方程的增根。
[x]
【归纳】：求解关于x的分式方程，若还含有其他字母，将这些字母当做常数对待参与运算，解方程的过程与不含参数的解法相同。
【归纳】：x=2使得分式方程的分母为0，将x=2称为分式方程的增根.
【归纳】：分式方程有增根，说明分式方程无解，而去分母之后的整式方程有解。在这道题目中，分式方程的分母为x-2,这也就是说整式方程的根（解）使得x-2=0,此时x=2,所以将x=2代入整式方程就可以求出参数m的值.
【归纳】：分式方程有增根，说明分式方程无解，而去分母之后的整式方程有解。在这道题目中，分式方程的分母为(x-2)(x+2),这也就是说整式方程的根（解）使得(x-2)(x+2)=0,此时x=2或者x=-2,所以将x=2或者x=-2代入整式方程就可以求出参数a的值.
【归纳】：分式方程无解分为两种情形：①分式方程有增根；②去分母之后的整式方程无解。对于第①种情形类比（3）问即可；对于第②种情形，要将去分母之后的整式方程整理成“ax=b”的形式，当a-1=0时，整式方程无解，原分式方程也无解。
[x]
【归纳】：分式方程的解为非负数说明分式方程一定有解，分式方程有解则去分母之后的整式方程也一定有解。因此，求参数的取值范围必须同时满足三个条件：①整式方程有解；②整式方程的解不能使得分式方程的分母为0；③整式方程的解大于等于0.就这道题而言，去分母整理之后整式方程为x=4-3a，说明整式方程一定有解；当x=2时，分式方程无解，所以4-3a≠2；又因分式方程的解为非负数，所以4-3a≥0.
拓展：若去分母整理之后整式方程为(a-1)x=4-3a,则需要考虑a-1≠0的情形,在此种情形下，分式方程才可能有解.
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[FREE] Prove the identity: \frac{\sin(\pi/2 + x)}{\sin(\pi - x)} = \cot x Note that each statement must be - brainly.com
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Prove the identity:
sin(π−x)sin(π/2+x)=cot x
Note that each statement must be based on a rule chosen from the rule menu.
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The trigonometric identity sin(π/2 + x)/sin(π - x) = cot(x) is true and is verified below
How to prove the trigonometric identity
From the question, we have the following parameters that can be used in our computation:
sin(π/2 + x)/sin(π - x) = cot(x)
By definition,
sin(π - x) = sin(x) and sin(π/2 + x) = cos(x)
So, we have the following equation
cos(x)/sin(x) = cot(x)
Evaluate
cot(x) = cot(x)
The above equation is true
This is so because both sides of the equation are equal
Hence, the trigonometric identity is true
Read more about the trigonometric ratio at
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We proved the identity s i n(π−x)s i n(2 π+x)=cot x by using trigonometric identities for sine functions. By substituting the sine values, we found that both sides of the equation are indeed equal. Therefore, the identity holds true.
Explanation
To prove the identity s i n(π−x)s i n(2 π+x)=cot x, we will use some fundamental trigonometric identities.
Using the sine addition formula: The sine of the sum of two angles can be expressed as:
sin(2 π+x)=cos(x)
Using the sine subtraction formula: The sine function has the property:
sin(π−x)=sin(x)
Substituting these values into the left side of our original equation:
sin(π−x)sin(2 π+x)=sin(x)cos(x)
Recognizing the right side: The right side of the equation, cot(x), is defined as the ratio of the cosine of x to the sine of x:
cot(x)=sin(x)cos(x)
Equating both sides:
Since both sides of the equation are equall, we get:
sin(π−x)sin(2 π+x)=cot(x)
Thus, we have shown that the identity is true. The steps taken confirm that both sides are equal under the definitions of sine and cotangent.
Examples & Evidence
For example, if x=0, then s i n(π)s i n(2 π)=0 1 which is undefined, while cot(0) is also undefined, maintaining equality. For x=4 π, both sides simplify to 1 showing equality.
This proof is based on established trigonometric identities which are widely accepted in mathematics.
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Sequence problem : Find |a 1||a 1|
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Given a sequence such that
a 1 a 2=1;a 2 a 3=2;a 3 a 4=3;…a 1 a 2=1;a 2 a 3=2;a 3 a 4=3;…
and
lim n→∞a n a n+1=1 lim n→∞a n a n+1=1
find |a 1||a 1|.
My attempts :
We can deduce from this a 1 a 2=1;a 2 a 3=2;a 3 a 4=3;…a 1 a 2=1;a 2 a 3=2;a 3 a 4=3;… that :
∏n=k n+1 a n=n∏n=k n+1 a n=n
Thus :
a n×a n+1 a n+1=n=n a n a n×a n+1=n a n+1=n a n
Therefore :
lim n→∞a n a n+1=lim n→∞a n n a n=lim n→∞a 2 n n=1 lim n→∞a n a n+1=lim n→∞a n n a n=lim n→∞a n 2 n=1
Is this can lead us to say that :
|a 2 n||a n|∼n∼n−−√|a n 2|∼n|a n|∼n
If this is true we can say that :
|a n|=1|a n|=1
But I don't know if this is true or not. Any tips or hints ?
Thanks in advance !
calculus
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edited Jun 27, 2021 at 9:12
user26857
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asked Jun 8, 2020 at 16:38
user784692 user784692
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The equivalence is for large n n, with a choice of error ε>0 ε>0. This is no use to try to compute an exact value at n=1 n=1 with it.Didier –Didier 2020-06-08 16:48:10 +00:00 Commented Jun 8, 2020 at 16:48
@DIdier_ Yeah, That's why I thought that it's false ! I used btw the definiton and I didn't solved it :'( !user784692 –user784692 2020-06-08 16:52:35 +00:00 Commented Jun 8, 2020 at 16:52
to me, it seems like there is no solution as many sequences can have the asymptotic behavior a n a n+1→1 a n a n+1→1 with prescribed values of a 1 a 2 a 1 a 2, a 2 a 4 a 2 a 4 and a 3 a 4 a 3 a 4.Didier –Didier 2020-06-08 16:59:25 +00:00 Commented Jun 8, 2020 at 16:59
Why do you have |a 1||a 1|? Do you suspect it can be negative?Alex –Alex 2020-06-08 17:26:53 +00:00 Commented Jun 8, 2020 at 17:26
@Alex , that was the given question user784692 –user784692 2020-06-08 17:35:00 +00:00 Commented Jun 8, 2020 at 17:35
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We find a 1/a 3=1/2 a 1/a 3=1/2, a 1 a 4=1⋅3/2 a 1 a 4=1⋅3/2, a 1/a 5=1⋅3/(2⋅4)a 1/a 5=1⋅3/(2⋅4). By induction, it one shows that
a 1 a 2 n=1⋅3⋅⋯⋅(2 n−1)2⋅4⋅⋯⋅(2 n−2)a 1 a 2 n=1⋅3⋅⋯⋅(2 n−1)2⋅4⋅⋯⋅(2 n−2)
for integer n≥2 n≥2, whereas
a 1/a 2 n+1=1⋅3⋅⋯⋅(2 n−1)2⋅4⋅⋯⋅(2 n)a 1/a 2 n+1=1⋅3⋅⋯⋅(2 n−1)2⋅4⋅⋯⋅(2 n)
for integers n≥1 n≥1.
Since a 2 n/a 2 n+1→1 a 2 n/a 2 n+1→1 by assumption, we have
a 2 1=lim n→∞(1⋅3⋅⋯⋅(2 n−1))2(2⋅4⋅⋯⋅(2 n−2))2(2 n)=lim n→∞(2 n−1)!2 2 4 n−3(n−1)!4 n=lim n→∞(2 n)!2 n 2 4 n−1 n!4.a 1 2=lim n→∞(1⋅3⋅⋯⋅(2 n−1))2(2⋅4⋅⋯⋅(2 n−2))2(2 n)=lim n→∞(2 n−1)!2 2 4 n−3(n−1)!4 n=lim n→∞(2 n)!2 n 2 4 n−1 n!4.
A calculation using Stirling's formula shows that a 2 1=2/π a 1 2=2/π and hence
|a 1|=2 π−−√.|a 1|=2 π.
Edit: a) Let me complete the solution by a calculation of the limit using Stirling's formula twice:
2 log((2 n)!)−4 log(n!)−(4 n−1)log 2+log n==2[(2 n+1 2)(log n+log 2)−2 n+1 2 log(2 π)]−4[(n+1 2)log n−n+1 2 log(2 π)]+log n−(4 n−1)log 2+O(1 n)2 log 2−log(2 π)+O(1 n)→log(2/π).2 log((2 n)!)−4 log(n!)−(4 n−1)log2+logn=2[(2 n+1 2)(logn+log2)−2 n+1 2 log(2 π)]−4[(n+1 2)logn−n+1 2 log(2 π)]+logn−(4 n−1)log2+O(1 n)=2 log2−log(2 π)+O(1 n)→log(2/π).
b) The solution also works in the complex domain. Then we obtain a 1=±2 π−−√a 1=±2 π which is more precise in the complex domain.
c) For arbitrary a 1 a 1, the solution shows that
a 2 n a 2 n+1→2 π a 2 1.a 2 n a 2 n+1→2 π a 1 2.
As a 2 n a 2 n+2=2 n 2 n+1→1 a 2 n a 2 n+2=2 n 2 n+1→1, this implies that a 2 n+1 a 2 n+2→π a 2 1 2.a 2 n+1 a 2 n+2→π a 1 2 2. Therefore a n/a n+1→1 a n/a n+1→1 if and only if a 2 1=2/π a 1 2=2/π.
d) One can also write using the first two formulas of the solution and a 2 n+2/a 2 n+1→1 a 2 n+2/a 2 n+1→1
a 2 1=lim n→∞a 2 1 a 2 n+2 a 2 n+1=lim n→∞(1⋅3⋅⋯⋅(2 n−1))2(2 n+1)(2⋅4⋅⋯⋅(2 n−2)(2 n))2=1⋅3 2 2⋅3⋅5 4 2⋯=2 π a 1 2=lim n→∞a 1 2 a 2 n+2 a 2 n+1=lim n→∞(1⋅3⋅⋯⋅(2 n−1))2(2 n+1)(2⋅4⋅⋯⋅(2 n−2)(2 n))2=1⋅3 2 2⋅3⋅5 4 2⋯=2 π
by the Wallis product.
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edited Jun 9, 2020 at 17:33
answered Jun 8, 2020 at 17:44
HelmutHelmut
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Thank you, this is a great answer !user784692 –user784692 2020-06-08 18:05:22 +00:00 Commented Jun 8, 2020 at 18:05
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Since a n+1=n/a n a n+1=n/a n the sequence a n a n is fully determined from the starting value a 1 a 1.
In fact it turns out that
a n=n−1 a n−1=n−1 n−2 a n−2=⋯=(n−1)!!(n−2)!!a(−1)n+1 1 a n=n−1 a n−1=n−1 n−2 a n−2=⋯=(n−1)!!(n−2)!!a 1(−1)n+1
Where N!!N!! denotes the semifactorial. From this you get that
a n a n+1=((n−1)!!)2(n−2)!!n!!(a 1)2(−1)n+1 a n a n+1=((n−1)!!)2(n−2)!!n!!(a 1)2(−1)n+1
In order to analyze asymptotic behaviour you may want distinguish even and odd n n
a 2 n a 2 n+1=((2 n−1)!!)2(2 n)!!(2 n−2)!!a−2 1=((2 n−1)!!(2 n)!!)2 2 n a 2 1=((2 n)!(2 n)!!2)2 2 n a 2 1 a 2 n a 2 n+1=((2 n−1)!!)2(2 n)!!(2 n−2)!!a 1−2=((2 n−1)!!(2 n)!!)2 2 n a 1 2=((2 n)!(2 n)!!2)2 2 n a 1 2
a 2 n+1 a 2 n+2=((2 n)!!)2(2 n+1)!!(2 n−1)!!a 2 1=((2 n)!!(2 n−1)!!)2 a 2 1 2 n+1=((2 n)!!2(2 n)!)2 a 2 1 2 n+1 a 2 n+1 a 2 n+2=((2 n)!!)2(2 n+1)!!(2 n−1)!!a 1 2=((2 n)!!(2 n−1)!!)2 a 1 2 2 n+1=((2 n)!!2(2 n)!)2 a 1 2 2 n+1
Passing to logarithms you get
log(a 2 n a 2 n+1)=2(log((2 n)!)−2 log((2 n)!!))+log(2 n)−2 log(a 1)log(a 2 n a 2 n+1)=2(log((2 n)!)−2 log((2 n)!!))+log(2 n)−2 log(a 1)
log(a 2 n+1 a 2 n)=2(2 log((2 n)!!)−log((2 n)!))−log(2 n+1)+2 log(a 1)log(a 2 n+1 a 2 n)=2(2 log((2 n)!!)−log((2 n)!))−log(2 n+1)+2 log(a 1)
Recall the Stirling approximation
log(n!)=n log(n)−n+1 2 log(n)+log(2 π−−√)+o(1)log(n!)=n log(n)−n+1 2 log(n)+log(2 π)+o(1)
from this you get approximations
log((2 n)!!)=log(2 n n!)=n log(n)+(log(2)−1)n+1 2 log(n)+log(2 π−−√)+o(1)log((2 n)!!)=log(2 n n!)=n log(n)+(log(2)−1)n+1 2 log(n)+log(2 π)+o(1)
log((2 n)!)=2 n log(n)+(2 log(2)−2)n+1 2 log(n)+log(2 π−−√)+o(1)log((2 n)!)=2 n log(n)+(2 log(2)−2)n+1 2 log(n)+log(2 π)+o(1)
Subtracting you have 2 log((2 n)!!)−log((2 n)!)=1 2 log(n)+log(π−−√)+o(1)2 log((2 n)!!)−log((2 n)!)=1 2 log(n)+log(π)+o(1), so
log(a 2 n a 2 n+1)=log(2)−log(π)−2 log(a 1)+o(1)log(a 2 n a 2 n+1)=log(2)−log(π)−2 log(a 1)+o(1)
log(a 2 n+1 a 2 n)=log(π)−log(2)+2 log(a 1)+o(1)log(a 2 n+1 a 2 n)=log(π)−log(2)+2 log(a 1)+o(1)
Hence there is a value of a 1 a 1 for which a n/a n+1→1 a n/a n+1→1 and it has to satisfy 2 log(a 1)=log(2/π)2 log(a 1)=log(2/π), that is a 1=2/π−−−√a 1=2/π.
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answered Jun 8, 2020 at 22:23
Dunnò000Dunnò000
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For me this should be the right answer, 1 upvote from me !euler_med –euler_med 2020-06-09 09:51:02 +00:00 Commented Jun 9, 2020 at 9:51
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You may notice that a celebrated sequence has very similar properties: by defining
I n=∫π/2 0(sin θ)n d θ I n=∫0 π/2(sinθ)n d θ
we have
I 2 n=π 2⋅4 n(2 n n),I 2 n+1=4 n(2 n+1)(2 n n)I 2 n=π 2⋅4 n(2 n n),I 2 n+1=4 n(2 n+1)(2 n n)
by integration by parts. {I n}n≥1{I n}n≥1 is obviously decreasing to zero and fulfills I n I n+1=π 2(n+1)I n I n+1=π 2(n+1).
Moreover, by Euler's Beta function (or just from the previous line) we have
I n=π−−√Γ(n+1 2)2 Γ(n+2 2)∼π 2 n−−−√.I n=π Γ(n+1 2)2 Γ(n+2 2)∼π 2 n.
It follows that the sequence
a n=π/2−−−√∫π/2 0(sin θ)n−1 d θ a n=π/2∫0 π/2(sinθ)n−1 d θ
meets the hypothesis of our problem.
Provided that the solution is unique, this proves a 1=2 π−−√a 1=2 π without invoking Stirling's approximation.
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edited Jun 12, 2020 at 14:04
answered Jun 11, 2020 at 10:10
Jack D'AurizioJack D'Aurizio
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I n I n is used in the proof of the Wallis product formula - without even using Euler's Beta function.Helmut –Helmut 2020-06-12 10:49:52 +00:00 Commented Jun 12, 2020 at 10:49
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2Check my solution: Find lim x→+∞((x+a 1)(x+a 2)−−−−−−−−−−−−−√−x)lim x→+∞((x+a 1)(x+a 2)−x)
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1Recursive sequence problem...
2If a 1 a 2=1,a 2 a 3=2,a 3 a 4=3⋯a 1 a 2=1,a 2 a 3=2,a 3 a 4=3⋯ and lim n→∞a n a n+1=1 lim n→∞a n a n+1=1, find |a 1||a 1|
4Prove ∑n p+1 a 1+2 p a 2+⋯+n p a n∑n p+1 a 1+2 p a 2+⋯+n p a n is convergent.
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160
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https://www.egrammatics.com/it-where-to-use-it/
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Where to use ‘IT’ – 7 uses
It is a pronoun used for nouns of neuter gender. The possessive case of ‘It’ is ‘Its‘.
The most general idea about this pronoun is that it is used for non-living things. But this a very narrow use of It as a pronoun. It is used in many different scenarios. It can be used as both subject and object. Let us look at it with some examples.
1. For things without life:
This is the most basic use. Nouns that are lifeless or noun that belongs to neuter gender are replaced using pronoun ‘it’.
Examples:
Court has passed a new privacy law. It was really the need of the hour.
She gifted me a coffee mug. It is a personalized mug with my picture on it.
My nephew broke the flower vase. It was an antique.
She sold her old car as it was just standing in her parking lot for years.
2. For animals when we do not clearly specify their gender:
When someone specifically talks with reference to the gender of an animal, they use gender specific personal pronouns. But when an animal is referred in general or we are not aware of the gender, we use personal pronoun ‘it’ which is used for neuter gender.
Examples:
We found a tiny cat crying in our backyard, it was all wet in the rain.
That is her favorite horse to ride on, it is very strong.
It was a german shepherd pup.
Our dog crashed into the car and hit its head.
3. For young child or babies when we do not specify their gender:
Similar to the case of animals, when someone specifically talks with reference to the gender of a baby, they use gender specific personal pronouns. But when a baby is referred in general or we are not aware of the gender, we use personal pronoun ‘it’ which is used for neuter gender.
Examples:
The child started crying, it is scared of the clown.
Her baby is sick, it caught cold.
4. Used as OBJECT to refer to an already stated action or information (to refer to previous sentence):
Many times, we state something, usually an action or information. And then we wish to restate the same action or information later to some other person. We can use ‘it’ for the phrase talking about that information to avoid repetition.
Examples:
Lilly spilt the milk and she is sorry about it.
I told my mother about the theft, but she already knew it.
Don’t call him now, we have already discussed it.
5. Used as temporary SUBJECT in passive sentences when real subject follows:
Certain passive sentences have no immediate subject. Either the subject here is understood or followed after pronoun ‘it’.
Examples:
It was appreciated that the entire nation stood together in the atrocity.
It is advised to exercise daily.
It is easy to blame others.
It has been adopted as an universal principal.
6. Used before a NOUN or PRONOUN to emphasis on it:
Sometimes, we want to emphasis or highlight the noun or pronoun. This is done by adding ‘it’ followed by a ‘to be’form of verb and the noun or pronoun.
Examples:
It was I who called you.
It was in Mumbai, where the event took place.
It is he who won the championship twice.
It was they who are my teammates.
7. Used as SUBJECT of Impersonal Verbs:
Impersonal Verbs is the verb that has no definite subject. For these verbs subject is not explicit though the subject is quite comprehensible. For such verbs ‘it’ is used as subject and here it is known as Impersonal Pronoun.
Examples:
It is extremely hot today. (the weather)
It rained quite heavily. (the rain rained heavily)
It is 12 o’clock in the afternoon. (the time is 12 o’clock)
It has been adopted as an universal principal.
These are the scenarios where ‘It’ is used as a pronoun. Here we saw that this pronoun has vast usage.
Also Read:
Personal Pronoun
Personal Pronoun: The word ‘personal’ means belonging to a particular...
Read More
Types of Pronouns
Types of Pronouns: English grammar has nine types of pronouns....
Read More
Noun Rules
NOUN : RULES Let us look at some of the...
Read More
Pronoun
What is a Pronoun? Pronouns are a grammatical unit used...
Read More
Practice Test
Practice what you have learned and keep it with you forever. These tests are with proper soloutions and explainations.
Interrogative pronoun
Learn about interrogative pronouns with examples.
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161
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https://www.mathbootcamps.com/graphing-linear-equations/
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Graphing linear equations
There are three ways you can graph linear equations: (1) you can find two points, (2) you can use the y-intercept and the slope, or (3) you can use the x- and y-intercepts. In the following guide, we will look at all three. Graphing linear equations doesn’t have to be difficult, and can even be fun once you have these methods down![adsenseWide]
Method 1: using two points to graph a linear equation
The graph of any linear equation like (y = 3x + 2) or (y = -x + 9) is a line, and only two points are needed to define a line. The idea with this method is to find two points on the line by picking values of (x).
Example
Graph the linear equation:(y = \dfrac{1}{3}x – 2)
Solution
To find two points on the line, pick any two values of (x) that would be easy to work with and then find the corresponding value of (y). Two easy values here would be 0 and 3 (Since the 3 will cancel with the 3 in the fraction.)
Let (x = 0):
(\begin{align}y &= \dfrac{1}{3}x – 2 \ &= \dfrac{1}{3}(0) – 2\ &= -2\end{align})
So, one point on the line is ((x,y) = (0, –2)).
Let (x = 3):
(\begin{align}y &= \dfrac{1}{3}x – 2\ &= \dfrac{1}{3}(3) – 2\ &= 1 – 2\ &= -1\end{align})
So, another point on the line is ((x,y) = (3,–1)).
Once you have your two points, you can plot them and then sketch the line. It is best to use a ruler or something similar to make sure that you draw the best representation of the graph as possible.
TIP: You might find it useful to find three or four points on the graph so that you can sketch it more accurately. To do this, just pick more x-values to find points!
Method 2: Use the slope and y-intercept
A linear equation written in the form (y = mx + b) is said to be in slope-intercept form. This form shows the slope (m) and the y-intercept (b) of the graph. Knowing these two values will let you quickly draw the graph of the linear equation, as you can see in the example below.
Example
Graph the linear equation:(y = \dfrac{2}{3}x + 4)
Solution
Since this equation is in the form (y = mx + b), you know that:
The slope is:(m=\dfrac{2}{3})
The y-intercept is:(b = 4)
Let’s look at the y-intercept first.
The y-intercept is the point where the graph crosses the y-axis (the vertical axis). So, you can plot this point as:
Now consider the slope. Slope can be viewed as a rate of change: it represents the change in (y) over the change in (x). Sometimes, this is called “rise over run”.
(\text{slope} = \dfrac{\text{rise}}{\text{run}} = \dfrac{\text{change in }y}{\text{change in }x})
For this example:
(\text{slope} = \dfrac{2}{3} = \dfrac{\text{change in }y}{\text{change in }x})
This can be translated to:
(\dfrac{2}{3} = \dfrac{\text{up 2 units}}{\text{for every 3 units right}})
Therefore, to find another point on the line, start with the y-intercept and go 3 unit right and 2 units up. Do this again, and you will find another point. In fact, you can keep doing this to find as many points as you think you will need to sketch a good graph.
Connect the points, and you have the graph of your linear function!
This method for graphing linear equations can be applied even when the slope is negative or when the slope is not a fraction, even if it doesn’t seem that way. The next example will show you how that works!
Example
Graph the linear equation:(y = -2x + 1)
Solution
The y-intercept here is 1, so plot this point first.
The slope is –2. While this isn’t a fraction, it can be viewed as one if you let the denominator equal 1.
(\text{slope} = -2 = \dfrac{-2}{1})
In other words:
(\dfrac{-2}{1} = \dfrac{\text{down two units}}{\text{for every 1 unit right}})
Now this can be applied to find points on the graph.
Finally, connect the points to sketch the graph of the linear equation.
TIP: If the slope is in decimal form, see if you can convert it to a fraction to apply this method. Otherwise, method 1 might be best.
Method 3: Using the x- and y-intercepts
When graphing linear equations that are given in the form (y = mx + b), it is easiest to just apply method 2. But sometimes, linear equations are given in standard form: (Ax + By = C), where (A), (B), and (C) are positive or negative whole numbers. In this case, using the x- and y-intercept may be the quickest approach.
Example
Graph the linear equation:(-3x + 2y = 6)
Solution
To find the x-intercept, which is the point where the graph crosses the x-axis, let (y = 0) and solve for (x):
(\begin{align}-3x + 2y &= 6\ -3x + 2(0) &= 6\-3x &= 6\x &= -2\end{align})
To find the y-intercept, which is the point where the graph crosses the y-axis, let (x = 0) and solve for (y):
(\begin{align}-3(0) + 2y &= 6\2y &= 6\y &= 3\end{align})
This gives you two points on the line, which you can plot and then connect to graph the linear equation.
To read more about x-intercepts and y-intercepts, check out the articles Understanding x-intercepts and Understanding y-intercepts.
[adsenseLargeRectangle]
Summary
When graphing linear equations, there are many possible approaches. Three common approaches are:
Finding points:pick simple values of (x) and find the corresponding values of (y). Plot these points and use these to graph your line.
Using the slope and y-intercept:use the concept of “rise over run” and the y-intercept to find points on the graph. This method is especially useful is the line is in slope-intercept form.
Using the x- and y-intercepts:let (x = 0) and (y = 0) to find the intercepts of the graph. Then, use these points to plot the line. This method is useful when the linear equation is in standard form.
Which method you use depends on the form of the linear equation you have and which method you are most comfortable with. No matter what, you can always find points if you get stuck.
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https://www.geeksforgeeks.org/dsa/generating-all-divisors-of-a-number-using-its-prime-factorization/
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Generating all divisors of a number using its prime factorization - GeeksforGeeks
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Generating all divisors of a number using its prime factorization
Last Updated : 12 Jul, 2025
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Given an integer N, the task is to find all of its divisors using its prime factorization.
Examples:
Input: N = 6
Output: 1 2 3 6
Input: N = 10
Output: 1 2 5 10
Approach: As every number greater than 1 can be represented in its prime factorization as p 1 a 1p 2 a 2......p k a k, where p i is a prime number, k ? 1 and a i is a positive integer.
Now all the possible divisors can be generated recursively if the count of occurrence of every prime factor of n is known. For every prime factor p i, it can be included x times where 0 ? x ? a i. First, find the prime factorization of n using this approach and for every prime factor, store it with the count of its occurrence.
Below is the implementation of the above approach:
C++
```cpp
// C++ implementation of the approach
include "iostream"
include "vector"
using namespace std;
struct primeFactorization {
// to store the prime factor
// and its highest power
int countOfPf, primeFactor;
};
// Recursive function to generate all the
// divisors from the prime factors
void generateDivisors(int curIndex, int curDivisor,
vector& arr)
{
// Base case i.e. we do not have more
// primeFactors to include
if (curIndex == arr.size()) {
cout << curDivisor << ' ';
return;
}
for (int i = 0; i <= arr[curIndex].countOfPf; ++i) {
generateDivisors(curIndex + 1, curDivisor, arr);
curDivisor = arr[curIndex].primeFactor;
}
}
// Function to find the divisors of n
void findDivisors(int n)
{
// To store the prime factors along
// with their highest power
vector<primeFactorization> arr;
// Finding prime factorization of n
for (int i = 2; i i <= n; ++i) {
if (n % i == 0) {
int count = 0;
while (n % i == 0) {
n /= i;
count += 1;
}
// For every prime factor we are storing
// count of it's occurrenceand itself.
arr.push_back({ count, i });
}
}
// If n is prime
if (n > 1) {
arr.push_back({ 1, n });
}
int curIndex = 0, curDivisor = 1;
// Generate all the divisors
generateDivisors(curIndex, curDivisor, arr);
}
// Driver code
int main()
{
int n = 6;
findDivisors(n);
return 0;
}
```
// C++ implementation of the approach
include "iostream"#include "vector"using namespace std;struct primeFactorization { // to store the prime factor // and its highest power int countOfPf, primeFactor;};// Recursive function to generate all the// divisors from the prime factors void generateDivisors(int curIndex, int curDivisor, vector& arr){ // Base case i.e. we do not have more // primeFactors to include if (curIndex == arr.size()) { cout << curDivisor << ' '; return; } for (int i = 0; i <= arr[curIndex].countOfPf; ++i) { generateDivisors(curIndex + 1, curDivisor, arr); curDivisor = arr[curIndex].primeFactor; }}// Function to find the divisors of n void findDivisors(int n){ // To store the prime factors along // with their highest power vector arr; // Finding prime factorization of n for (int i = 2; i i <= n; ++i) { if (n % i == 0) { int count = 0; while (n % i == 0) { n /= i; count += 1; } // For every prime factor we are storing // count of it's occurrenceand itself. arr.push_back({ count, i }); } } // If n is prime if (n > 1) { arr.push_back({ 1, n }); } int curIndex = 0, curDivisor = 1; // Generate all the divisors generateDivisors(curIndex, curDivisor, arr);}// Driver code int main(){ int n = 6; findDivisors(n); return 0;}
Java
```java
// Java implementation of the approach
import java.util.;
class GFG
{
static class primeFactorization
{
// to store the prime factor
// and its highest power
int countOfPf, primeFactor;
public primeFactorization(int countOfPf,
int primeFactor)
{
this.countOfPf = countOfPf;
this.primeFactor = primeFactor;
}
}
// Recursive function to generate all the
// divisors from the prime factors
static void generateDivisors(int curIndex, int curDivisor,
Vector arr)
{
// Base case i.e. we do not have more
// primeFactors to include
if (curIndex == arr.size())
{
System.out.print(curDivisor + " ");
return;
}
for (int i = 0; i <= arr.get(curIndex).countOfPf; ++i)
{
generateDivisors(curIndex + 1, curDivisor, arr);
curDivisor = arr.get(curIndex).primeFactor;
}
}
// Function to find the divisors of n
static void findDivisors(int n)
{
// To store the prime factors along
// with their highest power
Vector<primeFactorization> arr = new Vector<>();
// Finding prime factorization of n
for (int i = 2; i i <= n; ++i)
{
if (n % i == 0)
{
int count = 0;
while (n % i == 0)
{
n /= i;
count += 1;
}
// For every prime factor we are storing
// count of it's occurrenceand itself.
arr.add(new primeFactorization(count, i ));
}
}
// If n is prime
if (n > 1)
{
arr.add(new primeFactorization( 1, n ));
}
int curIndex = 0, curDivisor = 1;
// Generate all the divisors
generateDivisors(curIndex, curDivisor, arr);
}
// Driver code
public static void main(String []args)
{
int n = 6;
findDivisors(n);
}
}
// This code is contributed by Rajput-Ji
Python3python3
Python3 implementation of the approach
Recursive function to generate all the
divisors from the prime factors
def generateDivisors(curIndex, curDivisor, arr):
# Base case i.e. we do not have more
# primeFactors to include
if (curIndex == len(arr)):
print(curDivisor, end = ' ')
return
for i in range(arr[curIndex] + 1):
generateDivisors(curIndex + 1, curDivisor, arr)
curDivisor = arr[curIndex]
Function to find the divisors of n
def findDivisors(n):
# To store the prime factors along
# with their highest power
arr = []
# Finding prime factorization of n
i = 2
while(i i <= n):
if (n % i == 0):
count = 0
while (n % i == 0):
n //= i
count += 1
# For every prime factor we are storing
# count of it's occurrenceand itself.
arr.append([count, i])
# If n is prime
if (n > 1):
arr.append([1, n])
curIndex = 0
curDivisor = 1
# Generate all the divisors
generateDivisors(curIndex, curDivisor, arr)
Driver code
n = 6
findDivisors(n)
This code is contributed by SHUBHAMSINGH10
C#csharp
// C# implementation of the approach
using System;
using System.Collections.Generic;
class GFG
{
public class primeFactorization
{
// to store the prime factor
// and its highest power
public int countOfPf, primeFactor;
public primeFactorization(int countOfPf,
int primeFactor)
{
this.countOfPf = countOfPf;
this.primeFactor = primeFactor;
}
}
// Recursive function to generate all the
// divisors from the prime factors
static void generateDivisors(int curIndex, int curDivisor,
List arr)
{
// Base case i.e. we do not have more
// primeFactors to include
if (curIndex == arr.Count)
{
Console.Write(curDivisor + " ");
return;
}
for (int i = 0; i <= arr[curIndex].countOfPf; ++i)
{
generateDivisors(curIndex + 1, curDivisor, arr);
curDivisor = arr[curIndex].primeFactor;
}
}
// Function to find the divisors of n
static void findDivisors(int n)
{
// To store the prime factors along
// with their highest power
List<primeFactorization> arr = new List<primeFactorization>();
// Finding prime factorization of n
for (int i = 2; i i <= n; ++i)
{
if (n % i == 0)
{
int count = 0;
while (n % i == 0)
{
n /= i;
count += 1;
}
// For every prime factor we are storing
// count of it's occurrenceand itself.
arr.Add(new primeFactorization(count, i ));
}
}
// If n is prime
if (n > 1)
{
arr.Add(new primeFactorization( 1, n ));
}
int curIndex = 0, curDivisor = 1;
// Generate all the divisors
generateDivisors(curIndex, curDivisor, arr);
}
// Driver code
public static void Main(String []args)
{
int n = 6;
findDivisors(n);
}
}
// This code is contributed by PrinciRaj1992
JavaScriptjavascript
// Javascript implementation of the approach
// Recursive function to generate all the
// divisors from the prime factors
function generateDivisors(curIndex, curDivisor, arr)
{
// Base case i.e. we do not have more
// primeFactors to include
if (curIndex == arr.length) {
document.write(curDivisor + " ");
return;
}
for (var i = 0; i <= arr[curIndex]; ++i) {
generateDivisors(curIndex + 1, curDivisor, arr);
curDivisor = arr[curIndex];
}
}
// Function to find the divisors of n
function findDivisors(n)
{
// To store the prime factors along
// with their highest power
arr = [];
// Finding prime factorization of n
for (var i = 2; i i <= n; ++i) {
if (n % i == 0) {
var count = 0;
while (n % i == 0) {
n /= i;
count += 1;
}
// For every prime factor we are storing
// count of it's occurrenceand itself.
arr.push([ count, i ]);
}
}
// If n is prime
if (n > 1) {
arr.push([ 1, n ]);
}
var curIndex = 0, curDivisor = 1;
// Generate all the divisors
generateDivisors(curIndex, curDivisor, arr);
}
// driver code
var n = 6;
findDivisors(n);
// This code contributed by shubhamsingh10
```
Output1 3 2 6
Time Complexity: O(sqrt(n))
Auxiliary Space:O(sqrt(n))
Comment
More info
B
BhupendraYadav
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Article Tags :
Combinatorial
Recursion
DSA
prime-factor
divisors
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Eigenvalues and Eigenvectors of $2 \times 2$ Matrix
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$\begingroup$
Let's say I have a $2 \times 2$ matrix (actually the structure tensor of a discrete image - I): $$ \begin{bmatrix} \frac{\partial I}{\partial x}\frac{\partial I}{\partial x} & \frac{\partial I}{\partial x}\frac{\partial I}{\partial y} \\ \frac{\partial I}{\partial y}\frac{\partial I}{\partial x} & \frac{\partial I}{\partial y}\frac{\partial I}{\partial y} \end{bmatrix}$$
It has 2 properties:
Symmetric.
Positive Semidefinite.
Given those properties, what would be the easiest method to numerically compute its eigenvectors (orthogonal) and eigenvalues?
linear-algebra
matrices
eigenvalues-eigenvectors
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edited May 30, 2013 at 17:18
J. M. ain't a mathematician
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asked Nov 2, 2010 at 17:51
RoyiRoyi
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$\begingroup$ You sometimes need extra backslashes, which has something to do with Markdown. I put in the third backslash to fix the matrix. $\endgroup$
Jonas Meyer
– Jonas Meyer
2010-11-02 17:56:23 +00:00
Commented Nov 2, 2010 at 17:56
$\begingroup$ Jonas, Thank You. Noted my self. $\endgroup$
Royi
– Royi
2010-11-02 21:48:00 +00:00
Commented Nov 2, 2010 at 21:48
$\begingroup$ @Jonas, Since I can't edit the message could you fix a mistake I made? It shouldn't be the second derivative. It should be a multiplication of the first derivatives: Ix Ix, Ix Iy, Iy Ix, Iy Iy. Thanks. $\endgroup$
Royi
– Royi
2010-11-03 09:53:38 +00:00
Commented Nov 3, 2010 at 9:53
$\begingroup$ You should be able to edit all of your own questions and answers. Look to the bottom left of your answer, and click where it says "edit" (if you're logged in). $\endgroup$
Jonas Meyer
– Jonas Meyer
2010-11-03 16:53:03 +00:00
Commented Nov 3, 2010 at 16:53
$\begingroup$ I assumed it is like comments. Fixed it. Thank you. $\endgroup$
Royi
– Royi
2010-11-03 23:55:51 +00:00
Commented Nov 3, 2010 at 23:55
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3 Answers 3
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As a two-by-two matrix, applying Jacobi's method in fact gives the answer at once!
It is known that given a two-by-two symmetric matrix $\mathbf A$, one can construct an orthogonal matrix $\mathbf V=\bigl(\begin{smallmatrix}c&s\\-s&c\end{smallmatrix}\bigr)$ such that $\mathbf V^\top \mathbf A\mathbf V$ is diagonal, where the two numbers $c$ and $s$ satisfy $c^2+s^2=1$.
If the off-diagonal elements are not zero (why?), computing $c$ and $s$ can be done like so:
$$\begin{align} \tau&=\frac{a_{22}-a_{11}}{2a_{12}}\ t&=\frac{\mathrm{sgn}(\tau)}{|\tau|+\sqrt{1+\tau^2}}\ c&=\frac1{\sqrt{1+t^2}}\ s&=ct \end{align}$$
from which the eigendecomposition easily follows.
(See Golub and Van Loan's excellent book for further details.)
Here is a Mathematica demonstration of Jacobi's method:
a = N[{{3, -1}, {-1, 2}}, 20]; tau = (a - a)/a/2; t = Sign[tau]/(Abs[tau] + Sqrt[1 + tau^2]); {c, s} = {1, t}/Sqrt[1 + t^2] {0.85065080835203993218, 0.5257311121191336060} {l1, l2} = {a - t a, a + t a} {3.6180339887498948482, 1.3819660112501051518} Eigenvalues[a] == {l1, l2} True Eigenvectors[a] == {{c, -s}, {s, c}} True a.{c, -s} == l1 {c, -s} True a.{s, c} == l2 {s, c} True
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edited May 11, 2012 at 17:52
answered Nov 2, 2010 at 23:20
J. M. ain't a mathematicianJ. M. ain't a mathematician
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3
$\begingroup$ I leave as an exercise the derivation of the following formulae for the diagonal elements of $\mathbf{V}^T\mathbf{A}\mathbf{V}$: $a_{11}-ta_{12}$ and $a_{22}+ta_{12}$ $\endgroup$
J. M. ain't a mathematician
– J. M. ain't a mathematician
2010-11-02 23:25:04 +00:00
Commented Nov 2, 2010 at 23:25
$\begingroup$ @J.M., On the book you referred me to it says choosing between 2 solutions of t. You made a trick, could you explain it? Because my MATLAB code gets one of the Eigen Vectors in the wrong direction (Multiplied by 1). Thank You. $\endgroup$
Royi
– Royi
2014-10-12 19:53:19 +00:00
Commented Oct 12, 2014 at 19:53
$\begingroup$ @Drazick, "gets one of the eigenvectors in the wrong direction (multiplied by 1)." - you are aware that if $\mathbf x$ is an eigenvector of $\mathbf A$, then any nonzero multiple of $\mathbf x$ is also an eigenvector? $\endgroup$
J. M. ain't a mathematician
– J. M. ain't a mathematician
2015-05-01 13:45:15 +00:00
Commented May 1, 2015 at 13:45
Add a comment |
6
$\begingroup$
Since you work with a $2\times2$ matrix, the corresponding characteristic polynomial is quadratic so the eigenvalues can be expressed in closed form in terms of the matrix elements. As soon as you get the eigenvalues all you have to do is to solve two systems of linear equations with two unknowns (i.e. the coordinates of the corresponding eigenvector in the standard basis). Again, everything is nice and explicit, no need to apply a fancy numerical method.
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edited Nov 2, 2010 at 18:16
answered Nov 2, 2010 at 18:03
Andrey RekaloAndrey Rekalo
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$\endgroup$
3
$\begingroup$ But beware that the formula for the roots of a quadratic equation needs to be implemented with a little thought in order to avoid cancellation! See en.wikipedia.org/wiki/…. $\endgroup$
Hans Lundmark
– Hans Lundmark
2010-11-02 21:38:10 +00:00
Commented Nov 2, 2010 at 21:38
$\begingroup$ I want to implement it on GPU. Hence simplicity is very important both because of performance and the lack of libraries. Hence even solving the linear system isn't "Straightforward" (I'm spoiled by Matlab). $\endgroup$
Royi
– Royi
2010-11-02 21:54:28 +00:00
Commented Nov 2, 2010 at 21:54
3
$\begingroup$ @Drazick: solving a two-by-two homogeneous linear system is straightforward. $\endgroup$
Mariano Suárez-Álvarez
– Mariano Suárez-Álvarez
2010-11-02 23:24:39 +00:00
Commented Nov 2, 2010 at 23:24
Add a comment |
5
$\begingroup$
Despite other answers, I thought it might benefit the impatient to see the explicit answer below.
Let $$M = \left( \begin{array}{cc} a & b \ b & c \end{array} \right),$$ be the input matrix.
Define the discriminant: $\Delta = \sqrt{a^2+4 b^2-2 a c+c^2}$
Then, the eigenvalues of $M$ are given by:
$\lambda_1 = 0.5(a+c-\Delta)$ and $\lambda_2 = 0.5(a+c+\Delta)$
Now, you can find a matrix $V$ such that $$M = V^{-1} \begin{pmatrix} \lambda_1 & 0\ 0 & \lambda_2 \end{pmatrix}V.$$
Mathematica says that the matrix $V$ is given by $$ V = \begin{pmatrix} \frac{a-c-\Delta}{2b} & 1\ \frac{a-c+\Delta}{2b} & 1 \end{pmatrix} $$
If you are looking for orthogonal $V$, then the above calculations need some changes.
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answered Nov 3, 2010 at 21:34
user1709user1709
$\endgroup$
3
$\begingroup$ "Explicit" is great :-). Could you give the explicit solution for Orthogonal V? I don't have access to Mathematica. Thank You! $\endgroup$
Royi
– Royi
2010-11-03 23:23:21 +00:00
Commented Nov 3, 2010 at 23:23
$\begingroup$ @Drazick: The solution I gave returns orthogonal eigenvectors. $\endgroup$
J. M. ain't a mathematician
– J. M. ain't a mathematician
2010-11-04 00:02:37 +00:00
Commented Nov 4, 2010 at 0:02
$\begingroup$ @Drazick: You can use Maxima instead of Mathematica for symbolic solving. $\endgroup$
Samik R
– Samik R
2011-04-01 15:37:21 +00:00
Commented Apr 1, 2011 at 15:37
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Warm-up: Estimation Exploration: Large Denominators (10 minutes)
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Estimation Exploration
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The purpose of this estimation exploration is for students to reason about the size of a complex fraction sum with large denominators. Students can see that 1 is a good estimate because one fraction is small and the other is close to 1. In the synthesis they refine this estimate to explain why the value of the sum is a little larger than 1.
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Student Response
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Activity Synthesis
“How do you know that the sum is greater than 1?” ( is short of a whole. Since 17ths are bigger than 19ths, adding makes it greater than 1.)
Activity 1: Priya’s Salad Dressing (20 minutes)
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Addressing
5.NF.A.2
Routines and Access
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MLR6
Narrative
The purpose of this activity is for students to add and subtract fractions and estimate sums and differences of fractions using the context of a recipe. Students may have different responses and reasoning for the estimation questions. In both cases, they can calculate and compare fractions but they may have different thoughts about how these differences would affect the recipe or what exactly it means for the recipe to make “about cups.” In the synthesis, students discuss the reasonableness of the estimates and how to make precise calculations (MP6). When students relate their calculations to Priya's salad dressing they reason abstractly and quantitatively (MP2).
Reading: MLR6 Three Reads. Keep books or devices closed. Display only the problem stem, without revealing the questions. “We are going to read this question 3 times.” After the 1st Read: “Tell your partner what this situation is about.” After the 2nd Read: “List the quantities. What can be counted or measured?” Reveal the question(s). After the 3rd Read: “What strategies can we use to solve this problem?”
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“What kind of ingredients do you like to put in your salad?” (lettuce, cabbage, beans, seeds, beets, tomatoes, cheese)
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estimate to determine that Priya’s recipe will make about cups of dressing
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Priya’s Salad Dressing Recipe
cup olive oil
cup lemon juice
cup mustard
Pinch of salt and pepper
Priya has cup of olive oil. She is going to borrow some more from her neighbor. How much olive oil does she need to borrow to have enough to make the dressing?
1 tablespoon is equal to of a cup. Priya decides that 1 tablespoon of olive oil is close enough to what she needs to borrow from her neighbor. Do you agree with Priya? Explain or show your reasoning.
Priya says her recipe will make about cups of dressing. Do you agree? Explain or show your reasoning.
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“If Priya borrows a tablespoon of olive oil from her neighbor and uses it to make dressing, will she be putting in more or less olive oil than the recipe calls for?” ( is smaller than so she will be putting in less olive oil.)
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Poll the class.
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Ask previously selected students to share their estimates for the amount of salad dressing in the given order.
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“Does the recipe make more or less than cups? How do you know?” (More because is more than .)
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Representation
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The purpose of this activity is for students to solve multi-step problems involving the addition and subtraction of fractions with unlike denominators. Students work with both fractions and mixed numbers and can use strategies they have learned such as adding on to make a whole number.When students connect the quantities in the story problem to an equation, they reason abstractly and quantitatively (MP2).
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3–5 minutes: partner discussion
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Choose a problem to solve.
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Jada is baking protein bars for a hike. She adds cup of walnuts and then decides to add another cup. How many cups of walnuts has she added altogether?
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Problem B:
Kiran and Jada hiked miles and took a rest. Then they hiked another mile before stopping for lunch. How many miles have they hiked so far?
If the trail they are hiking is a total of miles, how much farther do they have to hike? Explain or show your reasoning.
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Revise your work if necessary.
Student Response
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“How were the problems the same? How were they different?” (For both problems I had to add fractions first and then subtract that total from another number. There were mixed numbers in both problems.)
“How did you use equivalent fractions to solve these problems?” (All the fractions we worked with had different denominators so we had to find equivalent fractions with the same denominators in order to add or subtract.)
Lesson Synthesis
Lesson Synthesis
“Today we solved problems that required adding and subtracting fractions.”
Display Priya‘s salad dressing recipe.
“What strategy did you use to find out how much salad dressing Priya’s recipe makes?” (The denominators for the fractions are 2, 3 and 4 so I used 12 because I know that it is a multiple of 2, 3, and 4. I put the half and fourths together first since I could use 4 as a common denominator and then I used 12 to add the fourths and third.
Display:
“What strategy did you use to find this difference for the olive oil?” (I knew that 48 is and so I used that as a common denominator. I used as a common denominator.)
“How do you decide which strategy to use?” (It depends on the numbers. If I know a small common multiple of the denominators, I use that. If I don’t, I can always use the product of the denominators.)
Cool-down: Evaluate Expressions (5 minutes)
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5.NF.A.2
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Inequality solving
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Asked 9 years, 1 month ago
Modified9 years, 1 month ago
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How must I proceed to solve inequalities of the following form:
−1≤r−1/r≤1−1≤r−1/r≤1
This is the final step for another problem that I'm attempting to solve and thus far, this is where I've gotten:
By taking the L.C.M and multiplying both sides by r r (given to take strictly positive (or zero) values), I got-
−r≤r ²−2≤2 r−r≤r ²−2≤2 r
Taking the left inequality separately, I found that
(r−1)(r+2)≥0(r−1)(r+2)≥0
which gives me
r∈(−∞,−2]∪[1,∞)r∈(−∞,−2]∪[1,∞)
(Clearly, the only valid region is [1,∞)[1,∞), since r r can only take strictly positive/zero values but let me leave that there until the very end, making a mental note of it.)
The right half of the inequality gives me r ²−2 r−2≤0 r ²−2 r−2≤0 and I have no insights as to how to solve this.
Please help! Thanks in advance :) Regards.
inequality
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edited Aug 23, 2016 at 10:43
Tom
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asked Aug 23, 2016 at 10:12
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Let's consider two cases, r>0 r>0 and r<0 r<0.
Case 1
Suppose r>0 r>0. Multiplying both sides of the inequality by r r,
−r≤r 2−1≤r.−r≤r 2−1≤r.
Rearranging the left-side inequality,
r 2+r−1≥0.r 2+r−1≥0.
Using the quadratic equation to find the roots of r 2+r−1 r 2+r−1,
r=−1±1 2−4(1)(−1)−−−−−−−−−−−√2(1)=−1±5–√2.r=−1±1 2−4(1)(−1)2(1)=−1±5 2.
Thus, our inequality becomes,
(r−−1+5–√2)(r−−1−5–√2)≥0.(r−−1+5 2)(r−−1−5 2)≥0.
From here, we get r∈(−∞,−1−5√2]r∈(−∞,−1−5 2] or r∈[−1+5√2,∞).r∈[−1+5 2,∞). However, we assumed r>0 r>0, thus we only have r∈[−1+5√2,∞).r∈[−1+5 2,∞).
Now let's rearrange the right hand side of the inequality to get
r 2−r−1≤0.r 2−r−1≤0.
The roots of r 2−r−1 r 2−r−1 are,
−(−1)±(−1)2−4(1)(−1)−−−−−−−−−−−−−−√2(1)=1±5–√2.−(−1)±(−1)2−4(1)(−1)2(1)=1±5 2.
Thus, the right hand side inequality becomes,
(r−1+5–√2)(r−1−5–√2)≤0.(r−1+5 2)(r−1−5 2)≤0.
From here, we get r∈[1−5√2,1+5√2]r∈[1−5 2,1+5 2]. But remember r>0 r>0, so this means r∈(0,1+5√2]r∈(0,1+5 2].
Case 2
Let's multiply both sides of the original inequality by r r, but remember to flip the direction of the inequalities,
−r≥r 2−1≥r.−r≥r 2−1≥r.
Rearranging the left and right side inequalities,
r 2+r−1≤0 and r 2−r−1≥0.r 2+r−1≤0 and r 2−r−1≥0.
These inequalties are the same as before except with the direction of the inequalties switched around so I'm not going to go through the same level as detail as in case 1.
For the left-hand inequality, we end up with r∈[−1−5√2,−1+5√2]r∈[−1−5 2,−1+5 2]; however, since r<0 r<0, this becomes r∈[−1−5√2,0)r∈[−1−5 2,0). For the right-hand inequality, we end up with r∈(−∞,1−5√2]r∈(−∞,1−5 2] or r∈[1+5√2,∞).r∈[1+5 2,∞). However; since r<0 r<0, this becomes just r∈(−∞,1−5√2].r∈(−∞,1−5 2].
Putting everything together
Looking at the result of everything, we have,
(r∈[−1+5–√2,∞)AND r∈[0,1+5–√2])OR(r∈[−1−5–√2,0)AND r∈(−∞,1−5–√2]).(r∈[−1+5 2,∞)AND r∈[0,1+5 2])OR(r∈[−1−5 2,0)AND r∈(−∞,1−5 2]).
More succintly,
r∈([−1+5–√2,∞)∩[0,1+5–√2])∪([−1−5–√2,0)∩(−∞,1−5–√2])r∈([−1+5 2,∞)∩[0,1+5 2])∪([−1−5 2,0)∩(−∞,1−5 2])
=[−1+5–√2,1+5–√2]∪[−1−5–√2,1−5–√2]=[−1+5 2,1+5 2]∪[−1−5 2,1−5 2]
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edited Aug 23, 2016 at 11:29
answered Aug 23, 2016 at 11:03
benguinbenguin
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Oh my God, thanks so so much! Now I can solve any inequality of this kind!user361896 –user361896 2016-08-23 11:04:30 +00:00 Commented Aug 23, 2016 at 11:04
No problem! :) This method is a bit long-winded (it took awhile to write up) but it gets the job done!benguin –benguin 2016-08-23 11:08:25 +00:00 Commented Aug 23, 2016 at 11:08
Yes, but I wonder why this solution doesn't match with that given by Josy Etxezarreta Martinez...user361896 –user361896 2016-08-23 11:09:57 +00:00 Commented Aug 23, 2016 at 11:09
I believe his answer is only addressing the part of the question where you have r 2−2 r−2≤0 r 2−2 r−2≤0. His solution to THAT inequality is correct. However, that inequality doesn't follow from the original inequality...I believe you might have accidentally added instead of subtracting somewhere or vice versa when rearranging your inequalities.benguin –benguin 2016-08-23 11:13:35 +00:00 Commented Aug 23, 2016 at 11:13
Oh, OK, I'll check :)user361896 –user361896 2016-08-23 11:15:16 +00:00 Commented Aug 23, 2016 at 11:15
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If you solve the equation with an equality, you get 1+3–√1+3 and 1−3–√1−3, so, and as 1 gives the output 1−2−2=−3 1−2−2=−3, then the region that you are looking for in the inequality is [1−3–√,1+3–√][1−3,1+3].
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answered Aug 23, 2016 at 10:43
Josu Etxezarreta MartinezJosu Etxezarreta Martinez
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What are the steps to find out this solution region?user361896 –user361896 2016-08-23 10:52:33 +00:00 Commented Aug 23, 2016 at 10:52
First you solve the equation r 2−2 r−2=0 r 2−2 r−2=0 to see when the function is crossing zero. Then as the inequality given is a parabola, you have to find when the parabola is negative and when it is positive, have doen that using the fact that as r=1 is between the solutions of the equation, if it is negative, then the region inside the solved values will be the region satisfying the inequality, and if it is bigger than zero, then the region outside the values would be the one you are looking for.Josu Etxezarreta Martinez –Josu Etxezarreta Martinez 2016-08-23 10:57:01 +00:00 Commented Aug 23, 2016 at 10:57
What about other possible regions of solution?user361896 –user361896 2016-08-23 10:57:46 +00:00 Commented Aug 23, 2016 at 10:57
There are three regions: (−∞,1−3–√)(−∞,1−3), where the function is positive, [1−3–√,1+3–√][1−3,1+3], where the function is negative or zero, and (1+3–√,∞)(1+3,∞) where the function is positive again. That's why I chose that region, because it is the only negative one.Josu Etxezarreta Martinez –Josu Etxezarreta Martinez 2016-08-23 11:00:57 +00:00 Commented Aug 23, 2016 at 11:00
Oh, OK, thanks :) I'll keep this in mind.user361896 –user361896 2016-08-23 11:01:25 +00:00 Commented Aug 23, 2016 at 11:01
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The simplest way to solve is to note first the domain of this inequality is r≠0 r≠0, and to replace it with the equivalent:
(r−1 r)2<1⟺(r 2−1)2<r 2(r−1 r)2<1⟺(r 2−1)2<r 2
Now set R=r 2 R=r 2. We have to solve for (R−1)20(R−1)20. So this comes down to finding the positive solutions of a quadratic inequation.
This quadratic inequation rewrites as R 2−3 R+1<0 R 2−3 R+1<0. The roots of the quadratic polynomial are 3±5–√2=(1±5–√2)2 3±5 2=(1±5 2)2. Both are positive, hence the required solutions are given by
3−5–√2<R<3+5–√2⟺⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪1−5–√2<r<1+5–√2 or−1+5–√2<r<−1+5–√2 3−5 2<R<3+5 2⟺{1−5 2<r<1+5 2 or−1+5 2<r<−1+5 2
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answered Aug 23, 2016 at 11:23
BernardBernard
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Logarithms
Try the free Interactive e-book for iPhone, iPad, and OS X. Chapter: Section: Home | Previous Section|Next Section Video Logarithms Author(s) David M. Lane Prerequisites Distributions Learning Objectives 1. Compute logs using different bases 2. Perform basic arithmetic operations using logs 3. State the relationship between logs and proportional change The log transformation reduces positive skew "Definition in a new window."). This can be valuable both for making the data more interpretable and for helping to meet the assumptions of inferential statistics. Basics of Logarithms (Logs) Logs are, in a sense, the opposite of exponents. Consider the following simple expression: 10 2 = 100 Here we can say the base of 10 is raised to the second power. Here is an example of a log: Log 10(100) = 2 This can be read as: The log base ten of 100 equals 2. The result is the power that the base of 10 has to be raised to in order to equal the value (100). Similarly, Log 10(1000) = 3 since 10 has to be raised to the third power in order to equal 1,000. These examples all used base 10, but any base could have been used. There is a base which results in "natural logarithms" and that is called e and equals approximately 2.718. It is beyond the scope of this book to explain what is "natural" about it. Natural logarithms can be indicated either as: Ln(x) or log e(x). Changing the base of the log changes the result by a multiplicative constant. To convert from Log 10 to natural logs, you multiply by 2.303. Analogously, to convert in the other direction, you divide by 2.303. Taking the antilog of a number undoes the operation of taking the log. Therefore, since Log 10(1000) = 3, the antilog 10 of 3 is 10 3 = 1,000. Taking the antilog of a number simply raises the base of the logarithm in question to that number. Logs and Proportional Change A series of numbers that increase proportionally will increase in equal amounts when converted to logs. For example, the numbers in the first column of Table 1 increase by a factor of 1.5 so that each row is 1.5 times as high as the preceding row. The Log 10 transformed numbers increase in equal steps of 0.176. Table 1. Proportional raw changes are equal in log units. | Raw | Log | --- | | 4.0 6.0 9.0 13.5 | 0.602 0.778 0.954 1.130 | As another example, if one student increased their score from 100 to 200 while a second student increased theirs from 150 to 300, the percentage change (100%) is the same for both students. The log difference is also the same, as shown below. Log 10(100) = 2.000 Log 10(200) = 2.301 Difference: 0.301 Log 10(150) = 2.176 Log 10(300) = 2.477 Difference: 0.301 Arithmetic Operations Rules for logs of products and quotients are shown below. Log(AB) = Log(A) + Log(B) Log(A/B) = Log(A) - Log(B) For example, Log 10(10 x 100) = Log 10(10) + Log 10(100) = 1 + 2 = 3. Similarly, Log 10(100/10) = Log 10(100) - Log 10(10) = 2 - 1 = 1. Question 1 out of 4. Changing the base changes the log by a multiplicative constant. an additive constant. an unpredcitable amount. Previous Section|Next Section
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17: Electrochemistry
Thermodynamics and Chemical Equilibrium (Ellgen)
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17.3: Defining Oxidation States
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Contents
1. Home
2. Bookshelves
3. Physical & Theoretical Chemistry
4. Thermodynamics and Chemical Equilibrium (Ellgen)
5. 17: Electrochemistry
6. 17.3: Defining Oxidation States
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Thermodynamics and Chemical Equilibrium (Ellgen)
Front Matter
1: Introduction - Background and a Look Ahead
2: Gas Laws
3: Distributions, Probability, and Expected Values
4: The Distribution of Gas Velocities
5: Chemical Kinetics, Reaction Mechanisms, and Chemical Equilibrium
6: Equilibrium States and Reversible Processes
7: State Functions and The First Law
8: Enthalpy and Thermochemical Cycles
9: The Second Law - Entropy and Spontaneous Change
10: Some Mathematical Consequences of the Fundamental Equation
11: The Third Law, Absolute Entropy, and the Gibbs Free Energy of Formation
12: Applications of the Thermodynamic Criteria for Change
13: Equilibria in Reactions of Ideal Gases
14: Chemical Potential - Extending the Scope of the Fundamental Equation
15: Chemical Potential, Fugacity, Activity, and Equilibrium
16: The Chemical Activity of the Components of a Solution
17: Electrochemistry
18: Quantum Mechanics and Molecular Energy Levels
19: The Distribution of Outcomes for Multiple Trials
20: Boltzmann Statistics
21: The Boltzmann Distribution Function
22: Some Basic Applications of Statistical Thermodynamics
23: The Ensemble Treatment
24: Indistinguishable Molecules - Statistical Thermodynamics of Ideal Gases
25: Bose-Einstein and Fermi-Dirac Statistics
26: Appendices
Back Matter
17.3: Defining Oxidation States
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17.2: Electrochemical Cells
17.4: Balancing Oxidation-reduction Reactions
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We introduce oxidation states to organize our thinking about oxidation–reduction reactions and electrochemical cells. When we define oxidation states, we create a set of rules for allocating the electrons in a molecule or ion to the individual atoms that make it up. The definition of oxidation states is therefore an accounting exercise. The definition of oxidation states predates our ability to estimate electron densities through quantum mechanical calculations. As it turns out, however, the ideas that led to the oxidation state formalism are directionally correct; atoms that have high positive oxidation states according to the formalism also have relatively high positive charges by quantum mechanical calculation. In general, the absolute values of oxidation states are substantially larger than the absolute values of the partial charges found by quantum-mechanical calculation; however, there is no simple quantitative relationship between oxidation states and the actual distribution of electrons in real chemical moieties. It is a serious mistake to think that our accounting system provides a quantitative description of actual electron densities.
It is a serious mistake to think that the Oxidation State system provides a quantitative description of actual electron densities.
The rules for assigning oxidation states grow out of the primitive (and quantitatively incorrect) idea that oxygen atoms usually acquire two electrons and hydrogen atoms usually lose one electron in forming chemical compounds and ionic moieties. The rest of the rules derive from a need to recognize some exceptional cases and from applying the basic ideas to additional elements. The rules of the oxidation state formalism are these:
For any element in any of its allotropic forms, the oxidation state of its atoms is zero.
In any of its compounds, the oxidation state of an oxygen atom is 2–, except in compounds that contain an oxygen–oxygen bond, where the oxidation state of oxygen is 1–. The excepted compounds are named peroxides. Examples include sodium peroxide, Na 2O 2, and hydrogen peroxide, H 2O 2.
In any of its compounds, the oxidation state of a hydrogen atom is 1+, except in compounds that contain a metal–hydrogen bond, where the oxidation state of hydrogen is 1–. The excepted compounds are named hydrides. Examples include sodium hydride, NaH, and calcium hydride, CaH 2.
In any of their compounds, the oxidation states of alkali metal atoms (lithium, sodium, potassium, rubidium, cesium, and francium) are 1+. (There are exceptional cases, but we do not consider them.)
In any of their compounds, the oxidation states of halogen atoms (fluorine, chlorine, bromine, iodine, and astatine) are 1–, except in compounds that contain a halogen–oxygen bond.
The oxidation states of any other atoms in a compound are chosen so as to make the sum of the oxidation states in the chemical moiety equal to its charge. So, for a neutral molecule, the oxidation states sum to zero. For a monovalent anion, they sum to 1–, etc.
This page titled 17.3: Defining Oxidation States is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Paul Ellgen via source content that was edited to the style and standards of the LibreTexts platform.
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Balance Reduction and Oxidation (Redox) Reactions
Oxidation States (Oxidation Numbers)This page explains what oxidation states (oxidation numbers) are and how to calculate and use them.
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Introductory Statistics
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## 10.1 Two Population Means with Unknown Standard Deviations
Use the following information to answer the next 15 exercises: Indicate if the hypothesis test is for
independent group means, population standard deviations, and/or variances known
independent group means, population standard deviations, and/or variances unknown
matched or paired samples
single mean
two proportions
single proportion
1.
It is believed that 70% of males pass their drivers test in the first attempt, while 65% of females pass the test in the first attempt. Of interest is whether the proportions are in fact equal.
A new laundry detergent is tested on consumers. Of interest is the proportion of consumers who prefer the new brand over the leading competitor. A study is done to test this.
3.
A new windshield treatment claims to repel water more effectively. Ten windshields are tested by simulating rain without the new treatment. The same windshields are then treated, and the experiment is run again. A hypothesis test is conducted.
