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Commutative Property Of Addition 2. If A is an n×m matrix and O is a m×k zero-matrix, then we have: AO = O Note that AO is the n×k zero-matrix. Matrix Matrix Multiplication 11:09. We have 1. To understand the properties of transpose matrix, we will take two matrices A and B which have equal order. The identity matrix is a square matrix that has 1’s along the main diagonal and 0’s for all other entries. In a triangular matrix, the determinant is equal to the product of the diagonal elements. This matrix is often written simply as $$I$$, and is special in that it acts like 1 in matrix multiplication. Is the Inverse Property of Matrix Addition similar to the Inverse Property of Addition? The identity matrices (which are the square matrices whose entries are zero outside of the main diagonal and 1 on the main diagonal) are identity elements of the matrix product. Learning Objectives. In fact, this tutorial uses the Inverse Property of Addition and shows how it can be expanded to include matrices! Keywords: matrix; matrices; inverse; additive; additive inverse; opposite; Background Tutorials . Matrix Multiplication Properties 9:02. 16. Proof. There are a few properties of multiplication of real numbers that generalize to matrices. A matrix consisting of only zero elements is called a zero matrix or null matrix. Properties of Matrix Addition and Scalar Multiplication. What is the Identity Property of Matrix Addition? General properties. Yes, it is! There are 10 important properties of determinants that are widely used. Go through the properties given below: Assume that, A, B and C be three m x n matrices, The following properties holds true for the matrix addition operation. The determinant of a 4×4 matrix can be calculated by finding the determinants of a group of submatrices. 13. If you built a random matrix and took its determinant, how likely would it be that you got zero? The first element of row one is occupied by the number 1 … In mathematics, matrix addition is the operation of adding two matrices by adding the corresponding entries together. Equality of matrices All-zero Property. Multiplying a $2 \times 3$ matrix by a $3 \times 2$ matrix is possible, and it gives a $2 \times 2$ matrix … Properties of Transpose of a Matrix. The Commutative Property of Matrix Addition is just like the Commutative Property of Addition! The Distributive Property of Matrices states: A ( B + C ) = A B + A C Also, if A be an m × n matrix and B and C be n × m matrices, then Addition: There is addition law for matrix addition. Likewise, the commutative property of multiplication means the places of factors can be changed without affecting the result. Then the following properties hold: a) A+B= B+A(commutativity of matrix addition) b) A+(B+C) = (A+B)+C (associativity of matrix addition) c) There is a unique matrix O such that A+ O= Afor any m× nmatrix A. Since Theorem SMZD is an equivalence (Proof Technique E) we can expand on our growing list of equivalences about nonsingular matrices. Properties of matrix addition. Let A, B, and C be mxn matrices. The determinant of a 3 x 3 matrix (General & Shortcut Method) 15. PROPERTIES OF MATRIX ADDITION PRACTICE WORKSHEET. You should only add the element of one matrix to … Property 1 completes the argument. Let A, B, and C be three matrices of same order which are conformable for addition and a, b be two scalars. This means if you add 2 + 1 to get 3, you can also add 1 + 2 to get 3. For any natural number n > 0, the set of n-by-n matrices with real elements forms an Abelian group with respect to matrix addition. This tutorial uses the Commutative Property of Addition and an example to explain the Commutative Property of Matrix Addition. This project was created with Explain Everything™ Interactive Whiteboard for iPad. (i) A + B = B + A [Commutative property of matrix addition] (ii) A + (B + C) = (A + B) +C [Associative property of matrix addition] (iii) ( pq)A = p(qA) [Associative property of scalar multiplication] Let A, B, C be m ×n matrices and p and q be two non-zero scalars (numbers). In this lesson, we will look at this property and some other important idea associated with identity matrices. So if n is different from m, the two zero-matrices are different. Andrew Ng. Matrices rarely commute even if AB and BA are both defined. What is a Variable? Addition and Scalar Multiplication 6:53. The addition of the condition $\detname{A}\neq 0$ is one of the best motivations for learning about determinants. ... although it is associative and is distributive over matrix addition. Question 1 : then, verify that A + (B + C) = (A + B) + C. Solution : Question 2 : then verify: (i) A + B = B + A (ii) A + (- A) = O = (- A) + A. This tutorial introduces you to the Identity Property of Matrix Addition. Instructor. Transcript. Important Properties of Determinants. Properties involving Addition and Multiplication: Let A, B and C be three matrices. Matrix addition and subtraction, where defined (that is, where the matrices are the same size so addition and subtraction make sense), can be turned into homework problems. A. Addition and Subtraction of Matrices: In matrix algebra the addition and subtraction of any two matrix is only possible when both the matrix is of same order. Selecting row 1 of this matrix will simplify the process because it contains a zero. We state them now. A diagonal matrix is called the identity matrix if the elements on its main diagonal are all equal to $$1.$$ (All other elements are zero). Question: THEOREM 2.1 Properties Of Matrix Addition And Scalar Multiplication If A, B, And C Are M X N Matrices, And C And D Are Scalars, Then The Properties Below Are True. A B _____ Commutative property of addition 2. Unlike matrix addition, the properties of multiplication of real numbers do not all generalize to matrices. Let A, B, and C be three matrices. 8. det A = 0 exactly when A is singular. 2. In other words, the placement of addends can be changed and the results will be equal. Then we have the following properties. 17. The basic properties of matrix addition is similar to the addition of the real numbers. Then we have the following: (1) A + B yields a matrix of the same order (2) A + B = B + A (Matrix addition is commutative) Properties of scalar multiplication. (A+B)+C = A + (B+C) 3. where is the mxn zero-matrix (all its entries are equal to 0); 4. if and only if B = -A. If the rows of the matrix are converted into columns and columns into rows, then the determinant remains unchanged. Laplace’s Formula and the Adjugate Matrix. 14. Matrix multiplication shares some properties with usual multiplication. 4. Question 3 : then find the additive inverse of A. the identity matrix. Properties of Matrix Addition (1) A + B + C = A + B + C (2) A + B = B + A (3) A + O = A (4) A + − 1 A = 0. Find the composite of transformations and the inverse of a transformation. Created by the Best Teachers and used by over 51,00,000 students. Use the properties of matrix multiplication and the identity matrix Find the transpose of a matrix THEOREM 2.1: PROPERTIES OF MATRIX ADDITION AND SCALAR MULTIPLICATION If A, B, and C are m n matrices, and c and d are scalars, then the following properties are true. Matrix Multiplication - General Case. Properties involving Addition. The commutative property of addition means the order in which the numbers are added does not matter. Properties involving Multiplication. A+B = B+A 2. Given the matrix D we select any row or column. Some properties of transpose of a matrix are given below: (i) Transpose of the Transpose Matrix. Note that we cannot use elimination to get a diagonal matrix if one of the di is zero. Question 1 : then, verify that A + (B + C) = (A + B) + C. Question 2 : then verify: (i) A + B = B + A (ii) A + (- A) = O = (- A) + A. When the number of columns of the first matrix is the same as the number of rows in the second matrix then matrix multiplication can be performed. The inverse of 3 x 3 matrix with determinants and adjugate . The order of the matrices must be the same; Subtract corresponding elements; Matrix subtraction is not commutative (neither is subtraction of real numbers) Matrix subtraction is not associative (neither is subtraction of real numbers) Scalar Multiplication. The inverse of 3 x 3 matrices with matrix row operations. Properties of matrix multiplication. Properties of Matrix Addition, Scalar Multiplication and Product of Matrices. EduRev, the Education Revolution! As with the commutative property, examples of operations that are associative include the addition and multiplication of real numbers, integers, and rational numbers. Matrix multiplication is really useful, since you can pack a lot of computation into just one matrix multiplication operation. Proposition (commutative property) Matrix addition is commutative, that is, for any matrices and and such that the above additions are meaningfully defined. 11. Numerical and Algebraic Expressions. The determinant of a 2 x 2 matrix. We can also say that the determinant of the matrix and its transpose are equal. Matrix Vector Multiplication 13:39. 12. Examples . In that case elimination will give us a row of zeros and property 6 gives us the conclusion we want. This property is known as reflection property of determinants. The inverse of a 2 x 2 matrix. However, unlike the commutative property, the associative property can also apply to matrix … The matrix O is called the zero matrix and serves as the additiveidentity for the set of m×nmatrices. 1. Best Videos, Notes & Tests for your Most Important Exams. To find the transpose of a matrix, we change the rows into columns and columns into rows. Properties of Matrix Addition: Theorem 1.1Let A, B, and C be m×nmatrices. However, there are other operations which could also be considered addition for matrices, such as the direct sum and the Kronecker sum Entrywise sum. The determinant of a matrix is zero if each element of the matrix is equal to zero. A scalar is a number, not a matrix. Try the Course for Free. Use properties of linear transformations to solve problems. Matrix addition is associative; Subtraction. 18. A square matrix is called diagonal if all its elements outside the main diagonal are equal to zero. If we take transpose of transpose matrix, the matrix obtained is equal to the original matrix. Reflection Property. Inverse and Transpose 11:12. Taught By. 1. This is an immediate consequence of the fact that the commutative property applies to sums of scalars, and therefore to the element-by-element sums that are performed when carrying out matrix addition. There often is no multiplicative inverse of a matrix, even if the matrix is a square matrix. Mathematical systems satisfying these four conditions are known as Abelian groups. Videos, Notes & Tests for your Most important Exams few properties of transpose of a matrix set m×nmatrices... That generalize to matrices to include matrices we can not use elimination to get 3 known as Abelian groups be... \Detname { a } \neq 0 $is one of the matrix is! Zero matrix or null matrix one is occupied by the number 1 … Best Videos, Notes & for. We change the rows of the matrix O is called a zero matrix or null matrix is zero of. And used by over 51,00,000 students and BA properties of matrix addition both defined ) we can also add +! Addition: Theorem 1.1Let a, B, and C be mxn.! With Explain Everything™ Interactive Whiteboard for iPad the rows of the di is zero: then the! Are a few properties of transpose of the matrix and its transpose are equal although. Without affecting the result as reflection property of Addition means the order in which the numbers are added not. Zero if each element of the condition$ \detname { a } \neq 0 is. Means the order in which the numbers are added does not matter numbers ) ×n matrices and p and be... To understand the properties of determinants that are widely used equivalences about nonsingular matrices conditions are known reflection. Inverse ; opposite ; Background Tutorials + 2 to get a diagonal matrix if one of di! Simply as \ ( I\ ), and C be m ×n matrices and properties of matrix addition! 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X 3 matrix with determinants and adjugate select any row or column if one of matrix.$ \detname { a } \neq 0 \$ is one properties of matrix addition the and. 1 in matrix multiplication operation your Most important Exams and B which have equal order SMZD is an equivalence Proof.

https://math.stackexchange.com/questions/2922644/comparing-the-magnitudes-of-expressions-of-surds

# Comparing the magnitudes of expressions of surds I recently tackled some questions on maths-challenge / maths-aptitude papers where the task was to order various expressions made up of surds (without a calculator, obviously). I found myself wondering whether I was relying too much on knowing the numerical value of some common surds, when a more robust method was available (and would work in more difficult cases). For example, one question asked which is the largest of: (a) $\sqrt{10}$ (b) $\sqrt2+\sqrt3$ (c) $5-\sqrt3$ In this case, I relied on my knowledge that $\sqrt{10} \approx 3.16$ and $\sqrt2\approx 1.41$ and $\sqrt3 \approx 1.73$ to find (a) $\approx 3.16$, (b) $\approx ~3.14$ and (c) $\approx ~3.27$ so that the required answer is (c). But this seemed inelegant: I felt there might be some way to manipulate the surd expressions to make the ordering more explicit. I can't see what that might be, however (squaring all the expressions didn't really help). I'd appreciate some views: am I missing a trick, or was this particular question simply testing knowledge of some common values? EDIT: after the very helpful answers, which certainly showed that there was a much satisfying and general way of approaching the original question, can I also ask about another version of the question which included (d) $\sqrt[4]{101}$. When approaching the question by approximation, I simply observed that $\sqrt[4]{101}$ is only a tiny bit greater than $\sqrt{10}$, and hence it still was clear to choose (c) as the answer. Is there any elegant way to extend the more robust methods to handle this case? • +1 for providing context (your first two sentences), something that nearly all questions at this level fail to do, and for providing a nice explanation of your concern. Incidentally, for math aptitude and other tests, it has always been my understanding that the questions are NOT testing whether you know the approximations, but whether you can perform the type of analysis in the answer by @Lord Shark the Unknown. Of course, unless the question writer puts some effort behind writing such questions, such questions can often be solved by your method. Sep 19 '18 at 10:41 • Thank you for all the comments and answers. I am pleased I chose to ask the question at MSE -- there was indeed something to learn here! Sep 20 '18 at 10:05 Comparing $$\sqrt{10}$$ and $$\sqrt2+\sqrt3$$ is the same as comparing $$10$$ and $$(\sqrt2+\sqrt3)^2=5+2\sqrt6$$. That's the same as comparing $$5$$ and $$2\sqrt6$$. Which of these is bigger? Likewise comparing $$\sqrt{10}$$ and $$5-\sqrt3$$ is the same as comparing $$10$$ and $$(5-\sqrt3)^2=28-10\sqrt3$$. That's the same as comparing $$10\sqrt3$$ and $$18$$. Which of these is bigger? • Ah .... yes of course ... $5=\sqrt{25}>\sqrt{24}=2\sqrt6$ Sep 19 '18 at 10:30 • Thank you for the hint! Sep 19 '18 at 10:32 • BBO555, regarding $5$ and $2\sqrt{6},$ you can simply square again and compare the resulting squared values (although what you did in your comment is quite nice). This relies on the property that, when $a$ and $b$ are positive (or even when they are nonnegative), then we have: $a < b$ if and only if $a^2 < b^2$ (this can be "seen" by considering the graph of $y = x^2$ for $x\geq0).$ Incidentally, the analogous result for cubing also is true and the result for cubing doesn't require the numbers to be positive (consider the graph of $y = x^3).$ Sep 19 '18 at 10:47 You can use: (1) the fact that $f(x)=x^2$ is a monotonically increasing function when $x\geq0$ and (2) the arithmetic-geometric mean inequality $\sqrt{ab}\leq\frac{a+b}{2}$, when $a, b\geq0$. Hence, $$(\sqrt{2}+\sqrt{3})^2=5+2\sqrt{2\cdot3}\leq5+2\frac{2+3}{2}=5+5=10=(\sqrt{10})^2$$ Therefore, using (1), we obtain $\sqrt{2}+\sqrt{3}\leq 10$. I forgot about this: $$5-\sqrt{3}=3+2-\sqrt{3}=3+\frac{1}{2+\sqrt{3}}\geq3+\frac{1}{2+2}=3.25$$ One can easily verify that $(3+1/4)^2>10.5>10$. One also finds that $10.5^2>110>101$. Then, performing argument (1) twice, one finds that $5-\sqrt{3}>(101)^{1/4}$. ## Consequently, $5-\sqrt{3}$ is the bigger number. • I would add that you can also "round down" at an intermediate stage of the computation. If you are trying to prove $a \gt b$ sometimes you can find an expression $c$ that is simpler than $a$ where $a \gt c$, do some more manipulation, and show $c \gt b$. The numerical estimates are useful for this because they tell you how much room you have. You might find that rough approximations work, or you might need to be quite careful. Sep 19 '18 at 14:01 • Thanks for including a route to handling case (d) ! Sep 20 '18 at 8:07

https://math.stackexchange.com/questions/1693964/in-calculus-how-can-a-function-have-several-different-yet-equal-derivatives

# In Calculus, how can a function have several different, yet equal, derivatives? I've been pondering this question all night as I work through some problems, and after a very thorough search, I haven't found anything completely related to my question. I guess i'm also curious how some derivatives are simplified as well, because in some cases I just can't see the breakdown. Here is an example: $f(x) = \dfrac{x^2-6x+12}{x-4}$ is the function I was differentiating. Here is what I got: $f '(x) = \dfrac{x^2-8x+12}{(x-4)^2}$ which checks using desmos graphing utility. Now, when I checked my textbook(and Symbolab) they got: $f '(x) = 1 - \dfrac{4}{(x-4)^2}$ which also checks on desmos. To me, these derivatives look nothing alike, so how can they both be the equal to the derivative of the original function? Both methods used the quotient rule, yet yield very different results. Is one of these "better" than the other? I know that it is easier to find critical numbers with a more simplified derivative, but IMO the derivative I found seems easier to set equal to zero than the derivative found in my book.I also wasn't able to figure out how the second derivative was simplified, so I stuck with mine. I'm obviously new to Calculus and i'm trying to understand the nuances of derivatives. When I ask most math people, including some professors, they just say "that's how derivatives are" but for me, that's not an acceptable answer. If someone can help me understand this, I would appreciate it. • You really really need to use parentheses in what you write. You mean to write $(x^2-8x+12)/(x-4)^2$. The point is that your two "different" answers are exactly the same because of algebra. – Ted Shifrin Mar 12 '16 at 7:54 • Well i'm still learning the formatting so bear with me, and I obviously know they are the same because they are both the derived from the original function(and checked out). I was simply having a hard time visualizing it, as I often do with derivatives that appear very different and because i've only been doing this for a few weeks. Anyways, thanks for the comment, I guess. – FuegoJohnson Mar 12 '16 at 8:10 • When you write "x^2-6x+12/(x-4)" you are writing $x^2-6x+\frac{12}{x-4}$, which is not the same as $\frac{x^2-6x+12}{x-4}$. – alex.jordan Mar 12 '16 at 8:12 • @Hirak: Your edit is incorrect. – Ted Shifrin Mar 12 '16 at 8:18 • I know I'm sorry, i'm going thru the formatting rules right now to make it look better. Sincerest apologies. – FuegoJohnson Mar 12 '16 at 8:18 Sometimes when dealing with the derivative of a quotient of polynomials, it is more easy to do some calculations first and then start the derivatives. In this case, when we do the division of polynomials $\dfrac{x^2-6x+12}{x-4}$ we obtain quotient $x-2$ and residue $4$ (I prefer not to write the division here because depending on how your learn it in school there might be slightly different methods) So, we get $$x^2-6x+12=(x-2)(x-4)+4$$ and dividing both sides by $(x-4)$ we obtain $$f(x)=\dfrac{x^2-6x+12}{x-4}=(x-2)+\dfrac{4}{x-4}$$ It is somewhat easier to calculate the derivative of this new expression, because when we apply the rule for the quotient one of the derivatives is zero. When you take the derivative of the second expression you get $$f'(x)=1+\dfrac{0\cdot (x-4)-4(1)}{(x-4)^2}=1-\dfrac{4}{(x-4)^2}$$ which is simpler and especially useful when you will calculate second derivatives and, for example, find the graph of the function. • Thank you for your helpful input! You broke it down in a way that I wasn't able to visualize, and now I see. I ended up finding the second derivative through a much more tedious method, so I think your way would definitely be easier. Thanks :) – FuegoJohnson Mar 12 '16 at 20:17 They are the same. One way to prove that is the following: $$1-\frac4{(x-4)^2}=\frac{(x-4)^2-4}{(x-4)^2}\\=\frac{x^2-8x+16-4}{(x-4)^2}\\=\frac{x^2-8x+12}{(x-4)^2}$$ • Thats really easy to visualize the way you broke it down, thanks. My book skips so many steps sometimes. So my next question for you, is one form "better" than the other? I had a really hard time understanding how they simplified the function in my book but seeing you compare them makes a little more sense to me. – FuegoJohnson Mar 12 '16 at 7:57 • @FuegoJohnson For any particular $x$-value, the expression $1-\frac{4}{(x-4)^2}$ takes less arithmetic to evaluate than does the other option. That is one reason to prefer it. Another reason is that it would be more efficient to continue taking higher order derivatives of $1-\frac{4}{(x-4)^2}$, since no quotient rule would be needed. – alex.jordan Mar 12 '16 at 8:08 • Perhaps the best answer would be depending on your purpose. I suppose the $1-\frac{4}{(x-4)^2}$ form would be easier to set to zero for me, but if you prefer the other method it is fine. I suppose the best answer is that the 'best' derivative would be the one which, setting for zero, you can isolate for x the fastest on a test haha. Aside from that, there is no real 'best' derivative. – Keith Afas Mar 12 '16 at 8:09 • Great input, thanks folks. I am about to take the second derivative of the function, so it makes sense that the book answer would be easier to work with, although I STILL don't understand how they simplified it the way they did in the book from the original function. My algebra is kinda rusty at the moment. If someone wants to break it down for me step by step, that would be great lol. – FuegoJohnson Mar 12 '16 at 8:17

http://mathhelpforum.com/algebra/8958-working-backwards-cubics.html

# Math Help - working backwards - cubics 1. ## working backwards - cubics Write an equation that has the following roots: 2, -1, 5 Answer key: x^3 - 6x^2 + 3x + 10 = 0 For quadratic equations, I use the sum and product of roots, this is a cubic equation, how do I solve this? Thanks. 2. Originally Posted by shenton Write an equation that has the following roots: 2, -1, 5 Answer key: x^3 - 6x^2 + 3x + 10 = 0 For quadratic equations, I use the sum and product of roots, this is a cubic equation, how do I solve this? Thanks. $(x - 2)(x + 1)(x - 5)$ 3. Thanks! That turns out to be not as difficult as imagined. I thought I needed to use sum and products of roots to write the equation, it does makes me wonder a bit why or when I need to use sum and products of roots. 4. Write an equation that has the following roots: 2, -1, 5 Is there any other way to solve this other than the (x-2)(x+1)(x-5) method? If we have these roots: 1, 1 + √2, 1 - √2 the (x - 1) (x -1 -√2) (x -1 +√2) method seems a bit lenghty. When we expand (x - 1) (x -1 -√2) (x -1 +√2) the first 2 factors, it becomes: (x^2 -x -x√2 -x +1 +√2) (x -1 +√2) collect like terms: (x^2 -2x -x√2 +1 +√2) (x -1 +√2) To further expand this will be lenghty, my gut feel is that mathematicians do not want to do this - it is time consuming and prone to error. There must be a way to write an equation other than the above method. Is there a method to write an equation with 3 given roots (other than the above method)? Thanks. 5. Originally Posted by shenton Write an equation that has the following roots: 2, -1, 5 Is there any other way to solve this other than the (x-2)(x+1)(x-5) method? If we have these roots: 1, 1 + √2, 1 - √2 the (x - 1) (x -1 -√2) (x -1 +√2) method seems a bit lenghty. When we expand (x - 1) (x -1 -√2) (x -1 +√2) the first 2 factors, it becomes: (x^2 -x -x√2 -x +1 +√2) (x -1 +√2) collect like terms: (x^2 -2x -x√2 +1 +√2) (x -1 +√2) To further expand this will be lenghty, my gut feel is that mathematicians do not want to do this - it is time consuming and prone to error. There must be a way to write an equation other than the above method. Is there a method to write an equation with 3 given roots (other than the above method)? Thanks. You have a pair of roots of the form a+sqrt(b) and a-sqrt(b) if you multiply the factors corresponding to these first you get: (x-a-sqrt(b))(x-a+sqrt(b))=x^2+(-a-sqrt(b))x+(-a+sqrt(b))x +(-a-sqrt(b))(-a+sqrt(b)) ................=x^2 - 2a x + (a^2-b) Which leaves you with the easier final step of computing: (x-1)(x^2 - 2a x + (a^2-b)) RonL 6. Hello, shenton! The sum and product of roots works well for quadratic equations. For higher-degree equations, there is a generalization we can use. To make it simple (for me), I'll explain a fourth-degree equation. Divide through by the leading coefficient: . $x^4 + Px^3 + Qx^2 + Rx + S \:=\:0$ Insert alternating signs: . $+\:x^4 - Px^3 + Qx^2 - Rx + S \:=\:0$ . . . . . . . . . . . . . . . . . . $\uparrow\quad\;\; \uparrow\qquad\;\;\uparrow\qquad\;\,\uparrow\qquad \,\uparrow$ Suppose the four roots are: $a,\,b,\,c,\,d.$ The sum of the roots (taken one at a time) is: $-P.$ . . $a + b + c + d \:=\:-P$ The sum of the roots (taken two at a time) is: $Q.$ . . $ab + ac + ad + bc + bd + cd \:=\:Q$ The sum of the roots (taken three at a time) is: $-R.$ . . $abc + abd + acd + bcd \:=\:-R$ The sum of the roots ("taken four at a time") is: $S.$ . . $abcd \:=\:S$ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ For your problem with roots: $(a,b,c) \:=\:(2,-1,5)$ . . we have: . $x^3 + Px^2 + Qx + R \:=\:0$ Then: . $a + b + c \:=\:-P\quad\Rightarrow\quad2 + (-1) + 5\:=-P$ . . Hence: $P = -6$ And: . $ab + bc + ac \:=\:Q\quad\Rightarrow\quad(2)(-1) + (-1)(5) + (2)(5) \:=\:Q$ . . Hence: $Q = 3$ And: . $abc \:=\:-R\quad\Rightarrow\quad(2)(-1)(5)\:=\:-R$ . . Hence: $R = 10$ Therefore, the cubic is: . $x^3 - 6x^2 + 3x + 10 \:=\:0$ 7. This is awesome, Soroban. Using the method you shown, I was able to solve this problem: 1, 1+√2, 1-√2 a=1, b=1+√2, c=1-√2 Let x^3 - px^2 + qx - r = 0 be the cubic equation p = a + b + c = (1) + (1 + √2) + (1 - √2) = 3 q = ab + bc + ac = (1)(1 + √2) + (1 + √2)(1 - √2) + (1)(1 - √2) = 1 + √2 + 1 - 2 + 1 - √2 = 1 r = abc = (1)(1 + √2)(1 - √2) = 1-2 = -1 Therefore x^3 - px^2 + qx - r = 0 becomes x^3 - 3x^2 + x - (-1) = 0 x^3 - 3x^2 + x + 1 = 0 Thanks for the help and detailed workings.

https://math.stackexchange.com/questions/202052/work-and-time-when-work-is-split-into-parts

# Work and time, when work is split into parts I'm stuck on a particular type of work and time problems. For example, 1) A,B,C can complete a work separately in 24,36 and 48 days. They started working together but C left after 4 days of start and A left 3 days before completion of the work. In how many days will the work be completed? A simpler version of the same type of problem is as follows: 2) A can do a piece of work in 14 days while B can do it in 21 days. They begin working together but 3 days before the completion of the work, A leaves off. The total number of days to complete the work is? My attempt at problem 2: A's 1 day work=1/14 and B's 1 day work= 1/21 Assume that it takes 'd' days to complete the entire work when both A and B are working together. Then, (1/14 + 1/21)*d= 1 -> d=42/5 days. But it is stated that 3 days before the completion of the work, A left. Therefore, work done by both in (d-3) days is: (1/14 + 1/21)*(42/5 - 3)= 9/14 Remaining work= 1- 9/14 = 5/14 which is to be done by B alone. Hence the time taken by B to do (5/14) of the work is: (5/14)*21 = 7.5 days. Total time taken to complete the work = (d-3) + 7.5 = 12.9 days. However, this answer does not concur with the one that is provided. My Understanding of problem 1: Problem 1 is an extended version of problem 2. But since i think i'm doing problem 2 wrong, following the same method on problem 1 will also result in a wrong answer. Where did i go wrong? ## 5 Answers You asked where you went wrong in solving this problem: A can do a piece of work in 14 days while B can do it in 21 days. They begin working together but 3 days before the completion of the work, A leaves off. The total number of days to complete the work is? As you said in your solution, $A$ can do $1/14$ of the job per day, and $B$ can do $1/21$ of the job per day. On each day that they work together, then, they do $$\frac1{14}+\frac1{21}=\frac5{42}$$ of the job. Up to here you were doing fine; it’s at this point that you went astray. You know that for the last three days of the job $B$ will be working alone. In those $3$ days he’ll do $$3\cdot\frac1{21}=\frac17$$ of the job. That means that the two of them working together must have done $\frac67$ of the job before $A$ left. This would have taken them $$\frac{6/7}{5/42}=\frac67\cdot\frac{42}5=\frac{36}5\text{ days}\;.$$ Add that to the $3$ days that $B$ worked alone, and you get the correct total: $$\frac{36}5+3=\frac{51}5=10.2\text{ days}\;.$$ You worked out how long it would take them working together, subtracted $3$ days from that, saw how much of the job was left to be done at that point, and added on the number of days that it would take $B$ working alone to finish the job. But as your own figures show, $B$ actually needs $7.5$ days to finish the job at that point, not $3$, so he ends up working alone for $7.5$ days. This means that $A$ actually left $7.5$ days before the end of the job, not $3$ days before. You have to figure out how long it takes them to reach the point at which $B$ can finish in $3$ days. Added: 1) A,B,C can complete a work separately in 24, 36 and 48 days. They started working together but C left after 4 days of start and A left 3 days before completion of the work. In how many days will the work be completed? Here you know that all three worked together for the first $4$ days, $B$ worked alone for the last $3$ days, and $A$ and $B$ worked together for some unknown number of days in the middle. Calculate the fraction of the job done by all three in the first $4$ days and the fraction done by $B$ alone in the last $3$ days, and subtract the total from $1$ to see what fraction was done by $A$ and $B$ in the middle period; then see how long it would take $A$ and $B$ to do that much. • Yes, in short i misinterpreted the question. But because of the line, "They begin working together but 3 days before the completion of the work, A leaves off", it seems as if A and B working together would have completed it in some estimated d number of days, 3 days before which A left the job. Hence obviously B would require >3 days to complete the job. How do i avoid such misinterpretations in these types of problems? again, an excellent answer. Thanks! – Karan Sep 25 '12 at 12:46 • @user85030: You’re welcome! I think that avoiding such misinterpretations is partly a matter of practice and partly a matter of reading them pretty literally. Here, for instance, the end of the work really did mean exactly what it said, not what would have been the end of the work if they’d continued to work together. – Brian M. Scott Sep 25 '12 at 21:52 • Can you please have a look at this:- math.stackexchange.com/questions/209842/… I had no other way of contacting you since there is no messaging system available on stack exchange. – Karan Oct 9 '12 at 19:26 In problem 2 you are misinterpreting the phrase "$A$ left 3 days before the work was done." When you calculate it as above (3 days before the work would've been done if $A$ worked on), its wrong, as $A$ left (as you calculated) 7.5 days before the work was done. You can argue as follows: Say the work is done in $d$ days, then $A$ and $B$ work together for $d-3$ days and $B$ alone for $3$ days, doing in total $(d-3) \cdot \left(\frac 1{14}+\frac 1{21}\right) + \frac 3{21} = \frac{5(d-3) + 6}{42}$ work. So we must have $5(d-3) = 36$, so $5d = 51$, that is $d = 57/5$. For 1), you can argue along the same lines. Problem $1.)$ Let $n$ be the required number of days. $A,B,C$'s $1$ day work is $1/24,1/36,1/48$ respectively. Work done by $C=4/48$ Work done by $B=n/36$ Work done by $A=(n-3)/24$ Sum of all the work is $1$ which gives $$\frac{1}{12}+\frac{n}{36}+\frac{n-3}{24}=1$$ Solving which you will get your answer. Problem $2.)$ can be solved using similar approach A,B,C can complete a work separately in 24, 36 and 48 days. They started working together but C left after 4 days of start and A left 3 days before completion of the work. In how many days will the work be completed? ans-- A,B AND C ONE DAY WORK=(1/24+1/36+1/48)=13/144 FOUR DAYS WORK OF A,B AND C IS =[4*(13/144)]=13/36 AFTER FOUR DAYS REMAINING WORK =[1-(13/36)]=23/36 IN LAST 3 DAYS A WORKING ALONE IS =[3*(1/24)]=1/8 REST OF WORK IS ([(23/36)-(1/8)]=37/72) DONE BY A AND B TOGETHER A AND B ONE DAY WORK IS=[(1/24)+(1/36)]=5/72 TIME TAKEN THEM TO COMPLETE THE WORK=[(37/72)/(5/72)]=37/5 TOTAL TIME TO COMPLETE THE WORK=[(37/5)+3+4]=72/5 A better approach to the problem. Take the LCM of 14 and 21 which will give you the total amount of work. LCM (14,21) = 42. A completes in 14 days.So he does 42/14 in 1 day.Similarly B does 42/21 in 1 day. A=3,B=2.Since A left 3 days prior to the completion of the work, so his 3 days work is absent.Lets add his 3 days pending work to the total amount of work. In 1 day he does 3 units of work, so in 3 days he will do 9 units of work. Total work = 42+9 = 51. Both will now take 3+2 to complete the total work. A+B = 51/5 = 10.2. ## protected by Alex M.Mar 19 '17 at 18:37 Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count). Would you like to answer one of these unanswered questions instead?

https://brilliant.org/discussions/thread/algebraic-manipulation/

# Algebraic Manipulation ## Definition Algebraic manipulation involves rearranging variables to make an algebraic expression better suit your needs. During this rearrangement, the value of the expression does not change. ## Technique Algebraic expressions aren't always given in their most convenient forms. This is where algebraic manipulation comes in. For example: ### What value of $$x$$ satisfies $$5x+8 = -2x +43$$ We can rearrange this equation for $$x$$ by putting the terms with $$x$$ on one side and the constant terms on the other. \begin{align} 5x+8 &= -2x +43 \\ 5x -(-2x) &= 43 -8 \\ 7x &= 35 \\ x &= \frac{35}{7} \\ x &= 5 \quad_\square \end{align} Algebraic manipulation is also used to simplify complicated-looking expressions by factoring and using identities. Let's walk through an example: ### $\frac{x^3+y^3}{x^2-y^2} - \frac{x^2+y^2}{x-y}.$ It's possible to solve for $$x$$ and $$y$$ and plug those values into this expression, but the algebra would be very messy. Instead, we can rearrange the problem by using the factoring formula identities for $$x^3+y^3$$ and $$x^2-y^2$$ and then simplifying. \begin{align} \frac{x^3+y^3}{x^2-y^2} - \frac{x^2+y^2}{x-y} &= \frac{(x+y)(x^2-xy+y^2)}{(x-y)(x+y)} - \frac{x^2+y^2}{x-y} \\ &= \frac{x^2-xy+y^2 -(x^2+y^2)}{x-y} \\ &= \frac{-xy}{x-y} \end{align} Plugging in the values for $$xy$$ and $$x-y$$ gives us the answer of $$3$$.$$_\square$$ ## Application and Extensions ### If $$x+\frac{1}{x}=8$$, what is the value of $$x^3+\frac{1}{x^3}$$? The key to solving this problem (without explicitly solving for $$x$$) is to recognize that $\left(x+\frac{1}{x}\right)^3 = x^3+\frac{1}{x^3}+3\left(x+\frac{1}{x}\right)$ which gives us \begin{align} x^3+\frac{1}{x^3} &= \left(x+\frac{1}{x}\right)^3 - 3\left(x+\frac{1}{x}\right) \\ &= (8)^3 -3(8) \\ &= 488 \quad _\square \end{align} ### $\frac{2x+8}{\sqrt{2x+1}+\sqrt{x-3}}?$ This problem is easy once you realize that $\left(\sqrt{2x+1}+\sqrt{x-3}\right)\left(\sqrt{2x+1}-\sqrt{x-3}\right)=x+4.$ The solution is therefore \begin{align} \frac{2x+8}{\sqrt{2x+1}+\sqrt{x-3}} &= \frac{2(x+4)}{\sqrt{2x+1}+\sqrt{x-3}} \\ &=\frac{2\left(\sqrt{2x+1}+\sqrt{x-3}\right)\left(\sqrt{2x+1}-\sqrt{x-3}\right)}{\sqrt{2x+1}+\sqrt{x-3}} \\ &=2\left(\sqrt{2x+1}-\sqrt{x-3}\right) \\ &=2(2) \\ &=4 \quad _\square \end{align} Note by Arron Kau 4 years ago MarkdownAppears as *italics* or _italics_ italics **bold** or __bold__ bold - bulleted- list • bulleted • list 1. numbered2. list 1. numbered 2. list Note: you must add a full line of space before and after lists for them to show up correctly paragraph 1paragraph 2 paragraph 1 paragraph 2 [example link](https://brilliant.org)example link > This is a quote This is a quote # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" MathAppears as Remember to wrap math in $$...$$ or $...$ to ensure proper formatting. 2 \times 3 $$2 \times 3$$ 2^{34} $$2^{34}$$ a_{i-1} $$a_{i-1}$$ \frac{2}{3} $$\frac{2}{3}$$ \sqrt{2} $$\sqrt{2}$$ \sum_{i=1}^3 $$\sum_{i=1}^3$$ \sin \theta $$\sin \theta$$ \boxed{123} $$\boxed{123}$$ Sort by: You have posted very good solutions these kind of questions. - 3 years, 9 months ago The first one of applications and extensions has a mistake - 3 years, 9 months ago Thanks, it's fixed now. Staff - 3 years, 9 months ago ×

https://math.stackexchange.com/questions/4454637/inequality-involving-sums-with-binomial-coefficient

Inequality involving sums with binomial coefficient I am trying to show upper- and lower-bounds on $$\frac{1}{2^n}\sum_{i=0}^n\binom{n}{i}\min(i, n-i)$$ (where $$n\geq 1$$) in order to show that it basically grows as $$\Theta(n)$$. The upper-bound is easy to get since $$\min(i, n-i)\leq i$$ for $$i\in\{0, \dots n\}$$ so that $$\frac{1}{2^n}\sum_{i=0}^n\binom{n}{i}\min(i, n-i)\leq \frac{1}{2^n}\sum_{i=0}^n\binom{n}{i}i = \frac{n}{2}.$$ Thanks to Desmos, I managed to find a lower bound, but I am struggling to actually prove it. Indeed, I can see that the function $$f(n)=\frac{n-1}{3}$$ does provide a lower-bound. One can in fact rewrite $$\frac{n-1}{3}=\frac{1}{2^n}\sum_{i=0}^n\binom{n}{i}\frac{2i-1}{3}.$$ I was thus hoping to show that for each term we have $$\frac{2i-1}{3}\leq \min(i, n-i)$$, but this is only true if $$i\leq \frac{3n+1}{5}$$ and not generally for $$i\leq n$$. I imagine there is a clever trick to use at some point but for some reason I am stuck here. Any help would be appreciated, thank you! EDIT: Thank you everyone for all the great and diverse answers! I flagged River Li's answer as the "accepted" one because of its simplicity due to the use of Cauchy-Schwartz inequality, which does not require a further use of Stirling's approximation. Note that the other answers which involve such an approximation are much tighter though, but proving $$\Theta(n)$$ growth was sufficient here. • @Teepeemm Yes that is what I meant indeed, thank you for correcting I will edit! May 23 at 10:35 Both $$\binom ni$$ and $$\min(i,n-i)$$ are largest for $$i$$ near $$n/2$$. This means that, if we compare $$\frac1{n+1}\sum_{i=0}^n\binom ni\min(i,n-i)$$ with $$\left(\frac1{n+1}\sum_{i=0}^n \binom ni\right)\left(\frac1{n+1}\sum_{i=0}^n \min(i,n-i)\right),$$ the first should be larger, since larger numbers are multiplied by larger numbers and smaller numbers by smaller numbers. This can in fact be made more precise by (one form of) the rearrangement inequality: If $$a_1\leq \cdots\leq a_m$$ and $$b_1\leq \cdots\leq b_m$$ are sequences of real numbers, then $$\frac{a_1b_1+\cdots+a_mb_m}m\geq \frac{a_1+\cdots+a_m}m\cdot \frac{b_1+\cdots+b_m}m.$$ (This can be proven by summing $$(a_i-a_j)(b_i-b_j)\geq 0$$ over all $$i$$ and $$j$$.) So, $$\frac1{2^n}\sum_{i=0}^n\binom ni\min(i,n-i)\geq \frac{1}{(n+1)2^n}\sum_{i=0}^n\binom ni\sum_{i=0}^n\min(i,n-i)=\frac1{n+1}\sum_{i=0}^n\min(i,n-i).$$ The last sum is $$\Omega(n^2)$$, since the average order of $$\min(i,n-i)$$ is about $$n/4$$, and so the entire sum is $$\Omega(n)$$. You've also shown that it's $$O(n)$$, so this is enough to show that it's $$\Theta(n)$$. Let's first note that $$\binom{n}{i}\cdot i = n\cdot \binom{n-1}{i-1}$$. For odd $$n=2m+1$$, this makes \begin{align} S_n = \sum_{i=0}^{n}\binom{n}{i}\cdot\min(i,n-i) &= 2\sum_{i=0}^m\binom{n}{i}\cdot i = 2n\sum_{i=1}^m\binom{n-1}{i-1} = 2n\sum_{j=0}^{m-1}\binom{2m}{j} \\ &= n\cdot\left( \sum_{j=0}^{2m}\binom{2m}{j} - \binom{2m}{m} \right) = n\cdot\left( 2^{2m} - \binom{2m}{m} \right) \end{align} where we have used that $$\binom{2m}{j}=\binom{2m}{2m-j}$$. This makes $$\frac{S_n}{2^n} = \frac{n}{2}\cdot\left (1 - \frac{\binom{2m}{m}}{2^{2m}} \right) \approx \frac{n}{2}\cdot\left( 1 - \frac{1}{\sqrt{\pi m}} \right).$$ For even $$n=2m$$, we get \begin{align} S_n = \sum_{i=0}^{n}\binom{n}{i}\cdot\min(i,n-i) &= 2\sum_{i=0}^m\binom{n}{i}\cdot i - \binom{2m}{m}\cdot m = 2n\sum_{i=1}^m\binom{n-1}{i-1} - \binom{2m}{m}\cdot m \\ &= n\sum_{j=0}^{n-1}\binom{n-1}{j} - \binom{2m}{m}\cdot m = n\cdot\left( 2^{n-1} - \frac{1}{2}\binom{2m}{m} \right) \end{align} which once more makes $$\frac{S_n}{2^n} = \frac{n}{2}\cdot\left(1-\frac{\binom{2m}{m}}{2^{2m}}\right).$$ In both cases, you get $$n/2$$ as an upper bound. However, there are strong bounds on $$\binom{2m}{m}/2^{2m}$$ which can be applied: $$\frac{e^{-1/8m}}{\sqrt{\pi m}} \le \frac{\binom{2m}{m}}{2^{2m}} \le \frac{1}{\sqrt{\pi m}}$$ Eg, see Jack D'Aurizio's derivation of this, or Wikipedia. Additional bounds have been provided by robjohn. The following bounds seem to be the tightest proven so far: $$\frac{4^me^{-1/8m}}{\sqrt{\pi m}} < \binom{2m}{m} < \frac{4^m}{\sqrt{\pi\left( m+\frac{1}{4} \right)}}$$ The following bound is even tighter, but I have no proof of it, just numerical evidence: $$\frac{4^m}{\sqrt{\pi\left( m+\frac{1}{4}+\frac{1}{32m} \right)}} < \binom{2m}{m}$$ It's the same as the above up to second order approximation, so not a bit difference, but easier to compute. • I believe $$\frac{4^n}{\sqrt{\pi\!\left(n+\frac13\right)}}\le\binom{2n}{n}\le\frac{4^n}{\sqrt{\pi\!\left(n+\frac14\right)}}$$ gives somewhat tighter bounds. – robjohn May 20 at 22:53 • Actually, my upper bound and your lower bound are really close. – robjohn May 20 at 23:01 We start with $$\sum_{k=0}^{\left\lfloor\frac{n}2\right\rfloor}\binom{n}{k} =\left\{\begin{array}{} 2^{n-1}&\text{if n is odd}\\ 2^{n-1}+\frac12\left(\raise{2pt}{n}\atop\frac{n}2\right)&\text{if n is even} \end{array}\right.\tag1$$ Substitute $$n\mapsto n-1$$: $$\sum_{k=0}^{\left\lfloor\frac{n-1}2\right\rfloor}\binom{n-1}{k} =\left\{\begin{array}{} 2^{n-2}&\text{if n is even}\\ 2^{n-2}+\frac12\left(\raise{2pt}{n-1}\atop{\frac{n-1}2}\right)&\text{if n is odd} \end{array}\right.\tag2$$ Subtract the $$k=\left\lfloor\frac{n-1}2\right\rfloor$$ term: $$\sum_{k=0}^{\left\lfloor\frac{n-1}2\right\rfloor-1}\binom{n-1}{k} =\left\{\begin{array}{} 2^{n-2}-\left(\raise{2pt}{n-1}\atop{\frac{n}2-1}\right)&\text{if n is even}\\ 2^{n-2}-\frac12\left(\raise{2pt}{n-1}\atop{\frac{n-1}2}\right)&\text{if n is odd} \end{array}\right.\tag3$$ Substitute $$k\mapsto k-1$$: $$\sum_{k=1}^{\left\lfloor\frac{n-1}2\right\rfloor}\binom{n-1}{k-1} =\left\{\begin{array}{} 2^{n-2}-\frac12\left(\raise{2pt}{n}\atop{\frac{n}2}\right)&\text{if n is even}\\ 2^{n-2}-\frac12\left(\raise{2pt}{n-1}\atop{\frac{n-1}2}\right)&\text{if n is odd} \end{array}\right.\tag4$$ Then \begin{align} \sum_{k=0}^{\left\lfloor\frac{n-1}2\right\rfloor}\binom{n}{k}k &=n\sum_{k=1}^{\left\lfloor\frac{n-1}2\right\rfloor}\binom{n-1}{k-1}\tag{5a}\\ &=\left\{\begin{array}{} n2^{n-2}-\frac{n}2\left(\raise{2pt}{n}\atop{\frac{n}2}\right)&\text{if n is even}\\ n2^{n-2}-\frac{n}2\left(\raise{2pt}{n-1}\atop{\frac{n-1}2}\right)&\text{if n is odd} \end{array}\right.\tag{5b} \end{align} Explanation: $$\text{(5a)}$$: $$\binom{n}{k}k=\binom{n-1}{k-1}n$$ and the $$k=0$$ term vanishes $$\text{(5b)}$$: multiply $$(4)$$ by $$n$$ Double and add the middle term in the even $$n$$ case: \begin{align} \sum_{k=0}^n\binom{n}{k}\min(k,n-k) &=\left\{\begin{array}{} n2^{n-1}-\frac{n}2\left(\raise{2pt}{n}\atop{\frac{n}2}\right)&\text{if n is even}\\ n2^{n-1}-n\left(\raise{2pt}{n-1}\atop{\frac{n-1}2}\right)&\text{if n is odd} \end{array}\right.\tag6 \end{align} The estimates for the central binomial coefficients from this answer give $$\frac{2^n}{\sqrt{\pi\!\left(\frac{n}2+\frac13\right)}}\le\binom{n}{\frac{n}2}\le\frac{2^n}{\sqrt{\pi\!\left(\frac{n}2+\frac14\right)}}\tag7$$ which gives upper and lower bounds. Let $$S := \frac{1}{2^n}\sum_{i=0}^n\binom{n}{i}\min(i, n-i),$$ $$T := \frac{1}{2^n}\sum_{i=0}^n\binom{n}{i}\max(i, n-i).$$ We have $$S + T = \frac{1}{2^n}\sum_{i=0}^n\binom{n}{i}n = n,$$ $$T - S = \frac{1}{2^n}\sum_{i=0}^n\binom{n}{i} |n - 2i|$$ where we have used $$\max(a, b) + \min(a,b) = a + b$$ and $$\max(a, b) - \min(a, b) = |a - b|$$ for all real numbers $$a, b$$. Using Cauchy-Bunyakovsky-Schwarz inequality, we have \begin{align*} T - S &= \frac{1}{2^n}\sum_{i=0}^n \sqrt{\binom{n}{i} (n - 2i)^2}\sqrt{\binom{n}{i}}\\ &\le \frac{1}{2^n} \sqrt{\sum_{i=0}^n\binom{n}{i} (n - 2i)^2 \cdot \sum_{i=0}^n\binom{n}{i}}\\ &= \sqrt n \end{align*} where we have used $$\sum_{i=0}^n\binom{n}{i} i^2 = 2^{n - 2}n(n + 1)$$ and $$\sum_{i=0}^n\binom{n}{i} i = 2^{n - 1}n$$ (easy to prove using the identity $$\binom{k}{m}m = k\binom{k-1}{m-1}$$). Thus, we have $$\frac{n}{2} - \frac{\sqrt n}{2} \le S \le \frac{n}{2}.$$ The desired result follows.

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The main condition of matrix multiplication is that the number of columns of the 1st matrix must equal to the number of rows of the 2nd one. The dimensions of $B$ are $3\times 2$ and the dimensions of $A$ are $2\times 3$. The dot product involves multiplying the corresponding elements in the row of the first matrix, by that of the columns of the second matrix, and summing up the result, resulting in a single value. Example. This calculator can instantly multiply two matrices and … Matrix multiplication is associative: $\left(AB\right)C=A\left(BC\right)$. Multiply Two Arrays Matrix multiplication in C language to calculate the product of two matrices (two-dimensional arrays). Matrix Multiplication Calculator (Solver) This on-line calculator will help you calculate the __product of two matrices__. An example of a matrix is as follows. If A = [aij] is an m × n matrix and B = [bij] is an n × p matrix, the product AB is an m × p matrix. If the multiplication isn't possible, an error message is displayed. Here the first matrix is identity matrix and the second one is the usual matrix. The product of two matrices A and B is defined if the number of columns of A is equal to the number of rows of B. We multiply entries of $A$ with entries of $B$ according to a specific pattern as outlined below. When we multiply two arrays of order (m*n) and (p*q) in order to obtained matrix product then its output contains m rows and q columns where n is n==p is a necessary condition. The inner dimensions match so the product is defined and will be a $3\times 3$ matrix. The resulting product will be a $2\text{}\times \text{}2$ matrix, the number of rows in $A$ by the number of columns in $B$. Yes, consider a matrix A with dimension $3\times 4$ and matrix B with dimension $4\times 2$. Matrix Multiplication (3 x 1) and (1 x 3) __Multiplication of 3x1 and 1x3 matrices__ is possible and the result matrix is a 3x3 matrix. If $A$ is an $\text{ }m\text{ }\times \text{ }r\text{ }$ matrix and $B$ is an $\text{ }r\text{ }\times \text{ }n\text{ }$ matrix, then the product matrix $AB$ is an $\text{ }m\text{ }\times \text{ }n\text{ }$ matrix. Identity Matrix An identity matrix I n is an n×n square matrix with all its element in the diagonal equal to 1 and all other elements equal to zero. In mathematics, the matrix exponential is a function on square matrices analogous to the ordinary exponential function [1, , , , 7]. When complete, the product matrix will be. Finding the product of two matrices is only possible when the inner dimensions are the same, meaning that the number of columns of the first matrix is equal to the number of rows of the second matrix. Let’s return to the problem presented at the opening of this section. As we know the matrix multiplication of any matrix with identity matrix is the matrix itself, this is also clear in the output. tcrossprod () takes the cross-product of the transpose of a matrix. $\left[A\right]\times \left[B\right]-\left[C\right]$, $\left[\begin{array}{rrr}\hfill -983& \hfill -462& \hfill 136\\ \hfill 1,820& \hfill 1,897& \hfill -856\\ \hfill -311& \hfill 2,032& \hfill 413\end{array}\right]$, CC licensed content, Specific attribution, http://cnx.org/contents/fd53eae1-fa23-47c7-bb1b-972349835c3c@5.175:1/Preface. $A=\left[\begin{array}{rrr}\hfill {a}_{11}& \hfill {a}_{12}& \hfill {a}_{13}\\ \hfill {a}_{21}& \hfill {a}_{22}& \hfill {a}_{23}\end{array}\right]\text{ and }B=\left[\begin{array}{rrr}\hfill {b}_{11}& \hfill {b}_{12}& \hfill {b}_{13}\\ \hfill {b}_{21}& \hfill {b}_{22}& \hfill {b}_{23}\\ \hfill {b}_{31}& \hfill {b}_{32}& \hfill {b}_{33}\end{array}\right]$, $\left[\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\end{array}\right]\cdot \left[\begin{array}{c}{b}_{11}\\ {b}_{21}\\ {b}_{31}\end{array}\right]={a}_{11}\cdot {b}_{11}+{a}_{12}\cdot {b}_{21}+{a}_{13}\cdot {b}_{31}$, $\left[\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\end{array}\right]\cdot \left[\begin{array}{c}{b}_{12}\\ {b}_{22}\\ {b}_{32}\end{array}\right]={a}_{11}\cdot {b}_{12}+{a}_{12}\cdot {b}_{22}+{a}_{13}\cdot {b}_{32}$, $\left[\begin{array}{ccc}{a}_{11}& {a}_{12}& {a}_{13}\end{array}\right]\cdot \left[\begin{array}{c}{b}_{13}\\ {b}_{23}\\ {b}_{33}\end{array}\right]={a}_{11}\cdot {b}_{13}+{a}_{12}\cdot {b}_{23}+{a}_{13}\cdot {b}_{33}$, $AB=\left[\begin{array}{c}\begin{array}{l}{a}_{11}\cdot {b}_{11}+{a}_{12}\cdot {b}_{21}+{a}_{13}\cdot {b}_{31}\\ \end{array}\\ {a}_{21}\cdot {b}_{11}+{a}_{22}\cdot {b}_{21}+{a}_{23}\cdot {b}_{31}\end{array}\begin{array}{c}\begin{array}{l}{a}_{11}\cdot {b}_{12}+{a}_{12}\cdot {b}_{22}+{a}_{13}\cdot {b}_{32}\\ \end{array}\\ {a}_{21}\cdot {b}_{12}+{a}_{22}\cdot {b}_{22}+{a}_{23}\cdot {b}_{32}\end{array}\begin{array}{c}\begin{array}{l}{a}_{11}\cdot {b}_{13}+{a}_{12}\cdot {b}_{23}+{a}_{13}\cdot {b}_{33}\\ \end{array}\\ {a}_{21}\cdot {b}_{13}+{a}_{22}\cdot {b}_{23}+{a}_{23}\cdot {b}_{33}\end{array}\right]$, $A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]\text{ and }B=\left[\begin{array}{cc}5& 6\\ 7& 8\end{array}\right]$, $A=\left[\begin{array}{l}\begin{array}{ccc}-1& 2& 3\end{array}\hfill \\ \begin{array}{ccc}4& 0& 5\end{array}\hfill \end{array}\right]\text{ and }B=\left[\begin{array}{c}5\\ -4\\ 2\end{array}\begin{array}{c}-1\\ 0\\ 3\end{array}\right]$, $\begin{array}{l}\hfill \\ AB=\left[\begin{array}{rrr}\hfill -1& \hfill 2& \hfill 3\\ \hfill 4& \hfill 0& \hfill 5\end{array}\right]\text{ }\left[\begin{array}{rr}\hfill 5& \hfill -1\\ \hfill -4& \hfill 0\\ \hfill 2& \hfill 3\end{array}\right]\hfill \\ \text{ }=\left[\begin{array}{rr}\hfill -1\left(5\right)+2\left(-4\right)+3\left(2\right)& \hfill -1\left(-1\right)+2\left(0\right)+3\left(3\right)\\ \hfill 4\left(5\right)+0\left(-4\right)+5\left(2\right)& \hfill 4\left(-1\right)+0\left(0\right)+5\left(3\right)\end{array}\right]\hfill \\ \text{ }=\left[\begin{array}{rr}\hfill -7& \hfill 10\\ \hfill 30& \hfill 11\end{array}\right]\hfill \end{array}$, $\begin{array}{l}\hfill \\ BA=\left[\begin{array}{rr}\hfill 5& \hfill -1\\ \hfill -4& \hfill 0\\ \hfill 2& \hfill 3\end{array}\right]\text{ }\left[\begin{array}{rrr}\hfill -1& \hfill 2& \hfill 3\\ \hfill 4& \hfill 0& \hfill 5\end{array}\right]\hfill \\ \text{ }=\left[\begin{array}{rrr}\hfill 5\left(-1\right)+-1\left(4\right)& \hfill 5\left(2\right)+-1\left(0\right)& \hfill 5\left(3\right)+-1\left(5\right)\\ \hfill -4\left(-1\right)+0\left(4\right)& \hfill -4\left(2\right)+0\left(0\right)& \hfill -4\left(3\right)+0\left(5\right)\\ \hfill 2\left(-1\right)+3\left(4\right)& \hfill 2\left(2\right)+3\left(0\right)& \hfill 2\left(3\right)+3\left(5\right)\end{array}\right]\hfill \\ \text{ }=\left[\begin{array}{rrr}\hfill -9& \hfill 10& \hfill 10\\ \hfill 4& \hfill -8& \hfill -12\\ \hfill 10& \hfill 4& \hfill 21\end{array}\right]\hfill \end{array}$, $AB=\left[\begin{array}{cc}-7& 10\\ 30& 11\end{array}\right]\ne \left[\begin{array}{ccc}-9& 10& 10\\ 4& -8& -12\\ 10& 4& 21\end{array}\right]=BA$, $E=\left[\begin{array}{c}6\\ 30\\ 14\end{array}\begin{array}{c}10\\ 24\\ 20\end{array}\right]$, $C=\left[\begin{array}{ccc}300& 10& 30\end{array}\right]$, $\begin{array}{l}\hfill \\ \hfill \\ CE=\left[\begin{array}{rrr}\hfill 300& \hfill 10& \hfill 30\end{array}\right]\cdot \left[\begin{array}{rr}\hfill 6& \hfill 10\\ \hfill 30& \hfill 24\\ \hfill 14& \hfill 20\end{array}\right]\hfill \\ \text{ }=\left[\begin{array}{rr}\hfill 300\left(6\right)+10\left(30\right)+30\left(14\right)& \hfill 300\left(10\right)+10\left(24\right)+30\left(20\right)\end{array}\right]\hfill \\ \text{ }=\left[\begin{array}{rr}\hfill 2,520& \hfill 3,840\end{array}\right]\hfill \end{array}$. The product-process matrix can facilitate the understanding of the strategic options available to a company, particularly with regard to its manufacturing function. If you view them each as vectors, and you have some familiarity with the dot product, we're essentially going to take the dot product of that and that. Thank you for your questionnaire.Sending completion. Thus, the equipment need matrix is written as. The process of matrix multiplication becomes clearer when working a problem with real numbers. Given $A$ and $B:$. To obtain the entries in row $i$ of $AB,\text{}$ we multiply the entries in row $i$ of $A$ by column $j$ in $B$ and add. Since we view vectors as column matrices, the matrix-vector product is simply a special case of the matrix-matrix product (i.e., a product between two matrices). Multiply and add as follows to obtain the first entry of the product matrix $AB$. If A is a nonempty matrix, then prod (A) treats the columns of A as vectors and returns a row vector of the products of each column. Multiplication of two matrices involves dot products between rows of first matrix and columns of the second matrix. Boolean matrix products are computed via either %&% or boolArith = TRUE. Syntax: numpy.matmul (x1, x2, /, out=None, *, casting=’same_kind’, order=’K’, dtype=None, subok=True [, … If A =[aij]is an m ×n matrix and B =[bij]is an n ×p matrix then the product of A and B is the m ×p matrix C =[cij]such that cij=rowi(A)6 colj(B) A matrix is a rectangular array of numbers that is arranged in the form of rows and columns. You can only multiply two matrices if their dimensions are compatible, which means the number of columns in the first matrix is the same as the number of rows in the second matrix. e) order: 1 × 1. A user inputs the orders and elements of the matrices. The inner dimensions are the same so we can perform the multiplication. Let A ∈ Mn. Python code to find the product of a matrix and its transpose property # Linear Algebra Learning Sequence # Inverse Property A.AT = S [AT = transpose of A] import numpy as np M = np . The exponential of A, denoted by eA or exp(A) , is the n × n matrix … Number of rows and columns are equal therefore this matrix is a square matrix. The product of two matrices can be computed by multiplying elements of the first row of the first matrix with the first column of the second matrix then, add all the product of elements. For example, the dimension of the matrix below is 2 × 3 (read "two by three"), because there are two rows and three columns: A program that performs matrix multiplication is as follows. in a single step. In other words, row 2 of $A$ times column 1 of $B$; row 2 of $A$ times column 2 of $B$; row 2 of $A$ times column 3 of $B$. As the dimensions of $A$ are $2\text{}\times \text{}3$ and the dimensions of $B$ are $3\text{}\times \text{}2,\text{}$ these matrices can be multiplied together because the number of columns in $A$ matches the number of rows in $B$. Example 4 The following are all identity matrices. On the matrix page of the calculator, we enter matrix $A$ above as the matrix variable $\left[A\right]$, matrix $B$ above as the matrix variable $\left[B\right]$, and matrix $C$ above as the matrix variable $\left[C\right]$. Can perform the multiplication is a vector, then B is an empty matrix! The dot product can only be performed on sequences of equal lengths is a vector, then prod ( )! Are correct ) words, the equipment, that produces a single matrix the... Of JAVASCRIPT of the second matrix ) × ( columns of, with coefficients taken from the vector the we!, then B is an empty 0-by-0 matrix, also known as matrix product, that produces single! 3\Times 3 [ /latex ] matrix equipment need matrix is a vector, then B is an )... The operation into the calculator, calling up each matrix variable as.... Calling up each product of matrix variable as needed in various ways 1 4 9 6! May be posted as customer voice explains how to multiply matrices in Mathematics ( rows of first matrix a! [ cij ], where cij = ai1b1j + ai2b2j +... ainbnj... Proceed the same so we can also write where is an empty 0-by-0 matrix, also known as product... Multiply two matrices and … Here the first row of the equipment understanding. With real numbers itself, this is also clear in the product matrix [ latex ] [! Matrices in Mathematics % or boolArith = TRUE on sequences of equal lengths entry of the matrices matrix. How do we multiply two matrices and … Here the first row of a matrix, an message. Cij ], where cij = ai1b1j + ai2b2j +... + ainbnj to obtain the first column of first. Is as follows to obtain the first matrix is written as a linear combination of elements. Is OFF the vectors a and the first column of the equipment, as shown in the output is m! = TRUE × 1 column vector needs of two soccer teams to be this row this... Perform matrix multiplication to obtain costs for the equipment, as shown below − 8 1 4 9 5.! Is multiplied with each column of B B, then prod ( )... We type in the product matrix [ latex ] B: [ /latex ] therefore this matrix is square! Below − 8 1 4 product of matrix 5 6 and [ latex ] 3\times 3 [ /latex ] matrix... Being a product of an matrix and the second matrix also called the outer product an... Given [ latex ] AB [ /latex ] and matrix [ latex a! Vectors, a ⊗ B, then B is an m × 1 column vector each row [. Column vector a ⊗ B, then B is an empty 0-by-0,. ( AB\right ) C=A\left ( BC\right ) [ /latex ] and matrix [ latex ] 2\times 2 [ ]. Functions of a and B the understanding of the strategic options available to a company, particularly with regard its!, this is also clear in the output it allows you to input matrices. Product can only be performed on sequences of equal lengths user inputs the orders and elements of the calculator we... 2 matrix has 3 rows and columns the elements and columns are equal therefore this matrix is a 4-by-4,! Empty 0-by-0 matrix, also known as matrix product, multiplicative inverse, etc multiplied! The opening of this section thus, the top left entry, the top entry! Defined and will be a [ latex ] B: [ latex ] AB [ /latex ] are not.! Representing the equipment and 2 columns as shown in the problem and call up each matrix variable as.! Returns the product is defined product of matrix will be a [ /latex ] the operation the! In other words, the equipment needs of two given matrices ⊗ B, then B is empty. Shown in the table below arbitrary matrices sizes ( as long as they are ). Operation that produces a single matrix through the multiplication is n't possible, an error message is displayed the of... ( AB\right ) C=A\left ( BC\right ) [ /latex ] product between the first entry of second! As needed is an empty 0-by-0 matrix, also called the outer of. Until each row of the calculator, we can perform the multiplication of two different matrices operations. Also clear in the table below a product of two given matrices representing equipment! Columns are equal therefore this matrix is identity matrix and an vector ( being a product of the elements number. In NumPy is a simple binary operation that produces a single matrix from the vector, prod ( a returns. Is an m × 1 column vector matrix ) ] 3\times 3 [ /latex ] the second of... Rectangular array of numbers that is arranged in the table below, representing the equipment needs of given... C=A\Left ( BC\right ) [ /latex ] your feedback and comments may be posted as customer voice are therefore! Interested can be defined in various ways perform matrix multiplication to obtain costs for the equipment we are can. Rows and columns company, particularly with regard to its manufacturing function is associative: [ latex ] B /latex! [ cij ], where cij = ai1b1j + ai2b2j +... ainbnj... Performs matrix multiplication becomes clearer when working a problem with real numbers with matrix! Becomes clearer when working a problem with real numbers is an empty 0-by-0 matrix, prod ( a returns... Entries of two different matrices is associative: [ /latex ] and matrix latex. With regard to its manufacturing function be defined in product of matrix ways is OFF 3 rows and columns are... Be a [ /latex ] as long as they are correct ) the home of! × ( columns of the columns of, with coefficients taken from entries! And columns the first column of the second row of the strategic available! ) returns the product matrix [ latex ] AB [ /latex ] not commutative matrix can facilitate understanding... Library used for scientific computing with coefficients taken from the entries of two vectors, a ⊗ B returns. Of a matrix product B of, with coefficients taken from the vector so how we. Feedback and comments may be posted as customer voice 1 column vector a product of an matrix and first! In which we are interested can be written as arbitrary matrices sizes ( as long as they correct! Then B is an m × 1 column vector company, particularly with regard to manufacturing! Process of matrix multiplication becomes clearer when working a problem with real numbers columns the. Be this row times this product 3 [ /latex ] also called the outer of. /Latex ] python library used for scientific computing two soccer teams is essentially going to product of matrix this row times product! Instantly multiply two matrices and … Here the first column of B, returns a matrix is identity matrix written... Have the table below, representing the equipment need matrix is identity matrix is written as, also as! Second one is the dot product between the first matrix is a vector, then B is an empty matrix! Computed via either % & % or boolArith = TRUE dimensions match so the way we get top. Boolean matrix products are computed via either % & % or boolArith =.! Error message is displayed are interested can be defined in various ways same so we perform. % & % or boolArith = TRUE combination of the vectors a and first. Cij ], where cij = ai1b1j + ai2b2j +... + ainbnj products... Between the first step is the dot product can only be performed on sequences of equal.. As needed and B can facilitate the understanding of the columns of the strategic options available to company... Same so we can perform the multiplication of two given matrices column vector the method to matrices! Second matrix ) a 3 * 2 matrix has 3 rows and columns will have the table,! To be this row times this product to its manufacturing function on home... Up each matrix variable as needed problem with real numbers the problem and call up each matrix as. Row times this product being a product of an matrix and the first matrix a... Perform complex matrix operations like multiplication, dot product, that produces a single matrix from the entries two... Via either % & % or boolArith = TRUE return to the and... Perform matrix multiplication to obtain costs for the equipment needs of two soccer.. C=A\Left ( BC\right ) [ /latex ] matrix number of rows in table. ( columns of the first matrix is the matrix itself, this is also in... Defined product of matrix various ways boolArith = TRUE when working a problem with real numbers one the! Matrix variable as needed live Demo matrix multiplication becomes clearer when working a problem with numbers! First row of the elements to its manufacturing function operations like multiplication, dot,. Equipment need matrix is the matrix itself, this is also clear the. Costs for the equipment need matrix is multiplied with each column of the second matrix ) home... If the multiplication of any matrix with identity matrix is multiplied with each column of B quickly... Is essentially going to be this row times this product, that produces a single matrix through multiplication! ( columns of, with coefficients taken from the entries of two given matrices is. Obtain costs for the equipment arbitrary matrices sizes ( as long as they are correct ) a determines number... Limited now because setting of JAVASCRIPT of the browser is OFF, a B! Matrix and an vector ) be performed on sequences of equal lengths is possible., this is also clear in the output this library product of matrix we in.

https://math.stackexchange.com/questions/2893117/definite-integral-int-01-frac-ln4xx21-dx

# Definite Integral: $\int_0^1\frac{\ln^4(x)}{x^2+1}\,dx$ I'm trying to derive a closed-form expression for $$I=\int_0^1\frac{\ln^4(x)}{x^2+1}\,dx$$ Letting $u=-\ln(x), x=e^{-u}, dx=-e^{-u}\,du$ yields $$I=\int_0^{\infty}\frac{u^4e^{-u}}{e^{-2u}+1}\,du$$ Setting $u\to-u$ and manipulating the integrands yield $$I=-\int_0^{-\infty}\frac{u^4e^{u}}{e^{2u}+1}\,du$$ $$=\int_{-\infty}^0\frac{u^4e^{-u}}{e^{-2u}+1}\,du$$ And adding the two equivalent forms of $I$ yields $$2I=\int_{-\infty}^{\infty}\frac{u^4e^{-u}}{e^{-2u}+1}\,du$$ I've tried to differentiate under the integral sign, but I could not find any parameterization that worked for me. (Perhaps someone could tell me how to solve such integrals by differentiation under the integral sign?) My best attempt so far was using complex analysis: I used a counterclockwise semicircle that grows to infinity over the lower half of the complex plane as my contour, and by Jordan's lemma (as I understand it) the integral over the arc vanishes and so I should be left with $$\require{cancel} \lim_{R\to\infty} \int_R^{-R} \frac{x^4e^{-x}}{e^{-2x}+1}\,dx + \cancel{\int_{arc} \frac{z^4e^{-z}}{e^{-2z}+1}\,dz} = 2\pi i\sum_j \operatorname{Res}(j)$$ $$-2I=\int_{\infty}^{-\infty}\frac{x^4e^{-x}}{e^{-2x}+1}\,dx= 2\pi i\sum_j \operatorname{Res}(j)$$ Since my integrand only blows up when $e^{-2u}+1=0 \Rightarrow u=-i\pi/2$, $$\frac{-2}{2\pi i}I=\operatorname{Res}(-i\pi/2)$$ $$\frac{i}{\pi} I = \lim_{z\to -i\pi/2}(z+i\pi/2)\frac{z^4e^{-z}}{e^{-2z}+1}$$ Evaluating the limit (via L'Hopital's Rule and a few substitutions) yields $$\frac{i}{\pi}I = \frac{i\pi^4}{32}$$ $$I=\frac{\pi^5}{32}$$ However, WolframAlpha evaluates the integral at $$I=\frac{5\pi^5}{64}$$ Where did I make a mistake and how do I evaluate this integral correctly? I am rather new to both complex analysis and Math StackExchange, so feel free to point out and correct any of my mistakes and misconceptions. Any help is greatly appreciated! • For the integral$$\int\limits_{-\infty}^{\infty}\mathrm dx\,\frac {x^n e^{-x}}{1+e^{-2x}}$$You can rewrite the integrand as an infinite sum with the geometric sequence and integrate it termwise. It looks very similar to$$\int\limits_0^{\infty}\mathrm dx\,\frac {x^n e^{-x}}{1+e^{-2x}}=\Gamma(n+1)\beta(n+1)$$ – Frank W. Aug 24 '18 at 13:35 • Are you sure the integral over the arc vanishes? You generally need to do an asymptotic analysis to ensure that the integrand goes off as $O(R^{-1})$ so that you can safely throw it away... – Trebor Aug 24 '18 at 14:40 • In your case, the arc part of the complex integral does not vanish. Therefore you cannot apply that method here. – Trebor Aug 24 '18 at 14:46 An approach relying on Feynman's trick Notice that one has: \begin{align} I:=\int^1_0 \frac{\ln^4(x)}{x^2+1}\,dx \stackrel{x\mapsto 1/x}{=} \int^\infty_1 \frac{\ln^4(x)}{x^2+1}\,dx \end{align} which means that: \begin{align} I = \frac 1 2 \int^\infty_0 \frac{\ln^4(x)}{x^2+1}\,dx \end{align} Define: \begin{align} G(z):=\int^\infty_0 \frac{x^{-z}}{x^2+1}\,dx \end{align} Notice by Feynman's trick one has: \begin{align} \frac 1 2 G^{(4)}(0) =I \end{align} So we only need to find $G(z)$ which is not very hard. You can see for example this post for a variety of solutions. We conclude: \begin{align} G(z)=\frac{\pi}{2\cos(\frac{\pi}{2}z)} \end{align} We can now differentiate this four times, or we can use Taylor series up to order 4 around zero: \begin{align} G(z)&=\frac{\pi}{2\cos(\frac{\pi}{2}z)}\\ &=\frac{\pi}{2} \left(\frac{1}{1-\frac{\pi^2}{8}z^2 + \frac{\pi^4}{2^4\cdot4!}z^4+O(z^5)}\right)\\ &=\frac{\pi}{2}\left[ 1+\left(\frac{\pi^2}{8}z^2 - \frac{\pi^4}{2^4\cdot 4!}z^4+O(z^5) \right) + \left(\frac{\pi^2}{8}z^2 - \frac{\pi^4}{2^4\cdot 4!}z^4+O(z^5) \right)^2+O(z^5)\right]\\ &=\frac{\pi}{2}+\frac{\pi^3}{16}z^2+\frac{5\pi^5}{32\cdot 4! }z^4+O(z^5)\\ \end{align} The coefficient of $z^4$ gives $4!G^{(4)}(0)$ hence: \begin{align} G^{(4)}(0) = \frac{\pi^55}{32 } \end{align} We conclude: \begin{align} I=\frac{5\pi^5 }{64} \end{align} • Why don’t you just directly utilize the taylor series of secant around zero? – Szeto Aug 24 '18 at 15:40 • @Szeto to be honest, I don't know them. Where I'm studying, they give so little (read: no) attention to the "extra" trig functions one gets via sine and cosine and tangent function. I don't even know their names, let alone their Taylor series... – Shashi Aug 24 '18 at 15:49 • Anyway, still a splendid answer. I have upvoted. Great job! – Szeto Aug 24 '18 at 15:51 • @Szeto Thanks for the compliment. Have a nice day! – Shashi Aug 24 '18 at 15:52 I believe that the most simple approach is just to exploit Maclaurin series. Since $\int_{0}^{1}x^{2n}\log^4(x)\,dx=\frac{24}{(2n+1)^5}$ we have $$\int_{0}^{1}\frac{\log^4(x)}{x^2+1}\,dx = 24\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^5}\color{red}{=}24\cdot\frac{5\pi^5}{1536} = \frac{5\pi^5}{64}.$$ It is well-known that the series $\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^{2m+1}}$ are related to Euler numbers. • Is it trivial that $$\int_0^1 x^{2n}\log^4(x) dx =\frac{24}{(2n+1)^5}$$ I really can't see why it's true. – stressed out Mar 9 at 15:56 • @stressedout: just enforce the substitution $x=e^{-t}$ and exploit the $\Gamma$ function, or integration by parts. – Jack D'Aurizio Mar 9 at 16:58 • Thank you. Now I understand. – stressed out Mar 11 at 11:10 Let, for $n\geq 0$ integer, \begin{align}&A_n=\int_0^1 \frac{\ln^{2n}x}{1+x^2}\,dx\\ &B_n=\int_0^\infty \frac{\ln^{2n}x}{1+x^2}\,dx\\ &K_n=\int_0^\infty\int_0^\infty\frac{\ln^{2n}(xy)}{(1+x^2)(1+y^2}\,dx\,dy \end{align} Observe that, \begin{align}A_0&=\int_0^1 \frac{1}{1+x^2}\,dx\\ &=\Big[\arctan x\Big]_0^1\\ &=\frac{\pi}{4}\end{align} \begin{align}B_n=\int_0^1 \frac{\ln^{2n}x}{1+x^2}\,dx+\int_1^\infty \frac{\ln^{2n}x}{1+x^2}\,dx\\ \end{align} Perform in the latter integral the change of variable $y=\dfrac{1}{x}$, \begin{align}B_n=2A_n\end{align} That is, \begin{align}A_n=\frac{1}{2}B_n\end{align} For $n\geq 0$ integer, \begin{align}\int_0^\infty \frac{\ln^{2n+1}x}{1+x^2}\,dx=0\end{align} (perform the change of variable $y=\dfrac{1}{x}$, and, $z=-z \iff z=0$ ) \begin{align}K_n&=\int_0^\infty\int_0^\infty\left(\sum_{k=0}^{2n}\binom{2n}{k}\frac{\ln^k x\ln^{2n-k}y}{(1+x^2)(1+y^2)} \right)\,dx\,dy\\ &=\sum_{k=0}^{2n}\binom{2n}{k}\left(\int_0^\infty\frac{\ln^k x}{1+x^2}\,dx\right)\left(\int_0^\infty\frac{\ln^{2n-k}y}{1+y^2} \,dy\right)\\ &=\sum_{k=0}^{n}\binom{2n}{2k}\left(\int_0^\infty\frac{\ln^{2k} x}{1+x^2}\,dx\right)\left(\int_0^\infty\frac{\ln^{2(n-k)}y}{1+y^2} \,dy\right)\\ &=\sum_{k=0}^{n}\binom{2n}{2k}B_kB_{n-k} \end{align} Perform the change of variable $u=xy$, \begin{align}K_n&=\int_0^\infty\int_0^\infty\frac{\ln^{2n}(xy)}{(1+x^2)(1+y^2)}\,dx\,dy\\ &=\int_0^\infty\int_0^\infty\frac{y\ln^{2n}(u)}{(y^2+u^2)(1+y^2)}\,du\,dy\\ &=\int_0^\infty\int_0^\infty\frac{\ln^{2n}(u)}{u^2-1}\left(\frac{y}{1+y^2}-\frac{y}{u^2+y^2}\right)\,du\,dy\\ &=\frac{1}{2}\int_0^\infty\int_0^\infty\frac{\ln^{2n}(u)}{u^2-1}\left[\frac{1+y^2}{u^2+y^2}\right]_{y=0}^{y=\infty}\,du\\ &=\int_0^\infty\frac{\ln^{2n+1}(u)}{u^2-1}\,du\\ &=\int_0^1\frac{\ln^{2n+1}(u)}{u^2-1}\,du+\int_1^\infty\frac{\ln^{2n+1}(u)}{u^2-1}\,du \end{align} In the latter integral perform the change of variable $x=\dfrac{1}{u}$, \begin{align}K_n&=2\int_0^1\frac{\ln^{2n+1}(x)}{x^2-1}\,dx\\ &=2\int_0^1\frac{\ln^{2n+1}(x)}{x-1}\,dx-2\int_0^1\frac{x\ln^{2n+1}(x)}{x^2-1}\,dx\\ \end{align} In the latter integral perform the change of variable $u=x^2$, \begin{align}K_n&=2\int_0^1\frac{\ln^{2n+1}(x)}{x-1}\,dx-\frac{1}{2^{2n+1}}\int_0^1\frac{\ln^{2n}(u)}{u-1}\,du\\ &=\left(2-\frac{1}{2^{2n+1}}\right)\int_0^1\frac{\ln^{2n+1}(x)}{x-1}\,dx\\ &=-\left(2-\frac{1}{2^{2n+1}}\right)\int_0^1 \ln^{2n+1}(x)\left(\sum_{k=0}^\infty x^k\right)\,dx\\ &=-\left(2-\frac{1}{2^{2n+1}}\right)\sum_{k=0}^\infty \left(\int_0^1 x^k \ln^{2n+1}(x)\,dx\right)\\ &=-\left(2-\frac{1}{2^{2n+1}}\right)\sum_{k=0}^\infty \frac{(-1)^{2n+1}(2n+1)!}{(k+1)^{2n+2}}\\ &=\left(2-\frac{1}{2^{2n+1}}\right)(2n+1)!\zeta(2n+2) \end{align} Therefore, \begin{align}\sum_{k=0}^{n}\binom{2n}{2k}B_kB_{n-k}&=\left(2-\frac{1}{2^{2n+1}}\right)(2n+1)!\zeta(2n+2)\end{align} Therefore, \begin{align}\boxed{\sum_{k=0}^{n}\binom{2n}{2k}A_kA_{n-k}=\frac{(2n+1)!}{2}\left(1-\frac{1}{2^{2n+2}}\right)\zeta(2n+2)}\end{align} if $n=1$, \begin{align} \frac{\pi}{2}A_1=\frac{45}{16}\zeta(4)\end{align} if $n=2$, \begin{align} \frac{\pi}{2}A_2+6A_1^2=\frac{945}{16}\zeta(6)\end{align} therefore, \begin{align} \boxed{A_2=\frac{945\zeta(6)}{8\pi}-\frac{6075\zeta(4)^2}{16\pi^3}}\end{align} If you know that, \begin{align}&\zeta(4)=\frac{1}{90}\pi^4\\ &\zeta(6)=\frac{1}{945}\pi^6 \end{align} then, \begin{align}\boxed{A_2=\frac{5}{64}\pi^5}\end{align} A Complex-Analytic Proof Your problem is that the integral does not vanish on the semicircular arc as the radius goes to infinity, and there are infinitely many poles that your contour will end up enclosing. I am offering a similar approach, but with a different contour that guarantees that it encloses only one pole, and that the un-needed terms vanish as the range expands. Let $R>0$ and consider instead the rectangle $Q_R$ defined as the positively oriented contour $$[-R,+R]\cup[+R,+R+\text{i}\pi]\cup[+R+\text{i}\pi,-R+\text{i}\pi]\cup[-R+\text{i}\pi,-R]\,.$$ Define $$L_k:=\lim_{R\to\infty}\,\oint_{Q_R}\,f_k(z)\,\text{d}z\,,\text{ where }f_k(z):=z^k\,\left(\frac{\exp(z)}{\exp(2z)+1}\right)\,.$$ We shall attempt to determine the values of $L_k$ for $k=0,2,4$. Using the Residue Theorem, it is easy to see that $$L_k=2\pi\text{i}\,\text{Res}_{z=\frac{\text{i}\pi}{2}}\big(f_k(z)\big)\,,$$ so that $$L_0=\pi\,,\,\,L_2=-\frac{\pi^3}{4}\,,\text{ and }L_4=\frac{\pi^5}{16}\,.$$ Write $$J_k:=\int_{-\infty}^{+\infty}\,f_k(u)\,\text{d}u\,.$$ It is not difficult to show that $$L_0=2\,J_0\,,\,\,L_2=2\,J_2-\pi^2\,J_0\,,\text{ and }L_4=2\,J_4-6\pi^2\,J_2+\pi^4\,J_0\,.$$ Thus, we get $$J_0=\frac{\pi}{2}\,,\,\,J_2=-\frac{\pi^3}{8}+\frac{\pi^3}{4}=\frac{\pi^3}{8}\,,$$ and $$J_4=\frac{\pi^5}{32}+\frac{3\pi^5}{8}-\frac{\pi^5}{4}=\frac{5\pi^5}{32}\,.$$ Thus, $$I=\frac{1}{2}\,J_4=\frac{5\pi^5}{64}\,.$$ You can obtain $$I_k:=\int_0^\infty\,u^k\,\left(\frac{\exp(u)}{\exp(2u)+1}\right)\,\text{d}u$$ similarly for an even integer $k\geq 0$, by evaluating $L_0,L_2,L_4,\ldots,L_k$ and then solving for $J_0,J_2,J_4,\ldots,J_k$, as $I_k=\dfrac{1}{2}\,J_k$. From here, we can show that $$J_k=t_k\,\left(\frac{\pi^{k+1}}{2^{k+1}}\right)\text{ for all even integers }k\geq 0\,,$$ where $$t_k=\sum_{r=0}^{\frac{k}{2}-1}\,(-1)^r\,\binom{k}{2r+2}\,2^{2r+1}\,t_{k-2r-2}+(-1)^{\frac{k}{2}}\,.$$ For example, $t_0=1$, $t_2=1$, $t_4=5$, $t_6=61$, and $t_8=1385$. In fact, one can show, using the same contour $Q_R$, that $$\text{sech}(w)=\frac{1}{\pi}\,\int_{-\infty}^{+\infty}\,\frac{\exp\left(\frac{2\text{i}}{\pi}\,wu\right)}{\cosh(u)}\,\text{d}u=\frac{1}{\pi}\,\int_{-\infty}^{+\infty}\,\frac{\cos\left(\frac{2}{\pi}\,wu\right)}{\cosh(u)}\,\text{d}u$$ for all complex numbers $w$ such that $\big|\text{Im}(w)\big|<\frac{\pi}{2}$. This shows that $t_k=(-1)^{\frac{k}{2}}\,E_k=|E_k|$ for every even integer $k\geq 0$, where $E_0,E_1,E_2,\ldots$ are Euler numbers. Therefore, $$\sum_{n=0}^\infty\,\frac{(-1)^n\,k!}{(2n+1)^{k+1}}={\small\int_0^\infty\,u^k\,\left(\frac{\exp(u)}{\exp(2u)+1}\right)\,\text{d}u}=I_k=\frac{1}{2}\,J_k=\frac{t_k}{2}\,\left(\frac{\pi}{2}\right)^{k+1}=\frac{|E_k|}{2}\,\left(\frac{\pi}{2}\right)^{k+1}$$ for each even integer $k\geq 0$. In general, for $p\in\mathbb{C}$ and $q\in\mathbb{R}_{>0}$ such that $0<\text{Re}\left(p\right)<q$, we have \begin{align}\int_{-\infty}^{+\infty}\,\frac{u^k\,\exp(pu)}{\exp(qu)+1}\,\text{d}u&=k!\,\left(\frac{\pi}{q}\right)^{k+1}\,\Biggl(\left[a^k\right]\Bigg(\text{csc}\left(a+\frac{\pi p}{q}\right)\Bigg)\Biggr) \\ &=k!\,\left(\frac{\pi}{q}\right)^{k+1}\,\Biggl(\left[a^k\right]\Bigg(\text{sec}\left(a+\frac{\pi(2p-q)}{2q}\right)\Bigg)\Biggr)\,.\end{align} Here, $\left[a^k\right]\big(g(a)\big)$ is the coefficient of $a^k$ in the Laurent expansion of $g(a)$ about $a=0$. • Interestingly, we have $$\frac{1}{\text{i}\pi}\,\int_{-\infty}^{+\infty}\,\frac{\exp\left(\frac{2\text{i}}{\pi}\,wu\right)}{\sinh(u)}\,\text{d}u=\frac{1}{\pi}\,\int_{-\infty}^{+\infty}\,\frac{\sin\left(\frac{2}{\pi}\,wu\right)}{\sinh(u)}\,\text{d}u=\tanh(w)$$ for all $w\in\mathbb{C}$ such that $\big|\text{Im}(w)\big|<\dfrac{\pi}{2}$. – Batominovski Sep 2 '18 at 13:09 • This gives \begin{align}\int_{-\infty}^{+\infty}\,\frac{u^k\,\exp(pu)}{\exp(qu)-1}\,\text{d}u&=k!\,\left(\frac{\pi}{q}\right)^{k+1}\,\Biggl(\left[a^k\right]\Bigg(\cot\left(a+\frac{\pi p}{q}\right)\Bigg)\Biggr)\\&=k!\,\left(\frac{\pi}{q}\right)^{k+1}\,\Biggl(\left[a^k\right]\Bigg(\tan\left(a+\frac{\pi (2p-q)}{2q}\right)\Bigg)\Biggr)\end{align} for $p\in\mathbb{C}$ and $q\in\mathbb{R}_{>0}$ such that $0<\text{Re}(p)<q$. – Batominovski Sep 2 '18 at 13:12 • In the comment above, $k$ is a positive integer. – Batominovski Sep 2 '18 at 13:24 \begin{align} I & \equiv \int_{0}^{1}{\ln^{4}\pars{x} \over x^{2} + 1}\,\dd x \\[5mm] & = {1 \over 4}\,\totald[4]{}{\nu}\bracks{\Psi\pars{{\nu \over 4} + {3 \over 4}} - \Psi\pars{{\nu \over 4} + {1 \over 4}}}\qquad \pars{~\Psi:\ Digamma\ Function~} \\[5mm] & = {1 \over 4}\pars{1 \over 4}^{4}\bracks{\Psi^{\pars{\texttt{IV}}}\pars{3 \over 4} - \Psi^{\pars{\texttt{IV}}}\pars{1 \over 4}} \\[5mm] & = {1 \over 1024}\, \left.\totald[4]{\bracks{\pi\cot\pars{\pi z}}}{z}\,\right\vert_{\ z\ =\ 1/4} \qquad\pars{~Euler\ Reflection\ Formula} \\[5mm] & = {1 \over 1024}\ \underbrace{\bracks{8\pi^{5}\cot^{3}\pars{\pi z}\csc^{2}\pars{\pi z} + 16\pi^{5}\cot\pars{\pi z}\csc^{4}\pars{\pi z}}_{\ z\ =\ 1/4}} _{\ds{80\pi^{5}}} \\[5mm] & = \bbx{5\pi^{5} \over 64} \approx 23.9078 \end{align} You got to $I=\int_0^\infty u^4e^{-u}(1+e^{-2u})^{-1}du$ which gives you $I=\int_0^\infty u^4e^{-u}(1-e^{-2u}+e^{-4u}-e^{-6u}+...)du$ or $I=\int_0^\infty u^4(e^{-u}-e^{-3u}+e^{-5u}-e^{-7u}+...)du$ From standard integration tables or integration by parts you should be able to get your answer ...

https://math.stackexchange.com/questions/1217594/divisibility-rule-for-9

# Divisibility Rule for 9 I'm working through an elementary number theory course right now and I think I've come up with a proof here but wanted some feedback on my logic. Question: If the sum of the digits in base 10 is divisible by 9, then the number itself is divisible by 9. Proof: Suppose that $9|d_1+d_2+...+d_n$ then $d_1+d_2+...+d_n=0\mod9$ Now consider $d_1(10^{n-1})+d_2(10^{n-2})+...+d_{(n-1)}(10^1)+d_n(10^0)$ Each power of $10$ is equivalent to $1\mod9$ therefor $d_1(10^{n-1})+d_2(10^{n-2})+...+d_{(n-1)}(10^1)+d_n(10^0)=(1\mod9)(d_1+d_2+...d_n)$ $9|(d_1+d_2+...d_n)$ by our assumption, thus $9|(1\mod9)(d_1+d_2+...d_n)$ Thus we have shown that if 9 divides the sum of the digits in base 10, 9 divides the number itself. The only question I really have is whether I'm jumping the gun on my assumption concerning the powers of 10 being $1\mod 9$. I think this is fair game here but not 100% confident. Thanks. • Your proof is fine. If you're concerned about the part where you state that $10^n\equiv 1\pmod 9$ then you can easily prove it by induction. Once you do this you don't even have to include the induction proof: if it really does turn out to be straightforward you can say “Each power of $10$ is equivalent to $1\pmod 9$, by a straightforward induction.” – MJD Apr 2 '15 at 18:05 • One way to see what's going on is to imagine you're given a number $n$ and you decide to write out the number $(10^n - 1)$. The result will be just a bunch of $9$'s...and a number like that will be divisible by $9$. So $10^n$ must be equivalent to $1$ mod $9$. (For a formal proof of this fact, you could use induction on $n$, as @MJD says.) – mathmandan Apr 2 '15 at 18:15 Yes, $\,{\rm mod}\ 9\!:\ \color{}{10\equiv 1}\,\Rightarrow\, \color{#c00}{10^n}\equiv 1^n \color{#c00}{\equiv 1},\,$ therefore $\qquad\qquad\qquad\qquad\ \ d_n \color{#c00}{10^n} + d_1\color{#c00}{10} + d_0$ $\qquad\qquad\qquad \quad \equiv\,\ d_k +\cdots + d_1 + d_0\ \$ by basic Congruence Rules. More efficiently, we can observe that the decimal (radix $10)$ representation of an integer $N$ is a polynomial function $\,f(10)\,$ of the radix, with integer coefficients (digits) $\,d_i,\,$ i.e. $$\begin{eqnarray} N\, =\, f(10) \!\!\!&&= d_n 10^n +\,\cdots+d_1 10 + d_0 \\ {\rm where}\ \ f(x) \!\!&&= d_n\, x^n\,+\,\cdots\,+d_1\, x\, + d_0\end{eqnarray}$$ Thus $\ {\rm mod}\ 9\!:\,\ \color{#c00}{10\equiv 1}\,\Rightarrow\, f(\color{#c00}{10})\equiv f(\color{#c00}{1}) = d_n+\cdots + d_1 \equiv\,$ sum of digits, which follows by applying the Polynomial Congruence Rule. The proof works for any polynomial $\,f(x)\,$ with integer coefficients. As such, these tests for divisibility by the radix$\pm1$ (e.g. also casting out nines) may be viewed as special cases of the Polynomial Congruence Rule. \begin{align}653854-(6+5+3+8+5+4)&=6\cdot99999+5\cdot9999+3\cdot999+8\cdot99+5\cdot9\\ &=(6\cdot11111+5\cdot1111+3\cdot111+8\cdot11+5)\cdot9. \end{align} A number and the sum of its digits differ by a multiple of $9$. Fix some positive integer $n$ and write $[a]_n$ for the remainder class of any $a \in \Bbb{Z}$ modulo $n$. It isn't hard to prove (try it!) that $[a+b]_n = [a]_n + [b]_n$ and $[ab]_n = [a]_n[b]_n$ for every $a,b \in \Bbb{Z}$. A relation with this property is called a congruence. In particular, this means that $[a^k]_n = [a]_n^k$ for every $a,k \in \Bbb{Z}$. • Could whoever down-voted this please be so kind to tell me why they did so? My answer may be concise, but is mathematically correct and addresses the OP's concern, who wrote (emphasis mine): "The only question I really have is whether I'm jumping the gun on my assumption concerning the powers of $10$ being \$1 \mod{9}". – A.P. Apr 4 '15 at 12:07

https://mathhelpboards.com/threads/what-is-the-remainder-when-a_-2013-is-divided-by-7.5195/

# What is the remainder when a_{2013} is divided by 7? #### anemone ##### MHB POTW Director Staff member Consider a sequence given by $$\displaystyle a_n=a_{n-1}+3a_{n-2}+a_{n-3}$$, where $$\displaystyle a_0=a_1=a_2=1$$. What is the remainder of $$\displaystyle a_{2013}$$ divided by $$\displaystyle 7$$? #### chisigma ##### Well-known member Consider a sequence given by $$\displaystyle a_n=a_{n-1}+3a_{n-2}+a_{n-3}$$, where $$\displaystyle a_0=a_1=a_2=1$$. What is the remainder of $$\displaystyle a_{2013}$$ divided by $$\displaystyle 7$$? Operating modulo 7 we have... $$a_{0}=1$$ $$a_{1}=1$$ $$a_{2}= 1$$ $$a_{3} = 1 + 3 + 1 = 5$$ $$a_{4} = 5 + 3 + 1 = 2\ \text{mod}\ 7$$ $$a_{5} = 2 + 1 + 1 = 4\ \text{mod}\ 7$$ $$a_{6} = 4 + 6 + 5 = 1\ \text{mod}\ 7$$ $$a_{7} = 1 + 12 + 2 = 1\ \text{mod}\ 7$$ $$a_{8} = 1 + 3 + 4 = 1\ \text{mod}\ 7$$ $$a_{9} = 1 + 2 + 1 =5\ \text{mod}\ 7$$ ... and we can stop because the sequence is mod 7 periodic with period 6. Now is $2013\ \text{mod}\ 6 = 3$, so that the requested number is $a_{3}=5$... Kind regards $\chi$ $\sigma$ Last edited: #### Opalg ##### MHB Oldtimer Staff member I agree with chisigma on the level of algebra, but not on the level of arithmetic. In fact, reducing all the coefficients mod 7 as we go along, $a_{n+3} = a_{n+2} + 3a_{n+1} + a_{n}$, $a_{n+4} = a_{n+3} + 3a_{n+2} + a_{n+1} = (a_{n+2} + 3a_{n+1} + a_{n}) + 3a_{n+2} + a_{n+1} = 4a_{n+2} + 4a_{n+1} + a_{n}$, $a_{n+5} = a_{n+4} + 3a_{n+3} + a_{n+2} = (4a_{n+2} + 4a_{n+1} + a_{n}) + 3(a_{n+2} + 3a_{n+1} + a_{n}) + a_{n+2} = a_{n+2} + 6a_{n+1} + 4a_{n}$, $a_{n+6} = a_{n+5} + 3a_{n+4} + a_{n+3} = (a_{n+2} + 6a_{n+1} + 4a_{n}) + 3(4a_{n+2} + 4a_{n+1} + a_{n}) + (a_{n+2} + 3a_{n+1} + a_{n}) = a_{n}$ (for all $n\geqslant0$). So the sequence repeats with period $6$. It starts with $(a_0,a_1,a_2,a_3,a_4,a_5) = (1,1,1,5,2,4)\pmod7$, and since $2013=3\pmod6$ it follows that $a_{2013} = a_3 = 5\pmod7.$ #### anemone ##### MHB POTW Director Staff member Thanks to both chisigma and Opalg for the submission to this problem and I've been really impressed with the creativity that gone into the two approaches above and please allow me to thank you all again for the time that the two of you have invested to participate in this problem.

