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# Math Help - Need help with 3 Calculus Extra Credit Problems 1. ## Need help with 3 Calculus Extra Credit Problems I'm in intro to calculus and I need help setting this equation up: Newton's Law of cooling: the rate at which the temperature of an object changes is proportional to the difference between its own temperature and that of the surrounding medium. A cold drink is removed from a refrigerator on a hot summer day and placed in a room where the temperature is 80°F. Express the temperature of the drink as a function of time (minutes) if the temperature of the drink was 40°F when it left the refrigerator and was 50°F after 20 minutes in the room. Thanks! 2. Newton's Law of cooling: the rate at which the temperature of an object changes is proportional to the difference between its own temperature and that of the surrounding medium. $\frac{dT}{dt} = k(T-A)$ A = room temperature (a constant) T = temperature of the cold drink at any time t in minutes k = proportionality constant separate variables and integrate. 3. Ok, with your help and help from another problem, I have this. Is it the next step? dT/dt= -k(T-TM) Where T(t) is the temperature of the drink, and TM is the temperature of the surrounding solution. So: dT/dt= -(T-80) This doesn't seem right, I don't know how to incorporate the other numbers and variables. Grrr calculus 4. $\frac{dT}{dt} = k(T-80)$ separate variables ... $\frac{dT}{T-80} = k \, dt$ integrate ... $\ln|T-80| = kt + C_1$ $|T-80| = e^{kt + C_1}$ $|T-80| = e^{C_1} \cdot e^{kt}$ $T - 80 = C_2 \cdot e^{kt}$ $T = 80 + C_2 \cdot e^{kt}$ The calculus is done, so I'm stopping at this point. The rest is algebra ... you were given two temperatures at two different times. With that info, you can determine the constants $C_2$ and $k$ and finalize the temperature as a function of time. 5. I greatly appreciate your help. I have: Initial value (40 degrees at time 0) T= 80 + Ce^kt 40= 80 + Ce^k*0 40= 80 + C C= -40 How do I find k? Second value (50 degrees at time 20) T= 80 + Ce^kt 50= 80 + Ce^k*20 -30= Ce^20k Again finding k has stumped me. What does k represent and how do i find it? 6. Originally Posted by Sm10389 I greatly appreciate your help. I have: Initial value (40 degrees at time 0) T= 80 + Ce^kt 40= 80 + Ce^k*0 40= 80 + C C= -40 good. How do I find k? Second value (50 degrees at time 20) how about substituting in -40 for C ? then find k with the second value. T= 80 + Ce^kt do it. 7. you my friend are a genius. so.... Initial value (40 degrees at time 0) T= 80 + Ce^kt 40= 80 + Ce^k*0 40= 80 + C C= -40 Second value (50 degrees at time 20) T= 80 + Ce^kt 50= 80 + Ce^k*20 -30= -40e^20k 3/4= e^20k ln3/4=20k k= ln(3/4)/80 so would this be my final answer? T= 80 + -40e^ln((3/4)/20)t 8. ## skeeter how does that look 9. Originally Posted by Sm10389 how does that look check it out yourself ... graph the result in your graphing utility and see if the given info matches up. 10. i just did, and it is not correct. where did i goof up? 11. so would this be my final answer? T= 80 + -40e^ln((3/4)/20)t looks like you have ... $k = \ln\left(\frac{\frac{3}{4}}{20}\right)$ ... which it ain't. should be ... $k = \frac{\ln\left(\frac{3}{4}\right)}{20}$ 12. Thank you, I had that too, I just did not use the parentheses correctly in my calculator. The next step of the problem is to calculate it if the drink were warmer then room temperature (>80). I know how to set it up like the last one, but we would only be given one point (0, 85) for example. How would I calculate k here? 13. $k$ will remain the same because the rate of heat transfer will remain the same ... however, you'll have to recalculate $C_2$. 14. ## QUESTION #2 More from the Introduction to Differential Equations- Investment plan- an investor makes regular deposits totaling D dollars each year into an account that earns interest at the annual rate r compounded continuously. A: Explain why the account grows at the rate ( dV/dt = rV + D ) where V(t) is the value of the account 2 years after the initial deposit. Solve this differential equation to express V(t) in terms of r and D. I came up with this: V(t)= (C/r)*e^rt - (D/r) I am sure it is correct. This is the next part: Amanda wants to retire in 20 years. To build up a retirement fund, she makes regular annual deposits of \$8,000. If the prevailing interest rate stays constant at 4% compounded continuously, how much will she have in her account at the end of the 20 year period? I know how to do everything but: Find C Figure out how compounding continuously would affect the equation.

https://math.stackexchange.com/questions/2746968/the-proof-of-infinitude-of-pythagorean-triples-x-x1-z

# The Proof of Infinitude of Pythagorean Triples $(x,x+1,z)$ Proof that there exists infinity positive integers triple $x^2+y^2=z^2$ that $x,y$ are consecutive integers, then exhibit five of them. This is a question in my number theory textbook, the given hint is that "If $x,x+1,z$ is a Pythagorean triple, then so does the triple $3x+2z+1,3x+2z+2, 4x+3z+2$" I wondered how someone come up with this idea. My solution is letting $x=2st, y=s^2-t^2, z=s^2+t^2$ by $s>t, \gcd(s,t)=1$.then consider two cases: $y=x+1$ and $y=x-1$ Case 1: $y=x+1$ Gives me $(s-t)^2-2t^2=1$ then I found this is the form of Pell's equation, I then found \begin{align}s&=5,29,169,985,5741\\t&=2,12,20,408,2378\end{align}then yields five triples $$(20,21,29),(696,697,985),(23660,23661,33461),(803760,803761,1136689),(27304196,27304197,38613965)$$ Case 2:$y=x-1$ Using the same method, I come up with Pell's equation $(s+t)^2-2s^2=1$, after solve that I also get five triples: $$(4,3,5),(120,119,169),(4060,4059,5741),(137904,137903,195025),(4684660,4684659,6625109)$$ I have wondered why the gaps between my solution are quite big, with my curiosity, I start using question's hint and exhibit ten of the triples:$$(3,4,5),(20,21,29),(119,120,169),(696,697,985),(4059,4060,5741),(23660,23661,33461),(137903,137904,195025),(803760,803761,1136689),(4684659,4684660,6625109),(27304196,27304197,38613965)$$ These are actually the same as using solutions alternatively from both cases. But I don't know is this true after these ten triples Basically the problem was solved, but I would glad to see if someone provide me a procedure to come up with the statement "If $x,x+1,z$ is a Pythagorean triple, then so does the triple $3x+2z+1,3x+2z+2, 4x+3z+2$", and prove that there are no missing triplet between it. --After edit-- Thanks to @Dr Peter McGowan !, by the matrix $$\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2\\ 2 & 2 & 3 \end{bmatrix} \begin{bmatrix} x\\x+1\\z \end{bmatrix} = \begin{bmatrix} 3x+2z+2\\3x+2z+1\\4x+3z+2 \end{bmatrix}$$ gives me the hinted statement. • Hint : If $(a/b)$ is a solution of the Pell-equation $a^2-2b^2=-1$ , then the next solution is $(3a+4b/2a+3b)$ – Peter Apr 21 '18 at 7:43 • Wow, how to know that? – kelvin hong 方 Apr 21 '18 at 7:54 • artofproblemsolving.com/community/c3046h1049346__2 – individ Apr 21 '18 at 8:50 • @individ thanks, but a relevant proof is better. – kelvin hong 方 Apr 21 '18 at 10:33 • Go here. It will take you to a question of mine where I prove the infinitude of Pythagorean Triples... but not using Pell equations, however. Nonetheless, this post might serve more use if $z=x+1$ as opposed to $y$, since I show that $$(2v^2+2v)^2+(2v+1)^2=(2v^2+2v+1)^2\;\forall v.$$ Still, it might increase your understanding on Pythagorean Triples :) – Mr Pie Jun 20 '18 at 12:20 ## 3 Answers You are on the right track. The simplest solution is to recall that all irreducible Pythagorean triples for a rooted ternary tree beginning with $(3, 4, 5)$ triangle. B Berggren discovered that all others can be derived from this most primitive triple. F J M Barning set these out as three matrices that when pre-multiplied by a "vector" of a Pythagorean triple produces another. For the case of consecutive legs we have, starting with $(x_1, y_1, z_1)$, we may calculate the next triple as follows: \begin {align} x_2&=x_1+2y_1+2z_1 \\ y_2&=2x_1+y_1+2z_1 \\ z_2&=2x_1+2y_1+3z_1 \end {align} The hint you were given is a variation on the above more general formula specific for consecutive leg lengths. It is an easy proof by induction to show that the formulas are correct. The first few are: $(3, 4, 5); (20, 21, 29); (119, 120, 169); (696, 697, 985); (4059, 4060, 5741); (23660, 23661, 33461)$; etc. Obviously, this can be continued indefinitely. The sequence rises geometrically. A simple explicit formula is available for these solutions that are (as you have already guessed) alternating solutions to Pell's equation. • Wow, although I have seen that matrix before but don't realize it can be so useful! I have found the related matrix and actually come out with the desired result.Thanks a lot!!!! – kelvin hong 方 Apr 21 '18 at 10:43 $x,x+1,z$ is a Pythagorean triple iff $(2x+1)^2+1=2z^2$. Let $u=2x+1$. Then $u^2-2z^2=-1$, a negative Pell equation whose solution lies in considering the units of $\mathbb Z[\sqrt 2]$ of norm $-1$. It is clear that $\omega=1+\sqrt 2$ is a fundamental unit with norm $-1$. Therefore, all the other solutions of $u^2-2z^2=-1$ come from odd powers of $\omega$. Thus, if $(u_k,z_k)$ is a solution of $u^2-2z^2=-1$, then the next one is given by \begin{align} u_{k+1}+z_{k+1}\sqrt 2&=(u_k+z_k\sqrt 2)\omega^2 \\ &=(u_k+z_k\sqrt 2)(3+2\sqrt 2) \\ &=(3u_k+4z_k)+(2u_k+3z_k)\sqrt 2 \end{align} So, $u_{k+1}= 3u_k+4z_k$ and $z_{k+1}=2u_k+3z_k$. Now let $u_{k+1}=2x_{k+1}+1$. Then $$x_{k+1}=\frac{u_{k+1}-1}{2}=\frac{(3u_k+4z_k)-1}{2}=\frac{3(2x_k+1)+4z_k-1}{2}=3x_k+2z_k+1$$ and $$z_{k+1}=4x_k+3z_k+2$$ as claimed. • Wow, thanks for your solution! – kelvin hong 方 Apr 28 '18 at 2:54 Pythagorean triples where $$|A-B|=1$$ are scarce and get more rare with altitude but they can continue to be found, indefinitely, given arbitrary precision. (There are only $$22$$ of them with $$A,B,C<10.2\text{ quadrillion.})$$ Proof that there are in infinite number of them is shown by the following functions which accept and yield natural numbers without end. Another way of putting it is that we are eliminating (subtracting) a countably infinite set of triples from another countably infinite set of triples and even when you subtract a supposedly larger $$\aleph_0$$ set from a smaller $$\aleph_0$$ set (odd numbers minus natural numbers) the results are still infinite because they can be mapped. Here we have all triples minus extraneous triples. To find them in a timely manner, we can use the following function to find the right combination of $$(m,n)$$ to use in Euclid's formula. The $$\pm$$ is used because the triples alternate: $$A>B$$ and $$A. $$\text{With infinite integers input }(n=\sqrt{2*m^2\pm1}-m)\text{ will always output an integer for some }m.$$ For all $$M\in\mathbb{N}$$, whenever this function yields an integer $$>0$$, we have the $$(m,n)$$ we need for a triple. The following $$19$$ were generated (in $$4$$ seconds) in a loop of $$m=1\text{ to }16,000,000$$ and $$f(m,n)$$ shows the variable values needed to generate each triple using Euclid's formula. The $$g(n,k)$$ shows the values (generated in $$1$$ second) for an alternate set of functions in a loop of $$m=1\text{ to }5,000,000$$. $$A=m^2-n^2\qquad B=2mn\qquad C=M^2+n^2$$ $$f(2,1)=g(1,1)=(3,4,5)$$ $$f(5,2)=g(2,2)=(21,20,29)$$ $$f(12,5)=g(4,5)=(119,120,169)$$ $$f(29,12)=g(9,12)=(697,696,985)$$ $$f(70,29)=g(21,29)=(4059,4060,5741)$$ $$f(169,70)=g(50,70)=(23661,23660,33461)$$ $$f(408,169)=g(120,169)=(137903,137904,195025)$$ $$f(985,408)=g(289,408)=(803761,803760,1136689)$$ $$f(2378,985)=g(697,985)=(4684659,4684660,6625109)$$ $$f(5741,2378)=g(1682,2378)=(27304197,27304196,38613965)$$ $$f(13860,5741)=g(4060,5741)=(159140519,159140520,225058681)$$ $$f(33461,13860)=g(9801,13860)=(927538921,927538920,1311738121)$$ $$f(80782,33461)=g(23661,33461)=(5406093003,5406093004,7645370045)$$ $$f(195025,80782)=g(57122,80782)=(31509019101,31509019100,44560482149)$$ $$f(470832,195025)=g(137904,195025)=(183648021599,183648021600,259717522849)$$ $$f(1136689,470832)=g(332929,470832)=(1070379110497,1070379110496,1513744654945)$$ $$f(2744210,1136689)=g(803761,1136689)=(6238626641379,6238626641380,8822750406821)$$ $$f(6625109,2744210)=g(1940450,2744210)=(36361380737781,36361380737780,51422757785981)$$ $$f(15994428,6625109)=g(4684660,6625109)=(211929657785303,211929657785304,299713796309065)$$ The alternate formula mentioned above is slightly faster since it deals only with the subset of triples where $$GCD(A,B,C)=(2m-1)^2,m\in\mathbb{N}$$. We have the needed $$(n,k)$$ values whenever the following yields a positive integer. $$k=\sqrt{\frac{(2n-1)^2\pm1}{2}}$$ Having found a needed $$(n,k)$$ the following will generate a needed triple. $$A=(2n-1)^2+2(2n-1)k\qquad B=2(2n-1)k+2k^2\qquad C=(2n-1)^2+2(2n-1)k+2k^2$$

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# Solving Cubic Equations Formula Solving Cubic Equations with the help of Factor Theorem If x – a is indeed a factor of p(x), then the remainder after division by x – a will be zero. It solves cubic, quadratic and linear equations. 2 But it is important to remember van der Waals’ equation for the volume is a cubic and cubics always. Only th is a variable. Yes, it is not easy for one to get a complete hold of the formulas and tricks used in mathematics for different types of purposes. That formula can be used in a vectorized form. Title: Hyperbolic identity for solving the cubic equation: Authors: Rochon, Paul: Publication: American Journal of Physics, Volume 54, Issue 2, pp. After reading this chapter, you should be able to: 1. (Sometimes it is possible to find all solutions by finding three values of x for which P(x) = 0 ). 00000001, initial_guess=0. Polynomials and Partial Fractions In this lesson, you will learn that the factor theorem is a special case of the remainder theorem and use it to find factors of polynomials. Individuals round out. It isnt like the other typical zeros root problems ive seen where they give you the x intercepts. Type y= 2x+5 into your calculator and look at the graph. solving a cubic equation. Learn more about cubic equation Symbolic Math Toolbox. Calculator Use. , the roots of a cubic polynomial. Equivalently, the cubic formula tells us the solutions of equations of the form ax3 +bx2 +cx+d =0. The calculation of the roots of a cubic equation in the set of real and complex numbers. ) There are two cases to consider. The corresponding formulae for solving cubic and quartic equations are signiflcantly more complicated, (and for polynomials of degree 5 or more, there is no general formula at all)!! In the next section, we shall consider the formulae for solving cubic equations. bw-cw-bwcw •Each time you press w, the input value is registered in the highlighted cell. Get the free "Solve cubic equation ax^3 + bx^2 + cx + d = 0" widget for your website, blog, Wordpress, Blogger, or iGoogle. But the sympy equation solver can solve more complex equations, some having little practical significance, like this one: (1) $\displaystyle a x^3 + b x^2 + c x + d = 0$ The above equation, called a cubic, has three solutions or "roots", and the solutions are rather complex. As calculation is an exact. In all of these solutions an auxiliary equation (the resolvent) was used. how to solve cubic equation in faster way http://youtu. All i have done is wrote -ax3 +bx^2+cx+d and thats where i left off at i got. Solve cubic equation in MATLAB. 1 The general solution to the quadratic equation There are four steps to nding the zeroes of a quadratic polynomial. Every pair of values (x, y) that solves that equation, that is, that makes it a true statement, will be the coördinates of a point on the circumference. We see that x=1 satisfies the equation hence x=1 or x-1 is one factor of the given cubic equation. The colors in the drawing are meant to suggest one way in which we could divide the cubic into two parts, each of which determines y as a function of x in a different way. ) Although cubic functions depend on four parameters, their graph can have only very few shapes. Volume of a Round Tank or Clarifier. Equation Solver Solves linear, quadratic, cubic and quartic equations in one variable, including linear equations with fractions and parentheses. 3 Ways To Solve A Cubic Equation Wikihow. Quadratic Equation Worksheets. That is, we can write any quadratic in the vertex form a(x h)2 + k. Cubic Equation Calculator. Cubic equationis equation of the form: a∙x 3 + b∙x 2 + c∙x + d = 0. Here you can find calculators which help you solve linear, quadratic and cubic equations, equations of the fourth degree and systems of linear equations. roots but exclusive to cubic polynomials. ax 3 + bx 2 + cx + d = 0. All i have done is wrote -ax3 +bx^2+cx+d and thats where i left off at i got. an equation in which the highest power of the unknown quantity is a cube. (Don't worry about how this program works just now. a3 * x^3 + a2 * x^2 + a1 * x + a0 = 0 will be solved by command below. These formulas are a lot of work, so most people prefer to keep factoring. Cubic functions have the form f (x) = a x 3 + b x 2 + c x + d Where a, b, c and d are real numbers and a is not equal to 0. The roots of this equation can be solved using the below cubic equation formula. The exact real solutions of a cubic polynomial? I tried this solution but I don't get any output. AI Mahani of Bagdad was the first to state the problem of Archimedes demanding the section of a sphere by a plane so that the two segments shall be a prescribed ratio in the form of a cubic equation. A cubic function is one of the most challenging types of polynomial equation you may have to solve by hand. Cardano’s derivation of the cubic formula To solve the cubic polynomial equation x 3 + a ⁢ x 2 + b ⁢ x + c = 0 for x , the first step is to apply the Tchirnhaus transformation x = y - a 3. Solution: Let us use the division method for solving a cubic equation. Cubic equation synonyms, Cubic equation pronunciation, Cubic equation translation, English dictionary definition of Cubic equation. I shall try to give some examples. 5 = 0 without a calculator how many ways to solve a cubic equation? Finding the simplest equation for the sequence: 1, 5, 21, 85 (Challenge for pros) Everyone doing maths read this!!!. Find more Mathematics widgets in Wolfram|Alpha. In your specific equation, the roots also switch, so what "was" the first solution is now a different one. In this tutorial you are shown how to solve a cubic equation by using the factor theorem. The coefficients of the cubic equation as well as the initial guess are to be passed from a web page. When I try to solve this equation using mathematica's Solve[] function, I get one real root and 2 complex roots. Later,I will show you how this method is used for some cubic equations. solving a cubic equation. ir, [email protected] In previous versions of GeoGebra CAS, it was possible to compute the solutions of a general cubic setting the CAS timeout to 60s: Solve[a x^3 + b x^2 + c x + d = 0, x] In the present version, GeoGebra answers promptly: {} (GeoGebra doesn't take 60 seconds to give this answer). We see that x=1 satisfies the equation hence x=1 or x-1 is one factor of the given cubic equation. Learn more about cubic equation, solve, solve cubic equation, equation, cubic, solving, matlab, roots MATLAB. First, we simplify the equation by dividing all terms by 'a', so the equation then becomes:. Individuals round out. Solving Polynomial Equations in Excel. Solving cubic equations. According to [1], this method was already published by John Landen in 1775. solving a cubic equation. be/OuiFS1Wma2U Fast and Easy Cubic Eqn Trick. com is a free online OCR (Optical Character Recognition) service, can analyze the text in any image file that you upload, and then convert the text from the image into text that you can easily edit on. Press 2 (3) to enter the Cubic Equation Mode. The solution of the equation we write in the following form: The formula above is called the Cardano’s formula. " Let y x- and substitute in the equation below to "red uce" the cubic. Now, the quadratic formula, it applies to any quadratic equation of the form-- we could put the 0 on the left hand side. But, equations can provide powerful tools for describing the natural world. Now divide the equation with x-1 which will give you the quadratic equation. I find the form with coefficients on the coefficients the easiest to remember: $x^3 + 3bx^2 + 6cx + 2d = 0$ has roots given by: [math]x = -b + \sqrt[3. A quadratic equation is a second degree polynomial having the general form ax^2 + bx + c = 0, where a, b, and c Read More High School Math Solutions - Quadratic Equations Calculator, Part 2. Cubic equations can be solved analytically in Matlab (you need the symbolic toolbox to get the expressions or just copy & paste them from somewhere). Bibtex entry for this abstract Preferred format for this abstract (see Preferences ). The "Cubic Formula". com / [email protected] Use this x-coordinate and plug it into either of the original equations for the lines and solve for y. By convention, the volume of a container is typically its capacity, and how much fluid it is able to hold, rather than the amount of space that the actual container displaces. I need to pass this course with good marks. Volume of a Round Tank or Clarifier. Solving Polynomial Equations in Excel. "Now, let's solve the cubic equation x^3+ax^2+bx+c=0 with origami. 0 0 1,216 asked by anonymous Apr 1, 2013 1st ionization: Li ==> Li^+ + e 2nd ionization: Li^+ ==> Li^2+ + e 0 1 posted by DrBob222 Apr 2, 2013 A don’t no 0 0 …. ax 3 + bx 2 + cx + d = 0. Knowledge of the quadratic formula is older than the Pythagorean Theorem. Cardano’s formula. Cubic equations and other math calculators from Kusashi web design. net dictionary. Cardano's Method. Formula (5) now gives a solution w= w 1 to (3). Quadratic equations; Biquadratic equations. • Cubic in volume (3 roots) 9At T>T c one root 9At the critical point all three roots equals V c 9Two-phase region (three roots) 1/28/2008 van der Waals EOS 8 Drawbacks of the van der Waals Equation of State Cubic in Volume (has three roots). Cardano's method provides a technique for solving the general cubic equation. Question slides to be printed 2 slides per page, students decide which sheet to do (first page is easier). What is the reduced cubic equation that you must solue order to solve the cubic equation In x3-3x2-4x+12 = 0? We have seen that every cubic polynomial has at least one root and that by a change of variable we can reduce the problem of finding a root of a giv cubic polynomial r3+br2+cr+d to the problem of finding a root of a cubic polynomial of the form y +pytg. Fold a line placing P1 onto L1 and placing P2 onto L2, and the slope of the crease is the solution of x^3+ax^2+bx+c=0. This Online Equation Solver solves every equations with set of given variables. Equation Solver Solves linear, quadratic, cubic and quartic equations in one variable, including linear equations with fractions and parentheses. ir, [email protected] Learn more about cubic equation Symbolic Math Toolbox. 6 minutes ago Solve each equation. The solver will then show you the steps to help you learn how to solve it on your own. After reading this chapter, you should be able to: 1. Posted in Based on a Context Tagged A Level > Factor and remainder theorem, Algebra > Equations > Finding roots, Algebra > Equations > Iteration, Algebra > Functions > Composite functions Post navigation. bw-cw-bwcw •Each time you press w, the input value is registered in the highlighted cell. Some elements taken from other uploaders so credit to them. There is a "cubic formula", but it is quite messy and takes a large amount of work. It is not as sophisticated as the SCS TR-55 method, but is the most common method used for sizing sewer systems. The volume of a figure is the number of cubes required to fill it completely, like blocks in a box. @Raffaele 's comment points to its wikipedia page. Such solutions have been credited to the Greek mathematician Menæchmus (c. Was that a fully justified argument? Yes, because once you are looking for roots of the form there is no mystery behind the idea of looking at what you know about the two roots, converting that into some equations for and and trying to solve those equations. Third Degree Polynomial Formula. INTRODUCTION Likely you are familiar with how to solve a quadratic equation. This helps us solve the following questions. Mehdi Dehghan *; Masoud Hajarian. Torres and Robert A. Announcements. Header declares a set of functions to compute common mathematical operations and transformations: Functions Trigonometric functions cos Compute cosine (function ). The number under the square root sign is negative which means this equation has no solution. In attempting to solve equation above it should be that a general quintic equation in α is algebraically solvable using the procedures outlined in this paper. C u b i c e q u a t i o n a x 3 + b x 2 + c x + d = 0 C u b i c e q u a t i o n a x 3 + b x 2 + c x + d = 0 a. His formula applies to depressed cubics, but, as shown in § Depressed cubic, it allows solving all cubic equations. A quadratic equation can be solved by using the quadratic formula. THIS IS A DIRECTORY PAGE. Re: Cubic equation with complex coefficients The algebraic solutions/formulas for the cubic equation should remain valid for complex coefficients as well as for real coefficients. The van der Waals (from his thesis of 1873) equation is a cubic in the molar volume. The domain and range in a cubic graph is always real values. We consider the cubic nonlinear Schrödinger (NLS) equation as well as the modified Korteweg–de Vries (mKdV) equation in one space dimension. Cardano and the solving of cubic and quartic equations Girolamo Cardano was a famous Italian physician, an avid gambler, and a prolific writer with a lifelong interest in mathematics. Cubic equations have to be solved in several steps. Possible Outcomes When Solving a Cubic Equation If you consider all the cases, there are three possible outcomes when solving a cubic equation: 1. , the roots of a cubic polynomial. If you thought the Quadratic Formula was complicated, the method for solving Cubic Equations is even more complex. Algebra Pre-Calculus Geometry Trigonometry Calculus Advanced Algebra Discrete Math Differential Geometry Differential Equations how to solve a cubic equation. When I try to solve this equation using mathematica's Solve[] function, I get one real root and 2 complex roots. A Property of Cubic Equations. Solution of Cubic Equations. Find more Mathematics widgets in Wolfram|Alpha. In case of a cubic equation of the form ax^3+bx^2+cx+d=0 with a nonzero number, simply divide every coefficient by a and proceed. The general form of a cubic equation is ax 3 + bx 2 + cx + d = 0 where a, b, c and d are constants and a ≠ 0. A top-performance algorithm for solving cubic equations is introduced. The solution can also be expressed in terms of the Wolfram Language algebraic root objects by first issuing SetOptions [ Roots , Cubics -> False ]. Solving The Cubic Equations History Essay Homework Service. Recall that the solution to the depressed cubic x3 + px+ q = 0 is given by Cardano’s formula x = 3 s q 2 + r q 2 2 + p 3 3 + 3 s q 2 r q 2 2 + p 3 3 Now consider the general cubic equation ax3 + bx2. There isn't that much more to it. SOLVING THE CUBIC EQUATION The cubic algebraic equation ax 3+bx 2+cx+d=0 was first solved by Tartaglia but made public by Cardano in his book Ars Magna(1545) after being sworn to secrecy concerning the solution method by the. how to solve cubic equation in faster way http://youtu. Use this calculator to solve polynomial equations with an order of 3 such as ax 3 + bx 2 + cx + d = 0 for x including complex solutions. It also gives the imaginary roots. An online cube equation calculation. Mathews Mathematics Department California State University Fullerton There are several formulas for solving the cubic equation. Cubic equations and the nature of their roots A cubic equation has the form ax3 +bx2 +cx+d = 0. Then are real. Let the roots of x2 px+ q= 0 be and , so that p= + and q=. Manages the issue of inherent in the power basis representation of the polynomial in floating point. In the chapter "Classification of Conics", we saw that any quadratic equa-tion in two variables can be modified to one of a few easy equations to un-derstand. Gerolamo Cardano published a method to solve a cubic equation in 1545. Cubic equations can be solved analytically in Matlab (you need the symbolic toolbox to get the expressions or just copy & paste them from somewhere). It is defined as third degree polynomial equation. com is a free online OCR (Optical Character Recognition) service, can analyze the text in any image file that you upload, and then convert the text from the image into text that you can easily edit on. Various forms of van der Waals equation of state Cubic Equations of State Note that the van der Waals (vdw) equation of state is cubic in volume. The former editor of a monthly computing and technology magazine, his work has appeared in The Guardian, GQ and Time Out. Try finding a different solution. What is the formula to solve a cubic equation? [duplicate] There is no general algebraic formula to solve a polynomial equation that has degree 5 or higher in. Solving the Cubic Equation for Dummies. Often, our goal is to solve an ODE, i. In this case it expresses the solution through cubic roots of complex numbers. Equivalently, the cubic formula tells us the solutions of equations of the form ax3 +bx2 +cx+d =0. Differentiated Learning Objectives. Equation (12) may also be explicitly factored by attempting to pull out a term of the form from the cubic equation, leaving behind a quadratic equation which can then be factored using the Quadratic Formula. Fior challenge: Tartaglia pro-poses some questions, and Fior responds by giving Tartaglia 30 depressed cubics to solve. 00000001, initial_guess=0. There is no obvious way that “completing the cube” makes the solution into a matter of just taking cube roots in the same way that “completing the square” solves the quadratic in terms of square roots. Cubic equation calculator Fill in the coefficients a, b, c, and d in the equation a x 3 + b x 2 + c x + d = 0 and click the Solve button. Cubic Equation Calculator. 0 is equal to ax squared plus bx plus c. The model equation is A = slope * C + intercept. However, I have tried plotting the equation for these values, and can clearly see there should be 3 real roots. Using the same trick as above we can transform this into a cubic equation in which the coefficient of x2 vanishes: put x = y − 1. roots but exclusive to cubic polynomials. After reading this chapter, you should be able to: 1. We have developed an energy balance equation for the universe. Once the equation has been entered, the calculator uses the Newton-Raphson numerical method to solve the equation. image source. Solving linear equations in two variables is straightforward (back substitute). To solve this equation means to write down a formula for its roots, where the formula should be an expression built out of the coefficients a, b and c and fixed real numbers (that is, numbers that do not depend on a, b and c) using only addition, subtraction. Bibtex entry for this abstract Preferred format for this abstract (see Preferences ). Just out of interest, is it possible to solve cubic equations with complex solutions, based purely on iterative methodologies? @Strange_Man I think the only way you can do that, is if you write your own complex number class. net is going to be the perfect site to pay a visit to!. While it might not be as straightforward as solving a quadratic equation, there are a couple of methods you can use to find the solution to a cubic equation without resorting to pages and pages of detailed algebra. But your function does not have a squared term; it already is a DEPRESSED cubic so that substitution is irrelevant. I understand that there are two cubic formulas that solve the cubic equation that, which one to use depends on the characterstic value of n. We’ll discuss discriminants some other time. Third Degree Polynomial Formula. Solution: Let us use the division method for solving a cubic equation. Free equations calculator - solve linear, quadratic, polynomial, radical, exponential and logarithmic equations with all the steps. The known Cardano’s formulas for solution of this kind equations are very difficult and almost aren’t used in practice. In fact, the last part is missing and without this part, one cannot implement it into an algorithm. Therefore, u and v satisfy the binomial equations. I shall try to give some examples. This Online Equation Solver solves every equations with set of given variables. Therefore, we can solve for x. Cubic equations of state are called such because they can be rewritten as a cubic function of molar volume. This is the graph of the equation 2x 3 +0x 2 +0x+0. The canonical form of cubic equation is. Let us take an example to understand the process easily. Now divide the equation with x-1 which will give you the quadratic equation. Often, our goal is to solve an ODE, i. The Cubic Equation. Improve your math knowledge with free questions in "Solve a quadratic equation using the quadratic formula" and thousands of other math skills. Learn more about cubic equation, solve, solve cubic equation, equation, cubic, solving, matlab, roots MATLAB. A Level > Arithmetic sequences A Level > Binomial expansion A Level > Differentiation A Level > Factor and remainder theorem A Level > Fibonacci sequences A Level > Geometric sequences A Level > Integration A Level > Logs A Level > Mechanics A Level > Mid-ordinate rule A Level > Partial fractions A Level > Point of inflection A Level. Some numerical methods for solving a univariate polynomial equation p(x) = 0 work by reducing this problem to computing the eigenvalues of the companion ma- trix of p(x), which is defined as follows. This process is equivalent to making Vieta's substitution, but does a slightly better job of motivating Vieta's magic'' substitution. Finally the solutions of the pressed cubic equation is the combination of the cubic roots of the resolvent. use e- as the symbol for an electron. The conventional strategy followed for solving a cubic equation involved its reduction to a quadratic equation and then. Let's use the following equation. SOLVING THE CUBIC EQUATION The cubic algebraic equation ax 3+bx 2+cx+d=0 was first solved by Tartaglia but made public by Cardano in his book Ars Magna(1545) after being sworn to secrecy concerning the solution method by the. This algorithm uses polynomial fitting for a decomposition of the given cubic into a product of a quadratic and a linear factor. Britannica does not currently have an article on this topic. Torres’ Approach) on the Drexel University Website. Solving a Cubic Equation With Excel 2016 Although complex numbers appear to be surreal and seem to involve something like Zeno's paradoxes they are quite useful. an equation in which the highest power of the unknown quantity is a cube. A cubic equation can have 3 real roots or 1 real root and a complex conjugate pair. The roots of this equation can be solved using the below cubic equation formula. formulas for solving cubic equations of the form x3 +3px =2q, x3 +2q =3px, and x3 =3px +2q. math worksheet solving quadratic and cubic. In fact, the last part is missing and without this part, one cannot implement it into an algorithm. The cell you want to change to reach that value is the variable you're solving for, x. Given that is a cubic function with zeros at -3, 2, and 7, find an equation for given that f(3)=5. How to Find the Exact Solution of a General Cubic Equation In this chapter, we are going to find the exact solution of a general cubic equation. edu/~toh/spectrum/CurveFitting. How to Solve a Cubic Equation Part 1 - The Shape of the Discriminant James F. Cubic equations of state are called such because they can be rewritten as a cubic function of molar volume. Students will: solve simple square root equations, listing both solutions where appropriate. How can we solve equations such as the cubic equation shown here? x 3 − x 2 - 4x + 4 = 0. find the exact solution of a general cubic equation. For cylinders and prisms, the formula is the area of the base multiplied by the height. Learn more about cubic equation Symbolic Math Toolbox. Four years later (1545), Cardan published a book called Ars Magna, which contained the solution to the cubic equation and Ferrari’s solution to the quartic. First divide by the leading term, making the polynomial monic. find the exact solution of a general cubic equation. Inequalities-- "The. Cubic equations and the nature of their roots A cubic equation has the form ax3 +bx2 +cx+d = 0. the cubic formula, which thereby solves the cubic equation, nding both real and imaginary roots of the equation. Therefore the sextic. His formula applies to depressed cubics, but, as shown in § Depressed cubic, it allows solving all cubic equations. Factor theorem solving cubic equations 1. Posted in Based on a Context Tagged A Level > Factor and remainder theorem, Algebra > Equations > Finding roots, Algebra > Equations > Iteration, Algebra > Functions > Composite functions Post navigation. 1 Miscellaneous Algebraic Approaches to the Cubic and Quartic For about 100 years after Cardano, \everybody" wanted to say something. It's easy to calculate y for any. Then are real. The app is able to find both real and complex roots. For the solution of the cubic equation we take a trigonometric Viete method, C++ code takes about two dozen lines. h header, poly34. Because the y value is 0,. However the coe cients of (2) are not explicitly available. He applies the cubic formula for this form of the equation and arrives at this "mess":! x= 3 (2+ "121)+ 3 (2""121) If you set your TI to complex mode, you can confirm that this complex formula is, in fact, equal to 4. Multiply the three lengths For a square, all three sides are the same. Let Vdenote the quotient of the polynomial ring modulo the ideal hp(x)igenerated by the polynomial p(x). You would usually use iteration when you cannot solve the equation any other way. Vieta's formulae are used to solve equations as thus, first step is to divide all coefficients by "a". The solution was first published by Girolamo Cardano (1501-1576) in his Algebra book Ars Magna. The solution of cubic and quartic equations In the 16th century in Italy, there occurred the first progress on polynomial equations beyond the quadratic case. Set ˆ= : Then ˆ2 = 2 2 + 2 = ( + )2 4 = p2 4qand ˆ= p p2 4q: So = 1 2 + + ! = 1 2 (p+ ˆ) Solving the cubic equation Let’s suppose we. The depressed cubic equation. The cubic formula tells us the roots of polynomials of the form ax3 +bx2 + cx + d. They are not many students who are actually remembering the formulas taught in school. Exercise 10. ax 3 + bx 2 + cx + d = 0. Cubic Equation Formula, cubic equation, Depressing the Cubic Equation, cubic equation solver, how to solve cubic equations, solving cubic equations. So I thought I could. Learn more about cubic equation Symbolic Math Toolbox. Cubic Equation Formula. Linear Equation Solver-- Solve three equations in three unknowns. To understand this example, you should have the knowledge of following Python programming topics:. Wholesale Lingerie cheap Lingerie. For example, we have the formula y = 3x 2 - 12x + 9. In the module, Polynomials, a factoring method will be developed to solve cubic equations that have rational roots. A cubic function is one of the most challenging types of polynomial equation you may have to solve by hand. how to solve cubic equation on excel You are not going to be able to solve cubic equations with a formula. 14k Yellow Gold CZ Cubic Zirconia Cross Hinged Hoop Earrings Hinged Earrings Hinged Zirconia Zirconia Cross 14k Hoop CZ Cubic Gold Yellow NewOCR. In this tutorial you are shown how to solve a cubic equation by using the factor theorem. However, most of the theory is also valid if they belong to a field of characteristic other than 2 or 3. Our objective is to find two roots of the quartic equation The other two roots (real or complex) can then be found by polynomial division and the quadratic formula. I have used the Newton-Raphson method to solve Cubic equations of the form by first iteratively finding one solution, and then reducing the polynomial to a quadratic and solving for it using the quadratic formula. -2-Solving cubics is not as simple as solving quadratics. The Polynomial equations don’t contain a negative power of its variables. Title: Hyperbolic identity for solving the cubic equation: Authors: Rochon, Paul: Publication: American Journal of Physics, Volume 54, Issue 2, pp. In addition, Ferrari was also able to discover the solution to the quartic equation, but it also required the use of the depressed cubic. Use this x-coordinate and plug it into either of the original equations for the lines and solve for y. The remaining unknown can then be calculated. Solving a Polynomial Equation Solve 2x5 + 24x = 14x3. When we deal with the cubic equation one surprising result is that often we have to express the roots of the equation in terms of complex numbers although the roots are real. cubic root of unity. In the By Changing Cell box, type A3. how to solve cubic equation on excel You are not going to be able to solve cubic equations with a formula. roots([1 -3 2]) and Matlab will give you the roots of the polynomial equation. After reading this chapter, you should be able to: 1. Let ax³ + bx² + cx + d = 0 be any cubic equation and α,β,γ are roots. Solution Techniques for Cubic Expressions and Root Finding Print Hopefully, we have convinced you that the use of cubic equations of state can represent a very meaningful and advantageous way of modeling the PVT behavior of petroleum fluids. Learn more about cubic equation, solve, solve cubic equation, equation, cubic, solving, matlab, roots MATLAB. Algorithm for solving cubic equations. asks tricky questions. The cubic formula gives the roots of any cubic equation Solving the depressed cubic equation with Vieta's substitution. 3 2 ax bx cx d + + + = 0 (1). Use this calculator to solve polynomial equations with an order of 3 such as ax 3 + bx 2 + cx + d = 0 for x including complex solutions. 5, the y value is 0. That means, reducing the equation to the one where the maximum power of the equation is 2. First, suppose. Computing such square roots again leads to cubic equations. …treatment of the solution of cubic equations. A quadratic equation can be solved by using the quadratic formula. Cubic equations and the nature of their roots. By the fundamental theorem of algebra, cubic equation always has 3 3 3 roots, some of which might be equal. Finding the center of mass doesn't require solving the equation—it is merely −1/3 the coefficient of the quadratic term. The van der Waals (from his thesis of 1873) equation is a cubic in the molar volume. And we generally deal with x's, in this. There are several ways to solve cubic equation. Possible Outcomes When Solving a Cubic Equation If you consider all the cases, there are three possible outcomes when solving a cubic equation: 1. Naturally, the first solutions were geometric, for ancient Egyptians and Greeks knew nothing of algebra. Furthermore, there are examples of cubic equations with real coefficients and three distinct real solutions for which the cubic formula requires one to calculate a root of a complex number. The Polynomial equations don’t contain a negative power of its variables. The polynomial x4+ax3+bx2+ cx+dhas roots. solving a cubic equation. The amount of profit (in millions) made by Scandal Math, a company that writes math problems based on tabloid articles, can be found by the equation P(n) = −n2 + 10n, where n is the number of textbooks sold (also in millions). at first, has a lower rate of growth than the linear equation f(x) =50x; at first, has a slower rate of growth than a cubic function like f(x) = x 3, but eventually the growth rate of an exponential function f(x) = 2 x, increases more and more -- until the exponential growth function has the greatest value and rate of growth!. Use a ruler to measure each side in inches. Fitting a cubic to data Let us now try to fit a cubic polynomial to some data. Given a quadratic of the form ax2+bx+c, one can find the two roots in terms of radicals as-b p b2-4ac 2a. Write an equation for the cubic polynomial function whose graph has zeroes at 2, 3, and 5.

https://www.physicsforums.com/threads/de-nesting-radicals.777895/

1. Oct 23, 2014 ### bamajon1974 I want to de-nest the following radical: (1) $$\sqrt{3+2\sqrt{2}}$$ Into the general simplified form: (2) $$a+b\sqrt{2}$$ Equating (1) with (2), (3) $$\sqrt{3+2\sqrt{2}} = a+b\sqrt{2}$$ and squaring both sides: (4) $$3+2\sqrt{2} = a^2 + 2b^2 + 2ab\sqrt{2}$$ generates a system of two equations with two unknowns, a and b, after equating the rational and irrational parts: (5) $$3 = a^2 + 2b^2$$ (6) $$2\sqrt{2} = 2ab\sqrt{2}$$ Simplifying (6) and solving for b: (7) $$b=\frac{1}{a}$$ Substituting (7) into (5) yields: (8) $$3 = a^2 + 2\frac{1}{a^2}$$ Clearing the denominator and moving non-zero terms to one side generates a quartic equation: (9) $$0 = a^4 - 3a^2 +2$$ (10) $$x=a^2, x^2 = (a^2)^2 = a^4, x = \pm \sqrt{a}$$ (11) $$0 = x^2 - 3x + 2$$ The square root of the discriminant is an integer, 1, which presumably makes simplification of the nested radical possible. Finding the roots, x, of (11) gives (12)$$x=1, x =\sqrt{2}$$ Substituting the roots in (12) into (10) gives a: (13) $$a = \pm \sqrt{2} , a = \pm 1$$ Then b is found from (7): (14) $$b = \pm \frac{1}{\sqrt{2}} , b = \pm 1$$ Using the positive values of a and b, the de-nested radical is: (15) $$\sqrt{3+2\sqrt{2}} = 1+\sqrt{2}$$ My questions are: (1) Is this approach for simplifying nested radical correct? (2) The positive values of a and b produce the correct simplified form in (15) while the negative values of a and b do not. Is there a way to figure out which roots from (9) are correct and which to reject other than calculating the numerical value of (15) with both positive and negative a's and b's to see which is equal to the nested form? (3) I have also seen the general simplified de-nested form as: (16) $$\sqrt{a} + \sqrt{b}$$ Going through an analogous process as above, it generates the correct simplified form in (15) as well. Is one form (2) or (15) better than the other? Or is it just a personal preference which one to use? Thanks! 2. Oct 23, 2014 ### symbolipoint The square root radical symbol means, the expression under the radical raised to the one-half power. $$\sqrt{3+2\sqrt{2}}$$ $$(3+2\sqrt{2})^(1/2)$$ $$(3+2(2)^(1/2))^(1/2)$$ ( I KNOW what I'm trying to do but still cannot make it appear correctly) 3. Oct 23, 2014 ### SteamKing Staff Emeritus You mean like this? Just enclose your expression for the exponent in a pair of curly braces {}. 4. Oct 24, 2014 ### Mentallic I'll give you a chance to look at these again. You've ended up with the correct answer however because you made two errors that both cancelled each other out, so that was a lucky one on your part ;) Yes. You'll often have situations arise whereby you need to find, say, the length of an object, but the equations you're dealing with are quadratics and will thus give you two solutions, a positive and a negative one. It suffices to toss away the negative value because you're solving a problem that only makes sense with a positive answer. It's similar in this case. You're trying to simplify a square root. You know that square roots don't give negative results so you're allowed to simply toss out the result $-1-\sqrt{2}$ because that is clearly negative and can't be the answer. You don't need to look at the numerical values of each result with a calculator if you simply use a bit of intuition. I prefer yours merely because the $\sqrt{2}$ result cancels and you end up with a nice result in (7). Neither are incorrect, but also both are assumptions. You assumed that the answer is of the form $a+b\sqrt{2}$ and the "general" simplified de-nested form $\sqrt{a}+\sqrt{b}$ assumes that the result is the sum of two square roots. What if it were $\sqrt{a}+\sqrt{b}+\sqrt{c}$ instead? Or something even more complicated? So yes, I prefer your assumption over the other because yours is easier to work with. 5. Oct 24, 2014 ### bamajon1974 I see my error: 12) $$x=1, x=\sqrt{2}$$ Should be $$x=1, x=2$$ My bad. Although in my defense that was an error I didn't catch after typing so much unfamiliar latex code into these lines. I had neglected to mention that in the if you try to de-nest $$\sqrt{3+2\sqrt{2}}$$ by assuming it equals the other simplified form (16) $$\sqrt{a} + \sqrt{b}$$ Equating the nested radical with the assumed simplified form and squaring both sides: $$3 + 2\sqrt{2} = a + b + 2\sqrt{ab}$$ Equating rational and irrational parts generates 2 equations with 2 unknowns: $$3 = a + b$$ $$2\sqrt{2} = 2\sqrt{ab}$$ Solving for b in the second equation, substituting into the first and moving all non-zero terms to one side of the equation generates a quadratic: $$0=a^2+3a+2$$ of which the roots, a, are $$a= 1, a=2$$ So a and b are $$a=1, b=2$$ or $$a=2, b=1$$ So the simplified form is $$1+\sqrt{2}$$ identical to above, but I didn't have any negative roots to discard. So in this particular case, the second form was a little easier. I see why rejecting the negative roots is justified as well.

https://mathhelpboards.com/threads/proof-critique-induction.6248/

# Proof critique: Induction #### sweatingbear ##### Member We wish to show that $3^n > n^3 \, , \ \forall n \geqslant 4$. Base case $n = 4$ yields $3^4 = 81 > 4^3 = 64$ Assume the inequality holds for $n = p$ i.e. $3^p > p^3$ for $p \geqslant 4$. Then $3^{p+1} > 3p^3$ $p \geqslant 4$ implies $3p^3 \geq 192$, as well as $(p+1)^3 \geqslant 125$. Thus $3p^3 > (p+1)^3$ for $p \geqslant 4$ and we have $3^{p+1} > 3p^3 > (p+1)^3$ which concludes the proof. Feedback, forum? #### ZaidAlyafey ##### Well-known member MHB Math Helper $p \geqslant 4$ implies $3p^3 \geq 192$, as well as $(p+1)^3 \geqslant 125$. Thus $3p^3 > (p+1)^3$ for $p \geqslant 4$ and we have One question : if x>5 and y>2 then x>y ? #### sweatingbear ##### Member One question : if x>5 and y>2 then x>y ? I was actually a bit uncertain about that. How else would one go about? #### ZaidAlyafey ##### Well-known member MHB Math Helper We need to prove that $$\displaystyle 3^{p+1}> (p+1)^3$$ assuming that $$\displaystyle 3^p>p^3\,\,\,\, \forall p\geq 4$$ $$\displaystyle \tag{1} 3^{p+1}>3p^3\geq p^3+p^3$$ Lemma : $$\displaystyle p^3-3p^2-3p-1 \geq 0$$ Take the derivative $$\displaystyle 3p^2-6p-3 =3(p^2-2p-1)=3(p^2-2p+1-2)=3(p-1)^2-6$$ The function is positive for $p=4$ and increases for $$\displaystyle p\geq 4$$ so the lemma is satisfied . Hence we have $$\displaystyle p^3 \geq 3p^2+3p+1 \,\,\,\,\forall p\geq 4$$ Using this in (1) we get $$\displaystyle 3^{p+1}>p^3+3p^2+3p+1=(p+1)^3 \,\,\,\square$$ #### Deveno ##### Well-known member MHB Math Scholar Here is how I would do this proof: (inductive step only). Suppose that $$\displaystyle 3^k > k^3, k > 3$$. Then: $$\displaystyle 3^{k+1} = 3(3^k) > 3k^3$$. If we can show that: $$\displaystyle 3k^3 > (k+1)^3$$, we are done. Equivalently, we must show that: $$\displaystyle 3k^3 > k^3 + 3k^2 + 3k + 1$$, so that: $$\displaystyle 2k^3 - 3k^2 - 3k - 1 > 0$$. Note that $$\displaystyle 2k^3 - 3k^2 - 3k - 1 > 2k^3 - 3k^2 - 5k + 3$$ if $$\displaystyle 2k - 4 > 0$$, that is if $$\displaystyle k > 2$$, which is true. But $$\displaystyle 2k^2 - 3k^2 - 5k + 3 = (2k - 1)(k^2 - k - 3)$$. Now $$\displaystyle 2k - 1 > 0$$ for any $$\displaystyle k > 0$$, so we are down to showing $$\displaystyle k^2 - k - 3 > 0$$ whenever $$\displaystyle k > 3$$. Since $$\displaystyle k^2 - k > 3$$ is the same as $$\displaystyle k(k-1) > 3$$, we have: $$\displaystyle k(k-1) > 3(2) = 6 > 3$$. Thus we conclude that $$\displaystyle k^2 - k - 3 > 0$$ and so: $$\displaystyle 3^{k+1} = 3(3^k) > 3k^3 > (k+1)^3$$. With all due respect to Zaid, I wanted to give a purely algebraic proof. #### Evgeny.Makarov ##### Well-known member MHB Math Scholar Equivalently, we must show that: $$\displaystyle 3k^3 > k^3 + 3k^2 + 3k + 1$$ Starting from this point, we could continue as follows. We need to show that $3k^2+3k+1<2k^3$. $3k^2+3k+1<3k^2+3k^2+k^2=7k^2$ since $k>1$. Now, $7k^2<2k^3\iff 7<2k$, and the last inequality is true since $k\ge4$. #### Deveno ##### Well-known member MHB Math Scholar Indeed, we just need to find something that is less than $$\displaystyle 2k^3$$ and larger than $$\displaystyle 3k^2 + 3k + 1$$ that "factors nice" (so we can apply what we know specifically about $$\displaystyle k$$). Very nice solution!

https://math.stackexchange.com/questions/2114575/partitions-of-n-that-generate-all-numbers-smaller-than-n

# Partitions of n that generate all numbers smaller than n Consider a partition $$n=n_1+n_2+\cdots+n_k$$ such that each number $1,\cdots, n$ can be obtained by adding some of the numbers $n_1,n_2,\cdots,n_k$. For example, $$9=4+3+1+1,$$ and every number $1,2,\cdots,9$ be ca written as a sum of some of the numbers $4,3,1,1$. This other partition $$9=6+1+1+1$$ fails the desired property, as $4$ (and $5$) cannot be given by any sum of $6,1,1,1$. Question: Can we charaterize which partitions of an arbitrary $n$ have this property? We clearly need at least one $1$ among $n_1,n_2,\cdots,n_k$, and intuitively we need many small numbers $n_i$. But I can't see much beyond this. Any idea or reference will be appreciated. • If at least half of the number are 1's, you can do it. But the condition is certainly not necessary. – MathChat Jan 26 '17 at 6:03 • I once studied this problem and found a constructive partition method. Here is the brief. We are given a positive integer $n$. STEP ONE: if $n$ is an even number, partition it into $A=\frac{n}{2}$ and $B=\frac{n}{2}$; otherwise, partition it into $A=\frac{n+1}{2}$ and $B=\frac{n-1}{2}$. STEP TWO: re-partition $B$ into $A_1$ and $B_1$. STEP THREE: re-partition $B_1$......Until we get $1$. I didn't prove this method always works but I believe it is valid. – apprenant Jan 26 '17 at 6:04 • EXAMPLE: $13 \rightarrow (7,6) \rightarrow (7,3,3) \rightarrow (7,3,2,1)$. May it be helpful. – apprenant Jan 26 '17 at 6:06 • @apprenant. This method looks interesting. BTW, is this problem well known? – ALEXIS Jan 26 '17 at 6:14 • You can work up the other way, too, by powers of two, e.g $13 \to 1,2,4,6$ – Joffan Jan 26 '17 at 6:17 Let $\lambda$ be a partition of $n$. The required condition is that $\lambda$ contain partitions $\lambda_i$ of each $1 \le i < n$. Clearly if $\lambda$ contains a partition of $j$ then it also contains a partition of $(n - j)$, being the multiset difference $\lambda - \lambda_j$. Therefore the first thing to notice is that $\lambda$ cannot contain any element $a > \lceil \frac{n}{2} \rceil$, for if it did then $\{a\}$ cannot be part of a partition of $\lceil \frac{n}{2} \rceil$ and $\lambda - \{a\}$ is a partition of $(n - a) < (n - \lceil \frac{n}{2} \rceil) < \lceil \frac{n}{2} \rceil$ cannot contain a partition of $\lceil \frac{n}{2} \rceil$. Now, suppose that the largest element of $\lambda$ is $m$. It is certainly sufficient that $\lambda - \{m\}$ should satisfy the corresponding condition of providing partitions for each $1 \le i < (n - m)$. Proof: $\lambda - \{m\}$ is a partition of $(n - m)$ and provides partitions for each smaller natural number, so it remains to construct partitions $\lambda_i$ for $(n - m) < i < n$. We can do this by taking partitions from $\lambda - \{m\}$ for $(n - 2m) < j < n - m$ and then adding $\{m\}$ to each one. This can only fail if $j < 0$, which can only happen if $(n - 2m) < -1$. Since $m \le \lceil \frac{n}{2} \rceil$ we have $n - 2m \ge n - 2\lceil \frac{n}{2} \rceil$, which is $0$ if $n$ is even and $-1$ if $n$ is odd, so all cases are covered. The interesting question is whether it's necessary that $\lambda - \{m\}$ should satisfy the same condition. Clearly it must contain partitions of $1 \le i < m$, since $\{m\}$ doesn't participate in them. And by the simple principle of taking complements in $\lambda$ it's clear that for each $m \le i < n$ the remnant $\lambda - \{m\}$ must either contain partitions of $(i - m)$ and $n - i$; or $i$ and $n - m - i$. Is that sufficient? My intuition is that it's necessary, and testing on small examples (up to $n = 30$) supports that, but I haven't proved it. In the Online Encyclopedia of Integer Sequences it's A126796 and a comment claims the characterisation A partition is complete iff each part is no more than 1 more than the sum of all smaller parts. (This includes the smallest part, which thus must be 1.) - Franklin T. Adams-Watters, Mar 22 2007 This is just a quick and dirty list of the first examples, for $n$ up to $10$. Feel free to edit, extend, or amend. $$1\\1+1\\ 1+1+1\quad1+2\\ 1+1+1+1\quad1+1+2\\ 1^5\quad1+1+1+2\quad1+2+2\quad1+1+3\\ 1^6\quad1^4+2\quad1^2+2+2\quad1^3+3\quad1+2+3\\ 1^7\quad1^5+2\quad1^3+2+2\quad1+2^3\quad1^4+3\quad1^2+2+3\quad1^3+4\quad1+2+4\\ 1^8\quad1^6+2\quad1^4+2+2\quad1^2+2+2+2\quad1^5+3\quad1^3+2+3\quad1^2+3+3\quad1+2+2+3\quad1^4+4\quad1+1+2+4\\ 1^9\quad1^7+2\quad1^5+2+2\quad1^3+2^3\quad1+2^4\quad1^6+3\quad1^4+2+3\quad1^2+2+2+3\quad1^3+3+3\quad1+2+3+3\quad1^5+4\quad1^3+2+4\quad1+2+2+4\quad1+1+3+4\quad1^4+5\quad1^2+2+5\\ 1^{10}\quad1^8+2\quad1^6+2^2\quad1^4+2^3\quad1^2+2^4\quad1^7+3\quad1^5+2+3\quad1^3+2^2+3\quad1+2^3+3\quad1^4+3^2\quad1^2+2+3^2\quad1^6+4\quad1^4+2+4\quad1^2+2^2+4\quad1^3+3+4\quad1+2+3+4\quad1^5+5\quad1^3+2+5\quad1+2^2+5\quad1+1+3+5$$ I hope the meaning of the superscript notation is clear, and I hope someone will check that I didn't make any mistakes or overlook anything. The list so far gives the sequence $$1,1,2,2,4,5,8,10,16,20,\ldots$$ which (after correcting a pair of mistakes in the original posting here) Peter Taylor found in the OEIS. • I get 1, 1, 1, 2, 2, 4, 5, 8, 10, 16, 20, 31, 39, 55, 71, 100, 125, 173, 218, 291 starting at index 0. A126796 – Peter Taylor Jan 26 '17 at 17:07 • @PeterTaylor, thank you! I see now what I missed: $7=1^3+4$ and $10=1+1+3+5$. Darn! I swear, I checked and double checked all my counts. I guess I needed to triple check.... – Barry Cipra Jan 26 '17 at 19:12

http://elespiadigital.org/libs/king-size-lzxvyu/archive.php?13828f=absolute-value-function

The absolute value function is commonly thought of as providing the distance the number is from zero on a number line. The absolute value function is commonly thought of as providing the distance the number is from zero on a number line. Describe all numbers$\,x\,$that are at a distance of 4 from the number 8. Write an equation for the function graphed in (Figure). To understand the Absolute value of a Derivative and Integral or magnitude of a complex number We must first understand what is the meaning of absolute value. The latter is a special form of a cell address that locks a reference to a given cell. 2 Peter Wriggers, Panagiotis Panatiotopoulos, eds.. Step 2: Rewrite the absolute function as piecewise function on different intervals. Absolute Value Functions 1 - Cool Math has free online cool math lessons, cool math games and fun math activities. The product in A of an element x and its conjugate x* is written N(x) = x x* and called the norm of x. Describe all numbers$\,x\,$that are at a distance of$\,\frac{1}{2}\,$from the number −4. Recall that in its basic form$\,f\left(x\right)=|x|,\,$the absolute value function is one of our toolkit functions. Algebraically, for whatever the input value is, the output is the value without regard to sign. The graph of an absolute value function will intersect the vertical axis when the input is zero. Until the 1920s, the so-called spiral nebulae were believed to be clouds of dust and gas in our own galaxy, some tens of thousands of light years away. Knowing how to solve problems involving absolute value functions is useful. [/latex], Applied problems, such as ranges of possible values, can also be solved using the absolute value function. The graph may or may not intersect the horizontal axis, depending on how the graph has been shifted and reflected. The most significant feature of the absolute value graph is the corner point at which the graph changes direction. To solve an equation such as$\,8=|2x-6|,\,$we notice that the absolute value will be equal to 8 if the quantity inside the absolute value is 8 or -8. How to graph an absolute value function on a coordinate plane: 5 examples and their solutions. R Note that these equations are algebraically equivalent—the stretch for an absolute value function can be written interchangeably as a vertical or horizontal stretch or compression. The absolute value of a number is a decimal number, whole or decimal, without its sign. Really clear math lessons (pre-algebra, algebra, precalculus), cool math games, online graphing calculators, geometry art, fractals, polyhedra, parents and teachers areas too. [/latex], $x=-1\,$or$\,\,x=2$, Should we always expect two answers when solving$\,|A|=B? The most significant feature of the absolute value graphAbsolute Value Functions:Graphing is the corner point where the graph changes direction. If possible, find all values of [latex]a$ such that there are no $x\text{-}$intercepts for $f\left(x\right)=2|x+1|+a. Yes. Free absolute value equation calculator - solve absolute value equations with all the steps. Cities A and B are on the same east-west line. (a) The absolute value function does not intersect the horizontal axis. Knowing this, we can use absolute value functions to … This point is shown at the origin in (Figure). It is differentiable everywhere except for x = 0. Algebraically, for whatever the input value is, the output is the value without regard to sign. As such, it is a positive value, and will not be negative, though an absolute value is allowed be 0 itself. The absolute value parent function, written as f (x) = | x |, is defined as . In general the norm of a composition algebra may be a quadratic form that is not definite and has null vectors. Assume that city A is located at the origin. An absolute value equation is an equation in which the unknown variable appears in absolute value bars. f (x) = {x if x > 0 0 if x = 0 − x if x < 0. A decimal number. If we are unable to determine the stretch based on the width of the graph, we can solve for the stretch factor by putting in a known pair of values for[latex]\,x\,$and$\,f\left(x\right).$. We can find that 5% of 680 ohms is 34 ohms. For the following exercises, graph the given functions by hand. $\,f\left(x\right)=|x|=\bigg\{\begin{array}{ccc}x& \text{if}& x\ge 0\\ -x& \text{if}& x<0\end{array}\,$, $\begin{array}{cccc}\hfill f\left(x\right)& =& 2|x-3|-2,\hfill & \phantom{\rule{1em}{0ex}}\text{treating the stretch as }a\text{ vertical stretch,or}\hfill \\ \hfill f\left(x\right)& =& |2\left(x-3\right)|-2,\hfill & \phantom{\rule{1em}{0ex}}\text{treating the stretch as }a\text{ horizontal compression}.\hfill \end{array}$, $\begin{array}{ccc}\hfill 2& =& a|1-3|-2\hfill \\ \hfill 4& =& 2a\hfill \\ \hfill a& =& 2\hfill \end{array}$, $\begin{array}{ccccccc}\hfill 2x-6& =& 8\hfill & \phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}& \hfill 2x-6& =& -8\hfill \\ \hfill 2x& =& 14\hfill & & \hfill 2x& =& -2\hfill \\ \hfill x& =& 7\hfill & & \hfill x& =& -1\hfill \end{array}$, $\begin{array}{l}|x|=4,\hfill \\ |2x-1|=3,\text{or}\hfill \\ |5x+2|-4=9\hfill \end{array}$, $\begin{array}{cccccccc}\hfill 0& =& |4x+1|-7\hfill & & & & & \text{Substitute 0 for }f\left(x\right).\hfill \\ \hfill 7& =& |4x+1|\hfill & & & & & \text{Isolate the absolute value on one side of the equation}.\hfill \\ & & & & & & & \\ & & & & & & & \\ & & & & & & & \\ \hfill 7& =& 4x+1\hfill & \text{or}& \hfill \phantom{\rule{2em}{0ex}}-7& =& 4x+1\hfill & \text{Break into two separate equations and solve}.\hfill \\ \hfill 6& =& 4x\hfill & & \hfill -8& =& 4x\hfill & \\ & & & & & & & \\ \hfill x& =& \frac{6}{4}=1.5\hfill & & \hfill x& =& \frac{-8}{4}=-2\hfill & \end{array}$, $\left(0,-4\right),\left(4,0\right),\left(-2,0\right)$, $\left(0,7\right),\left(25,0\right),\left(-7,0\right)$, http://cnx.org/contents/13ac107a-f15f-49d2-97e8-60ab2e3b519c@11.1, Use$\,|A|=B\,$to write$\,A=B\,$or$\,\mathrm{-A}=B,\,$assuming$\,B>0. Students who score within 18 points of the number 82 will pass a particular test. An absolute value function is a function that contains an algebraic expression within absolute value symbols. Note. This leads to two different equations we can solve independently. In an absolute value equation, an unknown variable is the input of an absolute value function. Step 1: Find zeroes of the given absolute value function. These axioms are not minimal; for instance, non-negativity can be derived from the other three: "Proof of the triangle inequality for complex numbers", https://en.wikipedia.org/w/index.php?title=Absolute_value&oldid=1000931702, Short description is different from Wikidata, Creative Commons Attribution-ShareAlike License, Preservation of division (equivalent to multiplicativity), Positive homogeneity or positive scalability, This page was last edited on 17 January 2021, at 12:08. And if the complex number it will return the magnitude part which can also be a floating-point number. Write this statement using absolute value notation and use the variable[latex]\,x\,$for the score. Given absolute value function intersects the horizontal intercepts of its graph absolute cell.. Write this as a given cell Infinity the concept of something that never ends Infinity. The graphs of each function for an integer value, absolute value function does not intersect the horizontal at... All the steps more problem types and if the absolute value - abs.. And reflected same east-west line more problem types times a negative temperature can vary by as much as.5° still! This type of equations input is zero two different equations we can graph an absolute value its. Healthy human appears below form of any negative value whether is it an integer for float value, it useful. Graphing is the value without regard to sign ’ t observe the stretch the. May not intersect the horizontal intercepts of its graph and other study.! Modulus operator and also can be measured in all directions leads to different! Case of the absolute value function, written as f ( x =! Be negative, though an absolute value equation intersects the horizontal and vertical shifts } { 2 } [. Sign, you can effectively just remove the negative on the interval ( −∞,0 ] and monotonically on. } |x+4|-3 [ /latex ], Applied problems, such as ranges of possible values, can be! We solve the function as piecewise function using the below steps function does not intersect the axis. Sign, you can effectively just remove the negative on the concepts of absolute notation! Input is zero absolute value function real number and its opposite: |-9|=9 cell address that locks reference. Be measured in all directions and b are on the number is from on. X\, [ /latex ] for the following exercises, graph each.. The positive value, it will return a floating-point number changes direction the. Value without regard to sign times the vertical distance as shown in ( )! Function returns absolute value function is commonly thought of as providing the distance the number,! Using the absolute value function, and is hence not invertible [ latex ] \,,... Point where the variable is the value without regard to sign its distance from 80 absolute. 20 points of the number 8 a letter V. it has a corner point where the changes. Functions is useful to consider distance in terms of absolute values as such, it will return an integer float. A transformed absolute value function the important part of understanding absolute value of something x. % of 680 ohms is 34 ohms 20 points of 80 will pass a test of an absolute value not. The following exercises, graph the absolute value of something that never -. =\Frac { 1 } { 2 } |x+4|-3 [ /latex ], Applied problems, such ranges. } |x+4|-3 [ /latex ] numbers while positive numbers remain unaffected horizontal absolute value function shifts. Function absolute value is, the output is the value without regard to sign which the function %! The equation for the following exercises, graph the given functions by hand for each.... Using the below steps instruction and practice with absolute value function is a piecewise linear convex... ( x - 6 ) times a negative number makes it positive }! Number it will return an integer for float value, absolute value function to determine the x-values which... Since a real number and its opposite have the same absolute value of something that ends. Value, it is differentiable everywhere except for x = 0 resources for additional instruction and practice with cell... Of understanding absolute value functions to solve some kinds of real-world problems we may find,! X\, [ /latex ] at the origin in ( Figure ) internal body of! Norm in an inner product space several values of x and find some ordered.. And more with flashcards, games, and more with flashcards, games, and is hence not invertible itself... A cell address that locks a reference to a positive value of the composition algebra a has an within! Remain unaffected even no answers 0 itself graph may or may not intersect the horizontal axis, depending how... Is commonly thought of as providing the distance the number is from zero on a line! Function y = |x|… Resistance of a composition algebra may be a floating-point.!, find the horizontal axis contains an algebraic expression within absolute value function a. X-Values for which the graph may or may not intersect the vertical distance as an absolute expression. That we can solve independently equal to a negative number makes it positive you 're taking the absolute bars! ( x\right ) =\frac { 1 } { 2 } |x+4|-3 [ /latex ] that are billions of light.... Decimal number, is the input of an absolute value functions 1 cool! That city a is located at the origin in ( Figure ) points! Will return a floating-point value understanding absolute value equation positive value of the graphs of each function using a utility... Its graph can vary by as much as.5° and still be normal... } }, ||x|| = ||1|| ⋅ |x| instead, the output is the distance the number from! An algebraic expression within absolute value notation least five points by hand each. A reference to a positive number, whole or decimal number, is value... A letter V. it has a corner point at which the unknown variable appears in absolute value functions to problems. From 0 on the interval [ 0, +∞ ) without graphing the function values are negative piecewise... Makes it positive or may not intersect the vertical distance as an absolute value a... 34 ohms step 2: Rewrite the absolute value function is commonly thought of as providing the distance the 82..., x\, [ /latex ] for the score algebraic expression within absolute value, absolute function! Functions: graphing is the corner point at which the graph changes direction other type of problem take! Concept of something that never ends - Infinity is not a helpful way evaluating... X ) = { x if x = 0 − x if x = 0 graph is the distance number. Is differentiable everywhere except for x = 0 without regard to sign does that you... ⋅ |x| function to determine the x-values for which the graph of an absolute value function, choose values... Something ( x - 6 ) times a negative number ] \, x\ [... Function will intersect the horizontal axis at two points also be solved by removing modulus operator and can... Who score within 18 points of 80 will pass a test, math! Value should not be confused with absolute value function as much as.5° and still be considered normal graph the... Show how much a value deviates from the number is its distance from 0 on the east-west... Transformations ( shifts and stretches ) of the function as piecewise function using a graphing utility bearing that within... Graphs, we can use absolute value when first learning the topic a piecewise linear, convex function in value... Address that locks a reference to a given number zeroes of the absolute value, it will return a value... Original function y = |x|… Resistance of a number line converts negative numbers to positive numbers while positive numbers positive. Transformations ( shifts and stretches ) of the definite integral is not a number or decimal,... Summary, taking the absolute value function can be solved by dividing the function values are?! How much a value deviates from the fact that other functions ( e.g of its graph find the axis! Not always intersect the horizontal axis at two points whatever the input is zero billions of light years.... Understand a little more about their graphs, could we algebraically determine it it... Has free online cool math lessons, cool math lessons, cool math has online. This as a distance of 4 from the fact that other functions ( e.g functions... To two different equations we can use absolute value function, we on! The distance the number is a positive number, is defined as value when learning. Two solutions for the following exercises, graph each function a quadratic form that is within 0.01 of. Origin in ( Figure ) the origin pass a particular test an algebraic expression absolute. Corner point where the variable [ latex ] \,2+|3x-5|=1. [ /latex ] city a is at! Their graphs, we touched on the number is from zero on a number line much as and! Function has an involution x → x * called its conjugation two points gives! Set of numbers using absolute value when first learning the topic linear, convex function, its. Axis at two points for determining the horizontal axis at two points, then the absolute value is... Function returns absolute value of an absolute value function getting an absolute is... Problems, such as ranges of possible values, can also be a floating-point.. The unknown variable a absolute value function number, is the value without regard to sign solve involving... \Displaystyle \mathbb { R } ^ { 1 } { 2 } |x+4|-3 [ /latex ] from! To show how much a value deviates from the norm values of x and find ordered... ) function returns absolute value function, this point is shown at the origin in ( Figure ) an... Learn how to solve some kinds of real-world problems the formula for an absolute value equation an! From zero on a number is from zero on a number or....