The known standard deviation in salary for all mid-level professionals in the financial industry is $11,000. Company A and Company B are in the financial industry. Suppose samples are taken of mid-level professionals from Company A and from Company B. The sample mean salary for mid-level professionals in Company A is $80,000. The sample mean salary for mid-level professionals in Company B is $96,000. Company A and Company B management want to know if their mid-level professionals are paid differently, on average.
5.
The average worker in Germany gets eight weeks of paid vacation.
According to a television commercial, 80% of dentists agree that Ultrafresh toothpaste is the best on the market.
7.
It is believed that the average grade on an English essay in a particular school system for females is higher than for males. A random sample of 31 females had a mean score of 82 with a standard deviation of three, and a random sample of 25 males had a mean score of 76 with a standard deviation of four.
The league mean batting average is 0.280 with a known standard deviation of 0.06. The Rattlers and the Vikings belong to the league. The mean batting average for a sample of eight Rattlers is 0.210, and the mean batting average for a sample of eight Vikings is 0.260. There are 24 players on the Rattlers and 19 players on the Vikings. Are the batting averages of the Rattlers and Vikings statistically different?
9.
In a random sample of 100 forests in the United States, 56 were coniferous or contained conifers. In a random sample of 80 forests in Mexico, 40 were coniferous or contained conifers. Is the proportion of conifers in the United States statistically more than the proportion of conifers in Mexico?
A new medicine is said to help improve sleep. Eight subjects are picked at random and given the medicine. The means hours slept for each person were recorded before starting the medication and after.
11.
It is thought that teenagers sleep more than adults on average. A study is done to verify this. A sample of 16 teenagers has a mean of 8.9 hours slept and a standard deviation of 1.2. A sample of 12 adults has a mean of 6.9 hours slept and a standard deviation of 0.6.
Varsity athletes practice five times a week, on average.
13.
A sample of 12 in-state graduate school programs at school A has a mean tuition of $64,000 with a standard deviation of $8,000. At school B, a sample of 16 in-state graduate programs has a mean of $80,000 with a standard deviation of $6,000. On average, are the mean tuitions different?
A new WiFi range booster is being offered to consumers. A researcher tests the native range of 12 different routers under the same conditions. The ranges are recorded. Then the researcher uses the new WiFi range booster and records the new ranges. Does the new WiFi range booster do a better job?
15.
A high school principal claims that 30% of student athletes drive themselves to school, while 4% of non-athletes drive themselves to school. In a sample of 20 student athletes, 45% drive themselves to school. In a sample of 35 non-athlete students, 6% drive themselves to school. Is the percent of student athletes who drive themselves to school more than the percent of nonathletes?
Use the following information to answer the next three exercises: A study is done to determine which of two soft drinks has more sugar. There are 13 cans of Beverage A in a sample and six cans of Beverage B. The mean amount of sugar in Beverage A is 36 grams with a standard deviation of 0.6 grams. The mean amount of sugar in Beverage B is 38 grams with a standard deviation of 0.8 grams. The researchers believe that Beverage B has more sugar than Beverage A, on average. Both populations have normal distributions.
Are standard deviations known or unknown?
17.
What is the random variable?
Is this a one-tailed or two-tailed test?
Use the following information to answer the next 12 exercises: The U.S. Center for Disease Control reports that the mean life expectancy was 47.6 years for White people born in 1900 and 33.0 years for non-White people. Suppose that you randomly survey death records for people born in 1900 in a certain county. Of the 124 White people, the mean life span was 45.3 years with a standard deviation of 12.7 years. Of the 82 non-White people, the mean life span was 34.1 years with a standard deviation of 15.6 years. Conduct a hypothesis test to see if the mean life spans in the county were the same for White and non-White people.
19.
Is this a test of means or proportions?
State the null and alternative hypotheses.
H0: __________
Ha: __________
21.
Is this a right-tailed, left-tailed, or two-tailed test?
In symbols, what is the random variable of interest for this test?
23.
In words, define the random variable of interest for this test.
Which distribution (normal or Student's t) would you use for this hypothesis test?
25.
Explain why you chose the distribution you did for Exercise 10.24.
Calculate the test statistic and p-value.
27.
Sketch a graph of the situation. Label the horizontal axis. Mark the hypothesized difference and the sample difference. Shade the area corresponding to the p-value.
Find the p-value.
29.
At a pre-conceived α = 0.05, what is your:
Decision:
Reason for the decision:
Conclusion (write out in a complete sentence):
Does it appear that the means are the same? Why or why not?
## 10.2 Two Population Means with Known Standard Deviations
Use the following information to answer the next five exercises. The mean speeds of fastball pitches from two different baseball pitchers are to be compared. A sample of 14 fastball pitches is measured from each pitcher. The populations have normal distributions. Table 10.18 shows the result. Scouters believe that Rodriguez pitches a speedier fastball.
| Pitcher | Sample Mean Speed of Pitches (mph) | Population Standard Deviation |
---
| Wesley | 86 | 3 |
| Rodriguez | 91 | 7 |
Table 10.18
31.
What is the random variable?
State the null and alternative hypotheses.
33.
What is the test statistic?
What is the p-value?
35.
At the 1% significance level, what is your conclusion?
Use the following information to answer the next five exercises. A researcher is testing the effects of plant food on plant growth. Nine plants have been given the plant food. Another nine plants have not been given the plant food. The heights of the plants are recorded after eight weeks. The populations have normal distributions. The following table is the result. The researcher thinks the food makes the plants grow taller.
| Plant Group | Sample Mean Height of Plants (inches) | Population Standard Deviation |
---
| Food | 16 | 2.5 |
| No food | 14 | 1.5 |
Table 10.19
Is the population standard deviation known or unknown?
37.
State the null and alternative hypotheses.
What is the p-value?
39.
Draw the graph of the p-value.
At the 1% significance level, what is your conclusion?
Use the following information to answer the next five exercises. Two metal alloys are being considered as material for ball bearings. The mean melting point of the two alloys is to be compared. 15 pieces of each metal are being tested. Both populations have normal distributions. The following table is the result. It is believed that Alloy Zeta has a different melting point.
| | Sample Mean Melting Temperatures (°F) | Population Standard Deviation |
---
| Alloy Gamma | 800 | 95 |
| Alloy Zeta | 900 | 105 |
Table 10.20
41.
State the null and alternative hypotheses.
Is this a right-, left-, or two-tailed test?
43.
What is the p-value?
Draw the graph of the p-value.
45.
At the 1% significance level, what is your conclusion?
## 10.3 Comparing Two Independent Population Proportions
Use the following information for the next five exercises. Two types of phone operating system are being tested to determine if there is a difference in the proportions of system failures (crashes). Fifteen out of a random sample of 150 phones with OS1 had system failures within the first eight hours of operation. Nine out of another random sample of 150 phones with OS2 had system failures within the first eight hours of operation. OS2 is believed to be more stable (have fewer crashes) than OS1.
Is this a test of means or proportions?
47.
What is the random variable?
State the null and alternative hypotheses.
49.
What is the p-value?
What can you conclude about the two operating systems?
Use the following information to answer the next twelve exercises. In the recent Census, three percent of the U.S. population reported being of two or more races. However, the percent varies tremendously from state to state. Suppose that two random surveys are conducted. In the first random survey, out of 1,000 North Dakotans, only nine people reported being of two or more races. In the second random survey, out of 500 Nevadans, 17 people reported being of two or more races. Conduct a hypothesis test to determine if the population percents are the same for the two states or if the percent for Nevada is statistically higher than for North Dakota.
51.
Is this a test of means or proportions?
State the null and alternative hypotheses.
H0: _________
Ha: _________
53.
Is this a right-tailed, left-tailed, or two-tailed test? How do you know?
What is the random variable of interest for this test?
55.
In words, define the random variable for this test.
Which distribution (normal or Student's t) would you use for this hypothesis test?
57.
Explain why you chose the distribution you did for the Exercise 10.56.
Calculate the test statistic.
59.
Sketch a graph of the situation. Mark the hypothesized difference and the sample difference. Shade the area corresponding to the p-value.
Figure 10.17
Find the p-value.
61.
At a pre-conceived α = 0.05, what is your:
Decision:
Reason for the decision:
Conclusion (write out in a complete sentence):
Does it appear that the proportion of Nevadans who are two or more races is higher than the proportion of North Dakotans? Why or why not?
## 10.4 Matched or Paired Samples
Use the following information to answer the next five exercises. A study was conducted to test the effectiveness of a software patch in reducing system failures over a six-month period. Results for randomly selected installations are shown in Table 10.21. The “before” value is matched to an “after” value, and the differences are calculated. The differences have a normal distribution. Test at the 1% significance level.
| Installation | A | B | C | D | E | F | G | H |
--- --- --- ---
| Before | 3 | 6 | 4 | 2 | 5 | 8 | 2 | 6 |
| After | 1 | 5 | 2 | 0 | 1 | 0 | 2 | 2 |
Table 10.21
63.
What is the random variable?
State the null and alternative hypotheses.
65.
What is the p-value?
Draw the graph of the p-value.
67.
What conclusion can you draw about the software patch?
Use the following information to answer next five exercises. A study was conducted to test the effectiveness of a juggling class. Before the class started, six subjects juggled as many balls as they could at once. After the class, the same six subjects juggled as many balls as they could. The differences in the number of balls are calculated. The differences have a normal distribution. Test at the 1% significance level.
| Subject | A | B | C | D | E | F |
--- --- ---
| Before | 3 | 4 | 3 | 2 | 4 | 5 |
| After | 4 | 5 | 6 | 4 | 5 | 7 |
Table 10.22
State the null and alternative hypotheses.
69.
What is the p-value?
What is the sample mean difference?
71.
Draw the graph of the p-value.
What conclusion can you draw about the juggling class?
Use the following information to answer the next five exercises. A doctor wants to know if a blood pressure medication is effective. Six subjects have their blood pressures recorded. After twelve weeks on the medication, the same six subjects have their blood pressure recorded again. For this test, only systolic pressure is of concern. Test at the 1% significance level.
| Patient | A | B | C | D | E | F |
--- --- ---
| Before | 161 | 162 | 165 | 162 | 166 | 171 |
| After | 158 | 159 | 166 | 160 | 167 | 169 |
Table 10.23
73.
State the null and alternative hypotheses.
What is the test statistic?
75.
What is the p-value?
What is the sample mean difference?
77.
What is the conclusion?
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What is the area bounded by the curves: y= x^2 − 4, y= 0, x= 4? - Quora
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What is the area bounded by the curves: y= x^2 − 4, y= 0, x= 4?
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Ernest Leung
B.Sc. (Hons.) in Chemistry Honors&Mathematics, The Chinese University of Hong Kong · Author has 11.9K answers and 5.8M answer views
·May 5
What is the area bounded by the curves: y = x² − 4, y = 0, x = 4?
Substitute y = 0 into y = x² - 4:
x² - 4 = (0)
x = ± 2
Hence, y = 0 cuts y = x² - 4 at (2, 0) and (-2, 0).
Substitute x = 4 into y = x² - 4:
y = (4)² - 4
y = 12
Hence, x = 4 cuts y = x² - 4 at (4, 12).
Refer to the following figure.
Continue Reading
What is the area bounded by the curves: y = x² − 4, y = 0, x = 4?
Substitute y = 0 into y = x² - 4:
x² - 4 = (0)
x = ± 2
Hence, y = 0 cuts y = x² - 4 at (2, 0) and (-2, 0).
Substitute x = 4 into y = x² - 4:
y = (4)² - 4
y = 12
Hence, x = 4 cuts y = x² - 4 at (4, 12).
Refer to the following figure.
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What is the area of the region bounded by the curve y=x-2 and x=0 to x=4?
What is the area bounded by the curve y = x^2 + 2, and the lines x = 0, y = 0 and x = 4?
What is the first quadrant area bounded by the following curves: y= x^2 + 2, y = 4 and x= 0?
What is the area bounded by the curve y = 4x - x^2 and the lines x = -2 and y=4?
What is the area bounded by x-axis and the curve [math]y = 4x-x^2[/math]?
Allen Ries
Math Major University of Alberta · Author has 25.1K answers and 9.7M answer views
·1y
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1y
To find the area bounded by the curves [math]y = x^2 - 4[/math], [math]y = 0[/math], and [math]x = 4[/math], we need to follow these steps:
Find the points of intersection: First, we set [math]y = x^2 - 4[/math] equal to [math]y = 0[/math] to find where the parabola intersects the x-axis.
[math]x^2 - 4 = 0[/math]
[math]x^2 = 4[/math]
[math]x = 2 \quad \text{and} \quad x = -2[/math]
The parabola intersects the x-axis at [math]x = -2[/math] and [math]x = 2[/math].
Determine the area: The area we want to find is between [math]x = 2[/math] and [math]x = 4[/math]. We will integrate the function [math]y = x^2 - 4[/math] from [math]x = 2[/math] to [math]x = 4[/math].
The area [math]A[/math] can be calculated using the integral:
[math]A = \int_{2}^{4} (x^2 - 4) \, dx[/math]
Calculate the integral:
[math]A = \int_{[/math]
Continue Reading
To find the area bounded by the curves [math]y = x^2 - 4[/math], [math]y = 0[/math], and [math]x = 4[/math], we need to follow these steps:
Find the points of intersection: First, we set [math]y = x^2 - 4[/math] equal to [math]y = 0[/math] to find where the parabola intersects the x-axis.
[math]x^2 - 4 = 0[/math]
[math]x^2 = 4[/math]
[math]x = 2 \quad \text{and} \quad x = -2[/math]
The parabola intersects the x-axis at [math]x = -2[/math] and [math]x = 2[/math].
Determine the area: The area we want to find is between [math]x = 2[/math] and [math]x = 4[/math]. We will integrate the function [math]y = x^2 - 4[/math] from [math]x = 2[/math] to [math]x = 4[/math].
The area [math]A[/math] can be calculated using the integral:
[math]A = \int_{2}^{4} (x^2 - 4) \, dx[/math]
Calculate the integral:
[math]A = \int_{2}^{4} (x^2 - 4) \, dx = \left[ \frac{x^3}{3} - 4x \right]_{2}^{4}[/math]
Now we compute the definite integral:
Evaluate at [math]x = 4[/math]:
[math]\frac{4^3}{3} - 4 \cdot 4 = \frac{64}{3} - 16 = \frac{64}{3} - \frac{48}{3} = \frac{16}{3}[/math]
Evaluate at [math]x = 2[/math]:
[math]\frac{2^3}{3} - 4 \cdot 2 = \frac{8}{3} - 8 = \frac{8}{3} - \frac{24}{3} = -\frac{16}{3}[/math]
Subtract the evaluations:
[math]A = \left( \frac{16}{3} \right) - \left( -\frac{16}{3} \right) = \frac{16}{3} + \frac{16}{3} = \frac{32}{3}[/math]
Therefore, the area bounded by the curves [math]y = x^2 - 4[/math], [math]y = 0[/math], and [math]x = 4[/math] is
[math]\boxed{\frac{32}{3}}.[/math]
Upvote ·
Brady West
BS in Civil Engineering, Michigan State University (Graduated 2020) · Author has 58 answers and 100.9K answer views
·4y
Here’s a graphical representation of the problem. As you can see, the area bounded by all three curves is that quasi-triangular section just right of center.
By integrating a function, we get the area below that curve down to the x-axis. Since that happens to be our lower boundary here, all we have to do is integrate the curve along the appropriate limits. If the lower bound was somewhere else, just subtract the integral of that lower bound curve from the integral of the upper bound to find the area.
So let’s get to it. As you can see from the graph, the area is bounded on top by y=x^2 - 4 and o
Continue Reading
Here’s a graphical representation of the problem. As you can see, the area bounded by all three curves is that quasi-triangular section just right of center.
By integrating a function, we get the area below that curve down to the x-axis. Since that happens to be our lower boundary here, all we have to do is integrate the curve along the appropriate limits. If the lower bound was somewhere else, just subtract the integral of that lower bound curve from the integral of the upper bound to find the area.
So let’s get to it. As you can see from the graph, the area is bounded on top by y=x^2 - 4 and on the bottom by the x-axis. This area extends from an x-value of 2 to 4.
(If not for the graph, we could solve for the intersection points by setting the two intersecting curves equal to each other and finding x.)
[math]\int_{2}^{4} (x^2 -4) dx = x^3/3 -4x|^4_2 = (4^3/3-4(4)) - (2^3/3-4(2)) = 32/3 = 10.666...[/math]
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Related questions
What is the area of the region bounded by the curve y=x-2 and x=0 to x=4?
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I will try to organize the work of this semester around several classical questions.
The first is, When is a prime p the sum of two squares? The question was raised by Fermat who gave the correct answer in 1640 and claimed he had a proof but never wrote it down; the first published proof, in 1747, is due to Euler. There is a beautiful article in Wikipedia which you can reach by Googling “sum of two squares”; it is well worth reading. Today there is a very short exceedingly clever proof which requires practically no preparation but which also gives you no idea of why the theorem should be true. Our approach here, however, will not be to get to the result as speedily as possible but to bring in the many modern algebraic ideas that lead to a conceptual understanding of why Fermat’s original assertion is correct.
1 First Theme: Sums of Squares When is an integer a sum of two squares, m = a2 + b2? First observation: if m, n are both sums of two squares then so is mn. For going over to complex numbers, m = a2 + b2 can be written as m = (a + bi)(a −bi). Similarly, if n = c2 + d2) then we can write n = (c + di)(c −di). It follows that mn = (a + bi)(c + di)(a −bi)(c −di) = [(ac −bd) + (ad + bc)i][(ad −bc) −(ac −bd)i], so we must have mn = (ac−bd)2 +(ad+bc)2. An easy calculation shows that this is correct. This suggests that we look first at the question of when a prime p is a sum of two squares. Knowing which primes are sums of two squares won’t completely answer the question, however. Certainly those numbers which can be written as products of primes each of which is a sum of two squares will, from what we have just seen, indeed be sums of two squares. It is still possible that some other numbers are, too. In fact, any number which is itself a square is trivially the sum of two squares, itself and zero. We will see that this really exhausts the possibilities: to be a sum of two squares a number must be a product of squares and of primes which are themselves a sum of two squares.
So which primes qualify?
Let’s try out some small primes: 2 = 12 + 12, 3 fails, 5 = 12 + 22, 7 fails, 11 fails, 13 = 22 + 32, 17 = 12 + 42, 19 fails. So far, except for 2, those primes p which are congruent 1 modulo 4, i.e., which leave a remainder of 1 when divided by 4 are sums of two squares. This is written p ≡1 mod 4 or simply p ≡1(4).
Those p with p ≡3(4) fail. And a number like 3 × 5 = 15 which has as a factor a prime that fails, namely 3, (but not 32), is not a sum of two squares. In fact we have the following Theorem 1 1. A prime p is a sum of two squares if and only if either p = 2 or p ≡1(4). 2. An integer m is a sum of two squares if and only if in the prime factorization of m those primes p ≡3(4) appear to an even power. 2 There are classical computational proofs of this which do not involve any higher algebra (cf. the Wikipedia article). The first few weeks of our course will be devoted to developing enough modern algebra so that we can understand a conceptual proof. Part of this will be understanding when prime factorizations 1 exist and are unique; it is implicit in the statement of the theorem that this is the case for the ordinary integers.
Passing to the complex numbers has been useful but before going further, let’s introduce some terminology and some notation. You have probably already been introduced to the concept of a group, G but let me review it very briefly: G consists of a set of elements together with a multiplication map G × G →G which is associative; moreover (i) there is a unit element, denoted 1 or e (or 1G if we want to emphasize that this is the unit element of the group G) with the property that 1 · x = x · 1 = x for all x ∈G, and (ii) for every x ∈G there is an inverse element x−1 such that x · x−1 = 1 = x−1 · x. This inverse is necessarily unique (as is the unit element). We say that x and y commute if xy = yx. A group in which all pairs of elements commute is called commutative or Abelian (in honor of N. H. Abel). In that case, multiplication is frequently written as addition and the ‘unit element’ is denoted by 0 and called the zero element; the group is then usually called an additive group. The set of all permutations of an arbitrary set S forms a group; it is non-Abelian whenever S has at least three elements. The most important Abelian group is the additive group of integers, here always denoted by Z. The integers, however, have more structure, there is an associative multiplication. This leads to the following definition: a ring R is an additive group with an associative multiplication that satisfies the distributive laws, i.e., where x(y + z) = xy + xz and (y + z)x = yx + zx. We need both of these statements since the multiplication need not be commutative. A good example of such a ring is set of all n × n matrices with real coefficients. We will always denote the real numbers by R and this ring by Mn(R). (This notation differs from that in the text but is more common.) Notice that Mn(R) has a unit element for multiplication. We will generally assume without mentioning it that the rings we deal with have a unit element but (unlike the text) not that they are commutative.
A field is a commutative ring in which every non-zero element has a mul-tiplicative inverse.
The most important examples are the rational numbers Q, the real numbers, R, and the complex numbers C.
There do exist non-commutative rings in which every non-zero element has a multiplicative inverse.
These are called division rings, or skew fields, a term which the elder Artin has contracted to sfield.
The most important example is the Hamiltonians, generally denoted by H after its discoverer. This algebra is a four-dimensional vector space over R with basis elements 1, i, j, k and multiplication defined by i2 = j2 = k2 = −1, ij = k = −ji. It follows that jk = i = −kj, ki = j = −ik.
The general quaternion thus has the form q = a+bi+cj+dk, where a, b, c, d ∈R.
To see that this is a division ring, define the conjugate ¯ q = a −bi −cj −dk.
Then q · ¯ q = ¯ q · q = a2 + b2 + c2 + d2. Since a, b, c, d are real, this can’t vanish unless all are zero, so we have q−1 = q/q¯ q. A ring which is a vector space over a field is frequently called an algebra.
This brings us to the concept of a morphism (older name homomorphism).
Recall that a morphism between groups f : G →H is a mapping which preserves the group multiplication, i.e., such that f(xy) = (fx)(fy). This implies that f(1g) = 1H and that f(x−1) = (fx)−1.
2 Exercise 1 Prove the preceding assertions.
In the newer terminology, a morphism which is one-to-one, i.e., where x ̸= y implies fx ̸= fy, is called a monomorphism although I will frequently use the older word as well. A morphism which is onto is now usually called an epimorphism, but here, too, I will frequently use the older term. One which is both is frequently called a bijection.
It is easy to check that if we have morphisms G f − − − − →H g − − − − →J then the composite G g◦f − − − − →J is again a morphism.
In elementary texts an isomorphism f : G →H is usually defined to be a morphism which is one-to-one and onto, a bijection. This will do for groups, rings, and all the structures you will encounter in this course, but the ‘categorical’ definition is this: f is an isomorphism if there is a morphism g : H →G such g ◦f is the identity map of g, denoted idG, and such that f ◦g = idH. The reason that the elementary definition works for groups and for rings is that if f : G →H is a bijection then there is a set mapping g : H →G sending every z ∈H back to the unique x ∈G with fx = z and this g is again a morphism.
Exercise 2 Prove the preceding assertion.
The concept of a morphism extends to rings: a morphism f : R →S is a mapping which preserves both the addition and the multiplication, i.e., such that f(x + y) = fx + fy and f(xy) = (fx)(fy).
While we certainly have f(0) = 0 it need no longer be the case that f(1R) = 1S. For example, let R just be R and S = M2(R). We can then define a morphism R →S by sending every real number x to the matrix µx 0 0 0 ¶ . This is certainly a morphism but the image of the unit element 1 ∈R is not the unit element of S. The image of 1 is, however, an element whose square is itself; such an element is called an idempotent. The general concept of a morphism is that it preserves all the structure there is. Having a unit element is not part of the definition of a ring, even though the rings we deal with here generally do have units. To be more precise, we define a unital ring to be one with a unit element and a unital morphism to be one preserving the unit element.
If you want a simple example where a morphism which is a bijection is not necessarily an isomorphism, considered partially ordered sets or posets, that is, sets in which for some but not necessarily all pairs of elements x, y there is a relation x ≺y. This has to satisfy the axioms that x ≺y and y ≺z imply x ≺z, x ≺x, and x ≺y together with y ≺x imply x = y. A morphism f : S →T of partially ordered sets is a set map such that x ≺y implies fx ≺fy. Now suppose that S carries a non-trivial partial order and that |S| is the same set with the partial order wiped out, i.e., in which there is no pair x, y with x ≺y unless x = y. Now consider the identity map |S| →S carrying every element to itself. This is trivially a morphism of partially ordered sets (since there are no conditions to satisfy) and is trivially a bijection, but the inverse map S →|S| is not a morphism of partially ordered sets.
3 The kernel K of a group morphism f : G →K is the set of all x ∈G such that fx = 1. It is a normal subgroup of G, i.e., a subgroup such that xKx−1 = K, where xKx−1 is the set of all elements xkx−1, k ∈K. The kernel of a ring morphism f : R →S is similarly the set I of all x ∈R such that fx = 0. This subset is an ideal of R, that is, (i) I is an additive subgroup of R (ii) with the additional property that if x ∈R and z ∈I then xz and zx are both in I. When we consider only the additive group structure of R (forget the multiplication for the moment) we generally write R+. Here is a simple but fundamental example. Suppose that R is commutative and pick an element a ∈R. Then the set of all multiples of a, i.e. elements of the form ax with x ∈R, forms an ideal. Such an ideal is called principal; it is the principal ideal generated by a, written aR or sometimes simply as (a) when the ring is understood. In particular, when R = Z if we pick an integer say 6, then the set of all multiples of 6 forms an ideal. As we will see, in Z every ideal is principal.
Rings with this special property are generally called principal ideal rings. In any ring, the entire ring and the subring reduced to the zero element alone are ideals but we generally don’t count these; all other ideals are called proper. It can happen that a ring has no proper ideals, in which case it is called simple.
The rings Mn(R) are simple.
Exercise 3 (i) Prove this. (ii) Prove that if I is an ideal of a ring R then Mn(I) is an ideal of Mn(R) and (iii) that every ideal of Mn(R) has this form.
Recall that if we have a normal subgroup K of a group G then we can form the quotient group G/K (the set of cosets xK). Every normal subgroup K is actually the kernel of a group morphism, namely of the canonical morphism G →G/K. Moreover, if we have a group morphism f : G →H and if the kernel of f contains K then we can define a new morphism ¯ f : G/K →H as follows: An element of G/K is a coset xK; set ¯ f(xK) = f(x). The “representative” x of the coset xK is not unique; it is just one of the elements of xK. However, if we have one representative x (that is, if we have in hand one element x of xK) then any other representative y must be a multiple xk of x. It follows that f(y) = f(x)f(k) = f(x), so the map ¯ f is well defined, and it is easy to see that it is a group morphism. With this, f can be factored: it can be written as the composite morphism G can − − − − →G/K ¯ f − − − − →H. The important thing now is that we can do the same for rings. Suppose that I is an ideal of R. Since I is an additive subgroup of R+ its cosets are written in the form x + I and we define R/I to be the set of these cosets. This is again a ring with addition and multiplication defined by (x+I)+(y+I) = (x+y)+I; (x+I)(y+I) = xy+I. In our simple example of the multiples of 6 in Z, the ring Z/(6) (sometimes written even more simply as Z/6 has exactly six elements, ¯ 0, ¯ 1, ¯ 2, . . . , ¯ 5, the cosets of 0 through 5, respectively. Addition and multiplication in Z/6 are easy but there are some peculiarities: ¯ 2 · ¯ 3 = ¯ 0 so here we have a pair of elements, neither of which is the zero element of the ring, but their product is zero; such elements are called zero divisors. Also, (¯ 3)2 = ¯ 3 · ¯ 3 = ¯ 3. This is another example of an idempotent.
4 The reason that Z/6 has zero divisors is that 6 is composite; 6 = 2 · 3. It is a basic theorem (a special case of a deeper one) that if p is a prime number then Z/p is a field. It is easy to check, for example, that Z/7 is a field: ¯ 2 · ¯ 4 = ¯ 3·¯ 5 = ¯ 6·¯ 6 = ¯ 1 (and, of course ¯ 1 is its own inverse), showing that every non-zero element of Z/7 has a multiplicative inverse. It follows that the non-zero elements form a group under multiplication. This group is cyclic; you can check that the powers of ¯ 3 give all the non-zero elements of Z/6. This is not an accident. We will prove the following Theorem 2 Any finite multiplicative subgroup of the multiplicative group of a field is cyclic. 2 Exercise 4 Give examples to show that this need not hold for (i) a skew field or (ii) an infinite group.
To prove this we shall need some information about the structure of Abelian groups. A group G (Abelian or not) is finitely generated if there is some finite subset of elements g1, . . . , gn such that every element of G can be written as a product of these elements and their reciprocals. A group G which can be generated by a single element g is cyclic.
If G is finite, say #G = n then G = {1, g, g2, . . . , n −1}, where gn = 1. In this case G is isomorphic (as an a group) to Z/n, the isomorphism being given by gm →¯ m. If G is infinite then G = {. . . , g−2, g−1, 1, g, g2, . . . }, where no power of G other then the zeroth is equal to the unit element.
In this case, G ∼ = Z, the isomorphism being given by gm →m.
Recall that the direct product or simply the product of two groups, G × H consists of all ordered pairs (g, h), g ∈G, h ∈H with group operation (g, h)(g′, h′) = (gg′, hh′). This construction extends in an obvious way to a product of any finite number of groups. With this we have the following fundamental theorem on finitely generated Abelian groups (a special case of a slightly more general theorem which we will prove) Theorem 3 Every finitely generated Abelian group is isomorphic to a unique group of the form Z/d1 × Z/d2 × · · · × Z/dr × Z × · · · × Z where d1|d2| · · · |dr. 2 The last condition means that d1 divides d2, d2 divides d3, etc.. If the group is finite, then the Z factors don’t appear. There are generally other ways to decompose an Abelian group, but this is the shortest (least number of factors).
The di are called the principal divisors of the group. Note that if the group is finite then the largest one, here denoted dr is the exponent of the group, i.e., the smallest integer e such that ge = 1 for every g ∈G. Notice that if a finite group is not cyclic, i.e., if r ≥2, then writing dr = e, the number of elements of the group satisfying the equation xe = 1 is greater than e. In a field the number of solutions to a polynomial equation can not exceed the degree of the equation. This is why a finite subgroup of the multiplicative group of a field must be cyclic. Sometimes, however, it is quite difficult to find a generator. For example, we will see that if p is a prime then Z/p is a field, usually denoted Fp. Since this field is finite and has exactly p elements, its multiplicative group 5 is of order p −1 and cyclic, but the difficulty of finding a generator when p is a large prime has sometimes been used in coding schemes. (The existence of finite fields was discovered by E. Galois.) The concept of direct product can also be extended to rings R and S except that there it is frequently called the direct sum and denoted R ⊕S. It is the set of ordered pairs (r, s) with r ∈R, s ∈S with addition and multiplication defined by (r, s) + (r′, s′) = (r + r′, s + s′), (r, s)(r′, s′) = (rr′, ss′).
Let’s get back (in a sophisticated way) to our problem of determining which primes p (and more generally, which integers) are sums of two squares, but first, one elementary observation: Any prime p ≡3(4) can not be a sum of two squares. For observe that if n is an even integer then n2 ≡0(4), while if n is odd, say n = 2m + 1 then n2 = 4m2 + 4m + 1 ≡1(4). Therefore, a sum of two squares can only be congruent to 0, 1, or 2 mod 4, hence never to 3.
For deeper results we will look at a generalization of the concept of integer.
A special case is the set of all complex numbers of the form a + bi where a and b are integers. These form a ring, the ring of Gaussian integers, denoted Z[i]. Ordinary primes (henceforth called rational primes because they are the primes in the field Q of rational numbers) may factor in Z[i]. For example, 2 = i(1+i)2, 5 = (1+2i)(1−2i). If p is a sum of two squares, p = a2 +b2, a, b ∈ Z, then p = (a + bi)(a −bi) in Z[i], so being a sum of two squares implies factorization in Z[i]. The converse is also true.
Lemma 1 A rational prime p factors in Z[i] if and only if it is a sum of two squares, in which case it has exactly two non-trivial factors, i.e., factors other than ±1, ±i.
Proof. Suppose that p is a prime which factors in a non-trivial way in Z[i] with p = αβ.
Taking conjugates we also have p = ¯ α¯ β.
Multiplying gives p2 = |α|2|β|2, a factorization of p2 in Z, but the only possibility for this is that |α|2 = |β|2 = p. It follows that neither α nor β can factor further since this would give a factorization of p in Z. Since both have the same absolute value and their product is real, one must be the conjugate of the other. So if α = a + bi, a, b ∈Z then β = a −bi and p = a2 + b2. ■ A complex number α which is a root of some polynomial f(x) = cnxn + cn−1xn−1 + · · · + c1x + c0 with integer coefficients is called an algebraic number.
We could make this equation monic, i.e., have leading coefficient equal to 1 by dividing by cn, but then the other coefficients would generally be rational numbers and not integers. If α is a root of a monic polynomial with integer coefficients then it is called an algebraic integer. If α is an algebraic number then there is a unique monic polynomial f(x) of minimum degree which it satisfies; that polynomial is called the minimum polynomial for α and its degree is called the degree of α. Before considering the general case, consider the special case where α = √ d, where d is some square-free integer (i.e., not divisible by the square of any integer other than 1) but which may be negative, a most important case being d = −1. This is obviously algebraic and even an algebraic integer since it satisfies the equation x2 −d = 0. In fact, all complex numbers of the form α = a + b √ d with a, b ∈Q are algebraic since α satisfies the equation 6 x2 −2ax + (a2 −b2d) = 0. The set of all these numbers is denoted Q( √ d) and we claim that they form a field: it is clear that the sums, differences, and products of two numbers of the form a + b √ d is again a number of the same form. To see that the inverse is also, observe first that since d is not a square the rational number a2 −b2d can not vanish whenever either a or b is not 0.
From x2 −2ax + (a2 −b2d) = 0 we have x(x −2a) = −(a2 −b2d) whence x−1 = −(x −2a)/(a2 −b2d), so Q √ d is indeed a field. It is also a vector space over Q with basis {1, √ d}, so it has dimension equal to 2 as a Q-space. Such a field is usually called a quadratic field. We will see that in fact the set of all algebraic numbers forms a field and that the set of all algebraic integers forms a subring of this field.
When is a number a + b √ d; a, b ∈Q an (algebraic) integer? Looking at the equation it satisfies, it is sufficient (and we will later show, necessary) that 2a and a2 −db2 be integers. If a is an ordinary or rational integer then db2 must also be an integer, and since d is square-free it follows that b is an integer, too. The only other possibility is that a be a “half-integer”, i.e., of the form m + 1/2, m ∈Z, in which case a2 = m2 + m + 1/4. But then a2 −db2 can only be an integer if b is also a half integer and d ≡1(4). So in this case the integers consist not just of all elements of the form m + n √ d, m, n ∈Z but also of the elements (m + 1 2) + (n + 1 2) √ d.
Exercise 5 Prove that the integers of Q √ d do form a ring.
Since −1 ̸≡1(4) we see finally that the ring of Gaussian integers is precisely the ring of algebraic integers inside the field Q(i).
We are now close to understanding why a prime p ≡1(4) must be a sum of two squares. Observe that if p = a2 + b2 = (a + bi)(a −bi), a, b ∈Z then the rational prime p has factored inside the ring of Gaussian integers Z[i]. This raises the question of whether Z[i] behaves like Z in that it has “primes” which can not be factored and where every element can be factored “uniquely” into a product of primes. Here is the reason for the quotes. Even in Z factorization is not strictly unique because we could introduce factors of −1 so we make the following definition: In a commutative ring R (with unit element) an element u which has an inverse is called a unit. (This may not be the best terminology, but it is the historical one.) The units form a group under multiplication, usually denoted R×. In Z the group of units consists only of {+1, −1}; in Z[i] it is {±1, ±i}. Elements x, y ∈R with y = ux where u is a unit are called associates; this is obviously an equivalence relation. In a commutative ring it is meaningful to say that y divides x if x = yz for some z but it generally does not follow that if x and y divide each other that they are associates, for there may be zero divisors. A commutative ring R in which there are no zero divisors is called an integral domain or simply a domain. (The older name was “domain of integrity”.) In a domain, if x and y divide each other, x = yz and y = xw, then we have x = xwz or x(1 −wz) = 0, so 1 −wz = 0. Thus w and z are units and x and y are associates. An element which has no divisors except itself and associates is called irreducible.
When we speak of “unique factorization” it means factorization into irreducibles which is unique up to the 7 order of the irreducible factors and multiplication by units (or replacement of factors by associates). It is quite possible in a domain for an element to have genuinely distinct factorizations into irreducibles. Consider, for example, the ring of integers of Q(√−5); it consists of all m + n√−5 with m, n ∈Z. In this ring the elements 3, 7, 1 + 2√−5, 1 −2√−5 are all irreducible.
Exercise 6 Prove this from first principles. (We will see more sophisticated reasons later.) Unfortunately unique factorization fails inside the ring of integers of Q(√−5 for we have the two distinct factorizations into irreducible factors 21 = 3 · 7 = (1 + 2√−5) · (1 −2√−5). A domain in which we have unique factorization into irreducibles is called a unique factorization domain, abbreviated UFD, or a factorial domain. In these we sometimes call irreducible elements “primes”, but bear in mind that with this definition +2 and −2 are both rational primes. It is a basic theorem (and not too difficult) that Z[i] is a factorial domain. Obviously in a factorial domain if an irreducible element divides a product then it must divide one of the factors; in fact, this is a crucial property that one must prove to show that a domain is factorial. If R is a factorial domain and π an irreducible element of R then the quotient ring R/π is again a domain and conversely. On the other hand Z[√−5]/3 is not a domain since (1+2√−5)·(1−2√−5) = 3·7 ≡0 mod 3. We will prove that Z[i] is factorial, but for the moment let’s accept it.
Lemma 2 A finite domain R is a field Proof.
We must show that every non-zero element x ∈R has an inverse.
Consider the set of all xy as y varies in R. No two of these can be identical, for if xy = xy′ then x(y−y′) = 0, contradicting the assumption that R is a domain.
Since R is finite, the set of all xy with fixed x must be all of R, so there is a y such that xy = 1. ■ Exercise 7 We really don’t need the commutativity in the preceding Lemma or even the existence of a unit element. Prove that if S is a finite set with an associative multiplication with both the left and right cancelation property, i.e., xy = xy′ implies y = y′ and yx = y′x also implies y = y′ then S is in fact a group (and in particular, there is a unit element for multiplication). (Hard) Suppose only one of the two cancelation properties holds. Is S still a group?
(Give a proof or a counterexample, but don’t spend too much time on it.) Finally (assuming some of the things we have not yet proven) we have the proof of Theorem 1: Proof. To prove assertion 1. we must show that if p ≡1(4) then p factors in Z[i]. Suppose to the contrary that it remained irreducible. Since Z[i] is a factorial domain it would follow that Z[i]/p is again a domain, with exactly p2 elements, and being finite it would be a field. The equation x4 = 1 could then have no more than four roots in Z[i]/p. But Z[i]/p ⊃Z/p and the latter is a field with exactly p elements. Its multiplicative group is therefore a cyclic 8 group with p −1 elements, and this is a multiple of 4, say p −1 = 4m. If a is any generator of this group then 1, am, a2m and a3m are four distinct elements satisfying x4 = 1. But (the classes of) i and −i are not amongst these and also satisfy the equation, so there are too many roots. Therefore Z[i]/p can not be a field, so p must factor. For 2., suppose that m is a sum of two squares, hence of the form m = (a + bi)(a −bi) for some a, b ∈Z and that a prime p ≡3(4) divides m. Since p is still irreducible in Z[i] it must divide one of the two factors.
Suppose pk||(a + bi) (meaning that k is the precise power to which p divides a+bi). Since pk is real, taking conjugates we see that also pk||(a−bi) so p2k||m.
■ So one way to understand the “if” (hard) part of Fermat’s original assertion, that an odd prime p is a sum of two squares if and only if p ≡1(4), is to say that such a prime must factor in Z[i]. There are many details that must be filled in; that will be our next project.
Exercise 8 Show that if a, b are integers then Z/(a + bi) always has exactly a2 + b2 elements, and if a, b are relatively prime then there is an isomorphism Z/(a2+b2) →Z/(a+bi) but not otherwise. What is the structure of the additive group of Z/(a + bi) when a and b are not relatively prime? (Hint: You might want to do the relatively prime case first.) 9
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Art of Problem Solving: Counting is as Easy as 1, 2, 3
Art of Problem Solving
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Art of Problem Solving's Richard Rusczyk explores a convenient method for counting lists of consecutive integers.
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Microbiology: Media Preparation
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HomeMicrobial Culture Media PreparationMicrobiology: Media Preparation
Microbiology: Media Preparation
What is Culture Media?
Microorganisms need nutrients, a source of energy, and certain environmental conditions in order to grow and reproduce. In the natural environment, microbes have adapted to the habitats most suitable for their needs. In the laboratory, however, these requirements must be met by culture media. A culture medium is an aqueous solution to which all the necessary nutrients have been added. Depending on the type and combination of nutrients, different categories of media can be made.
Section Overview
Types of Culture Media
Protein Hydrolysates
Aseptic Techniques
Media Supplements
Interaction of Media Components
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MicrObiological culture media: DIFFERENTIAL, SELECTIVE MEDIA, COMPLEX AND DEFINED MEDIA
Selective and differentialmediaare media based on either complex or defined media supplemented with growth-promoting or growth-inhibiting additives. The additives may be species- or organism-selective. e.g., a specific substrate, or an inhibitor such as cyclohexamide (artidione), which inhibits all eukaryotic growth and is typically used to prevent fungal growth in mixed cultures.
Complex mediaare rich in nutrients that contain water-soluble extracts of plant or animal tissue (e.g., enzymatically digested animal proteins such as peptone and tryptone). Usually, some sugar, often glucose, is added to serve as the main carbon and energy source. The combination of extracts and sugar creates a medium that is rich in minerals and organic nutrients, but since the exact composition is unknown, the medium is called complex.
Defined mediaare media composed of pure ingredients in carefully measured concentrations and dissolved in double distilled water (i.e., the exact chemical composition of the medium is known). Typically, they contain a simple sugar as the source of carbon and energy, a source of inorganic nitrogen, various mineral salts, and, if necessary, growth factors (purified amino acids, vitamins, purines, and pyrimidines).
Types of culture media and their importance
Culture Media Purpose
Complex Grow most heterotrophic organisms
Defined Grow specific heterotrophs and are often mandatory for chemoautotrophs, photoautotrophs and for microbiological assays
Selective Suppress unwanted microbes, or encourage desired microbes
Differential Distinguish colonies of specific microbes from others
Enrichment Simlar to selective media but designed to increase the numbers of desired microorganisms to a detectable level without stimulating the rest of the bacterial population
Reducing Growth of obligate anaerobes
The mixture of necessary nutrients can be used as a liquid medium, or a solidifying agent can be added. "Agar agar" is a natural polysaccharide produced by marine algae and is the most commonly used solidifying agent added to media (end concentration usually 1.5 % w/v). If hydrolysis of the agar is suspected, a silica gel is used as a replacement solidifying agent.
Explore our high-quality microbial culture media that can enhance reproducibility and reliability in microbiological testing in your laboratory. Visit Microbial Culture Media webpage.
Protein Hydrolysates
Complex media contain often protein hydrolysate which are are excellent natural sources of amino acids, peptides and proteins in growth media. It is the most important source for nitrogenous nutrients. They are most often obtained by enzymatic digestion or acid hydrolysis of natural products, such as animal tissues, milk, plants or microbial cultures. The number of available protein hydrolsate, also called peptones, is enormous and can promote and sustain the growth of most common organisms. For the enzymatic digestion often papain, pepsin, trypsin or a mixture of enzymes of the pancreatic juice are taken. Below is a list of often used expressions and the definitions.
Term Explanation
Trypic digested Protein hydrolysate was produced by protein digestion with trypsin
Peptic digested Protein was digested by pepsin
Pancreatic digested Protein was digested by a mixture of enzymes of the pancreatic juice
Proteose Peptonet A mixed enzymatically digestion of meat proteins. It is rich in peptides with the higher molecular weight.
Tryptone Casein which was tryptic digested
Tryptose/Tryplose A mixed enzymatically digestion of animal proteins. The digest conditions are such that it contains many different peptides including those of higher molecular weight (proteoses).
ASEPTIC TECHNIQUES USED in MICROBIOLOGICAL labs
Before inoculation, it is important to maintain the sterile condition of the microbiological media and all materials it encounters. During any subsequent handling of the bacterial cultures, unwanted or contaminant organisms must be excluded by employing aseptic techniques.
Sterilization implies the destruction of all microorganisms including spores, this is accomplished using heat, chemicals, radiation, and filtration.
Heat Sterilization
Heat sterilization causes denatures and coagulates vital proteins. It is simple to use, effective, easy to control, and cost-effective. There are various forms of heat sterilization.
Red Heat: Inoculating wires or loops are sterilized by holding them in a Bunsen flame until they are red hot.
Moist Heat: Bacteria are more readily destroyed by moist heat (steam) than dry heat. Usually used for the sterilization of culture media, aqueous solutions, and the destruction of discarded cultures. Air must first be removed in order to achieve the 121 °C necessary for successful sterilization. This is accomplished by the use of an autoclave (the technical version of a pressure cooker), which follows automatic cycles of heating under pressure for the required time.
Dry Heat: Usually employed for materials that could either be corroded by steam or must remain dry before use. These include metal instruments, glass Petri dishes, flasks and pipettes, and cotton wool. In practice, dry heat sterilization requires longer time intervals and higher temperatures than steam sterilization (e.g., steam sterilization 121 °C for 15 minutes or dry heat sterilization 160 °C for 120 minutes).
Chemical Sterilization
Chemical sterilization is usually employed for delicate equipment such as optical instruments and electrical devices, which would otherwise be damaged by heat. Due to the toxicity of the chemicals used, this is not the most popular form of sterilization. Chemicals employed include gaseous ethylene oxide, which alkylates amino, sulfhydryl, carboxyl, and hydroxyl groups of microbial cell compounds; formaldehyde, used as a fumigant; and hydrogen peroxide vapor used in aseptic packaging.
Sterilization by Radiation
Radiation is applied for heat-sensitive materials and for environmental samples such as soil and sediment where structural changes caused by heat need to be avoided. Two forms of radiation are used:
UV Sterilization initiates the excitation of atoms, which in nucleic acids leads to fatal mutations. UV light cannot penetrate materials so is used mainly for surface treatments (e.g., Laminar flow benches, and air and water).
Ionizing Radiation can penetrate samples, causing ionization within cells. Gamma radiation generated through a 60 Co is used to sterilize complex matrices such as soil and foodstuff. Microorganisms show increased resistance to radiation under anoxic conditions (2-5x) and in frozen samples.
Sterilization by Filtration
Filtration sterilization operates through the exclusion rather than the destruction of microorganisms. It is safe for the user and is employed for sensitive liquids and gases. Three types of filters are currently in use:
Depth Filters are made of columns packed with fibrous materials such as glass wool or cotton wool. The twisting and turning fibers entrap particles and act as filters; they show little resistance to flow and are used mainly for gases or as pre-filters for membrane filters which are easily clogged.
Membrane Filters act by screening out particles. Their effectiveness depends on the size of the membrane pores and the electrostatic attractions present. The most commonly used filters in microbiology are usually made of cellulose acetate or cellulose nitrate. Membrane filtration is usually employed for heat-sensitive substances, e.g., vitamin solutions; the filters are heat-sterilized before use.
Pore size of membrane filter for screening
Size of filter pores required to screen out:
Yeast### 0.45 -1.2 µm
Bacteria### 0.2 µm
Viruses and mycoplasmas### 0.01-0.1 µm
Nucleation Track (Nuclepore) Filters consist of very thin polycarbonate films that have been treated with nuclear radiation and then etched with a chemical to create very uniform vertical holes. They are applied for the same material as membrane filters but are more easily clogged.
Media Supplements
Table of the most commonly used media supplements, methods of sterilization, and solubility.
Media supplements, methods of sterilization, and solubility
Substance Solubility in Water at 25 °C Comments/Sterilisation
Actidione
(Cyclohexamide)2.1 g/100 mL Destroyed by boiling in aqueous solution at pH 7 for 1 hr.
Filter sterilise
L(+)-Arabinose 1 g/1 mL
Arginine15 g/100 mL Sat. solution bly alkaline, absorbs CO 2 pK 1, 2.18 pK 2 9.09 pK 3 13.2
Asparagine2.16 g/100 mL Stable in aq sol. at 100 °C, hydrolysed in acidic solutions
Biotin 22 mg/100 ml pH of 0.01% sol. = 4.5
acidic solutions can be heat-sterilised
Cysteinesoluble Neutral slightly alkaline solution is oxidised to cystine pK 1,
1.71 pK 2 8.33 pK 3 10.78
Dextrin soluble in 3 parts boiling water
Ehrlichs reagent Fructose soluble
soluble
Fuchsin 1 g/7 mL
Galactose a) soluble in 0.5 parts water, freely soluble in hot water
b) soluble in 1.7 parts water at 17 °C
Glucose 1 g/1 mL pH of 0.5 M aq solution = 5.9
Glycerol miscible
Glycogen soluble with opalescence
Lactose 21.6 g/100 mL a) 1 g/5 mL
b) 1 g/2.2 mL at 15 °C
Maltose soluble mp 102-103 °C
Mannitol soluble
Niacin soluble Stable to autoclaving at 120 °C for 20 mins
Ornithinesoluble Aqueous solution alkaline pK, 1.94 pK 2 8.65 pK 3 10.76
PhenylalanineL 29.6 g/L at 25 °C
D 1 g/35.5mL at 16 °C
DL 14.11 g/L at 25 °C
pK 1 2.58 pK 2 9.24
Resazurine insoluble in water
soluble in dil alkali hydroxides Indicator
0.1 g in 20mL 1 N NaOH + water up to 500 mL
pH 3.8 = orange, pH 6.5 dark violet
Ribose soluble
Citric acid soluble 59.2 % at 20 °C pH of 0.1 N solution = 2.2
Sorbitol soluble Up to83%mp 100/112 °C
Starch insoluble
Sucrose 1g/0.5mL
EDTA (disodium salt)soluble Used to complex iron in media
EDTA (acid)insoluble
All stock solutions of amino acids can be autoclaved at 120 °C for 20 mins.
Interaction of Media Components
When undertaking research where medium composition plays an important role, (e.g. toxicity studies), care must be taken to observe the various interactions which can take place between media components.
Media Components Type of Reaction
Yeast, peptone and
amino acids Bind large amounts of divalent Hg, Pb, Ag and Cu ions
Glucose exhibits non-selective metal binding which can cause metal deficiency of essential ions or decrease
the toxicity of toxic ions.
EDTA chelates essential trace metals causing deficiency; the effect can be reversed by the addition of
other metals e.g. Cu and Fe.
NaCI increases Zn toxicity (formation of a Zn-chloro complex), decreases Cd toxicity.
Agar in solid medium, Sn toxicity increases (formation of a soluble toxic tinagar complex).
Silica gel decreases toxicity of Sn, Cd, Pb, Ni and Zn divalent ions.
Phosphate precipitation of insoluble phosphates decreasing toxicity of toxic ions
(e.g Cd and Pb) or causing deficiency of essential ions (e.g. Fe)
Carbonate precipitation of insoluble carbonates
decreased Pb toxicity
Citrate non-selective metal binding
Tris non-selective metal binding
Can be avoided by addition of a HEPES buffer to reduce phosphate.
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https://corescholar.libraries.wright.edu/cgi/viewcontent.cgi?article=1004&context=math
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https://pmc.ncbi.nlm.nih.gov/articles/PMC9862925/
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Weighted Matrix Decomposition for Small Surface Defect Detection - PMC
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Micromachines (Basel)
. 2022 Dec 29;14(1):92. doi: 10.3390/mi14010092
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Weighted Matrix Decomposition for Small Surface Defect Detection
Zhiyan Zhong
Zhiyan Zhong
1 School of Automation, Guangdong Polytechnic Normal University, Guangzhou 510665, China
Conceptualization, Funding acquisition, Investigation, Methodology, Project administration, Resources, Software, Supervision, Validation, Visualization, Writing – original draft, Writing – review & editing
Find articles by Zhiyan Zhong
1, Hongxin Wang
Hongxin Wang
2 Machine Life and Intelligence Research Center, Guangzhou University, Guangzhou 510006, China
3 School of Computer Science, University of Lincoln, Lincoln LN6 7TS, UK
Conceptualization, Funding acquisition, Investigation, Methodology, Project administration, Resources, Software, Supervision, Validation, Visualization, Writing – original draft, Writing – review & editing
Find articles by Hongxin Wang
2,3,, Dan Xiang
Dan Xiang
1 School of Automation, Guangdong Polytechnic Normal University, Guangzhou 510665, China
4 Guangzhou Maritime University, Guangzhou 510725, China
Funding acquisition, Project administration, Validation, Writing – review & editing
Find articles by Dan Xiang
1,4
Editor: Tsz Ho Kwok
Author information
Article notes
Copyright and License information
1 School of Automation, Guangdong Polytechnic Normal University, Guangzhou 510665, China
2 Machine Life and Intelligence Research Center, Guangzhou University, Guangzhou 510006, China
3 School of Computer Science, University of Lincoln, Lincoln LN6 7TS, UK
4 Guangzhou Maritime University, Guangzhou 510725, China
Correspondence: howang@lincoln.ac.uk
Roles
Zhiyan Zhong: Conceptualization, Funding acquisition, Investigation, Methodology, Project administration, Resources, Software, Supervision, Validation, Visualization, Writing – original draft, Writing – review & editing
Hongxin Wang: Conceptualization, Funding acquisition, Investigation, Methodology, Project administration, Resources, Software, Supervision, Validation, Visualization, Writing – original draft, Writing – review & editing
Dan Xiang: Funding acquisition, Project administration, Validation, Writing – review & editing
Tsz Ho Kwok: Academic Editor
Received 2022 Nov 1; Revised 2022 Dec 17; Accepted 2022 Dec 23; Collection date 2023 Jan.
© 2022 by the authors.
Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (
PMC Copyright notice
PMCID: PMC9862925 PMID: 36677153
Abstract
Detecting small defects against a complex surface is highly challenging but crucial to ensure product quality in industry sectors. However, in the detection performance of existing methods, there remains a huge gap in the localization and segmentation of small defects with limited sizes and extremely weak feature representation. To address the above issue, this paper presents a weighted matrix decomposition model (WMD) for small defect detection against a complex surface. Firstly, a weighted matrix is constructed based on texture characteristics of RGB channels in the defect image, which aims to improve contrast between defects and the background. Based on the sparse and low-rank characteristics of small defects, the weighted matrix is then decomposed into low-rank and sparse matrices corresponding to the redundant background and defect areas, respectively. Finally, an automatic threshold segmentation method is used to obtain the optimal threshold and accurately segment the defect areas and their edges in the sparse matrix. The experimental results show that the proposed model outperforms state-of-the-art methods under various quantitative evaluation metrics and has broad industrial application prospects.
Keywords: defect detection, machine vision, weighted matrix decomposition
1. Introduction
Under the trend of intelligent manufacturing, automatic quality control is widely regarded as the top priority in industrial production [1,2]. Surface defect, the key factor of quality control, is generally defined as local anomalies embedded in homogeneous textures . Due to complexity of manufacturing processes and diversity of production environments, defects on product surfaces are always various and complex. For example, the manufacturing process of powder metallurgy is composed of casting, forging, rolling, machining, and extrusion, while its porosity, an important characteristic of powder metallurgy sintered materials, is influenced by multiple factors including temperature, sintering time, and pressure . Surface defects not only directly lower appearance quality, but also reduce product performance and commercial value . To effectively detect surface defects, automatic visual inspection methods with great advantage in non-destructive defect detection have been widely applied in rails , fabric , steel , thin-film-transistor , photovoltaic , and other flat products .
Conventional defect detection methods can be roughly divided into three categories: statistical-, spectral-, and model-based techniques . Statistical-based approaches compute local features at each pixel and then derives a set of statistics from distributions of the local features to discriminate defects . They generally contain eight representative algorithms, including thresholding , clustering , edge-based , fractal dimension , gray-level statistic , co-occurrence matrix , local binary pattern , and morphological . Among them, the most classical method is thresholding. Ostu presented a threshold selection method from gray level histograms, which is optimal for thresholding a histogram with bimodal or multimodal distribution but fails if the histogram is unimodal or close to unimodal . To address this issue, a number of modified methods were developed based on the work of Ostu . For example, Ng et al. proposed the valley-emphasis method (VE), which is suitable for images with unimodal and bimodal distributions . To obtain better and more stable thresholding results, the Gaussian valley-emphasis method (GVE) was developed . Truong et al. designed an entropy weighting scheme (EnOstu) to improve upon Otsu’s method . In addition, neighborhood valley-emphasis method (NVE) and improved valley-emphasis method (IVE) were proposed for small surface defect detection. Statistical-based approaches perform well on defect detection against steel surface whose pixel intensities exhibit remarkable regularity and apparent periodicity. However, they are vulnerable to illumination change , pseudo defects, and local noise .
To overcome the above limitation, the second method transforms the whole image from spatial domain to frequency domain where defects can be discriminated from defect-free regions more easily via comparison of their amplitudes . Spectral-based approaches are composed of six representative algorithms, such as Fourier transform , Gabor filters , optimized FIR filters , wavelet transform , multiscale geometric analysis , and hough transform . Tsai et al. proposed fabrics defects inspection method based on a global image restoration scheme using the Fourier transform, which does not rely on local features of textures . Similarly, Hu et al. combined Fourier analysis and wavelet shrinkage to detect textile defects . Aiger et al. presented a novel method based on the Phase Only Transform (PHOT) for detecting defects on textured surfaces. Its simplicity and generality enable it to work on various pattern without prior knowledge. Choi et al. proposed an unsupervised detection approach of surface defects by combining both global estimation and local refinement, which gives robust results even in noisy surface defect images . However, the spectral-based method is unable to effectively preserve local information after the transformation, which means it can only deal with defects that occupy a large amount of the whole image and exhibit significant difference to the background.
To achieve better performance for diverse defects, the third method projects original texture distribution of image blocks into low-dimensional space using different specialized models , such as Markov random field , Gaussian mixture entropy , and Weibull distribution . Model-based approaches accomplish defect detection by similarity measurement between the model which is established by the feature of image and the test image . Susanet et al. proposed a Gaussian mixture entropy model to automatically defect detection with no manual intervention . Han et al. developed a novel methodology for anomaly detection in noisy images with smooth backgrounds, named smooth-sparse decomposition (SSD) , which has superiority in terms of the detection accuracy as well as computation time. The following year, they proposed a spatio-temporal smooth sparse decomposition , which has the capability of identifying not only the time of process changes, but also the location of detected anomalies. Yu et al. proposed a coarse-to-fine model to identify rail defects at different scales . However, the model-based methods are designed for specific defects and difficult to generalize for various types of defects .
Most of the aforementioned approaches distinguish defects from background by extracting the textural features . The statistical and spectral methods can hardly deal with the situation when detect exhibit low contrast against complex background . The model-based approaches tend to consider the feature of the test images, which means the model is always specific and difficult to achieve universality. Convolutional neural networks (CNNs) have become a key element in the breakthrough of defect detection [50,51,52], but they may fail to detect small defects that are only a few pixels in size, since rich representations are difficult to learn from their poor-quality appearance and structure. Another obstacle presented by convolutional neural networks is that downsampling of the image always misses the opportunity to locate small defects; for example, YOLO apply a downsampling of to the input image, which means a defect with only a few pixels will disappear in the feature map . In view of these points, the above conventional methods are difficult to be applied directly for the inspection of small defects on surfaces with complex backgrounds. Although these conventional defect detection methods have accomplished grate success in the detection of specific defects, their performance still needs to be improved for small defects with extremely low gradient intensity or contrast to background while equate to only a few pixels in size within images.
Detecting small defects against complex surfaces has been an extremely challenging task for industrial manufacturing. Figure 1 gives an example of small surface defect . As shown, the resolution of the whole image is pixels where the defect and its surrounding region are enlarged in a square box with pixels in size. Figure 1b presents the ratio of the number of pixels in the defect region to the total pixel number of the original image. As can be seen, the proportion of the small defect to the whole image is quite low (less than ). In this case, it is difficult even for humans to notice such tiny defects. Specifically, difficulties for small defect detection are reflected in three aspects:
Figure 1.
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Examples of small surface defect dataset . (a) Typical challenging examples of small surface defect. (b) The ratio is calculated over the defective or defect-free regions to an image.
(1) Small defects always equate to a few pixels in size, revealing almost no other visual features, such as shape, texture, and color, which means their feature representations are extremely weak.
(2) Low contrast and unclear boundary between defect and defect-free regions of complex surface bring great uncertainties for defect feature extraction.
(3) Noise inevitably introduced in image acquisition processes is difficult to discriminate from small defects and also leads to inaccuracy of defect area segmentation.
To address these issues, a weighted matrix decomposition model (WMD) is proposed for small defect detection against complex surface. Specifically, a matrix is constructed by weighted summation of three channels from the original color image to improve the contrast between small defects and background. The weighted matrix is then decomposed into a sparse matrix and a low-rank matrix to identify defect regions while eliminating the interference of complex backgrounds. The pixel intensity of the defects identified by the sparse matrix are further analyszd. Finally, the binarization operation and the pixel value statistical analysis of the connected domain are performed to accurately segment defect regions and effectively filter out noise.
The rest of our paper is organized as follows. In Section 2, the weighted matrix decomposition model is introduced. In Section 3, experimental results on small surface defect detection are presented, together with the qualitative studies. Finally, the main idea and future work of this paper is summarized and concluded in Section 4.
2. Weighted Matrix Decomposition Model
2.1. Overall Network Architecture
The total framework of the proposed weighted matrix decomposition (WMD) model is shown in Figure 2. As can be seen, it is composed of three main steps, including weighted matrix construction, image decomposition, and defect region segmentation. Specifically, a weighted matrix is firstly constructed for a color image to improve contrast between defect and defect-free regions. The weighted matrix is further decomposed into a low-rank and sparse matrices to eliminate redundant background. Finally, defective regions in the latter matrix are located by an adaptive threshold. The following sections will elaborate on each step of the proposed WMD model.
Figure 2.
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Framework of the WMD model for defect detection.
2.2. Weighted Matrix Construction
Figure 3 shows the 2D views and 3D views of the RGB channels and gray-scale version of a steel surface image. As can be seen, pixel values of the defect are extremely close to those of the background. In this case, it is quite difficult to discriminate the defect and defect-free regions. To solve this problem, a weighted matrix is proposed to enhance contrast, which consists of three steps, including normalization, weighted summation, and linear transformation.
Figure 3.
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The first column shows R channel, G channel, B channel, grayscale version, and weighted matrix of a steel surface image, respectively, while the second and third columns present their corresponding 2D and 3D views. The weighted matrix is derived from the weighted summation of the R, G, and B channels.
(1) Normalization
The pixel values of the RGB channels are all normalized to for the convenience of calculation, that is,
(1)
where denotes input color image, represents spatial coordinates, and is the channel index corresponding to the R, G, B channels, respectively.
(2) Weighted Summation
To closely match human perception of lightness, weights of the RGB channels are assigned based on the CIELAB color space . Specifically, the higher contrast between defect and defect-free regions, the larger the weight. As shown in Figure 3, the contrast between the defect and defect-free regions in the three channels are arranged in descending order as G, R, and B channels. The weighted summation of the RGB channels is expressed as
(2)
where , , and are constant and satisfy , . To further enhance the contrast of the defect, the cube root of the weighted summation is adopted, that is
(3)
where denoted the contrast-enhanced matrix.
(3) Linear Transformation
The pixel values of are linearly transformed to integer values from 0 to 255, that is
(4)
where is a positive integer matrix. If there are decimals in Equation (4), we round them up to the nearest integer. In this stage, we obtain the weighted matrix , where the contrast of the defect is significantly enhanced, as can be seen from Figure 3. is further decomposed in the following sections to separate defects from the background.