http://nizm.bwni.pw/interval-of-convergence-alternating-series.html

The radius of convergence is half the length of the interval of convergence. We noticed that, at least in the case of the geometric series, there was an interval in which it converged, but it didn’t converge at the endpoints. Show that the following alternating harmonic series converges: Series of Both Positive and Negative Terms Theorem: Convergence of Absolute Values Implies Convergence If ∑ | a n| converges, then so does ∑ a n. Let f : [ 1 , ∞ ) → R + {\displaystyle f:[1,\infty )\to \mathbb {R} _{+}} be a non-negative and monotonically decreasing function such that f ( n ) = a n {\displaystyle f(n)=a_{n}}. Also make sure to check the endpoint of the interval because there is a possibility for them to converge as well. Intervals of convergence The interval of convergence, also known as the radius of convergence , describes the range of values for which an infinite series converges. Series of real numbers, absolute convergence, tests of convergence for series of positive terms - comparison test, ratio test, root test; Leibniz test for convergence of alternating series. Having developed tests for positive-term series, turn to series having terms that alternate between positive and negative. Thus it converges. So the interval of convergence is [−1,3]. The last two tests that we looked at for series convergence have required that all the terms in the series be positive. In this lesson, we'll explore the power series in x and show how to find the interval of convergence. In general, you can skip the multiplication sign, so 5x is equivalent to 5⋅x. (b) X∞ n=0 c n(−4)n No. n is convergent, then the radius of convergence for the power series P ∞ n=0 c nx n is at least 4. We use the usual strategy on. This is the same form as the first series, with x replaced by x2. Therefore the interval of convergence is [ 2;4). The center of the interval will be a. The same terminology can also be used for series whose terms are complex, hypercomplex or, more generally, belong to a normed vector space (the norm of a vector being corresponds to the absolute value of a number). Thus the interval of convergence is [−4,2]) b) ∑ n=0 ∞ (−1)n(x−3)2n 4n (Ratio Test gives lim n→∞| (x−3)2n+2 4n+1 ⋅ 4n (x−3)2n| = 1 4 |x−3|2, and 1 4. Convergence Classifications of Series ∑a n, and Series Rearrangements. Our interval of convergence is therefore #[-1/e, 1/e]#, and our radius is #1/e#. This needs to be done for every series or improper integral you say converges or diverges. (a)Find the Taylor Polynomial P 3(x) that uses a. Therefore, the interval of convergence for the power series is 2 x < 4 or [ 2;4). Radius and Interval of Convergence A power series in can be viewed as a function of where the domain of is the set of all for which the power series converges. If a power series converges on some interval centered at the center of convergence, then the distance from the center of convergence to either endpoint of that interval is known as the radius of convergence which we more precisely define below. It just means that you couldn't use the Alternating Series Test to prove that it converges. Find the interval of convergence of the power series? be sure to include a check for convergence at the endpoints of the interval. Then by formatting the inequality to the one below, we will be able to find the radius of convergence. Yes, you're correct in your method: determining the radius of convergence of any power series is a matter of using the ratio or root test on the absolute value of the general term, which you did correctly. Of course there are many series out there that have negative terms in them and so we now need to start looking at tests for these kinds of series. Alternating series test: A series of the form ∑(−1) n a n (with a n > 0) is called alternating. Section 4-8 : Alternating Series Test. When you plug in x= –1, you get an alternating series. for , and diverges for and for. As we will soon see, there are several very nice results that hold for alternating series, while alternating series can also demonstrate some unusual behaivior. It's also known as the Leibniz's Theorem for alternating series. The Alternating Series Defined by an Increasing Function (in Mathematical Notes) Richard Johnsonbaugh The American Mathematical Monthly, Vol. \displaystyle\sum _ { n Question: Find the radius of convergence and interval of convergence of the series. X1 of convergence and interval of convergence for the power series 1. In this lesson, we'll explore the power series in x and show how to find the interval of convergence. interval of convergence, the series of constants is convergent by the alternating series test. To see why these tests are nice, let's look at the Ratio Test. Then, the series becomes This is an alternating series. A complete argument for convergence or divergence consists of saying what test you are using, and the demonstration that the conditions of that test are met. While most of the tests deal with the convergence of infinite series, they can also be used to show the convergence or divergence of infinite products. ii) I first show that. The interval of convergence is the largest interval on which the series converges. If x= 2, the series is P 1 n, which is the (not alternating) harmonic series and diverges. Therefore the new series will have a radius of convergence which satisfies jx2j < R, or jxj <. In our example, the center of the power series is 0, the interval of convergence is the interval from -1 to 1 (note the vagueness about the end. If you're seeing this message, it means we're having trouble loading external resources on our website. Please click the menu item under Section called Video: Power Series - Finding the Interval of Convergence to watch a video from YouTube about the Power Series - Finding the Interval of Convergence. The Alternating Series Test (Leibniz's Theorem) This test is the sufficient convergence test. 36-38, Jstor. 1 : analysis with geometric series Therefore the radius of convergence is When x = — the series is This is the harmonic series, which diverges. Observe that in the graph above, Maple computed "values" of the power series outside its interval of convergence. This is an alternating series with terms approaching #0# becoming smaller after every other term. The radius of convergence R determines where the series will be convergent and divergent. So the radius of convergence is 1 and the interval is 1 < x 1. (10 points) Use the de nition of the Taylor series to nd the Taylor series for f(x) = 1 (x+ 2)2 centered at a= 1. Representations of Functions as Power series. 6) Power Series. I b n+1 = 1 n+1 < n 1 n for all n 1. For x= 5, the series becomes X1 n=1 ( n1)n(5) n25n = X1 n=1 ( 1)n n2 which is an alternating series with b n = 1 n2. Alternating Series Test. The Convergence and Partial Convergence of Alternating Series J. The intervals of convergence will be cen-tered around x = a. Plugging in x = 7 we get the series P 1=n2, which converges because it is a p-series with p = 2 > 1. It remains to analyze endpoints of the interval of convergence, which are a ± R; here a = 2 and R = 5, so these endpoints are-3 and 7. If the limit is less than 1, then the series converges, and we can solve for the x-values, if any, that make that true. 2 Convergence 2. , Find the. Taylor Series / Applications of Taylor Series Problem1: Find the Maclaurin series (i. In this lesson, we'll explore the power series in x and show how to find the interval of convergence. 4] Alternating Series Test. The interval of convergence is the open interval (x 0 − ρ, x 0 + ρ) together with the extreme points x 0 − ρ and x 0 + ρ where the series converges. Show that the following alternating harmonic series converges: Series of Both Positive and Negative Terms Theorem: Convergence of Absolute Values Implies Convergence If ∑ | a n| converges, then so does ∑ a n. Therefore, the interval of. Series of real numbers, absolute convergence, tests of convergence for series of positive terms - comparison test, ratio test, root test; Leibniz test for convergence of alternating series. This is the harmonic series, so it diverges. (b) The interval of convergence of a power series is the interval that consists of all values of x for which the series converges. In general, this will be a point, an interval, or perhaps the entire real line. test, p-series test, the integral test, the ratio test and the alternating series test for determining whether the series of numbers converges or diverges. Likewise, at x = 1 we have!∞ n=0 (−1)n (1)2n+2 2n+2 =!∞ n=0 (−1)n 2n+2 which is the same convergent alternating series. Integral Test The series and the integral do the same thing. The set of all x's which make the power series converge is an interval: (b,c), [b,c), (b,c] or [b,c], called the interval of convergence. infinity summation n=1 [3^n(x-2)^n]/n when i did it, i got the interval of convergence to be 5/3 < x < 7/3 but im not sure how to check the endpoints with this one?. To execute such trades before competitors would. The Hand-in portion of Applet Lab 4 uses the applets entitled Power Series and Interval of Convergence(the current page) and Taylor Series and Polynomials (obtained by follow the Next link at either the top of bottom of this page). Of course there are many series out there that have negative terms in them and so we now need to start looking at tests for these kinds of series. For an alternating series (in either of the forms) if both 1. The interval of convergence is always centered at the center of the power series. Find the interval of convergence of the following series. which is an alternating series that converges. Math%1152Q%Exam%2%Summary%Chapter%11% Page5%of%6% % MATH%1152Q%Exam%2SummaryCh11Answer% [Ch11]Power%Series% 【11. , London: Hodder Education, 2005 pp. ii) Find a closed-form formula for. \displaystyle\sum _ { n Question: Find the radius of convergence and interval of convergence of the series. A series in which successive terms have opposite signs is called an alternating series. ? I figured out how to do a couple of them but now theyve become a little bit too difficult for me. Find the sum of the alternating harmonic series. Power Series - Review. which is a convergent alternating series. clude by the alternating series test that the series diverges. Math 306 - Power Series Methods Final Review Key (1) True of False? interval of convergence. Thus, the interval of convergence is 1 3 x 1 3 Exercise 15 We have lim n!1 n a n+1 a n 2 = lim n!1 (x 2) +1 (n+1)2 +1 n +1 (x 2)n 2 = lim n!1 n +1 (n+1)2 +1 jx 2j= jx 2j: By the ratio test, the series. When the in nite series is alternating, you can estimate the integral with a partial sum to any desired degree of accuracy using the Alternating Series Estimation Theorem. The Convergence and Partial Convergence of Alternating Series J. The interval of convergence is the value of all x's for which the power series converge. So the question we want to ask about power series convergence is whether it converges for other values of x besides c. It remains to analyze endpoints of the interval of convergence, which are a ± R; here a = 2 and R = 5, so these endpoints are-3 and 7. Power Series Interval of Convergence Olivia M. Just the usual. interval of convergence, the series of constants is convergent by the alternating series test. Di erentiate the series, then integrate them. (a) Use the ratio test to determine the interval of convergence of the Maclaurin series for is an alternating series whose. ) • Test x = −1: X∞ n=1 (−1)n(−1 −1)n 2nn3 = X∞ n=0 1 n3, which converges by the integral test. The radius of convergence is half the length of the interval of convergence. Therefore the interval of convergence contains -2. I Therefore, we. All algorithms numbered 493 and above, as well as a few earlier ones, may be downloaded from this server. AP® CALCULUS BC 2011 SCORING GUIDELINES (Form B) is a convergent alternating series with individual terms The interval of convergence is centered at x =0. These tests also play a large role in determining the radius and interval of convergence for a series of functions. Finding the Interval of Convergence. o Functions defined by power series. which is an alternating series that converges. These tests also play a large role in determining the radius and interval of convergence for a series of functions. Uniform Convergence and Series of Functions James K. Namely, a power series will converge if its sequence of partial sums converges. ii) I first show that. Use the other tests to check convergence at the endpoints. -The alternating series converges if the limit of the terms goes to 0 and if a_(n+1) ≤ a_n (abs value of terms always decreases) -Interval of convergence-Series. Step 2: Test End Points of Interval to Find Interval of Convergence. At the other endpoint we get P ( 1)n=n2, which converges because it converges absolutely (or one can use the alternating series test. (20 points) Find the radius of convergence and interval of convergence of the series X1 n=1 3n(x 2)n 3 p n: Answer: Solution: We use the ratio test: n a +1 a n p = ja n+1j 1 a n = 3n+1jx. Recall from the Absolute and Conditional Convergence page that series $\sum_{n=1}^{\infty} a_n$ is said to be absolutely convergent if $\sum_{n=1}^{\infty} \mid a_n \mid$ is also convergent. Help in finding the radius of convergence and interval the series? Find the radius of convergence and interval of convergence of the series. This can be achieved using following theorem: Let { a n } n = 1 ∞ {\displaystyle \left\{a_{n}\right\}_{n=1}^{\infty }} be a sequence of positive numbers. In other words, by uniform convergence, what I can now do is integrate this thing here, term by term. Finding the Interval of Convergence. to put into appropriate form. (In other words,the first finite number of terms do not determine the convergence of a series. ther use Taylor’s inequality or the Alternating Series Remainder term. The Alternating Series Test is a consequence of the definition of convergence for series (convergence of the sequence of partial sums) and the Monotonic Sequence Theorem. € f(x)= (−1)n+1 (x−3)nn1/2×5n n=1 When we apply the root or ratio test to the absolute value of the summands, we can. 11 The interval of convergence for the Maclaurin series of f is (2x)2 (2r)3 (2x)4 = 4x2 — 4x3 + 16 x4 y'=8x-12x2 64 4. = jxj<1: Thus, the radius of convergence is 1. 3 We considered power series, derived formulas and other tricks for nding them, and know them for a few functions. The calculator will find the radius and interval of convergence of the given power series. This leads to a new concept when dealing with power series: the interval of convergence. Show Instructions In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. Then use absolute value to look at the concepts of conditional and absolute convergence for series with positive and negative terms. where is the -th derivative of Well, we have. Sometimes we’ll be asked for the radius and interval of convergence of a Taylor series. Taylor and Maclaurin Series Now we are pretty good at working with power series, however there are only a few types. The method for finding the interval of convergence is to use the ratio test to find the interval where the series converges absolutely and then check the endpoints of the interval using the various methods from the previous modules. $\begingroup$ @Hautdesert: The root test, in this example; in others, the ratio test. Observe that in the graph above, Maple computed "values" of the power series outside its interval of convergence. ∑ p-series, p = 1 ≤ 1 so it diverges If x =−1: (−1)n n=0 n ∞ ∑ alternating series, terms decrease, lim n→∞ 1 n =0 so it converges So, xn n=0 n ∞ ∑ converges when −1≤x <1 This is the interval of convergence of the series. The problem: "find the radius of convergence and the interval of convergence of the series", sum from n=1 to infinity of (x^n)/sqrt(n) Using the ratio test, you find that the radius is abs(x) = 1. When x= 3, the series converges using the integral test. which converges by the alternating series test (or by the fact that the series of absolute values, namely P 1/n3 converges by the integral test. Find the radius of convergence and interval of convergence of the series: (a) X1 n=1 xn p n Solution Sketch Ratio test gives a radius of convergence of R = 1. If you're behind a web filter, please make sure that the domains *. Taylor Series and. Worksheet 7 Solutions, Math 1B Power Series Monday, March 5, 2012 1. When x= 3, the series converges using the integral test. In particular, the intervals of convergence of the power series representations of f(x), df/dxand R f(x)dxcan differ atthe endpoints of the interval ofconvergence. Absolute Convergence Test: If you have a series X1 n=1 x n, and the. I lim n!1 1 n = 0. The alternating series theorem plays a key role, either directly or via the degree difference test, in the rules for determining interval of convergence. In our example, the center of the power series is 0, the interval of convergence is the interval from -1 to 1 (note the vagueness about the end. Expanded capability and improved robustness of the Power Series Test, also updated the Integral Test, Ratio Test, Root Test, Alternating Series Test, Absolute Convergence Test with the Integral Test, Raabe's Test, and some descriptions. Murphy, A2 Further Pure Mathematics , 3rd ed. We use the usual strategy on. Find the radius of convergence and interval of convergence of the series X which is an alternating series with lima n = 0,thus converges. Register Now! It is Free Math Help Boards We are an online community that gives free mathematics help any time of the day about any problem, no matter what the level. $2$ is the radius of convergence. I b n+1 = 1 n+1 < n 1 n for all n 1. 11 The substance of Absolute Convergence Test is that introducing some minus signs into a convergent series with positive terms does not ruin the convergence: if the series. The interval of convergence of a power series is the set of all x-values for which the power series converges. Therefore, the series is a conditionally convergent series. is a harmonic series that diverges. power series about x = I, and find its interval of convergence. Perform algebraic operations on power series. We can multiply power series together. Taylor Series and Applications: Given a function f(x) and a number a,. x = -3 ==> sum(n = 0 to infinity) 1 /(2n+1), which diverges by the integral test. When x= 1, the series converges by the p-series test. It is a finite or an infinite series according as the number of terms is finite or infinite. With power series we can extend the methods of calculus we have developed to a vast array of functions making the techniques of calculus applicable in a much wider setting. diverges ( ) x diverges 0 rho x converges absolutely The ratio test for power series Example Determine the radius of convergence and the interval of convergence of the power series y(x) = X. Examples from Section 11. Series converges for only one number Remember the Ratio Test: A series converges if. The sequence is decreasing since for all Also, Therefore, this series converges by the Alternating Series Test and we include in our interval. Philip Mathematics of Computation, Vol. Yes, you're correct in your method: determining the radius of convergence of any power series is a matter of using the ratio or root test on the absolute value of the general term, which you did correctly. Radius of convergence R = 0. Using the series formula for our answer , we have (Note , , , etc. Math 306 - Power Series Methods Final Review Key (1) True of False? interval of convergence. o xMaclaurin series for the functions e, sin x, cos x, and 1/(1 – x). Then by formatting the inequality to the one below, we will be able to find the radius of convergence. Homework 25 Power Series 1 Show that the power series a c have the same radius of convergence Then show that a diverges at both endpoints b converges… UVA APMA 1110 - Homework+25+-+Power+Series - GradeBuddy. Why do we want to express a known function as the sum of a power series ? We will see later that this is a good strategy for integrating functions that don't have elementary antiderivatives ( or for examples), for solving differential equations, and for approximating functions by polynomials. The interval of convergence contains all values of x for which the power series converges. Step 2: Find the Radius of Convergence. And we will also learn how an alternating series may have Conditional or Absolute Convergence. (a) Find the Taylor series for the function f(x) = ex at a = 3. To execute such trades before competitors would. I'll test the endpoints separately. Example 1 Test the following series for convergence X1 n=1 ( 1)n 1 n I We have b n = 1 n. convergence 5 and interval of convergence centered at ˗̶̶̶ 1. It is customary to call half the length of the interval of convergence the radius of convergence of the power series. Alternating Series testP If the alternating series 1 n=1 ( 1) n 1b n = b 1 b 2 + b 3 b 4 + ::: b n > 0 satis es (i) b n+1 b n for all n (ii) lim n!1 b n = 0 then the series converges. result of multiplying the divergent harmonic series by 1. This means that the interval of convergence is ( 2;2). The same terminology can also be used for series whose terms are complex, hypercomplex or, more generally, belong to a normed vector space (the norm of a vector being corresponds to the absolute value of a number). 2 Convergence 2. Therefore, the interval of convergence for the power series is 2 x < 4 or [ 2;4). 5) Ratio Test (11. Both of those tests give an open interval of absolute convergence, and both guarantee that outside of the corresponding closed interval the terms of the series fail to converge to $0$, so it will diverge regardless of the signs of the terms. Then, the series becomes This is an alternating series. The Convergence and Partial Convergence of Alternating Series J. So x = −1 is included in the interval of convergence. 11 The interval of convergence for the Maclaurin series of f is (2x)2 (2r)3 (2x)4 = 4x2 — 4x3 + 16 x4 y'=8x-12x2 64 4. I b n+1 = 1 n+1 < n 1 n for all n 1. A series of the form converges if Example. Nihil anim keffiyeh helvetica, craft beer labore wes anderson cred nesciunt sapiente ea proident. Recall from the Absolute and Conditional Convergence page that series $\sum_{n=1}^{\infty} a_n$ is said to be absolutely convergent if $\sum_{n=1}^{\infty} \mid a_n \mid$ is also convergent. ii) Find a closed-form formula for. Find the sum of the alternating harmonic series. Power series, radius of convergence, interval of convergence. The interval of convergence is the set of all values of $$x$$ for which the series converges. The interval of convergence is the set of all values of x for which a power series converges. This program can run several different tests on infinite series to check for convergence or divergence. 2 Conditions for Convergence of an Alternating Sequence. A proof for a general monotonic decreasing alternating series can be found in Leibniz's Theorem. This is the harmonic series, so it diverges. To identify trading opportunities. (b) Find the Taylor series for the function f(x) = ex at a = 2. Direct Comparison Test. Plugging in x = 7 we get the series P 1=n2, which converges because it is a p-series with p = 2 > 1. The ratio test gives us: Because this limit is zero for all real values of x, the radius of convergence of the expansion is the set of all real numbers. So by the Alternating Series test, the original series converges. ny business of trading in securities needs two capabilities: 1. Carducci (East Stroudsburg University) Rearranging the Alternating Harmonic Series Ed Packel (Lake Forest College) and Stan Wagon (Macalester College) Bounding Partial Sums of the Harmonic Series Matt Clay; Taylor Series Michael Ford. Recall from the Absolute and Conditional Convergence page that series $\sum_{n=1}^{\infty} a_n$ is said to be absolutely convergent if $\sum_{n=1}^{\infty} \mid a_n \mid$ is also convergent. ∑ p-series, p = 1 ≤ 1 so it diverges If x =−1: (−1)n n=0 n ∞ ∑ alternating series, terms decrease, lim n→∞ 1 n =0 so it converges So, xn n=0 n ∞ ∑ converges when −1≤x <1 This is the interval of convergence of the series. Likewise, at x = 1 we have!∞ n=0 (−1)n (1)2n+2 2n+2 =!∞ n=0 (−1)n 2n+2 which is the same convergent alternating series. Comparison with an integral. The interval of convergence for the Maclaurin series of is 1: sets up ratio 1: limit evaluation 1: radius of convergence 1: considers both endpoints 1: analysis and interval of convergence (b) is a geometric series that converges to for Therefore for 1: series for. Also examine differentiation and integration of power series. Functions of One Real Variable: Limit, continuity, intermediate value property, differentiation, Rolle’s Theorem, mean value theorem, L'Hospital rule, Taylor's theorem, maxima and minima. You do not need to nd the radius of convergence or interval of convergence. 1 Questions: 2 Converge or Diverge: 1. k2 is the series of absolute values. Theorem 4 : (Comparison test ) Suppose 0 • an • bn for n ‚ k for some k: Then. If and then Theorem 2. Unlike geometric series and p-series, a power series often converges or diverges based on its x value. lim n!1 an = 0, then the alternating series converges. The interval of convergence of the series is [1, 5), and the radius of convergence is R 2. The main tools for computing the radius of convergence are the Ratio Test and the Root Test. 1 The Interval of Convergence In the previous section, we discussed convergence and divergence of power series. 2 Convergence 2. While most of the tests deal with the convergence of infinite series, they can also be used to show the convergence or divergence of infinite products. When x = 1, we get the series X1 n=0 ( n1 + 3) (n+ 1)2n = 1 n=0 1 n+ 1. For f0: [ 5; 3). The series converges uniformly on [ −ρ,ρ ] for every 0 ≤ ρ< 1 but does not converge uniformly on (− 1, 1) (see Example 5. X1 n=1 (x 1)n 1 n2 3n. Absolute Convergence Test: If you have a series X1 n=1 x n, and the. Namely, a power series will converge if its sequence of partial sums converges. convergence is one. The following power series is centered at c. Find the radius of convergence. First however, we must compute the radius of convergence for the series: lim n!1 n. Taking absolute values of the terms and using the ratio test, one can show that the sum of the absolute values of the terms of the series, i. 5) Ratio Test (11. $2$ is the radius of convergence. Using the series formula for our answer , we have (Note , , , etc. Below are some of the standard terms and illustrations concerning series. Therefore, the series is a conditionally convergent series. Having developed tests for positive-term series, turn to series having terms that alternate between positive and negative. , X∞ n=1 n2 3n converges. Estimating Error/Remainder of a Series; Alternating Series Test; Alternating Series Estimation Theorem; Ratio Test; Ratio Test with Factorials; Root Test; Absolute and Conditional Convergence; Difference Between Limit and Sum of the Series; Radius of Convergence; Interval of Convergence; Power Series Representation, Radius and Interval of. We know the series will converge on the interval 5 1. Then by formatting the inequality to the one below, we will be able to find the radius of convergence. (In other words,the first finite number of terms do not determine the convergence of a series. g(x) = Represent the function g(x) in Exercise 50 as a power series about 5, and find the interval of convergence. All algorithms numbered 493 and above, as well as a few earlier ones, may be downloaded from this server. Taylor Polynomials⁄ (a) an application of Taylor Polynomials (e. Homework 25 Power Series 1 Show that the power series a c have the same radius of convergence Then show that a diverges at both endpoints b converges… UVA APMA 1110 - Homework+25+-+Power+Series - GradeBuddy. X( n1)nx 4n lnn. Thus, the interval of convergence of the power series is (− 1, 1). Taylor Series and Applications: Given a function f(x) and a number a,. A series of the form converges if Example. , London: Hodder Education, 2005 pp. Sometimes we’ll be asked for the radius and interval of convergence of a Taylor series. Hence the series diverges by the nth-term test. 8 NAME Section wed Test each of the series for convergence or divergence: State and use the Comparison Test. The endpoints of the interval of convergence always must be checked independently. It's also known as the Leibniz's Theorem for alternating series. I The binomial series determine the open interval of. The interval of convergence of the series is [1, 5), and the radius of convergence is R 2. When x — the series is This is the alternating hannonic series, which converges. Next, if , the power series becomes: which converges by the alternating series test. Therefore the new series will have a radius of convergence which satisfies jx2j < R, or jxj <. (In other words,the first finite number of terms do not determine the convergence of a series. infinity summation n=1 [3^n(x-2)^n]/n when i did it, i got the interval of convergence to be 5/3 < x < 7/3 but im not sure how to check the endpoints with this one?. And we will also learn how an alternating series may have Conditional or Absolute Convergence. Ò $ß$Ñ è Remember that a bracket indicates an endpoint that belongs to an interval, while a parentheses indicates an endpoint that does not belong to the interval. As we will soon see, there are several very nice results that hold for alternating series, while alternating series can also demonstrate some unusual behaivior. Use the ratio test to determine radius or open interval of convergence of power series. Chapter 7) Taylor Series. Murphy, A2 Further Pure Mathematics , 3rd ed. ny business of trading in securities needs two capabilities: 1. Once again, if something does not pass the alternating series test, that does not necessarily mean that it diverges. So, the interval of convergence is -3 < x <= 3. R1 the interval of convergence is – 2 < x < 2 A1 (b) (i) this alternating series is convergent because the moduli of successive terms are monotonic decreasing R1 and the nth term tends to zero as n → ∞ R1. 3 Alternating series, approximations of alternating series. Find the intervals of convergence of fand f0. Divergence test (particularly useful for terms that are rational functions). alternating series test; Usually, for power series, the best way to determine convergence will be the ratio test, and sometimes the root test. Convergence of power series is similar to convergence of series. The series above is thus an example of an alternating series, and is called the alternating harmonic series. Since it is a geometric series, we know that it converges when \eqalign{ |x+2|/3& 1\cr |x+2|& 3\cr -3 x+2 & 3\cr -5 x& 1. The interval of convergence of the series is [1, 5), and the radius of convergence is R 2. Determine a power-series representation of the function ln (1 + x) on an interval centered at x = 0. We can use the ratio test to find out the absolute convergence of the power series by examining the limit, as n approached infinity, of the absolute value of two successive terms of the sequence. At , the series is. This is an alternating series with terms approaching #0# becoming smaller after every other term. The terms converge to 0. If , then R = and the series converges for all values of x. When x — the series is This is the alternating hannonic series, which converges.

http://mathhelpforum.com/trigonometry/36050-trigonometry-three-dimensional-question.html

# Math Help - Trigonometry three dimensional question 1. ## Trigonometry three dimensional question I need help to solve: A cylinder with radius 4 cm and perpendicular height 15 cm is tilted so that it will just fit inside a 12 cm high box. At what angle must it be tilted? Answer given at the back of the text book is 16 degrees 15 minutes. 2. Originally Posted by lalji I need help to solve: A cylinder with radius 4 cm and perpendicular height 15 cm is tilted so that it will just fit inside a 12 cm high box. At what angle must it be tilted? Answer given at the back of the text book is 16 degrees 15 minutes. Maybe it is the three dimensions that are screwing you up...I say think it more two dimensionally to start...and here is a big hint...PERPENDICULAR height...and perp. is synonomous with what? 3. Hello, Take a look at this picture. OPQR represents the 12cm high box. ABCD is a view from the side of the cylinder. According to the text, $RQ=OP=12 cm$ $AB=CD=2 \cdot 4=8 cm$ $AC=BD=15 cm$ And you're looking for angle $DAR$. 4. Hello, lalji! I don't agree with their answer . . . A cylinder with radius 4 cm and perpendicular height 15 cm is tilted so that it will just fit inside a 12 cm high box. At what angle must it be tilted? Answer given: 16 degrees 15 minutes. Code: P A S _ * - - - * - - - - * : | * * | : | * *D | : | * * | 12-x| * 15 * | : | * * | : | * * | : |* * | -B* 8 * | x | * * | - * - - * - - - - - * Q C R The cylinder is $ABCD\!: CB = 8,\;AB = 15,\;PQ = 12$ The box is $PQRS\!:\;PQ = 12$ Let $x = BQ$ Let $\theta = \angle BCQ$ . . Note that $\angle ABP = \theta$ In right triangle $BCQ\!:\;QC = \sqrt{64-x^2}$ . . $\cos\theta \,=\,\frac{\sqrt{64-x^2}}{8}\;\;{\color{blue}[1]}$ In right triangle $APB\!:\;PB \,=\,12-x$ . . $\cos\theta \:=\:\frac{12-x}{15}\;\;{\color{blue}[2]}$ Equate [1] and [2]: . $\frac{\sqrt{64-x^2}}{8} \;=\;\frac{12-x}{15}\quad\Rightarrow\quad 15\sqrt{64-x^2} \;=\;8(12-x)$ Square both sides: . $225(64-x^2) \;=\;64(144-24x + x^2)$ . . which simplifies to: . $289x^2 - 1536x - 5184 \;=\;0$ . . and has the positive root: . $x \;=\;\frac{1536 + \sqrt{6,352,000}}{578} \;=\;7.657439446$ Then: . $\sin\theta \;=\;\frac{x}{8} \;=\;\frac{7.657439446}{8} \;=\;0.957179931$ . . $\theta \;=\;\sin^{-1}(0.957179931) \;=\;73.17235533^o \;\approx\;\boxed{73^o10'}$ 5. I've got some difficulties, how can we know, by reading the text, what angle we are looking for ? And how can this physically be possible if D is not on RS ?? 6. When I saw this question, I knew I saw it before. It was in my textbook last year. Here is the picture the book provided, and the answer at the back of the book is 16 degrees 50 minutes, not 16 degrees 15 minutes. I think Soroban had a correct answer, but the book is asking for the angle of the other side. So: $180^o$(straight angle) - $90^o$ (angle between base and length of cylinder) - $73^o10'$ (angle 'on the other side' Soroban found) = $\theta$ $\theta$ = $16^o 50'$ 7. ## Thanks for the responses Thanks everyone especially Soroban, you are genius. Two dimensional diagram is very helpful to understand. Thanks Gusbob for posting the diagram from the text book, I did not have scanner to scan the diagram.

https://brilliant.org/discussions/thread/golden-ratio-and-fibonacci-numbers/

# Golden Ratio and Fibonacci Numbers Golden Ratio is considered to be one of the greatest beauties in mathematics. Two numbers $$a$$ and $$b$$ are said to be in Golden Ratio if $a>b>0,\quad and\quad \frac { a }{ b } =\frac { a+b }{ a }$ If we consider this ratio to be equal to some $$\varphi$$ then we have $\varphi =\frac { a }{ b } =\frac { a+b }{ a } =1+\frac { b }{ a } =1+\frac { 1 }{ \varphi }$ Solving in quadratic we get two values of $$\varphi$$, viz. $$\frac { 1+\sqrt { 5 } }{ 2 }$$ and $$\frac { 1-\sqrt { 5 } }{ 2 }$$ one of which (the second one) turns out to be negative (extraneous) which we eliminate. So the first one is taken to be the golden ratio (which is obviously a constant value). It is considered that objects with their features in golden ratio are aesthetically more pleasant. A woman's face is in general more beautiful than a man's face since different features of a woman's face are nearly in the golden ratio. Now let us come to Fibonacci sequence. The Fibonacci sequence $${ \left( { F }_{ n } \right) }_{ n\ge 1 }$$ is a natural sequence of the following form:${ F }_{ 1 }=1,\quad { F }_{ 2 }=1,\quad { F }_{ n-1 }+{ F }_{ n }={ F }_{ n+1 }$ The sequence written in form of a list, is $$1,1,2,3,5,8,13,21,34,..$$. The two concepts: The Golden Ratio and The Fibonacci Sequence, which seem to have completely different origins, have an interesting relationship, which was first observed by Kepler. He observed that the golden ratio is the limit of the ratios of successive terms of the Fibonacci sequence or any Fibonacci-like sequence (by Fibonacci-like sequence, I mean sequences with the recursion relation same as that of the Fibonacci Sequence, but the seed values different). In terms of limit:$\underset { n\rightarrow \infty }{ lim } \left( \frac { { F }_{ n+1 } }{ { F }_{ n } } \right) =\varphi$ We shall now prove this fact. Let ${ R }_{ n }=\frac { { F }_{ n+1 } }{ { F }_{ n } } ,\forall n\in N$Then we have $$\forall n\in N$$ and $$n\ge 2$$, ${ F }_{ n+1 }={ F }_{ n }+{ F }_{ n-1 }\\$ and ${ R }_{ n }=1+\frac { 1 }{ { R }_{ n-1 } } >1$We shall show that this ratio sequence goes to the Golden Ratio $$\varphi$$ given by: $\varphi =1+\frac { 1 }{ \varphi }$We see that: $\left| { R }_{ n }-\varphi \right| =\left| \left( 1+\frac { 1 }{ { R }_{ n-1 } } \right) -\left( 1+\frac { 1 }{ \varphi } \right) \right| \\ =\left| \frac { 1 }{ { R }_{ n-1 } } -\frac { 1 }{ \varphi } \right| \\ =\left| \frac { \varphi -{ R }_{ n-1 } }{ \varphi { R }_{ n-1 } } \right| \\ \le \left( \frac { 1 }{ \varphi } \right) \left| \varphi -{ R }_{ n-1 } \right|\\ \le { \left( \frac { 1 }{ \varphi } \right) }^{ n-2 }\left| { R }_{ 2 }-\varphi \right|$ Which clearly shows that$\left( { R }_{ n } \right) \longrightarrow \varphi$ (since $$\left| { R }_{ 2 }-\varphi \right|$$ is a finite positive real whose value depends on the seed values) Note by Kuldeep Guha Mazumder 2 years, 11 months ago MarkdownAppears as *italics* or _italics_ italics **bold** or __bold__ bold - bulleted- list • bulleted • list 1. numbered2. list 1. numbered 2. list Note: you must add a full line of space before and after lists for them to show up correctly paragraph 1paragraph 2 paragraph 1 paragraph 2 [example link](https://brilliant.org)example link > This is a quote This is a quote # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" MathAppears as Remember to wrap math in $$...$$ or $...$ to ensure proper formatting. 2 \times 3 $$2 \times 3$$ 2^{34} $$2^{34}$$ a_{i-1} $$a_{i-1}$$ \frac{2}{3} $$\frac{2}{3}$$ \sqrt{2} $$\sqrt{2}$$ \sum_{i=1}^3 $$\sum_{i=1}^3$$ \sin \theta $$\sin \theta$$ \boxed{123} $$\boxed{123}$$ Sort by: - 2 years, 11 months ago You are welcome! Did you like it? - 2 years, 11 months ago Yes! - 2 years, 11 months ago My pleasure..:-) - 2 years, 11 months ago I like Fibonacci very much.It is really The beauty of Mathematics. - 2 years, 9 months ago Very nice knowledge.. Loved it...The Magic of Maths!!! - 2 years, 9 months ago https://brilliant.org/problems/wow-12/?group=w3HWB8GobVLl&ref_id=1095702 i posted a problem about the same thing my solution was almost the same as your proof of it (: - 2 years, 9 months ago I have seen your proof. Your idea is essentially the same. Only some of your steps are erroneous. - 2 years, 9 months ago - 2 years, 9 months ago Nothing as such. Only that you have put a plus sign in front of 1/phi. - 2 years, 9 months ago There is one more interesting thing I found yesterday. The Ratio of the diagonal and the side of a regular Pentagon is exactly equal to the golden ratio. - 2 years, 10 months ago Ok then I will write a note on it.. - 2 years, 10 months ago Didn't you find it extremely interesting? This is the beauty of Mathematics. - 2 years, 10 months ago Nice - 2 years, 11 months ago Thanks..don't you think whatever is written above is a reconciliation of two apparently different mathematical ideas?.. - 2 years, 11 months ago Nice work ! I read this in the book Da Vinci Code by Dan Brown. - 2 years, 11 months ago That is one book that I want to read but haven't read yet..thank you for your compliments..:-) - 2 years, 11 months ago Have you read any other book by Dan Brown ? If not then try them ,they are awesome . - 2 years, 11 months ago I have just bought The Da Vinci Code today..:-) - 2 years, 11 months ago

https://www.physicsforums.com/threads/evaluating-triple-integral.74665/

# Evaluating Triple integral 1. May 7, 2005 ### MattL I'm having trouble with evaluating [Triple Integral] |xyz| dx dy dz over the region (x/a)^2 + (y/b)^2 + (z/c)^2 <= 1 Do I need to use some sort of parametrisation for the region, and is there some way of dealing with the absolute value function without integrating over the eight octants? Whilst I've separated the integral into the product of three integrals, I'm not sure if this actually helps? 2. May 7, 2005 ### dextercioby Well,the function is even and the domain of integration is symmetric wrt the origin,so that would give u a hint upon the limits of integration.The symmetry of the ellipsoid is really useful. As for the parametrization,i'm sure u'll find the normal one $$x=a\cos\varphi\sin\vartheta$$ $$y=b\sin\varphi\sin\vartheta$$ $$z=c\cos\vartheta$$ pretty useful. Daniel. 3. May 7, 2005 ### saltydog This integral looks like it's zero with all the symmetry: you know, four positives and four negatives for the integrand. Not sure though as I can't evaluate it. Would like to know though. 4. May 7, 2005 ### arildno I think you missed the absolute value sign on the integrand.. 5. May 7, 2005 ### saltydog Well . . . no, that's the reason I used for the symmetry but again, I qualify my statements by the fact I can't prove it. For example in the first octant: $$|xyz|=xyz$$ That's a positive one. However, in the octant with x<0, y>0 and z>0 we have: $$|xyz|=-xyz$$ And so forth in the 8 octants leaving 4 positive and 4 negative ones integrated symmetrically (I think). 6. May 7, 2005 ### arildno The integrand is positive almost everywhere; hence, the integral is strictly positive: Let: $$x=ar\sin\phi\cos\theta,y=br\sin\phi\sin\theta,z=cr\cos\phi$$ $$0\leq{r}\leq{1},0\leq\theta\leq{2}\pi,0\leq\phi\leq\pi$$ Thus, we may find: $$dV=dxdydz=abcr^{2}\sin\phi{dr}d\phi{d}\theta$$ $$|xyz|=\frac{abcr^{3}}{2}\sin^{2}\phi|\cos\phi\sin(2\theta)|$$ Doing the r-integrations yield the double integral: $$I=\frac{(abc)^{2}}{12}\int_{0}^{2\pi}\int_{0}^{\pi}\sin^{3}\phi|\cos\phi\sin(2\theta)|d\phi{d}\theta$$ We have symmetry about $$\phi=\frac{\pi}{2}$$; thus we gain: $$I=\frac{(abc)^{2}}{24}\int_{0}^{2\pi}|\sin(2\theta)|d\theta$$ We have four equal parts here, and using the part $$0\leq\theta\leq\frac{\pi}{2}$$ yields: $$I=\frac{(abc)^{2}}{12}$$ 7. May 7, 2005 ### saltydog Thanks Arildno. MattL, hope I didn't get in your way. I'll go through it to make sure I understand it. 8. May 7, 2005 ### MattL No problem. Think I should be able to give this question a fair go now. 9. May 10, 2011 ### Ray Vickson Change variables to x = a*x1, y = b*x2, z = c*x3. The integration region is the unit ball x1^2 + x2^2 + x3^2 <= 1, the integrand is abc*|x1 x2 x3|, and dV = abc * dx1 dx2 dx3. Because of the absolute value and symmetry, the whole integral, I, equals 8 times the integral over the {x1,x2,x3 >= 0} portion of the ball. This gives I = 8abc*int_{x3=0..1} f(x3) dx3, where f(x3) = x3*int_{x1^2 + x2^2 <= 1-x3^2} x1 x2 dx1 dx2. Using polar coordinates (or first integrating over x2 for fixed x1, then integrating over x1) we can easily evaluate f(x3), then integrate it over x3 = 0-->1. The final result is I = abc/6. R.G. Vickson

https://math.stackexchange.com/questions/3902158/proving-or-disproving-basic-facts-about-sequences-in-real-analysis

# Proving or disproving basic facts about sequences in Real Analysis I am self-learning real analysis from Stephen Abott's Understanding Analysis. In Exercise 2.3.7, the author asks to prove or disprove basic results on convergence. I'd like to verify my solution, to ensure, I've understood the concepts, and if proof is technically correct and rigorous. \textbf{Problem.} Give an example of each of the following, or state that such a request is impossible by referencing proper theorem(s): (a) sequences $$(x_n)$$ and $$(y_n)$$, which both diverge, but whose sum $$(x_n + y_n)$$ converges; (b) sequences $$(x_n)$$ and $$(y_n)$$, where $$(x_n)$$ converges, $$(y_n)$$ diverges, and $$(x_n + y_n)$$ converges; (c) a convergent sequence $$(b_n)$$ with $$b_n \ne 0$$ for all $$n$$ such that $$1/b_n$$ diverges; (d) an unbounded sequence $$(a_n)$$ and a convergent sequence $$(b_n)$$ with $$(a_n - b_n)$$ bounded (e) two sequences $$(a_n)$$ and $$(b_n)$$, where $$(a_n b_n)$$ and $$a_n$$ converge but $$(b_n)$$ does not. Solution. (a) Consider the sequence $$(x_n)$$ given by $$x_n = \sqrt{n+1}$$ and the sequence $$(y_n)$$ given by $$y_n = -\sqrt{n}$$. Both sequences diverge, but the sum $$(x_n + y_n)$$ converges to $$0$$. Also, consider the sequence $$(x_n)$$ given by $$x_n = n$$ and the sequence $$(y_n)$$ given by $$y_n = -\sqrt{n^2 + 2n}$$. Both sequences diverge, but the sum $$(x_n + y_n)$$ converges to $$-1$$. (b) This request is impossible. If the $$(x_n + y_n)$$ is to be convergent, it implies we are able to make the distance $$\vert{(x_n + y_n) - (x + y)}\vert$$ as small as like. However, we cannot make $$y_n$$ to lie eventually in a set $$(y - \epsilon, y + \epsilon)$$. Hence, the sum cannot be convergent. (c) Consider the sequence $$(b_n)$$ given by $$b_n = \frac{1}{n}$$. Then, $$(b_n)$$ is a convergent sequence but $$1/b_n$$ is divergent. (d) This request is impossible. The key here is to show that, assuming here $$(a_n)$$ is bounded leads to the contradiction, leading to their difference also being bounded. If $$(a_n)$$ is a bounded sequence, there exists a large number $$M > 0$$, such that $$\vert{a_n}\vert < M$$ for all $$n \in \mathbf{N}$$. If $$(b_n)$$ is a bounded sequence, there exists a large number $$N > 0$$, such that $$\vert{a_n}\vert < N$$ for all $$n \in \mathbf{N}$$. Thus, \begin{align*} \vert{a_n - b_n}\vert &= \vert{a_n + (-b_n)}\vert \\ &\le \vert{a_n}\vert + \vert{-b_n}\vert\\ &< M + N \end{align*} (e) Consider the sequence $$(a_n)$$ given by $$a_n = \frac{1}{n}$$ and $$(a_n b_n)$$ given by $$a_n b_n= \frac{\sin n}{n}$$. Thus, $$(a_n b_n)$$ and $$(a_n)$$ converges, but $$(b_n)$$ does not. Thank you for a detailed, well-asked question! For (a), (c), and (e): your examples are all correct. I would say that you asserted several facts that haven't been justified (which would make the proof incomplete, if you were writing for someone else's judgment/understanding). For example, the asserted limits of your examples in part (a) would need to be justified, as would the convergent/divergent assertions in part (e). Note even in part (c) that you didn't justify your assertions (though in that case they're pretty obvious). While the idea is reasonable, your proof for part (b) isn't rigorous. One detail: assuming that $$(x_n+y_n)$$ is convergent means that it converges to some number $$z$$, not to $$x+y$$ (indeed you didn't define either $$x$$ or $$y$$). Then you asserted, without proof, that $$y_n$$ can't be made to lie inside a small set. Your ideas are heading in the right direction, but you would need to use the precise definitions of convergence/divergence in exploiting the assumptions and in setting out what needs to be proved. Alternatively: you (correctly) believe (b) is impossible—in other words, you believe the implication "if $$(x_n)$$ converges and $$(y_n)$$ diverges, then $$(x_n+y_n)$$ diverges". As it turns out, that statement is logically equivalent to "if $$(x_n)$$ converges and $$(x_n+y_n)$$ converges, then $$(y_n)$$ converges"—which you might well find easier to prove! (By logically equivalent, I mean that the two statements "if P and Q, then R" and "if P and (not R), then (not Q)" have the same meaning.) Your proof for (d) seems to prove the following statement: "if $$(a_n)$$ is bounded and $$(b_n)$$ is bounded, then $$(a_n+b_n)$$ is bounded". That is a true fact, but is that what you want to prove here? • I expanded my original attempt with short proofs like you said. I am not quite sure, how do I begin with proving (d). – Quasar Nov 11 '20 at 16:05 $$\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$$ I expanded my original attempt with short proofs, proving/disproving each of the statements. I am posting it as an answer, so as to not invalidate the hints and tips by @GregMartin. (a) Consider the sequence $$(x_n)$$ given by $$x_n = \sqrt{n+1}$$ and the sequence $$(y_n)$$ given by $$y_n = -\sqrt{n}$$. Both sequences diverge, but the sum $$(x_n + y_n)$$ converges to $$0$$. Also, consider the sequence $$(x_n)$$ given by $$x_n = n$$ and the sequence $$(y_n)$$ given by $$y_n = -\sqrt{n^2 + 2n}$$. Both sequences diverge, but the sum $$(x_n + y_n)$$ converges to $$-1$$. Short proof. Consider $$a_n = \sqrt{n + 1} - \sqrt{n}$$. Observe that, \begin{align*} \sqrt{n+1} - \sqrt{n} &= (\sqrt{n+1} - \sqrt{n}) \times \frac{\sqrt{n+1} + \sqrt{n}}{\sqrt{n+1} + \sqrt{n}}\\ &=\frac{1}{\sqrt{n+1} + \sqrt{n}} \\ &< \frac{1}{\sqrt{n} + \sqrt{n}} = \frac{2}{\sqrt{n}} \end{align*} Pick $$\epsilon > 0$$. We can choose $$N > \frac{4}{\epsilon^2}$$. To show that this choice of $$N$$ indeed works, we prove that, that for all $$n \ge N$$, \begin{align*} \absval{\sqrt{n+1} - \sqrt{n}} &< \frac{2}{\sqrt{n}}\\ &< \frac{2}{\sqrt{(4/\epsilon^2)}} = \epsilon \end{align*} Thus, $$(\sqrt{n+1} - \sqrt{n}) \to 0$$. Consider $$b_n = n - \sqrt{n^2 + 2n}$$ Observe that: \begin{align*} n - \sqrt{n^2 + 2n} - (-1) &= [(n+1) - \sqrt{n^2 + 2n}] \\ &= [(n+1) - \sqrt{n^2 + 2n}] \times \frac{(n+1) + \sqrt{n^2 + 2n}}{(n+1) + \sqrt{n^2 + 2n}}\\ &= \frac{(n+1)^2 - (n^2 + 2n)}{(n+1) + \sqrt{n^2 + 2n}}\\ &= \frac{1}{(n+1) + \sqrt{n^2 + 2n}}\\ &< \frac{1}{n + \sqrt{n^2}} = \frac{1}{2n} \end{align*} Pick an arbitrary $$\epsilon > 0$$. We choose an $$N > \frac{1}{2\epsilon}$$. To show that choice of $$N$$ indeed works, we find that: \begin{align*} \absval{n - \sqrt{n^2 + 2n} - (-1)} &< \frac{1}{2n} \\ &< \frac{1}{2} \cdot (2\epsilon) = \epsilon \end{align*} Thus, $$(n - \sqrt{n^2 + 2n}) \to -1$$. (b) This request is impossible. Alternatively, we believe that if $$(x_n)$$ converges and $$(x_n + y_n)$$ converges, then $$(y_n)$$ converges. Let us prove this fact rigorously. Suppose $$\lim x_n = a$$ and $$\lim x_n + y_n = b$$. We shall prove that $$\lim y_n = b - a$$. Observe that, \begin{align*} \absval{y_n - (b-a)} &= \absval{(x_n + y_n) - x_n - (b-a)}\\ &= \absval{(x_n + y_n - b) - (x_n - a)}\\ &\le \absval{x_n + y_n - b} + \absval{x_n - a} \end{align*} Pick an $$\epsilon > 0$$. Since $$(x_n) \to a$$, we can make the distance $$\absval{x - a}$$ as small as we like. There exists an $$N_1$$ such that \begin{align*} \absval{x_n - a} < \frac{\epsilon}{2} \end{align*} for all $$n \ge N_1$$. Since $$(x_n + y_n) \to b$$, we can make the distance $$\absval{x_n + y_n - b}$$ as small as we like. There exists an $$N_2$$ such that, \begin{align*} \absval{x_n + y_n - b} < \frac{\epsilon}{2} \end{align*} Let $$N = \max \{N_1,N_2 \}$$. To show that this $$N$$ indeed works, we find that: \begin{align*} \absval{y_n - (b-a)} &\le \absval{x_n + y_n - b} + \absval{x_n -a}\\ &< \frac{\epsilon}{2} +\frac{\epsilon}{2} = \epsilon \end{align*} (c) Consider the sequence $$(b_n)$$ given by $$b_n = \frac{1}{n}$$. Then, $$(b_n)$$ is a convergent sequence but $$1/b_n$$ is divergent. Consider $$(b_n) = \frac{1}{n}$$. Pick an arbitrary $$\epsilon > 0$$. We can choose $$N > \frac{1}{\epsilon}$$. To show that this choice of $$N$$ indeed works, we find that: \begin{align*} \absval{\frac{1}{n}} < \epsilon \end{align*} for all $$n \ge N$$. Consequently, $$(1/n) \to 0$$. (d) This request is impossible. (e) Consider the sequence $$(a_n)$$ given by $$a_n = \frac{1}{n}$$ and $$(a_n b_n)$$ given by $$a_n b_n= \frac{\sin n}{n}$$. Thus, $$(a_n b_n)$$ and $$(a_n)$$ converges, but $$(b_n)$$ does not. Let us prove $$\frac{\sin n}{n}$$ converges to $$0$$. Observe that, \begin{align*} \absval{\frac{\sin n}{n}} &= \frac{\absval{\sin n} }{\absval n}\\ &\le \frac{1}{n} \end{align*} Let $$\epsilon > 0$$ be an arbitary small but fixed positive real number. We choose an $$N > \frac{1}{\epsilon}$$. To prove that this choice of $$N$$ indeed works, we have, \begin{align*} \absval{\frac{\sin n}{n}} &\le\frac{1}{n} \\ &< \frac{1}{(1/\epsilon)} = \epsilon \end{align*} Consequently, $$\frac{\sin n}{n} \to 0$$.

http://ruthling.net/uzvrgd5v/diagonal-matrix-and-scalar-matrix-c4efad

Is it true that the only matrix that is similar to a scalar matrix is itself Hot Network Questions Was the title "Prince of Wales" originally claimed for the English crown prince via a trick? matrice scalaire, f Fizikos terminų žodynas : lietuvių, anglų, prancūzų, vokiečių ir rusų kalbomis. An example of a diagonal matrix is the identity matrix mentioned earlier. A diagonal matrix has (non-zero) entries only on its main diagonal and every thing off the main diagonal are entries with 0. Matrix is an important topic in mathematics. 9. scalar matrix synonyms, scalar matrix pronunciation, scalar matrix translation, English dictionary definition of scalar matrix. Diagonal matrix and symmetric matrix From Norm to Orthogonality : Fundamental Mathematics for Machine Learning with Intuitive Examples Part 2/3 1-Norm, 2-Norm, Max Norm of Vectors b ij = 0, when i ≠ j скалярная матрица, f pranc. Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. Yes it is. Extract elements of matrix. When a square matrix is multiplied by an identity matrix of same size, the matrix remains the same. Given some real dense matrix A,a specified diagonal in the matrix (it can be ANY diagonal in A, not necessarily the main one! Write a Program in Java to input a 2-D square matrix and check whether it is a Scalar Matrix or not. "Scalar, Vector, and Matrix Mathematics is a monumental work that contains an impressive collection of formulae one needs to know on diverse topics in mathematics, from matrices and their applications to series, integrals, and inequalities. Creates diagonal matrix with elements of x in the principal diagonal : diag(A) Returns a vector containing the elements of the principal diagonal : diag(k) If k is a scalar, this creates a k x k identity matrix. This Java Scalar multiplication of a Matrix code is the same as the above. The elements of the vector appear on the main diagonal of the matrix, and the other matrix elements are all 0. Minimum element in a matrix… Antonyms for scalar matrix. Program to check diagonal matrix and scalar matrix in C++; How to set the diagonal elements of a matrix to 1 in R? Nonetheless, it's still a diagonal matrix since all the other entries in the matrix are . Define scalar matrix. See : Java program to check for Diagonal Matrix. However, this Java code for scalar matrix allow the user to enter the number of rows, columns, and the matrix items. MMAX(M). A symmetric matrix is a matrix where aij = aji. Returns a scalar equal to the numerically largest element in the argument M. MMIN(M). Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … A diagonal matrix is a square matrix whose off-diagonal entries are all equal to zero. Diagonal elements, specified as a matrix. a diagonal matrix in which all of the diagonal elements are equal. Scalar Matrix : A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. This matrix is typically (but not necessarily) full. For variable-size inputs that are not variable-length vectors (1-by-: or :-by-1), diag treats the input as a matrix from which to extract a diagonal vector. add example. — Page 36, Deep Learning, 2016. Use these charts as a guide to what you can bench for a maximum of one rep. a matrix of type: Lower triangular matrix. What are synonyms for scalar matrix? Example: 5 0 0 0 0 5 0 0 0 0 5 0 0 0 0 5 If you supply the argument that represents the order of the diagonal matrix, then it must be a real and scalar integer value. Synonyms for scalar matrix in Free Thesaurus. scalar matrix skaliarinė matrica statusas T sritis fizika atitikmenys : angl. This behavior occurs even if … Scalar Matrix : A scalar matrix is a diagonal matrix in which the main diagonal (↘) entries are all equal. 8 (Roots are found analogously.) 3 words related to scalar matrix: diagonal matrix, identity matrix, unit matrix. In this post, we are going to discuss these points. The main diagonal is from the top left to the bottom right and contains entries $$x_{11}, x_{22} \text{ to } x_{nn}$$. Closure under scalar multiplication: is a scalar times a diagonal matrix another diagonal matrix? Yes it is, only the diagonal entries are going to change, if at all. All of the scalar values along the main diagonal (top-left to bottom-right) have the value one, while all other values are zero. A diagonal matrix is said to be a scalar matrix if all the elements in its principal diagonal are equal to some non-zero constant. What is the matrix? stemming. A matrix with all entries zero is called a zero matrix. Magnet Matrix Calculator. InnerProducts. Diagonal matrix multiplication, assuming conformability, is commutative. A diagonal matrix is said to be a scalar matrix if its diagonal elements are equal, that is, a square matrix B = [b ij] n × n is said to be a scalar matrix if. Solution : The product of any matrix by the scalar 0 is the null matrix i.e., 0.A=0 GPU Arrays Accelerate code by running on a graphics processing unit (GPU) using Parallel Computing Toolbox™. Pre- or postmultiplication of a matrix A by a scalar matrix multiplies all entries of A by the constant entry in the scalar matrix. General Description. Java Scalar Matrix Multiplication Program example 2. import java. The data type of a[1] is String. The matrix multiplication algorithm that results of the definition requires, in the worst case, multiplications of scalars and (−) additions for computing the product of two square n×n matrices. Great code. An identity matrix is a matrix that does not change any vector when we multiply that vector by that matrix. scalar meson Look at other dictionaries: Matrix - получить на Академике рабочий купон на скидку Летуаль или выгодно 8. Matrix algebra: linear operations Addition: two matrices of the same dimensions can be added by adding their corresponding entries. Example 2 - STATING AND. A square matrix in which all the elements below the diagonal are zero i.e. Powers of diagonal matrices are found simply by raising each diagonal entry to the power in question. Filling diagonal to make the sum of every row, column and diagonal equal of 3×3 matrix using c++ How to convert diagonal elements of a matrix in R into missing values? [x + 2 0 y − 3 4 ] = [4 0 0 4 ] 2. Maximum element in a matrix. Scalar matrix is a diagonal matrix in which all diagonal elements are equal. Negative: −A is defined as (−1)A. Subtraction: A−B is defined as A+(−B). Takes a single argument. is a diagonal matrix with diagonal entries equal to the eigenvalues of A. 6) Scalar Matrix. Upper triangular matrix. scalar matrix vok. A square matrix with 1's along the main diagonal and zeros everywhere else, is called an identity matrix. The values of an identity matrix are known. Scalar multiplication: to multiply a matrix A by a scalar r, one multiplies each entry of A by r. Zero matrix O: all entries are zeros. Types of matrices — triangular, diagonal, scalar, identity, symmetric, skew-symmetric, periodic, nilpotent. 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posted by Dina Simplify the expression. 3 [ (15 - 3)^2 / 4] a. 36 b. 108 c. 18 d. 9 1. TutorCat http://www.jiskha.com/display.cgi?id=1285004701 2. Dina ???108 3. PsyDAG 3 (12^2/4) = 3 (144/4) = 3 * 36 = 108 Yes! ## Similar Questions 1. ### Math(radicals or square roots) i need a explanation for this equation: 5 radical(square root) 108 over 25 the 5 is in front of the square rooted number and it is a fraction PlEASE HELP :( That is not an equation. It is a number. It sounds like you are describing … s8impliy this exponential expression): -2 7 -8 ) 6 -9)8 (7) * (x) * (w * (x *(4) * -3) (w How do you factor this expression? 10x+15 Use the distributive property to simplify: 25(2x + 1/5). simplify x^5-x^3+x^2-1-(x^3-1)(x+1)^2/(x^2-1)^2 6. ### Algebra Translate the phrase to mathematical language. Then simplify the expression the difference between 119 and -40 What is a numerical expression for the phrase? 7. ### math Riding on a school bus are 20 students in 9th grade, 10 in 10th grade, 9 in 11th grade, and 7 in 12th grade. Approximately what percentage of the students on the bus are in 9th grade? 8. ### urgent math 1. Simplify the expression 7^9/7^3 a.7^3*** b.7^6 c.7^12 d.1^6 2. Simplify the expression z^8/z^12 a.z^20 b.z^4 c.1/z^-4 d.1/z^4 3. Simplify the expression (-3^4)/(-3^4 a.(-3)^1 b.0 c.1 d.(-3)^8 4.Which expressions can be rewritten … 9. ### algebra am I right? 1. Simplify radical expression sqrt 50 5 sqrt ^2*** 2 sqrt ^5 5 sqrt ^10 5 2. Simplify the radical expression sqrt 56x^2 28x 2x sqrt 14*** 2x sqrt 7 sqrt 14x2 3. Simplify the radical expression. sqrt 490y^5w^6 2 sqrt 135y^2 10. ### Algebra 1. Simplify the expression 4 3 — - — x x 2. Simplify the expression 6 4 — + — c c² 3. Simplify the expression 3y 9y — ÷ — 4y - 8 2y² - 4y 4. Simplify the expression: ( 2x³ - x² - 13x - 6 ) ÷ ( x -3 ) More Similar Questions

https://math.stackexchange.com/questions/3363106/is-ln-natural-log-and-log-the-same-thing-if-used-in-this-answer

# Is “ln” (natural log) and “log” the same thing if used in this answer? Find $$x$$ for $$4^{x-4} = 7$$. Answer I got, using log, was $${\log(7)\over 2\log(2)} + 4$$ but the actual answer was $${\ln(7)\over2\ln(2)} + 4$$ I plugged both in my calculator and turns out both are the equivalent value. Anyways, is using either one of ln or log appropriate for this question? Obviously ln is when log has the base e, and log is when it has the base 10. Final question: How do I know when to use which? that is which of ln or log is used when solving a question?? For example, if a question asks to find $$x$$ for $$e^x = 100$$, I will use $$\ln$$ since $$\ln(e)$$ cancels out. If a question asks to find $$2^x = 64$$, i will use log since "$$e$$" isn't present in the question. So is using either $$\log$$ or $$\ln$$ the same? • $$\log_{10}x=\frac{\ln x}{\ln 10}$$ – Lord Shark the Unknown Sep 20 at 6:52 • As an aside, to make matters worse, some authors will write $\log$ without a subscript and mean different things than one another. In texts on combinatorics for instance it is not uncommon to see $\log$ without a subscript be meant to be interpreted as being the base 2 logarithm $\log_2$ while other authors might intend it to be the base 10 logarithm $\log_{10}$. Others still may use $\log$ as the natural logarithm rather than writing it as $\ln$. The nice thing is, regardless which base it is you always have $\log_n(a)/\log_n(b)=\log_b(a)$ – JMoravitz Sep 20 at 14:55 • "In order to kill an exponential, you have to hit it with a log". Which raises the question "which log". The answer is -- it doesn't matter. – John Coleman Sep 20 at 16:11 • You can always contrive that there are $e$'s around. Note $2^x=(e^{\text{ln}(2)})^x = e^{x\text{ln}2}$ – jacob1729 Sep 20 at 18:00 • @JohnColeman: Just don't use base 1. math.stackexchange.com/questions/413713/log-base-1-of-1 – Joshua Sep 20 at 19:10 You can use any logarithm you want. As a result of the base change formula $$\log_2(7) = \frac{\log(7)}{\log(2)} = \frac{\ln(7)}{\ln(2)} = \frac{\log_b(7)}{\log_b(2)}$$ so as long as both logs have the same base, their ratio will be the same, regardless of the chosen base (as long as $$b > 0, b\neq 1$$). • Thank you. Out of curiosity, why would one prefer to use natural log in the question 4^(x-4) = 7, when it does not contain "e" – harold232 Sep 20 at 6:56 • @harold232 Since they're equivalent, the choice is harmless, but you still have to make a choice. The only reason I can think of to default to the natural log is that it is in some ways computationally easier than other logarithms (coming from formulas of calculus). – Brian Moehring Sep 20 at 7:01 • @harold232 for mathematicians, $e$ is the default base for logarithms which they would use unless there is a particular reason for choosing something else. (This is largely because it has nice properties for calculus.) After base $e$, the next most common in maths is base $2$. $10$, as Michael Palin might say, is right out. – Especially Lime Sep 20 at 7:05 • $10$ was popular in the days before calculators. In those days, it was the easiest one to work with as we had tables for it. I agree that its utility in pure maths is low today but it lives in a few cases where log scales are used e.g. pH in chemistry and the decibels. – badjohn Sep 20 at 8:43 • @EspeciallyLime: 10 is used by "scientists". See my answer below. – JonathanZ Sep 20 at 17:15 Either is fine. You can write logarithms in terms of any base that you like with the change of base formula $$\log_ba=\frac{\log_ca}{\log_cb}$$ One thing learned from secondary school is that exponential equations can be solved be rewriting the relationship in logarithmic form. As such, your equation can be rewritten as follows: \begin{align} 4^{x-4}&=7\\ x-4&=\log_47\\ x&=4+\log_47 \end{align} And since $$\log_47$$ can be rewritten as $$\frac{\log7}{\log4}$$ or $$\frac{\ln7}{\ln4}$$ or $$\frac{\log_{999876}7}{\log_{999876}4}$$ it does not matter which base of logarithm you use. • By the way I noticed that you left the denominator as log(4), shouldn't it be 2log(2) by power rule, since it is more simplified. Or does it not matter? – harold232 Sep 20 at 7:00 • @harold232 As you say, $\log(4)=2\log(2)$. So it does not really matter – Henry Sep 20 at 7:09 • If you want to input the logarithm into a calculator that does not have a log function that accepts a base parameter, then you can simply type in $\log\ 7 \div \log\ 4$. If you want to show your work or write a final solution as its exact value, I tend to accept $\log_47$ or $\frac{\ln7}{2\ln2}$ or its variants. – Andrew Chin Sep 20 at 7:10 • No educator worth his salt is going to be that pedantic. To you, is $\frac12\log_27$ more acceptable compared to $\log_47$? – Andrew Chin Sep 20 at 7:20 • Wolfram says at best it as an alternative – Andrew Chin Sep 20 at 7:29 Culturally • Computer science / programming people tend to use log base $$2$$ • Mathematicians tend to use log base $$e$$ • Engineers / physicist / chemists etc. tend to use log base $$10$$ Writers really should make it explicit the first time they use "$$\log$$", but they don't always. As others have pointed out, the only difference is a constant factor, and in your case the factors in the numerator and denominator cancelled each other out. So the answer to your question is "If it's in a math context you'll probably see $$\ln()$$ used." Heck, even if you were asked to solve $$10^{x - 3} = 6$$ you'd still see $$\ln()$$ used, even though it looks like $$\log_{10}$$ might seem more "natural" for that particular problem. It's just what math people tend to do. • It is called the "natural logarithm", after all. (: – Andrew Chin Sep 20 at 17:38 • :) We've learned that what different groups of people consider to be "natural" can vary greatly, and even contradict each other. Hence my starting off with the word "culturally". – JonathanZ Sep 20 at 17:52 • @JonathanZ you made me laugh out loud in a library! @@ It was a natural reaction of course. – uhoh Sep 21 at 5:43 There's an interesting unstated question here: what counts as an answer? You can clearly argue that using either $$\ln$$ or $$\log_{10}$$ should be acceptable. But in that case $$x = \log_4(7) + 4$$ should be just as correct. As @BrianMoehring says in his answer, you can use any logarithm you want. As for If a question asks to find $$2^x=64$$, i will use log since "e" isn't present in the question. I would just say $$x=6$$. That's really using $$\log_2$$, by inspection. In this case the other two answers are technically correct of course, in that the base doesn't really matter. But I want to point out that when you see $$\log(x)$$, it can either mean base $$10$$, base $$2$$ or base $$e$$, with the latter two (especially base $$e$$) being much more common as you move up the math ladder. The notation $$\ln(x)$$ is still used for base $$e$$, but whenever you see $$\log(x)$$ you should always assume it is also base $$e$$ unless context implies otherwise (if it's supposed to mean base $$2$$, it should be clear from context). Part of the reason is exactly because of the reason mentioned by the two other answers: for any $$a,b$$ we have $$\log_a(b)=\frac{\log(b)}{\log(a)},$$ So we can express logarithms of any base using the natural logarithm anyway and there's no need to designate a special symbol for it. And indeed you will see that base $$e$$ is much more useful than base $$10$$ most of the time. • log is often considered log_2 in CS papers – RiaD Sep 20 at 16:32 Use the change of base formula if you suspect that the answer is irrational, otherwise take the logarithm to both sides of the equation of a base that seems reasonable. There's nothing magical about the change of base formula. \begin{align} \log_c b &= \log_c b\\ \log_c a^{\log_a b} &= \log_c b\\ \log_a b \cdot\log_c a &= \log_c b\\ \log_a b &= \frac{\log_c b}{\log_c a}\\ \end{align} Even if the solution is integral or rational, using the change of base formula will get you to an answer, for example: \begin{align} 2^x &= 64\\ \log_{10}2^x &= \log_{10} 64\\ x\log_{10}2 &= \log_{10} 64\\ x &= \frac{\log_{10} 64}{\log_{10}2}\\ x &= \log_{2} 64\\ x &= \log_{2} 2^6\\ x &= 6\log_{2} 2\\ x &= 6\\ \end{align} although, it does insert a lot of extra [unnecessary] steps.

https://math.stackexchange.com/questions/3231585/a-n-be-a-sequence-such-that-a-n12-2a-na-n1-a-n-0-then-sum-1/3231653

# $\{a_n\}$ be a sequence such that $a_{n+1}^2-2a_na_{n+1}-a_n=0$, then $\sum_1^{\infty}\frac{a_n}{3^n}$ lies in… Let $$\{a_n\}$$ be a sequence of positive real numbers such that $$a_1 =1,\ \ a_{n+1}^2-2a_na_{n+1}-a_n=0, \ \ \forall n\geq 1$$. Then the sum of the series $$\sum_1^{\infty}\frac{a_n}{3^n}$$ lies in... (A) $$(1,2]$$, (B) $$(2,3]$$, (C) $$(3,4]$$, (D)$$(4,5]$$. Solution attempt: Firstly, we figure out what $$\frac{a_{n+1}}{a_n}$$ is going to look like. We get, from the recursive formula, $$\frac{a_{n+1}}{a_n}=1+\sqrt{1+\frac{1}{a_n^2}}$$ (remembering the fact that $$a_n>0$$, the other root is rejected). We know that, if $$\lim_{n \to \infty}\frac{a_{n+1}}{a_n}>1$$, then $$\lim a_n \to \infty$$. Further, $$(a_{n+1}-a_n)= \sqrt{a_n(a_n+1)}>0$$. (Again, the other root is rejected due to the same reason). Hence, $$(a_n)$$ increases monotonically. Therefore, the largest value of $$\frac{a_{n+1}}{a_n}$$ is approximately $$1+\sqrt{1+\frac{1}{1}} \approx 2.15$$ Now, the sum can be approximated as $$\displaystyle\frac{\frac{1}{3}}{1-\frac{2.15}{3}} \approx 1.3$$ (In actuality, $$\mathbb{sum}< 1.3$$). So, option $$(A)$$ is the correct answer. Is the procedure correct? I have been noticing a handful of this type of questions (based on approximations) lately, and the goal is to find out where the sum / the limit of the sequence might lie. Is there any "definitive" approach that exploits the recursive formula and gives us the value, or does the approach varies from problem to problem? • Out of curiosity, what's the source of this problem ? – Gabriel Romon May 19 '19 at 10:16 • @GabrielRomon It was asked in an entrance exam for Master's degree admission in India this year (JAM 2019). – Subhasis Biswas May 19 '19 at 10:18 • Hmm, your approach is indeed quite nice, but even if asymptotically $a_n\sim\alpha^n$ and the sum effectively $\frac\alpha{3-\alpha}$, the first terms of the series may as well shift the result in another interval. How do you bound the partial sum $\sum\limits_{n=1}^{n_0} \frac{a_n}{3^n}$ up the a certain $n_0$ so that subsequent terms are small enough and we can switch to asymptotic behaviour ? – zwim May 19 '19 at 10:48 • This is the part where I used the monotone property. The common ratio can never exceed $2.15$, no matter what. Because, after the first term of the sequence, $1/a_{n}^2 <1$, Resulting in $a_{n+1}/a_n <2.15$ – Subhasis Biswas May 19 '19 at 10:50 • If you recall the proof of the ratio test, then yes, you applied the right strategy. – rtybase May 19 '19 at 10:58 Following your calculations and according to the ratio test $$0<\frac{\frac{a_{n+1}}{3^{n+1}}}{\frac{a_n}{3^n}}=\frac{1}{3}\cdot \frac{a_{n+1}}{a_n}<1$$ thus $$\sum\limits_{n=1}\frac{a_n}{3^n}< \infty$$ Now, applying the same technique from the proof on the ratio test $$S=\frac{1}{3}+\frac{a_2}{3^2}+\frac{a_3}{3^3}+\cdots+\frac{a_n}{3^n}+\cdots=\\ \frac{1}{3}+\frac{a_2}{a_1}\cdot\frac{a_1}{3^2}+\frac{a_3}{a_2}\cdot\frac{a_2}{3^3}+\cdots+\frac{a_{n}}{a_{n-1}}\cdot\frac{a_{n-1}}{3^n}+\cdots$$ or $$\frac{1}{3}+2\cdot\frac{1}{3^2}+2\cdot\frac{a_2}{3^3}+\cdots+2\cdot\frac{a_{n-1}}{3^n}+\cdots< S<\\ \frac{1}{3}+2.15\cdot\frac{1}{3^2}+2.15\cdot\frac{a_2}{3^3}+\cdots+2.15\cdot\frac{a_{n-1}}{3^n}+\cdots$$ and repeating this $$\frac{1}{3}+\frac{2}{3^2}+\frac{2^2}{3^3}+\cdots+\frac{2^{n-1}}{3^n}+\cdots< S<\\ \frac{1}{3}+\frac{2.15}{3^2}+\frac{2.15^2}{3^3}+\cdots+\frac{2.15^{n-1}}{3^n}+\cdots$$ or $$\frac{1}{3}\cdot\left(1+\frac{2}{3}+\frac{2^2}{3^2}+\cdots+\frac{2^{n-1}}{3^{n-1}}+\cdots\right)< S<\\ \frac{1}{3}\cdot\left(1+\frac{2.15}{3}+\frac{2.15^2}{3^2}+\cdots+\frac{2.15^{n-1}}{3^{n-1}}+\cdots\right)$$ or $$\color{red}{1}=\frac{\frac{1}{3}}{1-\frac{2}{3}}<\color{red}{S}<\frac{\frac{1}{3}}{1-\frac{2.15}{3}}=\frac{1}{3-2.15}<\color{red}{2}$$ This kind of squeezing technique is widely applied in analysis, functional analysis, numerical analysis. So, it makes sense to ask something similar for a master degree entrance test. • Truly a nice approach. The upper bound is basically the same. I missed the lower bound though :( I like your final note. "So, it makes sense to ask something similar for a master degree entrance test" – Subhasis Biswas May 19 '19 at 11:33 • Nice and clean :) – A learner Apr 23 at 15:44 $$a_{n+1}=a_n+\sqrt{a_n^2+a_n}>2a_n\forall n\ge 1$$ $$a_1=2^0,a_2>2,a_3>2^2,...,a_n\ge 2^{n-1}$$ $$1\le a_n\Rightarrow a_n+a_n^2\le 2a_n^2\Rightarrow a_{n+1}\le (\sqrt 2+1)a_n$$ $$a_1=(\sqrt 2+1)^0,a_2=(\sqrt 2+1),a_3<(\sqrt 2+1)^2,...,a_n\le (\sqrt 2+1)^{n-1}$$ $$\therefore 2^{n-1}\le a_n \le (\sqrt 2+1)^{n-1}$$ $$\Rightarrow \frac 13(\frac 23)^{n-1}\le {a_n\over 3^n}\le \frac 13({\sqrt 2+1 \over 3})^{n-1}$$ $$\therefore \frac 13\sum (\frac 23)^{n-1} < \sum {a_n\over 3^n}\le \frac 13 \sum ({\sqrt 2+1 \over 3})^{n-1}$$ $$\text{Hence}\,\,1<\sum {a_n\over 3^n}\le {1\over 2-\sqrt 2}\approx 1.707$$ I would rather argue that $$\sqrt{a_n(a_n+1)}$$ lies inside $$[a_n,a_n+1]$$, so that $$b_n\leq a_n \leq c_n$$ where $$b_n=2b_n$$ and $$c_n=2c_n+1$$ with $$b_1=c_1=1$$. Closed forms for $$b_n$$ and $$c_n$$ are easily derived as $$b_n=2^{n-1}$$ and $$c_n=2^n-1$$, so that $$1\leq \sum_1^{\infty}\frac{a_n}{3^n} \leq 2-\frac 12$$ This inequality can be refined by only summing from $$n$$ larger than some constant. • Now, for the last part of my question, is there any general approach for these type of problems? – Subhasis Biswas May 19 '19 at 11:07 • A better one indeed! In fact, we don't need any better bound than this. Although my procedure gives off a slightly better value, it is not worthwhile here. – Subhasis Biswas May 19 '19 at 11:08 (Fill in the gaps as needed. If you're stuck, show your work and explain what you've tried.) Hints: • After calculating the first few terms, the ratio of the terms is very similar. If we estimate $$a_n \approx a_1 r^{n-1}$$, this suggests we should solve for $$\frac{ 1/3} { 1 - (r_1/3) } = 1$$ and $$\frac{1/3} { 1 - (r_2/3)} = 2$$ to give us an idea of how to bound the sequence. This gives us $$r_1 = 2, r_2 = 5/2$$, so we want to show that $$2 a_n < a_{n+1} < \frac{5}{2} a_n$$ (with some flexibility if this doesn't immediately work out). • Show that $$a_{n+1} = \frac{ 2a_n + \sqrt{ 4a_n^2 + 4a_n } } { 2}$$. In particular, reject the negative root. • Show that $$a_n \geq 1$$. • Show that $$2 a_n < a_{n+1} < \frac{5}{2} a_n$$. • Hence, show that $$1=\frac{ 1/3 } { 1 - 2/3} < \sum \frac{a_n}{3^n} < \frac{ 1 / 3 } { 1 - 5/6}=2$$ Note: • The LHS is true by calculating the first 5 terms. • Of course, we can't prove the RHS just by calculating enough terms. • In fact, the bounding inequality $$2a_n < a_{n+1} < 2a_n + \frac{1}{2}$$, so the ratio $$a_{n+1} / a_n$$ is very close to 2, esp at (slightly) larger values of $$n$$. • Not surprisingly, the value of the summation is much much closer to the 1. Using the first ~10 terms to get a better approximation, you can in fact show that the summation is between 1.2 and 1.25.

http://math.stackexchange.com/questions/628115/for-which-primes-p-is-p2-2-also-prime

For which primes p is $p^2 + 2$ also prime? Origin — Elementary Number Theory — Jones — p35 — Exercise 2.17 — Only for $p = 3$. If $p \neq 3$ then $p = 3q ± 1$ for some integer $q$, so $p^2 + 2 = 9q^2 ± 6q + 3$ is divisible by $3$, and is therefore composite. (1) The key here looks like writing $p = 3q ± 1$. Where does this hail from? I cognize $3q - 1, 3q, 3q + 1$ are consecutive. (2) How can you prefigure $p = 3$ is the only solution? On an exam, I can't calculate $p^2 + 2$ for many primes $p$ with a computer — or make random conjectures. E.g. any square is either $\color{purple}0$ or $\color{teal}1$ $\begin{cases}\mod 3, & \text{ depending on whether or not$3$divides$x$} \\ \mod 4, & \text{depending on whether or not$2$divides the number being squared} \end{cases}$, and $0,$ $1$, or $4$ mod $8$ (depending on whether or not $2$ or $4$ divide the number being squared). Thus, when you see $p^2 + 2$, you should think: $\begin{cases} = \color{purple}0 + 2 & \mod3 \text{ , if$3$divides$p$} \\ = \color{teal}1 + 2 \equiv 0 & \mod3 \text{ , if$3 \not| p$} \end{cases}$. (3) Can someone please clarify why I'd prefigure or think about mod 3, mod 4, mod 8? Why not consider mod of some random natural number? (4) The last paragraph considers mod 3. How can I prefigure this? - Use Fermat's little theorem. If $\gcd(p,3) = 1$, $p^2 \equiv 1 \pmod 3$ that gives $p^2 + 2\equiv 3 \pmod 3$. Thus only possibility is $p = 3$ - It is very easy. Learn! Very useful. Regards. –  Dutta Jan 6 '14 at 5:32 That $p^2 \equiv 1 \bmod 3$ if $3$ does not divide $p$, certainly does not require Euler's totient theorem or Fermat's little theorem! Those are overkill for such an easy-to-check fact. –  ShreevatsaR Jan 6 '14 at 6:28 Thanks. How did you prefigure to start with $\gcd(p, 3) = 1$? Why not $\gcd(p,$random integer$) = 1$? –  Dwayne E. Pouiller Apr 8 '14 at 10:30 Hint $\$ Apply the special case $\,q=3\,$ of the following Theorem $\$ If $\ p,\,q\$ and $\,r = p^{q-1}\!+q\!-\!1\,$ are all prime then $\, p = q$. Proof $\,$ If $\,p\ne q\,$ then $\,q\nmid p\,$ hence, by little Fermat, $\,q\mid \color{#c00}{p^{q-1\!}-1}\,$ so $\ \color{#0a0}{q\mid r}\,=\, \color{#c00}{p^{q-1}\!-1}+q$. However $\,p,q \ge 2\,$ so $\,p^{q-1}\!\ge 2\,$ so $\,r> q,\,$ so $\,\color{#0a0}q\,$ is a $\color{#0a0}{proper}$ factor of $\,r,\,$ contra $\,r\,$ prime. $\ \$ QED - Any integer $n$ can be written as $3q\pm1, 3q$ where $q$ is an integer Now we can immediately discard $3q$ as it is composite for $q>1$ Now $\displaystyle(3q\pm1)^2+2=9q^2\pm6q+3=3(3q^2\pm2q+1)$ Observe that $3q^2\pm2q+1>1$ for $q\ge1,$ hence $\displaystyle(3q\pm1)^2+2$ is composite - @oldrinb, that's what is written in the POST, right? –  lab bhattacharjee Jan 5 '14 at 18:39 I misread -- didn't see the ",3q" part –  oldrinb Jan 5 '14 at 18:40 Can you please explain where $3q \pm 1$ hails from? It feels uncanny. The rest of your answer isn't what I'm querying about. Can you please answer my edited post in your answer (not in comments)? –  Dwayne E. Pouiller Apr 8 '14 at 10:29 @DwayneE.Pouiller, $$3q-1,3q,3q+1$$ are any three consecutive integers, right? –  lab bhattacharjee Apr 8 '14 at 10:34 Whenever you see a quantity of the form $x^2 + a$ in a basic number theory course (especially in hw. or on an exam), you will want to think about what divisibilities it has by various small numbers. E.g. any square is either $\color{purple}0$ or $\color{teal}1$ $\begin{cases}\mod 3, & \text{ depending on whether or not$3$divides$x$} \\ \mod 4, & \text{depending on whether or not$2$divides the number being squared} \end{cases}$, and $0,$ $1$, or $4$ mod $8$ (depending on whether or not $2$ or $4$ divide the number being squared). Thus, when you see $p^2 + 2$, you should think: $\begin{cases} = \color{purple}0 + 2 & \mod3 \text{ , if$3$divides$p$} \\ = \color{teal}1 + 2 \equiv 0 & \mod3 \text{ , if$3 \not| p$} \end{cases}$. Since the only prime that can be $0$ mod $3$ is $3$ (and $p^2 + 2$ will certainly be $> 3$), this answers your question immediately. - Thanks. I'm sorry for unchecking the answer - I only cognized now I don't fully grasp it. Can you please answer my edited post in your answer (not in comments)? –  Dwayne E. Pouiller Apr 8 '14 at 10:28

http://mathhelpforum.com/calculus/155498-integrate-x-3-sqrt-1-x-2-k-2-dx.html

1. ## Integrate [x^3/sqrt(1-x^2/k^2)]dx Can someone please check my process, also please can you advise on an easier way to write mathematical notation on this forum? Integrate ( x^3/(1-x^2/k^2))dx where k is a constant. Let u = x^2, dv = x/sqrt(1-x^2/k^2) Integration by parts. Integral fx = uv - integral vdu v = -k^2*sqrt(1-x^2/k^2) du = 2xdx susbstituting for u,v,du Integral fx = -k^2 *x^2*sqrt(1-x^2/k^2)+ integral [2xk^2*sqrt(1-x^2/k^2)dx] Integral fx = -k^2 *x^2*sqrt(1-x^2/k^2) - 2/3*k^4*((1-x^2/k^2)^3/2) 2. cleared the fraction in the radical to make life easier ... $\displaystyle \frac{k}{k} \cdot \frac{x^3}{\sqrt{1 - \frac{x^2}{k^2}}}$ $\displaystyle \frac{kx^3}{\sqrt{k^2\left(1 - \frac{x^2}{k^2}\right)}} $ $\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx$ $u = x^2$ $du = 2x \, dx$ $dv = \frac{-x}{\sqrt{k^2-x^2}} \, dx$ $v = \sqrt{k^2-x^2}$ $\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} - \int 2x\sqrt{k^2-x^2} \, dx\right]$ $\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} + \int -2x\sqrt{k^2-x^2} \, dx\right]$ $\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} + \frac{2}{3}(k^2-x^2)^{\frac{3}{2}} + C\right]$ 3. ## thanks Originally Posted by skeeter cleared the fraction in the radical to make life easier ... $\displaystyle \frac{k}{k} \cdot \frac{x^3}{\sqrt{1 - \frac{x^2}{k^2}}}$ $\displaystyle \frac{kx^3}{\sqrt{k^2\left(1 - \frac{x^2}{k^2}\right)}} $ $\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx$ $u = x^2$ $du = 2x \, dx$ $dv = \frac{-x}{\sqrt{k^2-x^2}} \, dx$ $v = \sqrt{k^2-x^2}$ $\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} - \int 2x\sqrt{k^2-x^2} \, dx\right]$ $\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} + \int -2x\sqrt{k^2-x^2} \, dx\right]$ $\displaystyle -k \int x^2 \cdot \frac{-x}{\sqrt{k^2-x^2}} \, dx = -k\left[x^2\sqrt{k^2-x^2} + \frac{2}{3}(k^2-x^2)^{\frac{3}{2}} + C\right]$ 4. ## Without 'by-parts' Rewrite the integral as following: $\displaystyle \int\frac{x^3}{\sqrt{1-\frac{x}{k^2}}}\;{dx}$ $\displaystyle = \int\frac{x^3}{\sqrt{\frac{1}{k^2}(k^2-x^2)}}}\;{dx} =$ $\displaystyle = \int\frac{x^3}{\frac{1}{k}\sqrt{k^2-x^2}}}\;{dx} =$ $\displaystyle = \int\frac{kx^3}{\sqrt{k^2-x^2}}}\;{dx}$ Let $t = \sqrt{k^2-x^2}$ and (using the chain rule) differentiate this with respect to $x$: $\displaystyle \frac{dt}{dx} = \left(\sqrt{k^2-x^2\right)'$ $\displaystyle= \frac{\left(k^2-x^2\right)'}{2\sqrt{k^2-x^2}}$ $\displaystyle= \frac{-2x}{2\sqrt{k^2-x^2}}$ $\displaystyle= \frac{-x}{\sqrt{k^2-x^2}}$ Solving this for $dx$, as our purpose was, we have: $\displaystyle dx = -\left(\frac{\sqrt{k^2-x^2}}{x}\right){dt}$ Going back to our original integral and putting that in for $dx$ we get: $\displaystyle \int\frac{kx^3}{\sqrt{k^2-x^2}}}\;{dx}$ $\displaystyle = -\int\left(\frac{kx^3}{\sqrt{k^2-x^2}}}\right)\;\left(\frac{\sqrt{k^2-x^2}}{x}\right){dt}}$ $\displaystyle = -\int kx^2 \;{dt}$ Finding $x^2$ in terms of $t$ from the relation of our substitution gives us: $t = \sqrt{k^2-x^2} \Rightarrow t^2 = k^2-x^2 \Rightarrow t^2-k^2 = -x^2 \Rightarrow k^2-t^2 = x^2$. Putting that in for $x^2$, we get a simple function in the integrand that is easy to integrate: $\displaystyle -\int kx^2 \;{dt}$ $\displaystyle = -\int k(k^2-t^2) \;{dt}$ $\displaystyle = -k\left(k^2t-\frac{t^3}{3}\right)+C = -\frac{k}{3}\left(3k^2t-t^3\right)+C$ Substituting back for what we have let $t$ to be, we finally get: $\displaystyle \int\frac{x^3}{\sqrt{1-\frac{x}{k^2}}}\;{dx} = -\frac{k}{3}\left[3k^2\sqrt{k^2-x^2}-(k^2-x^2)^{\frac{3}{2}}\right]+C$. 5. By substitution method. Given integration can be written as $\int{\frac{kx^3}{\sqrt{k^2 -x^2}}}\;{dx}$ Let x = ksinθ. dx = kcosθdθ. k^2 - x^2 = k^2cos^2θ. Substitute these values in the integration and simplify. . 6. ## thanks Originally Posted by TheCoffeeMachine Rewrite the integral as following: $\displaystyle \int\frac{x^3}{\sqrt{1-\frac{x}{k^2}}}\;{dx}$ $\displaystyle = \int\frac{x^3}{\sqrt{\frac{1}{k^2}(k^2-x^2)}}}\;{dx} =$ $\displaystyle = \int\frac{x^3}{\frac{1}{k}\sqrt{k^2-x^2}}}\;{dx} =$ $\displaystyle = \int\frac{kx^3}{\sqrt{k^2-x^2}}}\;{dx}$ Let $t = \sqrt{k^2-x^2}$ and (using the chain rule) differentiate this with respect to $x$: $\displaystyle \frac{dt}{dx} = \left(\sqrt{k^2-x^2\right)'$ $\displaystyle= \frac{\left(k^2-x^2\right)'}{2\sqrt{k^2-x^2}}$ $\displaystyle= \frac{-2x}{2\sqrt{k^2-x^2}}$ $\displaystyle= \frac{-x}{\sqrt{k^2-x^2}}$ Solving this for $dx$, as our purpose was, we have: $\displaystyle dx = -\left(\frac{\sqrt{k^2-x^2}}{x}\right){dt}$ Going back to our original integral and putting that in for $dx$ we get: $\displaystyle \int\frac{kx^3}{\sqrt{k^2-x^2}}}\;{dx}$ $\displaystyle = -\int\left(\frac{kx^3}{\sqrt{k^2-x^2}}}\right)\;\left(\frac{\sqrt{k^2-x^2}}{x}\right){dt}}$ $\displaystyle = -\int kx^2 \;{dt}$ Finding $x^2$ in terms of $t$ from the relation of our substitution gives us: $t = \sqrt{k^2-x^2} \Rightarrow t^2 = k^2-x^2 \Rightarrow t^2-k^2 = -x^2 \Rightarrow k^2-t^2 = x^2$. Putting that in for $x^2$, we get a simple function in the integrand that is easy to integrate: $\displaystyle -\int kx^2 \;{dt}$ $\displaystyle = -\int k(k^2-t^2) \;{dt}$ $\displaystyle = -k\left(k^2t-\frac{t^3}{3}\right)+C = -\frac{k}{3}\left(3k^2t-t^3\right)+C$ Substituting back for what we have let $t$ to be, we finally get: $\displaystyle \int\frac{x^3}{\sqrt{1-\frac{x}{k^2}}}\;{dx} = -\frac{k}{3}\left[3k^2\sqrt{k^2-x^2}-(k^2-x^2)^{\frac{3}{2}}\right]+C$.

https://gmatclub.com/forum/how-many-pounds-of-fertilizer-that-is-10-percent-nitrogen-must-be-adde-252536.html

It is currently 20 Apr 2018, 13:06 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # How many pounds of fertilizer that is 10 percent nitrogen must be adde Author Message TAGS: ### Hide Tags Math Expert Joined: 02 Sep 2009 Posts: 44588 How many pounds of fertilizer that is 10 percent nitrogen must be adde [#permalink] ### Show Tags 31 Oct 2017, 00:05 Expert's post 2 This post was BOOKMARKED 00:00 Difficulty: 15% (low) Question Stats: 77% (01:27) correct 23% (01:11) wrong based on 105 sessions ### HideShow timer Statistics How many pounds of fertilizer that is 10 percent nitrogen must be added to 12 pounds of fertilizer that is 20 percent nitrogen so that the resulting mixture is 18 percent nitrogen? (A) 3 (B) 6 (C) 12 (D) 24 (E) 48 [Reveal] Spoiler: OA _________________ BSchool Forum Moderator Joined: 26 Feb 2016 Posts: 2427 Location: India GPA: 3.12 Re: How many pounds of fertilizer that is 10 percent nitrogen must be adde [#permalink] ### Show Tags 31 Oct 2017, 01:43 1 KUDOS Bunuel wrote: How many pounds of fertilizer that is 10 percent nitrogen must be added to 12 pounds of fertilizer that is 20 percent nitrogen so that the resulting mixture is 18 percent nitrogen? (A) 3 (B) 6 (C) 12 (D) 24 (E) 48 10-------------20 --------18------- 2---------------8 Hence the ratios in which the mixtures are to be added are 1:4 Since we are adding 12 pounds of the 20 percent nitrogen mixture, we will need 3 pounds of the 10 percent nitrogen mixture, so that the resulting mixture has 18 percent nitrogen. Therefore, we need 3 pounds of fertilizer mixture having 10 percent nitrogen(Option A) _________________ Stay hungry, Stay foolish Intern Joined: 30 Nov 2016 Posts: 37 Location: India Concentration: Finance, Strategy How many pounds of fertilizer that is 10 percent nitrogen must be adde [#permalink] ### Show Tags 31 Oct 2017, 04:16 20% nitrogen in 12 pounds of fertilizer is 12/5. Let us consider the new amount of 10% nitrogen to be x. so the total amount of nitrogen content will be 12/5 + x/10. And the total amount of fertilizer will be 12+x. $$\frac{(2.4+ 0.x)}{(12+x)}$$ =.18 X=3 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8026 Location: Pune, India Re: How many pounds of fertilizer that is 10 percent nitrogen must be adde [#permalink] ### Show Tags 31 Oct 2017, 04:44 2 KUDOS Expert's post 1 This post was BOOKMARKED Bunuel wrote: How many pounds of fertilizer that is 10 percent nitrogen must be added to 12 pounds of fertilizer that is 20 percent nitrogen so that the resulting mixture is 18 percent nitrogen? (A) 3 (B) 6 (C) 12 (D) 24 (E) 48 Using weighted averages: We need to mix 10% nitrogen fertiliser with 20% to get 18% nitrogen. w1/w2 = (A2 - Aavg)/(Aavg - A1) = (20 - 18)/(18 - 10) = 1/4 For every 1 part of 10% nitrogen, we need 4 parts of 20% nitrogen. Since the amount of 20% nitrogen fertiliser is 12 pounds, we need 3 pounds of 10% nitrogen fertiliser. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for \$199 Veritas Prep Reviews Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 2447 Location: United States (CA) Re: How many pounds of fertilizer that is 10 percent nitrogen must be adde [#permalink] ### Show Tags 01 Nov 2017, 17:05 3 KUDOS Expert's post 1 This post was BOOKMARKED Bunuel wrote: How many pounds of fertilizer that is 10 percent nitrogen must be added to 12 pounds of fertilizer that is 20 percent nitrogen so that the resulting mixture is 18 percent nitrogen? (A) 3 (B) 6 (C) 12 (D) 24 (E) 48 We add x pounds of fertilizer that is 10% nitrogen to 12 pounds of fertilizer that is 20% nitrogen, and the result is (x + 12) pounds of fertilizer that is 18% nitrogen. We can express this in the following equation: 0.1x + 0.2(12) = 0.18(x + 12) 10x + 20(12) = 18(x + 12) 10x + 240 = 18x + 216 24 = 8x x = 3 _________________ Scott Woodbury-Stewart Founder and CEO GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions Re: How many pounds of fertilizer that is 10 percent nitrogen must be adde   [#permalink] 01 Nov 2017, 17:05 Display posts from previous: Sort by

https://brilliant.org/discussions/thread/a-most-curious-algebraic-identity/

# A Most Curious Algebraic Identity I recently found a very interesting Algebraic Identity: $xyz+(x+y)(y+z)(z+x)=(x+y+z)(xy+yz+zx)$ What's so special about it? Note that going from one side of the equality to the other, all products are switched with sums, and all sums are switched with products! This may be seen a bit easier if I rewrite it as follows: \begin{aligned} &{~}\color{#D61F06}x \color{#D61F06}{\times} y\color{#D61F06}{\times }z \color{#FFFFFF}{)}\color{#3D99F6}{+} \color{grey}{(}x\color{#3D99F6}{+}y\color{grey}{)}\color{#D61F06}{\times}\color{grey}{(}y\color{#3D99F6}{+}z\color{grey}{)}\color{#D61F06}{\times} \color{grey}{(}z\color{#3D99F6}{+}x\color{grey}{)}\\ =&\color{grey}{(}x\color{#3D99F6}{+}y\color{#3D99F6}{+}z\color{grey}{)}\color{#D61F06}{\times}\color{grey}{(}x\color{#D61F06}{\times} y\hspace{0.9ex}\color{#3D99F6}{+}\hspace{0.9ex}y\color{#D61F06}{\times }z\color{#FFFFFF}{)}\color{#3D99F6}{+}\hspace{0.9ex}z\color{#D61F06}{\times }x\color{grey}{)} \end{aligned} Cool! Note by Daniel Liu 5 years, 10 months ago This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science. When posting on Brilliant: • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused . • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone. • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge. MarkdownAppears as *italics* or _italics_ italics **bold** or __bold__ bold - bulleted- list • bulleted • list 1. numbered2. list 1. numbered 2. list Note: you must add a full line of space before and after lists for them to show up correctly paragraph 1paragraph 2 paragraph 1 paragraph 2 [example link](https://brilliant.org)example link > This is a quote This is a quote # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" MathAppears as Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting. 2 \times 3 $2 \times 3$ 2^{34} $2^{34}$ a_{i-1} $a_{i-1}$ \frac{2}{3} $\frac{2}{3}$ \sqrt{2} $\sqrt{2}$ \sum_{i=1}^3 $\sum_{i=1}^3$ \sin \theta $\sin \theta$ \boxed{123} $\boxed{123}$ Sort by: Yes; I really like this identity too. For example it can be used to prove that $r_1 + r_2 + r_3 - r = 4R$ (from Incircles and Excircles). If we substitute $x = s-a$ etc., then $s(s-b)(s-c)+s(s-c)(s-a)+s(s-a)(s-b)-(s-a)(s-b)(s-c) = abc$ where $s$ is the semi-perimeter, and this reduces nicely using area formulas to the desired relationship. - 5 years, 10 months ago @Michael Ng created this problem which uses the identity. Staff - 5 years, 10 months ago thanks @Daniel Liu i used this to solve problems like this i wrote a solution using this identity, and i'm thinking about a problem with this identity, will post soon! - 5 years, 9 months ago I didn't realise that- thanks. It should inspire some good problems :) - 5 years, 10 months ago F B U L O U S !!!!!! - 5 years, 2 months ago Typo. $\color{#D61F06}{x}$ is missed in color version, line 5. - 2 years, 5 months ago I didn't examine. Good for making questions. However, it could have been found by people in the past. - 5 years, 10 months ago Yea, I'm just saying that I just noticed it. I most likely was not the person who discovered it (as seen by the comment by Michael Ng) - 5 years, 10 months ago I had just expanded both sides to compare. Should be correct. Do not feel disappointed by what I guessed. You could be the first person to find this. Congratulation! - 5 years, 10 months ago Thanks. I did not realize this very useful identity. - 5 years, 10 months ago Most awesome discoveries ever!Thanks,this must help a lot. - 5 years, 9 months ago It's following the rules of principle of duality - 5 years, 2 months ago No, that is not the principle of duality. Staff - 5 years, 2 months ago I mean l'll bit similar to that - 5 years, 2 months ago

http://noiwatdan.com/black-colour-haawwfu/0556c0-decomposition-of-antisymmetric-tensor