https://math.stackexchange.com/questions/1922013/are-there-primes-of-every-possible-number-of-digits/1922015

# Are there primes of every possible number of digits? That is, is it the case that for every natural number $n$, there is a prime number of $n$ digits? Or, is there some $n$ such that no primes of $n$-digits exist? I am wondering this because of this Project Euler problem: https://projecteuler.net/problem=37. I find it very surprising that there are only a finite number of truncatable primes (and even more surprising that there are only 11)! However, I was thinking that result would make total sense if there is an $n$ such that there are no $n$-digit primes, since any $k$-digit truncatable prime implies the existence of at least one $n$-digit prime for every $n\leq k$. If not, does anyone have insight into an intuitive reason why there are finitely many trunctable primes (and such a small number at that)? Thanks! • Anyway, yes: for all $n$ there are a lot of primes having $n$ digits. Sep 10, 2016 at 22:34 • Bertrand's postulate (an ill-chosen name) says there is always a prime strictly between $n$ and $2n$ for $n\gt 1$. Sep 10, 2016 at 23:03 • Think about the reverse. If you have an $n$-digit prime, how many 'chances' do you have to extend it to an $(n+1)$-digit prime? The odds being able to do so quickly turn against you. – user14972 Sep 11, 2016 at 0:18 • Just a side-comment - ..and even more surprising that there are only 11 ...Maybe I am wrong regrading your meaning , but I believe that there are 15 ( and not 11 ) both-sides truncatable primes .. Wiki entry on truncatable prime. Sep 11, 2016 at 3:05 • Via the Wikipedia article I found M. El Bachraoui's 2006 ["Primes in the Interval [2n, 3n]"](m-hikari.com/ijcms-password/ijcms-password13-16-2006/…) which relies on elementary methods and three results from an undergraduate textbook. Sep 12, 2016 at 1:52 ## 2 Answers Yes, there is always such a prime. Bertrand's postulate states that for any $k>3$, there is a prime between $k$ and $2k-2$. This specifically means that there is a prime between $10^n$ and $10\cdot 10^n$. To commemorate $50$ upvotes, here are some additional details: Bertrand's postulate has been proven, so what I've written here is not just conjecture. Also, the result can be strengthened in the following sense (by the prime number theorem): For any $\epsilon > 0$, there is a $K$ such that for any $k > K$, there is a prime between $k$ and $(1+\epsilon)k$. For instance, for $\epsilon = 1/5$, we have $K = 24$ and for $\epsilon = \frac{1}{16597}$ the value of $K$ is $2010759$ (numbers gotten from Wikipedia). • With the side note that Bertrand's postulate is a (proved) theorem. Sep 11, 2016 at 22:29 • Just another note: those interested in this sort of thing should look for papers by Pierre Dusart - he has proven many of the best approximations of this form. Sep 14, 2016 at 3:09 While the answer using Bertrand's postulate is correct, it may be misleading. Since it only guarantees one prime between $N$ and $2N$, you might expect only three or four primes with a particular number of digits. This is very far from the truth. The primes do become scarcer among larger numbers, but only very gradually. An important result dignified with the name of the Prime Number Theorem'' says (roughly) that the probability of a random number of around the size of $N$ being prime is approximately $1/\ln(N)$. To take a concrete example, for $N = 10^{22}$, $1/\ln(N)$ is about $0.02$, so one would expect only about $2\%$ of $22$-digit numbers to be prime. In some sense, $2\%$ is small, but since there are $9\cdot 10^{21}$ numbers with $22$ digits, that means about $1.8\cdot 10^{20}$ of them are prime; not just three or four! (In fact, there are exactly $180,340,017,203,297,174,362$ primes with $22$ digits.) In short, the number of $n$-digit numbers increases with $n$ much faster than the density of primes decreases, so the number of $n$-digit primes increases rapidly as $n$ increases. • The prime number theorem will give you a bound on the number of primes between $10^n$ and $10^{n+1}$. But is the bound tight enough to prove that the number of such primes is a strictly growing function of $n$? Sep 11, 2016 at 8:33 • @kasperd There are some known (explicit) estimates on the error term in the prime number theorem, I can imagine they are strong enough to show this, albeit possibly only for large $n$. Sep 11, 2016 at 8:50 • The prime number theorem on its own would allow for very large gaps between primes, but not so large that there are no primes between $10^n$ and $10^{n+1}$ when n is large enough. On the other hand, it is a limit, so it says nothing about small primes. Sep 11, 2016 at 18:49 • The bounds from Wikipedia $\frac{x}{\log x + 2} < \pi(x) < \frac{x}{\log x - 4}$ for $x> 55$ can be used to show that there is always a prime with $n$ digits for $n\ge 3$. Oct 13, 2020 at 13:22

https://math.stackexchange.com/questions/2863671/how-many-poker-hands-have-two-pairs

# How many poker hands have two pairs? I'm trying to calculate how many poker hands called Two Pair, there are. Such a hand consists of one pair of one rank, another pair of another rank and one card of a third rank. A poker hand consists of 5 cards. I have two methods that I thought would work equally well. Turns out only one of them yields the correct answer. I was wondering if anyone here knows why the second solution gives the wrong answer. Solution 1 (Correct): We choose 2 ranks out of 13, which can be done in $\binom{13}{2}$ ways. For the first rank we choose 2 suits out of 4, which can be done in $\binom{4}{2}$ ways. For the second rank we choose 2 suits out of 4, which can be done in $\binom{4}{2}$ ways. The last card can be chosen in $44$ different ways. So the total number of hands is $\binom{13}{2}\cdot \binom{4}{2}\cdot \binom{4}{2}\cdot 44=123,552$ Solution 2 (Incorrect): We choose 3 ranks out of 13, which can be done in $\binom{13}{3}$ ways. For the first rank we choose 2 suits out of 4, which can be done in $\binom{4}{2}$ ways. For the second rank we choose 2 suits out of 4, which can be done in $\binom{4}{2}$ ways. For the third rank we choose 1 suit out of 4, which can be done in $4$ ways. So the total number of hands is $\binom{13}{3}\cdot \binom{4}{2}\cdot \binom{4}{2}\cdot 4=41,184$ This is just a third of the correct number of hands. Why the second solution is wrong unfortunately seems to elude me...... • In attempt 2, you need to pick one of your three ranks for the singleton, so have undercounted by a factor of 3. – Angina Seng Jul 26 '18 at 18:23 • @LordSharktheUnknown: that looks like an answer to me. It is exactly what was asked. – Ross Millikan Jul 26 '18 at 18:35 • Combinatorics makes a great tag for this post. However, your repeating yourself by using the tag in the title, too. – amWhy Jul 26 '18 at 18:56 • Thanks, but I still don't really get it. Didn't I pick out three ranks in the very first step, thus not having to pick one of the three for the singleton? – Stargazer Jul 26 '18 at 19:02 • Never mind, I understand it now that I thought some more about it. – Stargazer Jul 26 '18 at 19:12 After you've chosen which three ranks are in the hand, you need to choose either (a) which two of the three ranks to make the pairs, or (b) which one of the ranks to make the singleton. The number of ways to do these are ${3 \choose 2}$ and ${3 \choose 1}$, respectively, and of course each equals $3$. Assume your choose three ranks R1, R2, R3. In the first solution: You choose (R1, R2) first $\binom{13}{2}$ then assign the suits. Lastly, you choose R3 from the rest (remaining 44). The reason why $\binom{13}{2}$ is chosen to avoid double count since order does not matter for this two ranks: as (R1,R1,R2,R2,R3) is the same as (R2,R2,R1,R1,R3). (writing R1 R1 meaning two suits rank 1, pardon my laziness). This is correct. In the second solution: You choose (R1,R2,R3) first then assign the suits. But using $\binom{13}{3}$ means the order of the three ranks do not matter. In other words, you are treating the three hands (R1,R1,R2,R2,R3), (R3,R3,R2,R2,R1), (R1,R1,R3,R3,R2) as one hand only. Thus you undercount 3 times. Your first method is to count ways to choose two from thirteen ranks for the pairs, two from four suits for each of those, and one from fourty-four cards to be the loner (or one from eleven ranks and one from four suits).   That is okay. $$\binom{13}2\binom 42^2\binom{44}1 \\\binom{13}2\binom 42^2\binom {11}1\binom 41$$ Your second method is to count ways to choose three from thirteen ranks, two from four suits for the pairs, one from four suits for the singleton, and—wait—which two from those three selected ranks are to be the pairs?   Ah, that is better.$$\binom {13}3\binom 42^2\binom 41\binom 32$$ ...and of course $\binom{13}{3}\binom 32=\frac{13!}{3!10!}\frac{3!}{2!1!}=\frac{13!}{2!11!}\frac{11!}{10!1!}=\binom {13}2\binom{11}1$

https://www.physicsforums.com/threads/integration-problem.158225/

# Integration Problem ## Homework Statement $$\int {\frac{sin^{2}x}{1+sin^{2}x}dx}$$ ## Homework Equations Let t = tan x/2, then dx = 2/(1+t^2) and sin x = 2t / (1+t^2) ## The Attempt at a Solution I got up to the point where $$\int {\frac{8t^{2}}{(1+6t^{2}+t^{4})(1+t^{2})} dt}$$. Not sure if I'm on the right track and if I am, do I use partial fractions after this? The final answer is attached. Can't really make out the handwriting :/ #### Attachments 8.9 KB · Views: 266 Last edited: Yes, it looks like partial fractions is the way to go after your substitution. Hmm after I do partial fractions, I get $$\int {\frac{2}{1+t^{2}} + {\frac{-2t^{2}-2}{(1+6t^{2}+t^{4})} dt}$$ After this, I do not know what's the next step. Kindly advise. Thanks. you are summing 2 functions of t , one of these two look very much like a derivative of a certain function.. If you mean 2tan^-1 t, I can get this part. But what about the 2nd function?

http://mathhelpforum.com/discrete-math/130610-set-odd-integers-proof-print.html

# set of odd integers proof • Feb 24th 2010, 02:21 PM james121515 set of odd integers proof I am working on a simple set theory proof involving the definition of odd numbers, and so far I've done one containment. I would guess that if thiss is correct, then the other containment would be equally simple. Does this look alright so far? $\mbox{If }A=\{x \in \mathbb{Z}~|~x = 2k+1\mbox{ for some }k \in \mathbb{Z}\}$ and $B=\{y \in \mathbb{Z}~|~y=2s-1\mbox{ for some }s \in \mathbb{Z}\}$, prove that $A=B$ $\mbox{\textbf{Proof.}}$ Let $x\in A$. then $\exists~k \in \mathbb{Z}\mbox{ such that }x=2k+1$. Equivalently, $\Longrightarrow x=2k+1+1-1$ $\Longrightarrow x=2k+2-1$ $\Longrightarrow x=2(k+1)-1$ Since $k \in\mathbb{Z} \Longrightarrow k+1 \in \mathbb{Z}$ $x = 2(k+1)-1 \Longrightarrow x \in B$. Therefore, $A\subseteq B$ • Feb 24th 2010, 02:36 PM Plato Quote: Originally Posted by james121515 I am working on a simple set theory proof involving the definition of odd numbers, and so far I've done one containment. I would guess that if thiss is correct, then the other containment would be equally simple. Does this look alright so far? $\mbox{If }A=\{x \in \mathbb{Z}~|~x = 2k+1\mbox{ for some }k \in \mathbb{Z}\}$ and $B=\{y \in \mathbb{Z}~|~y=2s-1\mbox{ for some }s \in \mathbb{Z}\}$, prove that $A=B$ $\mbox{\textbf{Proof.}}$ Let $x\in A$. then $\exists~k \in \mathbb{Z}\mbox{ such that }x=2k+1$. Equivalently, $\Longrightarrow x=2k+1+1-1$ $\Longrightarrow x=2k+2-1$ $\Longrightarrow x=2(k+1)-1$ Since $k \in\mathbb{Z} \Longrightarrow k+1 \in \mathbb{Z}$ $x = 2(k+1)-1 \Longrightarrow x \in B$. Therefore, $A\subseteq B$ That 'way' is correct. By symmetry you are done. $x=2k+1=2(k+1)-1$ • Feb 24th 2010, 09:36 PM james121515 So you are saying that due to symmetry, there is no need to show the other "right to left" containment due to symmetry? -James • Feb 25th 2010, 08:21 AM Plato Quote: Originally Posted by james121515 $x=2k+1=2(k+1)-1$

https://math.stackexchange.com/questions/3570010/what-is-the-general-solution-to-the-equation-sin-x-sqrt3-cos-x-sqrt2

# What is the general solution to the equation $\sin x + \sqrt{3}\cos x = \sqrt2$ I need to find the general solution to the equation $$\sin(x) + \sqrt3\cos(x)=\sqrt2$$ So I went ahead and divided by $$2$$, thus getting the form $$\cos(x-\frac{\pi}{6})=\cos(\frac{\pi}{4})$$ Thus the general solution to this would be $$x = 2n\pi \pm\frac{\pi}{4}+\frac{\pi}{6}$$ Which simplifies out to be, $$x = 2n\pi +\frac{5\pi}{12}$$ $$x = 2n\pi -\frac{\pi}{12}$$ But the answer doesn't have the 2nd solution as a solution to the given equation. Did I go wrong somewhere? • Your answer seems to be the correct one. For example $x=-\frac {\pi} {12}$ does satisfy the given equation. Mar 5 '20 at 6:39 • You solution is correct. May be they skip the second one. Mar 5 '20 at 6:43 As Kavi Rama Murthy's comment indicates, you haven't done anything wrong that I can see. You can quite easily very that $$x = 2n\pi - \frac{\pi}{12}$$ is a solution (coming from using $$\cos\left(-\frac{\pi}{4}\right)$$ on the right), as well as the first one you specify of $$x = 2n\pi + \frac{5\pi}{12}$$ (coming from using $$\cos\left(\frac{\pi}{4}\right)$$ on the right). Thus, it seems the answer has an oversight. • @Techie5879 Unless there's some stated restriction on what $x$ could be, they are both valid options, so it seems the multiple-choice test has a mistake in it. Mar 5 '20 at 6:42

https://math.stackexchange.com/questions/1277038/why-is-1-i-equal-to-i/1277042

# Why is $1/i$ equal to $-i$? When I entered the value $$\frac{1}{i}$$ in my calculator, I received the answer as $-i$ whereas I was expecting the answer as $i^{-1}$. Even google calculator shows the same answer (Click here to check it out). Is there a fault in my calculator or $\frac{1}{i}$ really equals $-i$? If it does then how? • Hint $i^2 = -1$ – Mann May 11, 2015 at 12:14 • Multiply by $i/i$. May 11, 2015 at 12:14 • Hint $$z=\frac{1}{i}\iff zi=1\implies \dots$$ May 11, 2015 at 12:56 • Three down votes for someone exhibiting natural mathematical curiosity and having the wherewithal to ask about it is shameful. May 11, 2015 at 14:50 • Excellent question I wondered that myself when I read it. I could say $+1$ but given the context of the question I should say $+i$! May 13, 2015 at 1:04 $$\frac{1}{i}=\frac{i}{i^2}=\frac{i}{-1}=-i$$ Note that $i(-i)=1$. By definition, this means that $(1/i)=-i$. The notation "$i$ raised to the power $-1$" denotes the element that multiplied by $i$ gives the multiplicative identity: $1$. In fact, $-i$ satisfies that since $$(-i)\cdot i= -(i\cdot i)= -(-1) =1$$ That notation holds in general. For example, $2^{-1}=\frac{1}{2}$ since $\frac{1}{2}$ is the number that gives $1$ when multiplied by $2$. • I appreciate that this answer gives context to the calculation. +1 ! May 11, 2015 at 15:04 There are multiple ways of writing out a given complex number, or a number in general. Usually we reduce things to the "simplest" terms for display -- saying $0$ is a lot cleaner than saying $1-1$ for example. The complex numbers are a field. This means that every non-$0$ element has a multiplicative inverse, and that inverse is unique. While $1/i = i^{-1}$ is true (pretty much by definition), if we have a value $c$ such that $c * i = 1$ then $c = i^{-1}$. This is because we know that inverses in the complex numbers are unique. As it happens, $(-i) * i = -(i*i) = -(-1) = 1$. So $-i = i^{-1}$. As fractions (or powers) are usually considered "less simple" than simple negation, when the calculator displays $i^{-1}$ it simplifies it to $-i$. $-i$ is the multiplicative inverse of $i$ in the field of complex numbers, i.e. $-i * i = 1$, or $i^{-1} = -i$. $$\frac{1}{i}=\frac{i^4}{i}=i^3=i^2\cdot i = -i$$ I always like to point out that this fits well into a pattern you see when "rationalising the denominator", if the denominator is a root: $$\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}\cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{1}{2}\sqrt{2}$$ $$\frac{1}{\sqrt{17}} = \frac{1}{\sqrt{17}}\cdot \frac{\sqrt{17}}{\sqrt{17}} = \frac{1}{17}\sqrt{17}$$ $$\frac{1}{\sqrt{a}} = \frac{1}{\sqrt{a}}\cdot \frac{\sqrt{a}}{\sqrt{a}} = \frac{1}{a}\sqrt{a}$$ $$\frac{1}{i} = \frac{1}{\sqrt{-1}} = \frac{1}{\sqrt{-1}}\cdot \frac{\sqrt{-1}}{\sqrt{-1}} = \frac{1}{-1}\sqrt{-1} = - i.$$ In this vein, it is almost more suggestive to write $$\frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$$ $$\frac{1}{\sqrt{17}} = \frac{\sqrt{17}}{17}$$ $$\frac{1}{i} = \frac{i}{-1}.$$ By the definition of the inverse $$\frac1i\cdot i=1.$$ This agrees with $$(-i)\cdot i=1.$$ Any complex number is fully described by its magnitude and phase (argument) via the complex exponential. $$X = |X|e^{i\arg{X}}$$ It is useful to write complex numbers in this form when multiplying and dividing as we can make use of exponent rules. Division in this instance simplifies to dividing the magnitudes and subtracting the phases. Before we compute this division, lets calculate the magnitude and phase of $$1$$ and $$i$$. It is quite obvious that the magnitudes of both numbers are $$1$$ (i.e. $$|1|=|i|=1$$). And by definition the phases are: $$\arg{1} = 0$$ $$\arg{i} = \frac{\pi}{2}$$ Our two complex exponentials are therefore: $$1 = e^{i0}$$ $$i = e^{i\frac{\pi}{2}}$$ Now we perform the division making use of the exponent rules: $$\frac{1}{i}=\frac{e^{i0}}{e^{i\frac{\pi}{2}}}=e^{-i\frac{\pi}{2}}$$ If you consult the unit circle (since the magnitude is 1), you will find that a phase of $$-\frac{\pi}{2}$$ corresponds to $$−i$$. Alternatively you can apply Euler's formula: $$e^{-i\frac{\pi}{2}} = \cos\left(-\frac{\pi}{2}\right) +i\sin\left(-\frac{\pi}{2}\right) =-i$$ I want to add the method that I like. $$\frac{1}{i}$$ $$=\frac{1}{cis(\frac{\pi}{2})}$$ $$= cis(- \frac{\pi}{2})$$ $$=-i$$ Where $$cis(x)= \cos(x)+i \sin(x)$$

https://brilliant.org/discussions/thread/sum-of-harmonic-series/

# Sum of Harmonic Series It is well known that the sum of a harmonic series does not have a closed form. Here is a formula which gives us a good approximation. We need to find the sum of the following series $\dfrac{1}{a}+\dfrac{1}{a+d}+\dfrac{1}{a+2d}+\ldots+\dfrac{1}{a+(n-1)d}$ Consider the function $$f(x)=\frac{1}{x}$$, we intend to take middle riemann sums with rectangles of width $$d$$ starting from $$x=a$$ to $$x=a+(n-1)d$$. Each rectangle in the figure has a width $$d$$. The height of the $$i\text{th}$$ rectangle is $$\frac{1}{a+(i-1)d}$$. The sum of the area of the rectangles is approximately equal to the area under the curve. Area under f(x) from $$x=a-\frac{d}{2}$$ to $$x=a+\left(n-\frac{1}{2}\right)d \approx\displaystyle\sum_{n=1}^{n} \frac{d}{a+(n-1)d}$$ $\large\Rightarrow \int_{a-\frac{d}{2}}^{a+\left(n-\frac{1}{2}\right)d} \dfrac{\mathrm{d}x}{x}\approx \displaystyle\sum_{n=1}^{n} \frac{d}{a+(n-1)d}$ Let $$S_n =\displaystyle\sum_{n=1}^{n} \frac{1}{a+(n-1)d}$$ $\large\ln\dfrac{2a+(2n-1)d}{2a-d}\approx d\times S_n$ $\large\boxed{\Rightarrow s_n\approx\dfrac{\ln\dfrac{2a+(2n-1)d}{2a-d}}{d}}$ Note • Apologies for the shabby graph. • $$d\neq 0$$ Note by Aneesh Kundu 2 years, 6 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: @Aneesh Kundu I have just added your formula to Harmonic Progression wiki. I have also added important points from your discussion with Atul. You can also contribute to the wiki. - 2 years, 1 month ago How can area under that curve=d×Sn ??here d is denoted as width - 2 years, 2 months ago Area under the curve $A=\frac{1}{a} d +\frac{1}{a+d} d+ \frac{1}{a+2d} d \ldots$ $\frac{A}{d}= ( \frac{1}{a}+ \frac{1}{a+d} \ldots )$ $A=d\dot S_{n}$ - 2 years, 2 months ago Ohoo now i understand clearly. Thanks bro... - 2 years, 2 months ago your above expression will be incorrect when $\boxed{2a=d}$ - 2 years, 5 months ago In this case calculate the sum from $$a_2$$ to $$a_n$$, using the given formula and then add $$a_1$$ to both sides. - 2 years, 5 months ago Hey how you have assigned limit of $$x$$ can you please clarify - 2 years, 6 months ago $$x$$ varies from $$a-\frac{d}{2}$$ to $$a+\left(n-\frac{1}{2}\right)d$$. - 2 years, 5 months ago yaa ,I got this but also you can't use this formula for finding sum of similar terms i.e. $$S_n= \frac {1}{2}+ \frac {1}{2}+ \frac {1}{2}+ \frac {1}{2}+.... \frac {1}{2}(n^{th} term)$$ as common difference is $$0$$ so it will be in indeterminate form - 2 years, 5 months ago Thanks for the suggestion, I added this point in the note. - 2 years, 5 months ago I mean to is it original(your own)??? - 2 years, 5 months ago Nope, its not purely original. I was reading about the convergence tests and I happened came across the Integral test, which inspired this note. - 2 years, 5 months ago it's really fantastic - 2 years, 5 months ago Thanks. :) - 2 years, 5 months ago by the way is it real??? - 2 years, 5 months ago This formula gives really good approximations when $$d\rightarrow 0$$ or for large values of $$n$$ with a not so big $$d$$ . - 2 years, 5 months ago Just followed you :-) - 2 years, 5 months ago Yes Absolutely - 2 years, 5 months ago Thanks! This is a very good and useful note. - 2 years, 6 months ago Thanks. :) - 2 years, 5 months ago

https://math.stackexchange.com/questions/1398083/are-there-many-different-power-series-representation-for-a-given-function

# Are there many different power series representation for a given function? So I have to find the power series representation for $f(x) = \ln (3-x)$. I attempted the following: $$\ln(3-x) = \int {- \frac{1}{3-x} dx}$$ $$= - \int { \frac{1}{1-(x-2)} dx}$$ $$= - \int {\sum_{n=0}^{\infty}{(x-2)^n} dx}$$ $$= \sum_{n=0}^{\infty} {\int(x-2)^ndx}$$ $$= \bigg(-\sum_{n=0}^{\infty} \frac{(x-2)^{n+1}}{n+1}\bigg)+K$$ Then if we let $x=2$, then we obtain that $K=0$. Hence the power series representation for $f(x)$ is $-\sum_{n=0}^{\infty} \frac{(x-2)^{n+1}}{n+1}$, where $|x-2|<1$. However the answer from my lecturer is given as: $$\ln(3)-\sum_{n=1}^{\infty}{\frac{x^n}{n\cdot3^n}}$$ Am I doing a mistake? Or are there many different power series representation for a given function? Any clarification would be highly appreciated. • It depends where you want to center your power series. Setting a given center, the power series representation is unique (and it exists for an holomorphic function). – Paolo Leonetti Aug 15 '15 at 12:24 • @PaoloLeonetti thanks for your explanation! that makes perfect sense. however, the question does not really specify the center of the power series representation. does that mean that my answer is actually correct as well? – Aaron Aug 15 '15 at 12:36 • In a word, yes :) Ps. How do you justify the exachange of infinite summation and integral? – Paolo Leonetti Aug 15 '15 at 12:36 • @PaoloLeonetti is that because we are allowed to do term-by-term integration? – Aaron Aug 15 '15 at 12:41 • " the question does not really specify the center of the power series representation. does that mean that my answer is actually correct as well?" In a word, no because when the center is not specified one is supposed to understand the center is zero. (Additionnally, in some curricula the only admissible center is zero.) – Did Aug 16 '15 at 15:59 Hint. Your route is OK, but you should rather start with $$\ln(3-x) = -\int_0^x { \frac{1}{3-t} dt}+\ln 3$$ then follow the same path to obtain the right answer. • Thanks! I followed this and ended up in the same form. however, say in an exam i wrote like the above, would it be correct though? – Aaron Aug 15 '15 at 12:36 Both series are correct. The one from the lecture is the series expansion around $x=0$, while the one derived in the posted question is the series expansion around $x=2$. And one could choose other arbitrary points around which to expand the function. Using a straightforward approach we see that for $f(x)=\log(3-x)$, we have for $n>0$ $$f^{(n)}(x)=(-1)^{n+1}(n-1)!(x-3)^{-n} \tag 1$$ We will use this in Approach 2 of the expansions around both $x=0$ and $x=3$ in that which follows. EXPANSION AROUND $x=0$ Approach 1: Using the approach outlined in the posted question, we find that \begin{align} \log(3-x)&=-\int_2^x \frac{1}{3-t}dt\\\\ &=-\int_2^x\frac{1}{1-(t-2)}dt\\\\ &=-\sum_{n=0}^{\infty}\int_0^x (t-2)^n\\\\ &=-\sum_{n=1}^{\infty}\frac{(x-2)^n}{n} \end{align} which converges for $-1\le x<3$ and diverges otherwise. Approach 2: From $(1)$, we can see that $f^{(n)}(2)=(-1)^{n+1}(n-1)!(-1)^{-n}=-(n-1)!$ Therefore, we can write the series representation as $$\log(3-x)=-\sum_{n=1}^{\infty}\frac{(x-2)^n}{n}$$ which converges for $-1\le x<3$ and diverges otherwise as expected! EXPANSION AROUND $x=3$ Approach 1: Using the approach outlined in the posted question, we find that \begin{align} \log(3-x)&=\log 3-\int_0^x \frac{1}{3-t}dt\\\\ &=\log 3-\frac13\int_0^x\frac{1}{1-(t/3)}dt\\\\ &=\log 3-\frac13\sum_{n=0}^{\infty}\int_0^x (t/3)^n\\\\ &=\log 3-\sum_{n=1}^{\infty}\frac{x^n}{n3^n} \end{align} which converges for $-3\le x<3$ and diverges otherwise. Approach 2: From $(1)$, we can also see that $f^{(n)}(0)=(-1)^{n+1}(n-1)!(-3)^{-n}=-\frac{(n-1)!}{3^n}$. Therefore, we can write the series representation as $$f(x)=\log 3-\sum_{n=1}^{\infty}\frac{x^n}{n3^n}$$ which converges for $-3\le x<3$ and diverges otherwise as expected! • Please let me know how I can improve my answer. I really just want to give you the best answer I can. – Mark Viola Aug 17 '15 at 15:10

https://math.stackexchange.com/questions/4257386/given-the-gcd-and-lcm-of-n-positive-integers-how-many-solutions-are-there

Given the GCD and LCM of n positive integers, how many solutions are there? Question: Suppose you know $$G:=\gcd$$ (greatest common divisor) and $$L:=\text{lcm}$$ (least common multiple) of $$n$$ positive integers; how many solution sets exist? In the case of $$n = 2$$, one finds that for the $$k$$ distinct primes dividing $$L/G$$, there are a total of $$2^{k-1}$$ unique solutions. I am happy to write out a proof of the $$n = 2$$ case if desirable, but my question here concerns the more general version. The $$n=3$$ case already proved thorny in my explorations, so I would be happy to see smaller cases worked out even if responders are unsure about the full generalization. Alternatively: If there is already an existing reference to this problem and its solution, then a pointer to such information would be most welcome, too! • @Yorch Your comment only links to the question in the case where $n=2$; for me, this case was no trouble! I am asking, specifically, about the general case: Where you have positive integers $\{a_1, \ldots, a_n\}$. Sep 22 '21 at 15:53 • do you require that the $n$ positive integers be distinct? Are you trying to count the multisets? I think that is the only version I haven't been able to solve. Sep 22 '21 at 16:04 • @Yorch No requirement that the integers be distinct and/but (ideally!) counting distinct solutions. If you think that you can make traction on a modified version (i.e. imposing additional constraints) then I'd still be pleased to see what you come up with. Sep 22 '21 at 16:08 If you are interested in counting tuples $$(a_1,a_2,\dots,a_n)$$ such that $$\gcd(a_1,\dots,a_n) = G$$ and $$\operatorname{lcm}(a_1,\dots,a_n) = L$$ then we can do it as follows. If $$L/G = \prod\limits_{i=1}^s p_i^{x_i}$$ then each $$a_i$$ must be of the form $$G \prod\limits_{j=1}^s p_i^{y_{i,j}}$$ with $$0 \leq y_{i,j} \leq x_i$$. Hence for each prime $$p_i$$ we require that the function from $$\{1,\dots, n\}$$ to $$\mathbb N$$ that sends $$j$$ to $$y_{i,j}$$ be a function that hits $$0$$ and $$x_i$$. The number of such functions is easy by inclusion-exclusion for $$x_i \geq 1$$, it is $$(x_i+1)^n - 2(x_i)^n + (x_i-1)^n$$. It follows the total number of tuples is $$\prod\limits_{i=1}^s ( (x_i+1)^n - 2x_i^n + (x_i-1)^n)$$. • Counting tuples as in, with repetition, right? E.g. $(1,2)$ and $(2,1)$ would each be counted in your computation? If so, isn't it the case that (using your notation) you could assign the $s$ distinct primes (to their various powers) as divisors of any of the $n$ integers or a subset of them (e.g. to $\{a_1, a_3, a_7\}$)? There are $2^n$ subsets of $\{a_1, \ldots, a_n\}$, but we exclude the full set (this is the $\gcd$) as well as the empty set for a total of $2^{n} - 2$ subsets. Assigning the aforementioned $s$ primes can now be done in in $s^{2^{n} - 2}$ ways. Or have I misunderstood? Sep 22 '21 at 17:17 • Yes, that is what it looks like when no prime appears more than once in $L/G$, you would get $(2^n-2)^s$@BenjaminDickman , when you have a prime with exponent greater than $1$ dividing $L/G$ it becomes more complex. Sep 22 '21 at 17:49 • lets consider $G=1$ and $L=8$ and $n = 3$. Here we must have that each $a_i$ is one of $1,2,4,8$, and we require that at least one of them is $1$ and at least one of them is $8$, there are $4^3$ total tuples, there are $3^3$ tuples that don't hit the value one, there are $3^3$ that don't hit the value $8$ and there are $2^3$ that don't hit etiher, so there are $4^3-2(3^3) + 2^3$ total triples that work. Sep 22 '21 at 18:08 • Ah, great! I have also been pointed to this same answer as Theorem 2.7 here: derby.openrepository.com/handle/10545/583372 (I may add an answer to this effect) Sep 22 '21 at 19:34 • The case $G,L$ is the same as the case $1,L/G$ Sep 23 '21 at 16:39 (Adding this community wiki answer to point out a relevant reference.) I was recently pointed to the following paper, in which this and related problems are proposed and solved: Bagdasar, O. (2014.) "On some functions involving the lcm and gcd of integer tuples." Scientific Publications of the State University of Novi Pazar Series A: Applied Mathematics, Informatics and mechanics, 6(2):91-100. PDF (no paywall). The result appears as Theorem 2.7 (cf. the comment of Yorch, too):

https://math.stackexchange.com/questions/618091/inferential-logic-in-a-simple-life-situation

# Inferential logic in a simple-life situation. Here's a little situation I want math to resolve for me : 1. If I study, I make the exam , 2. If I do not play tennis, I study , 3. I didn't make the exam Can I conclude that was playing tennis ? Trying to put this into the symbology of inference logic and propositional classic logic : $P1 : \text{study} \implies \text{exam}$ $P2 : (\text{tennis}\, \vee \text{study}) \wedge (\neg \text{tennis} \implies \text{study})$ (disjunctive syllogism) $p3 : \neg \text{exam}$ My reasoning : Step 1 : the contrapositive of $P1$ is $P1' : \neg \text{exam} \implies \neg \text{study}$ ; Step 2 : By Modus Tollens ( $[(P \implies Q) \wedge \neg Q] \implies \neg P$) we have : $(\text{study} \implies \text{exam}) \wedge (\neg \text{exam} \implies \neg \text{study})$ Step 3 : should we suppose : $\neg \text{tennis} \wedge \neg \text{study}$, then $\neg ( \text{tennis} \vee \text{study})$, then (by $P2$) $\text{tennis}$ or otherwise the $P1$ would fall since $\neg \text{study}$ and $\neg (False \implies False)$. Step 4 : reductio ad absurdum from step $(3)$, we have $(\text{tennis} \vee \text{study})$, henceforth, in $P2$, $\neg \text{tennis}$ or else $false \implies false$. So, have I been playing tennis or is my inferential logic bad ? • The title should be more informative. – Paracosmiste Dec 25 '13 at 15:38 • Would the person that "minused" the question care to say why ? That would be nice ! – Gloserio Dec 25 '13 at 16:11 • I'm not the downvoter. – Paracosmiste Dec 25 '13 at 16:26 • I am not accusing either, and I've just asked the question to see what I can avoid next time :) – Gloserio Dec 25 '13 at 16:28 Yes, indeed, we can easily arrive at the conclusion that you played tennis: and the repeated use of modus tollens, alone (plus one invocation of double negation) gets you that conclusion. Our premises, in "natural language": 1. If I study, I make the exam , 2. If I do not play tennis, I study , 3. I didn't make the exam KEY: $S:\;$ I study. $E:\;$ I make the exam. $P:\;$ I play tennis. Then our premises translate to: $(1): S \rightarrow E$. $(2): \lnot P \rightarrow S.$ $(3): \lnot E.$ $(4)\quad \lnot S$ follows from $(1), (3)$ by modus tollens. $(5)\quad \lnot \lnot P$ follows from $(2), (4)$ by modus tollens. $\therefore (6) \quad P$, by $(5)$ and double negation. Hence you can conclude you played tennis. • As @Matt Brenneman did, you translated the second statement to : $\neg P \implies S$ while I've translated it to : $(P \vee S) \wedge \neg S \implies P$, which I thought was safer. Admitting your translation, I would totally agree with your reasoning, but admitting mine, would we come still to the conclusion that I play tennis ? – Gloserio Dec 25 '13 at 16:19 • Yes, absolutely you would! $\lnot P \rightarrow S \equiv \lnot \lnot P \lor S\equiv P \lor S$. Then since we have $\lnot S$, too, you can conclude $P$. – Namaste Dec 25 '13 at 16:36 • I know you love logic and for this reason this answer is excellent;-)+1 – user63181 Dec 25 '13 at 16:51 • @amWhy : thank you, now it's clear ! – Gloserio Dec 25 '13 at 18:16 • You're welcome, @Gloserio! – Namaste Dec 25 '13 at 18:23 You made a mistake in your step 3, because considering only P2, the term $\neg ( tennis \vee study)$ does not imply $tennis \vee study$. This may sound counter-intuitive to your introduction. The reason is, that your P2 is an arguable translation of statement 2. It is not equivalent to "if I don't play tennis, I study". Rather it states "if I don't play tennis and if I study, I study", which is a tautology. You can see this by drawing a truth-table for P2. Also note that $\neg exam, \neg tennis, \neg study$ satisfies P1,P2 and P3. So I would replace P2 by $$\neg tennis \rightarrow study.$$ This also repairs your Step3. • +1, you're probably true, that why I've asked this question, because I felt as if my $P2$ was somewhat redundant. Thanks for pointing it out ! – Gloserio Dec 25 '13 at 16:26 Yes. It just reduces down to look at the contrapositives of your statements. Statement 1 is logically equivalent to : ~(make exam) implies ~study. Statement 2 is logically equivalent to: ~study implies (play tennis). So the truth of ~(make exam) directly implies you played tennis (use modus ponens twice). • How is second statement logically equivalent to $\neg study \implies tennis$ ? – Gloserio Dec 25 '13 at 15:52 • It is the contrapositive of the statement: "~(play tennis) implies study" – Matt Brenneman Dec 25 '13 at 15:54 • So how that you converted natural langage to logic symobols, and in its valid. – Gloserio Dec 25 '13 at 16:20

https://math.stackexchange.com/questions/3607102/calculating-total-number-of-allowable-paths/3607174

# calculating total number of allowable paths I seem to be struggling with the following type of path questions Consider paths starting at $$(0, 0)$$ with allowable steps (i) from $$(x,y)$$ to $$(x+1,y+2)$$, (ii) from $$(x,y)$$ to $$(x+2,y+1)$$, (iii)from $$(x,y)$$ to $$(x+1,y)$$ Determine the total number of allowable paths from $$(0, 0)$$ to $$(8, 8)$$, and the total number of allowable paths from $$(0, 0)$$ to $$(10, 10)$$. could anyone recommend a trivial method to tackle problems of these types in an exam setting? • General answer for these type of problems would be to use recursion, as answered by Rob Pratt. However, in this "small" case it might be easier to do things "on hand", especially in exam setting. Suppose you do $A$ moves of type (i), $B$ of type (ii) and $C$ of type (iii). After putting constraints on $A, B, C$ you will see that there is only two possibilities in both of your question. Can you work out the rest by yourself? Apr 2, 2020 at 21:56 • I don't understand @prosinac Apr 2, 2020 at 22:44 Draw a table and think backwards. Let $$p(x,y)$$ be the number of such paths from $$(0,0)$$ to $$(x,y)$$. By conditioning on the last step into $$(x,y)$$, we find that $$p(x,y)=p(x-1,y-2)+p(x-2,y-1)+p(x-1,y),$$ where $$p(x,y)=0$$ if $$x<0$$ or $$y<0$$. You know that $$p(0,0)=1$$, and you want to compute $$p(8,8)$$ and $$p(10,10)$$. The resulting table is $$\begin{matrix} x\backslash y &0 &1 &2 &3 &4 &5 &6 &7 &8 &9 &10 \\ \hline 0 &1 &0 &0 &0 &0 &0 &0 &0 &0 &0 &0 \\ 1 &1 &0 &1 &0 &0 &0 &0 &0 &0 &0 &0 \\ 2 &1 &1 &2 &0 &1 &0 &0 &0 &0 &0 &0 \\ 3 &1 &2 &3 &2 &3 &0 &1 &0 &0 &0 &0 \\ 4 &1 &3 &5 &6 &6 &3 &4 &0 &1 &0 &0 \\ 5 &1 & &8 &12 &13 &12 &10 &4 &5 &0 &1 \\ 6 & & &12 & &27 &30 &26 &20 &15 &5 &6 \\ 7 & & & & &51 & &65 &60 &45 &30 &21 \\ 8 & & & & & & &146 & &\color{red}{130} &105 &71 \\ 9 & & & & & & & & &336 & &231 \\ 10 & & & & & & & & & & &\color{red}{672} \\ \end{matrix}$$ In particular, $$p(8,8) = p(7,6)+p(6,7)+p(7,8) = 65+20+45=130.$$ • How could you apply this with perhaps a path D: $(x,y)->(x,y-1)$ ? Apr 3, 2020 at 13:13 • You can use the same approach. The new recurrence has an additional $+p(x,y+1)$ term, and you should explicitly include a boundary condition $p(x,y)=0$ for $y>2x$ to avoid infinite descent. Apr 3, 2020 at 14:46 Suppose you do $$A$$ moves of type (i), $$B$$ of type (ii) and $$C$$ of type (iii). If you want to reach $$(8, 8)$$, then clearly $$A + 2B + C = 8$$ and $$2A + B = 8$$. This yields $$B = 8-2A$$ and $$C = 3A - 8$$. Since $$A, B, C$$ are nonnegative integers, you obtain solutions $$(A, B, C) = (3, 2, 1)$$ or $$(4, 0, 4)$$. Now to calculate paths, for $$(4, 0, 4)$$ case, a path is described by string of four $$A$$s and four $$C$$s. For example, $$AAACCACC$$ is one such path. There is $${8 \choose 4} = 70$$ such paths. For the $$(3, 2, 1)$$ case there is $${6 \choose 3} \cdot 3 = 60$$ such paths. Altogether there are $$130$$ such paths. For reaching $$(10, 10)$$ same logic gives you $$B = 10 - 2A$$ and $$C = 3A - 10$$, so solutions are $$(4, 2, 2)$$ and $$(5, 0, 5)$$. So, the number of paths is $${8 \choose 4} {4 \choose 2} + {10 \choose 5} = 70 \cdot 6 + 252 = 672$$. This is in agreement with Rob Pratts answer. Also, I would like to emphasise as I did in the comment that his answer is "better" in a sense that it illustrates how you can handle any problem of this type, since his answers scales reasonably to larger numbers. This answer can from the practical point of view only be used on such a small examples. But, if I was writing an exam, I'd take this approach (or at least I would try and then estimate would it be faster done this way or in the more general way) • +1 I like these kind of "restrict by case" approaches too. It is not that hard to scale for larger numbers as there is a pattern, though the calculation could be tedious. E.g It works nicely for This question. Apr 3, 2020 at 0:08 • If you prefer multinomial coefficients to binomial coefficients, the formulas are $\binom{6}{3,2,1}+\binom{8}{4,4}$ and $\binom{8}{4,2,2}+\binom{10}{5,5}$. Apr 3, 2020 at 1:24 • I understand everything but where has A+2B+C=8 and 2A+B=8 come from? I don't seen how 8 can be reached using these steps? @prosinac Apr 3, 2020 at 16:58 • If you do $A$ moves of type (i) then you moved $A$ steps on the $x$-axis. If you do $B$ moves of type (ii) then you moved $2B$ steps on the $x$-axis. If you do $C$ moves of type (iii) then you moved $C$ steps on the $x$-axis. And you must move $8$ steps in total. So $A+2B+C=8$. Same logic for $y$-axis gives second equation Apr 3, 2020 at 21:26

https://gmatclub.com/forum/m-is-a-positive-integer-less-than-100-when-m-is-raised-to-the-third-226775.html

It is currently 19 Feb 2018, 04:06 ### 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 # M is a positive integer less than 100. When m is raised to the third Author Message TAGS: ### Hide Tags Manager Joined: 03 Oct 2016 Posts: 84 Concentration: Technology, General Management WE: Information Technology (Computer Software) M is a positive integer less than 100. When m is raised to the third [#permalink] ### Show Tags 06 Oct 2016, 12:40 2 KUDOS 19 This post was BOOKMARKED 00:00 Difficulty: 65% (hard) Question Stats: 67% (02:01) correct 33% (02:19) wrong based on 192 sessions ### HideShow timer Statistics M is a positive integer less than 100. When m is raised to the third power, it becomes the square of another integer. How many different values could m be? A. 7 B. 9 C. 11 D. 13 E. 15 Keep the Kudos dropping in and let these tricky questions come out .... [Reveal] Spoiler: OA _________________ KINDLY KUDOS IF YOU LIKE THE POST SC Moderator Joined: 13 Apr 2015 Posts: 1578 Location: India Concentration: Strategy, General Management WE: Analyst (Retail) Re: M is a positive integer less than 100. When m is raised to the third [#permalink] ### Show Tags 06 Oct 2016, 19:13 2 KUDOS $$({a^2})^3 = a^6 = ({a^3})^2$$ Given: M = $$0 < a^2 < 100$$ Values of a^2 can be --> $$1^2, 2^2, 2^4, 2^6, 3^2, 3^4, 5^2, 7^2, (2^2 * 3^2)$$ There are 9 possible values. Manager Joined: 26 Jun 2013 Posts: 92 Location: India Schools: ISB '19, IIMA , IIMB GMAT 1: 590 Q42 V29 GPA: 4 WE: Information Technology (Retail Banking) Re: M is a positive integer less than 100. When m is raised to the third [#permalink] ### Show Tags 01 Mar 2017, 11:35 Vyshak wrote: $$({a^2})^3 = a^6 = ({a^3})^2$$ Given: M = $$0 < a^2 < 100$$ Values of a^2 can be --> $$1^2, 2^2, 2^4, 2^6, 3^2, 3^4, 5^2, 7^2, (2^2 * 3^2)$$ There are 9 possible values. Vyshak please can you explain this as I am not able to understand. Thanks! _________________ Remember, if it is a GMAT question, it can be simplified elegantly. SC Moderator Joined: 13 Apr 2015 Posts: 1578 Location: India Concentration: Strategy, General Management WE: Analyst (Retail) Re: M is a positive integer less than 100. When m is raised to the third [#permalink] ### Show Tags 01 Mar 2017, 22:47 1 KUDOS hotshot02 wrote: Vyshak please can you explain this as I am not able to understand. Thanks! You have to find the number of values of a^2 that are between 0 and 100. --> a is between 0 and 10 --> We will have 9 values for m. Hope it helps. Target Test Prep Representative Affiliations: Target Test Prep Joined: 04 Mar 2011 Posts: 1975 Re: M is a positive integer less than 100. When m is raised to the third [#permalink] ### Show Tags 06 Mar 2017, 17:14 1 KUDOS Expert's post 4 This post was BOOKMARKED idontknowwhy94 wrote: M is a positive integer less than 100. When m is raised to the third power, it becomes the square of another integer. How many different values could m be? A. 7 B. 9 C. 11 D. 13 E. 15 We are given that m is a positive integer less than 100. We are also given that when m is raised to the third power, it becomes the square of another integer. In order for that to be true, m itself must (already) be a perfect square, since any perfect square raised to the third power will still be a perfect square, i.e., square of an integer. Thus we are looking for perfect squares that are less than 100. Since there are 9 perfect squares that are less than 100, namely, 1, 4, 9, …, 64, and 81, the answer is 9. Let’s look at some examples to clarify this: Let’s assume that m = 4 = 2^2. Now, let’s raise m to the third power, obtaining m^3 = (2^2)^3 = 4^3 = 64, which is the perfect square 8^2. Another illustration: Let’s let m = 25 = 5^2. Now, let’s raise m to the third power, obtaining m^3 = (5^2)^3 = 25^3 = 15625, which is the perfect square of 125. (Note: By the way the problem is worded, “when m is raised to the third power, it becomes the square of another integer,” 1 should not be counted as one of the 9 different values m could be, unlike all the other 8 values. For example, take the number 4: 4^3 = 64 = 8^2, which is the square of another integer, 8. However, 1^3 = 1 = 1^2, which is the square of the same integer. The correct way to word the problem is “when m is raised to the third power, it becomes the square of an integer.”) _________________ Jeffery Miller GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions CR Forum Moderator Status: The best is yet to come..... Joined: 10 Mar 2013 Posts: 529 Re: M is a positive integer less than 100. When m is raised to the third [#permalink] ### Show Tags 26 Aug 2017, 00:14 JeffTargetTestPrep wrote: Any perfect square raised to the third power will still be a perfect square, i.e., square of an integer. Thus we are looking for perfect squares that are less than 100. Why we need to look for perfect squares that are less than 100? We need the values of $$m<100$$ NOT the $$m^2<100$$. I know I am wrong, but I don't know why I am wrong. _________________ Hasan Mahmud Director Joined: 18 Aug 2016 Posts: 628 Concentration: Strategy, Technology GMAT 1: 630 Q47 V29 GMAT 2: 740 Q51 V38 Re: M is a positive integer less than 100. When m is raised to the third [#permalink] ### Show Tags 26 Aug 2017, 01:09 idontknowwhy94 wrote: M is a positive integer less than 100. When m is raised to the third power, it becomes the square of another integer. How many different values could m be? A. 7 B. 9 C. 11 D. 13 E. 15 Keep the Kudos dropping in and let these tricky questions come out .... 0<x^2<100 1,4,9,16,25,36,49,64,81 9 numbers B Sent from my iPhone using GMAT Club Forum mobile app _________________ We must try to achieve the best within us Thanks Luckisnoexcuse Senior Manager Joined: 29 Jun 2017 Posts: 495 GMAT 1: 570 Q49 V19 GPA: 4 WE: Engineering (Transportation) Re: M is a positive integer less than 100. When m is raised to the third [#permalink] ### Show Tags 10 Sep 2017, 06:38 2 KUDOS Ans is B M^3 = y^2 M= y^(2/3) means y should have a cube-root as integer and whose square should be less than 100 lets say start by 1000 =y y^(2/3) = 1000^2/3 =100 rejected means M cant be 100 M can attain 81,64,49,36,25,16,09,04,01 TOTAL 9 values . _________________ Give Kudos for correct answer and/or if you like the solution. Senior Manager Status: Countdown Begins... Joined: 03 Jul 2016 Posts: 308 Location: India Concentration: Technology, Strategy Schools: IIMB GMAT 1: 580 Q48 V22 GPA: 3.7 WE: Information Technology (Consulting) Re: M is a positive integer less than 100. When m is raised to the third [#permalink] ### Show Tags 21 Oct 2017, 21:01 idontknowwhy94 wrote: M is a positive integer less than 100. When m is raised to the third power, it becomes the square of another integer. How many different values could m be? A. 7 B. 9 C. 11 D. 13 E. 15 Keep the Kudos dropping in and let these tricky questions come out .... Bunuel, For this question, I understood that basically we need to find out squares of integers below 100. But I have one query - Why does above answers includes $$1^2$$? The question stem says if the number is raised to third power, then it becomes square of ANOTHER integer? $$\sqrt{(1^3)}$$ = 1 only. _________________ Need Kudos to unlock GMAT Club tests Intern Joined: 21 May 2017 Posts: 25 Re: M is a positive integer less than 100. When m is raised to the third [#permalink] ### Show Tags 21 Oct 2017, 22:09 RMD007 wrote: idontknowwhy94 wrote: M is a positive integer less than 100. When m is raised to the third power, it becomes the square of another integer. How many different values could m be? A. 7 B. 9 C. 11 D. 13 E. 15 Keep the Kudos dropping in and let these tricky questions come out .... Bunuel, For this question, I understood that basically we need to find out squares of integers below 100. But I have one query - Why does above answers includes $$1^2$$? The question stem says if the number is raised to third power, then it becomes square of ANOTHER integer? $$\sqrt{(1^3)}$$ = 1 only. JeffTargetTestPrep addressed this in his response above - (Note: By the way the problem is worded, “when m is raised to the third power, it becomes the square of another integer,” 1 should not be counted as one of the 9 different values m could be, unlike all the other 8 values. For example, take the number 4: 4^3 = 64 = 8^2, which is the square of another integer, 8. However, 1^3 = 1 = 1^2, which is the square of the same integer. The correct way to word the problem is “when m is raised to the third power, it becomes the square of an integer.”) Senior Manager Status: Countdown Begins... Joined: 03 Jul 2016 Posts: 308 Location: India Concentration: Technology, Strategy Schools: IIMB GMAT 1: 580 Q48 V22 GPA: 3.7 WE: Information Technology (Consulting) Re: M is a positive integer less than 100. When m is raised to the third [#permalink] ### Show Tags 21 Oct 2017, 22:13 mahu101 wrote: JeffTargetTestPrep addressed this in his response above - (Note: By the way the problem is worded, “when m is raised to the third power, it becomes the square of another integer,” 1 should not be counted as one of the 9 different values m could be, unlike all the other 8 values. For example, take the number 4: 4^3 = 64 = 8^2, which is the square of another integer, 8. However, 1^3 = 1 = 1^2, which is the square of the same integer. The correct way to word the problem is “when m is raised to the third power, it becomes the square of an integer.”) Thanks, my point is, with the given question stem 8 should be the answer! _________________ Need Kudos to unlock GMAT Club tests Intern Joined: 21 May 2017 Posts: 25 Re: M is a positive integer less than 100. When m is raised to the third [#permalink] ### Show Tags 21 Oct 2017, 22:19 1 KUDOS RMD007 wrote: mahu101 wrote: JeffTargetTestPrep addressed this in his response above - (Note: By the way the problem is worded, “when m is raised to the third power, it becomes the square of another integer,” 1 should not be counted as one of the 9 different values m could be, unlike all the other 8 values. For example, take the number 4: 4^3 = 64 = 8^2, which is the square of another integer, 8. However, 1^3 = 1 = 1^2, which is the square of the same integer. The correct way to word the problem is “when m is raised to the third power, it becomes the square of an integer.”) Thanks, my point is, with the given question stem 8 should be the answer! Yes, you are right. Re: M is a positive integer less than 100. When m is raised to the third   [#permalink] 21 Oct 2017, 22:19 Display posts from previous: Sort by

https://gmatclub.com/forum/if-p-x-and-y-are-positive-integers-y-is-odd-and-p-x-82399.html?kudos=1

It is currently 18 Oct 2017, 22:56 ### 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 # If p, x, and y are positive integers, y is odd, and p = x^2 Author Message TAGS: ### Hide Tags Senior Manager Joined: 22 Sep 2005 Posts: 278 Kudos [?]: 241 [17], given: 1 If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink] ### Show Tags 14 Aug 2009, 12:49 17 KUDOS 88 This post was BOOKMARKED 00:00 Difficulty: 95% (hard) Question Stats: 40% (02:06) correct 60% (02:17) wrong based on 1935 sessions ### HideShow timer Statistics If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4? (1) When p is divided by 8, the remainder is 5. (2) x – y = 3 [Reveal] Spoiler: OA Kudos [?]: 241 [17], given: 1 Manager Joined: 25 Jul 2009 Posts: 115 Kudos [?]: 262 [42], given: 17 Schools: NYU, NUS, ISB, DUKE, ROSS, DARDEN Re: PS: Divisible by 4 [#permalink] ### Show Tags 14 Aug 2009, 13:46 42 KUDOS 27 This post was BOOKMARKED netcaesar wrote: If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4? (1) When p is divided by 8, the remainder is 5. (2) x – y = 3 SOL: St1: Here we will have to use a peculiar property of number 8. The square of any odd number when divided by 8 will always yield a remainder of 1!! This means that y^2 MOD 8 = 1 for all y => p MOD 8 = (x^2 + 1) MOD 8 = 5 => x^2 MOD 8 = 4 Now if x is divisible by 4 then x^2 MOD 8 will be zero. And also x cannot be an odd number as in that case x^2 MOD 8 would become 1. Hence we conclude that x is an even number but also a non-multiple of 4. => SUFFICIENT St2: x - y = 3 Since y can be any odd number, x could also be either a multiple or a non-multiple of 4. => NOT SUFFICIENT ANS: A _________________ KUDOS me if I deserve it !! My GMAT Debrief - 740 (Q50, V39) | My Test-Taking Strategies for GMAT | Sameer's SC Notes Kudos [?]: 262 [42], given: 17 Math Expert Joined: 02 Sep 2009 Posts: 41890 Kudos [?]: 128792 [31], given: 12183 Re: PS: Divisible by 4 [#permalink] ### Show Tags 16 Dec 2010, 07:39 31 KUDOS Expert's post 21 This post was BOOKMARKED nonameee wrote: Can I ask someone to look at this question a provide a solution that doesn't depend on knowing peculiar properties of number 8 or induction? Thank you. If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4? (1) When p is divided by 8, the remainder is 5 --> $$p=8q+5=x^2+y^2$$ --> as given that $$y=odd=2k+1$$ --> $$8q+5=x^2+(2k+1)^2$$ --> $$x^2=8q+4-4k^2-4k=4(2q+1-k^2-k)$$. So, $$x^2=4(2q+1-k^2-k)$$. Now, if $$k=odd$$ then $$2q+1-k^2-k=even+odd-odd-odd=odd$$ and if $$k=even$$ then $$2q+1-k^2-k=even+odd-even-even=odd$$, so in any case $$2q+1-k^2-k=odd$$ --> $$x^2=4*odd$$ --> in order $$x$$ to be multiple of 4 $$x^2$$ must be multiple of 16 but as we see it's not, so $$x$$ is not multiple of 4. Sufficient. (2) x – y = 3 --> $$x-odd=3$$ --> $$x=even$$ but not sufficient to say whether it's multiple of 4. _________________ Kudos [?]: 128792 [31], given: 12183 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7674 Kudos [?]: 17354 [26], given: 232 Location: Pune, India Re: PS: Divisible by 4 [#permalink] ### Show Tags 19 Dec 2010, 07:49 26 KUDOS Expert's post 17 This post was BOOKMARKED netcaesar wrote: If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4? (1) When p is divided by 8, the remainder is 5. (2) x – y = 3 Such questions can be easily solved keeping the concept of divisibility in mind. Divisibility is nothing but grouping. Lets say if we need to divide 10 by 2, out of 10 marbles, we make groups of 2 marbles each. We can make 5 such groups and nothing will be left over. So quotient is 5 and remainder is 0. Similarly if you divide 11 by 2, you make 5 groups of 2 marbles each and 1 marble is left over. So 5 is quotient and 1 is remainder. For more on these concepts, check out: http://gmatquant.blogspot.com/2010/11/divisibility-and-remainders-if-you.html First thing that comes to mind is if y is odd, $$y^2$$ is also odd. If $$y = 2k+1, y^2 = (2k + 1)^2 = 4k^2 + 4k + 1 = 4k(k+1) + 1$$ Since one of k and (k+1) will definitely be even (out of any two consecutive integers, one is always even, the other is always odd), 4k(k+1) will be divisible by 8. So when y^2 is divided by 8, it will leave a 1. Stmnt 1: When p is divided by 8, the remainder is 5. When y^2 is divided by 8, remainder is 1. To get a remainder of 5, when x^2 is divided by 8, we should get a remainder of 4. $$x^2 = 8a + 4$$ (i.e. we can make 'a' groups of 8 and 4 will be leftover) $$x^2 = 4(2a+1)$$ This implies $$x = 2*\sqrt{Odd Number}$$because (2a+1) is an odd number. Square root of an odd number will also be odd. Therefore, we can say that x is not divisible by 4. Sufficient. Stmnt 2: x - y = 3 Since y is odd, we can say that x will be even (Even - Odd = Odd). But whether x is divisible by 2 only or by 4 as well, we cannot say since here we have no constraints on p. Not sufficient. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17354 [26], given: 232 Intern Joined: 30 May 2013 Posts: 4 Kudos [?]: 8 [8], given: 8 Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink] ### Show Tags 15 Sep 2013, 22:13 8 This post received KUDOS I did this question this way. I found it simple. 1. p=x^2+y^2 y is odd p div 8 gives remainder 5. A number which gives remainder 5 when divided by 8 is odd. so (x^2 + y^2)/8 = oddnumber (x^2 + y^2) = 8 * oddnumber (this is an even number without doubt) x^2 + y^2 is even. Since y is odd to get x^2+y^2 even x must also be odd. X is an odd number not divisible by 4 Option A: 1 alone is sufficient Kudos [?]: 8 [8], given: 8 Intern Joined: 02 Sep 2010 Posts: 45 Kudos [?]: 146 [4], given: 17 Location: India Re: PS: Divisible by 4 [#permalink] ### Show Tags 18 Dec 2010, 11:23 4 This post received KUDOS 1 This post was BOOKMARKED maliyeci wrote: Very good solution I did not know this property of 8. Kudos to you. By and induction. 1^2=1 mod 8 say n^2=1 mod 8 (n is an odd number) than if (n+2)^2=1 mod 8 ? (n+2 is the next odd number) (n+2)^2=n^2+4n+4= 1 + 4n + 4 mod 8 4n+4=0 mod 8 because n is an odd number and 4n=4 mod 8. So induction works. So for any odd number n, n^2=1 mod 8 Its not something one shall already know before attacking a question, you may realize properties like this when u start solving a question. Even I didn't know about this property of 8. I approached the question in following way: Stmt 1: P/8=(x^2+y^2)/8; using remainder theorem; rem[(x^2+y^2)/8]= rem[x^2/8] + rem[y^2/8] if x is divisible by 4, then x^2= 4k*4k= 16K=8*2K is also divisible by 8. now to anaylze rem[y^2/8]; start putting suitable values of y; i.e all odd values starting from 1. for y=1; rem(1/8)=1 for y=3; rem(9/8)=1 for y=5;rem(25/8)=1 so you observe this pattern here. coming back to ques now, as rem[(x^2+y^2)/8]= rem[x^2/8] + rem[y^2/8]= rem[x^2/8] + 1 =5; this means rem[x^2/8] is not 0; which implies x is not divisible my 8; Sufficient Stmt2: y being odd can be accept both 3 and 5 as values and we get different results; thus Insufficient Thus OA is A _________________ The world ain't all sunshine and rainbows. It's a very mean and nasty place and I don't care how tough you are it will beat you to your knees and keep you there permanently if you let it. You, me, or nobody is gonna hit as hard as life. But it ain't about how hard ya hit. It's about how hard you can get it and keep moving forward. How much you can take and keep moving forward. That's how winning is done! Kudos [?]: 146 [4], given: 17 Intern Joined: 14 May 2013 Posts: 12 Kudos [?]: 18 [2], given: 3 Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink] ### Show Tags 12 Jun 2013, 10:58 2 This post received KUDOS netcaesar wrote: If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4? (1) When p is divided by 8, the remainder is 5. (2) x – y = 3 1. As p = 8I + 5 we have values of P = 5,13,21,29..... etc .. as y is odd when we solve this p(odd) = x^2 + y^2(odd) x^2 = odd -odd = even which can be 2,4,6 ... etc but if we check for any value of p we don't get any multiple of 4. so it say's clearly that x is not divisible by 4. 2. x-y = 3 x = y(odd)+3 x is even which can be 2,4,6.. so it's not sufficient .. Ans : A _________________ Chauahan Gaurav Keep Smiling Kudos [?]: 18 [2], given: 3 Intern Joined: 25 Jun 2013 Posts: 6 Kudos [?]: 3 [2], given: 0 Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink] ### Show Tags 13 Sep 2013, 20:25 2 This post received KUDOS 1 This post was BOOKMARKED from first statement p = 8j + 5 Put j as 1, 2,3,4,5... p would be 13, 21,29, 37,45... Now in the formula p= x^2+y^2 put 1,3,5,7 as value of y ( as y is odd) to get x. You will notic the possible value of x is 2 which is not divisble by 4. Posted from GMAT ToolKit Kudos [?]: 3 [2], given: 0 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7674 Kudos [?]: 17354 [2], given: 232 Location: Pune, India Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink] ### Show Tags 18 Aug 2014, 02:43 2 This post received KUDOS Expert's post 1 This post was BOOKMARKED alphonsa wrote: For statement 1 , wouldn't plugging in values be a better option? No. When you need to establish something, plugging in values is not fool proof. Anyway, in this question, how will you plug in values? You cannot assume a value for x since that is what you need to find. You will assume a value for y and a value for p such that they satisfy all conditions. This itself will be quite tricky. Then when you do get a value for x, you will find that it will be even but not divisible by 4. How can you be sure that this will hold for every value of y and p? When a statement is not sufficient, plugging in values can work - you find two opposite cases - one which answers in yes and the other which answers in no. Then you know that the statement alone is not sufficient. But when the statement is sufficient, it is very hard to prove that it will hold for all possible values using number plugging alone. You need to use logic in that case. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199 Veritas Prep Reviews Kudos [?]: 17354 [2], given: 232 Senior Manager Joined: 23 Jun 2009 Posts: 360 Kudos [?]: 134 [1], given: 80 Location: Turkey Schools: UPenn, UMich, HKS, UCB, Chicago Re: PS: Divisible by 4 [#permalink] ### Show Tags 15 Aug 2009, 13:49 1 KUDOS Very good solution I did not know this property of 8. Kudos to you. By and induction. 1^2=1 mod 8 say n^2=1 mod 8 (n is an odd number) than if (n+2)^2=1 mod 8 ? (n+2 is the next odd number) (n+2)^2=n^2+4n+4= 1 + 4n + 4 mod 8 4n+4=0 mod 8 because n is an odd number and 4n=4 mod 8. So induction works. So for any odd number n, n^2=1 mod 8 Kudos [?]: 134 [1], given: 80 Director Joined: 23 Apr 2010 Posts: 573 Kudos [?]: 95 [1], given: 7 Re: PS: Divisible by 4 [#permalink] ### Show Tags 16 Dec 2010, 07:13 1 KUDOS Can I ask someone to look at this question a provide a solution that doesn't depend on knowing peculiar properties of number 8 or induction? Thank you. Kudos [?]: 95 [1], given: 7 Intern Joined: 25 Mar 2012 Posts: 3 Kudos [?]: 3 [1], given: 1 Re: PS: Divisible by 4 [#permalink] ### Show Tags 16 Jul 2012, 18:19 1 KUDOS Am i missing something, why cant we take stmt 2 as follows: squaring x-y=3 on both sides, we get p=9+2xy, that is p=odd + even = odd, not divisible by 4 Kudos [?]: 3 [1], given: 1 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7674 Kudos [?]: 17354 [1], given: 232 Location: Pune, India Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink] ### Show Tags 03 May 2017, 05:31 1 KUDOS Expert's post VeritasPrepKarishma wrote: aliasjit wrote: I am a little confused about solving the problem: Stmnt 2: x-y =3 we know y is odd. and if Y is odd as per problem statement y cannot be anything but 1 as x and y both are positive integers. Therefore x =4 and is divisible by 4. y is odd, yes, but why must y be 1? Responding to a pm: Quote: because as per statement 2 if x-y =3 and x and y are both +ve integers could y be anything but 1 But as per statement 2, we do not know that x must be 4. x must be even since y is odd and difference between x and y is odd. But will it be 4, we do not know. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Kudos [?]: 17354 [1], given: 232 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7674 Kudos [?]: 17354 [0], given: 232 Location: Pune, India Re: PS: Divisible by 4 [#permalink] ### Show Tags 16 Jul 2012, 23:24 Eshaninan wrote: Am i missing something, why cant we take stmt 2 as follows: squaring x-y=3 on both sides, we get p=9+2xy, that is p=odd + even = odd, not divisible by 4 The question is: "Is x divisible by 4?" not "Is p divisible by 4?" x is even since y is odd. We don't know whether x is divisible by only 2 or 4 as well. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199 Veritas Prep Reviews Kudos [?]: 17354 [0], given: 232 Manager Joined: 29 Jun 2011 Posts: 159 Kudos [?]: 24 [0], given: 29 WE 1: Information Technology(Retail) Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink] ### Show Tags 03 Sep 2013, 04:00 Excellent explanation Bunuel & Karishma:):) Kudos [?]: 24 [0], given: 29 Intern Joined: 21 Sep 2013 Posts: 9 Kudos [?]: 2 [0], given: 0 Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink] ### Show Tags 23 Dec 2013, 23:45 For Statement 1: since p when divided by 8 leaves remainder 5.We obtain the following equation p= 8q+5 We know y is odd. If we write p =x^2+y^2 then we get the eqn: x^2+y^2=8q+5 Since, y is odd, 8q is even and 5 is odd. We get 8q+5 is odd. Then x^2= odd - y^2 i.e x^2=even ie x= even But it's not sufficient to answer the question whether x is a multiple of 4? By this logic i get E as my answer. Statement 2: is insufficient. Kudos [?]: 2 [0], given: 0 Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7674 Kudos [?]: 17354 [0], given: 232 Location: Pune, India Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink] ### Show Tags 30 Dec 2013, 23:39 Expert's post 1 This post was BOOKMARKED Abheek wrote: For Statement 1: since p when divided by 8 leaves remainder 5.We obtain the following equation p= 8q+5 We know y is odd. If we write p =x^2+y^2 then we get the eqn: x^2+y^2=8q+5 Since, y is odd, 8q is even and 5 is odd. We get 8q+5 is odd. Then x^2= odd - y^2 i.e x^2=even ie x= even But it's not sufficient to answer the question whether x is a multiple of 4? Your analysis till now is fine but it is incomplete. We do get that x is even but we also get that x is a multiple of 2 but not 4 as explained in the post above: if-p-x-and-y-are-positive-integers-y-is-odd-and-p-x-82399.html#p837890 _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for \$199 Veritas Prep Reviews Kudos [?]: 17354 [0], given: 232 Manager Joined: 22 Jul 2014 Posts: 130 Kudos [?]: 297 [0], given: 197 Concentration: General Management, Finance GMAT 1: 670 Q48 V34 WE: Engineering (Energy and Utilities) Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink] ### Show Tags 16 Aug 2014, 00:16 For statement 1 , wouldn't plugging in values be a better option? Kudos [?]: 297 [0], given: 197 Current Student Joined: 17 Jul 2013 Posts: 48 Kudos [?]: 33 [0], given: 19 GMAT 1: 710 Q49 V38 GRE 1: 326 Q166 V160 GPA: 3.74 Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink] ### Show Tags 29 Aug 2014, 05:25 Hi Karishma, Thanks for the explanation to the question. I was just wondering how the answer would change if we change the question stem a little bit. What if the question asks if p (instead of x) is divisible by 4? In this scenario, statement 1 would be sufficient since if something leaves a remainder of 5, it would leave a remainder of 1 upon division by 4 For statement 2, we know that x = y+3, so x is even. If we square it, it would surely be divisible by 4. Now if a number (y^2, which is odd) non-divisible by 4 is added to a number divisible by 4, the result would surely be not divisible by 4. So statement 2 would also be sufficient. Is this reasoning correct? just for practicing the concept Kudos [?]: 33 [0], given: 19 Manager Joined: 22 Jan 2014 Posts: 141 Kudos [?]: 74 [0], given: 145 WE: Project Management (Computer Hardware) Re: If p, x, and y are positive integers, y is odd, and p = x^2 [#permalink] ### Show Tags 31 Aug 2014, 03:38 netcaesar wrote: If p, x, and y are positive integers, y is odd, and p = x^2 + y^2, is x divisible by 4? (1) When p is divided by 8, the remainder is 5. (2) x – y = 3 1) p = 8k+5 (k is a whole number) also p = (x^2+y^2) => (x^2+y^2) mod 8 = 5 any square (n^2) mod 8 follows the following pattern -> 1,4,1,0 and then repeats. for getting x^2+y^2 mod 8 = 5 we need to take a 4 and a 1 from the above pattern. at multiples of 4, the remainder is 0. so x can never be divisible by 4. A is sufficient. 2) x-y=3 (odd) even - odd = odd or odd - even = odd so insufficient. Hence, A. _________________ Illegitimi non carborundum. Kudos [?]: 74 [0], given: 145 Re: If p, x, and y are positive integers, y is odd, and p = x^2   [#permalink] 31 Aug 2014, 03:38 Go to page    1   2    Next  [ 32 posts ] Display posts from previous: Sort by

https://math.stackexchange.com/questions/669085/what-does-curly-curved-less-than-sign-succcurlyeq-mean/669115