2.3. Matrix Decomposition
A weighted matrix is first partitioned into N nonoverlapping patches, and then decomposed into sparse matrix with defects and low-rank matrix with redundant backgrounds. To address the issues discussed in Section 1, a novel weighted matrix decomposition model is proposed as follows:
(5)
where is devoted to allow identification of the intrinsic feature subspace of the redundant background patches. is used to capture the spatial and feature relations of patches. Then, the defect detection task is transformed into a typical optimization problem of recovering low-rank and sparse components from a data matrix:
(6)
where the is weight coefficient (). r is a constant. represents the norm to count the number of non-zero values. k is the number of pixels in the defect area.
Figure 4 illustrate the matrix decomposition process for an image containing single defect, while Figure 5 present the matrix decomposition process for an image containing multiple defects. In each figure, the first column shows visual images of the weighted matrix, low-rank matrix, and sparse matrix after integerization while the second column presents their 3D views. In the following, the regularization of sparsity and low-rank matrices is verified, and their decomposition is implemented.
Figure 4.
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Matrix decomposition in single-defect images. The first column is a weighted matrix, which is decomposed into low-rank (in second column) and sparse matrices (in third column), respectively.
Figure 5.
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Matrix decomposition in multiple-defect images. The first column is a weighted matrix, which is decomposed into low-rank (in second column) and sparse matrices (in third column), respectively.
(1) Sparsity Regularization for Salient Defects
The contrast between defect and defect-free is glaringly obvious in the sparse matrices of Figure 4 and Figure 5. In their 3D view images, it is quite clear that most of the values are 0. Since the defect area is particularly smaller than the input image, the satisfies the following requirements:
(7)
where the image size is M × N, namely:
(8)
Equation (8) means that most of the elements in matrix are 0. That is, the assumption that the weight matrix of small surface defect images have sparsity, holds.
(2) Low-Rank Regularization for Image Background
As shown in Figure 4 and Figure 5, the texture of low-rank matrices is relatively smooth. If the image is divided into L blocks (where ), they are often similar and approximately lie in a low-dimensional subspace. The low-rank regularization on the background feature matrix is applied to pursue its intrinsic structure. Figure 6 presents three representative small surface defect images with the same size of , their corresponding weighted matrices in second column, and the singular value of the weighted matrix in third column where the patch size is and the vertical sliding steps are 8. Since the sizes of all corresponding patch-images are , the matrix has 60 singular values. The above analysis shows that the background feature matrix derived from the weighted matrix is low-rank, that is,
(9)
where r is a constant. Intrinsically, it constrains the complexity of the background image.
Figure 6.
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Low-rank matrix analysis. The first column shows three representative examples. In addition, the second and third one are weighted matrices and the singular values of the corresponding background patch-images.
(3) Sparse and Low-Rank Matrices Decomposition
In order to detect small surface defect of images, the defect detection task is intrinsically a typical problem of recovering a low-rank and sparse components from a data matrix. This problem can be effectively solved via Principal Component Pursuit and converted to solve the following convex optimization problem:
(10)
where is a positive constant. , replace in Equation (9) and in Equation (7), respectively, for tractable computation. Here, is the nuclear norm of a matrix (i.e., the sum of singular values); is the -norm, that is
(11)
The above optimization problem Equation (10) is convex and can be solved by applying the Accelerated Proximal Gradient approach proposed in . Through the above steps, the low-rank and sparse decomposition process of the weighted matrix has been completed, where the latter contains defect areas.
2.4. Pixels Segmentation
After obtaining the sparse matrix, the following features can be observed in combination with Figure 1, Figure 4 and Figure 5.
(1) From Figure 1, it can be seen that the percentage of small defects in the whole image is very low, which means that the background occupies most of the image, that is, the area difference between defect and defect-free is highly significant.
(2) Most pixels in the sparse matrix are zero (background regions), and the non-zero pixels correspond to defects or noises, as shown in the 3D view of Figure 4 and Figure 5 (the second row). For intuitive visualization, the pixel values of sparse matrix are readjusted to a value of positive integer between 0 and 255, as shown in Figure 4 and Figure 5 (the first row).
After the above analysis, combined with the sparse matrices of Figure 4 and Figure 5, we have two following conclusions. Firstly, the pixel values of the defect are less than 0, and the pixel value greater than 0 is interference such as noise. Secondly, after integer conversion, the mode in images is adopted as the threshold. If pixel values of an area are lower than the mode in images, then the area is regarded as a defect; otherwise, it is regarded as the background. The binarization image is expressed by
(12)
where the defect is represented as 1 and the background is 0.
In order to enhance the consistency of the defect images, for those images containing a small amount of noises or other interference factors, the connected domain pixel value statistics method is used to further instantly remove the interference and accurately segment the defect area.
3. Results and Discussions
3.1. Experimental Setup
3.1.1. Datasets
To guarantee the effectiveness of the algorithm, we use two publicly available datasets for experimentation. In order to convince the experimental effect, we use the existing methods and the proposed method in this paper for performance evaluation in the same dataset. The dataset used in this experiment is the silicon steel strip dataset ( accessed: 10 December 2022) , which contains a number of color images with a resolution of pixels. It is characterized by small defect areas, noise interference, complex background, and extremely low contrast between defect and defect-free regions. Another is the magnetic tile dataset ( accessed: 10 December 2022), consisting of numerous size-varied images. In the experiments, each image is cropped to uniform pixels. The contrast between the defect and the background is extremely low in each magnetic tile image, while interfering by stains and textures.
3.1.2. Parameter Settings
The parameters in the implementation of the proposed WMD model are set as follows. In weight matrix construction, to closely match human perception of lightness, we assign weights of the RGB channels based on the CIELAB color space, where = , = , = . In matrix decomposition, is related to the number of small defect pixel values, which is set as to . All experiments are tested on the MATLAB software platform under a machine that is equipped with Intel-i7 2.4-GHz CPU, 16-GB memory.
3.1.3. Comparison Algorithms
The proposed defect detection algorithm is compared with nine state-of-the-art solutions, including six statistical methods (EnOstu , VE , NVE , GVE , IVE , Ostu ), a spectral method (PHOT ), and a model-based method (SSD ). EnOstu is fully automatic and capable of detecting extremely small defect regions, and the images of low contrast between defect and defect-free. VE is widely used in defect detection, where the applicable defect range can be from no defect to small or large defects. NVE has accurate segmentation results in defect detection, such as a small defect image, a part image, and a number image. GVE resolves such problem of an optimal segmentation threshold selection in a image with unimodal or close to unimodal by introducing a Gaussian weighting scheme to enhance the weighting effect. IVE is suitable for the defect detection with uneven illumination, complex image texture, and relatively small defect area. Ostu is a classical automatic threshold segmentation method. PHOT can be applied to various patterns of defect detection, e.g., periodic texture defect, multiple textures defect of various size and regularities, arbitrary scene with synthetic defect, etc. SSD can detect various types of defects in a smooth background. With the popularization of deep learning methods, a defect classification method based on deep features, called MT , is selected as the comparison method of this experiment. In this method, SqueezeNet and MobileNetV2 models are adopted for feature extraction, while the ReliefF algorithm is used for feature selection.
3.1.4. Evaluation Metrics
To evaluate the performance of small surface defect detection methods, we employ nine comparison metrics including accuracy (), sensitivity (), specificity (), the Mean Intersection over Union (), the F-Measure () , the Mutual Information (), the Normalized Mutual Information (), the Structural Similarity () , and the E-Measure () , respectively. It should be noted that the recall rate and sensitivity are the same. The higher these nine indexes, the better the performance of the algorithm. The common evaluation indicators , , , , can be formulated as follows
(13)
(14)
(15)
(16)
(17)
(18)
where , , , and represent the number of True Positives, True Negatives, False Positives, and False Negatives, respectively. True Positive () denotes the number of correctly detected defect pixels mapping to the ground truth. True Negatives () represents the number of correctly detected background pixels mapping to the ground truth. False Positive () is the number of defect pixels that are false detected. False Negative () represents the number of undetected defect pixels. In general, the improvement of precision corresponds to the decrease of recall. For the metric, we use in this paper to balance precision and recall.
We also compare the detected result with ground truth by calculating their pixel and structure similarity. Specifically, and metrics compute pixels similarity between the test image and the ground truth, while the and compute the structural similarity.
(19)
where and are the mean values of x and y, respectively. and are the variances of x and y, respectively. is the covariance of x and y. and are constants. comprehensively evaluates image quality from pixel, region, boundary, and object level , that is
(20)
where w and h are the width and height of image, respectively. is an enhanced alignment matrix.
3.2. Comparison With the State-of-the-Art
The proposed WMD algorithm is evaluated on the small surface defect detection dataset and compared with nine recently proposed algorithms. The visual comparisons are shown in Figure 7, Figure 8, Figure 9, Figure 10, Figure 11 and Figure 12, and Table 1 shows some quantitative comparison. The results show that the proposed WMD, an unsupervised method, ranks first on the small surface defect detection dataset across different criteria.
Figure 7.
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Visual comparisons of detection results of the ten methods on the image with a single defect. Our detection result (WMD) is quite close to ground truth.
Figure 8.
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Visual comparisons of detection results of the ten methods on the image with multiple defects. The detection result of the proposed WMD method is quite close to ground truth, whereas the other competing methods are unable to either identify the defect location or completely separate the defect from the background.
Figure 9.
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Visual comparisons of detection results of the ten methods on the image with multiple defects. The detection result of the proposed WMD method is quite close to ground truth, whereas the other competing methods are unable to either identify the defect location or completely separate the defect from the background.
Figure 10.
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Visual comparisons of four methods on (a–e) five randomly selected single-defect images. The detection result of the proposed WMD method is quite close to ground truth (GT), whereas the other competing methods are unable to either identify the defect location or completely separate the defect from the background.
Figure 11.
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Visual comparisons of four methods on (a–e) five randomly selected multiple-defect images. The detection result of the proposed WMD method is quite close to ground truth (GT), whereas the other competing methods are unable to either identify the defect location or completely separate the defect from the background.
Figure 12.
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Quantitative comparison in terms of (a) (%), (b) (%), (c) (%), (d) , (e) , (f) , (g) , (h) , and (i) , where the x axis is the Image ID. Our method (WMD) significantly outperforms these evaluated defect detection algorithms across ten randomly selected surface defect detection images.
Table 1.
Results on the magnetic tile image in terms of (%), (%), (%), (%), , , , , , and .
| | (%) | (%) | (%) | (%) | | | | | |
:---: :---: :---: :---: :---: |
| EnOtsu | 0.45013 | 0.19990 | 100 | 0.25123 | 0.00399 | 0.00200 | 0.50008 | 0.86746 | 0.00130 |
| VE | 10.49514 | 0.22229 | 100 | 10.31631 | 0.00444 | 0.00222 | 0.50031 | 0.88316 | 0.23347 |
| NVE | 0.27837 | 0.19956 | 100 | 0.07913 | 0.00398 | 0.00200 | 0.50004 | 0.86711 | 0.02549 |
| GVE | 0.27837 | 0.19956 | 100 | 0.07913 | 0.00398 | 0.00200 | 0.50004 | 0.86711 | 0.02549 |
| IVE | 0.27837 | 0.19956 | 100 | 0.07913 | 0.00398 | 0.00200 | 0.50004 | 0.86711 | 0.02549 |
| PHOT | 99.66240 | 8.33333 | 6.93069 | 99.84768 | 0.07568 | 0.03933 | 0.50745 | 0.99984 | 0.58086 |
| SSD | 99.84009 | 79.41176 | 26.73267 | 99.98615 | 0.40000 | 0.25000 | 0.59021 | 0.99992 | 0.64337 |
| Ostu | 36.86725 | 0.30245 | 96.03960 | 36.74903 | 0.00603 | 0.00302 | 0.50046 | 0.92478 | 0.24855 |
| MT | 50.75417 | 0.38346 | 95.04950 | 50.66566 | 0.00764 | 0.00383 | 0.50072 | 0.94407 | 0.25117 |
| WMD | 99.86378 | 94.44444 | 33.66337 | 99.99604 | 0.49635 | 0.33010 | 0.62883 | 0.99994 | 0.66594 |
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3.2.1. Visual Comparison
Figure 7, Figure 8 and Figure 9 show the visualization of partial detection results. In Figure 7 and Figure 8, we can observe that EnOstu , VE , IVE , SSD , Ostu , and MT cannot identify the location of the defect. NVE , GVE , PHOT , and our method (WMD) are able to effectively separate the defect from the background. In Figure 9, EnOstu , VE , NVE , GVE , and IVE almost judge the background as a defect. PHOT and SSD identify background and a small number of defects. Ostu and MT are powerless against magnetic tile defects.
To further evaluate the performance of these methods, we conduct two experiments on five randomly selected single-defect images and multiple-defect images in a silicon steel strip dataset, as shown in Figure 10 and Figure 11, respectively. In these two experiments, the results from EnOstu , VE , IVE , SSD , Ostu , and MT methods are not present because they are incapable of detecting defects and their results contain a number of false positives. Similarly, there is no experimental result for the magnetic tile image because the traditional method is similar to Figure 9 in other images. In Figure 10, PHOT is unable to effectively identify defects on all the five images. The NVE and GVE methods can detect relatively salient defects in (a)–(c) and partial defect regions in (d), but fail to detect the defect that is highly similar to the background in (e). The method proposed (WMD) in this paper achieves the best detection results in the above five images. In Figure 11, we obtain a conclusion similar to that in Figure 10.
3.2.2. Performance Comparison
It can be seen that the methods of NVE and GVE are suitable for the detection of small surface defects by the comparison of the above experiments. Figure 12 shows some quantitative comparisons by the three methods (NVE, GVE, and WMD). As shown in Figure 12a,c–i, (%), (%), , , , , , and are clearly higher than those of the baseline methods. represents the probability of detected background pixels, and denotes the probability of detected defect pixels. However, it should be noted that recall is more indispensable than precision in small surface defect detection, in respect that a missed defect has a greater hazard than an error-checked noise. The performance comparison demonstrates that the proposed WMD model performs better than all the competitors in detecting the images with single defect and multiple defects.
To quantify the effect of Figure 9, the evaluation indicators of ten methods are shown in Table 1. As can be seen, the WMD works best in the (%), (%), (%), , , , , and indicators. The value of EnOstu , VE , NVE , GVE , and IVE respectively reaches 100%, because the five methods almost all judge the background as defects. However, their and values are extremely low, and the value for correct identification of background and defects is also quite small. The above results demonstrate the effectiveness of the proposed method in the magnetic tile image.
3.3. Verify the Effectiveness of Weighted Matrix Decomposition
In the experiment comparison in Section 3.2, the small surface defect images are converted to gray images, then the existing methods are used to perform defect detection. However, in Section 2.2, the contrast between the defect and the background of the weighted matrix is greater than that in the gray images (Figure 3). To verify the effectiveness of low-rank and sparse matrices decomposition in defect detection, all methods are simultaneously used in the weighted matrix. Visual comparisons are shown in Figure 13, Figure 14, Figure 15, Figure 16 and Figure 17, while some quantitative comparisons are presented in Figure 18 and Table 2.
Figure 13.
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Visual comparisons of detection results using the ten methods on weighed matrix. NVE, GVE, IVE, and PHOT do not have any output. EnOstu, VE, SSD, Ostu, MT, and our method (WMD) have the ability to separate defects from the background.
Figure 14.
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Visual comparisons of detection results using the ten methods on weighed matrix with multiple defects. NVE, GVE, and PHOT cannot effectively identify defects. EnOstu, VE, IVE, SSD, Ostu, MT and our method (WMD) can effectively provide the location of the defect.
Figure 15.
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Visual comparisons of detection results using the ten methods on weighed matrix of magnetic tile images. EnOstu, VE, NVE, GVE, IVE, PHOT, SSD, Ostu, and MT cannot effectively identify defects. VE only identifies some defects, and our method (WMD) can effectively provide the location of defect.
Figure 16.
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Visual comparisons of different methods on (a–e) five randomly selected single-defect images. The ground truth (GT) is shown in the last column. EnOstu, VE, SSD, and our method (WMD) have the ability to locate the defect.
Figure 17.
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Visual comparisons of different methods on (a–e) five randomly selected multiple-defect images. Compared with the baseline methods, the proposed WMD produces more discriminative models, which is capable of identifying the defect from backgrounds.
Figure 18.
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Quantitative comparison in terms of (a) (%), (b) (%), (c) (%), (d) , (e) , (f) , (g) , (h) , and (i) , where the x axis is the Image ID. Our method outperforms the baseline methods.
Table 2.
Results on weight matrix of magnetic tile image in terms of (%), (%), (%), (%), , , , , , and .
| Methods | (%) | (%) | (%) | (%) | | | | | |
:---: :---: :---: :---: :---: |
| EnOtsu | 0.55279 | 0.20011 | 100 | 0.35410 | 0.00399 | 0.00200 | 0.50009 | 0.86766 | 0.00655 |
| VE | 80.78457 | 0.63551 | 61.38614 | 80.82333 | 0.01258 | 0.00633 | 0.50085 | 0.97779 | 0.25874 |
| NVE | 0.28627 | 0.19957 | 100 | 0.08704 | 0.00398 | 0.00200 | 0.50004 | 0.86712 | 0.01858 |
| GVE | 0.28627 | 0.19957 | 100 | 0.08704 | 0.00398 | 0.00200 | 0.50004 | 0.86712 | 0.01858 |
| IVE | 0.28627 | 0.19957 | 100 | 0.08704 | 0.00398 | 0.00200 | 0.50004 | 0.86712 | 0.01858 |
| PHOT | 99.77691 | 0 | 0 | 99.97626 | 0 | 0 | 0.50001 | 0.99987 | 0.38060 |
| SSD | 99.60317 | 27.67857 | 61.38614 | 99.67953 | 0.38154 | 0.23574 | 0.57370 | 0.99986 | 0.66337 |
| Ostu | 62.84056 | 0.48664 | 91.08911 | 62.78412 | 0.00968 | 0.00486 | 0.50095 | 0.95823 | 0.25334 |
| MT | 49.19450 | 0.36789 | 94.05941 | 49.10486 | 0.00733 | 0.00368 | 0.50064 | 0.94140 | 0.25088 |
| WMD | 99.86378 | 94.44444 | 33.66337 | 99.99604 | 0.49635 | 0.33010 | 0.62883 | 0.99994 | 0.66594 |
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3.3.1. Visual Effect Verification
Figure 13, Figure 14 and Figure 15 show the visualization results. As can be seen, the detection effect of NVE , GVE , and IVE present great distinction between single-defect and multiple-defect images. In Figure 13, these methods cannot identify background and defects, whereas in Figure 14, they fail to separate the defect or only part of the region. Similar conclusions can be obtained in other small surface defect images. Ostu and MT are unable to distinguish defects and backgrounds in Figure 13 and Figure 14. PHOT only identifies one of the defects in a small number of multi-defect images, and performs poorly in single-defect images. The results are similar with those in Section 3.2. EnOstu , VE , SSD , and our method (WMD) are capable of discriminating the defect regions. In Figure 15, the visual results of the EnOstu , NVE , GVE , IVE , PHOT , SSD , Ostu , and MT in the weighted matrix are similar to those in Figure 9. The detection performance of the VE in the weighted matrix is better than that in the original image, but it only detects part of the defects, and also mistakenly judges some background as defects. Our method (WMD) is capable of separating defects from the background.
Figure 16 and Figure 17 show some visual comparisons of the best methods in the experiments. As can be seen from Figure 16, EnOstu , VE , and WMD successfully separate the defects from the background. However, SSD occasionally fails to identify the defect. In Figure 17a–d, EnOstu and VE correctly detect small defects, but are incapable of detecting defects similar to the background in Figure 17e. SSD is unable to detect the defects as small as a few pixels or similar to the background. The method proposed (WMD) in this paper successfully locates the defects in the above five images.
3.3.2. Performance Metrics Verification
Figure 18 shows some quantitative comparison of the best methods (EnOstu , VE , SSD , and WMD). It can be concluded that the proposed model WMD in this paper has the best performance in terms of eight evaluation metrics (Figure 18a,c–i). The quantitative evaluation results in Figure 15 are shown in Table 2, which yields similar conclusions to Table 1. Among them, the and values of PHOT are both 0, indicating that it judges all defects as background. However, the value reaches , because the proportion of defects in the image is extremely low, and the background is particularly large. The quantitative evaluation results in Figure 15 are shown in Table 2, which yields similar conclusions to Table 1. Among them, the and values of PHOT are both 0, indicating that it judges all defects as background. However, the value reaches , because the proportion of defects in the image is quite low, and the background is particularly large.
In summary, for the images whose defect and defect-free region share similar appearance, WMD successfully separates the defects from the background, while other methods often fail. These results illustrate the robustness of the WMD algorithm, and confirm the effectiveness of the proposed weighted matrix in separating the low-rank and sparse subspaces.
4. Conclusions
In this paper, we have presented a weighted matrix decomposition model which effectively solves the problem of detecting small surface defects. Firstly, a weighted matrix is established to improve the contrast between small defects and background. Then, the weighted matrix is decomposed into a low-rank matrix representing image background and a sparse matrix identifying defects, which eliminates the interference of complex backgrounds. Finally, the pixel intensity of the defects is analyzed to select the optimal threshold and accurately separate the defects from the background. Experiments on a small surface defect dataset have shown that, compared to the conventional methods, the proposed achieves encouraging performance in small defect detection against complex product surfaces on public benchmark datasets under various quantitative metrics. The proposed model provides an effective solution in industrial manufacturing for detecting small defects against complex product surfaces. In the future, we will further generalize our 2D patch model into 3D or more dimensions and investigate applications of the N-D patch model. We will also try multi-subspace cluster strategies to further improve the flexibility of our method in highly variant background cases.
Author Contributions
Conceptualization, Z.Z. and H.W.; methodology, Z.Z. and H.W.; software, Z.Z. and H.W.; validation, Z.Z., H.W. and D.X.; writing—original draft preparation, Z.Z. and H.W.; writing—review and editing, Z.Z., H.W. and D.X.; visualization, Z.Z. and H.W.; supervision, Z.Z. and H.W.; project administration, Z.Z., H.W. and D.X.; funding acquisition, Z.Z., H.W. and D.X. All authors have read and agreed to the published version of the manuscript.
Institutional Review Board Statement
Not applicable.
Informed Consent Statement
Not applicable.
Data Availability Statement
Not applicable.
Conflicts of Interest
The authors declare no conflict of interest.
Funding Statement
This research was funded by the National Natural Science Foundation of China (Grant No. 62103112), the China Postdoctoral Science Foundation (Grant No.2021M700921), Guangzhou Key Laboratory of Intelligent Building Equipment Information Integration and Control (Grant No. 202002010003), University scientific research project of Guangzhou Education Bureau (Grant No. 202234607), and Special projects in universities’ key fields of Guangdong Province (Grant No. 2020ZDZX2002).
Footnotes
Disclaimer/Publisher’s Note: The statements, opinions and data contained in all publications are solely those of the individual author(s) and contributor(s) and not of MDPI and/or the editor(s). MDPI and/or the editor(s) disclaim responsibility for any injury to people or property resulting from any ideas, methods, instructions or products referred to in the content.
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Abstract
1. Introduction
2. Weighted Matrix Decomposition Model
3. Results and Discussions
4. Conclusions
Author Contributions
Institutional Review Board Statement
Informed Consent Statement
Data Availability Statement
Conflicts of Interest
Funding Statement
Footnotes
References
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Convert between thermal conductivity units - table and diagram.
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| Btu ft / (h ft2 oF) | Btu in / (h ft2 oF) | Btu in / sec ft2 oF) | Cal / (cm s oC) | Kcal / (cm s oC) |
| Btu ft / (h ft2 oF) | 1 | 12 | 0.0033 | 0.0041 | 4.134×10-6 |
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| Kcal / (m h oC) | 0.672 | 8.06 | 0.0022 | 0.00278 | 2.778×10-6 |
| Erg / (cm s oC) | 5.78×106 | 6.93×10-5 | 1.92×10-8 | 2.39×10-8 | 2.389×10-11 |
| Joules / (m h oC) | 1.61×10-4 | 0.00193 | 5.35×10-7 | 6.64×10-7 | 6.635×10-10 |
| Watt / (ft oC) | 1.89 | 22.8 | 0.0078 | 0.0078 | 7.84×10-6 |
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| Btu ft / (h ft2 oF) | 1.49 | 173076 | 6230 | 0.527 | 1.73 |
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社会学研究 9 9 年第3 期 方法论背景中的问卷调查法 风 笑 天 本 作为一种具体的社会调查方法 , 问卷法有着其特定的方法论背景和基础 。本文认 为 : ①问卷调查法在本质土是一种实证的方法 , 这种实证的特征决定了问卷 、随机抽 样及统计分析三者之间密不可分的关系 ; ②成功的问卷调查必须有充分的理论分析 作基础 , 必须有明确的理论枢架作指导 ; ③效率是问卷调查法所面临的最大挑战之 一 , 而操作化的水平和质量则是决定调查效度的关健因素 ; ①人既作为研究的主体 , 又作为研究客体的现实 , 制约和影响着问卷调查的可靠性和适用性 ; ③不同的调查 内容也在一定程度上决定和影响着问卷调查法的适用范围 。 作者 : 风笑天 , 男 , 1 9 5 4 年生 , 华中师范大学政治系教授 。 冲 乍 间卷调查法 , 即以问卷为工具来收集资料的调查方法 , 是当前最常用的社会调查方法之 一 。正如英国著名社会学家莫泽( C . A . 材。s e r ) 所说 : “ 社会调查十有八九是采用问卷方法进行 的 ” 。①从目前情况看 , 在运用阿卷法收集资料的调查研究中 , 还存在着一定的问题 。 它不仅影 响到具体的社会调查研究的效果 , 还影响到人们对问卷调查法的总体信任感 。而造成这一现状 的原因 , 不仅有研究者对具体方法 、程序的掌握与运用问题 , 还有对与这些具体操作方法和程 序密切相关的方法论背景的认识间题 。 从一般意义上看 , 问卷调查法有着一套较为固定的程序和操作内容 。任何一个社会学研究 人员(或者社会科学研究人员) , 只要熟悉和掌握了这套程序及内容 , 就可以运用它来进行自己 的研究 。 然而 , 从更深一层的意义上看 , 问卷调查又有着其特定的方法论背景和基础 。 如果离 开了这种特定的背景和基础 , 就会在实际运用问卷调查法的时候产生出各种偏差 , 影响到调查 研究结果的可靠性和准确性 。大量实践表明 , 熟悉和掌握间卷调查法的具体程序和操作方法 , 对于成功地进行一项调查研究来说 , 仅仅只是必要的 , 而不是充分的 。 要提高问卷调查的水平 和质量 , 还必须熟悉和了解它的方法论背景 。本文拟就几个与问卷调查法有关的方法论问题作 一初步的探讨 。 一 、问卷调查法的本质 就象特定的社会学理论往往同特定的方法论以及特定的研究方法相联系一样 , 作为具体 方法的问卷调查法也有着与之密切相连的方法论背景或基础 。 正是这种方法论背景决定了问 ① C . A .
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K a l t o n : S u r v e y M e t h o d s 一n s o e i a l ln v e 、t i g a t i o n .
( e S e o n d E d it i o n ) H E B I t d , 19 7 1 , P 4 5 .
卷调查法的根本性质 , 也影响到问卷调查法的应用范围 , 甚至还影响到它的具体操作方式和步 骤 。 问卷调查法在本质上是一种实证的方法 。 这是它在方法论上区别于其他某些方法的一项 重要标志 。作为一种实证的方法 , 间卷调查法既有着与众多自然科学方法相类似的逻辑程序 , 又有着与它们相类似的内容结构 。在社会研究中 , 它是一种从宏观的角度 、采取定量的手段 、依 据客观的验证来认识和说明社会现象的调查研究方式 。这种方式与传统的 、以个别访谈和实地 观察为主要特征 , 多从微观的角度 、采用定性的方法 、依靠主观洞察和分析来认识社会现象的 调查研究方法 , 有着明显的区别 。 问卷调查法的这种实证性质 , 决定了它与随机抽样 、与统计分析密不可分 。或者说 , 问卷调 查法实际上意味着抽样 、问卷与统计分析三者之间存在某种必然的 、内在的联系 。 抽样解决的 是调查对象的问题 , 问卷则是进行变量测量和资料收集的工具 , 而统计分析则是处理这种虽来 自于样本却要反映总体的 、以间卷形式收集的大量资料的必由之路 。作为一种完整的社会调查 方法 , 问卷调查必须同时包括这三者 。也可以说 , 它们共同体现着现代社会调查研究方法的基 本特征 。 正是在这个意义上 , 我们说 , 问卷调查法是且仅仅只是社会研究方法中的一种具体方法 , 一种有着明显的实证色彩 , 适于进行大规模的 、定量研究的调查方法 。 在现代社会中 , 由于社会发展与社会变迁速度的日益加快 , 社会不同部分之间的差异日益 扩大 , 社会总体的异质性程度日益增强 , 社会生活现象也越来越复杂 。 因此 , 传统的 、以对少数 个案的深入观察和访问为主要手段 , 以定性分析和主观洞察为主要分析方式的社会调查方法 , 已不能满足人们认识社会现象的需要 。社会现实对新的调查方法的出现提出了要求 , 问卷调查 法的产生和广泛应用正是适应了这种要求 。 同时 , 电子计算机技术的发展 , 社会统计指标的建 立 , 多元统计分析方法的完善 , 也都为问卷调查法的广泛运用提供了必要的客观条件 。 而问卷 调查法的产生和广泛运用 , 又为社会学研究( 以及社会科学研究) 从微观走向宏观 , 从定性走向 定量 , 从思辩走向实证 , 提供了一条可行的途径 , 也为人们探索社会现象的奥秘 、认识社会现象 的规律性提供了一种新的工具 。 私 二 、理论与问卷调查法 理论作为人们对外部世界的一种理性认识 , 在各种经验的社会研究中有着其独特的作用 。 从方法论角度看 ,抽象的理论与具体的方法不是彼此分离 , 而是紧密相连的 。“ 科学方法 ’ 不能 简单地被理解成只是对细节的过份注重和不辞劳苦地收集 、分析数据资料 , 而是包含着对理论 问题的关注和对通过理论工作来解释社会现象的强调 。 ’心现实生活中一些问卷调查所反映出 来的缺陷和不足 , 正是由于它们未能对理论与问卷调查的关系 、理论在问卷调查中的作用等等 给予充分注意的缘故 。 任何一种形式和规模的问卷调查 , 都离不开明确的理论框架或理性分析的指引 。 正是理论 及其由理论导出的各种假设 , 引导着问卷调查走向特定的事实 。而如果 “ 没有假设的指引 , 我们 冷 ( 卫 M a r t 一n Blu m e r : o S e 一 o l o g i e a l R e s a r e h M e t h o d s , ( s e e o n d E d一t l o n ) M a e m 一 l l a n P u b ll s h e r s l 一t d , 19 8 4 , P 2 .
辛 , 就不知道观察什么 , 寻找什么 , 也不知道做什么样的实验来发现日常生活中的秩序 。” ①目前存 在于许多问卷调查中的一个主要问题 , 就是由于缺少理论框架 、缺乏理论分析而形成的目的性 差和盲目性大 。这种情况最集中地体现在调查问卷的设计及调查结果的分析和解释两个方面 。 我们知道 , 问卷设计是整个间卷调查过程中举足轻重的一环 。 问卷设计的水平和质量 , 直 接影响到整个调查研究工作的最终成果 , 决定其质量的高低 。 如果脱离了理论框架的指导 , 间 卷设计工作往往是漫无边际 、没有中心地提问题 。 其结果既可能缺少某些必要的资料 , 也可能 收集到太多与研究目的无关的资料 。一旦缺少必要的资料 , 则分析工作就难以进行 , 正确 、全面 的结论就难以得到 ; 而当与研究目的无关的资料太多 , 则又会浪费大量的人力 、物力和时间 , 并 且给资料的整理和分析工作带来许多不必要的麻烦 , 有时甚至会使研究者陷于资料的海洋难 以自拔 。 在对问卷调查资料进行分析时 , 有无理论框架的指引 , 其结果也大不一样 。 当理论框架及 研究假设十分明确 , 变量和指标都十分清楚时 , 分析阶段的任务也将是十分明确和具体的 。验 证假设将是资料分析的中心任务 , 而对结果的深入解释和讨论则将依据假设的证实与证伪来 展来 。 因此 , 研究的最终结论的获得 , 相对来说是比较容易的 , 其逻辑性 、条理性也都是十分清 楚的 。但是 , 当缺乏理论框架时 , 研究者往往只能被动地从所得资料中去寻找 , 即只能靠通常所 说的 “ 事后分析 ” 。这种 “ 不管三七二十一 , 打到网中便是鱼 ” 的指导思想和做法 , 必然造成资料 分析阶段那种在杂乱无章 、良芳混杂的资料里 “ 大海捞针 ” 的现象出现 。这也正是目前一些低质 量问卷调查所走过的道路 。因此 , 无论我们的问卷调查是描述性的 , 还是解释性的 , 也无论我们 的调查内容如何 , 规模大小 , 都必须以充分的理论分析作基础 , 都必须有明确的理论框架作指 导 。 三 、问卷调查法的效度 娜 对于任何一种研究方法来说 , 效度都是其面临的一个重要问题 。而效度问题对于问卷调查 法来说 ,似乎显得更为突出一些 。 这首羌 是由于各种社会调查方法都具有效度较低 , 信度较高 (也即准确性较低 , 可靠性较高) 的特点 , 问卷调查法自然也不例外 ; 而更重要的是 , 间卷调查法 的特定工具及操作方式 , 又在较大的程度上加重了这种状况 。 或者说 , 问卷调查法更经常地处 于低效度的境地中 ,更经常地面临效度问题的挑战 。 一方面 , 问卷调查法在工具及程序上的高度标准化要求 , 常常使得原本很复杂的问题流于 表面化 , 即形式的要求导致了内容的肤浅 。 尤其是问卷这种特定形式的人工化痕迹十分明显 , 研究者常常难以在问卷的形式中 , 恰如其分 、不偏不倚 、且又深入细致 、周到全面地去探测和度 量活生生的社会现实 。削足适履 、主观臆断 、似是而非 、含糊不清的现象 , 在实际调查所用的问 卷中是屡见不鲜的 。 这是导致一些问卷调查效度不高的一个主要原因 。 另一方面 , 问卷调查法完全依靠问卷 、依靠被调查者的自我报告来收集资料 。因此 , 研究者 同被调查者及其所处的社会生活背景之间 , 存在着一定的距离 , 或者说 , 是被问卷隔离开了 。研 究者同被调查者及其生活背景的接触是间接的 , 研究者往往很难对导致被调查者产生某种思 想 , 行为的背景原因 , 获得直接的 、切身的感受 。这同样是造成一些问卷调查效度较低的一个重 ① P e t e r H .
M a n n : M e t h o d s o f o S e i a l L n v e s t ig a t i o n , a B s i l B la e k w e ll I n c , 1 98 5 , P 4 6 .
要原因 。 无论是纷繁复杂的人类行为 , 还是千差万别的社会生活现象 , 当研究者投入其中时所 获得的感受 , 与他们面对一叠问卷表或一堆统计数据时的感受 , 显然是大不相同的 。 正是由于上述两方面的缺陷 , 使得问卷调查法在效度的挑战面前 , 常常是显得有些软弱无 力 。这也是间卷调查法经常受到批评 、责难和疑间的原因所在 。 出路何在呢? 笔者认为 , 要尽 可能改变这种状况 , 要在理想与现实之间架起一座沟通的桥梁 , 就不能不对间卷调查中的操作 化工作给予高度的重视 。因为这是提高间卷调查法效度的关健所在 , 也是更好地发挥间卷调查 法作用的出路所在 。一些问卷调变结果中所出现的各种偏差 , 大都与操作化过程中的不足紧密 相关 ; 而问卷调查法在实际应用过程中的许多缺陷 , 也可以通过对操作化过程的进一步改善来 得到某种程度的弥补 。 操作化的核心是把抽象变为具体 。从问卷调查的实际过程来看 , 它可以分为研究课题的操 作化 、假设的操作化 、以及概念的操作茸 亡三 个层次 。研究课题的操作化所要解决的 , 是如何把一 个笼统的 、高度概括的研究课题 , 变成一项(一套) 明确具体的研究方案和研究计划 , 所谓研究 设计指的就是这种过程 ; 假设的操作化所要解决的 , 是如何把来自于理论的抽象命题( 研究假 设) , 转化成或分解成若干个可验证的具体假设(工作假设) ; 而概念的操作化所要解决的 , 则是 如何将抽象的概念转化成一组具体的 、可实际观测的指标 。在这三个层次中 , 概念的操作化对 间卷调查的影响最为直接 、也最为明显 , 它与间卷设计的工作更是紧密相连 。同时 , 概念操作化 过程中的种种失误 , 也是导致众多问卷调查受到责难的突出原因 。 因此 , 要提高问卷调查的效 度 , 首先必须提高概念操作化的水平和质量 。 而这一目标的实现 , 在很大程度上依赖于研究者 对所涉及概念的理解 , 以及对所涉及的社会生活现象的熟悉和了解程度 , 特别是从本质上对它 们认识和体验的程度 。 在这方面 , 研究者值得花大功夫 , 下大力气 。 四 、人的因素与问卷调查 人既作为研究主体 , 又作为研究客体的现实 , 构成了众多社会科学的一个共同的 、带有根 本性的特征 。 这种特征使得这些社会科学明显地与自然科学区别开来 。 在社会调查乃至在社 会研究的过程中 , 人的因素的影响已是众所周知的事实 。 而在问卷调查中 , 这种影响则显得更 加直接 、更加普遍 , 也显得更加突出 、更加复杂一些 。 首先 , 作为客体的调查对象在年龄 、性别 、职业 、文化程度等方面的特征 , 以及他们对问卷 调查的态度和认识 , 都制约和影响着问卷调查的适用性和可靠性 。 由于不同的人们有着不同的社会背景 、不同的生活方式 、不同的价值观念和不同的社会阅 历 , 因此 , 他们对于同一种事物往往会有着不同的反应 。 而人们对于间卷调查这一事物的认识 和了解 , 特别是作为被调查者的人们对问卷调查本身的本度和反应 , 可以说是决定问卷调查能 否成功的关健因素 。任何一项社会调查都离不开被调查者的合作与支持 。对于问卷调查来说 , 这种合作与支持更是不可缺少 。因此 , 如果被调查者顾虑重重 、过份担心 , 或者是漫不经心 、过 份忽视 , 都意味着问卷调查的失败 。 被调查者自身的各种特征 , 会使得一项问卷调查的难易程度 、完满程度很不一样 。 比如说 在青年人中作一项间卷调查 , 比起在老年人中作同样的调查 , 在客观上往往要容易一些 , 但在 主观上却往往要困难一些 , 这是由于青年人在总体上比老年人具有高一些的文化程度 ,对新事 物( 比如问卷调查这样的新事物) 的了解往往更多一些 , 因而他们接受问卷调查 、完成间卷调查 · 1 6 · 玲 角 , 的客观条件相对好一些 。但另一方面又由于他们所具有的特定生理 、心理特点 , 以及他们的处 世态度 、行为方式 , 又可能使得他们作为被调查者来说 , 主观上往往不如老年人那么认真 、那么 负责 。 同样的道理 , 在很少与笔 、与纸打交道的工人 、农民中做问卷调查 , 与在常年只与笔墨书本 打交道的学生 、教师及其他知识分子中做同样的调查 , 情形显然是大不一样的 。 两种情况下所 遇到的困难会不同 , 问卷设计的要求 、最合适的调查方式等等也会不同 。 .
至于在不同文化程度 的人群中作问卷调查的差别 , 自然是最明显不过的了 。无论是阅读能力 、理解能力 、还是表达能 力 , 无不受着文化程度的制约 。 而我国社会中男性公民总体文体程度高于女性公民 , 城市居民 文化程度高于农村居民 , 文教科卫及行政干部的文化程度高于工人和商业人员 , 后者又高于服 务业人员及农民的现实 , 理所当然地又使得问卷调查的适用范围在上述几个方面的前者与后 者之间 , 呈现出一种由大逐渐变小的趋势 。 其次 , 作为主体的研究者在问卷调查中所具有的角色特点 , 也在很大程度上决定和影响着 问卷调查的成败优劣 。这种角色特点的表现方面之一 , 是研究者在问卷调查中往往既要从实证 性 、客观性的要求出发 , 保持 “ 价值中立 ” , 尽可能地减少人为因素的干扰 , 减少研究过程中的主 观成分和影响 , 以达到如实反映社会现象本来面目的目的 ; 同时 , 他又要从被调查对象是待定 社会背景中的人这一现实出发 , 充分发挥自己作为研究者的主体能动作用 , 通过深入体验 、主 观洞察 , 以及 “ 投入理解 ” 等等 , 去贴近调查对象所生活的人文背景 , 去熟悉和了解他们的心理 状况 、思维方式 、生活习惯和社区意识 , 以便更好地用间卷去测量他们的行为和态度 , 去收集相 关的资料和信息 。面对这样一种既相互对立 、又相互补充的角色要求 , 研究者要力求保持平衡 、 摆正关系 , 确非易事 。 这种角色特点的另一个表现方面 , 就是在问卷设计的过程中 , 研究者往往既要从研究的目 的和需要出发 , 尽可能多 、尽可能全面 、尽可能详细地提出间题 , 包括提出一些很复杂的 、准以 记忆 、难以回答的问题 ; 同时 , 他又要从被调查者的角度出发 , 去考虑他们填答问卷是否方便 , 是否容易 , 是否有利于他们理解和接受等等 。简言之 , 就是既要为自己着想 , 又要为被调查者着 想 。这是研究者在问卷设计中必将面对的另一种相互矛盾 、相互冲突的处境 。而只有妥善地处 理好这两方面的关系 , 问卷调查工作才能收到预期的效果 。 五 、问卷调查法与调查内容 一种具体的研究方法 , 如同工具箱中一件特定的工具 , 总是为着解决某类特定的问题而设 计或建立的 。对于不同的研究内容 , 它的适用性也不一样 。而各种不同的研究方法 , 在了解 、认 识和研究各种社会现象以及人们的各种社会行为方面 , 也是各有其用处的 。问卷调查法作为众 多方法中的一种 , 它的适用范围 , 或者说它的用武之地 , 与所调查的内容十分相关 。即它的适用 范围受着具体调查内容的限制和影响 。对于不同的调查内容 , 它的作用也大不一样 。认识到这 一点 , 可以使我们在实际调查研究中少走弯路 , 同时也可以充分发挥问卷调查法在认识社会中 的作用 。 一般来说 , 经验社会学所研究的现象大体上可分成社会的状况 、人们的行为等客观事实 , 以及人们的心理 、意向 、态度 、认识 、情感等主观因素两大方面 。从理论上看 , 任何一种方法都可 以用来研究这两个方面 。但由于每一种具体方法都受着其特定的方法背景( 实证的或人文的) 、 .
1 7 · 争 ǔ , .
特定的逻辑程序( 归纳的或演绎的 、特定的操作方式( 结构的或无结构的等因素的约束和限 制 , 因而在具体的研究领域中它们的运用情况往往不同 , 对不同研究内容的适用性也往往不一 样 。 具体地说 , 问卷调查法由于其明显的实证方法论背景 、以假设演绎为主的逻辑程序 、结构 化 、标准化的操作方式 , 加上它与抽样和统汁分析之间的内在联系 、以及以个人作为主要分析 单位 、以精心设计的间卷作为收集资料的工具等众多特点 , 使得它特别适用于描述一个大总体 的状况 、性质和特征 。 正如美国社会学家艾尔 · 巴比( E a lr B a b t ) 所说 : “ 一个认真抽取的概 率样本 , 加上一个标准化的间卷 , 可以提供对某一学生群体 、某个城市 、某个国家 、或其他大总 体的精确描述 ” 。①同时 , 也特别适用于了解 、分析和研究社会生活中具有不同背景的人们的行 为 。然而 , 对于了解和认识人们行为的动机 、人们的思想和感情 、人们对现实生活的各种主观感 受 、人们的各种心理状态等内容 , 一句话 , 对于了解人们的内心世界 , 以及了解人们相互之间各 种复杂微妙的关系来说 , 问卷调查法就不如参与观察 、个案研究 、深度访谈等方法那么可行 , 那 么有效 。 举例来说 , 在婚姻家庭研究领域中 , 离婚现象是一个重要问题 。对于离婚现象的研究 , 问卷 调查法可以较好地获得有关某一地区离婚者的基本状况 , 比如离婚者的年龄分布 、职业分布 、 文化程度分布 、结婚时间的长短 、结识的方式 、子女的数目 、家庭的规模等等 。即可以清楚地描 述出与离婚者有关的各种外部特征 , 并从中归纳出若干特点 , 使人们对离婚者的状况有一个总 体的和综合的认识 。尽管在这种问卷调查中 , 研究者同样也可以通过问卷收集到诸如离婚的主 要原因 、离婚前后的思想 、感情 、认识 、态度和感受等主观资料 , 但由于前述的 、与问卷调查法有 关的几个方面的原因 , 它在这些涉及个人隐私 、秘密和情感 , 并且通常是被自尊 、害羞 、戒备 、自 卑 、疑虑等多种敏感的心理因素所包围和掩饰的领域中 , 常常难以有很大的作为 。 它在这方面 的收效还远不尽如人意 , 其所得资料往往显得过于粗略 、肤浅和牵强 。 尤其难以使研究者真正 把握各种不同的离婚者在情感 、人际关系 、心理状态等方面所具有的个性特点和细微差别 。 总之 , 任何一种研究方法 , 总有其特别适用的领域和研究内容 , 也总有其不大适用的领域 和研究内容 。 对于社会研究中常用的问卷调查法来说 , 认识到它究竟能干些什么 , 又不能干些 什么 ; 或者说 , 它适于干些什么 , 不适于干些什么 , 是每位研究者应该明确的问题 。 这同样是提 高问卷调查质量和效果的一个重要方面 。 下 责任编辑 : 张死丽 冷 ’ ① E a r l a B b b i e : T h e P r a e t ie e o f o S e i a l R e s e a r e h .
( F if t h E d it i o n ) W a d s w o r t h , In e . 1 9 8 9 , P 2 54 .
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[FREE] If you roll a fair six-sided die twice, what is the probability of getting two even numbers? A. - brainly.com
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If you roll a fair six-sided die twice, what is the probability of getting two even numbers?
A. 4 1
B. 9 1
C. 3 1
D. 6 1
1
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The probability of getting two even numbers when rolling a fair six-sided die twice is 1/4.
Explanation
When rolling a fair six-sided die twice, there are a total of 6 x 6 = 36 possible outcomes. To find the probability of getting two even numbers, we need to determine the number of favorable outcomes.
Let's consider the even numbers on a die: 2, 4, and 6. Out of the six possible outcomes, half of them are even numbers. So, the probability of rolling an even number on the first roll is 1/2, and the same goes for the second roll.
To find the probability of two independent events occurring together, we multiply their individual probabilities. Therefore, the probability of getting two even numbers is (1/2) x (1/2) = 1/4.
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The probability of rolling two even numbers with a six-sided die is 4 1. This is calculated by determining the probability of rolling an even number on each die and multiplying them together. The correct answer is A. 4 1.
Explanation
To find the probability of rolling two even numbers when rolling a fair six-sided die twice, we first need to identify the even numbers on a six-sided die. The even numbers are 2, 4, and 6. Therefore, there are 3 even numbers out of a total of 6 possible outcomes on a single die roll.
Calculate the probability of rolling an even number on the first die:
The probability (P) of rolling an even number is:
P(e v e n)=6 3=2 1
Calculate the probability of rolling an even number on the second die:
Since the rolls are independent, the probability remains the same.
P(e v e n)=6 3=2 1
Multiply the probabilities of both rolls:
P(tw o e v e n n u mb ers)=P(e v e n o n f i rs t)×P(e v e n o n seco n d)
P(tw o e v e n n u mb ers)=2 1×2 1=4 1
This means the probability of getting two even numbers when rolling a fair six-sided die twice is 4 1. Therefore, the correct answer is A. 4 1.
Examples & Evidence
For example, if you roll the die twice, the following combinations represent rolling two even numbers: (2,2), (2,4), (2,6), (4,2), (4,4), (4,6), (6,2), (6,4), and (6,6). Each of these combinations contributes to the total count of outcomes that meet our criterion of rolling two even numbers.
The reasoning behind the calculations is based on probability theory, where the individual outcomes of rolling a die are equally likely, thus allowing for straightforward probability calculations through multiplication of independent events.
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### Free Mathematics solutions and answers
Community Answer When rolling a six-sided die twice, what is the probability of getting a sum of 6? a. 1/36 b. 1/12 c. 1/6 d. 1/4
Community Answer If you were to roll a fair, six-sided die, what is the probability of getting a 3? (A) 3/6 (B) 1/3 (C) 5/6 (D) 1/6
Community Answer 4.8 30 If you roll a single six-sided die, what is the probability of rolling an odd number? 1/6 , 1/3 , 1/2 , 3
Community Answer Probability of rolling a hard eight (two 4’s) using two fair six sided dice? 1/6 2/6 or 1/3 2/12 1/36 Probability of rolling two 6’s using two fair sided dice? 1/6 2/6 or 1/3 2/12 1/36 Probability of getting a 1 or a 4 from a SINGLE ROLL of a fair six sided die? 1/6 ½ ¼ 2/6 or 1/3
Community Answer 5.0 5 A single six-sided die is rolled twice. What is the probability of rolling an even number on the first roll and an odd number on the second roll? A. 1/12 B. 1/6 C. 1/4 D. 1/36
Community Answer When rolling a die twice, what is the probability of rolling a 3 first, followed by rolling an even number? Option a) 1/6 Option b) 1/4 Option c) 1/12 Option d) 1/9
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Community Answer 11 What is the present value of a cash inflow of 1250 four years from now if the required rate of return is 8% (Rounded to 2 decimal places)?
Community Answer 13 Where can you find your state-specific Lottery information to sell Lottery tickets and redeem winning Lottery tickets? (Select all that apply.) 1. Barcode and Quick Reference Guide 2. Lottery Terminal Handbook 3. Lottery vending machine 4. OneWalmart using Handheld/BYOD
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Limits of Rauzy graphs of low-complexity words Blair Drummond August 7, 2019 University of Ottawa The Benjamini-Schramm limit Limits of finite graphs The aim of the thesis is to show the convergence properties of particular sequences of finite graphs. Before doing this, it is natural to ask what it really means to take a “limit” of finite graphs, and what problems one might encounter in defining this.
So to begin, we motivate the definition of the Benjamini-Schramm limit by highlighting the complication with graphs.
1 Limits of finite graphs Graph limits: attempt #1 Consider the following sequence of graphs: take Bn(Z2, 0) to be the ball of radius n around 0 in the integer lattice (with the graph metric). Imagine coloring one of these balls the red, and color a second such ball blue, and then connect the two at (0, 0) with a line of length n. Call these graphs Gn.
2 Limits of finite graphs (Graph limits: attempt #1 (cont.)) This sequence Gn illustrates a particular problem. As n →∞, there are two growing graphs of different colors, which are moving away from eachother. Naively, in the “limit” one has two copies of Z2, with different colors, infinitely far away from eachother, with some line “connecting” them.
This is obviously a problem — the limit may be an infinite graph, but no two vertices should be infinitely far apart. What is it that goes wrong?
3 Limits of finite graphs (Graph limits: attempt #1 (cont.)) The issue is that the space of graphs simply cannot itself be given any reasonable topology. We are forced to think about rooted graphs.
The limit depends on how we choose the roots!
But, if one assigns the root into the red ball, then the blue ball “disappears offto infinity”, and conversely if one assigns the roots into the blue ball, then the red ball “disappears offto infinity”.
4 Limits of finite graphs (Graph limits: attempt #1 (cont.)) The issue is that the space of graphs simply cannot itself be given any reasonable topology. We are forced to think about rooted graphs.
The limit depends on how we choose the roots!
But, if one assigns the root into the red ball, then the blue ball “disappears offto infinity”, and conversely if one assigns the roots into the blue ball, then the red ball “disappears offto infinity”.
So how do we resolve this?
4 Limits of finite graphs So how do we resolve this?
We resolve it the way any good analyst would...
5 Limits of finite graphs So how do we resolve this?
We resolve it the way any good analyst would...
with measures!
5 Limits of finite graphs So how do we resolve this?
We resolve it the way any good analyst would...
with measures!
This leads us to the Benjamini-Schramm limit.
5 Limits of finite graphs Definition: G• We define G• to be the space of (isomorphism classes of) rooted, connected and locally finite directed graphs, equipped with the projective topology. [BS11, Kai15].
6 Limits of finite graphs Definition: G• We define G• to be the space of (isomorphism classes of) rooted, connected and locally finite directed graphs, equipped with the projective topology. [BS11, Kai15].
This space is locally compact and metrizable. And the subsets of G• of graphs with bounded vertex degree form compact subsets.
6 Limits of finite graphs The Benjamini-Schramm limit With G• so defined, every finite graph G yields a probability measure on G• by taking a random rooting (G, o) of G.
µ = 1 |G| X v∈G δ[(G, o)] That is, we can take a normalized sum of point masses of (isomorphism classes of) the different rootings of G 7 Limits of finite graphs The Benjamini-Schramm limit With G• so defined, every finite graph G yields a probability measure on G• by taking a random rooting (G, o) of G.
µ = 1 |G| X v∈G δ[(G, o)] That is, we can take a normalized sum of point masses of (isomorphism classes of) the different rootings of G We can then say that a sequence of finite graphs Gn converge in the Benjamini-Schramm sense, if the µn converge (in the weak-∗ topology) in M(G•) — the space of Borel probability measures on G•.
7 Limits of finite graphs This resolves our earlier problems. Effectively, the Benjamini-Schramm limit captures all of the different conceivable subsequential limits of rootings of Gn, forming a distribution over the limit set of different rootings.
8 Limits of finite graphs Sidenote: ...what is a graph?
Q: Everyone has their own definition of a “graph”; what definition are we using?
A: For our purposes, a graph will be a pair of vertices and oriented edges, (V , E) with E ⊆V × V . This would sometimes be called a digraph with no multiple edges. We will also consider edge-labelled graphs, where there is a map ℓ: E →A. Benjamini-Schramm convergence remains the same, we just modify the meaning of ∼ =.
9 Limits of finite graphs This solves our problem. Now what about these “particular sequences of graphs”?
10 Rauzy graphs Rauzy graphs What about these “Rauzy graphs” We will start with Rauzy graphs, because these are more closely related to the origin of this project.
11 Rauzy graphs What about these “Rauzy graphs” We will start with Rauzy graphs, because these are more closely related to the origin of this project.
We will start by talking about subshifts.
11 Rauzy graphs Subshifts Given a finite alphabet A with |A| ≥2, we can equip all An with the product-of-discrete topology, and then we define the space of (singly) infinite words with the projective topology AN := lim ← −An this space is compact, metrizable, and totally disconnected; we call it the space of infinite words. It comes with a continuous self-map, S, the shift.
S : ω0ω1ω2 · · · 7→ω1ω2ω3 . . .
We call compact S-closed subsets X ⊆AN subshifts.
12 Rauzy graphs Languages As the infinite words are themselves (projective) limits of finite words, it should be unsurprising that the subshifts X are determined by the permitted finite subwords [LM95]. For any subshift X, we can define the language, L(X) ⊂A∗(the free monoid on A) to be the set of all finite length words which appear in some infinite word in X. In fact, it is usually easiest to define a particular subshift by defining the language.
13 Rauzy graphs Example: the golden mean shift We can define the golden mean shift this way, by defining X to be the collection of infinite words which do not contain the subword 11. So the language of X is then: L(X) = {ϵ, 0, 1, 00, 01, 10, 001, 010, 100, 101, 0000, . . . } We can also define Ln(X) = L(X) ∩An to be the subwords of length n. In the above example, L3(X) = {000, 001, 010, 100, 101} 14 Rauzy graphs Rauzy graph With these definitions, the nth Rauzy graph of a subshift is a graph where the vertices are the length n subwords, and there is an (oriented) edge (u, v) between two length n words if u precedes v in a word w of length n + 1. That is, Rn(X) = (Ln(X), E) where (u, v) in E ⇔ u z }| { w1 w2 . . . wnwn+1 | {z } v , w ∈Ln+1(ω) 15 Rauzy graphs Labelled Rauzy graphs We can also add labels to the edges, by “coloring” the edge with the newly added letter of A. For every edge (u, v), (u, v) in E ⇔ u z }| { w1 w2 . . . wnwn+1 | {z } v , w ∈Ln+1(ω) and ℓ((u, v)) = wn+1.
We will denote these edge-labelled Rauzy graphs by ⇀ Rn(Xω).
16 Rauzy graphs Rauzy graphs of the golden mean shift labels: blue-thick = 0, red-dashed = 1 0 1 00 01 10 000 001 010 100 101 First three Rauzy graphs of the golden mean shift.
17 Rauzy graphs The high-complexity case The golden mean shift, and the full shift {0, 1}N are both examples of shifts of finite type, where the shifts are determined by a finite number of forbidden symbols. For the golden mean shift, {11} is forbidden, and for the full shift, nothing (∅). These subshifts are typically of high-complexity, having exponential growth in |Ln(X)|.
The only exceptions are degenerate cases where |X| < ∞, like when X is a periodic shift. This happens if {00, 11} are forbidden, or if {1} is forbidden, for example.
18 Rauzy graphs The high-complexity case In the case of high-complexity shifts, the Benjamini-Schramm limit of the associated Rauzy graphs has already been studied.
It was shown, for instance, that the labelled Benjamini-Schramm limit of the Rauzy graphs of the full shift (these are known as the de Bruijn graphs) converge to Cay (L2, {→, flip →}) where L2 is the lamplighter group, Z ≀Z2. The unlabelled graph limit yields the famous Diestel-Leader graph, DL(2, 2).
See [Lee16, GLN16, Kai18], and also unpublished work by Kaimanovich, Leeman, and Nagnibeda.
19 The low-complexity case The low-complexity case The low-complexity case The low-complexity case, however, is somewhat different, and that is what this thesis addresses.
20 The low-complexity case Low-complexity word Given an infinite word ω ∈AN, there is a smallest subshift Xω ⊆AN containing ω, which is easily seen to be Xω = { Skω : k ∈N } that is, the closure of the orbit of ω is a subshift (recall that S is the shift). We can say that ω is of low-complexity if for some K lim sup n |Ln(Xω)| n < K 21 The low-complexity case Low-complexity word Given an infinite word ω ∈AN, there is a smallest subshift Xω ⊆AN containing ω, which is easily seen to be Xω = { Skω : k ∈N } that is, the closure of the orbit of ω is a subshift (recall that S is the shift). We can say that ω is of low-complexity if for some K lim sup n |Ln(Xω)| n < K That is, low-complexity words have linearly many subwords of a given length, rather than exponentially many!
21 The low-complexity case A result of Cassaigne & Frid, [Fri01, Cas96] Cassaigne showed that the linear growth of subword complexity is equivalent to bounded differences between |Ln+1(Xω)| and |Ln(Xω)|. That is, ω is of low-complexity if and only if there is a k such that ∀n. |Ln+1(Xω)| −|Ln(Xω)| < k Frid interpreted this in terms of Rauzy graphs, and pointed out that this means that in any Rauzy graph, there must be a bounded number of vertices with in-degree or out-degree greater than 1.
22 Our first result Unlabelled convergence Unlabelled convergence for aperiodic low-complexity words By the result of Cassaigne & Frid, there are a bounded number of special vertices in any Rn(Xω): A special vertex is just a vertex where the in-degree or out-degree are not equal to 1.
• ⋆ ⋆ ⋆ ⋆ (•) is regular; the (⋆)s are special.
23 Unlabelled convergence Theorem 1 If ω is a low-complexity aperiodic word, then the unlabelled Rauzy graphs Rn(Xω) converge to a point mass concentrated on the line graph in G•.
Proof sketch: If ω is aperiodic then |Ln(Xω)| →∞, and so the special vertices can be made to occupy an arbitrarily small part of the graphs.
Adding into this the bounded vertex degree of Rauzy graphs (the vertex degree is always less than 2|A|), and using that in any Rauzy graph Rn(Xω) that there are at most K special vertices for some K, one gets that there are at most K + (K|A| −1)r vertices within distance r of a special vertex.
24 Unlabelled convergence Proof sketch (cont.): Since |Ln(Xω)| →∞, this means that for any r ∈N, an r-neighbourhood of (uniformly) randomly chosen vertex in Rn(Xω) will not include any special vertex with probability at least 1 −K + (K|A| −1)r |Ln(Xω)| →1 and since the (connected) r-neighborhood contains no special vertices, it must be a line1. Since every neighborhood randomly converges to a line, and since G• is equipped with the projective topology, the Benjamini-Schramm limit is shown to be the point mass of the line graph.
1It cannot be a cycle, or else Xω would be finite.
25 Unlabelled convergence Tidying up: the finite cases If ω is not aperiodic, then it is either eventually periodic or periodic. In these two cases, the Rauzy graphs both stabilize to either a finite cycle (the periodic case) or else a graph that resembles: The cycle has only one symmetry class, so the Benjamini-Schramm limit is just a point pass on the finite cycle. In the eventually periodic case, the above graph is rigid, so the limit is a uniform measure on all distinct rootings of the graph above (the length of the “handle” and the size of the “loop” may be different).
26 Unlabelled convergence This classifies the unlabelled limits. What about limits of labelled graphs?
27 The labelled case The labelled case Sequences of edges encode finite subwords For labelled Rauzy graphs, the labels of subsequent edges encode a sequence of letters wn+1wn+2 . . . wn+k, which are themselves a word in L(Xω). Since (by theorem 1) typical neighbourhoods are lines, this means that when we randomly sample a neighbourhood in the Rauzy graphs, we are really sampling finite words from ω.
abbab bbaba babaa abaab baabb aabba abbaa a a b b a a aabba abbaa 28 The labelled case This leads us to ask, how do we make sense of the “probability” or “frequency” of a subword in ω?
29 The labelled case Uniform frequencies It turns out that a finite word u only has well defined frequencies if the following limit exists uniformly in k 2: frequ(Skω) := lim n 1 n + 1#{ occurrences of u in ωkωk+1 . . . ωk+n } = ...
= lim n 1 n + 1 n X i=0 Iu Si+kω 2So require that for some c that limn supk 1 n+1 Pn i=0 Iu Si+kω −c = 0 30 The labelled case That is, it can be viewed as an ergodic average. Moreover, if the frequency of every subword is defined (we need it to be.), then since the Iu functions generate a dense subalgebra3 of C(AN, R), Oxtoby’s uniform ergodic theorem ([Oxt52]) gives us that (Xω, S) is uniquely ergodic, and we get that for the unique S-invariant measure µ, frequ(ω) = µ(u) 3The span of {Iu}u∈A∗separate points and contain the constant functions.
31 The labelled case The case of words with uniform frequencies In the case of words ω with uniform frequencies, we can identify the Benjamini-Schramm limit using the measure µ on Xω. We start with the aperiodic case.
Theorem 2 If ω is a low-complexity aperiodic word with uniform frequencies, then the labelled Rauzy graphs ⇀ Rn(Xω) converge to µ—viewing µ as a distribution on A-configurations of the (bi-infinite) line graph in ⇀ G•.
32 The labelled case Proof sketch: 1 ⃝If ω is aperiodic, then following the proof of Theorem 1 we get that for any fixed φ > 0 and size k, for sufficiently large n, a random neighbourhood of diameter k in ⇀ Rn(Xω) resembles a line with probability at least 1 −φ.
2 ⃝By the definition of uniform frequencies, for any u and error ϵ > 0, for sufficiently large diameter k, the frequency of u in ωiωi+1 . . . ωi+k is within ϵ of the true frequency of u, µ(u).