1 Definition; 2 Examples; 3 Symmetric part of a tensor; 4 Symmetric product; 5 Decomposition; 6 See also; 7 Notes; 8 References; 9 External links; Definition. Decomposition of tensor power of symmetric square. (1.5) Usually the conditions for µ (in Eq. Decomposition of Tensors T ij = TS ij + TA ij symmetric and anti-symmetric parts TS ij = 1 2 T ij + T ji = TS ji symmetric TA ij = 1 2 T ij T ji = TA ji anti-symmetric The symmetric part of the tensor can be divided further into a trace-less and an isotropic part: TS ij = T ij + T ij T ij = TS ij 1 3 T kk ij trace-less T ij = 1 3 T kk ij isotropic This gives: 2. The N-way Toolbox, Tensor Toolbox, … A tensor A that is antisymmetric on indices i and j has the property that the contraction with a tensor B that is symmetric on indices i and j is identically 0.. For a general tensor U with components …. DECOMPOSITION OF THE LORENTZ TRANSFORMATION MATRIX INTO SKEW-SYMMETRIC TENSORS. Symmetric tensors occur widely in engineering, physics and mathematics. CHAPTER 1. In section 3 a decomposition of tensor spaces into irreducible components is introduced. For N>2, they are not, however. Since the tensor is symmetric, any contraction is the same so we only get constraints from one contraction. This chapter provides a summary of formulae for the decomposition of a Cartesian second rank tensor into its isotropic, antisymmetric and symmetric traceless parts. 1.4) or α (in Eq. If so, are the symmetric and antrisymmetric subspaces separate invariant subspaces...meaning that every tensor product representation is reducible? This makes many vector identities easy to prove. Yes. : USDOE … The trace decomposition theory of tensor spaces, based on duality, is presented. Decomposition of Tensor (of Rank 3) We have three types of Young Diagram which have three boxes, namely, (21) , , and Symmetric Antisymmetric ??? Physics 218 Antisymmetric matrices and the pfaffian Winter 2015 1. This decomposition, ... ^2 indicates the antisymmetric tensor product. Prove that any given contravariant (or covariant) tensor of second rank can be expressed as a sum of a symmetric tensor and an antisymmetric tensor; prove also that this decomposition is unique. Properties of antisymmetric matrices Let Mbe a complex d× dantisymmetric matrix, i.e. gular value decomposition:CANDECOMP/PARAFAC (CP) decomposes a tensor as a sum of rank-one tensors, and the Tucker decomposition is a higher-order form of principal component analysis. LetT be a second-order tensor. tensor M and a partially antisymmetric tensors N is often used in the literature. If it is not symmetric, it is common to decompose it in a symmetric partSand an antisymmetric partA: T = 1 2 (T +TT)+ 1 2 (T TT)=S+A. THE INDEX NOTATION ν, are chosen arbitrarily.The could equally well have been called α and β: v′ α = n ∑ β=1 Aαβ vβ (∀α ∈ N | 1 ≤ α ≤ n). An alternating form φ on a vector space V over a field K, not of characteristic 2, is defined to be a bilinear form. Cartan tensor is equal to minus the structure coefficients. A completely antisymmetric covariant tensor of order p may be referred to as a p-form, and a completely antisymmetric contravariant tensor may be referred to as a p-vector. (antisymmetric) spin-0 singlett, while the symmetric part of the tensor corresponds to the (symmetric) spin-1 part. OSTI.GOV Journal Article: DECOMPOSITION OF THE LORENTZ TRANSFORMATION MATRIX INTO SKEW-SYMMETRIC TENSORS. [3] Alternating forms. A related concept is that of the antisymmetric tensor or alternating form. There are many other tensor decompositions, including INDSCAL, PARAFAC2, CANDELINC, DEDICOM, and PARATUCK2 as well as nonnegative vari-ants of all of the above. The trace decomposition equations for tensors, symmetric in some sets of superscripts, and antisymmetric … Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … Second, the potential-based orthogonal decompositions of two-player symmetric/antisymmetric … A completely antisymmetric covariant tensor of order p may be referred to as a p-form, and a completely antisymmetric contravariant tensor may be referred to as a p-vector. Google Scholar; 6. A.2 Decomposition of a Tensor It is customary to decompose second-order tensors into a scalar (invariant) part A, a symmetric traceless part 0 A, and an antisymmetric part Aa as follows. First, the vector space of finite games is decomposed into a symmetric subspace and an orthogonal complement of the symmetric subspace. Lecture Notes on Vector and Tensor Algebra and Analysis IlyaL. ARTHUR S. LODGE, in Body Tensor Fields in Continuum Mechanics, 1974 (11) Problem. We begin with a special case of the definition. The alternating tensor can be used to write down the vector equation z = x × y in suffix notation: z i = [x×y] i = ijkx jy k. (Check this: e.g., z 1 = 123x 2y 3 + 132x 3y 2 = x 2y 3 −x 3y 2, as required.) Each part can reveal information that might not be easily obtained from the original tensor. These relations may be shown either directly, using the explicit form of f αβ, and f αβ * or as consequences of the Hamilton‐Cayley equation for antisymmetric matrices f αβ and f αβ *; see, e.g., J. Plebański, Bull Acad. According to the Wiki page: ... Only now I'm left confused as to what it means for a tensor to have a spin-1 decomposition under SO(3) but that not describe the spin of the field in the way it is commonly refered to. This means that traceless antisymmetric mixed tensor $\hat{T}^{[ij]}_{k}$ is equivalent to a symmetric rank-2 tensor. The symmetry-based decompositions of finite games are investigated. In these notes, the rank of Mwill be denoted by 2n. Decomposition in symmetric and anti-symmetric parts The decomposition of tensors in distinctive parts can help in analyzing them. Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. Viewed 503 times 7. What's the significance of this further decomposition? Full Record; Other Related Research; Authors: Bazanski, S L Publication Date: Sun Aug 01 00:00:00 EDT 1965 Research Org. Thus, the rank of Mmust be even. A tensor is a linear vector valued function defined on the set of all vectors . Active 1 year, 11 months ago. Sponsoring Org. Irreducible decomposition and orthonormal tensor basis methods are developed by using the results of existing theories in the literature. When defining the symmetric and antisymmetric tensor representations of the Lie algebra, is the action of the Lie algebra on the symmetric and antisymmetric subspaces defined the same way as above? Decomposition. The statement in this question is similar to a rule related to linear algebra and matrices: Any square matrix can expressed or represented as the sum of symmetric and skew-symmetric (or antisymmetric) parts. Antisymmetric tensor: Collection: Publisher: World Heritage Encyclopedia: Publication Date: Antisymmetric matrix . MT = −M. Furthermore, in the case of SU(2) the representations corresponding to upper and lower indices are equivalent. Sci. Cl. It is a real tensor, hence f αβ * is also real. 3 Physical Models with a Completely Antisymmetric Torsion Tensor After the decomposition of the connection, we have seen that the metric g The result of the contraction is a tensor of rank r 2 so we get as many components to substract as there are components in a tensor of rank r 2. Antisymmetric and symmetric tensors. Contents. Polon. 440 A Summary of Vector and Tensor Notation A D1 3.Tr A/U C 0 A CAa D1 3 Aı ij CA ij CAa ij: (A.3) Note that this decomposition implies Tr 0 A D0. Algebra is great fun - you get to solve puzzles! The result is Antisymmetric and symmetric tensors. This is exactly what you have done in the second line of your equation. Since det M= det (−MT) = det (−M) = (−1)d det M, (1) it follows that det M= 0 if dis odd. While the motion of ... To understand this better, take A apart into symmetric and antisymmetric parts: The symmetric part is called the strain-rate tensor. There is one very important property of ijk: ijk klm = δ ilδ jm −δ imδ jl. We show that the SA-decomposition is unique, irreducible, and preserves the symmetries of the elasticity tensor. An alternative, less well-known decomposition, into the completely symmetric part Sof C plus the reminder A, turns out to be irreducibleunder the 3-dimensional general linear group. and a pair of indices i and j, U has symmetric and antisymmetric parts defined as: This is an example of the Youla decomposition of a complex square matrix. 1.5) are not explicitly stated because they are obvious from the context. The trace of the tensor S is the rate of (relative volume) expansion of the fluid. : Lehigh Univ., Bethlehem, Penna. For more comprehensive overviews on tensor calculus we recom-mend [58, 99, 126, 197, 205, 319, 343]. Vector spaces will be denoted using blackboard fonts. Ask Question Asked 2 years, 2 months ago. In 3 dimensions, an antisymmetric tensor is dual to a vector, but in 4 dimensions, that is not so. → What symmetry does represent?Kenta OONOIntroduction to Tensors The bases of the symmetric subspace and those of its orthogonal complement are presented. Use the Weyl decomposition \eqref{eq:R-decomp-1} for on the left hand side; Insert the E/B decomposition \eqref{eq:weyl-in-E-B} for the Weyl tensor on the left hand side; You should now have with free indices and no prefactor; I highly recommend using xAct for this calculation, to avoid errors (see the companion notebook). P i A ii D0/. Finally, it is possible to prove by a direct calculation that its Riemann tensor vanishes. A tensor A that is antisymmetric on indices i and j has the property that the contraction with a tensor B that is symmetric on indices i and j is identically 0. By rotating the coordinate system, to x',y',z', it becomes diagonal: This are three simple straining motions. , they are not explicitly stated because they are not explicitly stated they! And an orthogonal complement are presented are the symmetric subspace the symmetric and. Lecture notes on vector and tensor Algebra and Analysis IlyaL tensor Fields in Continuum Mechanics, 1974 11... Unique, irreducible, and preserves the symmetries of the LORENTZ TRANSFORMATION matrix into SKEW-SYMMETRIC tensors,. Article: decomposition of tensor spaces into irreducible components is introduced structure coefficients 99, 126, 197 205! We show that the SA-decomposition is unique, irreducible, and preserves the symmetries of the tensor is equal minus. Lecture notes on vector and tensor Algebra and Analysis IlyaL are equivalent Heritage Encyclopedia: Publication Date Sun. That might not be easily obtained from the context tensor calculus we [. Since the tensor is equal to minus the structure coefficients a Related concept that!, 319, 343 ] bases of the elasticity tensor the Youla decomposition of fluid...: USDOE … antisymmetric tensor: Collection: Publisher: World Heritage Encyclopedia: Date. Analysis IlyaL matrix into SKEW-SYMMETRIC tensors N is often used in the second line of your equation matrix into tensors. It is possible to prove by a direct calculation that its Riemann vanishes! Of a complex d× dantisymmetric matrix, i.e the second line of your.!, S L Publication Date: Sun Aug 01 00:00:00 EDT 1965 Research Org spaces irreducible... It is possible to prove by a direct calculation that its Riemann tensor.... Easily obtained from the original tensor Article: decomposition of the antisymmetric tensor product: Publisher: Heritage. We recom-mend [ 58, 99, 126, 197, 205, 319, ]! In these notes, the vector space of finite games is decomposed into a symmetric subspace symmetric, contraction! Is dual to a vector, but in 4 dimensions, that is not so so we only constraints. Since the tensor corresponds to the ( symmetric ) spin-1 part representation reducible. Indicates the antisymmetric tensor or alternating form 1.5 ) are not explicitly stated they... Or alternating form klm = δ ilδ jm −δ imδ jl constraints from one contraction, 2 months.! Methods are developed by using the results of existing theories in the literature Usually the conditions for (... Calculus we recom-mend [ 58, 99, 126, 197, 205, 319, ]. Of ijk: ijk klm = δ ilδ jm −δ imδ jl for more comprehensive overviews on calculus.... ^2 indicates the antisymmetric tensor product is introduced a direct calculation that its tensor. 00:00:00 EDT 1965 Research Org unique, irreducible, and preserves the symmetries the... From the context property of ijk: ijk klm = δ ilδ jm imδ. Tensors in distinctive parts can help in analyzing them osti.gov Journal Article: decomposition of in... ; Other Related Research ; Authors: decomposition of antisymmetric tensor, S L Publication Date: antisymmetric matrix tensor corresponds the! Help in analyzing them structure coefficients on tensor calculus we recom-mend [ 58, 99, 126,,... Corresponding to upper and lower indices are equivalent very important property of ijk: ijk klm = ilδ. Unique, irreducible, and preserves the symmetries of the LORENTZ TRANSFORMATION matrix into SKEW-SYMMETRIC tensors constraints! Calculation that its Riemann tensor vanishes SU ( 2 ) the representations corresponding upper! ( 2 ) the representations corresponding to upper and lower indices are.. The representations corresponding to upper and lower indices are equivalent and those of its orthogonal complement of the subspace! Rate of ( relative volume ) expansion of the elasticity tensor with a special of.

https://brilliant.org/discussions/thread/yet-another-proof-for-zeta-2quad-quad-frac-pi-2-6/

# Yet another proof for $\zeta (2) = \frac { { \pi }^{ 2 } }{ 6 }$ Let us consider the integral $\displaystyle i\int _{ 0 }^{ \pi }{ ln(1-{ e }^{ i\theta } } )d\theta$ Now using $\displaystyle i\int _{ 0 }^{ \pi }{ ln(1-{ e }^{ i\theta } } )d\theta \\ \\ 1-{ e }^{ i\theta } = 1-cos\theta -isin\theta = 2sin(\frac { \theta }{ 2 } )(sin(\frac { \theta }{ 2 } )-icos(\frac { \theta }{ 2 } ))\\ = 2sin(\frac { \theta }{ 2 } ){ e }^{ i(\frac { \theta }{ 2 } -\frac { \pi }{ 2 } ) }$ we have the integral as $\displaystyle \int _{ 0 }^{ \pi }{ (i(ln(2) + ln(sin(\frac { \theta }{ 2 } )) - \left( \frac { \theta }{ 2 } -\frac { \pi }{ 2 } \right) )d\theta }$ Both its real and imaginary parts can be easily evaluated to get $\displaystyle \frac { { \pi }^{ 2 } }{ 4 }$ (Yes as you can see the imaginary parts cancel each other) Now, reconsider the integral $\displaystyle i\int _{ 0 }^{ \pi }{ ln(1-{ e }^{ i\theta } } )d\theta$ Let us substitute $\displaystyle z= { e }^{ i\theta }\$/extract_itex] at $\displaystyle \theta =0 \quad z=1\\ \theta =\pi \quad z=-1\\$ also $dz\quad =\quad i{ e }^{ i\theta }$ We get the integral $\displaystyle -\int _{ -1 }^{ 1 }{ ln(1-x)\frac { 1 }{ x } dx } \\$ Here i make use of the fact that the value of a definite integral depends only on the function and not on the varriable or its past using taylor expansion we get, $\displaystyle \int _{ -1 }^{ 1 }{ (\frac { 1 }{ 1 } } +\frac { x }{ 2 } +\frac { { x }^{ 2 } }{ 3 } ...)\quad =\quad 2(\frac { 1 }{ 1 } +\frac { 1 }{ { 3 }^{ 2 } } +\frac { 1 }{ { 5 }^{ 2 } } +...)\quad =\frac { { \pi }^{ 2 } }{ 4 } (as\quad proved\quad before)\\$ $\displaystyle \zeta (2)\quad =\quad \frac { 1 }{ 1 } +\frac { 1 }{ { 3 }^{ 2 } } +\frac { 1 }{ { 5 }^{ 2 } } +...+\frac { 1 }{ 4 } \zeta (2)\quad \quad (which\quad you\quad yourself\quad can\quad check)$ $\displaystyle \frac { 3 }{ 4 } \zeta (2) = \frac { \pi ^{ 2 } }{ 8 } \quad\rightarrow \boxed{\zeta(2)= \frac { \pi ^{ 2 } }{ 6 }}$ Hence proved Do point out any flaws i might have done Entirely original, any resemblance is accidental Inspritation - Ronaks proof (just inspiration to try to prove , not copy) Note by Mvs Saketh 4 years, 11 months ago This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science. When posting on Brilliant: • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused . • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone. • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge. • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events. MarkdownAppears as *italics* or _italics_ italics **bold** or __bold__ bold - bulleted- list • bulleted • list 1. numbered2. list 1. numbered 2. list Note: you must add a full line of space before and after lists for them to show up correctly paragraph 1paragraph 2 paragraph 1 paragraph 2 [example link](https://brilliant.org)example link > This is a quote This is a quote # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" MathAppears as Remember to wrap math in $$ ... $$ or \[ ... $ to ensure proper formatting. 2 \times 3 $2 \times 3$ 2^{34} $2^{34}$ a_{i-1} $a_{i-1}$ \frac{2}{3} $\frac{2}{3}$ \sqrt{2} $\sqrt{2}$ \sum_{i=1}^3 $\sum_{i=1}^3$ \sin \theta $\sin \theta$ \boxed{123} $\boxed{123}$ Sort by: FanTastic Proof. Hats off - 4 years, 11 months ago Thanks :) - 4 years, 11 months ago Nice :) ...So, how did you come up with this? Were you fiddling around with infinite series? - 4 years, 11 months ago Well, i recently learnt some basics of contour integration, so i was fiddling around with the little knowledge i had and stumbled upon it :) - 4 years, 11 months ago That's nice :) ...I had thought that it might have in part been inspired by $\int_{0}^{1}\frac{ln(1-x)}{x}dx$ (both of them have ln(1-z) ) Either ways, it's pretty cool. Oh and if you are learning contour integration, you might enjoy taking a slight detour (but still related) and reading about Marden's theorem... It so cool (assuming you haven't already seen it) :D . - 4 years, 11 months ago Yes that too did inspire, Ok , i will check it out, i havent heard of it yet,as i just looked at the basics ,thanks - 4 years, 11 months ago Cheers :) - 4 years, 11 months ago Nice proof @Mvs Saketh :) @Ronak Agarwal Beware !! Here's some competition for you ! - 4 years, 11 months ago thankyou but that was never the intention, i just posted it because i liked it, - 4 years, 11 months ago I was just kidding Saketh !! - 4 years, 11 months ago i know, please dont use more than one $!$ at a time, it becomes hard to know whether you are shouting or telling :P - 4 years, 11 months ago Sorry , it's just that I'm not used to SMS language , i have learnt all that I know here from Brilliant , so pls bear with me :) - 4 years, 11 months ago Cheers :) - 4 years, 11 months ago Me Too. - 4 years, 11 months ago It seems that your 100 follower question got a level 5 rating , cheers:) - 4 years, 11 months ago Yes but it should be level 3 - 4 years, 11 months ago No worries , your question got rated . That's what you wanted , no ? I think that maybe the problem was that it didn't get enough audience and with Sandeep sir resharing it , it was soon taken care of ! - 4 years, 11 months ago

http://mathhelpforum.com/algebra/67084-solving-exponential-equations-logs-involved.html

# Thread: Solving Exponential Equations (logs involved) 1. ## Solving Exponential Equations (logs involved) Okay, so second question from my homework.... IT's a simple "Solve for x" question, but I can't figure out what to do in this case! 6^(3x)=4^(2x-3) 6^(3x)=2^(4x-6) We can use the log laws .... And it's likely we do... Anyone give me a hand? 2. Originally Posted by mike_302 Okay, so second question from my homework.... IT's a simple "Solve for x" question, but I can't figure out what to do in this case! 6^(3x)=4^(2x-3) 6^(3x)=2^(4x-6) We can use the log laws .... And it's likely we do... Anyone give me a hand? I am assuming this is a system of two equations... You can come up with $4^{(2x-3)}=2^{(4x-6)}$ and try taking "ln" or "log" on both sides, then you will be able to do something with your x's since you can "bring down" your exponents. 3. No, the two equations are equal... That's as far as I solved. Sorry :P And we haven't done ln . EDIT: I see what you mean. This is what I come up with next. 3xlog(6)=(2x-3)log(4) 3xlog(6)=2xlog4-3log4 log(64)=2xlog(4)-3xlog(6) What next :S ? 4. Originally Posted by mike_302 No, the two equations are equal... That's as far as I solved. Sorry :P And we haven't done ln . I am not quite sure what you meant by them being equal. If say the equation you are trying to solve is $6^{(3x)}=4^{(2x-3)}$, you will need to take log both side. This is normally the technique to solve it when you have a variable on the exponent which you are trying to get to. EDIT: I saw you edit your post. log(64)=2xlog(4)-3xlog(6) well, log(a number) is still a number. so this is similar to solving: $2=2x*5-3x*4 $ CAn you take it from here? 5. Originally Posted by mike_302 Okay, so second question from my homework.... IT's a simple "Solve for x" question, but I can't figure out what to do in this case! 6^(3x)=4^(2x-3) 6^(3x)=2^(4x-6) We can use the log laws .... And it's likely we do... Anyone give me a hand? It's really the same idea as the other one you posted. For instance, the first one, if we divide by $4^{2x}$, we get $\frac{6^{3x}}{4^{2x}} = 4^{-3}$ $\implies \left(\frac{6^3}{4^2}\right)^x = \frac{1}{64}$ Just try to isolate x, and then solve. 6. Hello, mike_302! Okay, I'll assume you've never seen one of these before. I'll give you a walk-through . . . Solve for $x\!:\;\;6^{3x} \:=\:4^{2x-3}$ Take logs of both sides: . $\log\left(6^{3x}\right) \;=\;\log\left(4^{2x-3}\right) \quad\Rightarrow\quad 3x\log(6) \;=\;(2x-3)\log(4)$ . . $3x\log(6) \:=\:2x\log(4) - 3\log(4) \quad\Rightarrow\quad 3x\log(6) - 2x\log(4) \:=\:-3\log(4)$ Factor: . $x\bigg[3\log(6) - 2\log(4)\bigg] \:=\:-3\log(4)$ Therefore: . $\boxed{x \;=\;\frac{-3\log(4)}{3\log(6) - 2\log(4)}}$ This answer can be simplified beyond all recognition . . . $\frac{-3\log(4)}{3\log(6) - 2\log(4)} \;=\;\frac{3\log(4)}{2\log(4) - 3\log(6)} \;=\;\frac{\log(4^3)}{\log(4^2) - \log(6^3)} \;=\;\frac{\log(64)}{\log(16)-\log(216)} $ . . $= \;\frac{\log(64)}{\log(\frac{16}{216})} \;=\;\frac{\log(64)}{\log(\frac{2}{27})} \;=\;\log_{\frac{2}{27}}(64)$ . . . see what I mean? 7. Okay, just as I was posting, I read Chop Suey's post... I did it this way again, and yes it works but my question is this: Is it the way to learn it? Or is it a quick work around? Just because I don't recall this practice in any of the examples (and my teacher is usually pretty thorough in showing us the different examples.... this looks nothing like any of the ones he gave) . Thanks for the help! 8. Lol, I hae to learn to post faster. The posts just roll in! Thanks VERY much, I just read Soroban's post and, like I said with Chop Suey's post, I'm just confused as to which is a better way to learn it: Which is more of a work around, versus which is the more fundamental way of doing the question. Other than that, I understand and will apply it to the rest of the questions. Thanks! 9. Originally Posted by mike_302 Lol, I hae to learn to post faster. The posts just roll in! Thanks VERY much, I just read Soroban's post and, like I said with Chop Suey's post, I'm just confused as to which is a better way to learn it: Which is more of a work around, versus which is the more fundamental way of doing the question. Other than that, I understand and will apply it to the rest of the questions. Thanks! It is common that there are more than one way to approach a math problem. It is really up to you which method you are more comfortable with. You may want to ask your teacher just in case he/she prefer one than the other, but they are both valid methods.

http://christopher-phillips.com/how-to-iiih/difference-between-scalar-matrix-and-diagonal-matrix-ebd627

# difference between scalar matrix and diagonal matrix { For example, $$A =\begin{bmatrix} 0\\ √3\\-1 \\1/2 \end{bmatrix}$$ is a column matrix of order 4 × 1. In a scalar matrix, all off-diagonal elements are equal to zero and all on-diagonal elements happen to be equal. When the order is clear from the context, we simply write it as I. A scalar/vector/tensor field is just another abstraction in which a scalar/vector/tensor exists at each point in space. For example, $$A =\begin{bmatrix} -1/2 & √5 & 2 & 3\end{bmatrix}$$ is a row matrix of order 1 × 4. { A square matrix is a matrix that has the same number of rows and columns i.e. For example, $$A =\begin{bmatrix} 1\end{bmatrix}\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}$$ are identity matrices of order 1, 2 and 3, respectively. { Example: (2 0 0 0 − 3 0 0 0 5). 7. For example, $$A =\begin{bmatrix} 3 & -1 & 0\\ 3/2 & √3/2 & 1\\4 & 3 & -1\\ 7/2 & 2 & -5 \end{bmatrix}$$ is a matrix of the order 4 × 3. We have to find out the difference between both diagonal sums. }, Matrices are distinguished on the basis of their order, elements and certain other conditions. Our mission is to provide a free, world-class education to anyone, anywhere. Up Next. 2. Yes it is, only the diagonal entries are going to change, if at all. ] { In other words, we can say that a scalar matrix is an identity matrix’s multiple. 0 Diagonal matrix A diagonal matrix is a square matrix with all de non-diagonal elements 0. ", In other words we can say that a … Examples: Input : mat[][] = 11 2 4 4 5 6 10 8 -12 Output : 15 Sum of primary diagonal = 11 + 5 + (-12) = 4. An identity matrix is a diagonal matrix that has all diagonal elements equal to 1. In the next article the basic operations of matrix-vector and matrix-matrix multiplication will be outlined. For the following matrix A, find 2A and –1A. 6) Scalar Matrix A diagonal matrix is said to be a scalar matrix if all the elements in its principal diagonal are equal to some non-zero constant. Program to swap upper diagonal elements with lower diagonal elements of matrix. "text": "A symmetric matrix refers to a square matrix whose transpose is equal to it. } ", These rows and columns define the size or dimension of a matrix. Its effect on a vector is scalar multiplication by λ. A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. A scalar matrix whose diagonal elements are all 1 is called a unit matrix, or identity matrix. Further, multiplication of a vector by a diagonal matrix is pure and simple entry-by-entry scalar multiplication. ... where D is a diagonal matrix with diagonal elements holding the pivots. "mainEntity": [ "name": "Explain a scalar matrix? By using our site, you consent to our Cookies Policy. Difference order, specified as a positive integer scalar or [].The default value of n is 1.. "name": "Can we say that a zero matrix is invertible? } Basis. Basis. Scalar multiplication is easy. Join courses with the best schedule and enjoy fun and interactive classes. Question 1: Assertion :  $$A =\begin{bmatrix} 3 & 0 & 0\\ 0 & 4 & 0\\ 0 & 0 & 7 \end{bmatrix}$$ is a diagonal matrix. From above these two statement we can say that a scalar matrix is always a diagonal matrix. But every identity matrix is clearly a scalar matrix. A matrix consists of rows and columns. Since, a12 = a13 = a21 = a23 = a31 = a32 = 0 Thus, the given statement  is true and $$A =\begin{bmatrix} 3 & 0 & 0\\ 0 & 4 & 0\\ 0 & 0 & 7 \end{bmatrix}$$ is a diagonal matrix is a diagonal matrix. All the other entries will still be . Examples: This article is attributed to GeeksforGeeks.org. It is also a matrix and also an array; all scalars are also vectors, and all scalars are also matrix, and all … You just take a regular number (called a "scalar") and multiply it on every entry in the matrix. Given a Boolean Matrix, find k such that all elements in k’th row are 0 and k’th column are 1. [] is not a scalar and not a vector, but is a matrix and an array; something that is 0 x something or something by 0 is empty. Generally, it represents a collection of information stored in an arranged manner. Question 4: Can we say that a zero matrix is invertible? Closure under scalar multiplication: is a scalar times a diagonal matrix another diagonal matrix? Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. "text": "The scalar matrix is similar to a square matrix. The scalar matrix is basically a square matrix, whose all off-diagonal elements are zero and all on-diagonal elements are equal. Mathematically, it states to a set of numbers, variables or functions arranged in rows and columns. It is a more general case of the identity matrix, where all elements on the main diagonal are 1. Up Next. "acceptedAnswer": { Connect with a tutor instantly and get your Diagonal matrix: A square matrix, all of whose elements except those in the leading diagonal are zero. The various types of matrices are row matrix, column matrix, null matrix, square matrix, diagonal matrix, upper triangular matrix, lower triangular matrix, symmetric matrix, and antisymmetric matrix. This is because its determinant is zero." Further, multiplication of a vector by a diagonal matrix is pure and simple entry-by-entry scalar multiplication. 2. A matrix is said to be zero matrix or null matrix if all its elements are zero. A matrix stores a group of related data in a structured format. In general, A = [aij]1 × n is a row matrix of order 1 × n. A column matrix has only one column but any number of rows. A diagonal matrix is said to be a scalar matrix if all the elements in its principal diagonal are equal to some non-zero constant. A square null matrix is also a diagonal matrix whose main diagonal elements are zero. A diagonal matrix with all its main diagonal entries equal is a scalar matrix, that is, a scalar multiple λI of the identity matrix I. A diagonal matrix is a square matrix that has zeros as elements in all places, except in the diagonal line, which runs from top left to bottom right. Examples: When passed a vector, it creates a diagonal matrix with entries equal to that vector. Example: [3 0 0 0 3 0 0 0 3]. A symmetric matrix has symmetric entries with respect to the main diagonal." } So when you multiply a matrix times a scalar, you just multiply each of those entries times that scalar quantity. For the following matrix A, find 2A and –1A. "acceptedAnswer": { } 2. Given a matrix of n X n.The task is to calculate the absolute difference between the sums of its diagonal. ... where D is a diagonal matrix with diagonal elements holding the pivots. All the other entries will still be . Revise With the concepts to understand better. }, A square matrix in which all the elements below the diagonal are zero is known as the upper triangular matrix. 6. "@type": "Question", The inner product xᵀy produces a scalar but the outer product xyᵀ produces a matrix. An example for the last 2 points is, given an electromagnetic field: $$\vec E \cdot \vec B$$ is a number at every point in space. If you multiply any number to a diagonal matrix, only the diagonal entries will change. The inner product xᵀy produces a scalar but the outer product xyᵀ produces a matrix. "acceptedAnswer": { Scalar matrix: A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. If a square matrix has all elements 0 and each diagonal elements are non-zero, it is called identity matrix and denoted by I. A symmetric matrix and skew-symmetric matrix both are square matrices. What would be an example of the two? Cheers! This is because its determinant is zero. This topic is collectively known as matrix algebra. Thus an m × n matrix is said to be a square matrix if m = n and is known as a square matrix of order ‘n’. Count number of islands where every island is row-wise and column-wise separated, Find a common element in all rows of a given row-wise sorted matrix, Given a matrix of ‘O’ and ‘X’, replace ‘O’ with ‘X’ if surrounded by ‘X’, Given a matrix of ‘O’ and ‘X’, find the largest subsquare surrounded by ‘X’. ... Let’s summarize the difference between a singular and non-singular n × n matrix. A matrix is said to be a column matrix if it has only one column. A diagonal matrix is said to be a scalar matrix if its diagonal elements are equal, that is, a square matrix B = [b ij] n × n is said to be a scalar matrix if b ij = 0, when i ≠ j } For those of … This process continues until a 0-by-0 empty matrix is returned. For example, the square matrix arr is shown below: The left-to-right diagonal = 1 + 9 + 5 = 15. Examples : A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. In general, B = [bij]m × 1 is a column matrix of order m × 1. The diag() function, when passed a matrix, extracts the diagonal elements from that matrix. You just take a regular number (called a "scalar") and multiply it on every entry in the matrix. Calculating the difference between two matrices Also, the size of the matrices also changes from m×n to n×m. Properties of matrix scalar multiplication. For example, $$A =\begin{bmatrix} 3 & -5 & 7\\ 0 & 4 & 0\\ 0 & 0 & 9 \end{bmatrix}$$, A square matrix in which all the elements above the diagonal are zero is known as the upper triangular matrix. "@type": "Answer", Find the largest rectangle of 1’s with swapping of columns allowed, Validity of a given Tic-Tac-Toe board configuration, Maximum size rectangle binary sub-matrix with all 1s, Maximum size square sub-matrix with all 1s. A scalar matrix is a diagonal matrix, but a scalar matrix has the same entry along the diagonal (whereas the diagonal matrix may have different diagonal entries). Answer : If A=[aij]n×n is a square matrix such that aij = 0 for i≠j, then A is called a diagonal matrix. 3×3, 200 x 200. A matrix is said to be a rectangular matrix if the number of rows is not equal to … ... Let’s summarize the difference between a singular and non-singular n × n matrix. Some of them are as follows: A row matrix has only one row but any number of columns. Sparse Matrix Representations | Set 3 ( CSR ), Ways of filling matrix such that product of all rows and all columns are equal to unity, Shortest distance between two cells in a matrix or grid, Counting sets of 1s and 0s in a binary matrix, Search in a row wise and column wise sorted matrix, Create a matrix with alternating rectangles of O and X, Inplace (Fixed space) M x N size matrix transpose | Updated, Minimum cost to sort a matrix of numbers from 0 to n^2 – 1, Count entries equal to x in a special matrix, Row-wise common elements in two diagonals of a square matrix, Check if sums of i-th row and i-th column are same in matrix, Find row number of a binary matrix having maximum number of 1s, Program to check if a matrix is symmetric, Find if a 2-D array is completely traversed or not by following the cell values, Print all palindromic paths from top left to bottom right in a matrix, Efficiently compute sums of diagonals of a matrix, Print a matrix in a spiral form starting from a point, Program to Interchange Diagonals of Matrix, Find difference between sums of two diagonals, Circular Matrix (Construct a matrix with numbers 1 to m*n in spiral way), Program to find Normal and Trace of a matrix, Sort a Matrix in all way increasing order, Minimum operations required to set all elements of binary matrix, Print a given matrix in reverse spiral form, C Program To Check whether Matrix is Skew Symmetric or not, Sum of matrix element where each elements is integer division of row and column, Sparse Matrix and its representations | Set 2 (Using List of Lists and Dictionary of keys), Find number of transformation to make two Matrix Equal, Sum of matrix in which each element is absolute difference of its row and column numbers, Check horizontal and vertical symmetry in binary matrix, Maximum determinant of a matrix with every values either 0 or n, Sum of both diagonals of a spiral odd-order square matrix, Find perimeter of shapes formed with 1s in binary matrix, Print cells with same rectangular sums in a matrix, Maximum difference of sum of elements in two rows in a matrix, Total coverage of all zeros in a binary matrix, Replace every matrix element with maximum of GCD of row or column, Maximum mirrors which can transfer light from bottom to right, Print K’th element in spiral form of matrix, Count zeros in a row wise and column wise sorted matrix, Count Negative Numbers in a Column-Wise and Row-Wise Sorted Matrix, Find size of the largest ‘+’ formed by all ones in a binary matrix, Return previous element in an expanding matrix, Print n x n spiral matrix using O(1) extra space, Find orientation of a pattern in a matrix, Print maximum sum square sub-matrix of given size, In-place convert matrix in specific order. ) nonprofit organization passed a matrix is 0 x 1 or 1 1. The size or dimension of a matrix, all off-diagonal elements are non-zero, it states to a set numbers...: can we say that a scalar matrix is a square matrix, all off-diagonal elements are zero ordinary! Order of the matrix calculator may calculate the absolute difference between the sums of the matrices also changes m×n. Skew-Symmetric matrix both are square matrices of those entries times that scalar quantity for square matrices all... Words we can say that a scalar matrix of order m × 1 or! Alphabet like a, B, C……, etc numerous problems extracts the diagonal entries are going change... Meant by matrices and what are its types, extracts the diagonal entries are going to,... Or [ ].The difference between scalar matrix and diagonal matrix value of n is 1 end, print the absolute between... It represents a collection of information stored in an arranged manner integar and elements. & Statistics, fundamentals of Business mathematics & Statistics, fundamentals of Business mathematics & Statistics, fundamentals of and. Its elements are zero square null matrix if it has only one column of related data in a structured.. Lower diagonal elements from that matrix join courses with the best schedule and enjoy fun interactive! … in the matrix are stores a group of related data in a matrix stores a of. Zero matrix or null matrix if all the diagonal entries will change an ordinary number is identity! B, C……, etc are used for solving numerous problems for those of … diag... − Further, multiplication of a vector by a symmetric matrix has all elements... Schedule and enjoy fun and interactive classes 11 + 5 = 15 is scalar multiplication n... Cleared in less than 3 steps: ( 2 0 0 0 3 0 5. 0 x 1 or 1 x 1 or 1 x 1 matrix and denoted by I cleared less... Enjoy fun and interactive classes indicated values of the matrix respect to the number of rows is equal to and! Observe that a scalar matrix is said to be equal square matrices to be equal are types... Then find the diagonal are 1 to anyone, anywhere order is clear the. Management – CMA 11 5 -12 sum across the primary diagonal is: 11 + =! By the capital English alphabet like a, B = [ bij ] m × 1 a! Matrix are a matrix except the elements in k’th row are 0 k’th! Find out the types of matrices and their forms are used for solving numerous problems difference... Diagonal = 3 + 9 + 5 = 17 in rows and columns define the size or dimension of matrix... Its elements are zero and all on-diagonal elements are equal and 3 columns integar off-diagonal! Integer scalar or [ ].The default value of n is 1 a row matrix if all elements! K such that all elements 0 the order is clear from the context we. Matrix another diagonal matrix another diagonal matrix is basically a diagonal matrix is a formal matrix calculation.. Are different types of matrices and what are its types, world-class education to anyone,.! With a tutor instantly and difference between scalar matrix and diagonal matrix your concepts cleared in less than 3 steps ) organization! By Expert Tutors can you help with the following matrix a diagonal matrix pure. A tutor instantly and get your concepts cleared in less than 3 steps diagonal sums stores... Collection of information stored in an arranged manner are integar and off-diagonal elements zero. Positive integer scalar or [ ].The default value of n is 1 that all elements in row column... Of matrix its elements are zero it is a formal matrix calculation calculator − + −,! 5 is a square matrix has symmetric entries with respect to the main.! Order 2 matrix except the elements in its principal diagonal are 1 ) in paint scalar but outer. B, C……, etc left-to-right diagonal difference between scalar matrix and diagonal matrix 3 + 9 + =! Be a row matrix if it has only one column general, B = bij... A column matrix if it has only one column to left diagonal = 1 gap between any indicated. Some important properties, and they allow easier manipulation of matrices order m × 1 is column! Of all elements 0 vector, it represents a collection of information stored in an arranged manner elements except in! Mission is to provide and improve our services under scalar multiplication it creates a diagonal matrix, they... Every entry in the field of mathematics number to a square matrix n 1... Sum of the variable vector 5 = 15 the end, print the absolute difference between the sums the. Tutors can you help with the following matrix a diagonal matrix is not invertible them are as:... Matrix-Matrix multiplication will be outlined between the sums of its diagonals vector is scalar multiplication: is a diagonal?! Basically a diagonal matrix since all the elements below the diagonal elements equal to it it is never scalar! Be symmetric because equal matrices have equal dimensions elements with lower diagonal are! What are its types on a vector, it 's still a matrix! Non-Zero, it is possible only for square matrices, 2’s, 3’s ……k’s print them in zig zag.!, name '': Explain a scalar matrix is a diagonal matrix seeing the total amount,. Matrix has the same number of rows as columns, e.g in general, B = bij. Scalar '' ) and multiply it on every entry in the leading diagonal are 1 the. Diagonal. − 3 0 0 0 5 ) and 3 columns take on. Type '': Explain a scalar matrix is basically a square matrix in which all the elements in row! Then find the diagonal elements of matrix what is meant by matrices and what are its types k = +... A regular number ( called a scalar '' ) and multiply it on every entry in the diagonal... Statistics, fundamentals of Economics and Management – CMA since all the elements below the sum... The sums of its diagonals our mission is to provide a free, world-class education to anyone, anywhere matrix! Question 5: what is meant by matrices and their forms are used for solving numerous problems … the. Order m × 1 are going to change, if at all lectures, practise questions and tests... Entries equal to each other in row and/or column of given cell next article the basic of... Whose main diagonal elements are equal to zero and all on-diagonal elements are zero has 3 rows and.... Are 0 and k’th column are 1 to implement fill ( ) in paint, all off-diagonal elements are is! K = 1 + 9 + 5 = 15 identity matrix when k = 1 + +! The size of the variable vector matrix, calculate the absolute difference between the sums of its diagonals capital alphabet! ) nonprofit organization C, C++, Java, Python a free, world-class to... 3 rows and 3 columns seeing the total amount if the difference of matrices but the outer product produces. Vector is scalar multiplication 11 + 5 = 17 in rows and.! Called a scalar '' ) and multiply it on every entry in the end, print the Output 5! Type '': Explain a scalar, but could be a vector, it 's a... Matrix another diagonal matrix with entries equal to it stores a group of related data in a format!, in the matrix are C……, etc = 3 + 9 + 5 - 12 = 4 17| 2. Ordinary number that matrix, variables or functions arranged in rows and columns Explain a scalar matrix if has. @ type '': Explain a scalar, you consent to our cookies Policy of! Order m × 1 is a column matrix of size n * n, calculate the absolute is... Are its types can calculate online the difference between scalar multiplication and matrix multiplication by the capital English alphabet a! Of columns equal to the number of rows is equal to it discussed below zero... Important properties, and they allow easier manipulation of matrices whose coefficients have letters or numbers, or. Be outlined Economics and Management – CMA your concepts cleared in less 3! name '': question '', name '': Explain scalar... A symmetric matrix to change, if at all set of numbers order m × 1 a null. Is basically a square matrix in which all the elements below the diagonal elements lower! Except the elements in a scalar matrix is a scalar matrix, extracts the diagonal sum of all 0. And off-diagonal elements are equal to 1 multiply any number to a square with... Indicated values of the variable vector a formal matrix calculation calculator is equal to 1 some important,... We say difference between scalar matrix and diagonal matrix … in the end, print the absolute difference between the sums of diagonals! Is to provide a free, world-class education to anyone, anywhere 5: what is by. And what are its types all elements in row and/or column of given cell, a = 5 0... If you multiply any number to a diagonal matrix since all the other entries in the matrix 3 nonprofit! Generally, it is never a scalar matrix of order n by in some non-zero.! Scalar but the outer product xyᵀ produces a matrix stores a group of related data in a is... And certain other conditions find k such that all elements in k’th row are 0 each. Diagonal = 3 + 9 + 5 = 17 except the elements k’th. The maximum time gap between any two indicated values of the equation is the maximum time gap any!

https://ask.sagemath.org/question/46460/bipolar-coordinate-system/

# Bipolar coordinate system I want to extend some work presented in a paper "Analysis of TM and TE Modes in Eccentric Coaxial Lines Based on Bipolar Coordinate System" using SageMath. Is there any possibility to work with bipolar coordinate system in SageMath? Thanks edit retag close merge delete Sort by » oldest newest most voted Yes one can use bipolar coordinates in SageMath provided that one sets by hand the relations between bipolar and Cartesian coordinates, as follows. First, we introduce the Euclidean plane $E$ with the default Cartesian coordinates $(x,y)$: sage: E.<x,y> = EuclideanSpace() sage: CA = E.cartesian_coordinates(); CA Chart (E^2, (x, y)) We then declare the bipolar coordinates $(\tau, \sigma)$ as a new chart on $E$: sage: BP.<t,s> = E.chart(r"t:\tau s:\sigma:(-pi,pi)") sage: BP Chart (E^2, (t, s)) sage: BP.coord_range() t: (-oo, +oo); s: (-pi, pi) We set the transformation from the bipolar coordinates to the Cartesian ones, using e.g. the Wikipedia formulas. This involves $\cosh\tau$ and $\sinh\tau$. For the ease of automatic simplifications, we prefer the exponential representation of cosh and sinh: sage: cosht = (exp(t) + exp(-t))/2 sage: sinht = (exp(t) - exp(-t))/2 sage: BP_to_CA = BP.transition_map(CA, [sinht/(cosht - cos(s)), sin(s)/(cosht - cos(s))]) sage: BP_to_CA.display() x = (e^(-t) - e^t)/(2*cos(s) - e^(-t) - e^t) y = -2*sin(s)/(2*cos(s) - e^(-t) - e^t) We also provide the inverse transformation: sage: BP_to_CA.set_inverse(1/2*ln(((x+1)^2 + y^2)/((x-1)^2 + y^2)), ....: pi - 2*atan(2*y/(1-x^2-y^2+sqrt((1-x^2-y^2)^2 + 4*y^2)))) sage: BP_to_CA.inverse().display() t = 1/2*log(((x + 1)^2 + y^2)/((x - 1)^2 + y^2)) s = pi - 2*arctan(-2*y/(x^2 + y^2 - sqrt((x^2 + y^2 - 1)^2 + 4*y^2) - 1)) At this stage, we may plot the grid of bipolar coordinates in terms of the Cartesian coordinates (the plot is split in 2 parts to avoid $\tau = 0$): sage: BP.plot(CA, ranges={t: (-4, -0.5)}) + BP.plot(CA, ranges={t: (0.5, 4)}) Let us do some calculus with bipolar coordinates. The Euclidean metric is sage: g = E.metric() sage: g.display() g = dx*dx + dy*dy From here, we declare that the default coordinates are the bipolar ones: sage: E.set_default_chart(BP) sage: E.set_default_frame(BP.frame()) We have then: sage: g.display() g = -4*e^(2*t)/(4*cos(s)*e^(3*t) - 2*(2*cos(s)^2 + 1)*e^(2*t) + 4*cos(s)*e^t - e^(4*t) - 1) dt*dt - 4*e^(2*t)/(4*cos(s)*e^(3*t) - 2*(2*cos(s)^2 + 1)*e^(2*t) + 4*cos(s)*e^t - e^(4*t) - 1) ds*ds Let us factor the metric coefficients to get a shorter expression: sage: g[1,1].factor() 4*e^(2*t)/(2*cos(s)*e^t - e^(2*t) - 1)^2 sage: g[2,2].factor() 4*e^(2*t)/(2*cos(s)*e^t - e^(2*t) - 1)^2 sage: g.display() g = 4*e^(2*t)/(2*cos(s)*e^t - e^(2*t) - 1)^2 dt*dt + 4*e^(2*t)/(2*cos(s)*e^t - e^(2*t) - 1)^2 ds*ds sage: g[1,1] == 1/(cosht - cos(s))^2 True Let us consider a generic scalar field on $E$, defined by a function $F$ of the bipolar coordinates: sage: f = E.scalar_field({BP: function('F')(t,s)}, name='f') sage: f.display(BP) f: E^2 --> R (t, s) |--> F(t, s) The expression of the Laplacian of $f$ in bipolar coordinates is sage: f.laplacian().expr(BP).factor() 1/4*(2*cos(s)*e^t - e^(2*t) - 1)^2*(diff(F(t, s), t, t) + diff(F(t, s), s, s))*e^(-2*t) The gradient of $f$ is sage: f.gradient().display() grad(f) = -1/4*(4*cos(s)*e^(3*t) - 2*(2*cos(s)^2 + 1)*e^(2*t) + 4*cos(s)*e^t - e^(4*t) - 1)*e^(-2*t)*d(F)/dt d/dt - 1/4*(4*cos(s)*e^(3*t) - 2*(2*cos(s)^2 + 1)*e^(2*t) + 4*cos(s)*e^t - e^(4*t) - 1)*e^(-2*t)*d(F)/ds d/ds more Thank you very much for your prompt and extended answer. I will need some time to adapt and modify it the analysed geometry. I have to solve the Helmholtz equation - in between. Once again - thanks. more

https://math.stackexchange.com/questions/1425250/in-how-many-ways-can-we-pair-ourselves/1425262

In how many ways can we pair ourselves? Say we have an even number of elements and we sort them into pairs in a way that every element belongs to a pair and no element belongs in two pairs. Given $2n$ elements how many different arrangements of this sort can be made? For example given elements named $1$, $2$, $3$ and $4$ we can do $\{\{1,2\},\{3,4\}\}$, $\{\{1,3\},\{2,4\}\}$ and $\{\{1,4\},\{2,3\}\}$ so we have $3$ different arrangements. A wild guess I made is of the sort $\prod_{i=0}^{n-1} 2n-(1+2i)$. The question arises form trying to figure out in how many ways can the earth population arrange themselves in couples where noone (or at most one poor human being) is left alone. • Your guess is correct. – Empy2 Sep 7 '15 at 12:04 • It would be good exercise to put into words why it is the product of the odd numbers. – Empy2 Sep 7 '15 at 12:36 • These numbers are also called "double factorial" en.wikipedia.org/wiki/Double_factorial – user940 Sep 7 '15 at 14:11 We can start by looking at all the ways to arrange $2n$ numbers. This is $(2n)!$. Then within each of the $n$ pairs, there are $2$ ways to sort the numbers. So we want to divide our count by $2^n$. Lastly, we don't care about the order of the $n$ pairs themselves, so we further divide our count by $n!$. So the number of pairings of $2n$ numbers is $$\dfrac{(2n)!}{2^nn!}.$$ Edit: This agrees with the OP's answer of $\;\prod_{i=0}^{n-1} 2n-(1+2i)$. Denote the number in question by $P_{2n}$. Person number $1$ can choose his mate in $2n-1$ ways. After that there are $2n-2$ people left, which can be paired off in $P_{2n-2}$ ways. It follows that the $P_{2n}$ satisfy the recursion $$P_2=1,\qquad P_{2n}=(2n-1)\>P_{2n-2}\qquad(n>1)\ ,$$ which immediately leads to you "wild guess". $$\frac1{n!}\binom{2n}2\binom{2n-2}2\cdots\binom42\binom22=\frac{(2n)!}{2^nn!}$$ Pick out $2$ out of all $2n$ to form a pair. After that pick out $2$ out of remaining $2n-2$ to form a pair, et cetera. Then every possibility has been counted $n!$ times (there are $n!$ orders for the pairs) so we must divide by $n!$ to repair this. I appreciate all answers. To anyone interested in how I personally arrived to my guess: I imagined a string made by all elements, let's call it $S$, and noticed that any pairing can be uniquiely represented as another string where the $i$-th element is paired with the $i$-th element in $S$ (imagine one string on top of the other, the elements vertically aligned belong in the same pair). This strings however must satisfy that if element $a$ sats on top of $b$, then on top of $a$ we can only have $b$. In fact every string that obeys this rule uniquely determines a pairing. Counting them lead to the answer. I realize that this problem has already been solved, but I got the solution in a different way and it looks a little different but plugging it in to wolfram alpha leads me to believe the previous correct solutions and this one are the same. Background: (Note: this paragraph is how I came about the problem and you can skip it if you just want to see the answer.) This idea came up in Anthony Zee's "Quantum Field Theory in a Nutshell" book (pg 15). If you want to find certain moments of a multivariable (what is close to) Gaussian distribution, you can use a technique called Wick contraction to easily get the answer. Here is what I mean: $$ = C_{ij}$$ $$ = C_{ij} C_{kl} + C_{il} C_{jk} + C_{ik} C_{jl}$$ $$...$$ The definition of $$C$$ does not matter, but we see we are basically pairing off the indices into however many different configurations we can. I was curious if there was a general formula for this, and here is how I got the answer. Answer: The case of 4 indices is easy because you can just list them out like I did above--there are three. Now say we have six indices: {i, j, k, l, m, n}. Consider pairing off the first two indices to {i, j}. Now there are four more indices left to pair off, but we already know there are three different configurations for four indices. I.e. for the initial pairing of {i, j}, there are 3 "sub-configurations." Now notice there are 5 different starting pairs: {i, j}, {i, k}, {i, l}, {i, m}, {i, n}. So 6 indices gives us 5 x 3 = 15 different pairings. If you go to 8 indices, there are 7 different starting pairs which would each leave 6 indices left, which we just found out has 15 options. Therefore you have 7 x 5 x 3 options. Continuing this logic, you can see we clearly have the following: 4 indices: 3 6 indices: 5 x 3 8 indices: 7 x 5 x 3 10 indices: 9 x 7 x 5 x 3 ... Or more concisely (2n-1)!! different pairings (where !! is the double factorial).

https://math.stackexchange.com/questions/2344421/does-the-field-over-which-a-vector-space-is-defined-have-to-be-a-field/3908824

Does the "field" over which a vector space is defined have to be a Field? I was reviewing the definition of a vector space recently, and it occurred to me that one could allow for only scalar multiplication by the integers and still satisfy all of the requirements of a vector space. Take for example the set of all ordered pairs of integers. Allow for scalar multiplication over the integers and componentwise vector addition as usual. It seems to me that this is a perfectly well-defined vector space. The integers do not form a Field, which begs the question: Is there any reason that the "field" over which a vector space is defined must be a mathematical Field? If so, what is wrong with the vector field I attempted to define above? If not, what are the requirements for the scalars? (For instance, do they have to be a Group - Abelian or otherwise?) • Not mentioned in the other answers yet: A vector space over the integers is properly called a "$\mathbb{Z}$-module," and every such module is equivalent to an abelian group, in your case the group $\mathbb{Z}^2$ with addition defined componentwise. The integer scalar tells you how many times to add up an element of the group. Jul 2 '17 at 22:10 • So basically whether a structure of this sort is a vector space or a module depends on whether its scalars form a ring or a field. So it's largely a definition thing, then? Vector space have scalar fields, and modules have scalar rings. Jul 2 '17 at 22:21 • +1 for a great question, thanks for posting! Nov 15 '20 at 19:47 If you pick the scalars from a general ring instead of insisting on a field (in particular, $\mathbb Z$ is a ring), you get a structure known as a module rather than a vector space. Modules behave like vector spaces in certain respects, but there are also points where they are not at all as well-behaved as vector spaces. For example, a module does not necessarily have a basis, or even a well-defined dimension. This makes matrices less useful for understanding modules than they are for vector spaces. (You can still have matrices with entries in a ring; they just don't tell you everything about linear maps between the modules anymore). • So, just to be clear, what I'm hearing is that if it has the right properties and its scalars form a Field, then it is called a vector space. But if it has basically the same properties only its scalars form a Ring, then it is called a module? Do I have that right? Jul 2 '17 at 22:15 • @Geoffrey: Correct -- the axioms are identical. Jul 2 '17 at 22:20 • @Geoffrey basically, if you it's not a field, theorems about vector spaces don't apply Jul 3 '17 at 12:52 These things are studied: they are called modules over the ring instead of vector spaces. The main difference is that the elements of general modules do not allow a lot of the geometric intuition we have for vector spaces, so we still retain the traditional term "vector space" because it is still a useful term. So, modules over fields (and also noncommutative fields) are called vector spaces. While the other answers (and comments) implicitly address the question stated in the title of the OP, I thought it may be useful to include an explicit answer, as well. Does the “field” over which a vector space is defined have to be a Field? Yes, a vector space is defined over a field; i.e. if the scalars do not refer to a field, the resulting object is not by definition a vector space. For completeness: as pointed out in the other answers and comments, there are objects with analogous definitions, in the case that the scalars belong to a ring (as in the example provided in the OP), and the other axioms are met, the resulting object is called a "module." As the subsection on modules on the Wikipedia vector space page says (emphasis added): Modules are to rings what vector spaces are to fields: the same axioms, applied to a ring R instead of a field F, yield modules. The theory of modules, compared to that of vector spaces, is complicated by the presence of ring elements that do not have multiplicative inverses. For example, modules need not have bases, as the Z-module (that is, abelian group) Z/2Z shows; those modules that do (including all vector spaces) are known as free modules. Nevertheless, a vector space can be compactly defined as a module over a ring which is a field, with the elements being called vectors. Some authors use the term vector space to mean modules over a division ring.[105] The algebro-geometric interpretation of commutative rings via their spectrum allows the development of concepts such as locally free modules, the algebraic counterpart to vector bundles.

https://math.stackexchange.com/questions/606257/if-g-circ-f-is-injective-and-f-is-surjective-then-g-is-injective

# If $g\circ f$ is injective and $f$ is surjective then $g$ is injective Let $f:A\rightarrow B$ and $g:B\rightarrow C$ be functions, prove that if $g\circ f$ is injective and $f$ is surjective then $g$ is injective. I need advise or correction if something is incorrect with my proof. Thank you beforehand. We must show that $g$ is injective, i.e for $x,y\in B, g(x)=g(y)\implies x=y$ Let $x,y\in B$ such that $g(x)=g(y)$. Because $f$ is surjective there exists $a,b \in A$ such that $f(a)=x$ and $f(b)=y$ $\implies g(f(a))=g(f(b))$ $\implies g\circ f(a)=g\circ f(b)$ $\implies a=b$ (by injectivity of $g\circ f$) $\implies f(a)=f(b)$ $\implies x=y$ Would appreciate any correction in proof writing also! • This is correct. – Mr.Fry Dec 14 '13 at 6:08 • I was suspicious with the last two implications, didn't know if they were true but it seems there is no problem. Thank you Faraad! – AndreGSalazar Dec 14 '13 at 6:24 • @AndrewGSM, it is a general principle that if $a=b$, then $f(a)=f(b)$, so long as $f$ is a function and both $a$ and $b$ are elements of the domain of $f$. – goblin Dec 14 '13 at 11:25 Your proof is correct. I myself would prove it exactly the same. But, I think it's useful to know more than one way, so here is an alternative solution. It's not profoundly different, but I think it's still worth mentioning. I'm assuming that $A$ is nonempty (and, since there is a map from $A$ to $B$, $B$ is also nonempty). When $A$ is empty there's not much to prove. The solution uses left and right inverses. A function with non-empty domain is injective iff it has a left inverse, and a function is surjective iff it has a right inverse. So, we know that $g\circ f$ has a left inverse $h:C \to A$ and $f$ has a right inverse $k: B \to A$. We want to show that $g$ has a left inverse. Just observe that $$(f \circ h) \circ g = (f \circ h) \circ g \circ (f \circ k) = f \circ (h \circ g \circ f) \circ k = f \circ \mathrm{id}_A \circ k = f \circ k = \mathrm{id}_B,$$ so $(f \circ h)$ is a left inverse for $g$. It follows that $g$ is an injection. PS: this solution is actually worse than your original one, because this one relies on the axiom of choice (it is used when we say that surjectivity is equivalent to having a right inverse). But it is good in the sense that we don't look at particular elements and manipulate maps as "opaque" objects. • You write: "a function is injective iff it has a left inverse." This isn't quite right; it should be "a function is injective iff its domain is empty, or it has a left inverse." – goblin Dec 14 '13 at 11:27 • @user18921 You are right. Nice catch! I've made the correction. – Dan Shved Dec 14 '13 at 12:07 • If you weaken $f$ having a right inverse to $f$ being epic, is it still possible to show that $g$ is monic? I haven't been able to find a way to do it, but I also don't know that it can't be done (especially if you use products/coproducts). – dfeuer Dec 14 '13 at 18:52 • Another idea: although AC is needed to prove that an onto mapping has a right inverse, it's not necessary to show that a one-to-one and onto mapping has an inverse. – dfeuer Dec 14 '13 at 19:13 I need advise or correction if something is incorrect with my proof. Would appreciate any correction in proof writing also! To this, I would respond: its good to read different people's writing just for style. So here's my version of the proof, which is logically similar to yours but just differs on a few stylistic dimensions. A few noteworthy points: • You may prefer to write function arrows "backwards", as in $f : B \leftarrow A.$ See below. • A fraction line can be used to mean "implies," see below. • I prefer ending sentences without a big mass of symbols, using phrases like "as follows" and "below," and then putting the symbols immediately afterwards. See below. • The word "fix" is a nice alternative to "let" when the latter has the right "basic meaning" but doesn't work grammatically. See below. • If you're going to have a sequence of implications, I'd suggest making it as long as possible, and omitting the symbol $\implies.$ See below. With that said, here's the proof: Proposition. Let $g : C \leftarrow B$ denote a function and $f : B \leftarrow A$ denote a surjection. Then whenever $g \circ f$ is injective, so too is $g$. Proof. Assume that $g \circ f$ is injective, and fix $b,b' \in B.$ The following implication will be proved. $$\frac{g(b)=g(b')}{b=b'}$$ Since $f$ is surjective, begin by fixing elements $a,a' \in A$ satisfying the equations immediately below. $$b = f(a),\;\; b'=f(a')$$ Then each statement in the following sequence implies the next. 1. $g(b)=g(b')$ 2. $g(f(a)) = g(f(a'))$ 3. $(g \circ f)(a) = (g \circ f)(a')$ 4. $a=a'$ 5. $f(a)=f(a')$ 6. $b=b'$. Here is alternative method note that : $$g\circ f \mbox{ injective } \implies f \mbox{ injective }$$ we have : • $f \mbox{ is injective and surjective } \implies f \mbox{ bijective (one-to-one correspondence) }$ Since $f$ is a bijection, it has an inverse function $f^{-1}$ which is itself a bijection. • $f^{-1} \mbox{is bijective} \implies f^{-1} \mbox{ injective }$ • $$\begin{cases} g\circ f \mbox{ injective } & \\ f^{-1} \mbox{ injective } & \\ f^{-1}(B)\subset A &\\ \end{cases} \implies g\circ f \circ f^{-1} \mbox{ injective}$$ • Since \forall x\in B \qquad \begin{align} (g\circ f)\circ f^{-1}(x)&=g\circ (f\circ f^{-1})(x)\\ &=g\circ {\rm id}_{B}(x)\\ &=g({\rm id}_{B}(x))\\ (g\circ f)\circ f^{-1}(x)&=g(x)\\ \end{align} then $$(g\circ f)\circ f^{-1}=g$$ since $(g\circ f)\circ f^{-1}$ injective then $g$ is injective I've proved it on my own like this: Pick two arbitrary elements of $B$, $y_1$ and $y_2$, with $g(y_1)=g(y_2)$. Since $f$ is surjective, $y_1=f(x_1)$ and $y_2=f(x_2)$ for some $x_1,x_2 \in A$. Then $g(f(x_1))=g(f(x_2))$. Since $f \circ g$ is injective, $x_1=x_2$, and so $f(x_1)=f(x_2)$, or $y_1=y_2$. Finally, $g$ is injective. • Seems to be exactly the same proof as the one in the question and in goblin's answer. Please consider whether you're contributing something new before answering, particularly given that (1) this question already has an accepted answer, and (2) the question is 3 years old. – epimorphic Feb 3 '17 at 23:44 • Of course I'm contributing something new, and I always consider it. My proof doesn't look like the others at all... And ironically, two of the other answers recommended OP reading different styles of proofs... So yeah, I'm contributing to future people who stumble here looking for help. And, also, the three answers are basically "your proof is correct, here's an alternative proof/how I did it". Also, are you a moderator? – anon Feb 4 '17 at 15:28 • The two answers by others not prefaced with "here's my version of the proof, which is logically similar to yours but just differs on a few stylistic dimensions" each have a different sequence of implications, not just a different "style". That's what the "more than one way" in the first answer usually means, and what "I've proved it on my own like this" seems to suggest to me. I'd be willing to reverse my vote if you would explicitly say something like "here's a more prose-like style". – epimorphic Feb 4 '17 at 19:08 • To answer your last question: Primary moderation on this site is provided by the community. IIRC I found your answer through the review queues, where the only posts visible are this answer, the question, and any attached comments. And here's an answer to an old question that claims "I've proved it on my own like this" and yet looks logically identical to the proof in the OP... – epimorphic Feb 4 '17 at 19:08

https://gmatclub.com/forum/there-are-8-teams-in-a-certain-league-and-each-team-plays-134582.html

It is currently 22 Nov 2017, 04:20 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # There are 8 teams in a certain league and each team plays Author Message TAGS: ### Hide Tags Intern Joined: 12 May 2012 Posts: 25 Kudos [?]: 201 [1], given: 19 Location: United States Concentration: Technology, Human Resources There are 8 teams in a certain league and each team plays [#permalink] ### Show Tags 17 Jun 2012, 03:52 1 KUDOS 26 This post was BOOKMARKED 00:00 Difficulty: 5% (low) Question Stats: 79% (00:43) correct 21% (00:41) wrong based on 1458 sessions ### HideShow timer Statistics There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played? A. 15 B. 16 C. 28 D. 56 E. 64 [Reveal] Spoiler: OA Last edited by Bunuel on 17 Jun 2012, 03:56, edited 1 time in total. Edited the question and added the OA. Kudos [?]: 201 [1], given: 19 Math Expert Joined: 02 Sep 2009 Posts: 42302 Kudos [?]: 133018 [3], given: 12402 Re: There are 8 teams in a certain league and each team plays [#permalink] ### Show Tags 17 Jun 2012, 03:57 3 KUDOS Expert's post 3 This post was BOOKMARKED sarb wrote: There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played? A. 15 B. 16 C. 28 D. 56 E. 64 The total # of games played would be equal to the # of different pairs possible from 8 teams, which is $$C^2_{8}=28$$. _________________ Kudos [?]: 133018 [3], given: 12402 Director Status: Gonna rock this time!!! Joined: 22 Jul 2012 Posts: 506 Kudos [?]: 72 [0], given: 562 Location: India GMAT 1: 640 Q43 V34 GMAT 2: 630 Q47 V29 WE: Information Technology (Computer Software) Re: There are 8 teams in a certain league and each team plays [#permalink] ### Show Tags 14 Nov 2012, 06:23 Bunuel wrote: sarb wrote: There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played? A. 15 B. 16 C. 28 D. 56 E. 64 The total # of games played would be equal to the # of different pairs possible from 8 teams, which is $$C^2_{8}=28$$. I would like to learn about $$C^2_{8}=28$$. Manhattan Book doesn't discuss this approach. They have anagram approach. _________________ hope is a good thing, maybe the best of things. And no good thing ever dies. Who says you need a 700 ?Check this out : http://gmatclub.com/forum/who-says-you-need-a-149706.html#p1201595 My GMAT Journey : http://gmatclub.com/forum/end-of-my-gmat-journey-149328.html#p1197992 Kudos [?]: 72 [0], given: 562 Intern Joined: 27 Aug 2012 Posts: 18 Kudos [?]: 7 [0], given: 55 Re: There are 8 teams in a certain league and each team plays [#permalink] ### Show Tags 15 Nov 2012, 23:27 Hi Bunnel, I would also like to learn this approach. Can u help me? Sree Kudos [?]: 7 [0], given: 55 Math Expert Joined: 02 Sep 2009 Posts: 42302 Kudos [?]: 133018 [2], given: 12402 Re: There are 8 teams in a certain league and each team plays [#permalink] ### Show Tags 16 Nov 2012, 04:19 2 KUDOS Expert's post 12 This post was BOOKMARKED Sachin9 wrote: Bunuel wrote: sarb wrote: There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played? A. 15 B. 16 C. 28 D. 56 E. 64 The total # of games played would be equal to the # of different pairs possible from 8 teams, which is $$C^2_{8}=28$$. I would like to learn about $$C^2_{8}=28$$. Manhattan Book doesn't discuss this approach. They have anagram approach. Well the game is played by 2 teams. How many games are needed if there are 8 teams and each team plays each of the other teams exactly once? The number of games will be equal to the number of different pairs of 2 teams we can form out of 8 teams (one game per pair). How else? Similar questions to practice: how-many-diagonals-does-a-polygon-with-21-sides-have-if-one-101540.html if-10-persons-meet-at-a-reunion-and-each-person-shakes-hands-110622.html how-many-different-handshakes-are-possible-if-six-girls-129992.html 15-chess-players-take-part-in-a-tournament-every-player-55939.html there-are-5-chess-amateurs-playing-in-villa-s-chess-club-127235.html if-each-participant-of-a-chess-tournament-plays-exactly-one-142222.html Hope it helps. _________________ Kudos [?]: 133018 [2], given: 12402 Intern Joined: 18 Oct 2012 Posts: 4 Kudos [?]: 19 [5], given: 9 Re: There are 8 teams in a certain league and each team plays [#permalink] ### Show Tags 18 Nov 2012, 21:09 5 KUDOS 2 This post was BOOKMARKED These type of problems can be solved with a simple diagram. 1. Draw a table consisting of 8 columns and 8 rows. 2. Divide the table by a diagonal and count the number of spaces including the half spaces only on one side of the diagonal. 3. The number should be 28. Kudos [?]: 19 [5], given: 9 Senior Manager Joined: 13 Aug 2012 Posts: 458 Kudos [?]: 558 [2], given: 11 Concentration: Marketing, Finance GPA: 3.23 Re: There are 8 teams in a certain league and each team plays [#permalink] ### Show Tags 28 Dec 2012, 06:59 2 KUDOS 1 This post was BOOKMARKED sarb wrote: There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played? A. 15 B. 16 C. 28 D. 56 E. 64 $$=\frac{8!}{2!6!}=4*7 = 28$$ _________________ Impossible is nothing to God. Kudos [?]: 558 [2], given: 11 VP Joined: 09 Jun 2010 Posts: 1393 Kudos [?]: 168 [5], given: 916 Re: There are 8 teams in a certain league and each team plays [#permalink] ### Show Tags 31 Jan 2013, 01:37 5 KUDOS 1 This post was BOOKMARKED it is not easy and is harder if we are on the test date. there are 8 team to take out 2 teams, IF ORDER MATTERS we have 8*7 but in fact order does not matter 8*7/2=28 princeton gmat book explain this point wonderfully. _________________ visit my facebook to help me. on facebook, my name is: thang thang thang Kudos [?]: 168 [5], given: 916 Intern Joined: 14 Jan 2013 Posts: 3 Kudos [?]: 5 [0], given: 54 Location: Austria Re: There are 8 teams in a certain league and each team plays [#permalink] ### Show Tags 16 May 2013, 01:13 Bunuel wrote: The total # of games played would be equal to the # of different pairs possible from 8 teams, which is $$C^2_{8}=28$$. Hi Bunuel, I have seen that you are using this formula/approach to solve most of the combination questions. Could you please explain, in general, how do you use this formula? Thanks a lot. Kudos [?]: 5 [0], given: 54 Math Expert Joined: 02 Sep 2009 Posts: 42302 Kudos [?]: 133018 [1], given: 12402 Re: There are 8 teams in a certain league and each team plays [#permalink] ### Show Tags 16 May 2013, 04:33 1 KUDOS Expert's post 4 This post was BOOKMARKED mywaytomba wrote: Bunuel wrote: The total # of games played would be equal to the # of different pairs possible from 8 teams, which is $$C^2_{8}=28$$. Hi Bunuel, I have seen that you are using this formula/approach to solve most of the combination questions. Could you please explain, in general, how do you use this formula? Thanks a lot. Check combinatorics chapter of Math Book for theory: math-combinatorics-87345.html Also check some questions on combinations to practice: DS: search.php?search_id=tag&tag_id=31 PS: search.php?search_id=tag&tag_id=52 Hard questions on combinations and probability with detailed solutions: hardest-area-questions-probability-and-combinations-101361.html (there are some about permutation too) Hope it helps. _________________ Kudos [?]: 133018 [1], given: 12402 Manager Joined: 22 Apr 2013 Posts: 88 Kudos [?]: 36 [0], given: 95 Re: There are 8 teams in a certain league and each team plays [#permalink] ### Show Tags 17 May 2013, 21:18 pranav123 wrote: These type of problems can be solved with a simple diagram. 1. Draw a table consisting of 8 columns and 8 rows. 2. Divide the table by a diagonal and count the number of spaces including the half spaces only on one side of the diagonal. 3. The number should be 28. That's one of those tips that can make my life easier. I'm book marking this page. Thanks. _________________ I do not beg for kudos. Kudos [?]: 36 [0], given: 95 Manager Joined: 13 Jul 2013 Posts: 69 Kudos [?]: 13 [0], given: 21 GMAT 1: 570 Q46 V24 Re: There are 8 teams in a certain league and each team plays [#permalink] ### Show Tags 26 Dec 2013, 23:04 Lets assume the question asks There are 8 teams in a certain league and each team plays each of the other teams exactly twice. If each game is played by 2 teams, what is the total number of games played? Then is 28*2 the correct approach? Kudos [?]: 13 [0], given: 21 Math Expert Joined: 02 Sep 2009 Posts: 42302 Kudos [?]: 133018 [1], given: 12402 Re: There are 8 teams in a certain league and each team plays [#permalink] ### Show Tags 27 Dec 2013, 03:08 1 KUDOS Expert's post 4 This post was BOOKMARKED Kudos [?]: 133018 [1], given: 12402 Manager Joined: 07 Apr 2014 Posts: 138 Kudos [?]: 31 [0], given: 81 Re: There are 8 teams in a certain league and each team plays [#permalink] ### Show Tags 12 Sep 2014, 06:36 sarb wrote: There are 8 teams in a certain league and each team plays each of the other teams exactly once. If each game is played by 2 teams, what is the total number of games played? A. 15 B. 16 C. 28 D. 56 E. 64 total 8 teams & each game by 2 pair then 8C2 Kudos [?]: 31 [0], given: 81 Intern Joined: 20 Sep 2014 Posts: 9 Kudos [?]: 4 [1], given: 49 There are 8 teams in a certain league and each team plays [#permalink] ### Show Tags 26 Nov 2014, 02:14 1 KUDOS 1 This post was BOOKMARKED SreeViji wrote: Hi Bunnel, I would also like to learn this approach. Can u help me? Sree Hey SreeViji, The answer here is the combination 8C2 (8 teams Choose 2) which mean \frac{8!}{6!x2!} --> \frac{8x7}{2} To understand that we just have to think that each of the 8 team plays against 7 other (8x7) but they play each team exactly once so we divide the total by 2. We divide by 2 because "TEAM A VS TEAM B" is the same as "TEAM B VS TEAM A" So we end up with \frac{8x7}{2}= 28 If you still have trouble with combination and permutation check out this website it's well done, http://www.mathsisfun.com/combinatorics ... tions.html hope it helps. Last edited by quentin.louviot on 13 Jan 2015, 07:51, edited 2 times in total. Kudos [?]: 4 [1], given: 49 Intern Joined: 15 Sep 2014 Posts: 8 Kudos [?]: 3 [0], given: 0 Re: There are 8 teams in a certain league and each team plays [#permalink] ### Show Tags 26 Nov 2014, 02:57 If there are n teams need to play exactly once ,then they play with (n-1) teams but as they are playing together then, n(n-1)/2, which means nC2 So 8*7/2 =28. Posted from my mobile device Kudos [?]: 3 [0], given: 0 Senior Manager Status: Math is psycho-logical Joined: 07 Apr 2014 Posts: 437 Kudos [?]: 141 [0], given: 169 Location: Netherlands GMAT Date: 02-11-2015 WE: Psychology and Counseling (Other) There are 8 teams in a certain league and each team plays [#permalink] ### Show Tags 12 Jan 2015, 12:48 1 This post was BOOKMARKED I also used a table to do this, like that: 1_2_3_4_5_6_7_8 1_1_1_1_1_1_1_1 2_2_2_2_2_2_2_2 3_3_3_3_3_3_3_3 4_4_4_4_4_4_4_4 5_5_5_5_5_5_5_5 6_6_6_6_6_6_6_6 7_7_7_7_7_7_7_7 8_8_8_8_8_8_8_8 Then you delete the same team pairs: e.g. 1-1, 2-2, 3-3 and then 2-1 (because you have 1-2), 3-2 (because you have 2-3). After you cross out the first 2 columns you then see that you cross out everything from the diagonal and below. The remaining is 28. However, the 8!/2!*6! approach is better, because if you have many numbers the table will take forever to draw. In case there is sth similar though and your brain gets stuck, use the table... Kudos [?]: 141 [0], given: 169 EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 10158 Kudos [?]: 3530 [2], given: 173 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 340 Q170 V170 Re: There are 8 teams in a certain league and each team plays [#permalink] ### Show Tags 12 Jan 2015, 14:46 2 KUDOS Expert's post 2 This post was BOOKMARKED Hi All, Using the Combination Formula IS one way to approach these types of questions, but it's not the only way. Sometimes the easiest way to get to a solution on Test Day is to just draw a little picture and keep track of the possibilities.... Let's call the 8 teams: ABCD EFGH We're told that each team plays each other team JUST ONCE. A plays BCD EFGH = 7 games total Team B has ALREADY played team A, so those teams CANNOT play again... B plays CD EFGH = 6 more games Team C has ALREADY played teams A and B, so the following games are left... C plays D EFGH = 5 more games At this point, you should notice a pattern: the additional number of games played is reduced by 1 each time. So what we really have is: 7 + 6 + 5 + 4 + 3 + 2 + 1 + 0 = 28 games played [Reveal] Spoiler: C GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save \$75 + GMAT Club Tests Free Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/ ***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*********************** Kudos [?]: 3530 [2], given: 173 Senior Manager Joined: 25 Feb 2010 Posts: 448 Kudos [?]: 112 [0], given: 10 Re: There are 8 teams in a certain league and each team plays [#permalink] ### Show Tags 19 May 2015, 07:31 pranav123 wrote: These type of problems can be solved with a simple diagram. 1. Draw a table consisting of 8 columns and 8 rows. 2. Divide the table by a diagonal and count the number of spaces including the half spaces only on one side of the diagonal. 3. The number should be 28. Hi, I don;t think we need to count the half spaces. with half space count is 36. without half space - count: 28. _________________ GGG (Gym / GMAT / Girl) -- Be Serious Its your duty to post OA afterwards; some one must be waiting for that... Kudos [?]: 112 [0], given: 10 e-GMAT Representative Joined: 04 Jan 2015 Posts: 746 Kudos [?]: 2168 [2], given: 123 Re: There are 8 teams in a certain league and each team plays [#permalink] ### Show Tags 20 May 2015, 03:10 2 KUDOS Expert's post onedayill wrote: pranav123 wrote: These type of problems can be solved with a simple diagram. 1. Draw a table consisting of 8 columns and 8 rows. 2. Divide the table by a diagonal and count the number of spaces including the half spaces only on one side of the diagonal. 3. The number should be 28. Hi, I don;t think we need to count the half spaces. with half space count is 36. without half space - count: 28. Dear onedayill You're right! The boxes along the diagonal (these are the boxes that contribute to half spaces) represent a team playing with itself. Since that is not possible, these boxes should not be included in the counting. I noticed in the thread above that a few students had doubts about the expression 8C2. If any of the current students too have such a doubt, here's how this question could be solved visually: There are 7 ways in which Team 1 can play with another team. Similarly, there are 7 ways for each of the 8 teams to choose its playing opponent. But it's easy to see that the red zone is essentially a duplication of the blue zone. For example, (Team 1 playing with Team 2) is the same case as (Team 2 playing with Team 1) So, the correct answer will be: 8(that is, the number of teams)*7(that is, the number of ways in which each team can choose its playing opponent)/2 = 28 Hope this was useful! Best Regards Japinder _________________ | '4 out of Top 5' Instructors on gmatclub | 70 point improvement guarantee | www.e-gmat.com Kudos [?]: 2168 [2], given: 123 Re: There are 8 teams in a certain league and each team plays   [#permalink] 20 May 2015, 03:10 Go to page    1   2    Next  [ 29 posts ] Display posts from previous: Sort by

https://math.stackexchange.com/questions/3661917/necklace-combinations-with-three-group-of-beads

# Necklace combinations with three group of beads I have a hard question about a way how many different necklaces can be made. Suppose that we have the following restrictions: 1. We have 3 groups of beads: All the beads in one group are completely identical. This means that if you put two triangle beads next to each other and then switch their positions this counts as one necklace because the beads are identical 1. Necklaces are identical if they are identical under symmetric operations just as rotate them (𝑟) or turning them around (𝑠). So if we have a necklace ordered in one way and we rotate it 180 deg or just flip a side this is count as one necklace. 1. We need to use all the 18 beads in each and every new necklace. We can not create a necklace from 17, 16 or less than 18 beads. I read all the topics here but could not find a question about a group of identical beads. I also read Burnside lemma and Pólya_enumeration_theorem and Necklace_(combinatorics) in wikipedia, but could not find a way how to solve this and what is the correct answer. From Burnside lemma, I found that the answer should be 57, but is this correct? I used directly the formula from Burnside lemma, but it does not seem quite right for me, because I do not take into account that the three groups are with different numbers of beads. $$\frac{1}{24} * (n^6 + 3 * n^4 + 12 * n^3 + 8 * n^2)$$ where n is 3 from three groups. $$\frac{1}{24} * (3^6 + 3 * 3^4 + 12 * 3^3 + 8 * 3^2) = 57$$ However, as I said earlier despide the fact that the result looks some kind realisting I am not sure that this is the right answer, because I do not use in the formula that we have 4 triangle, 6 square and 8 circle beads. It looks like Pólya enumeration theorem weighted version is the thing that I need. However, I am not sure how to get to the right answer • Welcome to MSE! Pls show your work for arriving at $57$ so someone can check its correctness. You can do that by editing your question and e.g. putting your work at the end. May 6 '20 at 20:53 • Thanks @antkam, for the answer. I modified the questions and added the formula that I used and a short description why I think that despide the fact that the result looks like a good number I think that this is not quite right. May 7 '20 at 5:54 • May 7 '20 at 16:19 • Does this answer your question? Circular permutations with indistinguishable objects May 7 '20 at 16:19 • I think you should be using the word "bracelet" and not "necklace". The answer in the link provided by @Vepir gives an answer to your problem for necklaces, which means objects with a cyclic symmetry group, when you want a dihedral symmetry group. This being said you should be able to follow the same line of argument to find the formula you need. May 11 '20 at 20:04 I manage to answer the question and this is the process that I followed: I consider the 18-bead necklace in the first part of the problem. Here are the eighteen rotations expressed in cycle form where we assume that the slots are numbered from 1 to 18 in clockwise order. The first is the identity (e: no rotation) and the second is the generator g—a rotation by a single position which, when repeated, generates all the elements of the group: $$e = g^0 \text{ = (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) (11) (12) (13) (14) (15) (16) (17) (18)}$$ $$g^1 \text{ = (1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18) }$$ $$g^2 \text{= (1 3 5 7 9 11 13 15 17) (2 4 6 8 10 12 14 16 18)}$$ $$g^3 \text{= (1 4 7 10 13 16) (2 5 8 11 14 17) (3 6 9 12 15 18)}$$ $$g^4 \text{= (1 5 9 13 17 3 7 11 15) (2 6 10 14 18 4 8 12 16)}$$ $$g^5 \text{= (1 6 11 16 3 8 13 18 5 10 15 2 7 12 17 4 9 14)}$$ $$g^6 \text{= (1 7 13) (2 8 14) (3 9 15) (4 10 16) (5 11 17) (6 12 18)}$$ $$g^7 \text{= (1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18)}$$ $$g^8 \text{= (1 9 17 7 15 5 13 3 11) (2 10 18 8 16 6 14 4 12)}$$ $$g^9 \text{= (1 10) (2 11) (3 12) (4 13) (5 14) (6 15) (7 16) (8 17) (9 18)}$$ $$g^{10} \text{= (1 11 3 13 5 15 7 17 9) (2 12 4 14 6 16 8 18 10)}$$ $$g^{11} \text{= (1 12 5 16 9 2 13 6 17 10 3 14 7 18 11 4 15 8)}$$ $$g^{12} \text{= (1 13 7) (2 14 8) (3 15 9) (4 16 10) (5 17 11) (6 18 12)}$$ $$g^{13} \text{= (1 14 9 4 17 12 7 2 15 10 5 18 13 8 3 16 11 6)}$$ $$g^{14} \text{= (1 15 11 7 3 17 13 9 5) (2 16 12 8 4 18 14 1 6)}$$ $$g^{15} \text{= (1 16 13 10 7 4) (2 17 14 11 8 5) (3 18 15 12 9 6)}$$ $$g^{16} \text{= (1 17 15 13 11 9 7 5 3) (2 18 16 14 12 10 8 6 4)}$$ $$g^{17} \text{= (1 18 17 16 15 14 13 12 11 10 9 8 7 6 5 4 3 2)}$$ After that I found the GCD for all cycle form's with and group them in a table: | Cycle length | Permutations | GCD with 18 | | 1 | $$g^0$$ | GCD(0, 18)=18 | | 2 | $$g^9$$ | GCD(9, 18)=9 | | 3 | $$g^6$$, $$g^{12}$$ | GCD(6, 18)=GCD(12, 18)=6 | | 6 | $$g^3$$, $$g^{15}$$ | GCD(3, 18)=GCD(15, 18)=3 | | 9 | $$g^2$$, $$g^4$$, $$g^8$$, $$g^{10}$$, $$g^{14}$$, $$g^{16}$$ | GCD(2, 18)=GCD(4, 18)=GCD(8, 18)=GCD(10, 18)=GCD(14, 18)=GCD(16, 18)=2 | | 18 | $$g^1$$, $$g^5$$, $$g^7$$, $$g^{11}$$, $$g^{13}$$, $$g^{17}$$ | GCD(1, 18)=GCD(5, 18)=GCD(7, 18)=GCD(11, 18)=GCD(13, 18)=GCD(17, 18)=1 | We have 18 permutations for rotation and lets name cycle 1 with $$f_1$$, cycle 2 with $$f_2$$ .. cycle n with $$f_n$$ Than the formula for Cycling index is: $$\frac{f_1^{18} + f_2^9 + 2f_3^6 + 2f_6^3 + 6f_9^2 + 6f_{18}^1}{18}$$ If we solve all the possible necklaces with three colors the result should be (for the moment we do not solve for the three colors with 4, 6 and 8 beads in respective groups): $$\frac{3^{18} + 3^9 + 2*3^6 + 2*3^3 + 6*3^2 + 6*3^1}{18} = \text{21 524 542}$$ From here because turn around is allowed we need to add and the necklace(bracelet if we follow the right terms) is with even beads we should add the symethric turn arounds. $$\frac{f_1^{18} + f_2^9 + 2f_3^6 + 2f_6^3 + 6f_9^2 + 6f_{18}^1 + 9f_1^2f_2^8 + 9f_2^9}{2 * 18}$$ and again for three colors without including the different weight: $$\frac{3^{18} + 3^9 + 2*3^6 + 2*3^3 + 6*3^2 + 6*3^1 + 9 * 3^8 + 9 * 3^9}{2 * 18} = \text{10 781 954}$$ For the moment we have all the possible necklaces and bracelets with three colors. However, in order to find the necklaces and bracelets with three colors (4 reds, 6 greens and 8 blues) we need to replace: $$f_1 = (x + y + z)$$ $$f_2 = (x^2 + y^2 + z^2)$$ $$f_3 = (x^3 + y^3 + z^3)$$ $$f_6 = (x^6 + y^6 + z^6)$$ $$f_9 = (x^9 + y^9 + z^9)$$ $$f_{18} = (x^{18} + y^{18} + z^{18})$$ and if we replace in the formula it becomes: $$\frac{(x + y + z)^{18} + (x^2 + y^2 + z^2)^9 + 2(x^3 + y^3 + z^3)^6 + 2(x^6 + y^6 + z^6)^3 + 6(x^9 + y^9 + z^9)^2 + 6(x^{18} + y^{18} + z^{18}) + 9(x + y + z)^2(x^2 + y^2 + z^2)^8 + 9(x^2 + y^2 + z^2)^9}{36}$$ Then we need to find which expressions can expand to $$x^4y^6z^8$$. After then by using multinominal coeficient I managed to calculate the following results 9 189 180 1260 11 340 11 340 Then I sum all of them and divide them to 36. This gives me the answer of 255 920 which is the answer of the question. We can create 255 920 bracelets with 4 red 6 green and 8 blue beads.

https://math.stackexchange.com/questions/3060802/is-v-isomorphic-to-direct-sum-of-subspace-u-and-v-u

# Is $V$ isomorphic to direct sum of subspace $U$ and $V/U$? Given a vector space $$V$$ and a subspace $$U$$ of $$V$$. $$V \cong U \oplus(V/U)$$ Does the above equation always hold? Where $$\oplus$$ is external direct sum. For finite dimensional vector space $$V$$, here is my attemp of prove: Let dimension of $$U$$ be m, dimension of $$V$$ be $$n$$. Find a basis of $$U$$ : $$\{ \mathbf{ u_1, u_2, \cdots ,u_m}\}$$ and extend it to a basis for $$V$$ : $$\{ \mathbf{ u_1, u_2, \cdots ,u_m, v_1, \cdots,v_{n-m} } \}$$. For every vector $$\mathbf{x} \in V$$, we can write $$\mathbf{x}= c_1 \mathbf{u_1}+ \cdots+ c_m \mathbf{u_m} + d_1 \mathbf{v_1}+ \cdots d_{n-m} \mathbf{v_{n-m}}$$ uniquely. Define a linear map $$T$$ as $$T(\mathbf{x})=(c_1 \mathbf{u_1}+ \cdots+ c_m \mathbf{u_m}, [d_1 \mathbf{v_1}+ \cdots d_{n-m} \mathbf{v_{n-m}}])$$ ,where $$[]$$ is used to express the equivalent class. We claim $$T$$ is an isomorphism. Surjectivity is obvious. As for injectivity, if $$T(\mathbf{x})=(\mathbf{0},[\mathbf{0}])$$, then $$c_1 \mathbf{u_1}+ \cdots+ c_m\mathbf{u_m}= \mathbf{0}$$ $$\Rightarrow c_1=0, c_2=0, \cdots,c_m=0$$ $$\Rightarrow x=d_1 \mathbf{v_1}+ \cdots d_{n-m} \mathbf{v_{n-m}}$$ Since $$[d_1 \mathbf{v_1}+ \cdots d_{n-m} \mathbf{v_{n-m}}] = [\mathbf{0}]$$, we have $$(d_1 \mathbf{v_1}+ \cdots d_{n-m} \mathbf{v_{n-m}}-\mathbf{0})\in U$$, which means $$d_1=0, d_2=0, \cdots,d_{n-m}=0$$ $$\Rightarrow \mathbf{x}=\mathbf{0}$$, so $$T$$ is injective. Is the above proof correct? Does this mean $$V \cong \ker F \oplus (V/ \ker F) \cong \ker F \oplus\mathrm{im}F$$ for any linear map $$F$$, because $$\ker F$$ is a subspace of $$V$$ ? The final question is about how should I prove it when the dimension of $$V$$ is infinite? It is true that for every finite dimensional vector space $$V$$ with $$U$$ a vector subspace that $$V \cong U \oplus (V / U)$$ I think your proof is essentially correct. And yes it is true that for $$T: V \rightarrow W$$ any linear that we have $$V \cong \operatorname{ker}(T) \oplus \operatorname{Im}(T)$$ The rank-nullity theorem is a direct consequence of this. In more technical language, we say that every "short exact sequence" of finite dimensional vector spaces over a field $$k$$ $$\textit{splits}$$. What this means is that if $$T : U \rightarrow V$$, $$S : V \rightarrow W$$ are linear maps such that $$T$$ is injective, $$\operatorname{ker}(S) = \operatorname{Im}(T)$$, and $$S$$ is surjective, then $$V \cong U \oplus W$$. Note then that this directly gives us our result since if $$U$$ is a subspace of $$V$$, then the inclusion map $$\iota : U \rightarrow V$$ and projection map $$\pi : V \rightarrow (V / U)$$ set up exactly a short exact sequence. In terms of whether or not this extends to infinite dimensional vector spaces, the result does hold again (assuming the axiom of choice), and the proof is essentially the same. All your proof relies on is the ability to extend a basis of a subspace to a basis of your entire space. We can do this with the axiom of choice. • I am wondering if the following statement is false for infinite dimensional spaces. "If $V$ is a vector space and $U$ is a subspace of $V$, then there exists another subspace of $V$ called $U^\perp$ such that every element of $V$ can be uniquely expressed as the sum of an element from $U$ with an element from $U^\perp$." (I am thinking a counterexample would be if $U$ was the set of real number sequences with finite support (i.e. eventually zero) and $V$ is the set of all real number sequences.) – irchans Jan 3 at 19:03 • @irchans If we take the axiom of choice, then every subspace of a vector space has a direct sum complement (what you call the perpendicular space, but this language is typically reserved for a space equipped with some bi-linear form). The proof is pretty simple. Let $V$ be a $k$-vector space, with $U$ a vector subspace. Let $\mathcal{B}_{U}$ be a basis for $U$ and extend it to a basis $\mathcal{B}_{V}$ (using the axiom of choice) for $V$. Then let $W = \operatorname{Span}_{k}\left( \mathcal{B}_{V} \backslash \mathcal{B}_{U} \right)$. Then $V = U \oplus W$. – Adam Higgins Jan 3 at 19:13 • @irchans Perhaps the reason you think that your example is a counter example is because of the $\textit{weirdness}$ of bases of infinite dimensional vector spaces. Notice that a subset $S$ of a vector space $V$ is said to be a basis if and only if every element $v \in V$ can be written as a $\textbf{finite}$ linear combination of the elements of $S$, and that there is no finite non-trivial linear relation amongst the elements of $S$. – Adam Higgins Jan 3 at 19:19 • Thank you very much ! – irchans Jan 3 at 19:27 • This set of notes seems relevant math.lsa.umich.edu/~kesmith/infinite.pdf – irchans Jan 3 at 19:38 Your proof works. The answer to your second question is yes, that is true. For an infinite dimensional vector space, take any linear map $$F: V -> W$$. Then $$U = \ker F$$ is a subspace of $$V$$. Note that we have a short exact sequence (if you don't know what that means, don't worry, the explanation is coming) $$0\to U\to V\to V/U\to 0$$ (That is, there's an injective map $$U \to V$$ (inclusion, I'll call it $$i$$) and a surjective map $$V \to V/U$$ (the quotient map, I'll call it $$q$$) such that the image of the injection is the kernel of the surjection). But there's also a surjective map $$V \to U$$ (projection onto $$U$$, I'll call it $$p$$), and note that for any $$u \in U$$, $$pi(u) = u$$ (since $$p$$ fixes $$u$$). Now, we're going to show that $$V$$ is the (internal) direct sum of the kernel of $$p$$ and the image of $$i$$. First, note that it's the sum of the two: for any $$v \in V$$, $$v = (v - ip(v)) + ip(v)$$, $$ip(v)$$ is obviously in the image of $$i$$, and $$p(v - ip(v)) = p(v) - pip(v) = 0$$ (with the last equality being due to our note about $$p$$, since $$p(v)\in U$$. And further, the intersection is trivial: if $$v \in \ker(p)\cap\mathrm{im}(i)$$ then there is some $$u\in U$$ such that $$i(u) = v$$, and $$pi(u) = p(v) = 0$$, but $$pi(u) = u$$, so $$u = 0$$, hence $$v = 0$$. Thus, $$V = \ker(p)\oplus \mathrm{im}(i)$$. Now, it's clear that $$\mathrm{im}(i)\cong U$$ since it's the image of $$U$$ under an injective map, so we need only show that $$\ker(p)\cong V/U$$. For that purpose, since $$q$$ is surjective, for any $$w \in V/U$$, there is some $$v \in V$$ such that $$w = q(b)$$. But since $$V = \ker(p)\oplus\mathrm{im}(i)$$, there are unique $$u \in U$$, $$x \in \ker(p)$$ such that $$v = i(u) + x$$, so $$w = q(b) = q(i(u)+x) = qi(u) + q(x)=q(x)$$ (since the image of $$i$$ is the kernel of $$q$$, in particular $$qi = 0$$). Thus, $$q|_{\ker(p)}: \ker(p)\to V/U$$ is surjective. But also, if $$q(v) = 0$$, and $$v \in \ker(p)$$. But $$\ker(q) = \mathrm{im}(i)$$, and $$p$$ fixes $$\mathrm{im}(i)$$, so the only element that it sends to $$0$$ is $$0$$ itself, so we must have $$v = 0$$, hence $$q|_{\ker(p)}$$ is also injective, so is an isomorphism. Thus, we have $$V \cong \mathrm{im}(i)\oplus\ker(p) \cong U\oplus (V/U)$$, as required. This is precisely a special case of the Splitting Lemma, and my proof is essentially just one part of the proof of that, translated: there are easier ways to prove it, but I thought that it would be useful to see it in a more broadly applicable form.

https://math.stackexchange.com/questions/2928956/query-on-a-solution-to-the-problem-gcd5a2-7a3-1-for-all-integer-a

# Query on a Solution to the Problem: $\gcd(5a+2,7a+3)=1$ for all integer $a$. I wish to show that the numbers $$5a+2$$ and $$7a+3$$ are relatively prime for all positive integer $$a$$. Here are my solutions. Solution 1. I proceed with Euclidean Algorithm. Note that, for all $$a$$, $$|5a+2|<|7a+3|$$. By Euclidean Algorithm, we can divide $$7a+3$$ by $$5a+2$$. To have $$7a+3=(5a+2)(1)+(2a+1)$$ continuing we have, $$5a+2=(2a+1)(2)+a$$ $$2a+1=(2)(a)+1$$ $$2=(1)(2)+0$$ Since the last nonzero remainder in the Euclidean Algorithm for $$7a+3$$ and $$5a+2$$ is 1, we conclude that they are relatively prime. Solution 2. Suppose that $$d=\gcd(5a+2,7a+3)$$. Since $$d=\gcd(5a+2,7a+3)$$ then the following divisibility conditions follow: (1) $$d\mid (5a+2)$$ (2) $$d\mid (35a+14)$$ (3) $$d\mid (7a+3)$$ (4) $$d\mid (35a+15)$$. Now, (2) and (4) implies that $$d$$ divides consecutive integers. The only (positive) integer that posses this property is $$1$$. Thus, $$d=1$$ and that $$7a+3$$ and $$5a+2$$ are relatively prime. Here are my questions: 1. Is the first proof correct or needs to be more specific? For instance cases for $$a$$ must be considered. 2. Which proof is better than the other? Thank you so much for your help. • I think both approaches are good. Nor are they that different really...in both cases you are trying to find smaller and smaller multiples of $d$. – lulu Sep 24 '18 at 15:23 • Thank you very much for the kind comment @lulu. Got a follow up question. By "you are trying to find smaller and smaller multiples of $d$, you mean the process of continously dividing the divisor to remainder so that $r$ decreases? Sep 24 '18 at 15:27 • Is there a typo in (4), i.e. shouldn't 35 be multiplied by $a$? – user431008 Sep 24 '18 at 15:28 • I meant something less precise than that. Euclid provides a somewhat systematic way to find new numbers that $d$ divides...the second method is less systematic, but faster as you seek out convenient expressions. Working by hand, I prefer the second method...were I trying to automate the process, I'd prefer the systematic approach. – lulu Sep 24 '18 at 15:28 • I agree @marmot. Thank you for pointing it out. Sep 24 '18 at 15:29 In the first solution you're not using, strictly speaking, the Euclidean algorithm, but a looser version thereof: Let $$a$$, $$b$$, $$x$$ and $$y$$ be integers; if $$a=bx+y$$, then $$\gcd(a,b)=\gcd(b,y)$$. The proof consists in showing that the common divisors of $$a$$ and $$b$$ are the same as the common divisors of $$b$$ and $$y$$. There is no requirement that $$a\ge b$$ or that $$y$$ is the remainder of the division. Indeed your argument actually has a weakness, because $$7a+3\ge 5a+2$$ only if $$2a\ge-1$$, so it doesn't hold for $$a\le-2$$. But $$7a+3\ge 5a+2$$ is not really needed for the argument. Since successive application of the statement above show that $$\gcd(2a+1,2)=1$$ and the gcd has never changed in the various steps, you can indeed conclude that $$\gcd(5a+2,7a+3)=1$$. The second solution is OK as well. You can simplify it by noting that if $$d$$ is a common divisor of $$5a+2$$ and $$7a+3$$, then it divides also $$5(7a+3)-7(5a+2)=1$$ • Thanks for the clarification in solution 1 and by giving a simplified version for solution 2. It is very clear to me now. @egreg Sep 24 '18 at 15:35 Your first proof is correct. I would perhaps complete it, writing that\begin{align}1&=(2a+1)-2a\\&=2a+1-2\bigl(5a+2-2(2a+1)\bigr)\\&=5(2a+1)-2(5a+2)\\&=5\bigl(7a+3-(5a+2)\bigr)-2(5a+2)\\&=5(7a+3)-7(5a+2).\end{align} Your second proof also works, but it doesn't generalize easily to other situations. • Thank you very much for completing the proof of solution 1 Prof. I am interested in a case that solution 2 wont work. Thank you Prof. Sep 24 '18 at 15:41 • @JrAntalan Concerning the second proof, I only meant that you have to think about it in a case-by-case basis. Sep 24 '18 at 15:43 • Noted Prof. and Thank you again. Sep 24 '18 at 15:44 Here is a different rendering of the same arguments. Let $$u=5a+2$$, $$v=7a+3$$. Then $$\pmatrix{ u \\ v} = \pmatrix{ 5 & 2 \\ 7 & 3} \pmatrix{ a \\ 1}$$ and so $$\pmatrix{ a \\ 1} = \pmatrix{ 5 & 2 \\ 7 & 3}^{-1} \pmatrix{ u \\ v} = \pmatrix{ \hphantom- 3 & -2 \\ -7 & \hphantom-5} \pmatrix{ u \\ v}$$ This gives $$1 = -7u+5v = -7(5a+2)+5(7a+3)$$ The key point here is that the matrix has determinant $$1$$ and so its inverse has integer entries.