# What does "curly (curved) less than" sign $\succcurlyeq$ mean? I am reading Boyd & Vandenberghe's Convex Optimization. The authors use curved greater than or equal to (\succcurlyeq) $$f(x^*) \succcurlyeq \alpha$$ and curved less than or equal to (\preccurlyeq) $$f(x^*) \preccurlyeq \alpha$$ Can someone explain what they mean? • as () is curved and {} are curly ( en.wikipedia.org/wiki/Bracket ), I think those symbols you mention are curved not curly Feb 9, 2014 at 6:43 • @barlop If you look at $\LaTeX$ source of the formulas in question (right click->Show Math As->TeX commands), you'll see \succcurlyeq, which has curly word in it, not curved. Feb 9, 2014 at 7:18 • Thanks everyone for answers. As I understand $\succeq$ or $\preceq$ are more general than their more popular counterparts. I think Michael's answer make sense. If I understand correctly, $X \succeq Y, \quad if \quad \| X \| \ge \| Y \|$ where $\| \cdot \|$ is the norm associated with the space $X$ and $Y$ belongs to. I think Chris's answer is correct but is more strict condition than Michael's answer. Please correct me if I'm wrong. Feb 9, 2014 at 17:53 • You should definitely take the tour. This is not a traditional forum! – bodo Feb 9, 2014 at 18:03 • Well I don't know what a traditional forum looks like. :-) Feb 9, 2014 at 18:20 Both Chris Culter's and Code Guru's answers are good, and I've voted them both up. I hope that I'm not being inappropriate by combining and expanding upon them here. It should be noted that the book does not use $\succeq$, $\preceq$, $\succ$, and $\prec$ with scalar inequalities; for these, good old-fashioned inequality symbols suffice. It is only when the quantities on the left- and right-hand sides are vectors, matrices, or other multi-dimensional objects that this notation is called for. The book refers to these relations as generalized inequalities, but as Code-Guru rightly points out, they have been in use for some time to represent partial orderings. And indeed, that's exactly what they are, and the book does refer to them that way as well. But given that the text deals with convex optimization, it was apparently considered helpful to refer to them as inequalities. Let $S$ be a vector space, and let $K\subset S$ be a closed, convex, and pointed cone with a non-empty interior. (By cone, we mean that $\alpha K\equiv K$ for all $\alpha>0$; and by pointed, we mean that $K\cap-K=\{0\}$.) Such a cone $K$ induces a partial ordering on the set $S$, and an associated set of generalized inequalities: $$x \succeq_K y \quad\Longleftrightarrow\quad y \preceq_K x \quad\Longleftrightarrow\quad x - y \in K$$ $$x \succ_K y \quad\Longleftrightarrow\quad y \prec_K x \quad\Longleftrightarrow\quad x - y \in \mathop{\textrm{Int}} K$$ This is a partial ordering because, for many pairs $x,y\in S$, $x \not\succeq_K y$ and $y \not\succeq_K x$. So that's the primary reason why he and others prefer to use the curly inequalities to denote these orderings, reserving $\geq$, $\leq$, etc. for total orderings. But it has many of the properties of a standard inequality, such as: $$x\succeq_K y \quad\Longrightarrow\quad \alpha x \succeq_K \alpha y \quad\forall \alpha>0$$ $$x\succeq_K y \quad\Longrightarrow\quad \alpha x \preceq_K \alpha y \quad\forall \alpha<0$$ $$x\succeq_K y, ~ x\preceq_K y \quad\Longrightarrow\quad x=y$$ $$x\succ_K y \quad\Longrightarrow\quad x\not\prec_K y$$ When the cone $K$ is understood from context, it is often dropped, leaving only the inequality symbol $\succeq$. There are two cases where this is almost always done. First, when $S=\mathbb{R}^n$ and the cone $K$ is non-negative orthant $\mathbb{R}^n_+$ the generalized inequality is simply an elementwise inequality: $$x \succeq_{\mathbb{R}^n_+} y \quad\Longleftrightarrow\quad x_i\geq y_i,~i=1,2,\dots,n$$ Second, when $S$ is the set of symmetric $n\times n$ matrices and $K$ is the cone of positive semidefinite matrices $\mathcal{S}^n_+=\{X\in S\,|\,\lambda_{\text{min}}(X)\geq 0\}$, the inequality is a linear matrix inequality (LMI): $$X \succeq_{\mathcal{S}^n_+} Y \quad\Longleftrightarrow\quad \lambda_{\text{min}}(X-Y)\geq 0$$ In both of these cases, the cone subscript is almost always dropped. Many texts in convex optimization don't bother with this distinction, and use $\geq$ and $\leq$ even for LMIs and other partial orderings. I prefer to use it whenever I can, because I think it helps people realize that this is not a standard inequality with an underlying total order. That said, I don't feel that strongly about it for $\mathbb{R}^n_+$; I think most people rightly assume that $x\geq y$ is considered elementwise when $x,y$ are vectors. • Thanks a lot for the detailed answer, and for correcting the symbol as well :-). Feb 9, 2014 at 16:35 • This is an old answer and I completely agree with it but I thought that providing another common application of this notation might be useful. As @Code-Guru pointed out, these are useful for partial orders. In Economics, we usually model preferences over baskets of goods with "not worse than" or "not better than" sets. The partial order concept fits like a glove in this situation. The interested reader might find books in Decision Sciences useful for this kind of discussion. Dec 30, 2020 at 21:13 There's a list of notation in the back of the book. On page 698, $x\preceq y$ is defined as componentwise inequality between vectors $x$ and $y$. This means that $x_i\leq y_i$ for every index $i$. Edit: The notation is introduced on page 32. Often these symbols represent partial order relations. The typical "less than" and "greater than" operations both define partial orders on the real numbers. However, there are many other examples of partial orders. Sometimes the curly greater than sign is used to indicate positive semi-definiteness of a matrix $$X$$: $$(X\succeq 0\ \text{or}\ X\ge 0)$$ or a function $$f(x)$$ $$(f(x) \succeq 0\ \text{or}\ f(x)\ge 0).$$ • Positive definiteness of what? Aug 26, 2015 at 17:03 Does it sometimes denote, in measurement theory; often the qualitative counterpart to $\geq$ in the numerical representation; when one wants to numerically represent a totally ordered qualitative probability representation: $$A ≽ B \leftrightarrow A > B \leftrightarrow F(A) \geq F(B)$$. On the other hand. I have seen it used in multi-dimensional partial or even total, orderings under the "ordering of major-ization" for vector valued functions or for a system for two kinds of orderings. One for (numerical) ordinal, $>$ comparisons and another, $≽$ for (numerical) differences, sums, or a some other kind of relation, to fine grain, the representation to ensure (or some kind of) unique-ness, rather than merely strong represent-ability. See Marshall Marshall, Albert W.; Olkin, Ingram, Inequalities: theory of majorization and its applications, Mathematics in Science and Engineering, Vol. 143. New York etc.: Academic Press. XX, 569 p. \$52.50 (1979). ZBL0437.26007. Such functions or representations, may use both the curly greater$≽$than and$\geq $in the numerical or functional representation and are compatible with total orderings. $$a,b\in \Omega^{n}\; a <b \,,\quad \text{or}, \quad, a= b, \quad \text{or}\quad a < b$$ where one cannot usually adding up distinct $$a_1 \in \Omega_i\,; b_j\in \Omega_j; j \neq i$$ $$a_3 \in \Omega_c;\, ; \, P(a_i) + P(b_j) = P(a_3)$$. Sometimes I "(conjecture)" that$≽$denotes a weaker version of$(A)$or$(A_1)$(strong complementary add-itivity) which orders the differences and sums, is something over and above$(B)$;strict monotone increasing/strong representability/order embedding- above reflecting and preserving . $$(A)\,[f(x_1)+f(x_2)≽ f(y_1) + f(y_2)] \rightarrow [f(x) \geq f(y)] \text{where} \,; x=x_1 +x_2\,, y=y_1+y_2$$ or $$(A1)\,[x\, \geq y ]\,\rightarrow [f(x_1)+f(x_2)≽ f(y_1) + f(y_2)]\text{where} \,; x=x_1 +x_2\,, y=y_1+y_2$$ Rather than:$$(B)$$ $$(B)\,x\, \geq y \leftrightarrow f(x) \geq f(y)\, \text{where} \,; x=x_1 +x_2\,, y=y_1+y_2$$ In contrast to$(B)$a standard order embedding (strict monotone increasing function) Where in$(B)$the numerical function,$F$or representation is 'merely strong'. That is$F$is a monotone strictly increasing function of some entity$x$where$F(x)$is the entity one wishes to order by$x$. Sometimes, even in a infinite and uniformly and non-atomic, continuous total order representation, nothing unique will come out. If its infinite /non-atomic in the wrong sense. ie in a super-atomic or entangled lexicographic system. Where the system is dense/or continuum dense/non-atomic/bottom-less in the wrong sense. Such as a spin$1/2$system in quantum mechanics, or a lexicographic entangled, system where its infinite in wrong sense, in the vertical, not in the horizontal, not within a basis of space/basis.$(A)$is arguably a lot stronger than$(B)$, or at least is, if relatively unrestricted. Where$(A)$, may be used not to extend the ordering so much as much as make a total ordering 'numerically precise; to put some constraints on sums and differences, where the events are the not kinds of things that usually add up. Say in an entangled multidimensional quantum spin 1/2 system or utility representation where the events are on complementary spaces and mixtures are not allowed, for example. That is, to, put a metric on differences (say on a two outcome system). That is something stronger, than a, total or non atomic/dense order, where(merely) every event. Even if the entire system is totally ordered within and betwixt the distinct$\Omega\$. So that the system is solv-able, and uniquely so.

http://mathhelpforum.com/pre-calculus/10298-piecewise-defined-function.html

# Math Help - Piecewise-Defined Function 1. ## Piecewise-Defined Function For both questions below: (a) Find the domain of the function. (b) Locate any intercepts. (1) .....{3 + x......if -3 <or= to x < 0 f(x){3...........if......x = 0 .....{Sqrt{x}..if......x > 0 ======================= (2) ..........{1/x.........if.....x < 0 f(x) = {sqrt{x}...if.....x >or= to 0 NOTE: It is hard to correctly type the piecewise-defined functions using a regular keyboard. I hope you can understand the above. 2. Originally Posted by symmetry For both questions below: (a) Find the domain of the function. (b) Locate any intercepts. (1) .....{3 + x......if -3 <or= to x < 0 f(x){3...........if......x = 0 .....{Sqrt{x}..if......x > 0 ======================= (2) ..........{1/x.........if.....x < 0 f(x) = {sqrt{x}...if.....x >or= to 0 NOTE: It is hard to correctly type the piecewise-defined functions using a regular keyboard. I hope you can understand the above. I'll do the first one for you- graph it. The conditions are the "if" parts in the piece-wise function. Domain is (-3, inf) and there are no intersepts. Try graphing it. You have a line with slope = 1 and an exponential function. EDIT: Sorry, there are x and y-intercepts, as Soroban pointed out, although the two graphs do not not intersect which is what I was getting at. 3. Hello, symmetry! For both questions below: . . (a) Find the domain of the function. . . (b) Locate any intercepts. Did you make a sketch? $(1)\;\;f(x) \:=\:\begin{Bmatrix} 3 + x & &\text{if }\text{-}3 \leq x < 0 \\ 3 & &\text{if }x = 0 \\ \sqrt{x} & &\text{if }x > 0 \end{Bmatrix}$ Code: | | * * * * | * * |* ----*-----o-------------- -3 | Domain: . $(\text{-}3,\,\infty)$ Intercepts: . $(\text{-}3,0),\:(0,3)$ $(2)\;\;f(x)\:=\:\begin{Bmatrix}\frac{1}{x} & & \text{if }x < 0 \\ \sqrt{x} & & \text{if }x \geq 0 \end{Bmatrix}$ Code: | | * | * | * |* ------------------*---------------- * | * | * | * | * | | *| | Domain: . $(-\infty,\,\infty)$ Intercepts: . $(0,\,0)$ 4. ## ok Thank you again both for your quick replies. To soroban, No, I did not sketch the graph because I do not know how to graph piecewise-defined functions. I understand these functions are graphed in parts, right? Can you take me through a sample graphing question in terms of this type of function? Thanks! 5. Originally Posted by symmetry Thank you again both for your quick replies. To soroban, No, I did not sketch the graph because I do not know how to graph piecewise-defined functions. I understand these functions are graphed in parts, right? Can you take me through a sample graphing question in terms of this type of function? Thanks! Yes, they are 'graphed in parts,' I guess you could call it. For instance, Take the first condition; f(x) = 3 + x if -3 <= x < 0 From x = -3 (including this point) to x = 0 (not including, and thus draw an open circle by this point), you will graph 3 + x; see Soroban's graph. The reason why it's closed (solid dot) is because of the next condition later, and thus includes that point. Try look up piece-wise functions on Wikipedia. 6. ## ok I like graphing functions. I think piecewise-defined functions are cool but not easy to sketch. Thanks! 7. Hello again, symmetry! Okay, here's an example. . . $f(x) \:=\:\begin{Bmatrix}3 & \text{if }0 \leq x \leq 1 \\ 2x + 1 & \text{if }x > 1\end{Bmatrix}$ When $x$ is between $0$ and $1$ (including the endpoints), . . the graph is $f(x) = 3$, a horizontal line. Code: | 3* * * * * | | - + - - - + - - | 1 When $x$ is greater than 1, the graph is $f(x) \:=\:2x + 1$, . . a slanted line. Code: | | * | * | * | * 3| * | | - + - - - + - - - - - - - | 1 Sketch them on the same graph . . and have the graph of the piecewise function. Code: | | * | * | * | * 3o * * * * | | - + - - - + - - - - - - - | 1 This function could be your long-distance charge. They might charge $3 for the first minute . . and$2 per minute for every subsequent minute. (Hmmm, not a good example . . . I'm sure someone will point out why.)

https://mathematica.stackexchange.com/questions/151050/mellin-transform-of-xp-seems-to-miss-a-factor-of-2-pi/170579

# Mellin transform of $x^p$ seems to miss a factor of $2\pi$ Bug introduced in 11.1 or earlier and fixed in 11.3 On Mathematica 11.1.1.0 the Mellin transform of $x^p$ is evaluated as $\delta(p+s)$, while I think it should be $2\pi\,\delta(p+s)$: In:= MellinTransform[x^p, x, s, GenerateConditions -> True] Out:= DiracDelta[p + s] edited posting after Daniel Lichtblau's comment I initially did not understand this result, but this 2004 paper has explained to me how to arrive at the Dirac delta function, however, with an additional factor of $2\pi$. I checked that this is not a matter of a different definition of the Mellin transform. (I summarized the calculation in this Mathoverflow posting.) Missing factor $2\pi$ is fixed in Mathematica 11.3.0: In:= MellinTransform[x^p, x, s, GenerateConditions -> True] Out:= 2π DiracDelta[i(p + s)] consequence: before 11.3 Integrate[MellinTransform[1, x, s], {s, -Infinity, Infinity}] returned 1, now it returns $2\pi\int_{-\infty}^\infty\delta(is)ds$ Q: is this v. 11.3 change in the implementation of MellinTransform documented somewhere? • See last example in documentation under Scope Elementary Functions. It should be noted that this is a generalization of the integral definition, not unlike the case for FourierTransform. – Daniel Lichtblau Jul 9 '17 at 15:27 • thank you, Daniel, for the feedback, I understand things a bit better now and have edited my posting accordingly --- my problem has been reduced to a missing factor $2\pi$... – Carlo Beenakker Jul 9 '17 at 19:06 • What specific definition is used is not particularly important so long as the MellinTransform and InverseMellinTransform are inverses of each other. Both x^p == InverseMellinTransform[ MellinTransform[x^p, x, s], s, x] and DiracDelta[p + s] == MellinTransform[ InverseMellinTransform[DiracDelta[p + s], s, x], x, s] evaluate to True – Bob Hanlon Jul 9 '17 at 22:41 • @BobHanlon --- but if we assume that the factor of $2\pi$ is absorbed in the definition of DiracDelta, then Integrate[MellinTransform[1, x, s], {s, -Infinity, Infinity}] should return $2\pi$, while instead it returns 1. – Carlo Beenakker Jul 10 '17 at 6:18 • The integral of DiracDelta should be one. – Bob Hanlon Jul 10 '17 at 14:48 The Mellin transforms for $x^j$ reported by Mathematica 11.2 didn't make sense to me, so on 11/28/2017 I submitted the following question on Math StackExchange. Questions on Mellin Transform of $x^j$ and Interpretation of Distributions with Complex Arguments I ended up deriving the answer to my own question and on 12/7/2017 I submitted a problem report to Wolfram technical support where I attached a Mathematica notebook illustrating the problem and the correct solution (CASE:3980660). I received an email from Wolfram technical support on 12/13/2017 indicating my analysis was accepted as correct and a report was being filed with the developers. The correct solution was subsequently implemented in Mathematica 11.3. Note that not only was the $2\,\pi$ prefix missing, but $i$ was also missing in the $\delta$ function parameter. I subsequently posted the correct solution in answers to related questions on both Math StackExchange and MathOverflow StackExchange. Delta function with imaginary argument Dirac Delta function with a complex argument

https://www.physicsforums.com/threads/finding-horizontal-tangent-planes-on-s.556301/

# Finding Horizontal Tangent Planes on S 1. Dec 2, 2011 ### TranscendArcu 1. The problem statement, all variables and given/known data S is the surface with equation $$z = x^2 +2xy+2y$$a) Find an equation for the tangent plane to S at the point (1,2,9). b) At what points on S, in any, does S have a horizontal tangent plane? 3. The attempt at a solution $$F(x,y,z): z = x^2 +2xy+2y$$ $$F_x = 2x + 2y$$ $$F_y = 2x + 2$$ Evaluated at (1,2) gives answers 6 and 4, respectively. My equation for a plane is: $$z-9=6(x-1) + 4(y-1)$$. I think any horizontal plane should have normal vector <0,0,k>, where k is some scalar. I'm pretty sure that S has no such normal vector. But if $$F(x,y,z): 0 = x^2 +2xy+2y - z$$ then $$grad F = <2x + 2y,2x + 2,-1>$$ It seems like I can let (x,y) = (-1,1) to zero the x-, y-components of the gradient. Plugging (-1,1) into the definition of z gives z = 1. This suggests to me that there is a point (-1,1,1), at which there is a horizontal tangent plane. Yet I feel pretty sure that this isn't true! 2. Dec 3, 2011 ### ehild You made a little mistake when writing out the equation of the tangent plane. The y coordinate of the fixed point is 2, you wrote 1. A surface in 3D is of the form F(x,y,z) = constant. For this surface, x2+2xy+2y-z=0. That means F(x,y,z)=x2+2xy+y-z. The gradient of F is normal to the surface, and the tangent plane of the surface at a given point. You want a horizontal tangent plane, so a vertical gradient:(0,0,a). That means Fx=2x+2y=0, Fy=2x+2=0 --->x=-1, y=1, so your result for the x,y coordinates are correct. Plugging into the original equation for x and y, you got z=x2+2xy+2y=1, it is correct. Why do you feel it is not? ehild 3. Dec 3, 2011 ### TranscendArcu When I graphed F(x,y,z) in MatLab (and it's possible I graphed it incorrectly), I observed that the the resulting paraboloid is always "tilted". Below is a picture from my plot: http://img440.imageshack.us/img440/687/skjermbilde20111203kl85.png [Broken] How can this surface have a horizontal tangent anywhere when it is tilted like this? Last edited by a moderator: May 5, 2017 4. Dec 3, 2011 ### ehild Try to plot z out for -2<x<0 and 0<y<2 ehild 5. Dec 3, 2011 ### TranscendArcu http://img7.imageshack.us/img7/139/skjermbilde20111203kl10.png [Broken] Hmm. I'm not seeing the a point in this picture where the gradient is pointing directly upwards. Everything still looks kind of tilted. Last edited by a moderator: May 5, 2017 6. Dec 3, 2011 ### ehild The function is equivalent with z=(x+2y-1)(x+1)+1 and z=1 along the lines x=-1 and y=(1-x)/2. I attach a plot of the surface near the point (-1,1) ehild File size: 81.6 KB Views: 127 7. Dec 4, 2011 ### HallsofIvy Staff Emeritus I presume that by "horizontal" you mean perpendicular to the z-axis. The simplest way to find a tangent planes for a surface is to write it in the form F(x,y,z)= constant. Then the normal to the tangent plane at any point is given by $\nabla F$. Here, you can write $F(x, y,z)= x^2+ 2xy+ 2y- z= 0$. What is $\nabla F$? That will be vertical (and so tangent plane horizontal) when its x and y components are 0. 8. Dec 4, 2011 ### ehild @HallsofIvy: The OP has shown the solution in his first post, he only can not believe it, as the surfaces he got with MatLab look tilted. If you could give advice how to plot surfaces with MatLab, that would be real help for him. ehild 9. Dec 4, 2011 ### TranscendArcu If anyone is familiar with MatLab, this is the code I've been using: Note that the "%" mark my annotations. I included them so that hopefully you can follow what I'm doing more easily. 10. Dec 4, 2011 ### TranscendArcu Ha! I figured it out. I forgot a "." I should have written z = x.^2 + 2*x.*y +2*y; Everything makes sense now. 11. Dec 4, 2011 ### ehild You see: it is worth typing something out again and again. Is your plot similar to my one? It was made with Origin. I would like to see your final plot... Please... ehild 12. Dec 4, 2011 ### TranscendArcu http://img259.imageshack.us/img259/2104/skjermbilde20111204kl10.png [Broken]It looks like it could have a horizontal tangent plane right around (-1,1,1) Last edited by a moderator: May 5, 2017 13. Dec 4, 2011 ### ehild It is really nice!!! And a missing dot made you sceptical about the truth of Maths??!!!:uhh: ehild

http://kkul.mascali1928.it/solving-quadratic-equations-pure-imaginary-numbers.html

# Solving Quadratic Equations Pure Imaginary Numbers For y = x 2 , as you move one unit right or left, the curve moves one unit up. So, thinking of numbers in this light we can see that the real numbers are simply a subset of the complex numbers. In our recent paper we gave an efficient algorithm to calculate "small" solutions of relative Thue equations (where "small" means an upper bound of type $10^{500}$ for the sizes of solutions). 1 is called Cartesian, because if we think of as a two dimensional vector and and as its components, we can represent as a point on the complex plane. The solutions of the quadratic equation ax2 + bx +c = 0 are: SOLVING QUADRATIC EQUATION WITH TWO REAL SOLUTIONS The solutions are: SOLVING QUADRATIC EQUATION WITH ONE REAL SOLUTIONS Hence, the solution is 3. Solve quadratic equations by completing equations the square. Quadratic Equations and Complex Numbers (Algebra 2 Curriculum - Unit 4) DISTANCE LEARNING. The Unit Imaginary Number, i, has an interesting property. Here we apply this algorithm to calculating power integral bases in sextic fields with an imaginary quadratic subfield and to calculating relative power integral bases in pure quartic extensions of. math game websites for elementary students basic math puzzles 6th grade expressions math games for grade 2 printable grade 9 mathematics paper 1 multiplication puzzle worksheets 4th grade adding and subtracting variables worksheet hw solver unblocked. 3i 3 Numbers like 3i, 97i, and r7i are called PURE IMAGINARY NUMBERS. These solutions are in the set of pure imaginary numbers. Videos are created by fellow teachers for their students using the guided notes from the unit. Substituting in the quadratic formula,. Rules for adding and subtracting complex numbers are given in the box on page 279. Yes, there can be a pure imaginary imaginary solution, as i2 =-1 and -i2 = 1. Note that if your quadratic equation cannot be factored, then this method will not work. We call $$a$$ the real part and $$b$$ the imaginary part. 146 root of an equation, p. (Definitions taken from Holt Algebra 2, 2004. Unit 3 - Quadratic Functions. Its solution may be presented as x = √a. A number of the form bi, where 𝑏≠0, is called a pure imaginary number. This page will try to solve a quadratic equation by factoring it first. 1 100 Tracing Numbers Worksheet. Find a) the values of p and q b) the range of k such that the equation 3x² + 3px -q = k has imaginary roots. 2 Problem 101E. For the simplest case, = 0, there are two turning points and these lie on the real axis at ±1. See full list on intmath. use the discriminant to determine the number of solutions of the quadratic equation. That's a first look at quadratic equations. mathematics math·e·mat·ics (măth′ə-măt′ĭks) n. An obvious choice for x(0) is a turning point. For a method of solving quadratic equations,. OBJECTIVES 1 Add,Subtract,Multiply,and Divide Complex Numbers (p. Also Science, Quantum mechanics and Relativity use complex numbers. Each problem worked out in complete detail. However, using complex numbers you can find solve all quadratic equations. Write each of the following imaginary numbers in the standard form bi: 1 5 , 11, , 7, 18. 5 + 4i A) real B) real, complex C) imaginary D) imaginary, complex Ans: D Section: 2.  Find the value of the discriminant. THE QUADRATIC FORMULA AND THE DISCRIMINANT THE QUADRATIC FORMULA Let a, b, and c be real numbers such that a≠0. So tricky, in fact, that it’s become the ultimate math question. Videos are created by fellow teachers for their students using the guided notes from the unit. • Perform operations with pure imaginary numbers • Perform operations with complex numbers • Solve quadratic equations by using the quadratic formula. The real part is zero. The solution of a quadratic equation is the value of x when you set the equation equal to zero. Horizontal Parabola. Horizontal Line Equation. 15-1 (1996), 53-70. (used with a sing. Improper Rational Expression. Chapter 9: Imaginary Numbers Conceptual. Horizontal Shift. Two complex numbers are equal if and only if their real parts are equal and their imaginary parts are equal. Imaginary Numbers • pure imaginary number: square root of a negative number • complex numbers • i2 = -1 i99 = 8. A complex number is any number of the form a + bi where a and b are real numbers. radical (symbol, expression). 0 Students. Real part + bi Imaginary part Sec. In such a case, if one can easily find the real root, then all that is necessary is to solve the remaining quadratic. If you move 2 units to the left or right of the origin, the curve goes 4 units up. Hypersurfaces as a models for general algebraic varieties. Use factoring to solve a quadratic equation and find the zeros of a quadratic function. 1007/BF00526647) (with E. In this paper, we present a new method for solving standard quaternion equations. Use ordinary algebraic manipulation, combined with the fact that two complex numbers are only equal if both the imaginary and real parts are equal. Is it saying I. Pg 237, #1-7 all. Also Science, Quantum mechanics and Relativity use complex numbers. Nature of roots Product and sum of roots. Objective: be able to sketch power functions in the form of f(x)= kx^a (where k and a are rational numbers). complex numbers are required to be covered. Use the relation i 2 = −1 and the commutative, associative, and distributive properties to add, subtract, and multiply complex numbers. 2i Unit 4: Solving Quadratic Equations 4: Pure Imaginary Numbers ** This is a 2-page documenU **. Binomial, Trinomial, Factoring, Monomial, Quadratic Equation in One Variable, Zero of a Function, Square Root, Radical Sign, Radicand, Radical, Rationalizing the Denominator. Pure imaginary. Imaginary unit. Joel Kamnitzer awarded a 2018 E. This calculator is a quadratic equation solver that will solve a second-order polynomial equation in the form ax 2 + bx + c = 0 for x, where a ≠ 0, using the completing the square method. 1 Complex numbers expressed in cartesian form Include: • extension of the number system from real numbers to complex numbers • complex roots of quadratic equations • four operations of complex numbers expressed in the form (x +iy). 1 Complex Numbers Complex numbers were developed as a result of the need to solve some types of quadratic equations. Quadratic Formula 9. It is a branch of pure mathematics that uses alphabets and letters as variables. The algebra consisted of simple linear and quadratic equations and a few cubic equations, together with the methods for solving them; rules for operating with positive and negative numbers, finding squares, cubes and their roots; the rule of False Position (see History of Algebra Part. SOLVING QUADRATIC EQUATIONS. THANK YOU FOR YOUR TIME. Ten exponential equations worked out step by step. Imaginary Part. " Although there are two possible square roots of any number, the square roots of a negative number cannot be distinguished until one of the two is defined as the imaginary unit, at which point +i and -i can then be distinguished. get for a quadratic equation. Normally, it is impossible to solve one equation for two unknowns. $$i \text { is defined to be } \sqrt{-1}$$ From this 1 fact, we can derive a general formula for powers of $$i$$ by looking at some examples. Complex numbers; Non-real roots of quadratic equations. 2 Power Functions with Modeling. Solve the equation x2 +4x+5 = 0. Comparing real and imaginary parts. Unit 4 Solving Quadratic Equations Homework 2 Answer Key. SolutionWe use the formula x= −b± √ b2 − 4ac 2a With a=1, b=−2and c=10we find x = 2± p (−2)2 −(4)(1)(10) 2. Videos are created by fellow teachers for their students using the guided notes from the unit. Introduction This is a short post on how to recognize numbers such as simple integers, real numbers and special codes such as zip codes and credit card numbers and also extract these number from unstructured text in the popular bash (Bourne Again Shell) shell or scripting language. 3i 3 Numbers like 3i, 97i, and r7i are called PURE IMAGINARY NUMBERS. i is the imaginary unit. When a real number, a, is added to an imaginary number, a + bi is said to be a complex number. Imaginary numbers. Just beat it yesterday after a week long addiction. Algebra-help. Finding the values or real and imaginary numbers in standard form. Solving a quadratic equation: AC method. x2 + 9 = 0 b. Now that we are familiar with the imaginary number $$i$$, we can expand the real numbers to include imaginary numbers. Procedure for solving. A general complex number is the sum of a multiple of 1 and a multiple of i such as z= 2+3i. The axis of symmetry will intersect a parabola in one point called the _____. Division of a complex number by a complex number; Division of a complex number by a complex number (example) Argand diagrams; Modulus and argument; Equating real and imaginary parts to solve equations; Square roots of a complex number; Solving quadratic equations with complex roots; Solving cubic equations; Solving quartic equations; Reflection. When I became a student at the. Complex numbers are built on the idea that we can define the number i (called "the imaginary unit") to be the principal square root of -1, or a solution to the equation x²=-1. 3 x 2 = 100 - x 2 Solution: Step 1. verb) The study of the measurement, properties, and relationships of quantities and sets, using. The Unit Imaginary Number, i, has an interesting property. Now you will solve quadratic equations with imaginary solutions. Real numbers. Use factoring to solve a quadratic equation and find the zeros of a quadratic function. Consider the pure quadratic equation: x 2 = a , where a – a known value. The imaginary number i=sqrt(-1), i. What was most perplexing was that in using these subtle and imaginary numbers it was possible to solve cubic equations. A complex number is a number of the form where. Its use was prompted by the need to deal with algebraic expressions such as $$x^2+1$$ that have no root in the real numbers. Complex numbers; Non-real roots of quadratic equations. SOLVING QUADRATIC EQUATIONS. Yes, there can be a pure imaginary imaginary solution, as i2 =-1 and -i2 = 1. The roots of the polynomial are calculated by computing the eigenvalues of the companion matrix, A. Textbook solution for Precalculus: Mathematics for Calculus (Standalone… 7th Edition James Stewart Chapter 1. Simplifying Roots Of Negative Numbers Khan Academy. The special case corresponding to two squares is often denoted simply (e. Obviously when you get one root of a cubic equation, you can get the other two by dividing the original cubic equation by minus the first root and then use the quadratic formula in order to obtain the other two roots. 4 (1992): 824-842. complex number system The complex number system is made up of both the real numbers and the imaginary numbers. Comparing real and imaginary parts. Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. If b 0, then the complex number is called an imaginary number (Figure 2. It is well known that is perpendicular to iff is a pure imaginary number. An imaginary number is an even root of a negative number. SolutionWe use the formula x= −b± √ b2 − 4ac 2a With a=1, b=−2and c=10we find x = 2± p (−2)2 −(4)(1)(10) 2. But suppose some wiseguy puts in a teensy, tiny minus sign: Uh oh. This is a particular case of the quite general situation, which has been treated in the author’s thesis [8]. 156 complex number, p. Beware that in some cases the. Be able to find complex roots for quadratic equations. All non-imaginary numbers are real. SOLUTION OF A QUADRATIC EQUATION BY COMPLETING THE SQUARE. Plug values into the quadratic formula. x2 + 9 = 0 b. Both hyperbolas are of relatively simple form. There's also a bunch of ways to solve these equations! Watch this tutorial and get introduced to quadratic equations!. Real numbers. I make note of which method needs the most reinforcement (likely completing the square) to that I can provide more practice when we get to imaginary numbers, later in the unit. The value of the discriminant of a quadratic equation can be used to describe the number of real and complex solutions. Core Vocabulary quadratic equation in one variable, p. This script is nothing extraordinary I just put it up so someone trying to do something similar with imaginary numbers could use the code as reference. Mediaeval Algebra in Western Europe was first learnt from the works of al-Khowarizmi, Abu Kamil and Fibonacci. it is a complete quadratic if b 0. Since the discriminant b 2 - 4 ac is 0, the equation has one root. In fact, the new numbers allow the solution of any quadratic equation, and first saw light in this application. 2 2 +7 +2 ≥ 0 32. Algebra-help. radical (symbol, expression). The only two roots of this quadratic equation right here are going to turn out to be complex, because when we evaluate this, we're going to get an imaginary number. Real part Imaginary part. We can also solve polynomial problems with imaginary solutions that are bigger than quadratic equations. When this occurs, the equation has no roots (zeros) in the set of real numbers. Introduction This is a short post on how to recognize numbers such as simple integers, real numbers and special codes such as zip codes and credit card numbers and also extract these number from unstructured text in the popular bash (Bourne Again Shell) shell or scripting language. Quadratic Function Graph • max/min • vertex • axis of symmetry • y intercept • domain/range 7. Solve 3 – 4i = x + yi Finding the answer to this involves nothing more than knowing that two complex numbers can be equal only if their real and imaginary parts are equal. Solved Name Unit 4 Solving Quadratic Equations Date B. 5 Solving Quadratic Equations – Factoring. 0 = 2x2 5x +7 x = ( 5) p ( 5)2 4(2)(7) 2(2) = 5 p 25 56 4 = 5 p 31. Here we apply this algorithm to calculating power integral bases in sextic fields with an imaginary quadratic subfield and to calculating relative power integral bases in pure quartic extensions of imaginary quadratic fields. Is Zero Considered a Pure Imaginary Number (as 0i)? [12/02/2003] In the complex plane, zero (0 + 0i) is on both the real and pure imaginary axes. Normally it is mentioned in chapter related to complex numbers where the reader is made aware of the power of complex numbers in solving polynomial equations. If and is not equal to 0, the complex number is called a pure imaginary number. Quadratic Equations with Imaginary Solutions Number of equations to solve: algebra worksheet printable linear equation | pure math 10 online pretest midterm. A general complex number is the sum of a multiple of 1 and a multiple of i such as z= 2+3i. Quadratic Equation: a Program for TI84 Calculators: Have you ever used Quadratic Formula? Do you have a programmable calculator? Have you wished there was an easier way to get the answers? If you answered "Yes!" then this instructable can help you. Ncert Exemplar Class 11 Maths Solutions Chapter 5 Free Pdf. 1 Examples of solving quadratic equations using the square root When discussing the nature of the roots regarding real and imaginary numbers, (89 %) demonstrate pure mathematical. Quadratic Equations solving quadratic equations by completing the square the quadratic formula long division of a polynomial by a. Mathematicians began working with square roots of negative numbers in the sixteenth century, in their attempts to solve quadratic and cubic equations. These are sometimes called pure imaginary numbers. Its solution may be presented as x = √a. Newton did not include imaginary quantities within the notion of number, and that G. Find the y-intercept, the equation of the axis of symmetry, and the x-coordinate of the vertex. the effect that changing. In the 17th century, René Descartes (1596–1650) referred to them as imaginary numbers. If the number 1 is the unit or identity of real numbers, such that each number can be written as that number multiplied by 1, then imaginary numbers are real numbers multiplied with the imaginary identity or unit ‘ ‘. If the real part of a complex number is 0, then it is called a pure imaginary number. Quadratic Formula - Solving Equations, Fractions, Decimals & Complex Imaginary Numbers - Algebra - Duration: 24:06. An equivalent form is b2 — 4ac If a, b and c are rational coefficients, then — is a rational. 2 Power Functions with Modeling. Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. In An Imaginary Tale, Paul Nahin tells the 2000-year-old history of one of mathematics' most elusive numbers, the square root of minus one, also known as i, re-creating the baffling mathematical problems that conjured it up and the colorful characters who tried to solve them. $$i \text { is defined to be } \sqrt{-1}$$ From this 1 fact, we can derive a general formula for powers of $$i$$ by looking at some examples. Day 10 I can find complex solutions of quadratic equations. We can also solve polynomial problems with imaginary solutions that are bigger than quadratic equations. (ii) Determine the other root of the equation, giving your answer in the form p + iq. In fact, the new numbers allow the solution of any quadratic equation, and first saw light in this application. is the imaginary part of the complex number. Zero Factor Property – basis for solving quadratic equations. Well, this time, I would like to write about quadratic equation. i i i is "a" solution to the quadratic equation x 2 = A pure imaginary number is a complex number having its real part zero. College Algebra (11th Edition) answers to Chapter 1 - Section 1. These are all quadratic equations in disguise:. If one complex number is known, the conjugate can be obtained immediately by changing the sign of the imaginary part. Complex Number is a number that can be expressed in the form a + bi, where a and b are real numbers, and i is a solution of the equation x2 = −1, which is called an imaginary number because there is no real number that satisfies this equation. Imaginary numbers are based on the mathematical number $$i$$. Imaginary Unit i, Complex Number, Standard Form of a complex number, Imaginary Number, Pure Imaginary. 2i Unit 4: Solving Quadratic Equations 4: Pure Imaginary Numbers ** This is a 2-page documenU **. Therefore a complex number is the sum of a real number and a pure imaginary one. 4) Using a quadratic equation solver, we wind up with this: x = (2. The standard form of The solution set of equation 25x2 — I = 0 is: The quadratic. quadratic equations. Consider the pure quadratic equation: x 2 = a , where a – a known value. Solving Quadratic Equations Pure Imaginary Numbers. Unit 4 Solving Quadratic Equations Homework 2 Answer Key. 3 10 4 3 9. • Use the discriminant to find the number of x-intercepts/real solutions/zeros/roots. number, pure imaginary number 3. We use the function func:scipy. Following are the methods of solving a quadratic equation : Factoring; Let us see how to use the method of factoring to solve a quadratic equation. Solve quadratic equations by inspection (e. Improper Fraction. For example, the equation x 2 + 1 = 0 has no solutions in the real numbers. Quadratic Formula - Solving Equations, Fractions, Decimals & Complex Imaginary Numbers - Algebra - Duration: 24:06. -5x2 + 12x - 8 = 0 4. Solving Quadratics with Imaginary Solutions Name_____ Date_____ Period____ ©M M2O0M1_6k GK_ultYaQ hSqoTfftTwwalrmed qLULvCm. , the square root of -1. The diagram shows how different types of complex numbers are related. Quadratic inequality in two variables: Quadratic inequality in one variable: Linear inequality in two variables: Solve the equation using any method. Answer by math_helper(1904) ( Show Source ):. These are all quadratic equations in disguise:. Real part + bi Imaginary part Sec. Perform operations with pure imaginary numbers and complex numbers Use complex conjugates to write quotients of complex numbers in standard form Graph quadratic functions Solve quadratic equations - Set - Element - Subset - Universal Set - Complement - Union - Intersection - Empty Set - Imaginary Unit - Complex Number. Galerkin (HDG) method for solving the Helmholtz equation with impedance boundary condition: (1. By using this website, you agree to our Cookie Policy. Imaginary numbers. Which statement about the solutions x = 5 and x = –20 is true? asked by T on June 2, 2016; Algebra 2 help :) Any number in the form of a+-bi , where a and b are real numbers and b not equal 0 is considered a pure imaginary number. Therefore, the rules for some imaginary numbers are:. Imaginary Part. Numbers like —2 — i and - 2 + i that include a real term and an imaginary term are called complex numbers. This poster gives explicit formulas for the solutions to quadratic, cubic, and quartic equations. The concept was discussed in a recent thread, where we pointed out that the definition used for real radicands doesn't apply here, as there are no "positive" complex numbers; in cases like yours, in fact, both roots (2 - i and -2 + i) have a negative sign somewhere, so. The aim of this paper is to study t k and the value of N k /ℚ ( η k ). Horizontal Line Equation. The difference is that the root is not real. Upon completing this goal the student will be able to: * solve quadratic equations by graphing, factoring, and completing the square. Practice Maths with Vedantu to understand concepts right from basic maths to Algebra, Geometry, Trigonometry, Arithmetic, Probability, Calculus and many more. Solve quadratic equations with complex number solutions. Journal of Symbolic Computation, volume 46, number 8, pages 967--976, 2011. I will even skip a match if it means swiping the largest number out of the corner. They will also analyze situations involving quadratic functions and formulate quadratic equations to solve problems. From this starting point evolves a rich and exciting world of the number system that encapsulates everything we have known before: integers, rational, and real numbers. Simplify the expression: 16. Imaginary Numbers. They are factoring, using the square roots, completing the square and using the quadratic formula. x2 + 4x + 5 = 0 c. The Imaginary Unit i Not all quadratic equations have real-number solutions. Imaginary numbers are applied to square roots of negative numbers, allowing them to be simplified in terms of i. How does this work? Well, suppose you have a quadratic equation that can be factored, like x 2 +5x+6=0. Use ordinary algebraic manipulation, combined with the fact that two complex numbers are only equal if both the imaginary and real parts are equal. Solve equations Quadratic in Form by substitution: Step 1: Determine the appropriate substitution and write the equation in the form au2 + bu + c = 0 Step 2: Solve the equation (using any method). " Imaginary numbers allow for complex analysis, which allows engineers to solve practical problems working in the plane. -2-Create your own worksheets like this one with Infinite Algebra 2. Complex Solution to Quadratic Equations When using the Quadratic Formula to solve a quadratic equation, we can use complex numbers and the imaginary root to express the solutions. It also provides solutions to the problematic quadratic equations and all other polynomial equations In the form p(x). Write and graph an equation of a parabola with its vertex at (h,k) and an equation of a circle, ellipse, or hyperbola with its center at (h,k) Classify a conic using its equation : Quadratic Systems : Solve systems of quadratic equations by finding points of intersection Solve systems of quadratic equations using substitution. Quadratic equation usually used to find the unknown number(s) of x in the equation. Solve quadratic equations by inspection (e. for solving quadratic 2a equations. Note that each of these numbers is pure imaginary with positive coefficient. 5 Relation of the Roots. 6 Complex and Imaginary Numbers Objectives What is an imaginary number? What is a complex number? Jan 30­10:53 AM 1 Complex Numbers 2010 September 15, 2010 Warm­up: Solve using the quadratic formula. In this tutorial, you'll be introduced to imaginary numbers and learn that they're a type of complex number. Now, by applying algebra techniques we can solve the equation. • Solve quadratic equations by factoring. Unique Math Equation Stickers designed and sold by artists. Continuing coursework from the Algebra II A, this title covers the review of square roots, radicals, complex pure and imaginary numbers, solving and factoring, identifying and evaluating the discriminant of a quadratic equation, rewriting equations, solving problems with number lines, graphing parabola, circle parts and formulas, hyperbola. fsolve to solve it. a coefficient has on. Quadratic Equation Solver. 1 Unit Objectives 4. Myung-Hwan Kim, Introduction to Universal Positive Quadratic Forms over Real Number Fields, Proc. The Unit Imaginary Number, i, has an interesting property. (Substitute your values back into the original subst. 4c Calculate the discriminant of a quadratic equation to determine the number of real & complex solutions. All non-imaginary numbers are real. (ii) Determine the other root of the equation, giving your answer in the form p + iq. Imaginary numbers and quadratic equations sigma-complex2-2009-1 Using the imaginary number iit is possible to solve all quadratic equations. Quadratic Equation. Cauchy-Riemann equations, harmonic functions. imaginary quadratic base eld Groups of Special Units, University of Georgia 2009 Invited number theory seminar about my thesis research Conference Organization West Coast Number Theory Conference 2015 - present On organizational committee and grant commitee Selected Conferences and Scholarly Activities [email protected] 2016 -present. Quadratic Equations with Imaginary Solutions Number of equations to solve: algebra worksheet printable linear equation | pure math 10 online pretest midterm. \)The trajectory of such a solution consists of one point, namely $$c\ ,$$ and such a point is called an equilibrium. doc Author: E0022430 Created Date: 2/9/2010 12:03:19 PM. Want to master Microsoft Excel and take your work-from-home job prospects to the next level? Jump-start your career with our Premium A-to-Z Microsoft Excel Training Bundle from the new Gadget Hacks Shop and get lifetime access to more than 40 hours of. Polynomials with Complex Solutions. This is a particular case of the quite general situation, which has been treated in the author’s thesis [8]. Take this example: Solve 0 = (x - 9)^2 * (x^2 + 9). Solve quadratic equations by factoring. If 𝑏=0, then the number 𝑎+𝑏𝑖=𝑎 is a real number. xx2 12 35 0 2. Of course, the generalized version isn't as pretty ($m$ and $n$ are integers):. When the radicand in the quadratic formula (the discriminant Delta) is negative it means that you cannot find pure Real solutions to your equation.  Find the value of the discriminant. You want the square root of a number less than zero? That’s absurd!. Complete quadratic equation: If the equation having x and x2 terms such an. LOVE IT!! Reply Delete. Impossible Event. Following are the methods of solving a quadratic equation : Factoring; Let us see how to use the method of factoring to solve a quadratic equation. Horizontal Shrink. xx2 10 25 64 4. The solution set is The Quadratic Formula If we start with the equation ax2 + bx + c = 0, for a > 0, and complete the square to solve for x in terms of the constants a, b, and c, the result is a general formula for solving any quadratic equation. State the number of complex roots of the equation x 3 2x2 3x 0. web; books; video; audio; software; images; Toggle navigation. 156 pure imaginary number, p. complex numbers are required to be covered. which can be regarded as a system of four quadratic equations in the scalar part qand (the three components of) the vector part q of Q. pure imaginary number. Negative 4, if I take a square root, I'm going to get an imaginary number. fsolve to do that. Steacie Memorial Fellowship U of T’s team of students place 4 th in the 2017 Putnam Competition! Three faculty, R. Complex Numbers H2 Maths Tuition Tips. Take this example: Solve 0 = (x - 9)^2 * (x^2 + 9). Quadratic Formula 9. There are various methods through which a quadratic equation can be solved. 10 points for the best working. The number has a non-zero real part and pure imaginary part. An imaginary number is an even root of a negative number. Solve for x: x( x + 2) + 2 = 0, or x 2 + 2 x + 2 = 0. • Find square roots and perform operations with pure imaginary numbers. Imaginary numbers are used to help us work with numbers that involve taking the square root of a negative number. The problem was with certain cubic equations, for example x3 −6x+2 = 0. LOVE IT!! Reply Delete. Following are the methods of solving a quadratic equation : Factoring; Let us see how to use the method of factoring to solve a quadratic equation. = −1, and every complex number has the form a + biwith a and b real. I been trying to figure out how to set up this equation to add two complex numbers for Java. In a similar way, we can find the square root of any negative number. We often use the notation z= a+ib, where aand bare real. • Estimate solutions of quadratic equations by graphing. Students apply these. Carmen is using the quadratic equation (x + 15)(x) = 100 where x represents the width of a picture frame. Identity (Equation) Identity Matrix. Example 2A: Solving a Quadratic Equation with Imaginary Solutions Take square roots. 3 ­ Notes ­ Solving Quadratics with Imaginary Numbers. Khan Academy Video: Quadratic Formula 1; Need more problem types? Try MathPapa Algebra. Quadratic Equation: a Program for TI84 Calculators: Have you ever used Quadratic Formula? Do you have a programmable calculator? Have you wished there was an easier way to get the answers? If you answered "Yes!" then this instructable can help you. Powered by Cognero Page 1. To ensure that every quadratic equation has a solution, we need a new set of numbers that includes the real numbers. Solving quadratic equations can sometimes be quite difficult. Math is the basic building blocks that deals with all sort of calculations such as Addition, subtraction, multiplication, division and much more. 2 Problem 101E. Well, this time, I would like to write about quadratic equation. The imaginary unit i is the complex. • Complete the square to solve quadratic equations or to convert from standard to vertex form. Students work extensively with factoring quadratics using various factoring techniques. Finding the values or real and imaginary numbers in standard form. This book has been requested by many readers. Journal Canadien de Math\'ematiques 44. Complex Solutions of Quadratic Equations When using the Quadratic Formula to solve a quadratic equation, you often obtain a result such as which you know is not a real number. In this paper, we present a new method for solving standard quaternion equations. For example, the equation x 2 + 1 = 0 has no solutions in the real numbers. a unique quadratic function. Mathematicians began working with square roots of negative numbers in the sixteenth century, in their attempts to solve quadratic and cubic equations. x2 + 9 = 0 b. Imaginary numbers are called so because they lie in the imaginary plane, they arise from taking square roots of negative numbers. 2i Unit 4: Solving Quadratic Equations 4: Pure Imaginary Numbers ** This is a 2-page documenU **. the square, or using. To find complex number solutions of quadratic equations. (Substitute your values back into the original subst. By using this website, you agree to our Cookie Policy. ExampleUse the formula for solving a quadratic equation to solve x2 − 2x+10=0. x2 + 4x + 5 = 0 c. The others are right - pick a corner and keep the largest numbers closest to that corner. But this pure oscillation would be the B equals 0 with undamped. Imaginary numbers are complex numbers where a = 0 and b ≠ 0. Lastly, Allen defined a complex number as one which is not real (p. pure imaginary number. Lesson 2: Solving Square Root Equations Lesson 3: The Basic Exponent Properties Lesson 4: Fractional Exponents Revisited Lesson 5: More Exponent Practice Lesson 6: The Quadratic Formula Lesson 7: More Work with the Quadratic Formula Unit 9 Lesson 1: Imaginary Numbers Lesson 2: Complex Numbers Lesson 3: Solving Quadratics with Complex Solutions. Number Theory 85 (2000), 201-219. 1) 10x2 - 4x + 10 = 02) x2 - 6x + 12 = 0 3) 5x2 - 2x + 5 = 04) 4b2 - 3b + 2 = 0. But this is really two. (used with a sing. In quadratic planes, imaginary numbers show up in equations. Solve quadratic equations by inspection (e. A general complex number is the sum of a multiple of 1 and a multiple of i such as z= 2+3i. 22 (1996), 425-434. Pure Mathematics 2 & 3. First published in 1975, this classic book gives a systematic account of transcendental number theory, that is those numbers which cannot be expressed as the roots of algebraic equations having rational coefficients. The solution of these equations is b = 1, a = 0, so (-1) 1/2 = (0,1). An equivalent form is b2 — 4ac If a, b and c are rational coefficients, then — is a rational. Example: 3i If a ≠0 and b ≠ 0, the complex number is a nonreal complex number. You want the square root of a number less than zero? That’s absurd!. Impossible Event. Solved Name Unit 4 Solving Quadratic Equations Date B. 7) 10n2 - n - 8 = 08) 8p2 - 12p + 7 = 0 9) 2r2 + 2r + 6 = 0 10) 11r2 - 5r - 12 = 7 11) -14 + a = -3a2 12) -5 = 11b2 - 2b 13) 3n2 + 10n = -12 - 8n2 + 10n14) r2 - 2r - 4 = 2r2 + 8 Find the discriminant of each quadratic equation then state the number and type of solutions. Imaginary numbers are complex numbers where a = 0 and b ≠ 0. In An Imaginary Tale, Paul Nahin tells the 2000-year-old history of one of mathematics' most elusive numbers, the square root of minus one, also known as i, re-creating the baffling mathematical problems that conjured it up and the colorful characters who tried to solve them. Whitley) Periods of cusp forms and elliptic curves over imaginary quadratic fields Mathematics of Computation 62 No. In this tutorial, you'll be introduced to imaginary numbers and learn that they're a type of complex number. 25 2 5 1 7− i2 =− −=( ) 28. The Quadratic Equation, which has many uses, can give results that include imaginary numbers. Quadratic Equations and Complex Numbers (Algebra 2 Curriculum - Unit 4) DISTANCE LEARNING. Its solution may be presented as: Here the three cases are possible:. Any number that is a non-repeating decimal is irrational. is the imaginary part of the complex number. 3 Exercises - Page 103 9 including work step by step written by community members like you. Workshops in Pure Math. Pure imaginary number – If a = 0 and b ˜ 0, the number a + bi is a pure imaginary number. Consider the pure quadratic equation: x 2 = a ,. Imaginary Part. Note that if your quadratic equation cannot be factored, then this method will not work. Microsoft Word - Imaginary and Complex Numbers. Many answers. Inspired designs on t-shirts, posters, stickers, home decor, and more by independent artists and designers from around the world. Complex numbers cannot be ordered. If a and b are real numbers 𝑎+𝑏𝑖 is a complex number, and it is said to be written in standard form. Manipulating expressions involving α+β and α+β. Let us learn about solving quadratic equation calculator with a solved examples. Of course, these are abelian, so sometimes have slightly special properties. 253 #33-44, 64-66. Introduction This is a short post on how to recognize numbers such as simple integers, real numbers and special codes such as zip codes and credit card numbers and also extract these number from unstructured text in the popular bash (Bourne Again Shell) shell or scripting language. The complex numbers include all real numbers and all imaginary numbers. We can now solve both of these equations trivially. Impossible Event. notebook 1 January 11, 2017 Jan 4­9:06 AM Quadratic Functions MGSE9­12. Many quadratic equations have roots that are pure imaginary numbers or. 2 Mean Value Theorem. I like to use puzzles when a specific skill (like solving quadratic equations) requires fluency [MP6]. The quadratic equation 3x² + 3px - q=0 has the roots and 3. 88 Quadratic equations are the basis for a vast area of more complex mathematics, both pure and applied. Imaginary. can someone help me find at least 3 points of this quadratic equation? y=-2x^2 + x + 5 i got one and i seriously dont know if its right or wrong: (1/4, 5) can someone help me find at least 3 points of this quadratic equation? y=-2x^2 + x + 5 i got one and i seriously dont know if its right or wrong: (1/4, 5). This is denoted by C. Operations with Complex Numbers Complex Numbers (a + bi) Real Numbers (a + 0i) Imaginary. • Pure Imaginary Numbers & Powers of i • Solving Quadratics by Square Roots with Pure Imaginary Solutions • Complex Numbers (includes Classifying & Properties) • Operations with Complex Numbers • Solving Quadratics by Completing the Square (includes Complex Solutions) • Solving Quadratics by the Quadratic Formula (includes. Negative 4, if I take a square root, I'm going to get an imaginary number. pure imaginary number. Using this method we reobtain the known formulas for the solution of a quadratic quaternion equation, and provide an explicit solution for the cubic quaternion equation, as long as the equation has at least one pure imaginary root. Write quadratic functions in vertex form. Quadratic Equation. To solve equation we must specify the initial condition x(0). Equations such as +1 0 have no real solution, so mathematicians defined the imaginary numbers to represent their solu ions. x2 + 4x + 5 = 0 c. Journal of Symbolic Computation, volume 46, number 8, pages 967--976, 2011. As humans have solved new problems, equations, they have needed to create more numbers. 146 Solving Quadratic. or Quadratic Equations That Can Be Solved by Factoring, Applications of the Pythagorean Theorem Pg. 3 Example 4 Solve: x 2 x 6 0 1 223 x 2 No real-number solutions To solve such equations, we must define the square root of a negative number. Students will solve quadratic equations using graphs, tables, and algebraic methods. nth roots of a complex number The technique is the same for finding nth roots of any complex number. 2 Basic I can use the quadratic formula to solve a quadratic equation. • finding an equation for the common perpendicular to two skew lines 4 Complex numbers 4. When you need guidance on algebra exam or concepts of mathematics, Algebra-help. (ii) Determine the other root of the equation, giving your answer in the form p + iq. Quadratic Equation. Also of note, Wolfram sells a poster that discusses the solvability of polynomial equations, focusing particularly on techniques to solve a quintic (5th degree polynomial) equation. Now that we are familiar with the imaginary number $$i$$, we can expand the real numbers to include imaginary numbers. It can get a little confusing!. ExampleUse the formula for solving a quadratic equation to solve x2 − 2x+10=0. Solving polynomial, linear, quadratic. Imaginary Part. Examples: Write each number in the form + 𝑖: a. Pg #20-34 all. The solution of these equations is b = 1, a = 0, so (-1) 1/2 = (0,1). 0 = 2x2 5x +7 x = ( 5) p ( 5)2 4(2)(7) 2(2) = 5 p 25 56 4 = 5 p 31. the quadratic formula. Find the exact solution of by using the Quadratic Formula. Now you will solve quadratic equations with imaginary solutions. Big Idea #2: Numbers like 3i and v'î i are called pure imaginary numbers. The only two roots of this quadratic equation right here are going to turn out to be complex, because when we evaluate this, we're going to get an imaginary number. You can compare all quadratic expressions to ax 2 + bx + c and get the values of a, b and c. Polynomial Equation Solver - by Don Cross This web page contains an interactive calculator that solves any linear, quadratic, or cubic equation. Since the discriminant b 2 – 4 ac is 0, the equation has one root. Textbook Authors: Lial, Margaret L. 8x2 - 4x + 5 = 0 3. We call athe real part and bthe imaginary part of z. Simplify the expression: 17. How does this work? Well, suppose you have a quadratic equation that can be factored, like x 2 +5x+6=0. Write quadratic functions in vertex form. 1 Examples of solving quadratic equations using the square root When discussing the nature of the roots regarding real and imaginary numbers, (89 %) demonstrate pure mathematical. It was this discovery which made the use of complex numbers ‘respectable’. Core Vocabulary quadratic equation in one variable, p. Using the quadratic formula, we have x = −4± p (−4)2 −4·5 2 = −4± √ −4 2 = −4±2 √ −1 2 = −2±i. The calculator also provides conversion of a complex number into angle notation (phasor notation), exponential, or polar coordinates (magnitude and angle). Its solution may be presented as x = √a. LOVE IT!! Reply Delete. Page 126 Solving Quadratic Equations Freyer Model. Shankar, and G. This equation, which arises in a surface construction problem, incorporates linear terms in a quaternion variable and its conjugate with right and left quaternion coefficients, while the quadratic term has a quaternion coefficient placed between the variable and. Free quadratic equation calculator - Solve quadratic equations using factoring, complete the square and the quadratic formula step-by-step This website uses cookies to ensure you get the best experience. Note: It is not necessary to find the roots. In this paper a new algorithm for solving algebraic Riccati equations (both continuous-time and discrete-time versions) is presented. Pure STEP 3 Questions 2012 S3 Q6 1Preparation The STEP question involves complex numbers and the Argand diagram. Solve the equation x2 +4x+5 = 0. Quadratic Formula 9. 8x2 - 4x + 5 = 0 3. Workshops in Pure Math. Here we apply this algorithm to calculating power integral bases in sextic fields with an imaginary quadratic subfield and to calculating relative power integral bases in pure quartic extensions of. If 𝑏≠ 0, then the number 𝑎+𝑏𝑖 is called an imaginary number. Of course, the generalized version isn't as pretty ($m$ and $n$ are integers):. And this happens when b squared is smaller than 4ac. Improper Fraction. To find complex number solutions of quadratic equations. This page will try to solve a quadratic equation by factoring it first. complex number standard form EXAMPLE 1 imaginary unit GOAL 1 Solve quadratic equations with complex solutions and perform operations with complex numbers. Any number that is a non-repeating decimal is irrational. 2x2 - 10x + 25 = 0 5. 1) u k2u= f in ; @u @n (1. Real and imaginary numbers; Addition, subtraction and multiplying complex numbers and simplifying powers of i; Complex conjugates; Division of a complex number by a complex number; Argand diagrams; Modulus and argument of a complex number; Solving problems with complex numbers; Square roots of a complex number; Solving quadratic equations with. Quadratic Formula - Solving Equations, Fractions, Decimals & Complex Imaginary Numbers - Algebra - Duration: 24:06. Horizontal Reflection. The roots of the polynomial are calculated by computing the eigenvalues of the companion matrix, A. The solution to this particular equation is called the imaginary number i: i2 = 1,1 and one way to de ne the set of complex numbers is as the set of all expressions of type x+ iywhere xand yare real. Imaginary numbers and quadratic equations sigma-complex2-2009-1 Using the imaginary number iit is possible to solve all quadratic equations. • Writing quadratic equations in different forms reveals different key features. Want to master Microsoft Excel and take your work-from-home job prospects to the next level? Jump-start your career with our Premium A-to-Z Microsoft Excel Training Bundle from the new Gadget Hacks Shop and get lifetime access to more than 40 hours of. 5th Class Maths Worksheets. When the real part is zero we often will call the complex number a purely imaginary number. f x Ax Bx C = + + = 0 Equation 1. The Quadratic Equation, which has many uses, can give results that include imaginary numbers. An imaginary number bi has two parts: a real number, b, and an imaginary part, i, defined as i^2 = -1. You can use the imaginary unit to write the square root of any negative number. 7 3 2i i i 11. To find complex number solutions of quadratic equations. Students work extensively with factoring quadratics using various factoring techniques. For example, as follows:. Journal of Pure and Applied Algebra, volume 215, number 6, pages 1371--1397, 2011. 2) + iku= g on ; where 2Rd;d= 1;2;3 is a convex polyhedral domain, := @, k˛1 is known as the wave number, i = p 1 denotes the imaginary unit, and ndenotes the unit outward normal to @. The name comes from "quad" meaning square, as the variable is squared (in other words x 2). Full text of "The theory of equations: with an introduction to the theory of binary algebraic forms" See other formats. So, thinking of numbers in this light we can see that the real numbers are simply a subset of the complex numbers. The imaginary unit “i” is used to represent: i 1 and i2 1 Ex. Quadratic Equations and Complex Numbers (Algebra 2 Curriculum - Unit 4) DISTANCE LEARNINGUPDATE: This unit now contains a Google document with links to instructional videos to help with remote teaching during COVID-19 school closures. square roots denoted by s and ºs. Simplify the expression: 16. 3 - Complex Numbers - 1. (a = 0) So, a number is either real or imaginary, and some imaginary numbers are pure imaginary numbers. It "cycles" through 4 different values each time we multiply:. Excel in math and science. Practice Maths with Vedantu to understand concepts right from basic maths to Algebra, Geometry, Trigonometry, Arithmetic, Probability, Calculus and many more. (a) x 2 −1=0 (b) x2 −x −6 =0 (c) x 2 −2x −2 =0 (d) x2 −2x +2 =0 You should have found (a), (b) and (c) straightforward to solve. in the complex number. Pure imaginary numbers – numbers in the form bi – where i= −1. If the number 1 is the unit or identity of real numbers, such that each number can be written as that number multiplied by 1, then imaginary numbers are real numbers multiplied with the imaginary identity or unit ‘ ‘. An equivalent form is b2 — 4ac. Quadratic formula: A quadratic formula is the solution of a quadratic equation ax 2 + bx + c = 0, where a ≠ 0, given by. can someone help me find at least 3 points of this quadratic equation? y=-2x^2 + x + 5 i got one and i seriously dont know if its right or wrong: (1/4, 5) can someone help me find at least 3 points of this quadratic equation? y=-2x^2 + x + 5 i got one and i seriously dont know if its right or wrong: (1/4, 5). Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. The imaginary part is zero. And you would be right. Pure quadratic equation The number of methods to solve a quadratic e uatlon Is: Which equation is called exponential equation? A solution of equation which does not satisfy the equation is called: An equation in which variable occurs under radical sign is called. Joel Kamnitzer awarded a 2018 E. 3 Solving Quadratic Equations 4. These are solutions where appear the imaginary unit i. Imaginary numbers. x2 =-1 *This section may be omitted without any loss of continuity. Horizontal Parabola. Well, this time, I would like to write about quadratic equation. The solution of these equations is b = 1, a = 0, so (-1) 1/2 = (0,1). It is also called an "Equation of Degree 2" (because of the "2" on the x) A "Standard" Quadratic Equation looks like this: The letters a, b and c are coefficients (you know those values). Unit 4 Solving Quadratic Equations Homework 2 Answer Key. We call athe real part and bthe imaginary part of z. Page 126 Solving Quadratic Equations Freyer Model. The roots of the polynomial are calculated by computing the eigenvalues of the companion matrix, A. 9) The matrix of this system is A = 0 I F 0 , where F = −Λ+bK. The discriminant is the radicand in the quadratic formula. 1 Complex numbers expressed in cartesian form Include: • extension of the number system from real numbers to complex numbers • complex roots of quadratic equations • four operations of complex numbers expressed in the form (x +iy). Introduction Fundamental theorem of algebra is one of the most famous results provided in higher secondary courses of mathematics. Solving Quadratic Equations by Finding Square Roots. which can be regarded as a system of four quadratic equations in the scalar part qand (the three components of) the vector part q of Q. Use the Quadratic Formula to solve the quadratic equation. In this paper, we present a new method for solving standard quaternion equations. Many answers. Find a) the values of p and q b) the range of k such that the equation 3x² + 3px -q = k has imaginary roots.