Combining 1 ⃝and 2 ⃝, we can get that at least a 1 −φ proportion of random k-diameter neighbourhoods are lines, and the frequency of u within these lines can be made within ϵ of µ(u). So the frequency of u in a large Rauzy graph can be bound between (1 −φ)(1 −ϵ)µ(u) and (1 + ϵ)µ(u).
33 The labelled case The finite case In the case of periodic or eventually periodic ω (which automatically have uniform frequencies), we can also identify the Benjamini-Schramm subsequential limits, however the limit only exists in the periodic (= minimal) case; in the eventually periodic case, there are p subsequential limits where Skω is periodic with period p.
34 The labelled case This happens because the labels on the “handle” (pictured below) cycle with n in ⇀ Rn(Xω), but all the graphs are rigid.
cca ccc bca abc cab a b c a ⃝ b ⃝ ccab ccca bcab abca cabc b c a b ⃝ c ⃝ ω = ccc(abc)∞ In the periodic case, the (unlabelled) graphs only have one symmetry class, so this doesn’t occur, and all Rauzy graphs (for n > p) are isomorphic. In this case, the measure is just determined by the chosen (from {1, . . . , p}) vertex.
35 The labelled case Remark All of the results of the earlier theorems also apply to bi-infinite words ω ∈AZ. The proofs are unaffected—one simply has to appropriately modify a few definitions.
36 The non-uniquely-ergodic case The non-uniquely-ergodic case The general situation is a mess.
By Theorem 1 for unlabelled Rauzy graphs, we know that any subsequential Benajmini-Schramm limit of ⇀ Rn(Xω) n is supported on what is basically a set of A-configurations of the bi-infinite line graph. The difference is that now, the space of S-invariant measures on Xω is not a singleton, {µ}, but some sort of (Choquet) simplex M(Xω, S).
[FM10] is a good reference for this theory for low-complexity words.
37 The non-uniquely-ergodic case A recent result of Cyr and Kra showed that low-complexity shifts have finitely many ergodic measures [CK19], generalizing older results of Boshernitzan which apply only to minimal shifts [Bos85].
Since we then know that E(Xω, S) ⊂M(Xω, S), the set of ergodic measures, is finite, we can show that (Prop 6.4.2) E(Xω) = [ Y ⊂Xω, Y minimal E(Y , S) But when there are two minimal subsystems, the behaviour can be complicated.
38 The non-uniquely-ergodic case Example without a limit Take t = 01101001 . . . to be the Thue-Morse word. Where σ : {0, 1}∗ is the substitution map σ : ( 0 7→01 1 7→10 we have that t = limn σn(0). Now, define ω = t × (ab)∞ 39 The non-uniquely-ergodic case Example without a limit (cont.) With ω = t × (ab)∞defined this way, the Rauzy graphs have the following structure ⇀ Rn(Xt) | {z } 3n≤size≤10 3 n tn−2 . . . 110a tn−3 . . . 10ab . . .
10aba . . .
0abab . . .
⇀ Rn(X(ab)∞) | {z } size = 2 size = n −1, labels from {a, b} With ⇀ Rn(Xt) n−1 oscillating between 3 and 10 3 , attaining both values as limit points.
40 The non-uniquely-ergodic case Example without a limit (cont.) One gets that, because the size of ⇀ Rn(Xt) oscillates in this way, the Benjamini-Schramm limit oscillates along with it. Where µ is the (unique) ergodic measure on Xt, and ν the uniform measure on X(ab)∞, the Benjamini-Schramm limit attains 3µ + ν 4 and 10/3µ + ν 13/3 as subsequential limits.
41 The non-uniquely-ergodic case Example without a limit (cont.) One gets that, because the size of ⇀ Rn(Xt) oscillates in this way, the Benjamini-Schramm limit oscillates along with it. Where µ is the (unique) ergodic measure on Xt, and ν the uniform measure on X(ab)∞, the Benjamini-Schramm limit attains 3µ + ν 4 and 10/3µ + ν 13/3 as subsequential limits.
And in fact, all possible subsequential limits are a convex combination of these two measures.
41 The non-uniquely-ergodic case Arrow reversal Now, consider the related word, ω′ = (ab)∞× t. Then the Rauzy graphs look like ⇀ Rn(X(ab)∞) | {z } size = 2 abab . . . 0 aba . . . 01 . . .
ab01 . . . tn−3 a011 . . . tn−2 ⇀ Rn(Xt) | {z } 3n≤size≤10 3 n size = n −1, labels from {0, 1} Here, by contrast, {a, b} only appear on two edges. The Benjamini-Schramm limit actually gives µ, the ergodic measure on (Xt, S)!
42 The non-uniquely-ergodic case Conclusion While in the non-uniquely-ergodic case the Benjamini-Schramm limit set can be viewed as a subset of M(Xω, S), the limit set itself does not necessarily contain any ergodic measures (though it can), and it may or may not be a singleton (so the limit may or may not exist).
Future work The unsatisfying loose end of this is the case of the minimal non-uniquely ergodic case. The constructions provided used non-minimality in order to use growth of distinct sub-Rauzy graphs to compute subsequential limits. For minimal words, where we cannot do this, it is less obvious (to me, at least) what can happen!
43 References i [Bos85] Michael Boshernitzan.
A unique ergodicity of minimal symbolic flows with linear block growth.
J. Analyse Math., 44:77–96, 1984/85.
[BS11] Itai Benjamini and Oded Schramm.
Recurrence of distributional limits of finite planar graphs [mr1873300].
In Selected works of Oded Schramm. Volume 1, 2, Sel.
Works Probab. Stat., pages 533–545. Springer, New York, 2011.
44 References ii [Cas96] Julien Cassaigne.
Special factors of sequences with linear subword complexity.
In Developments in language theory, II (Magdeburg, 1995), pages 25–34. World Sci. Publ., River Edge, NJ, 1996.
[CK19] Van Cyr and Bryna Kra.
Counting generic measures for a subshift of linear growth.
J. Eur. Math. Soc. (JEMS), 21(2):355–380, 2019.
45 References iii [FM10] S´ ebastien Ferenczi and Thierry Monteil.
Infinite words with uniform frequencies, and invariant measures.
In Combinatorics, automata and number theory, volume 135 of Encyclopedia Math. Appl., pages 373–409.
Cambridge Univ. Press, Cambridge, 2010.
[Fri01] A. E. Frid.
On factor graphs of DOL words [ MR1760731 (2001d:05156)].
Discrete Appl. Math., 114(1-3):121–130, 2001.
Discrete analysis and operations research.
46 References iv [GLN16] R. Grigorchuk, P.-H. Leemann, and T. Nagnibeda.
Lamplighter groups, de brujin graphs, spider-web graphs and their spectra, May 2016.
Preprint.
[Kai15] V. A. Kaimanovich.
Invariance, quasi-invariance and unimodularity for random graphs.
Zap. Nauchn. Sem. S.-Peterburg. Otdel. Mat. Inst.
Steklov. (POMI), 441(Veroyatnost’i Statistika.
22):210–238, 2015.
47 References v [Kai18] Vadim A. Kaimanovich.
Circular slider graphs: de Bruijn, Kautz, Rauzy, lamplighters and spiders.
In Unimodularity in randomly generated graphs, volume 719 of Contemp. Math., pages 129–154. Amer. Math.
Soc., Providence, RI, 2018.
[Lee16] Paul-Henry Leemann.
On subgroups and Schreier graphs of finitely generated groups.
PhD thesis, University of Geneva, 2016.
48 References vi [LM95] Douglas Lind and Brian Marcus.
An introduction to symbolic dynamics and coding.
Cambridge: Cambridge University Press, 1995.
[Oxt52] John C. Oxtoby.
Ergodic sets.
Bull. Amer. Math. Soc., 58:116–136, 1952.
49 Thank you!
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Cerclage for women with twin pregnancies: a systematic review and metaanalysis - PubMed
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Meta-Analysis
Am J Obstet Gynecol
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. 2019 Jun;220(6):543-557.e1.
doi: 10.1016/j.ajog.2018.11.1105. Epub 2018 Dec 7.
Cerclage for women with twin pregnancies: a systematic review and metaanalysis
Chunbo Li1,Jie Shen1,Keqin Hua2
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1 Department of Obstetrics and Gynecology, The Obstetrics & Gynecology Hospital of Fudan University, Shanghai, China.
2 Department of Obstetrics and Gynecology, The Obstetrics & Gynecology Hospital of Fudan University, Shanghai, China. Electronic address: huakeqinjiaoshou@163.com.
PMID: 30527942
DOI: 10.1016/j.ajog.2018.11.1105
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Meta-Analysis
Cerclage for women with twin pregnancies: a systematic review and metaanalysis
Chunbo Li et al. Am J Obstet Gynecol.2019 Jun.
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. 2019 Jun;220(6):543-557.e1.
doi: 10.1016/j.ajog.2018.11.1105. Epub 2018 Dec 7.
Authors
Chunbo Li1,Jie Shen1,Keqin Hua2
Affiliations
1 Department of Obstetrics and Gynecology, The Obstetrics & Gynecology Hospital of Fudan University, Shanghai, China.
2 Department of Obstetrics and Gynecology, The Obstetrics & Gynecology Hospital of Fudan University, Shanghai, China. Electronic address: huakeqinjiaoshou@163.com.
PMID: 30527942
DOI: 10.1016/j.ajog.2018.11.1105
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Abstract
Objective data: This study was conducted to estimate whether cerclage could extend the prolongation of pregnancy, reduce the risk of preterm birth, and improve perinatal outcomes in women with twin pregnancies.
Study eligibility criteria: We included randomized controlled trials and cohort studies comparing the efficacy of cerclage with no cerclage for women with twin pregnancies.
Study appraisal and synthesis methods: The following databases were searched for all published studies that compared cerclage placement with expectant management in twin pregnancies from inception to July 2018: Medline, EMBASE, Scopus, ClinicalTrials.gov, Web of Science, and Cochrane Library. Each report was reviewed for inclusion or exclusion standard, and data extraction was performed by 2 authors independently.
Results: A total of 16 studies with 1211 women that met the inclusion criteria were included in the final analysis. Our outcomes indicated that cerclage placement for twin pregnancies with a cervical length of <15 mm was associated with significant prolongation of pregnancy by a mean difference of 3.89 weeks of gestation (95% confidence interval, 2.19-5.59; P=.000; I 2=0%) and a reduction of preterm birth at <37 weeks of gestation (risk ratio, 0.86; 95% confidence interval, 0.74-0.99; P=.040; I 2=0%), <34 weeks of gestation (risk ratio, 0.57; 95% confidence interval, 0.43-0.75; P=.000; I 2=0%) and <32 weeks of gestation (risk ratio, 0.61; 95% confidence interval, 0.41-0.90; P=.010; I 2=0%), compared with those pregnancies in the control group. For women with a dilated cervix of >10 mm, cerclage placement was associated with significant prolongation of pregnancy by a mean difference of 6.78 weeks of gestation (95% confidence interval, 5.32-8.24; P=.000; I 2=0%); a reduction of preterm birth at <34 weeks of gestation (risk ratio, 0.56; 95% confidence interval, 0.45-0.69; P=.000; I 2=28%), <32 weeks of gestation (risk ratio, 0.50; 95% confidence interval, 0.38-0.65; P=.000; I 2=14%), <28 weeks of gestation (risk ratio, 0.41; 95% confidence interval, 0.20-0.85; P=.030; I 2=80%), and <24 weeks of gestation (risk ratio, 0.35; 95% confidence interval, 0.18-0.67; P=.001; I 2=24%), and improvement of perinatal outcomes compared with those in the control group. However, for twin pregnancies with a normal cervical length (eg, cerclage for an indication for women with a history of preterm birth or twin alone), the efficacy of cerclage placement was less certain because of the limited data.
Conclusion: Our metaanalysis indicates that cerclage placement is beneficial for the reduction of preterm birth and the prolongation of pregnancy in twin pregnancies with a cervical length of <15 mm or dilated cervix of >10 mm. However, the benefit of history-indicated or twin alone-indicated cerclage is less certain in twin pregnancies with normal cervical length according to current literature. Further high-quality studies were needed to confirm the findings.
Keywords: cervical cerclage; metaanalysis; preterm birth; twin pregnancy.
Copyright © 2018 Elsevier Inc. All rights reserved.
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Comment in
The placement of a cerclage in patients with twin pregnancies and a short cervix is associated with increased risk of preterm birth and adverse perinatal outcome.Sanchez-Ramos L.Sanchez-Ramos L.Am J Obstet Gynecol. 2020 Feb;222(2):194-196. doi: 10.1016/j.ajog.2019.09.038. Epub 2019 Oct 3.Am J Obstet Gynecol. 2020.PMID: 31586601 No abstract available.
Cerclage in twin pregnancies with short cervical length: more level 1 data are needed.Roman A, Berghella V.Roman A, et al.Am J Obstet Gynecol. 2020 Jun;222(6):637-638. doi: 10.1016/j.ajog.2020.01.048. Epub 2020 Feb 1.Am J Obstet Gynecol. 2020.PMID: 32014505 No abstract available.
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https://stackoverflow.com/questions/121847/where-can-i-learn-about-logarithms
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Where can I learn about logarithms? [closed]
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I hear logarithms mentioned quite a lot in the programming context. They seem to be the solution to many problems and yet I can't seem to find a real-world way of making use of them. I've read the Wikipedia entry and that, quite frankly, leaves me none the wiser.
So, where can I learn about the real-world programming problems that logarithms solve? Has anyone got any examples of problems they faced that were solved by implementing a logarithm?
math
logarithm
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edited Sep 23, 2008 at 17:48
tslocum
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asked Sep 23, 2008 at 15:44
Charles RoperCharles Roper
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I'm voting to close this question as off-topic because it is not about programming.
High Performance Mark
– High Performance Mark
2016-03-03 23:55:53 +00:00
Commented Mar 3, 2016 at 23:55
Can you explain why you think it isn't about programming? Note the sentence in bold in the second paragraph.
Charles Roper
– Charles Roper
2016-03-04 01:15:25 +00:00
Commented Mar 4, 2016 at 1:15
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19 Answers 19
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13
Logarithms in programming are also frequently used in describing the efficiency of an algorithm using Big O notation.
For instance, a binary search algorithm would have a worst case scenario of O(log(n)) (on a sorted set), whereas a linear search's worst case is O(n)
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answered Sep 23, 2008 at 15:52
Chris Marasti-GeorgChris Marasti-Georg
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Ben Hoffstein
Ben Hoffstein
Yes of course, this is a great example. Any time you have something being split in half after each iteration, think Log!
12
Say you've got $1000, and it's in a savings account with 2.4% interest.
How many years do you have to wait until you have $2000 to buy a new laptop?
1000 × 1.024x = 2000
1.024x = 2
x = log 1.024 2 = 29.23 years
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edited Feb 4, 2009 at 20:05
strager
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answered Sep 23, 2008 at 15:52
MendeltMendelt
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Chris Marasti-Georg
Chris Marasti-Georg
I think ^ is clearer than to denote an exponent... to me at least
Charles Roper
Charles Roper
This is very interesting and exactly what I meant by "real world". In trying to grok this answer I found the following very useful: math.about.com/od/formulas/a/compound.htm I ended up with the following to get back to 2000: 1000(1 + 0.024)29.2263362061568
Mendelt
Mendelt
Fixed the typo's, thanks for the feedback. @Charles Roper you're getting it now. In other words log's tell you how many times you have to multiply a number by itself to get to another number.
Charles Roper
Charles Roper
You're using inconsistent exponent operators now... :)
Mendelt
Mendelt
@Charles Roper: Damn, I shouldn't edit my posts late at night when drinking wine :-)
|
10
In my own research I came upon a few useful resources:
Khan Academy logarithms section
This is a terrific set of lessons on logarithms. This comment from a 6th grader sums it up nicely:
Thank you so much. This week, my math teacher told me to challenge myself, so I tried logarithms. At first I was like, 'I can't do this, it's too hard'. Then I watched the video, and now they're even fun! I'm in 6th grade, my math teacher is impressed. I can't thank you enough.
Ruby Quiz #105: Tournament Matchups
This article contains a good example of using a base 2 log to determine the number of rounds required to complete a knock-out tournament given x teams.
An Intuitive Guide To Exponential Functions & E
An excellent, intuitive (as you'd expect, given the title), guide to e, the base of the natural logarithm. Lots of illustrations and examples make this a gem of an article.
Demystifying the Natural Logarithm (ln)
This is the followup to the article to one about e and discusses the natural logarithm (ln) which, using the intuitive explanation given in the article, "gives you the time needed to reach a certain level of growth".
There's actually loads of good content on the Better Explained site. Truly, a splendid resource.
Another tool that I had actually come across before but since completely forgotten about is Instacalc. It seems to be by the same person - Kalid Azad - who authors the Better Explained site. It's a really useful tool when hacking about with maths.
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edited Feb 3, 2015 at 14:05
answered Sep 24, 2008 at 17:52
Charles RoperCharles Roper
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Yatharth Agarwal
Yatharth Agarwal
+1 for the Kalid Azad mention!
5
Logs are a type of meta-arithmetic. It's a way of thinking of every number as a (possibly fixed) base raised to an exponent. Operations are performed on the exponents alone. This means that you can do multiplication and division by doing addition and subtraction of the logs. In other words, you put your data into log space, perform a suite of arithmetic, and pull it back into non-log space. If the floating point precision loss and the overhead of transforming in or out of log space is cheap, then you may have an overall win in time.
One slick trick you can do with logs is calculate out the number of characters a number will take when printed by taking the log-base-2 of the number and divide it by the log-base-10(2), which is constant time compared with a set of multiplies.
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edited Sep 19, 2013 at 13:00
answered Sep 23, 2008 at 15:59
plinthplinth
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I've seen logarithms used for displaying tag clouds. This is a page that explains it:
Tag Cloud Font Distribution Algorithm
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answered Sep 23, 2008 at 15:52
brock.holumbrock.holum
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I assume you have heard about logarithms with contexts to time consumation.
A concrete example would be search algorithms. Given a set of ordered data (think a sorted array of int's), you want to find the index key to a value in that data. We can benefit from the fact that the array is sorted (1, 2, 6, 192, 404, 9595, 50000 for example). Let's say we we want to find the index to the value 2. We can minimize our search space by culling (ignoring) half the array each step. We start this search by testing the value at the middle of the array. There are 7 values in the array, we then make the index 7/2 = 3.5 = 3 as int. array is 192. The value we are looking for is 2, therefore we want to continue the search in the lower half of the search space. We completely ignore index 4, 5, 6 since they are all higher than 192, and in turn also higher then 2. Now we have a search space that looks like (1, 2, 6). We then index into middle again (repeat process), and we find the 2 instantly. The search is complete, the index to 2 is 1.
This is a very small example, but it shows how such an algorithm works.
For 16 values, you need to search at maximum 4 times. For 32 values, you search max 5 times, 64 values 6 times and so on.. 1048576 values are searched in 20 steps. This is far quicker than having to compare each item in the array separately. Of course, this only works for sorted collections of data.
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answered Sep 23, 2008 at 16:05
StatementStatement
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I recommend e: The Story of a Number for a good foundation of the importance of logarithms, their discovery and relevance to natural phenomena.
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answered Sep 23, 2008 at 16:16
Cade RouxCade Roux
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Another way of looking at it is looking at the number of base multipliers in a number. I am sure you can see how this all relates in the following examples.
Decimal (base 10):
log10 (1) = 0 , (10^0) = 1
log10 (10) = 1 , (10^1) = 10
log10 (100) = 2 , (10^2) = 100
log10 (1000) = 3 , (10^3) = 1000
log10 (10000) = 4 , (10^4) = 10000
log10 (100000) = 5 , (10^5) = 100000
Binary (base 2):
log2 (1) = 0 , (2^0) = 1
log2 (2) = 1 , (2^1) = 2
log2 (4) = 2 , (2^2) = 4
log2 (8) = 3 , (2^3) = 8
log2 (16) = 4 , (2^4) = 16
log2 (32) = 5 , (2^5) = 32
log2 (64) = 6 , (2^6) = 64
log2 (128) = 7 , (2^7) = 128
Hexadecimal (base 16):
log16 (1) = 0 , (16^0) = 1
log16 (16) = 1 , (16^1) = 16
log16 (256) = 2 , (16^2) = 256
log16 (4096) = 3 , (16^3) = 4096
log16 (65536) = 4 , (16^4) = 65536
If you want to think in variables:
log N (X) = Y
(N^Y) = X
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answered Sep 23, 2008 at 16:18
StatementStatement
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Many (many!) relationships in the real world are logarithmic. For instance, it would not surprise me if the distribution of reputation scores on Stack Overflow is log normal. The vast majority of users will have reputation scores of 1 and a handful of people will have unattainably high reputation. If you apply a logarithmic transformation to that distribution, it would likely be nearly a linear relation. A quick scan of shows this to be true.
You might be more familiar with The Long Tail concept, which is an application of the logarithmic distribution.
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edited May 23, 2017 at 12:02
CommunityBot
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answered Sep 23, 2008 at 16:27
Jon EricsonJon Ericson
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The only problem I can recall is having to calculate the product of a column in SQL. SQL Server does not have a PRODUCT() aggregate function, so this was accomplished using a sum of the logarithms (using the LOG10() function) of each value. The main drawback was that all numbers in the column had to be positive and non-zero (you cannot calculate a logarithm on a negative number or zero).
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answered Sep 23, 2008 at 15:49
Ben HoffsteinBen Hoffstein
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nlucaroni
nlucaroni
Definitely a great transformation you mention here that can be expanded past just SQL. When products are really small you can often take the sum of the log.
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The most obvious usage in every programming example is precision. Put simply, consider storing unsigned integers. How many bits do you need to store X? Well, the maximum value you can store in n bits is 2^n - 1, so you can need log_2 X + 1 bits to store X. Now you can pick short, int, word, long etc with ease.
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answered Sep 23, 2008 at 15:53
Adam WrightAdam Wright
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One example, out of many : Calculating compound interests at a very small rate with a large number of periods.
You can do it the most straightforward way, even using fast exponentiation, but accuracy may suffer, due to the way floats are stored and calculating s r power n still takes O(ln(n)) operations.
With logarithms, it's somewhat more accurate.
A = ln( s r power n ) = ln(s) + n ln(r)Two lookups in your logarithm database gives you ln(s) and ln(r), with ln(r) begin very small, and floats work at their best accuracy near 0 result = exp(A), a reverse lookup here.
It's also the only really efficient way if you work with non-integer exponents, to extract cubic roots for example.
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answered Sep 23, 2008 at 16:08
community wiki
Johan Buret
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Check out MIT's Open Courseware: Introduction to Algorithms. Free educations. Awesome.
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answered Sep 23, 2008 at 16:11
ZachZach
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Charles Roper
Charles Roper
Looks very interesting indeed, but does it cover logarithms?
2
One of the more "cool" applications of logarithms I've found is Spiral Storage. It's a hash table that allows you to split one bucket at a time as the table grows, relocating less than half of the records in that bucket to the same, new bucket. Unlike linear hashing, where the performance varies cyclically and all of the buckets tend to be split at around the same time, spiral hashing allows nice, smooth growth of the table.
It was published about 30 years ago by G. N. N. Martin, about whom I haven't been able to learn much besides the fact that he also invented Range Encoding. Seems like a smart guy! I haven't been able to get a copy of his original paper, but Per-Ã
ke Larson's paper "Dynamic hash tables" has a very clear description.
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answered Sep 23, 2008 at 16:41
ericksonerickson
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Logarithms are used quite often in charts and graphs, when one or both axes cover a large range of values.
Some natural phenomena are best expressed on a logarithmic scale; some examples are sound pressure levels (SPL in dB) and earthquake magnitude (Richter scale).
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answered Sep 23, 2008 at 16:06
Mark RansomMark Ransom
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As an example of what Chris is talking about, an algorithm that changes complexity based on the number of bits in a value is (probably) going to have an efficiency described by O(log(n)).
Another everyday example of exponents (and hence logarithms) is in the format of IEEE floating point numbers.
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edited May 23, 2017 at 10:29
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answered Sep 24, 2008 at 0:35
Andrew EdgecombeAndrew Edgecombe
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A logarithmic function is simply the inverse of an exponential function, in the same sense that subtraction is the inverse of addition. Just as this equation:
a = b + c
states the same fact as this equation:
a - c = b
this equation:
b p = x
(where is raising to a power) states the same fact as this equation:
log [base b] (x) = p
Although b can be any number (e.g. log [base 10] (10,000) = 4) the "natural" base for Mathematics is e (2.718281828...) about which see here.
"Common" logarithms, used more in engineering, use a base of 10. A quick-and-dirty (emphasis on dirty) interpretation of the common (base 10) logarithm of some number x is that it is one less than the number of decimal digits required to express numbers the size of x.
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answered Sep 24, 2008 at 2:14
joel.neelyjoel.neely
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Demystifying the Natural Logarithm (ln) at BetterExplained is the best i have found. It clears the concepts from the base and help you understand the underlying concepts. After that everything seems a cakewalk.
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answered Jan 17, 2009 at 13:12
NavneetNavneet
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Here are some sites that I have used:
I have used logarithms for calculating the yearly appreciation on a house to determine whether the seller was being fair.
House Appreciation Equations
Here is the basic equation:
Previous price = p
New price = n
Appreciation rate = r
Years of appreciation = y
p (1 + r)^y = n
So, if the price 6 years ago was $191,000 (by checking your couty auditor's site) and the asking price is $284,000, what is the appreciation rate (which would not take into account any one-time improvement costs)?
191,000 (1 + r)^6 = 284,000 (1 + r)^6 = 284,000 / 191,000 = 1.486 Using a property of exponents and logarithms¦ 6 ( log (1 + r) ) = log 1.486 log (1 + r) = (log 1.486) / 6 = 0.02866 Using another property of exponents and logarithms¦ 10 0.02866 = 1 + r 1.068 = 1 + r r = 1.068 1 = 0.068 = 6.8% (kind of high!)
To determine what a reasonable price would be¦use 4% and allow for whatever improvements they made (which should be listed on the web id they were major¦but it wouldnt include bathroom/kitchen remodeling, etc.)
191,000 (1 + 0.04)^6 = n n = 241,675 + reasonable cost of improvement which of course will depreciate over time and should not represent 100% of the cost of the improvement
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edited Feb 4, 2009 at 19:47
answered Jan 29, 2009 at 15:24
a bellea belle
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https://www.cuemath.com/ncert-solutions/ncert-solutions-class-10-maths-chapter-14-statistics/
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NCERT Solutions Class 10 Maths Chapter 14 Statistics
NCERT Solutions for Class 10 Maths Chapter 14 Statistics deals with the classification of ungrouped as well as grouped frequency distributions. In the previous classes, the students must have learned how to represent the data pictorially in various graphical formats such as bar graphs, histograms, and frequency polygons. They must be familiar with topics like numerical representatives of ungrouped data and measures of central tendencies such as mean, median, and mode. In the NCERT Solutions Class 10 Maths Chapter 14, learning will switch to all the three measures such as mean, median, mode from ungrouped data to the grouped data.
The students will understand some new topics of statistics such as cumulative frequency, cumulative frequency distribution, cumulative frequency curves or ‘ogives,’ etc. A detailed pdf file of class 10 maths NCERT solutions Chapter 14 Statistics can be found below and also you can find some of these in the exercises given below.
NCERT Solutions Class 10 Maths Chapter 14 Ex 14.1
NCERT Solutions Class 10 Maths Chapter 14 Ex 14.2
NCERT Solutions Class 10 Maths Chapter 14 Ex 14.3
NCERT Solutions Class 10 Maths Chapter 14 Ex 14.4
NCERT Solutions for Class 10 Maths Chapter 14 PDF
NCERT Solutions Class 10 Maths Chapter 14 Statistics enables the students to understand the concept of statistics by outlining its key points. This chapter also helps in the revision of concepts such as mean, median, mode. Apart from this, the students will be able to explore some new topics with the help of this chapter, such as measures of central tendency in the ungrouped data. The following links mentioned below will help understand the sections in this chapter further in detail :
☛ Download Class 10 Maths NCERT Solutions Chapter 14 Statistics
NCERT Class 10 Maths Chapter 14
NCERT Solutions for Class 10 Maths Chapter 14 Statistics
NCERT Solutions Class 10 Maths Chapter 14 Statistics forms the perfect study material for class 10 students to prepare for their board exams. With detailed explanations on the various methods of the step-deviation method, assumed mean method, direct method, etc., these solutions cover all the important points in a detailed step-by-step manner. A section-wise detailed analysis of NCERT Solutions class 10 chapter 14 Statistics can be seen below :
Class 10 Maths Chapter 10 Ex 14.1 - 9 Questions
Class 10 Maths Chapter 10 Ex 14.2 - 6 Questions
Class 10 Maths Chapter 10 Ex 14.3 - 7 Questions
Class 10 Maths Chapter 10 Ex 14.4 - 3 Questions
☛ Download Class 10 Maths Chapter 14 NCERT Book
Topics Covered: The Class 10 maths NCERT solutions Chapter 14 Statistics covers a variety of topics such as mean, median, and mode of grouped data, cumulative frequency, representation of cumulative frequency data in the graphical form, and obtaining the median of grouped data.
Total Questions: Class 10 Maths Chapter 14 Statistics consists of 25 questions. Out of these 25 problems, 15 are fairly easy to solve, 5 are of moderate level, while 5 are a bit complex in nature.
List of Formulas in NCERT Solutions Class 10 Maths Chapter 14
NCERT solutions class 10 maths Chapter 14 guides the students in studying the grouped data and finding their mean, median, and mode. Analyzing data is one of the most essential skills to have in today’s time hence this is the most important chapter of class 10 maths that has a lot of real-life applications. Statistics involves the use of formulas which are covered in detail in the chapter. Some of the important formulas that will help students study and analyze different data charts are as :
Class mark = (Upper class limit + Lower class limit)/ 2
Relationship between the three measures of central tendency : 3 Median = Mode + 2 Mean
Important Questions for Class 10 Maths NCERT Solutions Chapter 14
| CBSE Important Questions for Class 10 Maths Chapter 14 Exercise 14.1 |
| Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9 |
| CBSE Important Questions for Class 10 Maths Chapter 14 Exercise 14.2 |
| Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 |
| CBSE Important Questions for Class 10 Maths Chapter 14 Exercise 14.3 |
| Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 |
| CBSE Important Questions for Class 10 Maths Chapter 14 Exercise 14.4 |
| Question 1 Question 2 Question 3 |
Video Solutions for Class 10 Maths NCERT Chapter 14
| NCERT Video Solutions for Class 10 Maths Chapter 14 |
| Video Solutions for Class 10 Maths Exercise 14.1 |
| Question 1 | Question 2 |
| Question 3 | Question 4 |
| Question 5 | Question 6 |
| Question 7 | Question 8 |
| Question 9 | |
| Video Solutions for Class 10 Maths Exercise 14.2 |
| Question 1 | Question 2 |
| Question 3 | Question 4 |
| Question 5 | Question 6 |
| Video Solutions for Class 10 Maths Exercise 14.3 |
| Question 1 | Question 2 |
| Question 3 | Question 4 |
| Question 5 | Question 6 |
| Question 7 | |
| Video Solutions for Class 10 Maths Exercise 14.4 |
| Question 1 | Question 2 |
| Question 3 | |
FAQs on NCERT Solutions Class 10 Maths Chapter 14
What is the Importance of NCERT Solutions Class 10 Maths Chapter 14 Statistics?
NCERT Solutions Class 10 Maths Chapter 14 Statistics deals with the collection, analysis, presentation, and interpretation of data in different forms. This chapter enables the students to arrange data in a particular form in order to study the salient features of the same and present them in a way that can be easily understood by all. The modern world is highly inclined towards studying and analyzing data in order to make the right decisions, be it in any field, thereby these solutions become an important resource to study.
Do I Need to Practice all Questions Provided in NCERT Solutions Class 10 Maths Statistics?
There are a total of 25 questions in NCERT solutions class 10 maths Statistics. All the questions are curated to cover all the topics related to statistics, such as finding the mean, median, mode using different methods, cumulative frequency representation, etc. in detail. A consistent practice of all the examples and questions in the chapter will result in a better understanding and increased confidence in problem-solving. Thus, it is highly beneficial for the students to practice all the questions.
What are the Important Topics Covered in NCERT Solutions Class 10 Maths Chapter 14?
NCERT Solutions Class 10 Maths Chapter 14 will help you revise the numerical representation of ungrouped data, also known as the measures of central tendencies called mean, median, and mode. Along with that, these solutions also cover the concept of cumulative frequency, the cumulative frequency distribution, and steps involved in drawing ‘ogives’ or cumulative frequency curves.
How Many Questions are there in Class 10 Maths NCERT Solutions Chapter 14 Statistics?
NCERT Solutions Class 10 maths chapter 14 consists of a total of 25 questions distributed in 4 exercises. These questions challenge the students to think out of the box and solve complex problems based on the frequency distribution table, exclusive or continuous frequency, range, etc. Practicing all the questions in these NCERT solutions class 10 maths chapter 14 will result in an excellent preparation for board exams.
What are the Important Formulas in NCERT Solutions Class 10 Maths Chapter 14?
Some of the important formulas of this chapter is the mean of grouped data using the direct method, mean method, and step deviation method. Other formulas include median and mode for grouped data. These formulas can better be understood by solving questions in a stepwise manner. The derivation of these formulas is stated in the NCERT Solutions Class 10 Maths Chapter 14.
Why Should I Practice NCERT Solutions Class 10 Maths Statistics 14?
The NCERT Solutions Class 10 Maths Statistics 14 is a very important chapter not just for exams but for learning insights that have their applications in real-life situations as well. Through this chapter, students will learn how to arrange and analyze different kinds of data by arranging it in a definite or required order. This is one of the greatest skill-set in today’s time and can offer a life-long benefit to the students; hence they must practice NCERT solutions regularly.
| |
| NCERT Solutions Class 10 Maths Chapters |
| Chapter 1 Real Numbers | Chapter 2 Polynomials |
| Chapter 3 Pair of Linear Equations in Two Variables | Chapter 4 Quadratic Equations |
| Chapter 5 Arithmetic Progressions | Chapter 6 Triangles |
| Chapter 7 Coordinate Geometry | Chapter 8 Introduction to Trigonometry |
| Chapter 9 Some Applications Of Trigonometry | Chapter 10 Circles |
| Chapter 11 Constructions | Chapter 12 Areas Related to Circles |
| Chapter 13 Surface Areas and Volumes | Chapter 15 Probability |
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https://wayground.com/library/lessons/high-school/ela/reading/literature/literary-elements-and-devices/literary-techniques/advanced-devices/irony-verbal-situational-dramatic
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Irony (verbal, Situational, Dramatic) Lessons High School Ela | Wayground (formerly Quizizz)
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Irony (verbal, Situational, Dramatic)
High School Irony (verbal, Situational, Dramatic) Lessons
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8th Grade - 9th Grade
Irony Identification - Forms of Irony - Understanding Irony
Lesson
Irony Identification - Forms of Irony - Understanding Irony -----------------------------------------------------------
8th Grade - 9th Grade
English
Explore various examples of irony, focusing on the different forms such as verbal, situational, and dramatic irony. Learners will enhance their comprehension of ironic expressions, develop critical analysis skills, and gain a deeper understanding of how irony functions in literature and everyday communication.
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10th Grade - 11th Grade
Understanding Irony - Types & Examples
Lesson
Understanding Irony - Types & Examples --------------------------------------
10th Grade - 11th Grade
English
Explore irony by identifying and differentiating between dramatic, situational, and verbal types. Gain insight into recognizing irony in diverse contexts, enhancing analytical skills and comprehension of literary devices.
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7th Grade - 9th Grade
Understanding Irony - Types and Examples
Lesson
Understanding Irony - Types and Examples ----------------------------------------
7th Grade - 9th Grade
English
Explore the fascinating concept of irony through detailed examples of verbal, situational, and dramatic irony. Learners will enhance their ability to identify irony, comprehend its influence on storytelling, and differentiate between various ironic forms.
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7th Grade - 12th Grade
Identification and Classification - Types of Irony - Literary Devices
Lesson
Identification and Classification - Types of Irony - Literary Devices ---------------------------------------------------------------------
7th Grade - 12th Grade
English
Explore and classify types of irony, focusing on situational, dramatic, and verbal irony within literary contexts. Understand how irony functions in texts, enhancing analytical skills and the ability to identify subtle nuances in literature.
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9th Grade - 12th Grade
Irony Identification - Types of Irony - Literary Analysis
Lesson
Irony Identification - Types of Irony - Literary Analysis ---------------------------------------------------------
9th Grade - 12th Grade
English
Develop a deeper understanding of verbal, situational, and dramatic irony by analyzing diverse examples, which will enhance analytical skills and literary comprehension. Mastery in identifying irony allows for effective interpretation and discussion of literary works.
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8th Grade - 12th Grade
[Irony - Types and Applications - Literary Devices]
Lesson
[Irony - Types and Applications - Literary Devices] ---------------------------------------------------
8th Grade - 12th Grade
English
Exploration of irony highlights verbal, situational, and dramatic irony through engaging questions and scenarios. Learners will develop a deeper understanding of these literary devices and their applications, enhancing critical thinking and interpretative skills in identifying irony.
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8th Grade - 12th Grade
Understanding Satire - Applying Irony in Literature and Beyond
Lesson
Understanding Satire - Applying Irony in Literature and Beyond --------------------------------------------------------------
8th Grade - 12th Grade
English
Explore the world of satire, analyzing how literary devices such as irony, parody, and exaggeration serve to critique societal vices and follies. Gain insights into the workings and objectives of satire, learning to identify and interpret its application across literature and media.
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9th Grade
Understanding Irony - Types and Applications - Literary Devices
Lesson
Understanding Irony - Types and Applications - Literary Devices ---------------------------------------------------------------
9th Grade
English
Explore the concept of irony, including verbal, situational, and dramatic forms. Gain insights into how these literary devices function to enhance narrative depth and reader engagement, and develop the ability to identify and interpret irony in various texts.
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11th Grade - 12th Grade
Understanding Language Elements - Prepositions & Diction - Themes & Irony in Literature
Lesson
Understanding Language Elements - Prepositions & Diction - Themes & Irony in Literature ---------------------------------------------------------------------------------------
11th Grade - 12th Grade
English
Explore essential language components including prepositions, diction, themes, and irony. Gain valuable insights into literary analysis by learning to identify and analyze these elements within texts, thereby enhancing interpretative skills and appreciation for literature.
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8th Grade - 12th Grade
Understanding and Identifying Types of Irony - Literary Devices
Lesson
Understanding and Identifying Types of Irony - Literary Devices ---------------------------------------------------------------
8th Grade - 12th Grade
Performing Arts
Explore the forms of irony, focusing on verbal, situational, and dramatic irony through engaging examples. By mastering these literary devices, learners will enhance their ability to recognize and analyze irony in diverse contexts, enriching their interpretive skills and literary appreciation.
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9th Grade - 12th Grade
Drama Analysis - Key Dramatic Terms - Theater Studies
Lesson
Drama Analysis - Key Dramatic Terms - Theater Studies -----------------------------------------------------
9th Grade - 12th Grade
English
Explore key dramatic terms crucial for understanding theater, including monologue, soliloquy, and irony. Enhance comprehension of dramatic structures and improve the ability to critically analyze plays.
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9th Grade
Understanding Irony - Literary Devices - Night by Elie Wiesel
Lesson
Understanding Irony - Literary Devices - Night by Elie Wiesel -------------------------------------------------------------
9th Grade
English
Explore the concept of irony as demonstrated in Chapter 1 of "Night" by Elie Wiesel. Focus on understanding types such as situational, dramatic, and verbal irony. Gain insight into how these literary devices contribute to narrative depth while developing the ability to identify irony in literature critically.
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7th Grade - 9th Grade
Comprehension and Analysis - Plot and Themes - The Interlopers
Lesson
Comprehension and Analysis - Plot and Themes - The Interlopers --------------------------------------------------------------
7th Grade - 9th Grade
English
Investigate the intricate plot and themes of "The Interlopers" through a series of thought-provoking questions designed to enhance reading comprehension and critical analysis skills. This exploration provides insight into literary elements, helping learners improve their ability to interpret complex narratives.
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9th Grade - University
Techniques of Satire - Irony and Exaggeration - Literary Devices
Lesson
Techniques of Satire - Irony and Exaggeration - Literary Devices ----------------------------------------------------------------
9th Grade - University
English
Satire employs irony and exaggeration to critique societal issues and human behavior. Learners will develop the ability to identify and interpret these literary techniques, leading to a deeper understanding of satirical works.
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11th Grade
Irony Exploration - Types and Examples - Literary Techniques
Lesson
Irony Exploration - Types and Examples - Literary Techniques ------------------------------------------------------------
11th Grade
English
Investigate the concept of irony, examining its various forms and uses as a literary technique through thought-provoking questions and examples. Develop the ability to identify and distinguish between verbal, dramatic, and situational irony, thereby enhancing understanding of literary elements and critical thinking skills.
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7th Grade - 9th Grade
Key Literary Concepts - Dramatic Elements & Imagery - Suspense & Irony in Literature
Lesson
Key Literary Concepts - Dramatic Elements & Imagery - Suspense & Irony in Literature ------------------------------------------------------------------------------------
7th Grade - 9th Grade
English
Key literary concepts such as suspense, irony, and imagery are explored to create engaging narratives. Learners will gain a deeper understanding of these techniques and their applications in literature, enhancing their analytical skills and appreciation of storytelling.
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10th Grade
Identifying Prepositions - Language Use - English Grammar
Lesson
Identifying Prepositions - Language Use - English Grammar ---------------------------------------------------------
10th Grade
English
Focus on identifying prepositions within sentences and exploring different types of irony and literary devices. Enhance understanding of English grammar and improve skills in literary analysis.
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10th Grade - 12th Grade
Understanding Context Clues - Reading Comprehension Strategies
Lesson
Understanding Context Clues - Reading Comprehension Strategies --------------------------------------------------------------
10th Grade - 12th Grade
English
Focusing on reading comprehension, learners will enhance their skills in inferring meanings within texts by effectively using context clues. This development will lead to improved vocabulary and a deeper understanding of complex texts.
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9th Grade
Understanding and Identifying Irony - Types of Irony in Literary Contexts
Lesson
Understanding and Identifying Irony - Types of Irony in Literary Contexts -------------------------------------------------------------------------
9th Grade
English
Explore irony in literature by examining engaging examples and completing exercises focused on identifying verbal, situational, and dramatic irony. Enhance skills in recognizing different forms of irony and understanding their purposes in storytelling.
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9th Grade
Analyzing Language and Literary Devices - Understanding Irony
Lesson
Analyzing Language and Literary Devices - Understanding Irony -------------------------------------------------------------
9th Grade
English
Analyzing how authors utilize language and literary devices, such as irony, to serve specific purposes in English literature enhances students' ability to identify these tools, comprehend their meanings, and explore how they deepen textual understanding.
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https://artofproblemsolving.com/wiki/index.php/2014_USAMO_Problems/Problem_5?srsltid=AfmBOoqqqTFYJtTkTQ4RRW6kKzCNQQxNlhP8g5GB6_3Up3h5Xx_A9N66
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2014 USAMO Problems/Problem 5
Contents
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1 Problem
2 Solution 1
3 Solution 2
4 See also
Problem
Let be a triangle with orthocenter and let be the second intersection of the circumcircle of triangle with the internal bisector of the angle . Let be the circumcenter of triangle and the orthocenter of triangle . Prove that the length of segment is equal to the circumradius of triangle .
Solution 1
Let be the center of , be the center of . Note that is the reflection of across , so . Additionally so lies on . Now since are perpendicular to and their bisector, is isosceles with , and . Also But as well, and , so . Thus .
Solution 2
Since is a cyclic quadrilateral, . and , we find . That is, is a cyclic quadrilateral. Let be mid-point of . are collinear and . Let be second intersection of with circumcircle of the triangle . Let , . Since is mid-point of the arc , . Since is a cyclic quadrilateral, . Since is the orthocenter of triangle , . Thus, and is a cyclic quadrilateral. So, and . We will prove that is a parallelogram.
(figure link)
We see that is an isosceles triangle and . Also and . Then, . By spiral similarity, and . Hence, , . Since , we get is a parallelogram. As a result, .
(Lokman GÖKÇE)
See also
2014 USAMO (Problems • Resources)
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Problem 4Followed by
Problem 6
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All USAMO Problems and Solutions
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184
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https://people.math.wisc.edu/~duhlenbr/10p02Eccentricity.pdf
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10.2 Classifying Conic Sections by Eccentricity 9 October 2007 Eccentricity and Directrix Given two real quantities a > 0 and e > 0 with e ̸= 1, define the auxiliary quantities c = a · e and d = a e.
Define the focus as the point F(c, 0) and the directrix as the vertical line D with equation x = d .
Consider the set C = C(a, e) defined as the set of all points P(x, y) in the plane which satisfy the condition PF = e · PD .
(1) The points of C(a, e) are described by (PF)2 = e2 · (PD)2 (x −c)2 + (y −0)2 = e2 · (x −d)2 x2 −2cx + c2 + y2 = e2 x2 −2dx + d2 1 −e2 x2 + y2 + 2(de2 −c)x = e2d2 −c2 .
Since d e2 = c and d e = a we find that the points of C(a, e) are characterized by the equation 1 −e2 x2 + y2 = a2 −c2 .
Elliptic case: If 0 < e < 1 it follows that 0 < c < a < d .
Defining b > 0 by b2 = a2 −c2 the last equation above implies that 1 −e2 x2 + y2 = b2 .
Since b2 1−e2 = a2 we obtain the standard equation of an ellipse x2 a2 + y2 b2 = 1 .
Hyperbolic case: If 1 < e it follows that 0 < d < a < c .
Defining b > 0 by b2 = −a2 + c2 the last equation above im-plies that 1 −e2 x2 + y2 = −b2 .
Since −b2 1−e2 = a2 we obtain the standard equation of a hyperbola x2 a2 −y2 b2 = 1 .
Summary: Recall that the parabola was defined in terms of a focus F(p, 0) where p > 0 and the directrix D with equation x = −p in terms of the condition PF = 1 · PD .
Hence a common definition for the standard ellipse, parabola, hyperbola is provided by the focus-directrix equation PF = e · PD , where for fixed e > 0 the set of points in the plane satisfying this condition is an ellipse, if 0 < e < 1; a parabola, if e = 1; a hyperbola, if e > 1.
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185
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https://mathinsight.org/chain_rule_refresher
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A refresher on the chain rule - Math Insight
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A refresher on the chain rule
The chain rule is subtler than the previous rules, so if it seems trickier to you, then you're right. OK. But it is absolutely indispensable in general and later, and already is very helpful in dealing with polynomials.
The general assertion may be a little hard to fathom because it is of a different nature than the previous ones. For one thing, now we will be talking about a composite function instead of just adding or multiplying functions in a more ordinary way. So, for two functions f f and g g,
d d x(f(g(x))=f′(g(x))⋅g′(x)d d x(f(g(x))=f′(g(x))⋅g′(x)
There is also the standard notation
(f∘g)(x)=f(g(x))(f∘g)(x)=f(g(x))
for this composite function, but using this notation doesn't accomplish so very much.
A problem in successful use of the chain rule is that often it requires a little thought to recognize that some formula is (or can be looked at as) a composite function. And the very nature of the chain rule picks on weaknesses in our understanding of the notation. For example, the function
F(x)=(1+x 2)100 F(x)=(1+x 2)100
is really obtained by first using x x as input to the function which squares and adds 1 1 to its input. Then the result of that is used as input to the function which takes the 100 100 th power. It is necessary to think about it this way or we'll make a mistake. The derivative is evaluated as
d d x(1+x 2)100=100(1+x 2)99⋅2 x d d x(1+x 2)100=100(1+x 2)99⋅2 x
To see that this is a special case of the general formula, we need to see what corresponds to the f f and g g in the general formula. Specifically, let
f(input)=(input)100 f(input)=(input)100
g(input)=1+(input)2 g(input)=1+(input)2
The reason for writing ‘input’ and not ‘x’ for the moment is to avoid a certain kind of mistake. But we can compute that
f′(input)=100(input)99 f′(input)=100(input)99
g′(input)=2(input)g′(input)=2(input)
The hazard here is that the input to f f is not x x, but rather is g(x)g(x). So the general formula gives
d d x(1+x 2)100=f′(g(x))⋅g′(x)=100 g(x)99⋅2 x=100(1+x 2)99⋅2 x d d x(1+x 2)100=f′(g(x))⋅g′(x)=100 g(x)99⋅2 x=100(1+x 2)99⋅2 x
More examples:
d d x 3 x+2−−−−−√=d d x(3 x+2)1/2=1 2(3 x+2)−1/2⋅3 d d x 3 x+2=d d x(3 x+2)1/2=1 2(3 x+2)−1/2⋅3
d d x(3 x 5−x+14)11=11(3 x 5−x+14)10⋅(15 x 4−1)d d x(3 x 5−x+14)11=11(3 x 5−x+14)10⋅(15 x 4−1)
It is very important to recognize situations like
d d x(a x+b)n=n(a x+b)n−1⋅a d d x(a x+b)n=n(a x+b)n−1⋅a
for any constants a,b,n a,b,n. And, of course, this includes
d d x a x+b−−−−−√=1 2(a x+b)−1/2⋅a d d x a x+b=1 2(a x+b)−1/2⋅a
d d x 1 a x+b=−(a x+b)−2⋅a=−a(a x+b)2 d d x 1 a x+b=−(a x+b)−2⋅a=−a(a x+b)2
Of course, this idea can be combined with polynomials, quotients, and products to give enormous and excruciating things where we need to use the chain rule, the quotient rule, the product rule, etc., and possibly several times each. But this is not hard, merely tedious, since the only things we really do come in small steps. For example:
d d x(1+x+2−−−−√(1+7 x)33)=(1+x+2−−−−√)′⋅(1+7 x)33−(1+x+2−−−−√)⋅((1+7 x)33)′((1+7 x)33)2 d d x(1+x+2(1+7 x)33)=(1+x+2)′⋅(1+7 x)33−(1+x+2)⋅((1+7 x)33)′((1+7 x)33)2
by the quotient rule, which is then
(1 2(x+2)−1/2)⋅(1+7 x)33−(1+x+2−−−−√)⋅((1+7 x)33)′((1+7 x)33)2(1 2(x+2)−1/2)⋅(1+7 x)33−(1+x+2)⋅((1+7 x)33)′((1+7 x)33)2
because our observations just above (chain rule!) tell us that
d d x x+2−−−−√=1 2(x+2)−1/2⋅(x+2)′=1 2(x+2)−1/2 d d x x+2=1 2(x+2)−1/2⋅(x+2)′=1 2(x+2)−1/2
Then we use the chain rule again to take the derivative of that big power of 1+7 x 1+7 x, so the whole thing becomes
(1 2(x+2)−1/2)⋅(1+7 x)33−(1+x+2−−−−√)⋅(33(1+7 x)32⋅7)((1+7 x)33)2(1 2(x+2)−1/2)⋅(1+7 x)33−(1+x+2)⋅(33(1+7 x)32⋅7)((1+7 x)33)2
Although we could simplify a bit here, let's not. The point about having to do several things in a row to take a derivative is pretty clear without doing algebra just now.
Exercises
Find d d x((1−x 2)100)d d x((1−x 2)100)
Find d d x x−3−−−−√d d x x−3
Find d d x(x 2−x 2−3−−−−−√)d d x(x 2−x 2−3)
Find d d x(x 2+x+1−−−−−−−−√)d d x(x 2+x+1)
Find d d x(x 3+x 2+x+1−−−−−−−−−−−−√3)d d x(x 3+x 2+x+1 3)
Find d d x((x 3+x+1−−−−√)10)d d x((x 3+x+1)10)
Thread navigation
Calculus Refresher
Previous: The product rule
Next: Tangent and normal lines
Similar pages
The idea of the chain rule
Simple examples of using the chain rule
A refresher on the quotient rule
A refresher on the product rule
The quotient rule for differentiation
Introduction to the multivariable chain rule
Multivariable chain rule examples
Special cases of the multivariable chain rule
The idea of the derivative of a function
Derivatives of polynomials
More similar pages
See also
The idea of the chain rule
Simple examples of using the chain rule
Cite this as
Garrett P, “A refresher on the chain rule.” From Math Insight.
Keywords: chain rule, composition, derivative, derivative properties, ordinary derivative
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From Wikipedia, the free encyclopedia
This article is about the chemical element. For the nutrient commonly called sodium, see salt. For the use of sodium as a medication, see Saline (medicine). For other uses, see sodium (disambiguation).
"Natrium" redirects here. For other uses, see Natrium (disambiguation).
Chemical element with atomic number 11 (Na)
Sodium, 11Na
| | |
| Sodium | |
| Appearance | silvery white metallic |
| | |
| Standard atomic weight Ar°(Na) | |
| | 22.98976928±0.00000002 22.990±0.001 (abridged) |
| | |
| Sodium in the periodic table | |
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- | | Hydrogen | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | Helium | | Lithium | Beryllium | | | | | | | | | | | | | | | | | | | | | | | | | Boron | Carbon | Nitrogen | Oxygen | Fluorine | Neon | | Sodium | Magnesium | | | | | | | | | | | | | | | | | | | | | | | | | Aluminium | Silicon | Phosphorus | Sulfur | Chlorine | Argon | | Potassium | Calcium | | | | | | | | | | | | | | | Scandium | Titanium | Vanadium | Chromium | Manganese | Iron | Cobalt | Nickel | Copper | Zinc | Gallium | Germanium | Arsenic | Selenium | Bromine | Krypton | | Rubidium | Strontium | | | | | | | | | | | | | | | Yttrium | Zirconium | Niobium | Molybdenum | Technetium | Ruthenium | Rhodium | Palladium | Silver | Cadmium | Indium | Tin | Antimony | Tellurium | Iodine | Xenon | | Caesium | Barium | Lanthanum | Cerium | Praseodymium | Neodymium | Promethium | Samarium | Europium | Gadolinium | Terbium | Dysprosium | Holmium | Erbium | Thulium | Ytterbium | Lutetium | Hafnium | Tantalum | Tungsten | Rhenium | Osmium | Iridium | Platinum | Gold | Mercury (element) | Thallium | Lead | Bismuth | Polonium | Astatine | Radon | | Francium | Radium | Actinium | Thorium | Protactinium | Uranium | Neptunium | Plutonium | Americium | Curium | Berkelium | Californium | Einsteinium | Fermium | Mendelevium | Nobelium | Lawrencium | Rutherfordium | Dubnium | Seaborgium | Bohrium | Hassium | Meitnerium | Darmstadtium | Roentgenium | Copernicium | Nihonium | Flerovium | Moscovium | Livermorium | Tennessine | Oganesson | Li ↑ Na ↓ K neon ← sodium → magnesium | |
| Atomic number (Z) | 11 |
| Group | group 1: hydrogen and alkali metals |
| Period | period 3 |
| Block | s-block |
| Electron configuration | [Ne] 3s1 |
| Electrons per shell | 2, 8, 1 |
| Physical properties | |
| Phase at STP | solid |
| Melting point | 370.944 K (97.794 °C, 208.029 °F) |
| Boiling point | 1156.090 K (882.940 °C, 1621.292 °F) |
| Density (at 20° C) | 0.9688 g/cm3 |
| when liquid (at m.p.) | 0.927 g/cm3 |
| Critical point | 2573 K, 35 MPa (extrapolated) |
| Heat of fusion | 2.60 kJ/mol |
| Heat of vaporization | 97.42 kJ/mol |
| Molar heat capacity | 28.230 J/(mol·K) |
| Vapor pressure | P (Pa) | 1 | 10 | 100 | 1 k | 10 k | 100 k | --- --- --- | at T (K) | 554 | 617 | 697 | 802 | 946 | 1153 | | |
| Atomic properties | |
| Oxidation states | common: +1 −1, |
| Electronegativity | Pauling scale: 0.93 |
| Ionization energies | 1st: 495.8 kJ/mol 2nd: 4562 kJ/mol 3rd: 6910.3 kJ/mol (more) |
| Atomic radius | empirical: 186 pm |
| Covalent radius | 166±9 pm |
| Van der Waals radius | 227 pm |
| Spectral lines of sodium | |
| Other properties | |
| Natural occurrence | primordial |
| Crystal structure | body-centered cubic (bcc) (cI2) |
| Lattice constant | a = 428.74 pm (at 20 °C) |
| Thermal expansion | 69.91×10−6/K (at 20 °C) |
| Thermal conductivity | 142 W/(m⋅K) |
| Electrical resistivity | 47.7 nΩ⋅m (at 20 °C) |
| Magnetic ordering | paramagnetic |
| Molar magnetic susceptibility | +16.0×10−6 cm3/mol (298 K) |
| Young's modulus | 10 GPa |
| Shear modulus | 3.3 GPa |
| Bulk modulus | 6.3 GPa |
| Speed of sound thin rod | 3200 m/s (at 20 °C) |
| Mohs hardness | 0.5 |
| Brinell hardness | 0.69 MPa |
| CAS Number | 7440-23-5 |
| History | |
| Naming | possibly from Arabic suda, 'headache', for soda's use as an anticephalalgic |
| Discovery and first isolation | Humphry Davy (1807) |
| Symbol | "Na": from New Latin natrium, coined from German Natron, 'natron' |
| Isotopes of sodium v e | |
| | |
| | Main isotopes | | | Decay | | --- --- | Isotope | abundance | half-life (t1/2) | mode | product | | 22Na | trace | 2.6019 y | β+ | 22Ne | | 23Na | 100% | stable | | | | 24Na | trace | 14.956 h | β− | 24Mg | | |
| | |
| Category: Sodium view talk edit | references | |
Sodium is a chemical element; it has symbol Na (from Neo-Latin natrium) and atomic number 11. It is a soft, silvery-white, highly reactive metal. Sodium is an alkali metal, being in group 1 of the periodic table. Its only stable isotope is 23Na. The free metal does not occur in nature and must be prepared from compounds. Sodium is the sixth most abundant element in the Earth's crust and exists in numerous minerals such as feldspars, sodalite, and halite (NaCl). Many salts of sodium are highly water-soluble: sodium ions have been leached by the action of water from the Earth's minerals over eons, and thus sodium and chlorine are the most common dissolved elements by weight in the oceans.
Sodium was first isolated by Humphry Davy in 1807 by the electrolysis of sodium hydroxide. Among many other useful sodium compounds, sodium hydroxide (lye) is used in soap manufacture, and sodium chloride (edible salt) is a de-icing agent and a nutrient for animals including humans.
Sodium is an essential element for all animals and some plants. Sodium ions are the major cation in the extracellular fluid (ECF) and as such are the major contributor to the ECF osmotic pressure. Animal cells actively pump sodium ions out of the cells by means of the sodium–potassium pump, an enzyme complex embedded in the cell membrane, in order to maintain a roughly ten-times higher concentration of sodium ions outside the cell than inside. In nerve cells, the sudden flow of sodium ions into the cell through voltage-gated sodium channels enables transmission of a nerve impulse in a process called the action potential.
Characteristics
Physical
Sodium at standard temperature and pressure is a soft silvery metal that combines with oxygen in the air, forming sodium oxides. Bulk sodium is usually stored in oil or an inert gas. Sodium metal can be easily cut with a knife. It is a good conductor of electricity and heat.
The melting (98 °C) and boiling (883 °C) points of sodium are lower than those of lithium but higher than those of the heavier alkali metals potassium, rubidium, and caesium, following periodic trends down the group. These properties change dramatically at elevated pressures: at 1.5 Mbar, the color changes from silvery metallic to black; at 1.9 Mbar the material becomes transparent with a red color; and at 3 Mbar, sodium is a clear and transparent solid. All of these high-pressure allotropes are insulators and electrides.
In a flame test, sodium and its compounds glow yellow because the excited 3s electrons of sodium emit a photon when they fall from 3p to 3s; the wavelength of this photon corresponds to the D line at about 589.3 nm. Spin-orbit interactions involving the electron in the 3p orbital split the D line into two, at 589.0 and 589.6 nm; hyperfine structures involving both orbitals cause many more lines.
Isotopes
Main article: Isotopes of sodium
Twenty isotopes of sodium are known, but only 23Na is stable. 23Na is created in the carbon-burning process in stars by fusing two carbon atoms together; this requires temperatures above 600 megakelvins and a star of at least three solar masses. Two radioactive, cosmogenic isotopes are the byproduct of cosmic ray spallation: 22Na has a half-life of 2.6 years and 24Na, a half-life of 15 hours; all other isotopes have a half-life of less than one minute.
Two nuclear isomers have been discovered, the longer-lived one being 24mNa with a half-life of around 20.2 milliseconds. Acute neutron radiation, as from a nuclear criticality accident, converts some of the stable 23Na in human blood to 24Na; the neutron radiation dosage of a victim can be calculated by measuring the concentration of 24Na relative to 23Na.
Chemistry
Main article: Sodium compounds
Sodium atoms have 11 electrons, one more than the stable configuration of the noble gas neon. The first and second ionization energies are 495.8 kJ/mol and 4562 kJ/mol, respectively. As a result, sodium usually forms ionic compounds involving the Na+ cation.
Metallic sodium
Metallic sodium is generally less reactive than potassium and more reactive than lithium. Sodium metal is highly reducing, with the standard reduction potential for the Na+/Na couple being −2.71 volts, though potassium and lithium have even more negative potentials.
Salts and oxides
See also: Category:Sodium compounds
Sodium compounds are of immense commercial importance, being particularly central to industries producing glass, paper, soap, and textiles. The most important sodium compounds are table salt (NaCl), soda ash (Na2CO3), baking soda (NaHCO3), caustic soda (NaOH), sodium nitrate (NaNO3), di- and tri-sodium phosphates, sodium thiosulfate (Na2S2O3·5H2O), and borax (Na2B4O7·10H2O). In compounds, sodium is usually ionically bonded to water and anions and is viewed as a hard Lewis acid.
Most soaps are sodium salts of fatty acids. Sodium soaps have a higher melting temperature (and seem "harder") than potassium soaps.
Like all the alkali metals, sodium reacts exothermically with water. The reaction produces caustic soda (sodium hydroxide) and flammable hydrogen gas. When burned in air, it forms primarily sodium peroxide with some sodium oxide.
Aqueous solutions
Sodium tends to form water-soluble compounds, such as halides, sulfates, nitrates, carboxylates and carbonates. The main aqueous species are the aquo complexes [Na(H2O)n]+, where n = 4–8; with n = 6 indicated from X-ray diffraction data and computer simulations.
Direct precipitation of sodium salts from aqueous solutions is rare because sodium salts typically have a high affinity for water. An exception is sodium bismuthate (NaBiO3), which is insoluble in cold water and decomposes in hot water. Because of the high solubility of its compounds, sodium salts are usually isolated as solids by evaporation or by precipitation with an organic antisolvent, such as ethanol; for example, only 0.35 g/L of sodium chloride will dissolve in ethanol. A crown ether such as 15-crown-5 may be used as a phase-transfer catalyst.
Sodium content of samples is determined by atomic absorption spectrophotometry or by potentiometry using ion-selective electrodes.
Electrides and sodides
Like the other alkali metals, sodium dissolves in ammonia and some amines to give deeply colored solutions; evaporation of these solutions leaves a shiny film of metallic sodium. The solutions contain the coordination complex [Na(NH3)6]+, with the positive charge counterbalanced by electrons as anions; cryptands permit the isolation of these complexes as crystalline solids. Sodium forms complexes with crown ethers, cryptands and other ligands.
For example, 15-crown-5 has a high affinity for sodium because the cavity size of 15-crown-5 is 1.7–2.2 Å, which is enough to fit the sodium ion (1.9 Å). Cryptands, like crown ethers and other ionophores, also have a high affinity for the sodium ion; derivatives of the alkalide Na− are obtainable by the addition of cryptands to solutions of sodium in ammonia via disproportionation.