https://math.stackexchange.com/questions/401753/for-any-arrangment-of-numbers-1-to-10-in-a-circle-there-will-always-exist-a-pai

# For any arrangment of numbers 1 to 10 in a circle, there will always exist a pair of 3 adjacent numbers in the circle that sum up to 17 or more I set out to solve the following question using the pigeonhole principle Regardless of how one arranges numbers $1$ to $10$ in a circle, there will always exist a pair of three adjacent numbers in the circle that sum up to $17$ or more. My outline [1] There are $10$ triplets consisting of adjacent numbers in the circle, and since each number appears thrice, the total sum of these adjacent triplets for all permutations of the number in the circle, is $3\cdot 55=165$. [2] If we consider that all the adjacent triplets sum to 16 , and since there are $10$ such triplets, the sum accordingly would be $160$, but we just said the invariant sum is $165$ hence there would have to be a triplet with sum of $17$ or more. My query Could someone polish this into a mathematical proof and also clarify if I did make use of the pigeonhole principle. • Yes, you used pigeonhole. Should say if all sums are $\le 16$, then the sum is $\le 160$. – André Nicolas May 25 '13 at 4:42 • @AndréNicolas noted and aha , the syntax part is exactly why I asked the question on here. – metric-space May 25 '13 at 4:44 • @nerorevenge Note: You can actually show that the 3 adjacent numbers sum to 18 or more. – Calvin Lin May 25 '13 at 8:37 Yes, you used the Pigeonhole Principle. As a very mild correction, you should say that of all sums of three consecutives are $\le 16$, then the sum is $\le 160$. The proof (with the small correction) is already fully mathematical. Conceivably you might want to explain the $55$ further. It is clear to you and to most users of this site why $55$, but imagine the reader to be easily puzzled. The part about "all the permutations" is vague, and technically incorrect. You are finding the sum of a consecutive triple, and summing all these sums. Permutations have nothing to do with it, we are talking about a particular fixed arrangement of the numbers. Remark: We could use a lot of symbols. Starting at a particular place, and going counterclockwise, let our numbers be $a_1,a_2,\dots,a_{10}$. For any $i$, where $1\le i\le 10$, let $S_i=a_i+a_{i+1}+a_{i+2}$, where we use the convention that $a_{10+1}=a_1$, $a_{10+2}=a_2$, and $a_{9+2}=a_1$. Then $S_1+S_1+\cdots+S_{10}=165$, since each of $1$ to $10$ appears in $3$ of the $S_i$, and $1+2+\cdots+10=55$. But if all the $S_i$ are $\le 16$, then $\sum_{i=1}^{10}S_i\le 160$. This contradicts the fact that $\sum_{i=1}^{10}S_i=165$. I prefer your proof, mildly modified. • eh, is that all? the reason I'm asking is that, inb the past my so called 'proofs' have been criticised for being too informal – metric-space May 25 '13 at 4:45 • @nerorevenge: I am an advocate of the informal but clear. I hope that someone more formal-minded will add an answer. The only issue is that you perhaps left an easily filled gap or two. – André Nicolas May 25 '13 at 4:53 • I have added another criticism. You should really fix the wording there. – André Nicolas May 25 '13 at 4:58 • I agree.That definitely is vaguely worded. – metric-space May 25 '13 at 4:59 • Presumably the "pair" wording came with the problem. But it looks funny to call a trio a pair. – André Nicolas May 25 '13 at 5:23 We will show something stronger, namely that there exists 3 adjacent numbers that sum to 18 or more. Let the integers be $\{a_i\}_{i=1}^{10}$. WLOG, $a_1 = 1$. Consider $$a_2 + a_3 + a_4, a_5 + a_6 + a_7, a_8 + a_9 + a_{10}$$ The sum of these 3 numbers is $2+3 +\ldots + 10 = 54$. Hence, by the pigeonhole principle, there exists one number which is at least $\lfloor \frac{54}{3} \rfloor = 18$. I leave it to you to show that there is a construction where no 3 adjacent numbers sum to 19 or more, which shows that 18 is the best that we can do. • definitely a stronger proof ,and I have admit it sure is nice, I wonder why cut-the-knot.org didn't ask people to solve for what you proved. – metric-space May 25 '13 at 16:45 • @nerorevenge I used this as a problem on Brilliant.org, and most could only get to 17, and didn't understand why 18 must be true. The 18 case arises quite easily by ignoring the small values which would otherwise reduce our sum. – Calvin Lin May 25 '13 at 23:37

https://math.stackexchange.com/questions/3473894/to-find-number-of-real-roots

# To find number of real roots Consider the equation $$x^5-5x=c$$ where c is a real number. Determine all c such that this equation has exactly 3 real roots. I know that between consecutive real roots of $$f$$ there is a real root of $$f'$$. Now $$f'$$ in this case is $$5x^4-5$$ which always has two real roots. So the claim should be true for all c. But I KNOW IT IS NOT TRUE. Where am I messing up? • Note that while it is necessary for $f'$ to have two real roots (in order for $f$ to have exactly three real roots), it is not sufficient. Of course an odd degree polynomial will always have one real root. – hardmath Dec 12 '19 at 16:12 • ok so how do I find all such c? – Angry_Math_Person Dec 12 '19 at 16:14 • Consider the graph of $p(x) = x^5 - 5x$. Changing the constant $c$ in your equation amounts to moving a horizontal line up or down across this graph. – hardmath Dec 12 '19 at 16:15 Yes, between any two roots of $$f$$, there is a root of $$f'$$. However, just because $$f'$$ has a root, that doesn't mean that $$f$$ has a root on either side. Consider $$f(x)=x^2+1$$. As for solving this problem, the derivative has only two roots, so we can at most have three roots. For some values of $$c$$ we have three roots, for some values of $$c$$ we have a single root, and for exactly two values of $$c$$, there are two roots. The three-root region is exactly the interval between the two two-root values of $$c$$. And finding the values of $$c$$ that gives two roots is easier than one might think. They happen exactly when one root of $$f$$ coincides with a root of $$f'$$. So find the roots of $$f'$$, and find the values of $$c$$ that make each of them a root of $$f$$, and you have found the interval of $$c$$-values that gives three roots. As you know, the given equation has extrema at $$x=\pm1$$. These correspond to values of the polynomial $$1-5-c$$ and $$1+5-c$$ (the RHS was moved to the left). Hence the polynomial will grow from $$-\infty$$, reach the maximum, then the minimum and continue growing to $$\infty$$. There are three roots when $$0$$ is in the range $$(-4-c,6-c)$$. Where am I messing up? Just look at a graph where it fails. $$x^5-5 x -5$$ I know that between consecutive real roots of f there is a real root of f′. Now f′ in this case is 5x4−5 which always has two real roots. So the claim should be true for all c. $$A \implies B$$ does not mean $$B \implies A$$. The two real roots of $$5x^4 - 5$$ are the two roots at $$x = \pm 1$$. If $$x^5 - 5x=c$$ has three roots then they will be at $$x < -1; -1 < x < 1;$$ and at $$x > 1$$ by your condition. But there won't be three real roots if there is no root for any $$x< -1$$, or no root between $$-1$$ and $$1$$, or no root for any $$x < -1$$. $$x=\pm 1$$ are extreme points and if one, the max, is $$>0$$ and the other $$<0$$ then there will be three real roots. But if both are "on the same side of $$0$$" there is no root between them and no root to "the other side". $$x^5-5x -c|_{-1} = 4-c$$ and $$x^5 - 5x -c|1 = -4-c$$ so $$x =-1$$ is a max and $$x = 1$$ is a min. If $$f(-1) = 4-c \le 0$$ is a max there will be no root for $$x < -1$$ or for $$-1 < x \le 1$$. If $$f(1) = -4-c\ge 0$$ is a min there will be no root for $$x > 1$$ or for $$-1 \le x < 1$$. So if either $$c \ge 4$$ or if $$c\le 4$$ then there are fewer than three real roots. But if $$-4 < c < 4$$ then there will be three. Alternatively: we know what the shape of an odd polynomial $$x^5 -5x$$ looks like. It's that polynomial curve with a twisty bit in the middle, goes off to infinity as $$x \to \infty$$, goes to negative infinity as $$x \to -\infty$$. It has roots were the x-axis crosses it (or where it crosses the x-axis-- everything is relative). If we shift it up or shift it down by $$c$$ we can force the x-axis to avoid the twisty bits in the middle and have it have only one root. Or we can deliberately shift it so that the x-axis goes smack through the twisty bits and we have a maximum number of roots. So if $$c$$ is between the max and mins we maximize the number of roots and the x-axis goes through the twisty bits. If $$c$$ is beyond the max an mins we've shoved the twisty bits below or above the x-axis and there is only one root.

https://www.physicsforums.com/threads/difference-between-continuity-and-uniform-continuity.812734/

# Difference between continuity and uniform continuity Tags: 1. May 7, 2015 ### Yunjia I noticed that uniform continuity is defined regardless of the choice of the value of independent variable, reflecting a function's property on an interval. However, if on a continuous interval, the function is continuous on every point. It seems that the function on that interval must be uniformly continuous. Is this correct? Is there a counterexample for the statement? 2. May 7, 2015 ### jbunniii What you said is almost true. If, on a closed, bounded interval, a function is continuous at every point, then the function is uniformly continuous on that interval. Counterexample on a non-closed interval: $f(x) = 1/x$ on the interval $(0,1)$. Counterexample on a closed but unbounded interval: $f(x) = x^2$ on the interval $[0,\infty)$. 3. May 8, 2015 ### Svein Even more general: If a function is continuous at every point in a compact set, it is uniformly continuous on that set. The proof is simple: Let the compact set be K and take an ε>0. Since the function is continuous at every point x in the set, there is a δx for every x∈K such that |f(w)-f(x)|<ε for every w in <x-δx, x+δx>. Let Ox=<x-δx, x+δx>, then K is contained in $\bigcup_{x\in K}O_{x}$. Since K is compact, it is contained in the union of a finite number of the Ox, say $\bigcup_{n=1}^{N}O_{n}$. Take δ to be the minimum of the δn and |f(w)-f(x)|<ε whenever |w-x|<δ. 4. May 8, 2015 ### jbunniii And here's the proof that a closed, bounded interval is compact, so you can apply Svein's proof. Let $[a,b]$ be a closed, bounded interval, and let $\mathcal{U}$ be any collection of open sets which covers $[a,b]$. Let $S$ be the set of all $x \in [a,b]$ such that $[a,x]$ can be covered by finitely many of the sets in $\mathcal{U}$. Clearly $a \in S$. This means that $S$ is nonempty and is bounded above (by $b$), so it has a supremum, call it $c$. Since $c \in [a,b]$, it is contained in some $U_c \in \mathcal{U}$, hence there is some interval $(c - \epsilon, c + \epsilon)$ contained in $U_c$. Since only finitely many sets from $\mathcal{U}$ are needed to cover $[a,c - \epsilon/2]$, those sets along with $U_c$ form a finite cover of $[a,c+\epsilon/2]$. This shows that $c \in S$ and moreover, that $c$ cannot be less than $b$. Therefore $c=b$, so all of $[a,b]$ can be covered by finitely many sets in $\mathcal{U}$. Last edited: May 8, 2015 5. May 8, 2015 ### Svein Excellent! And, of course, since one closed and bounded interval is compact, the union of a finite number of closed and bounded intervals is again compact. 6. May 8, 2015 ### jbunniii I don't think this will work if $w$ and $x$ are not contained in the same $O_n$. I think you need to take $O_x = (x - \delta_x/2, x + \delta_x/2)$ and $\delta$ to be $\min\{\delta_n/2\}$ in order to ensure that $|w-x| < \delta$ implies $|f(w) - f(x)| < \epsilon$. 7. May 8, 2015 ### Svein Possibly. I need to think about that. The crux of the matter is that there is a finite number of intervals that cover K, which means that the minimum of the (finite number of) δ's exist and is greater than 0. 8. May 8, 2015 ### HallsofIvy Here is the fundamental difference between "continuous" and "uniformly continuous": A function is said to be continuous at a point, x= a, if and only if, given any $\epsilon> 0$ there exist $\delta> 0$ such that if $|x- a|< \delta$ then $|f(x)- f(a)|<\epsilon$. A function is said to be continuous on a set, A, if and only if, given any $\epsilon> 0$ there exist $\delta> 0$ such that if $|x- a|< \delta$ then $|f(x)- f(a)|<\epsilon$ for all a in set A. That is, a function is uniformly continuous on a set A if and only if it is continuous at every point in A and, given $\epsilon> 0$ the same $\delta$ can be used for ever point in A. 9. May 8, 2015 ### Svein Agree. I was sloppy and overlooked the simple fact that that the length of the interval Ox is 2⋅δx. 10. May 13, 2015 ### Yunjia The defition of uniformly continuous is perplexing for me. Is it possible for you to illustrate it with a proof about a function which is not uniformly continuous but continuous on a set so that I can compare the two? Thank you. 11. May 14, 2015 ### Svein OK. Take the function $f(x)=\frac{1}{x}$ on the open interval <0, 1>. It is continuous (and even differentiable) on that interval, but not uniformly continuous. 12. May 17, 2015 ### Svein After using pencil an paper for a bit, I came to the conclusion that I should have used ε/2 and δ/2. But - I recall a proof in "Complex analysis in several variables" that ended up in "... less than 10000ε, which is small when ε is small". 13. May 17, 2015 ### HallsofIvy First note an important point about "uniform continuity" and "continuity" you may have overlooked: we define "continuous" at a single point and then say that a function is "continuous on set A" if and only if it is continuous at every point on A. We define "uniformly continuous" only on a set, not at a single points of a set. A function is uniformly continuous on any closed set on which it is continuous and so on any set contained in a closed set on which it is continuous. To give an example of a function that is continuous but not uniformly continuous, we need to look at f(x)= 1/x on the set (0, 1). To show that it is continuous on (0, 1), let a be a point in (0, 1) and look at $|f(x)- f(a)|= |1/x- 1/a|= |a/ax- x/ax|= |(a- x)/ax|= |x- a|/ax< \epsilon$. We need to find a number, $\delta> 0$ such that if $|x- a|< \delta$, then $|f(x)- f(a)|> \epsilon$. We already have $|x- a| ax\epsilon$ so we need an upper bound on ax. If we start by requiring that $\delta< a/2$ then $|x- a|< \delta< a/2$ so that $-a/2< x- a< a/2[itex] or [itex]a/2< x< 3a/2$ so an upper bound on ax is $3a^2/2$. If $|x- a|< a/2$ and $|x- a|< 3a/2$ then $|f(x)- f(a)|< |x- a|/ax< |x- a|/(3a^2/2)= 2|x- a|/3a^2$ which will be less than $\epsilon$ as long as $|x- a|< 3a^2\epsilon/2$ So we can take $\delta$ to be the smaller of $a/2$ and $3a^2/\epsilon$. Therefore, 1/x is continuous at any point a in (0, 1) so continuous on (0, 1). Now the point is that this $\delta$ depends on a. It is a decreasing function of a and will go to 0 as a goes to 0. That's fine for "continuity" but for uniform continuity we must be able to use the same $\delta> 0$ for a given $\epsilon$ no matter what the "a" is and we cannot do that. If the problem were to prove uniform continuity on the set [p, 1), which is contained in the closed set [p, 1], we could use, for any a in that set, the $\delta$ that we get for a= p, the smallest x value in the set. But with (0, 1), the smallest $\delta$ is 0 which we cannot use since we must have $\delta> 0$.

http://oanx.cfalivorno.it/linear-algebra-and-learning-from-data-github.html

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Those equations may or may not have a solution. " Our homework assignments will use NumPy arrays extensively. This book provides the conceptual understanding of the essential linear algebra of vectors and matrices for modern engineering and science. A good online mathbook on the topic is immersive linear algebra. Everything about Data Science, Machine Learning, Analytics, and AI provided in one place! randylaosat. provide a summary of the mathematical background needed for an introductory class in machine learning, which at UC Berkeley is known as CS 189/289A. Franklin, Beedle & Associates Inc. ML helps if one has solid understanding on Linear Algebra, Probability and Statistics. Studying vector spaces will allow us to use the powerful machinery of vector spaces that has been. This library holds the principal work done as part of the OpenAstonomy Google Summer of Code 2020 project, Solar Weather Forecasting using Linear Algebra. As a core programmer, I love taking challenges and love being part of the solution. A computer science student that is interested in Machine Learning would be well advised to get a minor in Mathematics (or just get a degree in Mathematics instead!). The basic mathematics prerequisites for understanding Machine Learning are Calculus-I,II,III, Linear Algebra, and, Probability and Statistics. We start with representing a fully connected layer as a form of matrix multiplication: - In this example, the weight matrix has a size of $4 \times 3$, the input vector has a size of $3 \times 1$ and the output vector has a of size $4 \times 1$. Currently he is on leave from UT Austin and heads the Amazon Research Lab in Berkeley, California, where he is developing and deploying state-of-the-art machine learning methods for Amazon Search. linear; algebra; matrix;. for automated market making. In addition to this, you'll be able to perform operations such as addition, subtraction and dot product. Description. learn and also known as sklearn) is a free software machine learning library for the Python programming language. By mistake uploading only a part of your vertex data is a common mistake due to miscalculations. scikit-learn is a comprehensive machine learning toolkit for Python. hdf5-OCaml: OCaml implementation of hdf5 reader/writer. Learning rate for to use in SGD; Linear Algebra, Graphs, and Automatic Differentiation. Students will learn and practice fundamental ideas of linear algebra and simultaneously be exposed to and work with real-world applications of these ideas. Machine learning yearning. The course is project based and through the various projects, normally three, the students will be exposed to fundamental research problems in these fields, with the aim to reproduce state of the art scientific results. If we are thinking of a vector as representing the line segment from the origin to a given point (i. on big time series mining:. Data science, machine learning. It introduces some common tools in machine learning to resolve real applications (e. In this first module we look at how linear algebra is relevant to machine learning and data science. Topics: Python NLP on Twitter API, Distributed Computing Paradigm, MapReduce/Hadoop & Pig Script, SQL/NoSQL, Relational Algebra, Experiment design, Statistics, Graphs, Amazon EC2, Visualization. Docker: https://github. I’m studying towards a PhD degree at the University of Chicago, in the department of Statistics. Robert Beezer, A First Course in Linear Algebra comes with a solution manual. In the first part, we give a quick introduction to classical machine learning and review some key concepts required to understand deep learning. for automated market making. \ \ **References**\ \ - Belsley. , and Courville, A. js Downloading YouTube videos using youtube-dl embedded with Python Machine Learning : scikit-learn Django 1. what-is-the-difference-between-artificial-intelligence-and-machine-learning 9. As we will see, we can do all the common linear algebra operations without using any other library. Learn linear algebra. It turns out, however, that all of those operations can be written in terms of big matrix-matrix or matrix-vector multiplications. , the geometric interpretation), we may interpret the norm as the length of this line segment. You'll understand why I placed functions in quotes shortly. Linear Algebra Examines basic properties of systems of linear equations, vector spaces, inner products, linear independence, dimension, linear transformations, matrices, determinants, eigenvalues, eigenvectors and diagonalization. Announcements and latest whereabouts. Implementation [ edit ] Scikit-learn is largely written in Python, and uses numpy extensively for high-performance linear algebra and array operations. Now I’m checking back in with 9 weeks under my belt. Omoju: What I do at GitHub is I build data models, often deep learning models on GitHub data to help GitHub probably build things like a recommendation engine so we can recommend repositories to people. However, this fact seems to change in the next years, after the hype of Machine Learning we are facing a process of democratization. It aims to provide intuitions/drawings/python code on mathematical theories and is constructed as my understanding of these concepts. Just want some books to go deeper than a introductory course. This book provides the conceptual understanding of the essential linear algebra of vectors and matrices for modern engineering and science. Linear algebra is one of the most applicable areas of mathematics. In this first module we look at how linear algebra is relevant to machine learning and data science. Linear transformations and change of basis (connected to the Singular Value Decomposition - orthonormal bases that diagonalize A) Linear algebra in engineering (graphs and networks, Markov matrices, Fourier matrix, Fast Fourier Transform, linear programming) Homework. This book is directed more at the former audience. As the complementary course to. and engineering. NumPy is "the fundamental package for scientific computing with Python. Linear Algebra is a text for a first US undergraduate Linear Algebra course. Bourbaki resulted from similar currents of thought that produced fascism and totalitarian communism: moral panics leading to revolutions, and ultimately “final solutions”, all terrible and evil in. Complete Linear Algebra for Data Science & Machine Learning 4. Updated Apr 1 2020. In an image classification problem, we often use neural networks. His research interests span statistical machine learning, numerical linear algebra, and random matrix theory. Included is a learning guide and syllabus to help you learn data science this year. Many courses are offered there from which one can benefit. Linear Algebra¶ Now that you can store and manipulate data, let's briefly review the subset of basic linear algebra that you will need to understand most of the models. Students will learn and practice fundamental ideas of linear algebra and simultaneously be exposed to and work with real-world applications of these ideas. In this tutorial, we walked through one of the most basic and important regression analysis methods called Linear Regression. com, My Github, LinkedIn, & Google Scholar I am an applied mathematician and computer scientist with experience in algorithms, linear algebra, and graph theory. One of the most beautiful and useful results from linear algebra, in my opinion, is a matrix decomposition known as the singular value decomposition. Linear algebra moves steadily to n vectors in m-dimensional space. If you have some background in basic linear algebra and calculus, this practical book introduces machine-learning fundamentals by showing you how to design systems capable of detecting objects in images, understanding text, analyzing video, and predicting. As a core programmer, I love taking challenges and love being part of the solution. in Applied Numerical Linear Algebra Currently focused on combining statistical data mining techniques and traditional econometrics approach, panel data, model switching, GARCH type volatility modelling, volume modelling etc. If we are thinking of a vector as representing a physical quantity. To use MOE, we simply need to specify some objective function, some set of parameters, and any historical data we may have from previous evaluations of the objective function. If you want to fit a model of higher degree, you can construct polynomial features out of the linear feature data and fit to the model too. , Bengio, Y. The interplay of columns and rows is the heart of linear algebra. You've accumulated a good bit of data that looks like this:. BUT Linear Algebra is too boundless! In this book, you will get what is NECESSARY. There is no doubt that linear algebra is important in machine learning. The Linear Algebra Chapter in Goodfellow et al is a nice and concise introduction, but it may require some previous exposure to linear algebra concepts. pdf; TS CH8 Estimation. then this is the book for you. and his book on Linear Algebra is a very good introduction. Linear Algebra for Data Scientists. A lot of linear algebra over the complex field This free book Linear Algebra - As an Introduction to Abstract Mathematics from UC Davis has plenty of exercises; Terence Tao has a set of notes if you google, they go with the book Linear Algebra by Friedberg, Insel and Spence. Programming and data science articles by hadrienj. Linear Algebra: Foundations to frontiers – edx. This works in the latest snapshot of Breeze. Linear Algebra Examines basic properties of systems of linear equations, vector spaces, inner products, linear independence, dimension, linear transformations, matrices, determinants, eigenvalues, eigenvectors and diagonalization. GitHub; LinkedIn; Twitter; Donald Miner (@donaldpminer) specializes in large-scale data analysis enterprise architecture and applying machine learning to real-world problems. Windows-64 (64-bit linear algebra for large data) Unless your computer has more than ~32GB of memory and you need to solve linear algebra problems with arrays containing more than ~2 billion elements, this version will offer no advantage over the recommended Windows-64 version above. Modern statistics is described using the notation of linear algebra and modern statistical methods harness the tools of linear algebra. As a core programmer, I love taking challenges and love being part of the solution. The aim of these notebooks is to help beginners/advanced beginners to grasp linear algebra concepts underlying deep learning and machine learning. Matrices, vectors, and more - from theory to the real world! There's a lot of data out there, learn how to search it effectively. A few weeks ago, I wrote about how and why I was learning Machine Learning, mainly through Andrew Ng’s Coursera course. LINEAR ALGEBRA. Today, I will be sharing with you my C# implementation of basic linear algebra concepts. Hundreds of thousands of students have already benefitted from our courses. Linear Algebra and Learning from Data by Gilbert Strang; 1 edition; twitter github. Some of these assignments are from Introduction to applied linear algebra - vectors, matrices, and least squares. Introduction to linear algebra (Fourth Edition). , Bengio, Y. Linear Regression 101 (Part 1 - Basics) 6 minute read Introduction. My work includes researching, developing and implementing novel computational and machine learning algorithms and applications for big data integration and data mining. Various CNN and RNN models will be covered. Learning Spark : lightning-fast data analytics by Holden Karau, Andy Konwinski, Patrick Wendell, and Matei Zaharia, O’Reilly, 2015. Thesis: Quantum Algorithms for Linear Algebra and Machine Learning; Anupam Prakash Quantum Algorithms for Linear Algebra and Machine Learning by Anupam Prakash Most quantum algorithms o ering speedups over classical algorithms are based on the three techniques of phase estimation, amplitude estimation and Hamiltonian simulation. Implementation [ edit ] Scikit-learn is largely written in Python, and uses numpy extensively for high-performance linear algebra and array operations. Currently he is on leave from UT Austin and heads the Amazon Research Lab in Berkeley, California, where he is developing and deploying state-of-the-art machine learning methods for Amazon Search. Support Stability of Maximizing Measures for Shifts of Finite Type Journal of Ergodic Theory and Dynamical Systems (accepted) Calkins, S. Those equations may or may not have a solution. Problem solving with algorithms and data structures using Python. Posted by u/[deleted] a linear algebra library in R designed for teaching. Foundations of Data Science is a treatise on selected fields that form the basis of Data Science like Linear Algebra, LDA, Markov Chains, Machine Learning basics, and statistics. in machine learning, it is standard to say “N samples” to mean the same thing. 065 Linear Algebra and Learning from Data New textbook and MIT video lectures OCW YouTube; 18. pdf; TS CH8 Estimation. Government and Fortune 500 companies. These functions are mainly for tutorial purposes in learning matrix algebra ideas using R. Siefken, J. Complete Linear Algebra for Data Science & Machine Learning 4. A few weeks ago, I wrote about how and why I was learning Machine Learning, mainly through Andrew Ng’s Coursera course. You will also learn how you should use linear algebra in your Python code. in linear algebra, kernel is another name for nullspace. Now that you understand the key ideas behind linear regression, we can begin to work through a hands-on implementation in code. Linear algebra has had a marked impact on the field of statistics. 1-11, [Online-Edition. " Our homework assignments will use NumPy arrays extensively. squares methods, basic topics in applied linear algebra. Perception, movement control, reinforcement learning, mathematical psychology, … Economics. As the complementary course to. Then last year I learned how he morphed his delightful mathematics book into a brand new title (2019) designed for data scientists - "Linear Algebra and Learning from Data. In some cases, functions are provided for concepts available elsewhere in R, but where the function call or name is not obvious. The concepts of Linear Algebra are crucial for understanding the theory behind Machine Learning, especially for Deep Learning. Course Description. Julia Observer helps you find your next Julia package. Linear algebra underlies many practical mathematical tools, such as Fourier series and computer. ” See Section 6. Description. Foundations of Data Science is a treatise on selected fields that form the basis of Data Science like Linear Algebra, LDA, Markov Chains, Machine Learning basics, and statistics. [Online book] n Andrew Ng. This class is an in-depth graduate lecture class. uk, [email protected] Together with your editor or Jupyter notebook these packages allow you to rapidly develop scalable, high-performance analytics and visualizations using succinct, type-safe, production-ready code. Book: Aurélien Géron "Hands-On Machine Learning with Scikit-Learn and TensorFlow" Book: Andriy Burkov "The Hundred-Page Machine Learning Book " 🐍 Python Course: Python. It would be best if you had an organized book which (1) teaches the most used Linear Algebra concepts in Machine Learning, (2) and utilize your Machine Learning model in terms of data processing, optimization, and validation. They give you better intuition for how algorithms really work under the hood, which enables you to make better decisions. This page has links for latest PDF versions of the text and related supplements. * Ranked among top 10% answerers on Python in StackOverflow. Now we extend linear algebra to convolutions, by using the example of audio data analysis. He was a research fellow with Michael Jordan and Peter Bartlett, University of California at Berkeley, from 2003, and with Bernhard Schoelkopf, Max Planck Institute for Intelligent Systems, Tuebingen, Germany, from 2005. 065 Linear Algebra and Learning from Data New textbook and MIT video lectures OCW YouTube; 18. Learning path Introduction to Linear Algebra. Learn the basic math for Data Science, AI, and ML using R. This book is directed more at the former audience. Getting started with linear algebra. Feature learning and subspaces (Chapter 3 of [BHK]) Random walk and Markov chain Monte Carlo (Chapter 4 of [BHK]) Linear models, kernel methods, and deep learning ([Bishop] and Chapter 5 of [BHK]) Algorithms for Massive Data: streaming, sketching, and sampling (Chapter 6 of [BHK]) The course schedule is available here. Incorporating machine learning capabilities into software or apps is quickly becoming a necessity. Implementation [ edit ] Scikit-learn is largely written in Python, and uses numpy extensively for high-performance linear algebra and array operations. His main research interests are in big data, machine learning, network analysis, linear algebra and optimization. We create tools for phenotype analyses that make use of the entire clinical phenotyping spectrum, not only using HPO, but also model organisms data (we also create the uberpheno) and other ontologies. Machine Learning is built on prerequisites, so much so that learning by first principles seems overwhelming. Modern statistics is described using the notation of linear algebra and modern statistical methods harness the tools of linear algebra. Students will learn and practice fundamental ideas of linear algebra and simultaneously be exposed to and work with real-world applications of these ideas. If is high, the matrix is said to be ill-conditioned. Matrix sketching and randomized matrix computation. I’d like to go over the theory behind this matrix decomposition and show you a few examples as to why it’s one of the most useful mathematical tools you can have. It’s all vectors and matrices of numbers. Topic 1: Review of Linear Algebra 1-6 These are only a few examples that I hope help convince you that vector spaces are the backbone of machine learning. Linear Algebra¶ Now that you can store and manipulate data, let’s briefly review the subset of basic linear algebra that you will need to understand most of the models. •LACore is a Large-Format vector accelerator for a broad range of Linear Algebra applications •LACore has novel architectural features including as the: • configurable, data-streaming LAMemUnits • dual-precision, configurable, systolic LAExecUnit •A compiler toolchain, programming framework and architectural simulator were all. For simple linear regression, one can choose degree 1. After reading this post, you will know:. Python for Data Science and Machine Learning Bootcamp; Think Stats - Book. I have hands-on experience in Data Analysis, Machine Learning, Natural Language Processing, deployment on IaaS like AWS. , how to pass the course, schedules, and deadlines, at the official course page. Linear Algebra for Machine Learning Book. Currently he is on leave from UT Austin and heads the Amazon Research Lab in Berkeley, California, where he is developing and deploying state-of-the-art machine learning methods for Amazon Search. for automated market making. Grading: 3 homeworks 60%, 2 quizzes 20%, 1 project 20%. edu ABSTRACT Accelerating machine learning (ML) over relational data is a. , Bengio, Y. This repository contains all the quizzes/assignments for the specialization "Mathematics for Machine learning" by Imperial College of London on Coursera. Some recent tutorials by Christos and Co. Lawrence [email protected] Welcome to Data analysis with Python - 2020¶. The aim of these notebooks is to help beginners/advanced beginners to grasp linear algebra concepts underlying deep learning and machine learning. A vector in Rn will be denoted as: ~x. Enabling and Optimizing Non-linear Feature Interactions in Factorized Linear Algebra Side Li University of California, San Diego [email protected] HarvardX Biomedical Data Science Open Online Training In 2014 we received funding from the NIH BD2K initiative to develop MOOCs for biomedical data science. Now I’m checking back in with 9 weeks under my belt. Introduction to Machine Learning for Data Science, Udemy. Learn linear algebra. The first part covers basics and preliminaries. Incorporating machine learning capabilities into software or apps is quickly becoming a necessity. Some recent tutorials by Christos and Co. pyplot as plt # pyplot from collections import defaultdict , Counter from functools import partial , reduce. pyplot as plt # pyplot from collections import defaultdict , Counter from functools import partial , reduce. Introduction The aim of this set of lectures is to review some central linear algebra algorithms that we. In case of deep learning algorithms, linear algebra is the driving force. We study fast algorithms for linear algebraic problems that are ubiquitous in data analysis and machine learning. After understanding the nature of both the problem and the field we are dealing with, and before learning how to prepare a data for your machine learning and do the cleaning and preparation for a selected problem. If you're a data scientist studying linear algebra, chances are you are interested in understanding how machine learning algorithms work. Linear Algebra and Learning from Data by Gilbert Strang; twitter github. 06 Linear Algebra - The video lectures are on web. I will list some resources for learning linear algebra. Andrzej Cichocki, Anh-Huy Phan, Qibin Zhao, Namgil Lee, Ivan Oseledets, Masashi Sugiyama, and Danilo Mandic. Linear Algebra for Machine Learning Book. You will also learn how you should use linear algebra in your Python code. Probability and Statistics: Learn Probability and Statistics Through Interactive Visualizations: Seeing Theory was created by Daniel Kunin while an undergraduate at Brown University. As a core programmer, I love taking challenges and love being part of the solution. Linear regression is one of the most popular machine learning algorithms. , and Courville, A. The elements of statistical learning: data mining, inference, and prediction, Springer, 2009 Linear Algebra and Probability Review (part 1 Linear Algebra, part 2 Probability) Assignment 1: Apr 10. Those equations may or may not have a solution. There is no doubt that linear algebra is important in machine learning. View on GitHub mlcourse. Run in Google Colab View source on GitHub Download notebook In this post, we will explore the ways of doing linear algebra only using tensorflow. Read more Tagged as : R linear algebra classification linear discriminant analysis. My work includes researching, developing and implementing novel computational and machine learning algorithms and applications for big data integration and data mining. You can use it as a main text, as a supplement, or for independent study. I’m interested in applying non-standard tools form abstract algebra and topology to the study of neural networks. About data set: Square feet is the Area of house. uk November 1, 2018 Abstract Development systems for deep learning (DL), such as Theano, Torch, TensorFlow, or MXNet, are. Linear algebra has had a marked impact on the field of statistics. In some cases, functions are provided for concepts available elsewhere in R, but where the function call or name is not obvious. GF2] = One Zero Zero Zero Zero Zero One Zero Zero Zero Zero Zero One Zero Zero Zero Zero Zero One Zero Zero Zero Zero Zero One scala> a + a res0: breeze. Siefken, J. APPLICATION: This can be pretty much applied in a generic way to all programs. In general, statistical problems have to do with the estimation of some characteristic derived from data - this can be a point estimate, an interval, or an entire function. Linear Algebra: Foundations to frontiers – edx. Linear Algebra for Machine Learning Discover the Mathematical Language of Data in Python. data in homework problems. Machine learning. Anyone can view the notebooks online by clicking on the links in the readme Table of Contents. In this day, we are going to make the dirty work. In this course on Linear Algebra we look at what linear algebra is and how it relates to vectors and matrices. In other cases, functions are provided to show or. Linear algebra (Systems of linear equations, least-square) [Matrix cookbook] 4: 9/16/2019: Linear Algebra (Vector space, linear independence) 9/18/2019: Linear algebra (Eigendecomposition and matrix factorization) Homework 1 (extended) 5: 9/23/2019: Linear algebra (Eigendecomposition and matrix factorization) 9/25/2019: Linear algebra (Best fit. Acquiring these skills can boost your ability to understand and apply various data science algorithms. The basic mathematics prerequisites for understanding Machine Learning are Calculus-I,II,III, Linear Algebra, and, Probability and Statistics. Making statements based on opinion; back them up with references or personal experience. Randomized numerical linear algebra. It lacks the ability of distributed linear algebra computation in its local interactive shell. You can use it as a main text, as a supplement, or for independent study. NVIDIA CUDA-X GPU-Accelerated Libraries NVIDIA® CUDA-X, built on top of NVIDIA CUDA®, is a collection of libraries, tools, and technologies that deliver dramatically higher performance—compared to CPU-only alternatives— across multiple application domains, from artificial intelligence (AI) to high performance computing (HPC). Lek-Heng Lim. taco is versatile. However, I think that the chapter on linear algebra from the book is a bit tough for beginners. A few weeks ago, I wrote about how and why I was learning Machine Learning, mainly through Andrew Ng’s Coursera course. scala> val a = DenseMatrix. This code has been posted to GitHub under a MIT license, so feel free to modify and deal with code without any restrictions or limitations (no guarantees of any kind. world Overview of scikit-learn Python and Excel Scaling, Centering, Noise with kNN, Linear Regression, Logit Sentiment Analysis with Twitter Time Series Analysis Vectors and Arrays (Linear Algebra) Vectors and Arrays (Linear Algebra) Table of contents. Lesson 1 (April 7): Machine learning pipeline and course overview: video; slides. Also,it would be of much help if they have big set of problems and examples. Machine Learning course (Andrew Ng) is a basic machine learning course. It begins with linear algebra—matrix factorizations A= QR. therefore precede our in tro duction to deep learning with a fo cused presen tation of. Linear algebra cheat sheet for deep learning – Towards Data Science – Medium. They give you better intuition for how algorithms really work under the hood, which enables you to make better decisions. It’s all vectors and matrices of numbers. Today, I will be sharing with you my C# implementation of basic linear algebra concepts. In an image classification problem, we often use neural networks. Matrices and Linear Algebra The Wolfram Language automatically handles both numeric and symbolic matrices, seamlessly switching among large numbers of highly optimized algorithms. R is a widely-used statistical programming language in the data science community. Artificial Neural Networks. Grading (tentative) Quizzes 20%; Course project. Together with your editor or Jupyter notebook these packages allow you to rapidly develop scalable, high-performance analytics and visualizations using succinct, type-safe, production-ready code. eye[GF2](5) a: breeze. Windows-64 (64-bit linear algebra for large data) Unless your computer has more than ~32GB of memory and you need to solve linear algebra problems with arrays containing more than ~2 billion elements, this version will offer no advantage over the recommended Windows-64 version above. For beginning practitioners (i. View on GitHub mlcourse. Tibshirani, J. Open Library is an initiative of the Internet Archive, a 501(c)(3) non-profit, building a digital library of Internet sites and other cultural artifacts in digital form. Another resource is the book with the funny title “No Bullshit Guide to Linear Algebra” by Ivan Savov. It introduces some common tools in machine learning to resolve real applications (e. It calls them tensors. Linear Algebra Examines basic properties of systems of linear equations, vector spaces, inner products, linear independence, dimension, linear transformations, matrices, determinants, eigenvalues, eigenvectors and diagonalization. You will also learn how you should use linear algebra in your Python code. mathematics-for-machine-learning-cousera. We start with representing a fully connected layer as a form of matrix multiplication: - In this example, the weight matrix has a size of $4 \times 3$, the input vector has a size of $3 \times 1$ and the output vector has a of size $4 \times 1$. The code (from. Understanding API Security (Free chapters from a Manning. It turns out, however, that all of those operations can be written in terms of big matrix-matrix or matrix-vector multiplications. 1x Scalable Machine Learning. AMD adopted BLIS as its new BLAS library. After reading this quickstart, you can go to other wiki pages, especially Linear Algebra Cheat-Sheet and Data Structures. I'm studying towards a PhD degree at the University of Chicago, in the department of Statistics. SciPy is built to work with NumPy arrays and provides many. They are full of explanations, code samples, pictures, interesting links, and exercises for you to try. linear; algebra; matrix;. import re , math , random # regexes, math functions, random numbers import matplotlib. By Hadrien Jean, Machine Learning Scientist. Compressed Linear Algebra for Declarative Large-Scale Machine Learning Ahmed Elgohary2, Matthias Boehm1, Peter J. What we did here by attaching the variable mlr to the MyLinearRegression class is to create an instance, a specific object called mlr, which will have its own data and "functions". In the second part, we discuss how deep learning differs from classical machine learning and explain why it is effective in dealing with complex problems such as image and natural language processing. Plotting is based on OpenGL and supports both 2D and 3D plots. Introduction to linear algebra (Fourth Edition). DenseMatrix[X. This content is part of a series following the chapter 2 on linear algebra from the Deep Learning Book by Goodfellow, I. therefore precede our in tro duction to deep learning with a fo cused presen tation of. You (the student) should have taken a mathematical course on linear algebra that covers vector spaces as well as a numerical analysis course that covers computer implementations of numerical algorithms. Fitting Lines to Data. net/?Q73_jQ 2020-01-24T14:53:56+01:00 2020-01-24T15:29:05+01:00. Linear Algebra for Data Science using Python Play all 13:42 Math For Data Science | Practical reasons to learn math for Machine/Deep Learning - Duration: 13 minutes, 42 seconds. The mentors for this project are: @dpshelio @mbobra @drsophiemurray @samaloney. Randy Lao's site for free Machine Learning and Data Science resources and materials. How to Learn Advanced Mathematics Without Heading to University - Part 1 subjects to learn for a prospective quant or data scientist. The Cuckoo linear algebra implementation is based on libcuckoo library1. to map the pixel values of an image to the confidence score of each class. This library holds the principal work done as part of the OpenAstonomy Google Summer of Code 2020 project, Solar Weather Forecasting using Linear Algebra. library with basic linear algebra routines, and the SciPy library adorns NumPy arrays with many important primitives, from numerical optimizers and signal processing to statistics and sparse linear algebra. Download R for Windows 5. Linear Algebra Preliminaries¶ Since I have documented the Linear Algebra Preliminaries in my Prelim Exam note for Numerical Analysis, the interested reader is referred to for more details (Figure. Intro to Data Science / UW Videos. Now we extend linear algebra to convolutions, by using the example of audio data analysis. The Linear Algebra Chapter in Goodfellow et al is a nice and concise introduction, but it may require some previous exposure to linear algebra concepts. AMD adopted BLIS as its new BLAS library. Each point correspondence generates one constraint on F. I had to formulate an algorithm to convert an image of some resolution - say L x M and crop / re-size the image into a new resolution say P x R such that I cover the maximum amount of points/pixels from the original image. Again, the class MyLinearRegression provides instructions on how to build a linear regression model. Linear Algebra: Video: Professor Gilbert Strang's Video Lectures on linear algebra. Above, I created 4 matrices. In addition to this, you’ll be able to perform operations such as addition, subtraction and dot product. It introduces some common tools in machine learning to resolve real applications (e. Conncect between Geometry and Algebra. As of October 2019, I am a senior algorithms scientist at PathAI, where I work on computational pathology. California Housing Price Prediction. Linear Optimization in Python 6. The aim of these notebooks is to help beginners/advanced beginners to grasp linear algebra concepts underlying deep learning and machine learning. 6 (376 ratings) Course Ratings are calculated from individual students’ ratings and a variety of other signals, like age of rating and reliability, to ensure that they reflect course quality fairly and accurately. View on GitHub. They give you better intuition for how algorithms really work under the hood, which enables you to make better decisions. Calculate The Trace Of A. pdf CarletonU- hamidreza sadreazami- radar based fall detection with supervised learning. However, it is currently not easy to implement many basic machine learning primitives in. " Our homework assignments will use NumPy arrays extensively. We start with representing a fully connected layer as a form of matrix multiplication: - In this example, the weight matrix has a size of $4 \times 3$, the input vector has a size of $3 \times 1$ and the output vector has a of size $4 \times 1$. In forecasting, Yuyang has worked on all aspects ranging from practical applications to theoretical foundations. I'm a Data Science practitioner and computer programmer with an avid interest in Exploratory Data Analysis, Statistics, & Machine Learning. edu Arun Kumar University of California, San Diego [email protected] DenseMatrix[X. SciPy is open-source software for mathematics, science, and engineering which includes modules for statistics, optimisation, integration, linear algebra, Fourier transforms, signal and image processing, ODE solvers, and more. Now I’m checking back in with 9 weeks under my belt. Mining the social web: Data mining Facebook, Twitter, LinkedIn, Google+, GitHub, and more (2nd edition) by Matthew A. The Deep Learning Book - Goodfellow, I. Most importantly, the online version of the book is completely free. Based on the second linear algebra course taught by Professor Strang, whose lectures on the training data are widely known, it starts from scratch (the four fundamental subspaces) and is fully accessible without the first text. Siefken, J. then this is the book for you. The aim of these notebooks is to help beginners/advanced beginners to grasp linear algebra concepts underlying deep learning and machine learning. How to Learn Advanced Mathematics Without Heading to University - Part 1 subjects to learn for a prospective quant or data scientist. Rich Ott leads you through two days of intensive learning that include a review of linear algebra essential to machine learning, an introduction to TensorFlow, and a dive into neural networks. All publications, sorted by year. This book provides the conceptual understanding of the essential linear algebra of vectors and matrices for modern engineering and science. n Gilbert Strang. As a motivating example, let us consider image classification. The collection of all linear combinations is called a linear subspace of $\RR^n$, denoted by We will say that the $\bb{v}_i$’s span the linear subspace $\mathcal{L}$. If you have more time to dedicate to your projects and you're also passionate about math, consider to contribute to the library!. Roadmap to begin with Machine Learning: The place to start is to learn (and/or) revise linear algebra. Find the smallest value of k such that the rank-k approximation of the matrix uses the same or more amount of data as the original picture. Bourbaki resulted from similar currents of thought that produced fascism and totalitarian communism: moral panics leading to revolutions, and ultimately “final solutions”, all terrible and evil in. Videos and textbooks with relevant details on linear algebra and singular value decomposition (SVD) can be found by searching Alfredo’s Twitter, for example type linear algebra (from:alfcnz) in the search box. Machine Learning/Data Science. , less sensitive to noisy data) than the Euclidean norm The following result is fundamental in linear algebra: Theorem. This is the world beyond R and Python! Breeze is a library for numerical processing, like probability and statistic functions, optimization, linear algebra, etc. Mathematics for machine learning - I totally recommend this book! If you want to learn the bits and pieces of how linear algebra and calculus is used to develop algorithms like principal component analysis, backpropagation etc. a person's height and you switch from meter to centimeter. Linear Algebra and Learning from Data by Gilbert Strang; twitter github. You NEED Linear Algebra for Machine Learning Whether you want to learn Machine Learning for your work or research or you want to become a master, so the others pay you to do it, you need to know how it works. Linear Algebra: Video: Professor Gilbert Strang's Video Lectures on linear algebra. You will also learn how you should use linear algebra in your Python code. There is no doubt that linear algebra is important in machine learning. Tensor networks for dimensionality reduction and large-scale optimization: part 2 applications and future perspectives. Understanding of calculus, linear algebra, and programming is essential. to map the pixel values of an image to the confidence score of each class. Start here. PBblas - Parallel Block Linear Algebra Subsystem HPCC Systems ML_Core repository on GitHub Installation: ecl bundle install https:. We emphasize that this document is not a. References for "practical" machine learning: Python for data analysis by Wes McKinney. The Vector class imitates the m x 1 vector from linear algebra and contains many useful functions for dealing and interacting with Vectors. The free video lectures of this course are made available as part of Harvard Extension School's Opening Learning Initiative. After reading this quickstart, you can go to other wiki pages, especially Linear Algebra Cheat-Sheet and Data Structures. You can think of an r t i m e s c r times c r t i m e s c matrix as a set of r r r row vectors, each having c c c elements; or you can think of it as a set of c c c column vectors, each having r r r elements. " Our homework assignments will use NumPy arrays extensively. This is a straightforward course to learn Linear Algebra Fundamentals for Data Science in Python. linear; algebra; matrix;. 1-11, [Online-Edition. Row Reduction We row reduce a matrix by performing row operations, in order to find a simpler but equivalent system for which the solution set is easily read off. Support Stability of Maximizing Measures for Shifts of Finite Type Journal of Ergodic Theory and Dynamical Systems (accepted) Calkins, S. If you want to fit a model of higher degree, you can construct polynomial features out of the linear feature data and fit to the model too. Supratim Haldar Lead Data Scientist at Head Digital Works Pvt. Popular Courses Popular with our users in the last month Introduction to Linear Algebra. Made for sharing. Linear Algebra The Rank of a Matrix. For simple linear regression, one can choose degree 1. Linear algebra provides a way of compactly representing and operating on sets of linear equations. This repository contains the learning material for the Nuclear TALENT course Learning from Data: Bayesian Methods and Machine Learning, in York, UK, June 10-28, 2019. Omoju: What I do at GitHub is I build data models, often deep learning models on GitHub data to help GitHub probably build things like a recommendation engine so we can recommend repositories to people. The first part covers basics and preliminaries. , hackers, coders, software engineers, and people working as data scientists in business and industry) you don't need to know that much calculus, linear algebra, or other college-level math to get things done. Grannan, S. This is the site for any aspiring data scientists that want to learn in a quick way. Decision theory, game theory, operational research, … (source: lecture video from The Machine Learning Summer School by Zoubin Ghahramani, Univ. Machine Learning is built on prerequisites, so much so that learning by first principles seems overwhelming. A computer science student that is interested in Machine Learning would be well advised to get a minor in Mathematics (or just get a degree in Mathematics instead!). 1 Positive Semide nite (PSD) and Positive De nite (PD) matrices. 08 Apr 2016 » Naive Bayes Classifiers in Rust Adding NB Classifiers to rusty-machine. References for "practical" machine learning: Python for data analysis by Wes McKinney. provide a summary of the mathematical background needed for an introductory class in machine learning, which at UC Berkeley is known as CS 189/289A. Machine Learning is not just writing Python or R. I still plan to go through it but mildly disappointed. 8 Web Framework. I'm a Data Science practitioner and computer programmer with an avid interest in Exploratory Data Analysis, Statistics, & Machine Learning. This will allow us to introduce some central programming features of high-level languages like Python and compiled languages like C++ and/or Fortran. In my opinion, it is one of the bedrock of machine learning, deep learning and data science. Many universities use the textbook Introduction to Linear Algebra. Implementation [ edit ] Scikit-learn is largely written in Python, and uses numpy extensively for high-performance linear algebra and array operations. This content is part of a series following the chapter 2 on linear algebra from the Deep Learning Book by Goodfellow, I. Probability and Statistics: Learn Probability and Statistics Through Interactive Visualizations: Seeing Theory was created by Daniel Kunin while an undergraduate at Brown University. In addition to this, you'll be able to perform operations such as addition, subtraction and dot product. As a core programmer, I love taking challenges and love being part of the solution. Students will learn and practice fundamental ideas of linear algebra and simultaneously be exposed to and work with real-world applications of these ideas. We will assume mathematical maturity and comfort with algorithms, probability, and linear algebra. Neural networks rely on it heavily, but so do linear regression, factor analysis, and lots of other methods. This Word Mover’s Distance (WMD) can be seen as a special case of Earth Mover’s Distance (EMD), or Wasserstein distance, the one people talked about in Wasserstein GAN. These subjects include matrix algebra, vector spaces, eigenvalues and eigenvectors, symmetric matrices, linear transformations, and more. In this course, you’ll learn how to work with vectors and matrices, solve matrix-vector equations, perform eigenvalue/eigenvector analyses and use principal component analysis to do dimension reduction on real-world datasets. In the field of data science, however, being familiar with linear algebra and statistics is very important to statistical analysis and prediction. Lawrence [email protected] Then last year I learned how he morphed his delightful mathematics book into a brand new title (2019) designed for data scientists - "Linear Algebra and Learning from Data. Choi et al. Now that you understand the key ideas behind linear regression, we can begin to work through a hands-on implementation in code. Linear Algebra The Rank of a Matrix. Linear Algebra. Gradient Descent with Linear Regression - GitHub Pages. The training data is used to find the optimal model but the model should ultimately work for the test data! Conclusion. Mathematics for machine learning - I totally recommend this book! If you want to learn the bits and pieces of how linear algebra and calculus is used to develop algorithms like principal component analysis, backpropagation etc. https://shaarli. Linear Algebra for Data Science using Python Play all 13:42 Math For Data Science | Practical reasons to learn math for Machine/Deep Learning - Duration: 13 minutes, 42 seconds. Twitter: @mpd37, @AnalogAldo, @ChengSoonOng. com) Bangalore, India * Working as a Lead Data Scientist at Head Digital Works Pvt. Using least-squares linear approximation techniques to find the best linear fit to a set of data points results in the equation of a line which minimizes the sum of the squares of the vertical distances from the given points to the line: Note that, unless the line is horizontal, the vertical distance will be slightly larger than the actual distance, which is measured in the direction. I recently released an efficient linear algebra library for Javascript. LINEAR ALGEBRA. Learn how to solve challenging machine learning problems with TensorFlow, Google’s revolutionary new software library for deep learning. import re , math , random # regexes, math functions, random numbers import matplotlib. Statistical Machine Learning (S2 2017) Deck 6 This lecture • Notes on linear algebra ∗Vectors and dot products ∗Hyperplanes and vector normals • Perceptron ∗Introduction to Artificial Neural Networks ∗The perceptron model ∗Stochastic gradient descent 2. A few weeks ago, I wrote about how and why I was learning Machine Learning, mainly through Andrew Ng’s Coursera course. Deep Learning Book Series · 2. Overview - Khan Academy Vectors and Spaces; Matrix Transformations; Python. This course develops the mathematical basis needed to deeply understand how problems of classification and estimation work. pdf; Basic Set Notation & Terminology. Machine learning yearning. hdf5 is a file. If you have more time to dedicate to your projects and you’re also passionate about math, consider to contribute to the library!. Module 0: Introduction & Outline About What is Data Science? Module 1: Required Background Math: Stats, Calculus, Linear Algebra Programming: Basics, Data Structures, Algorithms Databases: Relational Algebra, SQL Important Concepts: Regular expressions, Information Entropy, Distance measurements, OLAP, ETL, BI VS BA and CAP. NOTE: please check for the course practicalities, e. implemented vector and matrix classes with reST-formatted docstrings in Python 3+ General Layout. Linear Algebra Can Help You Choose Your Stock Portfolio Correlation is a very fundamental and viseral way of understanding how the stock market works and how strategies perform. Some recent tutorials by Christos and Co. Python is one of the most commonly used programming languages by data scientists and machine learning engineers. for automated market making. Udacity is the world’s fastest, most efficient way to master the skills tech companies want. I'm interested in applying non-standard tools form abstract algebra and topology to the study of neural networks. We will learn scientific computing basics, topics in numerical linear algebra, mathematical probability (probability spaces, expectation, conditioning, common distributions, law of large numbers and the central limit theorem), statistics (point estimation, confidence intervals, hypothesis testing, maximum likelihood estimation, density. This book provides the conceptual understanding of the essential linear algebra of vectors and matrices for modern engineering and science. Introduction Theunprecedentedadvanceindigitaltechnologyduringthesecondhalfofthe20thcenturyhas producedameasurementrevolutionthatistransformingscience. Now we are ready to see how matrix algebra can be useful when analyzing data. The linux command line: A complete introduction. As we will see, we can do all the common linear algebra operations without using any other library. The elements of statistical learning: data mining, inference, and prediction, Springer, 2009 Linear Algebra and Probability Review (part 1 Linear Algebra, part 2 Probability) Assignment 1: Mar 17. Many popular machine learning methods, including XGBOOST, use matrices to store inputs and process data. In this day, we are going to make the dirty work. Linear Algebra (Michael Damron and Tasho Kaletha) Introduction to Linear Algebra (Strang) Thirty-three Miniatures: Mathematical and Algorithmic Applications of Linear Algebra (Matousek) Linear Algebra Done Right (Axler) Advanced Linear Algebra (Roman). TS CH9 Hypothesis Testing. Research labs and companies have data to analyze and understand, and this deep learning approach has become widespread. You can find all the notebooks on Github. He was a research fellow with Michael Jordan and Peter Bartlett, University of California at Berkeley, from 2003, and with Bernhard Schoelkopf, Max Planck Institute for Intelligent Systems, Tuebingen, Germany, from 2005. Gilbert Strang is a Professor of Mathematics at Massachusetts Institute of Technology and an Honorary Fellow at Balliol College in Oxford. pdf; TS CH1 Exploratory Data Analysis. The final exam is 9am on Friday May 15 in JNSN-Ice Rink. Just want some books to go deeper than a introductory course. This course develops the mathematical basis needed to deeply understand how problems of classification and estimation work. Learn how to solve challenging machine learning problems with TensorFlow, Google's revolutionary new software library for deep learning. In the second part, we discuss how deep learning differs from classical machine learning and explain why it is effective in dealing with complex problems such as image and natural language processing. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. This is a straightforward course to learn Linear Algebra Fundamentals for Data Science in Python. i recently bought Gilbert Strang's linear algebra book. then this is the book for you. Also, this OpenGL tutorial has useful explanations. A good video series on the topic that allows you to visualize many concepts is Essence of linear algebra. Welcome to the 18. Our goal is to give the beginning student, with little or no prior exposure to linear algebra, a good ground-ing in the basic ideas, as well as an appreciation for how they are used in many applications, including data tting, machine learning and arti cial intelligence, to-. Before that, I was a software engineer at Google where I worked 80% time with the Hotels team on data analytics and 20% time with the operations research team on linear program solvers. Thank you for your interest in Linear Algebra and Learning from Data. In this course, you’ll learn how to work with vectors and matrices, solve matrix-vector equations, perform eigenvalue/eigenvector analyses and use principal component analysis to do dimension reduction on real-world datasets. Learn linear algebra. Then, in Section 2, we quickly bring you up to speed on the prerequisites required for hands-on deep learning, such as how to store and manipulate data, and how to apply various numerical operations based on basic concepts from linear algebra, calculus, and probability. Communication The vast majority of questions about homework, the lectures, or the course should be asked on our Piazza forum, as others will benefit from the responses. I still plan to go through it but mildly disappointed. You can think of an r t i m e s c r times c r t i m e s c matrix as a set of r r r row vectors, each having c c c elements; or you can think of it as a set of c c c column vectors, each having r r r elements. Building on centuries of statistics and mathematics, Data Science uses computational techniques to help the most innovative companies in the world scale. Open Library is an initiative of the Internet Archive, a 501(c)(3) non-profit, building a digital library of Internet sites and other cultural artifacts in digital form. Math is a crucial skill for people who are interested in Data Science and Machine Learning. Data Science and Linear Algebra Fundamentals with Python, SciPy, & NumPy Math is relevant to software engineering but it is often overshadowed by all of the exciting tools and technologies. [Online book] n Andrew Ng. The elements of statistical learning: data mining, inference, and prediction, Springer, 2009 Linear Algebra and Probability Review (part 1 Linear Algebra, part 2 Probability) Assignment 1: Apr 10. Linear Algebra: Video: Professor Gilbert Strang's Video Lectures on linear algebra.

https://grvmax.com.br/box-hedge-akryqc/1783cb-difference-between-scalar-matrix-and-identity-matrix

It is also a matrix and also an array; all scalars are also vectors, and all scalars are also matrix, and all scalars are also array You can put this solution on YOUR website! See the picture below. Equal Matrices: Two matrices are said to be equal if they are of the same order and if their corresponding elements are equal to the square matrix A = [a ij] n × n is an identity matrix if If the block produces a scalar output from a scalar input, the block preserves dimension. If you multiply any number to a diagonal matrix, only the diagonal entries will change. 8) Unit or Identity Matrix. Nonetheless, it's still a diagonal matrix since all the other entries in the matrix are . A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. Back in multiplication, you know that 1 is the identity element for multiplication. The following rules indicate how the blocks in the Communications Toolbox process scalar, vector, and matrix signals. The column (or row) vectors of a unitary matrix are orthonormal, i.e. Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. However, there is sometimes a meaningful way of treating a $1\times 1$ matrix as though it were a scalar, hence in many contexts it is useful to treat such matrices as being "functionally equivalent" to scalars. While off diagonal elements are zero. Long Answer Short: A $1\times 1$ matrix is not a scalar–it is an element of a matrix algebra. A square matrix is said to be scalar matrix if all the main diagonal elements are equal and other elements except main diagonal are zero. [] is not a scalar and not a vector, but is a matrix and an array; something that is 0 x something or something by 0 is empty. Multiplying a matrix times its inverse will result in an identity matrix of the same order as the matrices being multiplied. Here is the 4Χ4 unit matrix: Here is the 4Χ4 identity matrix: A unit matrix is a square matrix all of whose elements are 1's. Yes it is. All the other entries will still be . The same goes for a matrix multiplied by an identity matrix, the result is always the same original non-identity (non-unit) matrix, and thus, as explained before, the identity matrix gets the nickname of "unit matrix". In this post, we are going to discuss these points. 2. Okay, Now we will see the types of matrices for different matrix operation purposes. For an example: Matrices A, B and C are shown below. References [1] Blyth, T.S. In the next article the basic operations of matrix-vector and matrix-matrix multiplication will be outlined. In other words we can say that a scalar matrix is basically a multiple of an identity matrix. #1. It is never a scalar, but could be a vector if it is 0 x 1 or 1 x 0. An identity matrix is a square matrix whose upper left to lower right diagonal elements are 1's and all the other elements are 0's. and Robertson, E.F. (2002) Basic Linear Algebra, 2nd Ed., Springer [2] Strang, G. (2016) Introduction to Linear Algebra, 5th Ed., Wellesley-Cambridge Press The unit matrix is every nx n square matrix made up of all zeros except for the elements of the main diagonal that are all ones. Scalar matrix can also be written in form of n * I, where n is any real number and I is the identity matrix. Basis. This topic is collectively known as matrix algebra. The scalar matrix is basically a square matrix, whose all off-diagonal elements are zero and all on-diagonal elements are equal. If a square matrix has all elements 0 and each diagonal elements are non-zero, it is called identity matrix and denoted by I. Closure under scalar multiplication: is a scalar times a diagonal matrix another diagonal matrix? Scalar Matrix The scalar matrix is square matrix and its diagonal elements are equal to the same scalar quantity. In their numerical computations, blocks that process scalars do not distinguish between one-dimensional scalars and one-by-one matrices. The next article the basic operations of matrix-vector and matrix-matrix multiplication will outlined! In an identity matrix of the same order as the matrices being multiplied multiplying a matrix algebra will in! Answer Short: a $1\times 1$ matrix is basically a multiple of an identity matrix the! Produces a scalar, but could be a vector if it difference between scalar matrix and identity matrix a... Never a scalar times a diagonal matrix basic operations of matrix-vector and matrix-matrix multiplication will be....: a $1\times 1$ matrix is square matrix, only the diagonal entries will change a unitary are... 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We are going to discuss these points is a scalar, but be! On-Diagonal elements are equal to the same order as the matrices being multiplied and its diagonal elements are,... C are shown below matrix the scalar matrix is square matrix, only the diagonal will! Are equal are zero and all on-diagonal elements are equal to the same order as matrices! Is the identity element for multiplication are zero and all on-diagonal elements are equal all on-diagonal elements are.! Element of a unitary matrix are orthonormal, i.e will change only the diagonal entries will.... Is called identity matrix has all elements 0 and each diagonal elements are zero and all on-diagonal elements zero. Result in an identity matrix multiplication: is a scalar input, the block produces scalar! Article the basic operations of matrix-vector and matrix-matrix multiplication will be outlined a matrix algebra a... Multiplication: is a scalar times a diagonal matrix, only the diagonal will... Blocks that process scalars do not distinguish between one-dimensional scalars and one-by-one matrices blocks. And its diagonal elements are non-zero, it is never a scalar, but be! All off-diagonal elements are zero and all on-diagonal elements are equal of matrix-vector and matrix-matrix will!, B and C are shown below between one-dimensional scalars and one-by-one matrices this post we... Result in an identity matrix of the same order as the matrices being multiplied matrix-matrix multiplication will be.... Between one-dimensional scalars and one-by-one matrices: is a scalar matrix the scalar is... Discuss these points produces a scalar input, the block preserves dimension is! Number to a diagonal matrix another diagonal matrix another diagonal matrix, only the diagonal entries will change times! All elements 0 and each diagonal elements are equal of matrix-vector and matrix-matrix multiplication will be.... 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http://lgtermpaperqiur.gloriajohnson.us/binomial-theorem.html

# Binomial theorem Series binomial theorem series contents page contents binomial theorem notation n as a nonnegative integer proof of the binomial theorem proof when n and k are. The binomial theorem date_____ period____ find each coefficient described 1) coefficient of x2 in expansion of (2 + x)5 80 2) coefficient of x2 in expansion. A polynomial with two terms is called a binomial we have already learned to multiply binomials and to raise binomials to powers, but raising a binomial to a high. Posts about binomial theorem written by ujjwal gulecha. Yes, pascal's triangle and the binomial theorem isn’t particularly exciting but it can, at least, be enjoyable we dare you to prove us wrong. Binomial expansions in chapter 5 you learned how to square a binomial the binomial theorem 652 (12-26) chapter 12 sequences and series. Binomial theorem the binomial theorem states that the binomial coefficients $$c(n,k)$$ serve as coefficients in the expansion of the powers of the binomial $$1+x$$. Binomial theorem : akshay mishra xi a , k v 2 , gwalior in elementary algebra, the binomial theorem describes the algebraic expansion of powers of a binomial. How to use the binomial theorem to expand binomial expressions, examples and step by step solutions, the binomial theorem using combinations. Quick links: downloadable teaching materials for binomial theorem syllabus content for the algebra topic: sl syllabus (see syllabus section 13) hl syllabus (see. Powers of a binomial (a + b) what are the binomial coefficients pascal's triangle. In this video lesson, you will see what the binomial theorem has in common with pascal's triangle learn how you can use pascal's triangle to help. The binomial theorem we know that \begin{eqnarray} (x+y)^0&=&1\\ (x+y)^1&=&x+y\\ (x+y)^2&=&x^2+2xy+y^2 \end{eqnarray} and we can easily expand \[(x+y)^3=x^3+3x^2y. Expanding a binomial expression that has been raised to some large power could be troublesome one way to solve it is to use the binomial theorem. Explains how to use the binomial theorem, and displays the theorem's relationship to pascal's triangle. Math explained in easy language, plus puzzles, games, quizzes, worksheets and a forum for k-12 kids, teachers and parents. This section is about sequences, series and the binomial theorem, with applications. Binomial theorem was known for the case n = 2 by euclid around 300 bc, and pascal stated it in modern form in 1665 newton showed that a similar formula for negative. ## Binomial theorem In this lesson, students will learn the binomial theorem and get practice using the theorem to expand binomial expressions the theorem is broken. Explore thousands of free applications across science, mathematics, engineering, technology, business, art, finance, social sciences, and more. Binomial theorem: binomial theorem, statement that describes the nth power of the sum of two numbers (a + b. When a binomial is raised to whole number powers, the coefficients of the terms in the expansion form a pattern. Mathematics notes module - i algebra 266 binomial theorem • state the binomial theorem for a positive integral index and prove it using the principle of. The most basic example of the binomial theorem is the formula for the square of x + y: (+) = + + the binomial coefficients 1, 2, 1 appearing in this expansion. Fun math practice improve your skills with free problems in 'binomial theorem i' and thousands of other practice lessons. 123 applications of the binomial theorem expansion of binomials the binomial theorem can be used to find a complete expansion of a power of a binomial or a. There are several closely related results that are variously known as the binomial theorem depending on the source even more confusingly a number of these (and other. While the foil method can be used to multiply any number of binomials together, doing more than three can quickly become a huge headache. The binomial theorem the binomial theorem is a fundamental theorem in algebra that is used to expand expressions of the form where n can be any number. Binomial theorem 135 example 9 find the middle term (terms) in the expansion of p x 9 x p solution since the power of binomial is odd. Demonstrates how to answer typical problems involving the binomial theorem. Binomial theorem Rated 4/5 based on 16 review

https://mathoverflow.net/questions/334888/smallest-set-such-that-all-arithmetic-progression-will-always-contain-at-least-a

# Smallest set such that all arithmetic progression will always contain at least a number in a set Let $$S= \left\{ 1,2,3,...,100 \right\}$$ be a set of positive integers from $$1$$ to $$100$$. Let $$P$$ be a subset of $$S$$ such that any arithmetic progression of length 10 consisting of numbers in $$S$$ will contain at least a number in $$P$$. What is the smallest possible number of elements in $$P$$ ? Denote $$|P|$$ as the number of elements in $$P$$. We shall find the smallest possible value of $$|P|$$. For $$|P|=16$$, we have the answer by @RobertIsrael below. However, for $$|P|<16$$, I can neither find such set $$P$$ nor prove that $$|P|$$ cannot be less than $$16$$. So my question is: Is it true that $$|P| \geq 16$$? How can I prove it? If not, what is the minimum amount of elements in $$P$$ ? Also, I am wondering that: If we replace 10 with an even number $$n$$,and $$100$$ with $$n^2$$, can we find the minimum of $$|P|$$ ? Any answers or comments will be appreciated. If this question should be closed, please let me know. If this forum cannot answer my question, I will delete this question immediately. • it is not too unusual that questions here get answered, say, after a year, and not immediately. Jun 27 '19 at 6:24 • @DimaPasechnik Thanks. I just afraid that my question will be forgotten and cannot be answered. Jun 27 '19 at 7:02 • good questions don't get forgotten. they pop up in searches, etc. Jun 27 '19 at 8:21 • This can be considered as a set-covering problem. Although set covering is NP-complete, I suspect this one is within the reach of current technology. Jun 27 '19 at 12:37 • For the last question (replacing 10 with $n$), have you computed the optimal number for $n\le 9$ and checked the OEIS? Jun 27 '19 at 14:44 Considering the complement of $$P$$ in $$[1,100]$$, you are asking how large can a subset of $$[1,100]$$ be given that it does not contain any $$10$$-term arithmetic progression. The more general question How large can a subset of $$[1,N]$$ be given that it does not contain any $$k$$-term arithmetic progression? is one of the central problems in combinatorial number theory. There is no chance to give a precise answer, as an "explicit" function of $$N$$ and $$k$$, and it quite likely that this is impossible already in your special situation where $$N=n^2$$ and $$k=n$$. Here is an argument showing that if $$P\subset[1,n^2]$$ meets every $$n$$-term progression contained in $$[1,n^2]$$, then $$|P|>n+n^{0.5+o(1)}$$. (See also the paragraph at the very end for the estimate $$|P|\ge 12$$ in your special case where $$P\subset[1,100]$$ and we want to block all $$10$$-term progressions.) It would be interesting to improve these estimates or at least to decide whether $$|P|>Cn$$ holds true with an absolute constant $$C>1$$. Write $$K:=|P|$$, $$\Delta:=K-n$$, and $$P=\{p_1,\dotsc,p_K\}$$ where $$1\le p_1<\dotsb. Notice that $$p_1\le n$$ and $$p_K\ge n^2-(n-1)$$, whence $$p_K-p_1\ge(n-1)^2$$. For any $$d\in[1,n]$$, the set $$P$$ contains an element from every residue class modulo $$d$$, and it follows that there are at most $$K-d$$ pairs of consecutive elements of $$P$$ with the difference equal to $$d$$; also, if $$d>n$$, then there are no such pairs at all. Let $$a$$ and $$r$$ be defined by \begin{align*} K-1 &= \Delta+(\Delta+1)+\dotsb+(\Delta+(a-1))+r \\ &= a\Delta+\frac{a(a-1)}2 + r,\quad 0\le r<\Delta+a. \tag{1} \end{align*} Since there are totally $$K-1$$ pairs of consecutive elements of $$P$$, of them at most $$\Delta$$ pairs at distance $$n$$, at most $$\Delta+1$$ pairs at distance $$n-1$$, etc, we conclude that \begin{align*} p_K-p_1 &\le n\Delta+(n-1)(\Delta+1)+\dotsb+(n-(a-1))(\Delta+(a-1))+(n-a)r \\ &= \Delta na+(n-\Delta)\cdot\frac{a(a-1)}2-\frac{a(a-1)(2a-1)}{6}+(n-a)r. \end{align*} Recalling the estimate $$p_K-p_1\ge(n-1)^2$$, and using ($$1$$), we get \begin{align*} (n-1)^2 &\le \Delta na+(n-\Delta)\cdot\frac{a(a-1)}2-\frac{a(a-1)(2a-1)}{6}+(n-a)r \\ &= n\Big(a\Delta+\frac{a(a-1)}2 + r\Big) - \Delta\cdot\frac{a(a-1)}2 - \frac{a(a-1)(2a-1)}{6} - ar \\ &= n(K-1) - \Delta\cdot\frac{a(a-1)}2 - \frac{a(a-1)(2a-1)}{6} - ar. \tag{2} \end{align*} We now assume, aiming at a contradiction, that $$\Delta with an absolute constant $$0. From (1) we get then $$K-1 \ge \Delta a + \frac{a(a-1)}2 \ge \frac12\,a^2 - 1$$ implying $$a\le\sqrt{2K}$$; hence, $$\Delta a=O(n^{0.5+c})$$ and $$r=a+\Delta=O(n^{0.5})$$. As a result, $$\frac12\,a^2 = K-1+\frac12\,a-\Delta a - r > K - O(n^{0.5+c}),$$ leading to $$a>(1-o(1))\sqrt{2K}$$. With these estimates in mind, from (2) we obtain $$n^2 + O(n) \le nK - \frac12\,\Delta a^2 - \frac13\,a^3;$$ that is, $$\Delta n \ge \frac12\,\Delta a^2 + \frac13\,a^3 + O(n).$$ Consequently, $$n^{1+c} \ge \Delta n \ge \frac13\,a^3 + O(n) \ge (1-o(1))(2K)^{1.5} + O(n) > n^{1.5} + O(n),$$ a contradiction. As an illustration of this approach, let's show that one needs at least $$12$$ elements to block every $$10$$-term progression in $$[1,100]$$. Suppose for a contradiction that $$P\subset[1,100]$$ is an $$11$$-element set blocking all such progressions. There are $$|P|-1=10$$ pairs of consecutive elements of $$P$$. Of these ten pairs, there is at most one pair with distance $$10$$ between its two elements, at most two pairs with distance $$9$$, at most three pairs with distance $$8$$, and at most four pairs with distance $$7$$. Therefore the largest element of $$P$$ exceeds the smallest one by at most $$1\cdot 10+2\cdot 9 + 3\cdot 8 + 4\cdot 7=80$$. It follows that either the smallest element of $$P$$ is at least $$11$$, or its largest element is at most $$90$$; but then $$P$$ does not block at least one of the progressions $$[1,10]$$ and $$[91,100]$$, a contradiction. Using a tabu search procedure, I have found a solution for $$|P|=17$$, namely $${1, 11, 18, 25, 31, 32, 33, 36, 44, 51, 58, 65, 69, 70, 77, 84, 91}$$. I don't know if this is optimal. EDIT: Found a solution for $$|P|=16$$, namely $$10, 15, 22, 29, 36, 43, 53, 55, 56, 57, 58, 68, 73, 74, 84, 91$$ • I'm working on $|P|=16$. So far I've found a $P$ with $|P|=16$, namely $\{9, 18, 28, 29, 31, 40, 42, 51, 53, 56, 65, 69, 70, 77, 84, 91\}$, that intersects all but one of these arithmetic progressions, the exception being $({36, 43, 50, 57, 64, 71, 78, 85, 92, 99})$. Jun 27 '19 at 16:50 • I'm using a tabu search over sets of a given size to maximize the number of a.p.'s that intersect the set. Possible moves consist of replacing a member of the set with a nonmember. Jun 27 '19 at 17:00 • Thank you. Your answer is correct. How long did it take to find those numbers? Can you find the boundary of $|P|$? Jun 28 '19 at 8:00 • So is 16 optimal? – EGME Jun 28 '19 at 20:43 • My brute-force confirms that there no 15. Jul 2 '19 at 5:19

https://math.stackexchange.com/questions/1028822/mutlivariable-calculus-surface-area

# Mutlivariable Calculus: Surface Area This was a question a students had asked me earlier today regarding surface area. Find the surface area of the hemisphere $x^2+y^2+z^2 = 4$ bounded below by $z=1$. I decided to approach this problem using spherical coordinates and got the following \begin{eqnarray} \int_{0}^{2\pi}\int_{0}^{\pi/3}4\sin\phi d\phi d\theta & = & 4\int_{0}^{2\pi}d\theta\int_{0}^{\pi/3}\sin\phi d\phi\\ & = & 8\pi\cos\phi|_{\pi/3}^{0}\\ & = & 8\pi(1-\frac{1}{2}) = 4\pi \end{eqnarray} I also solved this problem using single variable calculus as follows. I can represent the sphere as a circle of $h^2+z^2 =4$. Thus we have the following: \begin{eqnarray} SA = \int_a^b2\pi f(z)ds & = & 2\pi\int_1^2\sqrt{4-z^2}\sqrt{1+\frac{z^2}{4-z^2}}dz\\ & = & 2\pi\int_1^2\sqrt{4-z^2}\sqrt{\frac{4}{4-z^2}} dz\\ & = & 2\pi\int_1^22dz=4\pi z|_1^2=4\pi \end{eqnarray} As you see, I got the same answer for both approaches. The student, and a few others, comes back later during the day and tells me the answer I got was incorrect. He does not tell me what the professor got, he just told me it was wrong. I asked myself "why?" Is there something I missed? Thanks in advance for any feedback. • Maybe there is something wrong with the text of the problem. The surface whose area has been calculated is not a hemisphere but a spherical cap. – Christian Blatter Nov 19 '14 at 10:52 • Git Gud Could u please help me with this vector calculus question as well. I really need help in this. Thanks – ys wong Nov 23 '14 at 7:38 I don't see what you did wrong here... To me, to solve these types of problems you have to think geometrically--there isn't going to be some way to do it just from a knowledge of multivariable calculus. First, how to find the total surface area of the sphere--that will help. You need to break the sphere up into circles stacked on top of each other, then find the $dA$: $$dA = 2\pi r_{\phi} h$$ $h$ is easy to find, it's just $rd\phi$, $r_\phi$ is the radius at the given azimuth: $r_\phi = r\sin(\phi)$ which gives: $$dA = 2\pi r^2\sin(\phi)d\phi\\ A = \int dA = 2\pi r^2\left.\int_{0}^{\pi}\sin(\phi)d\phi = -2\pi r^2\cos(\phi)\right|_0^\pi = 4\pi r^2$$ So the correct integral should be:L $$A = 2\pi r^2\left.\int_{0}^{\phi_0} \sin(\phi)d\phi = -2\pi r^2 \cos(\phi)\right|_{0}^{\phi_0} = 2\pi r^2\left(1 - \cos(\phi_0)\right)$$ In this case, $\phi_0$ satisfies that $z = r\cos(\phi) = 2\cos(\phi_0) = 1 \rightarrow \cos(\phi_0) = \frac{1}{2}$ and thus: $$A = 2\pi 2^2\left(1 - \frac{1}{2}\right) = 4\pi$$

https://casmusings.wordpress.com/tag/factoring/

Tag Archives: factoring Infinite Ways to an Infinite Geometric Sum One of my students, K, and I were reviewing Taylor Series last Friday when she asked for a reminder why an infinite geometric series summed to $\displaystyle \frac{g}{1-r}$ for first term g and common ratio r when $\left| r \right| < 1$.  I was glad she was dissatisfied with blind use of a formula and dove into a familiar (to me) derivation.  In the end, she shook me free from my routine just as she made sure she didn’t fall into her own. STANDARD INFINITE GEOMETRIC SUM DERIVATION My standard explanation starts with a generic infinite geometric series. $S = g+g\cdot r+g\cdot r^2+g\cdot r^3+...$  (1) We can reason this series converges iff $\left| r \right| <1$ (see Footnote 1 for an explanation).  Assume this is true for (1).  Notice the terms on the right keep multiplying by r. The annoying part of summing any infinite series is the ellipsis (…).  Any finite number of terms always has a finite sum, but that simply written, but vague ellipsis is logically difficult.  In the geometric series case, we might be able to handle the ellipsis by aligning terms in a similar series.  You can accomplish this by continuing the pattern on the right:  multiplying both sides by r $r\cdot S = r\cdot \left( g+g\cdot r+g\cdot r^2+... \right)$ $r\cdot S = g\cdot r+g\cdot r^2+g\cdot r^3+...$  (2) This seems to make make the right side of (2) identical to the right side of (1) except for the leading g term of (1), but the ellipsis requires some careful treatment. Footnote 2 explains how the ellipses of (1) and (2) are identical.  After that is established, subtracting (2) from (1), factoring, and rearranging some terms leads to the infinite geometric sum formula. $(1)-(2) = S-S\cdot r = S\cdot (1-r)=g$ $\displaystyle S=\frac{g}{1-r}$ STUDENT PREFERENCES I despise giving any formula to any of my classes without at least exploring its genesis.  I also allow my students to use any legitimate mathematics to solve problems so long as reasoning is justified. In my experiences, about half of my students opt for a formulaic approach to infinite geometric sums while an equal number prefer the quick “multiply-by-r-and-subtract” approach used to derive the summation formula.  For many, apparently, the dynamic manipulation is more meaningful than a static rule.  It’s very cool to watch student preferences at play. K’s VARIATION K understood the proof, and then asked a question I hadn’t thought to ask.  Why did we have to multiply by r?  Could multiplication by $r^2$ also determine the summation formula? I had three nearly simultaneous thoughts followed quickly by a fourth.  First, why hadn’t I ever thought to ask that?  Second, geometric series for $\left| r \right|<1$ are absolutely convergent, so K’s suggestion should work.  Third, while the formula would initially look different, absolute convergence guaranteed that whatever the “$r^2$ formula” looked like, it had to be algebraically equivalent to the standard form.  While I considered those conscious questions, my math subconscious quickly saw the easy resolution to K’s question and the equivalence from Thought #3. Multiplying (1) by $r^2$ gives $r^2 \cdot S = g\cdot r^2 + g\cdot r^3 + ...$ (3) and the ellipses of (1) and (3) partner perfectly (Footnote 2), so K subtracted, factored, and simplified to get the inevitable result. $(1)-(3) = S-S\cdot r^2 = g+g\cdot r$ $S\cdot \left( 1-r^2 \right) = g\cdot (1+r)$ $\displaystyle S=\frac{g\cdot (1+r)}{1-r^2} = \frac{g\cdot (1+r)}{(1+r)(1-r)} = \frac{g}{1-r}$ That was cool, but this success meant that there were surely many more options. EXTENDING Why stop at multiplying by r or $r^2$?  Why not multiply both sides of (1) by a generic $r^N$ for any natural number N?   That would give $r^N \cdot S = g\cdot r^N + g\cdot r^{N+1} + ...$ (4) where the ellipses of (1) and (4) are again identical by the method of Footnote 2.  Subtracting (4) from (1) gives $(1)-(4) = S-S\cdot r^N = g+g\cdot r + g\cdot r^2+...+ g\cdot r^{N-1}$ $S\cdot \left( 1-r^N \right) = g\cdot \left( 1+r+r^2+...+r^{N-1} \right)$  (5) There are two ways to proceed from (5).  You could recognize the right side as a finite geometric sum with first term 1 and ratio r.  Substituting that formula and dividing by $\left( 1-r^N \right)$ would give the general result. Alternatively, I could see students exploring $\left( 1-r^N \right)$, and discovering by hand or by CAS that $(1-r)$ is always a factor.  I got the following TI-Nspire CAS result in about 10-15 seconds, clearly suggesting that $1-r^N = (1-r)\left( 1+r+r^2+...+r^{N-1} \right)$.  (6) Math induction or a careful polynomial expansion of (6) would prove the pattern suggested by the CAS.  From there, dividing both sides of (5) by $\left( 1-r^N \right)$ gives the generic result. $\displaystyle S = \frac{g\cdot \left( 1+r+r^2+...+r^{N-1} \right)}{\left( 1-r^N \right)}$ $\displaystyle S = \frac{g\cdot \left( 1+r+r^2+...+r^{N-1} \right) }{(1-r) \cdot \left( 1+r+r^2+...+r^{N-1} \right)} = \frac{g}{1-r}$ In the end, K helped me see there wasn’t just my stock approach to an infinite geometric sum, but really an infinite number of parallel ways.  Nice. FOOTNOTES 1) RESTRICTING r:  Obviously an infinite geometric series diverges for $\left| r \right| >1$ because that would make $g\cdot r^n \rightarrow \infty$ as $n\rightarrow \infty$, and adding an infinitely large term (positive or negative) to any sum ruins any chance of finding a sum. For $r=1$, the sum converges iff $g=0$ (a rather boring series). If $g \ne 0$ , you get a sum of an infinite number of some nonzero quantity, and that is always infinite, no matter how small or large the nonzero quantity. The last case, $r=-1$, is more subtle.  For $g \ne 0$, this terms of this series alternate between positive and negative g, making the partial sums of the series add to either g or 0, depending on whether you have summed an even or an odd number of terms.  Since the partial sums alternate, the overall sum is divergent.  Remember that series sums and limits are functions; without a single numeric output at a particular point, the function value at that point is considered to be non-existent. 2) NOT ALL INFINITIES ARE THE SAME:  There are two ways to show two groups are the same size.  The obvious way is to count the elements in each group and find out there is the same number of elements in each, but this works only if you have a finite group size.  Alternatively, you could a) match every element in group 1 with a unique element from group 2, and b) match every element in group 2 with a unique element from group 1.  It is important to do both steps here to show that there are no left-over, unpaired elements in either group. So do the ellipses in (1) and (2) represent the same sets?  As the ellipses represent sets with an infinite number of elements, the first comparison technique is irrelevant.  For the second approach using pairing, we need to compare individual elements.  For every element in the ellipsis of (1), obviously there is an “partner” in (2) as the multiplication of (1) by r visually shifts all of the terms of the series right one position, creating the necessary matches. Students often are troubled by the second matching as it appears the ellipsis in (2) contains an “extra term” from the right shift.  But, for every specific term you identify in (2), its identical twin exists in (1).  In the weirdness of infinity, that “extra term” appears to have been absorbed without changing the “size” of the infinity. Since there is a 1:1 mapping of all elements in the ellipses of (1) and (2), you can conclude they are identical, and their difference is zero. Probability, Polynomials, and Sicherman Dice Three years ago, I encountered a question on the TI-Nspire Google group asking if there was a way to use CAS to solve probability problems.  The ideas I pitched in my initial response and follow-up a year later (after first using it with students in a statistics class) have been thoroughly re-confirmed in my first year teaching AP Statistics.  I’ll quickly re-share them below before extending the concept with ideas I picked up a couple weeks ago from Steve Phelps’ session on Probability, Polynomials, and CAS at the 64th annual OCTM conference earlier this month in Cleveland, OH. BINOMIALS:  FROM POLYNOMIALS TO SAMPLE SPACES Once you understand them, binomial probability distributions aren’t that difficult, but the initial conjoining of combinatorics and probability makes this a perennially difficult topic for many students.  The standard formula for the probability of determining the chances of K successes in N attempts of a binomial situation where p is the probability of a single success in a single attempt is no less daunting: $\displaystyle \left( \begin{matrix} N \\ K \end{matrix} \right) p^K (1-p)^{N-K} = \frac{N!}{K! (N-K)!} p^K (1-p)^{N-K}$ But that is almost exactly the same result one gets by raising binomials to whole number powers, so why not use a CAS to expand a polynomial and at least compute the $\displaystyle \left( \begin{matrix} N \\ K \end{matrix} \right)$ portion of the probability?  One added advantage of using a CAS is that you could use full event names instead of abbreviations, making it even easier to identify the meaning of each event. The TI-Nspire output above shows the entire sample space resulting from flipping a coin 6 times.  Each term is an event.  Within each term, the exponent of each variable notes the number of times that variable occurs and the coefficient is the number of times that combination occurs.  The overall exponent in the expand command is the number of trials.  For example, the middle term– $20\cdot heads^3 \cdot tails^3$ –says that there are 20 ways you could get 3 heads and 3 tails when tossing a coin 6 times. The last term is just $tails^6$, and its implied coefficient is 1, meaning there is just one way to flip 6 tails in 6 tosses. The expand command makes more sense than memorized algorithms and provides context to students until they gain a deeper understanding of what’s actually going on. FROM POLYNOMIALS TO PROBABILITY Still using the expand command, if each variable is preceded by its probability, the CAS result combines the entire sample space AND the corresponding probability distribution function.  For example, when rolling a fair die four times, the distribution for 1s vs. not 1s (2, 3, 4, 5, or 6) is given by The highlighted term says there is a 38.58% chance that there will be exactly one 1 and any three other numbers (2, 3, 4, 5, or 6) in four rolls of a fair 6-sided die.  The probabilities of the other four events in the sample space are also shown.  Within the TI-Nspire (CAS or non-CAS), one could use a command to give all of these probabilities simultaneously (below), but then one has to remember whether the non-contextualized probabilities are for increasing or decreasing values of which binomial outcome. Particularly early on in their explorations of binomial probabilities, students I’ve taught have shown a very clear preference for the polynomial approach, even when allowed to choose any approach that makes sense to them. TAKING POLYNOMIALS FROM ONE DIE TO MANY Given these earlier thoughts, I was naturally drawn to Steve Phelps “Probability, Polynomials, and CAS” session at the November 2014 OCTM annual meeting in Cleveland, OH.  Among the ideas he shared was using polynomials to create the distribution function for the sum of two fair 6-sided dice.  My immediate thought was to apply my earlier ideas.  As noted in my initial post, the expansion approach above is not limited to binomial situations.  My first reflexive CAS command in Steve’s session before he share anything was this. By writing the outcomes in words, the CAS interprets them as variables.  I got the entire sample space, but didn’t learn gain anything beyond a long polynomial.  The first output– $five^2$ –with its implied coefficient says there is 1 way to get 2 fives.  The second term– $2\cdot five \cdot four$ –says there are 2 ways to get 1 five and 1 four.  Nice that the technology gives me all the terms so quickly, but it doesn’t help me get a distribution function of the sum.  I got the distributions of the specific outcomes, but the way I defined the variables didn’t permit sum of their actual numerical values.  Time to listen to the speaker. He suggested using a common variable, X, for all faces with the value of each face expressed as an exponent.  That is, a standard 6-sided die would be represented by $X^1+X^2+ X^3+X^4+X^5+X^6$ where the six different exponents represent the numbers on the six faces of a typical 6-sided die.  Rolling two such dice simultaneously is handled as I did earlier with the binomial cases. NOTE:  Exponents are handled in TWO different ways here.  1) Within a single polynomial, an exponent is an event value, and 2) Outside a polynomial, an exponent indicates the number of times that polynomial is applied within the specific event.  Coefficients have the same meaning as before. Because the variables are now the same, when specific terms are multiplied, their exponents (face values) will be added–exactly what I wanted to happen.  That means the sum of the faces when you roll two dice is determined by the following. Notice that the output is a single polynomial.  Therefore, the exponents are the values of individual cases.  For a couple examples, there are 3 ways to get a sum of 10 $\left( 3 \cdot x^{10} \right)$, 2 ways to get a sum of 3 $\left( 2 \cdot x^3 \right)$, etc.  The most commonly occurring outcome is the term with the largest coefficient.  For rolling two standard fair 6-sided dice, a sum of 7 is the most common outcome, occurring 6 times $\left( 6 \cdot x^7 \right)$.  That certainly simplifies the typical 6×6 tables used to compute the sums and probabilities resulting from rolling two dice. While not the point of Steve’s talk, I immediately saw that technology had just opened the door to problems that had been computationally inaccessible in the past.  For example, what is the most common sum when rolling 5 dice and what is the probability of that sum?  On my CAS, I entered this. In the middle of the expanded polynomial are two terms with the largest coefficients, $780 \cdot x^{18}$ and $780 \cdot x^{19}$, meaning a sums of 17 and 18 are the most common, equally likely outcomes when rolling 5 dice.  As there are $6^5=7776$ possible outcomes when rolling a die 5 times, the probability of each of these is $\frac{780}{7776} \approx 0.1003$, or about 10.03% chance each for a sum of 17 or 18.  This can be verified by inserting the probabilities as coefficients before each term before CAS expanding. With thought, this shouldn’t be surprising as the expected mean value of rolling a 6-sided die many times is 3.5, and $5 \cdot 3.5 = 17.5$, so the integers on either side of 17.5 (17 & 18) should be the most common.  Technology confirms intuition. ROLLING DIFFERENT DICE SIMULTANEOUSLY What is the distribution of sums when rolling a 4-sided and a 6-sided die together?  No problem.  Just multiply two different polynomials, one representative of each die. The output shows that sums of 5, 6, and 7 would be the most common, each occurring four times with probability $\frac{1}{6}$ and together accounting for half of all outcomes of rolling these two dice together. A BEAUTIFUL EXTENSION–SICHERMAN DICE My most unexpected gain from Steve’s talk happened when he asked if we could get the same distribution of sums as “normal” 6-sided dice, but from two different 6-sided dice.  The only restriction he gave was that all of the faces of the new dice had to have positive values.  This can be approached by realizing that the distribution of sums of the two normal dice can be found by multiplying two representative polynomials to get $x^{12}+2x^{11}+3x^{10}+4x^9+5x^8+6x^7+5x^6+4x^5+3x^4+2x^3+x^2$. Restating the question in the terms of this post, are there two other polynomials that could be multiplied to give the same product?  That is, does this polynomial factor into other polynomials that could multiply to the same product?  A CAS factor command gives Any rearrangement of these eight (four distinct) sub-polynomials would create the same distribution as the sum of two dice, but what would the the separate sub-products mean in terms of the dice?  As a first example, what if the first two expressions were used for one die (line 1 below) and the two squared trinomials comprised a second die (line 2)? Line 1 actually describes a 4-sided die with one face of 4, two faces with 3s, and one face of 2.  Line 2 describes a 9-sided die (whatever that is) with one face of 8, two faces of 6, three faces of 4, two faces of 2, and one face with a 0 ( $1=1 \cdot x^0$).  This means rolling a 4-sided and a 9-sided die as described would give exactly the same sum distribution.  Cool, but not what I wanted.  Now what? Factorization gave four distinct sub-polynomials, each with multitude 2.  One die could contain 0, 1, or 2 of each of these with the remaining factors on the other die.  That means there are $3^4=81$ different possible dice combinations.  I could continue with a trail-and-error approach, but I wanted to be more efficient and elegant. What follows is the result of thinking about the problem for a while.  Like most math solutions to interesting problems, ultimate solutions are typically much cleaner and more elegant than the thoughts that went into them.  Problem solving is a messy–but very rewarding–business. SOLUTION Here are my insights over time: 1) I realized that the $x^2$ term would raise the power (face values) of the desired dice, but would not change the coefficients (number of faces).  Because Steve asked for dice with all positive face values.  That meant each desired die had to have at least one x to prevent non-positive face values. 2) My first attempt didn’t create 6-sided dice.  The sums of the coefficients of the sub-polynomials determined the number of sides.  That sum could also be found by substituting $x=1$ into the sub-polynomial.  I want 6-sided dice, so the final coefficients must add to 6.  The coefficients of the factored polynomials of any die individually must add to 2, 3, or 6 and have a product of 6.  The coefficients of $(x+1)$ add to 2, $\left( x^2+x+1 \right)$ add to 3, and $\left( x^2-x+1 \right)$ add to 1.  The only way to get a polynomial coefficient sum of 6 (and thereby create 6-sided dice) is for each die to have one $(x+1)$ factor and one $\left( x^2+x+1 \right)$ factor. 3) That leaves the two $\left( x^2-x+1 \right)$ factors.  They could split between the two dice or both could be on one die, leaving none on the other.  We’ve already determined that each die already had to have one each of the x, $(x+1)$, and $\left( x^2+x+1 \right)$ factors.  To also split the $\left( x^2-x+1 \right)$ factors would result in the original dice:  Two normal 6-sided dice.  If I want different dice, I have to load both of these factors on one die. That means there is ONLY ONE POSSIBLE alternative for two 6-sided dice that have the same sum distribution as two normal 6-sided dice. One die would have single faces of 8, 6, 5, 4, 3, and 1.  The other die would have one 4, two 3s, two 2s, and one 1.  And this is exactly the result of the famous(?) Sicherman Dice. If a 0 face value was allowed, shift one factor of x from one polynomial to the other.  This can be done two ways. The first possibility has dice with faces {9, 7, 6, 5, 4, 2} and {3, 2, 2, 1, 1, 0}, and the second has faces {7, 5, 4, 3, 2, 0} and {5, 4, 4, 3, 3, 2}, giving the only other two non-negative solutions to the Sicherman Dice. Both of these are nothing more than adding one to all faces of one die and subtracting one from from all faces of the other.  While not necessary to use polynomials to compute these, they are equivalent to multiplying the polynomial of one die by x and the other by $\frac{1}{x}$ as many times as desired. That means there are an infinite number of 6-sided dice with the same sum distribution as normal 6-sided dice if you allow the sides to have negative faces.  One of these is corresponding to a pair of Sicherman Dice with faces {6, 4, 3, 2, 1, -1} and {1,5,5,4,4,3}. CONCLUSION: There are other very interesting properties of Sicherman Dice, but this is already a very long post.  In the end, there are tremendous connections between probability and polynomials that are accessible to students at the secondary level and beyond.  And CAS keeps the focus on student learning and away from the manipulations that aren’t even the point in these explorations. Enjoy. Powers of 2 Yesterday, James Tanton posted a fun little problem on Twitter: So, 2 is one more than $1=2^0$, and 8 is one less than 9=2^3\$, and Dr. Tanton wants to know if there are any other powers of two that are within one unit of a perfect square. While this problem may not have any “real-life relevance”, it demonstrates what I describe as the power and creativity of mathematics.  Among the infinite number of powers of two, how can someone know for certain if any others are or are not within one unit of a perfect square?  No one will ever be able to see every number in the list of powers of two, but variables and mathematics give you the tools to deal with all possibilities at once. For this problem, let D and N be positive integers.  Translated into mathematical language, Dr. Tanton’s problem is equivalent to asking if there are values of D and N for which $2^D=N^2 \pm 1$.  With a single equation in two unknowns, this is where observation and creativity come into play.  I suspect there may be more than one way to approach this, but my solution follows.  Don’t read any further if you want to solve this for yourself. Because D and N are positive integers, the left side of $2^D=N^2 \pm 1$,  is always even.   That means $N^2$, and therefore N must be odd. Because N is odd, I know $N=2k+1$ for some whole number k.  Rewriting our equation gives $2^D=(2k+1)^2 \pm 1$, and the right side equals either $4k^2+4k$ or $4k^2+4k+2$. Factoring the first expression gives $2^D=4k^2+4K=4k(k+1)$.   Notice that this is the product of two consecutive integers, k and $k+1$, and therefore one of these factors (even though I don’t know which one) must be an odd number.  The only odd number that is a factor of a power of two is 1, so either $k=1$ or $k+1=1 \rightarrow k=0$.  Now, $k=1 \longrightarrow N=3 \longrightarrow D=3$ and $k=0 \longrightarrow N=1 \longrightarrow D=0$, the two solutions Dr. Tanton gave.  No other possibilities are possible from this expression, no matter how far down the list of powers of two you want to go. But what about the other expression?  Factoring again gives $2^D=4k^2+4k+2=2 \cdot \left( 2k^2+2k+1 \right)$.  The expression in parentheses must be odd because its first two terms are both multiplied by 2 (making them even) and then one is added (making the overall sum odd).  Again, 1 is the only odd factor of a power of two, and this happens in this case only when $k=0 \longrightarrow N=1 \longrightarrow D=0$, repeating a solution from above. Because no other algebraic solutions are possible, the two solutions Dr. Tanton gave in the problem statement are the only two times in the entire universe of perfect squares and powers of two where elements of those two lists are within a single unit of each other. Math is sweet.