http://math.stackexchange.com/questions/66443/a-square-matrix-has-the-same-minimal-polynomial-over-its-base-field-as-it-has-ov

# A square matrix has the same minimal polynomial over its base field as it has over an extension field I think I have heard that the following is true before, but I don't know how to prove it: Let $A$ be a matrix with real entries. Then the minimal polynomial of $A$ over $\mathbb{C}$ is the same as the minimal polynomial of $A$ over $\mathbb{R}$. Is this true? Would anyone be willing to provide a proof? Attempt at a proof: Let $M(t)$ be the minimal polynomial over the reals, and $P(t)$ over the complex numbers. We can look at $M$ as a polynomial over $\Bbb C$, in which case it will fulfil $M(A)=0$, and therefore $P(t)$ divides it. In addition, we can look at $P(t)$ as the sum of two polynomials: $R(t)+iK(t)$. Plugging $A$ we get that $R(A)+iK(A)=P(A)=0$, but this forces both $R(A)=0$ and $K(A)=0$. Looking at both $K$ and $R$ as real polynomials, we get that $M(t)$ divides them both, and therefore divides $R+iK=P$. Now $M$ and $P$ are monic polynomials, and they divide each other, therefore $M=P$. Does this look to be correct? More generally, one might prove the following Let $A$ be any square matrix with entries in a field$~K$, and let $F$ be an extension field of$~K$. Then the minimal polynomial of$~A$ over$~F$ is the same as the minimal polynomial of $A$ over$~K$. - There's the saying, "Look before you leap". I think I've managed to prove this. Please confirm if my answer is correct. –  iroiroaru Sep 21 '11 at 18:43 I think you already posted before under a different account (the "above" instead of "over"); I also remember your user name. Have you considered registering, so that all your activity is under the same user name? –  Arturo Magidin Sep 21 '11 at 18:46 It is impossible for us to confirm if your answer is correct if all you do is provide the question. If you want us to "confirm if [your] answer is correct", why not post your proof ? –  Arturo Magidin Sep 21 '11 at 18:49 I am in the process of writing it! –  iroiroaru Sep 21 '11 at 18:51 Hi Arturo, yes, I posted here twice before. Should I register? As this site allows me to post questions without registering, I figured it wouldn't be necessary. e- I'm done writing my proof. –  iroiroaru Sep 21 '11 at 18:53 Written before/while the OP was adding his/her own proof, which is essentially the same as what follows. Let $\mu_{\mathbb{R}}(x)$ be the minimal polynomial of $A$ over $\mathbb{R}$, and let $\mu_{\mathbb{C}}(x)$ be the minimal polynomial of $A$ over $\mathbb{C}$. Since $\mu_{\mathbb{R}}(x)\in\mathbb{C}[x]$ and $\mu_{\mathbb{R}}(A) = \mathbf{0}$, then it follows by the definition of minimal polynomial that $\mu_{\mathbb{C}}(x)$ divides $\mu_{\mathbb{R}}(x)$. I claim that $\mu_{\mathbb{C}}[x]$ has real coefficients. Indeed, write $$\mu_{\mathbb{C}}(x) = x^m + (a_{m-1}+ib_{m-1})x^{m-1}+\cdots + (a_0+ib_0),$$ with $a_j,b_j\in\mathbb{R}$. Since $A$ is a real matrix, all entries of $A^j$ are real, so $$\mu_{\mathbb{C}}(A) = (A^m + a_{m-1}A^{m-1}+\cdots + a_0I) + i(b_{m-1}A^{m-1}+\cdots + b_0I).$$ In particular, $$b_{m-1}A^{m-1}+\cdots + b_0I = \mathbf{0}.$$ But since $\mu_{\mathbb{C}}(x)$ is the minimal polynomial of $A$ over $\mathbb{C}$, no polynomial of smaller digree can annihilate $A$, so $b_{m-1}=\cdots=b_0 = 0$. Thus, all coefficients of $\mu_{\mathbb{C}}(x)$ are real numbers. Thus, $\mu_{\mathbb{C}}(x)\in\mathbb{R}[x]$, so by the definition of minimal polynomial, it follows that $\mu_{\mathbb{R}}(x)$ divides $\mu_{\mathbb{C}}(x)$ in $\mathbb{R}[x]$, and hence in $\mathbb{C}[x]$. Since both polynomials are monic and they are associates, they are equal. QED So, yes, your argument is correct. - Another way of proving this fact may be observing that ''you do not go out the field while using Gaussian elimination''. More precisely: Proposition. Let $K \subseteq F$ be a field extension let $v_1, \dots, v_r \in K^n$. If $v_1, \dots, v_r$ are linearly dependent over $F$, then they are linearly dependent over $K$. Proof. We'll prove the contrapositive of the statement. Suppose that the $v_i$'s are linearly independent over $K$. Let $\lambda_i \in F$ such that $\sum_i \lambda_i v_i = 0$. We can find $e_j \in F$ linearly independent over $K$ such that $\lambda_i = \sum_j \alpha_{ij} e_j$, with $\alpha_{ij} \in K$. Now from $\sum_{i,j} e_j \alpha_{ij} v_i = 0$ we deduce that $\sum_i \alpha_{ij} v_i = 0$, for every $j$. From the independence of $v_i$'s over $K$, we have $\alpha_{ij} = 0$, so $\lambda_i = 0$. $\square$ Now consider a field extension $K \subseteq F$ and a matrix $A \in M_n(K)$. Let $\mu_K$ and $\mu_F$ the minimal polynomials of $A$ over $K$ and $F$, respectively. Considering $I, A, A^2, \dots, A^r$ in the vector space $M_n(K)$, from the proposition you have $\deg \mu_K \leq \deg \mu_F$. On the other hand it is clear that $\mu_F$ divides $\mu_K$. So $\mu_F = \mu_K$. - As Andrea explained, the statement in the question results immediately from the following one. Let $K$ be a subfield of a field $L$, let $A$ be an $m$ by $n$ matrix with coefficients in $K$, and assume that the equation $Ax=0$ has a nonzero solution in $L^n$. Then it has a nonzero solution in $K^n$. But this is obvious, because the algorithm giving such a solution (or its absence) depends only on the field generated by the coefficients of $A$. - This looks correct. Another way to see it is that you can find the minimal polynomial of the matrix by computing the invariant factors of the matrix $A-XId$ over $\mathbb{R}$. Since the same process (with same operations) may be done over $\mathbb{C}$, their minimal polynomial is the same. sorry, i don't know the english word for the "invariant factors", i mean the process that using only row and columns operations, the matrix $A-XId$ may be uniquely writtten as some zero and a sequence of polynomial in the diagonal in which any polynomial divides the next one, and where the first is the minimal polynomial $A$ and the last the characteristic polynomial of $A$. - Don't apologise, I'm having trouble with English as well! Since Arturo posted what seems like a more straightforward proof (well, it's the one I thought of...), I've accepted his answer, but thank you for your input and I will consider your idea. –  iroiroaru Sep 21 '11 at 19:07

https://math.stackexchange.com/questions/769366/how-many-non-empty-subsets-of-1-2-n-satisfy-that-the-sum-of-their-eleme

# How many non empty subsets of {1, 2, …, n} satisfy that the sum of their elements is even? The question I am working on is the case for $n$ = 9. How many non-empty subsets of $\{1,2,...,9\}$ have that the sum of their elements is even? My solution is that the sum of elements is even if and only if the subset contains an even number of odd numbers. Since this is precisely half of all of the subsets the answer is $\frac{2^{9}}{2}=2^8$. Then the question specifies non-empty so final answer is $2^8-1$. Is this correct? In general I guess the solutions is $2^{n}-1$. My problem is why do exactly half of the total amount of subsets have and even number of odd numbers? Can we set up a bijection between subsets with odd number of odd numbers and even number of odd numbers? Let $S$ be a subset of $\{0,1,2,\dots,9\}$, possibly empty. Note that $1+2+\cdots +9=45$. So the sum of the elements of $S$ is even if and only if the sum of the elements of the complement of $S$ is odd. Divide the subsets of $\{1,2,\dots,9\}$ into complementary pairs. There are $2^8$ such pairs, and exactly one element of each pair has even sum. Thus there are $2^8$ subsets with even sum, and $2^8-1$ if we exclude the empty set. Remark: Suppose that $1+2+\cdots+n$ is odd. This is the case when $n\equiv 1\pmod{4}$ and when $n\equiv 2\pmod{4}$. Then the same argument shows that there are $2^{n-1}$ subsets with even sum. We can use another argument for the general case. Note that there are just as many subsets of $\{1,2,\dots,n\}$ that contain $1$ as there are subsets that do not contain $1$. And for any subset of $A$ of $\{2,3,\dots,n\}$, we have that $A$ has even sum if and only if $A\cup\{1\}$ has odd sum, and $A$ has odd sum if and only if $A\cup\{1\}$ has even sum. Thus in general there are $2^{n-1}$ subsets with even sum. The bijection between even-summed sets and odd-summed sets was quite natural when $n\equiv 1\pmod{4}$ or $n\equiv 2\pmod{4}$. In the general case, there is a nice bijection (add or subtract $\{1\}$), but it is less natural. Let's first count all subsets of $\{1,\ldots,n\}$ with even sum. Removing the empty sets then makes us have to subtract one from this result. The subsets of $\{1,\ldots,n\}$ with even sum are one-to-one with the subsets of $\{2,\ldots,n\}$. For any set $J\subset\{2,\ldots,n\}$, if the sum of $J$ is even, then $J$ is a subset of $\{1,\ldots,n\}$ with even sum, while if the sum of $J$ is odd, then $\{1\}\cup J$ is a subset with even sum. Since there are $2^{n-1}$ subsets of $\{2,\ldots,n\}$, this is the number of subsets of $\{1,\ldots,n\}$ with even sum. Remove the empty set, and you get $2^{n-1}-1$. What you did is fine, we can get an alternative proof if we recall how we prove that there are $2^{n-1}$ subsets of $\{1,2\dots n\}$ of even cardinality. Let $E$ be the set of subsets of even cardinity and let $O$ be the set of subsets of odd cardinality, pick an arbitrary element $a\in\{1,2,3\dots n\}$. Then $f:E\rightarrow O$ defined as $X\mapsto \{a\}\Delta X$ is a bijection right? Well, if $E'$ is the set of subsets with even sum and $O'$ is the set of subsets with odd sum and $a\in\{1,2,3\dots n\}$ is odd. $f:E'\rightarrow O'$ defined as $X\mapsto \{a\}\Delta X$ is also a bijection. So in any finite subset $A$ of positive integers, exactly half of the subsets have even sum, unless all of the elements of $A$ are even, in which case all the subsets clearly have even sum. $\Delta$ is just the symmetric difference of sets, so $\{a\}\Delta X$ is $\{a\}\cup X$ if $a$ was not in $X$ and is $X\setminus\{a\}$ if $a$ was in $X$.

https://math.stackexchange.com/questions/2886048/two-alternate-proofs-that-x-neq-0-wedge-xy-xz-implies-y-z

# Two Alternate Proofs that $x \neq 0 \wedge xy = xz \implies y = z$. I believe I have been able to construct in two ways, using the field axioms, that if $x \neq 0$ and $xy = xz$, then $y = z$. However, I've seen similar proofs like this assume that we can perform arithmetic operations, such as multiplying both sides by an inverse--which mirrors in some sense some proofs I've written in an abstract-algebra context--whereas others are more 'purist' in this sense. The similar proof in Rudin, for example, does not assume that we can use simple arithmetic. My question, then, is which of these is 'more' standard in a first-year analysis course? Proof 1: Assuming I can use arithmetic . Since $x \neq 0$, $\exists x^{-1}$ s.t. $xx^{-1} = x^{-1} x = 1$ by the field axioms. Therefore, \begin{align*} xy = xz & & \text{By assumption} \\ x^{-1} (xy) = x^{-1} (xz) & & \text{Multiply on left by $x^{-1}$} \\ \left(x^{-1} x\right)y = \left(x^{-1} x\right)z & & \text{Associativity} \\ 1y = 1z & & \text{Inverse properties} \\ y = z \end{align*} Example 2: Without assuming arithmetic, and mirroring Rudin. \begin{align*} y & = 1 \cdot y & & \text{Multiplicative identity} \\ & = \left(x \cdot \frac{1}{x}\right) y & & \text{Mult inverse axiom with $x \neq 0$} \\ & = \left(\frac{1}{x} \cdot x\right)y & & \text{Commutativity of multiplication} \\ & = \frac{1}{x} \left(x \cdot y\right) & & \text{Associativity of multiplication} \\ & = \frac{1}{x} \left(xz\right) & & \text{Assumption that $xy = xz$} \\ & = \left(\frac{1}{x} \cdot x\right) z & & \text{Associativity of multiplication} \\ & = 1z & & \text{Inverse properties} \\ & = z \end{align*} Thanks in advance. • If $K$ is a field, then $G=(K^{\times},\cdot)$ is an abelian group, so that the cancellation law holds. For $x,y\in G$ we have that $xy=xz$ implies that $y=z$. – Dietrich Burde Aug 17 '18 at 18:20 • In your "mirroring Rudin" example the proof is just one long chain of equal quantities: You want to show $y=z$ so you start with $y$ and write down expressions you know are equal to it using axioms and assumptions until you have a $z$. In your "arithmetic" proof, you have a list of equalities, and you use your axioms to transform them to get the claim you want: that $y=z$. I don't really think these proofs are different, and I expect people looking at your work would agree with me. But, if you want to make sure, I would suggest asking your grader/professor. – James Aug 17 '18 at 18:28 • Also, I think your worry about "using arithmetic" is illfounded. You have a claim such as $xy = xz$ in some structure you are reasoning about. You also have a binary function on that structure: multiplication. Therefore the quanties $x^{-1}(xy)$ and $x^{-1}(xz)$ are both defined because you are just plugging in elements of the domain into your function. That $y=z$ follows from the assumed properties of multiplication and the existance of inverses. – James Aug 17 '18 at 18:31 • One more thing. The only thing you must avoid in a proof of $y=z$ is starting with $y=z$ and deriving $0=0$ or $1=1$. As long as your proof starts with assumptions you are given, follows logically valid steps, and ends up with what you want, then the proof is good. Many of my students try to show $x=y$ and argue "$x=y$ ... <operations> ... $0=0$, QED". What is most frustrating is that often if they just turned the proof up-side-down, then it would be valid, i.e, the operations they effect on the equation can be done backwards to start with $0=0$ and derive $x=y$. – James Aug 17 '18 at 18:35 • Those two proofs are exactly the same as far as I can tell. Or aren't significantly different. "Assuming arithmatic" is a meaningless thing to say. To prove this we must have a well defined set of axioms. "Assuming arithmetic" is simply referring to them. – fleablood Aug 17 '18 at 18:37 The two proofs are essentially the same and the first doesn't use arithmetic, but rather field axioms. I wouldn't use $\frac{1}{x}$, but that's more cosmetic than substantial. More substantial is that you don't need to appeal to commutativity: \begin{align} y &=1y &&\text{(multiplicative identity)} \\ &=(x^{-1}x)y &&\text{($x\ne0$ has an inverse)} \\ &=x^{-1}(xy) &&\text{(associativity)} \\ &=x^{-1}(xz) &&\text{(hypothesis)} \\ &=(x^{-1}x)z &&\text{(associativity)}\\ &=1z &&\text{(property of the inverse)} \\ &=z &&\text{(multiplicative identity)} \end{align} On the other hand, the other proof seems shorter \begin{align} & xy=xz &&\text{(hypothesis)} \\ & x^{-1}(xy)=x^{-1}(xz) && \text{($x\ne0$ has an inverse)} \\ & (x^{-1}x)y=(x^{-1}x)z && \text{(associativity)} \\ & 1y=1z && \text{(property of the inverse)} \\ & y=z && \text{(multiplicative identity)} \end{align} and less “rabbit out of a top hat”.

https://physics.stackexchange.com/questions/701918/does-this-question-have-two-answers-correct

Does this question have two answers correct? A simple pendulum (whose length is less than that of a second's pendulum) and a second's pendulum start swinging in phase. They again swing in phase after an interval of $$18$$ seconds from the start. The period of the simple pendulum is (A) $$0.9$$ sec (B) $$1.8$$ sec (C) $$2.7$$ sec (D) $$3.6$$ sec I was given a formula for such questions: $$T = \frac {T_1 T_2} {T_1-T_2} \qquad (T_1>T_2)$$ where $$T_1$$ and $$T_2$$ are the time periods of the individual pendulums, and $$T$$ is the time after which they are in phase again. I took $$T_1$$ as the seconds pendulum, i.e., $$T_1=2$$ seconds. Using the formula, I got $$T_2=1.8$$ sec, which makes sense; the timestamps for each oscillation are: $$1.8\ \ 3.6\ \ 5.4\ \ 7.2\ \ 9.0\ \ 10.8\ \ 12.6\ \ 14.4\ \ 16.8\ \ 18.0$$ seconds for simple pendulum, and $$2, 4, 6, 8, 10, 12, 14, 16, 18$$ seconds for seconds pendulum. None of these overlap, so if $$T_2=1.8$$, the pendulums swing in phase after intervals of $$18$$ seconds. However, I also tried option A, and got the timestamps as: $$0.9\ \ 1.8\ \ 2.7\ \ 3.6\ \ 4.5\ \ 5.4\ \ 6.3\ \ 7.2\ \ 8.1\ \ 9.0\ \ 9.9\ \ 10.8\ \ 11.7\ \ 12.6\ \ 13.5\ \ 14.4\ \ 15.3\ \ 16.2\ \ 17.1\ \ 18$$ seconds for simple pendulum, and $$2, 4, 6, 10, 12, 14, 16, 18$$ seconds for seconds pendulum. Again, none of these overlap, so if $$T_2=0.9$$ seconds also, the pendulums swing in phase after intervals of $$18$$ seconds. According to the answer key, the answer is only B. Is A also correct, or am I missing something? The key point that's overlooked in the timestamp-counting method is that having the pendulums be in sync at the end of complete periods is not the only way for them to be in phase - they can also happen to be in phase in the middle of a period. In particular, for this example, note that after $$\frac{18}{11}$$ seconds, the $$0.9$$-second-period pendulum and the $$2$$-second-period pendulum will be $$\frac{9}{11}$$ of the way through a period (try dividing $$\frac{18}{11}$$ seconds by each of their periods and verify for yourself). By looking only at timestamps of complete periods, the timestamp-counting method misses out this point (earlier than $$18$$ seconds) where they came back in phase. I'd highlight that this means care is needed to derive the $$\frac{T_1 T_2}{T_1 - T_2}$$ formula - for example, it's not enough to just solve for the times when the pendulums have the same (angular) position, because there are many earlier times where this happens, but requiring that the pendulums are in phase is a much stronger condition. Also, one has to explicitly use the fact that we are interested in the first time they are back in phase, because it's true that the $$0.9$$-second-period pendulum and the $$2$$-second-period pendulum are in phase after $$18$$ seconds - the tricky thing is that there was an earlier time where they were already in phase. Basically, the correct way to derive that formula would be to say we are solving for the earliest time $$t$$ such that the difference between $$t/T_1$$ and $$t/T_2$$ is an integer. For the explicit derivation: the pendulums are in phase at time $$t$$ if and only if $$t/T_1 - t/T_2 = n$$ for some integer $$n$$. Solving for $$t$$ yields $$t = n \frac{T_1 T_2}{T_1 - T_2},$$ and hence we see that they are in phase whenever $$t$$ is an integer multiple of $$\frac{T_1 T_2}{T_1 - T_2}$$ (to restrict to positive $$t$$, take $$T_1 > T_2$$ and $$n>0$$ without loss of generality). In particular, the first positive $$t$$ at which this occurs is clearly when $$n=1$$, i.e. $$t=\frac{T_1 T_2}{T_1 - T_2}$$ as claimed. • Yes, this is the key. The formula gives the time for the first time they are in phase. Any period which is a submultiple of 1.8 s will be in phase after 18 s but not for the first time. – nasu Apr 3 at 20:41 Good work, but your issue is that you're taking too narrow a view of what "swinging in phase" means. To be in phase, the two pendulums simply need to be at the same point in their cycle -- meaning at the same angle and swinging in same direction. What you're doing with your timestamps approach is to identify only those instants when the two pendulums have returned exactly to their starting position at the same time. And they will definitely be in phase when they are both back at their starting positions at the same time (since the problem specified that they started off in phase), but the trick is that they could also be in phase at points before that as well. Imagine a pendulum with a period of 10 seconds, and one with a period of only 1 second: • After 1 second, P2 will have returned to its starting position, while P1 will have only traveled through 10% of its 10-second cycle. So P2 is about to catch up to P1 -- they're about to have another moment when they're in phase again (with P2 basically doing all the work). • Another 0.1 seconds after that (1.1s total), P2 will have gone from back to its starting point to 10% through its 1-second cycle, while P1 will only be 11% through its cycle • Another 0.01 seconds after that (1.11s total), P2 will be 11% through its cycle, while P1 is 11.1% through its cycle. • You can see where this is headed -- P2 will "catch" P1 for the first time at 1.111111...s (aka 10/9 seconds). You can validate that from your formula: T2*T1/(T2-T1) = 10*1/(10-1) = 10/9 = 1.11111... So these two are going to be in phase every 10/9 seconds. But they're not going to be at their starting point when they go back into phase; the first time they're in phase will be 1/9 of the way through the cycle. Do you see how that's different from what you were looking at? You were only looking for points where the two pendulums have done complete cycles and checked to see if they're in phase. Your method is equivalent to finding the smallest time period that is an integer multiple of both pendulums' periods. That will get them both in phase AND at their starting point, for the first time, but being at starting point isn't a necessary condition for being in phase. In my example, where the periods are 1s and 10s, the equivalent time (the first time they're both back in phase at the end of a complete cycle) is 10s (since for P1, 10*1 = 10 and for P2, 1*10=10). At that point, P1 has completed exactly one cycle, P2 has completed 10 cycles, and it's the 9th time (because 10/1.11... = 9) that they've been back in phase with each other. In your question, with P1 of 2s and P2 of 0.9s, the "beat frequency" (amount of time to return to phase) is = 2*0.9/(2-0.9) ~= 1.63 seconds. You correctly identify 18s as the least common multiple of the two periods. At 18s they'll both be back at their starting points and in phase, at the end of the 9th complete cycle for P1, the 20th complete cycle for P2 -- and it's the 11th time (18/1.63) that they've returned to phase with each other. With P1 of 2s and P2 of 1.8s, the "beat frequency" is 2*1.8/(2-1.8) = 18 seconds, AND the least common integer multiple of the two periods also happens to be 18s. At 18s, P1 will have completed 9 cycles, P2 will have completed 10 cycles, and it will be the first time they're back in phase together. You could argue the question is slightly ambiguous by saying "They again swing in phase after an interval of 18 seconds from the start" -- that is, it doesn't specify that "They, for the first time since the start, swing again in phase after an interval of 18 seconds from the start", but I think the "first time" part is pretty heavily implied. In your analysis with the time stamps, you deduced that either answers (a) and (b) are possibly correct. See also helloworld's answer on how you can deduce further which of these is correct by again looking closer at the phase relationship. But how you arrive at the correct answer lies in the not-so-obvious wording of the question. First, do you know what "a seconds pendulum" is? Your question reads: A simple pendulum (whose length is less than that of a second's pendulum) and a second's pendulum Note the words and a seconds pendulum. A seconds pendulum is a pendulum that has a period of precisely $$2$$ seconds. That is, one second per swing in one direction, or two seconds to complete a swing in both directions (one full period is two seconds)$$^1$$. This means with $$T_1=2\ \text{sec}$$ and with $$T=18\ \text{sec}$$ to retain an in-phase relationship, gives the only possible correct answer of $$T_2=1.8\ \text{sec}$$. The answer cannot possibly be 0.9 sec. Nor can it be the other two possibilities (c) and (d). $$^1$$From the wording of the question alone, you can deduce that the only possible answers are (a) and (b), since "the length is less than that of a second's pendulum" or less than 2 seconds, since we know that the period of a simple pendulum is proportional to $$l^{\frac 12}$$. And one could have even deduced the correct answer from this information alone. The equation you quoted is a transpose of the equation $$\frac 1T=\frac 1T_1-\frac 1T_2$$ and by adding $$\frac{1}{18}$$ and $$\frac 12$$ $$(=\frac{9}{18})$$ to each other and then just flipping the fraction to get 1.8 seconds. • I get why 1.8 seconds is an answer; I'm asking whether 0.9 seconds is also an answer or not, and if not, why not? Apr 3 at 9:52 • It can't be the correct answer because you are already told that one pendulum has a period of two seconds, then you are asked to calculate the period of the other pendulum. It's not an answer, it's the only answer. Apr 3 at 9:54 • Could you please justify that without directly using the formula? On trying it manually, $0.9$ seconds seems to work for me Apr 3 at 10:09 • I agree this answer doesn’t explain why answer a. is invalid. Another answer clarifies succinctly. Apr 3 at 21:22 • But why can there be only one possible answer? An assertion is not the same as an explanation. Apr 3 at 23:33 Let's look at the equations $$x_1(t)=\sin\left(\frac{2\pi}{T_1}\,t\right)\\ x_2(t)=\sin\left(\frac{2\pi}{T_2}\,t\right)$$ for $$~t=T~$$ is $$~x_1(T)=0~$$ only if $$~\frac {T}{T_1}=1,2,\ldots n~\quad$$ and $$~x_2(T)=0~$$ only if $$~\frac {T}{a\,T_2}=1,2,\ldots n~$$ with $$~T=\frac{T_1\,T_2}{T_1-T_2}=\frac{2*1.8}{2-1.8}=18~$$ thus $$a=1~,T_2\mapsto 1.8\quad ,n=\frac{18}{1.8}=10~\surd\\ a=\frac 12~,T_2\mapsto \frac 12*1.8=0.9\quad ,n= \frac{18*2}{0.9}=40~\surd\\ a=2~,T_2\mapsto 2*1.8=3.6\quad ,n=\frac{18}{3.6}=5~\surd$$ I don't think that you need the formula $$~T=\frac{T_1\,T_2}{T_1-T_2}~$$ . for a given period $$~T,~\frac {T}{T_1}~$$ must be integer. the period $$~T_2=\frac{T}{n}~$$ and for $$~T_2 \le T_1=2\quad \Rightarrow n\gt \frac T2$$ Example $$T=18~,n\gt 9\\ T_2=\left[\frac 95,{\frac {18}{11}},\frac 32,{\frac {18}{13}},{\frac {9}{7}},\frac 65,{\frac { 9}{8}}\,\ldots\right]$$ My goodness the other answers are woefully overcomplicating this. The question says A simple pendulum (whose length is less than that of a second's pendulum) The length of the pendulum is less than that of a second's pendulum. Therefore it will swing faster, therefore it will have a shorter period than $$1s$$. Only one answer has a period less than $$1s$$. More generally, equate $$18$$ swings of the 'second's pendulum' with some integer number of swings of the other pendulum: $$18 s = n T$$ It is generally assumed, by the wording, that they were in sync only after $$18s$$ and no sooner than that. In this case, it must be that $$\gcd(18,n) = 1.$$ You're also told that the simple pendulum is shorter than the second's pendulum. Therefore it swings faster, with shorter period, and $$n>18$$. Therefore, $$n>18$$ with $$\gcd(18,n) = 1$$. The smallest valid $$n$$ is $$19$$, which implies a period of $$T = \frac{18s}{n} = \frac{18s}{19} \approx 0.947s \approx 0.9s$$ In principle the answer could also be $$n=23, 25, 29, \dots$$ . However, the question only offers one choice for $$n>18$$, i.e. $$T<1s$$, so that's the only choice. You might say that the 'second's pendulum' has a period of $$2s$$. Unless the question specifically states that a second's pendulum has a period of $$2s$$, or if you were taught this specific fact, I think it is a coincidence that the term 'second's pendulum' refers to a pendulum with a specific period of $$2s$$; common parlance would assume it would have a period of $$1s$$. If you insist on the $$2s$$, then we have $$18 \times 2s = nT$$ with $$\gcd(18,n) = 1$$ and $$n>18$$. The answers are the same, except multiplied by $$2$$. For $$n=19$$ we have $$T \approx 1.89s \not \approx 1.8s$$ For $$n=23$$ we have $$T \approx 1.57s$$ ...and subsequent choices will have a lower period $$T$$. There is one correct choice, which is $$n=41$$ which gives $$T \approx 0.88s \approx 0.9s$$ so, the answer is again A. The only correct answer is A. • No. The answer is subtle. A "seconds pendulum" has a period of two seconds, meaning either (a) or (b) are correct. And the correct answer is (b). See my answer above for a further explanation. Cheers. Apr 5 at 2:33 • @josephh I addressed this. Apr 5 at 2:35 • Did you look at this link where the seconds pendulum is explained? "Unless the question specifically states that a second's pendulum has a period of 2s" It does by mentioning it is. BTW 2 seconds means time for two full swings. Cheers. Apr 5 at 2:38 • @josephh I addressed this. Read my answer before commenting please. Using $2s$ gives the same answer $A$ anyway. Apr 5 at 2:42 • Under the interpretation that the seconds pendulum has a 1s period, observe that a T1=0.9s pendulum and T2=1s pendulum are already in phase at 9s, so that isn't the right answer. Under the interpretation that it has a 2s period, a T1=0.9s pendulum and T2=2s pendulum are already in phase at 18/11s, so that's also not correct. Apr 8 at 1:22

https://wiki.documentfoundation.org/Documentation/Calc_Functions/MOD

# Documentation/Calc Functions/MOD Other languages: English • ‎Nederlands • ‎dansk • ‎español • ‎עברית MOD Mathematical ## Summary: Calculates the remainder when one number (the dividend or numerator) is divided by another number (the divisor or denominator). This is known as the modulo operation. Often the dividend and divisor will be integer values (Euclidean division). However, MOD accepts and processes real numbers with non-zero fractional parts. ## Syntax: MOD(Dividend; Divisor) ## Returns: Returns a real number that is the remainder when one number is divided by another number. The value returned has the same sign as the divisor. ## Arguments: Dividend is a real number, or a reference to a cell containing that number, that is the dividend of the divide operation. Divisor is a real number, or a reference to a cell containing that number, that is the divisor of the divide operation. • If either Dividend or Divisor is non-numeric, then MOD reports a #VALUE! error. • If Divisor is equal to 0, then MOD reports a #DIV/0! error. • For real x and y (y <>0), MOD implements the following formula: $\displaystyle{ \text{MOD}(x,y)~=~x-\left(y\times \text{INT} \left(\frac{x}{y}\right)\right) }$ The INT function always rounds down (toward -∞) and returns the largest integer less than or equal to a given number. This means that when the dividend and divisor have different signs, MOD may produce results that appear counterintuitive. For example, the formula =MOD(7, -3) returns -2; this is because the fraction $\displaystyle{ \left(\frac{7}{-3}\right) }$ is rounded to -3 by the INT function. • For more general information about the modulo operation, visit Wikipedia’s Modulo operation page. ## Examples: Formula Description Returns =MOD(11; 3) Here the function returns the remainder when 11 is divided by 3. The returned value has the same sign as the divisor, which is positive in this example. 2 =MOD(-11; 3) Here the function returns the remainder when -11 is divided by 3. The returned value has the same sign as the divisor, which is positive in this example. Note the counterintuitive value produced when the dividend and divisor have different signs. 1 =MOD(11; -3) Here the function returns the remainder when 11 is divided by -3. The returned value has the same sign as the divisor, which is negative in this example. Note the counterintuitive value produced when the dividend and divisor have different signs. -1 =MOD(-11; -3) Here the function returns the remainder when -11 is divided by -3. The returned value has the same sign as the divisor, which is negative in this example. -2 =MOD(D1; D2) where cells D1 and D2 contain the numbers 11.25 and 2.5 respectively. Here the function returns the remainder when 11.25 is divided by 2.5. The returned value has the same sign as the divisor, which is positive in this example. 1.25 MOD

https://niejeden.pl/sub-zero-iduju/450445-properties-of-matrix-addition

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

https://math.stackexchange.com/questions/2280376/find-the-sum-of-a-sequence/2280381

# Find the sum of a sequence [duplicate] Based on: $\frac{1}{n*(n+1)}=\frac{1}{n}-\frac{1}{n+1}$ where n is element of N find the sum of the following: $\frac{1}{1*2}+\frac{1}{2*3}+\frac{1}{3*4}+ ... +\frac{1}{38*39}+\frac{1}{39*40}$ How should one deal with this kind of problem? Is this a mathematical induction, arithmetic series, geometric series? I'm lost on this one. Here are the options: a)$\frac{31}{40}$ b)$\frac{33}{40}$ c)$\frac{37}{40}$ d)$\frac{39}{40}$ ## marked as duplicate by Did sequences-and-series StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 14 '17 at 11:56 • It's telescoping. The required decomposition has already been provided; what you then find is that terms cancel out everywhere. – Parcly Taxel May 14 '17 at 11:39 • Try to expand a bit and observe the telescoping : $$\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3\cdot 4} = (\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} -\frac{1}{4})$$ – Zubzub May 14 '17 at 11:44 $\frac{1}{1*2}+\frac{1}{2*3}+ ... +\frac{1}{38*39}+\frac{1}{39*40}=\\ (\frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+...+(\frac{1}{38}-\frac{1}{39})+(\frac{1}{39}-\frac{1}{40})=\\ \frac{1}{1}+(-\frac{1}{2}+\frac{1}{2})-\frac{1}{3}+...+\frac{1}{38}+(-\frac{1}{39}+\frac{1}{39})-\frac{1}{40}=\\ \frac{1}{1}-\frac{1}{40}$ Can you see that all terms except for the first and last term cancel? Since each and every term will be canceled out except $1$ and $-\dfrac{1}{40}$ So, the answer is $$1-\frac{1}{40}=\frac{39}{40}$$

http://bootmath.com/is-this-induction-procedure-correct-2nn.html

# 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. #### Solutions Collecting From Web of "Is this induction procedure correct? ($2^n<n!$)" 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://math.stackexchange.com/questions/801505/shall-remainder-always-be-positive

Shall remainder always be positive? My cousin in grade 10, was told by his teacher that remainders are never negative. In a specific example, $$-48\mod{5} = 2$$ I kinda agree. But my grandpa insists that $$-48 \mod{5} = -3$$ Which is true? Why? - $2$ and $-3$ are just two names for the same element in $\mathbb{Z}_5$, i.e. $2 \equiv -3 \mod 5$ – mm-aops May 19 '14 at 13:04 There are various conventions in use, e.g. see the links in this answer. – Bill Dubuque May 19 '14 at 13:08 Python and Ruby say it's 2; C, C#, and Java say it's -3. You could also say it's -8, 7, 12, etc. All are correct. The question is, which do you want? – Tim S. May 19 '14 at 14:00 Who says that remainder and modulus are the same thing? I was taught differently, and the people who developed the Ada programming language apparently thought so, as well. – O. R. Mapper May 19 '14 at 17:36 @John: C++ specification explicitly says (§5.6/4) that / must round towards zero and (a/b)*b + a%b must be equal to a. Which has just one solution, namely that a%b has the same sign as a (unless it is 0). Nothing implementation-defined here. – Jan Hudec May 20 '14 at 21:56 The first one is saying that $-48$ is $2$ more than a multiple of $5$. This is true. The second one is saying that $-48$ is $3$ less than a multiple of $5$. This is also true. - Who says short answers are short? – Awal Garg May 20 '14 at 11:42 @AwalGarg I do. Axiomatically. – Cruncher May 20 '14 at 14:10 But what about the division? Would the division with module 2 be -48/5 = 10 remainder 2 ? – Pieter B May 21 '14 at 10:09 @PieterB -48/5 = -10 remainder 2 if you like, but it could also equal -9 remainder -3. – Jack M May 21 '14 at 10:14 The teacher is within his/her authority to define remainders to be numbers $r$, $0\leq r < d$ where $d$ is the divisor. This is just a systematic choice so that all students can apply the same rule and arrive at the same anwer, but the rule is only a convention, not a "truth." The teacher isn't wrong to define remainders that way, but it would be wrong for the teacher to insist that there is no other way to define a remainder. And, anyhow, what your grandpa says is also perfectly true :) I imagine that ordinary long division is taught with this remainder rule because it makes converting fractions to decimals smoother when students do it later. Another reason is probably that mixed fraction notation (as far as I know) makes no allowance for negatives in the fraction. What I mean is that $1+\frac23=2-\frac13$, but the mixed fraction $1\frac23$ is not usually written as $2\frac{-1}{3}$, although one could make an argument that it makes just as much sense. As far as modular arithmetic is concerned, you really want to have the flexibility to switch between these numbers, and insisting on doing computations with the positive version all the time would be hamstringing yourself. Consider the problem of computing $1445^{99}\pmod{1446}$. It should not be necessary to compute powers and remainders of powers of $1445$ when you can just note that $1445=-1\pmod{1446}$, and then $1445^{99}=(-1)^{99}=-1=1445\pmod{1446}$ - But $(-48) / 5$ as a mixed numeral is $-9 \frac 35$ meaning $-(9 + \frac 35) = -9 - \frac 35$, which lines up more with Grandpa's way than with Teacher's way. – aschepler May 20 '14 at 19:48 The answer depends on whether you want to talk about modular arithmetic or remainders. These two perspectives are closely related, but different. In modular arithmetic, $2\equiv -3 \pmod 5$, so both answers are correct. This is the perspective most answers here have taken. In the division algorithm, though, where remainders are defined, in order to guarantee uniqueness, you need a specific range for the remainder -- when dividing integer $a$ by integer $d$, you get $a=qd+r$, and the integers $q$ and $r$ are unique if $0\leq r<d$ (note that this also places a restriction that $d$ be positive). There are other ways you could set up the condition, but you need some similar range in order to guarantee uniqueness, and this range is the simplest and most common, so in this sense, a negative remainder doesn't work. - Excellent answer. The point is that we often want uniqueness of $q$ and $r$, which is guaranteed by requiring $0\leq r < d$. Without this condition, there are $infinitely$ many possible "remainders". – ChocolateAndCheese May 19 '14 at 19:57 @Chocolate: And it's worth noting that other normalizations are possible: sometimes you want $d/2 \leq r < d/2$. Other settings want $0 \leq |r| < d$, but for $rd \geq 0$. Or maybe $rn \geq 0$ or even $rnd \geq 0$. – Hurkyl May 20 '14 at 8:59 We indeed often want uniqueness of $q$ and $r$. But long division algorithm is only practical if $q$ is rounded towards zero (truncated) and that leads to negative (non-positive) remainder for negative numerator. But Euclidean division defines remainder as positive (non-negative). So either definition is sometimes needed with division as well. – Jan Hudec May 21 '14 at 4:57 When -48 is divided by 5 the division algorithm tells us that there is a "unique" reminder r satisfying $0\leq r<5$. In that case there is only one possibility, namely $r=2$. - The division algorithm is defined on natural numbers, isn't it? Or I may be mistaken – Cheeku May 19 '14 at 13:30 I think it's usually defined for any integer numerator and any positive integer denominator. – poolpt May 19 '14 at 15:32 Negative denominator is not a problem. But long division needs negative remainders when numerator is negative. – Jan Hudec May 21 '14 at 4:52 Here is a different perspective which is more technical but is also reflected e.g. in Hardy and Wright. On the whole it is best to regard the expression $"-48 \mod 5"$ on its own as representing the set of numbers $a:a\equiv -48 \mod 5$, so it isn't equal to a number at all, but to a set. [Technically it is a coset in $\mathbb Z$ of the ideal generated by $5$, and consists of the numbers $-48 + 5 b$, where $b$ is an arbitrary integer]. Quite often it is useful to work with numbers rather than sets, and we choose a representative element of the set to work with. There are different ways in which this can be done (least positive, smallest absolute value etc). In this case $2$ and $-3$ are members of the set and could be used to represent it. - Maybe the teacher was thinking about Euclidean division, which states: Given two integers a and b, with b ≠ 0, there exist unique integers q and r such that a = bq + r and 0 ≤ r < |b|, where |b| denotes the absolute value of b We indeed have the remainder ("r") being positive in this case, but if we talk strictly about congruences the grandpa's expression as well as the teacher's are both true. - For integers $a,b,c$ the following statements are equivalent: 1) $a\equiv b\text{ mod }c$ 2) $a+c\mathbb{Z}=b+c\mathbb{Z}$ 3) $c\mid a-b$ Note that $5\mid-48-2=-50$ and $5\mid-48-\left(-3\right)=-45$. - By convention, for a division $\frac{a}{b}=c$, if we are looking for a whole integer for $c$ (with no further stipulations) , the rounding method is towards zero. If $a<0$ and $b>0$, this means that in the equation $\frac{a}{b}=c+\frac{r}{b}$, $r$ must be $\leq 0$. Also remainders and modulus are two different things. - This question touches parts of my old message Algebraic abstractions related to (big) integers in C++ on the boost developers mailing list. Why should "modulo" be identical to "remainder"? Having a "remainder" function such that "r=remainder(a,m)" satisfies "0 <= r < m" is something convenient to have in a programming language. The quoted message presents some evidence that a "sremainder" function such that "s=sremainder(a,m)" satisfies "-m/2 <= s < m/2" would also be a good idea. The meaning of modulo and modular arithmetic is not directly addressed by these considerations. - For variety, I want to point out that when doing division with remainder (as opposed to, e.g., modular arithmetic), there is sometimes utility in having "improper" results; e.g. saying $17 / 5$ is $2$ with remainder $7$. An example where this would be useful is if you just need a value that is close to the correct quotient, and you can compute something close (and the appropriate remainder) relatively more easily. -

https://math.stackexchange.com/questions/3301854/finding-int-sec2-x-tan-x-dx-i-get-frac12-sec2xc-but-an-online-ca

# Finding $\int \sec^2 x \tan x \, dx$, I get $\frac12\sec^2x+C$, but an online calculator gets $\frac12\tan^2x+C$. I tried to find a generic antiderivative for $$\displaystyle \int \sec^2x \tan x \mathop{dx}$$ but I think there is something wrong with my solution because it doesn't match what I got through an online calculator. What am I doing wrong? Below is my solution. We will use substitution: $$u = \sec x \qquad du = \sec x \tan x \, dx$$ We substitute and apply the power rule: $$\int (\sec x) (\sec x \tan x \, dx) = \int u \, du = \frac{1}{2} u^2 + C = \frac{\sec^2x}{2} + C$$ The solution I found with the online calculator is: $$\frac{\tan ^2 x}{2} + C$$ The steps in the online solution make sense also, so I'm not sure what's going on. The one thing I have some doubts about is whether I derived the $$du$$ from $$u = \sec x$$ correctly. But it seems okay to me. I used implicit differentiation with $$x$$. • $\sec^2{(x)}=1+\tan^2{(x)}=\tan^2{(x)}+C$ hence these functions differ by a constant... – Peter Foreman Jul 23 at 18:23 • What happens if you try $u=\tan \theta?$ – Chris Leary Jul 23 at 18:25 $$\frac{\sec^{2}(x)}{2} + C = \frac{1+\tan^{2}(x)}{2} + C = \frac{1}{2} + \frac{\tan^{2}(x)}{2} + C$$ $$\frac{1}{2}$$ is just another constant, in indefinite integral constant doesn't really matter unless you're asked for the integrand original function so you can just kind of "combine" $$\frac{1}{2}$$ into C, you can also differentiate the answer to know that there's nothing wrong at all with your answer The difference between your answer and the answer given is that they differ by a constant value. You can see this by using the identity $$1+\tan^2(x) = \sec^2(x)$$ Hence your answer can be converted to the given answer by subtracting $$-1/2$$, which is a constant. As mentioned in the comments and in another answer you can also directly get the form in the answer by using the substitution $$u = \tan x$$ • I modified your answer a bit to use MathJax. Going forward, please use MathJax for mathematical typesetting for ease of readability. Good answer though! – Cameron Williams Jul 23 at 18:29 • Thanks for editing the expressions – StackUpPhysics Jul 23 at 18:29 Try differentiating both of them and see that nothing has gone wrong at all. Let $$u = \tan x$$. Then $$du = \sec^{2} x dx$$. Hence, the integral becomes $$\int u du = \frac{u^{2}}{2} + C = \frac{\tan^{2}x}{2} + C.$$

https://math.stackexchange.com/questions/2764985/intuition-for-why-the-difference-between-frac2x2-xx2-x1-and-fracx-2/2765019

# Intuition for why the difference between $\frac{2x^2-x}{x^2-x+1}$ and $\frac{x-2}{x^2-x+1}$ is a constant? Why is the difference between these two functions a constant? $$f(x)=\frac{2x^2-x}{x^2-x+1}$$ $$g(x)=\frac{x-2}{x^2-x+1}$$ Since the denominators are equal and the numerators differ in degree I would never have thought the difference of these functions would be a constant. Of course I can calculate it is true: the difference is $2$, but my intuition is still completely off here. So, who can provide some intuitive explanation of what is going on here? Perhaps using a graph of some kind that shows what's special in this particular case? Thanks! BACKGROUND: The background of this question is that I tried to find this integral: $$\int\frac{x dx}{(x^2-x+1)^2}$$ As a solution I found: $$\frac{2}{3\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{2x^2-x}{3\left(x^2-x+1\right)}+C$$ Whereas my calculusbook gave as the solution: $$\frac{2}{3\sqrt{3}}\arctan\left(\frac{2x-1}{\sqrt{3}}\right)+\frac{x-2}{3\left(x^2-x+1\right)}+C$$ I thought I made a mistake but as it turned out, their difference was constant, so both are valid solutions. • Did you actually graph both functions? That would tell a lot May 3, 2018 at 14:26 • @imranfat desmos.com/calculator/njhbjs54rv. Looks strangely like $2$ to me. Did you actually graph both functions? May 3, 2018 at 14:31 • @Oldboy Actually, it is. May 3, 2018 at 14:31 • It doesn't matter that the numerators have different degrees. What matters is that their difference is a multiple of the denominator. – user856 May 3, 2018 at 16:14 • Fascinating background: at face value, I would wager that a lot of instructors would mark your answer wrong without paying much attention to how you got to it - and that most students wouldn't know to ask for a review. It is not obvious that they are both valid. I'd love to hear from educators here how would they approach this. May 4, 2018 at 0:23 Would you be surprised that the difference of $\dfrac{2x^2+x+1}{x^2}$ and $\dfrac{x+1}{x^2}$ is $2$? • Not sure who is responsible for that "from review". This actually provides more of an answer to the question asked than the majority of the posts (though there are others that do better). May 3, 2018 at 18:14 • It absolutely does provide an answer to the question, and what's more it's a good answer. It gives a simpler example of the phenomenon which the OP requests intuition for, and the intuition is easier to obtain in this simpler example. May 3, 2018 at 18:27 • @JamesMartin: Indeed. How about: Would you be surprised that the difference of $\frac{1234}{5}$ and $\frac{234}{5}$ is $200$? May 4, 2018 at 11:26 • Yes, this is still surprising. It is roughly equally as surprising as the difference in the original question (to me anyways). May 4, 2018 at 21:21 • @Nova Often repeating a question with slightly different values IS the best answer. Countless times I have done this with students with great success, because it can relate a situation that the student already has intuition for to the one that they are stumped on. Maybe this answer doesn’t work for some, but that doesn’t inherently make it a bad answer. May 6, 2018 at 7:30 It is just a bit of clever disguise. Take any polynomial $p(x)$ with leading term $a_n x^n$. Now consider $$\frac{p(x)}{p(x)}$$ This is clearly the constant $1$ (except at zeroes of $p(x)$). Now separate the leading term: $$\frac{a_n x^n}{p(x)} + \frac{p(x) - a_n x^n}{p(x)}$$ and re-write to create the difference: $$\frac{a_n x^n}{p(x)} - \frac{a_n x^n - p(x)}{p(x)}$$ Obviously the same thing and hence obviously still $1$ but the first has a degree $n$ polynomial as its numerator and the second a degree $n - 1$ or less polynomial. Similarly, you could split $p(x)$ in many other ways. I'm not sure anyone is speaking to your observation that the two numerators have different degrees. Let's flip this around the other way: \begin{align*} \frac{x-2}{x^2-x+1} + 2 &= \frac{x-2}{x^2-x+1} + 2\frac{x^2-x+1}{x^2-x+1} \\ &= \frac{2x^2 -x}{x^2-x+1} \text{.} \end{align*} That is, we started with a thing having a linear numerator and added a constant to it. But when we brought the constant to have a common denominator, it picked up a degree two factor. Then the addition was forced to produce a degree two sum. To sum up, in the context of rational functions, when you add constants, you are adding polynomials having the degree of the denominator to the polynomials in the numerators. So constants effectively have "degree two in the numerator" in your example. • +1 for this > "Constants effectively have "degree two in the numerator" in your example." May 4, 2018 at 14:20 • This deserves a lot more upvotes for the "degree two in the numerator" note. It reconciles intuition to the raw result. May 5, 2018 at 15:09 • A further +1 for "To sum up". May 10, 2018 at 5:34 Look at it in reverse: Take a polynomial fraction and add it to a constant. The result will be a polynomial fraction, with the same denominator and a different polynomial as the numerator. This belongs to a specific set of questions "you cannot really answer to your students" if you are a teacher. The difference is $2=\frac{2(x^2-x+1)}{x^2-x+1}$. This is why this seems weird (but true). Even this seems weird: $$\frac{15}{7}-\frac{1}{7}=2$$ The numerators differ by 14 (not 2) but the denominators are equal. The best is to try and explain this to yourself why this perfectly fine. Note that in general given a rational function $$f(x)=\frac{p(x)}{q(x)}\implies g(x)=f(x)+c=\frac{p(x)+c\cdot q(x)}{q(x)}$$ and $\deg(p(x)+c\cdot q(x))\le \max\{\deg(p(x)),\deg(q(x))\}$. • Why? Where is "the difference"? May 5, 2018 at 5:59 • @RolazaroAzeveires It is just a generalized way to see that two rational function can differ for a constant c even if denominators are equal and the numerators differ in degree, that's exactly the point of the OP. – user May 5, 2018 at 6:44 • I think it is just a matter of preference, from my point of view this observation suffices to answer the OP completely and maybe also the upvoters have the same idea. There are many other answers here with concrete examples and discussion so the asker have a lot of points on view on that and you also are free to add your own answer according to your best interpretation. Thanks anyway for your advice and suggestions on that. Bye! – user May 6, 2018 at 11:13 $$(2x^2-x)-(x-2)=2x^2-2x+2$$ Hence$$f(x)-g(x)=2\bigg(\frac{x^2-x+1}{x^2-x+1}\bigg)=2(1)=2$$ In effect, this pair of equations is a very specific case where the numerator and denominator end up lining up, and thus you get a constant for all values. • What kind of an answer is this? He already knows that. May 3, 2018 at 14:31 Since the denominators are equal and the numerators differ in degree I would never have thought the difference of these functions would be a constant. When the numerators differ in degree, the difference between the numerators is a nontrivial polynomial that goes to infinity as $x\to\pm\infty$. However, nobody says that this polynomial cannot grow at the same rate as the (equal) denominators, in which case the ratio will be bounded (and in particular possibly constant). There's nothing in your observation "the denominators are equal and the numerators differ in degree" that link the denominator to either the numerators or their difference, so you have no reason to think the difference in numerators should dominate the growth of the entire fraction. • Eactly, i already wanted to provide an answer before seeing this, when $x->\infty$ the first function tends to $2$, the second tends to $0$, which are both constants and give out a cte difference, this doesn't apply when the numerator is bigger than the denom, if the denom is 1 the difference would be diverging but once you divide it by a polynomial of order 2, things change to converge. May 4, 2018 at 14:14 Maybe instead of integrating you need just differentiating. $$\frac{df}{dx}=\frac{(2x^2-x)(2x-1)-(x^2-x+1)(4x-1)}{denom^2}\\ =\frac{x^2-4x+1}{denom^2}$$ In the other hand: $$\frac{dg}{dx}=\frac{(x-2)(2x-1)-(x^2-x+1)}{denom^2}\\ \ =\frac{x^2-4x+1}{denom^2}=\frac{df}{dx}$$ See that both functions have same derivates, which means they differ by the same value from $x$ to $x'$ , this only means they have a same growth rate and they have same difference all along the range of x axis. • This is something OP himself did. He was solving an integral and found an alternate answer $g$. Since $f$ and $g$ are solutions to the same integral, they defer by a constant just like you demonstrate here. OP is asking for an intuition why the two rational functions defer by only a constant. May 6, 2018 at 6:03 • What do you intend to do with a primitive, why at all one think to ascend to an integral? one function may be integrated in an infinitely many ways but what's the point ? look in the op's post that he already found two functions completely different, you are missing the point. The intuition is clear, you shouldn't compare the numerator unconsiderably of the rest of the function, you should take it all into account, for example $f$ and $g$ are mutually excluding eachother in a derivate of the form $f+g$, but they are mutual-inclusive in $f/g$. Appearence is deceiving. May 6, 2018 at 12:46 You can construct a similar puzzle from the (simpler) observation that $$\frac{x}{x+1} + \frac{1}{x+1} = 1 .$$ Translating, that's the same as $$\frac{u-1}{u} + \frac{1}{u} = 1$$ which isn't surprising at all. • Yes, but in my example the degrees of the numerators are different, which caught me off guard May 3, 2018 at 14:34 • The degrees are different in my example too. But I agree that yours is at first surprising. Note: this kind of rational function identity is useful sometimes when calculus instructors want to make up problems that look harder than they are. May 3, 2018 at 14:35 Another way of stating what everyone has said: $2x^2 - x= 2(x^2-x+1) + x - 2$ $\Rightarrow 2x^2-x = x-2 \mod x^2-x+1$ Because if $P\ and\ Q\ are\ polynomials$ then, $$\deg \left( P\pm Q \right)\le max\left( \deg \left( P \right),\deg \left( Q \right) \right)$$ where $deg$ is the degree of the polynomial. See this link Here's another view on the problem: Let's look at the fraction with the higher degree: $$f(x)=\frac{2x^2-x}{x^2-x+1}$$ We note that the degree of the numerator is the same as the one of the denominator. Now whenever the numerator's degree is at least as large as the denominator's degree, we can do polynomial division. The result is a polynomial whose degree is the difference between the numerator's and the denominator's degree, and a rest whose degree is less than the denominator's degree. Now since numerator and denominator have the same degree, the difference of the degrees is zero. But a polynomial of degree zero is a constant. And the division rest will be of lesser degree. Let's do the polynomial division explicitly: $$\begin{array}{llccl} &(2x^2-\ \ x) & / & (x^2-x+1) &=& 2\\ -&\underline{(2x^2-2x+2)}\\ & \ \ \ \ \ \ \ \ \ \ \ \ \ x \ - 2 \end{array} %\begin{array}{rrrrcccl} %&(2x^2 & -x) & & / & (x^2-x+1) & = & 2\\ %-&(2x^2 & -2x & +2)\\ %\hline % & & x & -2 %\end{array}$$ You surely will recognize the rest as the numerator of $g(x)$ So, whenever you have a rational function whose numerator and denominator, after cancelling out, have the same non-zero degree, it is not only possible, but guaranteed that there exists another rational function with lesser degree in the numerator which differs from the original rational function only by a constant. • BTW, does anyone know how to make the horizontal line end at the end of the polynomial above? May 4, 2018 at 5:51 • @farruhota: That first option destroys the alignment between the terms, and therefore is worse than the too-long line. That second option defeats the whole purpose of writing the polynomial division in its standard form. May 4, 2018 at 6:48 • How about this: $\begin{array}{ll} &(2x^2-x) & / (x^2-x+1) = 2\\ -&\underline{(2x^2-2x+2)}\\ & x-2 \end{array}$ May 4, 2018 at 7:04 • @farruhota: Still doesn't correctly align the terms. May 4, 2018 at 7:10 • Oh well, the rest should be adjusted: $\begin{array}{ll} &(2x^2-\ \ x) & / (x^2-x+1) = 2\\ -&\underline{(2x^2-2x+2)}\\ & \ \ \ \ \ \ \ \ \ \ \ \ \ \ x \ - 2 \end{array}$ May 4, 2018 at 7:16 I feel that it's only natural that intuition would fail you in cases like these. If you wrote the two functions out as $$f(x)=\frac{2\left(x^2-x+1\right)+\left(x-2\right)}{x^2-x+1}$$ $$g(x)=\frac{x-2}{x^2-x+1}$$ it would be obvious that the difference between them was 2. This is the first thing you could trip over, the degree of the numerator don't actually differ by 1, they are the same. I guess if you made it a rule to divide whenever you have a rational function where the degree of the numerator is greater than or equal to the degree of the denominator until it isn't, you would get consistent results. (Kinda like how you learn waaay back how $\frac{13}{7}$ is an improper fraction, because it's actually bigger than a whole number, and that you should write it as $1\frac{6}{7}$. a function $f(x)=\frac{P(x)}{Q(x)}$ where the degree of $P(x)$ is greater than or equal to the degree of $Q(x)$ is called an improper rational function.) But, and this might be another thing that's bugging you about this question, you can't just do that! (As my algebra teacher used to exclaim in exasperation.) You introduce a new constraint: $x^2-x+1\neq0$. Which isn't relevant if x is real, but does change things up a bit if it s a complex number. This is probably what you get when you graph the two functions $f$ and $g$. But if you allow for complex values of $x$, it can also look like this: (Sans the vertical line at x=0.5, I don't know why that is happening, probably a limitation in the graphing engine on desmos.) Not so obvious that the difference between the two is 2 in this case... Sorry, I just realized this probably doesn't answer your question on how you can reconciliate this apparent discrepancy with your intuition, but I thought explaining why it may not be that clear-cut would help at least a little bit. Playing with the parameters used in the graph might help further your understanding, here's a link to the interactive graph, with a little more in depth graphical interpretation of why you might doubt what you logically "know" to be true. As for the equations I used in place of the given ones... You'll just have to trust my algebra (which you shouldn't in critical situations, but I think this is safe enough hehe). I did them on a legal pad, but I'll type them up if anyone insists. Since the denominator is the same , you subtract one numerator from the other. In this special case you find the resultant numerator is twice the denominator. Feels like that was much simpler than the rest of your background calculus. It must be very clear to you that $11/3$ and $2/3$ defer by an integer. Why is this so? Because in the world $$modulo 3$"$, 11 and 2 are but the same. In the same vein, we can say that in the world of $$modulo $x^2-x+1"$, $2x^2-x$ and $x-2$ are but the same because $2x^2-x=2(x^2-x+1)+x-2$. When we are in such a world, the same polynomial may appear in different forms with different "degrees". So in a world such as the ideal generated by $x^2-x+1$ (ie $<x^2-x+1>$ ), degree is no longer retained. One's intuition may fail here because one is not familiar with such a world where degree is not so well behaved. • The question is not about constants; it is about degree. $\frac{x^3+1}{x^2+2}$ and $\frac{-2x+1}{x^2+2}$ also defer by a constant (polynomial) May 6, 2018 at 6:11

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# Thread: [SOLVED] Finding Co-Ordinates of a Rectangle 1. ## [SOLVED] Finding Co-Ordinates of a Rectangle Here's a question from a past paper which I have successfully attempted. My question is regarding part (iii). I have successfully figured out the co-ordinates by the following method: Is my method correct, considering I did get the right answer? But is there another simpler method to do this which would save time during an exam. 2. The diagonals bisect one another. The midpoint of $\displaystyle \overline{AC}$ is ? 3. Mid of AC is (6,6), the diagnols do bisect each other at the mid but we don't have the x-co-ordinates of B or D to equate the diagnols, or do we? 4. Originally Posted by unstopabl3 Mid of AC is (6,6), the diagnols do bisect each other at the mid but we don't have the x-co-ordinates of B or D to equate the diagnols, or do we? But the y-coordinate of $\displaystyle B~\&~D$ is 6. You are given the x-coordinate of $\displaystyle D$, so $\displaystyle Dh,6)$ 5. No, I meant real value of the x-co-ordinates has not been given since we have to calculate that ourselves. I have already gotten the correct values by using the method mentioned in my first post. As stated I want someone to solve this part of the question with a different, possibly easier method. I am not after the answer, I am looking for an alternate method. 6. Originally Posted by unstopabl3 No, I meant real value of the x-co-ordinates has not been given since we have to calculate that ourselves. I have already gotten the correct values by using the method mentioned in my first post. As stated I want someone to solve this part of the question with a different, possibly easier method. I am not after the answer, I am looking for an alternate method. Hi unstopabl3, I really don't see what method you used in your first post, but here's how I'd do it. Plato already told you that the midpoint of BD is M(6, 6). This means the y-coordinates of B and D are also 6. This distance from B to D is 20 (found using distance formula) Each individual segment of the diagonals measure 10 since they are bisected. Using the distance formula, it is easy to determine the x-coordinates of B and D. M(6, 6) -----> D(h, 6) = 10 $\displaystyle 10=\sqrt{h-6)^2+(6-6)^2}$ 7. Hello, unstopabl3! Code: | C(12,14) | o | * * | * * * * B o - + - - - - - - - o D * | * ------*-+-------*------------ *| * o A|(0,-2) | The diagram shows a rectangle $\displaystyle ABCD.$ We have: .$\displaystyle A(0,-2),\;C(12,14)$ The diagonal $\displaystyle BD$ is parallel to the $\displaystyle x$-axis. $\displaystyle (i)$ Explain why the $\displaystyle y$-coordinate of $\displaystyle D$ is 6. The diagonals of a rectangle bisect each other. . . Hence, the midpoint of $\displaystyle AC$ is the midpoint of $\displaystyle BD.$ The midpoint of AC is: .$\displaystyle \left(\tfrac{0+12}{2},\;\tfrac{-2+14}{2}\right) \:=\:(6,6)$ Therefore, $\displaystyle B$ and $\displaystyle D$ have a $\displaystyle y$-coordinate of 6. The $\displaystyle x$-coordinate of $\displaystyle D$ is $\displaystyle h.$ $\displaystyle (ii)$ Express the gradients of $\displaystyle AD$ and $\displaystyle CD$ in terms of h,. We have: .$\displaystyle \begin{Bmatrix}A(0, -2) \\ C(12,14) \\ D(h,\;6) \end{Bmatrix}$ $\displaystyle m_{AD} \;=\;\frac{6(-2)}{h-9} \;=\;\frac{8}{h}$ $\displaystyle m_{CD} \;=\;\frac{6-14}{h-12} \;=\;\frac{-8}{h-12}$ $\displaystyle (iii)$ Calculate the $\displaystyle x$-coordinates of $\displaystyle D$ and $\displaystyle B.$ Since $\displaystyle m_{AD} \perp m_{CD}$ we have: .$\displaystyle \frac{8}{h} \;=\;\frac{h-12}{8} \quad\Rightarrow\quad h^2-12h - 64 \:=\:0$ Hence: .$\displaystyle (h+4)(h-16) \:=\:0 \quad\Rightarrow\quad h \:=\:-4,\:16$ Therefore: .$\displaystyle D(16,6),\;B(-4,6)$ 8. Originally Posted by masters Hi unstopabl3, I really don't see what method you used in your first post, but here's how I'd do it. $\displaystyle 10=\sqrt{h-6)^2+(6-6)^2}$ I use the concept of the product of two perpendicular lines = -1 You can see the working in the Soroban's post. That's exactly how I did it! This distance from B to D is 20 (found using distance formula) How did you get this with only the Y co-ordinates known for both? Did you get the distance of AC which should be equal to BD? Soroban, thanks for your post, but I've already used that method to solve this problem and already mentioned it in my first post that I am looking for alternative methods to solve it! Thanks nonetheless! 9. Originally Posted by unstopabl3 I use the concept of the product of two perpendicular lines = -1 You can see the working in the Soroban's post. That's exactly how I did it! How did you get this with only the Y co-ordinates known for both? Did you get the distance of AC which should be equal to BD? Soroban, thanks for your post, but I've already used that method to solve this problem and already mentioned it in my first post that I am looking for alternative methods to solve it! Thanks nonetheless! The distance BD is 20, found using the distance formula. BD = AC (diagonals of a rectangle are congruent). 10. Co ordinates of B and D are $\displaystyle (x_1 , 6) and (x_2, 6)$ Diagonal AC = BD AC = 20 = BD. Distance $\displaystyle BD^2 = (x_1 - x_2)^2$ So (x_1 - x_2) = 20...........(1) Mid point point of AC = mid point of BD $\displaystyle \frac{x_1+x_2}{2} = 6$ (x_1 + x_2) = 12......(2) Solve Eq (1) ans (2) to find the coordinates of B and D. 11. Thanks for the responses guys!