Organosodium compounds
Main article: Organosodium chemistry
Many organosodium compounds have been prepared. Because of the high polarity of the C-Na bonds, they behave like sources of carbanions (salts with organic anions). Some well-known derivatives include sodium cyclopentadienide (NaC5H5) and trityl sodium ((C6H5)3CNa). Sodium naphthalene, Na+[C10H8•]−, a strong reducing agent, forms upon mixing Na and naphthalene in ethereal solutions.
Intermetallic compounds
Sodium forms alloys with many metals, such as potassium, calcium, lead, and the group 11 and 12 elements. Sodium and potassium form KNa2 and NaK. NaK is 40–90% potassium and it is liquid at ambient temperature. It is an excellent thermal and electrical conductor. Sodium-calcium alloys are by-products of the electrolytic production of sodium from a binary salt mixture of NaCl-CaCl2 and ternary mixture NaCl-CaCl2-BaCl2. Calcium is only partially miscible with sodium, and the 1–2% of it dissolved in the sodium obtained from said mixtures can be precipitated by cooling to 120 °C and filtering.
In a liquid state, sodium is completely miscible with lead. There are several methods to make sodium-lead alloys. One is to melt them together and another is to deposit sodium electrolytically on molten lead cathodes. NaPb3, NaPb, Na9Pb4, Na5Pb2, and Na15Pb4 are some of the known sodium-lead alloys. Sodium also forms alloys with gold (NaAu2) and silver (NaAg2). Group 12 metals (zinc, cadmium and mercury) are known to make alloys with sodium. NaZn13 and NaCd2 are alloys of zinc and cadmium. Sodium and mercury form NaHg, NaHg4, NaHg2, Na3Hg2, and Na3Hg.
History
Because of its importance in human health, salt has long been an important commodity. In medieval Europe, a compound of sodium with the Latin name of sodanum was used as a headache remedy. The name sodium is thought to originate from the Arabic suda, meaning headache, as the headache-alleviating properties of sodium carbonate or soda were well known in early times.
Although sodium, sometimes called soda, had long been recognized in compounds, the metal itself was not isolated until 1807 by Sir Humphry Davy through the electrolysis of sodium hydroxide. In 1809, the German physicist and chemist Ludwig Wilhelm Gilbert proposed the names Natronium for Humphry Davy's "sodium" and Kalium for Davy's "potassium".
The chemical abbreviation for sodium was first published in 1814 by Jöns Jakob Berzelius in his system of atomic symbols, and is an abbreviation of the element's Neo-Latin name natrium, which refers to the Egyptian natron, a natural mineral salt mainly consisting of hydrated sodium carbonate. Natron historically had several important industrial and household uses, later eclipsed by other sodium compounds.
Sodium imparts an intense yellow color to flames. As early as 1860, Kirchhoff and Bunsen noted the high sensitivity of a sodium flame test, and stated in Annalen der Physik und Chemie:
In a corner of our 60 m3 room farthest away from the apparatus, we exploded 3 mg of sodium chlorate with milk sugar while observing the nonluminous flame before the slit. After a while, it glowed a bright yellow and showed a strong sodium line that disappeared only after 10 minutes. From the weight of the sodium salt and the volume of air in the room, we easily calculate that one part by weight of air could not contain more than 1/20 millionth weight of sodium.
Occurrence
The Earth's crust contains 2.27% sodium, making it the sixth most abundant element on Earth and the fourth most abundant metal, behind aluminium, iron, calcium, and magnesium and ahead of potassium.Sodium's estimated oceanic abundance is 10.8 grams per liter. Because of its high reactivity, it is never found as a pure element. It is found in many minerals, some very soluble, such as halite and natron, others much less soluble, such as amphibole and zeolite. The insolubility of certain sodium minerals such as cryolite and feldspar arises from their polymeric anions, which in the case of feldspar is a polysilicate. In the universe, sodium is the 15th most abundant element with a 20,000 parts-per-billion abundance, making sodium 0.002% of the total atoms in the universe.
Astronomical observations
Atomic sodium has a very strong spectral line in the yellow-orange part of the spectrum (the same line as is used in sodium-vapor street lights). This appears as an absorption line in many types of stars, including the Sun. The line was first studied in 1814 by Joseph von Fraunhofer during his investigation of the lines in the solar spectrum, now known as the Fraunhofer lines. Fraunhofer named it the "D" line, although it is now known to actually be a group of closely spaced lines split by a fine and hyperfine structure.
The strength of the D line allows its detection in many other astronomical environments. In stars, it is seen in any whose surfaces are cool enough for sodium to exist in atomic form (rather than ionized). This corresponds to stars of roughly F-type and cooler. Many other stars appear to have a sodium absorption line, but this is actually caused by gas in the foreground interstellar medium. The two can be distinguished via high-resolution spectroscopy, because interstellar lines are much narrower than those broadened by stellar rotation.
Sodium has also been detected in numerous Solar System environments, including the exospheres of Mercury and the Moon, and numerous other bodies. Some comets have a sodium tail, which was first detected in observations of Comet Hale–Bopp in 1997. Sodium has even been detected in the atmospheres of some extrasolar planets via transit spectroscopy.
Commercial production
Employed in rather specialized applications, about 100,000 tonnes of metallic sodium are produced annually. Metallic sodium was first produced commercially in the late nineteenth century by carbothermal reduction of sodium carbonate at 1100 °C, as the first step of the Deville process for the production of aluminium:
: Na2CO3 + 2 C → 2 Na + 3 CO
The high demand for aluminium created the need for the production of sodium. The introduction of the Hall–Héroult process for the production of aluminium by electrolysing a molten salt bath ended the need for large quantities of sodium. A related process based on the reduction of sodium hydroxide was developed in 1886.
Sodium is now produced commercially through the electrolysis of molten sodium chloride (common salt), based on a process patented in 1924. This is done in a Downs cell in which the NaCl is mixed with calcium chloride to lower the melting point below 700 °C. As calcium is less electropositive than sodium, no calcium will be deposited at the cathode. This method is less expensive than the previous Castner process (the electrolysis of sodium hydroxide).
If sodium of high purity is required, it can be distilled once or several times.
The market for sodium is volatile due to the difficulty in its storage and shipping; it must be stored under a dry inert gas atmosphere or anhydrous mineral oil to prevent the formation of a surface layer of sodium oxide or sodium superoxide.
Uses
See also: Sodium supplements
Though metallic sodium has some important uses, the major applications for sodium use compounds; millions of tons of sodium chloride, hydroxide, and carbonate are produced annually. Sodium chloride is extensively used for anti-icing and de-icing and as a preservative; examples of the uses of sodium bicarbonate include baking, as a raising agent, and sodablasting. Along with potassium, many important medicines have sodium added to improve their bioavailability; though potassium is the better ion in most cases, sodium is chosen for its lower price and atomic weight. Sodium hydride is used as a base for various reactions (such as the aldol reaction) in organic chemistry.
Metallic sodium is used mainly for the production of sodium borohydride, sodium azide, indigo, and triphenylphosphine. A once-common use was the making of tetraethyllead and titanium metal; because of the move away from TEL and new titanium production methods, the production of sodium declined after 1970. Sodium is also used as an alloying metal, an anti-scaling agent, and as a reducing agent for metals when other materials are ineffective.
Note the free element is not used as a scaling agent, ions in the water are exchanged for sodium ions. Sodium plasma ("vapor") lamps are often used for street lighting in cities, shedding light that ranges from yellow-orange to peach as the pressure increases. By itself or with potassium, sodium is a desiccant; it gives an intense blue coloration with benzophenone when the desiccate is dry.
In organic synthesis, sodium is used in various reactions such as the Birch reduction, and the sodium fusion test is conducted to qualitatively analyse compounds. Sodium reacts with alcohols and gives alkoxides, and when sodium is dissolved in ammonia solution, it can be used to reduce alkynes to trans-alkenes. Lasers emitting light at the sodium D line are used to create artificial laser guide stars that assist in the adaptive optics for land-based visible-light telescopes.
Heat transfer
Liquid sodium is used as a heat transfer fluid in sodium-cooled fast reactors because it has the high thermal conductivity and low neutron absorption cross section required to achieve a high neutron flux in the reactor. The high boiling point of sodium allows the reactor to operate at ambient (normal) pressure, but drawbacks include its opacity, which hinders visual maintenance, and its strongly reducing properties. Sodium will explode in contact with water, although it will only burn gently in air.
Radioactive sodium-24 may be produced by neutron bombardment during operation, posing a slight radiation hazard; the radioactivity stops within a few days after removal from the reactor. If a reactor needs to be shut down frequently, sodium-potassium alloy (NaK) is used. Because NaK is a liquid at room temperature, the coolant does not solidify in the pipes. The pyrophoricity of the NaK means extra precautions must be taken to prevent and detect leaks.
Another heat transfer application of sodium is in poppet valves in high-performance internal combustion engines; the valve stems are partially filled with sodium and work as a heat pipe to cool the valves.
Biological role
Main article: Sodium in biology
Biological role in humans
In humans, sodium is an essential mineral that regulates blood volume, blood pressure, osmotic equilibrium and pH. The minimum physiological requirement for sodium is estimated to range from about 120 milligrams per day in newborns to 500 milligrams per day over the age of 10.
Diet
Sodium chloride, also known as 'edible salt' or 'table salt' (chemical formula NaCl), is the principal source of sodium (Na) in the diet and is used as seasoning and preservative in such commodities as pickled preserves and jerky. For Americans, most sodium chloride comes from processed foods. Other sources of sodium are its natural occurrence in food and such food additives as monosodium glutamate (MSG), sodium nitrite, sodium saccharin, baking soda (sodium bicarbonate), and sodium benzoate.
The U.S. Institute of Medicine set its tolerable upper intake level for sodium at 2.3 grams per day, but the average person in the United States consumes 3.4 grams per day. The American Heart Association recommends no more than 1.5 g of sodium per day.
The committee to Review the Dietary Reference Intakes for Sodium and Potassium, which is part of the National Academies of Sciences, Engineering, and Medicine, has determined that there isn't enough evidence from research studies to establish Estimated Average Requirement (EAR) and Recommended Dietary Allowance (RDA) values for sodium. As a result, the committee has established Adequate Intake (AI) levels instead, as follows. The sodium AI for infants of 0–6 months is established at 110 mg/day, 7–12 months: 370 mg/day; for children 1–3 years: 800 mg/day, 4–8 years: 1,000 mg/day; for adolescents: 9–13 years – 1,200 mg/day, 14–18 years 1,500 mg/day; for adults regardless of their age or sex: 1,500 mg/day.
Sodium chloride (NaCl) contains approximately 39.34% of its total mass as elemental sodium (Na). This means that 1 gram of sodium chloride contains approximately 393.4 mg of elemental sodium. For example, to find out how much sodium chloride contains 1500 mg of elemental sodium (the value of 1500 mg sodium is the adequate intake (AI) for an adult), we can use the proportion:
: 393.4 mg Na : 1000 mg NaCl = 1500 mg Na : x mg NaCl
Solving for x gives us the amount of sodium chloride that contains 1500 mg of elemental sodium
: x = (1500 mg Na × 1000 mg NaCl) / 393.4 mg Na = 3812.91 mg
This mean that 3812.91 mg of sodium chloride contain 1500 mg of elemental sodium.
High sodium consumption
Main article: Health effects of salt
High sodium consumption is unhealthy, and can lead to alteration in the mechanical performance of the heart. High sodium consumption is also associated with chronic kidney disease, high blood pressure, cardiovascular diseases, and stroke.
High blood pressure
There is a strong correlation between higher sodium intake and higher blood pressure. Studies have found that lowering sodium intake by 2 g per day tends to lower systolic blood pressure by about two to four mm Hg. It has been estimated that such a decrease in sodium intake would lead to 9–17% fewer cases of hypertension.
Hypertension causes 7.6 million premature deaths worldwide each year. Since edible salt contains about 39.3% sodium—the rest being chlorine and trace chemicals; thus, 2.3 g sodium is about 5.9 g, or 5.3 ml, of salt—about one US teaspoon.
One scientific review found that people with or without hypertension who excreted less than 3 grams of sodium per day in their urine (and therefore were taking in less than 3 g/d) had a higher risk of death, stroke, or heart attack than those excreting 4 to 5 grams per day. Levels of 7 g per day or more in people with hypertension were associated with higher mortality and cardiovascular events, but this was not found to be true for people without hypertension. The US FDA states that adults with hypertension and prehypertension should reduce daily sodium intake to 1.5 g.
Physiology
The renin–angiotensin system regulates the amount of fluid and sodium concentration in the body. Reduction of blood pressure and sodium concentration in the kidney result in the production of renin, which in turn produces aldosterone and angiotensin, which stimulates the reabsorption of sodium back into the bloodstream. When the concentration of sodium increases, the production of renin decreases, and the sodium concentration returns to normal. The sodium ion (Na+) is an important electrolyte in neuron function, and in osmoregulation between cells and the extracellular fluid. This is accomplished in all animals by Na+/K+-ATPase, an active transporter pumping ions against the gradient, and sodium/potassium channels. The difference in extracellular and intracellular ion concentration, maintained by the sodium-potassium pump, produce electrical signals in the form of action potentials that supports cardiac muscle contraction and promote long-distance communication between neurons. Sodium is the most prevalent metallic ion in extracellular fluid.
In humans, unusually low or high sodium levels in the blood is recognized in medicine as hyponatremia and hypernatremia. These conditions may be caused by genetic factors, ageing, or prolonged vomiting or diarrhea.
Biological role in plants
In C4 plants, sodium is a micronutrient that aids metabolism, specifically in regeneration of phosphoenolpyruvate and synthesis of chlorophyll. In others, it substitutes for potassium in several roles, such as maintaining turgor pressure and aiding in the opening and closing of stomata. Excess sodium in the soil can limit the uptake of water by decreasing the water potential, which may result in plant wilting; excess concentrations in the cytoplasm can lead to enzyme inhibition, which in turn causes necrosis and chlorosis.
In response, some plants have developed mechanisms to limit sodium uptake in the roots, to store it in cell vacuoles, and restrict salt transport from roots to leaves. Excess sodium may also be stored in old plant tissue, limiting the damage to new growth. Halophytes have adapted to be able to flourish in sodium rich environments.
Safety and precautions
Sodium
| Hazards | |
| GHS labelling: | |
| Pictograms | |
| Signal word | Danger |
| Hazard statements | H260, H314 |
| Precautionary statements | P223, P231+P232, P280, P305+P351+P338, P370+P378, P422 |
| NFPA 704 (fire diamond) | 3 1 2 W |
Chemical compound
Sodium forms flammable hydrogen and caustic sodium hydroxide on contact with water; ingestion and contact with moisture on skin, eyes or mucous membranes can cause severe burns. Sodium spontaneously explodes in the presence of water due to the formation of hydrogen (highly explosive) and sodium hydroxide (which dissolves in the water, liberating more surface). However, sodium exposed to air and ignited or reaching autoignition (reported to occur when a molten pool of sodium reaches about 290 °C, 554 °F) displays a relatively mild fire.
In the case of massive (non-molten) pieces of sodium, the reaction with oxygen eventually becomes slow due to formation of a protective layer. Fire extinguishers based on water accelerate sodium fires. Those based on carbon dioxide and bromochlorodifluoromethane should not be used on sodium fire. Metal fires are Class D, but not all Class D extinguishers are effective when used to extinguish sodium fires. An effective extinguishing agent for sodium fires is Met-L-X. Other effective agents include Lith-X, which has graphite powder and an organophosphate flame retardant, and dry sand.
Sodium fires are prevented in nuclear reactors by isolating sodium from oxygen with surrounding pipes containing inert gas. Pool-type sodium fires are prevented using diverse design measures called catch pan systems. They collect leaking sodium into a leak-recovery tank where it is isolated from oxygen.
Liquid sodium fires are more dangerous to handle than solid sodium fires, particularly if there is insufficient experience with the safe handling of molten sodium. In a technical report for the United States Fire Administration, R. J. Gordon writes (emphasis in original)
Molten sodium is extremely dangerous because it is much more reactive than a solid mass. In the liquid form, every sodium atom is free and mobile to instantaneously combine with any available oxygen atom or other oxidizer, and any gaseous by-product will be created as a rapidly expanding gas bubble within the molten mass. Even a minute amount of water can create this type of reaction. Any amount of water introduced into a pool of molten sodium is likely to cause a violent explosion inside the liquid mass, releasing the hydrogen as a rapidly expanding gas and causing the molten sodium to erupt from the container.
When molten sodium is involved in a fire, the combustion occurs at the surface of the liquid. An inert gas, such as nitrogen or argon, can be used to form an inert layer over the pool of burning liquid sodium, but the gas must be applied very gently and contained over the surface. Except for soda ash, most of the powdered agents that are used to extinguish small fires in solid pieces or shallow pools will sink to the bottom of a molten mass of burning sodium – the sodium will float to the top and continue to burn. If the burning sodium is in a container, it may be feasible to extinguish the fire by placing a lid on the container to exclude oxygen.
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^ Stallings, Virginia A.; Harrison, Meghan; Oria, Maria, eds. (2019). Dietary Reference Intakes for Sodium and Potassium. National Academies of Sciences, Engineering, and Medicine. Washington, DC: The National Academies Press. doi:10.17226/25353. ISBN 978-0-309-48834-1. PMID 30844154.
^ Jump up to: a b "NaCl (Sodium Chloride) Molar Mass". Archived from the original on 18 March 2024. Retrieved 18 March 2024.
^ Jump up to: a b Patel, Yash; Joseph, Jacob (13 December 2020). "Sodium Intake and Heart Failure". International Journal of Molecular Sciences. 21 (24): 9474. doi:10.3390/ijms21249474. ISSN 1422-0067. PMC 7763082. PMID 33322108.
^ CDC (28 February 2018). "The links between sodium, potassium, and your blood pressure". Centers for Disease Control and Prevention. Archived from the original on 17 January 2021. Retrieved 5 January 2021.
^ Jump up to: a b Geleijnse, J. M.; Kok, F. J.; Grobbee, D. E. (2004). "Impact of dietary and lifestyle factors on the prevalence of hypertension in Western populations" (PDF). European Journal of Public Health. 14 (3): 235–239. doi:10.1093/eurpub/14.3.235. PMID 15369026.
^ Lawes, C. M.; Vander Hoorn, S.; Rodgers, A.; International Society of Hypertension (2008). "Global burden of blood-pressure-related disease, 2001" (PDF). Lancet. 371 (9623): 1513–1518. CiteSeerX 10.1.1.463.887. doi:10.1016/S0140-6736(08)60655-8. PMID 18456100. S2CID 19315480. Archived from the original (PDF) on 26 October 2015. Retrieved 25 October 2017.
^ Armstrong, James (2011). General, Organic, and Biochemistry: An Applied Approach. Cengage Learning. pp. 48–. ISBN 978-1-133-16826-3.
^ Table Salt Conversion Archived 23 September 2014 at the Wayback Machine. Traditionaloven.com. Retrieved on 11 November 2015.
^ Jump up to: a b "Use the Nutrition Facts Label to Reduce Your Intake of Sodium in Your Diet". US Food and Drug Administration. 3 January 2018. Archived from the original on 25 January 2018. Retrieved 2 February 2018.
^ Jump up to: a b Andrew Mente; et al. (2016). "Associations of urinary sodium excretion with cardiovascular events in individuals with and without hypertension: a pooled analysis of data from four studies". The Lancet. 388 (10043): 465–75. doi:10.1016/S0140-6736(16)30467-6. hdl:10379/16625. PMID 27216139. S2CID 44581906. Archived from the original on 6 November 2023. Retrieved 10 March 2023.
^ McGuire, Michelle; Beerman, Kathy A. (2011). Nutritional Sciences: From Fundamentals to Food. Cengage Learning. p. 546. ISBN 978-0-324-59864-3. OCLC 472704484.
^ Campbell, Neil (1987). Biology. Benjamin/Cummings. p. 795. ISBN 978-0-8053-1840-1.
^ Srilakshmi, B. (2006). Nutrition Science (2nd ed.). New Age International. p. 318. ISBN 978-81-224-1633-6. OCLC 173807260. Archived from the original on 1 February 2016. Retrieved 8 January 2016.
^
Pohl, Hanna R.; Wheeler, John S.; Murray, H. Edward (2013). "Sodium and Potassium in Health and Disease". In Astrid Sigel; Helmut Sigel; Roland K. O. Sigel (eds.). Interrelations between Essential Metal Ions and Human Diseases. Metal Ions in Life Sciences. Vol. 13. Springer. pp. 29–47. doi:10.1007/978-94-007-7500-8_2. ISBN 978-94-007-7499-5. PMID 24470088.
^ Kering, M. K. (2008). "Manganese Nutrition and Photosynthesis in NAD-malic enzyme C4 plants PhD dissertation" (PDF). University of Missouri-Columbia. Archived (PDF) from the original on 25 April 2012. Retrieved 9 November 2011.
^ Subbarao, G. V.; Ito, O.; Berry, W. L.; Wheeler, R. M. (2003). "Sodium—A Functional Plant Nutrient". Critical Reviews in Plant Sciences. 22 (5): 391–416. Bibcode:2003CRvPS..22..391S. doi:10.1080/07352680390243495. S2CID 85111284.
^ Zhu, J. K. (2001). "Plant salt tolerance". Trends in Plant Science. 6 (2): 66–71. Bibcode:2001TPS.....6...66Z. doi:10.1016/S1360-1385(00)01838-0. PMID 11173290.
^ Jump up to: a b "Plants and salt ion toxicity". Plant Biology. Archived from the original on 3 April 2012. Retrieved 2 November 2010.
^ "Sodium 262714". Sigma-Aldrich. Archived from the original on 15 January 2016. Retrieved 1 October 2018.
^ Hazard Rating Information for NFPA Fire Diamonds Archived 17 February 2015 at the Wayback Machine. Ehs.neu.edu. Retrieved on 11 November 2015.
^ Angelici, R. J. (1999). Synthesis and Technique in Inorganic Chemistry. Mill Valley, CA: University Science Books. ISBN 978-0-935702-48-4.
^ Jump up to: a b Gordon, Routley J. (25 October 1993). Sodium explosion critically burns firefighters, Newton, Massachusetts (Technical report). United States Fire Administration. 75.
^ Jump up to: a b c U.S. National Research Council Committee on Prudent Practices for Handling, Storage, and Disposal of Chemicals in Laboratories (1995). Prudent Practices in the Laboratory: Handling and Disposal of Chemicals. National Academies. p. 390. ISBN 978-0-309-05229-0.
^ An, Deukkwang; Sunderland, Peter B.; Lathrop, Daniel P. (2013). "Suppression of sodium fires with liquid nitrogen" (PDF). Fire Safety Journal. 58: 204–207. Bibcode:2013FirSJ..58..204A. doi:10.1016/j.firesaf.2013.02.001. Archived from the original (PDF) on 8 August 2017.
^ Clough, W. S.; Garland, J. A. (1 July 1970). Behaviour in the Atmosphere of the Aerosol from a Sodium Fire (Report). U.S. Department of Energy Office of Scientific and Technical Information. OSTI 4039364.
^ Ladwig, Thomas H. (1991). Industrial fire prevention and protection. Van Nostrand Reinhold, 1991. p. 178. ISBN 978-0-442-23678-6.
^ Jump up to: a b Günter Kessler (8 May 2012). Sustainable and Safe Nuclear Fission Energy: Technology and Safety of Fast and Thermal Nuclear Reactors (illustrated ed.). Springer Science & Business Media, 2012. p. 446. ISBN 978-3-642-11990-3.
Greenwood, Norman N.; Earnshaw, Alan (1997). Chemistry of the Elements (2nd ed.). Butterworth-Heinemann. doi:10.1016/C2009-0-30414-6. ISBN 978-0-08-037941-8.
External links
Sodium at The Periodic Table of Videos (University of Nottingham)
Etymology of "natrium" – source of symbol Na
The Wooden Periodic Table Table's Entry on Sodium
Sodium isotopes data from The Berkeley Laboratory Isotopes Project's
| Periodic table | |
--- |
| | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- --- | | 1 | 2 | | | | | | | | | | | | | | | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | | 1 | H | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | | He | | 2 | Li | Be | | | | | | | | | | | | | | | | | | | | | | | | | B | C | N | O | F | Ne | | 3 | Na | Mg | | | | | | | | | | | | | | | | | | | | | | | | | Al | Si | P | S | Cl | Ar | | 4 | K | Ca | | | | | | | | | | | | | | | Sc | Ti | V | Cr | Mn | Fe | Co | Ni | Cu | Zn | Ga | Ge | As | Se | Br | Kr | | 5 | Rb | Sr | | | | | | | | | | | | | | | Y | Zr | Nb | Mo | Tc | Ru | Rh | Pd | Ag | Cd | In | Sn | Sb | Te | I | Xe | | 6 | Cs | Ba | La | Ce | Pr | Nd | Pm | Sm | Eu | Gd | Tb | Dy | Ho | Er | Tm | Yb | Lu | Hf | Ta | W | Re | Os | Ir | Pt | Au | Hg | Tl | Pb | Bi | Po | At | Rn | | 7 | Fr | Ra | Ac | Th | Pa | U | Np | Pu | Am | Cm | Bk | Cf | Es | Fm | Md | No | Lr | Rf | Db | Sg | Bh | Hs | Mt | Ds | Rg | Cn | Nh | Fl | Mc | Lv | Ts | Og | | |
| | | | | | --- --- | | s-block | f-block | d-block | p-block | | |
| v t e Sodium compounds | |
--- |
| Inorganic | | | | --- | | Halides | NaF NaHF2 NaCl NaBr NaI NaAt | | Chalcogenides | NaO3 NaO2 Na2O Na2O2 NaOH NaOD Na2S NaSH Na2Se NaSeH Na2Te NaHTe Na2Po | | Pnictogenides | Na3N NaN3 NaNH2 NaN2H3 Na3P Na3As Na3Sb | | Oxyhalides | NaClO NaClO2 NaClO3 NaClO4 NaBrO NaBrO2 NaBrO3 NaBrO4 NaIO NaIO3 NaIO4 | | Oxychalcogenides | Na2SO3 Na2SO4 NaHSO3 NaHSO4 Na2S2O3 Na2S2O4 Na2S2O5 Na2S2O6 Na2S2O7 Na2S2O8 Na2S4O6 Na2SeO3 Na2SeO4 NaHSeO3 Na2TeO3 | | Oxypnictogenides | NaNO2 NaNO3 Na2N2O2 Na2N2O3 NaH2PO4 NaPO3 Na2HPO4 NaPO2H2 Na2HPO3 Na2PO3F Na3PS2O2 Na3PO4 Na5P3O10 Na4P2O7 Na2H2P2O7 Na3AsO3 Na3AsO4 Na2HAsO4 NaH2AsO4 NaSbO3 | | Metalates | NaAlH4 NaAlO2 Na3AlF6 NaAl(SO4)2 NaAuCl4 Na2TiF6 NaBiO3 NaCoO2 NaMnO4 NaReO4 NaTcO3 NaTcO4 NaVO3 Na2CrO4 Na2Cr2O7 Na2Cr3O10 Na2[Fe(CO)4] Na2MnO4 Na2MoO4 Na3IrCl6 Na2PtCl6 Na2TiO3 Na2U2O7 Na2WO4 Na2Zn(OH)4 Na3VO4 Na6V10O28 Na4Fe(CN)6 Na3Fe(CN)6 Na3Fe(C2O4)3 Na3[Co(NO2)6] Na2PdCl4 | | Others | NaSbF6 NaAsF6 NaBH4 NaBH3(CN) NaBO2 Na2B4O7 Na2B2O9 Na2B8O13 NaCN NaCNO NaH NaHCO3 Na4XeO6 NaHXeO4 NaOCN NaSCN Na2CO3 Na2C2O4 Na2C3S5 Na2GeO3 Na2He Na2xSi3yO2y+x (Na2SiO3 Na6Si2O7 Na4SiO4) Na2SiF6 NaNSi2(CH3)6 | |
| Organic | CH3ONa C2H5ONa HCOONa C2H5COONa C3H7COONa Na2C4H4O6 C4H5NaO6 NaCH3COO NaC6H5CO2 NaC6H4(OH)CO2 NaC12H23O2 NaC10H8 Na2[Fe[CN5]NO] C6H16AlNaO4 NaC6H7O6 C5H8NO4Na C6H5Na C4H9Na NaC5H5 C15H31COONa C17H33COONa C18H35O2Na C164H256O68S2Na2 |
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1 Background
2 Sample variance
3 Partition — analysis of variance
3.1 Sums of squared deviations
3.2 Example
3.3 Two-way analysis of variance
4 See also
5 References
Squared deviations from the mean
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Calculations in probability theory
Squared deviations from the mean (SDM) result from squaring deviations. In probability theory and statistics, the definition of variance is either the expected value of the SDM (when considering a theoretical distribution) or its average value (for actual experimental data). Computations for analysis of variance involve the partitioning of a sum of SDM.
Background
[edit]
An understanding of the computations involved is greatly enhanced by a study of the statistical value
: , where is the expected value operator.
For a random variable with mean and variance ,
:
(Its derivation is shown here.) Therefore,
From the above, the following can be derived:
Sample variance
[edit]
Main article: Sample variance
The sum of squared deviations needed to calculate sample variance (before deciding whether to divide by n or n − 1) is most easily calculated as
From the two derived expectations above the expected value of this sum is which implies
This effectively proves the use of the divisor n − 1 in the calculation of an unbiased sample estimate of σ2.
Partition — analysis of variance
[edit]
Main article: Partition of sums of squares
In the situation where data is available for k different treatment groups having size ni where i varies from 1 to k, then it is assumed that the expected mean of each group is
and the variance of each treatment group is unchanged from the population variance .
Under the Null Hypothesis that the treatments have no effect, then each of the will be zero.
It is now possible to calculate three sums of squares:
Individual
Treatments
Under the null hypothesis that the treatments cause no differences and all the are zero, the expectation simplifies to
Combination
Sums of squared deviations
[edit]
Under the null hypothesis, the difference of any pair of I, T, and C does not contain any dependency on , only .
: total squared deviations aka total sum of squares
: treatment squared deviations aka explained sum of squares
: residual squared deviations aka residual sum of squares
The constants (n − 1), (k − 1), and (n − k) are normally referred to as the number of degrees of freedom.
Example
[edit]
In a very simple example, 5 observations arise from two treatments. The first treatment gives three values 1, 2, and 3, and the second treatment gives two values 4, and 6.
Giving
: Total squared deviations = 66 − 51.2 = 14.8 with 4 degrees of freedom.
: Treatment squared deviations = 62 − 51.2 = 10.8 with 1 degree of freedom.
: Residual squared deviations = 66 − 62 = 4 with 3 degrees of freedom.
Two-way analysis of variance
[edit]
This section is an excerpt from Two-way analysis of variance.[edit]
In statistics, the two-way analysis of variance (ANOVA) is an extension of the one-way ANOVA that examines the influence of two different categorical independent variables on one continuous dependent variable. The two-way ANOVA not only aims at assessing the main effect of each independent variable but also if there is any interaction between them.
See also
[edit]
Absolute deviation
Algorithms for calculating variance
Errors and residuals
Least squares
Mean squared error
Residual sum of squares
Root mean square deviation
Variance decomposition of forecast errors
References
[edit]
^ Mood & Graybill: An introduction to the Theory of Statistics (McGraw Hill)
Retrieved from "
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Analysis of variance
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Mastering the Midpoint & Distance on the Number Line
In geometry, lines have starting and ending points. Between these points, we can find the exact middle, called the midpoint, and measure how long the line is, known as the distance. This guide will explain these two concepts in simple terms. So, if you're looking to understand how to find the middle of a line or how long it is, you're in the right place!
Step-by-step Guide: Distance and Midpoint on the Number Line
Understanding the Midpoint:
The midpoint between two numbers (or points) on a number line is the number that lies exactly halfway between them. The formula to find the midpoint 𝑀 M between two points 𝐴(𝑥 1)A(x 1) and 𝐵(𝑥 2)B(x 2) is:
𝑀=𝑥 1+𝑥 2 2 M=x 1+x 2 2
the Concept ofDistance:
The distance between two numbers (or points) on a number line is the absolute difference between them. The formula to find the distance 𝐷 D between two points 𝐴(𝑥 1)A(x 1) and 𝐵(𝑥 2)B(x 2) is:
𝐷=|𝑥 2–𝑥 1|D=|x 2–x 1|
Examples
Example 1: Midpoint
Find the midpoint between the numbers 4 4 and 10 10 on the number line.
Solution:
Using the midpoint formula:
𝑀=4+10 2 M=4+10 2
𝑀=14 2 M=14 2
𝑀=7 M=7
So, the midpoint is 7 7.
Example 2: Distance
Find the distance between the numbers −3−3 and 5 5 on the number line.
Solution:
Using the distance formula:
𝐷=|5–(−3)|D=|5–(−3)|
𝐷=|5+3|D=|5+3|
𝐷=8 D=8
So, the distance is 8 8 units.
Practice Questions:
Find the midpoint between the numbers 2 2 and 8 8 on the number line.
Calculate the distance between the numbers −7−7 and 3 3 on the number line.
If the midpoint between two numbers 𝑎 a and 𝑏 b is 5 5, what is the value of 𝑏 b given 𝑎=3 a=3?
Answers:
The midpoint is 5 5.
The distance is 10 10 units.
The value of 𝑏 b is 7 7.
by: Effortless Math Team about 2 years ago (category: Articles)
Effortless Math Team
3 weeks ago
Effortless Math Team
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What is frequency density?
How to calculate frequency density
Frequency density worksheet
Frequency density examples
Example 1: calculating frequency density Example 2: calculating frequency density Example 3: calculating frequency density Example 4: finding frequencies from the frequency density Example 5: finding the frequencies from the frequency density Example 6: finding the frequencies from the frequency density
Common misconceptions
Practice frequency density questions
Frequency density GCSE questions
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Introduction
What is frequency density?
How to calculate frequency density
Frequency density worksheet
Frequency density examples
↓
Example 1: calculating frequency density Example 2: calculating frequency density Example 3: calculating frequency density Example 4: finding frequencies from the frequency density Example 5: finding the frequencies from the frequency density Example 6: finding the frequencies from the frequency density
Common misconceptions
Practice frequency density questions
Frequency density GCSE questions
Next lessons
Still stuck?
GCSE Maths Statistics Histograms
Frequency Density
Frequency Density
Here we will learn about frequency density, including how to calculate it from a grouped frequency table.
There is also a histogramworksheet based on Edexcel, AQA and OCR exam questions, and further guidance on where to go next if you’re still stuck.
What is frequency density?
Frequency density is the frequency per unit for the data in each class. It is calculated by dividing the frequency by the class width ie the difference between the upper limit of the class interval and the lower limit of the class interval.
We use frequency density to plot histograms.
\text{Frequency density }=\frac{\text{Frequency}}{\text{Class width}}
Step-by-step guide: Frequency density formula (coming soon)
To work out frequency density we first need to identify the class width of the interval by subtracting the lower bound from the upper bound of the class interval.
Once we have identified the class width the frequency density can be calculated:
In a histogram:
the frequency density is the height of the bar
the frequency of a class interval is equal to the area of the bar
the total area of the bars is equal to the total frequency
What is frequency density?
How to calculate frequency density
In order to calculate frequency density:
Identify the upper and lower bounds of the class interval.
Find the class width of the class interval by finding the difference of the upper and lower bounds.
Divide the frequency of the class interval by its class width.
How to calculate frequency density
Histograms worksheet
Get your free histograms worksheet of 20+ questions and answers including frequency density and frequency density formula questions. With reasoning and applied questions.
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Histograms worksheet
Get your free histograms worksheet of 20+ questions and answers including frequency density and frequency density formula questions. With reasoning and applied questions.
DOWNLOAD FREE
Related lessons on histograms
Frequency density is part of our series of lessons to support revision on histograms. You may find it helpful to start with the main histograms lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:
Histogram
Frequency density formula
Frequency density examples
Example 1: calculating frequency density
The table shows information about the ages of people at a cinema.
Calculate the frequency densities.
Identify the upper and lower bounds of the class interval.
2Find the class width of the class interval by finding the difference of the upper and lower bounds.
3Divide the frequency of the class interval by its class width.
Example 2: calculating frequency density
The table shows information about the heights of pupils in a mathematics class.
Calculate the frequency densities.
Identify the upper and lower bounds of the class interval.
Find the class width of the class interval by finding the difference of the upper and lower bounds.
Divide the frequency of the class interval by its class width.
Example 3: calculating frequency density
The table shows information about the mass of fish in a lake.
Calculate the frequency densities.
Identify the upper and lower bounds of the class interval.
Find the class width of the class interval by finding the difference of the upper and lower bounds.
Divide the frequency of the class interval by its class width.
Example 4: finding frequencies from the frequency density
The table shows information about the heights of plants in a garden.
Calculate the missing frequencies and frequency densities.
Identify the upper and lower bounds of the class interval.
Find the class width of the class interval by finding the difference of the upper and lower bounds.
Divide the frequency of the class interval by its class width.
Here we have some missing frequencies as well as missing frequency densities.
We can calculate the frequency densities for the final two class intervals in the usual way.
To calculate the missing frequencies we need to work backwards.
\text{Frequency density} = \frac{\text{Frequency}}{\text{Class width}}
therefore
\text{Frequency} = \text{Frequency density} \times \text{Class width}
Multiply the frequency density by the class width to find the frequency.
Example 5: finding the frequencies from the frequency density
This table shows information about the ages of people playing bingo.
Calculate the missing frequencies and frequency densities.
Identify the upper and lower bounds of the class interval.
Find the class width of the class interval by finding the difference of the upper and lower bounds.
Divide the frequency of the class interval by its class width.
\text{Frequency density} = \frac{\text{Frequency}}{\text{Class width}}
therefore
\text{Frequency} = \text{Frequency density} \times \text{Class width}
Example 6: finding the frequencies from the frequency density
This table shows how far some children travel to school.
Work out the missing frequencies and frequency densities.
Identify the upper and lower bounds of the class interval.
Find the class width of the class interval by finding the difference of the upper and lower bounds.
Divide the frequency of the class interval by its class width.
\text{Frequency density} = \frac{\text{Frequency}}{\text{Class width}}
therefore
\text{Frequency} = \text{Frequency density} \times \text{Class width}
Common misconceptions
Midpoint of the group
A common error is to use the midpoint when calculating frequency densities rather than the class width. This is because the midpoint is used for estimating the mean from a frequency table and frequency polygons.Frequency density is the frequency per unit for the data in each class so it is important to use the width of the class in the calculation.
Practice frequency density questions
To calculate the frequency density we use the formula
\text{Frequency density }=\frac{\text{Class width}}{\text{Frequency}}
\text{Frequency density }=\frac{\text{Frequency}}{\text{Class width}}
\text{Frequency density }=\frac{\text{Frequency}}{\text{Midpoint}}
\text{Frequency density }=\frac{\text{Cumulative frequency}}{\text{Class width}}
Frequency density is the frequency per unit for the data in each class.
A class interval was given as 25 \leq x < 30. How would you calculate the class width?
25-30
30-25
(25+30) \div 2
(30-25) \div 2
Class width = upper bound – lower bound
A class interval given as 15 \leq x < 20. has a frequency density of 1.6. What is its frequency?
16
3.125
17.5
8
We need to multiply the frequency density by the class width.
1.6 \times 5=8
The table shows information about the heights of a group of children.
Which table shows the correct frequency densities?
\text{Frequency density }=\frac{\text{Frequency}}{\text{Class width}}
The table shows the frequency densities for the mass of some stones in a garden.
Which table shows the correct frequencies?
We need to multiply the frequency density by the class width.
This table shows information about the heights of trees in a wood.
Which table shows the correct missing values?
To find the missing frequencies, we need to multiply frequency density by class width.
To find the missing frequency densities we need to divide the frequency by the class width.
Frequency density GCSE questions
The table shows information about the heights of sunflowers in an allotment.
Complete the frequency density column.
(3 marks)
Show answer
Attempt to divide frequency by class width
(1)
Three correct
(1)
All correct
(1)
The table shows the frequency densities for the mass of sheep in a field.
Complete the frequency column.
(3 marks)
Show answer
Attempt to multiply frequency density by class width
(1)
Three correct
(1)
All correct
(1)
The table shows information about the ages of guests at a hotel.
Complete the missing values.
(4 marks)
Show answer
Attempt to multiply frequency density by class width
(1)
Attempt to divide frequency by class width
(1)
Three correct
(1)
All correct
(1)
Learning checklist
You have now learned how to:
");--ub-list-item-fa-li-top:3px;--ub-list-item-spacing:0px;">
");">
Construct and interpret diagrams for grouped discrete data and continuous data, i.e. histograms with equal and unequal class intervals and cumulative frequency graphs, and know their appropriate use
The next lessons are
Types of sampling methods
Cumulative frequency
Frequency table
Still stuck?
Prepare your KS4 students for maths GCSEs success with Third Space Learning. Weekly online one to one GCSE maths revision lessons delivered by expert maths tutors.
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Introduction
What is frequency density?
How to calculate frequency density
Frequency density worksheet
Frequency density examples
↓
Example 1: calculating frequency density Example 2: calculating frequency density Example 3: calculating frequency density Example 4: finding frequencies from the frequency density Example 5: finding the frequencies from the frequency density Example 6: finding the frequencies from the frequency density
Common misconceptions
Practice frequency density questions
Frequency density GCSE questions
Next lessons
Still stuck?
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Absolute value
The absolute value of a real number, denoted , is the unsigned portion of . Geometrically, is the distance between and zero on the real number line.
The absolute value function exists among other contexts as well, including complex numbers.
Contents
[hide]
1 Real numbers
2 Complex numbers
3 Examples
4 Problems
5 See Also
Real numbers
When is real, is defined as For all real numbers and , we have the following properties:
(Alternative definition)
(Non-negativity)
(Positive-definiteness)
(Multiplicativeness)
(Triangle Inequality)
(Symmetry)
Note that
and
Complex numbers
For complex numbers, the absolute value is defined as , where and are the real and imaginary parts of , respectively. It is equivalent to the distance between and the origin, and is usually called the complex modulus.
Note that , where is the complex conjugate of .
Examples
If , for some real number , then or .
If , for some real numbers , , then or , and therefore or .
Problems
Find all real values of if .
Find all real values of if .
(AMC 12 2000) If , where , then find .
See Also
Magnitude
Norm
Valuation
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Four Useful Centrifugal Pump Calculators for Pump Professionals - Empowering Pumps and Equipment
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Four Useful Centrifugal Pump Calculators for Pump Professionals
Home » Four Useful Centrifugal Pump Calculators for Pump Professionals
Energy EfficiencyNovember 25, 2017
Four Useful Centrifugal Pump Calculators for Pump Professionals
By: Cieana Detloff
If you are required to perform centrifugal pump calculations in your current job assignment, we think you are going to find this website useful. The site contains numerous on-line calculators. At this time, we want to introduce you to four handy, easy-to-use centrifugal pump calculators:
The Brake Horsepower calculator can be used to determine the required centrifugal pump brake horsepower (BHP) for a given flow rate (Q), head (H), specific gravity (S.G.), and efficiency (Eff). When you enter any four of these variables as inputs, the calculator will automatically determine the fifth variable. This is a handy calculator for sizing centrifugal pump driver requirements.
ThePump Efficiency with Electric Motor Driveris a unique calculator that can be used to determine the centrifugal pump efficiency (eff p) of a motor driven pump. If you know these five inputs:
Motor KW,
Discharge pressure (P d),
Suction pressure (P s)
Flowrate (Q), and
Electric motor efficiency (eff m).
the calculator will automatically determine the current pump efficiency. This is a handy calculator for evaluating a centrifugal pump’s hydraulic condition in the field.
The Specific Speed (US) calculator can be used to determine a centrifugal pump’s specific speed (N s) if you know the flow rate (Q), head (H), rotational speed (N). If you enter any three of these variables as inputs, the calculator will determine the fourth variable. A pumps specific speed is an indicator of the basic geometry of the impeller, i.e. low specific speed pumps tend to have large outside diameters and narrow, radial flow passages while high specific speed pumps tend to have small outside diameters and large, more axial flow passages. Pumps of the same specific speed can be scaled up or down for greater or less flow and head requirements.
The Suction Specific Speed calculator can be used to determine a centrifugal pump’s suction specific speed (N ss) for a given flow rate (Q), suction head required (NPSHr), rotational speed (N). If you enter any three of these variables as inputs, the calculator will determine the fourth variable. A pump’s suction specific speed is an indicator of an impellers flow stability at low flows.
Give these calculators a try and let us know what you think.
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Basic performance, cavitation and runaway investigations of a high specific speed Francis turbine with emphasis on cavitation analysis: A case study - ScienceDirect
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Highlights
Abstract
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1. Introduction
2. Flow system of the high specific speed model Francis turbine, research stand, devices and instrumentation used in experimental research
3. Determination of the basic quantities measured and determined in laboratory tests
4. Basic performance investigations of the model Francis turbine
5. Cavitation investigations of the model Francis turbine
6. Runaway investigations of the model Francis turbine
7. Summary and conclusions
CRediT authorship contribution statement
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Energy Conversion and Management
Volume 314, 15 August 2024, 118653
Research Paper
Basic performance, cavitation and runaway investigations of a high specific speed Francis turbine with emphasis on cavitation analysis: A case study
Author links open overlay panel Zbigniew Krzemianowski, Janusz Steller
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Highlights
•High specific speed Francis turbine as an alternative to low specific speed Kaplan/semi-Kaplan/Propeller turbine.
•Comprehensive experimental studies of the high specific speed Francis turbine.
•Multi-aspect experimental approach to determining the Thoma cavitation number.
•Comprehensive vortex visualization for various cavitation states in the draft tube.
•Saddles were found on the performance characteristics for high guide vane opening angles.
Abstract
The paper presents comprehensive experimental research on a high specific speed model Francis turbine with a characteristic runner diameter of 250 mm under the head of 12 m. Essentially, the turbine was designed for use at heads between 10 and 50 m, which are typical for installations with low specific speed Kaplan turbines. The basic performance, cavitation and runaway characteristics of the machine, which were determined as a result of the work, are quite rarely available in the literature in the case of high specific speed Francis turbines. The turbine achieved the maximum efficiency of 91%, kinematic specific speed of 82, and relatively wide operation range, so that it can be attractive as an alternative to the low specific speed Kaplan, semi-Kaplan or propeller turbines.
In result of the basic laboratory tests, the performance characteristics of the machine operation in a wide range of guide vane opening angles between 9° and 36° as well as the shell efficiency characteristics were determined. On the characteristics plotted vs. rotational speed, the saddle effect was identified for the guide vane opening angles with values greater than the optimal one (22°). During the operation of the machine in the area of saddles, the change in the flow rate took quite surprising patterns. The share of the turbine thrust torque in the total torque was also determined, the characteristics of which are also burdened with a similar saddle effect.
In the case of cavitation tests, a multi-aspect approach to cavitation diagnostics was used to characterize the cavitation phenomenon from the viewpoint of various criteria. Following this approach, the signals of the draft tube wall acceleration, acoustic emission at the draft tube cone upper flange, and pressure fluctuations inside the draft tube were used. The determined cavitation characteristics indicate little cavitation susceptibility of the Francis turbine blade system, despite the presented turbine having been designed to operate at higher flow rates than the classic high head Francis turbines with long interblade channels. The cavitating flow in the draft tube, directly below the runner, was also visualized during the cavitation tests. The paper presents the results of visualization for all cavitation states at five selected settings of the guide vane opening angle (two states in Partial Load, two states in High Load area and one state in the Best Efficiency Point), thanks to which it was possible to track the development of vortices during pressure reduction in the flow system.
The runaway coefficient was determined as a result of the runaway tests. The value of 1.497 was reached at the optimum guide vane opening angle.
Additionally, the work also presents in detail the efficiency scale effects to be expected at prototype turbines, i.e. those intended for operation at a higher head and with larger runner diameters. The efficiency of the machine with a runner diameter of 1.5 m and working at a head of 50 m reaches∼93%. The influence of water viscosity on the change in efficiency of the prototype turbine is also shown.
The presented methodology can be used for the proper selection of turbines intended for hydropower plants and for the proper installation of the runner in respect to the tailwater level in order to avoid the phenomenon of cavitation erosion.
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Keywords
High specific speed Francis turbine
Model hydraulic turbine tests
Draft tube cavitation analysis
Runaway characteristics
Cavitation visualisation
Scale effect
Nomenclature
_g_
Gravitational acceleration [m/s 2]
_n_
Rotational speed [rpm]
_n red_
Rotational speed reduced on reference head [rpm]
_n 1 I_
Double-reduced rotational speed [rpm]
_n sQ_
Kinematic specific speed [_not defined_]
_p_
Pressure [Pa]
_p p_
Water vapour pressure [Pa]
_p s_
Pressure at lower cross-section of turbine [Pa]
_t w_
Water temperature [˚C]
_A_
Area [m 2]
_D_
Characteristic runner diameter [m]
_H_
Head [m]
_H n_
Net head [m]
_H_ _dyn_
Dynamic head [m]
_H_ _stat_
Static head [m]
_H ref_
Reference head [m]
_P_
Power on the shaft [W]
_P r_
Shaft torque power with runner in the air [W]
_P w_
Windage power [W]
_P raw_
Raw power [W]
_P red_
Power reduced on reference head [W]
_Q_
Volumetric flow rate [m 3/s]
_Q red_
Volumetric flow rate reduced on reference head [m 3/s]
_Q 1 I_
Double-reduced volumetric flow rate [m 3/s]
_Re_
Reynolds number
_T_
Total torque [Nm]
_T f_
Mechanical friction torque [Nm]
_T r_
Shaft torque with runner in the air [Nm]
_T t_
Torque measured by means of torque meter [Nm]
_T w_
Windage torque [Nm]
_V in_
Velocity at turbine spiral inlet [m/s]
_V out_
Velocity at draft tube outlet [m/s]
_α GV_
Guide vane opening angle [°]
_δ_
Measurement uncertainty [%]
_η_
Efficiency [%][-]
_ρ_
Water density [kg/m 3]
_σ_
Cavitation number (Thoma number) [-]
_ω_
Angular velocity [rad/s]
_BEP_
Best Efficiency Point
_M_
Model
_P_
Prototype
_Δ_
Difference
_in_
Inlet
_out_
Outlet
_opt_
Optimal
_max_
Maximum
1. Introduction
Designing hydraulic turbines with high efficiency, exceeding 88% (taking into account the power on the shaft and without mechanical losses) is nowadays generally expected from the designers. In the case of low specific speed turbines, this is generally not a very difficult task due to the significantly smaller flows through the machine, associated with lower velocities, which causes less disturbance in operation due to less intense secondary flows, lower vibrations and pressure fluctuations in the blade system or in the draft tube. In the case of high specific speed turbines, it is more difficult, because relatively higher flow rates are required in the machine, which is associated with the acceleration of water in the blade channels and the draft tube. The above phenomena have a negative impact on the operation of the entire machine. The turbine requires a comprehensive analysis to assess the efficiency of its operation, not only in terms of basic performance parameters, but also in terms of cavitation, in order to prevent the occurrence of cavitation and, more importantly, the phenomenon of cavitation erosion, which gradually destroys the flow system of the machine.
The subject of the study under consideration is an experimental analysis of a high specific speed model Francis turbine (_n_ sQ=82), which is intended for installations using the lowest heads dedicated to Francis turbines. The scope of application of the presented machine is in the range of heads from about 8 m to about 50 m. In this range, the efficiency of such a machine should have very high peak efficiencies, significantly exceeding 90%, also as a model machine, i.e. without taking into account the scale effect. Lower specific speed Francis turbines are more efficient and their efficiency can reach over 93%, but their design differs from that of the high specific speed ones because they pass relatively much lower flows at lower rated speeds. By using the appropriate geometry of the machine’s flow system (mainly the runner), it is possible to reduce the size of a high specific speed turbine and thus to reduce installation costs. There is also the possibility of eliminating the gearbox, which further reduces the cost of the machine and installation, as well as the size of the generator (higher rated rotational speed). However, the most significant argument for using high specific speed Francis turbines is the possibility of avoiding the use of a Kaplan turbine whenever it leads to excessive costs of civil works. It is also related to significant financial savings resulting from the elimination of the runner blade adjustment system during the rotation of a Kaplan turbine runner. The above-mentioned arguments became the motivation to start the development of high specific speed Francis turbines at the Institute of Fluid-Flow Machinery (IFFM) of the Polish Academy of Sciences (in Polish: IMP PAN).
In general, it can be concluded from the literature that since Kaplan turbines came to a widespread use, little research studies have been conducted on high specific speed Francis turbines. On the other hand, there is a lot of works on low- and medium specific speed Francis turbines, intended for high and medium heads, respectively. If such a machine is installed on a low head, it should be considered that its blade system is not adequate for this head and therefore has a lower efficiency than could be achieved using a high specific speed machine. Their significant size makes them also uncompetitive in comparison to Kaplan turbines of similar performance parameters in terms of installation costs. The works on the design and analysis of high specific speed Francis turbines (this also concerns Francis pump-turbines) that can be found in the literature are relatively new. It can be concluded that the design and analysis of high specific speed Francis turbines is a relatively young research field, which also explains why it is worthy of interest. High specific speed turbines are becoming more and more attractive due to the fact that lower and lower water dammings are being considered for development, which requires higher specific speeds of the machines. On the other hand, low specific speed Francis turbines are often installed on low-head dammings, which is not relevant at the current state of technology, as low specific speed turbines require more space and have lower efficiency parameters than the high specific speed ones. Therefore, undertaking work on high specific speed Francis turbines becomes justified from the economic point of view.
From the literature review of high specific speed Francis turbines a few papers can be quoted. For example, Enomoto et al. presented a high specific speed Francis turbine with a newly designed and optimized runner shape, thanks to which it was possible to increase efficiency compared to the old solution. Nakamura and Kurosawa using Multi-Objective Genetic Algorithm, and Kawajiri et al. using Particle Swarm Optimization also designed new runners, thanks to which the efficiencies of their machines were increased as well. In turn, Lee et al. , as a result of the use of CFD tools, optimized the high specific speed prototype Francis turbine, thanks to which they achieved high efficiency and stable operation of the machine. Kubo et al. used CFD to rehabilitate high specific Francis turbine (spiral case, whole blade system and draft tube) to increase the efficiency of the machine, also achieving noise and cavitation risk reduction.
The limitation of the applicability of any turbine, including Francis turbines, is the occurrence of the cavitation phenomenon, which can lead to cavitation erosion in the machine and other unwanted effects, which in turn may pose a significant threat to machine structural integrity due to the adverse effect of the resulting fluctuations in the flow. Eventually, erosive damage may lead to the loss of original flow system geometry and result in the degradation of performance parameters , . Cavitation diagnostics in model tests has been known for a very long time , . Laboratory tests of Francis turbines allow for a deep analysis of the cavitation risk. This results in the proper recommendation of the runner elevation in respect to the tailwater level, and thus safe operation of the machine under field conditions , . Cavitation tests are so important because the impact of cavitation on the hydrounit is complex and has a negative impact on turbine elements, such as the runner (both the blades, the hub and the shroud) and the draft tube. The level of cavitation phenomenon impact depends on the flow velocity and the load on the runner blades , as well as on parameters such as atmospheric pressure, suction head, temperature, gas content in the liquid and turbine operating time . In addition, cavitation tests are a source of knowledge about the flow behaviour under various machine load conditions (especially in partial load conditions). During operation of the machine, downstream the runner, especially under the hub, a flow field with a large pressure reduction is created. This generates undesirable flows in the draft tube in the form of vortex ropes flowing mainly from the runner hub or vortices flowing from the runner blades (e.g. interblade vortex is typically observed in Part Load machine operation) , , , , . This all leads to dangerous fluctuations in the draft tube , , and the phenomenon of cavitation surge, which induces large values of pressure fluctuations , , . Additionally, it leads to a significant loss in static pressure recovery and influences the overall machine efficiency . Summarising, the negative impact of cavitation may lead to resonances in the structure and its destruction through the formation of cavitation ropes that permanently affect the walls of the draft tube, which may cause its systematic destruction. To prevent this, the modern trend is to add breaker fins to the draft tube to prevent the development of a cavitation rope and its negative impact on the flow . Nevertheless, based on the literature review, one can get the impression that there is a lack of comprehensive studies of high specific speed Francis turbines.
This paper presents a comprehensive experimental analysis of a model high specific speed Francis turbine carried out at a head of 12 m, with a kinematic specific speed n sQ of 82, which is defined according to the following formula:(1)n sQ=n Q 0.5 H 0.75 where: _n_ – the rotation speed [rpm], _Q_ – the optimal volumetric flow rate [m 3/s], and _H_ – the head[m].
Achieving this value allows Francis turbines to compete with low specific speed Kaplan turbines, i.e. machines with double-adjustable blades (guide vanes and runner blades), both in terms of maximum efficiency and, more importantly, a wide range of high efficiency characteristics. In general, Kaplan turbines are more expensive, so the approach of using less expensive Francis turbines instead seems reasonable.
The turbine has been designed to work at the lowest heads recommended for Francis turbines (in such a case, it is extremely important to choose the right meridional shape of the runner). The preliminary runner design was made using a quasi three-dimensional method implemented in the ADT/TurboDesign™ software, based on the Inverse Problem Method solution. The generated shape of the runner blade was used to perform the blade shape optimization process using the ANSYS/DesignXplorer™ environment with the use of the ANSYS/CFX™ computational solver. The runner (its meridional cross-section) was divided into 6 subdomains using 7 axisymmetric stream surfaces. At each intersection of the meridional blade view and the flow surface, 5 points in the streamline direction were determined (excluding the point on the leading edge). In this way, 35 (7x5) free points were defined and their position in the circumferential direction was allowed to change. The allowable range of blade angle changes was±15°. Further, the created runner geometry was meshed using ANSYS/TurboGrid™. Releasing only 35 points resulted in a large number (over 660) of runners with different geometry of blade shapes. Due to the large number of calculation cases, the computations were performed for one runner channel and one guide vane channel. The simplification of the computational domain forced setting the periodic boundary conditions in domain, which had to have some influence on the computed flow field distribution, since the guide vanes and runner were positioned in specific circumferential location relative to the spiral casing. In this case, secondary flows may not be sufficiently modelled. However, it should be kept in mind that the process of optimization is carried out in the area of optimal performance. This means that the impact of undesirable phenomena is minimized and the influence on operational parameters can be considered that is small. At the end, from a large number of runner blade shapes, the final shape was determined using an optimization procedure based on Multi-Objective Genetic Algorithm (MOGA).
Therefore, it should be emphasized that the flow system with the runner designed in the above-mentioned way is a relatively modern one and consistent with the current state of knowledge, which was confirmed by the conducted experimental research. From this point of view, the results of experimental research are a novelty and it was difficult to find similar comprehensive results in the literature. The analysis included basic performance tests, cavitation tests and runaway tests, which are presented in detail in the paper. The focus was put on the analysis of the cavitation phenomenon in the draft tube under the runner. Various cavitation states were achieved by lowering the pressure in the entire hydraulic loop. Thanks to this, it was possible to determine the efficiency characteristics depending on the cavitation number (Thoma number). To improve the quality of the results cavitation tests were performed using several methods. In particular, the acceleration of the draft tube wall, the acoustic emission at the draft tube cone upper flange, and pressure fluctuations inside the tube were monitored. Cavitation development was observed visually and documented photographically via the Plexiglass draft tube cone wall. The detailed results for several selected guide vane opening angles are presented in this paper. Additionally, a method of estimating the measurement uncertainty as well as the friction torque is presented. In the case of the friction torque, the method of measuring and calculating the torque components and the dependence of the total torque on the double-reduced rotational speed and for constant guide vane opening angles of the turbine are also presented.
It should be added that the paper is a continuation of the IMP PAN studies on the comprehensive design and analysis of high efficiency Francis turbines with high or very high kinematic specific speeds, exceeding the value of 80. The first results were presented in the reference . The design of the 11-blade runner (this paper concerns a Francis turbine with a completely different 13-blade runner) with a very high kinematic specific speed (94) was shown there, designed on the basis of the Inverse Problem Method, using the method based on a two-dimensional axisymmetric model developed in a curvilinear system. As a result, a completely different runner shape was obtained, even if all other flow system components (spiral case, 13 stationary guide vanes, 20 guide vanes and axisymmetric conical draft tube) remained the same. The change of the runner resulted in a complete change in the parameters of the machine, with a significantly different specific speed.
2. Flow system of the high specific speed model Francis turbine, research stand, devices and instrumentation used in experimental research
For the model high specific speed Francis turbine, a 13-blade runner with a characteristic diameter of _D_=250 mm was designed for laboratory testing under head of 12 m. The runner design method will not be discussed in this paper (a brief description is provided in Introduction), as the design details and cavitation analysis of its operation will be the subject of a separate work. Other elements of the flow system, such as the inlet pipeline, the spiral case, the stationary and adjustable guide vanes and the draft tube were designed on the basis of the literature analysis and taking into account the conditions for the installation of the entire turbine at the laboratory test stand. Table. 1 presents the main geometric and rated flow parameters describing the high specific speed model Francis turbine. The two-dimensional meridional section and three-dimensional elements of the flow system (spiral, stationary and adjustable vanes, runner and draft tube) are shown in Fig. 1.
Table 1. Geometrical and rated flow parameters of the model Francis turbine.
| Empty Cell | Value |
| Number of stationary guide vanes (_N_ SGV) | 13 |
| Number of adjustable guide vanes (_N_ GV) | 20 |
| Number of runner blades (_N_ R) | 13 |
| Diameter of the inlet pipeline | 393 mm |
| Diameter of the draft tube outlet | 567 mm |
| Draft tube (diffuser) angle | 10° |
| Draft tube length | ∼1000 mm |
| Optimal guide vane opening | 22° |
| Rated volumetric flow rate | 0.2091 m 3/s |
| Rated rotational speed | 1155 rpm |
| Rated head | 12 m |
| Rated power | 22.36 kW |
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Fig. 1. Meridional geometry and 3D flow domains (A – spiral case, B – stationary and adjustable guide vanes, C – runner, and D – draft tube) of the high specific speed model Francis turbine.
Basic performance, cavitation and runaway tests of the model Francis turbine were carried out in the laboratory of the IMP PAN Gdańsk (Poland). The hydraulic turbine test stand can be used to test Kaplan and Francis turbines , , , with the following maximum parameters: flow rate _Q_ max = ∼0.45 m 3/s and head _H_ max = ∼30 m.
The water circulation loop at the stand is shown in Fig. 2, while the installation of the model Francis turbine is shown in Fig. 3. The water circulation can be briefly described as follows. The water flow is forced by two centrifugal circulation pumps P1 and P2, pumping water to the reservoir 2. Next, water is supplied to the bottom part of the upper water reservoir 3 via two pipelines. Electromagnetic flowmeters are located on these pipelines. The upper reservoir is divided into two parts by an internal vertical wall. After flowing into the first part of the reservoir, the water is dammed up to the height of the wall. Then, it pours into the second part, which allows permanent water degassing. In this part, in the middle of the height of the reservoir, there is an outlet stub. A knife gate valve is connected to the stub pipe, behind which a straight axial pipeline with a diameter of 400 mm is installed, supplying water to the model turbine. Directly in front of the turbine, the pipeline deviates by an angle of 7° and connects to the turbine’s spiral case (Fig. 3 – bottom view). After passing the blade system the water flows into the draft tube and then into the lower water reservoir 1. From the lower reservoir, water flows directly to the circulating pumps through two pipelines and the water circuit closes. The stand is equipped with devices for lowering the pressure under the runner (increasing the suction height), which also allows, as already mentioned, to conduct cavitation tests. It is also possible to pressurize the airbag in the upper reservoir.