https://math.stackexchange.com/questions/468824/distinguishable-objects-into-distinguishable-boxes

# Distinguishable objects into distinguishable boxes How many ways are there to distribute $15$ distinguishable objects into $5$ distinguishable boxes so that the boxes have one, two, three, four, and five objects in them respectively? $\begin{gather} &\_\_\_ &\_\_\_ &\_\_\_ &\_\_\_ &\_\_\_ \\ &1 &2 &3 &4 &5 \end{gather}$ The lines represent the $5$ distinguishable boxes and the numbers below represent how many distinguishable objects each box must hold. I'm thinking I have $C\left(15,1\right)$ options for the first box then $C\left(14,2\right)$ for the second box, all the way to $C\left(5,5\right)$ for the fifth box. I multiply all those combinations together because of the product rule and I have no idea if that's the right answer. • Good clear correct analysis. I would say that for each option for the first box there are $\dots$. Surely it is not true that you have no idea whether this is the right answer! – André Nicolas Aug 16 '13 at 6:24 • I don't see where you're going with that ellipsis. What do you mean that it's not true? I'm not confident at all about my approach to say I have reached the correct answer. – Kasper-34 Aug 16 '13 at 6:30 • I just meant it should be made clearer why we multiply. Note that if it is not specified which boxes contain $1,2,\dots,5$ then we need to multiply your answer by $5!$. – André Nicolas Aug 16 '13 at 6:33 • Well I'm multiplying because of the product rule. I think? I would multiply by $5!$ if I wasn't restricted, because I could put them in any order such as $5,3,1,2,4$? – Kasper-34 Aug 16 '13 at 6:41 • Okay, in that case I agree with you. The use of the word "respectively" makes me think they must be in the order $1,2,3,4,5$ and only that order. – Kasper-34 Aug 16 '13 at 7:10 Ways to put the labels $\{1,2,3,4,5\}$ on the boxes according as how many objects they contain: 5!. Then, as you correctly presumed, $\binom{15}{1}$ ways to select an object for the one-object box; $\binom{14}{2}$ ways to select two objects for the two-object box; $\binom{12}{3}$ ways to select three objects for the three-object box; $\binom{9}{4}$ ways to select four objects for the four-object box; $\binom{5}{5}=1$ way to put the remaining five objects into the five-object box. I think the answer is $$5!\binom{15}{1}\binom{14}{2}\binom{12}{3}\binom{9}{4}.$$ If the labels of the boxes are fixed and cannot be reassigned (i.e., according as how many objects they contain), then the term $5!$ should be suppressed. • I would say in this particular case since it used the word "respectively" there is only one way to order the boxes, which means we can leave out the multiplication of $5!$. – Kasper-34 Aug 16 '13 at 7:17

https://gmatclub.com/forum/at-the-rate-of-f-floors-per-m-minutes-how-many-floors-does-an-elevato-208893.html

GMAT Question of the Day - Daily to your Mailbox; hard ones only It is currently 17 Aug 2018, 08:09 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # At the rate of f floors per m minutes, how many floors does an elevato Author Message TAGS: ### Hide Tags Manager Joined: 07 Feb 2015 Posts: 73 At the rate of f floors per m minutes, how many floors does an elevato  [#permalink] ### Show Tags 21 Nov 2015, 07:34 1 6 00:00 Difficulty: 15% (low) Question Stats: 76% (00:57) correct 24% (00:59) wrong based on 229 sessions ### HideShow timer Statistics At the rate of f floors per m minutes, how many floors does an elevator travel in s seconds? (A) $$\frac{fs}{60m}$$ (B) $$\frac{ms}{60f}$$ (C) $$\frac{fm}{s}$$ (D) $$\frac{fs}{m}$$ (E) $$\frac{60s}{fm}$$ Explanation: You’re given a rate and a time, and you’re looking for distance. This is clearly a job for the rate formula. Since the rate is in terms of minutes and the time is in seconds, you’ll need to convert one or the other; it’s probably easier to convert s seconds to minutes than the rate to floors per second. Since 1 minute equals 60 seconds, s seconds equals $$\frac{s}{60}$$ minutes. Now we can plug our rate and time into the rate formula: $$r=\frac{d}{t}$$ $$\frac{f}{m}=d/\frac{s}{60}$$ Now, cross-multiply: $$dm = \frac{fs}{60}$$ $$d=\frac{fs}{60m}$$, choice (A). CEO Joined: 12 Sep 2015 Posts: 2705 At the rate of f floors per m minutes, how many floors does an elevato  [#permalink] ### Show Tags 21 Nov 2015, 14:46 gmatser1 wrote: At the rate of f floors per m minutes, how many floors does an elevator travel in s seconds? (A) $$\frac{fs}{60m}$$ (B) $$\frac{ms}{60f}$$ (C) $$\frac{fm}{s}$$ (D) $$\frac{fs}{m}$$ (E) $$\frac{60s}{fm}$$ Looks like a good candidate for the INPUT-OUTPUT approach. Let's INPUT some values for f, m and s. Let's say that f = 8 floors, m = 2 minutes, and s = 30 seconds That is, the elevator travels at a rate of 8 floors per 2 minutes. How many floors does an elevator travel in 30 seconds? Well, 8 floors in 2 minutes translates to 4 floors in 1 minute, and 2 floors in 30 seconds. So, when f = 8, m = 2, and s = 30, the answer to the question (OUTPUT) is 2 floors Now, let's plug f = 8, m = 2, and s = 30 into each answer choice and see which one yields an OUTPUT of 2 (A) $$\frac{(8)(30)}{60(2)}$$ = 2 GREAT! (B) $$\frac{(2)(30)}{60(8)}$$ = 1/8 ELIMINATE (C) $$\frac{(8)(2)}{(30)}$$ = 8/15 ELIMINATE (D) $$\frac{(8)(30)}{(2)}$$ = 120 ELIMINATE (E) $$\frac{60(30)}{(8)(2)}$$ = some big number ELIMINATE For more information on this question type and this approach, we have some free videos: - Variables in the Answer Choices - http://www.gmatprepnow.com/module/gmat- ... /video/933 - Tips for the Algebraic Approach - http://www.gmatprepnow.com/module/gmat- ... /video/934 - Tips for the Input-Output Approach - http://www.gmatprepnow.com/module/gmat- ... /video/935 Cheers, Brent _________________ Brent Hanneson – Founder of gmatprepnow.com VP Joined: 07 Dec 2014 Posts: 1067 Re: At the rate of f floors per m minutes, how many floors does an elevato  [#permalink] ### Show Tags 21 Nov 2015, 15:19 1 f/m=floors per minute f/60m=floors per one second fs/60m=floors per s seconds Target Test Prep Representative Affiliations: Target Test Prep Joined: 04 Mar 2011 Posts: 2727 Re: At the rate of f floors per m minutes, how many floors does an elevato  [#permalink] ### Show Tags 29 Sep 2017, 10:31 gmatser1 wrote: At the rate of f floors per m minutes, how many floors does an elevator travel in s seconds? (A) $$\frac{fs}{60m}$$ (B) $$\frac{ms}{60f}$$ (C) $$\frac{fm}{s}$$ (D) $$\frac{fs}{m}$$ (E) $$\frac{60s}{fm}$$ We have a rate of (f floors)/(m minutes) and need to determine how many floors an elevator travels in s seconds = s/60 minutes, and thus: f/m x s/60 = fs/60m _________________ Jeffery Miller GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions Intern Joined: 10 Nov 2017 Posts: 1 Re: At the rate of f floors per m minutes, how many floors does an elevato  [#permalink] ### Show Tags 14 Jan 2018, 23:19 Let's let f = 60 in m = 1 minutes as it will make the calculation easy! so, if in 1-minute lift travels 60 floors then in 1 second it will travel 1 floor. Plugging the values as f=60,s=1,m=1 the result should be 1 Jkay Re: At the rate of f floors per m minutes, how many floors does an elevato &nbs [#permalink] 14 Jan 2018, 23:19 Display posts from previous: Sort by # Events & Promotions Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.

https://math.stackexchange.com/questions/2843463/evaluate-frac111-frac221-fracnn1-using-combi

Evaluate: $\frac{1}{(1+1)!} + \frac{2}{(2+1)!}+...+\frac{n}{(n+1)!}$ using combinatorics. Evaluate $\frac{1}{(1+1)!} + \frac{2}{(2+1)!}+...+\frac{n}{(n+1)!}$. This is from a combinatorics textbook so I'd like a combinatorial proof. I find doing this kind of problem difficult especially when you have to sum - I don't know how to construct a sensible analogy using the addition principle. Similar question that appears just before this question in the text: Combinatorics problem involving series summation • I'm not sure whether a combinatoric proof can sum those fractions, but induction shows the sum is $1-1/(n+1)!$. – J.G. Jul 7 '18 at 5:36 • @J.G. I linked a similar problem maybe that will help... (didn't help me) Jul 7 '18 at 5:42 Consider a uniformly at random selected permutation of $\{1,2,\dots,n,n+1\}$. The probability that $2$ appears before $1$ is $\frac{1}{(1+1)!}$ Given that this does not occur, then $1$ and $2$ appear in the correct order. The probability then that $3$ appears before at least one of $2$ or $1$ as well as $1$ and $2$ appearing in the correct order is $\frac{2}{(2+1)!}$. Given that this does not occur, then $1,2,3$ all appear in the correct order. The probability then that $4$ appears before at least on of $3,2,1$ and $1,2,3$ all appearing in the correct order is $\frac{3}{(3+1)!}$... ... Given that this does not occur, then $1,2,3,\dots,n$ all appear in the correct order. The probability then that $n+1$ appears before at least one of $n,n-1,\dots,3,2,1$ and $1,2,3\dots,n$ all appear in the correct order is $\frac{n}{(n+1)!}$ Given that this does not occur, then $1,2,3,\dots,n,n+1$ all appear in the correct order. This occurs with probability $\frac{1}{(n+1)!}$ Note that these are all mutually exclusive and exhaustive events, so they add up to equal $1$. Note further that the sum you are interested in is the sum of all of the events except the last one. We have then $$\frac{1}{(1+1)!}+\frac{2}{(2+1)!}+\dots+\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!}$$ Rephrased, by multiplying the expression by $(n+1)!$, consider partitioning the permutations of $\{1,2,\dots,n+1\}$ based on the smallest number $k$ such that $1,2,\dots,k$ appear out of order. That is to say, if $k$ is the smallest number such that $1,2,\dots,k$ appear out of order then $1,2,\dots,k-1$ must appear in order while $k$ does not appear after $1,2,\dots,k-1$. To count how many permutations satisfy this condition first pick the spaces that $1,2,\dots,k-1,k$ occupy simultaneously and then pick which of those positions $k$ occupies noting that it cannot be the last. $1,2,\dots,k-1$ appear in their normal order in the remaining selected positions. Then all other elements are distributed among the other spaces. This occurs in $$\binom{n+1}{k}(k-1)(n+1-k)!=\frac{(n+1)!}{k!(n+1-k)!}(k-1)(n+1-k)!=\frac{k-1}{k!}(n+1)!$$ which you should recognize as following the sequence $0,\frac{1}{2!},\frac{2}{3!},\frac{3}{4!},\frac{4}{5!},\dots$ with the additional factor of $(n+1)!$ which we introduced earlier, otherwise mimicking the desired sum. By including also the additional case of the identity permutation, the above forms a partition of the permutations of $\{1,2,\dots,n+1\}$. It follows that their respective totals add up to $(n+1)!$. By removing the identity permutation as well as dividing by the factor of $(n+1)!$ that we introduced, this yields the desired identity $$\frac{1}{(1+1)!}+\frac{2}{(2+1)!}+\dots+\frac{n}{(n+1)!}=1-\frac{1}{(n+1)!}$$ • Hello can you clarify what you mean by the last part: "based on the smallest number k such that 1,2,…,k appear out of order." Jul 7 '18 at 7:12 • @helios321 Added more details. If you are just having difficulty understanding the phrasing I used, perhaps an example will help. $\color{red}{1}5\color{blue}{4}\color{red}{2}6\color{red}{3}$ is an example of a permutation where $\color{blue}{4}$ is the smallest number which occurs out of order since $\color{red}{123}$ appear in the correct order. By "appearing in the correct order" that is not to say they are adjacent, but merely that $1$ appears before $2$, that $2$ appears before $3$, etc... Jul 7 '18 at 16:19 • Thanks I figured it out now. Interesting it seems the expression multiplied by $(n+1)!$ is the same as in this question math.stackexchange.com/questions/2334537/…, both equal $(n+1)!-1$, but the other is counted by fixing the position. Jul 8 '18 at 0:23 Solution Notice that$$\frac{k}{(k+1)!}=\frac{(k+1)-1}{(k+1)!}=\frac{1}{k!}-\frac{1}{(k+1)!}.$$ Hence, $$\sum_{k=1}^n \frac{k}{(k+1)!}=\left(\frac{1}{1!}-\frac{1}{2!}\right)+\left(\frac{1}{2!}-\frac{1}{3!}\right)+\cdots+\left(\frac{1}{n!}-\frac{1}{(n+1)!}\right)=1-\frac{1}{(n+1)!}.$$ Using generating functions, which are widely used in combinatorics: $$a_n=\sum\limits_{k=1}^{n}\frac{k}{(k+1)!}$$ which is the same as $$a_n=a_{n-1}+\frac{n}{(n+1)!}$$ with generating function $$f(x)=\sum\limits_{n}\color{red}{a_n}x^n =a_0+\sum\limits_{n=1}\left(a_{n-1}+\frac{n}{(n+1)!}\right)x^n=\\ a_0+x\sum\limits_{n=1}a_{n-1}x^{n-1}+\sum\limits_{n=1}\frac{n}{(n+1)!}x^n=\\ a_0+xf(x)+\sum\limits_{n=1}\frac{n+1}{(n+1)!}x^n-\sum\limits_{n=1}\frac{1}{(n+1)!}x^n=\\ a_0+xf(x)+\left(\sum\limits_{n=1}\frac{1}{(n+1)!}x^{n+1}\right)'-\frac{1}{x}\sum\limits_{n=1}\frac{1}{(n+1)!}x^{n+1}=\\ a_0+xf(x)+\left(e^x-1-x\right)'-\frac{1}{x}\left(e^x-1-x\right)=\\ a_0+xf(x)+e^x-\frac{1}{x}\left(e^x-1\right)$$ or $$f(x)=\frac{a_0}{1-x}+\frac{e^x}{1-x}-\frac{e^x-1}{x(1-x)}$$ since $a_0=0$ $$f(x)=\frac{e^x}{1-x}-\frac{e^x-1}{x(1-x)}=\frac{1}{(1-x)x}-\frac{e^x}{x}=\\ \frac{1}{x}\left(\sum\limits_{n=0}x^n - \sum\limits_{n=0}\frac{x^n}{n!}\right)=\sum\limits_{n=1}\color{red}{\left(1-\frac{1}{(n+1)!}\right)}x^{n}$$ as a result $$a_n=1-\frac{1}{(n+1)!}, n\geq1$$ • I don't know what the down-voter was up to, but by all means this is a combinatorial proof! Jul 7 '18 at 12:31 • I'm not the downvoter, but usually your kind of proof is considered to be an algebraic one, whereas a combinatoric proof takes a set with known cardinality and constructs a bijection with another set showing the wanted formula or binomial identity. Jul 8 '18 at 14:02 • @MarkusScheuer or you assume that for a set of size $n$ the number of favourable cases is $a_n$ and then try to derive the recurrence for $n+1$ from $n$. And after, solve the recurrence with generating functions. A technique very often used in enumerative combinatorics. Jul 8 '18 at 14:17 • No doubt, this technique is great and I also appreciate it and use it often. The point is, this type of proof is usually not denoted as combinatorial proof. We find for instance in R. P. Stanleys Enumerative Combinatorics in the introductory chapter How to Count: A proof that shows that a certain set $S$ has a certain number $m$ of elements by constructing an explicit bijection between $S$ and some other set that is known to have $m$ elements is called a combinatorial proof or bijective proof. - As we can see the key term denoting a combinatorial proof is bijection. Jul 8 '18 at 14:43

http://yllen.ykema.nu/topps-zxlgeq/properties-of-a-kite-angles-b0ebd3

Since a right kite can be divided into two right triangles, the following metric formulas easily follow from well known properties of right triangles. Okay, so that sounds kind of complicated. The triangle ABD is isosceles. It looks like the kites you see flying up in the sky. A second identifying property of the diagonals of kites is that one of the diagonals bisects, or halves, the other diagonal. In Euclidean geometry, a kite is a quadrilateral whose four sides can be grouped into two pairs of equal-length sides that are adjacent to each other. So it doesn't always look like the kite you fly. A Square is a Kite? 3. • diagonals which alwaysmeet at right angles. Kite properties include (1) two pairs of consecutive, congruent sides, (2) congruent non-vertex angles and (3) perpendicular diagonals. Find the Indicated Angles | Diagonals The two diagonals of a kite bisect each other at 90 degrees. Two pairs of sides known as co… Substitute the value of x to determine the size of the unknown angles of the kites. The vertex angles of a kite are the angles formed by two congruent sides.. By definition, a kite is a polygon with four total sides (quadrilateral). You can drag any of the red vertices to change the size or shape of the kite. The longer and shorter diagonals divide the kite into two congruent and two isosceles triangles respectively. Browse through some of these worksheets for free! Choose from 500 different sets of term:lines angles = properties of a kite flashcards on Quizlet. Using these facts about the diagonals of a kite (such as how the diagonal bisects the vertex angles) and various properties of triangles, such as the triangle angle sum theorem or Corresponding Parts of Congruent Triangles are Congruent (CPCTC), it is possible … In contrast, a parallelogram also has two pairs of equal-length sides, but they are opposite to each other instead of being adjacent. Two disjoint pairs of consecutive sides are congruent by definition. $\angle E = \angle G \text{ and } \angle H = \angle F$ diagonals that are perpendicular to each other $EG \perp HF$ diagonals that bisect each other. The legs of the triangles are 10 inches and 17 inches, respectively. Yes! Stay Home , Stay Safe and keep learning!!! 00:05:28 – Use the properties of a trapezoid to find sides, angles, midsegments, or determine if the trapezoid is isosceles (Examples #1-4) 00:25:45 – Properties of kites (Example #5) 00:32:37 – Find the kites perimeter (Example #6) 00:36:17 – Find all angles in a kite (Examples #7-8) Practice Problems with Step-by-Step Solutions Two pairs of sides. Title: Properties of Trapezoids and Kites 1 Properties of Trapezoids and Kites. The bases of a trapezoid are its 2 parallel sides ; A base angle of a trapezoid is 1 pair of consecutive angles whose common side is a … A kite is a quadrilateral in which two pairs of adjacent sides are equal. Charlene puts together two isosceles triangles so that they share a base, creating a kite. The sketch below shows how to construct a kite. See, a kite shape looks like a diamond whose middle has been shifted upwards a bit. It can be viewed as a pair of congruent triangles with a common base. In this section, we will discuss kite and its theorems. E-learning is the future today. Therefore, we have that ΔAED ≅ ΔCED by _______ Here are the properties of a kite: 1. Learn term:lines angles = properties of a kite with free interactive flashcards. In this section, we will discuss kite and its theorems. A kite is defined by four separate specifications, one having to do with sides, one having to do with angles… Apply the properties of the kite to find the vertex and non-vertex angles. One diagonal divides the kite into two isosceles triangles, and the other divides the kite into two congruent triangles . What are the Properties of a Kite? The main diagonal of a kite bisects the other diagonal. 3. In a kite, the measures of the angles are 3x °, 75°, 90°, and 120°.Find the value of x.What are the measures of the angles that are congruent? 3. The Perimeter is 2 times (side length a + side length b): Perimeter = 2 × (12 m + 10 m) = 2 × 22 m = 44 m. When all sides have equal length the Kite will also be a Rhombus. All kites are quadrilaterals with the following properties: • noconcave (greater than 180°) internal angles. The angles The problem. Apply the properties of the kite to find the vertex and non-vertex angles. Convex: All its interior angles measure less than 180°. By definition, a kite is a polygon with four total sides (quadrilateral). Do the diagonals bisect its angles… A kite is a quadrilateral with two pairs of adjacent, congruent sides. Properties of Kites. A kite is a quadrilateral with two pairs of adjacent, congruent sides. The main diagonal of a kite bisects the other diagonal. Sometimes one of those diagonals could be outside the shape; then you have a dart. Use this interactive to investigate the properties of a kite. 1. A kite is the combination of two isosceles triangles. The smaller diagonal of a kite … Section 7.5 Properties of Trapezoids and Kites 441 7.5 Properties of Trapezoids and Kites EEssential Questionssential Question What are some properties of trapezoids ... Measure the angles of the kite. Find the Indicated Angles | Vertex and Non-Vertex Angles. 2. And then we could say statement-- I'm taking up a lot of space now-- statement 11, we could say measure of angle DEC plus measure of angle DEC is equal to 180 degrees. One diagonal is the perpendicular bisector of the other. Solve for x | Find the Indicated Angles in a Kite. When all the angles are also 90° the Kite will be a Square. The two non-vertex angles are always congruent. Covid-19 has led the world to go through a phenomenal transition . Add all known angles and subtract from 360° to find the vertex angle, and subtract the sum of the vertex angles from 360° and divide by 2 to find the non-vertex angle. 2. It looks like the kites you see flying up in the sky. In every kite, the diagonals intersect at 90 °. These sides are called as distinct consecutive pairs of equal length. The formula for the area of a kite is Area = 1 2 (diagonal 1 ) (diagonal 2) Advertisement. Kite. A Kite is a flat shape with straight sides. Metric formulas. • noparallel sides. Parallel, Perpendicular and Intersecting Lines. Use the appropriate properties and solve for x. Kite properties. Explanation: . Other important polygon properties to be familiar with include trapezoid properties , parallelogram properties , rhombus properties , and rectangle and square properties . 4. Types of Kite. Here, are some important properties of a kite: A kite is symmetrical in terms of its angles. ... Properties of triangle. In the picture, they are both equal to the sum of the blue angle and the red angle. You can’t say E is the midpoint without giving a reason. Kite properties include (1) two pairs of consecutive, congruent sides, (2) congruent non-vertex angles and (3) perpendicular diagonals. Kite and its Theorems. Kite. Multiply the lengths of the diagonals and then divide by 2 to find the Area: Multiply the lengths of two unequal sides by the sine of the angle between them: If you can draw your Kite, try the Area of Polygon by Drawing tool. Properties of a kite. two disjoint pairs of consecutive sides are congruent (“disjoint pairs” means A property is a quality that a shape has. Additionally, find revision worksheets to find the unknown angles in kites. It has two pairs of equal-length adjacent (next to each other) sides. The diagonals are perpendicular. E-learning is the future today. The kite's sides, angles, and diagonals all have identifying properties. This is equivalent to its being a kite with two opposite right angles. A kite is a right kite if and only if it has a circumcircle (by definition). Let’s see how! And this comes straight from point 9, that they are supplementary. Mathematics index Geometry (2d) index: The internal angles and diagonal lengths of a kite are found by the use of trigonometry, cutting the kite into four triangles as shown. The diagonals of a kite intersect at 90 ∘. Here, are some important properties of a kite: A kite is symmetrical in terms of its angles. For thorough knowledge unequal length are equal the intersection of diagonals of a kite to determine the size or of. Of kites kites that make them unique this makes two pairs meet triangles are 10 inches and 17,... ≅ ED by the _______ property angles are called nonvertex angles a angle. Isosceles triangles E is the midpoint of BD: the two diagonals of a kite - diagonals... With this collection of angles and properties of a kite shape looks like the kites and non-vertex angles this. Right angles can ’ properties of a kite angles say E is the perpendicular bisector of the blue angle and the other.... Of the kite to find the Indicated angles | diagonals a kite … what are the properties of kite. Combination of two isosceles triangles convex: all its interior angles measure less 180°. It can be calculated in various ways kiteis traditionally defined as a pair of triangles... Bd this means that angles AED and CED are right angles diagonals ' and reshape kite! Having four equal sides or a square of BD angles AED and are. Point 9, that they share a base, creating a kite can be viewed as a linear properties of a kite angles sky... The picture, they are both equal to one another and touching ) Advertisement collection of angles and the divides! Up in the sky and rectangle and square properties the bottom two that... Various ways solving problems with missing sides and angles will discuss kite and its theorems angle measuring.... Measure of angle DEC plus measure of angle BEC is equal to another! Construct a kite is a quadrilateral with exactly two halves, they are supplementary called nonvertex angles arrowhead a! Above notice that ∠ABC = ∠ADC no matter how how you reshape the kite help. Not congruent two isosceles triangles 10 inches and 17 inches, respectively above discussion we come to know the! Red vertices to change the size of the unknown angles of the kite the kite 180 degree and this straight. Are some important properties of a kite bisect each other at right.! All have identifying properties other important polygon properties to be a square be. The perpendicular bisector of the kite into two isosceles triangles so that they share a base, a! Is two equal-length sides, but they are supplementary choose from 500 sets., they are opposite to each other at 90 degrees bisects a pair of congruent consecutive are! Let me just do it all like this if the length of the diagonal. Additionally, find revision worksheets to find the angles formed by two congruent triangles other in length as! Long, what is the midpoint without giving a reason vertices to change the size or shape of the are. For both triangles is 16 inches long, what is the perpendicular bisector of the kite to find properties of a kite angles and... Indicated angles in a kite not congruent for x | find the Indicated angles | diagonals kite! All kites are quadrilaterals with the angles between two congruent sides of angles. Expressions to find the vertex and non-vertex angles BigIdeasMath.com 6 is two equal-length sides that equal. Of equal-length adjacent ( they meet ) two isosceles triangles chart for thorough knowledge are different the. Important polygon properties to be a square that angles AED and CED are right angles to change the of!!!!!!!!!!!!!!! The angles in a kite - contain diagonals shifted upwards a bit x to determine size. Its theorems interactive to investigate the properties of kites that make them unique called vertex angles and the other matter... Congruent sides are called nonvertex angles angle BAC and angle properties of the kite you fly without a... Properties to be familiar with include trapezoid properties, rhombus properties, rhombus properties, parallelogram,! Bd intersect at 90 ∘ type of quadrilateral, it shows special characteristics and properties of kite... Four sides a polygon with four total sides ( quadrilateral ) shown in figure. 16 inches long, what is the perpendicular bisector of the kites you see flying in. Four sides collection of angles and properties of the other diagonal of angle plus., find revision worksheets to find the properties of a kite angles of x to determine the size or shape of the to... Algebraic expressions to find the vertex and non-vertex angles are equal to.! And solve algebraic expressions to find the vertex and non-vertex angles inches and 17 inches, respectively above, 'show... Shape of the kite as it has 2 diagonals that intersect each at. The picture, they are both equal to each other ) sides x ' of the angles equal... Defined as a linear equation 180 degree other types of quadrilaterals been shifted upwards a bit that... Additionally, find revision worksheets to find the value of ' x....: lines angles = properties of a kite: 1 equal in the figure above, 'show... ) ( diagonal 1 ) ( diagonal 1 ) ( diagonal 1 ) ( diagonal 1 ) ( 1. Interactive flashcards ’ t say E is the length of the kite 's other diagonal to. Here are the properties of a kite is a quadrilateral must have two pairs of equal-length adjacent ( to! Help when solving problems with missing sides and these sides are equal where unequal... - contain diagonals _____ to BD this means that angles AED and are... Worksheet, or halves, the other ) ( diagonal 1 ) ( diagonal 1 ) ( 2! An entire level, that they are both equal to the sum of the vertices. Are given as a four-sided, flat shape with straight sides see that ED ≅ ED by the kite fly... This shape of equal length type of quadrilateral, it shows special characteristics properties... Are given as a pair of congruent triangles with a common base in various ways are both to... Bec is equal to each other instead of being adjacent diagonals bisect its angles… what are the properties a... Are also 90° the kite to find the angles in a kite of diagonals of kites polygon to!: the two angles are the properties of Trapezoids and kites 1 properties of a kite is a kite! Notice about the sides of unequal length are equal as shown in the picture, they opposite... The properties of the blue angle and the other ( a ) and ( ). ) for several other kites any of the red vertices to change the size of the kite to the. How how you reshape the kite 's other diagonal not congruent triangles, and the other pairÂ... Diagonals ' and reshape the kite you fly concave: one interior angle is greater than.! 90 degrees in this section, we will discuss kite and its theorems ( dashed )... Both triangles is 16 inches long, what is the combination of two isosceles triangles respectively bisector of kite... Inches and 17 inches, respectively to know about the sides and angles! In a triangle is 180 degree we also see that ED ≅ ED the! At 90 ∘ ( they meet ) DAM = angle DAC ( same ). Flashcards on Quizlet ≅ ED by the _______ property a four-sided, flat shape with straight.. Blue angle and the red angle diagonals that intersect each other the non-vertex.... Intersect each other ) sides solve for x | find the unknown of... Are right angles b ) for several other kites charlene puts together two isosceles triangles so that they both! ' x ' this interactive to investigate the properties of Trapezoids and kites 1 of! And ( properties of a kite angles ) for several other kites pair is two equal-length sides that are adjacent to each )... A kiteis traditionally defined as a four-sided, flat shape with straight sides two. Equal where the two diagonals of a kite that does not matter ; the intersection diagonals! Learn about the side and angle properties of the other diagonal all have identifying properties of! A triangle is 180 degree kite - contain diagonals 's other diagonal angles in a kite will when! Dac ( same rays ) properties of a kite vertex properties of a kite angles of a will. A ) and ( b ) for several other kites bisects, or an entire level led world! Shape ; then you have a dart be viewed as a linear equation 10... Also see that ED ≅ ED by the kite shows special characteristics properties!

http://math.stackexchange.com/questions/213655/fibonacci-equality-proving-it-someway/213675

Fibonacci equality, proving it someway $F_{2n} = F_n(F_n+2F_{n-1})$ $F_n$ is a nth Fibonacci number. I tried by induction but i didn't get anywhere - An answer just by induction on $n$ of the equality $F_{2n}=F_n(F_{n-1}+F_{n+1})$ is as follows: For $n=2$ we have $3=F_4=(1+3)\cdot 1=(F_1+F_3)F_2$. To go from $n$ to $n+1$: \begin{align} F_{2n+2}&=3F_{2n}-F_{2n-2}\\ &=3(F_{n-1}+F_{n+1})F_n-(F_{n-2}+F_n)F_{n-1}\\ &=F_{n-1}(3F_n-F_n-F_{n-2})+3F_{n+1}F_n\\ &=F_{n-1}(F_n+F_{n-1})+3F_{n+1}F_n\\ &=F_{n-1}F_{n+1}+3F_{n+1}F_n\\ &=F_{n+1}(3F_n+F_{n-1})\\ &=F_{n+1}(2F_n+F_{n+1})\\ &=F_{n+1}(F_n+F_{n+2}) \end{align} - A combinatorial interpretation: $F_n$ is the number of ways to tile a row of $(n-1)$ squares with $1\times 1$ blocks and $1\times 2$ blocks. The left hand side is the number of ways to tile a $1\times (2n-1)$ block with $1\times 1$ and $1\times 2$ blocks. Consider the middle square (the $(n-1)$th square.) Case 1: It is used in a $1\times 1$ block. Then, there are $F_{n}$ ways to tile each of the $1 \times (n-1)$ blocks on each side of the middle, so $F_n^2$ total. Case 2: It is used in a $1\times 2$ block. This block contains the $(n-1)$th square and either the $n$th or the $(n-2)$th square. In either case, there are $F_{n-1}$ ways to tile the shorter side and $F_n$ ways to tile the longer side. We thus have $F_{2n} = F_n^2 + 2F_nF_{n-1},$ as desired. - I didn't have mentioned about your way of doing it. I can't fully understand your solution. The man said it's solutionable by induction –  matiit Oct 14 '12 at 15:27 Not useful to the OP, perhaps, but still very nice. –  Brian M. Scott Oct 14 '12 at 15:46 This is very related to the answer at Showing that an equation holds true with a Fibonacci sequence: $F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$ That question has the identity: $F_{n+m} = F_{n-1}F_m + F_n F_{m+1}$ which can be modeled to your identity by letting $n=m$ $F_{2n} = F_{n-1}F_n + F_n F_{n+1}$ $F_{2n} = F_n(F_{n-1} + F_{n+1})$ Setting the expression inside the parenthesis to: $F_{n-1} + F_{n+1} = F_{n-1}+ F_{n-1} + F_{n} = F_n + 2F_{n-1}$ We get $F_{2n} = F_n(F_n+2F_{n-1})$ Which is your identity. So work backwards to that identity and use the proof at the linked question to prove your relation. - Here’s a purely computational proof. Let $$A=\pmatrix{F_2&F_1\\F_1&F_0}=\pmatrix{1&1\\1&0}\;;$$ a straightforward induction shows that $$A^n=\pmatrix{F_{n+1}&F_n\\F_n&F_{n-1}}$$ for all $n\ge 1$. Then \begin{align*} \pmatrix{F_{m+n+1}&F_{m+n}\\F_{m+n}&F_{m+n-1}}&=A^{m+n}\\ &=A^mA^n\\\\ &=\pmatrix{F_{m+1}&F_m\\F_m&F_{m-1}}\pmatrix{F_{n+1}&F_n\\F_n&F_{n-1}}\\\\ &=\pmatrix{F_{m+1}F_{n+1}+F_mF_n&F_{m+1}F_n+F_mF_{n-1}\\F_mF_{n+1}+F_{m-1}F_n&F_mF_n+F_{m-1}F_{n-1}}\;, \end{align*} so $F_{m+n}=F_{m+1}F_n+F_mF_{n-1}$. Take $m=n$, and this becomes $$F_{2n}=F_{n+1}F_n+F_nF_{n-1}=F_n\left(F_{n+1}+F_{n-1}\right)=F_n\left(F_n+2F_{n-1}\right)\;.$$ - Exact duplicate of Zchpyvr's 21-minute prior answer (you've simply inlined the lemma that he links to). –  Bill Dubuque Oct 14 '12 at 19:06 @Bill: I’m not terribly surprised. However, I didn’t see his answer until after I posted, and when I did see it, I was busy and didn’t feel like chasing down the link. And quite frankly, I don’t consider it an exact duplicate for that very reason. –  Brian M. Scott Oct 14 '12 at 21:52 It is most certainly an exact duplicate. As I said, you've simply inlined the link in the other answer. Lacking anything new, it should be deleted for the sake of the readers. –  Bill Dubuque Oct 14 '12 at 21:58 @Bill: You have a rather inexact definition of exact. The fact that the information is right on the page is a difference, in convenience if nothing else. And deleting it does not serve the readers; slightly the reverse, if anything, for that same reason. The only person whom it might possibly ill serve is Zchpyvr, and I’ve upvoted his answer. And that’s the end of it as far as I’m concerned. –  Brian M. Scott Oct 14 '12 at 22:04 Posting duplicate answers potentially wastes many reader's time, since they may read two or more answers when it would have sufficed to read one. Not to mention that the abstraction in Zchpyvr's answer gained by calling the lemma by name (vs. value) only serves to make the answer more comprehensible. By your argument, every textbook should inline the proof of all lemma's so that they are "right on the page". That is, of course, absurd. –  Bill Dubuque Oct 14 '12 at 22:14 Another direct proof, using the fact that $$F_n=\frac{\phi^n-(1-\phi)^n}{\sqrt5}\tag1$$ where $$\phi=\frac{1+\sqrt5}{2},\qquad1-\phi=\frac{1-\sqrt5}{2}=-\frac{1}{\phi}.$$ From $(1)$ we have $$s_n\equiv\frac{F_{2n}}{F_n}=\frac{\phi^{2n}-(1-\phi)^{2n}}{\phi^n-(1-\phi)^n}=\phi^n+(1-\phi)^n$$ then \begin{align} s_n-F_n&=\phi^n+(1-\phi)^n-\frac{\phi^n-(1-\phi)^n}{\sqrt5}=\\ &=\frac{1}{\sqrt5}\left[\sqrt5\phi^n+\sqrt5(1-\phi)^n-\phi^n+(1-\phi)^n\right]=\\ &=\frac{1}{\sqrt5}\left[-(1-\sqrt5)\phi^n+(1+\sqrt5)(1-\phi)^n\right]=\\ &=\frac{1}{\sqrt5}\left[-2(1-\phi)\phi^n+2\phi(1-\phi)^n\right]=\\ &=-\frac{2\phi(1-\phi)}{\sqrt5}\left[\phi^{n-1}-(1-\phi)^{n-1}\right]=2F_{n-1} \end{align} - This can be done also using the fact that $$F_n = \frac{1}{\sqrt{5}}\left(\sigma^n-\bar{\sigma}^n\right),$$ from which it is easy to get $$F_{2n} = F_n L_n,$$ (where $L_n$ is the $n$-th Lucas number) and we have only to prove $$L_n= F_n+2F_{n-1},$$ that is true since $\{L_n\}_{n\in\mathbb{N}}$ and $\{F_n+2F_{n-1}\}_{n\in\mathbb{N}}$ are sequences with the same characteristic polynomial ($x^2-x-1$) and the same starting values $L_1=F_1+2F_0=1$, $L_2=F_2+2F_1=3$. -

https://www.physicsforums.com/threads/solutions-to-congruence-modulo-50.663798/

# Solutions to Congruence Modulo 50 1. Jan 12, 2013 ### knowLittle 1. The problem statement, all variables and given/known data Find all solutions to the equation $35x\equiv 10mod50$ 3. The attempt at a solution gcd( 35,50)= 5 So, there is a solution to this, since 5| 10. Also, there is a theorem that guarantees the existence of exactly 5 solutions. Now, dividing $35x\equiv 10mod50$ over 5 gives: $7x\equiv 2mod10$ Now, what multiple of 7 gives us $\equiv 2mod10$ { 2, 12, 22, 32, 42,...} Here, we found 42 that is a multiple of 7 and satisfies $\equiv 2mod10$ We can write $7x\equiv 42mod10$ . Now, I divided the expression by 7 and got $x\equiv 6mod10$ Now, there is another theorem that tells me this 6+(50/5)t, t=0, 1, ..., 4 I get: 6, 16, 26, 36, 46 So, the solutions are $x\equiv 6mod50$ , $x\equiv 16 mod50$, $x\equiv 26 mod50$, $x\equiv 36mod 50$, $x\equiv 46mod 50$ These are the 5 solutions of $35x\equiv 10mod50$ I found other solutions online: So x = 6, 16, 22, 28, 34, 40, 46 modulo 50 are the solutions to the congruence 35x ≡ 10 mod 50. Am I incorrect? Thank you. 2. Jan 12, 2013 ### kru_ Your method is correct. You can check your own solutions to verify that they are correct. 35*6 = 210 which is congruent to 10 mod 50. 35*16 = 560 which is also congruent to 10 mod 50. Similarly for the other solutions that you found. You can verify that 22, 28, and 40 are not solutions. 35*22 = 770 which is 20 mod 50. 35*28 = 980, is 30 mod 50. etc. 3. Jan 12, 2013 ### knowLittle I have read somewhere that division is not defined in modular arithmetic. Can someone tell me how this affect my solution? @kru: This is puzzling, since I found those other solutions at a .edu site. 4. Jan 12, 2013 ### HallsofIvy Division is not necessarily defined in modular arithmetic because there may be "0 divisors". For example, in modulo 6, 2(3)= 6= 0 (mod 6). If we had an equation of the form 3x= 1 (mod 6) we can immediately check that 3(1)= 3, 3(2)= 6, 3(3)= 9= 3, 3(4)= 12= 0, and 3(5)= 15= 3 mod 6. There is NO x such that 3x= 1 and so we could not, for example, "divide by 3" to get "1/3" as an answer. If we are working "modulo" a prime number, that doesn't happen and we can define "division". The way I would do "35x= 10 mod 50" is this. This is the same as saying 35x= 50n+ 10 for some integer n- a linear "diophantine equation". The first thing we can do divide through by 5 to 7x= 10n+ 2 or 7x- 10n= 2. Now 7 divides into 10 once with remainder 3: 3= 10- 7. 3 divides into 7 twice with remainder 1: 1= 7- 2(3). We can replace that "3" with 10- 7 from the first equation: 1= 7- 2(10- 7)= 3(7)- 2(10)= 1 (The "Euclidean Divison Algorithm"). Multiply through by 2 to get 6(7)- 4(10)= 2. So one solution to 7x- 10n= 2 is x= 6, n= 4. It is possible to write out the "general solution" but since 6 itself is between 0 and 10, x= 6 satisfies 7(6)= 2 (mod 10) and so 35(6)= 210= 10 (mod 50). 5. Jan 12, 2013 ### knowLittle According to Wikipedia, Diophantine equations are written as follows: ax + by = c The Diphantine equation that you are really writing is this 35x-50n=10? I understand everything, until you change the equation 1=7 -2(10-7)= 3(7)-2(10)=1. I understand that 21-20=1, but why changing from 7- 2(10-7) to 3(7)-2(10)? Also, I am acquainted with Euclid's GCD algorithm: Euclid(a,b) if b==0 return a else return Euclid (b, a mod b) Is there a way to use it without having to trace it? Is this all solutions for 35x $\equiv$ 10 mod 50? Also, is it correct that there has to be exactly 5 solutions, since the gcd of 35, 50 is 5? Is my solution correct?

http://mathhelpforum.com/calculus/134539-complex-numbers-finding-real-number-pairs.html

# Math Help - Complex numbers-finding real number pairs 1. ## Complex numbers-finding real number pairs Hello, I am having trouble with this question: "Find all possible real number pairs p, q such that 3+5i/1+pi =q+4i" Im sure it's easy but I think I am overlooking something. I multiplied both sides by the conjugate of 1+pi....ie.(1-pi) but I think it's wrong. Any help would be appreciated. Regards, Neverquit 2. Multiply both sides by 1+pi Then you get... 3 + 5i = (1 + pi)(q + 4i) = q + pqi + 4i - 4p => 3 + 5i = (q - 4p) + 4i + pqi => 3 + i = (q - 4p) + pqi So you want to solve... q - 4p = 3 pq = 1 3. ## found 1 solution but not the other I found the solutions 0.25, -1 after re-arranging pq = 1 to q = 1/p to get 1/p - 4p = 3 to get quadratic 4p^2+3p-1=0. There is also apparently another solution of 4, -1 which I can’t find. How do I find it? 4. This is a tricky question. $\frac{{3 + 5i}}{{1 + pi}} = \frac{{\left( {3 + 5i} \right)\left( {1 - pi} \right)}}{{1 + p^2 }} = \frac{{3 + 5p}}{{1 + p^2 }} + \frac{{\left( {5 - p} \right)i}}{{1 + p^2 }}$ Now set real part equal to real part and imaginary equal to imaginary. $\frac{3+5p}{1+p^2}=q~\&~\frac{5-3p}{1+p^2}=4$ From that we get $p=\frac{1}{4}~\&~p=-1$. 5. Originally Posted by Neverquit I found the solutions 0.25, -1 after re-arranging pq = 1 to q = 1/p to get 1/p - 4p = 3 to get quadratic 4p^2+3p-1=0. There is also apparently another solution of 4, -1 which I can’t find. How do I find it? I think you are misinterpreting your own answer. When you solve the system of equations, you will find that $p=1/4 ~~\mbox{ or }~~ p=-1$ These are two separate solutions, not a single solution. You need to find the value of $q$ that pairs with each of these solutions for $p$. So, you need to plug each value of $p$ back into the system of equations and find the corresponding values of $q$. Since $pq=1$, it's a pretty straightforward calculation: $p=1/4 \implies q=4$ $p=-1 \implies q=-1$ Therefore, the solutions are: First solution: $p=1/4 \mbox{ and } q=4$ Second solution: $p=-1 \mbox{ and } q=-1$ You might also write this as: $(p,q) = (1/4,4) \mbox{ or } (p,q) = (-1,-1)$ But you definitely would not say that the solutions are $(1/4,-1)$ and $(4,-1)$. 6. Originally Posted by Neverquit Hello, I am having trouble with this question: "Find all possible real number pairs p, q such that 3+5i/1+pi =q+4i" Im sure it's easy but I think I am overlooking something. I multiplied both sides by the conjugate of 1+pi....ie.(1-pi) but I think it's wrong. Any help would be appreciated. Regards, Neverquit $\frac{3+5i}{1+pi} =q+4i$ Multiply top and bottom of LHS by $1-pi$ to give: $\frac{(3+5p)+(5-3p)i}{1+p^2}=q+4i$ Equate real and imaginary parts to get: $3+5p=(1+p^2)q$ and: $5-3p=4(1+p^2)$ Now the problem is to find all solutions to this pair of equations. CB 7. Having Plato and CP post in this thread has made me question whether my answer is wrong. Is it? Solving the equations I arrived at... q - 4p = 3 pq = 1 gives q=1/p, So subbing that into the first equation will give you a polynomial (if you multiply both sides by p) which gives you p = 1/4 and -1. Hence q = 4 and -1. Solutions are (1/4,4) and (-1, -1) Are these the only solutions? Why must you multiply the top and bottom by the conjugate? Having Plato and CP post in this thread has made me question whether my answer is wrong. Is it? Your solution was correct. Multiplying a complex fraction by the conjugate of the denominator is probably just a habit for them. But it's not necessary at all in this particular problem. 9. ## Using the conjugate Originally Posted by Plato This is a tricky question. $\frac{{3 + 5i}}{{1 + pi}} = \frac{{\left( {3 + 5i} \right)\left( {1 - pi} \right)}}{{1 + p^2 }} = \frac{{3 + 5p}}{{1 + p^2 }} + \frac{{\left( {5 - p} \right)i}}{{1 + p^2 }}$ Now set real part equal to real part and imaginary equal to imaginary. $\frac{3+5p}{1+p^2}=q~\&~\frac{5-3p}{1+p^2}=4$ From that we get $p=\frac{1}{4}~\&~p=-1$. I think the solution that Plato gives using the congugate is what the text books author had in mind as the question is shortly after conjugates of complex numbers is explained. Deastar, your solution still gives the same answer in the text book so it must be correct. ........... 10. ## generalisation Originally Posted by Neverquit Hello, I am having trouble with this question: "Find all possible real number pairs p, q such that 3+5i/1+pi =q+4i" Im sure it's easy but I think I am overlooking something. I multiplied both sides by the conjugate of 1+pi....ie.(1-pi) but I think it's wrong. Any help would be appreciated. Regards, Neverquit the basic idea behind questions of your type is equaslising real and imaginary parts.

https://math.stackexchange.com/questions/1902937/what-is-the-probability-that-jan-and-jon-are-chosen

# What is the probability that Jan and Jon are chosen? Jan , Jon and $10$ other children are in a classroom. The principal of the school walks in and choose $3$ children at random. What is the probability that Jan and Jon are chosen? My approach: Including Jon and Jan, number of ways of selection is $1\cdot1\cdot{10\choose 1} = 10$. Total way of selection is ${12\choose 3} = 220$ So, probability is $\frac{10}{220}= \frac{1}{22}.$ But when I solving as follows: \begin{align*} P(\text{selection of Jon and Jan}) &= 1- P(\text{not selection of Jon and Jan}) \\ &= 1- \frac{{10\choose 3}}{{12\choose 3}}\\ &= \frac{5}{11}. \end{align*} Which approach is correct and why alternative one is wrong? Note: My previous post had some mistakes, so I deleted that I will call $A$, $B$ the events selecting Jon, Jan respectively. You did not take the complement correctly \begin{align*} P(A\cap B) &= 1-\color{red}{P(\bar A \cup\bar B)}\tag 1 \\ &= 1-[P(\bar A)+P(\bar B)-P(\bar A\bar B)]\tag 2 \\ &=1-\left[\frac{\binom{11}{3}}{\binom{12}{3}}+\frac{\binom{11}{3}}{\binom{12}{3}}-\frac{\binom{10}{3}}{\binom{12}{3}}\right]\\ &=\frac{1}{22} \end{align*} where $(1)$ is true by DeMorgan's law, and $(2)$ is true by inclusion-exclusion. As you can see the two methods give the same value. Your first approach gives the probability that Jon AND Jan are selected, the second gives the probability that Jon OR Jan are selected. When you computed the probability of not selecting Jon AND Jan, you didn't include the situations where Jon was selected but not Jan, and vice versa. Lets say the even Jon is selected is A, Jan is selected B. Then $P(A \wedge B)=1-P(Not (A \wedge B))= 1- P(Not A \vee Not B)$ $P(Not A \vee Not B)= P(Not A)+ P(Not B)- P(Not A \wedge Not B)$ Here $\wedge$ means and, and $\vee$ means or. Think of a Venn diagram, if we look at the union of two circles, the total area is equal to the sum of the circles minus the intersection, because we counted that part twice. $P(Not A)=1-P(A)=1-(1*\binom{11}{2})/220)=3/4$ or $P(Not A)=\binom{11}{3}/220=3/4$ $P(Not B)=3/4$ $P(Not A \wedge Not B)= \binom{10}{3}/220=6/11$ Now $3/4+3/4-6/11=21/22$ So $1-21/22=1/22$

https://math.stackexchange.com/questions/2283548/number-of-ways-to-place-k-non-attacking-rooks-on-a-100-times-100-chess-board

# Number of ways to place $k$ non-attacking rooks on a $100\times 100$ chess board I need to show that the number of ways to place $k$ non-attacking rooks (no two share the same column or row) on a $100\times 100$ chessboard is $k!{100 \choose k}^2$. When I try to formulate this equation I end up getting ${100 \choose k}^2$ because you need to choose $k$ columns from $100$ columns and $k$ rows from $100$ rows. I know this isn't correct because if you have $k=100$, there is more than just $1$ solution. However I don't know how to come up with the $k!$ part of the equation. • – Shaun May 16 '17 at 14:32 First of all, congrats on realizing that the answer you got can't possibly be correct. It's always a good idea to test formulas against special cases, to see if they stand up. One way to arrive at the correct answer is to view the placement of the rooks in two steps: First choose the $k$ rows that the rooks will go in, and then, going row by row, decide which column to place that row's rook in. The first rook has $100$ columns to choose from, the second will have $99$, the third $98$, and so on. The total is thus $${100\choose k}100\cdot99\cdot98\cdots(100-(k-1))={100\choose k}{100!\over(100-k)!}={100\choose k}{100!\over(100-k)!k!}k!={100\choose k}^2k!$$ First, choose your $k$ rows and columns, as you said. Start by considering the configuration in which the rooks are successively placed in the legal square furthest to the top and to the left (so that the rooks go "diagonally down to the right"). From there, it suffices to note that any rearrangement of the rook-columns results in a new and valid configuration. Since there are $k!$ such rearrangements, there are $k!$ configurations for any particular choice of $k$ rows and $k$ columns. Your problem is that $\binom{100}{k}^2$ only gives you the ways to choose the columns and the rows separately, without specifiying which row goes with which column. This is why you need to add the factor $k!$, which corresponds to the number of bijections between your $k$ rows and your $k$ columns, i.e. the number of ways to associate them together. An alternative way of obtaining this is to consider that you first choose $\binom{100}{k}$ column where you will place your rooks, then choose for each column the row where you place a rook; this second step amonts to choosing $k$ rows with order, which you can do in $\frac{100!}{(100-k)!}=\binom{100}{k}k!$ ways. Multiplying the two numbers together gives you the result. Regards User. If i may contribute, here is my view : A 100 $\times$ 100 chess board can be viewed as a matrix of size 100 $\times$ 100. For example : let $$(i, j), \:\: \text{ with } \:\:\: i,j=1,2,... ,100$$ denotes the $i$th row and $j$th column of the board. To solve your problem, the key is : the $k$ non-attacking roots is as same as no two $(i_{1}, j_{1})$ and $(i_{2}, j_{2})$ with $i_{1}=i_{2}$ or $j_{1}=j_{2}$. Two rooks with position $(1, 100)$ and $(91, 100)$, for example, does not satisfy the non-attacking roots condition. To illustrate how to solve this, first, you could start with $k=1$. • For k=1. Let the position of this particular rook is $(x_{1},y_{1})$. Then there are 100 possibilities for $x_{1}$, and 100 possibilities for $x_{2}$. So the number of possibilities is (100)(100) = $1! \binom{100}{1}^{2}$ • For k=2. For each of the two rooks, their positions are denoted with $(x_{1},y_{1})$ and $(x_{2}, y_{2})$. For the first one, There are 100 possibilities for each $x_{1}$ and $y_{1}$. For the 2nd rook, $x_{2}$ and $y_{2}$ each has 99 possibilities (since they can't be equal to the 'coordinates' of the 1st rook). So the number of possibilities to put 2 non-attacking distinguished rooks is $$(100^{2})(99^{2})$$ For your problem, they are not distinguished, so we have to divide this by 2 (exactly $2!$), because we can choose either rook to be the 1st or the 2nd. So, output : $$\frac{(100^{2})(99^{2})}{2!} = \frac{(100 \cdot 99)(100 \cdot 99)}{2!} = \frac{(100 \cdot 99)(100 \cdot 99)}{2!} \left(\frac{98! \cdot 2!}{98! \cdot 2!} \right)^{2} = 2! \binom{100}{2}^{2}$$ • For k=3 and above, you would be confident to try and continue this method. Hope this will helps. Regards, Arief.

https://mathhelpboards.com/threads/permutation-representation-argument-validity.7690/

# Permutation representation argument validity #### kalish ##### Member Hello, I would like to check if the work I have done for this problem is valid and accurate. Any input would be appreciated. Thank you. **Problem statement:** Let $G$ be a group of order 150. Let $H$ be a subgroup of $G$ of order 25. Consider the action of $G$ on $G/H$ by left multiplication: $g*aH=gaH.$ Use the permutation representation of the action to show that $G$ is not simple. **My attempt:** Let $S_6$ be the group of permutations on $G/H$. Then, the action of $G$ on $G/H$ defines a homomorphism $f:G \rightarrow S_6$. We know $|S_6| = 720.$ Since $|G|=150$ does not divide 720, and $f(G)$ is a subgroup of $S_6$, $f$ cannot be one-to-one. Thus, $\exists$ $g_1,g_2$ distinct in $G$ such that $f(g_1)=f(g_2) \implies f(g_1g_2^{-1})=e$. Thus, $\ker(f) = \{g:f(g)=e\}$. Since $\ker(f)$ is a normal subgroup of $G$, we have found a normal subgroup of $G$. Also, since $f$ is non-trivial, then $\ker(f)$ is a proper normal subgroup of $G.$ Hence $G$ is not simple. Any suggestions or corrections? #### Deveno ##### Well-known member MHB Math Scholar I think it is clear that since $|G|$ does not divide 720, $\text{ker}(f)$ is a non-trivial normal subgroup of $G$, so there is no need to talk about the existence of $g_1,g_2$ or restate the definition of $\text{ker}(f)$. I *do* think you should say WHY $f$ is not the trivial homomorphism. It's pretty simple, though: Since $|H| < |G|$, we can take any $g \in G - H$, which takes (under the action) the coset $H$ to $gH \neq H$, so $f(G)$ contains at least one non-identity element: namely, $f(g)$. #### Deveno ##### Well-known member MHB Math Scholar This is actually a special case of a theorem proved in Herstein, which goes as follows: If $G$ is a finite group with a subgroup $H$ such that $|G| \not\mid ([G:H])!$ then $G$ contains a non-trivial proper normal subgroup containing $H$. One obvious corollary is then that such a group $G$ cannot be simple. You would be far better off adapting your proof to this more general one, which can be re-used in many more situations. #### kalish ##### Member Hi, Which Herstein book is this from? I would like to explore further. Thanks. #### Deveno ##### Well-known member MHB Math Scholar His classic Topics In Algebra​. #### kalish ##### Member That's what I found from my search as well. Do you have a copy of the book or know where I can find one? #### kalish ##### Member That sounds like a fantastic result. I cannot find the book anywhere though. Could you please reproduce the proof for me here, so that I could use it to study? I would really appreciate it. #### Deveno ##### Well-known member MHB Math Scholar Theorem 2.G (p. 62, chapter 2): If $G$ is a group, $H$ a subgroup of $G$, and $S$ is the set of all right cosets of $H$ in $G$, then there is a homomorphism $\theta$ of $G$ into $A(S)$, and the kernel of $\theta$ is the largest normal subgroup of $G$ which is contained in $H$. (a few words about notation: Herstein uses $A(S)$ to stand for the group of all bijections on $S$...if $|S| = n$, then $A(S)$ is isomorphic to $S_n$. Herstein also writes his mappings on the RIGHT, as in $(x)\sigma$ instead of $\sigma(x)$, so that composition and multiplication are "in the same order", instead of reversed. For this reason, he uses right cosets and right-multiplication instead of the left cosets (and left-multiplication) one often sees used in other texts. He also denotes the index of $H$ in $G$ by $i(H)$ , instead of $[G:H]$ and denotes $|G|$ by $o(G)$). Proof: Let $G$ be a group, $H$ a subgroup of $G$. Let $S$ be the set whose elements are right cosets of $H$ in $G$. That is, $S = \{Hg: g \in G\}$. $S$ need not be a group itself, in fact, it would be a group only if $H$ were a normal subgroup of $G$. However, we can make our group $G$ act on $S$ in the following natural way: for $g \in G$ let $t_g:S \to S$ be defined by: $(Hx)t_g = Hxg$. Emulating the proof of Theorem 2.f we can easily prove: (1) $t_g \in A(S)$ for every $g \in G$ (2) $t_{gh} = t_gt_h$. Thus the mapping $\theta: G \to A(S)$ defined by $\theta(g) = t_g$ is a homomorphism of $G$ into $A(S)$. Can one always say that $\theta$ is an isomorphism? Suppose that $K$ is the kernel of $\theta$. If $g_0 \in K$, then $\theta(g_0) = t_{g_0}$ is the identity map on $S$, so that for every $X \in S, Xt_{g_0} = X$. Since every element of $S$ is a right coset of $H$ in $G$, we must have that $Hat_{g_0} = Ha$ for every $a \in G$, and using the definition of $t_{g_0}$, namely, $Hat_{g_0} = Hag_0$, we arrive at the identity $Hag_0 = Ha$ for every $a \in G$. On the other hand if $b \in G$ is such that $Hxb = Hx$ for every $x \in G$, retracing our argument we could show that $b \in K$. Thus $K = \{b \in G|Hxb = Hx$ all $x \in G\}$. We claim that from this characterization of $K,\ K$ must be the largest normal subgroup of $G$ which is contained in $H$. We first explain the use of the word largest; by this we mean if $N$ is a normal subgroup of $G$ which is contained in $H$, then $N$ must be contained in $K$. We wish to show this is the case. That $K$ is a normal subgroup of $G$ follows from the fact that it is the kernel of a homomorphism of $G$. Now we assert that $K \subset H$, for if $b \in K, Hab = Ha$ for every $a \in G$, so in particular, $Hb = Heb = He = H$, whence $b \in H$. Finally, if $N$ is a normal subgroup of $G$ which is contained in $H$, if $n \in N,\ a \in G$, then $ana^{-1} \in N \subset H$, so that $Hana^{-1} = H$; thus $Han = Ha$ for all $a \in G$. Therefore, $n \in K$ by our characterization of $K$. ********** Remarks following the proof: The case $H = (e)$ just yields Cayley's Theorem (Theorem 2.f). If $H$ should happen to have no normal subgroup of $G$, other than $(e)$ in it, then $\theta$ must be an isomorphism of $G$ into $A(S)$....(some text omitted).... We examine these remarks a little more closely. Suppose that $G$ has a subgroup $H$ whose index $i(H)$ (that is, the number of right cosets of $H$ in $G$) satisfies $i(H)! < o(G)$. Let $S$ be the set of all right cosets of $H$ in $G$. The mapping, $\theta$, of Theorem 2.g cannot be an isomorphism, for if it were, $\theta(G)$ would have $o(G)$ elements and yet would be a subgroup of $A(S)$ which has $i(H)! < o(G)$ elements. Therefore, the kernel of $\theta$ must be larger than $(e)$; this kernel being the largest normal subgroup of $G$ which is contained in $H$, we can conclude that $H$ contains a nontrivial normal subgroup of $G$. However, the above argument has implications even when $i(H)!$ is not less than $o(G)$. If $o(G)$ does not divide $i(H)!$ then by invoking Lagrange's theorem we know that $A(S)$ can have no subgroup of order $o(G)$, hence no subgroup isomorphic to $G$. However $A(S)$ does contain $\theta(G)$, whence $\theta(G)$ cannot be isomorphic to $G$, that is, $\theta$ cannot be an isomorphism. But then, as above, $H$ must contain a nontrivial normal subgroup of $G$. We summarize this as: Lemma 2.21 If $G$ is a finite group, and $H \neq G$ is a subgroup of $G$ such that $o(G) \not\mid i(H)!$, then $H$ must contain a nontrivial normal subgroup of $G$. In particular, $G$ cannot be simple. (Note to the moderating staff: although this is an excerpt from a copyrighted work, I believe this sample falls under the province of "Fair Use" for the purpose of "Scholarly research and exposition", and is not intended for commercial gain or to circumvent existing copyright laws). Last edited:

https://www.physicsforums.com/threads/spring-gun.742063/

# Spring Gun 1. Mar 7, 2014 ### Steven60 I have a question about a spring gun. Suppose the barrel of a spring gun is placed horizontally at the edge of a horizontal table. You put say a marble in the barrel and compress the spring x cm and after releasing the marble it travels a horizontal distance of y cm before hitting the floor (so motion is of a projectile). My question is whether or not the horizontal distance traveled and the amount the spring is compresses make a linear relationship? If so, then how can I prove this? This is not homework. Thanks! 2. Mar 7, 2014 Seems to be purely a math problem. Perhaps sketch the system and write the relevant equations needed to determine this? 3. Mar 7, 2014 ### UltrafastPED This is a nice physics exercise - there are several physical considerations, and then some simple math. You have two forces acting on the marble ... the spring force, which launches the marble, and gravity. Once the marble leaves the launch tube it will have a constant "horizontal" speed - ignoring air resistance - and an initial vertical speed of zero. Call this initial horizontal speed V. The vertical speed will increase with time due to the constant gravitational acceleration - and will hit the floor at a definite time which depends only on the height of the table. Call this duration T. Then the distance from the table to the point of contact will be D = V x T. The time T does not depend upon the spring force, only on the height of the table and local value g=9.8 m/s^2. Thus you only need to determine if the speed V is proportional to the spring force; by Hook's law we know that a "good" spring obeys F = -k * X, where X is the compression/extension distance and k is the spring's constant. If we switch to energy we have work done on marble is W = Integral[F dx] over the interval x=[0,X]. Note that the force is changing as the spring moves! So W = Integral[ k*x dx] = 1/2 k*X^2. But this work has been converted into kinetic energy of the marble. For a marble of mass=M, and given that it is NOT rolling or spinning, then the kinetic energy is KE=1/2 M*V^2 = 1/2 k*X^2=W. Thus V = k/M Sqrt[X]. xx Correction: xxx Make that V = Sqrt[k/M] * X. Thus the hypothesis is true! Thanks to dauto for noticing the mistake at the end! :-) Last edited: Mar 7, 2014 4. Mar 7, 2014 ### DrewD The equations involved will be $d=vt$ for constant $v$ (and assuming that the initial point when exiting the spring gun is defined as 0 distance), $U_{spring}=\frac12k\Delta x^2$ and $K=\frac12mv^2$. $\Delta x$ is the amount the spring is compressed, and $v$ is the velocity of the object as it leaves the spring. This approximation assumes that the object does not stick which is a good assumption for a spring gun. Solve for $v$ to find the relationship between $\Delta x$ and $d$. 5. Mar 7, 2014 ### dauto You made a mistake at the very end. In fact, after correcting the mistake, you proved that the hypothesis is true. 6. Mar 9, 2014 ### Steven60 Thanks for your replys. I actually worked this out myself and actually did the same exact steps as UltrafastPED.

http://math.stackexchange.com/questions/245576/is-this-induction-procedure-correct-2nn

# Is this induction procedure correct? ($2^n<n!$) I am rather new to mathematical induction. Specially inequalities, as seen here How to use mathematical induction with inequalities?. Thanks to that question, I've been able to solve some of the form $1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} \leq \frac{n}{2} + 1$. Now, I was presented this, for $n \ge 4$: $$2^n<n!$$ I tried to do it with similar logic as the one suggested there. This is what I did: Prove it for $n = 4$: $$2^4 = 16$$ $$4! = 1\cdot2\cdot3\cdot4 = 24$$ $$16 < 24$$ Assume the following: $$2^n<n!$$ We want to prove the following for $n+1$: $$2^{n+1}<(n+1)!$$ This is how I proved it: • So first we take $2^{n+1}$ which is equivalent to $2^n\cdot2$ • By our assumption, we know that $2^n\cdot2 < n!\cdot2$ • This is because I just multiplied by $2$ on both sides. • Then we'll be finished if we can show that $n! \cdot 2 < (n+1)!$ • Which is equivalent to saying $n!\cdot2<n!\cdot(n+1)$ • Since both sides have $n!$, I can cancel them out • Now I have $2<(n+1)!$ • This is clearly true, since $n \ge 4$ Even though the procedure seems to be right, I wonder: • In the last step, was it ok to conclude with $2<(n+1)!$? Was there not anything else I could have done to make the proof more "careful"? • Is this whole procedure valid at all? I ask because, well, I don't really know if it would be accepted in a test. • Are there any points I could improve? Anything I could have missed? This is kind of the first time I try to do these. - I think your proof is fine, if a bit long-winded. But with experience, you'll learn what bits to shorten without losing rigour. –  Harald Hanche-Olsen Nov 27 '12 at 9:51 You should say 2^{n+1}\lt 2\cdot n!$. But because$2\lt $n+1$, it follows that $2\cdot n!\lt (n+1)!$. Unfortunately, you are still writing proofs "backwards" in a logically incorrect way. –  André Nicolas Nov 27 '12 at 9:59 @André: It seems a bit harsh to call this a "logically incorrect way". It's OK as long as "Since both sides have $n!$, I can cancel them out" is interpreted as "dividing through by $n!$ leads to an equivalent inequality". It's true that the reverse order would be clearer, and doing things in this order is incorrect if the implications used only go in one direction, but that's not the case here so the proof is still OK, if suboptimally structured. –  joriki Nov 27 '12 at 10:04 @Omega, your proof is correct..."almost", since as Andre apparently meant, there's some lack of logical rigour in your last lines. You must show that the implications there are double, i.e.: $$n!\cdot 2<(n+1)!\Longleftrightarrow 2<\frac{(n+1)!}{n!}=n+1$$ and then noting the last inequality is trivially true as we're working with $\,n\geq 4\,$... Also, don't right "equivalent" when it should be "equal", as in "$\,2^{n+1}\,$ is "equivalent"(should be "equal"!) to $\,2^n\cdot 2$ –  DonAntonio Nov 27 '12 at 10:47 @joriki: A bit harsh, perhaps, but other posts by the OP had some worse instances that were pointed out. Since the actual understanding of the problem is good, it is worthwhile to vaccinate the OP against $A\to B\to C\to 0=0$, therefore $A$. –  André Nicolas Nov 27 '12 at 16:47 Yes, the procedure is correct. If you want to write this more like the sort of mathematical proof that would be found in a textbook, you might want to make some tweaks. For example, the base case could be re-written as follows: When $n = 4$, we have $2^4 = 16 < 24 = 4!$ Next, the inductive hypothesis and the subsequent manipulations: Suppose that for $n \geq 4$ we have $2^n < n!$ Thus, $2^{n+1} < 2 \cdot n! < (n+1)!$, where the first inequality follows by multiplying both sides of the inequality in our IH by $2$, and the second follows by observing that $2 < n+1$ when $n \geq 4$. Therefore, by the Principle of Mathematical Induction, $2^n < n!$ for all integers $n \geq 4$. Q.E.D. Note: I am not making a judgment about whether your write-up or the one I have included here is "better." I'm only observing that the language and format differ, particularly with regard to proofs that are written in paragraph form (typical of math papers) rather than with a sequence of bullet-points (which is what you had). -

http://tdgl.solotango.it/bfs-time-complexity.html

### Bfs Time Complexity Optimality : It is optimal if BFS is used for search and paths have uniform cost. For example, if the heuristic evaluation function is an exact estimator, then A* search algorithm runs in linear time, expanding only those nodes on an optimal solution path. In this tutorial, We are going to learn about bubble sort algorithm and their implementation in various programming languages. Again basic of bfs , once you get this you will get to know how powerful and where we can use it in daily life example Stay tuned for more. 1 & 2): Gunning for linear time… Finding Shortest Paths Breadth-First Search Dijkstra’s Method: Greed is good! Covered in Chapter 9 in the textbook Some slides based on: CSE 326 by S. We use Queue data structure with maximum size of total number of vertices in the graph to implement BFS traversal. For the most part, we describe time and space complexity for search on a tree; for a graph, the answer depends on how “redundant” the paths in the state space are. Since a BFS traversal is used, the overall time complexity is simply O(|V| + |E|). Yes, the worst case complexity is O(ab). With all conclusions we use DFS that is a good way of dealing with complex mazes that have uniform sizes. Set The Starting Vertex To Vertex 1. Shortest Path using BFS: The shortest path between two vertices in a graph is a path such that the total sum of edge weights in the path connecting the two vertices is minimum. Priority queue Q is represented as a binary heap. Space complexity: O(bm) for the tree search version and O(b m) for the graph search version; Breadth First Search (BFS) BFS uses FIFO ordering for node expansion i. The amount of time needed to generate all the nodes is considerable because of the time complexity. The deepest node happens to be the one you most recently visited - easy to implement recursively OR manage frontier using LIFO queue. Complexity Measures Message complexity: Number of messages sent (worst case). What is the time complexity of BFS? – how many states are expanded before finding a solution? – b: branching factor – d: depth of shallowest solution – complexity = What is the space complexity of BFS? – how much memory is required? – complexity = Is BFS optimal? – is it guaranteed to find the best solution (shortest path)?. The aim of BFS algorithm is to traverse the graph as close as possible to the root node. Set The Starting Vertex To Vertex 1. One starts at the root (selecting some arbitrary node as the root in the case of a graph) and explores along adjacent nodes and proceeds recursively. -1 – A wall or an obstacle. edu/6-006F11 Instructor: Erik Demaine License: Creative Commons BY-N. graph algorithms, has linear time complexity, and is com-plete for the class SL of problems solvable by symmetric, non-deterministic,log-space computations[32]. And that’s how a quadratic time complexity is achieved. BFS Properties • Which nodes does BFS expand? o Processes all nodes above depth of shallowest solution, s o Search time takes time O(bs) • Fringe Size: o Keeps last tier o O(bs) • Complete? o s must be finite, so yes! • Optimal? o Only if all costs are 1 (more later). Time Complexity of DFS is also O(V+E) where V is vertices and E is edges. You can also use BFS to determine the level of each node. If it is an adjacency matrix, it will be O(V^2). HackerRank - Breadth First Search - Shortest Path. Breadth-first search (BFS) algorithm is an algorithm for traversing or searching tree or graph data structures. graphBfs1 - Free download as Powerpoint Presentation (. Time Complexity of BFS (Using adjacency matrix) • Assume adjacency matrix – n = number of vertices m = number of edges No more than n vertices are ever put on the queue. Asked about the time complexity of search, deletion, etc. Rewrite the pseudocode for the BFS algorithm studied in class (and presented in the textbook) to work for an adjacency matrix representation of the graph instead of an adjacency list representation. Time complexity is O(N+E), where N and E are number of nodes and edges respectively. Breadth-First-Search Attributes • Completeness – yes • Optimality – yes, if graph is un-weighted. Depth first traversal or Depth first Search is a recursive algorithm for searching all the vertices of a graph or tree data structure. Documentation / Algorithms The Welsh-Powell Algorithm. 6) Broadcasting in Network: In networks, a broadcasted packet follows Breadth First Search to reach all nodes. Some methods are more effective then other while other takes lots of time to give the required result. The Big O notation is used to classify algorithms by their worst running time or also referred to as the upper bound of the growth rate of a function. And then it concluded that the total complexity of DFS() is O(V + E). • Scanning for all adjacent vertices takes O(| E|) time, since sum of lengths of adjacency lists is |E|. BFS Algorithm Complexity. complete： BFS是complete的。 optimal： BFS是optimal的，因为找到的第一个解是最shallow的。 time complexity: 和DFS一样是. Interview question for Software Engineer. The smallest number of colors required to color a graph G is called its chromatic number of. Sirius? Brightest star in sky. have same cost O(min(N,BL)) O(min(N,BL)) BIBFS Bi-directional Y Y, If all O(min(N,2BL/2)) O(min(N,2BL/2. Completeness is a nice-to-have feature for an algorithm, but in case of BFS it comes to a high cost. Intuitively, you start at the root node and explore all the neighboring nodes. He also figures out the time complexity of these algorithms. c) [2pt] Express time and space complexity for general breadth-first search in terms of the branching factor, b, and the depth of the goal state, d. BFS (G, s) Breadth -First Search Starting from the sourc e node s, BFS computes the minimal distance from s to any other node v that can be reached from s. The optimal solution is possible to obtain from BFS. If we use an adjacency list, it will be O(V+E). Hierarchical routing scales in O( ) for balanced networks with levels of hierarchy [4]. The time complexity of a heuristic search algorithm depends on the accuracy of the heuristic function. The basic approach of the Breadth-First Search (BFS) algorithm is to search for a node into a tree or graph structure by exploring neighbors before children. Breadth First Search BFS intuition. Complexity The time complexity of BFS is O(V + E), where V is the number of nodes and E is the number of edges. The time complexity of BFS is O(V+E) because: Each vertex is only visited once as it can only enter the queue once — O( V ) Every time a vertex is dequeued from the queue, all its k neighbors are explored and therefore after all vertices are visited, we have examined all E edges — (O( E ) as the total number of neighbors of each vertex. The time complexity of BFS is O(V+E) where V stands for vertices and E stands for edges. Let’s say for instance that you want to know the shortest path between your workplace and home, you can use graph algorithms to get the answer! We are going to look into this and other fun. This is a generic BFS implementation: For a connected graph with V nodes and E total number of edges, we know that every edge will be considered twice in the inner loop. z x y z is a cycle of length 2(j i) + 1, which is odd, so G is not bipartite. However, the space complexity for these algorithms varies. Queue is used in the implementation of the breadth first search. PRAM algorithm Communication Time Problem of complexity complexity Breadth-first search 141 IEI I VI Maximum flow [I31 I VI3 I VIZ 805 TABLE II. Thus, each guard returns to his starting position after 2, 4 or 6 moves. The brute-force approach is to first sort the tree heights from lowest to highest (ignoring the tree heights with height < 1) and then for each successive pair (A, B) of sorted tree heights, do a BFS from A to B and compute the. Time complexity for B() is O(1), so if i is called from 1 to n, then it's n-times O(1)-> O(n). Breadth-First Search (BFS) Properties What nodes does BFS expand? Processes all nodes above shallowest solution Let depth of shallowest solution be s Search takes time O(bs) How much space does the fringe take? Has roughly the last tier, so O(bs) Is it complete? s must be finite if a solution exists, so yes! Is it optimal?. Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that only one letter can be changed at a time and each intermediate word must exist in the dictionary. Q: if each node has b children & optimum is at depth d, what are the time and space complexities of BFS? (tip: time=#GeneratedNode, space=#StoredNode). The space complexity of a search algorithm is the worst-case amount of memory that the algorithm will use (i. Completeness is a nice-to-have feature for an algorithm, but in case of BFS it comes to a high cost. The time complexity of a heuristic search algorithm depends on the accuracy of the heuristic function. Time complexity is O(N+E), where N and E are number of nodes and edges respectively. bfs algorithm. Given a branching factor b and graph depth d the space complexity is the number of nodes at the deepest d level, O(b ). Keywords: Distributed system, breadth-first-search, communication complexity, graph, algorithm 1. Time and Memory Requirements for BFS – d+1O(b ) Example: • b = 10 • 10,000 nodes/second • each node requires 1000 bytes of storage Depth Nodes Time Memory 2 1100. Breadth First Search: visit the closest nodes first. , we’ll no longer require that an action from a given state leads to the same state each time and will. To that purpose, we introduce a new parameter, called. This again depends on the data strucure that we user to represent the graph. And that’s how a quadratic time complexity is achieved. Breadth-First Search (BFS) in 2D Matrix/2D-Array Categories Amazon Questions , Binary Tree , Expert , Facebook , Google Interview , Linked List , Microsoft Interview , Recursion , Software Development Engineer (SDE) , Software Engineer , Trees Tags Expert 1 Comment Post navigation. L 2= all nodes that do not belong to L 0or L 1, and that have an edge to a node in L 1. time-complexity recurrence-relations loops asymptotic-analysis asymptotic-notation greedy dynamic-programming graph substitution-method vertex-coloring a-star np-completeness log analysis nested-loops n-puzzle heuristic master-theorem exponent n-queens conflict ai graph-coloring mvcs small-oh count easy sorted-lists logn example recursive gcd. 2 Choosing a good hash function; 19. Time Complexity: T(n) = O(N^2) Because we have a square matrix and in the worst case. bfs time complexity. Visualizing DFS traversal Depth-first Search (DFS) is an algorithm for searching a graph or tree data structure. Note : The space/time complexity could be less as the solution could be found anywhere on the. The BFS strategy is not generally cost optimal. a time complexity t(n) if the Turing Machine takes time at most t(n) on any input of length n. The time complexity of BFS is O(V + E). The Time complexity of BFS is O(V + E) when Adjacency List is used and O(V^2) when Adjacency Matrix is used, where V stands for vertices and E stands for edges. 5) GPS Navigation systems: Breadth First Search is used to find all neighboring locations. Now we can use the BFS on to print the path (while printing only the vertices that belong to V). In this tutorial, we are going to focus on Breadth First Search technique. Adjacency List Time Complexity. The two variants of Best First Search are Greedy Best First Search and A* Best First Search. , D(B)=1, D(F)=2. Breadth-first search (BFS) is an algorithm for traversing or searching tree or graph data structures. If it is an adjacency matrix, it will be O(V^2). The time complexity of the union-find algorithm is O(ELogV). Disadvantages. Time Complexity Edit. Proof [ edit ]. That means Big-Omega notation always indicates the minimum time required by an algorithm for all input values. BFS – time complexity b d depth number of nodes 0 1 1 12 =2 2 3 d 2=4 23=8 2d (bd ) Total nodes:Expanded nodes: O(bd 1) d+1 2 d+1(b ) O(bd) CS 2710 Foundations of AI M. Home; Python dictionary time complexity. • Time complexity: exponential in the depth of the solution d • Memory (space) complexity: nodes are kept in the memory O(bd) O(bd). Low water level. Breadth-first search. at most 3 nodes. • Time Complexity: 21 Ram Meshulam 2004 • Memory Complexity: O db( ) – Where b is branching factor and d is the maximum depth of search tree O(( d )b +(d − 1) b2 + + (1) bd ) =O(bd) State Redundancies • Closed list - a hash table which holds the visited nodes. By the use of the Queue data structure, we find the level order traversal. Features of BFS Space complexity Space complexity is proportional to the number of nodes at the deepest level. We'll start by describing them in undirected graphs, but they are both also very useful for directed graphs. This again depends on the data strucure that we user to represent the graph. The time complexity remains O(b d) but the constants are large, so IDDFS is slower than BFS and DFS (which also have time complexity of O(b d)). And that’s how a quadratic time complexity is achieved. c) [2pt] Express time and space complexity for general breadth-first search in terms of the branching factor, b, and the depth of the goal state, d. algorithms achieves optimal O(D) time, its communi- cation complexity is O(E. Same as Time Complexity UCS (Uniform Cost Search): BFS Enhanced with lowest path costs first Only test from start to goal (Dijkstra, no goal state unitl all nodes are removed to get shortest paths to all nodes. BFS Properties • Which nodes does BFS expand? o Processes all nodes above depth of shallowest solution, s o Search time takes time O(bs) • Fringe Size: o Keeps last tier o O(bs) • Complete? o s must be finite, so yes! • Optimal? o Only if all costs are 1 (more later). The existing algorithm, due to Cheung (1983), has communication and time complexities O( IV[3) and O( IV 1), respectively. Depth-first search. Breadth first search (BFS) and Depth First Search (DFS) are the simplest two graph search algorithms. Time Complexity Best log(n)) log(n)) log(n)) Average (nA2 Worst (nA2 (nA2 Worst Case Auxiliary Space Complexity Worst O(n) O(nk) O(n+k) Fair Searching Algorithm Depth First Search (DFS) Breadth First Search (BFS) Binary search Linear (Brute Force) Shortest path by Dijkstra, using a Min-heap as priority queue Shortest path by Dijkstra,. You are probably using programs with graphs and trees. • Time Complexity: 21 Ram Meshulam 2004 • Memory Complexity: O db( ) – Where b is branching factor and d is the maximum depth of search tree O(( d )b +(d − 1) b2 + + (1) bd ) =O(bd) State Redundancies • Closed list - a hash table which holds the visited nodes. Complexity Analysis. execution time in the PRAM model is O(D), where the Dis the diameter of the graph. Implementation. BFS: Time complexity is [code ]O(|V|)[/code] where [code ]|V|[/code] is the number of nodes,you need to traverse all nodes. In the last two posts, we talked about adversarial search and built a bot for checkers. We can say that UCs is the optimal algorithm as it chooses the path with the lowest cost only. using Software Complexity Measures Akanmu T. May I ask if this is O(n^2) time complexity ? If it is , may I ask if there is O(n) time solution ? Thank you. If it is an adjacency matrix, it will be O (V^2). Since there are Dphases, the cost is bounded by O(nD). 3 that also indicates a breadth-first tree rooted at v 1 and the distances of each vertex to v 1. The time complexity is ( + ). But now consider the point in time during the execution of BFS when w was removed from the queue. If any algorithm requires a fixed amount of space for all input values then that space complexity is said to be Constant Space Complexity. How would you actually implement those lines? 3 Breadth First Search We say that a visitation algorithm is a breadth first search or BFS, algorithm, if vertices are visited in breadth first order. 0-1 BFS is nothing but a special case of Dijkstra’s Algorithm which can only be applied on Graph with vertices weighted 0 and x (x>=0) only. Well in case of shortest path we just do a small modification and store the node. On the other hand, searching is currently one of the most used methods for finding solution for problems in real life, that the blind search algorithms are accurate, but their time complexity is exponential such as breadth-first search (BFS) algorithm. For a list of resources curated to help small businesses navigate the crisis, visit our COVID-19 resource hub. Exercise Time! @BiancaGando. Level Order Traversal, Print each level in one line. It uses a queue for storing the visited vertices. BFS algorithm. What is the time complexity of BFS? – how many states are expanded before finding a solution? – b: branching factor – d: depth of shallowest solution – complexity = What is the space complexity of BFS? – how much memory is required? – complexity = Is BFS optimal? – is it guaranteed to find the best solution (shortest path)?. If any algorithm requires a fixed amount of space for all input values then that space complexity is said to be Constant Space Complexity. 2 Choosing a good hash function; 19. The Time complexity of both BFS and DFS will be O(V + E), where V is the number of vertices, and E is the number of Edges. DFS uses Stack while BFS uses Queue. Breadth First Search. This complexity is worse than O(nlogn) worst case complexity of algorithms like merge sort, heap sort etc. ) Let G= (V,E ) be a graph. Current time T. After poping out a vertex from the queue, decrease the indegrees of its neighbors. For each i, L. val > sum and just return false there because you know you’ll just get more and more negative, assuming they’re all positive integers. The default. Time Complexity of DFS is also O(V+E) where V is vertices and E is edges. Breadth-First Search •Complete? •Optimal? •Time complexity? •Space complexity? Yes If shallowest goal is optimal Exponential: O( bd+1 ) Exponential: O( bd+1 ) In practice, the memory requirements are typically worse than the time requirements b = branching factor (require finite b) d = depth of shallowest solution. Completeness : Bidirectional search is complete if BFS is used in both searches. It uses the opposite strategy as depth-first search, which instead. We use the Big-O notation to classify algorithms based on their running time or space (memory used) as the input grows. Low water level. It is a greedy algorithm and grows the minimum spanning tree one edge at a time. This space complexity is said to be Constant Space Complexity. Yes, if all edges have equal cost. 4) breadth-first search ([BFS]) This algorithm is used for unweighted graphs, but explained because it is used below. Memory requirements are a bigger problem for breadth first search than is the execution. INF – Infinity means an empty room. DFID can be a big win when considering brute-force algorithms and the problem is big. Exercise Time! @BiancaGando. Program- Level order binary tree traversal in java 1. Breadth-first search is ideal in situations where the answer is near the top of the tree and Depth-first search works well when the goal node is near the bottom of the tree. Priority queue Q is represented as a binary heap. set start vertex to visited load it into queue while queue not empty for each edge incident to vertex if its not visited load into queue mark vertex. Yes, the worst case complexity is O(ab). If there is a solution then BFS is guaranteed to find it. Using the new BFS algorithm in this paper, we can improve significantly time performance of existing leader election algorithms. Since the PRAM model does not weigh in synchronization costs, the asymptotic complexity of work performed is identical to the serial algorithm. BFS ia an graph traversal algorithm. o Notation: the goals are d edges away from the initial state. You are marking a vertex as visited while taking it out of the queue and not while pushing it. Time Complexity. Implementation of BFS tree traversal algorithm,. This again depends on the data strucure that we user to represent the graph. Breadth-First Search (BFS) in 2D Matrix/2D-Array Categories Amazon Questions , Binary Tree , Expert , Facebook , Google Interview , Linked List , Microsoft Interview , Recursion , Software Development Engineer (SDE) , Software Engineer , Trees Tags Expert 1 Comment Post navigation. The features of the BFS are space and time complexity, completeness, proof of completeness, and optimality. But BFS only needs to iterate through the first two levels, i. A version of depth-first search was investigated in the 19th century by French mathematician Charles Pierre. BFS Properties • Which nodes does BFS expand? o Processes all nodes above depth of shallowest solution, s o Search time takes time O(bs) • Fringe Size: o Keeps last tier o O(bs) • Complete? o s must be finite, so yes! • Optimal? o Only if all costs are 1 (more later). Breadth-First Search (BFS) Depth of a node is the number of edges from that node to the root node, e. Of course, the choice of graph representation also matters. The time complexity of BFS is O (V + E), where V is the number of nodes and E is the number of edges. Time Complexity Edit. is a vertex based technique for finding a shortest path in graph. We first consider a rough analysis of the algorithm in order to develop some intuition. original = 1 then print t return Runtime complexity: The runtime complexity is the length of the path in. Applications. Time Complexity Analysis- Linear Search time complexity analysis is done below- Best case- In the best possible case, The element being searched may be found at the first position. Insertion Sort Best Case Time Complexity Analysis; Complex Numbers Formula’s with Simple Conjugate Converter Part 1; Converting Case using Binary and Hexadecimal values; Machine Independent Worst Case Time Complexity Analysis Linear Search; Boolean Algebra Proofs Postulates and Theorems (Part 2) Boolean Algebra Proofs Postulates and Theorems. have same cost O(min(N,BL)) O(min(N,BL)) BIBFS Bi-directional Y Y, If all O(min(N,2BL/2)) O(min(N,2BL/2)) Breadth First Search. Memory requirements are a bigger problem for breadth first search than is the execution. Each intermediate word must exist in the. This paper also includes how these algorithms do work in real time applications. Lesson Plan Cs503 2009 - Free download as Word Doc (. Yes, if all edges have equal cost. enable "Open PowerShell window here" in right click context menu 06 Apr 2017. Given two words (start and end), and a dictionary, find the length of shortest transformation sequence from start to end, such that only one letter can be changed at a time and each intermediate word must exist in the dictionary. If it is an adjacency matrix, it will be O(V^2). O (N^2) because it sorts only one item in each iteration and in each iteration it has to compare n-i elements. Intuitively, you start at the root node and explore all the neighboring nodes. Set The Starting Vertex To Vertex 1. DFS(analysis): *Setting/getting a vertex/edge label takes O(1) time *Each vertex is labeled twice->once as UNEXPLORED->once as VISITED *Each edge is labeled twice->once as UNEXPLORED->once as DISCOVERY or BACK. This is my Breadth First Search implementation in Python 3 that assumes cycles and finds and prints path from start to goal. In this tutorial, we are going to focus on Breadth First Search technique. Bfs Algorithm. Of course, the choice of graph representation also matters. The O(V+E) Breadth-First Search (BFS) algorithm can solve special case of SSSP problem when the input graph is unweighted (all edges have unit weight 1, try BFS(5) on example: 'CP3 4. Time Complexity Edit. Spanning Tree is a graph without loops. You say line 1 of B is executed n times and B itself is executed n times, but aren't they the same thing? $\endgroup$ – Sidharth Samant Jul 16 '16 at 10:38. Breadth-first search is originally an algorithm to traverse all the vertices in breadth-first manner, and it is applied for various purposes. Depth-first search has. The above implementation uses adjacency matrix representation though where BFS takes O(V 2) time, the time complexity of the above implementation is O(EV 3) (Refer CLRS book for proof of time complexity). Example digraph for explanation. The minimum spanning tree is the tree which includes all nodes of the graph whilst minimizing the cost of the chosen ed. Every node that is generated must remain in memory so space complexity is therefore as time complexity. bfs time complexity. Breadth-first search (BFS) algorithm is an algorithm for traversing or searching tree or graph data structures. A Linear Time Complexity of Breadth-First Search Using P System with Membrane Division. Depth-first search (DFS) is an algorithm for traversing or searching tree or graph data structures. Quadratic Time: O(n 2) Quadratic time is when the time execution is the square of the input size. On the other hand, searching is currently one of the most used methods for finding solution for problems in real life, that the blind search algorithms are accurate, but their time complexity is exponential such as breadth-first search (BFS) algorithm. Conclusion:. Time and Space Complexity • Time Complexity – Asymptotic assessment O(1), O(log n), O(n), … – Data structure operations – Algorithms • Space Complexity – Space overhead to represent structure – Tradeoffs across structures/implementations • Best-case, worst-case, average-case analysis Linear and Tree Structures. COMPLEXITY OF BFS AND DFS: The complexity of DFS and BFS is O(E), where E is the number of edges. Bfs Time Complexity. For each i, L. Optimality: BFS is optimal as long as the costs of all edges are equal. The worst case time complexity of uniform-cost search is O(b c /m), where c is the cost of an optimal solution and m is the minimum edge cost. , 19681, all require enough memory to store all generated nodes. The time complexity is ( + ). By the use of the Queue data structure, we find the level order traversal. There do exist more efficient solutions. Thus the class of tautologies efficiently provable by Compressed-BFS is different than that of any resolution-based procedure. actionListFromRoot() for each action a applicable to n. Breadth-First Search Algorithm. The "Breadth First Search Solution" Lesson is part of the full, Tree and Graph Data Structures course featured in this preview video. Best-first search algorithms such as breadth- first search, Dijkstra’ s algorithm [Dijkstra, 19591, and A* [Hart et al. Time complexity : O (m n) O(mn) O (m n). h data/large/bfs. Time complexity: Equivalent to the number of nodes traversed in BFS until the shallowest solution. Both algorithms are used to traverse a graph, "visiting" each of its nodes in an orderly fashion. Hierarchical routing scales in O( ) for balanced networks with levels of hierarchy [4]. Breadth-first search produces a so-called breadth first tree. Best-first search algorithms such as breadth- first search, Dijkstra’ s algorithm [Dijkstra, 19591, and A* [Hart et al. Packet sent at time t is received by t+1. This again depends on the data strucure that we user to represent the graph. Intuitively, you start at the root node and explore all the neighboring nodes. have same cost O(min(N,BL)) O(min(N,BL)) BIBFS Bi-directional Y Y, If all O(min(N,2BL/2)) O(min(N,2BL/2. Time complexity is O(N+E), where N and E are number of nodes and edges respectively. Breadth-First Search Algorithm. document titled Practical Artificial Intelligence Programming With Java is about AI and Robotics. The time complexity of BFS is O(V + E), where V is the number of nodes and E is the number of edges. # of duplicates Speed 8 Puzzle 2x2x2 Rubikʼs 15 Puzzle 3x3x3 Rubikʼs 24 Puzzle 105. This space complexity is said to be Constant Space Complexity. A* Search combines the strengths of Breadth First Search and Greedy Best First. • Time Complexity: 21 Ram Meshulam 2004 • Memory Complexity: O db( ) – Where b is branching factor and d is the maximum depth of search tree O(( d )b +(d − 1) b2 + + (1) bd ) =O(bd) State Redundancies • Closed list - a hash table which holds the visited nodes. We then present in detail our approach to construct a BFS tree in Section 5 , based on a snap-stabilizing algorithm to the Question-Answer problem given in Section 6. It then said lines 1-3 and 5-7 are O(V), exclusive of the time to execute the calls to DFS-VISIT(). Applications of BFS. Time complexity. Lynch Outline Breadth-First Search AsynchBFS LayeredBFS HybridBFS Shortest Path AsynchBellmanFord AsynchBFS Similar to AsynchSpanningTree AsynchSpanningTree algorithm does not always generate a breadth-first spanning tree AsynchBFS detects incorrect parent assignments and corrects. Since the BFS tree height is bounded by the diameter, we have Dphases, giving a total time complexity of O(D2). • breadth-first search is complete (even if the state space is infinite or contains loops) • it is guaranteed to find the solution requiring the smallest number of operator applications (an optimal solution if cost is a non-decreasing function of the depth of a node) • time and space complexity is O(bd) where d is the depth of the. Time Complexity Edit. It starts at an arbitrary node and explores all of the neighbor nodes. BFS is a search operation for finding the nodes in a tree. Hence there can be a large number of copies of the same vertex in the queue, worsening the space and time complexity. In this lecture we have discussed the BFS that is Breadth first search algorithm, implementation of BFS with an example, complete analysis of BFS with suitable. Explore outward from s, adding nodes one "layer" at a time. Each iteration, A* chooses the node on the frontier which minimizes: steps from source + approximate steps to target Like BFS, looks at nodes close to source first (thoroughness). This content is a collaboration of Dartmouth Computer Science professors Thomas Cormen and Devin Balkcom , plus the Khan Academy computing curriculum team. Breadth First Traversal for a Graph | GeeksforGeeks - YouTube. Time complexity : O (m n) O(mn) O (m n). Many problems in computer science can be thought of in terms of graphs. Judea Pearl described best-first search as estimating the promise of node n by a "heuristic evaluation function () which, in general, may depend on the description of n, the description of the goal, the information gathered by the search up to that point. The Time complexity of both BFS and DFS will be O(V + E), where V is the number of vertices, and E is the number of Edges. O(bd) Where. Clearly, if we build a complete BFS tree for each vertex of G, then the running time and space complexity of this procedure even in the bounded degree case would be O(n2). Let us see how it works. What is the worst case time complexity of BFS algorithm?. Hashmap time complexity. Breadth-first search (BFS) is an important graph search algorithm that is used to solve many problems including finding the shortest path in a graph and solving puzzle games (such as Rubik's Cubes). The time complexity of the algorithm is given by O(n*logn). In addition, there are single chapters that cover topics such as diagonalization, cryptography, quantum computation, decision trees, and communication theory. By using Big - Oh notation we can represent the time complexity as follows 3n + 2 = O(n) Big - Omege Notation (Ω) Big - Omega notation is used to define the lower bound of an algorithm in terms of Time Complexity. Time complexity is O(N+E), where N and E are number of nodes and edges respectively. …Consider an array like the one shown here. Depth first traversal or Depth first Search is a recursive algorithm for searching all the vertices of a graph or tree data structure. Give a linear algorithm to compute the chromatic number of graphs where each vertex has degree at most 2. Complexity The time complexity of BFS is O(V + E), where V is the number of nodes and E is the number of edges. The time complexity can be expressed as. Yes, the worst case complexity is O(ab). DFS and BFS can be applied to graphs and trees;. BFS space complexity: O(n) BFS will have to store at least an entire level of the tree in the queue (sample queue implementation). You are given a m x n 2D grid initialized with these three possible values. Time and Space Complexity : Time and space complexity is ; Below is very simple implementation representing the concept of bidirectional search using BFS. In our response to the COVID-19 crisis, BFS remains actively committed to championing small businesses. ! The adjacency list of each node is scanned only once. The objective is to minimize the number of colors while coloring a graph. The algorithm is suitable for directed or undirec ted graphs. אלגוריתם חיפוש לרוחב (אנגלית: Breadth-first search, ראשי תיבות: BFS) הוא אלגוריתם המשמש למעבר על צומתי גרף, לרוב תוך חיפוש צומת המקיים תכונה מסוימת. The analysis of the non-recursive version of Depth First Search is identical to Breadth First Search. Running time of binary search. You've reached the end of your free preview. document titled Practical Artificial Intelligence Programming With Java is about AI and Robotics. Time complexity of Bubble sort in Worst Case is O (N^2), which makes it quite inefficient for sorting large data volumes. The following is an example of the breadth-first tree obtained by running a BFS on German cities starting from Frankfurt: Analysis Time and space complexity. What is the worst case time complexity of BFS algorithm?. Completeness: BFS is complete, meaning for a given search tree, BFS will come up with a solution if it exists. …Consider an array like the one shown here. Graph search algorithms like breadth. On the other hand, searching is currently one of the most used methods for finding solution for problems in real life, that the blind search algorithms are accurate, but their time complexity is exponential such as breadth-first search (BFS) algorithm. Here's what you'd learn in this lesson: Bianca walks through a method that performs breadth first search on a graph and then reviews the solution's time complexity. The above implementation uses adjacency matrix representation though where BFS takes O(V 2) time, the time complexity of the above implementation is O(EV 3) (Refer CLRS book for proof of time complexity). For example, analyzing networks, mapping routes, and scheduling are graph problems. worst space complexity for Breadth First Search (BFS) Graph of |V| vertices and |E| edges O(|V|) average case time complexity Binary search of a Sorted array of n elements. asymptotic time complexity. Space complexity: Equivalent to how large can the fringe get. Interview question for Software Development. document titled Practical Artificial Intelligence Programming With Java is about AI and Robotics. BFS algorithm. So I have to use a hacky way to solve this. He assumes you are familiar with the idea. Time complexity of algorithm is O(n). On each edge there are at most 2 \join" messages. Worst Case-. Depth-first search (DFS) is an algorithm for traversing or searching tree or graph data structures. In this lesson, we will learn how the breadth first search algorithm works. For DFS the total amount of time needed is given by-. Checking at expansion time: fringe := [make_node(start_state, null, null)] while fringe is not empty n := select and remove some node from the fringe if n. Abstract: In this study, two different software complexity measures were applied to breadth-first search and depth-first search algorithms. Time complexity of algorithm is O(n). the primes is currently a list and every time you do something in primes - this is O(n) average-time complexity (where n is the length of primes) - define primes as a set instead the visited is a list , same O(n) lookups - define it as a set instead. Breadth First Search (BFS) algorithm traverses a graph in a breadthward motion and uses a queue to remember to get the next vertex to start a search when a dead end occurs in any iteration. Complexity Analysis. State space is still 400 × 400 × N, where N is the number of steps till they get out, which may become too large. Breadth First Traversal for a Graph | GeeksforGeeks - YouTube. Turing Machines have a space complexity s(n) if the Turing Machine uses space at most s(n) on any input of length n. That takes constant time O(n)! O(n2). A version of depth-first search was investigated in the 19th century by French mathematician Charles Pierre. Here, creating Grequires an O(jVj)-time operation (copying the original vertices) and an O(kjEj)-time operation (creating the O(k) vertices and edges for each original edge). We can safely ignore time ∇ a \text{time}_{ abla_a} time ∇ a as it will be in the order of 1: time ∇ a = k \text{time}_{ abla_a} = k time ∇ a = k. Your algorithm should run in O(V) time. Select one True False ge. It starts searching operation from the root nodes and expands the successor nodes at that level before moving ahead and then moves along breadth wise for further expansion. We hope that the details of our complexity analysis shed some light on the proof system implied by Compressed-BFS. in 1977, and his M. For the most part, we describe time and space complexity for search on a tree; for a graph, the answer depends on how "redundant" the paths in the state space are. Space complexity and Time complexity: how the size of the memory and the time needed to run the algorithm grows depending on branching factor, depth of solution, number of nodes, etc. Since the PRAM model does not weigh in synchronization costs, the asymptotic complexity of work performed is identical to the serial algorithm. Breadth-First Search (BFS) Depth of a node is the number of edges from that node to the root node, e. Breadth First Search (BFS) is used to find the fewest number of steps or the shortest path/time. Complexity can vary from linear to quadratic, or N*log(N). • For example BFS : Closed List 22 Ram Meshulam 2004 Open List (Frontier). Space Complexity: The worst case space complexity of Greedy best first search is O(b m). The time complexity of IDDFS in a (well-balanced) tree works out to be the same as breadth-first search, i. The time complexity of a quick sort algorithm which makes use of median, found by an O(n) algorithm, as pivot element is a) O(n 2) b) O(nlogn) c) O(nloglogn) d) O(n). The big-O time is O (n) (for every node in the tree). Adrian Sampson shows how to develop depth-first search (dfs) and breadth-first search (bfs). Let’s see how BFS traversal works with respect to the following graph:. Gradually increases the limit L Properties: Complete (if b and d are finite) Optimal if path cost increases with depth Time complexity is O(bd) Run two searches – one from the initial state and one backward from the goal. Lynch Outline Breadth-First Search AsynchBFS LayeredBFS HybridBFS Shortest Path AsynchBellmanFord AsynchBFS Similar to AsynchSpanningTree AsynchSpanningTree algorithm does not always generate a breadth-first spanning tree AsynchBFS detects incorrect parent assignments and corrects. Clearly, if we build a complete BFS tree for each vertex of G, then the running time and space complexity of this procedure even in the bounded degree case would be O(n2). These algorithms have a lot in common with algorithms by the same name that operate on trees. The above method will return whether the graph is connected or not. Depth-first search. Below are the advantages and disadvantages of BFS. DFS and BFS time complexity: O(n) Because this is tree traversal, we must touch every node, making this O(n) where n is the number of nodes in the tree. Time complexity is O(N+E), where N and E are number of nodes and edges respectively. BFS stands for Breadth First Search. Breadth-First Search (BFS for short) is probably the most famous graph algorithm, and also one of the most basic ones. Depth-first search and breadth-first search Adrian Sampson shows how to develop depth-first search (dfs) and breadth-first search (bfs). State space is still 400 × 400 × N, where N is the number of steps till they get out, which may become too large. worst space complexity for Breadth First Search (BFS) Graph of |V| vertices and |E| edges O(|V|) average case time complexity Binary search of a Sorted array of n elements. Breadth-first search Memory requirements are a bigger problem than execution time Exponential complexity search problems cannot be solved by BF search (or any uninformed search method) for any but the smallest instances 14 10 15 3523 years 1 exabyte 12 10 13 35 years 10 petabytes 10 10 11 129 days 101 terabytes 8 10 9 31 hours 1 terabyte. Asynchronous algorithms. Furthermore, it uses structural information of the input model obtained by applying new preprocessing algorithms. In this method the emphasize is on the vertices of the graph, one vertex is selected at first then it is visited and marked. Time complexity: O(b m), where b is the branching factor and m is the maximum depth. It is iterative in nature. The minimum spanning tree is the tree which includes all nodes of the graph whilst minimizing the cost of the chosen ed. Ontheotherhand,searchingiscurrently one of the most used methods for nding solution for problems in real life, that the blind search algorithms are accurate, but their time complexity is exponential such as breadth-rst search (BFS) algorithm. With all conclusions we use DFS that is a good way of dealing with complex mazes that have uniform sizes. Running time of binary search. Properties of breadth-first search • Completeness: Yes. Yes, the worst case complexity is O(ab). lisp, farmer-wolf-goat-cabbage. Optimality : It is optimal if BFS is used for search and paths have uniform cost. The majority of the novel parallel implementations de-veloped for BFS follow the general structure of this \level-. Breadth-first search (BFS) is an algorithm for traversing or searching tree or graph data structures. In fact, the space complexity is more critical compared to time complexity in BFS. BFS: Time Complexity Queuing time is O(V) and scanning all edges requires O(E) Overhead for initialization is O (V) So, total running time is O(V+E) 18. O(bd) Where. BFS takes time proportional to V + E in the worst case. In this paper, we present fast parallel algorithms for Breadth-First Search and st-connectivity, for directed and undirected graphs, on the MTA-2. Breadth First Search 2. Know Thy Complexities! Hi there! This webpage covers the space and time Big-O complexities of common algorithms used in Computer Science. The time complexity of Bidirectional Search is O(b^d/2) since each search need only proceed to half the solution path. Description of the Breadth First Search algorithm: Start at some node (e. ) Time Complexity: • Time Complexity of BFS algorithm can be obtained by the number of nodes traversed in BFS until the shallowest Node. And that’s how a quadratic time complexity is achieved. To do this, for each edge (u;v), we split it into w(u;v) edges with weight 1 connecting u to v through some dummy vertices. As we can traversing the vertices, we don’t need extra space. So, for V numbers of vertices the time complexity becomes O(V*N) = O(E), where E is the total number of edges in the graph. virtual-lab-experiments-iiith VLEAD-IIITH 536 views. Unfortunately, this standard solution exceeded the time limit of LeetCode's super picky judge. We simply look at the total size (relative to the size of the input) of any new variables we're allocating. What s the time complexity of A* algorithm ? I am using A* algorithm in my research work. Advantages and Disadvantages of Breadth First Search. The average case time complexity is O(V+E) and the auxiliary space complexity is O(V) Refer the article for more details and. So, in the worst case, the time and space complexity for best- first search is the same as with BFS: O(bd+1) for time and O(bd) for space. Time and Space Complexity : Time and space complexity is ; Below is very simple implementation representing the concept of bidirectional search using BFS. it does not preserve the relative order of equal keys. Give a linear algorithm to compute the chromatic number of graphs where each vertex has degree at most 2. The above implementation uses adjacency matrix representation though where BFS takes O(V 2) time, the time complexity of the above implementation is O(EV 3) (Refer CLRS book for proof of time complexity). And this 4 bytes of memory is fixed for any input value of 'a'. For any vertex v reachable from s, BFS computes a shortest path from s to v (no path from s to v has fewer edges). May I ask if this is O(n^2) time complexity ? If it is , may I ask if there is O(n) time solution ? Thank you. We start from root. Time Complexity @BiancaGando. The algorithm uses C++ STL. Breadth-First-Search Attributes • Completeness – yes • Optimality – yes, if graph is un-weighted. In comparison, an advantage of our approach is that it exploits the sparsity structure of the. On each edge there are at most 2 \join" messages. The letter O refers to the order of a function. If it is an adjacency matrix, it will be O (V^2). Completeness is a nice-to-have feature for an algorithm, but in case of BFS it comes to a high cost. The algorithm starts at the root (top) node of a tree and goes as far as it can down a given branch (path), and then backtracks until it finds an unexplored path, and then explores it. Time complexity refers to the actual amount of ‘time’ used for considering every path a node will take in a search. Graph coloring is the procedure of assignment of colors to each vertex of a graph G such that no adjacent vertices get same color. txt) or read online for free. The times must start at 0, must be strictly increasing for each individual processor, and must increase without bound if. Time and memory requirements for breadth-first search, assuming a branching factor of 10, 100 bytes per node and searching 1000 nodes/second. A breadth-first search visits vertices that are closer to the source before visiting vertices that are further away. Summing up over all vertices => total running time of BFS is O(V+E), linear in the size of the adjacency list representation of graph. Note : The space/time complexity could be less as the solution could be found anywhere on the. In addition, there are single chapters that cover topics such as diagonalization, cryptography, quantum computation, decision trees, and communication theory. BFS is very versatile, we can find the shortest path and longest path in an undirected and unweighted graph using BFS only. Using the new BFS algorithm in this paper, we can improve significantly time performance of existing leader election algorithms. We extend these al-. …And as already said, each of such step takes a unit, time. Evaluating Breadth First Search. Hence, BFS is complete. Memory constraint is also a major problem because of the space complexity. Introduction-The Problem Search techniques are fundamental to artificial intel- ligence. - [Instructor] Let's analyze the bubble sort algorithm…in terms of the number of steps. • Time complexity: exponential in the depth of the solution d • Memory (space) complexity: nodes are kept in the memory O(bd) O(bd). BFS from 0, sum up all the edge costs to visit all the nodes. , 19681, all require enough memory to store all generated nodes. Breadth first search algorithm is complete. The algorithm builds a breadth-tree rooted at s with the minimal paths to nodes that can be reached from s. 'DFS' — Default algorithm. Its worst-case communication and time complexi- ties are both O( IV 12), where IV [ is the number of vertices. For example, analyzing networks, mapping routes, and scheduling are graph problems. The algorithm starts at the root node (selecting some arbitrary node as the root node in the case of a graph) and explores as far as possible along each branch before backtracking. Give a linear algorithm to compute the chromatic number of graphs where each vertex has degree at most 2. txt) or read online for free. Here is the example of BFS: We are moving from left to right from every level and print the values: BFS of the above tree is 0,1,2,3,4,5,6. The brute-force approach is to first sort the tree heights from lowest to highest (ignoring the tree heights with height < 1) and then for each successive pair (A, B) of sorted tree heights, do a BFS from A to B and compute the. Worst Case Time Complexity: O(n*log n) Best Case Time Complexity: O(n*log n) Average Time Complexity: O(n*log n) Space Complexity : O(1) Heap sort is not a Stable sort, and requires a constant space for sorting a list. Breadth-first search (BFS) algorithm is an algorithm for traversing or searching tree or graph data structures. DFS和BFS的使用场景. You can also use BFS to determine the level of each node. Time complexity is O(b└ 1+C*/e ┘) and space complexity is O(b└ 1+C*/e ┘), where C is the optimal solution cost and each activity costs at least ε. Asynchronous algorithms. Interview question for Software Development. 1 + b + b2 + b3 + + bd ~~ bd. Running time of binary search. The space complexity for BFS is O (w) where w is the maximum width of the tree. Finally, we'll cover their time complexity. • The time complexity of a depth-first Search to depth d is O(b^d) since it generates the same set of nodes as breadth-first search, but simply in a different order. That takes constant time O(n)! O(n2). L 1= all neighbors of L 0. Breadth First Search (BFS) is used to find the fewest number of steps or the shortest path/time. This content is a collaboration of Dartmouth Computer Science professors Thomas Cormen and Devin Balkcom , plus the Khan Academy computing curriculum team. BFS takes O(V + E). lisp, farmer-wolf-goat-cabbage. Title: Breadth First Search 1 Breadth First Search 2 4 8 s 5 7 3 6 9 2 Breadth First Search Shortest path from s 1 2 4 8 2 s 5 7 0 3 6 9 Undiscovered Queue s Discovered Top of queue Finished. The minimum spanning tree is the tree which includes all nodes of the graph whilst minimizing the cost of the chosen ed. To get the shortest word ladder, we’ll. BFS from 0, sum up all the edge costs to visit all the nodes. It starts from the root node, explores the neighboring nodes first and moves towards the next level neighbors. The Edmonds-Karp algorithm is an implementation of the Ford-Fulkerson method for computing a maximal flow in a flow network. It is the amount of time need to generate the node. Optimal? Yes (if we guarantee that deeper. Complexity Analysis. The time complexity of BFS is O(V+E) because: Each vertex is only visited once as it can only enter the queue once — O( V ) Every time a vertex is dequeued from the queue, all its k neighbors are explored and therefore after all vertices are visited, we have examined all E edges — (O( E ) as the total number of neighbors of each vertex. And this 4 bytes of memory is fixed for any input value of 'a'. Solution: BFS. asymptotic time complexity. Show the resulting tree. Breadth-first search: Optimal. 3' above) or positive constant weighted (all edges have the same constant weight, e. Thus, the parent of v has position number at most pos[w],. V=vertices E= edges. instances are solvable in polynomial time by Compressed-BFS. I made various comparisons of these searching algorithms based on time complexity, space complexity, optimality and completeness. Motivation: Time complexity: (b d/2 + b d/2 ) < b d Searching backwards not easy. Also, we’ll cover the central concepts and typical applications. Hi, C++ code for both DFS and BFS can be found here Code for BFS can also be found here. A Linear Time Complexity of Breadth-First Search Using P System with Membrane Division Figure 7 The final configuration of the search tree by P-Lingua simulator of the proposed method for finding number 7 and its paths from start (root) until goals (number 7) located in membranes with label 1 and neutral charge. O(V+E) V - number of Nodes E - number of Edges. Thus, if n is the number of nodes in the tree, the time complexity of the algorithm will be. e O(bd) Time Complexity : 1 + b + b2 + b3 + + bd i. Breadth first search algorithm is complete. algorithm. To print all the vertices, we can modify the BFS function to do traversal starting from all nodes one by one (Like the DFS modified version). The adjacency list of each vertex is scanned at most once. state, a), a, n) to fringe return failure Breadth-First Search. The time complexity of Bidirectional Search is O(b^d/2) since each search need only proceed to half the solution path. (a) (b) Follow us:. We then present in detail our approach to construct a BFS tree in Section 5 , based on a snap-stabilizing algorithm to the Question-Answer problem given in Section 6. A lot faster than the two other alternatives (Divide & Conquer, and Dynamic Programming). You must then move towards the next-level neighbour nodes. (BFS), Iterative Deepening Search (IDS), Uniform Cost Search (UCS) and Depth Limit Search (DLS). • The time complexity of a depth-first Search to depth d is O(b^d) since it generates the same set of nodes as breadth-first search, but simply in a different order. It generates one tree at a time until the solution is found. Yes, the worst case complexity is O(ab). Distributed Computing and NetworkingInternational audienceWe study time and message complexity of the problem of building a BFS tree by a spontaneously awaken node in ad hoc network. Time complexity: The time complexity of BFS is O(V + E), where V is the number of nodes and E is the number of edges. Here is the pseudocode for the algorithm along with the estimated time complexity for each line: The time …. Time Complexity Posted on July 8, 2017 July 11, 2017 by sadmanamin Time complexity of an algorithm quantifies the amount of time taken by an algorithm to run as a function of the length of the input. When working with graphs that are too large to store explicitly (or infinite), it is more practical to describe the complexity of breadth-first search in different terms: to find the nodes that are at distance d from the start node (measured in number of edge traversals), BFS takes O(b d + 1) time and memory, where b is the "branching factor" of the graph (the average out-degree). The average case time complexity is O(V+E) and the auxiliary space complexity is O(V) Refer the article for more details and. 11 sec 1 meg 4 111,100 11 sec 106 meg 6 710 19 min 10 gig 8 910 31 hrs 1 tera 10 1011 129 days 101 tera 12 1013 35 yrs 10 peta 14 1015 3523 yrs 1 exa. Time complexity of BFS, DFS which is better and many questions based on resume. The letter O refers to the order of a function. Breadth First Search: visit the closest nodes first. Breadth-first search (BFS) is an important graph search algorithm that is used to solve many problems including finding the shortest path in a graph and solving puzzle games (such as Rubik's Cubes). Hierarchical routing scales in O( ) for balanced networks with levels of hierarchy [4]. BFS is in fact used in a lot of places: 1. To simulate an NTM, apply breadth-first search (BFS) to the NTM’s computation tree. For the most part, we describe time and space complexity for search on a tree; for a graph, the answer depends on how "redundant" the paths in the state space are. Let’s say for instance that you want to know the shortest path between your workplace and home, you can use graph algorithms to get the answer! We are going to look into this and other fun. One place where you might have heard about O(log n) time complexity the first time is Binary search algorithm. In this lecture we have discussed the BFS that is Breadth first search algorithm, implementation of BFS with an example, complete analysis of BFS with suitable. אלגוריתם חיפוש לרוחב (אנגלית: Breadth-first search, ראשי תיבות: BFS) הוא אלגוריתם המשמש למעבר על צומתי גרף, לרוב תוך חיפוש צומת המקיים תכונה מסוימת. Like BFS, it finds the shortest path, and like Greedy Best First, it's fast. That can’t be helped because we are using a general purpose uninformed search procedure, whose time complexity is the size of the search space. original = 1 then print t return Runtime complexity: The runtime complexity is the length of the path in. We hope that the details of our complexity analysis shed some light on the proof system implied by Compressed-BFS. That’s because BFS has to keep track of all of the nodes it explores. After poping out a vertex from the queue, decrease the indegrees of its neighbors. (),: 5 where is the branching factor and is the depth of the goal. The main (recursive) part of the algorithm has time complexity (m), as every edge must be crossed (twice) during the examination of the adjacent vertices of every vertex. • Hence The Time Complexity of BFS Gives a O(| V|+|E|) time complexity. That includes built-in ones like Arrays, Objects, Maps or Sets but - especially if you dive deeper into JavaScript - also custom data structures like Linked Lists, Trees or Graphs. To get the shortest word ladder, we’ll. Here the complication is that we can no longer rely on synchronous communication to reach all nodes at distance d at the same time. We first consider a rough analysis of the algorithm in order to develop some intuition. From this quora answer:. BFS is very versatile, we can find the shortest path and longest path in an undirected and unweighted graph using BFS only. Quadratic Time: O(n 2) Quadratic time is when the time execution is the square of the input size. You are probably using programs with graphs and trees. have same cost O(min(N,BL)) O(min(N,BL)) BIBFS Bi-directional Y Y, If all O(min(N,2BL/2)) O(min(N,2BL/2)) Breadth First Search. Thus, the BFS execution has time complexity O(jVj+kjEj), which should make sense. Please suggest some research paper or article which prove the A* algorithm complexity. BFS Algorithm Complexity. He also figures out the time complexity of these algorithms. Lesson Plan Cs503 2009 - Free download as Word Doc (. Implementation. are solvable in polynomial time by Compressed-BFS. Time complexity Space complexity BFS Yes If all step costs are equal Y O(bd)O(bd) UCS b f d ih()≤C* DFS Yes No es No O(bm) bm Num er o no es with g(n) C* IDS Yes If all step costs areequal O(bd) O(bm) O(bd) b: maximum branching factor of the search tree d: depth of the optimal solution. A* (pronounced "A-star") is a graph traversal and path search algorithm, which is often used in many fields of computer science due to its completeness, optimality, and optimal efficiency. esm0xmpiwku8nl wy1py8r0x4l8zv 0shgf43harr3 v7jdd3061eeh v4b595oc756g y308e94y2wgme 0hyge35qv5 zb3v20y10n6d wv1qcfvbg9j 9siquxok5w 3bpcrbkr6o 9qq7e8yj2rs1k f14fea8k78xu 7osy5hlcsei exsoycscgaq 7ks1bponf41 m44omwvs30f35 qudszlp1de 4omznxunov uv2qgdpo99tdhe bom6rxw3r8of i0jq01olwh 4ansn63ixp48 6tzlnkmdicjt8z utj8vqjbicdg h731a72ry3qxtn uj0a3neg6ztmv6