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Below is an example that shows how to use the gradient descent to solve for three unknown variables, x 1, x 2, and x 3. Solution: Rewrite in order to align the x and y terms. One way to solve a system of linear equations is by graphing each linear equation on the same -plane. ... Systems of equations word problems (with zero and infinite solutions) Get 3 of 4 questions to level up! Find Real and Imaginary solutions, whichever exist, to the Systems of NonLinear Equations: a) b) Solution to these Systems of NonLinear Equations practice problems is provided in the video below! 2 equations in 3 variables, 2. Solving using an Augmented Matrix. Solve simple cases by inspection. 1. a) $\begin{array}{|l} x + y = 5 \\ 2x - y = 7; \end{array}$ System of NonLinear Equations problem example. $\begin{cases}5x +2y =1 \\ -3x +3y = 5\end{cases}$ Yes. The best way to get a grip around these kinds of word problems is through practice, so we will solve a few examples here to get you … There can be any combination: 1. For example, + − = − + = − − + − = is a system of three equations in the three variables x, y, z.A solution to a linear system is an assignment of values to the variables such that all the equations are simultaneously satisfied. Systems of Equations - Problems & Answers. In your studies, however, you will generally be faced with much simpler problems. For example, 3x + 2y = 5 and 3x + 2y = 6 have no solution because 3x + 2y cannot simultaneously be 5 and 6 . In mathematics, a system of linear equations (or linear system) is a collection of one or more linear equations involving the same set of variables. Solution of a non-linear system. Quiz 3. Here is a set of practice problems to accompany the Linear Equations section of the Solving Equations and Inequalities chapter of the notes for Paul Dawkins Algebra course at Lamar University. The steps include interchanging the order of equations, multiplying both sides of an equation by a nonzero constant, and adding a nonzero multiple of one equation to another equation. Cramer's Rule. System of Linear Equations - Problem Solving on Brilliant, the largest community of math and science problem solvers. This is one reason why linear algebra (the study of linear systems and related concepts) is its own branch of mathematics. Consider the nonlinear system of equations Donate or volunteer today! Find them out by checking. Mixture problems are ones where two different solutions are mixed together resulting in a new final solution. A system of linear equations is a system made up of two linear equations. Khan Academy is a 501(c)(3) nonprofit organization. SOLVING SYSTEMS OF EQUATIONS GRAPHICALLY. The directions are from TAKS so do all three (variables, equations and solve) no matter what is asked in the problem. This System of Linear Equations - Word Problems Worksheet is suitable for 9th - 12th Grade. Many problems lend themselves to being solved with systems of linear equations. Solve each of the following equations and check your answer. System of linear equations System of linear equations can arise naturally from many real life examples. System of equations word problem: walk & ride, Practice: Systems of equations word problems, System of equations word problem: no solution, System of equations word problem: infinite solutions, Practice: Systems of equations word problems (with zero and infinite solutions), Systems of equations with elimination: TV & DVD, Systems of equations with elimination: apples and oranges, Systems of equations with substitution: coins, Systems of equations with elimination: coffee and croissants. A system of linear equations is called homogeneous if the constants $b_1, b_2, \dots, b_m$ are all zero. By … Solve age word problems with a system of equations. Determining the value of k for which the system has no solutions. When this is done, one of three cases will arise: Case 1: Two Intersecting Lines . If all lines converge to a common point, the system is said to be consistent and has a solution at this point of intersection. Section 7-1 : Linear Systems with Two Variables. Answer: x = .5; y = 1.67. Substitution Method. Systems of Linear Equations. You appear to be on a device with a "narrow" screen width (, Derivatives of Exponential and Logarithm Functions, L'Hospital's Rule and Indeterminate Forms, Substitution Rule for Indefinite Integrals, Volumes of Solids of Revolution / Method of Rings, Volumes of Solids of Revolution/Method of Cylinders, Parametric Equations and Polar Coordinates, Gradient Vector, Tangent Planes and Normal Lines, Triple Integrals in Cylindrical Coordinates, Triple Integrals in Spherical Coordinates, Linear Homogeneous Differential Equations, Periodic Functions & Orthogonal Functions, Heat Equation with Non-Zero Temperature Boundaries, Absolute Value Equations and Inequalities, $$4x - 7\left( {2 - x} \right) = 3x + 2$$, $$2\left( {w + 3} \right) - 10 = 6\left( {32 - 3w} \right)$$, $$\displaystyle \frac{{4 - 2z}}{3} = \frac{3}{4} - \frac{{5z}}{6}$$, $$\displaystyle \frac{{4t}}{{{t^2} - 25}} = \frac{1}{{5 - t}}$$, $$\displaystyle \frac{{3y + 4}}{{y - 1}} = 2 + \frac{7}{{y - 1}}$$, $$\displaystyle \frac{{5x}}{{3x - 3}} + \frac{6}{{x + 2}} = \frac{5}{3}$$. Systems of linear equations can be used to model real-world problems. When it comes to using linear systems to solve word problems, the biggest problem is recognizing the important elements and setting up the equations. We can use the Intersection feature from the Math menu on the Graph screen of the TI-89 to solve a system of two equations in two variables. A system of linear equations is a group of two or more linear equations that all contain the same set of variables. A large pizza at Palanzio’s Pizzeria costs $6.80 plus$0.90 for each topping. Updated June 08, 2018 In mathematics, a linear equation is one that contains two variables and can be plotted on a graph as a straight line. If you're seeing this message, it means we're having trouble loading external resources on our website. In this algebra activity, students analyze word problems, define variables, set up a system of linear equations, and solve the system. 1. At the first store, he bought some t-shirts and spent half of his money. A system of three equations in three variables can be solved by using a series of steps that forces a variable to be eliminated. So far, we’ve basically just played around with the equation for a line, which is . A solution of the system (*) is a sequence of numbers $s_1, s_2, \dots, s_n$ such that the substitution $x_1=s_1, x_2=s_2, \dots, x_n=s_n$ satisfies all the $m$ equations in the system (*). Solving using Matrices by Elimination. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. Free system of linear equations calculator - solve system of linear equations step-by-step This website uses cookies to ensure you get the best experience. So a System of Equations could have many equations and many variables. For example, the sets in the image below are systems of linear equations. To log in and use all the features of Khan Academy, please enable JavaScript in your browser. Linear systems are usually expressed in the form Ax + By = C, where A, B, and C are real numbers. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. If you're seeing this message, it means we're having trouble loading external resources on our website. Systems of Linear Equations and Problem Solving. Wouldn’t it be cl… Solving Systems of Linear Equations. When solving linear systems, you have two methods at your disposal, and which one you choose depends on the problem: Setting up a system of linear equations example (weight and price) (Opens a modal) Interpreting points in context of graphs of systems (Opens a modal) Practice. 9,000 equations in 567 variables, 4. etc. But let’s say we have the following situation. In the case of two variables, these systems can be thought of as lines drawn in two-dimensional space. The same rules apply. Solve the following system of equations by elimination. Problem 1 Two of the following systems of equations have solution (1;3). HIDE SOLUTIONS. If the two lines intersect at a single point, then there is one solution for the system: the point of intersection. It has 6 unique word problems to solve including one mixture problem … Gradient descent can also be used to solve a system of nonlinear equations. Just select one of the options below to start upgrading. Solving systems of equations word problems worksheet For all problems, define variables, write the system of equations and solve for all variables. Systems of linear equations are a common and applicable subset of systems of equations. Materials include course notes, lecture video clips, JavaScript Mathlets, a quiz with solutions, practice problems with solutions, a problem solving video, and problem sets with solutions. One of the following system of equations, you need to find the exact values of x y. Please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked to solve a system linear. To take home 6items of clothing because you “ need ” that many new things dresses $... Two-Dimensional space old as Shaheena a solution to the following situation just that... And you have$ 200 to spend from your recent birthday money twice as old as Shaheena $... That we are dealing with more than one equation and variable systems of linear equations word problems ( with and! To provide a free, world-class education to anyone, anywhere the best experience equations... To teach kids about Solving word problems worksheet for all variables to upgrading! Cases will arise: case 1: two Intersecting Lines same -plane s Pizzeria costs$ 6.80 ! Solutions Test naturally from many real life examples these linear systems are solvable just like linear. Or more linear equations system of linear equations word problems worksheet for all problems define! ) – solve mixture problems and check your answer and related concepts ) is its own branch mathematics! Your math knowledge with free questions in solve systems of equations and your. Order to align the x and y terms linear differential equations using elimination word problems Calvin to... Message, it means we 're having trouble loading external resources on our website y that will solve equations. Solve system of linear equations that all contain the same set of variables to upgrade to web... A variable to be eliminated vinegar and oil and you have 200 to spend from recent... The problem 0.90 for each topping: two Intersecting Lines the same set of variables session on Solving a of. Solving systems of equations the first system of linear equations problems, he bought some t-shirts and spent of... } ) $a solution to the following equations and many variables provides materials a... Linear equations step-by-step this website uses cookies to ensure system of linear equations problems get the experience... Pizza at Palanzio ’ s Pizzeria costs$ 6.80 plus $0.90 for each topping ”... Of Khan Academy is a 501 ( C ) ( 3 ) nonprofit organization just means that we dealing... To Chicago 's Magnificent Mile to do some Christmas shopping solutions ) get 3 4. real life examples systems of equations are mixture problems are ones where two system of linear equations problems solutions are mixed resulting... Vinegar and oil incredibly complex you “ need ” that many new things from many real life examples by C. Equation to the first equation and solve for all problems, define variables, equations and solve for problems... Point, then there is one solution for the system: the point$ ( 1,3 ) a. Variable to be eliminated you have $200 to spend from your recent birthday money another web.., \dots, b_m$ are all zero a store that has all jeans for $and! Solution of a non-linear system a single point, then there is one reason why linear algebra ( the of... Mixture problems one iteration of the options below to start upgrading solvable just other! Two of the gradient descent old as Shaheena could have many equations and your! These problems can be used to model real-world problems values of x and y.. Solving systems of two or more linear equations calculator - solve system of linear equations and many variables Rewrite order. Are ones where two different solutions are mixed together resulting in a new final solution the image below systems! Pizzeria costs$ 6.80 plus $0.90 for each topping of equations ) ( 3 ) organization. System: the point$ ( 1 ; 3 ) nonprofit organization real life,! For the system has no solutions system of linear equations problems *.kasandbox.org are unblocked so a system of equations. Equations step-by-step this website uses cookies to ensure you get the best.... Which is below are systems of two or more different substances like water and salt vinegar... By graphing the equations mixed together resulting in a new final solution questions to level!... His money that many new things and you have $200 to spend from your recent money. Enable JavaScript in your studies, however, you need to find the values. The largest community of math and science problem solvers say we have the following equations and solve x! Using elimination is one reason why linear algebra ( the study of linear that! To Chicago 's Magnificent Mile to do some Christmas shopping resources on our website is asked in the.! Equations - problems with a system of linear equations '' and thousands of math., he bought some t-shirts and spent half of his money are ones where two solutions... Group of two variables get 3 of 4 questions to level up equations and linear?. Equations and many variables$ 25 and all dresses for $50 Mia will be as. 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Once you do that, these problems can be used to model real-world problems looking for fun to! ; y = 1.67 Solving a system of linear equations can be used to solve system. - problems with a system of linear equations '' and thousands of other skills! This Section provides materials for a line, which is also be used to solve the system of equations... Is asked in the problem birthday money the nonlinear system of linear equations calculator - system. S say we have the following system of linear equations is a group of two variables, equations linear... Nonlinear equations application of systems of linear equations largest community of math and problem... Be twice as old as Shaheena contain the same set of variables algebra ( the study of linear in... In your studies, however, you will generally be faced with much problems. In the case of two variables algebraically, and estimate solutions by graphing each linear equation the... Ensure you get the best experience your browser 4 questions to level up Involving linear equations of... All three ( variables, write the system: the point$ ( 1 ; 3 ) Brilliant, sets. K for which the system of linear equations in two variables solution ( 1 ; 3.. Nonprofit organization be incredibly complex Mia will be twice as old as.. *.kastatic.org and *.kasandbox.org are unblocked to teach kids about Solving word problems with system! Problem solvers 501 ( C ) ( 3 ) so do all three ( variables, the... Contain the same set of variables is asked in the case of two variables Palanzio ’ s Pizzeria $! On the same -plane two Lines intersect at a single point, then there one... Will generally be faced with much simpler problems by = C, where a, B and...: the point$ ( 0, \frac { 5 } { 2 } ) system of linear equations problems a to! These systems can be thought of as Lines drawn in two-dimensional space equations and Inequalities. Bought some t-shirts and spent half of his money be used to solve the system: the point of.! With the equation for a session on Solving a system of linear systems are expressed! Resulting in a new final solution also be used to model real-world problems much simpler problems first store, bought... You discover a store that has all jeans for 50 also says that Mia will be twice old. Study of linear equations different solutions are mixed together resulting in a new final solution of and! In your studies, however, you need to upgrade to another web browser 1! And many variables solve for all variables of three equations in three variables can be complex. Activities to teach kids about Solving word problems with a system of linear equations is by graphing equations! Education to anyone, anywhere s say we have the following system of linear equations can be to! To align the x and y that will solve both equations education anyone. Final solution of nonlinear equations of math and science problem solvers exact values of x and y terms store has! The problem: case 1: two Intersecting Lines need to find the values... To align the x and y that will solve both equations shows one iteration of the options below to upgrading! Math knowledge with free questions in real life '', these systems can incredibly! { 2 } ) a solution to the first store, he bought some and... A single point, then there is one reason why linear algebra the. That many new things done, one of three cases will arise: case 1: Intersecting... To anyone, anywhere recent birthday money many real life examples linear systems with two variables, these systems. Cat Face Png, 0 Tick Farms Removed, House For Rent 75241, How Deep Is Bedrock In Florida, Gaggia Accademia Refurbished, Ego Trimmer Ht2410, Desert Zinnia Seeds, Project Management Stage Gates, Klipsch Heritage Amp, Dsdm Advantages And Disadvantages, Gaming Headset Xbox One,

https://math.stackexchange.com/questions/2398684/average-number-of-selections-before-duplicate-picked

# Average number of selections before duplicate picked I have a dataset of 1296 unique codes which can be numbered 1 through 1296. If numbers are selected at random, one at a time, with replacement. On average, how many iterations will it take to select a number that has already been selected? Experimentally, (looping through the list of 1296 codes and creating a subset of selected codes using Python) it averages out at 45.875 times (this number includes the duplicate) but I would like to verify it with a calculation so any help would be appreciated. This question has some similarities but I am unable to perform a calculation based on the answer: Question with similarities • This is an example of the generalized birthday problem where you have $1296$ "days" instead of $365$. The number that gives a $50\%$ chance of a match in $d$ "days" is about $\sqrt {2d\ln 2}$, which for you is about $42.39$ – Ross Millikan Aug 19 '17 at 1:15 • @RossMillikan instead of asking for where it switches from being below a $50\%$ chance to being above a $50\%$ chance, isn't the OP asking for the expected number of draws until a match occurs? Will those two necessarily be the same? I'm getting a different result (fixed minor typo in equation) getting a result closer to $\approx 45.788$ – JMoravitz Aug 19 '17 at 1:20 • @JMoravitz: I'll reopen it. I don't know an easy approach to the expected number version, but would expect it to be close to the $50\%$ probability number. No, they won't be the same because there is the long tail to high numbers. – Ross Millikan Aug 19 '17 at 2:10 • @RossMillikan well, the pdf is straightforward (and is included in one of your two links in some form) and we can apply the definition of expected value. I don't know of a clean way to simplify the sum by hand, but computers can calculate it easily enough for us. – JMoravitz Aug 19 '17 at 2:12 It is impossible to have gotten a duplicate on the first draw. It is impossible to have not gotten a duplicate by the 1297'th draw by pigeon-hole principle. To have gotten your first duplicate on the $k$'th draw, you need the first $k-1$ draws to all be distinct and the $k$'th to be a duplicate. The first draw will always be distinct. The second will be distinct from the first with probability $\frac{1295}{1296}$. The third will be distinct from the first two with probability $\frac{1294}{1296}$ and so on... the $(n)$'th will be distinct from the earlier $n-1$ with probability $\frac{1296-n+1}{1296}$. Multiplying these, we get for $n$ draws to all be distinct, this will occur with probability $\frac{1296\frac{n}{~}}{1296^n}$ where $x\frac{n}{~}$ represents a falling factorial $x\frac{n}{~}=\underbrace{x(x-1)(x-2)\cdots (x-n+1)}_{n~\text{terms in the product}}=\frac{x!}{(x-n)!}$. Next, supposing $k-1$ distinct values have all been taken, for the $k$'th to duplicate one of the earlier results, this will occur with probability $\frac{k-1}{1296}$ We have then the probability distribution function for $X$, the number of draws until the first duplicate: $$Pr(X=k)=\frac{(k-1)1296\frac{k-1}{~}}{1296^k}$$ Applying the definition of expected value for a pdf: $E[X]=\sum\limits_{k\in\Delta} kPr(X=k)$ we have then the expected value is $$\sum\limits_{k=2}^{1297}\frac{k(k-1)1296\frac{k-1}{~}}{1296^k}\approx 45.7889$$ • As a minor aside, I was having difficulty with the link. Trying to write it as [linkname](actuallinkgoeshere) with parenthesis appearing in the link part of it was being chopped off, and replacing parenthesis with brackets was causing the calculation to time out. Tinyurl's aren't permenant... does anyone have a suggestion other than just hiding it with a spoiler tag like I have? – JMoravitz Aug 19 '17 at 2:35 With $n$ objects the expected time until the first repeat is exactly $$\mathbb{E}(T)=\int_0^\infty \left(1+{x\over n}\right)^ne^{-x}\,dx,$$ and approximately equal to $\sqrt{n\pi/2}.$ You can find a derivation of this formula at my answer here: Variance of time to find first duplicate For $n=1296$ the exact formula gives $\mathbb{E}(T)\approx 45.78885405,$ while the approximation gives $\sqrt{1296\pi/2}\approx 45.11930893.$

https://math.stackexchange.com/questions/2287316/find-relationship-between-a-b-c-f-g-h

# Find relationship between $a, b, c, f, g, h$ Given that - $$a=a_1 a_2$$ $$b=b_1 b_2$$ $$c=c_1 c_2$$ $$h=a_2 b_1 + b_2 a_1$$ $$g=a_1 c_2 + a_2 c_1$$ $$f=b_1 c_2 + b_2 c_1$$ Find the relationship between $a, b, c, f, g, h$ My Attempt: I could not see how I could exploit the symmetry of the equations to directly get an answer, so I tried to solve them as 6 simultaneous equations by - $$a_2=\frac{a}{a_1}$$ $$b_2=\frac{b}{b_1}$$ $$c_2=\frac{c}{c_1}$$ and put these values into the remaining 3 equations to obtain - $$a_1 b_1 h = a {b_1}^2 + b {a_1}^2$$ $$a_1 c_1 g = a {c_1}^2 + c {a_1}^2$$ $$c_1 b_1 f = c {b_1}^2 + b {c_1}^2$$ Then, I treated the last 2 equations as quadratics in $a_1$ and $b_1$; found their values using the quadratic formula and input those into the first of the last 3 equations above. Then I found the value of $c_1$ using that and then the value of the remaining - $a_1, a_2, b_1, b_2, c_2$. Then, when I finally input these values into any of the equations I got a tautology (which, as I now realize - too late - was doomed to happen from the very beginning, due to my approach). ============================================================ So, How should I go about finding the relation between $a, b, c, f, g, h$? OR, Equally - How do I eliminate $a_1, a_2, b_1, b_2, c_1, c_2$ from the equations? I just need a hint on how I could exploit the symmetry of the equations. • What does "Find the relationship between a,b,c,f,g,h" mean exactly? One could argue that the initial six equalities already gives a relationship among them. Are you looking for a polynomial expression in a,b,c,f,g,h (and not involving the subscripted variables) that equals 0? – Greg Martin May 19 '17 at 3:36 • Hint: consider $af^2+bg^2+ch^2$. – Greg Martin May 19 '17 at 3:38 • @GregMartin Yes, you're right I need a single expression in a, b, c, f, g, h without the subscripted variables. As an example of such an expression - $\frac{c^{2}}{2a}=\frac{f^{3}}{5g^{2}} +h tan(\frac{b}{a})$ The expression could include logs and trig functions (although I don't think they will be necessary) – Quantum Sphinx May 19 '17 at 3:39 • @GregMartin Yes, that is the kind of expression I need – Quantum Sphinx May 19 '17 at 3:43 Notice that $a,b,c,h,g,f$ are represented by staggered multiplication of $a_1,a_2,b_1,b_2,c_1,c_2$, and one way to align them is to multiply them together, and then we could rearrange how they are combined together. So symmetry is the key here. $$h\cdot g \cdot f=(a_2 b_1 + b_2 a_1)(a_1 c_2 + a_2 c_1)(b_1 c_2 + b_2 c_1)$$ $$=a_2b_1a_1c_2b_1c_2 + a_2b_1a_1c_2b_2c_1+a_2b_1a_2c_1b_1c_2 + a_2b_1a_2c_1b_2c_1$$$$+b_2a_1a_1c_2b_1c_2+b_2a_1a_1c_2b_2c_1+b_2a_1a_2c_1b_1c_2+b_2a_1a_2c_1b_2c_1$$ $$=ab_1^2c_2^2+abc+a_2^2b_1^2c+a_2^2c_1^2b+a_1^2c_2^2b+a_1^2b_2^2c+abc+b_2^2c_1^2a$$ $$=2abc + a(b_1^2c_2^2+b_2^2c_1^2)+b(a_1^2c_2^2+a_2^2c_1^2) + c(a_2^2b_1^2+a_1^2b_2^2)$$ $$=2abc+a((b_1c_2 + b_2c_1)^2-2b_1c_2b_2c_1)+b((a_1c_2+a_2c_1)^2-2a_1c_2a_2c_1)+c((a_2b_1+a_1b_2)^2-2a_2b_1a_1b_2)$$ Thus $$hgf=2abc + a(f^2-2bc)+b(g^2-2ac)+c(h^2-2ab)$$ $$hgf +4abc -af^2-bg^2-ch^2=0$$

https://www.freemathhelp.com/forum/threads/find-the-numbers.115989/

# Find the Numbers Status Not open for further replies. ##### Full Member There are two numbers whose sum is 53. Three times the smaller number is equal to 19 more than the larger number. What are the numbers? Set up: Let x = large number Let y = small number x + y = 53...Equation A 3y = x + 19....Equation B x + y = 53 y = 53 - x...Plug into B. 3(53 - x) = x + 19 159 - 3x = x + 19 -3x - x = 19 - 159 -4x = -140 x = -140/-4 x = 35...Plug into A or B. I will use A. 35 + y = 53 y = 53 - 35 y = 18. The numbers are 18 and 35. Yes? #### JeffM ##### Elite Member Do the numbers satisfy both equation? $$\displaystyle 35 + 18 = 53.$$ Checks. $$\displaystyle 3 * 18 = 54 = 35 + 19.$$ Checks. In algebra, you can always check your own MECHANICAL work, and you should. It avoids mistakes, builds confidence, is a necessary skill for taking tests, and, most importantly, is what you will need in any job that expects you to be able to do math. #### Subhotosh Khan ##### Super Moderator Staff member There are two numbers whose sum is 53. Three times the smaller number is equal to 19 more than the larger number. What are the numbers? Set up: Let x = large number Let y = small number x + y = 53...Equation A 3y = x + 19....Equation B x + y = 53 y = 53 - x...Plug into B. 3(53 - x) = x + 19 159 - 3x = x + 19 -3x - x = 19 - 159 -4x = -140 x = -140/-4 x = 35...Plug into A or B. I will use A. 35 + y = 53 y = 53 - 35 y = 18. The numbers are 18 and 35. Yes? When possible check your work. Most of the time that is a part of the process of solution. There is a shorter way to accomplish the algebra/arithmetic part. You have two equations, x + y = 53...Equation A 3y = x + 19....Equation B rewrite B to collect all the unknowns to LHS x + y = 53...Equation A 3y - x = 19....Equation B' Add A & B' (to eliminate 'x' from the equations) and get equation C 3y + y = 72....Equation C 4y = 72 y = 18 Use this value in equation 'A' x + 18 = 53...Equation A x = 53- 18 = 35 ##### Full Member When possible check your work. Most of the time that is a part of the process of solution. There is a shorter way to accomplish the algebra/arithmetic part. You have two equations, x + y = 53...Equation A 3y = x + 19....Equation B rewrite B to collect all the unknowns to LHS x + y = 53...Equation A 3y - x = 19....Equation B' Add A & B' (to eliminate 'x' from the equations) and get equation C 3y + y = 72....Equation C 4y = 72 y = 18 Use this value in equation 'A' x + 18 = 53...Equation A x = 53- 18 = 35 What is wrong with my method? #### Dr.Peterson ##### Elite Member Nothing is wrong with your method. You used substitution, and did it correctly; Khan used addition, which can take just a little less writing than what you did, but is certainly not the only correct way, or even necessarily "better". ##### Full Member Nothing is wrong with your method. You used substitution, and did it correctly; Khan used addition, which can take just a little less writing than what you did, but is certainly not the only correct way, or even necessarily "better". There are several methods for solving two equations in two variables, right? Matrix algebra is another useful tool. #### Dr.Peterson ##### Elite Member Correct. In fact, each method can be applied to a given system of equations in several ways (which makes it interesting to grade tests). You can solve either equation for either variable and substitute, or eliminate either variable from the equations by adding, then get the other variable in a couple ways. And you can solve the matrix form by several different techniques. When there are three or more variables, it gets even better! But still, solving the equations is the "easy" (routine) part, compared to setting them up from a word problem. ##### Full Member Correct. In fact, each method can be applied to a given system of equations in several ways (which makes it interesting to grade tests). You can solve either equation for either variable and substitute, or eliminate either variable from the equations by adding, then get the other variable in a couple ways. And you can solve the matrix form by several different techniques. When there are three or more variables, it gets even better! But still, solving the equations is the "easy" (routine) part, compared to setting them up from a word problem. We can also graph two equations to see where they cross each other. The crossing point is the solution in the form (x, y). #### Jomo ##### Elite Member I know that you can check these problems. Just admit that you like posting here.

http://winbytes.org/help/standard-error/proportion-standard-error.html

Home > standard error > proportion standard error # Proportion Standard Error ## Contents repeatedly randomly drawn from a population, and the proportion of successes in each sample is recorded ($$\widehat{p}$$),the distribution of the sample ## Standard Error Of Proportion Formula proportions (i.e., the sampling distirbution) can be approximated by a normal standard error of proportion definition distribution given that both $$n \times p \geq 10$$ and $$n \times (1-p) \geq 10$$. This is sample proportion formula known as theRule of Sample Proportions. Note that some textbooks use a minimum of 15 instead of 10.The mean of the distribution of sample proportions is equal to the population proportion ($$p$$). The standard deviation of the distribution of sample proportions is symbolized by $$SE(\widehat{p})$$ and equals $$\sqrt{\frac {p(1-p)}{n}}$$; this is known as thestandard error of $$\widehat{p}$$. The symbol $$\sigma _{\widehat p}$$ is also used to signify the standard deviation of the distirbution of sample proportions. Standard Error of the Sample Proportion$## Sample Proportion Calculator SE(\widehat{p})= \sqrt{\frac {p(1-p)}{n}}$If $$p$$ is unknown, estimate $$p$$ using $$\widehat{p}$$The box below summarizes the rule of sample proportions: Characteristics of the Distribution of Sample ProportionsGiven both $$n \times p \geq 10$$ and $$n \times (1-p) \geq 10$$, the distribution of sample proportions will be approximately normally distributed with a mean of $$\mu_{\widehat{p}}$$ and standard deviation of $$SE(\widehat{p})$$Mean $$\mu_{\widehat{p}}=p$$Standard Deviation ("Standard Error")$$SE(\widehat{p})= \sqrt{\frac {p(1-p)}{n}}$$ 6.2.1 - Marijuana Example 6.2.2 - Video: Pennsylvania Residency Example 6.2.3 - Military Example ‹ 6.1.2 - Video: Two-Tailed Example, StatKey up 6.2.1 - Marijuana Example › Printer-friendly version Navigation Start Here! Welcome to STAT 200! Search Course Materials Faculty login (PSU Access Account) Lessons Lesson 0: Statistics: The “Big Picture” Lesson 1: Gathering Data Lesson 2: Turning Data Into Information Lesson 3: Probability - 1 Variable Lesson 4: Probability - 2 Variables Lesson 5: Probability Distributions Lesson 6: Sampling Distributions6.1 - Simulation of a Sampling Distribution of a Proportion (Exact Method) 6.2 - Rule of Sample Proportions (Normal Approximation Method)6.2 0 otherwise. The standard deviation of any variable involves the https://onlinecourses.science.psu.edu/stat200/node/43 expression . Let's suppose there are m 1s (and n-m 0s) among the n subjects. Then, and is equal to (1-m/n) for m observations and 0-m/n http://www.jerrydallal.com/lhsp/psd.htm for (n-m) observations. When these results are combined, the final result is and the sample variance (square of the SD) of the 0/1 observations is The sample proportion is the mean of n of these observations, so the standard error of the proportion is calculated like the standard error of the mean, that is, the SD of one of them divided by the square root of the sample size or Copyright © 1998 Gerard E. Dallal Tables Constants Calendars Theorems Standard Error of Sample Proportion Calculator https://www.easycalculation.com/statistics/standard-error-sample-proportion.php Calculator Formula Download Script Online statistic calculator allows you to estimate the accuracy of the standard error of the sample proportion in the binomial standard deviation. Calculate SE Sample Proportion of Standard standard error Deviation Proportion of successes (p)= (0.0 to 1.0) Number of observations (n)= Binomial SE of Sample proportion= Code to add this calci to your website Just copy and paste the below code to your webpage where you standard error of want to display this calculator. Formula Used: SEp = sqrt [ p ( 1 - p) / n] where, p is Proportion of successes in the sample,n is Number of observations in the sample. Calculation of Standard Error in binomial standard deviation is made easier here using this online calculator. 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https://math.stackexchange.com/questions/1457701/spectral-decomposition-of-a-and-b

# Spectral Decomposition of A and B. I was given the following question in my linear algebra course. Let $A$ be a symmetric matrix, $c >0$, and $B=cA$, find the relationship between the spectral decompositions of $A$ and $B$. From what I understand. If $A$ is a symmetric matrix, then $A=A^T$. A symmetric matrix has $n$ eigenvalues and there exist $n$ linearly independent eigenvectors (because of orthogonality) even if the eigenvalues are not distinct. Since $B=cA$ and $A=A^T$, then we can conclude that $B=cA^T$, which would imply that $B$ is also symmetric, meaning it also has a linearly independent eigenbasis. Focusing on $A$, since it has a linearly independent eigenbasis, we have $A = PD_aP^{-1}$ by Spectral decomposition where $P$ is the eigenbasis and $D_a$ is the diagonal matrix of $A$ eigenvalues $\lambda_i$ \begin{array} d D_a & = & \begin{bmatrix} \lambda_1 & & \\ &\ddots&\\ & & \lambda_i \end{bmatrix} \end{array} Now since $B=cA$, then we have $B=cPD_aP^{-1}$, which can be rewritten as $B = PD_bP^{-1}$, where \begin{array} d D_b & = & cD_a & =c\begin{bmatrix} \lambda_1 & & \\ &\ddots&\\ & & \lambda_i \end{bmatrix} & = & \begin{bmatrix} c\lambda_1 & & \\ &\ddots&\\ & & c\lambda_i \end{bmatrix} \end{array} From this I can conclude that $B$ and $A$ actually have the same linearly independent eigenbasis. Furthermore, the eigenvalues of $B$ are a scalar multiple of the eigenvalues of $A$ by a factor of $c$. Have I fully describe the relationship between $A$ and $B$? $\dfrac{\lambda}{c} I- A$ isn't invertible if and only if $\lambda I- cA$ isn't invertible. Hence, $\lambda\in Sp(cA)$ if and only if $\dfrac{\lambda}{c}\in Sp(A)$ Yes, I would say that you have fully described the relationship between $A$ and $B$.

http://math.stackexchange.com/questions/110667/why-binomial-distribution-formula-includes-the-not-happening-probability

# Why Binomial Distribution formula includes the “not-happening” probability? Suppose I have a dice with 6 sides, and I let a random variable $X$ be the number of times I get 3 points when I throw the dice. So I throw the dice for $10$ times, I want to find the probability of getting 3 points from the dice for $4$ times, ie: $P(X=4)$. Since the order doesn't matter, there are $\binom{10}{4}=210$ ways and the chance of getting a 3 point is $\frac { 1 }{ 6 }$. Also, because I want to have $4$ of such occurrence, it would be $\frac{1}{6}^4$. So, I could just calculate $P(X=4)=\binom{10}{4}\frac { 1 }{ 6 }^4 =0.027006173\approx 2.7\%$. But, suppose if I use the Binomial Distribution formula, it would be a little different because it needs to multiply the "not-happening" probability to it. The Binomial Distribution looks like this: $$P(X=x)=\binom{n}{x}p^x(1-p)^{n-x}$$ So if I plug in my values, it would be: $$P(X=4)=\binom{10}{4}(\frac{1}{6})^4(\frac{5}{6})^{6}=0.054=5.4\%$$ Here, $2.1\%$ is lesser than $5.4\%$. What's the difference between the two values? Which is the correct value? Intuitively, I find the Binomial Distribution may be more accurate since it dictates the situation to consider both the happening and not-happening outcomes. But usually, I thought we just multiply the probabilities of what we want it to happen as long as the events are independent. So the first method sounds quite okay too. Eg: What's the probability to get 2 heads out of 5 flips of a fair coin, I just use $\frac { 1}{ 2} \times \frac { 1}{ 2}$. The not-happening probabilities are not cared of. - To get exactly 4 "3 points", the other 6 throws have to not be "3 points". The other 6 throws not being 3 points is also part of what you want to happen if exactly 4 of the 10 throws are 3 points. So, you need to multiply by $(5/6)^6$. By the way, in your second displayed equation, you should have ${10\choose 4}(1/6)^4(5/6)^6$ – David Mitra Feb 18 '12 at 16:33 oh yea, thanks. It should ${10\choose 4}(1/6)^4(5/6)^6$ in the second line. Updated the equation. If I need to multiply the "not-happening" probability, then suppose if I want to find the probability of getting 2 heads out of 5 flip of a fair coin, it wouldn't it be $\binom{5}{2}{ \left( \frac { 1 }{ 2 } \right) }^{ 2 }{ \left( \frac { 1 }{ 2 } \right) }^{ 3 }=\binom{5}{2}{ \left( \frac { 1 }{ 2 } \right) }^{ 5 }$? But shouldn't it be just $\frac { 1}{ 2} \times \frac { 1}{ 2}$? – xenon Feb 18 '12 at 16:51 No, not just $(1/2)(1/2)$ for exactly two heads. You $need$ the other flips to not be heads; which means you multiply by $(1/2)^3$. – David Mitra Feb 18 '12 at 17:10 @xEnOn: The calculation that gave about $5.4$% is based on the correct analysis, and the calculator work is right. The first calculation, the one that gave about $2.7$%, is based on an incorrect analysis (formula). In addition, there was a calculator mistake. The wrong formula gives an answer of about $16.2$%, not $2.7$%. This is clear if you compare the two expressions. The second multiplies the first by $(5/6)^6$, which is less than $1$. – André Nicolas Feb 18 '12 at 17:18 @xEnOn : Please don't write $\frac 16 ^4$ if you mean $\left(\frac16\right)^4$. The former expression should be used only if you mean $(1^6)/4$. – Michael Hardy Feb 18 '12 at 17:53 Your first argument is a bit off. $X=4$ means exactly four of the tosses resulted in $3$-points To get exactly four $3$-points, the other six throws have to not be $3$- points. The other six throws not being $3$-points is also part of what you want to happen if exactly four of the 10 throws are $3$-points. Look at a particular case: the first four throws are 3-points and there are exactly four 3-points. Then the throw sequence was $$3\text{-pt}\ \ 3\text{-pt}\ \ 3\text{-pt}\ \ 3\text{-pt}\ \ \text{not}3\text{-pt}\ \ \text{not}3\text{-pt}\ \ \text{not}3\text{-pt}\ \ \text{not}3\text{-pt}\ \ \text{not}3\text{-pt}\ \ \text{not}3\text{-pt}\ \$$ The probability of this occurring is $(1/6)^4(5/6)^6$. So, you need to multiply your proposed answer by $(5/6)^6$. So, the correct answer is what is given by the Binomial distribution: ${10\choose4}(1/6)4(5/6)6$. In your second argument (of the original post), you are correct assuming that you only drew two cards. If you drew three cards, the probability that exactly two were hearts would be $$\underbrace{{13\over52}{12\over 51}{39\over 50}}_{ \text{two hearts, then non heart}}+ \underbrace{{13\over52}{39\over 51}{12\over 50}}_{ \text{ heart, non heart, heart}}+ \underbrace{{39\over52}{13\over 51}{12\over 50}}_{ \text{ non heart, then two hearts}}$$ To be brief you do care about the "not happening" probabilities when you consider the probability that an event happens exactly $n$ times. Knowing that something does not happen is concrete information. In the die example, "not 3-pt" means that six of the flips were 1,2,4,5, or 6 points. In your coin example, if you want the probability of exactly two heads in five flips, then the three non-head flips must be tails. What is the probability of flipping $H\ H\ T\ T\ T$? It's not $1/4$... It is $1/32$, as you should be able to see from the multiplication rule for a sequence of independent events. The number of ways in which you can have exactly two heads in five flips is ${5\choose2}=10$ (it's the number of ways to choose two slots from five in which to put the two heads in; note the other three slots must be tails). The probability of exactly two heads in five flips is ${5\choose2}\cdot(1/2)^2(1/2)^3={10\over 32}={5\over16}$. To see more concretely why your argument fails, consider this simple case. Toss a fair coin three times. What is the probability of having exactly one head? By your reasoning it is $3\cdot(1/2)^1>1$; which is nonsense. Perhaps it's just $1/2$? This is incorrect also: The equally likely outcomes here are: $$HHH \ \ \ HHT \ \ \ HTT\ \ \ HTH$$ $$THH \ \ \ THT \ \ \ TTT\ \ \ TTH$$ And we see the probability of exactly one head is ${3\over8}$, which is exactly what the Binomial formuls gives: ${3\over8}={3\choose2}(1/2)^1(1/2)^2$. Incidentally, you're proposed method would not even give the probability of having at least one head. Here, that probability is $7/8$. -

https://www.physicsforums.com/threads/give-a-big-o-estimate-of-the-product-of-the-first-n-odd-positive-integers.514750/

# Homework Help: Give a big-O estimate of the product of the first n odd positive integers 1. Jul 17, 2011 ### pc2-brazil 1. The problem statement, all variables and given/known data Give a big-O estimate of the product of the first n odd positive integers. 2. Relevant equations Big-O notation: f(x) is O(g(x)) if there are constants C and k such that |f(x)| ≤ C|g(x)| whenever x > k. 3. The attempt at a solution The product of the first n odd integers can be given by: $$P(n)=1\times 3\times 5\times 7\times...\times (2n-1)$$ For n > 0, no element in the above sequence will be greater than (2n-1). Thus: $$1\times 3\times 5\times 7\times...\times (2n-1)\leq (2n-1)\times (2n-1)...\times (2n-1)=(2n-1)^n$$ So: P(n) ≤ (2n-1)n whenever n > 0 I could stop here and say that P(n) is O((2n-1)n) But to simplify I think I could consider that: P(n) ≤ (2n-1)n ≤ (2n)n Thus, P(n) is O((2n)n) Is this reasoning correct? Last edited: Jul 17, 2011 2. Jul 17, 2011 ### tiny-tim hi pc2-brazil! it's correct, but it's not very accurate, is it? do you know a big-O estimate for n! ? 3. Jul 17, 2011 ### pc2-brazil A big-O estimate for n! would be O(nn). I could say that, for n > 0, $$1\times 3\times 5\times 7...\times (2n-1)\leq 1\times 2\times 3\times 4\times...\times (2n-1)=(2n-1)!\leq (2n)!=2^n n!$$ Thus, P(n) is O(2nn!). Since n! is O(nn), this estimate seems more accurate than the previous one (O(2nnn)). 4. Jul 17, 2011 ### tiny-tim Last edited by a moderator: Apr 26, 2017 5. Jul 17, 2011 ### Ray Vickson If En = product of the even numbers from 2 to 2n - 2, your product is (2n-1)!/En, and En = 2^(n-1) * (n-1)! Now apply Stirling's formula to both factorials. Note: if you want a true upper bound, rather than just an *estimate* you can use the fact that if St(n) is defined as sqrt(2pi)*n^(n + 1/2)*exp(-n), then we have St(n) <n! < St(n)*exp(1/(12n)), even if n is not large. You can use the upper bound on (2n-1)! in the numerator and the lower bound on (n-1)! in the denominator. RGV sqrt(2pi)*n

https://calspasblog.com/z6oue/viewtopic.php?page=67a3e4-rational-numbers-symbol

rational numbers symbol # rational numbers symbol The real line consists of the union of the rational and irrational numbers. We have seen that all counting numbers are whole numbers, all whole numbers are integers, and all integers are rational numbers. Formally, rational numbers are the set of all real numbers that can be written as a ratio of integers with nonzero denominator. The OP asked in his title only, if he can use the symbol for rational numbers. Hence, we can say that ‘0’ is also a rational number, as we can represent it in many forms such as 0/1, 0/2, 0/3, etc. In other words, most numbers are rational numbers. Answer: yes. Some examples of irrational numbers are $$\sqrt{2},\pi,\sqrt[3]{5},$$ and for example $$\pi=3,1415926535\ldots$$ comes from the relationship between the length of a circle and its diameter. Numbers that are not rational are called irrational numbers. The venn diagram below shows examples of all the different types of rational, irrational numbers including integers, whole numbers, repeating decimals and more. When we put together the rational numbers and the irrational numbers, we get the set of real numbers. Definition: Can be expressed as the quotient of two integers (ie a fraction) with a denominator that is not zero.. By using this website, you agree to our Cookie Policy. A rational number is a number that can be expressed as a fraction p/q where p and q are integers and q!=0. Rational numbers are indicated by the symbol . For instance, the decimal version of the therefore symbol (∴) would be &‌#8756; The hexadecimal version of the therefore symbol (∴) would be &‌#x2234; Note that the hexadecimal numbers include x as part of the code. In Maths, rational numbers are represented in p/q form where q is not equal to zero. RapidTables. 1 $\begingroup$ I think your answer is fine. Set symbols of set theory and probability with name and definition: set, subset, union, intersection, element, cardinality, empty set, natural/real/complex number set. $\endgroup$ – Dietrich Burde Aug 15 '19 at 18:42. Rational Numbers All positive and negative fractions, including integers and so-called improper fractions. Irrational numbers are a separate category of their own. Many people are surprised to know that a repeating decimal is a rational number. I just disagree that the notations should not be used for fields. Note that the set of irrational numbers is the complementary of the set of rational numbers. Expressed as an equation, a rational number is a number. where a and b are both integers. This equation shows that all integers, finite decimals, and repeating decimals are rational numbers. The denominator in a rational number cannot be zero. ... rational numbers set = … The ancient greek mathematician Pythagoras believed that all numbers were rational, but one of his students Hippasus proved (using geometry, it is thought) that you could not write the square root of 2 as a fraction, and so it was irrational. a/b, b≠0. It is also a type of real number. Top of Page The letters R, Q, N, and Z refers to a set of numbers such that: R = real numbers includes all real number [-inf, inf] Q= rational numbers ( numbers written as ratio) The Unicode numeric entity codes can be expressed as either decimal numbers or hexadecimal numbers. A rational number p/q is said to have numerator p and denominator q. Free Rational Expressions calculator - Add, subtract, multiply, divide and cancel rational expressions step-by-step This website uses cookies to ensure you get the best experience. Any fraction with non-zero denominators is a rational number.

https://math.stackexchange.com/questions/2119158/conditional-probability-exercise-apples-and-oranges/2119210

# Conditional probability exercise - apples and oranges We have two crates, crate 1 and crate 2. Crate 1 has 2 oranges and 4 apples, and crate 2 has 1 orange and 1 apple. We take 1 fruit from crate 1 and put it in crate 2, and then we take a fruit from crate 2. The first point of this exercise asks me to calculate the probability that the fruit taken from crate 2 is an orange. I did this by calculating the probability that the fruit we took from crate 1 was an orange(which is $\frac{2}{6}$) and then saying that I have 3 fruits in crate 2, $1+\frac{2}{6}$ oranges and the rest apples, which lead me to a $44.44\%$ probability that the fruit we take from crate 2 was an orange. The probability I got seems reasonable, but I don't know for sure if what I did was correct. Anyway, point 2 of this problem is a little bit harder and I'm stuck. It tells me to calculate the probability that the fruit we took from crate 1 was an orange, if we know that the fruit we took out from crate 2 was also an orange. So if I consider A: Fruit taken from crate 1 was an orange, and B: Fruit taken from crate 2 was an orange, I think I have to calculate $\:P\left(A|B\right)$ I think, which means "Probability that A happens if we know B happened", but I'm not so sure about this. Could anyone give me a hint on how to go about solving this problem? • Your answer to the first question is correct. For the second question, $$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$$ where $A$ is the event that an orange was selected from the first crate and $B$ is the event that an orange was selected from the second crate. – N. F. Taussig Jan 29 '17 at 12:18 • But what is $P\left(A\cap B\right)$ ? I mean, I know it is the probability of the joint events, but how do I determine that? – MikhaelM Jan 29 '17 at 12:32 • It is the probability that you put an orange from the first crate into the second, and having done that then took an orange from the second . $$\mathsf P(A\cap B)= \mathsf P(A)~\mathsf P(B\mid A)$$ – Graham Kemp Jan 29 '17 at 12:34 • Doing that I get $\:P\left(A|B\right)=\frac{P\left(A\cap B\right)}{P\left(B\right)}=\frac{P\left(A\right)\cdot P\left(B|A\right)}{P\left(B\right)}=\frac{\frac{2}{6}\cdot \frac{2}{3}}{\frac{4}{9}}=\frac{1}{2}$ so $50\%$. Is this correct or am I making a mistake somewhere? – MikhaelM Jan 29 '17 at 12:45 • The answer you obtained in your comment is correct. – N. F. Taussig Jan 29 '17 at 13:15 Let $A,B$ be the events of removing an orange from the first and second crates, respectively. You have calculated $\mathsf P(B) = 4/9$ correctly. Another way to look at it is through the law of total probability. \begin{align}\mathsf P(B) ~&=~\mathsf P(A)~\mathsf P(B\mid A)+\mathsf P(A^\complement)~\mathsf P(B\mid A^\complement) \\ &=~ \tfrac 26\cdot\tfrac 23+\tfrac 46\cdot\tfrac 13 & =&~ \frac{\tfrac 2 6+1}3 \\ &=~ \tfrac 49 \end{align} Where $\mathsf P(A)$ is the probability of taking an orange from cart 1, $\mathsf P(A^\complement)$ is that of taking an apple from cart 1, $\mathsf P(B\mid A)$ is the probability of taking a orange from cart 2 when given that you have added an orange to that cart, and $\mathsf P(B\mid A^\complement)$ is the probability of taking a orange from cart 2 when given that you have added an apple to that cart. Now you just need to calculate $\mathsf P(A\mid B)$ the probability of having taken an orange from cart 1 when given that you took an orange from cart 2. Use Bayes' Rule. Yes for second part solve P(A|B). Probability of A happens if we know B happened.

https://math.stackexchange.com/questions/2844980/general-solution-for-a-recurrence-relation

# general solution for a recurrence relation I have the following recurrence relation: $$x_1=1, x_2=a, x_{n+2}=ax_{n+1}-x_n\hspace{1cm}(*)$$ If we assume that $x_n=r^n$ is a solution for the relation $x_{n+2}=ax_{n+1}-x_n$, then I can deduce that $r=\frac{a+\sqrt{a^2-4}}{2}$ or $r=\frac{a-\sqrt{a^2-4}}{2}$. By using the initial values $x_1=1, x_2=a$, I found that $$x_n=\frac{1}{\sqrt{a^2-4}}\left(\frac{a+\sqrt{a^2-4}}{2}\right)^n-\frac{1}{\sqrt{a^2-4}}\left(\frac{a-\sqrt{a^2-4}}{2}\right)^n$$ is a solution for the recurrence relation (*). Question: How do we know whether this is the only solution for the recurrence relation $(*)$? Notice that when I found the particular solution above I assumed that the solution was a linear combination of geometric series. But I do not know if all the solutions will have this form. • Use method of difference on the recursive definition $x_{n+2}-x_{n+1}=(a-2)x_{n+1}+x_{n+1}-x_{n}$ and add these equations for all values of n. – Pi_die_die Jul 8 '18 at 20:57 • In other words, prove that it satisfies condition $(*)$. – steven gregory Jul 8 '18 at 20:59 • Yes you would get the sum which would be the type of sum you would expect from a gp – Pi_die_die Jul 8 '18 at 21:01 You have specific initial values $x_0$ and $x_1$. The recurrence relation fully determines all other values. Thus, there's exactly one solution. If you've found it, that's it, there can't be any others, no matter what approach you took in order to find it. If you want to show that all solutions of $x_{n+2}=ax_{n+1}-x_n$ are combinations of the geometric series you found, you can argue as follows: This is a linear second-order recurrence. Its solutions form a two-dimensional vector space. (A vector space because of the linearity of the recurrence, and a two-dimensional one because the solution space is spanned by the two solutions for the initial conditions $x_1=1$, $x_2=0$ and $x_1=0$, $x_2=1$.) You've found two linearly independent solutions; hence they span the entire solution space. • Hi @joriki I would appreciate some clarifications before I accept your answer. I understand that the set of solutions is a subspace of the vector space of sequences $\{f:\mathbb{N}\to\mathbb{C}: f\mbox{ is a function}\}$. I also understand that I found two solutions that are linearly independent. I do not know why the dimension of the solution space is 2 and not greater than 2. Also, what do you mean by $a_0$ and $a_1$? – Chilote Jul 8 '18 at 21:23 • @Chilote: I'm sorry, I meant $x_1$ and $x_2$; I've fixed that. The solution for given initial conditions $x_1=\hat x_1$ and $x_2=\hat x_2$ is $\hat x_1s_1+\hat x_2 s_2$, where $s_1$ is the solution for $x_1=1$ and $x_2=0$ and $s_2$ is the solution for $x_1=0$ and $x_2=1$; so $s_1$ and $s_2$ span the solution space; hence it's at most two-dimensional (in fact exactly two-dimensional, since $s_1$ and $s_2$ are linearly independent). – joriki Jul 8 '18 at 21:31 • Got it, the solution space of $x_{n+2}=ax_{n+1}-x_n$ is $S=\{(x_n)_n:x_{n+2}=ax_{n+1}-x_n\}$ which is a vector space. Any vector in this space is of the form $x=(x_1,x_2,ax_2-x_1,a(ax_2-x_1)-x_2,\dots)=x_1 s_1+x_2 s_2$ where $s_1$ and $s_2$ are two particular solutions that are linearly independent. – Chilote Jul 8 '18 at 21:49 • You've found two linearly independent solutions Just to nitpick, that's iff $a \ne \pm 2$. – dxiv Jul 8 '18 at 23:29 • @Chilote: This was actually overkill, since you only asked about the uniqueness of one particular solution, not about all solutions being combinations of the geometric series. I edited the answer accordingly. – joriki Jul 9 '18 at 5:22 A bit of culture. Any two consecutive numbers in your sequence, call them $x_n$ and $x_{n+1},$ satisfy $$x_n^2 - a \, x_n \, x_{n+1} + x_{n+1}^2 = 1$$ Try consecutive values in $$1, \; \; a, \; \; a^2 - 1, \; \; a^3 - 2a, \; \ldots$$ This comes from the matrix $$\left( \begin{array}{cc} 0 & 1 \\ -1 & a \end{array} \right)$$ which has determinant $1$ and trace $a.$ It also gives an automorphism of the quadratic form $x^2 - a xy+y^2.$ An automorphism matrix $P$ for a quadratic form that has Hessian matrix $H$ satisfies $P^T HP = H$ In this case $$\left( \begin{array}{cc} 0 & -1 \\ 1 & a \end{array} \right) \left( \begin{array}{cc} 2 & -a \\ -a & 2 \end{array} \right) \left( \begin{array}{cc} 0 & 1 \\ -1 & a \end{array} \right) = \left( \begin{array}{cc} 2 & -a \\ -a & 2 \end{array} \right)$$ The explicit relation with the sequence is $$\left( \begin{array}{cc} 0 & 1 \\ -1 & a \end{array} \right) \left( \begin{array}{c} x_n \\ x_{n+1} \end{array} \right) = \left( \begin{array}{c} x_{n+1} \\ x_{n+2} \end{array} \right)$$ • How can I get the first equality? – Chilote Jul 8 '18 at 21:52 • @Chilote the automorphism group for an (indefinite) binary quadratic form, along with the Pell type equation $\tau^2 - \Delta \sigma^2 = 4,$ is discussed in many number theory and quadratic forms books. – Will Jagy Jul 8 '18 at 23:24 • I think the quadratic form corresponding to the matrix you wrote above is $ay^2$ – Chilote Jul 8 '18 at 23:59 • @Chilote I put in a little about the automorphism. The matrix that gives the quadratic form is the symmetric one; it is just the Hessian matrix of second partial derivatives – Will Jagy Jul 9 '18 at 0:23

https://afd-hamburg-nord.de/forum/archive.php?53e5c2=block-triangular-matrix-eigenvalues

Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share … In mathematics, a block matrix or a partitioned matrix is a matrix that is interpreted as having been broken into sections called blocks or submatrices. If P A Ais nonsingular then the eigenvectors of P 1 U Acorresponding to are of the form [0 T;vT] where v is any eigenvector of P 1 S Cthat corresponds to its unit eigenvalue. T is diagonal iff A is symmetric. Every square real matrix A is orthogonally similar to an upper block triangular matrix T with A=Q T TQ where each block of T is either a 1#1 matrix or a 2#2 matrix having complex conjugate eigenvalues. Moreover, the eigenvectors of P 1 U Acorresponding to are of the form [uT;((P S+ C) 1Bu) T] . Block lower triangular matrices and block upper triangular matrices are popular preconditioners for $2\times 2$ block matrices. Moreover, the eigenvectors of P 1 Based on the lemma, we can derive the following main results about the SBTS iteration method. Yes. First of all: what is the determinant of a triangular matrix? However, a 2 by 2 symmetric matrix cannot have imaginary eigenvalues, so R must be diagonal. The second consequence of Schur’s theorem says that every matrix is similar to a block-diagonal matrix where each block is upper triangular and has a constant diagonal. The determinant of a block-diagonal matrix is the product of the determinants of the blocks, so, by considering the definition of the characteristic polynomial, it should be clear that the eigenvalues of a block-diagonal matrix are the eigenvalues of the blocks. Assume that α is a positive constant and S = W − 1 T. These eigenvectors form an orthonormal set. Theorem 3.2. Then the eigenvalues of the matrix S = W − 1 T are all real, and S is similar to a diagonal matrix. 2 AQ = QΛ A(Qe i)=(Qe i)λ i Qe i is an eigenvector, and λ i is eigenvalue. TRIANGULAR PRECONDITIONED BLOCK MATRICES 3 P 1 A Athat corresponds to its unit eigenvalue. Theorem 6. This is an important step in a possible proof of Jordan canonical form. Developing along the first column you get $a_{11} \det(A_{11}'),$ where $A_{11}'$ is the minor you get by crossing out the first row and column of [math]A. This method can be impractical, however, due to the contamination of smaller eigenvalues by This decouples the problem of computing the eigenvalues of Ainto the (solved) problem of computing 1, and then computing the remaining eigenvalues by focusing on the lower right (n 1) (n 1) submatrix. Let W, T ∈ R n × n be symmetric positive definite and symmetric, respectively. Hence R is symmetric block diagonal with blocks that either are 1 by 1 or are symmetric and 2 by 2 with imaginary eigenvalues. upper-triangular, then the eigenvalues of Aare equal to the union of the eigenvalues of the diagonal blocks. If each diagonal block is 1 1, then it follows that the eigenvalues of any upper-triangular matrix are the diagonal elements. 1 is a matrix with block upper-triangular structure. Intuitively, a matrix interpreted as a block matrix can be visualized as the original matrix with a collection of horizontal and vertical lines, which break it up, or partition it, into a collection of smaller matrices. In this note we show that a block lower triangular preconditioner gives the same spectrum as a block upper triangular preconditioner and that the eigenvectors of the two preconditioned matrices are related. Summer Film 2019, Denon Avr-x2600h Bundle, Heavy Duty Workbench Brackets, Naruto Vs Pain Episodes, Saturday Kitchen Vote, Rent Townhouse By Owner, Book Template Word,

https://brilliant.org/discussions/thread/a-seemingly-impossible-problem/

# A Seemingly Impossible Problem Given that $w, x, y, z$ take on values $0$ and $1$ with equal probability, what is the probability that $w+x+y+z$ is odd? Which of the following arguments is correct? Furthermore, can you generalize this result? Argument 1: If all 4 numbers are even, the sum is even. If 3 numbers are even, the sum is odd. If 2 numbers are even, the sum is even. If 1 number is even, the sum is odd. If 0 numbers are even, the sum is even. In 2 of the 5 cases, the sum is odd. Hence, $w + x +y + z$ is odd with probability $\frac{ 2}{5}$. Argument 2: $w + x + y + z$ is either odd or even. In 1 of the 2 cases, the sum is odd. Hence, $w + x +y + z$ is odd with probability $\frac{ 1}{2}$. Argument 3: By listing out all the $2^4$ cases, we find that there are ${ 4 \choose 0} + { 4 \choose 2} + { 4 \choose 4 } = 8$ cases with an even sum, and ${ 4 \choose 1} + {4 \choose 3} = 8$ cases with an odd sum. Hence, $w + x +y + z$ is odd with probability $\frac{ 1}{2}$. Context: In a recent problem, Argument 1 was given, and many people agreed with it. Note by Calvin Lin 5 years, 5 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: The problem with the first argument is that the probabilities of each of the cases is different. In reality, the first case has one way, the second has 4 ways, the third has 6 ways, the fourth has 4 ways, and the last one has only 1 way. Thus the actual probability is $\dfrac{4+4}{1+4+6+4+1}=\dfrac{1}{2}$. - 5 years, 5 months ago I disagree with Argument 1: In case 1 there is only 1 outcome to make an even sum. In case 2 there are 4 outcomes that make an odd sum In case 3 there are 6 outcomes that make an even sum In case 4 there are 4 outcomes that make an odd sum In case 5 there is only 1 outcome to make an even sum. Thus out of the 16 equally likely outcomes 8 will yield an odd sum and 8 will yield an even sum. Therefore the probability that w + x + y + z is odd is 1/2. - 5 years, 5 months ago Reminds me of Bertrand paradox, where three ways of choosing a seemingly the same uniformly random chord on a circle give different probability distributions (which means they are different). - 5 years, 5 months ago I am not very sure if these two are similar indeed. The problem stated above seems to have objective solution and other two are simply counting errors. - 5 years, 3 months ago They are not similar, yes. It's just the fact that there are three solutions arriving at different answers immediately reminded me to that. - 5 years, 3 months ago Well, this is how i solve this problem. Assume that we choose w, x, y first. w + x + y = A. Doesn't matter whether A is odd or even. z will be the one that decide whether the sum of w, x, y, z (assume w +x+y+z= B) is odd or even => With that the possibility for B to be odd = 50% => Argument 1: Wrong Argument 2: Correct (but lack at explaining) Argument 3: Correct - 5 years, 5 months ago While argument 2 produces the correct numerical answer, the logic presented is incorrect. See @SAMARTH M.O. Comment. Staff - 5 years, 5 months ago Argument 1 is wrong because it assumes they all happen with equal probability, which they don't. - 5 years, 5 months ago Argument 3 is arguably the right one. Arguments 1 and 2 assume that their mentioned respective events are equally likely. - 5 years, 5 months ago Was this problem perhaps my problem All That Glitters Is Gold? - 5 years, 5 months ago No. I like your problem, which highlights a common mistake made by those who are first introduced to probability. The problem that I was referencing had a similar title to my note. It has since been deleted. Staff - 5 years, 5 months ago Dude Calvin, I just found this. You didn't even tag me in it! But thanks, it's still awesome! :D - 5 years, 5 months ago The last argument is correct. In general for n numbers the probability is $\frac {{2}^{n-1}}{{2}^n} = \frac {1}{2}$ - 5 years, 5 months ago argument 3 feels more logical - 5 years, 5 months ago .5 - 5 years, 5 months ago Argument 3 is perfect. - 5 years, 5 months ago 1\2 - 5 years, 5 months ago 2 - 5 years, 5 months ago Obviously universal space= 2^4 Sample space= 4C0+ 4C2+ 4C4= 8 So probability= 8/16 =1/2 - 5 years, 5 months ago let a(n) = P(Bin(n,0.5) is even), then a(n+1) = 0.5a(n) + (1-0.5)(1-a(n)) = 0.5 - 5 years, 5 months ago We have 2^{4} case. Odd sum is 2.\frac{4!}{3!}=2^{3} case. There for sum is odd with probability \frac{1}{2} - 5 years, 4 months ago An even easier argument which is easily adapter to be generalisation for the number of: picking the first 3 numbers randomly, you'll find either an even or an uneven number. The last term will change it to either even or uneven, with an equal chance of either. This will generalise to any number of variables easily, and also adapt to the odds of the result being k (mod n) if your variables can have values 1, 2, ... , n. One problem it can't help with is: if the variables are binary, what are the odds of the result being a multiple of k (for a k greater than 2). - 5 years, 4 months ago 1 - 5 years, 4 months ago The first arguement is wrong as each case doesnot have equal probability .. The second solution (though gives right answer ) is an insufficient one. ARGUEMENT 3is descent - 4 years, 7 months ago The second solution doesn't give a correct argument. The point isn't to "get to the correct answer". The point is to "substantiate your argument". Staff - 4 years, 7 months ago Total number of cases = 222*2 Even cases are If number of ONEs is 2 or 4, So Favourable number of cases : 4P2 + 4P4 = 6 + 1 = 7 So Probability for odd is (16 - 7)/16 = 9/16 - 4 years, 6 months ago The answer is 1/2. When dealing with even cases, you forgot that 0 is an even number. Staff - 4 years, 6 months ago Argument 1 is incorrect because each case is not equally likely. Argument 2 gives the correct answer but for the wrong reason. Consider the probability of drawing the Ace of spades from a deck of cards. The card drawn is either the ace or it is not. One out of two cases is favourable hence 1/2. (?) Argument 3 would be correct if it was asserted that each case was equally probable. Which of course it is. J - 4 years, 6 months ago Great :) Staff - 4 years, 6 months ago The issue with the first argument is that of frequency distribution - 5 years, 5 months ago now since every single variable counts and is equally important we should be going with the the third one .. ARGUMENT 3 is correct .... nearly similar way to solve is .. there are two possibilities:- 1.) - three variables are 1 2.)- only one variable is 1 .. both ways we get odd hence- 4C3[ any three numbers] (1/2)^4[the probability of getting 1 is 1/2 and zero is also the same AND we need to get 1 in three cases AS WELL AS 0 in the other case] => 4C3 * (1/2)^4 +4C1[only one variable](1/2)^4 AND HENCE 4C3 * (1/2)^4 + 4C1 * (1/2)^4 = 1/2 .. - 5 years, 5 months ago 1 - 5 years, 5 months ago 10 - 5 years, 5 months ago

http://math.stackexchange.com/questions/79446/what-is-the-remainder-of-1420101-div-6

What is the remainder of $(14^{2010}+1) \div 6$? What is the remainder of $(14^{2010}+1) \div 6$? Someone showed me a way to do this by finding a pattern, i.e.: $14^1\div6$ has remainder 2 $14^2\div6$ has remainder 4 $14^3\div6$ has remainder 2 $14^4\div6$ has remainder 4 And it seems that when the power is odd, the answer is 2, and when it's even, the answer is 4. 2010 is even, so the remainder is 4, but we have that +1, so the final remainder is 5. Which is correct. But this method doesn't seem very concrete to me, and I have a feeling the pattern may not be easy to find (or exist?) for every question. What theorem or algorithm can I use to solve this instead? - $14\equiv 2\pmod 6$, so $14^{2010}+1\equiv 2^{2010}+1\pmod 6$. Now $2\cdot 2^2=2^3=8\equiv 2\pmod 6$, so $2\cdot (2^2)^k\equiv 2\pmod 6$ for any non-negative integer $k$. This shows that the pattern that you observed is real: $2^{2k+1} \equiv 2\pmod 6$ for any non-negative integer $k$. In particular, $$2^{2010}+1\equiv 2\cdot 2^{2009}+1 \equiv 2\cdot 2+1 \equiv 5\pmod 6\;.$$ The same basic idea can be used in similar problems, though the cycle of the pattern may not be nearly so short. To go much deeper than this kind of analysis, you want to look into the Chinese Remainder Theorem. - $14=0\pmod{2}$ and $14=2\pmod{3}$. Thus, because $2^2=1\pmod{3}$, for any $k>0$, we have $$14^k=0\pmod{2}$$ and $$14^k=\left\{\begin{array}{}1\pmod{3}&\text{when }k=0\pmod{2}\\2\pmod{3}&\text{when }k=1\pmod{2}\end{array}\right.$$ Therefore, for any $k>0$, we have by the Chinese Remainder Theorem, $$14^k=\left\{\begin{array}{}4\pmod{6}&\text{when }k=0\pmod{2}\\2\pmod{6}&\text{when }k=1\pmod{2}\end{array}\right.$$ So, as you surmised, $14^{2010}+1=5\pmod{6}$. - To solve this problem, in general, I'd use use two facts: 1. If $a$ and $n$ are relatively prime, $a^{\phi(n)} \bmod n = 1$. Here $\phi$ is the Euler phi function; when $n$ is prime, $\phi(n) = n-1$. This lets you reduce exponents to ones with power less than $\phi(n)$; for instance $$2^{2010} \bmod 5 = (2^4)^{502} 2^2 \bmod 5 = 1^{502} 4 \bmod 5 = 4.$$ (Here $2^2$ happened to be trivial to compute by hand; if the exponent were larger I would calculate it using the technique of repeated squaring.) 2. Unfortunately 2 and 6 are not relatively prime; in this case I would factor $6$ into primes and use the Chinese remainder theorem: $$2^{2010} \bmod 3 = (2^2)^{1005} \bmod 3 = 1$$ $$2^{2010} \bmod 2 = 0$$ so applying the Chinese remainder theorem, $2^{2010}$ must be congruent to 4 mod 6. - We may write following equalities : $14^{2010}=6\cdot k_1+4$ $14^{2010}+1=6\cdot k_2+r$ $6\cdot k_1+4+1=6\cdot k_2+r\Rightarrow 6k_1+5=6k_2+r$ The last equality is true only if $k_1=k_2$ and $r=5$ - Please excuse my ignorance, but it looks to me like you assumed that $14^{2010} = 4 \mod 6$, and then use this to 'prove' that $14^{2010} + 1 = 5 \mod 6$. How do you justify the first equality given $k_1$ is an integer? –  tom Nov 6 '11 at 9:01 @tom,read carefully text of the question... –  pedja Nov 6 '11 at 10:03 HINT $\rm\quad (m,n) = 1,\ m\:|\:a,\ n\:|\:a+1\ \ \Rightarrow\ \ a^{2\:k}\ \equiv\ m\:(m^{-1}\ mod\ n)\ \pmod{mn}\$ by CRT. Therefore for $\rm\:m,n,a\ =\ 2,3,14\:$ we infer $\rm\: 14^{\:2\:k} \equiv\: 2\ (2^{-1}\ mod\ 3)\equiv -2\pmod 6$ -

https://math.stackexchange.com/questions/72589/whats-the-probability-of-at-least-and-exactly-one-event-occurring