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Fig. 2. A simplified schematic of the stand for model hydraulic turbine tests (IMP PAN, Gdańsk, Poland). (Main markings: 1 – low pressure reservoir; 2 – collector of water from circulation pumps; 3 – high pressure reservoir; 4 – model turbine; P1,P2 – circulation pumps; P3,P4 – vacuum pumps; P5 – water drainage pump; Zz1,Zz2,Zz3,Zz4 – gate valves; Zm1,Zm2 – butterfly valve; Q1,Q2 – electromagnetic flowmeters; Zr1 – regulating valve; S1,S2 – compressors; Zp – water draining valve).
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Fig. 3. The model Francis turbine at the hydraulic turbine test stand - a simplified assembly drawing.
A view of the model Francis hydraulic unit installed on the experimental stand is shown in Fig. 4 (top left view). The figure shows also the system for visualizing the flow at the runner blading outlet and in the draft tube directly below the runner (top right view). The lower pictures show the inlet to the spiral case (bottom left view), the guide vane adjustment apparatus (bottom middle view) and the outlet from the runner (bottom right view).
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Fig. 4. Top left view: the model Francis turbine mounted on the hydraulic turbine test stand; top right view: turbine prepared to PIV and flow visualisation; bottom left view: inlet to the spiral; bottom middle view: guide vane adjustment apparatus; bottom right view: runner outlet.
The instruments and devices used to measure relevant performance parameters were as follows:
•
Head. The turbine head was determined using pressure difference (Δ _p_) between control cross-sections upstream and downstream of the turbine, measured by means of a Deltabar S intelligent differential pressure transducer with an Endress+Hauser DelTabar S PMD75-2RJ4/0 valve manifold (Fig. 5-B). The pressure was tapped by means of collective manifolds, each of which was connected to four holes drilled circumferentially, every 90° in the X pattern. The high pressure cross-section was located before the inlet to the spiral (Ø393) – (Fig. 5-C). The low pressure cross-section was located at the outlet of the draft tube (Ø567) – (Fig. 5-A). Measurement accuracy: 0.05%.
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Fig. 5. Sensors and devices used during the model Francis turbine tests.
•Torque. The Hottinger-Baldwin-Messtechnik (HBM) torque meter T22 with a range of 500 Nm was used to measure the shaft torque (Fig. 5-D). Measurement accuracy: 0.5%.
•Flow rate. The flow rate was measured using two electromagnetic Siemens flowmeters (Fig. 5-E) installed at DN200 and DN350 pipelines. Type of flowmeters: Sitrans FM MAG 5100 W+MAG 6000 W. Measurement accuracy: 0.25%.
•Rotational speed. The rotational speed was measured using the Sick DFS-60 encoder (Fig. 5-F) with a resolution of 1024 pulses per revolution. Measurement accuracy: 0.2%.
•Opening angle of the guide vane apparatus. The Introl PFK-75 rotary code converter was used to measure the opening angle of the guide vane apparatus (Fig. 5-D). The parameters of the device were as follows: number of graduations 4096 and angular resolution 5.3′. Measurement accuracy 0.5%. In addition, a linear displacement transducer Peltron Piz-100 (Fig. 5-G) was used.
•Pressure measurement. TPS-05A pressure transducers by Arex were used to measure the pressure signal in the low-pressure manifold (Fig. 5-B). The pressure transducer installed on the low-pressure manifold was used to record the pressure (_p_ s), which was used to calculate the cavitation number (_σ_). Measurement accuracy: 0.2%.
•Generator. A separately excited DC generator G16.08S by MultiMoto (Poznań, Poland) was used to produce electricity (Fig. 5-F). This is a device with a power of 42 kW and rated rotational speed of 1500 rpm.
•Electric power reception. The power from the generator was received by a water-cooled electrolytic resistor DFME (Wrocław, Poland) (Fig. 5-H) with the rated power of 150 kW.
3. Determination of the basic quantities measured and determined in laboratory tests
3.1. Definitions of energy parameters
Below are the definitions of the quantities used in the model tests.
The net head _H_ n was calculated as the sum of the static and the dynamic heads. The level difference between the measurement sections (_z_ in − _z_ out) was omitted because when measuring static pressure using a differential pressure transducer it is zero.(2)H n=Δ p ρ g⏟H stat+V in 2-V out 2 2 g⏟H dyn+z in-z out⏟0 where: Δ _p_ – the static pressure difference between the high- and low-pressure measuring cross-sections measured using a pressure difference transducer, _ρ_ – the water density depending on its temperature (the water temperature was measured using a meter located in the upper reservoir), _g_ – the gravitational acceleration (9.8145 m/s 2), _V_ in – the average flow velocity in the high pressure cross-section, _V_ out – the average flow velocity in the low pressure cross-section.
The velocities were calculated based on the measured flow rate _Q_ and areas of the high- (_A_ in) and low-pressure (_A_ out) measurement cross-sections:(3a)V in=Q A in=Q π D in 2 4(3b)V out=Q A out=Q π D out 2 4 where: _D in_ – the inlet diameter to the spiral (0.393 m), _D out_ – the outlet diameter of the conical draft tube (0.567 m).
The power on the shaft _P_ (the product of the total torque _T_ and the angular velocity _ω_) and the raw power _P raw_ (the power carried by the inflowing water) were determined using the following formulas, respectively:(4)P=T ω(5)P raw=ρ g Q red H ref where: _H ref_ – the reference net head (12 m), _Q red_ – the flow rate reduced on the reference net head (defined below).
The water evaporation pressure _p_ v is a quantity necessary to determine the cavitation number _σ_. During cavitation tests, it was determined depending on the current water temperature based on the formula taken from the ISO 13788:2012 standard , which is defined as follows:(6)p v=A e C t w B+t w where: _t w_ – the water temperature [˚C], _A_, _B_, _C_ – the constants: _A_=610.5 Pa, _B_=237.3˚C, _C_=17.269.
The rotational speed, the volumetric flow rate, and the power reduced on the reference head _n_ red (_H_ ref=12 m) were determined as follows:(7)n red=n H n H ref 0.5(8)Q red=Q H n H ref 0.5(9)P red=P H n H ref 1.5
The double reduced rotational speed _n I I_ (rotational speed of the turbine operating at a head of 1 m and having a runner diameter of 1 m) and the double reduced volumetric flow rate _Q I I_ (flow rate of the turbine operating at a head of 1 m and having a runner diameter of 1 m) were determined as follows:(10)n I I=nD H n 0.5(11)Q I I=Q H n 0.5 D 2
where: _D_ – the characteristic runner diameter (0.25 m).
The cavitation number _σ_ used in cavitation tests was calculated from the following formula:(12)σ=p s-p v ρ g+V out 2 2 g-H z H n
where: _p s_ – the absolute pressure in the low-pressure cross-section (manifold pressure) [Pa], _p v_ – the water evaporation pressure [Pa], _V out_ – the flow velocity in the low-pressure cross-section [m/s], _H n_ – the net head[m], _H z_=1.17 m – the height between the spiral axis and the level of the pressure sensor in the low-pressure cross-section (Fig. 1).
Turbine efficiency _η_ was calculated from the following formula:(13)η=P red P raw
where: _P red_ – the reduced power [W], _P raw_ – the raw power [W].
3.2. Determination of the total torque
The total torque _T_ was calculated as follows:(14)T=T t+T f=T t+T r-T w⏟T f
where: _T t_ – the torque measured on the turbine shaft using a torque meter [Nm], _T f_ – the friction torque [Nm]. The _T_ f torque is the difference between the torque measured during the runner operation in the air _T_ r and the theoretically estimated windage torque of the runner wading in the air _T_ w.
The methods of determining all the listed torques are shown below.
•Shaft torque with runner in the air
The torque _T r_ was measured using a torque meter during the operation of the turbine runner in the air (generator run in the motor mode). The generator was set in motor motion using the Ward-Leonard system, and the tests were carried out in 10 measurement series (5 series from the lowest rotational speed to the highest one and 5 series vice versa). In each of these series, the rotational speed was changed in the range from∼400 to∼1700 rpm and vice versa, every 100 rpm. The test results were obtained in form of the shaft torque dependence on the rotational speed. As a result, the average representative series of shaft torque was determined, which is presented in Table.2. The values of the average series range from 2.08 Nm at the lowest speed of 406 rpm to 4.92 Nm at the highest speed of 1712 rpm. On its basis, a trend line was determined, which, after analysing the course of points, was described by a linear equation. This equation was used further on to estimate the shaft torque depending on the turbine rotational speed. All measurement series along with the average series and the trend line (and its equation) are shown in Fig. 6.
•Windage torque
Table 2. Determination of the shaft torque _T r_ of the model Francis turbine runner in the range of rotational speed changes from∼400 to∼1700 rpm – average series parameters.
| _No._ | _n_ | _n s_ | _ω_ | _T r_ | _P r_ |
--- --- --- |
| rpm | rps | rad/s | Nm | W |
| 1 | 406.6 | 6.777 | 42.579 | 2.08 | 88.5 |
| 2 | 500.4 | 8.340 | 52.402 | 2.35 | 122.9 |
| 3 | 600.3 | 10.004 | 62.859 | 2.65 | 166.4 |
| 4 | 699.0 | 11.651 | 73.203 | 3.00 | 219.4 |
| 5 | 799.7 | 13.329 | 83.746 | 3.11 | 260.1 |
| 6 | 899.8 | 14.997 | 94.227 | 3.26 | 307.3 |
| 7 | 998.1 | 16.634 | 104.516 | 3.48 | 363.9 |
| 8 | 1099.8 | 18.330 | 115.169 | 3.71 | 427.7 |
| 9 | 1196.2 | 19.936 | 125.262 | 3.85 | 482.8 |
| 10 | 1298.6 | 21.643 | 135.988 | 4.05 | 550.3 |
| 11 | 1397.0 | 23.283 | 146.294 | 4.25 | 621.4 |
| 12 | 1500.0 | 25.001 | 157.084 | 4.47 | 701.5 |
| 13 | 1602.4 | 26.706 | 167.800 | 4.68 | 785.8 |
| 14 | 1712.7 | 28.544 | 179.348 | 4.92 | 881.6 |
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Fig. 6. The equation of the trend line, all measurement series and the average series determined to obtain the dependence of the resistance torque _T r_ on the rotational speed in the range between∼400 and∼1700 rpm, acting on the model Francis turbine runner rotating in the air.
The windage torque _T w_ is the torque caused by aerodynamic forces acting on the runner rotating in the air. The existence of this torque increases the measured shaft torque necessary to keep the runner rotating in the air. Its value results from estimating the power of winding loss. The method of determining this power was taken, for the case of the Francis turbine, from the EC 60041:1994 standard. The windage power is described by the following formula:(15)P w=4.6 60 n s 3 D outer 4 B 10-7=100 n s 3 B D outer 4
where: _n s_ – the rotational speed [rps], _B_ – the runner height at the inlet [m], _D outer_ – the outer runner diameter[m].
The height of the runner inlet _B_ and the outer diameter of the runner _D outer_ are determined according to the scheme taken from the above-mentioned standard, as shown in Fig. 7.
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Fig. 7. Scheme for determining the characteristic dimensions of a model Francis runner required to determine the windage power P w.
Based on the height of the runner inlet, as defined in Fig. 7, was _B_=0.085 m. The outer diameter of the runner was determined as the arithmetic mean of the diameters located at the points on the inlet line at the runner hub and the shroud, equal to: 0.2 m and 0.258 m, respectively, which gives the final runner diameter of _D outer_=0.229 m.
The cited standard states that the width of the uncertainty band of formula (15) is±50%. Therefore, the calculated winding power _P w_ should be treated as a very rough estimation. Fortunately, the value of this power is negligible in comparison with the total torque. The winding torque _T w_ can be determined from the formula:(16)T w=P w ω
where: _ω_ – angular velocity [rad/s],(17)ω=2 π n 60=π n 30
where: _n_ – rotational speed [rpm].
In addition, the standard limits the application of the formula to the case when the rate of leakage through the labyrinth seals is less than 0.8·1e-3 of the flow rate at the operating point with the highest efficiency (with the uncertainty band of the given value also being±50%). Since it is known from the basic performance tests that the flow rate at the BEP was∼0.209 m 3/s, then 0.8·1e-3 of this value was calculated to be∼0.00017 m 3/s. As a result of visual observation, it was assumed that the leakage flow rate through the labyrinth seals did not exceed the given value. On this basis, the winding torque was then estimated from the given formula in all measurement series of the basic performance, cavitation and runaway tests.
In Table 3 and Fig. 8, the range of changes in the winding torque in the rotational speed range between 400 and 1800 rpm, every 100 rpm, is presented.
•Relative friction torque
Table 3. Determination of the windage torque _T w_ of the model Francis turbine runner in the range of changes in the rotational speed between 400 and 1800 rpm.
| _No._ | _n_ | _n s_ | _ω_ | _T w_ | _P w_ |
--- --- --- |
| − | rpm | rps | rad/s | Nm | W |
| 1 | 400 | 6.667 | 41.888 | 0.17 | 6.9 |
| 2 | 500 | 8.333 | 52.360 | 0.26 | 13.5 |
| 3 | 600 | 10.000 | 62.832 | 0.37 | 23.4 |
| 4 | 700 | 11.667 | 73.304 | 0.51 | 37.1 |
| 5 | 800 | 13.333 | 83.776 | 0.66 | 55.4 |
| 6 | 900 | 15.000 | 94.248 | 0.84 | 78.9 |
| 7 | 1000 | 16.667 | 104.720 | 1.03 | 108.2 |
| 8 | 1100 | 18.333 | 115.192 | 1.25 | 144.0 |
| 9 | 1200 | 20.000 | 125.664 | 1.49 | 187.0 |
| 10 | 1300 | 21.667 | 136.136 | 1.75 | 237.8 |
| 11 | 1400 | 23.333 | 146.608 | 2.03 | 297.0 |
| 12 | 1500 | 25.000 | 157.080 | 2.33 | 365.2 |
| 13 | 1600 | 26.667 | 167.552 | 2.65 | 443.3 |
| 14 | 1700 | 28.333 | 178.024 | 2.99 | 531.7 |
| 15 | 1800 | 30.000 | 188.496 | 3.35 | 631.1 |
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Fig. 8. Dependence of the windage torque _T w_ of the model Francis runner in the range of rotational speed variation between 400 and 1800 rpm.
Fig. 9 shows the dependence of the relative total friction torque (_T f_/_T_) on the double-reduced rotational speed _n_ 1 1. The highest relative values of the total friction torque, rising from 1.75% to 3.83% in the direction of increasing _n_ 1 1 values, have been reached at the lowest guide vane opening angle of 9°. They decreased with the increase of the guide vane opening angle. The lowest total friction torque values ranged from 0.8% to 1.4% with increasing speed _n_ 1 1 in case of the highest guide vane opening angle of 36°. However, it can be seen that further opening of the guide vanes does not reduce the total friction torque (asymptotic trend), so it is no longer possible to reduce it.
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Fig. 9. Results of basic tests of the model Francis turbine – relative total friction torque curves _T f_ depending on the double-reduced rotational speed _n_ I I for constant guide vane opening angles.
3.3. Determination of measurement uncertainty
The systematic uncertainty of the efficiency measurement _δ Syst_ was estimated on the basis of the relative uncertainties of measurement of differential pressure (expressed as a head _H_) _δ H_, torque _δ T_, flow rate _δ Q_, rotational speed _δ n_, and uncertainty related to the measurement data acquisition system _δ K_. The systematic uncertainty was calculated according to the formula:(18)δ Syst=δ H 2+δ T 2+δ Q 2+δ n 2+δ K 2=0.05 2+0.5 2+0.25 2+0.2 2+0.1 2=±0.61%
The measurement uncertainty associated with the data acquisition system was estimated on the basis of the absolute accuracy specified by a manufacturer of the measurement card, related to the part of the measurement range used, and taking into account the accuracy of precision resistors used to convert current signals into voltage. Based on the performed estimations, it was assumed that it is not worse than±0.1%. Finally, based on the adopted values of individual components, the systematic uncertainty of the measurement results of the efficiency of the tested model turbine was not worse than±0.61%.
To determine the random uncertainty, the dependence of the efficiency on the double-reduced rotational speed _n_ I I was used. For this purpose, it was necessary to select the curve characterized by the worst approximation of the approximated points. The efficiency curve _η_=_η_(_n_ I I) was selected for the guide vane opening angle of 10.5° (see Fig. 10). The exact method is shown below.
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Fig. 10. Results of basic tests of the Francis model turbine – efficiency curves _η_ depending on the double-reduced rotational speed _n_ I I for constant guide vane opening angles.
The random uncertainty at the 95% confidence level is determined using the Student's _t_-distribution:(19)δ η R a n d=±t S y n
where the _S_ y estimator of the standard deviation is calculated as follows:(20)S y=∑y aproks-y pomiar 2 n-m-1=∑Δ 2 n-m-1
and where: _n_ – the number of measurement points, _t_ – the parameter of the Student's _t_-distribution, _m_ – the degree of the approximating polynomial.
For the selected _η_=_η_(_n_ I I) curve, the results of fitting the _m_-th degree curve to the efficiency measurement points are shown in Table 4. The characteristic values required for the Student's _t_-distribution are as follows: _t_=2.263, _n_=10, _m_=5 (the efficiency function for a guide vane opening angle 10.5° is a 5-degree polynomial), _S_ y=0.5045721. The _t_-value is calculated using the formula:(21)t=1.96+2.36 ν+3.2 ν 2+5.2 ν 3.84 where: _ν_=_n_-1 (_ν_=9).
Table 4. Determination of the random uncertainty of a model Francis turbine (the curve _η_=_η_ (_n_ I I) was used for a guide vane opening angle of 10.5°).
| No. | Double reduced rotational speed | Measured efficiency | Approximated efficiency | Efficiency difference | Δ 2 |
--- --- --- |
| _n 1 1_ | _η measured_ | _η approx_ | Δ _=η approx-η measured_ |
| [-] | [rpm] | [%] | [%] | [-] | [-] |
| 1 | 46.8 | 71.75 | 71.63 | −0.12 | 0.015 |
| 2 | 50.8 | 74.84 | 75.19 | 0.35 | 0.125 |
| 3 | 55.3 | 77.63 | 77.46 | −0.17 | 0.029 |
| 4 | 60.2 | 79.60 | 79.17 | −0.42 | 0.180 |
| 5 | 64.4 | 80.14 | 80.44 | 0.30 | 0.089 |
| 6 | 70.1 | 81.16 | 81.61 | 0.45 | 0.198 |
| 7 | 74.8 | 81.79 | 81.41 | −0.39 | 0.149 |
| 8 | 79.8 | 79.38 | 79.09 | −0.29 | 0.083 |
| 9 | 83.8 | 74.75 | 75.13 | 0.38 | 0.143 |
| 10 | 89.2 | 66.41 | 66.32 | −0.09 | 0.007 |
| Total: | 0.00 | 1.018 |
Finally, it was found that the random uncertainty δ R a n d for the presented results was not worse than±0.36% (Eq.(19)). This value was considered the highest random uncertainty for the basic performance tests of a model Francis turbine.
The total uncertainty _δ_ was determined from the equation:(22)δ=δ Syst 2+δ Rand 2=0.61 2+0.36 2≈±0.71%
Hence, the total uncertainty was assumed to be not worse than ∼±0.71%.
4. Basic performance investigations of the model Francis turbine
4.1. Description of experimental investigations
Basic performance laboratory tests were carried out in accordance with the applicable standard for testing model hydraulic turbines . They were conducted at the reference net head _H n_=12 m, for 15 settings of the guide vane opening angle _α_ GV, namely: 9°, 10.5°, 12°, 13.5°, 15°, 16.5°, 18°, 20°, 22°, 24°, 25.5°, 27°, 30°, 33°, and 36°, resulting in 15 performance characteristics of the model turbine, which were developed based on test results at 195 measurement points. At a given setting of the guide vane, the tests were carried out by changing the rotational speed of the turbine from the lowest to the highest values. The adjustment of the net head was carried out by changing the rotational speed of the circulation pumps, each motor of which, in turn, was regulated using a frequency converter. The range of rotational speed changes was in the range (depending on the guide vane opening angle) from∼630 to∼1650 rpm, every∼50 rpm. At each point, the measurement lasted about 3 min (with 2-minute stabilization of the operating state). The head was maintained within the±1% band around the reference head.
4.2. Results of experimental investigations
The results of basic performance tests are presented in Fig. 10 and Fig. 11. These are efficiency characteristics depending on the double-reduced rotational speed _n I I_ and flow rate _Q I I_, respectively. Fig. 13 also shows the relationship between double-reduced rotational speed and flow rate.
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Fig. 11. Results of basic tests of the Francis model turbine – efficiency curves _η_ depending on the double-reduced flow rate _Q_ I I for constant guide vane opening angles.
The parameters of the Best Efficiency Point (BEP) achieved by the model Francis turbine are as follows:
•turbine efficiency: 91.0%,
•kinematic specific speed: _n sQ_=82 (Eq.(1)),
•double-reduced rotational speed: _n I I_=83 rpm,
•double-reduced volumetric flow rate: _Q I I_=0.9659 m 3/s,
•guide vane opening angle: _α_ GV=22°,
•rotational speed: _n_=1155 rpm,
•volumetric flow rate: _Q_=0.2091 m 3/s,
•torque: _T_=186 Nm,
•power: _P_=22.36 kW.
In the range of the guide vane opening angles from 9° to 25.5°, the efficiency curves plotted vs. _n I I_ show a convex course (Fig. 10). However, this changes with the further increase of guide vane opening. In the range between the opening angle of 27° up to the largest opening angle of 36°, saddles are formed in the middle of the curves. The largest saddle is located on the _η_(_n I I_) characteristics at the guide vane opening angle of 30°. Analysing the efficiency curves depending on Q _I I_ (Fig. 11), from the guide vane opening angle 9° to 25.5°, the flow rate decreases with the increase of rotational speed _n I I_. However, starting from the opening angle of 27° (for the angle of 25.5°, the beginnings of changes are already visible), up to the largest opening angle of 36°, there is a clear change in the shape of the _η_(_Q I I_) curves at the saddles.
Fig. 12 shows a detailed course of changes in the flow rate curves for constant angles of guide vane opening. In these cases, the flow rate increases – Fig. 12 (detail A). The most likely reason seems to lie in the reordering of the flow separation zones and other recirculation flow structures with the change of angle of flow incidence on the runner blade profiles. The above mentioned phenomena show direct impact on the increase in flow instability and the corresponding pressure fluctuations in the draft tube. This phenomenon is well known in Kaplan turbines, but still little research has been done for Francis turbines with high specific speeds, i.e. such as the presented model Francis turbine. The figure of the flow rate dependence on the rotational speed (Fig. 13) confirms the existence of saddles on the curves for high guide vane opening angles. Again, intense recirculation in the inter-balade passages seems to be the most probable reason.
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Fig. 12. The course directions of efficiency curves _η_ depending on the double-reduced flow rate _Q_ I I for increasing values of the double-reduced rotational speed _n_ I I.
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Fig. 13. Results of basic tests of the Francis model turbine – the flow rate _Q_ I I depending on the double-reduced rotational speed _n_ I I for constant guide vane opening angles.
On the basis of the performance characteristics, the universal characteristics shown in Fig. 14 was also determined, i.e. the dependence of efficiency on the double-reduced values _n I I_ and _Q I I_ (the figure also includes curves of constant openings of the guide vanes _α_ GV). The upper right portion of the efficiency hill shows a ridge extending towards higher speeds and discharges. The efficiency characteristics along this area is the flattest and the turbine operation is the most stable. Model tests have also confirmed the low noise level of flowing water in this part of efficiency hill. A Francis turbine operating at a high load is characterized by better operating dynamics than in case of partial load operation. Therefore, this is the area recommended for the selection of turbines for an on-site facility. This is also consistent with the conclusion that was noticed during the work on the high specific speed Kaplan turbine, the results of which are shown in , i.e. the turbine designed for the lowest heads (up to 3 m) for Kaplan turbines. High specific speed turbines (Kaplan and Francis) operate very smoothly at high loads, sometimes even better than in their BEPs, although it is worth mentioning the risk of allowable cavitation number exceeding the installation value in a result of too high overload (see Section 5). The presented turbine is also recommended for the lowest heads, which are appropriate for installations in the case of Francis turbines, i.e. heads from about 8 to 50 m. Under these heads, the presented turbine should reach the best possible efficiency and ensure minimization of the machine structure size. The use of Francis turbines with lower specific speeds means that a turbine of larger size must be used for the same discharge.
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Fig. 14. Shell characteristics of the model Francis turbine (the dependence of the efficiency _η_ on the double-reduced rotational speed _n_ I I and the double-reduced flow rate _Q_ I I).
4.3. Impact of scale effect on a prototype Francis turbine efficiency
The procedure for scaling the energy parameters of the model Francis turbine to an industrial facility (prototype) according to is presented below. The estimation of the efficiency gain of a prototype machine is based on the measured maximum efficiency and flow conditions (Reynolds number) of the model Francis turbine. The general formula for the change in efficiency _Δη_, taking into account the scale effect can be written as follows:(23)Δ η=δ ref Re ref R e M 0.16-Re ref R e P 0.16
where: _Re ref_ – the reference Reynolds number, (_Re_)_M_ – the Reynolds number of a model turbine [-], (_Re_)_P_–the Reynolds number of a prototype turbine [-], _δ ref_ – the relative losses scaled for the reference Reynolds number[-].
The relative losses _δ ref_ are expressed as follows:(24)δ ref=1-η max M Re ref R e M 0.16+1-V ref V ref
where: _V ref_ – the loss distribution coefficient (for Francis turbines: _V ref_=0.7), η max M=0.91 – the maximum turbine efficiency obtained from model tests [-].
The reference Reynolds number _Re ref_ has a fixed value equal to Re ref=7 e 6 _._ In turn, the Reynolds number of the turbine model (_Re_)_M_ is given by the following formula:(25)R e M=D ν 2 g H M
where: _D_=0.25 m – the characteristic diameter of the model runner, _H_=12 m – the turbine head in model tests, _ν_=0.946 1e-6 m 2/s – the kinematic water viscosity (for the average circulating water temperature in basic performance tests: 23°C), _g_=9.8145 m/s 2 – the gravitational acceleration for the city of Gdańsk (Poland).
The Reynolds number (_Re_)_P_ of the turbine prototype is defined by an analogous formula to that used in the case of Reynolds number for a model:(26)R e P=D ν 2 g H P
where: _D_ – the characteristic diameter of the prototype runner, _H_ – the head under which the prototype turbine operates, _ν_=1.306·1e-6 m 2/s – the kinematic viscosity of water flowing through the prototype turbine (for the assumed average water temperature: 10°C).
The volumetric flow rate in the BEP of the prototype turbine _Q opt_ was determined using the formula:(27)Q o p t P=Q I I M D 2 H 0.5 P
where: Q I I M=0.9659 m 3/s – the double-reduced volumetric flow rate in the BEP of the model turbine.
The rotational speed of the prototype turbine n P was determined using the formula:(28)n P=n I I M H 0.5 D P
where: n I I M=83 rpm – the double-reduced rotational speed in the BEP of the model turbine.
The power at the optimal operating point N opt P of the prototype turbine was determined using the formula:(29)N opt P=η max ρ g H Q opt P
where: _η max_ – the maximum efficiency of the prototype turbine [-], _ρ_ – the density of water for the assumed working conditions of the prototype [kg/m 3], _g_ – the acceleration of gravity [m/s 2], _H_ – the head [m], _Q opt_ – the volumetric flow rate in the BEP of the prototype [m 3/s].
For calculations according to the formulas presented above, the following 6 values of heads were assumed, which are highly realistic values for high specific speed applications of Francis turbines: H P=12,15,20,30,40,50 m, and the following 10 values of characteristic runner diameters: D P=0.25,0.3,0.4,0.5,0.6,0.7,0.8,1.0,1.2,1.5 m.
Calculation results for the above-mentioned heads and diameters of the prototype turbine are shown in Table 5.
Table 5. Performance parameters of the prototype Francis turbines series based on model tests taking into account the scale effect.
| Head | Reynolds number of the prototype turbine | Max. efficiency increase of the prototype turbine | Max. efficiency of the prototype turbine | The volumetric flow rate of the prototype turbine at the Best Efficiency Point | The power of the prototype turbine at the Best Efficiency Point | Rotational speed of the prototype turbine |
--- --- ---
| _H_ | (Re)P | (Δ _η_)P | (_η_)P | (_Q opt_)_P_ | (_P opt_)P | (_n_)P |
| m | − | % | % | m 3/s | kW | rpm |
| 1.Runner diameter _D_=0.25 _m_ |
| 12 | 2,937,897 | −0.34 | 90.66 | 0.2091 | 22.3 | 1150.1 |
| 15 | 3,284,669 | −0.22 | 90.78 | 0.2338 | 31.2 | 1285.8 |
| 20 | 3,792,809 | −0.07 | 90.93 | 0.2700 | 48.2 | 1484.7 |
| 30 | 4,645,223 | 0.14 | 91.14 | 0.3307 | 88.7 | 1818.4 |
| 40 | 5,363,841 | 0.28 | 91.28 | 0.3818 | 136.8 | 2099.8 |
| 50 | 5,996,957 | 0.39 | 91.39 | 0.4269 | 191.4 | 2347.6 |
| 2.Runner diameter _D_=0.3 _m_ |
| 12 | 3,525,476 | −0.15 | 90.85 | 0.3011 | 32.2 | 958.4 |
| 15 | 3,941,602 | −0.03 | 90.97 | 0.3367 | 45.1 | 1071.5 |
| 20 | 4,551,370 | 0.12 | 91.12 | 0.3888 | 69.5 | 1237.3 |
| 30 | 5,574,267 | 0.32 | 91.32 | 0.4761 | 128.0 | 1515.4 |
| 40 | 6,436,609 | 0.46 | 91.46 | 0.5498 | 197.3 | 1749.8 |
| 50 | 7,196,348 | 0.57 | 91.57 | 0.6147 | 276.1 | 1956.3 |
| 3.Runner diameter _D_=0.4 _m_ |
| 12 | 4,700,635 | 0.15 | 91.15 | 0.5354 | 57.5 | 718.8 |
| 15 | 5,255,470 | 0.26 | 91.26 | 0.5985 | 80.4 | 803.6 |
| 20 | 6,068,494 | 0.40 | 91.40 | 0.6911 | 124.0 | 928.0 |
| 30 | 7,432,356 | 0.60 | 91.60 | 0.8465 | 228.2 | 1136.5 |
| 40 | 8,582,146 | 0.73 | 91.73 | 0.9774 | 351.9 | 1312.3 |
| 50 | 9,595,131 | 0.83 | 91.83 | 1.0928 | 492.3 | 1467.2 |
| 4.Runner diameter _D_=0.5 _m_ |
| 12 | 5,875,794 | 0.37 | 91.37 | 0.8365 | 90.0 | 575.0 |
| 15 | 6,569,337 | 0.48 | 91.48 | 0.9352 | 125.9 | 642.9 |
| 20 | 7,585,617 | 0.62 | 91.62 | 1.0799 | 194.1 | 742.4 |
| 30 | 9,290,446 | 0.80 | 91.80 | 1.3226 | 357.4 | 909.2 |
| 40 | 10,727,682 | 0.93 | 91.93 | 1.5272 | 551.0 | 1049.9 |
| 50 | 11,993,914 | 1.03 | 92.03 | 1.7075 | 770.9 | 1173.8 |
| 5.Runner diameter _D_=0.6 _m_ |
| 12 | 7,050,952 | 0.55 | 91.55 | 1.2046 | 129.8 | 479.2 |
| 15 | 7,883,204 | 0.65 | 91.65 | 1.3467 | 181.7 | 535.8 |
| 20 | 9,102,740 | 0.78 | 91.78 | 1.5551 | 280.1 | 618.6 |
| 30 | 11,148,535 | 0.97 | 91.97 | 1.9046 | 515.6 | 757.7 |
| 40 | 12,873,219 | 1.09 | 92.09 | 2.1992 | 794.8 | 874.9 |
| 50 | 14,392,696 | 1.19 | 92.19 | 2.4588 | 1112.0 | 978.2 |
| 6.Runner diameter _D_=0.7 _m_ |
| 12 | 8,226,111 | 0.69 | 91.69 | 1.6395 | 177.0 | 410.7 |
| 15 | 9,197,072 | 0.79 | 91.79 | 1.8330 | 247.6 | 459.2 |
| 20 | 10,619,864 | 0.92 | 91.92 | 2.1166 | 381.8 | 530.3 |
| 30 | 13,006,624 | 1.10 | 92.10 | 2.5923 | 702.8 | 649.4 |
| 40 | 15,018,755 | 1.22 | 92.22 | 2.9934 | 1083.4 | 749.9 |
| 50 | 16,791,479 | 1.31 | 92.31 | 3.3467 | 1515.6 | 838.4 |
| 7.Runner diameter _D_=0.8 _m_ |
| 12 | 9,401,270 | 0.81 | 91.81 | 2.1414 | 231.5 | 359.4 |
| 15 | 10,510,939 | 0.91 | 91.91 | 2.3942 | 323.9 | 401.8 |
| 20 | 12,136,987 | 1.04 | 92.04 | 2.7646 | 499.3 | 464.0 |
| 30 | 14,864,713 | 1.21 | 92.21 | 3.3859 | 919.0 | 568.3 |
| 40 | 17,164,292 | 1.33 | 92.33 | 3.9097 | 1416.7 | 656.2 |
| 50 | 19,190,262 | 1.42 | 92.42 | 4.3712 | 1981.9 | 733.6 |
| 8.Runner diameter _D_=1.0 _m_ |
| 12 | 11,751,587 | 1.01 | 92.01 | 3.3460 | 362.5 | 287.5 |
| 15 | 13,138,674 | 1.11 | 92.11 | 3.7409 | 507.1 | 321.5 |
| 20 | 15,171,234 | 1.23 | 92.23 | 4.3196 | 781.8 | 371.2 |
| 30 | 18,580,891 | 1.40 | 92.40 | 5.2905 | 1438.8 | 454.6 |
| 40 | 21,455,365 | 1.51 | 92.51 | 6.1089 | 2218.0 | 524.9 |
| 50 | 23,987,827 | 1.60 | 92.60 | 6.8299 | 3102.7 | 586.9 |
| 9.Runner diameter _D_=1.2 _m_ |
| 12 | 14,101,905 | 1.17 | 92.17 | 4.8182 | 522.9 | 239.6 |
| 15 | 15,766,409 | 1.26 | 92.26 | 5.3869 | 731.5 | 267.9 |
| 20 | 18,205,481 | 1.38 | 92.38 | 6.2203 | 1127.6 | 309.3 |
| 30 | 22,297,069 | 1.54 | 92.54 | 7.6183 | 2075.2 | 378.8 |
| 40 | 25,746,438 | 1.65 | 92.65 | 8.7968 | 3198.8 | 437.4 |
| 50 | 28,785,393 | 1.74 | 92.74 | 9.8351 | 4474.6 | 489.1 |
| 10.Runner diameter _D_=1.5 _m_ |
| 12 | 17,627,381 | 1.35 | 92.35 | 7.5284 | 818.6 | 191.7 |
| 15 | 19,708,011 | 1.44 | 92.44 | 8.4171 | 1145.2 | 214.3 |
| 20 | 22,756,851 | 1.56 | 92.56 | 9.7192 | 1765.3 | 247.5 |
| 30 | 27,871,337 | 1.71 | 92.71 | 11.9035 | 3248.5 | 303.1 |
| 40 | 32,183,047 | 1.82 | 92.82 | 13.7450 | 5007.2 | 350.0 |
| 50 | 35,981,741 | 1.90 | 92.90 | 15.3674 | 7004.0 | 391.3 |
As expected, the highest efficiency increase of 1.9% has been stated in the case of a prototype with the largest runner diameter (1.5 m) and head (50 m). The maximum absolute efficiency of the prototype in this case would be 92.9%. The predicted power would be then over 7 MW with a flow rate of∼15.4 m 3/s. Considering that the presented model turbine is a high specific speed turbine, it should be stated that this efficiency value is very high. Modern Francis turbines achieve higher efficiencies. However, this applies to turbines with a much lower kinematic specific speed, showing values of 40, 50 and 60 (according to Eq.(1)), which are assumed to be more efficient turbines due to lower flow rates in the flow system. It has been proven that the presented high specific speed Francis turbine is not able to effectively compete with turbines with _n_ sQ=40, 50, and 60. This principle has been proven by a statistical analysis (see the chart prepared by the American Society of Mechanical Engineers ) – Fig. 15. Statistically, the best-designed Francis turbines have the specific speed of 60. Hence, for the specific speed of the presented machine equal to 82, the relative efficiency loss will be greater than 0.7%. Therefore, it can be concluded that the obtained efficiency parameters from the tests of the presented turbine are very promising in terms of efficiency and cavitation (the cavitation tests conducted show that the turbine is characterized by relatively high resistance to the cavitation phenomenon and, consequently, to cavitation erosion).
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Fig. 15. Dependence of relative efficiency on the kinematic specific speed for various types of hydro turbines .
It is worth emphasizing that in the case where the head of the prototype turbine is equal to the head of the model turbine H M=H P=12 m and the characteristic diameter of the prototype turbine is equal to the diameter of the model turbine D P=D M=0.25 m, the calculations show that the increase in the maximum efficiency of the prototype turbine is negative and amounts to −0.34% – Fig. 16 (top view − red arrow). This may seem quite surprising and, at first glance, illogical. However, the analysis shows that the reason for this is the assumed different viscosities used in the model tests and assumed for the prototype. If one assumes the same average viscosity (as used in the model tests), then the efficiency increase is 0% and the efficiency obtained by the prototype turbine is 91% – Fig. 16 (bottom view − red arrow). This value of the absolute increase in efficiency as a result of assuming water temperature in the field increased to that under lab conditions seems to be quite significant. It also applies to the maximum efficiencies for the remaining cases presented. This indicates an important aspect, which is the influence of viscosity on the efficiency results achieved by a turbine installed in a field facility, which are not trivial when the parameters of the installed turbine are required by an investor. A detailed analysis of the impact of viscosity shows that the presented case is extreme in terms of efficiency increase, and for larger diameters of the prototype turbine from 0.25 m and a head of 12 m, the efficiency increase as a result of the change in viscosity will not be higher.
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Fig. 16. Scale effect − efficiency increase depending on the change in head and runner diameter − top diagram: for water viscosity in the prototype turbine higher than water viscosity in the model turbine resulting from the temperature difference of 13°C (23°-10°); bottom diagram: for equal water viscosities in the prototype and model turbine resulting from the same temperatures.
5. Cavitation investigations of the model Francis turbine
5.1. Methodology
Typically, the primary purpose of hydraulic machinery cavitation tests conducted under laboratory conditions is to determine the allowable cavitation number and check cavitation performance of the machine within the so called controlled cavitation band of operational parameters. In the case of large units with hydraulic machines designed for installation conditions specified by the client, the key task is to check machine performance under these conditions including establishing the safety margin and the controlled cavitation band allowed conditionally (within a rather small time limit) for commercial operation . Deep analysis of cavitation development in this area is generally a topic of numerous scientific studies.
The outcome point of any hydraulic turbine cavitation test procedure is establishing the classic cavitation characteristics representing efficiency curves plotted vs. decreasing cavitation number _σ_ and featured by an abrupt breakdown close to some critical _σ_ value. The mentioned critical value can be established with reasonable precision if the efficiency drop by a certain amount, e.g. 1 or 0.5%, is taken as a criterion. However, as the efficiency curve breakdown takes place generally at highly advanced cavitation development stage, the inception of cavitation impact on the efficiency curve is often used as an auxiliary, although imprecise, criterion as well. Furthermore, visual observation of the runner suction side by means of a stroboscope lamp is always recommended during laboratory tests in order to get better insight into the cavitation phenomena taking place downstream the runner. Records in form of photos and/or video clips are essential for proper correlation and interpretation of performance and diagnostic measurement data acquired during cavitation tests.
The abovementioned diagnostic measurements are of fundamental significance for monitoring potential threat to the structural integrity of the hydraulic machine due to the cavitation vortex rope in the draft tube and erosive load occurring mainly within the runner. While dynamic loads of the draft tube are rather easy to follow by monitoring vibration acceleration and/or pressure fluctuation at the draft tube wall, the assessment of erosive loads to the blading system is much more difficult. The most frequently used approach is based on multi-sensor acoustic emission (AE) techniques, developed step-by-step since the early 90-ies, e.g. , , , . In the authors’ practice the essential criterion used to determine the critical cavitation number referring to a selected diagnostic signal is an abrupt rise in its relevant descriptor value, e.g. peak-peak value, which is not followed by any valley and repeated rise with falling _σ_ value. Only single sensor AE techniques have been used so far by the IMP PAN team.
In this context, it is necessary to emphasise that recording several signals at the same operation point is of fundamental significance for the proper interpretation of the results, and in particular to ensure satisfactory reliability of final conclusions. The main reason lies in high vulnerability of diagnostic signals to external disturbances and the change of sensor positioning. The last remark refers especially to the acoustic emission signal.
Cavitation laboratory tests of the turbine under consideration , were carried out at the reference net head _H n_=12 m (the same as in the case of basic performance tests), for the following 11 settings of the guide vane opening angle _α_ GV: 9°, 12°, 15°, 18°, 20°, 22°, 24°, 27°, 30°, 33°, and 36°. The head was maintained within the range of±1% of the reference head.
For all basic performance characteristics (_η_=_f_(_n I I_)), characterized by a constant guide vane opening angle (Fig. 10), 5 constant values of the double-reduced rotational speed _n I I_ were selected, for which cavitation curves were subsequently determined. Firstly, the _n I I_ value corresponding to the BEP was selected, i.e.: _n I I_=83 rpm. Then, 5 rpm and 11 rpm were subtracted from or added to this value. Hence, the following values of the double-reduced rotational speed were obtai01 ned: _n I I_=72, 78, 83, 88, and 94 rpm (shown in Fig. 10). This corresponds to the absolute rotational speeds of the Francis runner, respectively: 997.7, 1080.8, 1150.1, 1219.4, and 1302.5 rpm. The number of points measured during cavitation tests was 436, located on 51 measurement characteristics. The settings of _n I I_ were maintained within±1 rpm. The cavitation number _σ_ at the starting points of the cavitation characteristics (_η_=_f_(_σ_)) was set to a value higher than 1 (_σ_>1), which was determined after basic performance tests. This procedure ensured starting each measurement series in non-cavitation conditions. Subsequently, by lowering the pressure in the flow system, this number was reduced in order to gradually induce cavitation in the turbine. As in the case of basic performance tests, the measurement at each operation point lasted about 3 min, with the stabilization of operating conditions preceding it about 2 min.
Video recording of cavitation in the draft tube cone was carried out using a high-speed camera from Dantec – Fig. 17.
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Fig. 17. High-speed camera used to visualize the flow in the draft tube.
The following diagnostic signals were recorded:
1.housing acceleration signal of a) the guide bearing and b) the draft tube;
2.pressure pulsation signal directly under the turbine runner;
3.acoustic emission signal of the draft tube flange.
Data acquisition and recording in the DDF format were conducted using dedicated measurement sets with swift A/D converters and DASYLAB® software. DASYLAB software was used also to conduct most of the digital signal processing, including filtering in time domain as well as determining peak-to-peak and rms values. Filtering in the time domain was conducted using the 8-th order Butterworth filter. As explained in Section 5.3 only the acceleration signals have been used eventually for detailed analysis and determining the allowable cavitation number value.
The acceleration signals of the draft tube wall and the turbine guide bearing housing were delivered via signal conditioning and data acquisition system from the PCB Piezotronics 357B03 piezoelectric accelerometers. The accelerometers were mounted on the conical draft tube wall about 30 cm below the flange connecting the metal part with the part made of Plexiglass (Fig. 18 – green arrow), and at the guide bearing housing (Fig. 1).
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Fig. 18. Mounting locations of the accelerometer on the draft tube wall (green arrow), acoustic emission sensor (red arrow) and piezoelectric pressure transducer ICP under the turbine runner (blue arrow). (For interpretation of the references to colour in this figure legend, the reader is referred to the web version of this article.)
The acoustic emission signal was provided by the Kistler 8152A broadband sensor. The sensor was mounted at the upper flange of the transparent draft tube cone segment whereas the adjacent runner chamber flange was used for a pressure tap (Fig. 18 – red arrow). A 113A22 PCB Piezotronics ICP pressure transducer was connected to the drilled tap conduit via a short impulse tube (Fig. 18 – blue arrow).
5.2. Results of experimental investigations
As mentioned, cavitation curves were determined for all guide vane opening angles using five above-mentioned values of the double-reduced rotational speed _n I I_. Due to the limited length of this paper, the results of cavitation tests will be presented only for 5 selected guide vane opening angles, namely: 9°, 15°, 22° (optimal opening), 27°, and 36°. In the case of the smallest opening angle 9°, the double-reduced rotational speeds were limited to 72 and 78 rpm (the remaining values are outside the curve – see Fig. 10). The selection of these values was dictated by the intention to conduct cavitation tests of the turbine operation, in which its efficiency was not worse than 70%. Another reason for limiting the scope of research was the need to avoid operating machine in areas where, due to very low efficiency, there is a risk of machine damage due to its dynamic condition (vibrations and pressure pulsations).
At operating points far from optimal conditions, a vortex rope was observed quite early. The breakdown of the efficiency curve was usually accompanied by acoustic effects, which resembled the sound of strong fabric flapping in the wind. As the cavitation number was lowered further, the noise became an intense but steady hum. In this state, the observed flow space was already completely filled with a vapour-gas mixture. When working in some non-typical conditions, i.e. highly closed guide vanes and relatively low rotational speeds, the strong acoustic effects were accompanied by a temporary increase in efficiency – even by 2 to 3%.
Below is a detailed description of the individual operating states of the machine, including performance and visual aspects. Non-cavitation states in the diagrams below are marked with "0". They correspond to the flow conditions during basic performance tests, which results are shown in Fig. 10, Fig. 11, and Fig. 13.
•Guide vane opening angle 9°
The results of the cavitation tests and the visualization of the flow at consecutive cavitation numbers are shown in Fig. 19. The Francis turbine at this opening was operating under a deep partial load conditions.
•_n I I_=72 rpm. The visualization of the initial operating state (state no. 0) did not show any signs of cavitating swirl in the flow under the runner (_σ_=1.02), despite relatively low efficiencies of∼78.4%. The same happened in states no. 1 (_σ_=0.68), no. 2 (_σ_=0.415), and no. 3 (_σ_=0.255). In state no. 4 (_σ_=0.2), bubbles appeared in the flow throughout the entire visualized volume. Efficiency increased by 0.4%. In subsequent state no. 5, the appearance of one large vortex flowing from the middle part of runner hub and blades (with the half runner outlet diameter – 0.5 _D_) was observed. In states no. 6 and no. 7, the vortex increased its diameter, reaching the full runner diameter. The structural nature of the vortex appears to be very chaotic, i.e. not having a constant or monotonically varying diameter along its length, but having a disorderly varying thickness. In state no. 7, the highest efficiency of the machine was recorded, which reached the value of 79.76%, with the cavitation number _σ_=0.09. The increase in efficiency related to the initial state without cavitation (no. 0) was∼1.3%. From state no. 7, there was a rapid loss of efficiency – states no. 8 and no. 9. Pictures in these states indicate the disappearance of the above-mentioned vortex and the appearance of cavitation in the entire flow volume. In the last state (no.9), the cavitation number was measured with the value _σ_ = ∼0.04 with the turbine efficiency of 74.84%.
•_n I I_=78 rpm. The initial operating condition (state no. 0) did not show, similarly to the previous case, any symptoms of cavitation in the flow (_σ_=1.14), despite a much lower efficiency of 74.79%. No cavitation signs could be noticed visually also in the subsequent states no. 1 (_σ_=0.76), no. 2 (_σ_=0.43), and no. 3 (_σ_=0.26). From state no. 4 (_σ_=0.205) to state no. 7 (_σ_=0.108), the formation of thin vortex structures without an ordered character could be clearly observed. Between states no. 3 and no. 6 (_σ_=0.135) there was a slight increase in machine efficiency∼0.3%. After state no. 6, a sharp increase in efficiency was observed, up to state no. 7, in which the efficiency reached 77.19%. Compared to the initial state no. 0, the increase in efficiency was 2.4%. Since state no. 7 up to the last state no. 11, there was a rapid loss of efficiency of about 3%, with the cavitation number of 0.04. The flow in these states was characterized by cavitation in the entire flow volume.
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Fig. 19. Results of cavitation tests of the Francis model turbine – efficiency characteristics depending on the cavitation number and visualization of the flow in the draft tube (under the runner) for the guide vane opening angle 9°.
The efficiency rise shortly before the efficiency curve breakdown is extremely high in both cases discussed.The essential reason can be seen in a far going pressure redistribution along the runner blade profiles, which in a 2D approach results in the lift coefficient rise, a phenomenon well known from numerous experimenal investigations conducted in the past, e.g. , .
•Guide vane opening angle 15°
The results of the cavitation tests and the visualization of the flow at lowering cavitation numbers are shown in Fig. 20. The Francis turbine is also operating at partial load at this opening.
•_n I I_=72 rpm. Despite the extreme value of _n I I_, the efficiency of points on the curve shows relatively high values (in state no. 0 the efficiency was 87.33%). Visualization of the initial operating state (no. 0) did not show any symptoms of cavitation (_σ_=1.06). This also happened in states no. 1 (_σ_=0.705) and no. 2 (_σ_=0.408). Only in states no. 3 (_σ_=0.29) and no. 4 (_σ_=0.23), the formation of disordered vortices in the form of threads, stochastically descending from the runner was observed. In states no. 5 (_σ_=0.195) and no. 6 (_σ_=0.13) the vortices clearly strengthened their structure by slightly increasing their diameter and moved deeper into the draft tube. However, they still retained their thread-like character. In states no. 7 (_σ_ = ∼0.105) and no. 8 (_σ_ = ∼0.08), one main thread vortex was clearly formed coming down from the runner hub, which expanded its circumferential diameter of the rotation quite rapidly to the value of about 0.5 _D_ of the runner at a distance of about 1 _D_ from the runner, after that the vortex disappeared. Between states no. 4 and no. 8, a slight increase in efficiency of about 0.4% was observed. In state no. 8, the highest efficiency was 87.8% and was higher than the initial efficiency by 1.1%. This is the only curve for the guide vane opening angle 15° that has a local increase in efficiency (the other curves did not show this feature). From state no. 8, a rapid decrease in efficiency was observed, up to the value of 83.93% (state no. 10), with the cavitation number _σ_ = ∼0.022. The flow in states no. 9 and no. 10 showed volumetric cavitation.
•_n I I_=78 rpm. This is the rotational speed at which the highest efficiency was obtained in the conditions of no cavitation in the flow (in state no. 0 it was 87.7%, with the cavitation number _σ_=1.044). Flow visualization in states no. 0 and no. 1 (_σ=_ 0.72) did not indicate the presence of cavitation structures. State no.2 (_σ_=0.412) shows the beginnings of the vortices formation with a thread-like structure, moving disorderly to the draft tube, but without a clear circumferential movement. In the following states no. 3 (_σ=_ 0.25), no. 4 (_σ=_ 0.19), and no. 5 (_σ=_ 0.14), one can see the strengthening of the vortices that rotate in the flow volume and penetrate each other. There was a clear increase in the diameter of the vortices in state no. 5 (_σ_ = ∼0.14, _η_=87.66%), which tend to merge and form a cavitation cloud. From state no. 5 there was a loss of efficiency up to state no. 8 (_σ_ = ∼0.04, _η_=83.13%). In states no. 6 and no. 7, the structure of a vortex with a very large diameter, peripherally extending into the draft tube was still visible. In state no. 8, the flow already showed cavitation cloud throughout the entire flow volume.
•_n I I_=83 rpm. In the initial states no. 0 (_σ_=1.01, _η_=86.37%) and no. 1 (_σ_=0.71), no traces of vortices coming down from the runner were observed. Only starting from state no. 2 (_σ_ = ∼0.41) one can see the beginnings of the formation of a peripheral vortex coming off the runner hub and rotating around the entire circumference of the draft tube. In contrary to the previous values of _n I I_, in this case the vortex is characterized by a clearly large internal diameter, which increases as the cavitation number decreases to states no. 3 (_σ_=0.25), no. 4 (_σ_=0.18), and no. 5 (_σ_=0.13). In state no. 5 the beginning of the loss of machine efficiency occurred from the value of 86.81% to the value of 82.53%, in the last state no. 7 (_σ_=0.055). In state no. 7, the flow was characterised by cavitation cloud throughout the entire flow volume.
•_n I I_=88 rpm. In the initial states no. 0 (_σ_=1.01, _η_=85.18%), 1 (_σ_=0.71) and no. 2 (_σ_=0.42), no traces of cavitation bubbles were observed. Starting from state no. 3 (_σ_ = ∼0.25), a cloud of bubbles was observed throughout the entire flow volume, which intensified as the cavitation number decreased. It is worth noting that no clear vortex structures characteristic for states with lower _n I I_ values were developed in the flow. From state no. 4 (_σ_ = ∼0.18) there was a clear decrease in efficiency from 85.15% to 80.66% in the last measured state no. 6 (_σ_ = ∼0.06). In state no. 6, the flow was characterised by cavitation cloud throughout the entire flow volume.
•_n I I_=94 rpm. This is the curve with clearly the lowest efficiency of∼80%. Despite this, no traces of cavitation bubbles were observed in initial states no. 0 (_σ_=1.03, _η_=80.08%), no. 1 (_σ_=0.705), and no.2 (_σ_=0.41). It is only starting from state no. 3 (_σ_=0.3) that the bubble vortex structures of small diameter appear in the entire flow volume. As the cavitation number decreases, the bubbles increase in diameter, causing the formation of cavitation fog in the entire flow volume. From state no. 5 (_σ_ = ∼0.19) a clear decrease in efficiency was observed up to state no. 7 (_σ_ = ∼0.072, _η_=76.14%). In state no. 7, as previously, the flow was characterised by a cavitation cloud in the entire volume of the flow.
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Fig. 20. Results of cavitation tests of the Francis model turbine – efficiency characteristics depending on the cavitation number and visualization of the flow in the draft tube (under the runner) for the guide vane opening angle 15°.
In conclusion, it is worth emphasizing that an increase in the double-reduced rotational speed (and the related absolute rotational speed) causes the rapid loss of machine efficiency, due to more intense swirl beneath the runner and the flow is more prone to develop cavitating vortex ropes at a faster rate.
•Guide vane opening angle 22°
The results of the cavitation tests and the visualization of the flow in individual states depending on the cavitation number for the opening angle of 22° are shown in Fig. 21. This is the optimal opening of the guide angle of the model Francis turbine.
•_n I I_=72 rpm. Visualization of the initial state of operation no. 0 (_σ_ = ∼1.02, _η_=87.82%) and in the next state no. 1 (_σ_=0.715) did not show any cavitating vortex-like structures. From state no. 2 (_σ_ = ∼0.505) a fairly weak thread vortex with a small circumferential rotation diameter (of an almost rectilinear shape) was observed, coming down from the runner hub. From this point, along with the decrease in the cavitation number, the diameter of the vortex increased in states from no. 2 to no. 9 (_σ_ = ∼0.09). In the last recorded state no. 10 (_σ_=0.055), cavitation bubbles with a visible circumferential whirl of water (a helical shape) were observed in the entire flow volume. Starting from state no. 7 (_σ_=0.155, _η_=87.67%) there was a sharp decrease in efficiency to the value of 83.96% in state no. 10.
•_n I I_=78 rpm. Visualization of states no. 0 (_σ_=1.01, _η_=90.08%) and no. 1 (_σ_=0.71) did not show the presence of cavitation bubbles. The subsequent states no. 2 (_σ_ = ∼0.41) and no. 3 (_σ_ = ∼0.3) indicated the formation of a very poorly visible thin vortex rope coming down from the runner hub. The rope increases in diameter as the cavitation number decreases. From state no. 4 (_σ_ = ∼0.24) to state no. 7 (_σ_ = ∼0.1) there was a clear vortex rope with increasing diameter as the cavitation number decreases. From state no. 5 (_σ_ = ∼0.19, _η_=90.13%) there was a decrease in efficiency and was characterized by a rather mild curvature in the initial phase until state no. 7 (_σ_ = ∼0.1, _η_=89.27%). From state no. 7 to state no.8 there was a sharp decrease in efficiency to the value of 88.73% (_σ_ = ∼0.06).
•_n I I_=83 rpm. At initial state no. 0, this is the Best Efficiency Point (_σ_=1.015, _η_=91.21%). No distinct vortex was observed in all visualized operating states. In the last state no. 8 cavitation bubbles circumferentially with a helical shape are visible. A gentle decrease in efficiency occurs between state no.5 (_σ_ = ∼0.185, _η_=91.20%) and state no. 7 (_σ_ = ∼0.1, _η_=89.67%). After that, a rapid decrease in efficiency to the final state no. 8 is visible (_σ_ = ∼0.06, _η_=85.90%).
•_n I I_=88 rpm. The nature of the flow in all states was similar to the states for the double-reduced rotational speed _n I I_=83 rpm. From state no. 5 (_σ_ = ∼0.185, _η_=90.17%), the efficiency decreases to final state no. 8 (_σ_ = ∼0.07, _η_=85.40%). In the final state, helically rotating cavitation bubbles were observed.
•_n I I_=94 rpm. In the initial state no. 0 (_σ_ = ∼1.04, _η_=87.58%) and no. 1 (_σ_ = ∼0.707), no traces of cavitation bubbles were found. From state no. 2 (_σ_ = ∼0.42), the formation of a thin thread-like vortex was observed, rotating over the entire diameter of the inlet draft tube. This vortex grows in the next states no. 3 (_σ_=0.295), no. 4 (_σ_=0.245), no. 5 (_σ_=0.19), and no. 6 (_σ_=0.16). In the following state no. 7 (_σ_ = ∼0.13) one vortex can be seen coming down from the runner hub with an internal diameter of∼0.2 _D_. In the last state no. 8, similarly to other cases of _n I I_, helically rotating cavitation bubbles were observed. The efficiency decrease starts from state no. 5 (_σ_ = ∼0.19, _η_=87.72%) to final state no. 8 in a fairly rapid manner (_σ_ = ∼0.095, _η_=84.07%).
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Fig. 21. Results of cavitation tests of the Francis model turbine – efficiency characteristics depending on the cavitation number and visualization of the flow in the draft tube (under the runner) for the guide vane opening angle 22°.
Summing up, at this guide vane opening angle no significant increase in efficiency was observed on any of the analysed cavitation curves. However, it can be noticed that in states followed by a sharp decrease in efficiency, the efficiency curves show slight waviness, indicating some pressure redistribution effects at the runner blades, which further lead to a radical change in the nature of the flow.
•Guide vane opening angle 27°
The results of the cavitation test and the visualization of the flow in individual states depending on the cavitation number for a guide vane opening angle of 27° are presented in Fig. 22.
•_n I I_=72 rpm. The visualization showed that already in the initial state no. 0 (_σ_ = ∼1.04, _η_=86.7%), a straight vortex rope was formed under the runner hub. The rope gradually increased its diameter in subsequent states until the state no. 6 (_σ_ = ∼0.11, _η_=84.49%), in which its diameter reached∼0.6 _D_. In the final visualized state no. 7 (_σ_ = ∼0.07, _η_=82.58%) the vortex was completely eliminated and a mixture of helically rotating bubbles was formed throughout the flow volume. The efficiency decreased quite quickly from state no. 3 (_σ_ = ∼0.25, _η_=86.53%) to the final state no. 7 (_σ_ = ∼0.07, _η_=82.58%).
•_n I I_=78 rpm. The course of flow changes in this case is quite similar to that described above. The difference is that the formation of the vortex under the runner hub is delayed. In state no. 1 (_σ_ = ∼0.72, _η=_ 87.54%) the vortex is barely noticeable. In state no. 2 (_σ_ = ∼0.405) it already appears very clearly. In all states, the vortex has a smaller diameter compared to _n I I_=72 rpm. In state no. 6 (_σ_ = ∼0.11) it reaches a value of approximately 0.3 _D_, and then disappears in the final state no. 7, becoming a mixture of helically rotating cavitation bubbles. The efficiency decreased quite rapidly from state no. 3 (_σ_ = ∼0.27, _η_=87.54%) to the final state no. 7 (_σ_ = ∼0.07, _η_=83.66%).
•_n I I_=83 rpm. The course of changes in the flow pattern is also quite similar to the two ones described above. The formation of the vortex rope changes as the cavitation number decreases. In state no. 2 (_σ_ = ∼0.41, _η_=88.23%), a trace of vortex rope formation was observed. In states no. 3 (_σ_=0.25), no. 4 (_σ_=0.2), no. 5 (_σ_=0.15), and no. 6 (_σ_ = ∼0.105) it was already very clearly visible, but with a smaller diameter compared to the rotational speeds presented above, reaching a value of∼0.25 _D_ under the runner. In the last state no. 7, the vortex disappeared becoming a mixture of helically rotating cavitation bubbles. The efficiency decreased quite rapidly in state no. 4 (_σ_ = ∼0.195, _η_=88.31%) to final state no. 7 (_σ_ = ∼0.07, _η_=83.78%).
•_n I I_=88 rpm. Vortex formation was observed consistently from state no. 2 (_σ_ = ∼0.42, _η_=88.8%). The diameter increased in states no. 3 (_σ_=0.246), no. 4 (_σ_=0.19), no. 5 (_σ_=0.15), up to state no. 6 (_σ_= ∼0.12, _η_=87.23%). However, the diameter was much smaller than in the previous states described for double-reduced rotational speeds. In state no. 6, the vortex had a diameter of∼0.15 _D_. In the last state no. 7 (_σ_ = ∼0.085, _η_=84.2%), the vortex disappeared, becoming a mixture of helically rotating cavitation bubbles. The decrease in efficiency was initially gentle, and then sharp from state no. 3 (_σ_ = ∼0.25, _η_=88.89%) to the final state no. 7.
•_n I I_=94 rpm. In this case, there is a clear difference in the flow patterns compared to the previously described double-reduced rotational speeds. The vortex rope that existed at that time practically ceased to exist. Only in state no. 5 (_σ_ = ∼0.165) and also in state no. 6 (_σ_ = ∼0.14), a slight trace of the vortex was visible. In the last state no. 7, the vortex disappeared becoming a helically rotating cavitation bubbles. The decrease in efficiency was initially gentle, and then sharp from state no. 3 (_σ_ = ∼0.26, _η_=89.16%) to the final state no. 7 (_σ_ = ∼0.11, _η_=84.53%).
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Fig. 22. Results of cavitation tests of the Francis model turbine – efficiency characteristics depending on the cavitation number and visualization of the flow in the draft tube (under the runner) for the guide vane opening angle 27°.
•Guide vane opening angle 36°
The cavitation test results and the visualization of the flow in individual states depending on the cavitation number for a guide vane opening angle of 36° are presented in Fig. 23. Generally, the flow was characterized by the occurrence of a straight single vortex rope under the runner hub, which quickly formed at the beginning in the state (no. 0) and gradually took form of an hourglass-like periodic vortex. The vortices did not disappear in close proximity to the runner, but passed deeper into the draft tube. Decreases in efficiency began at higher cavitation numbers than at lower guide vane opening angles.
•_n I I_=72 rpm. Since the initial state no. 0 (_σ_=1.025, _η_=81.77%), a straight vortex rope with a diameter of∼0.15 _D_ was formed. As the cavitation number decreased, the vortex increased in diameter, transforming into a periodic hourglass-shaped vortex from state no. 2 (_σ_ = ∼0.44) to state no. 6 (_σ_ = ∼0.235, _η_=80.17%), occupying almost the entire diameter of the runner, and then disappeared in state no. 7, in which the mixture of bubbles covered the entire flow volume. This had an impact on the flow efficiency, because the efficiency decrease began at a relatively high cavitation number in state no.2 (_σ_ = ∼0.44, _η_=81.52%) and proceeded quite gently until the final state no. 7 (_σ_ = ∼0.18, _η_=78.93%).