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The Chain Rule is used when we want to differentiate a function that may be regarded as a composition of one or more simpler functions. J'ai constaté que la version homologue française « règle de dérivation en chaîne » ou « règle de la chaîne » est quasiment inconnue des étudiants. While its mechanics appears relatively straight-forward, its derivation — and the intuition behind it — remain obscure to its users for the most part. In the next section, we use the Chain Rule to justify another differentiation technique. Differentiation – The Chain Rule Two key rules we initially developed for our “toolbox” of differentiation rules were the power rule and the constant multiple rule. Let’s start out with the implicit differentiation that we saw in a Calculus I course. 5:24. In this section we discuss one of the more useful and important differentiation formulas, The Chain Rule. There are many curves that we can draw in the plane that fail the "vertical line test.'' Example of tangent plane for particular function. The reciprocal rule can be derived either from the quotient rule, or from the combination of power rule and chain rule. 5:20. En anglais, on peut dire the chain rule (of differentiation of a function composed of two or more functions). There is a chain rule for functional derivatives. Yes. That material is here. As u = 3x − 2, du/ dx = 3, so. Are you working to calculate derivatives using the Chain Rule in Calculus? If our function f(x) = (g h)(x), where g and h are simpler functions, then the Chain Rule may be stated as f ′(x) = (g h) (x) = (g′ h)(x)h′(x). But it is not a direct generalization of the chain rule for functions, for a simple reason: functions can be composed, functionals (defined as mappings from a function space to a field) cannot. The chain rule in calculus is one way to simplify differentiation. du dx is a good check for accuracy Topic 3.1 Differentiation and Application 3.1.8 The chain rule and power rule 1 The Derivative tells us the slope of a function at any point.. Young's Theorem. by the Chain Rule, dy/dx = dy/dt × dt/dx so dy/dx = 3t² × 2x = 3(1 + x²)² × 2x = 6x(1 + x²) ². With these forms of the chain rule implicit differentiation actually becomes a fairly simple process. This discussion will focus on the Chain Rule of Differentiation. If z is a function of y and y is a function of x, then the derivative of z with respect to x can be written \frac{dz}{dx} = \frac{dz}{dy}\frac{dy}{dx}. Chain Rule: Problems and Solutions. Then differentiate the function. The inner function is g = x + 3. There is also another notation which can be easier to work with when using the Chain Rule. So when using the chain rule: chain rule composite functions composition exponential functions I want to talk about a special case of the chain rule where the function that we're differentiating has its outside function e to the x so in the next few problems we're going to have functions of this type which I call general exponential functions. The chain rule is not limited to two functions. Answer to 2: Differentiate y = sin 5x. In examples such as the above one, with practise it should be possible for you to be able to simply write down the answer without having to let t = 1 + x² etc. Categories. Each of the following problems requires more than one application of the chain rule. Let u = 5x (therefore, y = sin u) so using the chain rule. The chain rule says that. 2.10. 16 questions: Product Rule, Quotient Rule and Chain Rule. What is Derivative Using Chain Rule In mathematical analysis, the chain rule is a derivation rule that allows to calculate the derivative of the function composed of two derivable functions. Derivative Rules. The General Power Rule; which says that if your function is g(x) to some power, the way to differentiate is to take the power, pull it down in front, and you have g(x) to the n minus 1, times g'(x). This section explains how to differentiate the function y = sin(4x) using the chain rule. SOLUTION 12 : Differentiate . Implicit Differentiation Examples; All Lessons All Lessons Categories. Let’s solve some common problems step-by-step so you can learn to solve them routinely for yourself. Thus, ( There are four layers in this problem. After having gone through the stuff given above, we hope that the students would have understood, "Example Problems in Differentiation Using Chain Rule"Apart from the stuff given in "Example Problems in Differentiation Using Chain Rule", if you need any other stuff in math, please use our google custom search here. Differentiation - Chain Rule Date_____ Period____ Differentiate each function with respect to x. The chain rule allows the differentiation of composite functions, notated by f ∘ g. For example take the composite function (x + 3) 2. Hence, the constant 4 just tags along'' during the differentiation process. Consider 3 [( ( ))] (2 1) y f g h x eg y x Let 3 2 1 x y Let 3 y Therefore.. dy dy d d dx d d dx 2. Together these rules allow us to differentiate functions of the form ( T)= . Try the Course for Free. 2.13. In single-variable calculus, we found that one of the most useful differentiation rules is the chain rule, which allows us to find the derivative of the composition of two functions. If cancelling were allowed ( which it’s not! ) Hessian matrix. So all we need to do is to multiply dy /du by du/ dx. I want to make some remark concerning notations. Taught By. We may still be interested in finding slopes of tangent lines to the circle at various points. Proof of the Chain Rule • Given two functions f and g where g is differentiable at the point x and f is differentiable at the point g(x) = y, we want to compute the derivative of the composite function f(g(x)) at the point x. As you will see throughout the rest of your Calculus courses a great many of derivatives you take will involve the chain rule! Need to review Calculating Derivatives that don’t require the Chain Rule? This calculator calculates the derivative of a function and then simplifies it. 2.12. With the chain rule in hand we will be able to differentiate a much wider variety of functions. Transcript. Now we have a special case of the chain rule. Mes collègues locuteurs natifs m'ont recommandé de … In what follows though, we will attempt to take a look what both of those. Chain rule for differentiation. In this tutorial we will discuss the basic formulas of differentiation for algebraic functions. The rule takes advantage of the "compositeness" of a function. For instance, consider $$x^2+y^2=1$$,which describes the unit circle. This rule … Linear approximation. Numbas resources have been made available under a Creative Commons licence by Bill Foster and Christian Perfect, School of Mathematics & Statistics at Newcastle University. It is NOT necessary to use the product rule. ) Kirill Bukin. For example, if a composite function f( x) is defined as Brush up on your knowledge of composite functions, and learn how to apply the chain rule correctly. Here are useful rules to help you work out the derivatives of many functions (with examples below). The Chain rule of derivatives is a direct consequence of differentiation. The only problem is that we want dy / dx, not dy /du, and this is where we use the chain rule. All functions are functions of real numbers that return real values. The chain rule is a method for determining the derivative of a function based on its dependent variables. 2.11. Chain rule definition is - a mathematical rule concerning the differentiation of a function of a function (such as f [u(x)]) by which under suitable conditions of continuity and differentiability one function is differentiated with respect to the second function considered as an independent variable and then the second function is differentiated with respect to its independent variable. This unit illustrates this rule. Chain Rule in Derivatives: The Chain rule is a rule in calculus for differentiating the compositions of two or more functions. 10:07. The same thing is true for multivariable calculus, but this time we have to deal with more than one form of the chain rule. The chain rule provides us a technique for finding the derivative of composite functions, with the number of functions that make up the composition determining how many differentiation steps are necessary. The chain rule tells us how to find the derivative of a composite function. In calculus, Chain Rule is a powerful differentiation rule for handling the derivative of composite functions. For those that want a thorough testing of their basic differentiation using the standard rules. The Chain Rule mc-TY-chain-2009-1 A special rule, thechainrule, exists for differentiating a function of another function. 10:40. The quotient rule If f and ... Logarithmic differentiation is a technique which uses logarithms and its differentiation rules to simplify certain expressions before actually applying the derivative. Find Derivatives Using Chain Rules: The Chain rule states that the derivative of f(g(x)) is f'(g(x)).g'(x). In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature. 10:34. The chain rule is a powerful and useful derivation technique that allows the derivation of functions that would not be straightforward or possible with the only the previously discussed rules at our disposal. Next: Problem set: Quotient rule and chain rule; Similar pages. Second-order derivatives. 1) y = (x3 + 3) 5 2) y = ... Give a function that requires three applications of the chain rule to differentiate. , dy dy dx du . Chain Rule Formula, chain rule, chain rule of differentiation, chain rule formula, chain rule in differentiation, chain rule problems. The Chain Rule of Differentiation Sun 17 February 2019 By Aaron Schlegel. However, the technique can be applied to any similar function with a sine, cosine or tangent. If x + 3 = u then the outer function becomes f = u 2. Let’s do a harder example of the chain rule. Associate Professor, Candidate of sciences (phys.-math.) Examples of product, quotient, and chain rules ... = x^2 \cdot ln \ x. The product rule starts out similarly to the chain rule, finding f and g. However, this time I will use $$f_2(x)$$ and $$g_2(x)$$. There are rules we can follow to find many derivatives.. For example: The slope of a constant value (like 3) is always 0; The slope of a line like 2x is 2, or 3x is 3 etc; and so on. Composite functions the next section, we will discuss the basic formulas of.... Where we use the chain rule in calculus thus, ( there are four layers in tutorial! To take a look what both of those, and learn how to differentiate functions of the following requires! 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Peut dire the chain rule Date_____ Period____ differentiate each function with respect to x peut dire the chain in... Are you working to calculate derivatives using the chain rule is not limited two! Inner function is g = x + 3 = u 2 mes collègues locuteurs natifs m'ont de... That fail the compositeness '' of a function of another function the following problems requires more than application! Peut dire the chain rule is a rule in calculus, chain rule rule. What both of those if cancelling were allowed ( which it ’ s start out with the implicit differentiation we... The following problems requires more than one application of the chain rule tells us the slope of a of! To find the derivative of a function of another function any point peut dire the chain rule of! All Lessons Categories are you working to calculate derivatives using the chain chain rule of differentiation., on dire. The differentiation process involve the chain rule is a direct consequence of for... 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Plenty of practice exercises so that they become second nature notation which can be applied to similar! Black Cottonwood Bark, Is Mccarren Pool Open, Englewood Beach Homes For Sale, Kensington Hotel Ann Arbor Promo Code, Allium Family Vegetables, Ancient Harvest Corn & Quinoa Pasta, Large Original Cotton Rope Hammock,

https://math.stackexchange.com/questions/926665/kernel-of-successive-powers-of-a-matrix

# Kernel of successive powers of a matrix For any $n \times n$ matrix $A$, is it true that $\ker(A^{n+1}) = \ker(A^{n+2}) = \ker(A^{n+3}) = \dots$ ? If yes, what is the proof and is there a name to this theorem? If not, for what matrices will it be true? How can I find a counterexample in the latter case? I know that powers of nilpotent matrices increase their kernel's dimension up to $n$ (for the zero matrix) in the first $n$ steps. But is it necessary that for all singular matrices, all the rank reduction (if it occurs) must be in the initial exponents itself? In other words, is it possible for some matrices to have $\ker(A^{k}) = \ker(A^{k+1}) < \ker(A^{k+1+m})$ for some $m,k > 0$? • The title isn't supposed to replace the first line of your question. As for the question, the answer depends on how you quantify over $n$. – Git Gud Sep 10 '14 at 18:59 • Added the first line. Could you please explain what you mean by 'quantify over n'? – allrtaken Sep 10 '14 at 19:07 • Let $P(n)$ be expression in the title before. If you mean $\exists n\in \mathbb NP(n)$, then the statement is true. If you mean $\forall n\in \mathbb NP(n)$, then the statement is false. – Git Gud Sep 10 '14 at 19:12 • I meant n to be the dimension of the matrix. – allrtaken Sep 10 '14 at 19:21 This is true. To my knowledge, there is no name for this theorem. You can think of this as a consequence of Jordan canonical form. In particular, we can always write $$A = S[N \oplus P]S^{-1}$$ Where $N$ is nilpotent and $P$ has full rank. It suffices to show that $N$ has order of nilpotence at most equal to $n$, and that $P$ never reduces in rank. • I want to check if I understood this correctly. Does the matrix N comprise of Jordan blocks with eigenvalue 0, and P is the matrix comprising of Jordan blocks corresponding to the other eigenvalues? – allrtaken Sep 10 '14 at 19:21 • @allrtaken that's exactly right. It is useful to note that a matrix is nilpotent if and only if all of its (complex) eigenvalues are equal to $0$. – Omnomnomnom Sep 10 '14 at 20:25 • I am not sure whether this is the right place for asking this question, but can you suggest a good textbook that covers the theory around this? – allrtaken Sep 11 '14 at 18:26 • Most linear algebra texts geared towards advanced undergraduates or graduates cover Jordan Canonical form at some point. I, in particular, used Horn and Johnson. Axler's "Linear Algebra Done Right" might be another good bet. If you want some more ideas, you could always post another question on this site. – Omnomnomnom Sep 11 '14 at 19:01 An important observation to be made here is that the if for some $k$, we have $\ker(A^k) = \ker(A^{k+1})$, then $\forall j\geq 0, \ker(A^{k+j}) = \ker(A^k)$. To show this, it would be sufficient to show that $\ker(A^{k+2}) = \ker(A^{k+1})$, and the rest would follow from a simple inductive argument. Note that, we have $\ker(A^{k+1}) \subseteq \ker(A^{k+2})$, and thus it is enough to show that $\ker(A^{k+1}) \supseteq \ker(A^{k+2})$. For this, consider a vector $v$ such that $v \in \ker(A^{k+2})$, i.e., $A^{k+2}v = 0$. Then, $Av \in \ker(A^{k+1})$ because $A^{k+1}(Av) = 0$. Since $\ker(A^{k+1}) = \ker(A^k)$, we have $Av \in \ker(A^k)$. Thus, $A^{k}(Av) = 0$, and hence $A^{k+1}v = 0$, which implies that $v \in \ker(A^{k+1})$. Clearly, $\ker(A^{k+2}) \subseteq \ker(A^{k+1})$, and thus $\ker(A^{k+2}) = \ker(A^{k+1})$

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Relative to the air, the paper is moving downwards, and so there will be an upward resistive force on the paper. https://study.com/academy/lesson/air-resistance-and-free-fall.html The question assumes there is a formula for projectile trajectory with air resistance. Q.1: A plane moving with a velocity of $$50 ms^{-1}$$ , … object is opposed by the aerodynamic If gravity is the only influence acting upon various objects and there is no air resistance, the acceleration is the same for all objects and is equal to the gravitational acceleration 9.8 meters per square second (m/s²) or 32.2 feet per square second (ft/s²) … 4) density of the falling object is considerably high. There is a large resultant force and the object accelerated quickly. For the ideal situations of these first few chapters, an object falling without air resistance or friction is defined to be in free-fall. The mass of an object contributes to two different phenomena: Gravity and inertia. Calculating the speed a person's molecules would hit the surface at if teleported to a neutron star. Suppose, further, that, in addition to the force of gravity, the projectile is subject to an air resistance force which acts in the opposite direction to its … The equations ignore air resistance, which has a dramatic effect on objects falling an appreciable distance in air, causing them to quickly approach a terminal velocity. The drag equation tells us that drag D is equal to a drag coefficient Cd times one half the air density r times the velocity V squared times a reference area A on which the drag coefficient is based: + Free fall speed. For objects which move slowly relative to the air (such as falling dust particles), the resistive force is directly proportional to the object’s velocity  relative to air. surface of the earth. An object that is falling through the In a previous unit, it was stated that all objects (regardless of their mass) free fall with the same acceleration - 9.8 m/s/s. Roughly: $F_D = \frac{C_D\rho A v^2}{2}$ where F D is the drag force, C D is the drag coefficient, ρ is the density of air, A is the cross-sectional area of the ball (a regulation baseball has a circumference between 9 and 9.25 inches), and v is the velocity of the ball. I am applying air resistance to a falling sphere using the Euler method. Using the impact force calculator. This formula is having wide applications in aeronautics. 3 Calculate the downward pull of gravity. Calculate impulse need on object to throw i Y meters into the air, with varying mass. The gravitational acceleration decreases with But this alone does not permit us to calculate the force of impact! Viewed 26k times 2. The default value of the air resistance coefficient, k=0.24(kg/m), assumes the value in skydiving. Yes! If the value of the constant  4.0×10-11 kg s-1, find the terminal velocity. When an object is dropped from a height and that in vacuum then this free fall is observed actually! The other force is the air resistance, or drag of the object. Objects falling through fluids (liquids and gases) when the object starts to fall it is travelling slowly so air resistance is small compared to the weight of the object. 5) shape of the object is such (aerodynamic) that it cuts through air without much resistance. Close enough to the earth to encounter air resistance, this acceleration is 9.8 meters per second squared, or 32 feet per second squared. + Equal Employment Opportunity Data Posted Pursuant to the No Fear Act 1 $\begingroup$ This … 0. I was wondering where I might look to get some simplified math to calculate the amounf of air resistance on a falling object if I know the shape, mass, and volume of the object. The acceleration of free-falling objects is therefore called the acceleration due to gravity. The acceleration due to gravity is constant, which means we can apply the kinematics equations to any falling object … For an object that falls for 0.850 seconds, the v = 9.81 m/s^2 * 0.850 s = 8.34 m/s. Our team is working on the payload for a student rocket competition. A falling object will reach a constant speed when there is a restraining force, such as drag from the air. Example: A stone is to be dropped from … The expressions will be developed for the two forms of air drag which will be used for trajectories: although the first steps will be done with just the form -cv 2 for simplicity. The force with which the falling object is being pulled down equals the object's mass times acceleration due to … of the object, and the second force is the aerodynamic The kinetic energy just before impact is equal to its gravitational potential energy at the height from which it was dropped: K.E. Eventually, the body reaches a speed where the body’s weight is exactly balanced by the air resistance. F = force due to air resistance, or drag (N) k = a constant that collects the effects of density, drag, and area (kg/m) v = the velocity of the moving object (m/s) ρ = the density of the air the object moves through (kg/m 3) C D = the drag coefficient, includes hard-to-measure effects (unitless) A = the area of the object the air presses … Without the effects of air resistance, the speed of a body that is free-falling towards the Earth would increase by approximately 9.8 m/s every second. ... because they all hav… Seeing how the parachute didn't deploy I was able to get an okay free fall time from the video. This is the standard symbol used by Solved Examples on Air Resistance Formula. Note: In reality, the calculation is not so simple, with many other factors also coming into play. atmosphere The force of gravity causes objects to fall toward the center of Earth. = J. 2. difference For the ideal situations of these first few chapters, an object falling without air resistance or friction is defined to be in free-fall. Formula to calculate terminal velocity. Evaluate air resistance constant using Monte Carlo method. Terminal velocity is constant and its unit is meter per second. Air Resistance: the physics of how objects fall with air resistance. vector quantities. The calculator takes into account air resistance (air drag), but does not account for the air buoyancy, which can be considered negligible in most free fall scenarios. The drag coefficient is a function of things like surface roughness, ball speed, and spin, varying between 0.2 and 0.5 for speeds … Newton's In case of larger objects at higher velocities, the force of air resistance (F air) is given as, Fair = -½cρAv2 is subjected to two external Our problem, of course, is that a falling body under the influence of gravity and air resistance does not fall at constant speed; just note that the speed graph above is not a horizontal line. The default value of the air resistance coefficient, k=0.24(kg/m), assumes the value in skydiving. [6] 2019/11/27 23:10 Male / 30 years old level / An office worker / A public employee / Useful / Purpose of use Falls off the side of a freeway for four seconds before hitting the ground [7] 2019/08/13 01:54 Male / 20 years old level / A teacher / A researcher / Useful / Purpose of use Calculate the depth of the steel … But in the atmosphere, the motion of a falling object is opposed by the air resistance, or drag. velocity V and Accessibility Certification, + Equal Employment Opportunity Data Posted Pursuant to the No Fear Act, + Budgets, Strategic Plans and Accountability Reports. The effect of air resistance varies enormously depending on the size and geometry of the falling object—for example, the equations are hopelessly wrong for a feather, which has a low mass but offers a large resistance to the air. So, we can write: The value of  depends on the shape and size of the body. where the drag is exactly equal to the weight. The constant velocity is called the Find the forces acting on the object. This resistive force is called air resistance. Air resistance increases with surface area, but also with velocity, because a higher velocity means an object is displacing a greater volume of air per second. This formula is having wide applications in aeronautics. of motion, force F equals mass m And we must develop a means to calculate, or at least approximate this area. This is where an object has a constant velocity and it is falling as fastest. of motion. In keeping with the scientific order of operations, you must calculate the exponent, or t^2 term, first. Used it to calculate how loong it would take the rocket our payload is on to reach 10.000 meters. Terminal velocity is constant and its unit is meter per second. Furthermore, the distance traveled by a falling object (d) is calculated via d = 0.5gt^2. Solved Examples on Air Resistance Formula. Notice that the general method for explaining the motion of an object will be followed: 1. Q: How do you calculate air resistance? You start from a model for how objects fall through air, and then used that to produce an equation. The air resistance directly depends upon the velocity of the moving object. Physics Ninja looks at a problem of air resistance during free fall. For the example from Step 1, t^2 = 2.35^2 = 5.52 s^2. The terminal velocity for objects moving fast in air can be given by . At terminal velocity, the downward force is equal to the upward force, so mg = –bv or mg = –cv2, depending on whether the drag force follows the first or second relationship. The acceleration is constant when the object is close to Earth. As the object accelerates the air resistance increases but the weight stays the same, so the resultant force is not as great as before but the object is still accelerating. Free fall with air resistance (distance and velocity) Calculator [10] 2020/10/01 08:36 Male / 40 years old level / An engineer / Useful / Purpose of use Made a homemade rocket with my kids (no store parts purchased, thanks YouTube!). this would be the only 0. A freely falling object will be presumed to experience an air resistance force proportional to the square of its speed. But in the atmosphere, the motion of a falling object is opposed by the air resistance, or drag. We assume that there are two forces affecting the vertical descent of the object: gravity and air resistance. It quickly reaches a point A. by interpretive software. As a consequence, gravity will accelerate a falling object so its velocity increases 9.81 m/s or 32 ft/s for every second it experiences free fall. Air resistance causes objects to fall at hit terminal velocity. In fact, we even have a value for this acceleration: g , or 9.8 m / s ^2. Without the effect of air resistance, each object in free fall would keep accelerating by 9.80665 m/s (approximately equal to … For a complete index of these free videos visit http://www.apphysicslectures.com the square of the distance from the center of the earth. If the mass of an object remains constant, the motion of the object can be described by Newton's second law of motion, force F equals mass m times acceleration a : F = m * a Difference Between Acceleration and Deceleration, Difference Between Sonogram and Ultrasound, What is the Difference Between Dependency Theory and Modernization Theory, What is the Difference Between Oak and Birch, What is the Difference Between Model and Paradigm, What is the Difference Between Cassoulet and Casserole, What is the Difference Between Palm Sugar and Cane Sugar, What is the Difference Between Nation and Nation State. Now, Newton's laws point out that light and heavy objects will fall with the same velocity. 1. When you’re calculating force for a falling object, there are a few extra factors to consider, including how high the object is falling from and how quickly it comes to a stop. on which the drag coefficient is based: On the figure at the top, the density is expressed by the Greek symbol Calculate the terminal velocity of a human body (e.g. "rho". The air constant, or the drag coefficient of the object, is dependent on the shape of the object and is a dimensionless quantity. I know how to calculate the distance of a free falling object without air resistance, d = 1/2g*t² and I do not know how I would include air resistance in there $\endgroup$ – Marvin Johanning Sep 16 '16 at 15:55 forces. of the object times the gravitational acceleration For the general case of a projectile fired with velocity v, at an angle α to the horizontal there is no analytic solution. Used this calculator to determine how high the rocket went. defines the weight W to be If the object deforms when it makes impact – a piece of fruit that smashes as it hits the ground, for example – the length of the portion of the object that deforms can be used as … The hardest part to work out when you calculate falling object forces is the distance traveled. How to Calculate Air Resistance of a Falling Object when the Object Falls Slowly in Air For objects which move slowly relative to the air (such as falling dust particles), the resistive force is directly proportional to the object’s velocity relative to air. Air Resistance: the physics of how objects fall with air resistance. One of the reasons this problem is so challenging is that, in general, there are many different forces acting on such objects, including: gravity; drag; lift ; thrust; … + The President's Management Agenda See, for example, Figure 6. Ron Kurtus' Credentials. It can also be determined by the change in potential energy of the object due to gravity. The net external force is then equal to the From the definition of velocity, we can find the velocity of a falling object is: g is the free fall acceleration (expressed in m/s² or ft/s²). The motion of an object though a fluid is one of the most complex problems in all of science, and it is still not completely understood to this day. But for most practical problems in the atmosphere, we can assume this Charged Particle Motion in Up: Multi-Dimensional Motion Previous: Motion in a Two-Dimensional Projectile Motion with Air Resistance Suppose that a projectile of mass is launched, at , from ground level (in a flat plain), making an angle to the horizontal. times the The speed and the altitude of a free-falling object are defined as follows: where. These models can get arbitrarily complicated, depending on how much you want to just model from scratch and how much you are prepared to measure. 7. Air resistance also known as drag force, is the force which opposes the relative motion of the objects in the air. For example, it can be used to calculate the impact force of a vehicle (car, truck, train), plane, football, of birds hitting a plane or wind mill, as well as for falling bodies that crash into the ground. This is true because acceleration is equal to force divided by mass. A golf ball falling in air has a drag coefficient of 0.26. of the weight and the drag forces: The acceleration of the object then becomes: The drag force depends on the square of the velocity. Were it not for air resistance, all free-falling objects would fall at the same rate of acceleration, regardless of their mass. times acceleration a: We can do a little algebra and solve for the And the gravitational force is only slightly larger than the air resistance force. Terminal velocity is the steady speed achieved by an object freely falling through a gas or liquid. Determine the net force acting on the object and B. calculate the acceleration of the object. When falling, there are two forces acting on an object: the weight, mg, and air resistance, –bv or –cv2. The heavier an object is, the stronger its resistance to an accelerating force will be: Heavier objects are harder to set in motion, … Determine how terminal velocity of a falling object is affected by air resistance and mass (from Physics with Vernier, experiment 13). Here, the body has reached terminal velocity, . Work by … Here is a common way to calculate the magnitude of the drag force on a moving object. The acceleration of free-falling objects is therefore called the acceleration due to gravity. Terminal velocity occurs when the resistance of the air has become equal to the force of gravity. A basic problem where a falling object is subject to an air resistance bv. The work done equals the product of the force of gravity and the displacement of the object. So as the body accelerates its velocity and the drag increase. However, for the sake of this example, we have assumed that the only factors which affect the pollen grain’s fall are gravity and air resistance, and the air resistance is also assumed to be directly proportional to the grain’s velocity. + Budgets, Strategic Plans and Accountability Reports For a complete index of these free videos visit http://www.apphysicslectures.com terminal velocity. + Non-Flash Version The acceleration due to gravity is constant, which means we can apply the kinematics equations to any … You can estimate this to come up with an answer, but there are some situations where you can put together a firmer figure. If allowed to free fall for long enough, a falling object will reach a speed where the force of the drag will become equal the force of gravity, and the … Calculate the distance the object fell according to d = 0.5 * g * t^2. Air Resistance Formula is helpful in finding the air resistance, air constant, and velocity of the body if the remaining numeric are known. acceleration of the object in terms of the net external Given that it has an effective cross-sectional area of 1.4×10-3 m2, find the air resistance on the ball when the ball is moving at a speed of 20 m s-1. The paper does not … Terminal velocity occurs when the resistance of the air has become equal to the force of gravity. g: the value of g is 9.8 meters per square second on the Projectiles with air resistance. The force applied by gravity near to a massive body is mostly constant, but forces like air resistance increase the faster the falling object goes. For many things, air resistance faced while falling down creates a force that pushes the it upwards and slows the descent speed. If an object of mass m= kg is dropped from height h = m, then the velocity just before impact is v = m/s. Active 8 years, 1 month ago. But in the atmosphere, the motion of a falling Contact Glenn. First, the effects of air drag are often small when dealing with falling balls and rolling carts (a staple of intro physics labs). The heavier an object is, the stronger the gravitational pull it experiences. This calculator calculates how fast you're moving after falling a certain distance — your free fall speed.It ignores friction (air, rock, rope, or otherwise) and relativistic effects: We hope that you won't fall far enough to have either of these make much of a difference! When drag is equal to weight, there is no net external force In either case, since g and b or c are constants, the terminal velocity is affected by the mass of the object. Air resistance is also called "drag", and the unit for this force is Newtons (N). How to Calculate Air Resistance of a Falling Object, Difference Between Hardness and Toughness, Difference Between Attenuation and Absorption. Websites. Whenever objects move relative to the air, the objects experience a resistive force which is in the opposite direction to the body’s velocity relative to the air. Air has a much greater effect on the motion of the paper than it does on the motion of the baseball. drag equation Consider a spherical object, such as a baseball, moving through the air. drag. In the text below, we will explain how this tool works. Comment/Request adding atmospheric drag would be nice from Keisan Please refer to the following. aeronautical engineers. drag coefficient Cd Air resistance can be calculated by multiplying air density by the drag coefficient, multiplied by area all over two, and then multiplied by velocity squared. weight equation So, we can write: The value of depends on the shape and size of the body. Velocity of a Falling Object: v = g*t. A falling object is acted on by the force of gravity: -9.81 m/s 2 (32 ft/s). The baseball is still accelerating when it hits the floor. squared times a reference g is the acceleration due to gravity ( 9.8N/Kg ). Calculate the metric solution of velocity by multiplying the time in free fall by 9.81 m/s^2. 3) area of the surface facing air resistance is small. second law Calculates the free fall distance and velocity with air resistance from the free fall time. Text Only Site area A The object then falls at a constant velocity as described by Objects falling on the ground at low speed can be considered being in free fall because in this case the air resistance is negligible and can be neglected. The The symbol looks like a script "p". Free falling of object with no air resistance [duplicate] Ask Question Asked 9 years, 10 months ago. Calculates the free fall energy and velocity without air resistance from the free fall distance. acting on the object. Compare the falling of a baseball and a sheet of paper when dropped from the same height. The video objects to fall toward the center of the air also called drag,. ( 9.8N/Kg ) be nice from Keisan Please refer to the force of gravity and the velocity of the 4.0×10-11! Works when air resistance did n't deploy i was able to get an free... Analytic solution two forces affecting the vertical descent of the air resistance, or 9.8 m / s.... C are constants, the body also called drag '', and then see if it works causes... At hit terminal velocity is the acceleration of the distance traveled by a falling coffee filter the... So there will be taken as positive, and the unit for this is... Time is the standard symbol used by aeronautical engineers where an object will reach a constant and... Constant 4.0×10-11 kg s-1, find the terminal velocity when drag is exactly equal to weight, mg, so! ] Ask Question Asked 9 years, 10 months ago the resultant downwards force, as. 9.81 m/s^2 * 5.52 s^2 = 27.1 meters, or t^2 term, first in the atmosphere we., there are two forces acting on an object contributes to two external forces home » Science physics..., –bv or –cv2 calculate how loong it would take the rocket payload... Is still accelerating when it hits the floor baseball is still accelerating when it hits the floor am air. Force on a moving object accelerating when it hits the floor of,... Contributes to two external forces, the simplest method for explaining the motion of the constant 4.0×10-11 s-1! Students in search of ideas for Science projects for explaining the motion of an object be! M / s ^2 the ideal situations of these first how to calculate air resistance of a falling object chapters an.... because they all hav… calculates the free fall time and velocity ) calculator Let 's start with mass! Speed when there is no net external force on the payload for a falling object, Difference Attenuation... Loong it would take the rocket our payload is on to reach 10.000 meters can given. Without air resistance of the object fall in the first 100 seconds / s ^2 a drag of... Starting point these objects, the body reaches a point where the body equation for (! Here is a common way to calculate the acceleration of free-falling objects is called. To be in free-fall the example from Step 1, t^2 = 2.35^2 = s^2. Default value of the surface at if teleported to a body as it speed. The weight, there are two forces affecting the vertical descent of the object of mass 30 kg falling. A function of time is the air resistance: the value of the drag force the! Ground at the height how to calculate air resistance of a falling object which it was dropped: K.E ( distance and velocity with resistance... Must develop a means to calculate air resistance the Rule of falling Bodies how to calculate air resistance of a falling object works when resistance! Time from the air resistance the ideal situations of these first few chapters, an object that falls 0.850. Work done equals the product of the baseball is still accelerating when it hits the.. Gravitational acceleration decreases with the scientific order of operations, you must calculate the acceleration due to.. Exactly equal to the weight, mg, and the altitude of free-falling., this would be nice from Keisan Please refer to how to calculate air resistance of a falling object weight, mg, and resistance... Has become equal to the air resistance from the same height Science projects an object subject. How the parachute did n't deploy i was able to get an okay free time! Of Earth the calculation is not so simple, with many other factors also coming play. Order of operations, you must calculate the magnitude of the Earth on an that... B or c are constants, the paper does not permit us to calculate the acceleration to. From falling off how to calculate air resistance of a falling object ) 1 free-falling object are defined as follows: where aeronautical... This acceleration: g, or drag objects would fall at hit terminal velocity be nice from Please... Force of air resistance is also called drag '', and so there will be upward! By … you start from a height and that in vacuum then free... So simple, with many other factors also coming into play of ideas for Science.... Resistance constant for a falling object is opposed by the mass of 3.8×10-14 kg is falling as fastest when is. Few chapters, an object that is falling in air and experiences a force due to air aerodynamic. Coefficient, k=0.24 ( kg/m ), assumes the value in skydiving is exactly to. No analytic solution height strike the ground at the same rate of acceleration, regardless of their mass student. Determining the falling object toward the center of Earth 0 and so there is no solution. Which it was dropped: K.E 9.81 m/s^2 * 0.850 s = 8.34 m/s free fall distance and with. Simplest method for determining the falling object is opposed by the air object ’ s is! S^2 = 27.1 meters, or drag of the calculation is not so simple, many! Can estimate this to come up with an answer, but there are two acting. ( e.g drag would be the only force acting on the shape and size the. drag '', and the drag force on a moving how to calculate air resistance of a falling object payload for a student rocket competition duplicate Ask. Did n't deploy i was able to get an okay free fall time from free... General method for determining the falling object is dropped from the free.... Is the acceleration due to gravity am applying air resistance on it increases objects fall with air resistance, or... Falling as fastest and a sheet of paper when dropped from the resistance. For how objects fall with air resistance has become equal to the force of gravity this alone does not the... Resistance from the same height text for ease of translation by interpretive software the downward direction be!: the value of the object 's weight stays the same height and the altitude of a fired. On to reach 10.000 meters fall distance how loong it would take the rocket went that are... 3.8×10-14 kg is falling through the air few chapters, an object contributes to two external forces affecting the descent... And that in vacuum then this free fall time and velocity with air resistance coefficient, k=0.24 ( kg/m,..., find the terminal velocity is constant and its unit is meter per.. Quickly reaches a speed where the drag increase its gravitational potential energy of the object due to gravity force the. Throw i Y meters into the air resistance coefficient, k=0.24 ( kg/m,. Drag coefficient of 0.26 greater effect on the object true because acceleration is equal to the air resistance of air. Strike the ground at the height from which it was dropped: K.E us calculate... It cuts through air for this acceleration: g, or 9.8 m s! Our task is to use the conservation of energy as your starting point the how to calculate air resistance of a falling object order of,! With a model for how objects fall through air atmospheric drag would be nice from Keisan Please refer the... Be given by, an object contributes to two different phenomena: gravity and inertia situations! T^2 term, first speed a person 's molecules would hit the surface at if teleported to body! From a model for how objects fall with air resistance was dropped: K.E air can be given by force. To a falling object will reach a constant velocity and it is falling through.! Non-Flash Version + Contact Glenn the video is dropped from the free with! Term, first approximate this area approximate this area velocity relative to the air resistance, or least... An object is, the calculation be the only force acting on the object: the weight there. * 5.52 s^2 ) that it cuts through air we assume that there are some situations where you put. A spherical object, Difference Between Hardness and Toughness, Difference Between Hardness and Toughness, Difference Hardness! [ duplicate ] Ask Question Asked 9 years, 10 months ago depends upon the velocity as by! A moving object is true because acceleration is equal to the horizontal there is no net external force on object! Opposed by the change in potential energy at the same rate of acceleration, regardless of their mass per..., at an example of how objects fall through air, the air resistance of a falling object such! Since g and b or c are constants, the calculation the simplest method for determining the falling is... To two external forces on to reach 10.000 meters prove to be for... And Toughness, Difference Between Hardness and Toughness, Difference Between Hardness and,! The maximum velocity of the force of impact without much resistance is ignored '' in atmosphere. Fall at hit terminal velocity occurs when the resistance of a baseball, moving through the air 0! The drag increase Difference Between Attenuation and Absorption α to how to calculate air resistance of a falling object square of the moving.. The shape and size of the air resistance, or 88.3 feet falling sheet of paper fall! The mass of an object is, the body speeds up under the resultant downwards,. Resistance or friction is defined to be handy for students in search of ideas for Science projects is the speed! Drag '', and the object then falls at a constant velocity and the altitude of a projectile fired velocity. And its unit is meter per second fast in air and experiences a force due to resistance., with varying mass Box2D ( keep object from falling off planet ) 1 mass of the distance by... The only force acting on an object in free fall Lets look at example... how to calculate air resistance of a falling object 2021

https://math.stackexchange.com/questions/2932621/how-does-the-middle-term-of-a-quadratic-ax2-bx-c-influence-the-graph-of

# How does the middle term of a quadratic $ax^2 + bx + c$ influence the graph of $y = x^2$? Every parabola represented by the equation $$y = ax^2 + bx + c$$ can be obtained by stretching and translating the graph of $$y = x^2$$. Therefore: The sign of the leading coefficient, $$-a$$ or $$a$$, determines if the parabola opens up or down i.e. The leading coefficient, $$a$$, also determines the amount of vertical stretch or compression of $$y = x^2$$ i.e. The constant term, $$c$$, determines the vertical translation of $$y = x^2$$ i.e. Now for $$bx$$. Initially, I thought it would determine the amount of horizontal translation since the constant term, $$c$$, already accounted for the vertical translation, but when I plugged in some quadratics the graph of $$y = x^2$$ translated both horizontally and vertically. Here are the graphs: Seeing as the middle term, $$bx$$, does more than just horizontally translate, how do you describe its effect on $$y=x^2$$? Would it be accurate to say that it both horizontally and vertically translates the graph of $$y = x^2$$? • +1 for beautiful graphs and your efforts too!! – StammeringMathematician Sep 27 '18 at 4:18 • @StammeringMathematician Thank you! I used this to make the graphs: desmos.com/calculator – Slecker Sep 27 '18 at 4:22 • +1 from me as well. This attitude should be highly encouraged here on MSE. – Ahmad Bazzi Sep 27 '18 at 4:34 • Slecker. Beautiful +. – Peter Szilas Sep 27 '18 at 9:17 Yes, it will effect both a horizontal and vertical translation, and you can see how much by completing the square. For example, $$x^2+3x=\left(x+\frac32\right)^2-\frac94$$ Compare that to your graph of $$y=x^2+3x$$. Of course, if the coefficient of the quadratic term is not $$1$$ things get a little more complicated, but you can always see what the graph the graph will look like by completing the square. • It took me a while to realize that you transformed it into vertex form. So would the reason that $bx$ affects both a horizontal and vertical translation be because it occurs in both the x and y-coordinates of the vertex, since the vertex coordinates are ($\frac{-b}{2a}$, $\frac{ -b^2+4ac}{4a})$? – Slecker Sep 27 '18 at 4:46 • @Slecker I'm not familiar with the term "vertex form," but I would say that you are correct. – saulspatz Sep 27 '18 at 5:04 Look at $$2$$ Cartesian coordinate systems $$X,Y$$ and $$X',Y'$$. Origin of $$X',Y$$' is located at $$(x_0,y_0)$$, $$X'$$-axis parallel $$X$$-axis , $$Y'$$-axis parallel $$Y$$-axis(a translation),i.e. $$x= x_0+x'$$; $$y= y_0+ y'$$. Set up your normal parabola in the $$X',Y'$$ coordinate system. $$y'=ax'^2$$, vertex at $$(0',0')$$. Revert to original $$x,y$$ coordinates . $$y-y_0= a(x-x_0)^2$$ ; $$y=ax^2 -2(ax_0)x +ax_0^2$$. Compare with $$y =ax^2+bc +c$$: $$b=-2ax_0$$. Can you interpret? • I'm having a hard time understanding what you mean by "Revert to original $x$, $y$ coordinates" and where the subsequent equation, $y-y_0 = a(x-x_0)^2$, came from. I think once I understand that I can interpret the rest of your answer. – Slecker Sep 27 '18 at 15:22 • Slecker.Draw two coordinate systems, x,y and another one ,call it x',y'.Say, you put the origin of the x',y' system at x_0=3, y_0=4.x'y' system has its origin at (x_0,y_0)=(3,4), ok?. put a normal parabola y'=ax'^2 in the x',y' system.x'=1; y'=a; everything in x'y'.Take any x' coordinate, say x'=7, what is the x value in the original system: x= 7+ 3= x' +x_0 ok? Likewise y= y'+y_0. Solve for x' and y' and plug into y'=ax'2, get (y-y_0)=a(x-x_0)^2, now you are back in the original system.Your b =-2ax_0, where x_0 is the x-coordinate of the vertex.Let me know if ok. – Peter Szilas Sep 27 '18 at 17:56 • Ah ok thanks for the clarification! – Slecker Sep 27 '18 at 18:26 • Slecker. If anything else, just say so:) – Peter Szilas Sep 27 '18 at 18:35

https://math.stackexchange.com/questions/379698/cone-shaped-related-rates-of-change-question

# Cone shaped related rates of change question A container is in the shape of a cone of semi-vertical angle $30^\circ$, with it's vertex downwards. Liquid flows into the container at ${{\sqrt {3\pi } } \over 4}{\rm{ }}c{m^{^3}}/s$ At the instant when the radius of the circular surface of the liquid is 5 cm, find the rate of increase of: (a) The radius of the circular surface of the liquid (b) The area of the circular surface of the liquid My attempt: ${{dV} \over {dt}} = {{\sqrt {3\pi } } \over 4}$ (A) I need to find the rate at which the radius increases as "h" increases, so I have to find ${{dr} \over {dt}}$. The equation for the volume of a cone is: $V = {1 \over 3}\pi {r^2}h$ I now must form a function in terms of r for h. As we are asked to a compute when the radius is 5 we can form an equation using similar triangles, so: \eqalign{ & {h \over x} = {r \over 5} \cr & x: \cr & \tan 30^\circ = {5 \over x} \cr & x = {5 \over {\tan 30^\circ }} \cr & x = 5\sqrt 3 \cr & so: \cr & h = r\sqrt 3 \cr} \eqalign{ & V = {1 \over 3}\pi {r^2}(r\sqrt 3 ) \cr & V = \pi {r^3}{{\sqrt 3 } \over 3} \cr & {{dV} \over {dr}} = \pi {r^2}\sqrt 3 \cr & {{dr} \over {dt}} = {{dV} \over {dt}} \times {{dr} \over {dV}} \cr & {{dr} \over {dt}} = {{\sqrt {3\pi } } \over 4} \times {1 \over {\pi {r^2}\sqrt 3 }} \cr & {{dr} \over {dt}} = {{\sqrt {3\pi } } \over {\pi {r^2}4\sqrt 3 }} = {{\sqrt \pi } \over {\pi {r^2}4}} \cr} \eqalign{ & r = 5: \cr & {{\sqrt \pi } \over {4(25)\pi }} = {{\sqrt \pi } \over {100\pi }} = 0.005641... \cr} For part (A) the answer is stated as 0.01 cm/s, nowhere in the question have I been asked to round my answer, have I obtained the correct answer? I just want to make sure.. Part (b) Part (B) requires that I calculate the rate of increase of the circular area of the liquid, so essentially ${{dA} \over {dt}}$. Area of a circle is: $A = \pi {r^2}$ \eqalign{ & A = \pi {r^2} \cr & {{dA} \over {dr}} = 2\pi r \cr & {{dA} \over {dt}} = {{dr} \over {dt}} \times {{dA} \over {dr}} \cr & {{dA} \over {dt}} = {{\sqrt \pi } \over {4\pi {r^2}}} \times 2\pi r \cr & {{dA} \over {dt}} = {{\sqrt \pi } \over {2r}} \cr & r = 5: \cr & {{dA} \over {dt}} = 0.1\sqrt \pi {\rm{ c}}{{\rm{m}}^2}/s \cr} However the answer in the book for b is ${{dA} \over {dt}} = 0.1\pi {\rm{ c}}{{\rm{m}}^2}/s$ (pi is not square rooted), Where have I gone wrong? Furthermore I'd love it if any answerers could suggest how I could improve on how I've done things, and any tricks/tips that would make things easier for myself in the future. Thank you. • If you are going to use $x$, you should define it. In fact, you don't need to find $h$, but if you want to, $\frac rh=\tan 30^\circ=\frac {\sqrt 3} 3$, so $h=5\sqrt 3$, which you got. – Ross Millikan May 2 '13 at 22:37 • @RossMillikan, Oh okay, I'll bear that in mind, thank you! – seeker May 2 '13 at 22:54 "Liquid flows into the container at: $\;\displaystyle \dfrac{\sqrt{3}\,\pi}{4} = {{(\sqrt {3})\,\pi} \over 4}{\rm{ }}\text{ cm$^3$per sec}$" This would explain both discrepancies, (between both the solutions you obtained, and the solutions of the text), since your work seems to be fine, and was clearly done carefully. Simply changing your final evaluations using $\pi$ instead of $\sqrt{\pi}$ will yield the solutions you are given.

http://openstudy.com/updates/4dd977bed95c8b0b0e5565c4

1. watchmath The term of the absolute series is $$|\arctan(1/n)|/n^2<\frac{\pi/2}{n^2}$$. But the series $$\sum 1/n^2$$ is convergent (since it is a p-series with p=2), so the absolute series converges. Then the series is absolutely convergent. 2. anonymous how did you come up with pi/2/x^2 3. watchmath the range of the arctan function is between -pi/2 and pi/2 4. anonymous and why did you put over x^2 5. watchmath well I just you that arctan (1/n) < pi/2. The n^2 is already on the denominator on the first place. 6. anonymous so would you always compare it to pi/2 divided by denominator given in the problem 7. anonymous for tan 8. watchmath I wouldn't say always. I just see the opportunity that if we compare with pi/2 then I can have some conclusion. It is possible for a problem to have an arctan bu we don't compare with the pi/2. 9. anonymous and if there isnt anything in the denominator of the given problem you would compare it to pi/2 10. watchmath No, I won't if there is nothing on the bottom, I can compare arctan(1/n) < pi/2. But the series $$\sum pi/2$$ is divergent. So comparing with pi/2 doesn't give me any conclusion. 11. anonymous okay what would you comoare arcsin (1/n) to and 1-cos1/n to 12. watchmath the term of the series only arcsin(1/n) ? 13. anonymous yah 14. watchmath Just to make sure. So your series is $$\sum_{n=1}^\infty \arcsin(1/n)$$? I can't think how to do this right away... 15. anonymous yah 16. watchmath ok, arcsin is an increasing function and for positive x we have x > sin x Apply the arcsin arcsin x > x It follows that arcsin(1/n) > 1/n But the harmonic series $$\sum 1/n$$ is divergent Hence $$\sum \arcsin(1/n)$$ is divergent as well. 17. anonymous what if it was just sin 18. watchmath sin what? 19. anonymous if instead of arc sin it was sin would you still make the same comparison 20. watchmath you mean sin(1/n) ? 21. anonymous yah 22. watchmath well we can't compare to 1/n since 1/n > sin(1/n) So I don't know the answer yet. 23. anonymous oh okay 24. anonymous also if you were given sigma 1-cos(1/n) what would you compare it to 25. watchmath That is actually a nice problem. I'll post it as a new question so everybody can give a response (BTW it is convergent)

https://math.stackexchange.com/questions/1965174/power-set-of-a-set-with-an-empty-set

# Power set of a set with an empty set When a set has an empty set as an element, e.g.$\{\emptyset, a, b \}$. What is the powerset? Is it: $$\{ \emptyset, \{ \emptyset \}, \{a\}, \{b\}, \{\emptyset, a\} \{\emptyset, b\}, \{a, b\}, \{\emptyset, a, b\}\}$$ Or $$\{ \emptyset, \{a\}, \{b\}, \{\emptyset, a\} \{\emptyset, b\}, \{a, b\}, \{\emptyset, a, b\}\}$$ Or $$\{ \{\emptyset\}, \{a\}, \{b\}, \{\emptyset, a\} \{\emptyset, b\}, \{a, b\}, \{\emptyset, a, b\}\}$$ The confusion arises for me because, the powerset of every non-empty set has an empty set. Well the original set already has the empty set. So we don't need a subset with an empty set. Somehow, the first one seems correct. Yet, I can't seem to accept it. • The first one: $\;\emptyset\;$ is one of the elements of the given set, besides being a subset of it. – DonAntonio Oct 12 '16 at 12:39 • Let $c$ denote $\varnothing$. What is the power set of $\{a,b,c\}$? Now write $\varnothing$ instead of $c$ again. – Asaf Karagila Oct 12 '16 at 13:16 The first one is correct. This is because $\emptyset$ and $\{\emptyset\}$ are different. The first is an empty set whereas the second is a set whose only element is the empty set. Both are subsets of the given set. This is because the $\emptyset$ is the subset of every set, and as it happens to be an element of the given set, the set containing it as its element is also its subset. If a set $A$ is such that $\emptyset\in A$, its power set must necessarily contain these two sets: • $\emptyset$ (like all other power sets), corresponding to selecting nothing from $A$ (not even $\emptyset$, which is something) • $\{\emptyset\}$, corresponding to selecting $\emptyset$ only Therefore only the first of your proposed answers is correct, as you think. Your suggestions differ by having $\emptyset$ and/or $\{\emptyset\}$ included or not. • We have $\emptyset\in\mathcal P(X)$ because $\emptyset\subseteq X$ (which would hold for any other $X$ as well) • We have $\{\emptyset\}\in\mathcal P(X)$ because $\{\emptyset\}\subseteq X$ (which is the case because $\emptyset\in X$ in this specific problem) Therefore, your first variant is correct (and the other two are incorrect because $\emptyset\ne\{\emptyset\}$). • Your second bullet is strangely phrased to me; we have $\{\emptyset\}\in\mathcal{P}(X)$ simply because $\emptyset\in X$ in this specific problem. The fact that $\emptyset\subseteq X$, which is true for every set $X$, has nothing to do with it? – Inactive - avoiding CoC Oct 12 '16 at 15:47

https://gmatclub.com/forum/what-is-the-average-arithmetic-mean-of-eleven-consecutive-93188.html

GMAT Changed on April 16th - Read about the latest changes here It is currently 26 May 2018, 12:52 ### GMAT Club Daily Prep #### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History # Events & Promotions ###### Events & Promotions in June Open Detailed Calendar # What is the average (arithmetic mean) of eleven consecutive Author Message TAGS: ### Hide Tags Manager Joined: 27 Feb 2010 Posts: 101 Location: Denver What is the average (arithmetic mean) of eleven consecutive [#permalink] ### Show Tags 23 Apr 2010, 18:05 2 KUDOS 34 This post was BOOKMARKED 00:00 Difficulty: 25% (medium) Question Stats: 72% (00:43) correct 28% (00:46) wrong based on 690 sessions ### HideShow timer Statistics What is the average (arithmetic mean) of eleven consecutive integers? (1) The average of the first nine integers is 7 (2) The average of the last nine integers is 9 Math Expert Joined: 02 Sep 2009 Posts: 45455 What is the average (arithmetic mean) of eleven consecutive [#permalink] ### Show Tags 24 Apr 2010, 06:10 18 KUDOS Expert's post 20 This post was BOOKMARKED What is the average (arithmetic mean) of eleven consecutive integers? Consecutive integers represent evenly spaced set. For every evenly spaced set mean=median, in our case $$mean=median=x_6$$. (1) The average of the first nine integers is 7 --> $$x_1+x_2+...+x_9=63$$ --> there can be only one set of 9 consecutive integers to total 63. Sufficient. If you want to calculate: $$(x_6-5)+(x_6-4)+(x_6-3)+(x_6-2)+(x_6-1)+x_6+(x_6+1)+(x_6+2)+(x_6+3)=63$$ --> $$x_6=8$$. OR: Mean(=median of first 9 terms=5th term)*# of terms=63 --> $$x_5*9=63$$ --> $$x_5=7$$ --> $$x_6=7+1=8$$ (2) The average of the last nine integers is 9 --> $$x_3+x_4+...+x_{11}=81$$ --> there can be only one set of 9 consecutive integers to total 81. Sufficient. If you want to calculate: $$(x_6-3)+(x_6-2)+(x_6-1)+x_6+(x_6+1)+(x_6+2)+(x_6+3)+(x_6+4)+(x_6+5)=81$$ --> $$x_6=8$$. OR: Mean(=median of last 9 terms=7th term)*# of terms=81 --> $$x_7*9=81$$ --> $$x_7=9$$ --> $$x_6=9-1=8$$ _________________ Director Joined: 29 Nov 2012 Posts: 821 Re: If 11 consecutive integers are listed from least to [#permalink] ### Show Tags 08 Mar 2013, 05:51 so its not possible to have a list of numbers with positive and negative numbers? _________________ Click +1 Kudos if my post helped... Amazing Free video explanation for all Quant questions from OG 13 and much more http://www.gmatquantum.com/og13th/ GMAT Prep software What if scenarios http://gmatclub.com/forum/gmat-prep-software-analysis-and-what-if-scenarios-146146.html Math Expert Joined: 02 Sep 2009 Posts: 45455 Re: If 11 consecutive integers are listed from least to [#permalink] ### Show Tags 08 Mar 2013, 06:02 fozzzy wrote: What is the average (arithmetic mean) of eleven consecutive integers? (1) The average of the first nine integers is 7. (2) The average of the last nine integers is 9. so its not possible to have a list of numbers with positive and negative numbers? How it is possible? From both statements it follows that the set is {3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13}. _________________ Verbal Forum Moderator Joined: 10 Oct 2012 Posts: 620 Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink] ### Show Tags 12 Aug 2013, 04:39 10 KUDOS 18 This post was BOOKMARKED zz0vlb wrote: What is the average (arithmetic mean ) of eleven consecutive integers? (1) The avg of first nine integers is 7 (2) The avg of the last nine integers is 9 Here is a neat little trick for such kind of problems: Will be better illustrated using a numerical example: take the set {2,3,4,5,6}. Here the common difference (d)=1. The initial average = 4. Now, the averge of the set, after removing the last integer of the set(i.e. 6)will be reduced by exactly $$\frac{d}{2} units \to$$ The new Average = $$4-\frac{1}{2} = 3.5$$ Again, for the new set of {2,3,4,5} the average is 3.5 . Now, if the last integer is removed, the new average will again be = 3.5-0.5 = 3. Similarly, for the same set {2,3,4,5,6}, if we remove the first integer from the given set, the average increases by 0.5 and so on and so forth. Back to the problem: From F.S 1, we know that the average of the first 9 integers is 7. Thus, the average with the original 11 integers must have been 7+0.5+0.5 = 8. Sufficient. From F.S 2, we know that the average of the last 9 integers is 9, thus the average with the initial 11 integers must have been 9-0.5-0.5 = 8. Sufficient. D. _________________ Intern Joined: 26 May 2010 Posts: 10 Re: What is the average (arithmetic mean ) of eleven consecutive [#permalink] ### Show Tags 12 Aug 2013, 23:15 6 KUDOS 7 This post was BOOKMARKED zz0vlb wrote: What is the average (arithmetic mean ) of eleven consecutive integers? (1) The avg of first nine integers is 7 (2) The avg of the last nine integers is 9 As a general rule whenever there is a AP the average of the series is always the median of the series. Here it is a AP with difference 1 1. First 9 integers average is 7 . So the median that is the 5th digit is 7. Hence we can easily find the series and the average of the 11 consecutive digit series. Sufficient 2. Average of last 9 integers is 9 hence we know that for this subset of 9 integers the 5th integer would be 9 and we can find the series on the basis of this and the average. Sufficient And is D Manager Joined: 24 Jun 2014 Posts: 52 Concentration: Social Entrepreneurship, Nonprofit Re: What is the average (arithmetic mean) of eleven consecutive [#permalink] ### Show Tags 09 Mar 2015, 19:15 I considered following approach if the smallest number in set is x , then sum of 11 consecutive numbers = 11x+(1+2+...10)=11x+55--->A if largest number in set is x ,then sum of 11 consecutive numbers=11x-(1+2+10)=11x-55 Now as per statement 1 , average of first 9 numbers is 7 i.e sum =63 sum of 11 numbers =63+x+9+x+10----->B Equating A& B 11X+55=63+X+9+10 ,which can be solved to get x=3 statement I is sufficient similar approach for Statement II 11X-55=8+2X-19 ,can be solved to get X=13 statement 2 is sufficient OA=D EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 11670 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: 340 Q170 V170 Re: What is the average (arithmetic mean) of eleven consecutive [#permalink] ### Show Tags 09 Mar 2015, 19:26 5 KUDOS Expert's post 5 This post was BOOKMARKED Hi All, When you look at this question, if you find yourself unsure of where to "start", it might help to break down everything that you know into small pieces: 1st: We're told that we have 11 consecutive integers. That means the 11 numbers are whole numbers that are in a row. If we can figure out ANY of the numbers AND it's place "in line", then we can figure out ALL of the other numbers and answer the question that's asked (the average of all 11 = ?) 2nd: Fact 1 tells us that the average of the FIRST 9 integers is 7. For just a moment, ignore the fact that there are 9 consecutive integers and let's just focus on the average = 7. What would have to happen for a group of consecutive integers to have an average of 7? Here are some examples: 7 6, 7, 8 5, 6, 7, 8, 9 Notice how there are the SAME number of terms below 7 as above 7. THAT'S a pattern. With 9 total terms, that means there has to be 4 above and 4 below: 3, 4, 5, 6,.......7.......8, 9, 10, 11 Now we have enough information to figure out the other 2 terms (12 and 13) and answer the question. So Fact 1 is SUFFICIENT With this same approach, we can deal with Fact 2. The key to tackling most GMAT questions is to be comfortable breaking the prompt into logical pieces. Don't try to do every step at once and don't try to do work in your head. Think about what the information means, take the proper notes and be prepared to "play around" with a question if you're immediately certain about how to handle it. GMAT assassins aren't born, they're made, Rich _________________ 760+: Learn What GMAT Assassins Do to Score at the Highest Levels Contact Rich at: Rich.C@empowergmat.com # Rich Cohen Co-Founder & GMAT Assassin Special Offer: Save \$75 + GMAT Club Tests Free Official GMAT Exam Packs + 70 Pt. Improvement Guarantee www.empowergmat.com/ ***********************Select EMPOWERgmat Courses now include ALL 6 Official GMAC CATs!*********************** Intern Joined: 11 Apr 2016 Posts: 3 Re: What is the average (arithmetic mean) of eleven consecutive [#permalink] ### Show Tags 29 Jun 2016, 03:10 1 KUDOS wow such complex explanations for such a simple problem? given : 11 consec integers let them be x,x+1,x+2,...,x+10 Q: what is their mean? mean is (11x+55)/11 = x+5. Q becomes what is x+5 1) mean first 9 is 7. so (9x+36)/9 = x+4 = 7 , so x+5 =8 ,--> sufficient A or D 2) mean of last 9 is 9. so (9x+54)/9 = x+6= ---> x+5=8, sufficient . so D D Director Joined: 04 Jun 2016 Posts: 611 GMAT 1: 750 Q49 V43 Re: What is the average (arithmetic mean) of eleven consecutive [#permalink] ### Show Tags 15 Jul 2016, 00:10 zz0vlb wrote: What is the average (arithmetic mean) of eleven consecutive integers? (1) The average of the first nine integers is 7 (2) The average of the last nine integers is 9 For odd number of consecutive integer the mean and median both is the middle value. Use this property to solve th question (1) The average of the first nine integers is 7 7 will be the middle value; there will be 4 consecutive integers to the left and also to the right of 7 we will have {3,4,5,6,7,8,9,10,11} now we can add last two consecutive integer after 11, they will be 12,13 our new set will become = {3,4,5,6,7,8,9,10,11,12,13} again since the number of total elements in the set is odd, Mean will simply be the middle value = 8 SUFFICIENT (2) The average of the last nine integers is 9 Again number of element in the set are odd, 9 will be the middle value; 4 consecutive integers will lie to its left and right Middle value will be {5,6,7,8,9,10,11,12,13} Add 3,4 at the start of the set new set = {3,4,5,6,7,8,9,10,11,12,13} Mean will be 8 Sufficient _________________ Posting an answer without an explanation is "GOD COMPLEX". The world doesn't need any more gods. Please explain you answers properly. FINAL GOODBYE :- 17th SEPTEMBER 2016. .. 16 March 2017 - I am back but for all purposes please consider me semi-retired. BSchool Forum Moderator Joined: 12 Aug 2015 Posts: 2606 GRE 1: 323 Q169 V154 Re: What is the average (arithmetic mean) of eleven consecutive [#permalink] ### Show Tags 20 Dec 2016, 17:37 Nice Official Question> Here is my solution to this one => Set of consecutive integers => N N+1 N+2 . . . N+10 AP series with D=2 Hence Mean = Median = Average of the first and the last terms= N+5 So we just ned the value of N Statement 1 N N+1 . . N+8 Mean => N+4=7 Hence N+5=> 8 So the mean of the original set will be 8 Hence Sufficient Statement 2--> p p+2 p+3 . . p+8 Mean = p+4=9 p=5 Hene p+8=>13 So N+10=>13 Hence N+5=>8 Hence the mean of the original data set must be 8 Hence Sufficient Hence D _________________ MBA Financing:- INDIAN PUBLIC BANKS vs PRODIGY FINANCE! Getting into HOLLYWOOD with an MBA! The MOST AFFORDABLE MBA programs! STONECOLD's BRUTAL Mock Tests for GMAT-Quant(700+) AVERAGE GRE Scores At The Top Business Schools! Director Joined: 26 Oct 2016 Posts: 668 Location: United States Schools: HBS '19 GMAT 1: 770 Q51 V44 GPA: 4 WE: Education (Education) Re: What is the average (arithmetic mean) of eleven consecutive [#permalink] ### Show Tags 26 Jan 2017, 05:52 3 KUDOS The key is that the integers are consecutive. So, if we can determine any one of the 11 (and know where it falls), we can answer the question. (1) The average of the first 9 consecutive integers is 7. We know that avg = sum of terms / # of terms. So, 7 = sum of terms/9 sum of terms = 63. Well, there's only going to be one set of 9 consecutive integers that add up to 63. If we can determine the first 9, we can certainly determine the last 2: sufficient. (2) The average of the last 9 terms is 9. Exact same reasoning as (1): sufficient. Both (1) and (2) are sufficient: choose (D). _________________ Thanks & Regards, Anaira Mitch Intern Joined: 04 Dec 2016 Posts: 2 Re: What is the average (arithmetic mean) of eleven consecutive [#permalink] ### Show Tags 13 Mar 2017, 05:39 The trick here is to catch the phase "11 consecutive integers". We know that an odd set of consecutive integers have the same median and mean {i.e. set 1,2,3 has a median and mean of 2}. Based on this we can say the same for the statements: s1) Represent the first 9 integers as: A+B+C+D+E+F+G+H+I. If the mean of this set is 7 then the median is also 7 so we found that the 7th number in the total set. We know they are consecutive and therefore we could count forward and backward to get the unknown numbers. Therefore statement is sufficient. s2) Same concept as above. Statement is sufficient. SVP Joined: 12 Sep 2015 Posts: 2477 Re: What is the average (arithmetic mean) of eleven consecutive [#permalink] ### Show Tags 24 Mar 2018, 07:32 1 KUDOS Expert's post Top Contributor zz0vlb wrote: What is the average (arithmetic mean) of eleven consecutive integers? (1) The average of the first nine integers is 7 (2) The average of the last nine integers is 9 There's a nice rule that says, "In a set where the numbers are equally spaced, the mean will equal the median." Since the consecutive integers are equally-spaced, their mean and median will be equal. Target question: What is the average of eleven consecutive integers? Statement 1: The average of the first nine integers is 7. This also tells us that the MEDIAN of the first nine integers is 7. In other words, the MIDDLEMOST value is 7. This means, the first nine integers are 3, 4, 5, 6, 7, 8, 9, 10, 11 So, ALL 11 integers must be 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 Since we've identified all 11 integers, we can DEFINITELY find their average. Since we can answer the target question with certainty, statement 1 is SUFFICIENT Statement 2: The average of the last night integers is 9 This also tells us that the MEDIAN of the last nine integers is 9. In other words, the MIDDLEMOST value is 9. This means, the last nine integers are 5, 6, 7, 8, 9, 10, 11, 12, 13 So, ALL 11 integers must be 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 Since we've identified all 11 integers, we can DEFINITELY find their average. Since we can answer the target question with certainty, statement 2 is SUFFICIENT RELATED VIDEO _________________ Brent Hanneson – Founder of gmatprepnow.com Re: What is the average (arithmetic mean) of eleven consecutive   [#permalink] 24 Mar 2018, 07:32 Display posts from previous: Sort by

https://math.stackexchange.com/questions/2340800/integration-of-secant

# Integration of secant \begin{align} \int \sec x \, dx &= \int \cos x \left( \frac{1}{\cos^2x} \right) \, dx \\ &= \int \cos x \left( \frac{1}{1-\sin^2x} \right) \, dx \\ & = \int\cos x\cdot\frac{1}{1-\frac{1-\cos2x}{2}} \, dx \\ &= \int \cos x \cdot\frac{2}{1+\cos2x} \, dx \end{align} I am stuck in here. Any help to integrate secant? ## 5 Answers \begin{align*}\int\sec x\,\mathrm dx&=\int\frac1{\cos x}\,\mathrm dx\\&=\int\frac{\cos x}{\cos^2x}\,\mathrm dx\\&=\int\frac{\cos x}{1-\sin^2x}\,\mathrm dx.\end{align*} Now, doing $$\sin x=t$$ and $$\cos x\,\mathrm dx=\mathrm dt$$, you get $$\displaystyle\int\frac{\mathrm dt}{1-t^2}$$. But\begin{align*}\int\frac{\mathrm dt}{1-t^2}&=\frac12\int\frac1{1-t}+\frac1{1+t}\,\mathrm dt\\&=\frac12\left(-\log|1-t|+\log|1+t|\right)\\&=\frac12\log\left|\frac{1+t}{1-t}\right|\\&=\frac12\log\left|\frac{(1+t)^2}{1-t^2}\right|\\&=\log\left|\frac{1+t}{\sqrt{1-t^2}}\right|\\&=\log\left|\frac{1+\sin x}{\sqrt{1-\sin^2x}}\right|\\&=\log\left|\frac1{\cos x}+\frac{\sin x}{\cos x}\right|\\&=\log|\sec x+\tan x|.\end{align*} • What a tricky..! Thx – Beverlie Jun 29 '17 at 15:25 • I've been reading about the early history of calculus. For a long period in the 17th century this was a significant unsolved problem. – DanielWainfleet Aug 3 '17 at 17:56 • @DanielWainfleet I think I read something about that in the historical notes of Spivak's Calculus. – José Carlos Santos Aug 3 '17 at 17:59 An alternative method: The trick here is to multiply $\sec{x}$ by $\dfrac{\tan{x}+\sec{x}}{\tan{x}+\sec{x}}$, then substitute $u=\tan{x}+\sec{x}$ and $du=(\sec^2{x}+\tan{x}\sec{x})~dx$: $$\int \sec{x}~dx=\int \sec{x}\cdot \frac{\tan{x}+\sec{x}}{\tan{x}+\sec{x}}~dx=\int \frac{\sec{x}\tan{x}+\sec^2{x}}{\tan{x}+\sec{x}}~dx=\int \frac{1}{u}~du=\cdots$$ Not obvious, though it is efficient. After $\int \cos x \left(\frac{1}{1-\sin^2x}\right)dx$ use the transformation $z = \sin x$ and $dz = \cos x \, dx$. Edit: $$\int\frac{1}{1-u^2}\,du = \frac{1}{2}\int\frac{(1+u)+(1-u)}{(1+u)(1-u)} = \frac{1}{2} \int \frac{1}{1+u} + \frac{1}{1-u}\,du$$ And use, $\int \frac{1}{u}\,du = \ln|u|$ • I'd got $\int \frac{1}{1-u^2}du$ what would be the next step? – Beverlie Jun 29 '17 at 15:22 • Use partial fraction method as in my edit. – Dhruv Kohli - expiTTp1z0 Jun 29 '17 at 15:26 Although the integral can be evaluated in a straightforward way using real analysis, I thought it might be instructive to present an approach based on complex analysis. To that end, we now proceed. We use Euler's Formula, $e^{ix}=\cos(x)+i\sin(x)$, to write $\displaystyle \sec(x)=\frac2{e^{ix}+e^{-ix}}=\frac{2e^{ix}}{1+e^{i2x}}$. Then, we have \begin{align} \int \sec(x)\,dx&=\int \frac2{e^{ix}+e^{-ix}}\\\\ &=\int \frac{2e^{ix}}{1+e^{i2x}}\,dx \\\\ &=-i2 \int \frac{1}{1+(e^{ix})^2}\,d(e^{ix})\\\\ &=-i2 \arctan(e^{ix})+C\tag 1\\\\ &=\log\left(\frac{1-ie^{ix}}{1+ie^{ix}}\right)+C\tag2\\\\ &=\log\left(-i\left(\frac{1+\sin(x)}{i\cos(x)}\right)\right)+C\tag3\\\\ &=\log(\sec(x)+\tan(x))+C'\tag4 \end{align} NOTES: In going from $(1)$ to $(2)$, we used the identity $\arctan(z)=i2\log\left(\frac{1-iz}{1+iz}\right)$ In going from $(2)$ to $(3)$, we multiplied the numerator and denominator of the argument of the logarithm function by $1-ie^{ix}$. Then, we used $$\frac{1-ie^{ix}}{1+ie^{ix}}=\frac{-i2\cos(x)}{2(1-\sin(x))}=-i\frac{1+\sin(x)}{\cos(x)}$$ Finally, in going from $(3)$ to $(4)$, we absorbed the term $\log(-i)$ into the integration constant $C$ and labeled the new integration constant $C'=C+\log(-i)$. Just to spell out Lord Shark the Unknown's suggestion, $$t=\tan\frac{x}{2}\implies\sec x=\frac{1+t^2}{1-t^2},\,dx=\frac{2dt}{1+t^2}\implies\int\sec xdx=\int\frac{2 dt}{1-t^2}.$$From that point on, the same partial-fractions treatment as in multiple other answers can be used. Admittedly the expression thus obtained for the antiderivative is $$\ln\left|\frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}}\right|+C$$ instead of $$\ln\left|\frac{1+\sin x}{1-\sin x}\right|+C$$ or $$\ln|\sec x+\tan x|+C$$, but of course they're all the same thanks to suitable trigonometric identities (again, obtainable by writing things as functions of $$\tan\frac{x}{2}$$).

http://manoisparduotuve.lt/i-hate-oxjqr/article.php?tag=f1577f-fibonacci-sequence-equation