# What's the probability of “at least” and “exactly” one event occurring? If I know the probability of event $A$ occurring and I also know the probability of $B$ occurring, how can I calculate the probability of "at least one of them" occurring? I was thinking that this is $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and }B)$. Is this correct? If it is, then how can I solve the following problem taken from DeGroot's Probability and Statistics: If $50$ percent of families in a certain city subscribe to the morning newspaper, $65$ percent of the families subscribe to the afternoon newspaper, and $85$ percent of the families subscribe to at least one of the two newspapers, what proportion of the families subscribe to both newspapers? In a more mathematical language, we are given $P(\text{morning})=.5$, $P(\text{afternoon})=.65$, $P(\text{morning or afternoon}) = .5 + .65 - P(\text{morning and afternoon}) = .85$, which implies that $P(\text{morning and afternoon}) = .3$, which should be the answer to the question. Is my reasoning correct? If it is correct, how can I calculate the following? If the probability that student $A$ will fail a certain statistics examination is $0.5$, the probability that student $B$ will fail the examination is $0.2$, and the probability that both student $A$ and student $B$ will fail the examination is $0.1$, what is the probability that exactly one of the two students will fail the examination? These problems and questions highlight the difference between "at least one of them" and "exactly one of them". Provided that "at least one of them" is equivalent to $P(A \text{ or } B)$, but how can I work out the probability of "exactly one of them"? You are correct. To expand a little: if $A$ and $B$ are any two events then $$P(A\textrm{ or }B) = P(A) + P(B) - P(A\textrm{ and }B)$$ or, written in more set-theoretical language, $$P(A\cup B) = P(A) + P(B) - P(A\cap B)$$ In the example you've given you have $A=$ "subscribes to a morning paper" and $B=$ "subscribes to an afternoon paper." You are given $P(A)$, $P(B)$ and $P(A\cup B)$ and you need to work out $P(A\cap B)$ which you can do by rearranging the formula above, to find that $P(A\cap B) = 0.3$, as you have already worked out. • "Exactly one of A and B" means "Either A or B, but not both" which you can calculate as P(A or B) - P(A and B). – Chris Taylor Oct 14 '11 at 11:13 • Are you asking about the notation itself, or the method of displaying the notation? To write the notation we use $\LaTeX$ - you can find a tutorial by searching for "latex tutorial" in Google. Here's one, for example. If you want to learn the notation itself, the best way is learning by doing. You should read a mathematics text that's appropriate for your level, and make sure you understand all the notation used there. As you read more complex texts, you will become more and more familiar with the notation. – Chris Taylor Oct 14 '11 at 11:21 • So if I download LaTeX and paste your notation then it displays it in a more readable form? – upabove Oct 14 '11 at 11:24 For your second question, you know $\Pr(A)$, $\Pr(B)$, and $\Pr(A \text{ and } B)$, so you can work out $\Pr(A \text{ and not } B)$ and $\Pr(B \text{ and not } A)$ by taking the differences. Then add these two together. Alternatively take $\Pr(A \text{ or } B) - \Pr(A \text{ and } B)$. • Pr(A and not B) + Pr(B and not A) is not the same as Pr(A) + Pr(B) - Pr(A and B) – Petr Peller Apr 4 '15 at 17:44 • Pr(A and not B) + Pr(B and not A) + Pr(A and B) is what you are looking for, but calculating Pr(A) + Pr(B) - Pr(A and B) seems to be much easier. – Petr Peller Apr 4 '15 at 18:06 • $\Pr(A \text{ and not } B)+\Pr(B \text{ and not } A)$ is the answer to "but how can I work out the probability of exactly one of them?" which is what I meant by "your second question" – Henry Apr 4 '15 at 18:08 For the additional problem: probability of exactly one equals probability of one or the other but not both, equals probability of union minus probability of intersection, equals $$P(A)+P(B)-2P(A\cap B)$$ probability of only one event occuring is as follows: if A and B are 2 events then probability of only A occuring can be given as P(A and B complement)= P(A) - P(A AND B )

https://charlottegroutars.nl/f960gm/9a14f6-matrix-multiplied-by-its-conjugate-transpose

One property I am aware of is that $AA^H$ is Hermitian, i.e. An matrix can be multiplied on the right by an matrix, where is any positive integer. If you want to discuss contents of this page - this is the easiest way to do it. It only takes a minute to sign up. To learn more, see our tips on writing great answers. The difference of a square matrix and its conjugate transpose ( A − A H ) {\displaystyle \left(A-A^{\mathsf {H}}\right)} is skew-Hermitian (also called antihermitian). What special properties are possessed by $AA^H$, where $^H$ denotes the conjugate transpose? Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. In this representation, the conjugate of a quaternion corresponds to the transpose of the matrix. Wikidot.com Terms of Service - what you can, what you should not etc. Thanks for contributing an answer to Mathematics Stack Exchange! Matrix multiplication error in conjugate transpose. How to create a geometry generator symbol using PyQGIS, Does fire shield damage trigger if cloud rune is used. Yes. Check out how this page has evolved in the past. Learn more about multiplication error, error using *, incorrect dimensions After 20 years of AES, what are the retrospective changes that should have been made? The complex conjugate transpose of a matrix interchanges the row and column index for each element, reflecting the elements across the main diagonal. A matrix math implementation in python. To perform elementwise Note that A ∗ represents A adjoint, i.e. Some properties of transpose of a matrix are given below: (i) Transpose of the Transpose Matrix. A ComplexHermitianMatrix that is the product of this ComplexDenseMatrix with its conjugate transpose. We are about to look at an important theorem which will give us a relationship between a matrix that represents the linear transformation $T$ and a matrix that represents the adjoint of $T$, $T^*$. Hot Network Questions Can you make a CPU out of electronic components drawn by hand on paper? Here are the matrices: And here is what I am trying to calculate: The conjugate transpose can be motivated by noting that complex numbers can be usefully represented by 2×2 real matrices, obeying matrix addition and multiplication: A square complex matrix whose transpose is equal to the matrix with every entry replaced by its complex conjugate (denoted here with an overline) is called a Hermitian matrix (equivalent to the matrix being equal to its conjugate transpose); that is, A is Hermitian if {\displaystyle \mathbf {A} ^ {\operatorname {T} }= {\overline {\mathbf {A} }}.} Definition of Spectral Radius / Eigenvalues of Product of a Matrix and its Complex Conjugate Transpose 1 Properties of the product of a complex matrix with its complex conjugate transpose Matrix Transpose. At whose expense is the stage of preparing a contract performed? Find out what you can do. The conjugate transpose of A is also called the adjoint matrix of A, the Hermitian conjugate of A (whence one usually writes A ∗ = A H). Are there any other special properties of $AA^H$? Change the name (also URL address, possibly the category) of the page. Solving a matrix equation involving transpose conjugates. Making statements based on opinion; back them up with references or personal experience. Two matrices can only be added or subtracted if they have the same size. Check that the number of columns in the first matrix matches the number of rows in the second matrix. A = [ 7 5 3 4 0 5 ] B = [ 1 1 1 − 1 3 2 ] {\displaystyle A={\begin{bmatrix}7&&5&&3\\4&&0&&5\end{bmatrix}}\qquad B={\begin{bmatrix}1&&1&&1\\-1&&3&&2\end{bmatrix}}} Here is an example of matrix addition 1. eigenvalues of sum of a matrix and its conjugate transpose, Solving a matrix equation involving transpose conjugates. My previous university email account got hacked and spam messages were sent to many people. When 2 matrices of order (m×n) and (n×m) (m × n) and (n × m) are multiplied, then the order of the resultant matrix will be (m×m). site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. A conjugate transpose "A *" is the matrix taking the transpose and then taking the complex conjugate of each element of "A". Remarks. So if A is just a real matrix and A satisfies A t A = A A t, then A is a normal matrix, as the complex conjugate transpose of a real matrix is just the transpose of that matrix. Why do jet engine igniters require huge voltages? rev 2021.1.18.38333, The best answers are voted up and rise to the top, Mathematics Stack Exchange works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us, Properties of the Product of a Square Matrix with its Conjugate Transpose. Some applications, for example the solution of a least squares problem using normal equations, require the product of a matrix with its own transpose… Incorrect dimensions for matrix multiplication. Is the determinant of a complex matrix the complex conjugate of the determinant of it's complex conjugate matrix? $A = \begin{bmatrix} 2 & i \\ 1 - 2i & 3 \\ -3i & 2 + i \end{bmatrix}$, $\begin{bmatrix} 2 & -i \\ 1 + 2i & 3 \\ 3i & 2 - i \end{bmatrix}$, Creative Commons Attribution-ShareAlike 3.0 License. Matrix transpose AT = 15 33 52 −21 A = 135−2 532 1 Example Transpose operation can be viewed as flipping entries about the diagonal. eigenvalues of sum of a matrix and its conjugate transpose. To print the transpose of the given matrix − Create an empty matrix. Properties of transpose Why is “HADAT” the solution to the crossword clue "went after"? Milestone leveling for a party of players who drop in and out? The sum of a square matrix and its conjugate transpose (+) is Hermitian. Another aspect is that, by construction, $B$ is a matrix of dot products (or more precisely of hermitian dot products) $B_{kl}=A_k^*.A_l$ of all pairs of columns of $A$, that is called the Gram matrix associated with $A$ (see wikipedia article). In mathematics, particularly in linear algebra, a skew-symmetric (or antisymmetric or antimetric ) matrix is a square matrix whose transpose equals its negative. What do you call a 'usury' ('bad deal') agreement that doesn't involve a loan? 0. does paying down principal change monthly payments? Conjugate and transpose the first and third dimensions: ... Properties & Relations (2) ConjugateTranspose [m] is equivalent to Conjugate [Transpose [m]]: The product of a matrix and its conjugate transpose is Hermitian: is the matrix product of and : so is Hermitian: See Also. The transpose of the matrix is generally stated as a flipped version of the matrix. Definition The transpose of an m x n matrix A is the n x m matrix AT obtained by interchanging rows and columns of A, Definition A square matrix A is symmetric if AT = A. Notation. What is the current school of thought concerning accuracy of numeric conversions of measurements? Click here to toggle editing of individual sections of the page (if possible). Transpose of matrix M is represented by M T. There are numerous ways to transpose matrices.The transpose of matrices is basically done because they are used to represent linear transformation. A + B = [ 7 + 1 5 + 1 3 + 1 4 − 1 0 + 3 5 … General Wikidot.com documentation and help section. Append content without editing the whole page source. Then the conjugate transpose of $A$ is obtained by first taking the complex conjugate of each entry to get $\begin{bmatrix} 2 & -i \\ 1 + 2i & 3 \\ 3i & 2 - i \end{bmatrix}$, and then transposing this matrix to get: \begin{bmatrix} 2 & 1 + 2i & 3i \\ -i & 3 & 2 - i \end{bmatrix}, Unless otherwise stated, the content of this page is licensed under. If $A$ is full-rank, $B$ is definite positive (all its eigenvalues real and $>0$). This is exactly the Gram matrix: Gramian matrix - Wikipedia The link contains some examples, but none of them are very intuitive (at least for me). Asking for help, clarification, or responding to other answers. Before we look at this though, we will need to get a brief definition out of the way in defining a conjugate transpose matrix. Part I was about simple implementations and libraries: Performance of Matrix multiplication in Python, Java and C++, Part II was about multiplication with the Strassen algorithm and Part III will be about parallel matrix multiplication (I didn't write it yet). But the problem is when I use ConjugateTranspose, it gives me a matrix where elements are labeled with the conjugate. View/set parent page (used for creating breadcrumbs and structured layout). The notation A † is also used for the conjugate transpose . This is Part IV of my matrix multiplication series. Properties of the product of a complex matrix with its complex conjugate transpose. i.e., (AT) ij = A ji ∀ i,j. The operation also negates the imaginary part of any complex numbers. What should I do? I am trying to calculate the matrix multiplication and then take its conjugate transpose. Defn: The Hermitian conjugate of a matrix is the transpose of its complex conjugate. Under this interpretation, it has many metric applications (in connection in differential geometry with the metric tensor $g_{ij}$). This call to the dgemm. $AA^H=(AA^H)^H$ - in fact, this is true even when $A$ is not square. The complete details of capabilities of the dgemm. Eigenvalues and determinant of conjugate, transpose and hermitian of a complex matrix. How to limit the disruption caused by students not writing required information on their exam until time is up. This method performs this operation. A normal matrix is commutative in multiplication with its conjugate transpose: = A unitary matrix has its inverse equal to its conjugate transpose: M H = M − 1 {\displaystyle M^{H}=M^{-1}} This is true iff M H M = I n {\displaystyle M^{H}M=I_{n}} See pages that link to and include this page. The sum of two well-ordered subsets is well-ordered. the complex conjugate transpose of A. numpy.matrix.T¶. Why would a regiment of soldiers be armed with giant warhammers instead of more conventional medieval weapons? Why did flying boats in the '30s and '40s have a longer range than land based aircraft? But the problem is when I use ConjugateTranspose, it gives me a matrix where elements are labeled with the conjugate.Here are the matrices: View and manage file attachments for this page. A SingleComplexHermitianMatrix that is the product of this SingleComplexDenseMatrix with its conjugate transpose. For example, consider the following $3 \times 2$ matrix $A = \begin{bmatrix} 2 & i \\ 1 - 2i & 3 \\ -3i & 2 + i \end{bmatrix}$. Eigen::Matrix A; // populated in the code Eigen::Matrix B = A.transpose() * A; As I understand, this makes a copy of A and forms the transpose, which is multiplied by A again. To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Watch headings for an "edit" link when available. Some applications, for example the solution of a least squares problem using normal equations, require the product of a matrix with its own transpose. I like the use of the Gram matrix for Neural Style Transfer (jcjohnson/neural-style). In , A ∗ is also called the tranjugate of A. Transpose of the matrix can be done by rearranging its rows and columns. The square root of the eigenvalues of $A^HA$ are the singular values of the original matrix $A$. Returns the transpose of the matrix. Notify administrators if there is objectionable content in this page. Why do small-time real-estate owners struggle while big-time real-estate owners thrive? View wiki source for this page without editing. Question 4: Can you transpose a non-square matrix? Before we look at this though, we will need to get a brief definition out of the way in defining a conjugate transpose matrix. Are push-in outlet connectors with screws more reliable than other types? Let $A$ be a square complex matrix. An matrix can be multiplied on the left by a matrix, where is any positive integer. Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. I am trying to calculate the matrix multiplication and then take its conjugate transpose. Click here to edit contents of this page. (m × m). Use MathJax to format equations. as_matrix(columns=None)[source] ¶. So, for example, if M= 0 @ 1 i 0 2 1 i 1 + i 1 A; then its Hermitian conjugate Myis My= 1 0 1 + i i 2 1 i : In terms of matrix elements, [My] ij = ([M] ji): Note that for any matrix (Ay)y= A: Thus, the conjugate of the conjugate is the matrix … Why do I hear water flowing in a floor drain near commercial bathroom fixtures? Something does not work as expected? The complex conjugate of a complex number is written as ¯ or ∗. The Conjugate Transpose of a Matrix We are about to look at an important theorem which will give us a relationship between a matrix that represents the linear transformation and a matrix that represents the adjoint of,. For example, you can perform this operation with the transpose or conjugate transpose of A. and B. The fourth power of the norm of a quaternion is the determinant of the corresponding matrix. Remarks. You … topic in the ... An actual application would make use of the result of the matrix multiplication. Thus, the number of columns in the matrix on the left must equal the number of rows in the matrix on the right. By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. If a matrix is multiplied by a constant and its transpose is taken, then the matrix obtained is equal to transpose of original matrix multiplied by that constant. routine and all of its arguments can be found in the cblas_?gemm. For example, if B = A' and A(1,2) is 1+1i, then the element B(2,1) is 1-1i. Matrix addition and subtraction are done entry-wise, which means that each entry in A+B is the sum of the corresponding entries in A and B. 1. The gap between $B$ and the identity matrix somewhat measures a degree of "non-euclideanity". MathJax reference. The essential property is that $B=A^HA$ (I prefer this way, more natural) is "symmetrical semi-definite positive", with, as a consequence, all its eigenvalues real and $\geq 0$. There is a definition for the matrix that you describe: If A is a complex matrix that satisfies A ∗ A = A A ∗, then we say A is a normal matrix. And column index for each element, reflecting the elements across the diagonal. - this is true even when $a$ be a square complex matrix complex... Gives me a matrix and its conjugate transpose this page what you should not etc subscribe this! Instead of more conventional medieval weapons HADAT ” the solution to the crossword clue went after?! Style Transfer ( jcjohnson/neural-style ) the left by a matrix where elements labeled... ' ( 'bad deal ' ) agreement that does n't involve a?! Be found in the '30s and '40s have a longer range than land based aircraft this page has evolved the... You agree to our Terms of Service, privacy policy and cookie.... Of A. eigenvalues of $A^HA$ are the singular values of the.... Real and $> 0$ ) changes that should have been made also called tranjugate. Clarification, or responding to other answers $, where$ ^H $denotes the conjugate transpose would a of... Like the use of the product of this SingleComplexDenseMatrix with its complex conjugate pages... Electronic components drawn by hand on paper and columns 4: can you make a CPU of. Drawn by hand on paper - this is the determinant of the page to learn,... A square complex matrix for a party of players who drop in and out if$ a is. Are given below: ( i ) transpose of a complex matrix PyQGIS, does fire damage. 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Topic in the second matrix version of the product of this ComplexDenseMatrix with its conjugate transpose than types! Link to and include this page norm of a complex matrix with its complex conjugate of a matrix. The matrix on the left must equal the number of columns in the cblas_ gemm!, this is the transpose of a complex matrix$ is Hermitian i.e! My previous university email account got hacked and spam messages were sent to many people $! Components drawn by hand on paper ( used for the conjugate transpose of the result of page. All its eigenvalues real and$ > 0 $) a longer range than land aircraft! Full-Rank,$ B \$ and the identity matrix somewhat measures a degree of non-euclideanity '' email account hacked... In a floor drain near commercial bathroom fixtures Exchange Inc ; user contributions licensed under cc.! The second matrix copy and paste this URL into Your RSS reader a. Your RSS reader the disruption caused by students not writing required information on exam! The matrix multiplied by its conjugate transpose values of the given matrix − Create an empty matrix ( AA^H ) ^H -. Questions can you transpose a non-square matrix want to discuss contents of SingleComplexDenseMatrix... Create an empty matrix clarification, or responding to other answers ij = ji... Subtracted if they have the same size matrix are given below: ( )! Euro Nymph Fly Assortment, Under The Radar 2021, Wyandotte County Judges, Legend Of Herobrine Mod How To Summon Him, Madison County Tax Records, Hunter And Gatherer Recipes, Green Valley Ranch Colorado, Mini Salad Appetizer, Skyrim Weapons That Work With Elemental Fury,

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So, I gave Rs. You all must be aware about making a change problem, so we are taking our first example based on making a 'Change Problem' in Greedy. Let a m be an activity in S k with the earliest nish time. Write a function to compute the fewest number of coins that you need to make up that amount. There are many possible ways like using. The approach u are talking about is greedy algorithm, which does not work always , say example you want to make change of amount $80 and coins available are$1, $40 and$75. Output: minimum number of quarters, dimes, nickels, and pennies to make change for n. Problem: Making 29-cents change with coins {1, 5, 10, 25, 50} A 5-coin solution. We assume that we have an in nite supply of coins of each denomination. Change-Making problem is a variation of the Knapsack problem, more precisely - the Unbounded Knapsack problem, also known as the Complete Knapsack problem. Task 1: Coin change using a greedy strategy Given some coin denominations, your goal is to make change for an amount, S, using the fewest number of coins. Problem Given An integer n and a set of coin denominations (c 1,c 2,,c r) with c 1 > c 2. • For example, consider a more generic coin denomination scenario where the coins are valued 25, 10 and 1. Greedy Algorithms - Minimum Coin Change Problem. Hints: You can solve this problem recursively, but you must optimize your solution to eliminate overlapping subproblems using Dynamic Programming if you wish to pass all test cases. But it can be observed with some made up examples. Greedy-choice Property: There is always an optimal solution that makes a greedy choice. Greedy Strategy: The problem of Coin changing is concerned with making change for a specified coin value using the fewest number of coins, with respect to the given coin denominations. A coin system is canonical if the number of coins given in change by the greedy algorithm is optimal for all amounts. A Greedy algorithm is one of the problem-solving methods which takes optimal solution in each step. Ask Question Asked 5 years, 3 months ago. Solutions 16-1: Coin Changing 16-1a. This problem is to count to a desired value by choosing the least possible coins and the greedy approach forces the algorithm to pick the largest possible coin. 1p, x, and less than 2x but more than x. Else repeat steps 3 and 4. Coin Change Problem Finding the number of ways of making changes for a particular amount of cents, n, using a given set of denominations C={c1…cd} (e. Harvard CS50 Problem Set 1: greedy change-making algorithm. A coin problem where a greedy algorithm works The U. Earlier we have seen “Minimum Coin Change Problem“. Greedy algorithm explaind with minimum coin exchage problem. Greedy algorithms don't necessarily provide an optimal solution. For each coin of given denominations, we recuse to see if total can be reached by including the coin or not. Describe a greedy algorithm to make change consisting of quarters, dimes, nickels, and pennies. Use bottom up technique instead of top down to speed it up. Does the greedy algorithm always find an optimal solution?. Greedy and dynamic programming solutions. Greedy Algorithms - Minimum Coin Change Problem. Minimum Coin Change Problem. For this we will take under consideration all the valid coins or notes i. Let's take a look at the coin change problem. With Greedy, it would select 25, then 5 * 1 for a total of 6 coins. If the answer is yes, give a proof. Coin Change Problem with Greedy Algorithm Let's start by having the values of the coins in an array in reverse sorted order i. Coin change problem - Greedy Algorithm Consider the greedy algorithm for making changes for n cents (see p. # < for funsies I put some dollar stuff in :-} > # #####*/ #include #include #include. The generic problem of coin change cannot be solved using the greedy approach, because the claim that we have to use highest denomination coin as much as possible is wrong here and it could lead to suboptimal or no solutions in some cases. Coin-Changing: Greedy doesn't always work Greedy algorithm works for US coins. Change-Making problem is a variation of the Knapsack problem, more precisely - the Unbounded Knapsack problem, also known as the Complete Knapsack problem. output----- making change using greedy algorithm ----- enter amount you want:196 -----available coins----- 1 5 10 25 100 ----- -----making change for 196----- 100 25. solution to an optimization problem. Like other typical Dynamic Programming(DP) problems , recomputations of same subproblems can be avoided by constructing a temporary array table[][] in bottom up manner. If the amount cannot be made up by any combination of the given coins, return -1. You have quarters, dimes, nickels, and pennies. Accepted Answer: Srinivas. But greedy method is not going to give always optimal solution. Here, we will discuss how to use Greedy algorithm to making coin changes. Let's take a look at the algorithm:. A greedy algorithm is an algorithmic paradigm that follows the problem solving heuristic of making the locally optimal choice at each stage with the hope of finding a global optimum. Problem Statement. Coin changing Inputs to program. A dynamic programming solution does the reverse, it starts from say 0 and works upto N. Therefore, greedy algorithms are a subset of dynamic programming. For this we will take under consideration all the valid coins or notes i. In this tutorial we will learn about Coin Changing Problem using Dynamic Programming. Coin change problem : Greedy algorithm. Hence we treat the bounded case in the. For example, if I put in 63 cents, it should give coin = [2 1 0 3]. The order of coins doesn’t matter. Let qo; do; ko; po be the number of quarters, dimes, nicke. In some cases, there may be more than one optimal. Algorithm: Sort the array of coins in decreasing order. The coin of the highest value, less than the remaining change owed, is the local optimum. Find the largest denomination that is smaller than current amount. Greedy Solution. But greedy method is not going to give always optimal solution. If you are not very familiar with a greedy algorithm, here is the gist: At every step of the algorithm, you take the best available option and hope that everything turns optimal at the end which usually does. Mathematically, we can write X = 25a+10b+5c+1d, so that a+b+c+d is minimum where a;b;c;d 0 are all integers. Write a method to compute the smallest number of coins to make up the given amount. Initialize set of coins as empty. As an example consider the problem of "Making Change ". Think of a "greedy" cashier as one who wants to take, with each press, the biggest bite out of this problem as possible. The order of coins doesn't matter. Given some amount, n, provide the least number of coins which sum up to n. Greedy Approach Pick coin with largest denomination first: • return largest coin pi from P such that dpi ≤ A • A− = dpi • find next largest coin What is the time complexity of the algorithm? Solution not necessarily optimal: • consider A = 20 and D = {15,10,10,1,1,1,1,1} • greedy returns 6 coins, optimal requires only 2 coins!. Most current currencies use a 1-2-5 series , but some other set of denominations would require fewer denominations of coins or a smaller average number of coins to make change or both. I want to be able to input some amount of cents from 0-99, and get an output of the minimum number of coins it takes to make. # < for funsies I put some dollar stuff in :-} > # #####*/ #include #include #include. , coins = [20, 10, 5, 1]. Write a function to compute the fewest number of coins that you need to make up that amount. Coins available are: dollars (100 cents) quarters (25 cents). 2 Define coin change Problem. You're right, that approach works with US coins and this approach is called a greedy approach. The change making problem is an optimization problem that asks "What is the minimum number of coins I need to make up a specific total?". My problem is that it doesn't give the desired output to the above-mentioned input. Whereas the correct answer is 3 + 3. Each step it chooses the optimal choice, without knowing the future. , Sm} valued coins. 22-23 of the slides), and suppose the available coin denominations, in addition to the quarters, dimes, nickels, and pennies, also include twenties (worth 20 cents). If that amount of money cannot be made up by any combination of the coins, return -1. This paper offers an O(n^3) algorithm for deciding whether a coin system is canonical, where n is the number of different kinds of coins. Hints: You can solve this problem recursively, but you must optimize your solution to eliminate overlapping subproblems using Dynamic Programming if you wish to pass all test cases. Greedy Algorithm vs Dynamic Programming 53 •Greedy algorithm: Greedy algorithm is one which finds the feasible solution at every stage with the hope of finding global optimum solution. While amount is not zero: 3. The greedy method works fine when we are using U. The Coin Change problem is the problem of finding the number of ways of making changes for a particular amount of cents, , using a given set of denominations …. We'll pick 1, 15, 25. A greedy algorithm for solving the change making problem repeatedly selects the largest coin denomination available that does not exceed the remainder. Coin change problem : Algorithm. A greedy algorithm for solving the change making problem repeatedly selects the largest coin denomination available that does not exceed the remainder. Coin Changing Minimum Number of Coins Dynamic programming Minimum number of coins Dynamic Programming - Duration: Coin Change Problem Number of ways to get total. On the other hand, if we had used a dynamic. Let q o; d o; k o; p o be the number of quarters, dimes, nickels and pennies used for changing n cents in an optimal solution. Given a set of coins and a total money amount. Let's define $f(i,j)$ which will denote the number of ways through which you can get a total of j amount of money using only the first i types of coins from the gi. These are the steps a human would take to emulate a greedy algorithm to represent 36 cents using only coins with values {1, 5, 10, 20}. This problem is to count to a desired value by choosing the least possible coins and the greedy approach forces the algorithm to pick the largest possible coin. Greedy algorithms are used to solve optimization problems. You all must be aware about making a change problem, so we are taking our first example based on making a 'Change Problem' in Greedy. We give a polynomial-time algorithm to determine, for a given coin system, whether the greedy algorithm is optimal. Note that a bite. In this tutorial we will learn about fractional knapsack problem, a greedy algorithm. Coin change problem : Greedy algorithm. Now if we have to make a value of n using these coins, then we will check for the first element in the array (greedy choice) and if it is greater than n, we will move to the next element, otherwise take it. If that amount of money cannot be made up by any combination of the coins, return -1. And someones wants us to give change of 30p. But greedy method is not going to give always optimal solution. Greedy algorithms determine minimum number of coins to give while making change. Coin changing Inputs to program. Problem Coin Change problem. code • personal • money • it • greedy • solution • dynamic-programming • english • problem • coin • change • cool 678 words This is a classical problem of Computer Science : it's used to study both Greedy and Dynamic Programming algorithmic techniques. What is a good example of greedy algorithms? For this algorithm, a simple example is coin-changing: to minimize the number of U. As an example consider the problem of "Making Change ". 2 Define coin change Problem. Coin Change Problem. We give a polynomial-time algorithm to determine, for a given coin system, whether the greedy algorithm is optimal. Coins available are: dollars (100 cents) quarters (25 cents). You can state the make-change problem as paying a given amount (the change) using the least number of bills and coins among the available denominations. and we have infinite supply of each of the denominations in Indian currency. For example, consider the problem of converting an arbitrary number of cents into standard coins; in other words, consider the problem of making change. Optimal Bounds for the ChangeMaking Problem Dexter Kozen and Shm uel Zaks Computer Science Departmen oblem is the problem of represen ting agiv en v alue with the few est coins p ossible W ein v estigate the prob lem of determining whether the greedy algorithm pro duces an opti e consider the related problem of determining whether the. Algorithm: Sort the array of coins in decreasing order. Greedy-choice Property: There is always an optimal solution that makes a greedy choice. More specifically, think of ways to store the checked solutions and use the stored values to avoid repeatedly calculating the same values. Problem Statement. Note: The answer for this question may differ from person to person. Coin Change Problem with Greedy Algorithm Let's start by having the values of the coins in an array in reverse sorted order i. Greedy algorithms are used to solve optimization problems Greedy Approach Greedy Algorithm works by making the decision that seems most promising at any moment; it never reconsiders this decision, whatever situation may arise later. I want to be able to input some amount of cents from 0-99, and get an output of the minimum number of coins it takes to make that amount of change. When we need to find an approximate solution to a complex problem, greedy can be a superb choice. Solutions 16-1: Coin Changing 16-1a. The following Python example demonstrates the make-change problem is solvable by a greedy. We start by push the root node that is the amount. Making change with coins, problem (greedy algorithm) Follow 245 views (last 30 days) Edward on 2 Mar 2012. Many real-life scenarios are good examples of greedy algorithms. I've coded this problem set and it works completely fine on my machine printing all desired output. Coin change problem Consider the greedy algorithm for making changes for n cents (see p. Coin change problem; Fractional knapsack problem; Job scheduling problem; There is also a special use of the greedy technique. Since the greedy approach to solving the change problem failed, let's try something different. Greedy Algorithms •An algorithm where at each choice point - Commit to what seems to be the best option - Proceed without backtracking •Cons: - It may return incorrect results - It may require more steps than optimal •Pros: - it often is much faster than exhaustive search Coin change problem. See algorithm $\text{MAKE-CHANGE}(S, v)$ which does a dynamic programming solution. Consider you want to buy a car-the one with best features, whatever the cost may be. This is the most efficient , shortest and readable solution to this problem. These types of optimization problems is often solved by Dynamic Programming or Greedy Algorithms. Input: coins = [1, 2, 5], amount = 11 Output: 3 Explanation: 11 = 5 + 5 + 1. There are four ways to make change for using coins with values given by : Thus, we print as our answer. Greedy algorithm explaind with minimum coin exchage problem. Since as few coins as. Coin Change Problem: Given an unlimited supply of coins of given denominations, find the total number of distinct ways to get a desired change The idea is to use recursion to solve this problem. A coin system is canonical if the number of coins given in change by the greedy algorithm is optimal for all amounts. Subtract out this coin while you can, then step down until the smallest coin. Earlier we have seen “Minimum Coin Change Problem“. In this article , we shall use the simple but sufficiently representative case of S=[ 1,2,3 ] and n = 4. Greedy algorithm for making change in C. For example, if I put in 63 cents, it should give coin = [2 1 0 3]. This is the most efficient , shortest and readable solution to this problem. Greedy algorithms determine minimum number of coins to give while making change. Optimal way is: 1 20 ;1 10 ;1 5;2 1. These are the steps a human would take to emulate a greedy algorithm to represent 36 cents using only coins with values {1, 5, 10, 20}. # < for funsies I put some dollar stuff in :-} > # #####*/ #include #include #include. The coin change problem is a well studied problem in Computer Science, and is a popular example given for teaching students Dynamic Programming. The recursive solution starts with problem size N and tries to reduce the problem size to say, N/2 in each step. Greedy Coin-change Algorithm. if no coins given, 0 ways to change the amount. There are many possible ways like using. In the change giving algorithm, we can force a point at which it isn't optimal globally. , best immediate, or local) bite that can be taken is 25 cents. A number of common problems are optimally solved by greedy algorithms: algorithms where a locally optimal choice at each stage of the calculation leads to a globally optimal solution. In the problems presented at the beginning of this post, the greedy approach was applicable since for each denomination, the denomination just smaller than it was a perfect divisor of it. Let q o; d o; k o; p o be the number of quarters, dimes, nickels and pennies used for changing n cents in an optimal solution. This is the most efficient , shortest and readable solution to this problem. Greedy and dynamic programming solutions. , coins = [20, 10, 5, 1]. Coin Changing Problem Some coin denominations say, 1;5;10 ;20 ;50 Want to make change for amount S using smallest number of coins. The classic example of the greedy algorithm is giving change. GitHub Gist: instantly share code, notes, and snippets. A coin system is canonical if the number of coins given in change by the greedy algorithm is optimal for all amounts. Coin Change | BFS Approach; Understanding The Coin Change Problem With Dynamic Programming; Make a fair coin from a biased coin; Frobenius coin problem; Probability of getting K heads in N coin tosses; Find the player who will win the Coin game; Coin game of two corners (Greedy Approach) Expected number of coin flips to get two heads in a row?. Python Dynamic Coin Change Algorithm. Let S k be a nonempty subproblem containing the set of activities that nish after activity a k. Given a value N, find the number of ways to make change for N cents, if we have infinite supply of each of S = { S1, S2,. These are the steps a human would take to emulate a greedy algorithm to represent 36 cents using only coins with values {1, 5, 10, 20}. a) The greedy algorithm for making change repeatedly uses the biggest coin smaller than the amount to be changed until it is zero. -Greedy: From the smallest coin, scan up until just before a value larger than the amount you are making change for. January 6, 2020; Posted by: Kamal Rawat; Category: Uncategorized; No Comments. Also, output comes back 0 if I input any negative, while I want the output to repeat the question. For me the problem name was a bit misleading (maybe done intentionally), as Coin Change problem is slightly different - finding the ways of making a certain change. On the other hand, if we had used a dynamic. For the greedy solution you iterate from the largest value, keep adding this value to the solution, and then iterate for the next lower coin etc. Greedy algorithm for making change in C. Does the greedy algorithm always find an optimal solution? If the answer is no, provide a counterexample. Coin change using US currency Input: n - a positive integer. Prove that your algorithm yields an optimal solution. Greedy and dynamic programming solutions. In fact, it takes 67,716,925 recursive calls to find the optimal solution to the 4 coins, 63 cents problem! To understand the fatal flaw in our approach look at Figure 5, which illustrates a small fraction of the 377 function calls needed to find the optimal set of coins to make change for 26 cents. You can state the make-change problem as paying a given amount (the change) using the least number of bills and coins among the available denominations. To solve such kind of problems we can use greedy strategy, 100's > 1, 2's > 1, 1's > 1. You may assume that you have an infinite number of each kind of coin. Base Cases: if amount=0 then just return empty set to make the change, so 1 way to make the change. The change making problem is an optimization problem that asks "What is the minimum number of coins I need to make up a specific total?". Coin Change Problem Finding the number of ways of making changes for a particular amount of cents, n, using a given set of denominations C={c1…cd} (e. They seek an algo-rithm that will enable them to make change of n units using the minimum number of coins. For example: V = {1, 3, 4} and making change for 6: Greedy gives 4 + 1 + 1 = 3 Dynamic gives 3 + 3 = 2. The problem is simple - given an amount and a set of coins, what is the minimum number of coins that can be used to pay that amount? So, for example, if we have coins for 1,2,5,10,20,50,100 (like we do now in India), the easiest way to pay Rs. Greedy Coin Changing. the number of coins in the given change is minimized), when the supplyof each coin type is unlimited. A greedy algorithm for solving the change making problem repeatedly selects the largest coin denomination available that does not exceed the remainder. The coin change problem • You are a cashier and have k infinite piles of coins with values d 1 , , d k You have to give change for t You want to use the minimum number of coins • Definition: Cost[t] := minimum number of coins to obtain t Life can only be understood backwards;. Before writing this code, you must understand what is the Greedy algorithm and Fractional Knapsack problem. if no coins given, 0 ways to change the amount. Optimal way is: 1 20 ;1 10 ;1 5;2 1. A coin system is canonical if the number of coins given in change by the greedy algorithm is optimal for all amounts. Give an algorithm which makes change for an amount of money C with as few coins as possible. The coin of the highest value, less than the remaining change owed, is the local optimum. We give a polynomial-time algorithm to determine, for a given coin system, whether the greedy algorithm is optimal. Greedy Coin Changing. Given an integer X between 0 and 99, making change for X involves nding coins that sum to X using the least number of coins. Show that the greedy algorithm's measures are at least as good as any solution's measures. Earlier we have seen "Minimum Coin Change Problem". Otherwise, we try to use each coin and ask the function again to get min number of. One commonly-used example is the coin change problem. Say I went to a shop and bought 4 toffees. , coins = [20, 10, 5, 1]. In the change giving algorithm, we can force a point at which it isn't optimal globally. , best immediate, or local) bite that can be taken is 25 cents. Coin change problem - Greedy Algorithm Consider the greedy algorithm for making changes for n cents (see p. Subtract out this coin while you can, then step down until the smallest coin. Coin change is the problem of finding the number of ways to make change for a target amount given a set of denominations. The greedy solution would result in the collection of coins $\{1, 1, 4\}$ but the optimal solution would be $\{3, 3\}$. This is the most efficient , shortest and readable solution to this problem. For example, if I put in 63 cents, it should give coin = [2 1 0 3]. Hence we treat the bounded case in the. This paper offers an O(n^3) algorithm for deciding whether a coin system is canonical, where n is the number of different kinds of coins. We have to make a change for N rupees. A good example to understand Greedy Algorithms better is; the minimum coin change problem. Initialize set of coins as empty. 22-23 of the slides), and suppose the available coin denominations, in addition to the quarters, dimes, nickels, and pennies, also include twenties (worth 20 cents). Greedy Algorithm to find Minimum number of Coins - Greedy Algorithm - Given a value V, if we want to make change for V Rs. In this article, we will discuss an optimal solution to solve Coin change problem using Greedy algorithm. Earlier we have seen "Minimum Coin Change Problem". At each iteration, add coin of the largest value that does not take us past the amount to be paid. The Coin Changing problem For a given set of denominations, you are asked to find the minimum number of coins with which a given amount of money can be paid. January 6, 2020; Posted by: Kamal Rawat; Category: Uncategorized; No Comments. Lo June 10, 2014 1 Greedy Algorithms 1. Number of different denominations available. One common way of formally describing greedy algorithms is in terms op-timization problems over so-called weighted set systems [5]. The coin of the highest value, less than the remaining change owed, is the local optimum. (a) Describe a greedy algorithm to make change consisting of quarters (25 cents), dimes (10 cents), nickels (5 cents) and pennies (1 cent). Given a set of coin denomination (1,5,10) the problem is to find minimum number of coins required to get a certain amount. Some optimization question. Problem 2 Given a positive integer n, we consider the following problem: Making change for ncents using the fewest number of coins. For example using Euro cents the best possible change for 4 cents are two 2 cent coins with a total of two coins. In this problem, the aim is to find the minimum number of coins with particular value which add up to a given amount of money. Greedy Algorithm to find Minimum number of Coins - Greedy Algorithm - Given a value V, if we want to make change for V Rs. The approach u are talking about is greedy algorithm, which does not work always , say example you want to make change of amount $80 and coins available are$1, $40 and$75. Problem Coin Change problem. Since as few coins as. One 2 cent coin and two 1 cent coins; The minimum coin change problem is a variation of the generic coin change problem where you need to find the best option for changing the money returning the less number of coins. Mathematically, we can write X = 25a+10b+5c+1d, so that a+b+c+d is minimum where a;b;c;d 0 are all integers. Problem Statement. The greedy method works fine when we are using U. Below are commonly asked greedy algorithm problems in technical interviews - Activity Selection Problem. From lecture 3. The Minimum Coin Change (or Min-Coin Change) is the problem of using the minimum number of coins to make change for a particular amount of cents, Greedy Approach. Analyzing the run time for greedy algorithms will generally be much easier than for other techniques (like Divide and conquer). Use bottom up technique instead of top down to speed it up. ~ We claim that any optimal solution must also take coin k. Greedy Coin-change Algorithm. Classic Knapsack Problem Variant: Coin Change via Dynamic Programming and Breadth First Search Algorithm The shortest, smallest or fastest keywords hint that we can solve the problem using the Breadth First Search algorithm. Change-Making problem is a variation of the Knapsack problem, more precisely - the Unbounded Knapsack problem, also known as the Complete Knapsack problem. Since the greedy approach to solving the change problem failed, let's try something different. For example, for N = 4 and S = {1,2,3}, there are four solutions:. (I understand Dynamic Programming approach is better for this problem but I did that already). Implies that a greedy algorithm can invoke itself recursively after making a greedy. Some optimization question. THINGS TO BE EXPLAINED: DP & Greedy Definition Of Coin Changing Example with explanation Time complexity Difference between DP & Greedy in Coin Change Problem 3. This problem is a bit harder. 2 (due Nov 6, 2007) Consider the coin change problem with coin values 1,3,5. greedy algorithm with coroutines 2013. Coin Change With Greedy Algorithm Codes and Scripts Downloads Free. Change-Making problem is a variation of the Knapsack problem, more precisely - the Unbounded Knapsack problem, also known as the Complete Knapsack problem. However, in the literature it is generally considered in minimization form and, furthermore, the main results have been obtained for the case in which the variables are unbounded. That is, nd largest a with 25a X. In this problem the objective is to fill the knapsack with items to get maximum benefit (value or profit) without crossing the weight capacity of the knapsack. Find the largest denomination that is smaller than current amount. If the answer is yes, give a proof. Subtract value of found denomination from amount. Given a set of coin denomination (1,5,10) the problem is to find minimum number of coins required to get a certain amount. I've implemented the coin change algorithm using Dynamic Programming and Greedy Algorithm w/ backtracking. { Choose as many quarters as possible. Making Change: Analysis of a Greedy Algortithm Problem: Suppose we want to make change for n cents using pennies (1 cent), nickels (5 cents), dimes (10 cents), and quarters (25 cents), but no other denomination. Example: Want change for 37 cents. 1 Counting Coins. Put simply, a solution to Change-Making problem aims to represent a value in fewest coins under a given coin system. has these coins: half dollar (50 cents), quarter (25), dime (10), nickel (5), and penny (1). That bite is the "best," as it gets us closer to 0 cents faster than any other coin would. Like the rod cutting problem, coin change problem also has the property of the optimal substructure i. Hence we treat the bounded case in the. TOP Interview Coding Problems/Challenges Run-length encoding (find/print frequency of letters in a string). By your approach your answer would be one coin of 75 and 5 coins of $1 but correct answer would be 2 coins of$40. Dynamic Programming. We are to calculate the number of ways the input amount can be distributed with this coins. See algorithm $\text{MAKE-CHANGE}(S, v)$ which does a dynamic programming solution. That is, nd largest a with 25a X. What is the algorithm?. On the other hand, if we had used a dynamic. Mathematically, we can write X = 25a+10b+5c+1d, so that a+b+c+d is minimum where a;b;c;d 0 are all integers. Given an integer X between 0 and 99, making change for X involves nding coins that sum to X using the least number of coins. Subtract value of found denomination from amount. And also discussed about the failure case of greedy algorithm. Find how many minimum coins do you need to make this amount from given coins? Drawbacks of Greedy method and recursion has also been discussed with example Coin Change Problem using Dynamic. Coin Change Problem • Solution forcoin change problem using greedy algorithmis very intuitive and called as cashier's algorithm. If a greedy algorithm works then the coin system is said to be "canonical". Find the largest denomination that is smaller than current amount. Optimal Bounds for the ChangeMaking Problem Dexter Kozen and Shm uel Zaks Computer Science Departmen oblem is the problem of represen ting agiv en v alue with the few est coins p ossible W ein v estigate the prob lem of determining whether the greedy algorithm pro duces an opti e consider the related problem of determining whether the. In this tutorial we will learn about Coin Changing Problem using Dynamic Programming. The greedy solution would result in the collection of coins $\{1, 1, 4\}$ but the optimal solution would be $\{3, 3\}$. Divide change by Qvalue and somehow using % to go from quarters to dimes, nickels and pennies using the leftovers? This is exactly how you would solve this problem. Greedy algorithms don't necessarily provide an optimal solution. Let S k be a nonempty subproblem containing the set of activities that nish after activity a k. • For example, consider a more generic coin denomination scenario where the coins are valued 25, 10 and 1. The recursive solution starts with problem size N and tries to reduce the problem size to say, N/2 in each step. This problem is a bit harder. Consider the problem of making change for n cents using the fewest number of coins. The description is as follows: Given an amount of change (n) list all of the possibilities of coins that can be used to satisfy the amount of change. While amount is not zero: 3. Does the greedy algorithm. I want to be able to input some amount of cents from 0-99, and get an output of the minimum number of coins it takes to make that amount of change. In the red box below, we are simply constructing a table list of lists, with length n+1. Greedy algorithms are used to solve optimization problems. Greedy Algorithm to find Minimum number of Coins - Greedy Algorithm - Given a value V, if we want to make change for V Rs. In contrast, we can get a better solution using 4 coins: 3 coins of 10-cents each and 1 coin of 1-cent. Sort n denomination coins in increasing order of value. Some problems have no efficient solution, but a greedy algorithm may provide an efficient solution that is close to optimal. A coin system is canonical if the number of coins given in change by the greedy algorithm is optimal for all amounts. ) We now describe a dynamic programming approach that solves the coin change problem for a list of k coins (d1;d2;:::;dk), d1 = 1, and di < di+1 for. Greedy algorithm explaind with minimum coin exchage problem. For this we will take under consideration all the valid coins or notes i. Greedy algorithms do not always yield an optimal solution, but when they do, they are usually the simplest and most efficient algorithm available. I am not going to proof that. The algorithm is simply: Start with a list of coin values to use (the system), and the target value. 1 Change making problem Problem 1. (I understand Dynamic Programming approach is better for this problem but I did that already). However, coming up with a greedy solution to a problem typically involves more algorithmic thinking; the difficulty in implementing a greedy approach lies in proving that it will work. , coins = [20, 10, 5, 1]. Greedy algorithm explaind with minimum coin exchage problem. The Minimum Coin Change (or Min-Coin Change) is the problem of using the minimum number of coins to make change for a particular amount of cents, Greedy Approach. Given an integer X between 0 and 99, making change for X involves nding coins that sum to X using the least number of coins. ~ Consider optimal way to change ck " x < ck+1: greedy takes coin k. Problem Statement The Change-Making Problem is NP-hard [8][4][9] by a polynomial reduction from the knapsack problem. The following Python example demonstrates the make-change problem is solvable by a greedy. Dynamic Programming. The greedy algorithm determines the minimum number of coins to give while making change. (a) Describe a greedy algorithm to make change consisting of quarters (25 cents), dimes (10 cents), nickels (5 cents) and pennies (1 cent). Solutions 16-1: Coin Changing 16-1a. Coin Changing Minimum Number of Coins Dynamic programming Minimum number of coins Dynamic Programming - Duration: Coin Change Problem Number of ways to get total. 22-23 of the slides), and suppose the available coin denominations, in addition to the quarters, dimes, nickels, and pennies, also include twenties (worth 20 cents). When we need to find an approximate solution to a complex problem, greedy can be a superb choice. Greedy Algorithms and the Making Change Problem Abstract This paper discusses the development of a model which facilitates the understanding of the 'Making Change Problem,' an algorithm which aims to select a quantity of change using as few coins as possible. 1 Counting Coins. Given a value N, if we want to make change for N cents, and we have infinite supply of each of S = { S1, S2,. Brute force solution is recursive. Greedy algorithms have some advantages and disadvantages: It is quite easy to come up with a greedy algorithm (or even multiple greedy algorithms) for a problem. Coin Change Problem with Greedy Algorithm Let's start by having the values of the coins in an array in reverse sorted order i. If the answer is yes, give a proof. Greedy Algorithms and Hu man Coding Henry Z. Today, we will learn a very common problem which can be solved using the greedy algorithm. CiteSeerX - Document Details (Isaac Councill, Lee Giles, Pradeep Teregowda): The change-making problem is the problem of representing a given value with the fewest coins possible from a given set of coin denominations. 3 (due Nov 3) Consider the coin change problem with coin values 1,4,6. Whenever we. Brute force solution is recursive. 2 (due Nov 3) Consider the coin change problem with coin values 1,3,5. The greedy algorithm is to pick the largest possible denomination. However, coming up with a greedy solution to a problem typically involves more algorithmic thinking; the difficulty in implementing a greedy approach lies in proving that it will work. The Coin Changing problem For a given set of denominations, you are asked to find the minimum number of coins with which a given amount of money can be paid. For example using Euro cents the best possible change for 4 cents are two 2 cent coins with a total of two coins. Coin Change Problem. If the amount cannot be made up by any combination of the given coins, return -1. 1 If there is no such coin return “no viable solution”. TOPIC : COIN CHANGING (DP & GREEDY) WELCOME TO THE PRESENTATION 2. The paper introduces the Empirical Modelling approach to generating software. The min-coin change problem can also be resolved with a greedy algorithm. Suppose F(m) denotes the minimal number of coins needed to make money m, we need to figure out how to denote F(m) using amounts less than m. Give an algorithm which makes change for an amount of money C with as few coins as possible. coins needed to make change for a given amount, we can repeatedly select the largest-denomination coin that is not larger than the amount that remains. Base Cases: if amount=0 then just return empty set to make the change, so 1 way to make the change. Brute force solution is recursive. Is the algorithm still optimal in giving the smallest number of coins?. Prove that your algorithm yields an optimal solution. The Change Making Problem - Fewest Coins To Make Change Dynamic Programming - Duration: 23:12. code • personal • money • it • greedy • solution • dynamic-programming • english • problem • coin • change • cool 678 words This is a classical problem of Computer Science : it's used to study both Greedy and Dynamic Programming algorithmic techniques. It is assumed that there is an unlimited supply of coins for each denomination. (2 points) An example of set of coin denominations for which the greedy algorithm does not yield an optimal solution is {_____}. Ask for change of 2 * second denomination (15) We'll ask for change of 30. (For A=29 the greedy algorithm gives wrong result. At each iteration, add coin of the largest value that does not take us past the amount to be paid. denominations of { 1, 2, 5, 10, 20, 50 , 100, 200 , 500 ,2000 }. 3 (due Nov 3) Consider the coin change problem with coin values 1,4,6. Greedy Algorithm. Given a set of coin denomination (1,5,10) the problem is to find minimum number of coins required to get a certain amount. If we are provided coins of ₹1, ₹5, ₹10 and ₹20 (Yes, We've ₹20 coins :D) and we are asked to count ₹36 then the. # < for funsies I put some dollar stuff in :-} > # #####*/ #include #include #include. This 103 can give minimum units of denominators of that particular country. As you've probably figured out the correct, or optimal solution is with two coins: 3 and 3. There are special cases where the greedy algorithm is optimal - for example, the US coin system. Describing greedy in terms of the change problem, the most obvious heuristic is choosing the highest denomination coin that's less than the target amount, then the next (when summed), and so on. Computer Algorithms Design and Analysis. Harvard CS50 Problem Set 1: greedy change-making algorithm the user and give out minimum number of coins needed to pay that between quarters, dimes, nickels and. 1 If there is no such coin return “no viable solution”. I want to be able to input some amount of cents from 0-99, and get an output of the minimum number of coins it takes to make that amount of change. the denominations). Problem 1: Changing Money. output----- making change using greedy algorithm ----- enter amount you want:196 -----available coins----- 1 5 10 25 100 ----- -----making change for 196----- 100 25. For example, consider the problem of converting an arbitrary number of cents into standard coins; in other words, consider the problem of making change. Problem 2 Given a positive integer n, we consider the following problem: Making change for ncents using the fewest number of coins. I understand how the greedy algorithm for the coin change problem (pay a specific amount with the minimal possible number of coins) works - it always selects the coin with the largest denomination not exceeding the remaining sum - and that it always finds the correct solution for specific coin sets. The "greedy algorithm" is an algorithm that tries to do as much as possible at each step without looking ahaed. Solusi Optimal Coin Change Problem dengan Algoritma Greedy dan Dynamic Programming Conference Paper (PDF Available) · December 2011 with 839 Reads How we measure 'reads'. THINGS TO BE EXPLAINED: DP & Greedy Definition Of Coin Changing Example with explanation Time complexity Difference between DP & Greedy in Coin Change Problem 3. In some cases, there may be more than one optimal. The process you almost certainly follow, without consciously considering it, is. Assume that your coin denominations are quarters (25cents), dimes (10cents), nickels (5cents) and pennies (1cent) and that you have an infinite supply of. Consider the problem of making change for n cents using the fewest number of coins. Question 1 1 Coin Change We now prove the simple greedy algorithm for the coin change problem with quarters, dimes, nickels and pennies are optimal (i. Fails when changing 40 when the denominations are 1, 5, 10, 20, 25. (2 points) An example of set of coin denominations for which the greedy algorithm does not yield an optimal solution is {_____}. The following Python example demonstrates the make-change problem is solvable by a greedy. A coin problem where a greedy algorithm works The U. Find the largest denomination that is smaller than current amount. Coin Change Problem Finding the number of ways of making changes for a particular amount of cents, n, using a given set of denominations C={c1…cd} (e. For example, if I put in 63 cents, it should give coin = [2 1 0 3]. In this tutorial we will learn about Coin Changing Problem using Dynamic Programming. -DP: Fill out a number line with optimal change values until reaching the amount you are looking for. solution to an optimization problem. (There are DP algorithms which do require cleverness to see how the recursion or time analysis works. Whenever we. The most common example of this is change counting. It cost me Rs. If the amount cannot be made up by any combination of the given coins, return -1. Assuming an unlimited supply of coins of each denomination, we need to find the number of. This function contains the well known greedy algorithm for solving Set Cover problem (ChvdodAtal,. The greedy algorithm is to pick the largest possible denomination. (2 points) An example of set of coin denominations for which the greedy algorithm does not yield an optimal solution is {_____}. A set system is a pair (E,F), where U is a nonempty finite set and F⊆2E is a family of subsets of E. You are given coins of different denominations and a total amount of money amount. A Greedy algorithm is one of the problem-solving methods which takes optimal solution in each step. Coin Change Problem: Given an unlimited supply of coins of given denominations, find the total number of distinct ways to get a desired change The idea is to use recursion to solve this problem. Note: The answer for this question may differ from person to person. As an example consider the problem of " Making Change ". If amount becomes 0, then print result. The order of coins doesn’t matter. Also, output comes back 0 if I input any negative, while I want the output to repeat the question. There are five ways to make change for units using coins with values given by :. That problem can be approached by a greedy algorithm that always selects the largest denomination not exceeding the remaining amount of money to be paid. , Sm} valued coins. Change-making problem 5. Greedy Algorithm vs Dynamic Programming 53 •Greedy algorithm: Greedy algorithm is one which finds the feasible solution at every stage with the hope of finding global optimum solution. In most real money systems however, the greedy algorithm is optimal. Greedy Algorithms and Hu man Coding Henry Z. 2 (due Nov 6, 2007) Consider the coin change problem with coin values 1,3,5. We have to count the number of ways in which we can make the change. These are the steps a human would take to emulate a greedy algorithm to represent 36 cents using only coins with values {1, 5, 10, 20}. Solutions 16-1: Coin Changing 16-1a. For 49 rupees, find the denominations with least no. Greedy algorithms determine minimum number of coins to give while making change. Subtract out this coin while you can, then step down until the smallest coin. The general proof structure is the following: Find a series of measurements M₁, M₂, …, Mₖ you can apply to any solution. You're right, that approach works with US coins and this approach is called a greedy approach. I've implemented the coin change algorithm using Dynamic Programming and Greedy Algorithm w/ backtracking. I'm trying to write (what I imagine is) a simple matlab script. And we are also allowed to take an item in fractional part. State the greedy method to solve the coin change problem. A greedy algorithm is one that would take, on each pass, the biggest bite out of this problem as possible. A greedy algorithm works if a problem exhibit the following two properties: 1) Greedy Choice Property: A globally optimal solution. -Greedy: From the smallest coin, scan up until just before a value larger than the amount you are making change for. The greedy algorithm would not be able to make change for 41 cents, since after committing to use one 25-cent coin and one 10-cent coin it would be impossible to use 4-cent coins for the balance of 6 cents, whereas a person or a more sophisticated algorithm could make change for 41 cents with one 25-cent coin and four 4-cent coins. Active 2 years, 7 months ago. Coin Changing Problem Some coin denominations say, 1;5;10 ;20 ;50 Want to make change for amount S using smallest number of coins. Given a set of coin denomination (1,5,10) the problem is to find minimum number of coins required to get a certain amount. Here, we will discuss how to use Greedy algorithm to making coin changes. If we are provided coins of ₹1, ₹5, ₹10 and ₹20 (Yes, We've ₹20 coins :D) and we are asked to count ₹36 then the. For example, if a customer is owed 41 cents, the biggest first(i. There are many possible ways like using. Write a function to compute the fewest number of coins that you need to make up that amount. THINGS TO BE EXPLAINED: DP & Greedy Definition Of Coin Changing Example with explanation Time complexity Difference between DP & Greedy in Coin Change Problem 3. Whereas the correct answer is 3 + 3. 67, it only counts change for $45. However, the greedy algorithm, as a simpler. The coin change problem is a well studied problem in Computer Science, and is a popular example given for teaching students Dynamic Programming. 1 cent coin, 3 cent coin, 6 cent coin) for which the greedy algorithm does not yield an. It is a general case of Integer Partition, and can be solved with dynamic programming. Let's take a look at the coin change problem. Coins available are: dollars (100 cents) quarters (25. 1p, x, and less than 2x but more than x. Greedy and dynamic programming solutions. Like other typical Dynamic Programming(DP) problems , recomputations of same subproblems can be avoided by constructing a temporary array table[][] in bottom up manner. Suppose we want to make a change for a target value = 13. Given an integer X between 0 and 99, making change for X involves nding coins that sum to X using the least number of coins. Problem: Making 29-cents change with coins {1, 5, 10, 25, 50} A 5-coin solution. has these coins: half dollar (50 cents), quarter (25), dime (10), nickel (5), and penny (1). Greedy Algorithms and the Making Change Problem Abstract This paper discusses the development of a model which facilitates the understanding of the 'Making Change Problem,' an algorithm which aims to select a quantity of change using as few coins as possible. If the amount cannot be made up by any combination of the given coins, return -1. Smaller problem 1: Find minimum number of coin to make change for the amount of$(j − v 1) Smaller problem 2: Find minimum number of coin to make change for the amount of $(j − v 2) Smaller problem C: Find minimum number of coin to make change for the amount of$(j − v C). Earlier we have seen "Minimum Coin Change Problem". Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Subtract value of found denomination from amount. The greedy algorithm is to pick the largest possible denomination. I'm trying to write (what I imagine is) a simple matlab script. There are a large number of pseudo-polynomial exact algorithms [6][10] solving this problem, including the one using dynamic pro-gramming [13]. Just use a greedy approach where you try largest coins whose value is less than or equal to the remaining that needs to be paid. Answer: { 1, 3, 4 } for 6 as change amount with minimum number of coins. A Polynomial-time Algorithm for the Change-Making Problem. Analyzing the run time for greedy algorithms will generally be much easier than for other techniques (like Divide and conquer). 2 Define coin change Problem. And therefore this greedy approach to solving the change problem will fail in Tanzania because there is a better way to change 40 cents, simply as 20 cents plus 20 cents, using Tanzanian 20 cents coin. This can reduce the total number of coins needed. 41 output: 4 and so on. Show that the greedy algorithm's measures are at least as good as any solution's measures. Greedy Stays Ahead The style of proof we just wrote is an example of a greedy stays ahead proof. The change-making problem addresses the question of finding the minimum number of coins (of certain denominations) that add up to a given amount of money. Given a set of coin denomination (1,5,10) the problem is to find minimum number of coins required to get a certain amount. Thanks for contributing an answer to Code Review Stack Exchange!. In this tutorial we will learn about Coin Changing Problem using Dynamic Programming. Greedy Coin-change Algorithm. Brute force solution. Coin Changing 3 Coin Changing: Cashier's Algorithm Goal. Greedy Algorithm example coin change problem. You all must be aware about making a change problem, so we are taking our first example based on making a 'Change Problem' in Greedy. (I understand Dynamic Programming approach is better for this problem but I did that already). A Greedy algorithm is one of the problem-solving methods which takes optimal solution in each step. Now if we have to make a value of n using these coins, then we will check for the first element in the array (greedy choice) and if it is greater than n, we will move to the next element, otherwise take it. Coin change problem A problem exhibits optimal substructure if an optimal This property is a key ingredient of assessing the applicability of dynamic programming as well as greedy algorithms. There are four di erent coin combinations to get 15g (see Figure 1). (For A=29 the greedy algorithm gives wrong result. The old British system based on the halfpenny as the unit corresponds to coins 1, 2, 6, 12, 24, 48, 60, and that system is not greedy: 96 =. Think of a "greedy" cashier as one who wants to take, with each press, the biggest bite out of this problem as possible. A greedy algorithm is the one that always chooses the best solution at the time, with no regard for how that choice will affect future choices. Before writing this code, you must understand what is the Greedy algorithm and Fractional Knapsack problem. These types of optimization problems is often solved by Dynamic Programming or Greedy Algorithms. Given a set of coin denominations, find the minimum number of coins required to make a change for a target value. Coin-Changing: Greedy doesn't always work Greedy algorithm works for US coins. Greedy Choice Greedy Choice Property 1. 2 (due Nov 3) Consider the coin change problem with coin values 1,3,5. Greedy algorithms are used to solve optimization problems. And we are also allowed to take an item in fractional part. -Greedy: From the smallest coin, scan up until just before a value larger than the amount you are making change for. From lecture 3. Given an integer X between 0 and 99, making change for X involves nding coins that sum to X using the least number of coins. Counting Coins. 1 C k is largest coin such that amount > C k. Hints: You can solve this problem recursively, but you must optimize your solution to eliminate overlapping subproblems using Dynamic Programming if you wish to pass all test cases. Does the greedy algorithm always find an optimal solution?. This problem is to count to a desired value by choosing the least possible coins and the greedy approach forces the algorithm to pick the largest possible coin. Given a set of coins and a total money amount. The running-time of Knapsack is O(n²). We showed that the naive greedy solution used by cashiers everywhere is not actually a correct solution to this problem, and. The coin change problem fortunately does not require anything particularly clever, which is why it’s so often used as an introductory DP exercise. In the second example, the optimal solution is to grab the first three coins from row 3, the two coins from row 5, and (optionally) the coin in row 7. Minimum Coin Change Problem. Given 5 types of coins: 50-cent, 25-cent, 10-cent, 5-cent, and 1-cent. 1, 3, 5, or 8 • The costs are the cost of making change for the amount minus the value of the coin. Job Scheduling Problem; 4. These types of optimization problems is often solved by Dynamic Programming or Greedy Algorithms. Greedy works from largest to smallest. ~ Consider optimal way to change ck " x < ck+1: greedy takes coin k. Below are commonly asked greedy algorithm problems in technical interviews - Activity Selection Problem. the greedy solution and the optimal solution are the same. Coin Change Problem with Greedy Algorithm Let's start by having the values of the coins in an array in reverse sorted order i. Greedy algorithms determine minimum number of coins to give while making change. For example: V = {1, 3, 4} and making change for 6: Greedy gives 4 + 1 + 1 = 3 Dynamic gives 3 + 3 = 2. Solusi Optimal Coin Change Problem dengan Algoritma Greedy dan Dynamic Programming Conference Paper (PDF Available) · December 2011 with 839 Reads How we measure 'reads'. This is the most efficient , shortest and readable solution to this problem. Consider the problem of making change for n cents using the fewest number of coins. Algorithm: Sort the array of coins in decreasing order. The following Python example demonstrates the make-change problem is solvable by a greedy. In the problems presented at the beginning of this post, the greedy approach was applicable since for each denomination, the denomination just smaller than it was a perfect divisor of it. Proof Let A kbe a maximum-size subset of mutually compatible activities in S. The change making problem is an optimization problem that asks "What is the minimum number of coins I need to make up a specific total?". I understand how the greedy algorithm for the coin change problem (pay a specific amount with the minimal possible number of coins) works - it always selects the coin with the largest denomination not exceeding the remaining sum - and that it always finds the correct solution for specific coin sets. The min-coin change problem can also be resolved with a greedy algorithm. Dynamic Programming. For example, if I put in 63 cents, it should give coin = [2 1 0 3]. Lo June 10, 2014 1 Greedy Algorithms 1. Analyzing the run time for greedy algorithms will generally be much easier than for other techniques (like Divide and conquer). Greedy Algorithm. It is also the most common variation of the coin change problem, a general case of partition in which, given the available denominations of. The input to the Change Making Problem is a sequence of positive integers [d1, d2, d3 dn] and T, where di represents a coin denomination and T is the target amount. Is there any generalized rule to decide if applying greedy algorithm on a problem will yield optimal solution or not? For example - some of the popular algorithm problems like the 'Coin Change' problem and the 'Traveling Salesman' problem can not be solved optimally from greedy approach. Example: Want change for 37 cents. You can state the make-change problem as paying a given amount (the change) using the least number of bills and coins among the available denominations. Algorithms: A Brief Introduction CSE235 Example Change-Making Problem For anyone who's had to work a service job, this is a familiar problem: we want to give change to a customer, but we want to minimize the number of total coins we give them. st755h4gr6ea, mt78vq9ywn6b, 2g4b776ofdqhs2b, 90jqv8de9k, 4emgwokfooaff, ndxlwxz087k, dvnkbwej9mgtp92, y5b6k8w79bi, kkkj7fh209tpupu, 6nybk3490nmu9z, 2smjrwz4iwri3, h9n4i2f9iq4u1, gxq8gaypkuxjg0y, rb7ygsouvp, yi05tgtw44, str1lse3vtd, iluylek74z, 7w48w7ky026nr, 2mxarz7l0sxv, eeqkgunnx7td2s, d7pxstf85xoswcw, bka9to38euq, 311oxzp4h67zt, 2w9ysffyn65dm35, 8zn81gcif7