•_n I I_=78 rpm. The course of flow changes is quite similar to that described above. The runner hub vortex rope was also in the initial state no. 0 (_σ_=1.005, _η_=82.58%) and was characterized by a smaller diameter in all states than in the case of the double-reduced rotational speed of 72 rpm. From state no.2 (_σ_ = ∼0.5), the vortex took a periodical hourglass shape, up to state no. 6 (_σ_ = ∼0.21, _η_=80.75%), in which the vortex diameter reached a value of about 0.7 _D_ under the runner. In the final state no. 7, the vortex disappeared and was replaced by a mixture of helically rotating cavitation bubbles. The decrease in efficiency started at a relatively high cavitation number in state no. 2 (_σ_ = ∼0.5, _η_=82.52%) and proceeded gently until the final state no. 7 (_σ_ = ∼0.16, _η_=79.37%).
•_n I I_=83 rpm. The course of changes in flow pattern is also quite similar to the two cases described above. The vortex rope under the runner hub also occurred in the initial state no. 0 (_σ_ = ∼1.01, _η_=82.24%) and was characterized by a smaller diameter in all states compared to the cases of the double-reduced rotational speed of 72 rpm and 78 rpm. From state no. 2 (_σ_ = ∼0.495) it took a periodic hourglass shape until state no. 6 (_σ_ = ∼0.19). In state no. 6, the vortex diameter under the runner was approximately 0.6 _D_. In the final state no. 7, the vortex disappeared and was replaced by a mixture of helically rotating cavitation bubbles. The efficiency decrease began with a relatively high cavitation number in state no.3 (_σ_ = ∼0.38, _η_=82.12%) and proceeded quite gently until the final state no. 7 (_σ_ = ∼0.115, _η_=77.49%).
•_n I I_=88 rpm. The straight vortex rope did not occur in the initial state no. 0 (_σ_ = ∼1.005, _η_=81.58%), although the beginnings of the formation of a vortex structure, which was not fully formed were visible under the runner hub. From state no. 1 (_σ_ = ∼0.7) to state no. 5 (_σ_ = ∼0.185), the situation was similar to the cases described above. In state no. 5, the vortex diameter reached approximately 0.5 _D_ under the runner. In the final state no.6 (_σ_ = ∼0.13, _η_=78.32%), the vortex disappeared and was replaced by a mixture of helically rotating cavitation bubbles. The efficiency decrease started in state no. 4 (_σ_ = ∼0.28, _η_=81.54%) and progressed quite quickly to final state no. 6.
•_n I I_=94 rpm. The straight vortex rope did not occur in the initial state no. 0 (_σ_=1.01, _η_=81.36%). The beginnings of its formation could already be seen in state no. 1 (_σ_ = ∼0.705), but it did not have a clear straight course into the draft tube. Only starting from state no. 2 (_σ_=0.506) the vortex under the runner hub was visible. With lowering _σ_ the vortex split into two smaller vortices, which then very quickly dispersed in the flow at a distance of approximately 0.5 _D_ downstream the runner. In the next state no. 3 (_σ_ = ∼0.385), the situation took the form described in the previous cases of rotational speeds, i.e. from state no. 4 (_σ_ = ∼0.3, η=81.57%), up to state no. 6 (_σ_ = ∼0.19, _η_=80.21%) an hourglass-shaped vortex were observed, which increased in diameter with decreasing cavitation number. In state no. 6, the vortex diameter reached a value of about 0.3 _D_ under the runner. In the final state no. 7 (_σ_ = ∼0.12, _η_=76.47%), the vortex disappeared and was replaced by a mixture of helically rotating cavitation bubbles. The efficiency decrease started in state no. 4 and ended in state no. 7.
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Fig. 23. Results of cavitation tests of the Francis model turbine – efficiency characteristics depending on the cavitation number and visualization of the flow in the draft tube (under the runner) for the guide vane opening angle 36°.
•Critical cavitation numbers
By using the obtained curves, it was possible to determine the cavitation number σ 3 at which a 3% decrease in efficiency occurred. It is one of parameters characterizing hydraulic turbines at which fully developed volume cavitation occurs in the flow. The lowest cavitation numbers∼0.02, for a 3% decrease in efficiency, were obtained for the guide vane opening angle 12°. For the largest one, the lowest cavitation numbers at a 3% decrease were∼0.16. The description of the efficiency changes and the corresponding cavitation numbers are presented in Table.6.
Table 6. Results of cavitation tests of the model Francis turbine – specific cavitation numbers corresponding to measured decreases and increases in efficiency.
| Guide vane opening angle (_α_ GV) | Double-reduced rotational speed (_n_ I I) | Cavitation number (_σ_) for a 3% decrease in efficiency | Maximum measured decrease in efficiency | Cavitation number (_σ_) for maximum measured decrease in efficiency | Maximum measured increase in efficiency | Cavitation number (_σ_) for maximum measured increase in efficiency | Critical cavitation number (_σ_ cr) |
--- --- --- --- |
| ° | rpm | − | % | − | % | − | − |
| 9 | 72 | 0.048 | 3.74 | 0.039 | 1.31 | 0.100 | 0.28 |
| 78 | Not attained | 0.72 | 0.038 | 2.42 | 0.109 | 0.24 |
| 12 | 72 | 0.018 | 3.32 | 0.014 | 1.40 | 0.095 | 0.28 |
| 78 | 0.056 | 3.11 | 0.052 | 0.17 | 0.203 | 0.20 |
| 83 | 0.057 | 4.08 | 0.037 | 0.18 | 0.167 | 0.26 |
| 88 | 0.032 | 4.03 | 0.019 | 0.31 | 0.274 | 0.28 |
| 15 | 72 | 0.028 | 3.40 | 0.023 | 0.49 | 0.079 | 0.18 |
| 78 | 0.060 | 4.59 | 0.038 | Not occurred | − | 0.18 |
| 83 | 0.075 | 4.36 | 0.053 | Not occurred | − | 0.18 |
| 88 | 0.095 | 4.55 | 0.061 | Not occurred | − | 0.21 |
| 94 | 0.100 | 3.96 | 0.073 | Not occurred | − | 0.22 |
| 18 | 72 | 0.050 | 4.04 | 0.032 | Not occurred | − | 0.18 |
| 78 | 0.055 | 3.70 | 0.051 | Not occurred | − | 0.18 |
| 83 | 0.065 | 3.48 | 0.059 | Not occurred | − | 0.18 |
| 88 | 0.090 | 4.60 | 0.068 | Not occurred | − | 0.25 |
| 94 | 0.105 | 4.21 | 0.084 | Not occurred | − | 0.28 |
| 20 | 72 | 0.050 | 3.65 | 0.042 | 0.40 | 0.151 | 0.20 |
| 78 | 0.065 | 3.95 | 0.050 | Not occurred | − | 0.18 |
| 83 | 0.070 | 3.11 | 0.067 | Not occurred | − | 0.18 |
| 88 | 0.090 | 4.16 | 0.075 | Not occurred | − | 0.18 |
| 94 | 0.110 | 5.20 | 0.083 | Not occurred | − | 0.20 |
| 22 | 72 | 0.065 | 3.86 | 0.054 | Not occurred | − | 0.16 |
| 78 | 0.068 | 3.36 | 0.062 | Not occurred | − | 0.18 |
| 83 | 0.083 | 5.33 | 0.058 | Not occurred | − | 0.23 |
| 88 | 0.090 | 4.82 | 0.070 | Not occurred | − | 0.24 |
| 94 | 0.102 | 3.52 | 0.095 | 0.29 | 0.246 | 0.28 |
| 24 | 72 | 0.068 | 3.22 | 0.063 | Not occurred | − | 0.26 |
| 78 | 0.078 | 3.83 | 0.063 | Not occurred | − | 0.24 |
| 83 | 0.088 | 4.94 | 0.067 | Not occurred | − | 0.24 |
| 88 | 0.103 | 3.03 | 1.030 | Not occurred | − | 0.26 |
| 94 | 0.100 | 4.12 | 0.097 | Not occurred | − | 0.28 |
| 27 | 72 | 0.095 | 4.13 | 0.073 | Not occurred | − | 0.28 |
| 78 | 0.085 | 3.90 | 0.069 | Not occurred | − | 0.27 |
| 83 | 0.090 | 4.46 | 0.069 | Not occurred | − | 0.24 |
| 88 | 0.102 | 4.60 | 0.085 | Not occurred | − | 0.25 |
| 94 | 0.125 | 4.58 | 0.110 | Not occurred | − | 0.28 |
| 30 | 72 | 0.135 | 4.12 | 0.105 | Not occurred | − | 0.26 |
| 78 | 0.115 | 3.67 | 0.095 | Not occurred | − | 0.24 |
| 83 | Not attained | 2.69 | 0.097 | Not occurred | − | 0.24 |
| 88 | 0.115 | 3.90 | 0.100 | Not occurred | − | 0.26 |
| 94 | 0.135 | 4.92 | 0.104 | Not occurred | − | 0.30 |
| 33 | 72 | 0.155 | 3.53 | 0.138 | Not occurred | − | 0.32 |
| 78 | 0.140 | 4.12 | 0.110 | Not occurred | − | 0.30 |
| 83 | 0.130 | 3.67 | 0.107 | Not occurred | − | 0.28 |
| 88 | 0.110 | 3.65 | 0.103 | Not occurred | − | 0.26 |
| 94 | 0.135 | 4.96 | 0.105 | Not occurred | − | 0.32 |
| 36 | 72 | Not attained | 2.84 | 0.182 | Not occurred | − | 0.38 |
| 78 | 0.160 | 3.21 | 0.156 | Not occurred | − | 0.36 |
| 83 | 0.160 | 4.75 | 0.114 | Not occurred | − | 0.32 |
| 88 | 0.140 | 3.26 | 0.130 | Not occurred | − | 0.36 |
| 94 | 0.150 | 4.90 | 0.120 | Not occurred | − | 0.38 |
A clear increase in efficiency close to the efficiency curve breakdown occurred only for the two lowest guide vane opening angles, i.e. 9° and 12°. At the remaining opening angles, only single cases with small efficiency increases not exceeding 0.5% were observed.
The table also presents the values of critical cavitation number _σ cr_=_σ_ 0, i.e. the number at which the change in efficiency begins (decrease or local increase and then decrease). The “sigma dell” as formed by the _σ_ 0-isolines is shown in Fig. 24. It is to be noticed that the bottom of the dell (best cavitation properties) corresponds to slightly lower discharges of the BEP. It was also noticed that the resulting minimum _σ_ 0 value is quite close to that following from the relationship _σ_ allowable=_σ_ allowable(_n_ sQ) according to Arndt (Fig. 25).
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Fig. 24. The isolines of critical cavitation number σ 0 in the (_n_ 1 I, _Q_ 1 I) plane – “sigma dell”.Assessment based on the zero-efficiency drop criterion.
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Fig. 25. Approximate values of the allowable cavitation number on the kinematic specific speed of hydraulic turbines together with typical controllable cavitation bands .
5.3. Diagnostic signals and allowable cavitation number
Cavitation affected all diagnostic parameters, which does not mean that this impact was unambiguous. The last remark refers in particular to the pressure fluctuations beneath the runner (guide vane opening angles: 9° and 27°), which have generally appeared too weak for reliable diagnostic purposes (Fig. 26) even if applying the procedure specified at the beginning of Section 5.1. Though, it usually leads to reasonable assessments of the critical cavitation number. Therefore, the tests were conducted with the pressure tap shifted to the draft tube wall .
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Fig. 26. Results of cavitation tests of the model Francis turbine – pressure fluctuations amplitude (peak-peak) versus cavitation number.
The reliability question also arose in the case of acoustic emission (guider vane opening angles: 9° and 27°), which, after several unsuccessful attempts with the spiral case/runner chamber block, was used on the upper flange of the draft tube (Fig. 27). Satisfactory results with tests conducted with another model suggested that the problem lied in failure to find the optimum final AE sensor location.
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Fig. 27. Results of cavitation tests of the model Francis turbine – acoustic emission (peak-peak) versus cavitation number.
Eventually, it was decided to analyse cavitation load to turbine structure based on the high-frequency (>5 kHz) acceleration component of the draft tube wall (Fig. 28) and the turbine guide bearing (Fig. 29) (guide vane opening angles: 9°, 15°, 22°, 27°, and 36°). The peak-to-peak amplitude of the guide bearing housing was used to determine the allowable cavitation numbers for each operation point under consideration.. The procedure consisted of two stages. In the first stage, the critical cavitation number _σ_ vib=_σ_ vib(_n_ I I) curves were plotted for each guide vane angle (Fig. 30). In the second stage, the whole curve family was cut by a series of _σ_ allowable=const levels. The curve cutting points determined co-ordinates to be used when plotting the _σ_ allowable isolines in the (_n_ I I, _Q_ I I) plane (Fig. 31). The curves in both graphs were smoothed taking into account other characteristics and visual observation results.
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Fig. 28. Results of cavitation tests of the model Francis turbine – draft tube wall accelerations (peak-peak) versus cavitation number.
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Fig. 29. Results of cavitation tests of the model Francis turbine – guide bearing accelerations (peak-peak) versus cavitation number.
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Fig. 30. Guide bearing acceleration based on critical cavitation number versus double-reduced rotational speed as established for various guide vane angles.
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Fig. 31. The isolines of allowable cavitation number in the (_n_ 1 I, _Q_ 1 I) plane. Assessment based on the guide bearing vibration criterion.
It is to be noticed that while acceleration signals taken at the draft tube wall and the turbine guide bearing may show almost the same critical cavitation numbers, the differences in proportions between curves corresponding to different speeds are quite apparent. The highest values are attained during the partial load operation at 9° and 15° guide vane angles. In particular, they are much higher than those under overload conditions, which is a feature consistent with the findings of Gerich and Raabe .
An important feature of all plotted diagnostic curves are bumps and valleys appearing at relatively high cavitation number values, quite often at states with no cavitation visually discernible, sometimes at cavitation number close to 1.0. These diagnostic curve fluctuations do not always coincide with each other showing that at this stage some cavitation effects are not global.
Typical examples of single major bumps can be observed in the acceleration signals recorded at extreme partial load (guide vane opening angle 9°). As it can be seen from Fig. 28, the draft tube wall acceleration signal at _n_ I I=78 rpm starts to rise with lowering _σ_ already at _σ_ close to 0.8. One may suppose that the main reason lies in large amounts of microbubbles – still invisible, but already affecting the physical properties of the liquid, which gets more compressible – especially in vortex cores. One possible mechanism is also development of invisible cavitating threads inside the secondary flow and partial load vortex cores, which could make them more stiff and powerful. Whatever would be the mechanism, at some stage this effect gets compensated and damped by rising amount of vapour in the vortex core, not only in form of a single threads but also surrounding small bubbles. Finally the tread gets probably decomposed. As it can be seen from Fig. 19, only at _σ_=0.2 cavitation is easily discernible as a foggy bubble structure. However, at this point the acceleration of the draft tube amplitude is 10-fold lower than the maximum at _σ_=0.42. It is interesting that no such curve course can be observed at speed _n_ I I=72 rpm. Cavitation bubbles are present in a much larger volume, probably preventing strong interaction of the vortex rope (or thread) with the draft tube wall. As the bumps still occur at the guide bearing acceleration curve, one may suppose that the presence of even invisible cavitation structures inside the vortex rope core can increase substantially the force acting on the runner hub due to the vortex precession.
With rising guide vane angle the flow gets more swift and better ordered, which may explain lower amplitudes of high-frequency vibrations under conditions of invisible cavitation. In this context, it is worthwhile to notice a profound difference between the acceleration diagrams under overload conditions. As it can be seen from Fig. 28 and Fig. 29, cavitation developing at cavitation numbers below 0.3 only dampens the vortex rope radial force acting on the hub or even helps to position the rotating assembly along the rotation axis. This effect may disappear at too low rotation speeds.
At the same time, an abrupt rise in the draft tube wall acceleration may be noticed under the cavitation number fall below some critical value. The effect seems to follow from radial fluctuation of the vortex rope cavitating core rather than from precession, which is always present under partial load conditions.
The main purpose of cavitation tests described in this paper was to establish the allowable cavitation number above which the machine may be safely operated. The procedure was based on a multi-criterion approach, typically applied in hydraulic machinery labs since mid of the previous century, e.g. , but rarely described in literature in any detail. The phenomena responsible for diagnostic signal waviness occurring already at relatively high cavitation numbers are generally not mentioned even in a hypothetical way.
In addition to efficiency drop and visual observation results, the abrupt rise of relevant diagnostic signal estimate with falling Thoma number was used in this study in order to identify critical cavitation numbers according to different criteria. Rational and consistent explanation of cavitation development basing on test outcomes was the core of detailed analysis conducted. Four diagnostic signals were studied, including that of Acoustic Emission (AE) – the signal used in cavitation diagnostics under different name at least since seventies of the previous century, e.g. , and having entered into wide use at the turn of 80-ies and 90-ies, e.g. . The authors find their approach efficient enough for the modest goal of their investigation, but insufficient for an in-depth study of cavitation phenomenon, which usually requires not only spectral analysis of the pressure fluctuation beneath the runner, but also wavelet analysis of the signal coming from the runner blades, advanced AE signal correlation studies as well as advanced techniques used in order to decompose the signals and identify separate cavitation mechanisms, e.g. , , , . Typically, much higher effort, including higher number of signals has to be used for such purposes.
6. Runaway investigations of the model Francis turbine
Runaway tests were carried out with the reference net head the same as in the case of the basic performance and cavitation tests, _H n_=12 m. As a result, the characteristics of reduced rotational speed were created depending on the guide vane opening angle: _n red_=_f_(_α_ GV). 15 approximate guide vane angle openings were assumed for the tests, namely: 9°, 10.5°, 12°, 13.5°, 15°, 16.5°, 18°, 20°, 22°, 24°, 25.5°, 27°, 30°, 33°, and 36°. The head was maintained within a range of changes below±1% of the reference head. It should be emphasized, that its regulation was much more difficult than in the case of basic performance and cavitation tests due to less stable operation resulting from the lack of electrical load.
The results of the runaway tests are shown in Fig. 32 (the runaway curve was also plotted on the shell characteristics shown in Fig. 14). These are the following 2 characteristics: 1) rotational speed depending on the guide vane opening angle and 2) rotational speed depending on the double-reduced flow rate. An interesting fact is that the characteristics have a slight concavity (saddle) with an inflection point corresponding to the guide vane opening angle of∼17°. Fixed lines of the optimal rotational speed _n opt_=1155 rpm (in the BEP), the optimal guide vane opening angle _α_ GV=22°, and the optimal double-reduced flow rate _Q_ 1 1=0.9659 m 3/s were also drawn on the graphs.
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Fig. 32. Results of runaway tests of the model Francis turbine – dependence of the rotational speed _n_ on the guide vane opening angle _α_ GV and the double-reduced flow rate _Q_ 1 1.
Let us assume that the rotational speed of the turbine operating on the facility will be equal to the optimal rotational speed (_n_=_n_ opt). Then, the coefficients of increase in the runaway rotational speed (_n_ runaway) to the optimal rotational speed of the basic performance tests (_n_ opt) with respect to the guide vane opening angle can be presented as shown in Table.7 and in Fig. 33. The runaway coefficient at the optimal guide vane opening angle (22°) is∼1.497, while at the largest guide vane opening angle (36°) it is∼1.737.
Table 7. Results of runaway tests of the Francis model turbine – rotational speed factor increase above the optimal rotational speed.
| No. | Guide vane opening angle (_α_ GV) | Runaway rotational speed (_n_ runaway) | Runaway coefficient (_n_ runaway / _n_ opt) |
--- --- |
| − | ° | rpm | − |
| 1 | 9.1 | 1399.3 | 1.212 |
| 2 | 10.5 | 1452.5 | 1.258 |
| 3 | 12.1 | 1516.7 | 1.313 |
| 4 | 13.7 | 1545.6 | 1.338 |
| 5 | 15.0 | 1571.4 | 1.360 |
| 6 | 16.5 | 1597.1 | 1.383 |
| 7 | 18.0 | 1624.5 | 1.406 |
| 8 | 20.0 | 1671.5 | 1.447 |
| 9 | 22.1 | 1728.6 | 1.497 |
| 10 | 24.2 | 1797.7 | 1.556 |
| 11 | 25.6 | 1843.8 | 1.596 |
| 12 | 27.1 | 1885.2 | 1.632 |
| 13 | 30.1 | 1949.9 | 1.688 |
| 14 | 33.1 | 1990.1 | 1.723 |
| 15 | 36.1 | 2006.1 | 1.737 |
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Fig. 33. Results of runaway tests of the model Francis turbine – dependence of the runaway speed factor _n runaway/n opt_ on the guide vane opening angle _α_ GV.
7. Summary and conclusions
Development studies aimed at understanding the flow through a high specific speed Francis turbine and accompanying phenomena are of interest to many researchers. The reason is continuous demand for high quality designs of turbine flow systems, including increase of efficiency. However, nowadays it should be said that the efficiency of hydraulic turbines is already reaching the upper limit of what is really possible to achieve. On the other hand, the phenomenon of cavitation, especially in the runner and draft tube, and the associated cavitation erosion still require special attention. Diagnostics of flow behaviour in order to mitigate undesirable effects still takes much time in many research centres. This issue is addressed in the presented paper, which concerns a comprehensive analysis of the flow (using basic performance, cavitation and runaway tests) through a high specific speed model Francis turbine (with the diameter of 0.25 m) operating at a head of 12 m. This type of Francis turbines still remains largely unexplored, which was the main motivation to report on our study.
On the basis of the comprehensive tests carried out for the high specific speed model Francis turbine, the following can be concluded.
1.The aim of the turbine runner design was to obtain a flow system of a high specific speed machine, i.e. intended for the lowest heads in the Francis turbine applicability range. As part of this, the commercial TurboDesign™ program was used to design the runner. As a result, a turbine with the kinematic specific speed of _n_ sQ=82 was developed. This value allows concluding that the goal has been achieved, and the obtained characteristics allow for safe operation with high efficiency in a wide range of machine operation parameters (the tests were performed in the range of guide vane opening angles from 9° to 36°). The maximum efficiency of the model machine reached a value of∼91%, which also seems to be a high value, taking into account the fact that it should always be expected that such a turbine will never be as efficient as model turbines with lower specific speeds, intended for high heads. The analysis of the impact of the scale effect shows that it is possible to achieve efficiency of up to 92.9% (1.9% increase) under field conditions for the largest assumed head height of 50 m, and the largest assumed runner diameter of 1.5 m. Additionally, it was found that the effect of the temperature difference (13°C) of water flowing through the model turbine (average measured temperature 23°C) and the prototype turbine (average assumed temperature 10°C) causes a change in efficiency by 0.34%, with the same head and runner diameter parameters. Operation of a prototype turbine with water of higher viscosity reduces its efficiency.
2.Basic performance tests indicate the presence of saddles on the machine operating characteristics for guide vane opening angles greater than optimal. The size of the saddles is not large, but it can cause a threat in the form of a tendency to uneven operation of the machine.
3.The values of the contribution of the friction torque in the total torque ranged from 0.8%, at the highest opening of the guide vane opening angle (36°) and the double-reduced rotational speed (_n_ 1 1 = ∼60 rpm) to 3.8%, at the lowest angle (9°) and the rotational speed (_n_ 1 1 = ∼82 rpm). The contribution value of 0.8% is the lowest value one can get. Further reduction of mechanical friction losses is not possible.
4.Cavitation tests of the model turbine have shown highly satisfactory cavitation properties with allowable cavitation numbers below the cavitation limit following statistical analyses . Achieving such a result at the upper limit of the turbine type applicability range can be considered a reason for satisfaction. In the case of the Best Efficiency Point, the flow visualization showed no disturbing behaviour of the flow in the draft tube.
5.During the runaway tests, it was found that the runaway coefficient at the optimal guide vane opening angle (22°) was∼1.497, while in the case of the largest angle (36°) it was∼1.737. The runaway tests also indicated the presence of a saddle on the characteristics, although in the case of the runaway condition it has no significance for the safe operation of the machine.
6.In the paper, a multi-aspect approach to cavitation diagnostics was used. Acceleration of the draft tube wall, and guide-bearing as well as acoustic emission and pressure fluctuations were used to analyse development of the cavitation phenomenon. In addition, visual techniques were applied for the same purpose. The analysis made it possible to determine the beginning of dangerous machine operating states. At off-design points of operation, the main flow pattern downstream the runner is characterized by a vortex flow dominated by a braided (hourglass-shaped) vortex descending from the runner hub. As the cavitation number decreases, the vortex expands in diameter and then turns into a swirled mixture of bubbles throughout the entire flow volume in the draft tube.
7.The diagnostic measurements as conducted consecutively with lowering cavitation number have confirmed again that visually detectable signs of cavitation are often preceded by dynamic effects (pressure fluctuations, mechanical vibrations, acoustic emission) showing a fluctuating character against the _σ_ value. Especially high amplitude of these fluctuations as encountered in the case of off-design points of operation may be due to intense secondary flows. The mentioned fluctuations are expected to be linked with the growth of microbubbles, which can affect the liquid compressibility even before initiating the cavitation development. According to this hypothesis, water quality may be a significant factor. Further studies in this direction are planned in the future.
8.Due to initial problems with identifying the optimum location for the Acoustic Emission (AE) transducer and some ambiguity in interpreting the pressure fluctuation signal only vibration signals were used eventually to determine the critical and permissible cavitation number values. The above-mentioned problems with both signals were removed later on during tests of another Francis turbine with a different runner . Nevertheless, the multiyear experience of the authors learns that while broad band AE signal can provide a lot of valuable information, it is also extremely vulnerable to any possible external disturbances and depends strongly on sensor location as well as on the signal route from its source. On the other hand, vibration signals – especially those from the draft tube – provide usually a highly reliable information even if not allowing for very detailed analysis. The last one can be obtained by analysing the low frequency pressure fluctuation signal (e.g. those due to cavitating vortex rope). The high frequency analysis (>20 kHz) is usually recommended for much more detailed studies including statistical and/or wavelet techniques as applied to cavitation pulsatile signals. Statistical techniques employing the notion of ME cavitation intensity factor or vector have been used for decades by the IMP PAN team and other researchers both under field and laboratory conditions – also in order to describe cavitation erosive load , . It can also confirm the opinion of various cavitation experts that increasing the number of suitably selected signals and their estimates generally contributes to the quality of final assessments. While the number of signals used for determining the allowable cavitation number should be kept below some reasonable limit, it should also not fall below 2 or 3.
The aim of the research on Francis turbines was to develop turbine structure for the lowest possible heads, dedicated for the installation of Francis turbines (i.e. from 8 to 30 m), which are turbines with high or very high kinematic specific speed (from 80 to 100). The presented structure was also tested with another runner with a higher specific speed (with a different shape of the runner blades and the blade number reduced to 11 and a kinematic specific speed of 94). Thanks to this, it is now possible to propose turbines to install on the lowest heads for Francis turbines, for operation in hydroelectric power plants. Both structures were tested in terms of their efficiency and cavitation safety. This allows predicting parameters for safe operation, thanks to which the lifetime of these machines can be extended. Additionally, the high specific speed feature of these machines allows for the elimination of gears and reduction of the size of generators, therefore reducing the costs of a hydrounit.
The presented turbine has already had one implementation in field conditions in Poland, in the urban network for energy recovery, i.e. as a pressure reducer (power: ∼90 kW, head: ∼23 m). However, it has not been tested in terms of efficiency and cavitation risk. Therefore, the goal is still to check the operation of the prototype in field conditions, which will allow assessing the correctness of the research and diagnostic methods used.
The future research directon will aim to expand cavitation diagnostics. We hope that for cavitation diagnostics we will be able to use the so-called wavelet analysis method, which is quite a popular method in cavitation diagnostics and should increase the accuracy of cavitation prediction.
CRediT authorship contribution statement
Zbigniew Krzemianowski: Writing – review & editing, Writing – original draft, Supervision, Software, Project administration, Methodology, Investigation, Formal analysis, Conceptualization. Janusz Steller: Writing – review & editing, Writing – original draft, Methodology, Formal analysis.
Declaration of competing interest
The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper.
Acknowledgement
The works were co-financed by the (Polish) National Centre for Research and Development and the (Polish) National Fund for Environmental Protection and Water Management under project no. GEKON1/04/214228/27/2015 co-ordinated jointly by ZRE Gdańsk (industrial turbine manufacturer) and IMP PAN Gdańsk (scientific centre).
The authors would like to express their gratitude to Dr. T. Przybyliński, Dr. J. Karwacki, and MSc. A. Tomaszewski for their contribution to the documentation of photographic research. Many thanks are also due to the other members of the IMP PAN research team, including Dr. W. Janicki, who took care of the data acquisition system, as well as MSc. A. Góralczyk and MSc. A. Domański, who helped carry out laboratory tests and MSc. A.Henke, who drew the shell characteristics. Final thanks are due to Prof. A. Adamkowski for verification of the final internal reports.
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The data that has been used is confidential.
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Cited by (6)
### Experimental research of a high specific speed low-head model Francis turbine – A case study with emphasis on flow visualisation under runner using PIV method and energy production analysis
2025, Energy Show abstract Currently, there are many low-head facilities in hydropower equipped with Francis hydrounits, whose flow systems are inadequate for the heads under which they should operate. The paper presents performance investigations of a model Francis turbine with a kinematic specific speed of 94. Experimental investigations were carried out in a wide range of changes of the guide vane opening angle and the rotational speed. The efficiency reached 89% at high values of the double reduced flow rate _Q_ 1 1=1.033 m 3/s and the double reduced rotational speed _n_ 1 1=93 rpm. In particular, the paper presents the shell characteristics, as well as the results on friction losses generated during machine operation. The presence of saddles on the torque and flow rate characteristics was found at high guide vane opening angles, which seems to be a novelty in the case of Francis turbines. Based on the shell characteristics, the machine selection areas are also indicated, in which the energy production is higher than at the optimum operating point despite the reduction in turbine efficiency. Additionally, a photographic visualization of the flow under the runner and the velocity fields obtained by the PIV method for different guide vane angles are presented.
### Cavitation control in Francis turbines by design optimization across variable operating conditions
2025, Next Energy Show abstract Cavitation is a critical challenge which significantly affects performance of hydro turbines leading to inefficient operation. Cavitation primarily arises due to improper design and frequent variations in operating conditions. Prediction or maintaining cavitation within controlled limits is crucial and challenging task in turbines yet it is essential for employing effective mitigation strategies to ensure the efficient and reliable operation of turbines. In this study, computational fluid dynamics(CFD) analysis of Francis turbine runner employed at DulHasti Power Station (DPS) was conducted using Ansys CFX for 3 different loading conditions (Underload, Full load and Over load) to identify and mitigate cavitation prone areas. Blade loading profiles were analysed to detect the sudden pressure drops responsible for manifesting cavitation and Elliptic Ratio, the key blade design parameter was optimized to control sudden pressure drops and enhance cavitation resistance of runner. The effectiveness of the modified design was validated using Thoma’s cavitation number to ensure improved resistance against cavitation. Our study concluded that the modified runner exhibited reduced cavitation intensity demonstrating the potential for improved operational reliability. This novel framework, optimizing the elliptic ratio of blade to mitigate cavitation, establishes a benchmark for cavitation control in Francis turbine and can be extended to other reaction turbines as well.
### Cavitation erosion characterization of Cu-Ni alloy laser cladding layer under the action of high-pressure water jet
2025, Journal of Alloys and Compounds Show abstract The large number of applications of Cu-Ni alloys in the marine field and the prevalence of cavitation erosion (CE) necessitate a thorough examination and improvement of the CE characterization of Cu-Ni alloys. In this study, Cu-Ni alloy cladding layers were prepared by laser cladding technology and compared with the Cu-Ni casting alloy of the same composition. It was found that the grains of the Cu-Ni cladding layer were more uniform, with an average size of only 5μm, while the grain size of the casting alloy varied greatly, ranging from approximately 20 to 40μm. The cladding layer with fine grains and uniform structure reduced the weight loss of the Cu-Ni sample by 16.8 % during the CE stabilization period. Additionally, the influence of Ni content on the CE resistance of Cu-Ni alloy cladding layers was investigated. The results showed that during the CE stabilization period, the average weight loss rates of the three Cu-Ni cladding layers (Ni25Cu, Ni35Cu, and Ni45Cu) were 81.07 mg/h, 77.25 mg/h and 61.21 mg/h, respectively. With the increase in Ni content, the weight loss at each stage of CE decreased. The doping of Ni into the Cu lattice caused lattice distortion, promoted dislocation formation and grain refinement, thereby increasing the surface hardness. That made it more difficult for bubbles to destroy the surface, and limited the internal expansion of defects caused by CE.
### Machine learning-based efficiency prediction of Francis type hydraulic turbines through comprehensive performance testing
2025, Proceedings of the Institution of Mechanical Engineers Part A Journal of Power and Energy
### A review of condition monitoring in Francis turbines for predictive maintenance
2025, JVC Journal of Vibration and Control
### A comprehensive investigation of hydraulic performance and internal flow characteristics of Francis turbine model at part load cavitating flow for various turbulence models
2025, Engineering Applications of Computational Fluid Mechanics
© 2024 The Author(s). Published by Elsevier Ltd.
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Master Water PPM: Meaning, Ideal Levels, and How to Reduce
You may have been in this situation before: you take a glass of water and swallow it, but the taste is somewhat different than expected, or on the other hand, the dishes are clean but washed in the dishwasher without a trace of the detergent. These are all indications that the water that you are using may not be of good quality. One of the most critical parameters in evaluating water quality is water PPM (parts per million). But what does it even entail, and why should it matter to you? In this blog, we shall explain in detail the ppm water meaning, the types of PPM in water that are acceptable for drinking purposes, and, most importantly, how you can get rid of it when necessary.
Table of Contents
1. What is Water PPM?
2. How does PPM affect water Quality?
3. How to Test Water PPM?
4. Ideal PPM for Drinking Water: What is a Good PPM?
5. How to Reduce Water PPM?
1. What is Water PPM?
Water PPM (Parts Per Million) is a term used to express the amount of some constituents, such as minerals, salts, or chemicals in water. Simply, if you have one part of the base solution in one million volumes of water, that is in PPM equal to one. For instance, if your water consists of 300 parts per million of dissolved solids, that is, minerals, salts, etc, then its PPM level is 300.
So why should this concern you? Quantifying the amount of PPM present in water is important since it will help assess the level of contamination in the given water. If the pollution is high and the water PPM is above the average health standard, degenerative diseases could result due to over-intake of dissolved minerals. On the contrary, low-PPM hydrated water is considered to be safe for drinking.
It is possible to obtain PPM values in water through a TDS (Total Dissolved Solids) meter, which only gives the average water quality for a sample. The TDS meter quickly displays the concentration of total dissolved solids in a water sample, which is the straightforward definition of the instrument's operational principle. Measuring the TDS for drinking water is a simple and cost-effective method to determine water's safety for consumption.
PPM Water Meaning
Before discussing PPM meaning water, let’s understand its correlation with TDS regarding drinking water. TDS (Total Dissolved Solids) is a gauge of dissolved substances in the water, such as minerals, salts, and metals. PPM is just the unit for expressing the measurement of TDS. Put differently, TDS is the ‘what’, whereas PPM is the ‘how many.’
As an example, the answer is presented in PPM when measuring TDS in water. So when a TDS meter reads 150 PPM, it means that in one million parts of water, there are 150 parts of some dissolved solids.
2. How Does PPM Affect Water Quality?
As we have established the meaning of PPM in water, let us discuss its PPM and the quality of the water affected by it. The level of PPM determines the following properties of your water, to mention a few.
Taste and odor: An increase in the hard water ppm can produce a salty, bitter, and metallic taste. However, water with a low PPM is colorless, tasteless, and quite refreshing.
Potential health impacts: High PPM drinking water may contain minerals that pose health threats when taken at high levels over time. TDS water for drinking purposes demonstrated above the recommended guidelines can be associated with certain impurities, such as heavy metals, salts, etc.
Effects on appliances and piping: Water that is hard or rich in calcium and magnesium has a high ppm, meaning it contains many dissolved minerals. High levels of hard water ppm can clog appliances and pipelines, making them inefficient and expensive to maintain.
3. How To Test Water PPM?
Monitoring water's PPM is easy and straightforward. Let’s focus on the more frequently used techniques.
You may employ the following water PPM testing methods to check PPM level of water at home.
TDS Meters: These are simply hand-held meters that can give an almost instant reading of the water PPM. They are cheap, simple, and quick devices that help one gauge the quality of water available.
Water Quality Test Kits: These kits also come with strips containing certain reactants that color depending on the amount of dissolved solids in the water, thus approximating the PPM levels.
If there is any doubt about the results achieved, an individual may opt for assistance from a water testing service. They will offer you a close examination of what constituents exist within the water and how to deal with them.
4. Ideal PPM for Drinking Water: What is a Good PPM?
Knowing the acceptable PPM levels in drinking water for health and taste perspectives is essential. Here is how to assess the PPM levels of your water source.
Interpreting Your Water's PPM
Low PPM (0-50): This shows the availability of pure water. However, water with these PPM levels does not have essential elements that other waters have, hence the problem of taste and also adding nutritional value.
Moderate PPM (51-300): This range is regarded as the best for drinking water. It is normally enough to avoid health problems associated with both excess minerals and too much purification.
High PPM (301-500): This water may be hard, and the high mineral concentration may cause an unpalatable taste. It may be non-toxic and still be considered for use, but it is not recommended.
Very High PPM (500+): The water containing above 500 PPM is marked unfit for human consumption by the EPA. It can get into your body and equipment as poison.
Recommended TDS Levels for Drinking Water
The World Health Organization (WHO) recommends that the TDS for drinking water levels preferably range from 50 to 300 PPM for optimal taste and health. The Environmental Protection Agency (EPA) however places a threshold of 500 PPM and considers any water below that value safe for human consumption (Read about What TDS is Safe to Drink?). Beyond this, if your water is above this threshold, it would be wise to consider filtration methods to reduce water TDS levels.
5. How to Reduce Water PPM?
If you have tested the water and verified that the levels of PPM as set are higher for your water, do not panic. There are useful importations that can help you cut it down.
Filtration Methods
Reverse osmosis systems are very effective in eliminating the dissolved solids in water, hence the drastic reduction of the levels of PPM. They work by passing water through a screen that allows only water particles to pass while leaving all other impurities behind. You may want to purchase reliable RO systems from SimPure.
SimPure reverse osmosis systems are intended to efficiently eliminate all dissolved solids, including minerals, salts, and unwanted elements such as lead and chlorine, specific bacteria, etc. Thus, they considerably reduce the TDS levels of treated water, ultimately delivering pure and pleasant drinking water. For many householders, a SimPure reverse osmosis system would be the most recommended option as it helps to minimize hard water PPM levels and enhance the overall quality of water.
Distillation: Here, water is boiled, causing steam to be produced. The steam produced is then cooled to give back water, leaving behind the other components. Thus, very little PPM containing water can be achieved.
Ion Exchange: This process reduces the PPM of hard water by softening it using calcium, magnesium, sodium, and potassium fibrous beads, which are common in water softeners.
Conclusion
Knowing the water ppm level is one of the basic requirements if you want to maintain the quality of your water. Too low a PPM level may mean your water has been de-mineralized, while too high a PPM may present health risks and damage some appliances. Testing is recommended, along with filtration techniques when appropriate, to maintain water quality and safety for drinking. Test your water right now and take action to avoid unhealthy levels of water PPM!
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Further split graphs known to be Class 1 and a characterization of subgraph-overfull split graphs - ScienceDirect
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Discrete Applied Mathematics
Volume 345, 15 March 2024, Pages 114-124
Further split graphs known to be Class 1 and a characterization of subgraph-overfull split graphs
Author links open overlay panel Cintia Izabel Cararo a, Sheila Morais de Almeida a, Cândida Nunes da Silva b
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Highlights
•Correction of a flaw in the proof that odd maximum degree split graphs are Class 1.
•Proof that certain split graphs with even maximum degree are Class 1.
•Characterization of the split graphs which are subgraph-overfull, and thus Class 2.
Abstract
The chromatic index, χ′(G), is the smallest integer k for which a graph G has a proper edge coloring. The Classification Problem involves determining whether a graph G is Class 1 (χ′(G)=Δ(G)) or Class 2 (χ′(G)=Δ(G)+1). It is known that subgraph-overfull graphs must be Class 2. In this paper we are concerned with the Classification Problem for split graphs G[Q∪S] where Q is a clique and S is an independent set. When Δ(G) is odd, G is known to be Class 1. The original proof presented by Chen et al. (1995) has a minor flaw which we detail in this paper while also clarifying that it does not compromise their result. We prove that their technique can be adapted in a non-trivial way to show that some split graphs with even Δ(G) are also Class 1. We show that to solve the Classification Problem for split graphs it suffices to consider that all vertices in Q have degree Δ(G). Considering the subset X of S of the vertices of degree at most Δ(G)/2, we show that if the neighborhood of X has at least ⌊|Q|/2⌋ vertices, then G is Class 1; in the remaining cases we characterize the subgraph-overfull split graphs.
Introduction
Throughout this paper, G denotes a simple, finite, undirected, and connected graph with vertex set V(G) and edge set E(G). For any v in V(G), the neighborhood of v, denoted by N G(v), is the set of vertices adjacent to v. The closed neighborhood of v, denoted by N G[v], is the set N G(v)∪v. For X⊆V(G), the set N G(X) is the union of all N G(v) such that v∈X. The degree of a vertex v is denoted by d G(v). A graph is chordal if it has no induced cycle with more than three vertices. A clique is a set of pairwise adjacent vertices and a stable set is a set of pairwise nonadjacent vertices. A split graph G is a graph whose vertices can be partitioned into a clique Q G and a stable set S G. Alternatively, a graph G is split if and only if G and G¯ are chordal graphs. When there is no ambiguity, we hide the reference to the graph in the notations.
The Classification Problem consists of classifying G as Class 1, if G has chromatic index equal to Δ(G), or Class 2, otherwise. In the early 80s, Holyer proved that this problem is NP-complete. In the same decade, Johnson selected some graph theory problems known to be NP-complete and graph classes that he considered “having broad algorithmic significance” for these problems. For each problem and class, he either gave the known complexity status or, when it was an open problem, he made an educated guess as to how easy it seemed to identify the complexity of that problem in that class. Among these problems was the Classification Problem, which for most classes the complexity status was open. This increased the interest in this problem which was intensely studied ever since. Almost four decades after Johnson’s publication, the Classification Problem is the one with more open classes among those chosen by Johnson. Table 1 shows a clipping of the known and open results for subclasses of perfect graphs listed by Johnson. The 1985 column refers to the complexity of the problem in the year of Johnson’s publication and the 2022 column refers to an update presented in. For cases where the complexity is known, the value “N” is used for NP-complete, and the value “P” for polynomial. For the other cells, the symbol “O?” means that Johnson believed to be “apparently open, but possibly easy to resolve”. Even though Johnson thought it could be easy to determine the complexity of the open classes of Table 1, the little progress made since 1985 is an evidence of the difficulty of the problem.
In it is proved that the limit of the fraction of chordal graphs with n vertices that are split approaches one as n goes to infinity; so almost all chordal graph is a split graph. This emphasizes the importance of studying the Classification Problem in the class of split graphs.
Given a graph G with an edge coloring, a color class is a maximal set of edges that are colored the same. A set M of pairwise non-adjacent edges is called a matching. In an edge coloring, each color class is a matching. Given a graph G, the maximum number of edges in a matching is |V(G)|/2. Hence, when |E(G)|>Δ(G)|V(G)|/2, G is called overfull and must clearly be Class 2. A vertex v such that N G(v)=V(G)∖{v} is a universal vertex. Theorem 1 gives the chromatic index of graphs with a universal vertex.
Theorem 1
Let G be a graph with a universal vertex, then G is Class 1 if and only if G is not overfull. □
Let H be a subgraph of G with Δ(H)=Δ(G). If H is overfull, then G is subgraph-overfull. The Overfull Conjecture, , states that every graph G with Δ(G)>|V(G)|/3 is Class 2 if and only if G is subgraph-overfull. Some partial results that are evidence for the validity of this conjecture are presented in, , . Let v be a vertex of G with d(v)=Δ(G). If G[N[v]] is overfull, then G is neighborhood-overfull. There is also the following well-known conjecture concerning chordal graphs:
Conjecture 2
Any Class 2 chordal graph is neighborhood-overfull.
Note that every neighborhood-overfull graph is subgraph-overfull and every subgraph-overfull graph is Class 2. For split graphs, being subgraph-overfull is equivalent to being neighborhood-overfull, as stated below.
Theorem 3
Let G be a split graph. If G is subgraph-overfull then G is neighborhood-overfull. □
A Latin square is a square matrix where in each row and column the elements are pairwise distinct. In 1995, Chen, Fu and Ko proved the theorem below by showing that if G is a split graph with odd Δ(G), then it is possible to use a suitable Latin square of order Δ(G) to obtain an edge coloring for G in polynomial time.
Theorem 4
If G is a split graph with odd Δ(G), then G is Class 1. □
The technique developed by Chen, Fu and Ko to prove Theorem 4 involves decomposing G into three edge-disjoint subgraphs: a complete graph and two bipartite graphs. They show that for each subgraph in this decomposition there is a subset of the cells of a Latin square of order Δ(G), which they call a color diagram (defined later), that can be used to obtain an edge coloring of the subgraph. Their proof relies on the fact that the color diagrams chosen have the monotonicity property, as they call it. We observed a minor flaw in their claim as it is not true that the diagrams they define are indeed always monotonic. On the other hand, we also observed that the diagrams actually needed in their proof do satisfy the monotonicity property, thus ensuring the correctness of the authors’ final result. A detailed discussion of these technical issues regarding Chen et al.’s proof for Theorem 4 is presented in Section 3. In fact, we present in this paper Chen et al.’s proof in order to be able to explain in the detail their minor flaw and why it does not compromise their conclusion. Such detailed discussion regarding Chen et al.’s proof allow us to clarify why the technique cannot be directly applied to graphs with even Δ(G) as well as to identify some graphs with even Δ(G) for which Chen et al.’s decomposition technique can be adapted (in a non-trivial way) to obtain an edge coloring with Δ(G) colors. Let G be a split graph with an even Δ(G), where every v∈Q has a degree d(v)=Δ(G), and let X be the subset consisting of all vertices in S whose degree is at most Δ(G)/2. Specifically, we prove that if there are at least ⌊|Q|/2⌋ vertices in the neighborhood of the vertices in X, then G is Class 1. We also show that this result implies that split graphs that are neighborhood-overfull must satisfy certain structural properties, thus obtaining a characterization for such graphs, presented in Section 4. For the sake of clarity, before presenting all the detailed and technical discussions regarding the proof of Theorem 4 and our contributions in this paper, we provide several definitions and properties relative to Latin squares and color diagrams in Section 2. These concepts are essential for understanding the proofs presented in Section 3.
Section snippets
Latin square and monotonic color diagrams
A Latin square of order k, denoted M k,k, is a square matrix k×k whose cells contain elements of a set of order k such that every element appears in each row and column exactly once. A Latin square M k,k is commutative if m i,j=m j,i, for 0≤i,j≤k−1, and is idempotent if the elements of the main diagonal are pairwise distinct. Latin squares are often used to obtain edge colorings for graphs by relating the Latin square to the adjacency matrix of the graph, specifying the color of each existing edge.
Edge coloring of split graphs
In the previous section we presented counterexamples to statements made by Chen et al. in the proof of Theorem 4. In this section, we present a proof for this theorem which relies on Lemma 20, Lemma 21 proved earlier in this paper, thus guaranteeing that the edge-coloring problem is solved when the maximum degree is odd. Since Lemma 20, Lemma 21 do not depend on the parity of k, we are able to show that the decomposition technique can be generalized (in a non-trivial way) to prove that
Characterization of subgraph-overfull split graphs
In the previous section, we presented structures that guarantee that a split graph is Class 1. In this sense, it is interesting to study the structure of split graphs with open chromatic index. In this section, we present structures that are forbidden for split graphs with undefined chromatic index, that is, we characterize split graphs that are subgraph-overfull and hence Class 2. Lemma 30 presents a necessary condition for a split graph G to be subgraph-overfull. This Lemma is used in
Final remarks
This paper contributes in several ways to the solution of the Classification Problem in split graphs; a long standing open problem. We observe a minor flaw in Chen et al.’s proof for Theorem 4 while also showing that it does not compromise their conclusion that split graphs with odd Δ(G) are Class 1. We extended the decomposition technique in a non-obvious way to show that certain split graphs with even Δ(G) are Class 1 as well. According to Theorem 28, the main contribution of this paper,
Acknowledgments
This study was financed in part by UTFPR, the Coordenação de Aperfeiçoamento de Pessoal de Nível Superior - Brasil – CAPES (Finance Code 001), and CNPq - National Council for Scientific and Technological Development - Brazil (grants 300934/2006-8, 482521/2007-4, 140709/2008-8, 428941/2016-8, and 420079/2021-1).
Recommended articles
References (15)
Cai L. et al.
NP-completeness of edge-colouring some restricted graphs
Discrete Appl. Math. (1991)
de Figueiredo C.M. et al.
Revising Johnson’s table for the 21st century
Discrete Appl. Math. (2022)
Johnson D.S.
The NP-completeness column: an ongoing guide
J. Algorithms (1985)
Perkovic L. et al.
Edge coloring regular graphs of high degree
Discrete Math. (1997)
Bender E.A. et al.
Almost all chordal graphs split
J. Aust. Math. Soc. A (1985)
Chen B.L. et al.
Total chromatic number and chromatic index of split graphs
J. Combin. Math. Combin. Comput. (1995)
Chetwynd A.G. et al.
The chromatic index of graphs of even order with many edges
J. Graph Theory (1984)
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https://en.wikipedia.org/wiki/Experiment_(probability_theory)
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1 Experiments and trials
2 Mathematical description
3 See also
4 References
5 External links
Experiment (probability theory)
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From Wikipedia, the free encyclopedia
Procedure that can be infinitely repeated, with a well-defined set of outcomes
This article is about the probabilistic model used in actual experiments. For a discussion about actual experiments, see experiment.
| |
| Part of a series on statistics |
| Probability theory |
| Probability + Axioms Determinism + System Indeterminism Randomness |
| Probability space Sample space Event + Collectively exhaustive events + Elementary event + Mutual exclusivity + Outcome + Singleton Experiment + Bernoulli trial Probability distribution + Bernoulli distribution + Binomial distribution + Exponential distribution + Normal distribution + Pareto distribution + Poisson distribution Probability measure Random variable + Bernoulli process + Continuous or discrete + Expected value + Variance + Markov chain + Observed value + Random walk + Stochastic process |
| Complementary event Joint probability Marginal probability Conditional probability |
| Independence Conditional independence Law of total probability Law of large numbers Bayes' theorem Boole's inequality |
| Venn diagram Tree diagram |
| v t e |
In probability theory, an experiment or trial (see below) is the mathematical model of any procedure that can be infinitely repeated and has a well-defined set of possible outcomes, known as the sample space. An experiment is said to be random if it has more than one possible outcome, and deterministic if it has only one. A random experiment that has exactly two (mutually exclusive) possible outcomes is known as a Bernoulli trial.
When an experiment is conducted, one (and only one) outcome results— although this outcome may be included in any number of events, all of which would be said to have occurred on that trial. After conducting many trials of the same experiment and pooling the results, an experimenter can begin to assess the empirical probabilities of the various outcomes and events that can occur in the experiment and apply the methods of statistical analysis.
Experiments and trials
[edit]
Random experiments are often conducted repeatedly, so that the collective results may be subjected to statistical analysis. A fixed number of repetitions of the same experiment can be thought of as a composed experiment, in which case the individual repetitions are called trials. For example, if one were to toss the same coin one hundred times and record each result, each toss would be considered a trial within the experiment composed of all hundred tosses.
Mathematical description
[edit]
Main article: Probability space
A random experiment is described or modeled by a mathematical construct known as a probability space. A probability space is constructed and defined with a specific kind of experiment or trial in mind.
A mathematical description of an experiment consists of three parts:
A sample space, Ω (or S), which is the set of all possible outcomes.
A set of events , where each event is a set containing zero or more outcomes.
The assignment of probabilities to the events—that is, a function P mapping from events to probabilities.
An outcome is the result of a single execution of the model. Since individual outcomes might be of little practical use, more complicated events are used to characterize groups of outcomes. The collection of all such events is a sigma-algebra . Finally, there is a need to specify each event's likelihood of happening; this is done using the probability measure function, P.
Once an experiment is designed and established, ω from the sample space Ω, all the events in that contain the selected outcome ω (recall that each event is a subset of Ω) are said to “have occurred”. The probability function P is defined in such a way that, if the experiment were to be repeated an infinite number of times, the relative frequencies of occurrence of each of the events would approach agreement with the values P assigns them.
As a simple experiment, we may flip a coin twice. The sample space (where the order of the two flips is relevant) is {(H, T), (T, H), (T, T), (H, H)} where "H" means "heads" and "T" means "tails". Note that each of (H, T), (T, H), ... are possible outcomes of the experiment. We may define an event which occurs when a "heads" occurs in either of the two flips. This event contains all of the outcomes except (T, T).
See also
[edit]
Probability space
References
[edit]
^ Albert, Jim (21 January 1998). "Listing All Possible Outcomes (The Sample Space)". Bowling Green State University. Archived from the original on 16 October 2000. Retrieved June 25, 2013.
^ Papoulis, Athanasios (1984). "Bernoulli Trials". Probability, Random Variables, and Stochastic Processes (2nd ed.). New York: McGraw-Hill. pp. 57–63.
^ "Trial, Experiment, Event, Result/Outcome - Probability". Future Accountant. Retrieved 22 July 2013.
External links
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Convert n to kN - Conversion of Measurement Units
Convert n to kN - Conversion of Measurement Units
Convert newton to kilonewton
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How many n in 1 kN? The answer is 1000. We assume you are converting between newton and kilonewton. You can view more details on each measurement unit: n or kN The SI derived unit for force is the newton. 1 newton is equal to 0.001 kN. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between newtons and kilonewtons. Type in your own numbers in the form to convert the units!
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Definition: Newton
In physics, the newton (symbol: N) is the SI unit of force, named after Sir Isaac Newton in recognition of his work on classical mechanics. It was first used around 1904, but not until 1948 was it officially adopted by the General Conference on Weights and Measures (CGPM) as the name for the mks unit of force.
Definition: Kilonewton
The SI prefix "kilo" represents a factor of 10 3, or in exponential notation, 1E3.
So 1 kilonewton = 10 3 newtons.
The definition of a newton is as follows:
In physics, the newton (symbol: N) is the SI unit of force, named after Sir Isaac Newton in recognition of his work on classical mechanics. It was first used around 1904, but not until 1948 was it officially adopted by the General Conference on Weights and Measures (CGPM) as the name for the mks unit of force.
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https://allen.in/jee/chemistry/alkane
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Alkane
1.0Introduction
Alkanes are a group of chemical compounds made up only of carbon and hydrogen atoms, and all the atoms in alkanes are joined by single covalent bonds. Because they contain only single bonds, they are called saturated hydrocarbons.
Alkanes follow a general molecular formula:+CnH2n+2
This means that for every ‘n’ carbon atom, there will be 2n+2 hydrogen atoms.
2.0Structure and Bonding
In an alkane molecule:
This results in a stable, saturated structure. Chemists often use line-angle formulas to represent alkanes because they're quick to draw and easy to interpret.
For example, the simplest alkane, methane, contains just one carbon atom and four hydrogen atoms.
Its formula is CH4, and its structure looks like a central carbon bonded to four hydrogen atoms.
Long-Chain Alkanes
In larger alkanes, carbon atoms link to one another in a chain using single bonds. Each carbon is also bonded to enough hydrogen atoms so that it makes four total bonds.
An example of this is octane, which has eight carbon atoms.
Its molecular formula is C₈H₁₈.
List of the first ten alkanes, along with their molecular formulas:
| | |
--- |
| Name | Molecular Formula |
| Methane | CH₄ |
| Ethane | C₂H₆ |
| Propane | C₃H₈ |
| Butane | C₄H₁₀ |
| Pentane | C₅H₁₂ |
| Hexane | C₆H₁₄ |
| Heptane | C₇H₁₆ |
| Octane | C₈H₁₈ |
| Nonane | C₉H₂₀ |
Name
Molecular Formula
Methane
CH₄
Ethane
C₂H₆
Propane
C₃H₈
Butane
C₄H₁₀
Pentane
C₅H₁₂
Hexane
C₆H₁₄
Heptane
C₇H₁₆
Octane
C₈H₁₈
Nonane
C₉H₂₀
3.0Physical Properties of Alkanes
4.0Structural Formulas of Alkanes
Here are the condensed structural formulas for the first five straight-chain alkanes:
| | | |
---
| Name | Molecular Formula | Condensed Structure |
| Methane | CH₄ | CH₄ |
| Ethane | C₂H₆ | CH₃–CH₃ |
| Propane | C₃H₈ | CH₃–CH₂–CH₃ |
| Butane | C₄H₁₀ | CH₃–CH₂–CH₂–CH₃ |
| Pentane | C₅H₁₂ | CH₃–CH₂–CH₂–CH₂–CH₃ |
Name
Molecular Formula
Condensed Structure
Methane
CH₄
CH₄
Ethane
C₂H₆
CH₃–CH₃
Propane
C₃H₈
CH₃–CH₂–CH₃
Butane
C₄H₁₀
CH₃–CH₂–CH₂–CH₃
Pentane
C₅H₁₂
CH₃–CH₂–CH₂–CH₂–CH₃
Alkanes can be represented in different ways:
Sometimes, different compounds may share the same molecular formula but have different structures. These are called structural isomers.
5.0Types of Alkanes
Alkanes can exist in three main forms:
Even if the number of carbon atoms is the same, the structure can be different. For example, a molecule with 8 carbon atoms can exist as:
6.0Alkyl Groups and Substitution Reactions in Alkanes
When a substituent—such as a halogen—attaches itself to an alkane molecule, it does so by replacing one of the hydrogen atoms bonded to a carbon. This results in the formation of a new bond between the carbon atom and the substituent.
The reaction between methane (CH₄) and chlorine (Cl₂) as an example. When methane reacts with chlorine (in the presence of light or heat), one hydrogen atom from methane is substituted by a chlorine atom. This gives rise to a new compound called chloromethane (CH₃Cl). In this compound, a CH₃ group (methyl group) is bonded to a chlorine atom.
Alkyl Groups
When an alkane loses a hydrogen atom, the resulting fragment is called an alkyl group. This group still contains carbon and hydrogen atoms, but it's now ready to bond with other atoms or groups.
For convenience, chemists often represent an alkyl group with the symbol R—just like halogens are often denoted by the letter X.
For example, if we remove one hydrogen atom from methane (CH₄), we get the methyl group (CH₃–). This methyl group can now form a bond with another atom, such as chlorine, to make chloromethane (CH₃Cl).
This substitution reaction can be generalized as:
RH + X₂ → RX + HX
Where:
Table of Contents
Frequently Asked Questions
Why are alkanes called saturated hydrocarbons?
Alkanes are called saturated because all the carbon-carbon bonds are single bonds, and each carbon atom is fully "saturated" with hydrogen atoms—meaning it has formed the maximum number of bonds possible.
What is an alkyl group?
An alkyl group is formed when one hydrogen atom is removed from an alkane. It is a part of a molecule that can bond with other atoms or groups and is usually represented by the symbol R.
What is the difference between an alkane and an alkyl group?
An alkane is a complete, stable hydrocarbon molecule (e.g., CH₄), while an alkyl group is a fragment of an alkane that results from the loss of a hydrogen atom (e.g., CH₃–). The alkyl group is reactive and can form new bonds.
What is a substitution reaction in alkanes?
In a substitution reaction, one or more hydrogen atoms in an alkane are replaced by another atom or group—commonly a halogen (like chlorine or bromine). For example: CH₄ + Cl₂ → CH₃Cl + HCl
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Hum Reprod Update
. 2015 Aug 7;21(6):748–761. doi: 10.1093/humupd/dmv038
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Menstrual physiology: implications for endometrial pathology and beyond
Jacqueline A Maybin
Jacqueline A Maybin
1 MRC Centre for Reproductive Health, University of Edinburgh, The Queen's Medical Research Institute, 47 Little France Crescent, Edinburgh EH16 4TJ, UK
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1, Hilary OD Critchley
Hilary OD Critchley
1 MRC Centre for Reproductive Health, University of Edinburgh, The Queen's Medical Research Institute, 47 Little France Crescent, Edinburgh EH16 4TJ, UK
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1,
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1 MRC Centre for Reproductive Health, University of Edinburgh, The Queen's Medical Research Institute, 47 Little France Crescent, Edinburgh EH16 4TJ, UK
Correspondence address. E-mail: hilary.critchley@ed.ac.uk
Received 2015 Apr 14; Revised 2015 Jul 3; Accepted 2015 Jul 8; Issue date 2015 Nov.
© The Author 2015. Published by Oxford University Press on behalf of the European Society of Human Reproduction and Embryology.
This is an Open Access article distributed under the terms of the Creative Commons Attribution License ( which permits unrestricted reuse, distribution, and reproduction in any medium, provided the original work is properly cited.
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PMCID: PMC4594618 PMID: 26253932
Abstract
BACKGROUND
Each month the endometrium becomes inflamed, and the luminal portion is shed during menstruation. The subsequent repair is remarkable, allowing implantation to occur if fertilization takes place. Aberrations in menstrual physiology can lead to common gynaecological conditions, such as heavy or prolonged bleeding. Increased knowledge of the processes involved in menstrual physiology may also have translational benefits at other tissue sites.
METHODS
Pubmed and Cochrane databases were searched for all original and review articles published in English until April 2015. Search terms included ‘endometrium’, ‘menstruation’, ‘endometrial repair’, ‘endometrial regeneration’ ‘angiogenesis’, ‘inflammation’ and ‘heavy menstrual bleeding’ or ‘menorrhagia’.
RESULTS
Menstruation occurs naturally in very few species. Human menstruation is thought to occur as a consequence of preimplantation decidualization, conferring embryo selectivity and the ability to adapt to optimize function. We highlight how current and future study of endometrial inflammation, vascular changes and repair/regeneration will allow us to identify new therapeutic targets for common gynaecological disorders. In addition, we describe how increased knowledge of this endometrial physiology will have many translational applications at other tissue sites. We highlight the clinical applications of what we know, the key questions that remain and the scientific and medical possibilities for the future.
CONCLUSIONS
The study of menstruation, in both normal and abnormal scenarios, is essential for the production of novel, acceptable medical treatments for common gynaecological complaints. Furthermore, collaboration and communication with specialists in other fields could significantly advance the therapeutic potential of this dynamic tissue.
Keywords: endometrium, inflammation, angiogenesis, progesterone, hypoxia
Introduction
The phenomenon of human menstruation has been shrouded in mystery throughout history. Many questions regarding menstrual physiology remain unanswered, not least ‘why does it happen?’ Historically, menstruation has been regarded negatively. Historia Naturalis states ‘Wine sours if they pass, vines wither, grass dies, and buds are blasted. Should a menstruating woman sit under a tree, the fruit will fall. A looking glass will discolour at her glance, and a knife turn blunt’ (Pliny, AD 77–79). Aristotle viewed menstruation as an outward sign of female inferiority, a view that persisted into the nineteenth century and beyond. A leading British psychiatrist in 1874 wrote ‘with one week of the month more or less sick and unfit for hard work she is intellectually handicapped’. A pioneering nineteenth century Scottish gynaecologist claimed, ‘young girls should not play music or read serious books because it makes much mischief with their menstrual cycle’. Hence menstruation was regarded as incapacitating and, in turn, intellect dangerous for menstrual physiology.