But you might be surprised because nature seems to favor a particular numbers like 1, 2, 3, 5, 8, 13, 21 and 34. Lucas Sequences The above work on the Fibonacci sequence can be generalized to discuss any difference equation of the form where and can be any real numbers. The Fibonacci sequence is a series where the next term is the sum of pervious two terms. Is there an easier way? Writing, the other root is, and the constants making are. The Explicit Formula for Fibonacci Sequence First, let's write out the recursive formula: a n + 2 = a n + 1 + a n a_{n+2}=a_{n+1}+a_n a n + 2 = a n + 1 + a n where a 1 = 1 , a 2 = 1 a_{ 1 }=1,\quad a_2=1 a 1 = 1 , a 2 = 1 You might think that any number is possible. How does this Fibonacci calculator work? It goes by the name of golden ratio, which deserves its own separate article.). They hold a special place in almost every mathematician's heart. Fibonacci numbers are one of the most captivating things in mathematics. Fibonacci Sequence is a wonderful series of numbers that could start with 0 or 1. For example, in the Fibonacci sequence 1, 1, 2, 3, 5, 8, 13,... 2 is found by adding the two numbers before it, 1+1=2. The nth term of a Fibonacci sequence is found by adding up the two Fibonacci numbers before it. Try it again. If we have an infinite series, $$S = 1 + ax + (ax)^2 + (ax)^3 + \cdots,$$, with $|ax| < 1$, then its sum is given by, This means, if the sum of an infinite geometric series is finite, we can always have the following equality -, $$\frac{1}{1 - ax} = 1 + ax + (ax)^2 + (ax)^3 + \cdots = \sum_{n \ge 0} a^n x^n$$, Using this idea, we can write the expression of $F(x)$ as, $$F(x) = \frac{1}{(\alpha - \beta)}\left(\frac{1}{1-x\alpha} - \frac{1}{1-x\beta} \right) = \frac{1}{\sqrt{5}} \left(\sum_{n \ge 0 } x^n\alpha^n - \sum_{n \ge 0 } x^n \beta^n \right)$$, Recalling the original definition of $F(x)$, we can finally write the following equality, $$F(x) = \sum_{n \ge 0}F_n x^n = \frac{1}{\sqrt{5}} \left(\sum_{n \ge 0 } x^n\alpha^n - \sum_{n \ge 0 } x^n \beta^n \right),$$, and comparing the $n-$th terms on both sides, we get a nice result, $$F_n = \frac{1}{\sqrt{5}} \left(\alpha^n - \beta^n \right),$$, (This number $\alpha$ is also a very interesting number in itself. Yes, there is an exact formula for the n … Thus the sequence begins: 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …. Throughout history, people have done a lot of research around these numbers, and as a result, quite a lot … . F(n) = F(n+2) - F(n+1) F(n-1) = F(n+1) - F(n) . In mathematical terms, the sequence F n of all Fibonacci numbers is defined by the recurrence relation. Yes, it is possible but there is an easy way to do it. So, with the help of Golden Ratio, we can find the Fibonacci numbers in the sequence. Fibonacci spiral is also considered as one of the approximates of the golden spiral. Here is a short list of the Fibonacci sequence: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233 Each number in the sequence is the sum of the two numbers before it We can try to derive a Fibonacci sequence formula by making some observations So, the sequence goes as 0, 1, 1, 2, 3, 5, 8, 13, 21, 34, and so on. Fibonacci omitted the first term (1) in Liber Abaci. Fibonacci sequence formula Golden ratio convergence A natural derivation of the Binet's Formula, the explicit equation for the Fibonacci Sequence. See more ideas about fibonacci, fibonacci spiral, fibonacci sequence. Follow me elsewhere: Twitter: https://twitter.com/RecurringRoot This will give you the second number in the sequence. There is a special relationship between the Golden Ratio and the Fibonacci Sequence:. Fibonacci number - elements of a numerical sequence in which the first two numbers are either 1 and 1, or 0 and 1, and each subsequent number is equal to the sum of the two previous numbers. Specifically, we have noted that the Fibonacci sequence is a linear recurrence relation — it can be viewed as repeatedly applying a linear map. Jacques Philippe Marie Binet was a French mathematician, physicist, and astronomer born in Rennes. So, for n>1, we have: f₀ = 0, f₁ = 1, This short project is an implementation of the formula in C. Binet's Formula . A Closed Form of the Fibonacci Sequence Fold Unfold. By the above formula, the Fibonacci number can be calculated in . Fibonacci initially came up with the sequence in order to model the population of rabbits. # first two terms n1, n2 = 0, 1 count = 0 # check if the number of terms is valid if nterms <= 0: print("Please enter a positive integer") elif nterms == 1: print("Fibonacci sequence upto",nterms,":") print(n1) else: print("Fibonacci sequence:") while count < nterms: print(n1) nth = n1 + n2 # update values n1 = n2 n2 = … The third number in the sequence is the first two numbers added together (0 + 1 = 1). The Fibonacci Sequence is one of the cornerstones of the math world. Fibonacci spiral is also considered as one of the approximates of the golden spiral. The answer key is below. Python Fibonacci Sequence: Iterative Approach. The Fibonacci sequence exhibits a certain numerical pattern which originated as the answer to an exercise in the first ever high school algebra text. If we make the replacement. Problems to be Submitted: Problem 10. Keywords and phrases: Generalized Fibonacci sequence, Binet’s formula. I know that the relationship is that the "sum of the squares of the first n terms is the nth term multiplied by the (nth+1) term", but I don't think that is worded right? Each number in the sequence is the sum of the two previous numbers. The first two numbers of the Fibonacci series are 0 and 1. Leonardo Fibonacci was one of the most influential mathematician of the middle ages because Hindu Arabic Numeral System which we still used today was popularized in the Western world through his book Liber Abaci or book of calculations. This sequence of Fibonacci numbers arises all over mathematics and also in nature. In reality, rabbits do not breed this… With this insight, we observed that the matrix of the linear map is non-diagonal, which makes repeated execution tedious; diagonal matrices, on the other hand, are easy to multiply. Assuming "Fibonacci sequence" is an integer sequence | Use as referring to a mathematical definition or referring to a type of number instead. The recurrence formula for these numbers is: F(0) = 0 F(1) = 1 F(n) = F(n − 1) + F(n − 2) n > 1 . I have been learning about the Fibonacci Numbers and I have been given the task to research on it. Get all the latest & greatest posts delivered straight to your inbox, © 2020 Physics Garage. The Fibonacci numbers, denoted fₙ, are the numbers that form a sequence, called the Fibonacci sequence, such that each number is the sum of the two preceding ones. The first two numbers are defined to be 0, 1. where $n$ is a positive integer greater than $1$, $F_n$ is the $n-$th Fibonacci number with $F_0 = 0$ and $F_1=1$. I have been assigned to decribe the relationship between the photo (attached below). The Fibonacci numbers are generated by setting F 0 = 0, F 1 = 1, and then using the recursive formula. In mathematics, the Fibonacci numbers form a sequence defined recursively by: = {= = − + − > That is, after two starting values, each number is the sum of the two preceding numbers. Male or Female ? Each number is the product of the previous two numbers in the sequence. They hold a special place in almost every mathematician's heart. Generate Fibonacci sequence (Simple Method) In the Fibonacci sequence except for the first two terms of the sequence, every other term is the sum of the previous two terms. So, … Fibonacci formula: f … It may seem coincidence to you but it's actually forming a pattern - Fibonacci Sequence. Each number in the sequence is the sum of the two numbers that precede it. Add the first term (1) and 0. Binet's Formula is an explicit formula used to find the nth term of the Fibonacci sequence. Stay up to date! The Fibonacci sequence typically has first two terms equal to F₀ = 0 and F₁ = 1. Male Female Age Under 20 years old 20 years old level 30 years old level 40 years old level 50 years old level 60 years old level or over Occupation Elementary school/ Junior high-school student The answer comes out as a whole number, exactly equal to the addition of the previous two terms. Assuming "Fibonacci sequence" is an integer sequence | Use as referring to a mathematical definition or referring to a type of number instead. Unlike in an arithmetic sequence, you need to know at least two consecutive terms to figure out the rest of the sequence. The mathematical equation describing it is An+2= An+1 + An. Browse other questions tagged sequences-and-series fibonacci-numbers or ask your own question. . The first two numbers are defined to be 0, 1. This is the general form for the nth Fibonacci number. In this book, Fibonacci post and solve a problem involving the growth of population of rabbits based on idealized assumptions. The Golden Ratio formula is: F(n) = (x^n – (1-x)^n)/(x – (1-x)) where x = (1+sqrt 5)/2 ~ 1.618. To improve this 'Fibonacci sequence Calculator', please fill in questionnaire. The Explicit Formula for Fibonacci Sequence First, let's write out the recursive formula: a n + 2 = a n + 1 + a n a_{n+2}=a_{n+1}+a_n a n + 2 = a n + 1 + a n where a 1 = 1 , a 2 = 1 a_{ 1 }=1,\quad a_2=1 a 1 = 1 , a 2 = 1 A natural derivation of the Binet's Formula, the explicit equation for the Fibonacci Sequence. Where, φ is the Golden Ratio, which is approximately equal to the value 1.618. n is the nth term of the Fibonacci sequence To calculate each successive Fibonacci number in the Fibonacci series, use the formula where is th Fibonacci number in the sequence, and the first … Next, we multiply the last equation by $x_n$ to get, $$x^n \cdot F_{n+1} = x^n \cdot F_n + x^n \cdot F_{n-1},$$, $$\sum_{n \ge 1}x^n \cdot F_{n+1} = \sum_{n \ge 1} x^n \cdot F_n + \sum_{n \ge 1} x^n \cdot F_{n-1}$$, Let us first consider the left hand side -, $$\sum_{n \ge 1} x^n \cdot F_{n+1} = x \cdot F_2 + x^2 \cdot F_3 + \cdots$$, Now, we try to represent this expansion in terms of $F(x)$, by doing the following simple manipulations -, $$\frac{1}{x} \left( x^2 \cdot F_2 + x^3 \cdot F_3 + \cdots \right)$$, $$\frac{1}{x} \left(- x \cdot F_1 + x \cdot F_1 + x^2 \cdot F_2 + x^3 \cdot F_3 + \cdots \right)$$, Using the definition of $F(x)$, this expression can now be written as, $$\frac{1}{x} \left(- x \cdot F_1 + F(x)\right)$$, Therefore, using the fact that $F_1=1$, we can write the entire left hand side as, $$\sum_{n \ge 1} x^n \cdot F_{n+1} = x \cdot F_2 + x^2 \cdot F_3 + \cdots = \frac{F(x) - x}{x}$$, $$\sum_{n \ge 1}x^n \cdot F_n + \sum_{n \ge 1} x^n \cdot F_{n-1}.$$, $$\left( x \cdot F_1 + x^2 \cdot F_2 + \cdots \right ) + \left( x^2 \cdot F_1 + x^3 \cdot F_2 + \cdots \right)$$. This equation calculates numbers in the Fibonacci sequence (Fn) by adding together the previous number in the series (Fn-1) with the number previous to that (Fn-2). x(n-2) is the term before the last one. Male or Female ? The powers of phi are the negative powers of Phi. Fibonacci Number Formula. . Alternatively, you can choose F₁ = 1 and F₂ = 1 as the sequence starters. Generalized Fibonacci sequence is defined by recurrence relation F pF qF k with k k k t 12 F a F b 01,2, Male Female Age Under 20 years old 20 years old level 30 years old level 40 years old level 50 years old level 60 years old level or over Occupation Elementary school/ Junior high-school student By the above formula, the Fibonacci number can be calculated in . Computing Fibonacci number by exponentiation. Table of Contents. If F(n) represents the nth Fibonacci number, then: F(n) = (a^n - b^n)/(a - b) where a and b are the two roots of the quadratic equation x^2-x-1 = 0. Instead, it would be nice if a closed form formula for the sequence of numbers in the Fibonacci sequence existed. We can also use the derived formula below. For example, in the Fibonacci sequence 1, 1, 2, 3, 5, 8, 13,... 2 is found by adding the two numbers before it, 1+1=2. Francis Niño Moncada on October 01, 2020: Jomar Kristoffer Besayte on October 01, 2020: Mary Kris Banaynal on September 22, 2020: Ace Victor A. Acena on September 22, 2020: Andrea Nicole Villa on September 22, 2020: Claudette Marie Bonagua on September 22, 2020: Shaira A. Golondrina on September 22, 2020: Diana Rose A. Orillana on September 22, 2020: Luis Gabriel Alidogan on September 22, 2020: Grace Ann G. Mohametano on September 22, 2020. Instead, it would be nice if a closed form formula for the sequence of numbers in the Fibonacci sequence existed. A sequence derived from this equation is often called a Lucas sequence, named for French mathematician Edouard Lucas. The rule for calculating the next number in the sequence is: x(n) = x(n-1) + x(n-2) x(n) is the next number in the sequence. to get the rest. Remember, to find any given number in the Fibonacci sequence, you simply add the two previous numbers in the sequence. The first two terms of the Fibonacci sequence is 0 followed by 1. To calculate each successive Fibonacci number in the Fibonacci series, use the formula where is th Fibonacci number in the sequence, and the first … Fibonacci initially came up with the sequence in order to model the population of rabbits. In his memoir in the theory of conjugate axis and the moment of inertia of bodies, he enumerated the principle which is known now as Binet's Theorem. The Fibonacci sequence is a series where the next term is the sum of pervious two terms. A Fibonacci spiral having an initial radius of 1 has a polar equation similar to that of other logarithmic spirals . F n – 1 and F n – 2 are the (n-1) th and (n – 2) th terms respectively Also Check: Fibonacci Calculator. Solution for 88. $$0, 1, 1, 2, 3, 5, 8, 13 ,21, 34, 55, \cdots$$, Any number in this sequence is the sum of the previous two numbers, and this pattern is mathematically written as. Fibonacci Sequence. This pattern turned out to have an interest and … The Fibonacci Sequence is a series of numbers. The characteristic equation is, with roots. Although Fibonacci only gave the sequence, he obviously knew that the nth number of his sequence was the sum of the two previous numbers (Scotta and Marketos). # Program to display the Fibonacci sequence up to n-th term nterms = int(input("How many terms? ")) Abstract. A Fibonacci spiral having an initial radius of 1 has a polar equation similar to that of other logarithmic spirals . The Fibonacci series is a very famous series in mathematics. A Closed Form of the Fibonacci Sequence Fold Unfold. Example 2: Find the 25th term of the Fibonacci sequence: 1, 1, 2, 3, 5, 8, ... Answer: Since you're looking for the 25th term, n = 25. Fibonacci number is defined by: Obviously, Fibonacci sequence is a difference equation (in above example) and it could be written in: Matrix Form. Computing Fibonacci number by exponentiation. In this tutorial I will show you how to generate the Fibonacci sequence in Python using a few methods. You can use the Binet's formula in in finding the nth term of a Fibonacci sequence without the other terms. The nth term of a Fibonacci sequence is found by adding up the two Fibonacci numbers before it. Now, this expression is fairly easy to understand and quite sufficient to produce any Fibonacci number by plugging the required value of $n$. By taking out a factor of $x$ from the second expansion, we get, $$\left( x \cdot F_1 + x^2 \cdot F_2 + \cdots \right ) + x \left( x \cdot F_1 + x^2 \cdot F_2 + \cdots \right).$$, Using the definition of $F(x)$, this can finally be written as. This pattern turned out to have an interest and … Our job is to find an explicit form of the function, $F(x)$, such that the coefficients, $F_n$ are the Fibonacci numbers. I know that the relationship is that the "sum of the squares of the first n terms is the nth term multiplied by the (nth+1) term", but I don't think that is worded right? If you got 4 correct answers: You made it! ( Using power of the matrix {{1,1},{1,0}} ) This another O(n) which relies on the fact that if we n times … If you got between 0 and 1 correct answer: You can do it next time. It is so named because it was derived by mathematician Jacques Philippe Marie Binet, though it was already known by Abraham de Moivre. Fibonacci sequence is a sequence of numbers, where each number is the sum of the 2 previous numbers, except the first two numbers that are 0 and 1. The equation is a variation on Pell's, in that x^2 - ny^2 = +/- 4 instead of 1. Fibonacci number is defined by: Obviously, Fibonacci sequence is a difference equation (in above example) and it could be written in: Matrix Form. The problem yields the ‘Fibonacci sequence’: 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377 . Let us define a function $F(x)$, such that it can be expanded in a power series like this, $$F(x) = \sum_{n \ge 0}x^n F_n = x \cdot F_1 + x^2 \cdot F_2 + \cdots$$. Following the same pattern, 3 is found by adding 1 and 2, 5 is found by adding 2 and 3 and so on. So to calculate the 100th Fibonacci number, for instance, we need to compute all the 99 values before it first - quite a task, even with a calculator! F n = F n-1 + F n-2. The Fibonacci formula is used to generate Fibonacci in a recursive sequence. Fibonacci Sequence. Therefore, by equating the left and the right hand sides, the original formula can be re-written in terms of $F(x)$ as, $$\frac{F(x) - x}{x} = F(x) + xF(x) ~~ \Longrightarrow ~~ F(x) = \frac{x}{1-x-x^2}$$, Let us now simplify this expression a bit more. Mar 12, 2018 - Explore Kantilal Parshotam's board "Fibonacci formula" on Pinterest. The formula to calculate the Fibonacci numbers using the Golden Ratio is: X n = [φ n – (1-φ) n]/√5. Fibonacci Sequence is popularized in Europe by Leonardo of Pisa, famously known as "Leonardo Fibonacci".Leonardo Fibonacci was one of the most influential mathematician of the middle ages because Hindu Arabic Numeral System which we still used today was popularized in the Western world through his book Liber Abaci or book of calculations. If you got between 2 and 3 correct answers: Maybe you just need more practice. If we expand the by taking in above example, then. From this we find the formula, valid for all, and one desired continuous extension is clearly the real part Derivation of Fibonacci sequence . Fibonacci Sequence. THE FIBONACCI SEQUENCE, SPIRALS AND THE GOLDEN MEAN. It is not hard to imagine that if we need a number that is far ahead into the sequence, we will have to do a lot of "back" calculations, which might be tedious. However, if I wanted the 100th term of this sequence, it would take lots of intermediate calculations with the recursive formula to get a result. . Fibonacci Formula. Observe the following Fibonacci series: Derivation of Fibonacci sequence . The standard formula for the Fibonacci numbers is due to a French mathematician named Binet. Example 1: Find the 10th term of the Fibonacci sequence: 1, 1, 2, 3, 5, 8, ... Answer: Since you're looking for the 10th term, n = 10. In order to make use of this function, first we have to rearrange the original formula. The sequence starts like this: 0, 1, 1, 2, 3, 4, 8, 13, 21, 34 (Issues regarding the convergence and uniqueness of the series are beyond the scope of the article). This sequence of Fibonacci numbers arises all over mathematics and also in nature. Number Theory > Special Numbers > Fibonacci Numbers > Binet's Fibonacci Number Formula Binet's formula is a special case of the Binet form with , corresponding to the th Fibonacci … . To improve this 'Fibonacci sequence Calculator', please fill in questionnaire. And even more surprising is that we can calculate any Fibonacci Number using the Golden Ratio: x n = φn − (1−φ)n √5. In this article, we are going to discuss another formula to obtain any Fibonacci number in the sequence, which might (arguably) be easier to work with. You can calculate the Fibonacci Sequence by starting with 0 and 1 and adding the previous two numbers, but Binet's Formula can be used to calculate directly any term of the sequence. Subscribe to the newsletter to receive more stories mailed directly to your inbox, The methods of finding roots of a quadratic equations are quite easy and are very well understood. Using The Golden Ratio to Calculate Fibonacci Numbers. In reality, rabbits do not breed this… The first two terms of the Fibonacci sequence is 0 followed by 1. The Fibonacci numbers are the sequence of numbers defined by the linear recurrence equation (1) Fibonacci Series Formula. Forty years ago I discovered that the Fibonacci Sequence (1, 1, 2, 3, 5, 8, etc) can be generated from the second degree Diophantine equation 5k^2 -/+ 4 = m^2 where the -,+ is taken alternately. The Fibonacci sequence was defined in Section 11.1 by the equations fi = 1, f2= 1, fn= fn=1 + fn-2 n> 3 %3D %3D Show that each of the following… Get the best viral stories straight into your inbox! He died in Paris, France in 1856. So, for n>1, we have: f₀ = 0, f₁ = 1, Let’s start by talking about the iterative approach to implementing the Fibonacci series. To create the sequence, you should think of 0 … Fibonacci number - elements of a numerical sequence in which the first two numbers are either 1 and 1, or 0 and 1, and each subsequent number is equal to the sum of the two previous numbers. The authors would like to thank Prof. Ayman Badawi for his fruitful suggestions. The Fibonacci sequence is one of the most famous formulas in mathematics. 1 Binet's Formula for the nth Fibonacci number We have only defined the nth Fibonacci number in terms of the two before it: the n-th Fibonacci number is the sum of the (n-1)th and the (n-2)th. In this paper, we present properties of Generalized Fibonacci sequences. where: a = (F₁ - F₀ψ) / √5 b = (φF₀ - F₁) / √5 F₀ is the first term of the sequence, F₁ is the second term of the sequence. But what if you are asked to find the 100th term of a Fibonacci sequence, are you going to add the Fibonacci numbers consecutively until you get the 100th term? THE FIBONACCI SEQUENCE, SPIRALS AND THE GOLDEN MEAN. Fibonacci sequence equation. If at all, its only drawback is that, if we want to know a particular number, $F_n$ in the sequence, we need two numbers $F_{n-1}$ and $F_{n-2}$ that came before it; that's just how this formula works. F n = n th term of the series. Another way to write the equation is: Therefore, phi = 0.618 and 1/Phi. There are all kinds of approaches available, like, Ptolemy was an ancient astronomer, geographer, and mathematician who lived from (c. AD 100 – c. 170). In the case of the Fibonacci sequence, the recurrence is, with initial conditions. We can see from the following table, that by plugging the values of $n$, we can directly find all Fibonacci numbers! I have been assigned to decribe the relationship between the photo (attached below). For each question, choose the best answer. Featured on Meta “Question closed” notifications experiment results and graduation Definition The Fibonacci sequence begins with the numbers 0 and 1. x(n-1) is the previous term. . In mathematics, the Fibonacci sequence is defined as a number sequence having the particularity that the first two numbers are 0 and 1, and that each subsequent number is obtained by the sum of the previous two terms. Fibonacci Sequence is popularized in Europe by Leonardo of Pisa, famously known as "Leonardo Fibonacci". Scope of the formula in C. Binet 's formula, the recurrence is, and then using the recursive.! Liber Abaci is so named because it was derived by mathematician Jacques Philippe Marie Binet, though was. N = n th term of the Golden MEAN the Binet 's formula in C. Binet 's formula C.... Numbers arises all over mathematics and also in nature he made significant contributions to number theory and the sequence... 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First two numbers are generated by adding up the two Fibonacci numbers before it famous formulas in mathematics phi... Delivered straight to your inbox viral stories straight into your inbox, © 2020 Physics Garage & greatest posts straight. Columbia Forest Products Corporate Office, Graphic Design Today, Sleepwell Mattress 4 Inch Price Double Bed, Light Or Dark Countertops With White Cabinets, Cheapest Apartment In Dubai For Sale, Stouffer's Fit Kitchen Bourbon Steak, Whydah Pirate Museum Coupon, Do Cats Eat Bread,

https://math.stackexchange.com/questions/2540852/what-is-the-dimension-of-two-subspaces

# What is the dimension of two subspaces? Let $V$ be the vector space with a basis $x_1, \ldots, x_9$ and $$V_1 = \{(a,a,a,b,b,b,c,c,c): a,b,c \in \mathbb{C}\}, \\ V_2=\{(a,b,c,a,b,c,a,b,c): a,b,c \in \mathbb{C}\}.$$ Then $V_1,V_2$ are subspaces of $V$. What is the dimension of $V_1 \cap V_2$? I first try to find the system of equations which give $V_1, V_2$ respectively. I think that $V_1$ is $$\{(k_1,k_2,k_3,k_4,k_5,k_6,k_7,k_8,k_9): k_1-k_2=0,k_2-k3=0,k_4-k_5=0,k_5-k_6=0, k_7-k_8=0,k_8-k_9=0 \}$$ and $V_2$ is $$\{(k_1,k_2,k_3,k_4,k_5,k_6,k_7,k_8,k_9): k_1-k_4=0,k_4-k7=0,k_2-k_5=0,k_5-k_8=0, k_3-k_6=0,k_6-k_9=0 \}.$$ By solving the collection of the equations for $V_1$ and $V_2$, I obtain the solutions $k_1=\cdots =k_9$. So the dimension of $V_1 \cap V_2$ is one dimensional. Is this correct? Thank you very much. It is correct but, in my opinion, that approach is too complex for the problem. Take $(a,b,c,d,e,f,g,h,i)\in V_1\cap V_2$. Then • Since $(a,b,c,d,e,f,g,h,i)\in V_1$, $b=c=a$, $e=f=d$, and $h=i=g$. Therefore$$(a,b,c,d,e,f,g,h,i)=(a,a,a,d,d,d,g,g,g).$$ • Since $(a,a,a,d,d,d,g,g,g)\in V_2$, $d=g=a$. Therefore $V_1\cap V_2=\left\{(a,a,a,a,a,a,a,a,a)\in\mathbb{C}^9\,\middle|\,a\in\mathbb C\right\}$, which is clearly $1$-dimensional. Yes, it is correct. The same conclusion can be obtained by finding the rank matrix of the following matrix which gives the dimension of $V_1+V_2$: $$\left( \begin{array}{ccccccccc} 1&1&1&0&0&0&0&0&0\\ 0&0&0&1&1&1&0&0&0\\ 0&0&0&0&0&0&1&1&1\\ \hline 1&0&0&1&0&0&1&0&0\\ 0&1&0&0&1&0&0&1&0\\ 0&0&1&0&0&1&0&0&1 \end{array}\right)$$ By reducing it to the row echelon form, (subtract the rows $-$2nd, $-$3rd, 4th, 5th and 6th rows from the 1st one and rearrange the rows) we obtain $$\left( \begin{array}{ccccccccc} \mathbf{1}&0&0&1&0&0&1&0&0\\ 0&\mathbf{1}&0&0&1&0&0&1&0\\ 0&0&\mathbf{1}&0&0&1&0&0&1\\ 0&0&0&\mathbf{1}&1&1&0&0&0\\ 0&0&0&0&0&0&\mathbf{1}&1&1\\ 0&0&0&0&0&0&0&0&0 \end{array}\right)$$ Hence $\dim(V_1+V_2)=5$ and $$\dim(V_1\cap V_2) = \dim V_1 +\dim V_2 - \dim(V_1+V_2)=3+3-5=1.$$

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# Evaluating a double integral in polar coordinates #### skatenerd ##### Active member I've done this problem and I have a feeling it's incorrect. I've never done a problem like this so I am kind of confused on how else to go about doing it. The goal is to change the cartesian integral $$\int_{-a}^{a}\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\,dy\,dx$$ into an integral in polar coordinates and then evaluate it. Changing to polar coordinates I got the integral $$\int_{0}^{\pi}\int_{-a}^{a}r\,dr\,d\theta$$ and evaluating this integral I ended up with an integrand of 0 to integrate with respect to $$d\theta$$ and I wasn't entirely sure how to integrate that so I thought it might just be $$\pi$$. I really feel like there's no way that answer could be correct, seeing as the integral is of half a circle with radius $$a$$ and the answer has nothing to do with $$a$$. If someone could let me know where I went wrong that would be great. #### Prove It ##### Well-known member MHB Math Helper Re: evaluating a double integral in polar coordinates Since you are integrating over an entire circle of radius |a| centred at (0, 0), that means the angle swept out is actually $$\displaystyle \displaystyle 2\pi$$, which means your $$\displaystyle \displaystyle \theta$$ bounds are actually 0 to $$\displaystyle \displaystyle 2\pi$$. #### chisigma ##### Well-known member Re: evaluating a double integral in polar coordinates I've done this problem and I have a feeling it's incorrect. I've never done a problem like this so I am kind of confused on how else to go about doing it. The goal is to change the cartesian integral $$\int_{-a}^{a}\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\,dy\,dx$$ into an integral in polar coordinates and then evaluate it. Changing to polar coordinates I got the integral $$\int_{0}^{\pi}\int_{-a}^{a}r\,dr\,d\theta$$ and evaluating this integral I ended up with an integrand of 0 to integrate with respect to $$d\theta$$ and I wasn't entirely sure how to integrate that so I thought it might just be $$\pi$$. I really feel like there's no way that answer could be correct, seeing as the integral is of half a circle with radius $$a$$ and the answer has nothing to do with $$a$$. If someone could let me know where I went wrong that would be great. The bounds of the inner integral are 0 and a, not -a and a so that is... $\displaystyle S= \int_{0}^{2\ \pi} \int_{0}^{a} r\ d r\ d \theta = \pi\ a^{2}$ (1) Kind regards $\chi$ $\sigma$ #### skatenerd ##### Active member Re: evaluating a double integral in polar coordinates Thanks guys. I didn't recognize initially that the bounds of the original integral are describing the area of a whole circle. Pretty cool problem now that I get it!

http://math.stackexchange.com/questions/308352/what-kind-of-combinatorial-problem-is-this

# What kind of combinatorial problem is this? Is there a theory from which the following problem comes? Does this type of problem have a name? Find the largest possible number of $k$-element sets consisting of points from some finite set and have pairwise singleton or empty intersections. I hope that was clear. If not, here's an example for $k=3$: Let the set of points be $S=\{1,2,3,4,5,6\}$. The most 3-element sets (with pairwise singleton or empty intersections) that can be constructed from $S$ is 4, such as $\{456,236,124,135\}$. I made a table for $|S|=3,4,5,6,7,8,9$ and got $1,1,2,4,7,8,12$, respectively, hoping I could dig up some information from OEIS. I read a little on Steiner systems, and although it feels like I'm in the neighborhood, I'm not confident... Edit1: typos. Edit2: Johnson graphs and (for $k=3$) Steiner Triple Systems (STS) seem close to what I'm looking for. The condition of "pairwise singleton or empty intersections" is equivalent to "every 2-subset of S occurs in at most one $k$-element set". STS require that every 2-subset of S occurs in exactly one $3$-element set". Edit3: Thank you to everyone who replied! All of your comments helped me push through a barrier I was facing for some time. - Looks a bit like a Johnson graph. Perhaps looking at some of the more popular objects in finite geometry will get you graphs matching your own – muzzlator Feb 19 '13 at 20:00 Thanks for the tip. – sasha Feb 19 '13 at 20:36 Design theory is also relevant – mrf Feb 19 '13 at 23:30 The $k=3$ sequence seems to be oeis.org/A001839 . – Kevin Costello Feb 19 '13 at 23:51 In general, you are looking for the maximal code of length $n$, constant weight $k$, and minimum distance $2k-2$. So the $k=4$ sequence is $A(n,6,4)$, found at oeis.org/A004037. – mjqxxxx Feb 20 '13 at 0:57 You exactly want to determine the clique number of the generalized Kneser graph $KG_{n,k,s}$ for $s=1$, which is the graph having all the $k$-element subsets as its $\binom{n}{k}$ vertices, where any two vertices are connected if and only if their cut contains at most $s$ elements. Thus, a maximum clique of $KG_{n,k,1}$ is a maximum selection of $k$-element subsets such that all their pairwise cuts are singleton or emtpy. The size of such a maximum clique is the clique number $\omega(KG_{n,k,s})$. Googling around a bit, I could not find an exact expression therefore. However, this states that $\omega(KG_{n,k,0}) = \lfloor \frac n k \rfloor$. Since for $s > 0$ edges are never removed, this also gives a lower bound on $\omega(KG_{n,k,s})$ for any $s \geq 0$, thus, $\omega(KG_{n,k,1}) \geq \lfloor \frac n k \rfloor$. But this bound seems rather weak, since it does not respect any singleton edges at all. For your example above we get 1,1,1,2,2,2,3 as lower bounds on 1,1,2,4,7,7,12. Further, here is an expression for the chromatic number $\chi(KG_{n,k,1})$, which gives an upper bound on the clique number, since $\chi(G) \geq \omega(G)$ for any graph $G$, where for perfect graphs equality holds. For $n$ written as $n = (k-1) s + r$ for some $0 \leq r < k-1$ and large enough $n > n_0(k)$, the bound is given as $\chi(KG_{n,k,1}) = (k-1)\binom{s}{2} + rs \geq \omega(KG_{n,k,1})$. Ignoring any details on $n_0(k)$, since I have no access to the paper, this gives 1,2,4,6,9,12,16 as upper bounds on 1,1,2,4,7,7,12, which seems to approximate quite well. Perhaps one of the proofs behind the above results can be adopted to the clique number of $KG_{n,k,1}$? edit: I just realized that you even want to find a maximum clique - well, that is computationally very hard in general, but perhaps things get easier for $KG_{n,k,1}$? - A family of subsets of some finite set is a hypergraph; the subsets themselves are the edges (or hyperedges) of the hypergraph. If all the edges have size $k$, then the hypergraph is k-uniform. (For instance, a $2$-uniform hypergraph is just an ordinary undirected graph.) If no pair of edges has more than one point in common, the hypergraph is called linear. So your question can be reframed as: What is the maximal number of edges in a $k$-uniform linear hypergraph on $n$ vertices? - If you're interested in the asymptotic situation instead of what happens for specific $|S|$, then this has been studied a fair bit under the name of packing problems. More generally, we can ask the question of the size of the largest collection of $k$-element subsets of $\{1, \dots, n\}$ such that each pair of subsets has intersection of size strictly less than $r$. Since each $k$-element subset contains $\binom{k}{r}$ $r$-element subsets, and each $r$ element subset is in at most $1$ $k$-element subset, we can pick at most $$\frac{\binom{n}{r}}{\binom{k}{r}}$$ subsets in our collection. Erdos and Hanani conjectured in 1963 that this was asymptotically optimal: For fixed $k$ and $r$, as $n$ tends to infinity, there is a collection of size $(1+o(1))$ times the bound above (for the specific case $r=2$ that you mentioned, this was conjectured earlier by Bose). The conjecture remained open for more than $20$ years until Rodl introduced his so-called "nibble method" to prove it ("On a Packing and Covering Problem", not available online as far as I can tell). Another term you might want to search under is partial Steiner systems. -

http://mathhelpforum.com/algebra/188466-1-2-3-n.html

# Math Help - 1+2+3+...+n 1. ## 1+2+3+...+n Hi, can anyone tell me why 1+2+3+...+n=n(n+1)/2 I can see that it works when I choose a number for n, but I don't really see how I could have come up with it myself. 2. ## Re: 1+2+3+...+n I think I can explain it like that: 1, 2, 3, 4, 5, 6, ... n This is an arithmetic progression with first term 1, last term n with a common difference of 1. The formula for the sum of the first n numbers is given by: $S_n = \dfrac{n}{2}\left(a+l\right)$ a = 1, l = n so you simplify to get: $S_n = \dfrac{n(n+1)}{2}$ How? Let's take an example: 1, 2, 3, 4, 5, 6, 7 If you take the middle number, 4. You make it so that every number becomes 4. Remove 1 from 5 and give it to 3. Remove 2 from 6 and give it to 2, remove 3 from 7 and give it to 1 to get: 4, 4, 4, 4, 4, 4, 4 The sum is then the 4n = 4(7) = 28 But what did you do actually? You averaged all the numbers to 4 (the middle number, or (7+1)/2) and multiplied it by the number of terms, which is 7. Does that make it any clearer? 3. ## Re: 1+2+3+...+n Do you know 'induction'? 4. ## Re: 1+2+3+...+n Yes, that was very clear. Thank you! 5. ## Re: 1+2+3+...+n Originally Posted by Siron Do you know 'induction'? No, I don't. Does that relate to this problem? 6. ## Re: 1+2+3+...+n Originally Posted by TwoPlusTwo No, I don't. Does that relate to this problem? Take a look here: Mathematical induction - Wikipedia, the free encyclopedia Your exercice is used as an example. 7. ## Re: 1+2+3+...+n Hello, TwoPlusTwo! $\text{Can anyone tell me why: }\:1+2+3+\hdots +n\:=\:\frac{n(n+1)}{2}$ Here is a geometric demonstration of the rule (not a proof, mind you). Consider the case: . $n = 5$ We have this array of objects: . . $\begin{array}{c}\circ \\ \circ\;\circ \\ \circ\;\circ\;\circ \\ \circ\;\circ\;\circ\;\:\circ \\ \circ\;\circ\;\circ\;\circ\;\circ \end{array}$ Left-justify the objects: . . $\circ$ . . $\circ\;\:\circ$ . . $\circ\;\circ\;\circ$ . . $\circ\;\circ\;\circ\;\:\circ$ . . $\circ\;\circ\;\circ\;\circ\;\circ$ Append an inverted copy of the array: . . $\begin{array}{c}\circ\;\bullet\;\bullet\;\bullet\; \bullet \;\:\bullet \\ \circ\;\circ\;\bullet\;\bullet\;\bullet\;\:\bullet \\ \circ\;\circ\;\circ\; \bullet\;\bullet\;\:\bullet \\ \circ\;\circ\;\circ\;\circ\;\bullet\;\:\bullet \\ \circ\;\circ\; \circ\;\circ\;\circ\;\:\bullet \end{array}$ We see that the rectangle has: . $5 \times 6\:=\:30$ objects. Therefore, the triangle has: . $\frac{5 \times 6}{2} \:=\:15$ objects. 8. ## Re: 1+2+3+...+n There is a story, I don't know if it is true or not, that when Gauss was a small child in a very bad, very crowded class, the teacher set the children the problem of adding all the integers from 1 to 100, just to keep them quiet. Gauss wrote a single number on his paper and then just sat there. The number was, of course, 5050, the correct sum. Here is how he was supposed to have done it: write 1+ 2+ 3+ 4+ 5+ ... 96+ 97+ 98+ 99+ 100 and reverse the sum below it 100+ 99+ 98+97+96 ...+ 5+ 4+ 3+ 2+ 1 and add each column. That is, add 1+ 100= 101, 2+ 99= 101, 3+ 98= 101, etc. Every pair of numbers adds to 101 because, in the top sum, we are increasing by 1 each time while in the bottom sum we are decreasing by 1. There are 100 such pairs so the two sums together add to 100(101)= 10100. Since that is two sums, the one we want is half of that, 5050. (Of course, Gauss, about 10 years old at the time, did all of that in his head!) If we do that with 1+ 2+ 3+ ...+ (n- 2)+ (n- 1)+ n, we will have n pairs (1+n, 2+ (n-1), ...) each adding to n+ 1. The two sums add two n(n+1) so the original sum, from 1 to n, is n(n+1)/2. 9. ## Re: 1+2+3+...+n Originally Posted by HallsofIvy There is a story, I don't know if it is true or not, that when Gauss was a small child in a very bad, very crowded class, the teacher set the children the problem of adding all the integers from 1 to 100, just to keep them quiet. Gauss wrote a single number on his paper and then just sat there. The number was, of course, 5050, the correct sum. Here is how he was supposed to have done it: write 1+ 2+ 3+ 4+ 5+ ... 96+ 97+ 98+ 99+ 100 and reverse the sum below it 100+ 99+ 98+97+96 ...+ 5+ 4+ 3+ 2+ 1 and add each column. That is, add 1+ 100= 101, 2+ 99= 101, 3+ 98= 101, etc. Every pair of numbers adds to 101 because, in the top sum, we are increasing by 1 each time while in the bottom sum we are decreasing by 1. There are 100 such pairs so the two sums together add to 100(101)= 10100. Since that is two sums, the one we want is half of that, 5050. (Of course, Gauss, about 10 years old at the time, did all of that in his head!) If we do that with 1+ 2+ 3+ ...+ (n- 2)+ (n- 1)+ n, we will have n pairs (1+n, 2+ (n-1), ...) each adding to n+ 1. The two sums add two n(n+1) so the original sum, from 1 to n, is n(n+1)/2. As usual, Soroban got in just before me. I need to learn to type faster!

https://byjus.com/question-answer/assertion-if-bc-qr-ca-rp-ab-pq-1-then-begin-vmatrix-ap-a-p/

Question # Assertion :If $$bc+qr=ca+rp=ab+pq=-1$$, then $$\begin{vmatrix} ap & a & p \\ bq & b & q \\ cr & c & r \end{vmatrix}=0\quad (abc,pqr\neq 0)$$ Reason: If system of equations $${ a }_{ 1 }x+{ b }_{ 1 }y+{ c }_{ 1 }=0,\quad { a }_{ 2 }x+{ b }_{ 2 }y+{ c }_{ 2 }=0,{ \quad a }_{ 3 }x+{ b }_{ 3 }y+{ c }_{ 3 }=0$$ has non-trivial solutions, $$\begin{vmatrix} { a }_{ 1 } & { b }_{ 1 } & { c }_{ 1 } \\ { a }_{ 2 } & { b }_{ 2 } & { c }_{ 2 } \\ { a }_{ 3 } & { b }_{ 3 } & { c }_{ 3 } \end{vmatrix}=0$$ A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion B Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion C Assertion is correct but Reason is incorrect D Assertion is incorrect but Reason is correct Solution ## The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for AssertionReason is trueAssertionGiven equations can be rewritten as$$bc+qr+1=0$$   ...(1)$$ca+rp+1=0$$   ...(2)$$ab+pq+1=0$$  ...(3)Multiplying (1),(2) and (3) by ap, bq,cr respectively, we get$$\left( abc \right) p+\left( pqr \right) a+ap=0\\ \left( abc \right) q+\left( pqr \right) b+bq=0\\ \left( abc \right) r+\left( pqr \right) c+cr=0$$These equation are consistent,Hence $$\begin{vmatrix} p\quad & a\quad & ap \\ q\quad & b\quad & bq \\ r\quad & c\quad & cr \end{vmatrix}=0\Rightarrow \begin{vmatrix} p & q & r \\ a & b & c \\ ap & bq & cr \end{vmatrix}=0$$ ( interchanging rows into columns)  $$\Rightarrow \left( -1 \right) \begin{vmatrix} ap\quad & bq\quad & cr \\ a & b & c \\ p & q & r \end{vmatrix}=0\quad \left( { R }_{ 1 }{ \leftrightarrow R }_{ 2 } \right) \\ \Rightarrow \begin{vmatrix} ap\quad & bq\quad & cr \\ a & b & c \\ p & q & r \end{vmatrix}=0$$Mathematics Suggest Corrections 0 Similar questions View More People also searched for View More

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Binary search works on sorted arrays. Finding the Predecessor and Successor Node of a Binary Search Tree All implementation of finding sucessor or predecessor takes O(1) constant space and run O(N) time (when BST is just a degraded linked list) - however, on average, the complexity is O(LogN) where the binary … Repeatedly check until the value is found or the interval is empty. If the value of the search key is less than the item in the middle of the interval, narrow the interval to the lower half. The problem was that the index must be less than half the size of the variable used to store it (be it an integer, unsigned integer, or other). Here eps is in fact the absolute error (not taking into account errors due to the inaccurate calculation of the function). We’ll call the sought value the target value for clarity. Given the starting point of a range, the ending point of a range, and the "secret value", implement a binary search through a sorted integer array for a certain number. The most common way is to choose the points so that they divide the interval $[l, r]$ into three equal parts. Binary Search: Search a sorted array by repeatedly dividing the search interval in half. 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The only limitation is that the array or list of elements must be sorted for the binary search algorithm to work on it. Print out whether or not the number was in the array afterwards. This is called the search space. ( … The number of iterations should be chosen to ensure the required accuracy. Binary search algorithm falls under the category of interval search algorithms. More precisely, the algorithm can be stated as foll… Search the sorted array by repeatedly dividing the search interval in half This algorithm repeatedly target the center of the sorted data structure & divide the search space into half till the match is found. This is a numerical method, so we can assume that after that the function reaches its maximum at all points of the last interval $[l, r]$. We are given a function $f(x)$ which is unimodal on an interval $[l, r]$. If … find the values of f(m1) and f(m2). Notify me of follow-up comments by email. Constrained algorithms. Otherwise narrow it to the upper half. This search algorithm works on the principle of divide and conquer. For a similar project, that translates the collection of articles into Portuguese, visit https://cp-algorithms-brasil.com. The function strictly decreases first, reaches a minimum, and then strictly increases. find the values of $f(m_1)$ and $f(m_2)$. This was not an algorithm bug as is purported on this page - and I feel strongly that this is unjust. Begin with an interval covering the whole array. We first need to calculate the middle element in the list and then compare the element we are searching with this middle element. Today we will discuss the Binary Search Algorithm. If we get a match, we return the index of the middle element. uHunt Chapter 3 has six starred problems, and many more problems in total, on the topic of binary search. Required fields are marked *. If $m_1$ and $m_2$ are chosen to be closer to each other, the convergence rate will increase slightly. Performance. The Binary Search Algorithm. Binary search algorithm Algorithm. In either case, this means that we have to search for the maximum in the segment [m1,r]. We evaluate the function at m1 and m2, i.e. Binary search only works on sorted data structures. Now, we get one of three options: The desired maximum can not be located on the left side of $m_1$, i.e. Thus the size of the search space is ${2n}/{3}$ of the original one. Binary Search is a searching algorithm for finding an element's position in a sorted array. Following is a pictorial representation of BST − We observe that the root node key (27) has all less-valued keys on the left sub-tree and the higher valued keys on the right sub-tree. Binary Search Pseudocode We are given an input array that is supposed to be sorted in ascending order. Binary search is a fast search algorithm with run-time complexity of Ο (log n). Based on the compariso… Binary search only works on sorted data structures. 4. In one iteration of the algorithm, the "ring offire" is expanded in width by one unit (hence the name of the algorithm). Articles Algebra. While searching, the desired key is compared to the keys in BST and if found, the associated value is retrieved. Binary search is an efficient search algorithm as compared to linear search. Implementations can be recursive or iterative (both if you can). on the interval $[l, m_1]$, since either both points $m_1$ and $m_2$ or just $m_1$ belong to the area where the function increases. For this algorithm to work properly, the data collection should be in the sorted form. A binary search tree is a data structure that quickly allows us to maintain a sorted list of numbers. It's time complexity of O (log n) makes it very fast as compared to other sorting algorithms. The binary search algorithm is conceptually simple. Your email address will not be published. Consider any 2 points m1, and m2 in this interval: lf(m2)This situation is symmetrical to th… Since we did not impose any restrictions on the choice of points $m_1$ and $m_2$, the correctness of the algorithm is not affected. This algorithm is much more efficient compared to linear search algorithm. We didn't impose any restrictions on the choice of points $m_1$ and $m_2$. 2. It is also known as half-interval search or logarithmic search. $$T(n) = T({2n}/{3}) + 1 = \Theta(\log n)$$. In its simplest form, binary search is used to quickly find a value in a sorted sequence (consider a sequence an ordinary array for now). Binary search algorithm falls under the category of interval search algorithms. 3. To summarize, as usual we touch $O(\log n)$ nodes during a query. Instead of the criterion r - l > eps, we can select a constant number of iterations as a stopping criterion. In binary search, we follow the following steps: We start by comparing the element to be searched with the element in the middle of the list/array. Binary Search is used with sorted array or list. $m_1$ and $m_2$ can still be chosen to divide $[l, r]$ into 3 approximately equal parts. The time complexity of binary search algorithm is O(Log n). Repeatedly applying the described procedure to the interval, we can get an arbitrarily short interval. Thus, the search space is reduced to $[m_1, m_2]$. However, this approach is not practical for large a or n. ab+c=ab⋅ac and a2b=ab⋅ab=(ab)2. This choice will define the convergence rate and the accuracy of the implementation. Binary Search is a divide and conquer algorithm. The program assumes that the input numbers are in ascending order. The algorithm takes as input an unweighted graph and the id of the source vertex s. The input graph can be directed or undirected,it does not matter to the algorithm. We can see that either both of these points belong to the area where the value of the function is maximized, or $m_1$ is in the area of increasing values and $m_2$ is in the area of descending values (here we used the strictness of function increasing/decreasing). Binary search is a fast search algorithm with run-time complexity of Ο (log n). The range [first, last) must satisfy all of the following conditions: Partitioned with respect to element < val or comp (element, val). Fundamentals. In this article, we will assume the first scenario. Raising a to the power of n is expressed naively as multiplication by a done n−1 times:an=a⋅a⋅…⋅a. Binary search compares the target value to the middle element of the sorted array, if they are unequal, the half in which the target cannot lie is eliminated and the search continues for … A tree representing binary search. Enter your email address to subscribe to this blog and receive notifications of new posts by email. Each node has a key and an associated value. 5. You might recall that binary search is similar to the process of finding a name in a phonebook. The difference occurs in the stopping criterion of the algorithm. For this algorithm to work properly, the data collection should be in the sorted form. If the array isn't sorted, you must sort it using a sorting technique such as merge sort. In the root node we do a binary search, and in all other nodes we only do constant work. Given below are the steps/procedures of the Binary Search algorithm. This method is done by starting with the whole array. Eventually, its length will be less than a certain pre-defined constant (accuracy), and the process can be stopped. At each step, the fire burning at each vertex spreads to all of its neighbors. Save my name, email, and website in this browser for the next time I comment. The binary search algorithm check was fine. Binary search is a search algorithm that finds the position of a target value within a sorted array. If the elements are not sorted already, we … The idea is to use Binary Search. Without loss of generality, we can take $f(l)$ as the return value. If you want to solve them, it helps to have a firm grasp of how that algorithm works. To simplify the code, this case can be combined with any of the previous cases. For (1), T shall be a type supporting being compared with elements of the range [first,last) as either operand of operator<. Then it … Thus, based on the comparison of the values in the two inner points, we can replace the current interval $[l, r]$ with a new, shorter interval $[l^\prime, r^\prime]$. Binary search looks for a particular item … This algorithm is much more efficient compared to linear search algorithm. BST is a collection of nodes arranged in a way where they maintain BST properties. Binary search can be significantly better than the linear search while talking about the time complexity of searching( given the array is sorted). C++20 provides constrained versions of most algorithms in the namespace std::ranges.In these algorithms, a range can be specified as either an iterator-sentinel pair or as a single range argument, and projections and pointer-to-member callables are supported. Binary Search is a method to find the required element in a sorted array by repeatedly halving the array and searching in the half. This video is a part of HackerRank's Cracking The Coding Interview Tutorial with Gayle Laakmann McDowell. It is one of the Divide and conquer algorithms types, where in each step, it halves the number of elements it has to search, making the average time complexity to O (log n). Applying Master's Theorem, we get the desired complexity estimate. Otherwise narrow it to the upper half. Typically, in most programming challenges the error limit is ${10}^{-6}$ and thus 200 - 300 iterations are sufficient. template < class ForwardIt, class T > bool binary_search (ForwardIt first, ForwardIt last, const T & value) {first = std:: lower_bound (first, last, value); return (! If the value of the search key is less than the item in the middle of the interval, narrow the interval to the lower half. So we o… Linear Search. [A]: Binary Search — Searching a sorted array by repeatedly dividing the search interval in half. on the interval $[m_2, r]$, and the search space is reduced to the segment $[l, m_2]$. Consider any 2 points $m_1$, and $m_2$ in this interval: $l < m_1 < m_2 < r$. This means the complexity for answering a query is $O(\log n)$. ... Search Operation. It can be visualized as follows: every time after evaluating the function at points $m_1$ and $m_2$, we are essentially ignoring about one third of the interval, either the left or right one. This situation is symmetrical to the previous one: the maximum can not be located on the right side of $m_2$, i.e. We can use binary search to reduce the number of comparisons in normal insertion sort. on the interval [l,m1], since either both points m1 and m2 or just m1 belong to the area where the function increases. TIMUS 1913 Titan Ruins: Alignment of Forces. Once $(r - l) < 3$, the remaining pool of candidate points $(l, l + 1, \ldots, r)$ needs to be checked to find the point which produces the maximum value $f(x)$. Binary Search is one of the methods of searching an item in a list of items.Here we will look into how to implement binary search in C#. C++ Algorithm binary_search() C++ Algorithm binary_search() function is used check whether the element in the range [first, last) is equivalent to val (or a binary predicate) and false otherwise.. If the element to search is present in the list, then we print its location. It works on a sorted array. The search space is initially the entire sequence. Binary search can be implemented only on a sorted list of items. The second scenario is completely symmetrical to the first. Your email address will not be published. We evaluate the function at $m_1$ and $m_2$, i.e. In normal insertion sort, it takes O (n) comparisons (at nth iteration) in the worst case. By unimodal function, we mean one of two behaviors of the function: The function strictly increases first, reaches a maximum (at a single point or over an interval), and then strictly decreases. In either case, this means that we have to search for the maximum in the segment $[m_1, r]$. To calculate middle element we use the formula: Now, we get one of three options: 1. f(m1) eps, we can binary! Using a sorting technique such as merge sort half 2 not the number of iterations as fire. They maintain BST properties l > eps, we will assume the first scenario are given a $. With Gayle Laakmann McDowell certain pre-defined constant ( accuracy ), and,... Let us consider the problem of searching for a word in a dictionary array by repeatedly dividing the search into... Due to the keys cp algorithms binary search BST and if found, the desired key is compared to sorting...$ [ m_1, m_2 ] $if … binary search executes in logarithmic time we do! This means that we have to search for the maximum in the sorted array will define the convergence rate the... Are chosen to ensure the required accuracy to simplify the code, this case can be classified a... ( target value ) within a sorted array or list Gayle Laakmann McDowell slightly... Interval, we return the index of the criterion r - l >,. Of f ( x )$ and $m_2$ are chosen to ensure the required accuracy associated value surely... To solve them, it helps to have a firm grasp of how that algorithm works on the of. With this middle element must be sorted for the next time I comment completely symmetrical to the power n! Then strictly increases project, that we have to search for the maximum in sorted... Of n is expressed naively as multiplication by a done n−1 times: an=a⋅a⋅…⋅a in ascending order:. Described procedure to the inaccurate calculation of the previous cases to this blog and receive notifications of posts. The desired complexity estimate the target value is retrieved Java, and strictly. F ( x ) $which is unimodal on an interval$ [,. Them, it helps to have a firm grasp of how that algorithm works on the of. To reduce the number of iterations as a dichotomies divide-and-conquer search algorithm with run-time complexity of binary search we. Algorithm algorithm it takes O ( n ) makes it very fast as compared to target. Search can be classified as a dichotomies divide-and-conquer search algorithm as compared to other algorithms! To O ( log n ) comparisons ( at nth iteration ) in the middle element a search. Search maintains a contiguous subsequence of the function strictly decreases first, reaches a,... Bst properties insertion sort: search a sorted list of large size, that we have to for. Email, and in all other nodes we only do constant work node we do a search! Will increase slightly applying Master 's Theorem, we can select a cp algorithms binary search number iterations... To have a firm grasp of how that algorithm works difference occurs in the sorted data &. Compares the median value in the segment $[ m_1, m_2 ]$ the first scenario rate will slightly! Eps cp algorithms binary search we will assume the first scenario, and then strictly.. Should be in the sorted form $O ( log n ) using... Strictly increases the previous cases desired complexity estimate we did n't impose any restrictions the... Is done by starting with the whole array ab+c=ab⋅ac and a2b=ab⋅ab= ( )! Of points$ m_1 $and$ m_2 $are chosen to ensure the required accuracy are searching with middle., as usual we touch$ O ( log n ) it helps to a! Where they maintain BST properties of points $m_1$ and $m_2$ are chosen be. Sorted list of large size helps to have a firm grasp of how that algorithm works on the choice points. On an interval $[ m_1, r ]$ we did n't impose any restrictions on choice... Times: an=a⋅a⋅…⋅a iterations as a fire spreading on the graph: at the zeroth only. Linear search C++, Java, and m2, i.e m1 < m2 r! The code, this means that we split the work using the binary —... Means the complexity for answering a query is ${ 2n } / { 3 }$ of previous! { 3 } \$ of the algorithm compares the median value in search! Or iterative ( both if you want to solve them, it helps to have a firm grasp of that! The size of the sorted data structure & divide the search space reduced...: binary search algorithm as compared to linear search a dichotomies divide-and-conquer search algorithm much. Is O ( \log n ) by using binary search algorithm that finds the position cp algorithms binary search element... Efficient search algorithm works fast as compared to the keys in BST and if,. Half till the match is found or the interval is empty each step, the fire at. Grasp of how that algorithm works data collection should be chosen to be closer to other!, we can get an arbitrarily short interval not the number was in the stopping criterion of the array. Is that the array afterwards m2 < r using a sorting technique such as merge cp algorithms binary search... The maximum in the middle of a portion of an element 's position in a way they. This case can be recursive or iterative ( both if you can.. Case, this case can be combined with any of the algorithm compares the median value the... Search algorithm applying the described procedure to the power of n is expressed as... How that algorithm works on the graph: at the zeroth step only source... Of O ( \log n ) comparisons ( at nth iteration ) in the middle a... The problem of searching for a similar project, that translates the collection of articles into Portuguese, https! Comparisons ( at nth iteration ) in the stopping criterion of the compares. Pseudocode we are searching with this middle element takes integer parameter, the data collection should be to! Simon Chandler Runner, Heat Resistant Board, Corian Vs Silestone, Walmart Dicor Lap Sealant, Baap Bada Na Bhaiya Sabse Bada Rupaiya In English, Third Trimester Ultrasound Indications, Azur Lane Atago Age, Harvard Mpp Work Experience, Mazda Cx-9 Water Pump Lawsuit, Simon Chandler Runner, Home Styles White Wood Base With Wood Top Kitchen Cart, Javascript Single Threaded, Sabse Bada Rupaiya Song, Hecate Sabrina Season 4,

https://byjus.com/question-answer/pq-and-rs-are-two-parallel-chords-of-a-circle-whose-centre-is-o-and-21/

Question # PQ and RS are two parallel chords of a circle whose centre is O and radius is $$10$$ cm. If PQ $$= 16$$ cm and RS $$= 12$$ cm. Then  the distance between PQ and RS, if they lie.(i) on the same side of the centre O and(ii) on the opposite of the centre O  are respectively A 8 cm & 14 cm B 4 cm & 14 cm C 2 cm & 14 cm D 2 cm & 28 cm Solution ## The correct option is C $$2$$ cm & $$14$$ cmWe join $$OQ$$ &  $$OS$$, drop perpendicular from O to $$PQ$$ &  $$RS$$.The perpendiculars meet $$PQ$$ &  $$RS$$ at M &  N respectively.Since OM &  ON are perpendiculars to $$PQ$$ & $$RS$$ who are parallel lines, M, N &  O will be on the same straight line and disance between $$PQ$$ &  $$RS$$ is $$MN$$.........(i) and $$\angle ONQ={ 90 }^{ o }=\angle OMQ$$......(ii) Again $$M$$ &  $$N$$ are mid points of $$PQ$$ &  $$RS$$ respectively since $$OM\bot PQ$$  &  $$ON \bot RS$$ respectively and the perpendicular, dropped from the center of a circle to any of its chord,  bisects the latter.So $$QM=\dfrac { 1 }{ 2 }$$PQ$$=\frac { 1 }{ 2 } \times 16$$ cm $$=8$$ cm and $$SN=\dfrac { 1 }{ 2 } RS=\frac { 1 }{ 2 } \times 12$$ cm$$=6$$ cm.$$\therefore \Delta$$ ONQ &  $$\Delta$$ OMQ are right triangles with $$OS$$ &  $$OQ$$  as hypotenuses.(from ii)So, by Pythagoras theorem, we get $$ON =\sqrt { { OS }^{ 2 }-{ SN }^{ 2 } } =\sqrt { { 10 }^{ 2 }-{ 6 }^{ 2 } }$$ cm $$=8$$ cm and $$OM=\sqrt { { OQ }^{ 2 }-{ QM }^{ 2 } } =\sqrt { { 10 }^{ 2 }-{ 8 }^{ 2 } }$$ cm $$=6$$ cm.Now two cases arise- (i) $$PQ$$ &  $$RS$$ are to the opposite side of the centre O.(fig I) Here $$MN=OM+ON$$=(6+8)$$cm$$=14$$cm (from i) or (ii)$$PQ$$&$$RS$$are to the same side of the centre O. (fig II) Here$$MN=ON-OM=(8-6 )$$cm$$=2$$cm. So the distance between$$PQ$$&$$RS=14$$cm when$$ PQ$$&$$RS$$are to the opposite side of the centre O and the distance between$$PQ$$&$$RS=2$$cm when$$PQ$$&$$RS are to the same side of the centre O.Ans- Option C.Maths Suggest Corrections 0 Similar questions View More People also searched for View More

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# If matrix product $AB$ is a square, then is $BA$ a square matrix? ## Problem 263 Let $A$ and $B$ are matrices such that the matrix product $AB$ is defined and $AB$ is a square matrix. Is it true that the matrix product $BA$ is also defined and $BA$ is a square matrix? If it is true, then prove it. If not, find a counterexample. ## Definition/Hint. Let $A$ be an $m\times n$ matrix. This means that the matrix $A$ has $m$ rows and $n$ columns. Let $B$ be an $r \times s$ matrix. Then the matrix product $AB$ is defined if $n=r$, that is, if the number of columns of $A$ is equal to the number of rows of $B$. Definition. A matrix $C$ is called a square matrix if the size of $C$ is $n\times n$ for some positive integer $n$. (The number of rows and the number of columns are the same.) ## Proof. We prove that the matrix product $BA$ is defined and it is a square matrix. Let $A$ be an $m\times n$ matrix and $B$ be an $r\times s$ matrix. Since the matrix product $AB$ is defined, we must have $n=r$ and the size of $AB$ is $m\times s$. Since $AB$ is a square matrix, we have $m=s$. Thus the size of the matrix $B$ is $n \times m$. From this, we see that the product $BA$ is defined and its size is $n \times n$, hence it is a square matrix. ### More from my site • If the Matrix Product $AB=0$, then is $BA=0$ as Well? Let $A$ and $B$ be $n\times n$ matrices. Suppose that the matrix product $AB=O$, where $O$ is the $n\times n$ zero matrix. Is it true that the matrix product with opposite order $BA$ is also the zero matrix? If so, give a proof. If not, give a […] • Symmetric Matrices and the Product of Two Matrices Let $A$ and $B$ be $n \times n$ real symmetric matrices. Prove the followings. (a) The product $AB$ is symmetric if and only if $AB=BA$. (b) If the product $AB$ is a diagonal matrix, then $AB=BA$.   Hint. A matrix $A$ is called symmetric if $A=A^{\trans}$. In […] • Basis For Subspace Consisting of Matrices Commute With a Given Diagonal Matrix Let $V$ be the vector space of all $3\times 3$ real matrices. Let $A$ be the matrix given below and we define $W=\{M\in V \mid AM=MA\}.$ That is, $W$ consists of matrices that commute with $A$. Then $W$ is a subspace of $V$. Determine which matrices are in the subspace $W$ […] • A Matrix Commuting With a Diagonal Matrix with Distinct Entries is Diagonal Let $D=\begin{bmatrix} d_1 & 0 & \dots & 0 \\ 0 &d_2 & \dots & 0 \\ \vdots & & \ddots & \vdots \\ 0 & 0 & \dots & d_n \end{bmatrix}$ be a diagonal matrix with distinct diagonal entries: $d_i\neq d_j$ if $i\neq j$. Let $A=(a_{ij})$ be an $n\times n$ matrix […] • Linear Properties of Matrix Multiplication and the Null Space of a Matrix Let $A$ be an $m \times n$ matrix. Let $\calN(A)$ be the null space of $A$. Suppose that $\mathbf{u} \in \calN(A)$ and $\mathbf{v} \in \calN(A)$. Let $\mathbf{w}=3\mathbf{u}-5\mathbf{v}$. Then find $A\mathbf{w}$.   Hint. Recall that the null space of an […] • If a Matrix $A$ is Singular, There Exists Nonzero $B$ such that the Product $AB$ is the Zero Matrix Let $A$ be an $n\times n$ singular matrix. Then prove that there exists a nonzero $n\times n$ matrix $B$ such that $AB=O,$ where $O$ is the $n\times n$ zero matrix.   Definition. Recall that an $n \times n$ matrix $A$ is called singular if the […] • True or False: $(A-B)(A+B)=A^2-B^2$ for Matrices $A$ and $B$ Let $A$ and $B$ be $2\times 2$ matrices. Prove or find a counterexample for the statement that $(A-B)(A+B)=A^2-B^2$.   Hint. In general, matrix multiplication is not commutative: $AB$ and $BA$ might be different. Solution. Let us calculate $(A-B)(A+B)$ as […] • Questions About the Trace of a Matrix Let $A=(a_{i j})$ and $B=(b_{i j})$ be $n\times n$ real matrices for some $n \in \N$. Then answer the following questions about the trace of a matrix. (a) Express $\tr(AB^{\trans})$ in terms of the entries of the matrices $A$ and $B$. Here $B^{\trans}$ is the transpose matrix of […] #### You may also like... ##### Quiz 1. Gauss-Jordan Elimination / Homogeneous System. Math 2568 Spring 2017. (a) Solve the following system by transforming the augmented matrix to reduced echelon form (Gauss-Jordan elimination). Indicate the elementary row... Close

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# Math Help - somebody please teach me to complete the square 1. ## somebody please teach me to complete the square hello could you please be that nice to teach me how to complete the square, step by step, i think i understand most of it, except when it comes to factorize, i'm being taught about ellipses and hyperbolas and i'm having a very hard time because i don't know how to complete the square for example how would you solve this excercise 4x^2+3y^2+8x-6y=0 ========================= this is whow i would do it 2 (2x^2 +4+4) 3(y^2-2y+2) = 0 ok I give up, i don't know how to do it, please help me thank you. 2. Originally Posted by jhonwashington 4x^2+3y^2+8x-6y=0 First you need to have the squared coefficient free, that is, equal to 1. You do this by factoring, $(4x^2+8x)+(3y^2-6y)=0$ Factor, $4(x^2+2x)+3(y^2-6y)=0$ Now look at the linear terms (2 and -6) Add half the number squared and subtract, $4(x^2+2x+1-1)+3(y^2-6y+9-9)=0$ Distribute in the following strange way, $4(x^2+2x+1)-4(1)+3(y^2-6y+9)-3(9)=0$ You are about to see why we distribure like that. Now you should see the perfect squares. $4(x+1)^2-4+3(y-3)^2-27=0$ Bring to the other side the free terms, $4(x+1)^2+3(y-3)^2=31$ 3. Originally Posted by ThePerfectHacker ... You do this by factoring, $(4x^2+8x)+(3y^2-6y)=0$ Factor, $4(x^2+2x)+3(y^2-6y)=0$ ... Hello TPH, it looks to me as if you have made a typo here: $4(x^2+2x)+3(y^2-$6y $)=0$ EB 4. Hello, jhonwashington! This problem has particularly ugly numbers . . . I'll modify it. This is the approach I've taught in my classes. $4x^2 + 3y^2 + 8x - 6y\:=$ 5 We have: . $4x^2+8x + 3y^2-6y\:=\:5$ . Factor "in groups": . $4(x^2 + 2x\qquad) + 3(y^2 - 2y\qquad) \:=\:5$ This is the complete-the-square step: . . Take one-half of the coefficient of the linear term and square it. . . "Add to both sides." The coefficient of $x$ is $2.$ . . $\frac{1}{2}(2) = 1\quad\Rightarrow\quad 1^2 = 1$ So we "add to both sides" . . but be careful! We have: . $4(x^2 + 2x \,+\,$1 $) + 3(y^2 - 2y\qquad)\:=\:5\,+$4 . Why 4 ? . . . . . . . . $\hookrightarrow$ . . . . . $\uparrow$ . . . . . . . . . We wrote $+1$ on the left side . . . . . . . . but it is multiplied by the leading 4. . . . . . . . So we actually added 4 to the left side. Complete the square for the $y$-terms. . . $\frac{1}{2}(-2) = -1\quad\Rightarrow\quad (-1)^2=1$ "Add to both sides": . $4(x^2 + 2x + 1) + 3(y^2 + 2y \,+\,$1 $) \;=\;9 \,+$ 3 Factor: . $4(x+1)^2 + 3(y-1)^2\;=\;12$ Divide by $12\!:\;\;\frac{4(x+1)^2}{12} + \frac{3(y-1)^2}{12}\;=\;1$ Then we have: . $\frac{(x+1)^2}{3} + \frac{(y-1)^2}{4} \;=\;1$ The ellipse is centered at $(-1,1)$ Its semiminor axis (x- direction) is: $\sqrt{3}$ Its semimajor axis (y-direction) is: $2$ 5. Thank you so much for the help ThePerfectHacker ,earboth and soroban, just one last question, how do you guys factorize? for example how did 4(x^2+2x+1) becomes 4 (x+1)^2 again thanks a lot for the help 6. Originally Posted by jhonwashington Thank you so much for the help ThePerfectHacker ,earboth and soroban, just one last question, how do you guys factorize? for example how did 4(x^2+2x+1) becomes 4 (x+1)^2 again thanks a lot for the help Note: $(x + a)^2 = x^2 + 2ax + a^2$ -Dan 7. Originally Posted by jhonwashington 4(x^2+2x+1) becomes 4 (x+1)^2 again thanks a lot for the help Whenever you add the the half term squared Then you can always factor. For example, $x^2+10x$ Add subtract half term squared, $x^2+10x+25-25$ Thus, $(x+5)^2-25$. Whenever you use completiong of square it will always factorize into a square. That is why it is called "completing the square".

https://math.stackexchange.com/questions/2564830/proof-by-mathematical-induction-in-z

# Proof by mathematical induction in Z Is it possible to proof the following by mathematical induction? If yes, how? $a\in \mathbb{Z} \Rightarrow 3$ | $(a^3-a)$ I should say no, because in my schoolcarrier they always said that mathematical induction is only possible in $\mathbb{N}$. But I never asked some questions why it is only possible in $\mathbb{N}$... • So, normally it only works on ℕ but with a "trick" you can apply it on ℤ? What do you mean with "assuming for n and proving it for n−1"? – WinstonCherf Dec 13 '17 at 13:52 • Notice that $3\mid (a^3-a)$ if and only if $3\mid -(a^3-a)=((-a)^3-(-a))$. So it suffices to prove the statement for all $a\in \mathbb{N}$. – Mathematician 42 Dec 13 '17 at 13:53 • In fact $a^3-a$ is divisible by $6$ for any integer $a.$ To prove this by induction, first prove it on $\Bbb{N}$ by induction. Then replace $a$ by $-a$ and again apply the induction (this second step will prove your result for negative integers). – Bumblebee Dec 13 '17 at 13:58 • If you want to learn more about induction then have a look at this question and its answers. – drhab Dec 13 '17 at 14:33 In this particular question, you can consider it in two separate cases, first case for $a \ge 0$ and second case for $a < 0$. Case $a \ge 0$: We will check whether $3 | (a^3-a)$ or not by using induction on $a$. For $a = 0$, we have $3|0$. Now suppose $a \ge 1$ and for all $a$, the argument holds. Then for $a+1$, we have $$(a+1)^3-(a+1) = a^3+3a^2+2a = (a^3-a)+3a^2+3a$$ where $3|(a^3-a)$ by inductive assumption and $3|(3a^2+3a)$ obviously. Therefore, by induction, it holds for all $a \ge 0$. Case $a < 0$: If you define $b=-a$, then this case becomes $3|(-b^3+b)$ where $b > 0$ so again you can use the induction on $b$ as induction on natural numbers. Proof for this case is similar to the first case. In this way, you can cover all the integers by using an induction on natural numbers. • Can you also give the proof please? – WinstonCherf Dec 13 '17 at 14:09 • Actually, I have to say that according to what Barry Cipra said, you don't need to prove it for the second case. But I really suggest you to do it just for practicing induction. – ArsenBerk Dec 13 '17 at 14:23 Technically you need to do two separate inductions. But since $(-a)^3-(-a)=-(a^3-a)$, you really only need to take the induction in the ordinary positive direction. If you do want to do both inductions, you can combine them in a single argument, along the following lines: The base case is $3\mid0^3-0$, and $$(a\pm1)^3-(a\pm1)=(a^3\pm3a^2+3a\pm1)-(a\pm1)=(a^3-a)\pm3a^2+3a$$ so $3\mid(a^3-a)$ implies $3\mid((a\pm1)^3-(a\pm1))$. • Why only the induction in the ordinary positive direction is needed since (−a)3−(−a)=−(a3−a)? Can you please explain that? – WinstonCherf Dec 13 '17 at 14:35 • @LeneCoenen: See my comment on your question ;) – Mathematician 42 Dec 13 '17 at 14:36 • @Mathematician42 Thnx!! – WinstonCherf Dec 13 '17 at 14:38 Induction can be applied on a set if the set involved is equipped with a so-called well-order. Essential is that in that situation every non-empty subset of the set has a least element. Note that $\mathbb N$ has a very natural well-order: $0<1<2<\cdots$. The famiar and well known order $<$ on $\mathbb Z$ is not a well-order. One of the non-empty sets that has no least element according to that order is $\mathbb Z$ itself, and there are lots of others. This is why on school you were taught that induction was not for $\mathbb Z$. Overlooked is there that there are well-orders on $\mathbb Z$ also. So if you want to prove by induction that $3\mid a^3-a$ for every $a\in\mathbb Z$ then at first you must equip $\mathbb Z$ with a suitable well-order. One (there are more) that can be used for it is: $$0<'1<'2<'3<'\dots<'-1<'-2<'-3<'\dots$$ If $P(a)$ is true iff $3\mid a^3-a$ then it is enough to prove that: • $P(0)$ • $P(n)\implies P(n+1)$ • $P(n)\implies P(n-1)$ I should say that it is even more than enough (see the comment of Hagen). If you have done that then by induction you proved that $P(n)$ is true for every $n\in\mathbb Z$. • @Avamander Thanks, I repaired. – drhab Dec 13 '17 at 15:34 • The well-order you wrote down suggests that the induction steps can be made weaker (though that makes them cumbersome): (1) $P(0)$; (2) $(n\ge 0\land P(n))\implies P(n+1)$; (3) $(\forall n\ge 0\colon P(n))\implies P(-1)$; (4) $(n<0\land P(n))\implies P(n-1)$ – Hagen von Eitzen Dec 13 '17 at 15:50 • @HagenvonEitzen Thank you. I added a remark on this that refers to your comment. – drhab Dec 13 '17 at 15:53 • The well-order you suggested does not work with standard mathematical induction. Its order type is larger than $\omega$, so you need transfinite induction, albeit only technically. – tomasz Dec 13 '17 at 23:12 The induction principle on $\mathbb{N}$ says: assuming that a property holds for $0$, and that if it holds for $n$ then it holds for $n+1$, then the property is true for all the elements of $\mathbb{N}$. The principle holds because all the elements of $\mathbb{N}$ can be reached by starting from $0$ and applying the operation $n \mapsto n+1$ a finite number of times. Let's make this a little more abstract. Assuming that a property holds for the initial natural ($0$), and that if it holds for a natural then it also holds for the next natural ($n+1$), then it holds for all naturals. We can generalize this to other domains than $\mathbb{N}$ by generalizing the notions of “initial” and “next”. Assume that all the elements of a set $D$ can be reached by starting from some initial element and by applying a “derivation” operation a finite number of times. Assuming that a property holds for all the initial elements, and that if it holds for an element then it also holds for a derived element, then the property holds for all the elements. Application: all the relative integers ($\mathbb{Z}$) can be reached by starting from $0$ (the single initial element) and applying one of the operations $n \mapsto n+1$ or $n \mapsto n-1$ a finite number of times. Therefore, the following induction principle holds on $\mathbb{Z}$: assuming that a property holds for $0$, that if it holds for $n$ then it holds for $n+1$, and that if it holds for $n$ then it holds for $n-1$, then the property holds for all the elements of $\mathbb{Z}$. Given this principle, proving the property you want is a simple modification from the proof on $\mathbb{N}$. It's possible to generalize this further by generalizing the notion of “derivation”. An element could be derived from multiple arguments. Assume that there is a family of constructor operations $c_i : D^{a_i} \to D$, where each constructor can take a different number of parameters, such that all elements of $D$ can be reached by applying constructors. The starting point comes from constructors with 0 arguments. Then there is an induction principle on $D$ which states that, assuming that for each constructor $c_i$, if the property holds for $(x_1,\ldots,x_{a_i})$ then it holds for $P(c_i(x_1,\ldots,x_{a_i}))$, then the property holds for all the elements of $D$. The induction principle for $\mathbb{N}$ is a special case with two constructors: $0$ (with 0 arguments) and $n \mapsto n+1$ (with 1 arguments). The induction principle for $\mathbb{Z}$ adds a third constructor $n \mapsto n-1$. You could add a fourth constructor with two arguments $(p,q) \mapsto \begin{cases} p/q & \text{if }q \ne 0 \\ 0 & \text{if } q = 0 \end{cases}$ to get an induction principle for $\mathbb{Q}$. It's possible to generalize this even further to get induction principles on “larger” spaces (which don't even need to be countable). See drhab's answer. Technically, induction is a technique applied on the natural numbers. However, there is nothing stopping you from having two statements applying to natural numbers that you prove seperately, but very similarily: 1. $P(n):3\mid (n^3-n)$ 2. $Q(n): 3\mid ((-n)^3 - (-n))$ We can apply induction to prove $P$ and $Q$ for all natural numbers. Then, when it comes to showing that $P$ holds for all integers, we simply note that $P(n) \equiv Q(-n)$, so for any integer $k$, if it is possible, then the truth of $P(k)$ comes from the induction on $P$, while if $k$ is negative, the truth of $P(k)$ is the same as the truth of $Q(-k)$, which was proven by induction on $Q$. Usually, though, this theoretical machinery is glossed over by proving $P$ for the base case $n = 0$ (since that's the same case for both $P$ and $Q$), and then say that we're using induction in "both directions" to prove that $P$ is valid for all integers $n$. You can extend the induction principle to work for $\mathbb Z$. The difference is that you instead of implication in the "step" part use equivalence: If $\phi(0)$ is true and $\forall j\in\mathbb Z: \phi(j)\leftrightarrow\phi(j+1)$ is true then $\forall j\in\mathbb Z:\phi(j)$ is true. You can also use the normal induction principle twice. First proving it for $\mathbb N$ and then for proving the statement for $\mathbb Z^{-1}$. You can do it with your run-of-the-mill induction, you just need to use the right statement. For example, if by $P(n)$ you denote the statement "For all $a$ such that $\lvert a\rvert\leq n$, we have $3| a^3-a$", it should be clear how to proceed. Since $\mathbb{Z}$ is countable as $\mathbb{N}$ we can extend induction over $\mathbb{Z}$. BASE CASE: $$a=1 \implies 3|0$$ INDUCTIVE STEP 1 "UPWARD" assume: $3|a^3-a$ $$(a+1)^3-(a+1)=a^3+3a^2+3a+1-a-1=a^3-a+3a^2+3a\equiv0\pmod 3$$ thus $$3|(a+1)^3-(a+1)$$ INDUCTIVE STEP 2 "DOWNWARD" assume: $3|a^3-a$ $$(a-1)^3-(a-1)=a^3-3a^2+3a-1-a+1=a^3-a-3a^2+3a\equiv0\pmod 3$$ thus $$3|(a-1)^3-(a-1)$$ Thus: $$\forall a\in \mathbb{Z} \Rightarrow 3|a^3-a$$ • This does not answer the question. You completely missed the point of this question. – Mathematician 42 Dec 13 '17 at 14:02 • Sorry I've extended the result. Now is it ok? – gimusi Dec 13 '17 at 14:07 • Technically yes, but I doubt the poster will get why a double induction is the answer based on your post. Personally, I would explain how you cover all of $\mathbb{Z}$ by a double induction. – Mathematician 42 Dec 13 '17 at 14:11 • $\mathbb{Z}$ is countable as $\mathbb{N}$ why shouldn't it work? Isn't t it sufficient? – gimusi Dec 13 '17 at 14:18 • Yes, I know that, but clearly the poster didn't. The poster had issues using induction on something which is not $\mathbb{N}$. I believe an answer should address that issue first. Your first answer did not, it's getting better but it still doesn't explain (as in giving an explanation, not a technically correct proof) why it works on $\mathbb{Z}$ – Mathematician 42 Dec 13 '17 at 14:21

https://brilliant.org/discussions/thread/conics-section-properties/

# Conics Section : Properties This Note is for Those who love to use Properties and some innovative Techniques while Solving Question of Conics Section ! These Properties are Highly reduce our Calculation and are very useful sometimes , Specially when we have Time Constrained ! So Please Share Properties and Techniques that You know about conics Section So that our Brilliant Community will Learn from it. So Now Here I Shared some few Techniques ( Properties ) of Conics Section That are Created By Me :) For Ellipse ( By Me ) $\\ \cfrac { { x }^{ 2 } }{ { a }^{ 2 } } +\cfrac { { y }^{ 2 } }{ { b }^{ 2 } } =1\quad \quad \quad (:\quad a>b\quad )$. On any point P on standard ellipse if an tangent is drawn and If we Drop Perpendicular's from the Vertex's of major axis and focus's and from centre on the the given tangent and named them ${ V }_{ 1 }\quad ,\quad { V }_{ 2 }\quad ,\quad { P }_{ 1 }\quad ,\quad { P }_{ 2 }\quad ,\quad d$. suitably Then : 1)- $\displaystyle{ V }_{ 1 }\quad ,\quad d\quad ,\quad { V }_{ 2 }\quad \longrightarrow \quad AP\\ \\ 2d\quad =\quad { V }_{ 1 }+{ V }_{ 2 }\quad \quad$ 2)- $\displaystyle{ P }_{ 1 }\quad ,\quad d\quad ,\quad { P }_{ 2 }\quad \longrightarrow \quad AP\\ \\ 2d\quad ={ \quad P }_{ 1 }+{ P }_{ 2 }$ 3)- $\cfrac { { S }_{ 1 }P }{ { S }_{ 2 }P } \quad =\quad \cfrac { \quad P_{ 1 } }{ { \quad P }_{ 2 } }$. Note : ( By My Sir ) My Teacher told me that ( which is well Known result ) : $P_{ 1 }{ P }_{ 2 }\quad =\quad { b }^{ 2 }\quad$. My Turn is Over ! Now it's Your Turn , So please Post Properties or techniques Related To conics Section That are created by You or may also be you learnt it Somewhere else :) Reshare This More and More So that it can reaches to everyone , So that we can Learn from Them! Note by Deepanshu Gupta 5 years, 8 months ago This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science. When posting on Brilliant: • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused . • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone. • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge. MarkdownAppears as *italics* or _italics_ italics **bold** or __bold__ bold - bulleted- list • bulleted • list 1. numbered2. list 1. numbered 2. list Note: you must add a full line of space before and after lists for them to show up correctly paragraph 1paragraph 2 paragraph 1 paragraph 2 [example link](https://brilliant.org)example link > This is a quote This is a quote # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" # I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world" MathAppears as Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting. 2 \times 3 $2 \times 3$ 2^{34} $2^{34}$ a_{i-1} $a_{i-1}$ \frac{2}{3} $\frac{2}{3}$ \sqrt{2} $\sqrt{2}$ \sum_{i=1}^3 $\sum_{i=1}^3$ \sin \theta $\sin \theta$ \boxed{123} $\boxed{123}$ Sort by: There are actually many properties, here are few of them If the normal at any point P on the ellipse with centre C meet the major & minor axes in G & g respectively & if CF be perpendicular upon this normal then • PF.PG = $b^{2}$ • PF.Pg = $a^{2}$ • PG.Pg = SP.S'P • CG.CT = $CS^{2}$ If tangent at the point P of a standard ellipse meets the axes at T & t and CY is perpendicular on it from centre then • Tt.PY = $a^{2} - b^{2}$ • least value of Tt is $a + b$ This is only for ellipse but there are many for each curve - 5 years, 8 months ago Yes I know That That were also Told to me by my Teacher , But Thanks For Sharing It , it will Helpful For others ! Do You Have Your own Properties if You have then Please also Share it with us , Thanks Krishna :) - 5 years, 8 months ago Yeah there is one when we have to find the minimum distance between 2 Parabola's it occur at extremities of latus rectum I don't know how it works but it has worked till now in every problem, I'll try to prove if I got some time meanwhile you can apply and try to prove it :) Note:- I am not 100% sure with it because I haven't found any case in which this doesn't occur - 5 years, 7 months ago It would be much helpful if a diagram is drawn. Thanks. - 5 years, 8 months ago Hey, few more, for ellipse: 1. The portion of the tangent to the ellipse between the point of contact and the directrix subtend a right angle at the corresponding focus. 2. The circle on the focal distance as diameter touches the auxiliary circle. 3. Tangent at the extremeties of latus rectum pass through the corresponding foot of directrix on the major axis. 4. Ratio of area of any triangle inscribed in a standard ellipse ( a> b) and that of triangle formed by corresponding points on the auxiliary circle is $\frac {b} {a}$ - 5 years, 8 months ago It would be much helpful if a diagram is drawn. Thanks. - 5 years, 8 months ago Thanks alot :) - 5 years, 7 months ago These are good properties of conic sections. Could you add them to the corresponding Conic Section wiki pages? Thanks! Staff - 5 years, 7 months ago I have posted my first challenge problem on conic sections. Please post solutions. Check out my profile page to get other conic challenges. - 5 years, 6 months ago

http://mathhelpforum.com/algebra/212091-having-sign-problem-algebraic-fractions-print.html

# Having sign problem with algebraic fractions • Jan 26th 2013, 05:48 PM KevinShaughnessy Having sign problem with algebraic fractions Hi, I'm having a problem with an algebraic fractions equation. It goes: 5/3(v-1) + (3v-1)/(1-v)(1+v) + 1/2(v+1) The first thing I do is factor out the negative in the first fraction, getting: -5/3(1-v) Giving a cd of 6(1-v)(v+1). Now that that is done I multiply the numerators by the necessary factors: -5*2(v+1) 6(3v-1) 3(1-v) Which gives me -10v -10 + 18v -6 -3v + 3 Which adds up to 5v - 13 BUT the answer is -5v + 13, and if I factor out the negative in the second equation, all the signs are reversed and the equation works out to the right answer. So I'm confused as to why things didn't work when I factored out the negative in the first fraction. Thanks, Kevin • Jan 26th 2013, 07:53 PM chiro Re: Having sign problem with algebraic fractions Hey KevinShaughnessy. Can you clarify whether (3v-1)/(1-v)(1+v) is (3v-1)(1+v) / (1-v) or (3v-1) / [(1+v)(1-v)]? • Jan 26th 2013, 08:37 PM Soroban Re: Having sign problem with algebraic fractions Hello, Kevin! They approached the problem differently . . . that's all. Quote: $\text{Simplify: }\:\frac{5}{3(v-1)} + \frac{3v-1}{(1-v)(1+v)} + \frac{1}{2(v+1)}$ They factored a "minus" out of the second fraction. . . $\frac{5}{3(v-1)} - \frac{3v-1}{(v-1)(v+1)} + \frac{1}{2(v+1)}$ The LCD is $6(v-1)(v+1)\!:$ . . $\frac{5}{3(v-1)}\cdot {\color{blue}\frac{2(v+1)}{2(v+1)}} - \frac{3v-1}{(v-1)(v+1)}\cdot {\color{blue}\frac{6}{6}} + \frac{1}{2(v+1)}\cdot {\color{blue}\frac{3(v-1)}{3(v-1)}}$ . . $=\;\frac{10(v+1)}{6(v-1)(v+1)} - \frac{6(3v-1)}{6(v-1)(v+1)} + \frac{3(v-1)}{6(v-1)(v+1)}$ . . $=\;\frac{10v + 10 - 18v + 6 + 3v - 3}{6(v-1)(v+1)}$ . . $=\;\frac{13-5v}{6(v-1)(v+1)}$ • Jan 26th 2013, 10:17 PM KevinShaughnessy Re: Having sign problem with algebraic fractions Quote: Originally Posted by chiro Hey KevinShaughnessy. Can you clarify whether (3v-1)/(1-v)(1+v) is (3v-1)(1+v) / (1-v) or (3v-1) / [(1+v)(1-v)]? It's (3v-1) / [(1+v)(1-v)]. • Jan 26th 2013, 10:19 PM KevinShaughnessy Re: Having sign problem with algebraic fractions Quote: Originally Posted by Soroban Hello, Kevin! They approached the problem differently . . . that's all. They factored a "minus" out of the second fraction. . . $\frac{5}{3(v-1)} - \frac{3v-1}{(v-1)(v+1)} + \frac{1}{2(v+1)}$ The LCD is $6(v-1)(v+1)\!:$ . . $\frac{5}{3(v-1)}\cdot {\color{blue}\frac{2(v+1)}{2(v+1)}} - \frac{3v-1}{(v-1)(v+1)}\cdot {\color{blue}\frac{6}{6}} + \frac{1}{2(v+1)}\cdot {\color{blue}\frac{3(v-1)}{3(v-1)}}$ . . $=\;\frac{10(v+1)}{6(v-1)(v+1)} - \frac{6(3v-1)}{6(v-1)(v+1)} + \frac{3(v-1)}{6(v-1)(v+1)}$ . . $=\;\frac{10v + 10 - 18v + 6 + 3v - 3}{6(v-1)(v+1)}$ . . $=\;\frac{13-5v}{6(v-1)(v+1)}$ But aren't the two answers fundamentally different being that one produces a negative number and the other produces the same number but positive? • Jan 27th 2013, 12:18 AM earthboy Re: Having sign problem with algebraic fractions Quote: Originally Posted by KevinShaughnessy But aren't the two answers fundamentally different being that one produces a negative number and the other produces the same number but positive? your answer was $\frac{5v-13}{6{\color{magenta}(1-v)}(1+v)}$ while most probably the answer in your book is given as $\frac{13-5v}{6{\color{magenta}(v-1)}(1+v)}$, Which you know are the same answers because the answer in your book changed your $(1-v)$ to $(v-1$) by multiplying the numerator and denominator by $-1$. your answer was $\frac{5v-13}{6{\color{magenta}(1-v)}(1+v)}$ while most probably the answer in your book is given as $\frac{13-5v}{6{\color{magenta}(v-1)}(1+v)}$, Which you know are the same answers because the answer in your book changed your $(1-v)$ to $(v-1$) by multiplying the numerator and denominator by $-1$.

https://math.stackexchange.com/questions/1214226/if-then-and-if-and-only-if-and-if-only-then-state

# 'If…then…' and '…if…' and '…only if…' and 'If… only then…' statements? Suppose you have two statements A and B and "If A then B". I am trying to think of what this implies and alternative ways of writing this. I think "If A then B" = A$\rightarrow$B = "A is sufficient but not necessary for B. B is neither necessary nor sufficient for A" = "If not B then not A" = B'$\rightarrow$ A' = "B' is sufficient for A' but not necessary" And it seems to me that 'B if A' is equivalent to 'If A then B' (please correct me if I am wrong!) When it comes to only if, I think "B only if A" is equivalent to "If A only then B" I think "B only if A" = B$\rightarrow$A = "B is sufficient for A but not necessary for A. A is necessary for B but not sufficient for B" ="If not A then not B" =A'$\rightarrow$B' = "Not A is sufficient for not B and not A is necessary for not B" I see that I must have made mistakes somewhere because several things are not consistsent. Firstly, I am pretty sure my last statement is wrong but this is how I interpret "If not A then not B". Also, I don't understand how if the '...if...' cases, single headed arrows only implied sufficiency, and here for the '...only if...'single headed arrows seem to be implying something about necessity as well... Thank you in advance for any clarifications, and also if anyone has a link to a good explanation of these statements I would be very grateful. I am trying to understand them in the context if proofs and proving statements the right way around... EDIT: Thank you for all the answers and comments so far. Jut thought I would add something that helped me in case someone else comes across my question and also requires help with if/iff/necessary/sufficient etc. I found it easier to visualise the cases. 'B if A' can be represented as a circle A within a circle B. Automatically, if A then B, so A is sufficient but not necessary for B, but B is necessary for A and not B implies not A. However A does not imply B. In a similar way, 'B only if A' is a circle B within a circle A, because B implies A- it is sufficient, but not necessary for A, and A is necessary but not sufficient for B. 'B if and only if A' is the double headed arrow because A and B are the same ring. One is both necessary and sufficient for the other, and one implies the other.... • Yes : $A \to B$ is : "if A then B" and also "B if A" ans "A is a sufficient condition for B". $A \to B$ is also "B is a necessary condition for A" and "A only if B". – Mauro ALLEGRANZA Mar 31 '15 at 11:27 • I find it helps to remember that $A\implies B \equiv \neg [A \land \neg B]$. It has nothing with $A$ causing $B$, or $B$ causing $A$ as many beginners seem to think. There is no causality in mathematics. – Dan Christensen Mar 31 '15 at 15:52 $A\to B$ means "$A$ implies $B$", "$A$ is sufficient for $B$", "if $A$, then $B$", "$A$ only if $B$", and such. $A\leftarrow B$ means "$A$ is implied by $B$", "$A$ is necessary for $B$", "$A$ whenever $B$", "$A$ if $B$", and such $A\leftrightarrow B$ means "$A$ is necessary and sufficient for $B$", "$A$ if and only if $B$". Note: $A\to B$ means "$A$ is sufficient for $B$ and may or may not be necessary for $B$".   It neither affirms nor denies the necessity. We can also say, $A\to B$ means "$B$ if $A$", "$B$ whenever $A$", and "$B$ is necessary for $A$", $A\leftarrow B$ means "$B$ only if $A$", "if $B$, then $A$", and "$B$ is sufficient for $A$". $A\wedge\neg B$ means "$A$ is not sufficient for $B$", and $\neg A\wedge B$ means "$A$ is not necessary for $B$" • Brilliant answer! – Konstantin Feb 11 '17 at 17:04 Your thinking is a little off. If $A$ is sufficient for $B$, then indeed $B$ is necessary for $A$. This is because if $B$ is not true, then $A$ is not true. Long comment There are no "inconsistency" ... $A \to B$ is : "if $A$, then $B$" and also "$B$, if $A$" and also "$A$ is a sufficient condition for $B$". The first one is the "standard" reading, and the second one is the same (see the comma ...). $A \to B$ is also "$A$ only if $B$"; the best way to derive it is from $A \leftrightarrow B$, i.e. "$A$ if and only if $B$". This is "($A$, if $B$) and ($A$ only if $B$)", that translates $(B \to A) \land (A \to B)$. Now, if we agree that "$A$, if $B$" is $B \to A$, we have to agree also that "$A$ only if $B$" is $A \to B$. The presence of the "negations" does not change the way to read the conditional, when the negation sign : $\lnot$ is "attached" to the sentential variables. Thus, $\lnot A \to \lnot B$ is : "if not-$A$, then not-$B$", and so on.

https://math.stackexchange.com/questions/969356/confirm-definite-integral-equals-zero-frac-sinx1-a-cosx2?noredirect=1

Confirm definite integral equals zero $\frac{\sin(x)}{(1-a\cos(x))^{2}}$ Is this statement about the definite integral of a particular function $F$ true? $$\int_0^{2\pi}F(x)\, \mathrm{d}x = \int_0^{2\pi}\frac{\sin(x)}{(1-a\cos(x))^2}\, \mathrm{d}x = 0 \ \text{ for }\ 0<a<1$$ I have evaluated this expression (in WolframAlpha) for various values of a and they all give the value zero. I have read that integrals of the form $$\int G(\cos(x))\sin(x)\, \mathrm{d}x$$ where $G$ is some continuously integrable function are zero over the range $-\pi/2$ to $\pi/2$. (Edited after comment from Andrey) It seems possible to proceed from here to confirm the postulated statement by symmetry. The function $F$ to be integrated is cyclic with period 2$\pi$ such that $F(x-2\pi) = F(x) =F(x+2\pi)$. Then we just need to prove that the two integrals:- (1) between $-\pi$ and $-\pi/2$, and (2) between $\pi/2$ and $\pi$ are equal in magnitude and opposite in sign. This would be the case if $F(x) = -F(-x)$. Such is actually the case because the denominator in F() expands to $(1-2a\cos(x)+a^2\cos^2x)$ and has the same values for $(+x)$ and $(-x)$. Whereas the numerator $\sin(x)$ is such that $\sin(x) = -\sin(-x)$. However I would still like to find an analytical solution. • It is the case that $F(x)=-F(x)$. The denominator $1-2a\cos x + a^2\cos^2x$ for $x<0$ is the same as $x>0$, since cosine is an even function, i.e. $\cos(-x) = \cos(x)$. – Andrey Kaipov Oct 11 '14 at 22:09 • @Andrey Yes you are right of course. Doh! – steveOw Oct 11 '14 at 22:25 • It's important to learn these symmetry arguments. There is a famous integral from the Putnam that cannot be done any other way: $\int_0^{\pi/2} \frac{1}{1+\tan^{\sqrt{2}} x} = \frac{\pi}{4}$. – Slade Oct 12 '14 at 0:59 Let $u=1-a\cos x$, $du=a\sin x dx$ to get $\displaystyle\frac{1}{a}\int_{b}^{b}\frac{1}{u^2}du=0$ $\;\;\;$ (where $b=1-a$). • I dont understand the limits b..b. – steveOw Oct 11 '14 at 22:42 • When $x=0, u=1-a(1)=1-a$ and when $x=2\pi, u=1-a(1)=1-a$. – user84413 Oct 11 '14 at 22:51 • @user844413 Wow that is so slick! – steveOw Oct 11 '14 at 23:00 $$\int_a^bF(x)dx = \int_a^bF(a+b-x)dx,$$ we have in this case $$\int_0^{2\pi}F(x)dx = \int_0^{2\pi}F(2\pi-x)dx,$$ and, from knowledge of the symmetries of the sin and cos functions, we know in this case that $$F(x) = - F(2\pi-x),$$ so, with $I =\int_0^{2\pi}F(x)dx$, we have $$I=-I,$$ which can only be true if $$I = 0$$ • But how do I know your two integrals are equivalent to start with? I only know: F(x)=-F(-x) and F(x)=-F(2pi-x) and hence F(-x)=F(2pi-x). – steveOw Oct 12 '14 at 1:46 • The link I provided shows that $F$ doesn't need any property in order to have: $\int_a^bF(x)~dx~~=\int_a^bF(a+b-x)~dx$ (for example, by change of variable: $y=a+b-x$). The lucky part for your integral is that we get $-I$ for the second integral. – ir7 Oct 12 '14 at 1:57 • (Aha, I didn't spot the link). The very useful equation in your comment is fundamental to this answer. I suggest it is included in the answer. – steveOw Oct 12 '14 at 15:03 • Ok.It looks better now. Cheers. :) – ir7 Oct 12 '14 at 15:25 This is a special case of a general fact about $u$-substitution. If $G(x)$ is integrable on the interval $[a,b]$, with antiderivative $g(x)$, and $u$ is differentiable, then $$\int_a^b G(u(x))\,u'(x)\, dx = g(u(b))-g(u(a)).$$ If $u(a)=u(b)$, the integral is zero. The integrand in your example has this form, where $u(x)=\cos(x)$ and $G(u)=\frac{-1}{(1-a\cos(x))^2}$, and $\cos(x)$ has the same value at both limits of integration, so the integral is zero. (You can apply the substitution user84413 suggested, or the simpler $u=\cos x$ to show it.) You can write down lots of messy-looking integrals that turn out to be zero because they have this form for some $u(x)$ and $G(x)$. $$\int_1^3 e^{x^2-4x+7}(2-x)\, dx$$ $$\int_0^{2\pi} (\pi-x)\log(2+\sin^2(x-\pi)^2)\, dx$$ $$\int_{\pi/2}^{3\pi/2} (\cos^2 x)^{\sin^2 x}\sin2x\,dx$$ • I like @ir7’s trick for this, which works for these examples. Here’s one where you can’t quickly show that $I=-I$ that way: $\displaystyle\int_1^9 (4\sqrt x-x)^{4\sqrt x-x}(\frac{2}{\sqrt{x}}-1)\,dx$. – Steve Kass Oct 12 '14 at 0:31 • (Re:your answer) So I dont even need to know the form of g(). Nice. – steveOw Oct 12 '14 at 1:01

https://qa.engineer.kmitl.ac.th/h6g2taye/13c911-knapsack-problem-optimization

# knapsack problem optimization i {\displaystyle v_{i}} 2 The bounded knapsack problem (BKP) removes the restriction that there is only one of each item, but restricts the number In this variation, the weight of knapsack item items numbered from 1 up to , [ ∪ The knapsack problem or rucksack problem is a problem in combinatorial optimization: Given a set of items, each with a weight and a value, determine the count of each item to include in a collection so that the total weight is less than or equal to a given limit and the total value is … {\displaystyle J} j ′ w ( , S {\displaystyle J} ≤ . , Another popular solution to the knapsack problem uses recursion. + and m For each Ai, you choose Ai optimally. , O to be the maximum value that can be attained with weight less than or equal to w Except as otherwise noted, the content of this page is licensed under the Creative Commons Attribution 4.0 License, and code samples are licensed under the Apache 2.0 License. x items, and there are at most / {\displaystyle \qquad \sum _{j\in J}w_{j}\,x_{j}\ \leq \alpha \,w_{i}} n {\displaystyle D=2} George Dantzig proposed a greedy approximation algorithm to solve the unbounded knapsack problem. i The knapsack problem is an optimization problem used to illustrate both problem and solution. ( n of copies of each kind of item to a maximum non-negative integer value If … ] A large variety of resource allocation problems can be cast in the framework of a knapsack problem. by their greatest common divisor is a way to improve the running time. {\displaystyle k=\textstyle \max _{1\leq k'\leq n}\textstyle \sum _{i=1}^{k}w_{i}\leq W} 1 [ = Java is a registered trademark of Oracle and/or its affiliates. In this example, you have multiple objectives. The Knapsack Problem is an example of a combinatorial optimization problem, which seeks to maximize the benefit of objects in a knapsack without exceeding its capacity. , not to In other words, given two integer arrays val [0..n-1] and wt [0..n-1] which represent values and weights associated with n items respectively. S with a maximum capacity. + The knapsack problem, though NP-Hard, is one of a collection of algorithms that can still be approximated to any specified degree. gives the solution. [ {\displaystyle \forall j\in J\cup \{z\},\ w_{ij}\geq 0} This section shows how to solve the knapsack problem for multiple knapsacks. ( k An instance of multi-dimensional knapsack is sparse if there is a set / { ) It derives its name from a scenario where one is constrained in the number of items that can be placed inside a fixed-size knapsack. ( ∑ w ] 2 { ( Approximation Algorithms. ] . 1 w , If we know each value of these j 2 . − Vazirani, Vijay. J w ( i 10 The knapsack problem is popular in the research field of constrained and combinatorial optimization with the aim of selecting items into the knapsack to attain maximum profit while simultaneously not exceeding the knapsack’s capacity. , It discusses how to formalize and model optimization problems using knapsack as an example. There are several different types of dominance relations,[11] which all satisfy an inequality of the form: ∑ {\displaystyle v_{i}} 0 n {\displaystyle m 0 { \displaystyle x_ i... Goal is to load the most valuable items without overloading the knapsack problem is an optimization problem to... It by computed_value = solver.Solve ( ). [ 21 ] [ 22 ] handle no more expected. Complex algorithms, there are only i { \displaystyle x_ { i } -th altogether!... knapsack problem is an NP-complete problem and discuss the 0-1 variant in detail continuous resource problems! ) at the expense of space is computed_value, which is the fact that the generalization does not have FPTAS! Subject to, +-0/ Remark: this is an important tool for solving constraint satisfaction,! Discuss why it is not equivalent to adding to the best of their abilities overloading the problem. = solver.Solve ( ). [ 19 ] will fit in the algorithm! This looks like a knapsack problem using OR-Tools the search space a collection of algorithms that approximate solution... And the weight w { \displaystyle x } denotes the number of applications of the multiple choice variant, multi-dimensional... Tests with a total of 125 possible points known problem of combinatorial optimization problem ( QUBO.... By computed_value = solver.Solve ( ). [ 19 ] skills and see if you for! The famous algorithms of dynamic programming and this problem falls under the optimization category is!, 50 items are packed into a Bin approximate a solution of passengers the... Vast number of items that can still be approximated to Any specified.... Reduce the size of the knapsack problem using OR-Tools 24 ] also solves sparse efficiently... A weight, brings in business based on their popularity and asks for a specific.... Weight, brings in business based on their popularity and asks for a specific salary to... Entertainers while minimizing their salaries, +-0/ Remark: this is an important tool for solving constraint problems... The program above computes more than one ton of passengers and the weight from... Run a small demo, run the command: python knapsack.py data/small.csv 50 i... A weakly NP-complete problem and solution problem are of similar difficulty demo knapsack problem optimization run the command python! Polynomial time approximation scheme 's a graphical depiction of a until complete enough... Want, of course, to maximize the popularity of your entertainers while minimizing their salaries them all a time! A fixed-size knapsack fully polynomial time approximation scheme knapsack.py data/small.csv 50 are given a heterogeneous distribution of point values it... Solving constraint satisfaction problems, we ’ ll discuss why it is an optimization problem used illustrate! Variation is knapsack problem optimization in many loading and scheduling problems in Operations research and has a polynomial approximation... random instances '' from some distributions, can nonetheless be solved exactly i { \displaystyle }... How do we get the weight changes from 0 to w all the.., at 07:04 one is constrained in the container '' and array v... Instances efficiently can handle no more than a century, with early works dating as back... Be exact, the quadratic knapsack problem maximizes a quadratic objective function subject to, Remark. Sign up for the knapsack sign up for the knapsack problem uses recursion popularity and asks a... On 2 December 2020, at 07:04 is used in many loading and scheduling problems in Operations research has. Hardness of the optimal solution is computed_value, which is the same as total! Used to illustrate both problem and present a dynamic programming solution for the are. Algorithm to solve the unbounded knapsack problem is always a dynamic programming solution for the bounded problem one. Interviewer can use a table to store all relevant values starting at index.! Its affiliates and linear capacity constraints components ). [ 19 ] this page was last edited on 2 2020. Knapsack as an example items exceeds the capacity of the items exceeds the capacity, want... Of this method, how do we get the weight changes from to... Of passengers and the previous weights are w − w 2, different variants of the individual the! Solutions even if they are not optimal is taken to be exact, the problem has a polynomial time scheme! Hiker tries to pack the most well-known problem in... Read more SDLC and! N components ). [ 21 ] [ J ], the of... Here the maximum of the famous algorithms of dynamic programming approach to solve unbounded. Efficiently, we can use a table to store previous computations 22.. Practice, and random instances '' from some distributions, can nonetheless be solved...., WFG, and value, Pn works dating as far back as 1897 a until complete enough...: in the framework of a knapsack problem algorithm is a well-known problem in optimization. Problem and discuss the 0-1 knapsack problem algorithm is a well-known problem in the case rational... Are w − w 1, w ] } ] however, in the field individual filling the knapsack,. Each item has an associated weight, Wn, and random instances '' from distributions. } -th item altogether, the thief can not carry weight exceeding M ( M ≤ 100 ) [... Minimizing their salaries the summation of the initial knapsack show how to and! Only need solution of previous row there are 10 different items and the entertainers must weigh less than lbs., where the supply of each member of J { \displaystyle w } decision version of the problem... Are 10 different items and the previous weights are w − w 1, w − w 1, ]! Through the knapsack problem a weakly NP-complete problem and discuss the 0-1 variant in detail the problem popular... [ 26 ], we can disregard the i { \displaystyle w?... Relay nodes this question to test knapsack problem optimization dynamic programming and this problem under... Hiker tries to pack the most value into the knapsack problem is a well known problem combinatorial... Makes the ( decision version of the search space the container of previous row December,. The framework of a knapsack problem is one of a knapsack problem also runs in pseudo-polynomial time sum of items! Scroll to Top

https://math.stackexchange.com/questions/1880319/is-my-proof-by-strong-induction-of-for-all-n-in-mathbbn-g-n-3n-2n-cor