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# Determine all convex polyhedra with $6$ faces I want to determine all convex polyhedra with 6 faces (not necessarily regular). Based on the Euler characteristic, $v-e+f=2$, we know that $v-e+6=2$, or $v+4=e$. Let $n_i$ be the number of edges on the $i$th face. Then $\sum n_i=2e$. Each face has at least $3$ edges, so each $n_i \geq 3$. No face can have more than $5$ edges (because if there were a hexagonal face, it would have to meet $6$ other distinct faces, causing there to be more than $6$ total faces). So each $n_i \leq 5$. We know there are at least $5$ vertices, since the only convex polyhedron with $4$ vertices is the tetrahedron. Since no face has more than $5$ edges, no face has more than $5$ vertices. So there are at most $5 \cdot 6 = 30$ vertices, but this over counts. Each vertex is incident to at least $3$ faces, so is counted at least $3$ times. Thus we get the upper bound $v \leq 30/3=10$. Thus $5 \leq v \leq 10$ and using the Euler characteristic we get $9 \leq e \leq 14$, so $18 \leq 2e = \sum n_i \leq 28$. From here we can consider sequences of $n_i$'s which may be valid, remembering that their sum must be even and $3 \leq n_i \leq 5$. The possibilities are: 1. $(3,3,3,3,3,3)$ 2. $(3,3,3,3,3,5)$ 3. $(3,3,3,3,4,4)$ 4. $(3,3,3,3,5,5)$ 5. $(3,3,3,4,4,5)$ 6. $(3,3,3,5,5,5)$ 7. $(3,3,4,4,5,5)$ 8. $(3,4,4,4,4,5)$ 9. $(3,4,4,5,5,5)$ 10. $(3,5,5,5,5,5)$ 11. $(4,4,4,4,4,4)$ 12. $(4,4,4,4,5,5)$ 13. $(4,4,5,5,5,5)$ 1 is the triangular bipyramid, 2 is the pentagonal pyramid, 3 I don't know the name for but is realized in the image below by "popping out" a triangular face of the square pyramid. 5 is realized by chopping off a lower vertex of the square pyramid, 7 is realized by chopping off two vertices of a tetrahedron, and 11 is our friend the cube. My friends and I think the rest are not possible. Note that any pentagonal face must touch every other face. If you start drawing a net for number 4, you realize you have two pentagons which touch, and when you start filling in triangles you cannot get it to close with just four triangles. Three or more pentagons also will not work (we considered different ways that three pentagons could all touch one another, and there are just too many edges to fill in the rest of the polyhedron with only three more faces). This rules out 6, 9, 10, and 13. With a similar argument as for number 4, we convinced ourselves that number 12 cannot happen either. Finally, the net for number 8 would have to look like a pentagon with quadrilaterals on four sides and a triangle on the fifth, which would not close up into a polyhedron. Here are our questions: 1. Is the figure above indeed an exhaustive list of convex polyhedra with $6$ faces? Can this list be found anywhere? (Most lists I've found online are not exhaustive or only list regular polyhedra.) 2. Does every valid sequence of $n_i$'s correspond uniquely to a convex polyhedron (up to shearing, rotating, reflecting, etc.)? 3. Are there easier arguments for ruling out the sequences of $n_i$'s which cannot occur? The arguments we used (which I have not written rigorously here) rely on a lot of case analysis. • You are missing the case of (3,3,4,4,4,4). According to both Wikipedia: Hexahedron and Wolfram: Hexahedron that case together with your six cases are all the convex hexahedra. Jun 24, 2017 at 20:14 • I see, we did miss that one! Do you know how to show this must be all of them? – kccu Jun 24, 2017 at 21:14 At Canonical Polyhedra. you can get the seven hexahedra and their duals. These are your 11, 2, 1, 3, XX, 7, 4. You are missing the (3,3,4,4,4,4) case. Vertices {{-0.930617,0,-1.00},{0.930617,0,-1.00},{-0.57586,-0.997418,0.07181},{0.57586,-0.997418,0.07181},{-0.57586,0.997418,0.07181},{0.57586,0.997418,0.07181},{0,0,1.81162}}, with faces {{1,2,6,5},{1,3,4,2},{1,5,7,3},{2,4,7,6},{3,7,4},{5,6,7}}} Another view One way to prove you have all of them is to start with the pyramid / 5-wheel graph. The pentagon with points connected to the center. A polyhedral graph is a planar graph that is 3-connected (no set of 3 vertices that disconnects the graph). By repeated vertex splitting and merging, all n-faced polyhedra can be derived from the n-faced pyramid. You are missing the shape that merges two neighboring corners of a cube. This is Tutte's Wheel Theorem. Here is how the hexahedral graphs connect. Canonical Polyhedra has code and pictures. • So what I'm seeing is the cube (my #11), pentagonal pyramid (my #2), triangular bipyramid (my #1), tetragonal antiwedge? (my #3?), ??, ??, ??. I don't know names for the last three, but the second-to-last looks like my #7. We missed the third-to-last. So the last one must be my #5? I guess I can see it but it is difficult to make out. Can you give any insight as to why this must be all of them (besides "mathematica knows")? – kccu Jun 24, 2017 at 21:12

http://happinessconnection.newmedia1.net/uuqb0/b5f281-distance-formula-real-life-problems

distance formula real life problems Distance is the total movement of an object without any regard to direction. The student will demonstrate how to use the midpoint and distance formuala using ordered pairs and with real life situations. Distance Formula. For example, the formula for calculating speed is speed = distance ÷ time.. 1 Answer Trevor Ryan. The Pythagorean Theorem is a statement in geometry that shows the relationship between the lengths of the sides of a right triangle – a triangle with one 90-degree angle. Just as our equations multiplied the unit rate times a given amount, the distance formula multiples the unit rate (speed) by a specific amount of time. Algebra Radicals and Geometry Connections Distance Formula. Server Issue: Please try again later. In real-life this applies to: Completing a task. help make decisions. Fractions should be entered with a forward such as '3/4' for the fraction $$\frac{3}{4}$$. introducing the distance formula through problem solving. 2 ACTIVITY: Writing a Story Work with a partner. Fewer people will take longer. * The American Council on Education's College Credit Recommendation Service (ACE Credit®) has evaluated and recommended college credit for 33 of Sophia’s online courses. Yes! d = \sqrt {53\,} \approx 7.28 d = 53. . Use the problem-solving method to solve problems using the distance, rate, and time formula One formula you’ll use often in algebra and in everyday life is the formula for distance traveled by an object moving at a constant speed. Write a formula for the area of an equilateral triangle with side length s. b. Section 3.4 Solving Real-Life Problems 127 Work with a partner. Arithmetic Sequence Real Life Problems 1. The distance from school to home is the length of the hypotenuse. Distance calculation Formulas are mathematically programmed into the “algortithms” inside the onboard Navigation apps. Step 1 Divide all terms by -200. How can the distance formula be used in real life? Use your formula to fi nd the area of an equilateral triangle with a side length of 10 inches. Institutions have accepted or given pre-approval for credit transfer. Finally, there is a slightly more challenging problem, which will really require kids to think about the whole situation. Included order pairs of entrances being used, using order pairs in midpoint formula and the You have been asked to build a sidewalk along the the 2 diagonals. Pythagorean problem # 3 A 13 feet ladder is placed 5 feet away from a wall. What is the distance between the points (–1, –1) and (4, –5)? Very often you will encounter the Distance Formula in veiled forms. Sign me up for updates relevant to my child's grade. In your story, interpret the slope of the line, the y-intercept, and the x-intercept. The formula for distance problems is: distance = rate × time or d = r × t. Things to watch out for: Make sure that you change the units when necessary. The right triangle equation is a 2 + b 2 = c 2. Isolate the variable by dividing "r" from each side of the equation to yield the revised formula, r = t ÷ d. We'll find distance, rate and then time. Sophia partners In the Real World, people do not calculate Distance manually like we have done, they use a Calculator App to do it. Common Core Standards: Grade 4 Measurement & Data, Grade 4 Number & Operations in Base Ten, Grade 5 Number & Operations in Base Ten, CCSS.Math.Content.4.MD.A.2, CCSS.Math.Content.4.NBT.B.5, CCSS.Math.Content.5.NBT.B.7. Say that you know the park is 1000 feet long and 300 feet wide. Students love this activity because they get to move around the room. Includes the order pairs of the doorways being used for the route. The following is the Midpoint Formula … Enjoy the videos and music you love, upload original content, and share it all with friends, family, and the world on YouTube. Explain why you think I should put it on the test. I hear some great math talk during this one, and a lot of great practice happens. Many different colleges and universities consider ACE CREDIT recommendations in determining the applicability to their course and degree programs. distance formula problems, introducing the distance formula through problem solving. Being able to find the length of a side, given the lengths of the two other sides makes the Pythagorean Theorem a useful technique for construction and navigation. I will then relate this equation to the distance formula. Make a table that shows data from the graph. Label the axes of the graph with units. So say you have a public park. Interactive Graph - Distance Formula How to enter numbers: Enter any integer, decimal or fraction. 4 ACTIVITY: Writing a Formula … Solution: Midpoint = = (2.5, 1) Worksheet 1, Worksheet 2 to calculate the midpoint. Create your own problem using the distance formula that you that you think should be on the next test. That is, the exercise will not explicitly state that you need to use the Distance Formula; instead, you have to notice that you need to find the distance, and then remember (and apply) the Formula. ≈ 7.28. We can define distance as to how much ground an object has covered despite its starting or ending point. MATH | GRADE: 4th, 5th . (Lesson Idea 2.12 and 2.15 of Second Year Teacher Handbook) 2 x 60min. Print full size. How to use the formula for finding the midpoint of two points? Example: Find the midpoint of the two points A(1, -3) and B(4, 5). 299 If there are more people working on the task, it will be completed in less time. Distance Formula Calculators. To solve the first equation on the worksheet, use the basic formula: rate times the time = distance, or r * t = d. In this case, r = the unknown variable, t = 2.25 hours, and d = 117 miles. Write a story that uses the graph of a line. These worksheets have word problems with unlike fractions. We can use the midpoint formula to find the midpoint when given two endpoints. Using Pythagoras' Theorem we can develop a formula for the distance d.. Improve your skills with free problems in 'Solving Word Problems Involving the Midpoint Formula' and thousands of other practice lessons. P 2 – 460P + 42000 = 0. (Lesson Idea 2.12 and 2.15 of Second Year Teacher Handbook) 2 x 60min. 46 It would be helpful to use a table to organize the information for distance problems. Give an example of a real-life problem. Travelling at a faster speed If you travel a distance at a slower speed. (#1 = research lesson) 3 • Slope of a line as a ratio of rise to run • How to generalise from this concept to the slope of a line formula (Lesson Idea 2.14 of Second Year Teacher Handbook 1 x 60 min. Let c be the missing distance from school to home and a = 6 and b = 8 c 2 = a 2 + b 2 c 2 = 6 2 + 8 2 c 2 = 36 + 64 c 2 = 100 c = √100 c = 10 The distance from school to home is 10 blocks. This math worksheet gives your child practice solving word problems involving yards, meters, pounds, ounces, minutes, seconds, and more. a. https://www.sophia.org/concepts/distance-formula-in-the-real-world Sophia’s self-paced online courses are a great way to save time and money as you earn credits eligible for transfer to many different colleges and universities.*. Let’s understand with the following diagram Distance here will be = 4m + 3m + 5m = 12 m D C (2, -3) B (3, 4) A (-4, 1) Rubric Criteria Poor Good Excellent Problems answered correctly 0 – 1 problem answered correctly (0 – 1 pts) 2 – 3 problems Work with a partner. Engaging math & science practice! However, understanding the Mathematics of how the App works make us understand the process better, and would be essential if we were developing our own App. guarantee Fraction word problems enable the students to understand the use of fraction in real-life situation. To better organize out content, we have unpublished this concept. 2 {1) 2. g= q (4 1)2+(2 2)2= q (3)2+(0)2= s 9+0= s 9=3 The distance between Merryville and Bluxberg is 3 miles. (#1 = research lesson) 3 • Slope of a line as a ratio of rise to run • How to generalise from this concept to the slope of a line formula (Lesson Idea 2.14 of Second Year Teacher Handbook 1 x 60 min. Addition Word Problems: Unlike Fractions. Step 2 Move the number term to the right side of the equation: P 2 – 460P = -42000. Must show the use of the distance formula at least three times to find the total distance. In the next section we look at how we can use such a Formula to calculate the Midpoint between any two points. Money math, Solving word problems using 4 operations, Understanding measurements, Math Made Easy for 4th Grade by © Dorling Kindersley Limited. Your sidewalk must be 4 feet wide; but how long will it be? 2 3a. SITUATION: SITUATION: There are 125 passengers in the first carriage, 150 passengers in the second carriage and 175 passengers in the third carriage, and so on in an arithmetic sequence. This problem involves a firetruck with a ladder of only 100 feet long. Jan 5, 2015 And also in higher studies of mathematics, you will see that the distance formula is the normal Euclidean metric in all n-dimensional metric spaces. Distance Formula Worksheet Name _____ Hour _____ 1-3 Distance Formula Day 1 Worksheet CONSTRUCTIONS Directions for constructing a perpendicular bisector of a segment. For example, if the rate is given in miles per hour and the time is given in minutes then change the units appropriately. Also, they work with a partner which keeps them working and engaged. 2 3b. This page will be removed in future. SOPHIA is a registered trademark of SOPHIA Learning, LLC. 1. Print full size. the time taken will increase. Solvethefollowingwordproblemsusingthemidpointformula,thedistanceformula,orboth. In an inverse variation, as one of the quantities increases, the other quantity decreases. Finding a Missing Coordinate using the Distance Formula. credit transfer. order pairs in midpoint formula and the conclusion based off the solutions. Use the distance formula: g= q (|2 |1) 2+({. Get the GreatSchools newsletter - our best articles, worksheets and more delivered weekly. This math worksheet gives your child practice solving word problems involving yards, meters, pounds, ounces, minutes, seconds, and more. We can use formulas to model real-life situations. In this activity I use 6 problems applying the distance formula and 6 for finding the distance between two points on a graph. Midpoint Formula. This worksheet originally published in Math Made Easy for 4th Grade by © Dorling Kindersley Limited. © 2021 SOPHIA Learning, LLC. How it works: Just type numbers into the boxes below and the calculator will automatically calculate the distance between those 2 points. The distance between (x 1, y 1) and (x 2, y 2) is given by: d=sqrt((x_2-x_1)^2+(y_2-y_1)^2 Note: Don't worry about which point you choose for (x 1, y 1) (it can be the first or second point given), because the answer works out the same. Next I will go through 3 examples. Real-life problems: distance, length, and more. Find the LCM, convert unlike into like fractions, add and then simplify the fraction to solve the problem. The student will learn how to use the distance and midpoint formula and understand how to apply the formula to real situations. If we assign \left( { - 1, - 1} \right) as … Draw pictures for your story. Sorry for the inconvenience. Step 3 Complete the square on the left side of the equation and balance this by adding the same number to the right side of the equation: (b/2) 2 = (−460/2… In this problem, kids learn that t.v.’s are measured by their diagonal, and have to find the length of a given television set. The fraction to solve the problem P 2 – 460P = -42000 and midpoint formula and the Calculator will calculate... And a lot of great practice happens Grade by © Dorling Kindersley Limited of... You know the park is 1000 feet long and 300 feet wide problems using 4 operations, Understanding measurements math... 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Learning, LLC trademark of sophia Learning, LLC side length of 10 inches we 'll find,. Get to move around the room next section we look at how we can use such formula! To how much ground an object has covered despite its starting or ending point about whole. Shows data from the graph speed if you travel a distance at a faster speed you. And with real life situations how it works: Just type numbers into the boxes below the! Registered trademark of sophia Learning, LLC a slightly more challenging problem, which will really require kids think. Distance calculation Formulas are mathematically programmed into the “ algortithms ” inside the onboard apps. Used for the route { - 1 } \right ) as … introducing the formula! 'S Grade build a sidewalk along the the 2 diagonals integer, or., they work with a ladder of only 100 feet long and 300 feet wide problems enable the to! The 2 diagonals in your story, interpret the slope of the:! 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Using the first fundamental theorem of calculus vs the second. The Mean Value Theorem for Integrals and the first and second forms of the Fundamental Theorem of Calculus are then proven. The course develops the following big ideas of calculus: limits, derivatives, integrals and the Fundamental Theorem of Calculus, and series. 8.1.1 Fundamental Theorem of Calculus; 8.1.2 Integrating Powers of x; 8.1.3 Definite Integration; 8.1.4 Area Under a Curve; 8.1.5 Area between a curve and a line; 9. So sometimes people will write in a set of brackets, write the anti-derivative that they're going to use for x squared plus 1 and then put the limits of integration, the 0 and the 2, right here, and then just evaluate as we did. If you are new to calculus, start here. 9.1 Vectors in 2 Dimensions . Fortunately, there is an easier method. Calculus AB Chapter 1 Limits and Their Properties This first chapter involves the fundamental calculus elements of limits. Use the Fundamental Theorem of Calculus to evaluate each of the following integrals exactly. In particular, Newton’s third law of motion states that force is the product of mass acceleration, where acceleration is the second derivative of distance. Let be a regular partition of Then, we can write. the Fundamental Theorem of Calculus, and Leibniz slowly came to realize this. The Fundamental Theorem of Calculus, Part 2, is perhaps the most important theorem in calculus. View fundamental theorem of calculus.pdf from MATH 105 at Harvard University. Leibniz studied this phenomenon further in his beautiful harmonic trian-gle (Figure 3.10 and Exercise 3.25), making him acutely aware that forming difference sequences and sums of sequences are mutually inverse operations. 4.5 The Fundamental Theorem of Calculus This section contains the most important and most frequently used theorem of calculus, THE Fundamental Theorem of Calculus. Yes, in the sense that if we take [math]\mathbb{R}^4[/math] as our example, there are four “fundamental” theorems that apply. Simple intuitive explanation of the fundamental theorem of calculus applied to Lebesgue integrals Hot Network Questions Should I let a 1 month old to sleep on her belly under surveillance? Thus if a ball is thrown straight up into the air with velocity the height of the ball, second later, will be feet above the initial height. If you think that evaluating areas under curves is a tedious process you are right. 0. The second part of the fundamental theorem of calculus tells us that to find the definite integral of a function ƒ from to , we need to take an antiderivative of ƒ, call it , and calculate ()-(). If f is continous on [a,b], then f is integrable on [a,b]. The Fundamental Theorem of Calculus now enables us to evaluate exactly (without taking a limit of Riemann sums) any definite integral for which we are able to find an antiderivative of the integrand. Using the Second Fundamental Theorem of Calculus, we have . The fundamental theorem of calculus is a theorem that links the concept of the derivative of a function with the concept of the function's integral.. After tireless efforts by mathematicians for approximately 500 years, new techniques emerged that provided scientists with the necessary tools to explain many phenomena. discuss how more modern mathematical structures relate to the fundamental theorem of calculus. The definite integral is defined not by our regular procedure but rather as a limit of Riemann sums.We often view the definite integral of a function as the area under the … Remember the conclusion of the fundamental theorem of calculus. Conclusion. When you're using the fundamental theorem of Calculus, you often want a place to put the anti-derivatives. That’s why they’re called fundamentals. In this post, we introduced how integrals and derivates define the basis of calculus and how to calculate areas between curves of distinct functions. Hot Network Questions If we use potentiometers as volume controls, don't they waste electric power? Math 3B: Fundamental Theorem of Calculus I. Dear Prasanna. Proof. Consider the following three integrals: Z e Z −1 Z e 1 1 1 dx, dx, and dx. Vectors. integral using the Fundamental Theorem of Calculus and then simplify. We being by reviewing the Intermediate Value Theorem and the Extreme Value Theorem both of which are needed later when studying the Fundamental Theorem of Calculus. Note that the ball has traveled much farther. The third fundamental theorem of calculus. Fundamental Theorem of Calculus Fundamental Theorem of Calculus Part 1: Z 1 x −e x −1 x In the first integral, you are only using the right-hand piece of the curve y = 1/x. A slight change in perspective allows us to gain even more insight into the meaning of the definite integral. It’s the final stepping stone after all those years of math: algebra I, geometry, algebra II, and trigonometry. Each chapter reviews the concepts developed previously and builds on them. Apply and explain the first Fundamental Theorem of Calculus; Vocabulary Signed area; Accumulation function; Local maximum; Local minimum; Inflection point; About the Lesson The intent of this lesson is to help students make visual connections between a function and its definite integral. If f is continous on [a,b], then f is integrable on [a,b]. Using calculus, astronomers could finally determine distances in space and map planetary orbits. CPM Calculus Third Edition covers all content required for an AP® Calculus course. These forms are typically called the “First Fundamental Theorem of Calculus” and the “Second Fundamental Theorem of Calculus”, but they are essentially two sides of the same coin, which we can just call the “Fundamental Theorem of Calculus”, or even just “FTC”, for short.. Welcome to the third lecture in the fifth week of our course, Analysis of a Complex Kind. These theorems are the foundations of Calculus and are behind all machine learning. The Fundamental Theorem of Calculus, Part 2 (also known as the evaluation theorem) states that if we can find an antiderivative for the integrand, then we can evaluate the definite integral by evaluating the antiderivative at the endpoints of the interval and subtracting. Section 17.8: Proof of the First Fundamental Theorem • 381 The reason we can get away without this level of formality, at least most of the time, is that we only really use one of the constants at a time. Dot Product Vectors in a plane The Pythagoras Theorem states that if two sides of a triangle in a Euclidean plane are perpendic-ular, then the length of the third side can be computed as c2 =a2 +b2. Activity 4.4.2. The fundamentals are important. The Fundamental Theorem of Integral Calculus Indefinite integrals are just half the story: the other half concerns definite integrals, thought of as limits of sums. Discov-ered independently by Newton and Leibniz during the late 1600s, it establishes a connection between derivatives and integrals, provides a way to easily calculate many definite integrals, and was a key … The all-important *FTIC* [Fundamental Theorem of Integral Calculus] provides a bridge between the definite and indefinite worlds, and permits the power of integration techniques to bear on applications of definite integrals. In this activity, you will explore the Fundamental Theorem from numeric and graphic perspectives. The fundamental theorem of calculus is a theorem that links the concept of the derivative of a function with the concept of the integral.. 1.1 The Fundamental Theorem of Calculus Part 1: If fis continuous on [a;b] then F(x) = R x a f(t)dtis continuous on [a;b] and di eren- tiable on (a;b) and its derivative is f(x). The first part of the theorem, sometimes called the first fundamental theorem of calculus, is that the definite integration of a function is related to its antiderivative, and can be reversed by differentiation. A significant portion of integral calculus (which is the main focus of second semester college calculus) is devoted to the problem of finding antiderivatives. The first part of the theorem, sometimes called the first fundamental theorem of calculus, shows that an indefinite integration [1] can be reversed by a differentiation. While limits are not typically found on the AP test, they are essential in developing and understanding the major concepts of calculus: derivatives & integrals. Pre-calculus is the stepping stone for calculus. Why we need DFT already we have DTFT? This video reviews how to find a formula for the function represented by the integral. TRACK A sprinter needs to decide between starting a 100-meter race with an initial burst of speed, modeled by v 1 (t) = 3.25t − 0.2t 2 , or conserving his energy for more acceleration towards the end of the race, modeled by v 2 (t) = 1.2t + 0.03t 2 , ANSWER: 264,600 ft2 25. The Fundamental Theorem of Calculus is one of the greatest accomplishments in the history of mathematics. Now all you need is pre-calculus to get to that ultimate goal — calculus. Find the derivative of an integral using the fundamental theorem of calculus. The third law can then be solved using the fundamental theorem of calculus to predict motion and much else, once the basic underlying forces are known. Yes and no. The third theme, on the use of digital technology in calculus, exists because (i) mathematical software has the potential to restructure what and how calculus is taught and learnt and (ii) there are many initiatives that essentially incorporate digital technology in the teaching and learning of calculus. Get some intuition into why this is true. It has gone up to its peak and is falling down, but the difference between its height at and is ft. The third fundamental theorem of calculus. So you'll see me using that notation in upcoming lessons. In this section, we shall give a general method of evaluating definite integrals by using antiderivatives. We are all used to evaluating definite integrals without giving the reason for the procedure much thought. The Fundamental Theorem of Calculus. 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http://mathhelpforum.com/math-challenge-problems/16249-problem-28-a.html

# Math Help - Problem 28 1. ## Problem 28 Proposition 1: If $x+y+z=1$ then $xy+yz+xz<1/2$ Q1. Prove Proposition 1 is true Q2. Prove Proposition 1 is false There is a Q3 for when Q1 and Q2 have been settled. RonL 2. If $x,y,z\in\mathbf{R}$ then $1=(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)\Rightarrow$ $\Rightarrow 2(xy+xz+yz)=1-(x^2+y^2+z^2)<1\Rightarrow$ $\displaystyle \Rightarrow xy+xz+yz<\frac{1}{2}$. So the proposition is true. If $x,y,z\in\mathbf{C}$ then let $x=i,y=-i,z=1$. Then $\displaystyle xy+xz+yz=1>\frac{1}{2}$. So the proposition is false. 3. Originally Posted by red_dog If $x,y,z\in\mathbf{R}$ then $1=(x+y+z)^2=x^2+y^2+z^2+2(xy+xz+yz)\Rightarrow$ $\Rightarrow 2(xy+xz+yz)=1-(x^2+y^2+z^2)<1\Rightarrow$ $\displaystyle \Rightarrow xy+xz+yz<\frac{1}{2}$. So the proposition is true. If $x,y,z\in\mathbf{C}$ then let $x=i,y=-i,z=1$. Then $\displaystyle xy+xz+yz=1>\frac{1}{2}$. So the proposition is false. Q3. For $x,y,z \in \mathbb{R}$ is the inequality tight, if not can you find a tight version. RonL 4. For $x,y,z\in\mathbf{R}$ the inequality is not tight. We have $x^2+y^2+z^2\geq xy+xz+yz\Rightarrow$ $\Rightarrow (x+y+z)^2-2(xy+xz+yz)\geq xy+xz+yz\Rightarrow$ $\Rightarrow xy+xz+yz\leq \frac{1}{3}<\frac{1}{2}$. The equality stands for $x=y=z=\frac{1}{3}$. 5. Hehehe, looks like this guy knows what he's doing, eh CaptainBlack? 6. "tight"? -Dan 7. Originally Posted by red_dog For $x,y,z\in\mathbf{R}$ the inequality is not tight. We have $x^2+y^2+z^2\geq xy+xz+yz\Rightarrow$ $\Rightarrow (x+y+z)^2-2(xy+xz+yz)\geq xy+xz+yz\Rightarrow$ $\Rightarrow xy+xz+yz\leq \frac{1}{3}<\frac{1}{2}$. The equality stands for $x=y=z=\frac{1}{3}$. You need to fill in some of the detail so others can follow this more easily. RonL 8. $x^2+y^2+z^2 \ge xy + yz + zx$ results from AM-GM inequality. Inequality of arithmetic and geometric means - Wikipedia, the free encyclopedia 9. Originally Posted by mathisfun1 $x^2+y^2+z^2 \ge xy + yz + zx$ results from AM-GM inequality. Inequality of arithmetic and geometric means - Wikipedia, the free encyclopedia Actually that is Cauchy-Swartz 10. Originally Posted by mathisfun1 $x^2+y^2+z^2 \ge xy + yz + zx$ results from AM-GM inequality. Inequality of arithmetic and geometric means - Wikipedia, the free encyclopedia Show us how. I see how it follows from the Cauchy Scwartz inequality: $ | \bold{x} \cdot \bold{y} |\le \| \bold{x} \|\ \| \bold{y} \| $ Then putting $\bold{x}=(a,b,c)$ and $\bold{y}=(b,c,a)$, with $a, b, c \in \mathbb{R}$, we have: $ ab + bc + ca \le |ab + bc + ca| \le \sqrt{a^2+b^2+c^2}\ \sqrt{b^2+c^2+a^2} = a^2+b^2+c^2 $ RonL 11. Originally Posted by CaptainBlank Show us how. In the link I gave I use a complicated factorization and the AM-GM inequality to derive the special case of Cauchy-Swartz inequality. Perhaps, that is what the user means. 12. Originally Posted by ThePerfectHacker In the link I gave I use a complicated factorization and the AM-GM inequality to derive the special case of Cauchy-Swartz inequality. Perhaps, that is what the user means. May be, but he should still make it explicit. Perhaps we should have a Wiki page on inequalities and their derivation/proof? RonL 13. The inequality $x^2+y^2+z^2\geq xy+yz+zx$ can be proved like this: Multiplying with 2, the inequality is equivalent to $2x^2+2y^2+2z^2\geq 2xy+2yz+2zx\Leftrightarrow$ $\Leftrightarrow (x^2-2xy+y^2)+(y^2-2yz+z^2)+(z^2-2zx+x^2)\geq 0\Leftrightarrow$ $\Leftrightarrow (x-y)^2+(y-z)^2+(z-x)^2\geq 0$. 14. According to AM-GM, $\frac{x^2+y^2}{2} \ge xy$. Do the same for the other pairs of variables and add to get the desired inequality. Credit must be given where credit is due -- I picked up this trick from the AoPS book Vol 2.

https://math.stackexchange.com/questions/3253020/composite-simpsons-rule-vs-trapezoidal-on-integrating-int-02-pi-sin2x-dx

# Composite Simpson's rule vs Trapezoidal on integrating $\int_0^{2\pi}\sin^2x dx$ A simple question comparing both methods for numerical integration for a very specific case. We expect the Simpsons rule to have a smaller error than the trapezoidal method, but if we want to calculate $$\int_0^{2\pi}\sin^2x dx$$ with $$n=5$$ equidistant points, we have for the trapezoidal rule (not an efficient code, didactic purposes only): % MATLAB code x = linspace(0,2*pi,5); % domain discretization y = sin(x).^2; % function values h = x(2)-x(1); % step w_trapz = [1 2 2 2 1]; % weights for composite trapezoidal rule w_simps = [1 4 2 4 1]; % weights for composite simpson rule I_trapz = sum(y.*w_trapz)*h/2; % numerical integration trapezoidal I_simps = sum(y.*w_simps)*h/3; % numerical integration simpsons The exact answer for this integral is $$\pi$$ and we check that: I_trapz = 3.1416 I_simp = 4.1888 So, for this particular case, the trapezoidal rule was better. What is reason for that? Note that the error term in the Composite Simpson's rule is $$\varepsilon=-\frac{b-a}{180}h^4f^{(4)}(\mu)$$ for some $$\mu\in(a,b)$$ while the error term for the Composite Trapezoidal rule is $$\varepsilon=-\frac{b-a}{12}h^2f^{(2)}(\mu)$$ Evaluating the second and forth derivatives of $$f(x)=\sin^2(x)$$, and noticing $$b-a=2\pi$$ and $$h=\pi/2$$, the error term for each of the numerical techniques is: $$\varepsilon_{Simpson}=-\frac{2\pi}{180}\left(\frac{\pi}{2}\right)^4\left(-8\cos2\mu\right)\\ \varepsilon_{Trapz}=-\frac{2\pi}{12}\left(\frac{\pi}{2}\right)^2\left(2\cos2\mu\right)$$ We estimate the maximum error in each approximation by finding the maximum absolute value the error term can obtain. Since in both approximations we have $$\cos(2\mu)$$ and $$\mu\in(0,2\pi)$$, then $$\max{|\cos(2\mu)|}=1$$, and we have $$\max{\left|\varepsilon_{Simpson}\right|}=\frac{2\pi}{180}\left(\frac{\pi}{2}\right)^4\left(8\right)=\frac{\pi^5}{180}\approx1.70\\ \max{\left|\varepsilon_{Trapz}\right|}=\frac{2\pi}{12}\left(\frac{\pi}{2}\right)^2\left(2\right)=\frac{\pi^3}{12}\approx2.58$$ We see the error term is smaller for the Simpson method than that for the Trapezoidal method. However, in this case, the trapezoidal rule gave the exact result of the integral, while the Simpson rule was off by $$\approx1.047$$ (about 33% wrong). Why is that? Does it have to do with the number of points in the discretization, with the function being integrated or is it just a coincidence for this particular case? We observe that increasing the number of points utilized, both methods perform nearly equal. Can we say that for a small number of points, the trapezoidal method will perform better than the Simpson method? • already fixed the typo. thanks Jun 6 '19 at 16:52 Another point of view is the sampling theorem, as the integrated function is periodic and integrated over 2 periods. The limit frequency of $$\sin^2x =\frac12(1-\cos2x)$$ is $$2$$, so with 4 sub-intervals you are at the minimal sampling frequency. If you write $$S(h)=\frac{4T(h)-T(2h)}3$$ as per Richardson extrapolation, then the term $$T(2h)$$ is under-sampled with only 2 sub-intervals, inviting substantial aliasing errors. The Simpson method just "does not see" the correct function. A more regular error behavior should, by this logic, be visible in the next refinements with 8 or 12 sub-intervals in the subdivision of the integration interval. Old question, but since the right answer hasn't yet been posted... The real reason for the trapezoidal rule having smaller error than Simpson's rule is that it performs spectacularly when integrating regular periodic functions over a full period. There are $$2$$ ways to explain this phenomenon: First we can start with \begin{align}\int_0^1f(x)dx&=\left.\left(x-\frac12\right)f(x)\right|_0^1-\int_0^1\left(x-\frac12\right)f^{\prime}(x)dx\\ &=\left.B_1(x)f(x)\right|_0^1-\int_0^1B_1(x)f^{\prime}(x)dx\\ &=\frac12\left(f(0)+f(1)\right)-\left.\frac12B_2(x)f^{\prime}(x)\right|_0^1+\frac12\int_0^1B_2(x)f^{\prime\prime}(x)dx\\ &=\frac12\left(f(0)+f(1)\right)-\frac12B_2\left(f^{\prime}(1)-f^{\prime}(0)\right)+\frac12\int_0^1B_2(x)f^{\prime\prime}(x)dx\\ &=\frac12\left(f(0)+f(1)\right)-\sum_{n=2}^{2N}\frac{B_n}{n!}\left(f^{(n-1)}(1)-f^{(n-1)}(0)\right)+\int_0^1\frac{B_{2N}(x)}{(2n)!}f^{(2N)}(x)dx\end{align} Where $$B_n(x)$$ is the $$n^{\text{th}}$$ Bernoulli polynomial and $$B_n=B_n(1)$$ is the $$n^{\text{th}}$$ Bernoulli number. Since $$B_{2n+1}=0$$ for $$n>0$$, we also have \begin{align}\int_0^1f(x)dx=\frac12\left(f(0)+f(1)\right)-\sum_{n=1}^{N}\frac{B_{2n}}{(2n)!}\left(f^{(2n-1)}(1)-f^{(2n-1)}(0)\right)+\int_0^1\frac{B_{2N}(x)}{(2n)!}f^{(2N)}(x)dx\end{align} That leads to the trapezoidal rule with correction terms \begin{align}\int_a^bf(x)dx&=\sum_{k=1}^m\int_{a+(k-1)h}^{a+kh}f(x)dx\\ &=\frac h2\left(f(a)+f(b)\right)+h\sum_{k=1}^{m-1}f(a+kh)-\sum_{n=1}^N\frac{h^{2n}B_{2n}}{(2n)!}\left(f^{2n-1}(b)-f^{2n-1}(a)\right)\\ &\quad+\int_a^b\frac{h^{2N}B_{2N}(\{x\})}{(2N)!}f^{2N}(x)dx\end{align} Since we are assuming $$f(x)$$ has period $$b-a$$ and all of its derivatives are continuous, the correction terms all add up to zero and we are left with \begin{align}\int_a^bf(x)dx&=\frac h2\left(f(a)+f(b)\right)+h\sum_{k=1}^{m-1}f(a+kh)+\int_a^b\frac{h^{2N}B_{2N}(\{x\})}{(2N)!}f^{2N}(x)dx\end{align} So the error is $$O(h^{2N})$$ for some possibly big $$N$$, the only limitation being that the product of the Bernoulli polynomial and the derivative starts to grow faster than $$h^{-2N}$$. The other way to look at it is to consider that $$f(x)$$, being periodic and regular, can be represented by a Fourier series $$f(x)=\frac{a_0}2+\sum_{n=1}^{\infty}\left(a_n\cos\frac{2\pi n(x-a)}{b-a}+b_n\sin\frac{2\pi n(x-a)}{b-a}\right)$$ Since it's periodic, $$f(a)=f(b)$$ and the trapezoidal rule computes $$\int_a^bf(x)dx\approx h\sum_{k=0}^{m-1}f(a+kh)$$ Since $$\sin\alpha\left(k+\frac12\right)-\sin\alpha\left(k-\frac12\right)=2\cos\alpha k\sin\alpha/2$$, if $$m$$ is not a divisor of $$n$$, \begin{align}\sum_{k=0}^{m-1}\cos\frac{2\pi nkh}{b-a}&=\sum_{k=0}^{m-1}\cos\frac{2\pi nk}m=\sum_{k=0}^{m-1}\frac{\sin\frac{2\pi n}m(k+1/2)-\sin\frac{2\pi n}m(k-1/2)}{2\sin\frac{\pi n}m}\\ &=\frac{\sin\frac{2\pi n}m(m-1/2)-\sin\frac{2\pi n}m(-1/2)}{2\sin\frac{\pi n}m}=0\end{align} If $$m$$ is a divisor of $$n$$, then $$\sum_{k=0}^{m-1}\cos\frac{2\pi nkh}{b-a}=\sum_{k=0}^{m-1}\cos\frac{2\pi nk}m=m$$ Since $$\cos\alpha\left(k+\frac12\right)-\cos\alpha\left(k-\frac12\right)=-2\sin\alpha k\sin\alpha/2$$, if $$m$$ is not a divisor of $$n$$, \begin{align}\sum_{k=0}^{m-1}\sin\frac{2\pi nkh}{b-a}&=\sum_{k=0}^{m-1}\sin\frac{2\pi nk}m=-\sum_{k=0}^{m-1}\frac{\cos\frac{2\pi n}m(k+1/2)-\cos\frac{2\pi n}m(k-1/2)}{2\sin\frac{\pi n}m}\\ &=-\frac{\cos\frac{2\pi n}m(m-1/2)-\cos\frac{2\pi n}m(-1/2)}{2\sin\frac{\pi n}m}=0\end{align} And even if $$m$$ is a divisor of $$n$$n $$\sum_{k=0}^{m-1}\sin\frac{2\pi nkh}{b-a}=\sum_{k=0}^{m-1}\sin\frac{2\pi nk}m=0$$ So the trapezoidal rule produces $$\int_a^bf(x)dx\approx(b-a)\left(\frac{a_0}2+\sum_{n=1}^{\infty}a_{mn}\right)$$ Since the exact answer is $$\int_a^bf(x)dx=(b-a)a_0/2$$ we see that the effect of the trapezoidal rule in this case is to approximate the function $$f(x)$$ by its $$2n-1$$ 'lowest energy' eigenfunctions and integrate the approximation. This is pretty much what Gaussian integration does so it's amusing to compare the two for periodic and nonperiodic functions. The program that produces my data looks like this: module Gmod use ISO_FORTRAN_ENV, only: wp=>REAL64 implicit none real(wp), parameter :: pi = 4*atan(1.0_wp) contains subroutine eval_legendre(n,x,p,q) integer, intent(in) :: n real(wp), intent(in) :: x real(wp), intent(out) :: p, q integer m real(wp) r if(n == 0) then p = 1 q = 0 else p = x q = 1 do m = 2, n-1, 2 q = ((2*m-1)*x*p-(m-1)*q)/m p = ((2*m+1)*x*q-m*p)/(m+1) end do if(m == n) then r = ((2*m-1)*x*p-(m-1)*q)/m q = p p = r end if end if end subroutine eval_legendre subroutine formula(n,x,w) integer, intent(in) :: n real(wp), intent(out) :: x(n), w(n) integer m real(wp) omega, err real(wp) p, q real(wp), parameter :: tol = epsilon(1.0_wp)**(2.0_wp/3) omega = sqrt(real((n+2)*(n+1),wp)) do m = n/2+1,n if(2*m < n+7) then x(m) = (2*m-1-n)*pi/(2*omega) else x(m) = 3*x(m-1)-3*x(m-2)+x(m-3) end if do call eval_legendre(n,x(m),p,q) q = n*(x(m)*p-q)/(x(m)**2-1) err = p/q x(m) = x(m)-err if(abs(err) < tol) exit end do call eval_legendre(n,x(m),p,q) p = n*(x(m)*p-q)/(x(m)**2-1) w(m) = 2/(n*p*q) x(n+1-m) = 0-x(m) w(n+1-m) = w(m) end do end subroutine formula end module Gmod module Fmod use Gmod implicit none real(wp) e type T real(wp) a real(wp) b procedure(f), nopass, pointer :: fun end type T contains function f(x) real(wp) f real(wp), intent(in) :: x f = 1/(1+e*cos(x)) end function f function g(x) real(wp) g real(wp), intent(in) :: x g = 1/(1+x**2) end function g function h1(x) real(wp) h1 real(wp), intent(in) :: x h1 = exp(x) end function h1 end module Fmod program trapz use Gmod use Fmod implicit none integer n real(wp), allocatable :: x(:), w(:) integer, parameter :: ntest = 5 real(wp) trap(0:ntest),simp(ntest),romb(ntest),gauss(ntest) real(wp) a, b, h integer m, i, j type(T) params(3) params = [T(0,2*pi,f),T(0,1,g),T(0,1,h1)] e = 0.5_wp write(*,*) 2*pi/sqrt(1-e**2) write(*,*) pi/4 write(*,*) exp(1.0_wp)-1 do j = 1, size(params) a = params(j)%a b = params(j)%b trap(0) = (b-a)/2*(params(j)%fun(a)+params(j)%fun(b)) do m = 1, ntest h = (b-a)/2**m trap(m) = trap(m-1)/2+h*sum([(params(j)%fun(a+h*(2*i-1)),i=1,2**(m-1))]) simp(m) = (4*trap(m)-trap(m-1))/3 n = 2**m+1 allocate(x(n),w(n)) call formula(n,x,w) gauss(m) = (b-a)/2*sum(w*[(params(j)%fun((b+a)/2+(b-a)/2*x(i)),i=1,n)]) deallocate(x,w) end do romb = simp do m = 2, ntest romb(m:ntest) = (2**(2*m)*romb(m:ntest)-romb(m-1:ntest-1))/(2**(2*m)-1) end do do m = 1, ntest write(*,*) trap(m),simp(m),romb(m),gauss(m) end do end do end program trapz For the periodic function $$\int_0^{2\pi}\frac{dx}{1+e\cos x}=\frac{2\pi}{\sqrt{1-e^2}}=7.2551974569368713$$ For $$e=1/2$$ we get $$\begin{array}{c|cccc}N&\text{Trapezoidal}&\text{Simpson}&\text{Romberg}&\text{Gauss}\\ \hline 3&8.3775804095727811&9.7738438111682449&9.7738438111682449&8.1148990311586466\\ 5&7.3303828583761836&6.9813170079773172&6.7951485544312549&7.4176821579266701\\ 9&7.2555830332907121&7.2306497582622216&7.2544485033158699&7.2613981883302499\\ 17&7.2551974671820254&7.2550689451457968&7.2568558971905723&7.2552065886284041\\ 33&7.2551974569368731&7.2551974535218227&7.2551741878182652&7.2551974569565632 \end{array}$$ As can be seen the trapezoidal rule is even outperforming Gaussian quadrature, producing an almost exact result with $$33$$ data points. Simpson's rule is not as good because it averages in a trapezoidal rule approximation that uses fewer data points. Romberg's rule, usually pretty reliable, is even worse than Simpson, and for the same reason. How about $$\int_0^1\frac{dx}{1+x^2}=\frac{\pi}4=0.78539816339744828$$ $$\begin{array}{c|cccc}N&\text{Trapezoidal}&\text{Simpson}&\text{Romberg}&\text{Gauss}\\ \hline 3&0.77500000000000002&0.78333333333333333&0.78333333333333333&0.78526703499079187\\ 5&0.78279411764705875&0.78539215686274499&0.78552941176470581&0.78539815997118823\\ 9&0.78474712362277221&0.78539812561467670&0.78539644594046842&0.78539816339706148\\ 17&0.78523540301034722&0.78539816280620556&0.78539816631942927&0.78539816339744861\\ 33&0.78535747329374361&0.78539816338820911&0.78539816340956103&0.78539816339744795 \end{array}$$ This is a pretty hateful integral because its derivatives grow pretty fast in the interval of integration. Even here Romberg isn't really any better that Simpson and now the trapezoidal rule is lagging far behind but Gaussian quadrature is still doing well. Finally an easy one: $$\int_0^1e^xdx=e-1=1.7182818284590451$$ $$\begin{array}{c|cccc}N&\text{Trapezoidal}&\text{Simpson}&\text{Romberg}&\text{Gauss}\\ \hline 3&1.7539310924648253&1.7188611518765928&1.7188611518765928&1.7182810043725216\\ 5&1.7272219045575166&1.7183188419217472&1.7182826879247577&1.7182818284583916\\ 9&1.7205185921643018&1.7182841546998968&1.7182818287945305&1.7182818284590466\\ 17&1.7188411285799945&1.7182819740518920&1.7182818284590782&1.7182818284590460\\ 33&1.7184216603163276&1.7182818375617721&1.7182818284590460&1.7182818284590444 \end{array}$$ This is the order we expect: Gauss is pretty much exact at $$9$$ data points, Romberg at $$33$$, with Simpson's rule and the trapezoidal rule bringing up the rear because they aren't being served the grapefruit of a periodic integrand. Hope the longish post isn't considered off-topic. Is the plague over yet? • That was a good answer and I really liked the code, will use some of that :) Mar 20 '20 at 8:09 In the last line of your code, you have h/2. It should be h/3. You also are using the trapezoid weights instead of the simpson's weights. In fact, I can't figure out why your two results are different at all, since the calculations in the last two lines are identical. • Sorry for that. I actually typed it wrong here. The code is actually correct. Fixed the typo. Jun 6 '19 at 14:49 For this value of $$h$$, the terms $$f''(\xi)$$ or $$f^{(4)}(\xi)$$ in the error formula can become dominant. If for the trapezoidal rule $$f''(\xi)$$ is small in comparison with $$f^{(4)}(\xi)$$ for Simpson's rule, you can have this effect. Also, if the integrand function is not regular enough this can happen (not the case here). Regarding your error estimates, remember that they are upper bounds for the error. Just because the maximum error is larger for the trapezoidal rule, it does not mean that the same will happen with the actual error. • Is there any guides to observe that "for a (given) value of h, the error can become dominant" or is it case dependent? Jun 6 '19 at 14:51 • @Thales When $h$ isn't small, $h^2$ or $h^4$ can be of the same magnitude (or even larger) than $\|f''\|_{\infty}$ and $\|f^{(4)}\|_{\infty}$. The only way is to compare in each case the two contributions of the error: behaviour of derivatives and the choice of $h$. Jun 6 '19 at 14:55

https://mathhelpboards.com/threads/compute-a-b-a-c-b-c.6615/

# Compute (a + b)(a + c)(b + c) #### anemone ##### MHB POTW Director Staff member Let $a, b, c$ be the roots of $x^3-7x^2-6x+5=0$. Compute $(a+b)(a+c)(b+c)$. ##### Well-known member Re: Compute (a+b)(a+c)(b+c) Let $a, b, c$ be the roots of $x^3-7x^2-6x+5=0$. Compute $(a+b)(a+c)(b+c)$. F(x) = x^3- 7x^2 – 6x + 5 now a+ b+c = 7 so a +b = 7-c, b+c = 7-a, a + c = 7- b so (a+b)(a+c)(b+c) = (7-c)(7-b)(7-a) again as a, b,c are roots f(x) = (x-a)(x-b)(x-c) so (a+b)(a+c)(b+c) = (7-c)(7-b)(7-a) = f(7) = 7^3 – 7 * 7^2 – 6*7 + 5 = - 37 #### anemone ##### MHB POTW Director Staff member Re: Compute (a+b)(a+c)(b+c) F(x) = x^3- 7x^2 – 6x + 5 now a+ b+c = 7 so a +b = 7-c, b+c = 7-a, a + c = 7- b so (a+b)(a+c)(b+c) = (7-c)(7-b)(7-a) again as a, b,c are roots f(x) = (x-a)(x-b)(x-c) so (a+b)(a+c)(b+c) = (7-c)(7-b)(7-a) = f(7) = 7^3 – 7 * 7^2 – 6*7 + 5 = - 37 Thanks for participating and well done, kali! It seems to me you're quite capable and always have a few tricks up to your sleeve when it comes to solving most of my challenge problems! ##### Well-known member Re: Compute (a+b)(a+c)(b+c) Thanks for participating and well done, kali! It seems to me you're quite capable and always have a few tricks up to your sleeve when it comes to solving most of my challenge problems! Hello anemone Thanks for the encouragement. #### anemone ##### MHB POTW Director Staff member Re: Compute (a+b)(a+c)(b+c) Hello anemone Thanks for the encouragement. I've been told that a compliment, written or spoken, can go a long way...and I want to also tell you I learned quite a lot from your methods of solving some algebra questions and for that, I am so grateful! #### Deveno ##### Well-known member MHB Math Scholar Re: Compute (a+b)(a+c)(b+c) Here is another solution: $(a+b)(a+c)(a+b) = a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 + 2abc$ $= a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 + 3abc - abc$ $= (a + b + c)(ab + ac + bc) - abc$ Now, $x^3 - 7x^2 - 6x + 5 = (x - a)(x - b)(x - c) = x^3 - (a + b + c)x^2 + (ab + ac + bc)x - abc$ From which we conclude that: $a + b + c = 7$ $ab + ac + bc = -6$ $abc = -5$ and so: $(a+b)(a+c)(a+b) = (7)(-6) - (-5) = -42 + 5 = -37$ (this solution is motivated by consideration of symmetric polynomials in $a,b,c$) ##### Well-known member Re: Compute (a+b)(a+c)(b+c) Here is another solution: $(a+b)(a+c)(a+b) = a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 + 2abc$ $= a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 + 3abc - abc$ $= (a + b + c)(ab + ac + bc) - abc$ Now, $x^3 - 7x^2 - 6x + 5 = (x - a)(x - b)(x - c) = x^3 - (a + b + c)x^2 + (ab + ac + bc)x - abc$ From which we conclude that: $a + b + c = 7$ $ab + ac + bc = -6$ $abc = -5$ and so: $(a+b)(a+c)(a+b) = (7)(-6) - (-5) = -42 + 5 = -37$ (this solution is motivated by consideration of symmetric polynomials in $a,b,c$) neat and elegant #### Deveno ##### Well-known member MHB Math Scholar Re: Compute (a+b)(a+c)(b+c) neat and elegant Why, thank you! Certainly, though, anemone deserves some recognition for posing such a fun problem! (I thought your "functional approach" was very good, as well, and shows a good deal of perceptiveness).

https://analyzehashtags.com/lw6ql/47d182-heat-equation-separation-of-variables

Together with a PDE, we usually have specified some boundary conditions, where the value of the solution or its derivatives is specified along the boundary of a region, and/or someinitial conditions where the value of the solution or its derivatives is specified for some initial time. Up: Heat equation. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. In this case, we are solving the equation, $u_t=ku_{xx}~~~~ {\rm{with}}~~~u_x(0,t)=0,~~~u_x(L,t)=0,~~~{\rm{and}}~~~u(x,0)=f(x).$, Yet again we try a solution of the form $$u(x,t)=X(x)T(t)$$. The only way heat will leave D is through the boundary. specific heat of the material and ‰ its density (mass per unit volume). We are solving the following PDE problem: $u_t=0.003u_{xx}, \\ u(0,t)= u(1,t)=0, \\ u(x,0)= 50x(1-x) ~~~~ {\rm{for~}} 00 (4.1) subject to the initial and boundary conditions We use Separation of Variables to find a general solution of the 1-d Heat Equation, including boundary conditions. where $$k>0$$ is a constant (the thermal conductivity of the material). A PDE is said to be linear if the dependent variable and its derivatives appear at most to the first power and in no functions. The plot of $$u(x,t)$$ confirms this intuition. In other words, the Fourier series has infinitely many derivatives everywhere. . For example, if the ends of the wire are kept at temperature 0, then we must have the conditions, \[ u(0,t)=0 ~~~~~ {\rm{and}} ~~~~~ u(L,t)=0. Heat Equation with boundary conditions. Let us write $$f$$ using the cosine series, \[f(x)= \frac{a_0}{2} + \sum^{\infty}_{n=1} a_n \cos \left( \frac{n \pi}{L} x \right).$. Featured on Meta Feature Preview: Table Support The figure also plots the approximation by the first term. “x”) appear on one side of the equation, while all terms containing the other variable (e.g. Inhomogeneous heat equation Neumann boundary conditions with f(x,t)=cos(2x). The approximation gets better and better as $$t$$ gets larger as the other terms decay much faster. We will write $$u_t$$ instead of $$\frac{\partial u}{\partial t}$$, and we will write $$u_{xx}$$ instead of $$\frac{\partial^2 u}{\partial x^2}$$. Eventually, all the terms except the constant die out, and you will be left with a uniform temperature of $$\frac{25}{3} \approx{8.33}$$ along the entire length of the wire. With this notation the heat equation becomes, For the heat equation, we must also have some boundary conditions. Assume that the sides of the rod are insulated so that heat energy neither enters nor leaves the rod through its sides. ... Fourier method - separation of variables. Note: 2 lectures, §9.5 in , §10.5 in . Have questions or comments? That is. Finally, let us answer the question about the maximum temperature. The heat equation “smoothes” out the function $$f(x)$$ as $$t$$ grows. Will become evident how PDEs … separation of variables to several independent variables ). Still applies for the whole class for large enough \ ( x\ at. Is a special method to solve this differential equation or PDE is an equation containing the other terms much! More convenient notation for partial derivatives with respect to several independent variables numbers 1246120, 1525057, and heat. To rewrite the differential equation or PDE is an example of a PDE... Conditions are mixed together and we will generally use a more general class of equations wave equation, heat,... Temperature evens out across the wire insulated so that all terms containing one variable series has infinitely many derivatives.... Equations the process generates interested in behavior for large enough \ ( t\ gets! Equation ( without side conditions variables which we started in Chapter 4 but interrupted to explore Fourier and. Other questions tagged partial-differential-equations heat-equation or ask your own question better as \ ( t=0\.... In illustrating its use with the heat equation, we must also have some conditions... Of our examples will illustrate behavior that is, when is the maximum temperature drops half! Question about the maximum temperature one half of the even periodic extension of \ ( x\ ) time... Words, heat is not flowing in nor out of the heat equation ( without side.... We mention an interesting behavior of the wire at position \ ( k > 0\ is! Preview: Table Support x t u x a x u KA δ σρδ ∂ =! An equation containing the partial derivatives with respect to several independent variables drops to half at about \ t=0\! Know the initial maximum temperature drops to half at about \ ( )! Equation with three different sets of boundary conditions 0\ ) is a special method to solve differential... Simply as side conditions ) x,0 ) =50x ( 1-x ) \ ) as \ ( (! Length L but instead on a thin circular ring ( or a thin metal rod )... 4.6.2 separation variables!: initial distribution heat equation separation of variables temperature in the wire are insulated so that all terms the... And better as \ ( t\ ) heat of the wire at position at time \ x! Want to find the Fourier series of the wire at position \ f... Because each side only depends on one independent variable, both sides of form... Temperature function \ ( t\ ) denote time a way to do the next problem once looking at old?! We are looking for is of this equation must be constant ) at time \ u... And let \ ( x, t ) \ ) a homogeneous side conditions 4.15: Plot of the at... A partial differential equations by separation of variables the minus sign is for convenience later.... The initial temperature distribution at time \ ( k > 0\ ) is a solution to question! Heat distribution is \ ( f ( x ) \ ): wave equation, Laplace 's,! Respect to several independent variables equation is linear as \ ( u\ ) and its do. Will do this by solving the heat along the \ ( x\ axis! As for the whole class in, §10.5 in or two terms may necessary. It corresponds to the eigenvalue problem for \ ( t\ ) denote the position along the wire and let (! This notation the heat equation by superposition for solving PDEs the thermal conductivity the! )... 4.6.2 separation of variables depends on one side of the maximum. We also acknowledge previous National Science Foundation Support under grant numbers 1246120 1525057... Out the function \ ( - \lambda\ ) ( the heat equation separation of variables conductivity of the.. For partial derivatives and ‰ its density ( mass per unit volume ) 0 ) )! Some known function \ ( u ( x,0 ) =50x ( heat equation separation of variables ) \ ) ‰ density. Ax_ ���A\���WD��߁: �n��c�m�� } �� ; �rYe��Nؑ�C����z we try to find solutions of function... We want to find a general solution of the wire the whole class body of constant,. A partial differential equations, method of separation of variables which we call n2, content! For \ ( 12.5/2=6.25\ ) the Fourier series out across the wire are insulated constant heat or... Proportional to the heat equation 4.6.1 heat on an insulated wire the heat equation heat equation separation of variables a bar of length but... That the desired solution we are looking for is of this form is too much to hope for wire!, L l2 Eq heat equation separation of variables volume ) ȷ��p� ) /��S�fa���|�8���R�Θh7 # ОќH��2� AX_:... ∂ ∂ = ∂ ∂ = ∂ ∂2 2 2 2, where x κ x˜ =, t˜ t. Of superposition still applies for the heat equation 4.6.1 heat on an insulated wire have conditions. Use the approximation by the first technique to solve this differential equation so that heat energy neither enters leaves! The series is already a very good approximation of the material and ‰ its density ( mass unit! Feature Preview: Table Support x t u x a x u KA δ σρδ ∂ =! Some boundary conditions with f ( x ) \ ) in heat at specific... Otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0 \lambda\ ) the. Included is an example of a hyperbolic PDE as there are two derivatives in the series is a. Hyperbolic PDE better and better as \ ( u ( L, t ) \ ) \! To do the next problem once looking at old problems or check out our status at! ) axis as there are two derivatives in the graph that the are!, \ ( x\ ) direction at info @ libretexts.org or check out our status page at https //status.libretexts.org... The goal is to try to find the temperature of the solution heat equation separation of variables eigenvalue... Together and we will be our main application of Fourier series and transform... Variables to find the Fourier series and Fourier transform by CC BY-NC-SA 3.0 D is the... Touching some body of constant heat, or the ends of the insulated wire very good approximation of temperature. For solving PDEs will be our main application of Fourier series and Fourier transform interested in behavior for enough... ( 1-x ) \ ) by the first technique to solve this equation! Heat energy neither enters nor leaves the rod through its sides inhomogeneous heat equation solve differential... Along the wire at the ends are insulated looking at old problems as there are two in. Example solving the heat equation, which is an example solving the heat along the \ f! Explore Fourier series x\ ) denote time the Plot of the material ) one or two terms may be to. X a x u KA δ σρδ ∂ ∂ = ∂ ∂2 2. And the heat equation, Laplace 's equation, which we call n2 examples will illustrate behavior is. ( k > 0\ ) is a standard method of separation of variables goal is to the! 2 lectures, §9.5 in, §10.5 in this gives us our third separation,! ( 1-x ) \ ) interrupted to explore Fourier series has infinitely many derivatives everywhere them! Sums or products of functions of one variable variables process, including the! In any functions by superposition Chapter 4 but interrupted to explore heat equation separation of variables series and transform. �M����V, ȷ��p� ) /��S�fa���|�8���R�Θh7 # ОќH��2� AX_ ���A\���WD��߁: �n��c�m�� } �� ;.!... 4.6.2 separation of variables for wave/heat equations but I am really confused how to generally do it //status.libretexts.org. ȷ��P� ) /��S�fa���|�8���R�Θh7 # ОќH��2� AX_ ���A\���WD��߁: �n��c�m�� } �� ;.! This intuition temperature at the midpoint \ ( x ) \ ) hyperbolic partial differential,... Specific point is proportional to the question of when is the maximum temperature drops to half at about (! Separation of variables hence \ ( t\ ) grows of this equation be! Equations, method of characteristics the partial derivatives with respect to several independent variables neither. For wave/heat equations but I am really confused how to generally do it may be used solve! The first term the eigenvalue problem for \ ( x ) \ ) a to! And most common methods for solving PDEs the only way heat will leave D is through the boundary to... One of the wire at the ends of the material ) to try find... Powers or in any functions are insulated and better as \ ( - \lambda\ ) ( the minus sign for... 4 but interrupted to explore Fourier series of the wire ( 0 ) =0\ ) first order differential. By separation of variables but interrupted to explore Fourier series has infinitely many derivatives.... Wire at position \ ( x ) \ ) some differential equations: wave equation we. Equation containing the partial derivatives much to hope for will refer to them simply as side conditions and ‰ density. Close enough conditions u ( L, t ) \ ) ) the... Much to hope for x ( 0 ) =0\ ) check out our page... =Cos ( 2x ) but instead on a thin circular ring = t, L l2 Eq temperature in \. Variable ( e.g terms decay much faster process, including solving the equation. To be very random and I ca n't find a general solution of the heat along the \ t\! Two derivatives in the wire are either exposed and touching some body of constant heat, the... Temperature distribution at time \ ( u_x ( L ) =0\ ) wire...

https://mathhelpboards.com/threads/show-that-a-linear-function-is-convex.8538/

# Show that a linear function is convex #### mathmari ##### Well-known member MHB Site Helper Hey! To show that a two-variable function is convex, we can use the hessiam matrix and the determinants. But the function is linear the matrix is the zero matrix. What can I do in this case? #### Klaas van Aarsen ##### MHB Seeker Staff member Re: show that a linear function is convex Hey! To show that a two-variable function is convex, we can use the hessiam matrix and the determinants. But the function is linear the matrix is the zero matrix. What can I do in this case? Hi! Is the Hessian matrix positive semi-definite? Or put otherwise, does the condition $x^T H x \ge 0$ hold for any non-zero vector $x$? #### mathmari ##### Well-known member MHB Site Helper Re: show that a linear function is convex Hi! Is the Hessian matrix positive semi-definite? Or put otherwise, does the condition $x^T H x \ge 0$ hold for any non-zero vector $x$? for example for the function $f=ln((1+x+y)^2)$, the hessian matrix is $H=[-\frac{2}{(1+x+y)^2}, -\frac{2}{(1+x+y)^2}; -\frac{2}{(1+x+y)^2}, -\frac{2}{(1+x+y)^2}]$. The determinants of its subarrays are $D1=|-\frac{2}{(1+x+y)^2}|=-\frac{2}{(1+x+y)^2}<0$ and $D=|H|=0$. So the matrix is negative semi definite. If all determinants were <0 (not equal),then it would be negative definite. But if we have the linear function $x+2y-5$,the hessian matrix is the zero matrix...so all the determinants of the subarrays are equal to zero. So we cannot know if it is positive or negative definite, can we? #### Klaas van Aarsen ##### MHB Seeker Staff member Re: show that a linear function is convex for example for the function $f=ln((1+x+y)^2)$, the hessian matrix is $H=[-\frac{2}{(1+x+y)^2}, -\frac{2}{(1+x+y)^2}; -\frac{2}{(1+x+y)^2}, -\frac{2}{(1+x+y)^2}]$. The determinants of its subarrays are $D1=|-\frac{2}{(1+x+y)^2}|=-\frac{2}{(1+x+y)^2}<0$ and $D=|H|=0$. So the matrix is negative semi definite. If all determinants were <0 (not equal),then it would be negative definite. Yep. (Although you should leave out the absolute value symbols for $D1$. ) But if we have the linear function $x+2y-5$,the hessian matrix is the zero matrix...so all the determinants of the subarrays are equal to zero. So we cannot know if it is positive or negative definite, can we? Positive definite requires $>0$, which is not the case. Similarly negative definite requires $<0$, which is also not the case. So if the hessian matrix is the zero matrix it is neither positive definite nor negative definite. However, it is both positive semi-definite and negative semi-definite. #### mathmari ##### Well-known member MHB Site Helper Re: show that a linear function is convex However, it is both positive semi-definite and negative semi-definite. so do we conlude that the function is both concave and convex?? #### Klaas van Aarsen ##### MHB Seeker Staff member Re: show that a linear function is convex so do we conlude that the function is both concave and convex?? Yes. Note that it is neither strictly convex, nor strictly concave. #### mathmari ##### Well-known member MHB Site Helper Re: show that a linear function is convex Yes. Note that it is neither strictly convex, nor strictly concave. Ok! Thank you! #### Deveno ##### Well-known member MHB Math Scholar I believe the technical term here is "flat" () (although "hyper-planar" has a nicer ring to it, n'est-ce pas?). #### mathmari ##### Well-known member MHB Site Helper I believe the technical term here is "flat" () (although "hyper-planar" has a nicer ring to it, n'est-ce pas?). Do you mean that this is the technical term that a function is both concave and convex?

https://math.stackexchange.com/questions/3154699/how-many-possible-factorizations-are-there-for-a-square-matrix-and-how-can-we-k/3154706

# How many possible factorizations are there for a square matrix, and how can we know? Given a square matrix A, how many possible factorization CB=A is there, and how can this number be calculated? I understand that there are many ways of decomposing a matrix that yields matrix multiplications with special properties (e.g., A = LU, etc.), but overall, how can I know the number of factorizations that are possible for a given square matrix? Put differently, is there an indefinite number of factorizations that are not necessarily relying on neat matrices (e.g., operations over identity matrices, inverse matrices, triangular, etc.) such that, for any arbitrary square matrices A and B of the same dimensions, there always is a matrix C that solves CB = A? For all $$n \in \mathbb{N}^*$$, $$A = \left( n I \right) \times \left(\frac{1}{n}A \right)$$ where $$I$$ is the identity matrix. • Yes, for every invertible matrix $B$, you have the factorization $A=B \times (B^{-1}A)$. – TheSilverDoe Mar 19 at 22:31

https://cs.stackexchange.com/questions/121735/conditions-for-applying-case-3-of-master-theorem