These negative connotations of menstruation are inextricably linked to the lower social position of women in society. Currently, global differences in women's rights and status have a dramatic impact on reproductive health and consequently their morbidity and mortality. As women receive high quality education, begin working outside the home, gain the right to vote and have easy access to emergency healthcare and birth control, the ‘taboo’ of menstruation weakens. Therefore, some see the attitude of a society to menstruation as a barometer for civilization and equality. When in the USA in the 1960s, it was suggested that women lacked the ability to hold positions of responsibility and power due to their menstrual cycle, eminent US endocrinologist Estelle Ramage counteracted, ‘In man, the shedding of blood is always associated with injury, disease, or death. Only the female half of humanity is seen to have the magical ability to bleed profusely and still rise phoenix-like each month from the gore’. Despite this positive outlook, historical negative connotations of menstruation still have a significant impact in current society, including the perceptions and expectations of women and their healthcare providers.
However, as women undertake positions of responsibility in the workplace and home, abnormal menstruation can cause significant socio-economic problems. Abnormal menstrual bleeding affects 20–30% of premenopausal women (RCOG, 2011), and more than 800 000 women seek treatment annually in the UK (NICE, 2007). A US study demonstrated financial losses of >$2000 per patient each year due to work absence and home management costs (Frick et al., 2009). Although time has proven that physiological menstruation is not a barrier to female success; family and career responsibilities may become impossible if heavy or painful bleeding occurs. Due to advances in family planning, women in developed countries now can expect greater than 400 episodes of menstruation in their lifetime. This is in stark contrast to our ancestors and women in very underdeveloped countries, who have ∼40 menstrual bleeds due to multiple pregnancy and long spells of lactational amenorrhoea (Short, 1976). In this way, menstrual abnormalities are a relatively modern disorder.
As societies' view of menstruation changes for the better, the views of individual women suffering from common menstrual problems remain understandably negative. This review article aims to provide scientific evidence of both facets of menstrual physiology. First, how normal menstruation could contribute to scientific and clinical breakthroughs in all areas of health and disease and conversely, how aberrations in menstrual physiology can result in significant reproductive disorders with a severe impact on quality of life (Fig. 1). As we detail the physiology of menstruation, we aim to highlight the clinical applications of what we know, the key questions that remain and the scientific and medical possibilities for the future.
Figure 1.
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The relevance of menstrual physiology. The perimenstrual endometrium (centre) is exposed to inflammation and hypoxia. Stem cells and EMT are involved at menses to enable scar-free repair (light blue). Aberrations in these processes can lead to gynaecological disorders (mid-blue). Study of endometrial physiology may help delineate the pathogenesis of a number of disorders in other tissue sites (dark blue).
Methods
Pubmed and Cochrane databases were searched for all original and review articles published in English until April 2015. Search terms included ‘endometrium’, ‘menstruation’, ‘endometrial repair’, ‘endometrial regeneration’ ‘angiogenesis’, ‘decidualization’, ‘inflammation’, ‘heavy menstrual bleeding (HMB)’ and ‘menorrhagia’. We reviewed the manuscripts and included them as appropriate.
Results
To bleed or not to bleed?
Human females are one of the few species that menstruate, alongside old world primates, elephant shrews and fruit bats. The ovarian steroid hormones regulate endometrial function and human menstruation. After human ovulation, the corpus luteum secretes high levels of progesterone to maintain endometrial receptivity should fertilization occur. In the absence of pregnancy the corpus luteum regresses, causing a sharp decline in circulating progesterone levels. This triggers a local inflammatory response in the endometrium involving infiltration of leukocytes, cytokine release, oedema and activation of matrix metalloproteinases (Jabbour et al., 2006). The result is tissue breakdown and shedding of the upper two-thirds of the endometrium (the functional layer) during the menstrual phase of the cycle (see ‘What causes menstruation’ section). However, in non-menstruating species tissue breakdown and bleeding do not occur in response to progesterone withdrawal. Instead of shedding, considerable remodelling and reabsorption of the endometrium takes place.
Many theories for why women menstruate have been proposed, including defence against pathogens contained in sperm or energy efficiency of shedding versus endometrial maintenance. However, these theories do not account for differences between menstruating and non-menstruating mammals or the evolutionary basis of menstruation (Finn, 1996). Current evidence favours the spontaneous decidualization hypothesis. During the secretory phase (post ovulation until menstruation) of the menstrual cycle progesterone acts upon an estrogen primed endometrium. This causes decidualization; converting the elongated endometrial stromal cells into more spherical decidual cells and increasing their production of prolactin, insulin-like growth factor binding protein-1 (IGBP-1) and glycogen (Brosens et al., 1999; Dunn et al., 2003). Decidualization is initiated by cAMP and occurs in the perivascular stromal cells before spreading in an outward ‘wave’ across the stromal compartment. In women, and indeed all of the menstruating species, decidualization occurs ‘spontaneously’ prior to implantation. In contrast, the endometrium of non-menstruating mammals only undergoes decidualization when there is contact between the embryo and endometrium, i.e. at the time of implantation (Finn, 1998).
There is a strong correlation between the degree of trophoblast invasion during placental development and the extent of decidualization (Finn, 1996). Of note, the human endometrium undergoes the most extensive decidualization process and demonstrates the greatest degree of trophoblast invasion of all species (Ramsey et al., 1976). This extensive and spontaneous decidualization reaction is thought to confer maternal immunotolerance to the partially allogenic embryo, allowing controlled placental invasion (Emera et al., 2012). In addition, spontaneous decidualization may provide a maternal screen for genetically abnormal embryos. Many human pre-implantation embryos contain genetic aneuploidies and chromosomal imbalances, similar to those found in cancer cells. It seems prudent that the maternal environment should provide some selection over the embryos that will invade the endometrium. This hypothesis is supported by findings in women with recurrent miscarriage, where a higher proportion of poor quality embryos implant into a suboptimally decidualized endometrium (Salker et al., 2012). Therefore, menstruation is obligatory in the absence of pregnancy, as spontaneous decidualization of the endometrium has taken place. This may be viewed as an inevitable consequence of reproductive quality control but an additional benefit has also been proposed (Blanks and Brosens, 2013). Repeated shedding of the endometrium necessitates complete repair and regeneration of the denuded surface. Therefore, events that would only otherwise occur after parturition are repeated monthly. This may bestow upon the human endometrium an extraordinary ability to adapt to optimize function and would explain why most women suffering from recurrent miscarriage eventually achieve a successful pregnancy (Blanks and Brosens, 2013). Hence, there may be an evolutionary benefit to menstruation that explains its occurrence, and persistence, in women. So what can we learn from this physiological process that has been so carefully preserved in women and what are the consequences when aberrations occur?
Menstruation: a model of self-limiting inflammation?
The menstrual endometrium displays the classic hallmarks of inflammation, including tissue oedema and influx of immune cells. This inflammatory process that occurs in the endometrium at menstruation is entirely physiological and tightly regulated to prevent loss of function (Critchley et al., 2001). Outwith the reproductive tract, this physiological inflammation does not occur. However, the ovary and endometrium display repeated inflammation throughout a woman's reproductive lifespan at ovulation and menstruation, respectively (Rae and Hillier, 2005). Delineation of the factors involved and their regulation may lead to therapeutic benefits for gynaecological conditions such as heavy menstrual bleeding (HMB) and may be applicable to a host of inflammatory disorders at other tissue sites.
What causes menstruation?
Progesterone withdrawal
It is widely accepted that the sharp decline in circulating progesterone levels due to corpus luteum demise is the trigger for menstruation in women. Human studies using progesterone antagonists during the secretory phase have mimicked the events of menstruation, providing proof that progesterone withdrawal is important in menstrual physiology. Administration of mifepristone during the mid-secretory phase has increased our knowledge of local endometrial events during human menstruation, revealing an increase in endometrial inflammatory mediators, such as cyclo-oxygenase (COX-2), nuclear factor (NF)κB and interleukin (IL)-8 (also known as CXCL8) (Critchley et al., 1999a, 2003). Studies in the rhesus macaque have confirmed the importance of progesterone withdrawal in the induction of menstruation (McClellan et al., 1984; Nayak et al., 2000). Menstruation was artificially induced in macaques by surgical removal of ovaries followed by 14 days of estrogen priming prior to insertion of a progesterone capsule to mimic the secretory phase. Consistent with findings in women, removal of the progesterone implant resulted in menstruation, even when estradiol exposure was maintained. This finding emphasizes the dominance of progesterone withdrawal over estradiol withdrawal for menses induction. In contrast, when attempting to induce simulated menstruation in the scientifically versatile murine model, progesterone withdrawal was insufficient for induction of bleeding (Finn and Pope, 1984; Brasted et al., 2003; Menning et al., 2012; Rudolph et al., 2012; Cousins et al., 2014). This problem was surmounted by an injection of arachis oil into the uterine lumen when progesterone levels are high. This ‘induced injury’ resulted in pre-implantation decidualization of the murine endometrium, analogous to naturally occurring mid-secretory events in the macaque and human. Hence, the murine model of simulated menstruation reiterates the importance of decidualization prior to progesterone withdrawal in menstrual physiology.
Further support for the key role of the decidualized endometrial stromal cell in menstrual induction is derived from human in vivo and in vitro studies. Of note, the progesterone receptor (PR) has at least two isoforms, PRA and PRB, which act as transcriptional regulators of progesterone responsive genes (Graham et al., 1995; Graham and Clarke, 1997). Although the basal endometrial layer shows persistent PR expression throughout the menstrual cycle, PR has differing temporal and locational expression in the functional layer (Lessey et al., 1988; Snijders et al., 1992). PR is widely present during the proliferative phase, but there is a significant decline in glandular epithelial cells of the functional layer during the secretory phase. In contrast, PR persists in the stromal compartment of the functional layer throughout the secretory phase, particularly in the perivascular region. Immunohistochemical analysis of human tissue revealed that PRA is the predominant isoform during the secretory phase, with PRB declining in both stromal and glandular cells in the latter half of the cycle (Wang et al., 1998; Brosens et al., 1999; Mote et al., 2001). Hence, endometrial stromal cells remain responsive to progesterone throughout the secretory phase. Gene microarray-based studies have been reviewed in Dey et al. (2004) and showed that analysis of mid-secretory uterine tissue, ex vivo progesterone/PR antagonist-treated endometrium, treated in vitro decidualized stromal cells and uterine tissue from PR-deficient mice have identified a panel of progesterone responsive genes that may be important for implantation. Hence the mid-secretory phase decidualized stromal cells retain PR expression and confer maximal progesterone responsiveness, priming the endometrium to respond to progesterone withdrawal.
In 2001, Kelly et al. published the hypothesis that local endometrial events following progesterone withdrawal occur in two phases (Kelly et al., 2001). The initial phase involves an influx of cytokines and prostaglandins (PG) to the endometrium that is dependent on an efficient response of the perivascular decidualized stromal cells to decreasing levels of the anti-inflammatory hormone progesterone (Catalano et al., 2007; Evans and Salamonsen, 2014). The second phase occurs as a consequence of increased cytokine production and results in an influx of leukocytes to the endometrial environment, activation and release of matrixmetalloproteinases (MMPs) and destruction of the extracellular matrix (ECM). This lytic phase is thought to be independent of PR actions. This hypothesis was supported by an elegant study in ovariectomized macaques, where progesterone implants were removed as normal at the end of the simulated cycle but replaced at staggered time-points from 12 to 72 h after initial withdrawal (Slayden and Brenner, 2006). Replacement up to 24 h after withdrawal prevented menstruation and prevented increases in endometrial MMP1, 2 and 3. Replacement after 36 h had no effect on menstruation and partially blocked MMP production, with significantly less endometrial MMP2 expression. More recently, these findings have been replicated in the murine model of simulated menstruation (Wang et al., 2013). These studies demonstrate a temporal progesterone deprived threshold, over which menstruation becomes inevitable.
Endometrial inflammation and leukocyte traffic
Although progesterone withdrawal has an undeniable role in the initiation of menstruation and MMPs are widely accepted as the mediators of endometrial breakdown (Marbaix et al., 1996), the intermediate mechanisms of menstruation remain under investigation. Progesterone withdrawal regulates phase one of menstruation, by up-regulating local cytokine presence (Hannan et al., 2004; Jones et al., 2004). However, phase two occurs despite progesterone replacement after the critical threshold suggesting subsequent, independent regulation. Further evidence for these downstream regulatory factors comes from observational studies of MMP expression in human endometrial tissue. MMPs have the ability to degrade all components of the ECM and are up-regulated at the time of menstruation as a result of progesterone withdrawal (Marbaix et al., 1996; Vassilev et al., 2005). However, MMP expression in the perimenstrual phase is limited to the functional endometrial layer despite the global hormonal changes and persistent PR expression in the basal layer, suggesting a more local tissue site-specific regulation.
Gene microarray analysis of endometrial tissue biopsies collected from women during the mid-secretory phase when compared with those taken following progesterone withdrawal has identified potential gene candidates involved in the regulation of menstruation. These studies revealed an increase in CXCL8 and cyclo-oxygenase (COX)-2 following progesterone withdrawal (Critchley et al., 1999a; Catalano et al., 2007). COX is the rate-limiting enzyme in the synthesis of PG and is present in two isoforms. COX-1 is widely expressed in many tissues, whereas COX-2 is highly inducible. PGE 2 and F 2α have important reproductive functions (Critchley et al., 2006). Loss of EP2, a PGE 2 receptor, resulted in impaired ovulation and reduced litter size (Kennedy et al., 1999; Tilley et al., 1999). Gene ablation of the FP receptor, the receptor for PGF 2α, resulted in loss of parturition (Sugimoto et al., 1997). Both PGE2 and PGF2α concentrations are increased significantly in the human during the window of implantation in natural cycles and also in patients undergoing in vitro fertilization (IVF) and ovum donation. Interestingly, this profile is abrogated when the endometrium is refractory (Vilella et al., 2013).
In vitro studies of decidualized human stromal cells revealed that steroid hormone withdrawal increased a host of inflammatory mediators, including IL-6, CCL11, GM-CSF, CCL2, IL1-RA, CXCL10 and CXCL8, and this response was mediated by NF-κB (Evans and Salamonsen, 2014). NF-κB increases the transcription of a wide variety genes, including cytokines (IL-1, IL-6), chemokines (CXCL8, chemokine ligand 2/CCL-2) and adhesion molecules (intercellular adhesion molecule 1/ICAM, vascular cell adhesion molecule 1/VCAM) (Kayisli et al., 2004). Human endometrial biopsies have also been shown to express components of the NF-κB pathway, with evidence for activation of NF-κB during the perimenstrual phase (King et al., 2001). These findings have been replicated in the mouse menstrual-like model (Xu et al., 2013).
A recent study in the mouse model of simulated menstruation links NF-κB and COX-2 in the menstrual process. Inhibition of the COX enzymes or NF-κB at the time of progesterone withdrawal significantly decreased the amount of bleeding and endometrial breakdown in this murine model (Xu et al., 2013). Furthermore, there was a significant decrease in leukocyte influx after both interventions. Chromatin immunoprecipitation analysis revealed that NFkB binds to the COX-2 promoter, providing a mechanism of NFkB-mediated COX-2 up-regulation and subsequent inflammatory cell influx at menstruation. Progesterone is known to have inhibitory effects on NF-κB activity, mediated by increasing IκB production or by competing with NF-κB for recognition sites on relevant genes (Kelly et al., 2001). In this way, the steroid hormones modulate the local endometrial inflammatory environment by suppressing NF-κB activity until menstruation is required.
Following progesterone withdrawal, there is a dramatic rise in the endometrial leukocyte population (Bonatz et al., 1992; Salamonsen and Lathbury, 2000). Neutrophil numbers are negligible throughout most of the cycle but increase perimenstrually to comprise 6–15% of the total cell number (Salamonsen and Lathbury, 2000). As key mediators of the inflammatory response, neutrophils respond to inflammation by migrating rapidly to the site of injury to contain and clear any noxious stimuli. Circulating neutrophils have a lifespan of a few hours, but neutrophils residing in inflamed tissue can survive for days. This is due to decreased neutrophil apoptosis induced by pro-inflammatory mediators and hypoxia (Ward et al., 1999; Cross et al., 2006). The importance of this neutrophil influx at menstruation was shown in the mouse model, where neutrophil depletion using the antibody RB6-8C5 affected endometrial breakdown and markedly delayed endometrial repair (Kaitu'u-Lino et al., 2007a). Neutrophils contain high levels of MMPs and have the ability to activate resident MMPs to initiate endometrial breakdown (Gaide Chevronnay et al., 2011). In contrast, chronic inflammation is characterized by a persistent neutrophil response due to decreased apoptosis (Serhan and Savill, 2005). This prolonged neutrophil response results in tissue damage and loss of function. Therefore, tight regulation of neutrophil influx and apoptosis is required for normal menstruation. B cell lymphoma 2 (bcl-2) is an apoptosis regulator protein that is expressed in the human endometrium (Otsuki et al., 1994). Examination of human endometrial tissue revealed the presence of bcl-2 during the proliferative and early secretory phases with decreased levels in the late secretory and menstrual phases. These decreased levels correlated with the appearance of apoptotic cells in the perimenstrual phase. This cyclic pattern suggests that ovarian hormones regulate bcl-2 levels in the endometrium (Critchley et al., 1999b). In this way, progesterone withdrawal may increase bcl-2 to limit the lifespan of endometrial neutrophils at menstruation, preventing a chronic inflammatory response.
Macrophages also increase in number throughout the secretory phase to reach maximal numbers perimenstrually, during the luteo-follicular transition (Bonatz et al., 1992; Critchley et al., 1999a; Thiruchelvam et al., 2013). The regulation of the endometrial macrophage remains under investigation. Lacking PR, these cells may be recruited from the circulation due to increased endometrial chemoattractant production and/or may proliferate in situ (Guo et al., 2011; Davies et al., 2013b). These cells produce cytokines and proteases and are involved in tissue remodelling and debris removal. The classic M1 (pro-inflammatory) and M2 (anti-inflammatory) phenotypes represent simplified extremes of macrophage function. These complex cells have the ability to adapt and respond to the tissue environment in which they reside (Gordon and Martinez, 2010; Davies et al., 2013a). The phenotype of endometrial macrophages during the perimenstrual phase is yet to be fully delineated, but considering their known functions they are likely to have a significant impact in the endometrium at menstruation (Thiruchelvam et al., 2013). Furthermore, delineation of macrophage phenotype in this physiological model of tissue ‘injury’ and ‘repair’ may provide novel insights to pathological conditions, such as chronic inflammation or cancer, where resident macrophages are involved in aberrant function (Laoui et al., 2014). A direct comparison of the macrophage profile throughout the physiological inflammatory response of menstruation with the macrophage response in areas of chronic inflammation may lead to novel therapeutic targets to improve tissue function.
Taken together, the studies described above support the hypothesis that the decidualized stromal cell compartment can increase cytokine and chemokine production to attract leukocytes, or encourage their proliferation in the functional endometrial layer, during the perimenstrual phase. For summary of perimenstrual leukocyte traffic, see Fig. 2. In turn, endometrial leukocytes produce MMPs and have the potential to stimulate MMP production from adjacent cells (Jabbour et al., 2006) making them attractive candidates for the regulation of local endometrial MMP expression. In this way the decidualized stromal cells of the functional layer help determine their own fate, limiting the inflammatory reaction and tissue breakdown to the upper luminal portion of the endometrium. This compartmentalization of inflammation, with sparing of the basal layer, may be critical for efficient repair of the endometrium after shedding (menstruation). There is evidence that the amount of endometrium that is shed during menstruation varies between individuals, but it remains undetermined if the depth of shedding is associated with gynaecological pathologies (Ludwig and Spornitz, 1991; Fraser et al., 2001).
Figure 2.
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Leukocyte trafficking in the perimenstrual human endometrium (derived from data published and reviews by Bonatz et al., 1992; Salamonsen and Lathbury, 2000; Moffett-King, 2002; Thiruchelvam et al., 2013). Top panel: Sex steroid profiles in the luteo-follicular transition (perimenstrual ‘window’). Bottom panel: Overview of leukocyte traffic in the endometrium with transition from secretory phase through menses/endometrial repair to the proliferative phase of next cycle. Size of cell image reflects abundance.
What limits endometrial inflammation?
An excessive or prolonged inflammatory response at menstruation will lead to excessive tissue damage and may result in HMB (NICE, 2007). Studies examining endometrial tissue from women with objective measurement of their menstrual blood loss (MBL) have identified a significantly increased inflammatory response in women with HMB. Increased levels of the pro-inflammatory cytokine tumour necrosis factor α were identified in the menstrual effluent of women with HMB (MBL > 80 ml) compared with women with normal MBL (Malik et al., 2006). Endometrial COX-2 mRNA expression was also significantly elevated in women with HMB (Smith et al., 2007). In addition, increased levels of total PGs have been found in the endometrium of women with HMB (Smith et al., 1981a, b). Furthermore, increased signalling of PGE 2 through its EP2 and EP4 receptors has been suggested due to elevated production and decreased hydrolysis of cyclic AMP (Smith et al., 2007). In support of these findings, PG synthesis inhibitors are a popular treatment for HMB. Mefenamic acid is a non-steroidal anti-inflammatory compound which significantly decreases MBL (Cameron et al., 1990). However, although women treated with mefenamic acid have a significant decrease in their menstrual loss, 52% maintained a blood loss greater than 80 ml after 2 months of treatment, highlighting the need for more effective medical therapies for this condition (Cameron et al., 1990).
Glucocorticoids
The inflammatory response of physiological menstruation appears to be self-limiting. The pro-inflammatory cytokine IL-1 has been shown to increase the expression of 11β hydroxysteroid dehydrogenase-1 (11βHSD-1) (Rae et al., 2004; Rae and Hillier, 2005). This enzyme converts cortisone (compound E) to the anti-inflammatory steroid cortisol (compound F). Glucocorticoids alter the inflammatory response by limiting cytokine production, increasing macrophage phagocytosis, increasing transcription of anti-inflammatory genes and repressing pro-inflammatory transcription factors (Zhang et al., 2009).
Endometrial 11βHSD-1 mRNA levels are significantly increased at menstruation, consistent with a role in endometrial breakdown and repair (McDonald et al., 2006). In addition, the glucocorticoid receptor is present throughout the cycle in the stromal compartment, including endometrial leukocytes and endothelial cells (Bamberger et al., 2001; Henderson et al., 2003). In this way, local generation of glucocorticoids by inflammatory mediators may prevent an excessive inflammatory response in the menstrual endometrium. Studies of endometrium from women with HMB further highlight the importance of glucocorticoids in endometrial physiology. Secretory endometrium from women with a blood loss greater than 80 ml was found to have significantly elevated levels of 11βHSD-2 when compared with endometrium from women with normal loss (Rae et al., 2009). 11βHSD-2 converts cortisol back to cortisone and may explain the excessive local inflammation of the endometrium in women with HMB at menses. Decreased cortisol levels and loss of its anti-inflammatory effects may prolong menses, contributing to heavy blood loss. We are currently exploring whether ‘rescue’ of putative luteal phase endometrial glucocorticoid deficiency could reduce menstrual bleeding (Warner et al., 2015).
Control of the MMP response at menses
MMPs have the ability to degrade all components of the ECM and have been shown to have an integral role in endometrial breakdown at menses (Marbaix et al., 1996). Lack of control of MMP action at menstruation will lead to excessive tissue damage and may lead to abnormal bleeding. The control of MMP action occurs at a number of levels to prevent an abnormal response during menses and allow for tissue regeneration and remodelling at other phases of the cycle. A full review of these processes is beyond the scope of this review, and the reader is referred to Gaide Chevronnay et al. (2011) for a comprehensive overview. It is well established that progesterone inhibits MMP transcription to suppress their expression during the secretory phase of the cycle (Schatz et al., 1994; Salamonsen et al., 1997; Vassilev et al., 2005). The withdrawal of progesterone and the up-regulation of MMP levels during menstruation have been discussed above. Following endometrial breakdown, MMP activity can be inhibited by tissue inhibitors of metalloproteinases (TIMPS) or by the protease inhibitor α 2-macroglobulin. These factors are expressed in the endometrium throughout the menstrual cycle (Sayegh et al., 1995; Zhang and Salamonsen, 1997) suggesting that they are overwhelmed by an increase in MMP production at menstruation and that the ratio of MMPs to TIMPs may dictate the ability of MMPs to breakdown tissue. Additionally, active MMPs undergo endocytic clearance by low-density lipoprotein receptor-related protein-1 (LRP-1) during the proliferative and secretory phase of the cycle, initiating lysosomal degradation. At menstruation, the LRP-1 protein is not present due to tissue shedding (Selvais et al., 2009), enhancing MMP activity. This multifactorial regulation limits the MMP response to menstruation, ensuring tissue damage is not prolonged.
The endometrium: a model of vascular function
Menstruation as a physiological ischaemia-reperfusion injury
The first observations of endometrial architecture at menstruation were from intraocular endometrial transplants in the rhesus macaque (Markee, 1940). Direct observation of the explants following progesterone withdrawal revealed shrinkage of endometrial thickness, followed by vasoconstriction of spiral arterioles and focal bleeding. The vasoconstriction observed was transient but intense, consistent with an ischaemia-reperfusion injury. However, the presence and role of hypoxia in the endometrium remain controversial.
Ischaemia has not been detected in the human endometrium during menstruation to date. Laser Doppler fluximetry measures the number of red blood cells transiting a monitored volume per unit time. This method failed to detect ischaemia during menstruation (Gannon et al., 1997), but the limited spatial resolution of fluximetry may not detect focal or prolonged ischaemia-reperfusion episodes. There is some indirect evidence that hypoxia occurs at menstruation in human endometrial tissue. Markers of tissue hypoxia (CAIX and hypoxia inducible factor (HIF)-1α) have been detected immunohistochemically in the human endometrium at menses, with a distinct reduction in staining of both markers after cycle day 5 (Critchley et al., 2006; Punyadeera et al., 2006). In addition, hypoxia has been detected in the menstrual endometrium of the simulated mouse menstruation model (Fan et al., 2008). Pimonidazole is a marker of oxygen partial pressures less than 10 mmHg, and its expression was seen in the uppermost endometrial zones during the simulated menstrual phase. Negligible pimonidazole levels were observed by Day 5 after progesterone withdrawal. In contrast, hypoxia, pimonidazole and HIF-1α were not detected following ovarian hormone withdrawal in a xenograft menses model, where a fragment of human endometrial functional layer was grafted into immunodeficient mice (Coudyzer et al., 2013). These differences may be explained by disturbance of the full thickness endometrial architecture in the immunodeficient xenograft model, where spiral arteriole function and immune cell function will be modified, but definitive proof that hypoxia is present in the human endometrium at menses is still lacking.
Even if hypoxia is present in the endometrium, there remains debate about its function. Primary human endometrial stromal cells cultured in normoxic (21% O 2) and hypoxic (2% O 2) conditions for 24 and 48 h revealed that hypoxia decreased the secretion of membrane-type 1 MMP, active MMP-2, proMMP-1 and proMMP-3 (Zhang and Salamonsen, 2002). Similar decreases in MMPs were also observed in the culture supernatants from whole endometrial explants cultured in 0.1% O 2 for 24 h (Gaide Chevronnay et al., 2010). This suggests that hypoxia is not involved in endometrial breakdown by MMPs at menstruation, but does not exclude a role in repair of the denuded surface and limitation of the MMP response. In the xenograft model described previously, increases in MMP expression were observed in the human endometrial grafts and breakdown occurred within 96 h of ovarian hormone withdrawal. In addition, the xenografted endometrium underwent complete repair despite the absence of hypoxia. This suggests that hypoxia is not essential for endometrial breakdown or repair. However, in vivo human menstruation occurs 48–72 h after withdrawal of ovarian hormones (Catalano et al., 2007) and the mouse model of menstruation demonstrates bleeding within 8–12 h of hormone withdrawal (Brasted et al., 2003; Menning et al., 2012; Cousins et al., 2014). Hence 8 h post-progesterone withdrawal in the murine model is approximately equivalent to 48 h in the human. Therefore, it remains possible that, although hypoxia is not necessary for endometrial breakdown and repair, it is desirable for maximal efficiency of these processes. HIF-1 is a transcription factor known to be the master regulator of the cellular response to hypoxia (Iyer et al., 1998). In hypoxic conditions, this factor increases the transcription of a number of genes involved in angiogenesis, mitogenesis and metabolism. Its prolonged activation is observed in the tumour microenvironment, leading to aberrant angiogenesis and metastasis (Mazzone, 2010). However, transient activation appears necessary in physiological situations to instigate repair processes. For example, pharmacological activation of HIF-1 provided protection against development of colitis in a murine model (Cummins et al., 2008). The role of HIF-1 in menstruation, if any, remains to be determined.
Vasoconstriction
Regardless of the presence or absence of hypoxia in the menstrual endometrium, vasoconstriction of spiral arterioles is desirable at this time to limit blood flow. Poiseuille's equation states that the radius of a vessel is the major determinant of resistance to flow, meaning that a small increase in vessel radius will dramatically increase the amount of blood flowing through it (Maybin et al., 2011a). Therefore, decreased constriction of endometrial vessels at the time of menstruation will contribute significantly to increased menstrual blood loss. PGF 2α and endothelin-1 (ET-1) are two endometrial factors with known vasoconstrictive properties (Baird et al., 1996; Marsh et al., 1997). In contrast, PGE 2 is a known vasorelaxant. Women with heavy MBL have been shown to have a significantly decreased PGF 2α/PGE 2 ratio (Smith et al., 1981b) and decreased FP receptor expression (Smith et al., 2007). Excessive PGE 2 production at the expense of PGF 2α may result in less constriction of the spiral arterioles prior to menstruation. In addition, women with HMB have decreased endometrial expression of the potent vasoconstrictor ET-1 and increased expression of its metabolising enzyme, neural endopeptidase (Marsh et al., 1997). Increased metabolism of endothelin could explain its decreased endometrial expression and cause dilation of endometrial vessels at menstruation. Furthermore, altered spiral arteriole maturation may also contribute to inefficient spiral arteriole vasoconstriction at menstruation. Vessel wall circumference and focal discontinuities were noted to be larger in the endometrium of women with HMB than normal controls (Mints et al., 2007). Women with heavy bleeding had significantly reduced vascular smooth muscle cell proliferation in spiral arterioles during the mid-late secretory phase when compared with normal controls (Abberton et al., 1999b). In addition, smooth muscle myosin heavy chain, a contractile protein used as a marker of vascular smooth muscle cell maturation, was significantly decreased in spiral arterioles of women with HMB (Abberton et al., 1999a). The endothelial cell lining in endometrial tissue from women with HMB was found to have increased gaps, possibly due to increased expression of angiopoietin-2 during the secretory phase (Mints et al., 2010). This suggests that vessels in these women are pre-programmed during the proceeding cycle to be more fragile at menstruation. Taken together, the decreased levels of vasoconstrictive factors and immature vessels present in women with HMB will significantly increase MBL.
The endometrial coagulation system
Cessation of menstruation relies on an intact endometrial coagulation system to achieve haemostasis (Fig. 3). Endometrial endothelial injury initiates immediate activation and aggregation of platelets to form a plug. This takes place by two mechanisms (i) platelet glycoprotein interaction with von Willebrand factor (vWF) or (ii) tissue factor generation of thrombin (Davies and Kadir, 2012). The resulting platelet plug forms a barrier to prevent further blood loss. The subsequent stage of haemostasis involves the formation of fibrin via the coagulation cascade. The coagulation cascade is activated by two pathways; extrinsic and intrinsic. Each culminates in the conversion of factor X to Xa, which catalyses the conversion of pro-thrombin to thrombin, ultimately leading to the formation of a more stable fibrin clot to seal previously bleeding vessels. Disorders that interfere with systemic haemostasis have an impact on MBL. Von Willebrand disease is the most common of these disorders, with a prevalence of 13% in women with a complaint of HMB (Shankar et al., 2004).
Figure 3.
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Endometrial coagulation pathways. Immediate: A platelet plug forms rapidly, relying on interactions with tissue factor, vWF and collagen. Subsequent: intrinsic and/or extrinsic activation of coagulation pathways result in formation of a fibrin clot to ensure haemostasis. Fibrinolysis drives the degradation of the fibrin clot. t-PA and u-PA convert plasminogen to plasmin, which breaks down the fibrin clot. PAI converts plasmin back to plasminogen.
Fibrinolysis involves conversion of plasminogen to active plasmin, promoting the degradation of fibrin deposits. Tissue plasminogen activator (t-PA) and urokinase plasminogen activator (u-PA) drive the production of plasmin. In contrast, plasminogen activator inhibitor (PAI) inhibits fibrinolytic activity. The human endometrium contains t-PA and u-PA, as well as PAI and the uPA receptor (Gleeson et al., 1993; Nordengren et al., 2004). There is evidence that an overactive fibrinolytic system interferes with haemostasis and contributes to HMB. Women with HMB had raised levels of t-PA activity on the second day of bleeding compared with those with normal loss (Gleeson et al., 1993). The efficacy of tranexamic acid as a treatment for HMB provides further evidence for over activation of the fibrinolytic system in the endometrium of these women. This antifibrinolytic reduces t-PA and PAI levels in women with HMB and results in a 58% reduction in blood loss (Gleeson et al., 1994).
Angiogenesis
Vascular modification and new blood vessel growth are essential components of endometrial physiology. At menstruation, rapid repair of injured blood vessels must occur to stop bleeding, and this is usually completed by Day 5 of the cycle. This process occurs despite lack of ovarian hormone support, as observed in women following surgical ovariectomy who stop bleeding despite the lack of ovarian hormonal support. In addition, the murine model of menstruation displayed complete repair of the endometrium in the absence of estradiol (Kaitu'u-Lino et al., 2007b), suggesting vascular repair at menses (in this animal model) is not reliant on estrogen. The regulation of vascular repair at this stage is still to be fully delineated. Vascular endothelial growth factor (VEGF), a key mediator of vascular function, is increased in women at menses, and there is mounting evidence from human and murine studies that endometrial VEGF is regulated by hypoxia (Charnock-Jones et al., 1993; Sharkey et al., 2000; Fan et al., 2008; Maybin et al., 2011b).
During the proliferative phase, there is rapid growth of the functional layer of the endometrium, necessitating angiogenesis to maintain perfusion of new tissue (Girling and Rogers, 2005). This physiological angiogenic response is unusual in the human adult and provides an accessible human model for comparison to pathological situations such as the tumour microenvironment. Therefore, defining the control and mechanisms of this normal angiogenesis may identify new approaches to the control of tumour growth. Despite the significant changes in endometrial architecture across the cycle, it has been repeatedly demonstrated that levels of endothelial cell proliferation within the human endometrium do not show any consistent pattern across the menstrual cycle (Girling and Rogers, 2005). Interpretation of these results is challenging, as samples taken from women at the same stage of the menstrual cycle are extremely variable. This may be due to differences in hormone levels at the time of sampling or a variation in the region from which the biopsy was obtained. To overcome this, Nayak and Brenner utilized the macaque simulated menses model (Nayak and Brenner, 2002). Using Ki67 or bromodeoxyuridine (BrdU) to identify proliferating endothelial cells, the authors demonstrated a 6-fold increase in proliferation during the mid-proliferative stage (Days 8–10 after progesterone withdrawal). This peak was absent in the hormone-deprived animals, indicating endothelial cell proliferation at this stage in the cycle is estradiol dependent, unlike the vascular response at menses. No significant changes in proliferation were observed at other stages of this artificial cycle. This finding concurs with stereological analysis of human endometrial endothelial cell staining, where vascular length density was greatest during the mid-late proliferative phase (Gambino et al., 2002). By examining vessel length and branch points, the authors concluded that vessel elongation is the major mechanism by which endometrial angiogenesis occurs in mid-proliferative phase.
In the secretory phase, coiling and maturation of the spiral arterioles and growth of the subepithelial capillary plexus must take place. The consequence of impaired vascular maturation during this phase has been discussed above in regard to decreased vasoconstriction and its relation to HMB. It is likely that the uterine natural killer (uNK) cells play an important role in spiral arteriole maturation.
Uterine NK cells are CD56 bright, CD16-, CD3- and are phenotypically different from peripheral blood NK cells (Koopman et al., 2003). uNK cells increase in number during the mid-luteal phase and are located close to the endometrial glands and spiral arteries (Dosiou and Giudice, 2005), supporting a role in vascular remodelling. Evidence derived from early pregnancy studies supports a role for uNK cells in the early stages of spiral artery remodelling, where failure of this process is considered to contribute to pregnancy pathology (Moffett-King, 2002; Robson et al., 2012). The lack of spiral arteriole modification observed in mice deficient in uNK cells further supports this hypothesis (Greenwood et al., 2000; Ashkar et al., 2003).
Additional evidence for the impact of uNK cells on endometrial vasculature comes from studies of the action of selective progesterone receptor modulators (SPRMs) in the human endometrium. Analysis of endometrium from women administered the SPRM asoprisnil revealed a suppressed IL-15 pathway, which regulates uNK development and function. There was a marked reduction in uNK cells, an abnormal appearance of endometrial vasculature with increased α-SMA staining surrounding the spiral arterioles. Women taking this SPRM had significantly decreased menstrual bleeding, linking the PR, uNK cells, vascular structure and menstrual function (Wilkens et al., 2013). The inability to identify the PR on uNK cells (Henderson et al., 2003) suggests an indirect mechanism of hormonal regulation, via paracrine mediators such as chemokines (Salamonsen and Lathbury, 2000; Hannan and Salamonsen, 2007).
The importance of vascular normalization has recently become apparent in the field of cancer biology. Blockade of VEGF to prevent angiogenesis in the tumour microenvironment was logically introduced as a treatment for cancer (Carmeliet, 2005). Although initial results were encouraging, the mean survival of patients treated with these inhibitors disappointingly remained unchanged (Carmeliet and Jain, 2011). Recent research has highlighted the benefits of vessel normalization, rather than inhibition of angiogenesis, as a mechanism to reduce metastasis and hopefully increase survival (Mazzone et al., 2009; Carmeliet and Jain, 2011). Therefore, delineation of normal vascular processes and their regulation within the human endometrium, including physiological angiogenesis and vessel maturation, could have widespread clinical application.
The perimenstrual endometrium: a model of scarless tissue repair
After shedding its luminal portion, the endometrium must efficiently repair to ensure implantation can take place if fertilization occurs in the subsequent cycle. The processes involved in endometrial repair appear to be analogous to classic wound healing and include inflammation, its resolution, angiogenesis, tissue formation and tissue remodelling. The first three processes have been discussed above, and this section will concentrate on the latter two, with discussion of the former where necessary. The cross disciplinary benefits of studying this scar-free repair system are obvious, but incisive data on the factors involved and their regulation remain elusive and concerted efforts are necessary to maximize the translational benefits.
The regulation of endometrial repair and regeneration
Scanning electron microscopy of human menstrual endometrial samples revealed a ragged and torn surface with gland openings and a lack of epithelial covering (Ludwig and Spornitz, 1991). Subsequent regrowth of the epithelium occurred before stromal expansion, with epithelial cells growing from the necks of the glands to meet migrating cells from other glands, forming a new luminal surface. This began on menstrual Day 2, and full coverage of the lumen was achieved by Day 6. A more recent study found that the functional endometrial layer displays simultaneous shedding and repair in a piecemeal fashion during menstruation (Garry et al., 2009). Both of these studies suggest that initial re-epithelialization of the endometrium occurs during active bleeding in the absence of ovarian hormones, consistent with findings in the murine menstruation-like model (Kaitu'u-Lino et al., 2007b) and in women post-oophorectomy.
Tissue recombination studies in the mouse model suggest that uterine epithelialization is required before the stromal compartment can respond to ovarian steroids (Bigsby, 2002). Stromal cell mitoses first appear on Days 5–6 of the human menstrual cycle, when estradiol levels are rising and the epithelial layer has completely healed (Ferenczy et al., 1979). Unlike the initial repair phase, this endometrial regeneration is dependent on ovarian hormone support. VEGF, a potent mitogenic and angiogenic factor, was found to have three peaks of expression in the ovariectomized macaque model of menstruation (Nayak and Brenner, 2002). These increases in VEGF mRNA occurred in the surface epithelium during the early proliferative phase, in the stroma during the mid-proliferative phase and in the glands during the late secretory phase. Comparison of hormone-deprived and estrogen-exposed animals revealed that estrogen is not essential for the early proliferative phase peak but is necessary for VEGF mRNA up-regulation in mid-proliferative stromal cells. These findings support an estrogen-independent initial repair phase and estrogen-dependent regeneration of the endometrium.
Hence alongside initiation of menstruation, progesterone withdrawal is also likely to trigger endometrial repair. Support for this hypothesis is found in gene microarray analysis of differentially expressed transcripts from human endometrial explants cultured in vitro in the presence of ovarian hormones or in the absence of hormonal support. This hormone deprivation model revealed ‘wound healing and inflammation’ as a top scoring biological process (Gaide Chevronnay et al., 2010). The importance of VEGF for luminal re-epithelialization and angiogenesis at menstruation was demonstrated using VEGF Trap in the macaque and murine models (Fan et al., 2008). Progesterone withdrawal has been shown to increase the expression of VEGF in the macaque model (Nayak and Brenner, 2002), murine model (Fan et al., 2008) and in human endometrial explants (Maybin et al., 2011b). Hypoxia and PGs have been associated with these increases in VEGF expression (Fan et al., 2008; Maybin et al., 2011b) and may represent downstream mediators of progesterone withdrawal.
The importance of the vascular niche in tissue regeneration is further supported by studies of stromal-derived growth factor (SDF-1) and its receptors CXCL4 and CXCL7. SDF-1 is present throughout the menstrual cycle and CXCR4 expression peaks in the early proliferative phase and is present in epithelial cells and endothelial cells (Laird et al., 2011). SDF has been shown to increase endometrial epithelial cell proliferation in vitro (Tsutsumi et al., 2011). A recent study combining an inducible endothelial-cell-specific mouse gene deletion strategy and complementary models of acute and chronic liver injury revealed that differential recruitment of pro-fibrotic CXCR4 or pro-regenerative CXCR7 signalling determines if the liver regenerates or becomes fibrotic after injury. Hence, autocrine signals from the endothelium may influence the rate and nature of the repair process. The role and regulation of CXCR4 and CXCR7 in the normal menstrual endometrium, where scarring is absent, and in the rare syndrome of endometrial scarring, Asherman's, remains to be determined.
The cellular and molecular mechanisms governing epithelial cell proliferation and migration after menstruation have not been fully elucidated. At least three hypothesized mechanisms exist, including (i) proliferation of luminal epithelial cells from the base of the epithelial glands, (ii) mesenchymal to epithelial transition of residual stromal cells and (iii) regeneration of the luminal epithelium from endometrial stem cells.
Mesenchymal-to-epithelial transition
Previously, the ‘free-edge’ effect was thought to be responsible for endometrial re-epithelialization, where the absence of neighbouring cells at the wound margin acts as a growth signal (Heimark and Schwartz, 1985). However, scanning electron microscopy of menstrual endometrium revealed that epithelial cells appeared to arise from underlying stromal cells in denuded portions, rather than solely from the necks of epithelial glands (Ludwig and Spornitz, 1991; Garry et al., 2009). This suggests that endometrial stromal cells are reprogrammed at menstruation to lose their mesenchymal cell characteristics and gain epithelial cell traits, a process known as mesenchymal-to-epithelial transition (MET). Evidence for MET during endometrial repair comes from the murine model of simulated menses, where co-expression of the epithelial marker pancytokeratin and the stromal cell marker vimentin occurred in endometrial cells after 24 h of hormone withdrawal (Patterson et al., 2013). Gene microarray analysis of murine uterine tissue from the simulated menses model taken pre- and post-progesterone withdrawal revealed significant changes in genes known to be involved in MET such as cytokeratin, Wnt1, E-cadherin and osteopontin (Cousins et al., 2014). This study also identified actively proliferating cells in the stromal compartment, where there was loss of luminal epithelial coverage and proliferation of adjacent luminal epithelial cells, consistent with simultaneous MET and epithelial cell migration. In this way, the residual basal layer of the endometrium and the adjacent unshed functional layer can contribute to re-epithelialization of the denuded surface. The contribution of the functional endometrial layer to menstrual repair is supported by microarray study of stromal and glandular cells from the basal and functional layer obtained by laser capture microdissection (Gaide Chevronnay et al., 2009). This revealed that in addition to up-regulation of transcripts involved in tissue degeneration, stromal cells from the functional layer also displayed increased levels of genes associated with ECM biosynthesis, indicating an important contribution to repair of adjacent denuded areas.
The reverse process of epithelial-to-mesenchymal transition (EMT) is also important for wound healing, embryogenesis and fibrosis (Gonzalez and Medici, 2014). The loss of adhesion molecules and tight junctions alongside increased expression of mesenchymal cell markers allows migration into tissues. In the embryo, cycles of EMT and MET are necessary for development and highlight the reversibility of these processes (Nieto, 2013). The role of EMT, if any, in the endometrium remains to be determined, but it is likely that a balance of EMT and MET is important for repair processes. Excessive EMT has been implicated in fibrotic diseases of the kidney and lung (Kothari et al., 2014). This may be due to the generation of extreme myofibroblasts that are resistant to apoptosis. Synthesis and remodelling of the ECM by fibroblasts is essential for wound healing. Fibroblasts differentiate into myofibroblasts during the last phases of wound healing and increase their expression of smooth muscle actin (SMA). These myofibroblasts initiate wound contraction and secrete type I collagen. Persistence of myofibroblasts at an injury site results in scar formation (Hantash et al., 2008). Therefore, excessive EMT may induce scaring via aberrant myofibroblast differentiation causing persistence at the injury site. Cytokines, hypoxia, growth factors and components of the ECM have all been implicated in the regulation of EMT (Gonzalez and Medici, 2014). Strict control of these factors in the human endometrium may therefore underpin its exceptional ability to heal without scarring. Interestingly, normal human endometrial stromal cells have significantly less α-SMA expression and contractility when compared with endometriotic stromal cells (Yuge et al., 2007). This suggests endometrial cells have less myofibroblastic differentiation, leading to a reduction in scar formation. In this way, the balance of MET and EMT may influence endometrial repair at menstruation. Aberrations in their control could lead to pathology such as endometriosis, with its associated adhesions and scarring, or delayed endometrial repair and its consequent increased MBL.
Stem cells
An alternative, or perhaps complimentary, method of endometrial repair is regeneration of tissue from stem cells or progenitor cells. Evidence of their existence in the endometrium comes from colony-forming units derived from human endometrial samples (Gargett et al., 2009). These cells fulfilled the criteria of self-renewal, high proliferative potential and multilineage differentiation. In addition, the mouse model of simulated menstruation suggests that re-epithelialization of the uterine surface arises from progenitor cells residing in the glandular epithelial cells (Kaitu'u-Lino et al., 2010). Unlike human studies, it is possible to utilize the label retaining technique in the murine model, identifying stem cells due to their relative quiescence and comparatively slower proliferation than more differentiated cells. A pulse of BrdU is followed by a chase period, when slowly cycling cells retain the BrdU label and transient amplifying cells proliferate rapidly and dilute the label. Examination of BrdU and proliferating cell nuclear antigen immunofluorescence in this model revealed that glandular cells proliferated selectively during repair and BrdU labelling remained constant. In contrast, luminal cells showed rapid dilution of BrdU at menstruation. Both epithelial and stromal label retaining cells have been identified in this mouse model (Chan and Gargett, 2006).
For a comprehensive review of the contribution, derivation and application of endometrial stem cells, we refer the reader to a number of papers (Gargett and Masuda, 2010; Cervello et al., 2011, 2013; Deane et al., 2013). Many questions remain, but it is clear that the multipotent potential of cells within the endometrium can have widespread benefits. Endometrial biopsies are obtainable in an outpatient setting, usually without the need for anaesthetic. This is in contrast to the painful bone marrow biopsy used to obtain haematopoietic stem cells. Mesenchymal stem cells obtained from the endometrium are highly proliferative (Gargett et al., 2009) and are therefore attractive for in vitro expansion and use in cell-based therapies. Furthermore, multipotent cells have also been derived from menstrual effluent, negating the need for any biopsy (Ulrich et al., 2013). Increased understanding and utilization of these unique endometrial cells will benefit many gynaecological conditions. Endometriosis is caused by implantation and growth of endometrial deposits in other tissue sites and is thought to occur secondary to retrograde menstruation. However, although retrograde menstruation occurs in many women, only ∼10% have evidence of endometrial deposits (Gargett et al., 2014). The prevalence or activity of endometrial stem cells in the endometrial fragments spilling into the abdominal cavity may explain this discrepancy. An insufficiently thick endometrium can contribute to sub-fertility and failed IVF. Endometrial stem cell therapy is a potential treatment to regenerate the endometrium and increase fertility rates in the future (Cervello et al., 2013). In addition, menstrual derived cells have displayed regenerative properties at other tissue sites. They have incorporated into atrophied skeletal muscle fibres in a mouse model of Duchenne muscular dystrophy and have resulted in improved cardiac tissue function in an infarction model (Cui et al., 2007; Toyoda et al., 2007; Hida et al., 2008). Endometrial cells have differentiated into morphologically and functionally glucose-responsive insulin secreting cells, providing a potential therapeutic strategy for diabetes (Santamaria et al., 2011). Therefore, accessible multipotent cells from the endometrium could have widespread and significant future clinical applications.
Conclusions
Many advances have been made to increase our knowledge of menstrual physiology. However, why women menstruate and what starts, limits and stops menstrual blood loss remain key questions. The endometrium functions as a complex multicellular structure that involves interactions of immune, endocrine and vascular systems. The strict regulation of this tissue to allow cyclical ‘injury’ and ‘repair’ at menstruation results in a remarkable physiological response that allows pregnancy to occur. This accessible tissue, alongside robust animal models, provides a fantastic resource in which to study inflammation, angiogenesis and tissue repair (Fig. 1) to identify new therapeutic targets for gynaecological conditions and generate translational knowledge for application at a host of other tissue sites.
Authors’ roles
J.A.M wrote the manuscript with supervisory support from H.O.D.C.
Funding
We acknowledge the following funding from the Medical Research Council (G0000066, G0500047; G0600048; MR/J003611/1) and the Wellcome Trust (083908/Z/07/Z, 100646/Z/12/Z) for support of several studies wherein data derived have been described in this review. Funding to pay the Open Access publication charges for this article was provided by the Wellcome Trust.
Conflict of interest
J.M.: no conflict of interest. H.O.D.C.: research grant support from Bayer Pharma AG; and consultancy for Bayer Pharma AG, PregLem SA, Gedeon Richter, Vifor Pharma UK Ltd, and AbbVie Inc.
Acknowledgements
We acknowledge Mrs Sheila Milne for secretarial support, and Mr Ronnie Grant and Greg Armstrong for graphical assistance.
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https://www.quora.com/What-is-an-example-of-a-function-that-has-two-different-local-extrema-but-no-global-maximum-or-minimum
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What is an example of a function that has two different local extrema but no global maximum or minimum? - Quora
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Mathematics
Local Maxima
Types of Functions
Absolute Extrema
Global Maximum
Calculus 2
Calculus (Mathematics)
Operations and Functions
Local Minima
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What is an example of a function that has two different local extrema but no global maximum or minimum?
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Jörg Straube
M.Sc. in Computer Science, ETH Zurich (Graduated 1987) · Author has 6.3K answers and 1.7M answer views
·2y
One possible example is the cubic function f(x) = x^3 - 3x
No global max nor min, but local max and min at x = +-1
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One possible example is the cubic function f(x) = x^3 - 3x
No global max nor min, but local max and min at x = +-1
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I usually look at three things when I’m buying work shoes: comfort, cushioning and arch support; how sturdy the sole is; and whether I can actually afford to get more than one pair if I want them in different colours.
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What is a function that has a local minimum but no global minimum?
What are some continuous functions that do not have a global maximum or global minimum?
What is the difference between the global maximum and minimum and local maximum and minimum?
Can a function have a local minimum but no global minimum/maximum?
What are some examples of functions with two local maximums and one local minimum at each end?
Duff Howell
Plenty both in college and in on-the-job use in the decades since. · Author has 1.7K answers and 2.5M answer views
·2y
The key to unravelling a homework problem like this it to distinguish between local and global.
The function has no global maximum or minimum. It’s value goes from negative infinity to positive infinity.
The function has two different local extrema. Somewhere between the infinities, there are points where it comes to a maxima relative to nearby points, and one where it comes to a minima. One of each will get you exactly two.
A familiar example might be a simple cubic polynomial. Say you wanted there to be extremes at +-1. So a derivative (a quadratic) would have zeroes at +-1 (the extrema, wher
Continue Reading
The key to unravelling a homework problem like this it to distinguish between local and global.
The function has no global maximum or minimum. It’s value goes from negative infinity to positive infinity.
The function has two different local extrema. Somewhere between the infinities, there are points where it comes to a maxima relative to nearby points, and one where it comes to a minima. One of each will get you exactly two.
A familiar example might be a simple cubic polynomial. Say you wanted there to be extremes at +-1. So a derivative (a quadratic) would have zeroes at +-1 (the extrema, where the slope goes to zero.
x^2–1 = 0 is such a polynomial, so, integrating x^3-x = 0 should fit the bill, as would many other examples. Perhaps toss in a constant when you are transcribing to your homework.
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Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views
·1y
Originally Answered: Can a function have local maximum and minimum points without any overall minimum or maximum values? If so, what is an example of such a function? ·
I see what you mean with a polynomial-type function such as this…
…but a trigonometrical function such as y = sin(x) just...
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I see what you mean with a polynomial-type function such as this…
…but a trigonometrical function such as y = sin(x) just...
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9 7
Student
Doctorate in Computer Science, Washington State University (Graduated 1988)
·2y
You need two inflection points to get two local extrema, so a cubic polynomial is perfect. You also want the value to go to infinity when x -> infinity or -infinity. Finally, to guarantee that you don’t get a saddle and three extremes, you need the behavior at the two endpoints to be in opposite directions. Something like -x3 + 6x -2 is very simple and works. Try it and experiment with changes, like adding an x2 term. With graphing calculators, this is pretty simple.
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Related questions
More answers below
What are some reasons why we can't find global extrema for some functions?
What are some examples where a function has more than one maximum or minimum value, but only one local maximum or minimum value?
What is the significance of a function having a local minimum or maximum at a point, but no global ones (without using calculus)?
How do you tell if a function is maximum or minimum?
What are some examples of functions which have no local maximum or minimum but have global maximum or minimum?
Jared Teslow
Author has 68 answers and 31.1K answer views
·2y
Let a,b,c∈R a,b,c∈R s.t. a<b a<b and c≠0 c≠0
Now consider c(x−a)(x−b)=c x 2−c(a+b)x+a b c c(x−a)(x−b)=c x 2−c(a+b)x+a b c this has roots a a and b b
If we integrate that we get
p(x)=c 3 x 3−c(a+b)2 x 2+a b c x+D p(x)=c 3 x 3−c(a+b)2 x 2+a b c x+D
where D D can be any real number.
Then p(x)p(x) has no global max or min as it tends toward positive or negative infinity as x x tends toward positive or negative infinity (the exact pairing depends on the sign of c c). Also note that p(x)p(x) has two local extrema at x=a x=a and b b.
This function, p(x)p(x), has the properties desired.
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Francesco Amato
Studied at Polytechnic University of Turin (Graduated 1977) · Author has 4.5K answers and 1M answer views
·1y
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Can you provide an example of a function that has both a local minimum and local maximum but no global minimum or global maximum?
f(x)=x 3−x f(x)=x 3−x
lim x→±∞=±∞lim x→±∞=±∞
The stationary points are where
f′(x)=3 x 2–1=0→x±=±1√3 f′(x)=3 x 2–1=0→x±=±1 3
f(1√3)=−0.384 f(1 3)=−0.384 (relative minimum)
f(−1√3)=0.384 f(−1 3)=0.384 (relative maximum)
They are relative since there are smaller and greater function values respectively.
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f(x)=x 3−x f(x)=x 3−x
lim x→±∞=±∞lim x→±∞=±∞
The stationary points are where
f′(x)=3 x 2–1=0→x±=±1√3 f′(x)=3 x 2–1=0→x±=±1 3
f(1√3)=−0.384 f(1 3)=−0.384 (relative minimum)
f(−1√3)=0.384 f(−1 3)=0.384 (relative maximum)
They are relative since there are smaller and greater function values respectively.
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9 3
David Joyce
Professor Emeritus of Mathematics at Clark University · Upvoted by
Parag Kalita
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Jeremy Collins
, M.A. Mathematics, Trinity College, Cambridge · Author has 9.9K answers and 68.4M answer views
·4y
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How can I prove the function y=(x+1)2(x−2)y=(x+1)2(x−2) doesn't have any global extrema?
First prove that any polynomial of odd degree has no global extrema, then note that this is a cubic polynomial so it’s has odd degree.
So, how do you prove that any odd degree polynomial f(x)f(x) has no global extrema? You look at what happens for large values of x.x. If the leading coefficient is positive, then as x x approaches +∞,+∞, show f(x)f(x) approaches +∞,+∞, too, and as x x approaches −∞,−∞, so does f(x)f(x) approach −∞.−∞.
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Maurice Dupre
Ph. D. in Mathematics, University of Pennsylvania (Graduated 1972) · Author has 1.6K answers and 452.2K answer views
·1y
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Can you provide an example of a function that has both a local minimum and local maximum but no global minimum or global maximum?
f:R→R, where R is the whole real line
f(x)=x^3–3x, for all x in R.
This function has local maximum at -1, local minimum at +1, but is unbounded both above and below. It has an inflection point at 0 where f has negative slope.
For instance any cubic curve with positive highest degree coefficient and having negative slope at its inflection point, or negative highest degree coefficient with positive slope at its inflection point (every cubic curve has a single inflection point and if it has a local maximum at and a local minimum, then the inflection point is midway between the two).
More generally, a
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f:R→R, where R is the whole real line
f(x)=x^3–3x, for all x in R.
This function has local maximum at -1, local minimum at +1, but is unbounded both above and below. It has an inflection point at 0 where f has negative slope.
For instance any cubic curve with positive highest degree coefficient and having negative slope at its inflection point, or negative highest degree coefficient with positive slope at its inflection point (every cubic curve has a single inflection point and if it has a local maximum at and a local minimum, then the inflection point is midway between the two).
More generally, any odd degree polynomial of degree n+1 having n+1 different roots will have n local extremes with local minimums alternating with local maximums, but no global minimum and no global maximum.
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Marek Čtrnáct
Translator From English · Author has 821 answers and 938.4K answer views
·2y
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Can a function have a local minimum but no global minimum/maximum?
Sure! A good example is y=x sin(x)y=x sin(x):
Each “wave” is bigger than the one before. There are infinitely many minimums and maximums, but none of them is global.
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Sure! A good example is y=x sin(x)y=x sin(x):
Each “wave” is bigger than the one before. There are infinitely many minimums and maximums, but none of them is global.
Upvote ·
9 2
9 2
Alfred Dominic Vella
Life long educator in mathematics, computing and the sciences · Author has 4.7K answers and 4.4M answer views
·1y
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Can you provide an example of a function that has both a local minimum and local maximum but no global minimum or global maximum?
Any polynomial of odd degree other than a straight line!
You can easily construct others.
For example take a sin or cos and multiply by a non trivial odd monomial :
Eg x sin x
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Philip Lloyd
Specialist Calculus Teacher, Motivator and Baroque Trumpet Soloist. · Author has 6.8K answers and 52.8M answer views
·2y
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How do I investigate the following function? Does the following function f:R−>R,f(x)x x 2+1 f:R−>R,f(x)x x 2+1 local and global extrema ?
Let’s consider the equation…
So the graph has a horizontal asymptote of y = 0
On the right, it approaches 0 from above the x axis and on the left it approaches zero from underneath the x axis.
To find any max/min points we will need to differentiate…
Continue Reading
Let’s consider the equation…
So the graph has a horizontal asymptote of y = 0
On the right, it approaches 0 from above the x axis and on the left it approaches zero from underneath the x axis.
To find any max/min points we will need to differentiate…
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99 10
Jörg Straube
M.Sc. in Computer Science, ETH Zurich (Graduated 1987) · Author has 6.3K answers and 1.7M answer views
·1y
Related
Can you provide an example of a function that has both a local minimum and local maximum but no global minimum or global maximum?
Instead of the “boring” polynomials of degree 3,
I’ll give another example: y = (x^2 + 1)/x
It has a local min at x = 1 and a local max at x = -1, but there is no global max nor a global min.
Continue Reading
Instead of the “boring” polynomials of degree 3,
I’ll give another example: y = (x^2 + 1)/x
It has a local min at x = 1 and a local max at x = -1, but there is no global max nor a global min.
Upvote ·
Gopal Menon
B Sc (Hons) in Mathematics, Indira Gandhi National Open University (IGNOU) (Graduated 2010) · Author has 10.2K answers and 15.2M answer views
·Updated 3y
Related
What is the function such that has no local minimum and local maximum, but the points x=2 and x=4 are critical of the function?
What is the function such that has no local minimum and local maximum, but the points x=2 and x=4 are critical of the function?
A function for which x=2 x=2 and x=4 x=4 are critical points is one where d y d x=(x−2)(x−4).d y d x=(x−2)(x−4).
Integrating this, we get, y=x 3 3−3 x 2+8 x,y=x 3 3−3 x 2+8 x, ignoring the constant of integration.
In this function x=2 x=2 and x=4 x=4 are a local maximum and a local minimum respectively.
y(2)=8 3−12+16=20 3.y(2)=8 3−12+16=20 3.
y(4)=64 3−48+32=16 3.y(4)=64 3−48+32=16 3.
⇒y(2)−y(4)=4 3.⇒y(2)−y(4)=4 3.
If x=2 x=2 and x=4 x=4 are not to be local maxima or local minima, the following modifications are necessary.
The
Continue Reading
What is the function such that has no local minimum and local maximum, but the points x=2 and x=4 are critical of the function?
A function for which x=2 x=2 and x=4 x=4 are critical points is one where d y d x=(x−2)(x−4).d y d x=(x−2)(x−4).
Integrating this, we get, y=x 3 3−3 x 2+8 x,y=x 3 3−3 x 2+8 x, ignoring the constant of integration.
In this function x=2 x=2 and x=4 x=4 are a local maximum and a local minimum respectively.
y(2)=8 3−12+16=20 3.y(2)=8 3−12+16=20 3.
y(4)=64 3−48+32=16 3.y(4)=64 3−48+32=16 3.
⇒y(2)−y(4)=4 3.⇒y(2)−y(4)=4 3.
If x=2 x=2 and x=4 x=4 are not to be local maxima or local minima, the following modifications are necessary.
The domain has to be split into a union of three sub-domains, (−∞,2]∪(2,4)∪[4,∞).(−∞,2]∪(2,4)∪[4,∞).
In the first sub-domain the function has to be continuous in the neighbourhood of 2 2 and monotonically increasing or decreasing with the slope approaching 0 0 has x x tends to 2−.2−.
In the third sub-domain the function has to be continuous in the neighbourhood of 4 4 and monotonically increasing or decreasing with the slope approaching 0 0 as x x tends to 4+.4+.
The functions in the first sub-domain and the third sub-domain should tend to the same limit as x x tends to 2−2− and 4+4+ respectively and this limit should be the constant value of the function in the second sub-domain.
There are an infinite number of functions meeting the above requirements.
One such function is as under:
y=x 3 3−3 x 2+8 x y=x 3 3−3 x 2+8 x for x∈(−∞,2],x∈(−∞,2],
y=20 3 y=20 3 for x∈(2,4),x∈(2,4),
y=x 3 3−3 x 2+8 x+4 3 y=x 3 3−3 x 2+8 x+4 3 for x∈[4,∞).x∈[4,∞).
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Related questions
What is a function that has a local minimum but no global minimum?
What are some continuous functions that do not have a global maximum or global minimum?
What is the difference between the global maximum and minimum and local maximum and minimum?
Can a function have a local minimum but no global minimum/maximum?
What are some examples of functions with two local maximums and one local minimum at each end?
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What is the significance of a function having a local minimum or maximum at a point, but no global ones (without using calculus)?
How do you tell if a function is maximum or minimum?
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