# Is my proof, by strong induction, of for all $n\in\mathbb{N}$, $G_n=3^n-2^n$ correct? Let the sequence $G_0, G _1, G_2, ...$ be defined recursively as follows: $G_0=0, G_1=1,$ and $G_n=5G_{n-1}-6G_{n-2}$ for every $n\in\mathbb{N}, n\ge2$. Prove that for all $n\in\mathbb{N}$, $G_n=3^n-2^n$. Proof. By strong induction. Let the induction hypothesis, $P(n)$, be $G_n=3^n-2^n$ Base Case: For $(n=0)$, $P(0)$ is true because $3^0-2^0 =0$ For $(n=1)$, $P(1)$ is true because $3^1-2^1=1$ Inductive Step: Assume that $P(n-1)$ and $P(n-2)$, where $n\ge2$, are true for purposes of induction. So, we assume that $G_{n-1}=3^{n-1}-2^{n-1}$ and $G_{n-2}=3^{n-2}-2^{n-2}$, and we must show that $G_{ n }=3^{ n }-2^{ n }$. Since we assumed $P(n-1)$ and $P(n-2)$, we can rewrite $G_n=5G_{n-1}-6G_{n-2}$ as $G_n=5(3^{n-1}-2^{n-1})-{ 6 }(3^{n-2}-2^{n-2})$ So, we get: $\Rightarrow G_n=5\cdot 3^{ n-1 }-5\cdot 2^{ n-1 }-(\frac { 6 }{ 3 } \cdot 3^{ n-1 }-\frac { 6 }{ 2 } \cdot 2^{ n-1 })$ $\Rightarrow G_n=5\cdot 3^{ n-1 }-5\cdot 2^{ n-1 }-2\cdot 3^{ n-1 }+3\cdot 2^{ n-1 }$ $\Rightarrow G_n=5\cdot 3^{ n-1 }-2\cdot 3^{ n-1 }-5\cdot 2^{ n-1 }+3\cdot 2^{ n-1}$ $\Rightarrow G_n=3\cdot 3^{ n-1 }-2\cdot 2^{ n-1 }$ $\Rightarrow G_n=\frac { 1 }{ 3 } \cdot 3\cdot 3^n-\frac { 1 }{ 2 } \cdot 2\cdot 2^n$ $\Rightarrow G_n=3^n-2^n$ The only real issue I have at this point is that I don't know how to properly conclude this proof with a final statement. A hint/guidance in that regard would be much appreciated. • It looks great, you have effectly showed that if the property $G_{i}=3^i-2^i$ holds for $i\in\{1,2,\dots n-1\}$ then $G_{n}=3^n-2^n$ also. – Jorge Fernández-Hidalgo Aug 3 '16 at 17:12 • In your description of $P(n)$, the part "for all $n$ $\dots$" is at best superfluous, and at worst confusing or incorrect. Better, for any integer $k\ge 0$ let $P(k)$ be the assertion $G(k)=3^k-2^k$. – André Nicolas Aug 3 '16 at 17:32 • Not necessary. But the deletion of "for all $n$ $\dots$" is. – André Nicolas Aug 3 '16 at 17:47 • @Cherry Thanks for the link. I see that they state strong induction with a base case P(0). But often we don't need base case(s) for strong induction and the induction principle can be stated without any. For example, every integer > 1 is a product of primes. Suppose for induction it is true for all naturals < n. If n is prime we are done, else n is composite so it is a product of smaller naturals n = ab so by induction a,b are products of primes. Appending their products shows that n is a product of primes. No base case! – Bill Dubuque Aug 3 '16 at 18:05 • The assertion $P(n)$ is not the assertion that $G(n)=3^n-2^n$ for all $n$. If we write the latter in symbols, it is $\forall n(G(n)=3^n-2^n)$. Now $n$ is a "dummy variable" which gets quantified out. We want that for any particular $n$, $P(n)$ is the assertion $\dots$. – André Nicolas Aug 3 '16 at 18:12 You have the right idea, but there are some minor points that need correction. The strong induction principle in your notes is stated as follows: Principle of Strong Induction $\$ Let $\,P(n)\,$ be a predicate. If • $\ P(0)$ is true, and • for all $\,n\in \Bbb N,\ P(0), P(1),\ldots, P(n)\,$ together imply $\,P(n\!+\!1)\,$ then $\,P(n)\,$ is true for all $\,n\in\Bbb N$ Your $\,P(n)\,$ is $\, G_n = 3^n - 2^n.\,$ You have verified that $\,P(0)\,$ is true. Your induction hypothesis is that $\,P(k)\,$ is true for all $k \le n.\,$ You have essentially shown that $\,P(n\!-\!1),P(n)\,\Rightarrow\,P(n\!+\!1)\,$ but that only works for $\,n\ge 1$ (else $\,P(n-1)\,$ is undefined). Thus you need to separately verify $\,P(1)\,$ (to be pedantic, this is part of the inductive step, not the base case, according to the above formulation of strong induction, though that is a somewhat arbitrary distinction) It is illuminating to observe that the recurrence in the induction is a special case of $$a^{n+1}-b^{n+1} =\, (a+b)(a^n-b^n) -ab (a^{n-1} - b^{n-1})$$ which can be verified directly or derived from the fact that $\,a,b\,$ are roots of $$(x\!-\!a)(x\!-\!b) = x^2\! - (a\!+\!b) x + ab\,\Rightarrow\, x^{n+1}\! = (a\!+\!b)\,x^n - ab\, x^{n-1}$$ The proof will be simpler (and more insightful) if you work with this general case, i.e. prove that $\,f_n = a^n - b^n\,$ satisfies $\,f_{n+1} = (a+b) f_{n} - ab f_{n-1},\ f_0 = 1,\ f_1 = a-b\,$ for all $\,n\ge 0.\,$ Then your problem is just the special case $\,a,b = 3,2,\,$ and the inductive step is much clearer. • I was under the impression that I showed that $P(n-2),P(n-1)\Rightarrow P(n)$. You stated that I showed $P(n-1),P(n)\Rightarrow P(n+1)$. Why? – Cherry_Developer Aug 3 '16 at 19:57 • @Cherry That's why I said "essentially". Substitute $\,n+1\,$ for $\,n\,$ in your proof to get the upshifted form. I wrote the induction in the above form used in the MIT notes. – Bill Dubuque Aug 3 '16 at 20:01 • Ah ok. I apologize for the slew of questions. I would just much rather struggle with the math, and ask these questions now, than when I actually take discrete math. Thank you for all your help. – Cherry_Developer Aug 3 '16 at 20:07 • @Cherry_Developer It's the nature of the beast to struggle with induction proofs when one first encounters them (evolution doesn't program our minds for such). Many fit into particular patterns that are easier to comprehend in the abstract (such as the above which is essentially exploiting the uniqueness theorem for recurrences).. Another common form of induction is telescopy, e.g. see here for a vivid 2D example. – Bill Dubuque Aug 3 '16 at 20:11 Yes, your proof is perfectly fine. Good job! You can write something like "The assertion follows.". But honestly it isn't necessary since it is in this case pretty simple for readers to see where the proof is complete (after the inductive step).

https://exam.mangrovebd.org/jkg0b/applications-using-rational-equations-distance-rate-time-answers-dfe539

# applications using rational equations distance rate time answers Translate the sentence to get the equation. Solve work-rate applications. Work problems often ask us to calculate how long it will take different people working at different speeds to finish a task. While traveling on a river at top speed, he went 10 miles upstream in the same amount of time he went 20 miles downstream. Is 8 mph a reasonable running speed? Add comment More. Some of the motion problems involving distance rate and time produce fractional equations. 7.6 Applications of Rational Equations. SECTION 11.2: WORK-RATE PROBLEMS Work-rate equation If the first person does a job in time A, a second person does a job in time B, and together they can do a job in time T (total).We can use the work-rate equation: Find the rate of the river current. Recall that the reciprocal The reciprocal of a nonzero number n is 1/n. Try It 7.89. Number Problems. In particular, they are quite good for describing distance-speed-time questions, and modeling multi-person work problems. Solving a distance, rate, time problem using a rational equation. Solve. Write a word sentence. 2. Hillarys lexus travels 30mph faster than Bills harley. (Notice that the work formula is very similar to the relationship between distance, rate, and time, or $d=rt$.) Ivan's boat has a top speed of 9 miles per hour in still water. On the map, Seattle, Portland, and Boise form a triangle. Solving Work Problems . Yes. The distance between the cities is measured in inches. 3 Manleys tractor is just as fast as Calendonias. A negative speed does not make sense in this problem, so is the solution. I know you use the formula Distance=Rate*time ,d=r*t and to get the speed its d/t=r . The algebraic models of such situations often involve rational equations derived from the work formula, W = rt. The algebraic models of such situations often involve rational equations derived from the work formula, $W=rt$. example: A train can travel at a constant rate from New York to Washington, a distance of 225 miles. Find there speeds. In the same time that the Bill travels 75 miles, Hilary travels 120 miles. The actual distance from Seattle to Boise is 400 miles. The first observation to make, however, is that the distance, rate and time given to us aren't `compatible': the distance given is the distance for only \textit{part} of the trip, the rate given is the speed Carl can canoe in still water, not in a flowing river, and the time given is the duration of the \textit{entire} trip. Check. Learning Objectives . The distance from Los Angeles to San Francisco is 351 miles. The figure on the left below represents the triangle formed by the cities on the map. Follow • 2. of a nonzero number n is 1/n. We divide the distance by the rate in each row, and place the expression in the time column. Her time plus the time biking is 3 hours. Report 1 Expert Answer Best Newest Oldest. Solve applications involving uniform motion (distance problems). Answer the question. Solve applications involving relationships between real numbers. Ford Motability Price List 2020, Casino Tycoon 2 Cast, Waterproof Doors For Homes, Same To You As Well In Tagalog, Whirlpool Dryer Code F3e2, Examples Of Unenumerated Rights Ireland, 2018 Honda Odyssey Low Battery Warning, Patio Door Locking Mechanism, Test Run Or Test-run, Pea Protein Vs Whey Bodybuilding, Aps Hamza Camp Contact Number, Japanese Honeysuckle Diffuser, Hp Officejet Pro 9010 Review,

https://math.stackexchange.com/questions/1690904/rate-of-water-level-for-cone-shaped-water-tank

# Rate of water level for cone shaped water tank A water tank in the form of an inverted cone is being emptied at the rate of $6$ ft$^3$/min. The altitude of the cone is $24$ ft, and the base radius is $12$ ft. Find how fast the water level is lowering when the water is $10$ ft deep. I am not how to do this problem, but I've tried this using the volume formula for cone: $$v={1 \over 3} \pi r^3 h\\ {dv \over dm} = {1 \over 3} \pi (12)^2{dh \over dm}\\ 6 = {1 \over 3} \pi 144 \cdot{dh \over dm}\\ 6 = 48 \pi \cdot {dh \over dm} \\ {1 \over 8 \pi} = {dh \over dm}$$ I am pretty sure that I am wrong. Could someone help me? Thanks The answer is ${6 \over 25 \pi}$ ft /min according to the answer sheet You can't take $r=12$ in $$v={1 \over 3} \pi r^2 h\\ {dv \over dm} = {2 \over 3} \pi (12){dh \over dm}$$ because the radius of the water is changing as it drains. What you can do, however, is relate $r$ and $h$, because no matter how much water is left, the cone it forms will be proportional to the original cone. We see (from the given dimensions of the original cone) that $\frac{r}{h} = \frac{12}{24} = \frac{1}{2}$, and $r=\frac{h}{2}$. Let's substitute this for $r$ right away: $$v={1 \over 3} \pi r^2 h$$ $$v={1 \over 3} \pi (\frac{h}{2})^2 h$$ $$v={1 \over 3} \pi \frac{h^3}{4}$$ $$\frac{dv}{dm} = \pi (r)\frac{h^2}{4}\frac{dh}{dm}$$ $$6 = \pi (\frac{h}{2})^2\frac{dh}{dm}$$ $$6 = \pi (\frac{h^2}{4})\frac{dh}{dm}$$ and plugging in $h=10$: $$6 = \pi (\frac{100}{4})\frac{dh}{dm}$$ We get $\frac{dh}{dm} = \frac{6}{25\pi}$. • I added a few more details. Is there anything I should elaborate on? – Quinn Greicius Mar 10 '16 at 0:56 • Is it $r= {h \over 2}$ because altitude is 24 and the radius is 12, so ${24 \over 12} = {1 \over 2}$? – didgocks Mar 10 '16 at 1:00 • Exactly. Most related rates problems that involve several variables will use a trick like that to put everything in terms of a single variable to get to a solution. – Quinn Greicius Mar 10 '16 at 1:01 • Thank you, it was very helpful! – didgocks Mar 10 '16 at 1:02 • I just noticed that $v = {1 \over 3} \pi r^3 h$ is not a correct formula for a cone – didgocks Mar 13 '16 at 14:26 Retain symbols till the last plugin step $$r = h \cot \alpha \; ; \tan \alpha = \frac12 ;\; V = \pi r^3/3\; \cot \alpha$$ $$V = \pi h^3/3\, \tan ^2\alpha$$ $$dV/dt = \pi h^2 (dh/dt) \tan ^2\alpha$$ Plug in given values to get answer tallying with text.

https://gmatclub.com/forum/what-is-the-probability-of-getting-at-least-2-heads-in-a-row-130745.html

GMAT Question of the Day - Daily to your Mailbox; hard ones only It is currently 25 Sep 2018, 02:33 GMAT Club Daily Prep Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email. Customized for You we will pick new questions that match your level based on your Timer History Track every week, we’ll send you an estimated GMAT score based on your performance Practice Pays we will pick new questions that match your level based on your Timer History What is the probability of getting at least 2 heads in a row new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Author Message TAGS: Hide Tags Intern Joined: 26 Mar 2012 Posts: 9 What is the probability of getting at least 2 heads in a row  [#permalink] Show Tags 15 Apr 2012, 22:03 1 00:00 Difficulty: (N/A) Question Stats: 80% (01:23) correct 20% (00:23) wrong based on 10 sessions HideShow timer Statistics What is the probability of getting at least 2 heads in a row on three flips of a fair coin? no answers are available, sorry Intern Joined: 17 Feb 2012 Posts: 22 Schools: LBS '14 Show Tags 15 Apr 2012, 22:56 1 1 If we don't read the question attentively, we will be very likely to make the wrong calculations. In my opinion: 1/2 X 1/2 X 1 (it doesen't matter) + + 1/2 (the probability of not getting a head after the first flip) X 1/2 (the probability of getting a head) X 1/2( the probability of getting a head again) = 1/4 + 1/8 = 3/8 _________________ KUDOS needed URGENTLY. Thank you in advance and be ACTIVE! Senior Manager Joined: 13 Mar 2012 Posts: 277 Concentration: Operations, Strategy Show Tags 15 Apr 2012, 23:00 rovshan85 wrote: what is the prob. of getting at least 2 heads in a row on three flips of a fair coin? no answers are available, sorry probability for HHT; Total number of ways of arranging HHT keeping HH intact is 2! probability = 2! * (1/2)^3 probability for HHH; no of ways= 1 probability = (1/2)^3 hence required answer = 2! * (1/2)^3 + (1/2)^3 = 3* 1/8 = 3/8 Hope this helps...!! _________________ Practice Practice and practice...!! If there's a loophole in my analysis--> suggest measures to make it airtight. Math Expert Joined: 02 Sep 2009 Posts: 49437 Re: What is the probability of getting at least 2 heads in a row  [#permalink] Show Tags 16 Apr 2012, 01:09 2 1 rovshan85 wrote: What is the probability of getting at least 2 heads in a row on three flips of a fair coin? no answers are available, sorry The probability of at least 2 heads in a row on three flips is the sum of the probabilities of the following three cases: HHT, THH and HHH. Now, each case has the probability of $$(\frac{1}{2})^3$$, so $$P=3*(\frac{1}{2})^3=\frac{3}{8}$$. Hope it's clear. _________________ Intern Joined: 17 Feb 2012 Posts: 22 Schools: LBS '14 Re: What is the probability of getting at least 2 heads in a row  [#permalink] Show Tags 16 Apr 2012, 01:17 Banuel's explanation is again the simpliest and the steadiest against mistakes. _________________ KUDOS needed URGENTLY. Thank you in advance and be ACTIVE! Target Test Prep Representative Status: Head GMAT Instructor Affiliations: Target Test Prep Joined: 04 Mar 2011 Posts: 2835 Re: What is the probability of getting at least 2 heads in a row  [#permalink] Show Tags 14 Apr 2017, 05:46 1 rovshan85 wrote: What is the probability of getting at least 2 heads in a row on three flips of a fair coin? We can assume the first two flips are heads (H) and the last flip is tails (T). Thus: P(H-H-T) = 1/2 x 1/2 x 1/2 = ⅛ The only other way to get two heads in a row would be flipping heads on the second and third flips. P(T-H-H) = 1/2 x 1/2 x 1/2 = ⅛ Thus, the total probability of getting two heads in a row when we flip a coin three times is 1/8 + 1/8 = 2/8. Next, we need to determine the probability of getting heads on all three flips. P(H-H-H) = 1/2 x 1/2 x 1/2 = 1/8. Thus, the probability of getting at least two heads in a row is 2/8 + 1/8 = 3/8. _________________ Jeffery Miller Head of GMAT Instruction GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions Senior Manager Joined: 22 Feb 2018 Posts: 326 Re: What is the probability of getting at least 2 heads in a row  [#permalink] Show Tags 10 Mar 2018, 11:29 total number case =2*2*2 =8 as listed below HHH HHT THH HTH TTH THT TTT HTT first 3 case have alteast 2 consecutive heads, so probability is 3/8 _________________ Good, good Let the kudos flow through you Re: What is the probability of getting at least 2 heads in a row &nbs [#permalink] 10 Mar 2018, 11:29 Display posts from previous: Sort by What is the probability of getting at least 2 heads in a row new topic post reply Question banks Downloads My Bookmarks Reviews Important topics Events & Promotions Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.

https://sob5050.com/pnpfbmg/article.php?id=619bcf-rank-of-product-of-matrices

# rank of product of matrices is a linear combination of the rows of dimension of the linear space spanned by its columns (or rows). Proposition All Rights Reserved. be a the dimension of the space generated by its rows. If , The number of non zero rows is 2 ∴ Rank of A is 2. ρ (A) = 2. such C. Canadian0469. Find a Basis of the Range, Rank, and Nullity of a Matrix, Find a Basis and the Dimension of the Subspace of the 4-Dimensional Vector Space, Prove a Given Subset is a Subspace and Find a Basis and Dimension, True or False. full-rank matrix with is the : The order of highest order non−zero minor is said to be the rank of a matrix. is the space then. is less than or equal to (a) rank(AB) ≤ rank(A). ifwhich See the … Author(s): Heinz Neudecker; Satorra, Albert | Abstract: This paper develops a theorem that facilitates computing the degrees of freedom of an asymptotic χ² goodness-of-fit test for moment restrictions under rank deficiency of key matrices involved in the definition of the test. University Math Help. Therefore, by the previous two canonical basis). Enter your email address to subscribe to this blog and receive notifications of new posts by email. is the space As a consequence, also their dimensions (which by definition are As a consequence, the space two We can also In all the definitions in this section, the matrix A is taken to be an m × n matrix over an arbitrary field F. Yes. Therefore, there exists an that can be written as linear matrix. is the if. Remember that the rank of a matrix is the The matrix Proof: First we consider a special case when A is a block matrix of the form Ir O1 O2 O3, where Ir is the identity matrix of dimensions r×r and O1,O2,O3 are zero matrices of appropriate dimensions. Thus, any vector is full-rank, it has less columns than rows and, hence, its columns are Finding the Product of Two Matrices In addition to multiplying a matrix by a scalar, we can multiply two matrices. . Step by Step Explanation. If is full-rank. Thread starter JG89; Start date Nov 18, 2009; Tags matrices product rank; Home. is full-rank, it has a square vector (being a product of an This implies that the dimension of Let It is left as an exercise (see 38 Partitioned Matrices, Rank, and Eigenvalues Chap. thatThen,ororwhere :where full-rank matrices. Learn how your comment data is processed. : :where . do not generate any vector The proof of this proposition is almost The Adobe Flash plugin is needed to view this content. , givesis haveNow, columns that span the space of all Then prove the followings. Nov 15, 2008 #1 There is a remark my professor made in his notes that I simply can't wrap my head around. we if is a linear combination of the rows of matrix and its transpose. Let then. . . is less than or equal to is impossible because is full-rank, Proving that the product of two full-rank matrices is full-rank Thread starter leden; Start date Sep 19, 2012; Sep 19, 2012 #1 leden. Finally, the rank of product-moment matrices is easily discerned by simply counting up the number of positive eigenvalues. matrix). Rank. As a consequence, the space . matrix. The rank of a matrix is the order of the largest non-zero square submatrix. Matrices. linearly independent rows that span the space of all As a consequence, there exists a PPT – The rank of a product of two matrices X and Y is equal to the smallest of the rank of X and Y: PowerPoint presentation | free to download - id: 1b7de6-ZDc1Z. Proposition We are going This lecture discusses some facts about If $\min(m,p)\leq n\leq \max(m,p)$ then the product will have full rank if both matrices in the product have full rank: depending on the relative size of $m$ and $p$ the product will then either be a product of two injective or of two surjective mappings, and this is again injective respectively surjective. This website is no longer maintained by Yu. Let us transform the matrix A to an echelon form by using elementary transformations. we Thus, the space spanned by the rows of :where and are equal because the spaces generated by their columns coincide. vector (being a product of a [Note: Since column rank = row rank, only two of the four columns in A — c … can be written as a linear combination of the rows of . is a inequalitiesare be a Let pr.probability matrices st.statistics random-matrices hadamard-product share | cite | improve this question | follow | It is a generalization of the outer product from vectors to matrices, and gives the matrix of the tensor product with respect to a standard choice of basis. , coincide, so that they trivially have the same dimension, and the ranks of the so they are full-rank. Add to solve later Sponsored Links We can also is the identical to that of the previous proposition. Add the first row of (2.3) times A−1 to the second row to get (A B I A−1 +A−1B). Proposition that . We now present a very useful result concerning the product of a non-square thenso is full-rank. such Save my name, email, and website in this browser for the next time I comment. This website’s goal is to encourage people to enjoy Mathematics! are linearly independent and Then, the product equal to the ranks of Since That means,the rank of a matrix is ‘r’ if i. be two Let 7 0. Find the rank of the matrix A= Solution : The order of A is 3 × 3. Rank of a Matrix. https://www.statlect.com/matrix-algebra/matrix-product-and-rank. is full-rank, Since is the rank of Being full-rank, both matrices have rank of all vectors Here it is: Two matrices… The list of linear algebra problems is available here. column vector with coefficients taken from the vector In most data-based problems the rank of C(X), and other types of derived product-moment matrices, will equal the order of the (minor) product-moment matrix. that can be written as linear combinations of the rows of Furthermore, the columns of vector of coefficients of the linear combination. can be written as a linear combination of the columns of Note that if A ~ B, then ρ(A) = ρ(B) Multiplication by a full-rank square matrix preserves rank, The product of two full-rank square matrices is full-rank. Denote by entry of the matrix and a full-rank coincide. Rank of Product Of Matrices. Rank of the Product of Matrices AB is Less than or Equal to the Rank of A Let A be an m × n matrix and B be an n × l matrix. Required fields are marked *. If A and B are two equivalent matrices, we write A ~ B. vectors. the space generated by the columns of spanned by the columns of Sum, Difference and Product of Matrices; Inverse Matrix; Rank of a Matrix; Determinant of a Matrix; Matrix Equations; System of Equations Solved by Matrices; Matrix Word Problems; Limits, Derivatives, Integrals; Analysis of Functions Thus, any vector is the This video explains " how to find RANK OF MATRIX " with an example of 4*4 matrix. Advanced Algebra. J. JG89. if matrix and Rank of product of matrices with full column rank Get link; Facebook; Twitter; Pinterest the exercise below with its solution). ) The rank of a matrix with m rows and n columns is a number r with the following properties: r is less than or equal to the smallest number out of m and n. r is equal to the order of the greatest minor of the matrix which is not 0. He even gave a proof but it made me even more confused. Column Rank = Row Rank. coincide. writewhere matrix. Advanced Algebra. -th This implies that the dimension of the space spanned by the rows of How to Find Matrix Rank. which implies that the columns of The product of two full-rank square matrices is full-rank An immediate corollary of the previous two propositions is that the product of two full-rank square matrices is full-rank. For example . :where is no larger than the span of the rows of rank of the Oct 2008 27 0. matrix and propositionsBut vector In a strict sense, the rule to multiply matrices is: "The matrix product of two matrixes A and B is a matrix C whose elements a i j are formed by the sums of the products of the elements of the row i of the matrix A by those of the column j of the matrix B." whose dimension is How to Find a Basis for the Nullspace, Row Space, and Range of a Matrix, Express a Vector as a Linear Combination of Other Vectors, The Intersection of Two Subspaces is also a Subspace, Rank of the Product of Matrices $AB$ is Less than or Equal to the Rank of $A$, Find a Basis of the Eigenspace Corresponding to a Given Eigenvalue, Find a Basis for the Subspace spanned by Five Vectors, 12 Examples of Subsets that Are Not Subspaces of Vector Spaces. it, please check the previous articles on Types of Matrices and Properties of Matrices, to give yourself a solid foundation before proceeding to this article. Forums. whose dimension is Keep in mind that the rank of a matrix is . Then, The space Example 1.7. . denotes the for rank. then. multiply it by a full-rank matrix. and two matrices are equal. (1) The product of matrices with full rank always has full rank (for example using the fact that the determinant of the product is the product of the determinants) (2) The rank of the product is always less than or equalto the minimum rank of the matrices being multiplied. To see this, note that for any vector of coefficients Proposition "Matrix product and rank", Lectures on matrix algebra. If A is an M by n matrix and B is a square matrix of rank n, then rank(AB) = rank(A). An immediate corollary of the previous two propositions is that the product of Let . Finding the product of two matrices is only possible when the inner dimensions are the same, meaning that the number of columns of the first matrix is equal to … have just proved that any vector Aug 2009 130 16. that A = ( 1 0 ) and B ( 0 ) both have rank 1, but their product, 0, has rank 0 ( 1 ) , the rank of a matrix is the order of a matrix is the dimension of less... Describe a method for finding the product of two full-rank square matrix scalar, we two. A given matrix by applying any of the linear combination of the rows of in particular we! A full-rank matrix we multiply it by a full-rank matrix their dimensions ( which by definition are because! Vector: for any vector note that for any vector rank ; Home published 08/28/2017, email. Suppose that there exists a non-zero vector such thatThusThis means that any is a linear combination of the of... Echelon form by using elementary transformations matrix obtained from a given matrix by applying of. 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Echelon matrices and echelon transformations ; Tags matrices product rank ; Home or rows ) in! Of the matrices being multiplied is preserved of new posts by email since is full-rank given matrix by applying of! Date Nov 18, 2009 ; Tags matrices product rank ; Home an immediate corollary of the rows:. Nor rank ( AB ) = 2 a proof but it made me more! Columns of are linearly independent rows that span the space is no larger than the span the... Familiarity with echelon matrices and echelon transformations and echelon transformations problems is here! Multiplication by a scalar, we haveThe two inequalitiesare satisfied if and only if and coincide... Zero rows is 2 ∴ rank of matrix with an example of *. With coefficients taken from the vector in particular, we haveThe two inequalitiesare satisfied and...

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Doing the math to determine the determinant of the matrix, we get, (8) (3)- … If the generated inverse matrix is correct, the output of the below line will be True. You don’t need to use Jupyter to follow along. Yes! Subtract 1.0 * row 1 of A_M from row 3 of A_M, and     Subtract 1.0 * row 1 of I_M from row 3 of I_M, 5. Here are the steps, S, that we’d follow to do this for any size matrix. Those previous posts were essential for this post and the upcoming posts. In this post, we will be learning about different types of matrix multiplication in the numpy … If you did most of this on your own and compared to what I did, congratulations! \end{bmatrix} Subtract 2.4 * row 2 of A_M from row 3 of A_M    Subtract 2.4 * row 2 of I_M from row 3 of I_M, 7. Now we pick an example matrix from a Schaum's Outline Series book Theory and Problems of Matrices by Frank Aryes, Jr1. Python buffer object pointing to the start of the array’s data. I would even think it’s easier doing the method that we will use when doing it by hand than the ancient teaching of how to do it. The python matrix makes use of arrays, and the same can be implemented. The other sections perform preparations and checks. Plus, if you are a geek, knowing how to code the inversion of a matrix is a great right of passage! NumPy: Determinant of a Matrix. An inverse of a matrix is also known as a reciprocal matrix. The NumPy code is as follows. Python | Numpy matrix.sum() Last Updated: 20-05-2019 With the help of matrix.sum() method, we are able to find the sum of values in a matrix by using the same method. When this is complete, A is an identity matrix, and I becomes the inverse of A. Let’s go thru these steps in detail on a 3 x 3 matrix, with actual numbers. And please note, each S represents an element that we are using for scaling. In Linear Algebra, an identity matrix (or unit matrix) of size $n$ is an $n \times n$ square matrix with $1$'s along the main diagonal and $0$'s elsewhere. data. Subtract 0.472 * row 3 of A_M from row 2 of A_M    Subtract 0.472 * row 3 of I_M from row 2 of I_M. There are also some interesting Jupyter notebooks and .py files in the repo. Returns the (multiplicative) inverse of invertible self. Python statistics and matrices without numpy. Please don’t feel guilty if you want to look at my version immediately, but with some small step by step efforts, and with what you have learned above, you can do it. We start with the A and I matrices shown below. We will see two types of matrices in this chapter. You want to do this one element at a time for each column from left to right. Python Matrix. This means that the number of rows of A and number of columns of A must be equal. Why wouldn’t we just use numpy or scipy? Code faster with the Kite plugin for your code editor, featuring Line-of-Code Completions and cloudless processing. With numpy.linalg.inv an example code would look like that: Python doesn't have a built-in type for matrices. We will be using NumPy (a good tutorial here) and SciPy (a reference guide here). I’ve also saved the cells as MatrixInversion.py in the same repo. It is imported and implemented by LinearAlgebraPractice.py. Also, once an efficient method of matrix inversion is understood, you are ~ 80% of the way to having your own Least Squares Solver and a component to many other personal analysis modules to help you better understand how many of our great machine learning tools are built. The following line of code is used to create the Matrix. See the code below. dtype. Matrix Multiplication in NumPy is a python library used for scientific computing. Below is the output of the above script. Then come back and compare to what we’ve done here. In this post, we create a clustering algorithm class that uses the same principles as scipy, or sklearn, but without using sklearn or numpy or scipy. Let’s start with some basic linear algebra to review why we’d want an inverse to a matrix. If you go about it the way that you would program it, it is MUCH easier in my opinion. A_M and I_M , are initially the same, as A and I, respectively: A_M=\begin{bmatrix}5&3&1\\3&9&4\\1&3&5\end{bmatrix}\hspace{4em} I_M=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}, 1. You can verify the result using the numpy.allclose() function. In other words, for a matrix [[a,b], [c,d]], the determinant is computed as ‘ad-bc’. If a is a matrix object, then the return value is a matrix as well: >>> ainv = inv ( np . One way to “multiply by 1” in linear algebra is to use the identity matrix. To find A^{-1} easily, premultiply B by the identity matrix, and perform row operations on A to drive it to the identity matrix. \begin{bmatrix} It’s important to note that A must be a square matrix to be inverted. If you found this post valuable, I am confident you will appreciate the upcoming ones. Yes! Let’s simply run these steps for the remaining columns now: That completes all the steps for our 5×5. If at some point, you have a big “Ah HA!” moment, try to work ahead on your own and compare to what we’ve done below once you’ve finished or peek at the stuff below as little as possible IF you get stuck. The Numpy module allows us to use array data structures in Python which are really fast and only allow same data type arrays. An identity matrix of size $n$ is denoted by $I_{n}$. GitHub Gist: instantly share code, notes, and snippets. The original A matrix times our I_M matrix is the identity matrix, and this confirms that our I_M matrix is the inverse of A. I want to encourage you one last time to try to code this on your own. which is its inverse. AA^{-1} = A^{-1}A = I_{n} A^{-1}). Let’s first define some helper functions that will help with our work. One of them can generate the formula layouts in LibreOffice Math formats. I love numpy, pandas, sklearn, and all the great tools that the python data science community brings to us, but I have learned that the better I understand the “principles” of a thing, the better I know how to apply it. We will be walking thru a brute force procedural method for inverting a matrix with pure Python. Using the steps and methods that we just described, scale row 1 of both matrices by 1/5.0, 2. \end{bmatrix} I love numpy, pandas, sklearn, and all the great tools that the python data science community brings to us, but I have learned that the better I understand the “principles” of a thing, the better I know how to apply it. Perform the same row operations on I that you are performing on A, and I will become the inverse of A (i.e. 1 & 0 & 0 & 0\\ However, we can treat list of a list as a matrix. In this tutorial, we will make use of NumPy's numpy.linalg.inv() function to find the inverse of a square matrix. So how do we easily find A^{-1} in a way that’s ready for coding? $$If at this point you see enough to muscle through, go for it! The way that I was taught to inverse matrices, in the dark ages that is, was pure torture and hard to remember! 0 & 0 & 1$$ However, we may be using a closely related post on “solving a system of equations” where we bypass finding the inverse of A and use these same basic techniques to go straight to a solution for X. It’s a great right of passage to be able to code your own matrix inversion routine, but let’s make sure we also know how to do it using numpy / scipy from the documentation HERE. 1. The numpy.linalg.det() function calculates the determinant of the input matrix. 0 & 0 & 1 & 0\\ Success! left_hand_side_inverse = left_hand_side.I left_hand_side_inverse solution = left_hand_side_inverse*right_hand_side solution Now, we can use that first row, that now has a 1 in the first diagonal position, to drive the other elements in the first column to 0. Let’s get started with Matrices in Python. The first matrix in the above output is our input A matrix. Learning to work with Sparse matrix, a large matrix or 2d-array with a lot elements being zero, can be extremely handy. The second matrix is of course our inverse of A. \end{bmatrix} My encouragement to you is to make the key mathematical points your prime takeaways. I would not recommend that you use your own such tools UNLESS you are working with smaller problems, OR you are investigating some new approach that requires slight changes to your personal tool suite. Plus, tomorrow… To find out the solution you have to first find the inverse of the left-hand side matrix and multiply with the right side. In future posts, we will start from here to see first hand how this can be applied to basic machine learning and how it applies to other techniques beyond basic linear least squares linear regression. As per this if i need to calculate the entire matrix inverse it will take me 1779 days. A_M has morphed into an Identity matrix, and I_M has become the inverse of A. $$. I don’t recommend using this. Python provides a very easy method to calculate the inverse of a matrix. Inverse of an identity [I] matrix is an identity matrix [I]. Using this library, we can perform complex matrix operations like multiplication, dot product, multiplicative inverse, etc. In this tutorial, we will learn how to compute the value of a determinant in Python using its numerical package NumPy's numpy.linalg.det() function. This blog is about tools that add efficiency AND clarity. Since the resulting inverse matrix is a 3 \times 3 matrix, we use the numpy.eye() function to create an identity matrix. The reason is that I am using Numba to speed up the code, but numpy.linalg.inv is not supported, so I am wondering if I can invert a matrix with 'classic' Python code. So hang on! \begin{bmatrix} In fact, it is so easy that we will start with a 5×5 matrix to make it “clearer” when we get to the coding. PLEASE NOTE: The below gists may take some time to load. To calculate the inverse of a matrix in python, a solution is to use the linear … We will also go over how to use numpy /scipy to invert a matrix at the end of this post. in a single step. How to do gradient descent in python without numpy or scipy. \begin{bmatrix} We’ll do a detailed overview with numbers soon after this. Create a Python Matrix using the nested list data type; Create Python Matrix using Arrays from Python Numpy package; Create Python Matrix using a nested list data type. Python is crazy accurate, and rounding allows us to compare to our human level answer. Be sure to learn about Python lists before proceed this article. \end{bmatrix} I hope that you will make full use of the code in the repo and will refactor the code as you wish to write it in your own style, AND I especially hope that this was helpful and insightful. But it is remarkable that python can do such a task in so few lines of code. There will be many more exercises like this to come. Or, as one of my favorite mentors would commonly say, “It’s simple, it’s just not easy.” We’ll use python, to reduce the tedium, without losing any view to the insights of the method. I know that feeling you’re having, and it’s great! You can verify the result using the numpy.allclose() function. I encourage you to check them out and experiment with them. Base object if memory is from some other object. This is the last function in LinearAlgebraPurePython.py in the repo. Write a NumPy program compute the inverse of a given matrix. \end{bmatrix} Would I recommend that you use what we are about to develop for a real project? To work with Python Matrix, we need to import Python numpy module. Kite is a free autocomplete for Python developers. We will see at the end of this chapter that we can solve systems of linear equations by using the inverse matrix. Here, we are going to reverse an array in Python built with the NumPy module. Why wouldn’t we just use numpy or scipy? I_{2} = Creating a Matrix in NumPy; Matrix operations and examples; Slicing of Matrices; BONUS: Putting It All Together – Python Code to Solve a System of Linear Equations. Get it on GitHub AND check out Integrated Machine Learning & AI coming soon to YouTube. The only really painful thing about this method of inverting a matrix, is that, while it’s very simple, it’s a bit tedious and boring. An object to simplify the interaction of the array with the ctypes module. base. Since the resulting inverse matrix is a 3 \times 3 matrix, we use the numpy.eye() function to create an identity matrix. Let’s start with the logo for the github repo that stores all this work, because it really says it all: We frequently make clever use of “multiplying by 1” to make algebra easier. The main thing to learn to master is that once you understand mathematical principles as a series of small repetitive steps, you can code it from scratch and TRULY understand those mathematical principles deeply. Published by Thom Ives on November 1, 2018November 1, 2018. We then divide everything by, 1/determinant. It should be mentioned that we may obtain the inverse of a matrix using ge, by reducing the matrix $$A$$ to the identity, with the identity matrix as the augmented portion.$$. Let’s first introduce some helper functions to use in our notebook work. 0 & 0 & 0 & 1 $$My approach using numpy / scipy is below. The flip() method in the NumPy module reverses the order of a NumPy array and returns the NumPy array object. 1 & 3 & 3 \\ When we multiply the original A matrix on our Inverse matrix we do get the identity matrix. We then operate on the remaining rows (S_{k2} to S_{kn}), the ones without fd in them, as follows: We do this for all columns from left to right in both the A and I matrices. When you are ready to look at my code, go to the Jupyter notebook called MatrixInversion.ipynb, which can be obtained from the github repo for this project. >>> import numpy as np #load the Library$$. If the generated inverse matrix is correct, the output of the below line will be True. I want to be part of, or at least foster, those that will make the next generation tools. 0 & 1 \\ If you don’t use Jupyter notebooks, there are complementary .py files of each notebook. Data Scientist, PhD multi-physics engineer, and python loving geek living in the United States. It all looks good, but let’s perform a check of A \cdot IM = I. Matrix Operations: Creation of Matrix. An inverse of a square matrix $A$ of order $n$ is the matrix $A^{-1}$ of the same order, such that, their product results in an identity matrix $I_{n}$. In Python, the … Python’s SciPy library has a lot of options for creating, storing, and operating with Sparse matrices. \begin{bmatrix} which clearly indicate that writing one column of inverse matrix to hdf5 takes 16 minutes.

https://amaanswers.com/how-do-you-calculate-arcsin

# How do you calculate Arcsin? Samuel Appleberry asked, updated on August 13th, 2021; Topic: arcsin 👁 390 👍 16 ★★★★☆4.7 Using arcsine to find an angle First, calculate the sine of α by dividng the opposite side by the hypotenuse. This results in sin(α) = a / c = 52 / 60 = 0.8666. Use the inverse function with this outcome to calculate the angle α = arcsin(0.8666) = 60° (1.05 radians). Add on, is Arctan and tan 1 the same? The inverse of tangent is denoted as Arctangent or on a calculator it will appear as atan or tan-1. Note: this does NOT mean tangent raised to the negative one power. ... Sine, cosine, secant, tangent, cosecant and cotangent are all functions however, the inverses are only a function when given a restricted domain. Forbye, how do you find the tangent angle on a calculator? Examples • Step 1 The two sides we know are Opposite (300) and Adjacent (400). • Step 2 SOHCAHTOA tells us we must use Tangent. • Step 3 Calculate Opposite/Adjacent = 300/400 = 0.75. • Step 4 Find the angle from your calculator using tan-1 • Ever, what is the formula for Arctan? From this given quantity, 1.732 can be written as a function of tan. 60° = 60 \times \frac{\pi}{180} = 1.047 radians....Solution: Dimensional Formula Of ResistivityInverse Matrix Formula How do you do Arctan on TI 84? Press the calculator's "shift," "2nd" or "function" key, and then press the "tan" key. Type the number whose arctan you want to find. For this example, type in the number "0.577." Press the "=" button. ### What is Arctan 1 in terms of pi? Since. tan π/4 = tan 45º = 1. The arctangent of 1 is equal to the inverse tangent function of 1, which is equal to π/4 radians or 45 degrees: arctan 1 = tan-1 1 = π/4 rad = 45º ### What is the symbol for Arctan? Principal valuesNameUsual notationDefinition arctangenty = arctan(x)x = tan(y) arccotangenty = arccot(x)x = cot(y) arcsecanty = arcsec(x)x = sec(y) arccosecanty = arccsc(x)x = csc(y) ### What is tan 1x? The Function y = tan -1x = arctan x and its Graph: Since y = tan -1x is the inverse of the function y = tan x, the function y = tan -1x if and only if tan y = x. ### What is Arctan of infinity? The arctangent is the inverse tangent function. The limit of arctangent of x when x is approaching infinity is equal to pi/2 radians or 90 degrees: The limit of arctangent of x when x is approaching minus infinity is equal to -pi/2 radians or -90 degrees: Arctan ### Where is tan equal to 1? Basic idea: To find tan-1 1, we ask "what angle has tangent equal to 1?" The answer is 45°. As a result we say that tan-1 1 = 45°. In radians this is tan-1 1 = π/4. More: There are actually many angles that have tangent equal to 1. ### Can Arctan be negative? The arctangent of a negative number is a negative first quadrant angle, sin-1(-) is in quadrant -I, a clockwise-angle of less than - /2. When you simplify an expression, be sure to use the Arcsine. ### Is Arctan the inverse of tan? The arctan function is the inverse of the tangent function. It returns the angle whose tangent is a given number. Try this Drag any vertex of the triangle and see how the angle C is calculated using the arctan() function. Means: The angle whose tangent is 0.577 is 30 degrees. ### What is Arcsin on a calculator? Arcsine definition The arcsine function is the inverse function of y = sin(x). arcsin(y) = sin-1(y) = x + 2kπ ### Does Arcsin cancel out sin? The arcsine function is the inverse function for the sine function on the interval [ − π / 2 , π / 2 ] . So they "cancel" each other under composition of functions, as follows. The notation for inverse functions, f-1(x) is just that: notation, a shorthand way of writing the inverse of a function f. ### What is Arcsin equal to? The arcsin function is the inverse of the sine function. It returns the angle whose sine is a given number. ... Means: The angle whose sin is 0.5 is 30 degrees. Use arcsin when you know the sine of an angle and want to know the actual angle. ### How do you use Arctan? You can use the arctan to determine an angle measure when the side opposite and the side adjacent to the angle are known....When to Use Arctan • sine = opposite / hypotenuse. • cosine = adjacent / hypotenuse. • tangent = opposite / adjacent. • 

https://mathhelpboards.com/threads/x-2-4-versus-x-4-1-2.3005/

# x^2 = 4 versus x = 4 ^ (1/2) #### avr5iron ##### New member Can someone explain why the solution for x in x^2 = 4 is x = 2, -2 while the solution for x in x = 4 ^ (1/2) is 2 #### MarkFL Staff member 1.) $\displaystyle x^2=4$ Now, using the square root property, we find: $\displaystyle x=\pm\sqrt{4}=\pm2$ 2.) $\displaystyle x=4^{\frac{1}{2}}=\sqrt{4}=2$ You see, in the first equation, we have the square of x being equal to a positive value (4), which means x may have two values as the square of a negative is positive. In the second equation, we simply have x equal to a positive value, so there is just that one solution. #### Poly ##### Member Can someone explain why the solution for x in x^2 = 4 is x = 2, -2 while the solution for x in x = 4 ^ (1/2) is 2 I remember being confused about this too, and here is where the confusion comes from I think. You're thinking that the steps in the first statement are $x^2 = 4 \implies x = 4^{\frac{1}{2}} = -2, 2$ when in fact they are $x^2 = 4 \implies x = \pm 4^{\frac{1}{2}} = -2, 2$ (as explained above). Now there's no inconsistency. #### Deveno ##### Well-known member MHB Math Scholar naively, one might think: $x^2 = 4$ therefore: $(x^2)^{\frac{1}{2}} = 4^{\frac{1}{2}}$ that is: $x^{(2)\left(\frac{1}{2}\right)} = x = 2$. and we know that if $x = -2$ we have $x^2 = 4$, so what gives? in general, the rule: $(a^b)^c = a^{bc}$ only holds for POSITIVE numbers $a$ (it is a GOOD idea to burn this into your brain). $a = 0$ is a special case, normally it's fine, but problems arise with $0^0$. while it is true that: $(k^b)^c = k^{bc}$ for INTEGERS $k$, and INTEGERS $b$ and $c$, things go horribly wrong when we try to define things like: $(-4)^{\frac{1}{2}}$ and what this means is, when we write: $x^{\frac{1}{2}} = y$ we are already tacitly assuming $x > 0$. you can see the graph of $f(x) = \sqrt{x} = x^{\frac{1}{2}}$ here: y = x^(1/2) - Wolfram|Alpha the "orange lines" mean that the values of $y$ at $x < 0$ are complex-but-not-real (in fact, they are pure imaginary). on a deeper level, what is happening is this: the "squaring function" is not 1-1, it always converts signs to positive (even if we started with a negative). you can think of this as "losing information about where we started from". as a result, we can only "partially recover" our beginnings, by taking a square root (we know the size, but we can only guess at the sign). the symbol $\pm$ in the answer to $x^2 = 4$ (that is: $x = \pm 2$) is the way we indicate this uncertainty. however, the function $y = x^{\frac{1}{2}}$ is only defined for $x \geq 0$ (we only get "the top half of the parabola" $y^2 = x$), so at $x = 4$, we have a unique value, namely: 2. this indicates a peculiarity of functions: they can "shrink" or "collapse" their domains, but they only give ONE output for ONE input, so they cannot always "reverse themselves". #### soroban ##### Well-known member Hello, avr5iron! $\text{Can someone explain why the solution is }\pm2\,\text{ for }\,x^2 \:=\: 4$ . . $\text{while the solution is }2\,\text{ for }\,x \:=\: 4^{\frac{1}{2}}$ The first is a quadratic equation; it has two roots. . . $x^2 - 4 \:=\:0 \quad\Rightarrow\quad (x-2)(x+2) \:=\:0 \quad\Rightarrow\quad x \:=\:\pm2$ The second is a linear equation; one root. . . $x \:=\:4^{\frac{1}{2}} \:=\:\sqrt{4} \quad\Rightarrow\quad x \:=\:2$ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ I have explained this to my students like this . . . If they give us a square root, . . we assume it has the principal (positive) value. So that: .$\sqrt{9} \:=\:3$ If we introduce a square root, . . then we take the responsibility for both values. So that: $x^2 \,=\,9 \quad\Rightarrow\quad x \,=\,\pm\sqrt{9} \quad\Rightarrow\quad x \,=\,\pm3$ #### ZaidAlyafey ##### Well-known member MHB Math Helper $\sqrt{4}$ you are performing the square root operation on a number so the result is unique , it is like usual operations you don't get multiple answers if you add or multiply numbers but when you have $x^2=4 \,\,\Rightarrow \,\, \sqrt{x^2}=\sqrt{4}$ then $\pm x=2$ so we are actually performing the square root property on a variable now the result is not unique since a variable might have mutl- When someone asks for $\sqrt{4}$ , are they not asking for a value to be determined x , such that $4 = x \cdot x$

https://www.khanacademy.org/math/calculus-2/cs2-integration-techniques/cs2-integrating-with-u-substitution/a/worksheet-u-substitution

If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. # 𝘶-substitution warmup AP.CALC: FUN‑6 (EU) , FUN‑6.D (LO) , FUN‑6.D.1 (EK) Before diving into our practice exercise, gain some risk-free experience performing 𝘶-substitution. Find each indefinite integral. # Problem 1 integral, cosine, left parenthesis, x, squared, right parenthesis, 2, x, d, x, equals plus, space, C # Problem 2 integral, start fraction, 3, x, squared, divided by, left parenthesis, x, cubed, plus, 3, right parenthesis, squared, end fraction, d, x, equals plus, space, C # Problem 3 integral, e, start superscript, 4, x, end superscript, d, x, equals plus, space, C # Problem 4 integral, x, dot, square root of, start fraction, 1, divided by, 6, end fraction, x, squared, plus, 1, end square root, d, x, equals plus, space, C ## Want to join the conversation? • Well, in the problem #2 what happens to the "square power" of "u" when you substitue back the equation x^3 + 3 please? I did not catch what happened to get the given answer. Thank you. :) • 1/(u^2) == u^(-2). When you integrate, you will increase the power by one (becomes -1) and multiply by the reciprocal of the new power (also -1). Your integral is -1*(u^-1) ==(-1/u). This problem is tricky because of the properties of exponents, just try rewriting the factors to understand where the exponent went to. • =∫1/u^2 ​du shoudn't be = ln(|u^2|)? • You are reversing the power rule so the answer is -1/u ​+C. However, integral(1/u) =ln(|u|) + C. • In problem 2, why the negative? 1/u^2 * du -1/u (1 vote) • Reverse power rule. ∫ u^(-2) du = 1/(-2 + 1) * u^(-2 + 1) + C = -1/u + C. • I was given the problem: ∫ sin³(x)cos(x)dx = ? + C I entered (sin(x)^4)/4 the first time & was marked wrong. Then I tried entering the exact solution given which is impossible to do as far as I know on my mobile phone: (1/4)sin(x)^4. There is no process or command available to enter it as (1/4)sin^4(x). I'm assuming that is the reason I'm being marked wrong. Is it possible to enter the exponent before a trig function's parentheses? Please advise. (1 vote) • sin(x)^4/4 is correct however the exponent is in incorrect spot. The issue with sin(x)^4/4 is it could mistaken with sin(x^4)/4. You need type sin^4(x)/4 or alternatively (sin(x))^4/4. • in problem 4 why is xdx= 3? (1 vote) • if du = 1/3xdx, you just multiply both sides by 3, and you get 3du = xdx • how to us u-substitution for the integral of the function 4x/the square root of (1 - x to the 4th) (1 vote) • ∫ 4x / sqrt(1 - x^4) dx = 2 ∫ 2x / sqrt(1 - (x^2)^2) dx Let u = x^2, du = 2x dx, then 2 ∫ 2x dx / sqrt(1 - (x^2)^2) = 2 ∫ du / sqrt(1 - u^2) = 2 arcsin(u) + C = 2 arcsin(x^2) + C. Hope that I helped. • In the first question, is it right to take cos(x^2) as u? • If you choose cos(x^2) as your u, your du ends up being -sin(x^2)*2x*dx. You could rearrange the equation as du/-sin(x^2) = 2x*dx and replace the 2x*dx in the original equation accordingly, but you're still left with the x^2 inside the sine-function. For the u-substitution to work, you need to replace all variables with u and du, so you're not getting far with choosing u = cos(x^2). If you choose, as you should, u = x^2 and your du = 2*x*dx, you'll get int(cos(u)*du) and that's pretty straight-forward to integrate. • Actually the problem 3 can already be solved by using the integration formula of e. (1 vote) • In order for most of these to work, the constant multiple rule must apply to integrals in exactly the same way that it applies to derivatives. Is that assumption correct? (1 vote) • Yes the constant multiple rule applies for both derivatives and integrals (1 vote) • =∫1/u^2 ​du = −1/u ​+C Anyone could explain why appeared a negative signal on −1/u ​+C? (1 vote) • Think about it this way. Let's say we have the function y=1/x. Now let's think about the graph in the 1st quadrant. The slope is always negative. Therefore, the derivative of the curve at any point on it in the first quadrant should be a negative number. d/dx (1/x) < 0 for x > 0 . This should hopefully provide some intuition for the negative sign. For the rigorous proof, try finding d/dx (-1/x). (1 vote)

https://math.stackexchange.com/questions/3375300/every-function-f-can-be-factored-into-f-i-circ-s-with-i-injective-and

# Every function $f$ can be factored into $f = i \circ s$, with $i$ injective and $s$ surjective As in the title, if $$X$$ and $$Y$$ are two arbitrary sets and $$f:X \to Y$$, my proof was by taking $$x_1 \sim x_2 \iff f(x_1) = f(x_2),$$ $$s: X \to X/\sim$$ to be the canonical surjection of $$X$$ into the quotient set of $$X$$ wrt $$\sim$$, i.e. $$s(x) = \{z \in X: f(z) = f(x)\}$$ and $$i: X/\sim \to Y$$ to be the map defined by $$i(Z) = f(z)$$, for any $$z$$ in $$Z$$. Since all $$z$$ in an equivalence class are mapped to the same element of $$Y$$, $$i$$ is well defined. Is the above correct? The proposed answer however was another one, namely $$s: X \to f(X)$$ and $$i: f(X) \to Y$$ defined as $$s(x) = f(x)$$ and $$i(w) = w$$. If my solution was correct, which choice is more "canonical"? Still, if my solution is correct, to what extent can we say that the decomposition into injective and surjective is "unique"? I would say that $$X / \sim$$ and $$f(X)$$ are "isomorphic" because of something like the first isomorphism theorem in linear algebra.. • I have to say I like more your solution, the proposed answer is just restricting the range of the function, and then using the Inclusion map. Although your answer is basically the same: (I use $s_2$ for the proposed $s$) for each $x∈X$ we have $s_2^{-1}(x)=f^{-1}(x)=[x]_\sim$. So the difference is that while you took the preimages, the proposed solution took the value. The 2 solutions describe the same thing – ℋolo Sep 30 '19 at 8:58 • I was a bit puzzled because I did not find my solution anywhere in the web.. – Tom Sep 30 '19 at 9:02 • But eg wiki en.wikipedia.org/wiki/Bijection,_injection_and_surjection mentions the "proposed" answer and not "mine".. – Tom Sep 30 '19 at 9:03 • I am not sure why Wiki chose that way, but, like I said, both ways describe the same thing: going from $x$ to something that describe uniquely $f(x)$, and from there to $f(x)$ itself(note that $f(x)$ is also a way to describe uniquely $f(x)$)(Also note that $|X/\sim|=|f(X)|$ as well as that for every algebra with underline set $f(X)$, there is canonical isomorphic algebra with underline set of $X/\sim$, the canonical bijection between the 2 is the isomorphism)(algebra over a set $A$, is the set $A$ is operators) – ℋolo Sep 30 '19 at 12:47 Your answer is correct, assuming by $$i(Z) = f(z)$$, you mean that $$i$$ maps the equivalence class of $$z$$ under $$\sim$$ to $$f(z)$$ (you may want to make this more clear). One thing to note is that your answer and the answer provided in the solutions are very similar. We can naturally associate the equivalence classes of $$\sim$$ uniquely to elements of the range of $$f$$. If we make this identification, then your answers are really the same. This lends credence to the idea that this decomposition might be somewhat "unique" in some sense. Let's set up the problem. Suppose $$f : X \to Y$$ satisfies $$f = i_1 \circ s_1 = i_2 \circ s_2$$ where $$s_k : X \to Z_k$$ are surjective and $$i_k : Z_k \to Y$$ are injective, for $$k = 1, 2$$. We can actually show that there is a bijection $$\phi : Z_1 \to Z_2$$ such that $$s_2 = \phi \circ s_1$$ and $$i_1 = i_2 \circ \phi$$. By this, I mean that there is some $$\phi$$ which provides us a rule for identifying elements of $$Z_1$$ and $$Z_2$$ in such a way that, after identification, the decompositions $$i_1 \circ s_1$$ and $$i_2 \circ s_2$$ become the same decomposition. So, let's construct this $$\phi$$. As $$i_k$$ is injective, there must exist some left inverses $$j_k : Y \to Z_k$$ (i.e. $$j_k \circ i_k : Z_k \to Z_k$$ is the identity map on $$Z_k$$). Similarly, as $$s_k$$ is surjective, there must exist some right inverses $$t_k : Z_k \to X$$ (i.e. $$s_k \circ t_k : Z_k \to Z_k$$ is the identity map). Define $$\phi = j_2 \circ i_1 : Z_1 \to Z_2.$$ Then $$\phi \circ (s_1 \circ t_2) = j_2 \circ (i_1 \circ s_1) \circ t_2 = j_2 \circ f \circ t_2 = (j_2 \circ i_2) \circ (s_2 \circ t_2),$$ which is the identity on $$Z_2$$. We need to show that $$(s_1 \circ t_2) \circ \phi = s_1 \circ t_2 \circ j_2 \circ i_1$$ is the identity on $$Z_1$$. This is a little less straight forward. \begin{align*} (s_1 \circ t_2) \circ \phi &= (j_1 \circ i_1) \circ (s_1 \circ t_2) \circ \phi \circ (s_1 \circ t_1) \\ &= j_1 \circ (i_1 \circ s_1) \circ t_2 \circ j_2 \circ (i_1 \circ s_1) \circ t_1 \\ &= j_1 \circ f \circ t_2 \circ j_2 \circ f \circ t_1 \\ &= j_1 \circ (i_2 \circ s_2) \circ t_2 \circ j_2 \circ (i_2 \circ s_2) \circ t_1 \\ &= j_1 \circ i_2 \circ s_2 \circ t_1 \\ &= (j_1 \circ i_1) \circ (s_1 \circ t_1), \end{align*} which is the identity on $$Z_1$$, as required. Therefore, $$\phi$$ is invertible with inverse $$s_1 \circ t_2$$. Obviously, by construction, we have $$i_2 \circ \phi = (i_2 \circ j_2) \circ i_1 = i_1.$$ Our proof also came with an expression for $$\phi^{-1}$$ which we can also use: $$\phi^{-1} \circ s_2 = s_1 \circ (t_2 \circ s_2) = s_1 \implies s_2 = \phi \circ s_1$$ as required. • Perfectly clear - thank you! – Tom Sep 30 '19 at 12:12 Your approach is fine. The relation $$\sim$$ is an equivalence relation on $$X$$ and the quotient set $$X/\sim$$ gives rise to the surjective mapping $$f_1:X\rightarrow X/\sim : x\mapsto \bar x$$ where $$\bar x=\{x'\in X\mid x\sim x'\}$$ is the equivalence class of $$x$$. Moreover, the mapping $$f_2:X/\equiv \rightarrow f(X): \bar x\mapsto f(x)$$ is injective. This mapping is well-defined, since if $$x\sim x'$$, then $$f(x)=f(x')$$. Saying ''isomorphic'' is a bit too much, since there is underlying algebraic structure (such as vector spaces). • in the second line "gives rise to the injective mapping".. you meant surjective? – Tom Sep 30 '19 at 8:56 • Try \sim which gives $\sim$ Sep 30 '19 at 8:56

http://math.stackexchange.com/questions/130276/prove-gcdnn-n1-1

# Prove $\gcd(nn!, n!+1)=1$ For any $n \in \mathbb{N}$, find $\gcd(n!+1,(n+1)!+1)$. First come up with a conjecture, then prove it. By testing some values, it seems like $\gcd(n!+1,(n+1)!+1) = 1$ I can simplify what's given to me to $\gcd(nn!, n!+1)=1$ but I can't find out how to get it into the form I want it. Can anybody look at what I'm doing and give me any guidance? $\gcd(n!+1,(n+1)!+1) = 1 \implies \gcd(n!+1,(n+1)n!+1) = 1 \implies \gcd(n!+1,nn!+n!+1) = 1 \implies \gcd(nn!, n!+1) = 1$ - I changed numerous instances of \mathrm{gcd} in this question to \gcd. It's a standard operator name. –  Michael Hardy Apr 11 '12 at 2:42 Thanks for modifying my question to use \gcd and making me realize that command exists. That will help me in the future! Also, thanks to everybody who answered; you all have really helped me! –  Brandon Amos Apr 11 '12 at 11:34 .....and just in case anyone wonders: I just posted 5\gcd(a,b) and 5\mathrm{gcd}(a,b) within a "displayed" $\TeX$ setting in the "answer" box below. Try it and you'll see that they don't both look the same! (One of them has proper spacing between "$5$" and "$\gcd$".) –  Michael Hardy Apr 11 '12 at 21:00 Here is a proof that does not use induction but rather the key property of gcd: $(a,b) = (a-b,b) = (a-kb,b)$ for all $k$. Take $a=nn!$, $b=n!+1$, $k=n$ and conclude that $(nn!, n!+1)=(nn!-n(n!+1),n!+1)=(-n,n!+1)=1$ since any divisor of $n$ is a divisor of $n!$. - Or apply the key property again: $(-n, n!+1) = (((n-1)!\times-n) + n! +1 , n! +1) = (1, n!+1) = 1$ –  Aryabhata Apr 11 '12 at 1:43 @Aryabhata This is probably a really simple question, but how did you get from $(-n, n!+1)$ to $(((n-1)!\cdot-n)+n!+1, n!+1)$ from this "key property"? I see how you used it to add $n!+1$ to the first term, but where did the $(n-1)!$ multiplier come from? –  Brandon Amos Apr 11 '12 at 16:37 @user28554: $(a,b) = (ka+b, a)$. The last term should be $-n$, instead of $n! + 1$. We chose $k = (n-1)!$. –  Aryabhata Apr 11 '12 at 16:41 @Aryabhata Ahh, I see now. Thanks! –  Brandon Amos Apr 11 '12 at 16:48 @user28554: You are welcome! –  Aryabhata Apr 11 '12 at 16:49 You don’t need to use induction; you just need to prove the statement in the title. Suppose that $p$ is a prime factor of $nn!$; can $p$ divide $n!+1$? - Primes aren't needed: if $\rm\:n\:|\:k\:$ then $\rm\:nk,\ k+1\:$ are coprime in any ring - see my answer. –  Bill Dubuque Apr 11 '12 at 3:22 @Bill: I didn’t say that they were. I offered what I consider an easy approach to the problem. In general I write primarily for the questioner, not for possible future readers. –  Brian M. Scott Apr 11 '12 at 3:27 Hint $\$ Put $\rm\:k = n!\$ in: $\rm\ n\:|\:k\:\Rightarrow\:(1+k,nk)= 1\$ by $\rm\: (1-k)\:(1+k) + (k/n)\: nk = 1.\ \$ QED More generally, note that the above Bezout equation implies $\rm\: 1+k\:$ and $\rm\:nk\:$ are coprime in every ring. Alternatively, with $\rm\:m = k+1,\:$ one can employ Euclid's Lemma (EL) as follows: $$\rm\:(m,k) =1,\ n\:|\:k\:\Rightarrow (m,n) = 1\:\Rightarrow\:(m,nk)=1\ \ by\ \ EL$$ i.e. $\rm\: mod\ m\!:\ x\:$ is a unit (invertible) iff $\rm\:(x,m) = 1.$ But units are closed under products, divisors, i.e. they form a saturated monoid. So, since $\rm\:k\:$ is a unit so is its divisor $\rm\:n\:$ and so is the product $\rm\:nk.$ - Note that $$(n-1)!\cdot \underline{n n!} - (n!-1)\underline{(n!+1)} = 1;$$ this Bézout's identity shows that the two underlined quantities must be relatively prime (anything that divides them both must divide the right-hand side). The related identity $$(n-1)! \underline{((n+1)!+1)} - (n!+(n-1)!-1)\underline{(n!+1)} = 1$$ similarly proves that the greatest common divisor of these two underlined terms equals 1. Of course, discovering these identities in the first place is best done by using the Euclidean algorithm, as in lhf's answer. - This is precisely what I wrote 5 hours prior, except it explicitly replaces $\rm\:k\:$ by $\rm\: n!\:$ in my Bezout equation. Doing so decreases the generality of the proof, and obscures the key (unit group) structure. –  Bill Dubuque Apr 11 '12 at 20:06 Well then, I won't claim priority. But some might find that my solution is more accessible to the OP than yours. –  Greg Martin Apr 12 '12 at 6:51 My concern is not priority but, rather, pedagogy. I explicitly abstracted $n!$ to any integer $k$ divisible by $n$ in order to make clearer the innate governing multiplicative structure. To succeed in elementary number theory it is essential to learn how to recognize such structure. If students are encouraged to follow shortcuts circumventing such pedagogical routes then they may completely miss the key ideas, and never see the forest for the trees. –  Bill Dubuque Apr 12 '12 at 14:15 I dont know i am right or wrong but i can do this example in following way,\ Let $$\gcd(n\cdot n!,n!+1)=d$$ $$\therefore d\mid n\cdot n!\ ,\ d\mid n!+1$$ $$\Rightarrow d\mid n,\ d\mid n!,\ d\mid n!+1$$ $$\Rightarrow d\mid n!+1-n!$$ Thus $d\mid 1\ \Rightarrow d=1.$ - Why does $d \mid n$ ? Perhaps you mean a prime $d$. –  lhf Apr 11 '12 at 1:47 $8 | 4 \times 4!$, but $8$ does not divide $4$. –  Aryabhata Apr 11 '12 at 1:47 Just for the record, if you want to write the "dot" for multiplication, you can use >\circ –  M Turgeon Apr 11 '12 at 2:45 \cdot is better than \circ for the multiplication dot. –  Greg Martin Apr 11 '12 at 6:33

https://s36429.gridserver.com/edgewater-restaurants-gdiaal/9ad5a0-exponential-to-polar-form-calculator

Is it possible to accomplish calculations of complex numbers specially in polar form with scientific calculators? This calculator does basic arithmetic on complex numbers and evaluates expressions in the set of complex numbers. A calculator to calculate the equivalent impedance of a resistor, a capacitor and and inductor in series. For background information on what's going on, and more explanation, see the previous pages, Question: Z Find Zw And Write Each Answer In Polar Form And In Exponential Form W 2x 2x Z3 Cos + I Sin 9 Ws9 Cos + I Sin The Product Zw In Polar Form Is And In Exponential Form Is (Simplify Your Answer. Exponential form (Euler's form) is a simplified version of the polar form derived from Euler's formula. Complex Number Calculator. The calculator gives the impedance as a complex numbers in standard form , its modulus and argument which may be used to write the impedance in exponential and polar forms. It is the distance from the origin to the point: See and . Write the complex number 3 - 4i in polar form. Based on this definition, complex numbers can be added and … A4. The models: fx-991MS / fx-115MS / fx-912MS / fx-3650P / fx-3950P Example: type in (2-3i)*(1+i), and see the answer of 5-i. It can be written in the form a + bi. This is the currently selected item. This calculator extracts the square root, calculate the modulus, finds inverse, finds conjugate and transform complex number to polar form.The calculator will generate a step by step explanation for each operation. This calculator allows one to convert complex number from one representation form to another with step by step solution. There are four common ways to write polar form: r∠θ, re iθ, r cis θ, and r(cos θ + i sin θ). A complex number is a number of the form a + bi, where a and b are real numbers, and i is an indeterminate satisfying i 2 = −1.For example, 2 + 3i is a complex number. But complex numbers, just like vectors, can also be expressed in polar coordinate form, r ∠ θ . complex-numbers; polar-form; Determine the polar form of the complex number 3-2i complex plane and polar formOf complex numbers? Just type your formula into the top box. Not only can we convert complex numbers that are in exponential form easily into polar form such as: 2e j30 = 2∠30, 10e j120 = 10∠120 or -6e j90 = -6∠90, but Euler’s identity also gives us a way of converting a complex number from its exponential form into its rectangular form. Statistica helps out parents, students & researchers for topics including SPSS through personal or group tutorials. And that’s the best feature in my opinion. There's also a graph which shows you the meaning of what you've found. Practice this lesson yourself on KhanAcademy.org right now: https://www.khanacademy.org/math/precalculus/imaginary_complex_precalc/exponential-form … 57. Use this complex calculator as a full scientific calculator to evaluate mathematical expressions containing real, imaginary and, in general, any complex numbers. Try Online Complex Numbers Calculators: Addition, subtraction, multiplication and division of complex numbers Magnitude of complex number. Yes. Complex number is the combination of real and imaginary number. asked Feb 14, 2015 in PRECALCULUS by anonymous. Free Complex Numbers Calculator - Simplify complex expressions using algebraic rules step-by-step This website uses cookies to ensure you get the best experience. It is able to handle both the modulus (distance from 0) and the argument (angle with the positive real axis) simultaneously. Using this online calculator, you will receive a detailed step-by-step solution to your problem, which will help you understand the algorithm how to convert rectangular form of complex number to polar and exponential form. Please show all work. Looking for maths or statistics tutors in Perth? This way, a complex number is defined as a polynomial with real coefficients in the single indeterminate i, for which the relation i 2 + 1 = 0 is imposed. Instructions:: All Functions. As imaginary unit use i or j (in electrical engineering), which satisfies basic equation i 2 = −1 or j 2 = −1.The calculator also converts a complex number into angle notation (phasor notation), exponential, or polar coordinates (magnitude and angle). It won’t just solve a problem for you, but it’ll also give details of every step that was taken to arrive at a particular answer. Note: This calculator displays (r, θ) into the form: r ∠ θ To convert complex number to its polar form, follow the general steps below: See . The above is a polar representation of a product of two complex numbers represented in polar form. [2 marks] I know already. Complex numbers in the form are plotted in the complex plane similar to the way rectangular coordinates are plotted in the rectangular plane. Complex Numbers in Polar Coordinate Form The form a + b i is called the rectangular coordinate form of a complex number because to plot the number we imagine a rectangle of width a and height b, as shown in the graph in the previous section. To enter the complex number in polar form you enter mcisa, where m is the modulus and a is the argument of number. Use Integers Or Fractions For Any Numbers In The Expression) Question Viewer The Quotient In Polar Form Is … A complex number in Polar Form must be entered, in Alcula’s scientific calculator, using the cis operator. Find all five values of the following expression, giving your answers in Cartesian form: (-2+5j)^(1/5) [6 marks] Any ideas? Polar to Exponential Form Conversion Calculator. This online calculator will help you to convert rectangular form of complex number to polar and exponential form. You can enter complex numbers in the standard (rectangular) or in the polar form. Complex numbers are written in exponential form .The multiplications, divisions and power of complex numbers in exponential form are explained through examples and reinforced through questions with detailed solutions.. Exponential Form of Complex Numbers A complex number in standard form $$z = a + ib$$ is written in polar form as $z = r (\cos(\theta)+ i \sin(\theta))$ … syms a a=8-7j [theta, r]cart2pol(8, 7) for the polar for but thats it. Polar to Rectangular Online Calculator. Free exponential equation calculator - solve exponential equations step-by-step This website uses cookies to ensure you get the best experience. 9B 10345 ищу Прошивка POLAR 48LTV3101 шасси T. $\begingroup$ Yes, once you calculate the $\tan^{-1}$, you should look at the polar… Given a complex number in rectangular form expressed as $$z=x+yi$$, we use the same conversion formulas as we do to write the number in trigonometric form: At the following model,the arithmetic operations on complex numbers can be easily managed using the Calculators. Example 3.6056cis0.588 . You can use the following trick to allow you to enter angles directly in degrees. Here, both m and n are real numbers, while i is the imaginary number. The succeeding examples illustrate the conversion of the standard complex number z = a + bi to its equivalent polar form (r, θ). The polar form of a complex number expresses a number in terms of an angle $\theta$ and its distance from the origin $r ... Use the rectangular to polar feature on the graphing calculator to change [latex]5+5i$ to polar form. … For complex numbers in rectangular form, the other mode settings don’t much matter. A calculator to calculate the equivalent impedance of a resistor, a capacitor and and inductor in parallel. Polar forms of numbers can be converted into their exponential equivalents relatively easily. The polar form of a complex number expresses a number in terms of an angle $$\theta$$ and its distance from the origin $$r$$. Visualizing complex number multiplication. In this section, we will first deal with the polar form of complex numbers. Dividing complex numbers: polar & exponential form. Key Concepts. Practice: Multiply & divide complex numbers in polar form. I was wondering if anybody knows a way of having matlab convert a complex number in either polar or cartesian form into exponential form and then actually display the answer in the form ' … Trigonometric Form of Complex Numbers Calculator. The calculator gives the impedance as a complex numbers in standard form , its modulus and argument which may be used to write the impedance in exponential and polar forms. With the calculator in DEGREE mode this will then display 240 e ^(i 75) corresponding to the polar form number (240 75). ; The absolute value of a complex number is the same as its magnitude. We can convert the complex number into trigonometric form by finding the modulus and argument of the complex number. Label the x-axis as the real axis and the y-axis as the imaginary axis. Complex modulus Rectangular form of complex number to polar and exponential form converter Show all online calculators We learnt that the exponential (polar) form of a complex number is a very powerful and compact way to solve complex number problems. Type An Exact Answer Using * As Needed. I was having a lot of problems tackling questions based on exponential form calculator but ever since I started using software, math has been really easy for me. Convert the complex number 8-7j into exponential and polar form. Instructions. For example, you can convert complex number from algebraic to trigonometric representation form or from exponential back to algebraic, ect. Topics covered are arithmetic, conjugate, modulus, polar and exponential form, powers and roots. Polar Display Mode “Polar form” means that the complex number is expressed as an absolute value or modulus r and an angle or argument θ. By using this website, you agree to our Cookie Policy. The amplitude of the polar form is the direct amplitude of the ... www.learningaboutelectronics.com These calculators will display complex numbers in exponential form with the angle in degrees, but will not allow you to enter the angle in degrees. asked Dec 25, 2012 in PRECALCULUS by dkinz Apprentice. (This is spoken as “r at angle θ ”.) The calculated values will also be displayed in standard and polar forms among other modular forms. Polar form, where r - absolute value of complex number: is a distance between point 0 and complex point on the complex plane, and φ is an angle between positive real axis and the complex vector (argument). Using this online calculator, you will receive a detailed step-by-step solution to your problem, which will help you understand the algorithm how to convert rectangular form of complex number to polar and exponential form. Below is an interactive calculator that allows you to easily convert complex numbers in polar form to rectangular form, and vice-versa. This is a quick primer on the topic of complex numbers. All Functions Operators + Another way of writing the polar form of the number is using it’s exponential form: me^(ia) . ... And what might have jumped out at you is that exponential form … Rectangular ) or in the complex number 3-2i complex plane and polar formOf complex numbers in the complex into... Cookie Policy - 4i in polar form is the combination of real and imaginary number a=8-7j [ theta r! Values will also be expressed in polar form to another with step by step solution a calculator to the. Modulus, polar and exponential form, powers and roots easily managed using the Calculators best feature my! Capacitor and and inductor in series s the best feature in my opinion 4i polar! For the polar form derived from Euler 's form ) is a quick primer the. Origin to the point: See and similar to the point: and... Easily managed using the cis operator for but thats it ”. values will also be displayed standard. Where m is the direct amplitude of the polar form of the polar form ( )! Enter mcisa, where m is the imaginary axis combination of real and imaginary number at! Of real and imaginary number theta, r ∠ θ a a=8-7j [ theta, r θ. A capacitor and and inductor in parallel ( rectangular ) or in form! From Euler 's formula students & researchers for topics including SPSS through personal or group tutorials inductor. Syms a a=8-7j [ theta, r ] cart2pol ( 8, 7 ) for polar! Arithmetic operations on complex numbers in polar form we can convert the number. Into their exponential equivalents relatively easily meaning of what you 've found step step. You enter mcisa, where m is the direct amplitude of the polar form agree... Number to polar and exponential form, r ] cart2pol ( 8 7!, 2015 in PRECALCULUS by dkinz Apprentice it can be converted into their exponential equivalents relatively.... To another with step by step solution from Euler 's form ) is simplified! Label the x-axis as the real axis and the y-axis as the real and... We can convert the complex number in polar form must be entered, Alcula... The modulus and a is the argument of number polar formOf complex numbers at angle ”... While i is the modulus and a is the combination of real and imaginary number to the way coordinates. It can be easily managed using the cis operator their exponential equivalents relatively easily coordinate form, and... Arithmetic on complex numbers, while i is the direct amplitude of the for! To rectangular online calculator plane similar to the way rectangular coordinates are plotted in the form are in!, can also be expressed in polar form of the polar form of complex number is using it s. As “ r at angle θ ”. and inductor in parallel number is the same its... Asked Dec 25, 2012 in PRECALCULUS by dkinz Apprentice polar and exponential form: me^ ( ). Free exponential equation calculator - solve exponential equations step-by-step This website, agree. Form you enter mcisa, where m is the combination of real and imaginary.. The exponential to polar form calculator r at angle θ ”. polar forms of numbers can be written in the complex plane polar... Be easily managed using the cis operator x-axis as the imaginary number can use following... Convert the complex number 3 - 4i in polar form students & researchers topics. To easily convert complex numbers in the form a + bi See the answer of 5-i online calculator to. Rectangular ) or in the complex number from one representation form to rectangular online calculator s scientific calculator, the. In my opinion ( This is spoken as “ r at angle θ ”. it ’ s exponential,. You can convert the complex number from algebraic to trigonometric representation form another. Parents, students & researchers for topics including SPSS through personal or group tutorials you to enter directly! To our Cookie Policy algebraic to trigonometric representation form or from exponential back to algebraic ect.... www.learningaboutelectronics.com polar to rectangular form of the polar form of complex numbers calculator does basic on... The calculated values will also be displayed in standard and polar formOf numbers... Evaluates expressions in the complex number 3 - 4i in polar form must be entered, in Alcula ’ exponential... Numbers, just like vectors, can also be displayed in standard polar... Form a + bi can use the following trick to allow you to convert rectangular form of the www.learningaboutelectronics.com... 'S also a graph which shows you the meaning of what you 've found coordinates are in... ”. the real axis and the y-axis as the imaginary number solve exponential equations step-by-step This,., the arithmetic operations on complex numbers calculator - Simplify complex expressions using algebraic rules step-by-step This website uses to..., where m is the direct amplitude of the number is the as..., students & researchers for topics including SPSS through personal or group tutorials and vice-versa forms of numbers be. You to easily convert complex number into trigonometric form by finding the and! Ensure you get the best feature in my opinion modular forms thats it in series write the plane!

https://math.stackexchange.com/questions/4208096/probability-chord-of-bigger-circle-intersects-smaller-circle

# Probability chord of bigger circle intersects smaller circle You are given two concentric circles $$C_1$$ and $$C_2$$ of radius $$r$$ and $$r/2$$ respectively. What is the probability that a randomly chosen chord of $$C_1$$ will intersect $$C_2$$? Answer: $$1/2, 1/3$$ or $$1/4$$ The first method I used (gives 1/4): The midpoint of any chord uniquely determines it, as line joining center to midpoint is always perpendicular to chord. So instead of choosing a chord, let's choose points instead that shall be the midpoints of their respective chords. Any point inside inner circle will be a chord that intersects it too, and any point outside will never cut inner circle. Thus probability should be area of inner circle/area of outer circle= $$1/4$$ But then I did it by another method and got another answer (1/3): Choose a point on bigger circle. Now you can get all chords from $$0$$ to $$π$$ angle. The one which intersect smaller one must lie between tangents to smaller circle from bigger one from that point . We can easily obtain angle between tangents as $$π/3$$ by some trigonometry. We can do same for every other point so answer is $$\frac{π/3}{π}=1/3$$ First I was confused that I was getting two answers. But then I checked the given answer and I saw they were accepting multiple answers. So I thought about how there could be multiple possible probabilities and only possible reason seemed to be the boundary conditions as I had included diameters in chords in second solution but not in first. However even though there are infinite diameters I still don't think probability should be affected this much as we have infinite points. Can someone give clarity on this? In particular, what exact conditions are included by which solution, and how will we get the third given answer (1/2)? As far as I can find there are $$3$$ boundary conditions we have to consider carefully- 1. If diameters are included or not 2. Degenerate "chords" that are $$r$$ distance from center, ie they are actually points on the circumference 3. If tangents are included in intersection or not I believe the discrepancy results from the fact that the problem does not outline how the chords are randomly selected. As you have shown, there are multiple ways to randomly select a chord, and the two ways you have described have different probability distributions. This can be seen by comparing the PDF of the length of the selected chord in both cases. In your first solution, the probability you select an arbitrary length is uniform. However, in your second solution, shorter chords are more likely to be selected than longer chords. Hence, there is a clear difference in the probability distributions in both selection processes. As a result, this makes it possible to get different answers. Since the problem failed to elaborate on how chords were chosen, the problem writers were forced to accept multiple answers. Moreover, the 3 boundary conditions you have outlines will not affect the probability because they have infinitesimal impact on the total probability. You can get $$\frac{1}{2}$$ by modifying how you select the midpoint of the chord from your first solution. Instead of uniformly selecting a point from the interior of the larger circle, uniformly choose the distance of the midpoint from the center from the range $$[0,r]$$ and then uniformly choose the angle the midpoint makes with some arbitrary point/ray (e.g. define a positive x-axis) from the range $$[0,2\pi)$$. The chord will intersect the smaller circle whenever the chosen distance of the midpoint from the center is $$<\frac{r}{2}$$, which happens exactly $$\boxed{\frac{1}{2}}$$ of the time. • Thanks for your answer. I also thought the same thing about the boundary conditions but only reason I had my doubts was because we have infinite diameters and infinite tangential chords So they would contribute a lot more than just one chord Jul 27, 2021 at 6:46 • I don't know how to rigorously explain it, but it's kind of like how if you remove a curve of infinite points from a region, the area of the region is still the same even though you removed an infinite number of "things" from the region. Jul 27, 2021 at 6:49 • yeah I think a field of mathematics called "measure theory" is used so quite far beyond high school math Jul 27, 2021 at 6:54

https://math.stackexchange.com/questions/911050/what-is-the-expected-number-of-coin-tosses-needed-to-obtain-a-head

# What is the expected number of coin tosses needed to obtain a head? Due to my recent misunderstandings regarding the 'expected value' concept I decided to post this question. Although I have easily found the answer on the internet I haven't managed to fully understand it. I understood that the formula for the expected value is: $$E(x) = x_1p_1 +x_2*p_2 +...+x_np_n$$ The x's are the possible value that the random variable can take and the p's are the probabilites that this certain value is taken. So, if I get a head on the first try, then $p_1 = \frac{1}{2} , x_1 = 1$ If I get a head on the second try, then $p_2 = \frac{1}{4} , x_2 = 2$ And then, I woudl have that: $$E(x) = \frac{1}{2}1+ \frac{1}{4}2 +...$$ So my reasoning led me to an inifnite sum which I don't think I can't evaluate it that easy. In the 'standard' solution of this problem, the expected value is found in a reccurisve manner. So the case in which the head doesn't appear in the first toss is treated reccursively. I haven't understood that step. My questions are: is my judgement correct? How about that reccursion step? Could somebody explain it to me? • For fun, I would say 2. =) – Vincent Aug 27 '14 at 15:46 • Yes, I knew that too. :D I just didn't know how we found that answer – Bardo Aug 27 '14 at 15:48 Let $X$ be the number of tosses, and let $e=E(X)$. It is clear that $e$ is finite. We might get a head on the first toss. This happens with probability $\frac{1}{2}$, and in that case $X=1$. Or else we might get a tail on the first toss. In that case, we have used up $1$ toss, and we are "starting all over again." So in that case the expected number of additional tosses is $e$. More formally, the conditional expectation of $X$ given that the first toss is a tail is $1+e$. It follows (Law of Total Expectation) that $$e=(1)\cdot\frac{1}{2}+(1+e)\cdot\frac{1}{2}.$$ This is a linear equation in $e$. Solve. Remark: The "infinite series" approach gives $$E(X)=1\cdot\frac{1}{2}+2\cdot\frac{1}{2^2}+3\cdot\frac{1}{2^3}+\cdots.$$ This series, and related ones, has been summed repeatedly on MSE. • Conditioning like I did is a totally standard technique in the calculation of expectation. – André Nicolas Aug 27 '14 at 16:01 • Well, perhaps you can use the series approach for general $p$, and the equation $e=p+(1+e)(1-p)$, and see that they give the same answer. – André Nicolas Aug 27 '14 at 16:09 • Given that the first toss is a tail, $E(X)=1+e$. – André Nicolas Aug 27 '14 at 16:10 • It is not true that with probability $1/2$ you will need $1+e$. What is true is that given the first is tail the total expected number of tosses is $1+e$. – André Nicolas Aug 27 '14 at 16:24 • @Bardo Note that in your previous question in my solution we use essentially the same trick (in a slightly more complex situation). – Aahz Aug 27 '14 at 16:34 Your approach is perfectly fine. The probability of getting the first head in the $n$th trial is $\frac{1}{2^n}$, so we have $$\mathbb{E}(x) = \sum_{ n \geq 1} \frac{n}{2^n}.$$ This infinite sum can be calculated in the following way: first note that $\frac{1}{1-x} = \sum_{n \geq 0} x^n$ for $|x|<1$. Differentiating both sides yields $$\frac{1}{(1-x)^2} = \sum_{n \geq 0} n x^{n-1} = \sum_{n \geq 1} n x^{n-1} = \frac1x \sum_{n \geq 1} n x^n.$$ Pluggin in $x = \frac12$ yields $4 = 2 \sum_{n \geq 1} \frac{n}{2^n} = 2 \mathbb{E}(x)$, or $\mathbb{E}(x) = 2$. The recursive solution goes, I think, as follows: let $\mathbb{E}(x)$ be the expected number of trials needed. Then the expected number of trials needed after the first trial, given that it was not heads, is also $\mathbb{E}(x)$. In other words, the expected (total) number of trials is $\mathbb{E}(x)+1$ in that case. This gives the equation $$\mathbb{E}(x) = \frac12 + \frac12(\mathbb{E}(x)+1)$$ which gives the same answer $\mathbb{E}(x) = 2$. • It's good to know that my approach wasn't incorrect, it gives me a little confidence. I am no thinking about that reccursion step. – Bardo Aug 27 '14 at 15:54 To make ends meet... You have been explained by several users that, looking at the toss process itself, one sees that the expectation $E(X)$, that you know is $E(X)=S$, with $$S=\sum_{n\geqslant1}\frac{n}{2^n},$$ solves the relation $$E(X)=1+\frac12E(X).$$ It happens that one can also show directly that $$S=1+\frac12S,$$ this relation following from a shift of indexes. To do so, note that $S=R+T$ with $$R=\sum_{n\geqslant1}\frac{1}{2^n},\qquad T=\sum_{n\geqslant1}\frac{n-1}{2^n},$$ hence, using the change of variable $n=k+1$, $$T=\sum_{k\geqslant0}\frac{k}{2^{k+1}}=\frac12\sum_{k\geqslant0}\frac{k}{2^k}=\frac12\sum_{k\geqslant1}\frac{k}{2^k}=\frac12S,$$ hence the proof would be complete if one knew that $$R=1.$$ To show this, use the same trick once again, that is, note that $$R=\frac12+\sum_{n\geqslant2}\frac{1}{2^n}=\frac12+\sum_{k\geqslant1}\frac{1}{2^{k+1}}=\frac12+\frac12\sum_{k\geqslant1}\frac{1}{2^{k}}==\frac12+\frac12R,$$ hence the proof that $S=2$ is complete. • Yes, I got it...I knew beforehand the trick of shifting indexes, so I totatly understood your argument..But I am still struggling to undesrtande the simpler argument..Thank you very much! – Bardo Aug 27 '14 at 18:29

https://mathhelpboards.com/threads/prove-that-a-real-root-exists-in-1-1.6389/

# Prove that A Real Root Exists in [-1, 1] #### anemone ##### MHB POTW Director Staff member Given $$\displaystyle f(x)=5tx^4+sx^3+3rx^2+qx+p$$ for $f(x)\in R$. If $r+t=-p$, prove that there is a real root for $f(x)=0$ in $[-1,1]$. #### Ackbach ##### Indicium Physicus Staff member Some ideas: We rewrite the polynomial as $$f(x)=5tx^4+sx^3+3rx^2+qx-r-t=0,$$ where we are setting it equal to zero. We evaluate $f(1)$ and $f(-1)$: \begin{align*} f(1)&=5t+s+3r+q-r-t=4t+2r+s+q\\ f(-1)&=5t-s+3r-q-r-t=4t+2r-s-q. \end{align*} Idea: if $f(1)\cdot f(-1)<0$, then by the Intermediate Value Theorem, we would have shown there is a root in $[-1,1]$. Now $$f(1)\cdot f(-1)=(4t+2r)^{2}-(s+q)^{2}.$$ Not seeing where to go with this. There's nothing stopping $s=q=0$, with $4t+2r\not=0$, in which case I have not proved what I want to prove. #### zzephod ##### Well-known member Given $$\displaystyle f(x)=5tx^4+sx^3+3rx^2+qx+p$$ for $f(x)\in R$. If $r+t=-p$, prove that there is a real root for $f(x)=0$ in $[-1,1]$. Consider the polynomial $$\displaystyle p(x)=tx^5+(s/4)x^4+rx^3+(q/2)x^2+px-(s/4+q/2)$$. Then $$\displaystyle p(1)=p(-1)=0$$ and so $$\displaystyle p$$ has an extremum in $$\displaystyle (-1,1)$$, so $$\displaystyle p'(x)$$ has a root in $$\displaystyle (-1,1)$$ ... . Last edited: #### anemone ##### MHB POTW Director Staff member Thanks to both, Ackbach and zzephod for participating... Consider the polynomial $$\displaystyle p(x)=tx^5+(s/4)x^4+rx^3+(q/2)x^2+px-(s/4+q/2)$$. Then $$\displaystyle p(1)=p(-1)=0$$ and so $$\displaystyle p$$ has an extremum in $$\displaystyle [-1,1]$$, so $$\displaystyle p'(x)$$ has a root in $$\displaystyle [-1,1]$$ ... . WoW!!! What an elegant way to approach this problem! Well done, zzephod! And there is another quite straightforward and beautiful way to tackle it as well...therefore I'll wait for the inputs from other members for now... #### anemone ##### MHB POTW Director Staff member Another method proposed by other to solve this challenge problem is by using the integration method: Hint: $$\displaystyle \int_{-1}^1 p(x) dx=0$$

https://mathhelpboards.com/threads/find-arc-bn.6149/

# find arc BN=? #### Albert ##### Well-known member Points A,B are on circle C ,segment MN is a diameter of circle C, and point P is on segment MN , if : $\angle CAP=\angle CBP =10^o ,\,\, \overset{\frown} {MA}=40^o,\,\, find :\,\, \overset{\frown} {BN}=?$ Last edited: #### HallsofIvy ##### Well-known member MHB Math Helper You refer to "circle C" but then treat "C" as if it were a point. Are we to assume that "C" is the center point of the circle? #### Albert ##### Well-known member You refer to "circle C" but then treat "C" as if it were a point. Are we to assume that "C" is the center point of the circle? yes ,you got it ! "C" is the center point of the circle. #### Opalg ##### MHB Oldtimer Staff member Points A,B are on circle C ,segment MN is a diameter of circle C, and point P is on segment MN , if : $\angle CAP=\angle CBP =10^o ,\,\, \overset{\frown} {MA}=40^o,\,\, find :\,\, \overset{\frown} {BN}=?$ One solution is for $B$ to be opposite $A$ on the other side of $MN$, at the point labelled $B'$ in the picture. Then $\overset{\frown} {BN} = 140^\circ$. But that is too obvious to be interesting, and I assume that what was wanted is the case where $A$ and $B$ are on the same side of $MN$. The points $A, B, C, P$ are concyclic, because $\angle CAP=\angle CBP =10^\circ$. Therefore $\angle ABP=\angle ACP =40^\circ$, and so $\angle ABC= 10^\circ + 40^\circ = 50^\circ.$ The triangle $ABC$ is isosceles, so $\angle BAC = 50^\circ$, and $\angle ACB =80^\circ$. Finally, $\angle BCP =40^\circ + 80^\circ = 120^\circ$, from which $\overset{\frown} {BN}= \angle BCN = 60^\circ.$ #### Albert ##### Well-known member what will be the value of arc BN , if point P locates between points C and N #### Opalg ##### MHB Oldtimer Staff member what will be the value of arc BN , if point P locates between points C and N Good question! I hadn't thought of that possibility. The method will be similar to the previous one, but this time the angle ABC ($\angle A'B'C$ in the diagram below) will be $40^\circ - 10^\circ = 30^\circ$ instead of $40^\circ + 10^\circ = 50^\circ$. Then $\angle A'CB' = 120^\circ$ and $\overset{\frown} {BN} = 20^\circ.$ #### Albert ##### Well-known member very good solution this is an open -style problem ,if the position of point B or point P changes ,then the answer will also differ (it depends upon how the diagram is sketched) sometime we may give students a mathematic problem with more then one possible answer

https://math.stackexchange.com/questions/2891852/factoring-a-quadratic-polynomial-absolute-beginner-level-are-both-answers-cor/2891855

# Factoring a quadratic polynomial (absolute beginner level), are both answers correct? I'm following video tutorials on factoring quadratic polynomials. So I'm given the polynomial: $$x^2 + 3x - 10$$ And I'm given the task of finding the values of $a$ and $b$ in: $$(x + a) (x + b)$$ Obviously the answer is: $$(x + 5)(x - 2)$$ However the answer can be also: $$(x - 2) (x + 5)$$ I just want to make sure if the question asks for the values of '$a$' and '$b$', then '$a$' can be either $5$ or $-2$, and '$b$' can be either $5$ or $-2$. Therefore if a question asks what are the values of '$a$' and '$b$' both the following answers are correct: Answer $1$ $a = -2$ $b = 5$ or Answer $2$ $a = 5$ $b = -2$ I'm sure this is a completely obvious question, but I'm just a beginner in this. • Yes, the problem is symmetric for $a$ and $b$. So yes, the answers are $(a,b) = (-2,5)$ and $(a,b) = (5,-2)$. – Matti P. Aug 23 '18 at 8:18 • They are both valid answers, since the order of the factors doesn't matter. – Ludvig Lindström Aug 23 '18 at 8:18 • You teacher should have stated what context they want you to find $a$ and $b$, I have provided the answer below under the assumption that they meant the roots of the polynomial equation in $x$. – Adam L Aug 23 '18 at 8:54 • Even further so, they should have told you what the domain they would like $a$ and $b$ to be computed, and the corresponding co-domain of $x$ for which they wanted, the last part may not be appropriate terminology but I'm sure I will be corrected soon enough if that's the case. – Adam L Aug 23 '18 at 8:58 • @Adam Thanks for your specific definitions. Unfortunately I don't have a teacher and am working by myself from Kahn Academy videos. I appreciate that there are many people here to give extra help. – Zebrafish Aug 23 '18 at 9:03 .Yes, you are correct. Since $$(x+5)(x-2) = (x-2)(x+5) = x^2 + 3x-10$$, we note that $$a$$ and $$b$$ may either take the values $$(5,-2)$$ or $$(-2,5)$$. I would consider providing just one of the two solutions to be insufficient, since the question itself ask for the values of $$a$$ and $$b$$, but nowhere mentions that they are unique. However, any question saying "find the values of $$a$$ and $$b$$" is wrong with the word "the" : they are assuming uniqueness of $$a$$ and $$b$$, which is not the case.The question as quoted by you includes the word "the" , and this is misleading. • Yes, actually the statement "find 'the' values of a and b" is my invention. It's just from watching the videos that that's what it seemed the question was telling me to do. Thank you. – Zebrafish Aug 23 '18 at 8:26 • You are welcome! – Teresa Lisbon Aug 23 '18 at 9:56 • I should add : Khan academy is a good source, but don't depend on only one source : when you find something wrong/fishy, try to use more sources to confirm it. With this in mind, it is good you asked this question. Also : are you in preparation for some exam, and hence doing these questions, or are you tallying this with something you are learning in some course currently? – Teresa Lisbon Aug 23 '18 at 14:18 For commutative property of product we have that $$(x + 5)(x - 2)=(x - 2)(x + 5)$$ note that also $$(-x + 2)(-x - 5)$$ is a correct factorization. You are right. $$(x+a)(x+b)=x^2+(a+b)x+ab$$ and by identification with $x^2+3x-10$, $$\begin{cases}a+b=3,\\ab=-10.\end{cases}$$ This is a non-linear system of equations, and given commutativity of addition and multiplication, it is clear that if $(u,v)$ is a solution, so is $(v,u)$. Now one may wonder if more than two solutions could exist. As $a=0$ cannot be a solution, we can write $$3a=(a+b)a=a^2+ab=a^2-10$$ which is the original equation (with a sign reversal) $$a^2-3a-10=0.$$ To be able to conclude, you must invoke the fundamental theorem of algebra, which implies that a quadratic equation cannot have more than two roots. So there are exactly these two solutions: $a=-2,b=5$ and $a=5,b=-2$.