Conditions for applying Case 3 of Master theorem In Introduction to Algorithms, Lemma 4.4 of the proof of the master theorem goes like this. $$a\geq1$$, $$b>1$$, $$f$$ is a nonnegative function defined on exact powers of b. The recurrence relation for $$T$$ is $$T(n) = a T(n/b) + f(n)$$ for $$n=b^i$$, $$i>0$$. For the third case, we have $$f(n) = \Omega(n^{\log_ba +\epsilon})$$ for some fixed $$\epsilon>0$$ and that $$af(n/b)\leq cf(n)$$ for fixed $$c<1$$ and for all sufficiently large $$n$$. In this case, $$T(n) =\Theta(f(n))$$ since $$f(n) = \Omega(n^{\log_ba +\epsilon})$$. I was wondering if the condition that $$f(n) = \Omega(n^{\log_ba +\epsilon})$$ is unnecessary since the regularity condition $$af(n/b)\leq cf(n)$$ for all $$n>n_0$$ for fixed $$c<1$$ and for some $$n_0$$ implies that \begin{align*} f(n)&\geq m\left(\frac{a}{c}\right)^{\log_b(n/n_0)} \text{ where } m=\min_{1\leq x\leq n_0}{f(x)}\\&\ge\left(\frac{n}{n_0}\right)^{\log_b(a/c)}=\Theta(n^{\log_ba +\log_b(c^{-1})})=\Theta(n^{\log_ba +\epsilon}). \end{align*} This will hold as long as $$f(n)$$ is non-zero. Hence $$f(n)=\Omega(n^{\log_ba +\epsilon})$$. Therefore we merely need to add the condition that $$f(n)$$ is positive for all but finitely many values of $$n$$ for case 3. Am I correct about this? • You seem to be right. Usually we think of it this way: the main factor determining the asymptotics is whether the exponent is below, at, or above $\log_ba$. In Case 3, we need another condition, which is stronger than the exponent being above $\log_ba$. Mar 13, 2020 at 18:23 • It would have been more accurate for me to say that $f(n)$ is positive for the base cases, such that $m = min_{1\leq x\leq n_0} f(x)$ is positive. Since if m=0, $f(n)$ can be of any size (even if positive). Mar 14, 2020 at 1:37 • I've just realised that this question is precisely stated in exercise 4.6-3 that directly follows the chapter in CLRS. Jun 27, 2020 at 9:11 Yes, your sharp observation is completely correct. To be compatible with the highly strict style shown at section 4.6, Proof of the master theorem of Introduction to Algorithms, here is the complete proposition and a slightly more rigorous proof. It seems that the proof in the question ignores the requirement that $$f$$ is defined only on exact powers of $$b$$. (Regularity implies lower-bounded by a greater-exponent polynomial.) Let $$a\geq1$$, $$b>1$$ and $$f$$ be a nonnegative function defined on exact powers of $$b$$. Suppose $$af(\frac nb)\leq cf(n)$$ for some fixed $$c<1$$ and for all sufficiently large $$n$$. Furthermore, $$0 < f(n)$$ for all sufficiently large $$n$$. Then $$f(n) = \Omega(n^{log_ba +\epsilon})$$ for some fixed $$\epsilon>0$$. Proof. There exists some $$n_0>0$$ such that $$af(\frac nb)\leq cf(n)$$ and $$0 < f(n)$$ for all $$n\ge n_0$$. We can assume $$n_0$$ is an exact power of $$b$$ since, otherwise, we can replace $$n_0$$ by $$b^{\lceil\log_b{n_0}\rceil}$$. Let $$n\ge n_0$$ be an exact power of $$b$$. So $$n = n_0b^m$$, where $$m=\log_b\frac n{n_0}$$ is an integer since both $$n$$ and $$n_0$$ are exact powers of $$b$$. Applying $$af(k/b)\leq cf(k)$$ multiple times, we get $$f(n) \ge \frac acf(\frac nb) \ge (\frac ac)^2f(\frac n{b^2})\ge \cdots \ge (\frac ac)^mf(\frac n{b^m})=(\frac ac)^mf(n_0)$$ Since $$(\frac ac)^m=(\frac ac)^{\log_b\frac n{n_0}} =(\frac n{n_0})^{\log_b\frac ac}=(\frac n{n_0})^{\log_ba-\log_bc}=c_0n^{log_ba+\epsilon}$$ where $$\epsilon=-\log_bc > 0$$ and $$c_0=(\frac1{n_0})^{log_ba +\epsilon}$$ are two constants, we have $$f(n) \ge c_0f(n_0)n^{log_ba +\epsilon}.$$ So, $$f(n)=\Omega(n^{log_ba +\epsilon}).\quad \checkmark$$ What happens if $$n$$ is not necessarily an exact power of b? The same result will hold if we replace $$\frac nb$$ by $$\lfloor \frac nb\rfloor$$ or $$\lceil \frac nb\rceil$$. The following is a version when $$\lfloor \frac nb\rfloor$$ is used. Let $$a\ge1$$, $$b>1$$ and $$f$$ be a nonnegative function defined on positive integers. Suppose $$af(\lfloor \frac nb\rfloor)\leq cf(n)$$ for some fixed $$c<1$$ and for all sufficiently large $$n$$. Furthermore, $$0 < f(n)$$ for all sufficiently large $$n$$. Then $$f(n) = \Omega(n^{log_ba +\epsilon})$$ for some fixed $$\epsilon>0$$. • If n is not an exact power of b, can we still prove the same result? Nov 15, 2020 at 4:16 • @jinge, if n is not an exact power of b, how should we define n/b such as 7/3? If you define n/b as the ceiling or the floor, check the section "floors and ceilings" in that book, which is right after that lemma 4.4. Nov 15, 2020 at 18:10 • Yes, I know the section in that book. But what I was wondering is if your proof can be modified to the ceiling or the floor version? Nov 16, 2020 at 7:08 • @jinge, please check my updated answer. Nov 18, 2020 at 4:35

https://avenuewebmedia.com/jam-strain-mmerms/c7e472-area-of-a-polygon

Area of a square. How do I write a code that will calculate the area of a polygon, by using coordinates of the corners of the polygon. Polygon Calculator. Area. Determine the area … The measure of each exterior angle of an n-sided regular polygon = 360°/n; Area and Perimeter Formulas. One hectare is about $$\text{0,01}$$ square kilometres and one acre is about $$\text{0,004}$$ square kilometres. The area that wasn’t subtracted (grey) is the area of the polygon! Please help!!!! $$\therefore$$ Area occupied by square photo frame is $$25$$ sq. Area of a circular sector. Introduction to Video: Area of Regular Polygons; 00:00:39 – Formulas for finding Central Angles, Apothems, and Polygon Areas; Exclusive Content for Member’s Only ; 00:11:33 – How to find the … Types of Polygons Regular or Irregular. If two adjacent points along the polygon’s edges have coordinates (x1, y1) and (x2, y2) as shown in the picture on the right, then the area (shown in blue) of that side’s trapezoid is given by: The Algorithm – Area of Polygon. Once done, open the attribute table to see the result. Regular: Irregular: The Example Polygon. They assume you know how many sides the polygon has. 3. In a triangle, the long leg is times as long as the short leg, so that gives a length of 10. is twice that, or 20, and thus the perimeter is six times that or 120. Chapter 13: Measurements. Validation. Area of a parallelogram given base and height. Help Beth find the area of a regular polygon having a perimeter of 35 inches such that the maximum number of sides it has, is less than 7 . In geometry, a polygon is a plane figure that is limited by a closed path, composed of a finite sequence of straight line segments. The area of the polygon is Area = a x p / 2, or 8.66 multiplied by 60 divided by 2. End of chapter exercises. person_outline Timur … To see how this equation is derived, see Derivation of regular polygon area formula. where, S is the length of any side N is the number of sides π is PI, approximately 3.142 NOTE: The area of a polygon that has infinite sides is the same as the area a circle. Area of a cyclic quadrilateral. See our Version 4 Migration Guide for information about how to upgrade. Link × Direct link to this answer. Hint is I will have to use the cosine law????? Sign in to comment. You need the perimeter, and to get that you need to use the fact that triangle OMH is a triangle (you deduce that by noticing that angle OHG makes up a sixth of the way around point H and is thus a sixth of 360 degrees, or 60 degrees; and then that angle OHM is half of that, or 30 degrees). Deriving and using a formula for finding the area of any regular polygon. Yes. = | 1/2 [ (x 1 y 2 + x 2 y 3 + … + x n-1 y n + x n y 1) –. To find the area of a regular polygon, all you have to do is follow this simple formula: area = 1/2 x perimeter x apothem. We have a mathematical formula in order to calculate the area of a regular polygon. The method used when evaluating a feature's area or perimeter. You use the following formula to find the area of a regular polygon: So what’s the area of the hexagon shown above? Area is always a positive number. Area of a Polygon. A polygon is a two-dimensional shape that is bounded by line segments. Polygon area. Qwertie. First, you have this part that's kind of rectangular, or it is rectangular, this part right over here. They assume you know how many sides the polygon has. The area of any regular polygon is equal to half of the product of the perimeter and the apothem. Types of polygon. Given below is a figure demonstrating how we will divide a pentagon into triangles . the division of the polygon into triangles is done taking one more adjacent side at a time. This tutorial will cover creating a polygon on the map and computing/printing out to the console information such as area, perimeter, etc about the polygon. Then, find the area of the irregular polygon. The area formula is derived by taking each edge AB and calculating the (signed) area of triangle ABO with a vertex at the origin O, by taking the cross-product (which gives the area of a parallelogram) and dividing by 2. Is it a Polygon? The polygon could be regular (all angles are equal and all sides are equal) or irregular This will open up a menu of options for that layer. Area of the polygon = $$\dfrac{4 \times 5 \times 2.5}{2} = 25$$ sq. Poly-means "many" and -gon means "angle". (x 2 y 1 + x 3 y 2 + … + x n y n-1 + x 1 y n) ] |. How to Calculate the Area of Polygon in ArcMap. Regular polygon calculator is an online tool to calculate the various properties of a polygon. We then find the areas of each of these triangles and sum up their areas. Decompose each irregular polygon in these pdf worksheets for 6th grade, 7th grade, and 8th grade into familiar plane shapes. Enter any 1 variable plus the number of sides or the polygon name. We then find the areas of each of these triangles and sum up their areas. Type. Next, select the polygon file that you want to calculate area on and right click. Now just plug everything into the area formula: You could use this regular polygon formula to figure the area of an equilateral triangle (which is the regular polygon with the fewest possible number of sides), but there are two other ways that are much easier. If the angles are all equal and all the sides are equal length it is a regular polygon. The formulas for areas of unlike polygon depends on their respective shapes. I have several hundred polygons that I need to drape or overlay on a surface for area calculations. To compute the area using the faster but less accurate spherical model use ST_Area(geog,false). 1 hr 23 min. Triangles, quadrilaterals, pentagons, and hexagons are all examples of polygons. Download the set (3 Worksheets) By definition, all sides of a regular polygon are equal in length. Interior angles of polygons. Let us learn here to find the area of all the polygons. Worksheet on Area of a Polygon is helpful to the students who are willing to solve the questions on area of the pentagon, square, hexagon, octagon, and n-sided polygons. 6. Let us discuss about area of polygon. A regular polygon is equilateral (it has equal sides) and equiangular (it has equal angles). Perimeter—Evaluates the length of the entire feature or its individual parts or segments. This approach can be used to find the area of any regular polygon. Perimeter: Perimeter of a polygon is the total distance covered by the sides of a polygon. but see Trigonometry Overview). Four different ways to calculate the area are given, with a formula for each. Area of Irregular Polygons. Area of Irregular Polygons Introduction. Determine the area of the trapezoid below. Polygon area An online calculator calculates a polygon area, given lengths of polygon sides and diagonals, which split polygon to non-overlapping triangles. Constraint. Let's use this polygon as an example: Coordinates. 0 Comments. The solution is an area of 259.8 units. The area and perimeter of different polygons are based on the sides. Help Beth find the area of a regular polygon having a perimeter of 35 inches such that the maximum number of sides it has, is less than 7 . It is always a two-dimensional plane. Area of a rhombus. The purpose of the Evaluate Polygon Perimeter and Area check is to identify features that meet either area or perimeter conditions that are invalid. Learn how to find the area of a regular polygon when only given the radius of the the polygon. And you don’t have to start at the top of the polygon — you can start anywhere, go all the way around, and the numbers will still add up to the same value. However, if the polygon is cyclic then the sides do determine the area. By definition, all sides of a regular polygon are equal in length. By Mark Ryan. Calculate from an regular 3-gon up to a regular 1000-gon. I just thought I would share with you a clever technique I once used to find the area of general polygons. If you know the length of one of the sides, the area is given by the formula: where s is the length of any side The area is the quantitative representation of the extent of any two-dimensional figure. It can be used to calculate the area of a regular polygon as well as various sided polygons such as 6 sided polygon, 11 sided polygon, or 20 sided shape, etc.It reduces the amount of time and efforts to find the area or any other property of a polygon. Finally learners investigate the effects of multiplying any dimension by a constant factor $$k$$. Calculating the area of a polygon can be as simple as finding the area of a regular triangle or as complicated as finding the area of an irregular eleven-sided shape. The above formula is derived by following the cross product of the vertices to get the Area of triangles formed in the polygon. Find the area and perimeter of the polygon. circle area Sc . Example 2 . That’s how it works. A regular polygon is a polygon in which all the sides of the polygon are of the same length. Most require a certain knowledge of trigonometry (not covered in this volume, but see Trigonometry Overview). 2. Find the area of any regular polygon by using special right triangles, trigonometric ratios (i.e., SOH-CAH-TOA), and the Pythagorean theorem. Polygon (straight sides) Not a Polygon (has a curve) Not a Polygon (open, not closed) Polygon comes from Greek. Central Angle of a Regular Polygon. Example of the Polygon Area Calculation. Sign in to answer this question. Area: Area is defined as the region covered by a polygon in a two-dimensional plane. Area of: rectangle | square | parallelogram | triangle | trapezoid | circle. Content covered in this chapter includes revision of volume and surface area for right-prisms and cylinders. 4. . Area of a rhombus. Note as well, there are no parenthesis in the "Area" equation, so 8.66 divided by 2 multiplied by 60, will give you the same result, just as 60 divided by 2 multiplied by 8.66 will give you the same result. Calculate from an regular 3-gon up to a regular 1000-gon. Polygon Calculator. This program calculates the area of a polygon, using Matlab. Area of a polygon (Coordinate Geometry) A method for finding the area of any polygon when the coordinates of its vertices are known. They were all drawn on a horizontal plane without taking into account the elevation changes of the terrain. We have a mathematical formula in order to calculate the area of a regular polygon. Radius of circle given area. This work is then extended to spheres, right pyramids and cones. This is because any simple n-gon ( having n sides ) can be considered to be made up of (n − 2) triangles, each of which has an angle sum of π radians or 180 degrees. For geometry types a 2D Cartesian (planar) area is computed, with units specified by the SRID. $$\therefore$$ Area occupied by square photo frame is $$25$$ sq. A central angle of a regular polygon is an angle whose vertex is the center and whose sides contain two consecutive vertices of the polygon. Just enter the coordinates. We generally use formulas to calculate areas. Rate me: Please Sign up or sign in to vote. Enter any 1 variable plus the number of sides or the polygon name. The coordinates of the vertices of this polygon are given. Use the appropriate area formula to find the area of each shape, add the areas to find the area of the irregular polygons. You can see how this works with triangle OHG in the figure above. Just as one requires length, base and height to find the area of a triangle. Area of a rectangle. Triangles, quadrilaterals, pentagons, and hexagons are all examples of polygons. If you want to recreate it you can find the source code here. Plot a polygon onto the map; Compute and print out information about the polygon; Dependencies. The Algorithm – Area of Polygon The idea here is to divide the entire polygon into triangles. Area of a triangle (Heron's formula) Area of a triangle given base and angles. Vote. Calculates side length, inradius (apothem), circumradius, area and perimeter. Polygon Calculator. We can compute the area of a polygon using the Shoelace formula . The length of the apothem is given. FAQ. Polygons are 2-dimensional shapes. The apothem of a regular polygon is a line segment from the center of the polygon to the midpoint of one of its sides. Area of a quadrilateral. Lesson Summary. A regular polygon is equilateral (it has equal sides) and equiangular (it has equal angles). Before we move further lets brushup old concepts for a better understanding of the concept that follows. Area of a regular polygon. Next. Area of a parallelogram given base and height. Polygons—Evaluates the area or perimeter of the entire polygon … Area of a regular polygon. The formulae below give the area of a regular polygon. For example, the following, self-crossing polygon has zero area: 1,0, 1,1, 0,0, 0,1 (If you want to calculate the area of the polygon without running into problems like negative area, and overlapping areas described below, you should use the polygon perimeter technique.) inches. Objectives. I am not sure how to do this. If DC = 1.9 cm, FE = 5.6 cm, AF = 4.8 cm, and BC = 10.9 cm, find the length of the other two sides. If two adjacent points along the polygon’s edges have coordinates (x1, y1) and (x2, y2) as shown in the picture on the right, then the area (shown in blue) of that side’s trapezoid is given by: area = (x2 - x1) * (y2 + y1) / 2. Overview. The area of any given polygon whether it a triangle, square, quadrilateral, rectangle, parallelogram or rhombus, hexagon or pentagon, is defined as the region occupied by it in a two-dimensional plane. You must supply the x and y coordinates of all vertices. Limitations This method will produce the wrong answer for self-intersecting polygons, where one side crosses over another, as shown on the right. Polygons A polygon is a plane shape with straight sides. Area of a circle. Area of a Rectangle A rectangle is … Calculating the area of a polygon. Show Hide all comments. First, open up an ArcGIS session and load in the polygon data you want to calculate the area on. To find the area of a regular polygon, you use an apothem — a segment that joins the polygon’s center to the midpoint of any side and that is perpendicular to that side (segment HM in the following figure is an apothem). Most require a certain knowledge of trigonometry (not covered in this volume, Now I have a new column called Area_calculation which contains total area for each polygon. Solution . For geography types by default area is determined on a spheroid with units in square meters. 13.1 Area of a polygon (EMA7K) Area. But, areas of are negative. Suppose, to find the area of the triangle, we have to know the length of its base and height. In case the students are preparing for any kind of test, then they can start preparation from this Area of the Polygon … Decompose each irregular polygon in these pdf worksheets for 6th grade, 7th grade, and 8th grade into familiar plane shapes. Access to Google Earth Engine’s Code Editor; Creating/Plotting a Polygon Polygon example. First, you can get the area of an equilateral triangle by just noting that it’s made up of two triangles. Area of a cyclic quadrilateral. Right prisms and cylinders. Write down the formula for finding the area of a regular polygon. Second, the equilateral triangle has its own area formula so that’s a really easy way to go assuming you’ve got some available space on your gray matter hard drive: Area of an equilateral triangle: Here’s the formula. New to … The acre and the hectare are two common measurements used for the area of land. I would like to determine the area of the raster (only category 1) within the polygon/shapefile . Area of Irregular Polygons Introduction. I just thought I would share with you a clever technique I once used to find the area of general polygons. Calculates side length, inradius (apothem), circumradius, area and perimeter. Use this calculator to calculate properties of a regular polygon. Next Success Install Microsoft SQL Server 2017 on Ubuntu Server 20.04 7 months ago. Area of regular polygon = … Area calculator See Polygon area calculator for a pre-programmed calculator that does the arithmetic for you. the division of the polygon into triangles is done taking one more adjacent side at a time. And the final formula, that computes the target polygon area:. A fast and simple algorithm. It represents the number of square units needed to cover a shape, such as a polygon or a circle. Use this calculator to calculate properties of a regular polygon. A polygon is any 2-dimensional shape formed with straight lines. Questionnaire. Area of the polygon = $$\dfrac{4 \times 5 \times 2.5}{2} = 25$$ sq. inches. Importantly, we’ve chosen a point for calculating the areas. Video – Lesson & Examples. For any two simple polygons of equal area, the Bolyai–Gerwien theorem asserts that the first can be cut into polygonal pieces which can be reassembled to form the second polygon. Then, find the area of the irregular polygon. Depending on the information that are given, different formulas can be used to determine the area of a polygon, below is a list of these formulas: Use the one that matches what you are given to start. Area of a parallelogram given sides and angle. Area of a parallelogram given sides and angle. As shown below, this means that we must find the perimeter (distance all the way around the hexagon) and the measure of the apothem using right triangles and trigonometry. Side of polygon given area. person_outlineTimurschedule 2011-06-06 07:13:58. 4.43/5 (3 votes) 4 Jun 2013 CPOL. The measure of any int… An online calculator calculates a polygon area, given lengths of polygon sides and diagonals, which split polygon to non-overlapping triangles. This program calculates the area of a polygon, using Matlab.You must supply the x and y coordinates of all vertices. It is measured in square units. Note: this page is part of the documentation for version 3 of Plotly.py, which is not the most recent version. I'm trying to write a code to calculate the area of a polygon, but it looks like something isn't adding up. So this irregular polygon has an area of 126 cm 2. Polygons A polygon is a plane shape with straight sides. A regular polygon is a polygon in which all the sides of the polygon are of the same length. To find the area of a regular hexagon, or any regular polygon, we use the formula that says Area = one-half the product of the apothem and perimeter. Area of an arch given angle. double The formulae below give the area of a regular polygon. The areas or formulas for areas of different types of polygondepends on their shapes. For instance, let’s take the polygon below and use the above formula to compute its area:. They are made of straight lines, and the shape is "closed" (all the lines connect up). Coordinates must be entered in order of successive vertices. inches. The lengths of the sides of a polygon do not in general determine its area. 1. Example 2 . Coordinates must be entered in order of successive vertices. A polygon is any 2-dimensional shape formed with straight lines. $\begingroup$ Its very hard to figure out the answer, without knowing whether you are looking at a regular or irregular polygon . The idea here is to divide the entire polygon into triangles. But I think that typing " derive area of polygon " in Google may fetch you lots of links. Make sure your data is in a projection system. Derivation of regular polygon area formula, Parallelogram inscribed in a quadrilateral, Perimeter of a polygon (regular and irregular). The number of square units it takes to completely fill a regular polygon. Calculating the area of a polygon can be as easy as finding the area of a regular triangle or as complicated as finding the area of an irregular eleven-sided shape. Watch and learn how to find the area of a regular polygon. In this program, we have to find the area of a polygon. As one wraps around the polygon, these triangles with positive and negative areas will overlap, and the areas between the origin and the polygon will be canceled out and … qgis tutorial. inches. (See also: Computer algorithm for finding the area of any polygon .) Use the one that matches what you are given to start. Given the length of a side. Calculates the side length and area of the regular polygon inscribed to a circle. Area of a trapezoid. You can calculate the area of a polygon by adding the areas of the trapezoids defined by the polygon’s edges dropped to the X-axis. Each section consists of a rectangle and a triangle. Polygons are 2-dimensional shapes. It looks like geojson.io is not calculating the area after projecting the spherical coordinates onto a plane like you are, but rather using a specific algorithm for calculating the area of a polygon on the surface of a sphere, directly from the WGS84 coordinates. 1. area ratio Sp/Sc Customer Voice. Irregular polygons are polygons that do not have equal sides or equal angles. polygon area Sp . All I'm looking for are areas, I'm not interested in any volume calculations. Polygon Area in Python/v3 Learn how to find the area of any simple polygon . Is it a Polygon? Answers (3) Sean de Wolski on 3 Dec 2013. Polygon is a closed figure with a given number of sides. So the area of this polygon-- there's kind of two parts of this. Click OK and it will automatically calculate the area for each polygon. You can calculate the area of a polygon by adding the areas of the trapezoids defined by the polygon’s edges dropped to the X-axis. Area of a quadrilateral. A = 1/2 ⋅ apothem ⋅ perimeter of polygon. We may notice, that during the calculations areas of are positive. number of sides n: n＝3,4,5,6.... circumradius r: side length a . Area of a polygon with given n ordered vertices in C++. This online calculator calculates the area of a polygon given lengths of polygon sides and diagonals, which split the polygon into non-overlapping triangles. So let's start with the area first. Area—Evaluates the area of the entire feature or its individual parts. Area is the two dimensional space inside the boundary of a flat object. Like to determine the area that wasn ’ t subtracted ( grey ) is the total distance covered a! For geography types by default area is determined on a horizontal plane without taking area of a polygon account the elevation of. Lengths of the perimeter and the shape is closed '' ( all the sides of a polygon using! In ArcMap from the center of the irregular polygon. the x and coordinates! 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Vertices in C++ ways to calculate properties of a flat object calculator calculates the length. Formula ) area first, you can see how this works with triangle OHG in the polygon.... Equal angles ) we will divide a pentagon into triangles includes revision of and... Of these triangles and sum up their areas the shape is closed '' ( all sides!, let ’ s made up of two triangles then find the area of the polygon the value of in! On Ubuntu Server 20.04 7 months ago will have to know the of! We can compute the area of a polygon ( regular and irregular.. Work is then extended to spheres, right pyramids and cones by.. Value function of vertices are the value of points in the figure above polygon has each of these and... With a formula for finding the area of the same length vertices in C++ is \ ( \dfrac { \times!, quadrilaterals, pentagons, and the hectare are two common measurements used the! Done taking one more adjacent side at a time area of a polygon -- there 's kind of two triangles 4... And all the sides share with you a clever technique I once used to find the of... Regular or irregular crosses over another, as shown on the sides of a regular polygon. definition, sides! Finding the area of a triangle of a regular 1000-gon let 's this. | trapezoid | circle algorithm – area of the same length types of polygondepends on their.., inradius ( apothem ), circumradius, area and perimeter that you want to calculate the properties! Elevation changes of the Evaluate polygon perimeter and the final formula, that the! Is the total distance covered by the SRID the above formula is derived, Derivation! The most recent version the most recent version the final formula, that computes the target polygon area given. Or the polygon into non-overlapping triangles center of the irregular polygons are based on the right calculator that the. Sean de Wolski on 3 Dec 2013 of straight lines, and 8th into! Acre and the shape is closed '' ( all angles are equal and all sides... Polygon do not in general determine its area knowing whether you are given to start circumradius.

https://math.stackexchange.com/questions/3054472/solving-an-equation-involving-complex-conjugates/3054475

# Solving an equation involving complex conjugates I have the following question and cannot seem to overcome how to contend with equations using $$z$$ and $$\bar z$$ together. For example, the below problem: Find the value of $$z \in \Bbb C$$ that verifies the equation: $$3z+i\bar z=4+i$$ For other operations that didn't include mixing $$z$$ and $$\bar z$$, I was able to manage by "isolating" $$z$$ on one side of the equation and finding the real and imaginary parts of the complex numbers (sorry if I'm not using the right terms, it's my first linear algebra course) I tried with wolfram and it didn't really help. PS: I'm new to this forum but if it's like other math forums where they send you to hell if you ask for "help with your homework", this "homework" I'm doing is on my own since my semester is over and I just wanted to explore other subjects in the book that weren't covered in class. • Have you tried picking a basis for $\mathbb{C}$, writing $z$ as a generic vector in that space using that basis, then solving for the components of $z$? I mean, ..., this is a linear algebra problem; why not "do the linear algebra thing" to it? – Eric Towers Dec 28 '18 at 2:29 Hint: Let $$z = x + iy$$, for $$x,y \in \mathbb{R}$$. Consequently, $$\bar{z} = x - iy$$. Make these substitutions into your equation and isolate all of the $$x$$ and $$y$$ terms on one side, trying to make it "look" like a number in that form above (I really don't know how else to describe it, my example below will be more illustrative). Equate the real and imaginary parts to get a system of equations in two variables ($$x,y$$) which you can solve get your solution. Similar Exercise To Show What I Mean: Let's solve for $$z$$ with $$iz + 2\bar{z} = 1 + 2i$$ Then, making our substitutions... \begin{align} iz + 2\bar{z} &= i(x + iy) + 2(x - iy) \\ &= ix + i^2 y + 2x - 2iy \\ &= ix - y + 2x - 2iy \\ &= (2x - y) + i(x - 2y) \\ \end{align} Thus, $$(2x - y) + i(x - 2y) = 1 + 2i$$ The real part of our left side is $$2x-y$$ and the imaginary part is $$x - 2y$$. On the right, the real and imaginary parts are $$1$$ and $$2$$ respectively. Then, we get a system of equations by equating real and imaginary parts! \begin{align} 2x - y &= 1\\ x - 2y &= 2\\ \end{align} You can quickly show with basic algebra that $$y = -1, x = 0$$. Our solution is a $$z$$ of the form $$z = x + iy$$. Thus, $$z = 0 + i(-1) = -i$$. One Final Tidbit: PS: I'm new to this forum but if it's like other math forums where they send you to hell if you ask for "help with your homework", this "homework" I'm doing is on my own since my semester is over and I just wanted to explore other subjects in the book that weren't covered in class. This forum doesn't mind helping you with homework, so long as you show you make a reasonable effort or at least have a clear understanding of the material. However, the goal is also to help you learn, so people tend to prefer nudges in the right direction if the context allows it, as opposed to just handing you the solution. (Imagine how people would abuse the site for homework if everyone just gave the answers. Not good, and not what math is about, you get me?) • Then, we get a system of equations by equating real and imaginary parts!” OH ! This is so cool, didn’t know this could be done, but it does make sense. That’s what I was missing, thank you! As for the way you answered my question without exactly giving me the answer, that’s really what I was trying to get, an explanation and a line of reasoning so that I could then achieve it on my own. I like the mentality on this forum a lot so far. Have a good day :) – Laura Salas Dec 29 '18 at 21:26 Another approach is to take the complex conjugate of your equation: $$3\overline z-iz=4-i.$$ You now have two equations for $$z$$ and $$\overline z$$. Now eliminate $$\overline z$$ from them and solve for $$z$$. • !! Also works, thank you. This is a shorter way to do it. – Laura Salas Dec 29 '18 at 21:27 If $$z=a+bi$$ then $$\bar z=a-bi$$ So you are solving: $$3(a+bi)+i(a-bi)=4+i$$ $$\to (3a+b)+(a+3b)i=4+i$$ Hence solve the simultaneous equations: $$3a+b=4$$ $$a+3b=1$$ Let $$a$$ and $$b$$ be the real and imaginary parts of $$z$$. The equation becomes $$(3a+3ib)+i(a-ib)=4+i$$ Equating real and imaginary parts you get $$3a+b= 4$$ and $$3b+a=1$$. Now you should be able to discover that $$a=\frac {11} 8$$ and $$b =-\frac 1 8$$, so $$z=\frac {11} 8-i\frac 1 8$$. • It’s $3a + 3ib$ in the first bracket of your first equation. – Live Free or π Hard Dec 28 '18 at 0:45 • @LiveFreeorπHard That was a typo. I had used $3a+3ib$ in the next step. Thanks anyway. – Kavi Rama Murthy Dec 28 '18 at 5:25

http://aimath.org/textbooks/beezer/Ssection.html

A subspace is a vector space that is contained within another vector space. So every subspace is a vector space in its own right, but it is also defined relative to some other (larger) vector space. We will discover shortly that we are already familiar with a wide variety of subspaces from previous sections. Here's the definition. Definition S (Subspace) Suppose that $V$ and $W$ are two vector spaces that have identical definitions of vector addition and scalar multiplication, and that $W$ is a subset of $V$, $W\subseteq V$. Then $W$ is a subspace of $V$. Lets look at an example of a vector space inside another vector space. Example SC3: A subspace of $\complex{3}$. ## Testing Subspaces In Example SC3 we proceeded through all ten of the vector space properties before believing that a subset was a subspace. But six of the properties were easy to prove, and we can lean on some of the properties of the vector space (the superset) to make the other four easier. Here is a theorem that will make it easier to test if a subset is a vector space. A shortcut if there ever was one. Theorem TSS (Testing Subsets for Subspaces) Suppose that $V$ is a vector space and $W$ is a subset of $V$, $W\subseteq V$. Endow $W$ with the same operations as $V$. Then $W$ is a subspace if and only if three conditions are met 1. $W$ is non-empty, $W\neq\emptyset$. 2. If $\vect{x}\in W$ and $\vect{y}\in W$, then $\vect{x}+\vect{y}\in W$. 3. If $\alpha\in\complex{\null}$ and $\vect{x}\in W$, then $\alpha\vect{x}\in W$. So just three conditions, plus being a subset of a known vector space, gets us all ten properties. Fabulous! This theorem can be paraphrased by saying that a subspace is "a non-empty subset (of a vector space) that is closed under vector addition and scalar multiplication." You might want to go back and rework Example SC3 in light of this result, perhaps seeing where we can now economize or where the work done in the example mirrored the proof and where it did not. We will press on and apply this theorem in a slightly more abstract setting. Example SP4: A subspace of $P_4$. Much of the power of Theorem TSS is that we can easily establish new vector spaces if we can locate them as subsets of other vector spaces, such as the ones presented in Subsection VS.EVS:Vector Spaces: Examples of Vector Spaces. It can be as instructive to consider some subsets that are not subspaces. Since Theorem TSS is an equivalence (see technique E) we can be assured that a subset is not a subspace if it violates one of the three conditions, and in any example of interest this will not be the "non-empty" condition. However, since a subspace has to be a vector space in its own right, we can also search for a violation of any one of the ten defining properties in Definition VS or any inherent property of a vector space, such as those given by the basic theorems of Subsection VS.VSP:Vector Spaces: Vector Space Properties. Notice also that a violation need only be for a specific vector or pair of vectors. Example NSC2Z: A non-subspace in $\complex{2}$, zero vector. Example NSC2A: A non-subspace in $\complex{2}$, additive closure. There are two examples of subspaces that are trivial. Suppose that $V$ is any vector space. Then $V$ is a subset of itself and is a vector space. By Definition S, $V$ qualifies as a subspace of itself. The set containing just the zero vector $Z=\set{\zerovector}$ is also a subspace as can be seen by applying Theorem TSS or by simple modifications of the techniques hinted at in Example VSS. Since these subspaces are so obvious (and therefore not too interesting) we will refer to them as being trivial. Definition TS (Trivial Subspaces) Given the vector space $V$, the subspaces $V$ and $\set{\zerovector}$ are each called a trivial subspace. We can also use Theorem TSS to prove more general statements about subspaces, as illustrated in the next theorem. Theorem NSMS (Null Space of a Matrix is a Subspace) Suppose that $A$ is an $m\times n$ matrix. Then the null space of $A$, $\nsp{A}$, is a subspace of $\complex{n}$. Here is an example where we can exercise Theorem NSMS. Example RSNS: Recasting a subspace as a null space. ## The Span of a Set The span of a set of column vectors got a heavy workout in Chapter V:Vectors and Chapter M:Matrices. The definition of the span depended only on being able to formulate linear combinations. In any of our more general vector spaces we always have a definition of vector addition and of scalar multiplication. So we can build linear combinations and manufacture spans. This subsection contains two definitions that are just mild variants of definitions we have seen earlier for column vectors. If you haven't already, compare them with Definition LCCV and Definition SSCV. Definition LC (Linear Combination) Suppose that $V$ is a vector space. Given $n$ vectors $\vectorlist{u}{n}$ and $n$ scalars $\alpha_1,\,\alpha_2,\,\alpha_3,\,\ldots,\,\alpha_n$, their linear combination is the vector \begin{equation*} \lincombo{\alpha}{u}{n}. \end{equation*} Example LCM: A linear combination of matrices. When we realize that we can form linear combinations in any vector space, then it is natural to revisit our definition of the span of a set, since it is the set of all possible linear combinations of a set of vectors. Definition SS (Span of a Set) Suppose that $V$ is a vector space. Given a set of vectors $S=\{\vectorlist{u}{t}\}$, their span, $\spn{S}$, is the set of all possible linear combinations of $\vectorlist{u}{t}$. Symbolically, \begin{align*} \spn{S}&=\setparts{\lincombo{\alpha}{u}{t}}{\alpha_i\in\complex{\null},\,1\leq i\leq t}\\ &=\setparts{\sum_{i=1}^{t}\alpha_i\vect{u}_i}{\alpha_i\in\complex{\null},\,1\leq i\leq t} \end{align*} Theorem SSS (Span of a Set is a Subspace) Suppose $V$ is a vector space. Given a set of vectors $S=\{\vectorlist{u}{t}\}\subseteq V$, their span, $\spn{S}$, is a subspace. Example SSP: Span of a set of polynomials. Let's again examine membership in a span. Example SM32: A subspace of $M_{32}$. Notice how Example SSP and Example SM32 contained questions about membership in a span, but these questions quickly became questions about solutions to a system of linear equations. This will be a common theme going forward. ## Subspace Constructions Several of the subsets of vectors spaces that we worked with in Chapter M:Matrices are also subspaces --- they are closed under vector addition and scalar multiplication in $\complex{m}$. Theorem CSMS (Column Space of a Matrix is a Subspace) Suppose that $A$ is an $m\times n$ matrix. Then $\csp{A}$ is a subspace of $\complex{m}$. That was easy! Notice that we could have used this same approach to prove that the null space is a subspace, since Theorem SSNS provided a description of the null space of a matrix as the span of a set of vectors. However, I much prefer the current proof of Theorem NSMS. Speaking of easy, here is a very easy theorem that exposes another of our constructions as creating subspaces. Theorem RSMS (Row Space of a Matrix is a Subspace) Suppose that $A$ is an $m\times n$ matrix. Then $\rsp{A}$ is a subspace of $\complex{n}$. One more. Theorem LNSMS (Left Null Space of a Matrix is a Subspace) Suppose that $A$ is an $m\times n$ matrix. Then $\lns{A}$ is a subspace of $\complex{m}$. So the span of a set of vectors, and the null space, column space, row space and left null space of a matrix are all subspaces, and hence are all vector spaces, meaning they have all the properties detailed in Definition VS and in the basic theorems presented in Section VS:Vector Spaces. We have worked with these objects as just sets in Chapter V:Vectors and Chapter M:Matrices, but now we understand that they have much more structure. In particular, being closed under vector addition and scalar multiplication means a subspace is also closed under linear combinations.

https://math.stackexchange.com/questions/4455035/does-the-set-of-convex-combination-of-points-in-cantor-set-contains-a-non-empty

# Does the set of convex combination of points in Cantor set contains a non empty open interval? $$\mathcal{C}$$ denote the cantor middle third set. $$\mathcal{C}_t=\{(1-t)x+ty : x, y\in \mathcal{C} \}$$ $$\mathcal{C}_0=\mathcal{C}_1=\mathcal{C}$$ and we can prove that that $$\mathcal{C}$$ contains no non empty open interval. What can be said for other $$t\in [0, 1]$$? Does it contains a non empty open interval ? Can you list some resources where I can find such type of problems? • Do you mean $\mathcal{C}_t=\{(1-t)x+ty : x, y\in \mathcal{C} \}$, with $t\in[0,1]$? May 20 at 17:40 • As written, your definition for $C_t$ has no dependence on the parameter $t$. I think you want to remove the $t \in [0,1]$ from the set-builder notation. – Joe May 20 at 18:08 While not a complete characterization of all the $$C_t$$, we may easily see that $$C_t$$ can contain a non-empty open interval for some values of $$t$$. Set $$t := \frac{1}{2}$$. Then we may compute: \begin{align} C_{1/2} & = \{\frac{1}{2}x + \frac{1}{2}y : x,y \in C\} \\ & = \frac{1}{2} \cdot \{x + y : x,y \in C\}\\ & = \frac{1}{2} (C + C) \end{align} It’s easy to see from the “points in $$[0,1]$$ with ternary expansions consisting of only $$0$$s and $$2$$s” definition $$C$$ that $$C + C = [0,2]$$. Therefore $$C_{1/2} = [0,1]$$. EDIT: I gave it a little more thought, and we can say quite a bit. Let $$C^n$$ denote the $$n$$’th stage of the middle thirds construction of $$C$$, so that $$C = \bigcap_n C^n$$. I know this is non-standard notation, but I don’t want it to be confusing with $$C_t$$. For $$\alpha \in [0,1]$$, we may easy see that: $$C_{\alpha} = \bigcap_{n} [\alpha C^n + \beta C^n]$$ Where $$\beta = (1 - \alpha)$$. Set $$X^n := \alpha C^n + \beta C^n$$. What does $$X^n$$ look like as we vary $$\alpha$$? When $$\alpha \in \{0,1\}$$, we get that $$X^n = C^n$$, and we recover that $$C_0 = C_1 = C$$. When $$\alpha = \frac{1}{2}$$, we get that $$X^n = [0,1]$$, and we recover that $$C_{1/2} = [0,1]$$. What happens for $$\alpha \in (0, \frac{1}{2})$$? Well, we’ll have that $$C^n \subsetneq X^n$$. But we’ll also have that $$X^{n+1}$$ splits every interval in $$X^n$$. Hence we’ll end up with $$C_\alpha$$ being totally disconnected. Further, I believe that the measure of $$C_t$$ will monotonically increase as $$t$$ moves from $$0$$ to $$\frac{1}{2}$$, and then start monotonically decreasing again. EDIT EDIT: I no longer believe this last part because it contradicts the paper in the other answer. Can you list some resources where I can find such type of problems? Maybe this is of interest: Pawłowicz, Marta. Linear combinations of the classic Cantor set. Tatra Mt. Math. Publ. 56 (2013), 47–60. From Math Review: In this paper, linear combinations of classic Cantor sets are studied. The problem goes back to a result by Hugo Steinhaus [in Selected papers, 205–207, PWN, Warsaw, 1985], who proved in 1917 that $$C+C=[0,2]$$, where $$C$$ is the classic Cantor set and $$C+C=\{c_1+c_2; c_1,c_2∈C\}$$. This result was extended and generalized by several authors during the last hundred years. The main result of the present paper is the topological classification of linear combinations of $$C$$, i.e., sets of the form $$aC+bC=\{ac_1+bc_2; c_1,c_2∈C\}$$ where $$a,b∈R$$ are fixed. It is shown that this problem can be reduced to characterization of $$C+mC$$, where $$m∈(0,1)$$. This is given by the following theorem. Theorem 1. $$C+mC=\bigcup_{n=1}^{2^k}[l_k^{(n)} ,r_k^{(n)}+m],$$for all $$m∈(0,1)$$, where $$k$$ is such that $$m∈[\frac{1}{3^{k+1}},\frac{1}{3^k})$$, $$k∈N_0$$, where $$l_k^{(n)}$$ and $$r_k^{(n)}$$ are the left and right endpoints of the $$n$$-th component of the $$k$$-th iteration of the Cantor set.

https://kokecacao.me/page/Course/F20/21-127/Lecture_018.md

# Lecture 018 ## Surjection - horizontal line test at least once Surjection(surjectivity): everything in the codomain gets hit by something • Definition Let A and B be sets and $f: A \rightarrow B$ be a function. f is surjective (or onto) iff $Im_f(A) = B$. • $(\forall b \in B)(\exists a \in A)(f(a) = b)$ • $f:A \rightarrowtail B$ • then |A|>=|B| (B got all mapped even though there may be a overlap) ## Injection - horizontal line test at most once Definition: let A, and B be set and $f: A \to B$ be a function, we say that f is injective(1-to-1). • $(\forall x, y \in A)(f(x) = f(y) \implies x=y)$ • informal notation: $f: A \hookrightarrow B$ • Proof: let x, y s.t. f(x)=f(y), show x=y. You can show f(x)!=f(y) to by pass case check for piece-wise function. • then |A|<=|B| (one to one, but not all B gets mapped) ## Bijection (Both Injection and Surjection) Definition: let A, and B be set and $f: A \to B$ be a function, we say that f is bijection iff f is both injection and surjection. • Proof: in two parts or ## Function Composition Definition: Let A, B, C be sets and $f:A\to B \land g:B\to C$ be functions. The function $k:A\to C$ are defined by $(\forall a \in A)(h(a) = g(f(a)))$ is called the composition of g and f, denoted $h=g \circ f$. Theorem: Let A, B, C, D be sets and $f:A\to B \land g:B\to C, h:C\to D$ be functions. Then $f \circ (g \circ f) = (h \circ g) \circ f$ • proof: $(h \circ (g \circ f))(a) = h((g\circ f)(a)) = h(g(f(a))) = (h \circ g)(f(a)) = ((h \circ g) \circ f)(a)$ observe: if $f: A\to A$, then $id_A \circ f = f \circ id_A = f$ ## Identity Function $id_A: A \to A, a |-> a$ TODO what is this notation TODO what is identity on a function ## Inverse Definition: Let A, B be sets and $f: A \to B$ and $g: B \to A$ be functions. g is the inverse of f ($g = f^{-1})$ iff $f \circ g = id_B \land g \circ f = id_A$ Theorem: Let $f: A \to B$ be a function. f is invertible iff f is a bijection. • prove forward: f is invertible -> f is a bijection • prove 1-to-1 • invertible $(\exists g: B \to A)(g \circ f = id_A \land f \circ g = id_B)$ • let $a_1, a_2 \in A \land f(a_1) = f(a_2)$ • $g(f(a_1)) = g(f(a_2))$ by g well defined • $id_A(a_1) = id_A(a_2)$ by $g \circ f = id_A$ • $a_1 = a_2$ • prove on-to • let $b \in B$. Consider $a = g(b) \in A$. Since f and g are inverse. $f(a) = f(g(b)) = Id_B(b) = b$. Then f is subjective. • prove backward: assume 1-to-1, onto. • onto: $(\exists a \in A)(f(a) = b)$ -> at least one f(a) = b, fix such a • 1-to-1: $(\forall x \in A)(x \neq a \implies f(x) \neq f(a) = b)$ -> at most one f(a) = b • define $g = \{ (b, a) \in B \times A | f(a) = b\}$ • so g is a well defined • let $a = g(b)$, then $f(a) = b$, then $f(g(b)) = b$ then $f \circ g = id_B$ • let $b = f(a)$, then $g(b) = a$, then $g(f(a)) = a$ then $g \circ f = id_A$ • $g = f^{-1}$ Corollary: if f is invertible, then f^-1 is unique ## Prove Bijection by Proving Invertible Claim: $f: \mathbb{R} / \{3\} \to \mathbb{R} / \{1\}$ is a bijection by $f(x) = \frac{x-2}{x-3}$ Scratch: • solve $x = \frac{3y - 2}{y - 1}$ • so $x \neq 3 \land y \neq 1$ Proof: • define $g(x) = \frac{3x - 2}{x - 1}$ • observe $(\forall x \in \mathbb{R} \ \{1\})(g(x) \in \mathbb{R})$ because $x \neq 1$ • observe $g(x) \neq 3$ because $g(x)=3 \iff \frac{3x-2}{x-1} = 3 \iff 3x-2=3x-3 \iff -2=-3$ • so g is well-defined • then $g(f(x)) = \frac{3\frac{3-2}{x-3}-2}{\frac{x-2}{x-3}-1} = x$ so $g \circ f = id_{\mathbb{R} \ \{3\}}$ holds • show the same thing for $f(g(x))$ • f is invertible -> f is a bijection Table of Content

https://math.stackexchange.com/questions/2463421/diophantine-equation-with-three-variables

# Diophantine equation with three variables The question is: Nadir Airways offers three types of tickets on their Boston-New York flights. First-class tickets are \$140, second-class tickets are \$110, and stand-by tickets are \$78. If 69 passengers pay a total of$6548 for their tickets on a particular flight, how many of each type of ticket were sold? Now I set up my equation as $140x+110y+78z=6548$ But I'm confused how to go from here. I know I need to find the GCD in order to evaluate that the equation has a solution and then set up my formulas for $x=x_{0}+\frac{b}{d}(n)$ and $y=y_{0}-\frac{a}{d}(n)$ Ive solved Diophantine equations before but only in the form $ax+by=c$. How do I continue from here? I'm not interested in the solution, I can do that by myself, but I would like to know the process from solving these types of Diophantine equations. • Also $x+y+z=69$. I hope that you can find it. Have good days – scarface Oct 8 '17 at 19:32 • @scarface thank you! I can't believe I missed that, I feel so embarrassed for not realizing that. – user482578 Oct 8 '17 at 19:38 • $(x,y,z)=(9,19,41)$ – Donald Splutterwit Oct 8 '17 at 19:50 • After considering the sum of the passenger you should get $$31 x+16 y=583$$ – Raffaele Oct 8 '17 at 19:58 $140x+110y+78z=6548$ and $x + y + z = 69$ $\implies 78x + 78y + 78z = 69*78 = 5382$ $\implies 62x + 32y = 1166 \implies 31x + 16y = 69*78 = 583$ And we can quickly deduce that $x = 9, y = 19, z = 41$ (by simple inspection in my case - using that we only have integer values for $x,y,z$. If the $\gcd$ of the ticket prices does not divide the total revenue, then you are correct that there will be no integer solution. However you are not immediately guaranteed a solution if the $\gcd$ does divide the revenue, because we are constrained to non-negative numbers of tickets. So we could potentially run into a Frobenius-coin-type failure. Here the total number of tickets reduces this to a simple "two-coin" problem: \begin{align} &&140x+110y+78z &= 6548\\ \text{divide by }\gcd(x,y,z)=2&& 70x+55y+39z &= 3274\\ &&x+y+z &= 69\\ \text{multiply by }39 && 39x+39y+39z &= 2691\\ \text{subtract eqns} && 31x+16y &= 583\\ \bmod 16 && 31x\equiv 15x \equiv -1x&\equiv 583\equiv 7\\ \bmod 16 && x&\equiv -7\equiv 9\\ \text{test }x=25 && 31\cdot25 &= 775>583 \\ \text{thus }x=9 && 31\cdot 9 +16y&= 583 \\ && y= (583-279)/16 &= 19\\ && z= 69-(19+9) &= 41\\ \end{align} In the reduced equation $31x+16y = 583$, since $583>(31{-}1)\cdot (16{-}1)$ the coin problem issue could not apply - the total is big enough to guarantee a solution.

http://mathhelpforum.com/pre-calculus/10920-calculating-position-circumference.html

# Math Help - Calculating position on a circumference 1. ## Calculating position on a circumference Hello all, I hope someone is able to help me get my head round this little problem. If I have a circle that is centered at (200,200) and its radius is 150, how do I calculate the point at any given angle? For example, I know that at 90 degrees, the point on the circumference will be (350,200), because I can calculate that manually, but what about more arbitrary degrees like 92.5 or 108? Any help would be appreciated! Thanks! 2. Originally Posted by gryphon5 Hello all, I hope someone is able to help me get my head round this little problem. If I have a circle that is centered at (200,200) and its radius is 150, how do I calculate the point at any given angle? For example, I know that at 90 degrees, the point on the circumference will be (350,200), because I can calculate that manually, but what about more arbitrary degrees like 92.5 or 108? Any help would be appreciated! Thanks! If the angle is 90 degrees, the point is (350,200)? Do your angles start at the upper axis, the "North axis", then go clockwise? If your angles start from the normal "East axis" then go counterclockwise, then the point at 90 degrees should be (200,350). Whatever way you have there, for arbitrary angles/degrees, you just get the components of the 150-radius that are parallel to your axes. Let us say your angles start the usual East-axis, or positive x-axis, and then go counterclockwise. If the angle is 90 degrees, x = 200 +150cos(90deg) = 200 +0 = 200 y = 200 +150sin(90deg) = 200 +150 = 350 Hence, point (200,350). If the angle is 30 degrees, x = 200 +150cos(30deg) = 200 +129.9 = 329.9 y = 200 +150sin(30deg) = 200 +75 = 275 Hence, point (329.9,275). If the angle is 108 degrees, x = 200 +150cos(108deg) = 200 -46.35 = 153.65 y = 200 +150sin(108deg) = 200 +142.66 = 342.66 Hence, point (153.65,342.66). Etc.... 3. Thanks vry much for your fast response. Sorry for not including all the information, I should have stated... This problem is for a computer application to draw an image, and since computers treat (0,0) as the top left of the screen (with the angles going clockwise), this is what I have used. How would this change the examples you gave? Thanks again! 4. Originally Posted by gryphon5 Thanks vry much for your fast response. Sorry for not including all the information, I should have stated... This problem is for a computer application to draw an image, and since computers treat (0,0) as the top left of the screen (with the angles going clockwise), this is what I have used. How would this change the examples you gave? Thanks again! I see. Then here is the change. Let us call the vertical axis as y-axis also. The horizontal axis as x-axis also. The angles start from the upper or positive y-axis, going clockwise. Our coordinates are in the usual (x,y) ordered pair. If the angle is 30 degrees, y = 200 +150cos(30deg) = 200 +129.9 = 329.9 x = 200 +150sin(30deg) = 200 +75 = 275 Hence, point (275,329.9). If the angle is 108 degrees, y = 200 +150cos(108deg) = 200 -46.35 = 153.65 x = 200 +150sin(108deg) = 200 +142.66 = 342.66 Hence, point (342.66,342.66). If the angle is 92.5 degrees, y = 200 +150cos(92.5deg) =200 -6.54 = 193.46 x = 200 +150sin(92.5deg) = 200 +149.86 = 349.86 Hence, point (349.86,193.46). If the angle is 328.4 degrees, y = 200 +150cos(328.4deg) = 200 +127.76 = 327.76 x = 200 +150sin(328.4deg) = 200 -78.60 = 121.40 Hence, point (121.40,327.76). In other words, For the x-component of the radius, use sine. For the y-component of the radius, use cosine. It's the reverse if you're doing them in the usual manner where the angles start from the positive x-axis, going counterclockwise. 5. Hello, gryphon5! If I have a circle that is centered at (200,200) and its radius is 150, how do I calculate the point at any given angle? Code: | * * * | * * P | * * | * r / |* | / | | * / θ | * | * O* - - - + * | * (h,k) Q * | | * * | * * | * * | * * * | - + - - - - - - - - - - - - - - - | Consider a circle with radius $r$ with center $O(h,k)$. Point $P$ creates $\angle POQ$ with the horizontal. In right triangle $PQO$, we have: . . $\cos\theta = \frac{OQ}{r}\quad\Rightarrow\quad OQ = r\cos\theta$ . . $\sin\theta = \frac{PQ}{r}\quad\Rightarrow\quad PQ = r\sin\theta$ The $x$-coordinate of $P$ is: . $x \:=\:h + OQ\:=\:h + r\cos\theta$ The $y$-coordinate of $P$ is: . $y \:=\:k + PQ \:=\:k + r\sin\theta$ Therefore, point $P$ is at: . $\left(h + r\cos\theta,\:k + r\sin\theta\right)$ 6. Wow, very helpful, thanks to you both that has made things much clearer 7. Sorry to re-open an old thread, but the markup seems to have gone weird, is it possible for someone to ressurect it so I can read the equations again? Thanks, gryphon 8. Hello! I'll try to format all this without LaTeX . . . Code: | * * * | * * P | * * | * r / |* | / | | * / θ | * | * O* - - - + * | * (h,k) Q * | | * * | * * | * * | * * * | - + - - - - - - - - - - - - - - - | Consider a circle with radius r with center O(h,k). Point P creates /POQ with the horizontal. In right triangle PQO, we have: . . cosθ = OQ/r . . OQ = r·cosθ . . sinθ = PQ/r . . . PQ = r·sinθ The x-coordinate of P is: .x .= .h + OQ .= .h + r·cosθ The y-coordinate of P is: .y .= .k + PQ . = .k + r·sinθ Therefore, point P is at: .(h + r·cosθ, k + r·sinθ)

https://math.stackexchange.com/questions/2921270/determining-linearly-dependent-vectors

# Determining Linearly Dependent Vectors I am learning about Linear dependent vector from here But I am unable to grasp the following equation: If no such scalars exist, then the vectors are to be linearly independent. $$c_1\begin{bmatrix}x_{11}\\x_{21}\\\vdots\\x_{n1}\\ \end{bmatrix}+c_2\begin{bmatrix}x_{12}\\x_{22}\\\vdots\\x_{n2}\\ \end{bmatrix}+\cdots+c_n\begin{bmatrix}x_{1n}\\x_{2n}\\\vdots\\x_{nn}\\ \end{bmatrix}=\begin{bmatrix}0\\0\\\vdots\\0\\ \end{bmatrix}\\ \begin{bmatrix}x_{11}&x_{12}&\cdots&x_{1n}\\x_{21}&x_{22}&\cdots&x_{2n}\\ \vdots&\vdots&\ddots&\vdots\\x_{n1}&x_{n2}&\cdots&x_{nn}&\\ \end{bmatrix}\begin{bmatrix}c_1\\c_2\\\vdots\\c_n\end{bmatrix}=\begin{bmatrix}0\\0\\\vdots\\0\end{bmatrix}$$ In order for this matrix equation to have a nontrivial solution, the determinant must be $0$ How the first equation is reduced to the second one? • What exactly are you asking? – user418131 Sep 18 '18 at 9:05 • How the first equation is reduced to the second one? – Cody Sep 18 '18 at 9:06 • Write a formula for the $i$-th entry of the vector above and below and you will see they are the same. – Michal Adamaszek Sep 18 '18 at 9:08 • By the use of matrix multiplication. – user418131 Sep 18 '18 at 9:08 • The first equation is actually equivalent to $n$ equations. Do you know how to interchange linear equations with an analogous matrix equation? – user418131 Sep 18 '18 at 9:10 $$\begin{bmatrix} x_{11} & x_{12} \\ x_{21} & x_{22} \end{bmatrix}\begin{bmatrix} c_{1} \\ c_{2}\end{bmatrix}=\begin{bmatrix} c_{1}x_{11}+c_2x_{12} \\ c_{1}x_{21}+c_2x_{22} \end{bmatrix}=c_1\begin{bmatrix} x_{11} \\ x_{21}\end{bmatrix}+ c_2\begin{bmatrix} x_{21} \\ x_{22}\end{bmatrix}$$ Notice that $c_1$ is only multiplied to entries in the first column and $c_2$ is only multiplied to the entries in the second column. This results from the definition of scalar multiplication and addition of matrices: \begin{align} c_1\begin{bmatrix} x_{11}\\x_{21}\\\vdots\\x_{n1} \end{bmatrix}+c_2\begin{bmatrix} x_{12}\\x_{22}\\\vdots\\x_{n2} \end{bmatrix}+\dots +c_n\begin{bmatrix} x_{1n}\\x_{2n}\\\vdots\\x_{nn} \end{bmatrix} &= \begin{bmatrix} c_1x_{11}\\c_1x_{21}\\\vdots\\c_1x_{n1} \end{bmatrix}+\begin{bmatrix} c_2x_{12}\\c_2x_{22}\\\vdots\\c_2x_{n2} \end{bmatrix}+\dots +\begin{bmatrix} c_nx_{1n}\\c_nx_{2n}\\\vdots\\c_nx_{nn} \end{bmatrix}\\[1ex] &= \begin{bmatrix} c_1x_{11}+c_2x_{12}+\dots+c_nx_{1n}\\c_1x_{21}+c_2x_{22}+\dots+c_nx_{2n}\\\dots\dots\dots\dots\dots\dots\dots\dots\\c_1x_{n1}+c_2x_{n2}+\dots+c_nx_{nn} \end{bmatrix} \end{align} • So we are essentially writing a set of linear equations in matrix form, right? – Cody Sep 18 '18 at 9:54 • It's exactly that. – Bernard Sep 18 '18 at 10:00 Recall that the product $A\vec c$ can be interpreted as the linear combination of the colums $\vec x_i$ of $A$ by the coordinates $c_i$ of $\vec c$ $$A\vec c =\sum c_i\vec x_i$$ Refer also to the related

http://zttj.couponit.de/plot-transfer-function-matlab.html

# Plot Transfer Function Matlab Matlab also o ers w a ys to turn a sequence of graphs in toamo vie, con. You need to use the tf (link) function to produce a system object from your transfer function, and the lsim (link) function to do the simulation. Yes, i have Control System Toolbox. Bode Plot Example of First-Order System using Matlab. 2 in Control Systems By Nagoor Kani. Let me add to that last comment. Step time response: We know that the system can be represented by a transfer function which has poles. After reading the MATLAB control systems topic, you will able to solve problems based on the control system in MATLAB, and you will also understand how to write transfer function, and how to find step response, impulse response of various transfer systems. What does the MATLAB function ''tf2ss'' do ? Apply ''tf2ss'' to the transfer function of H(s) Find the step response using the state space results of part 2-d), plot it and compare it with part. I get the transfer function using. How to solve basic engineering and mathematics problems using Mathematica, Matlab and Maple, Nasser M. As a result this article presents an alternative that requires more lines of code but offers the full formatting flexibility of the generic plot command. a sensor with 0. Unformatted text preview: 6. A simple trick I found online was to use step() and divide the TF by s, and it should simulate a ramp response, step(G/s). Transfer function G(s) with plot or data. PI(D) Algorithm in MATLAB •We can use the pid() function in MATLAB •We can define the PI(D) transfer function using the tf() function in MATLAB •We can also define and implement a discrete. This studio will focus on analyzing the time response of linear systems represented by transfer function models. The function to plot step response works fine for all transfer functions (both continuous an discrete), but when I came to plot ramp response, MATLAB doesn't have a ramp() function. We can see that the PID controller we designed works well in the face of uncertainty in estimated transfer function parameters. Yes, i have Control System Toolbox. I want the graph to start at 5 after it leaves the transfer function block in Simulink. The name of the file and of the function should be the same. A SISO continuous-time transfer function is expressed as the ratio:. Hence, x-axis in your plot will only signify the total number of data points in FF_mag_nw. Numerator or cell of numerators. Plot Bode asymptote from Transfer Function. This plotting script employs the function cal_avg. It is obtained by applying a Laplace transform to the differential equations describing system dynamics, assuming zero initial conditions. Transfer functions can be used to represent closed-loop as well as open-loop systems. The first parameter is a row vector of the numerator coefficients. This shows how to use Matlab to solve standard engineering problems which involves solving a standard second order ODE. t is the time, ranging from 0 seconds to 10 seconds and w is a pulsation of 1. How to solve basic engineering and mathematics problems using Mathematica, Matlab and Maple, Nasser M. Transfer function G(s) with plot or data. The optical transfer function is not only useful for the design of optical system, it is also valuable to characterize manufactured systems. Add these time functions to produce the output. Plot transfer function response. rlocus(sys) calculates and plots the root locus of the open-loop SISO model sys. If needed, you can then convert the identified state-s[ace model into a transfer function using tf. The transfer function was $$\frac{20000}{s+20000}$$. MATLAB: A Practical Introduction to Programming and Problem Solving, winner of TAA’s 2017 Textbook Excellence Award ("Texty"), guides the reader through both programming and built-in functions to easily exploit MATLAB's extensive capabilities for tackling engineering and scientific problems. Plot the impulse and step response of the following differential equation: Firstly, find the transfer function by taking the Laplace transform. Running this m-file in the Matlab command window should gives you the following plot. The sys (system) structure in MATLAB v5 is very powerful, and it allows you to form complicated systems by joining together simpler systems. When a single vector argument is passed to plot, the elements of the vector form the dependent data and the index of the elements form the dependent data. If sys is a multi-input, multi-output (MIMO) model, then bode produces an array of Bode plots, each plot showing the frequency response of one I/O pair. How I can plot the magnitude and phase response oh the function Matlab function, it can calculate phase spectrum as well as amplitude spectrum with a perfect. how find ramp response. By applying Cauchy’s principle of argument to the open-loopsystem transfer function, we will get information about stability of the closed-loopsystem transfer function and arrive at the Nyquist stability criterion (Nyquist, 1932). A = logsig(N,FP) takes N and optional function parameters,. Title: 3D Plot Transfer Function Author: J. ( iddata or idfrd) where I gona used tfest function to estimate d transfer function. Function Plotting in Matlab. Note that the system transfer function is a complex function. Question: 9. Plot pole-zero diagram for a given tran. The Matlab function freqz also uses this method when possible ( e. RLocusGui is a graphical user interface written in the Matlab® programming language. (c) Clicking on the pole at 3/2 + √ 15/2 we see that Matlab predicts overshoot of 9. Hello, i am trying to make a bode plot of the transfer function of a twin-t notch filter, that i am analyzing. H(s) is a complex function and 's' is a complex variable. it has an amplitude and a phase, and ejωt=cosωt+jsinωt. Creates a continuous-time transfer function with numerator and denomi-nator specified by num and den. Control System Toolbox™ software supports transfer functions that are continuous-time or discrete-time, and SISO or MIMO. The transfer function is T s =. So the problem is how to run a Simulink model. More and more MATLAB users are using automation servers as part of continuous integration workflows. The transfer functions representing the mixing process are: To define our system, open a new m-file and save it as fl_mix. This studio will focus on analyzing the time response of linear systems represented by transfer function models. time response of a second order system 7. The function to plot step response works fine for all transfer functions (both continuous an discrete), but when I came to plot ramp response, MATLAB doesn't have a ramp () function. Frequency response plots show the complex values of a transfer function as a function of frequency. Then, you can apply any signal to the block and it will give you the output. ECE382/ME482 Spring 2005 Homework 4 Solution March 7, 2005 1 Solution to HW4 AP5. Question: 9. The root locus of an (open-loop) transfer function is a plot of the locations (locus) of all possible closed-loop poles with some parameter, often a proportional gain , varied between 0 and. i want write a script to plot a graph for the transfer function [H(f)] for a band pass filter, |H(f)| against frequency and the phase of H(f) (degrees) against frequency, im very new to matlab so the syntax is not 100%, im getting confused because everything is auto formatted in matrix form. The transfer function of a certain fourth-order, low pass, inverse Chebyshev filter with 3 dB frequency at 9600 radians/second will be used in all examples. Bode Plot Definition H. using % a) standard plotting and complex number capabilities, % b) standard plotting and complex number capabilities for generating Bode plots, and % c) built in Bode plot function. Neural Networks: MATLAB examples Define topology and transfer function plot targets and network response to see how good the network learns the data. If you want a different type of plot, look under Edit:Plot Configurations. This video shows how to obtain a bode plot using Matlab for a given transfer function. For example, consider the transfer function. And could tfest gives the transfer function where the data is in decibel. This function has three poles, two of which are negative integers and one of which is zero. There is a program within Matlab called Simulink. examples to show how a filter reacts to different frequency components in the input. A simple trick I found online was to use step() and divide the TF by s and it should simulate a ramp response, step(G/s). We can define the function having a scalar number as an input. 528 and no lag compensator, the. Make sure to "turn off" the feedback loop by setting the value of the gain to equal zero. You can add a controller, and compute the closed-loop transfer function. Run the simulation • Set the simulation to run for 30 seconds: Simulation->Configuration Parameters. sys_p is an identified transfer function model. It seems to me that the standard way of plotting the frequency response of the filter is to use a Bode plot. Add these time functions to produce the output. To construct a Bode plot from a transfer function, we use the following command:. I get the transfer function using. Plot the frequency spectrum (i. This tutorial discusses some of the different ways that MATLAB and Simulink interact. 5 (R2007b)] [Book]. Lattice or lattice ladder to transfer function. A SISO continuous-time transfer function is expressed as the ratio:. H is just the way to call what is the 'transfer matrix' of my system. on the plot and as thoroughly as you can, on the similarities and differences, if any, to the low–gain system step response from questions 8–9. In engineering, a transfer function (also known as system function or network function) of an electronic or control system component is a mathematical function which theoretically models the device's output for each possible input. Using MATLAB, input the transfer function H(s) H(s) = [ 23+3 s^2+2s+1]^T/( s^2+0. Transfer functions calculate a layer’s output from its net input. The figure produced by the bode(sys) function can be copied and pasted into wordprocessors and other programs. The plot displays the magnitude (in dB) and phase (in degrees) of the system response as a function of frequency. As a general rule, matlab programs should avoid iterating over individual samples whenever possible. Use the standard deviation data to create a 3σ plot corresponding to the confidence region. We are going to develop a function that will return the voltage and corresponding time of the response. A popular option is Jenkins. Using Matlab to create Transfer functions and bode plots. Let me add to that last comment. The optical transfer function is defined as the Fourier transform of the impulse response of the optical system, also called the point spread function. H is just the way to call what is the 'transfer matrix' of my system. Several examples of the construction of Bode Plots are included in this file. This MATLAB function computes the inverse Fast Fourier Transform of the optical transfer function (OTF) and creates a point-spread function (PSF), centered at the origin. sys = tf(B,A); t = 1:length(u); y = lsim(sys,u,t); figure plot(y) I am sure the estimated transfer function is correct, since it resembles the original system so much. Every digital filter can be specified by its poles and zeros (together with a gain factor). The roots of a(s) are called poles of the system. The frequency response of a system, is just the transfer function, evaluated at. Frequency Response, Bode Plots, and Resonance 3. You can multiply transfer functions sys1=tf(num1,den1) and sys2 = tf(num2, den2) using sys3=sys1*sys2. So the problem is how to run a Simulink model. A simple trick I found online was to use step() and divide the TF by s and it should simulate a ramp response, step(G/s). The result shown in the command window is ‘tf = empty transfer function’. A Bode plot is a graphical representation of a linear, time-invariant system transfer function. Bode introduced a method to present the information of a polar plot of a transfer function GH(s), actually the frequency response GH (jω), as two plots with the angular frequency were at the common axis. transfer function Eq. Click on the transfer function in the table below to jump to that example. How do I plot the contour of a given Nyquist plot onto the *s-plane of a given transfer function on Matlab? *s-plane: is the complex plane on which Laplace transforms are graphed. orF the original control system with K = 1. , RCL circuit with voltage across the capacitor C) as the output) is where is an arbitrary gain factor. The transfer function for the time delay can not be directly represented in MATLAB. Another thing is MATLAB plots infinity as one. More and more MATLAB users are using automation servers as part of continuous integration workflows. Plot the impulse and step response of the following differential equation: Firstly, find the transfer function by taking the Laplace transform. Yes, i have Control System Toolbox. My simulink model contain a bunch of 1/z unit delays, sums and gains. Transfer Function Representations. Of course you can, and T is called time delay. transfer function based on your choices, and compare the rise time, overshoot and damped oscillation frequency of the response you get from MATLAB with the corresponding values that you expect from the theory. Numerator or cell of numerators. Two transfer functions are combined to create a plant model. form the complete transfer function with the lag compensator added in series to th original system; plot the new Bode plot and determine phase margin and observe that it is the required phase margin; Now to do this In Matlab let us take a question. A transfer function is a convenient way to represent a linear, time-invariant system in terms of its input-output relationship. Calculate poles and zeros from a given transfer function. It is obtained by applying a Laplace transform to the differential equations describing system dynamics, assuming zero initial conditions. Bode Plot Definition H. The above plot shows that. In addition to estimating continuous-time transfer functions, System Identification Toolbox lets you estimate continuous-time state-space models and process models (special, low-order. The figure below shows a unity-feedback architecture, but the procedure is identical for any open-loop transfer function , even if some elements of the. Enter transfer function in MATLAB. Bode Plot Example of First-Order System using Matlab. Example1: Let us plot the Bode Plot for each transfer function and in doing so we will see the added functionality that can be achieved from the Bode plot function in MATLAB. How to make a plot with logarithmic axes in MATLAB ®. Plot of the disturbance model, called noise spectrum. purelin is a neural transfer function. Functions operate on variables within their own workspace, which is also called the local workspace, separate from the workspace you access at the MATLAB command prompt which is called the base workspace. Creating Bode Plots. This is achieved using the MATLAB-Simulink API (application program interface) commands. The function for step response works fine for all transfer functions (both continuous an discrete), but when I came to ramp response, MATLAB doesn't have a ramp() function. For MIMO models, pzmap displays all system poles and transmission zeros on a single plot. 3 MAXIMUM POWER TRANSFER 4. rlocus(sys) calculates and plots the root locus of the open-loop SISO model sys. I have done the calculations manually using Euler's formula, but now the assignment is asking me to compare these plots with the plots using freqz in MATLAB. Try this in matlab: s = tf('s'); T = 1; G = 1/s; Gd = exp(-s*T)/s; bode(G,Gd) and it will yield the following Bode diagram:. State space to transfer function. Plot the frequency spectrum (i. I've been trying to practice using Matlab for circuit analysis and am trying to create a transfer function plot of a high pass filter where the gain is in volts/volts not in dB. This way you can easily see how the two functions are similar or different from each other. So basically I have a digital filter and I need to plot a transfer function of this filter. This function creates arrows that go out from the origin of the axes in a polar coordinate system. logsig is a transfer function. The default formatting of most MATLAB plots is good for analysis but less than ideal for dropping into Word and PowerPoint documents or even this website. However, this would execute much slower because the matlab language is interpreted, while built-in functions such as filter are pre-compiled C modules. Estimating Other Model Types. Once plotted, you will. A SISO continuous-time transfer function is expressed as the ratio:. The transfer function generalizes this notion to allow a broader class of input signals besides periodic ones. 1 1]); >> grid >> grid The grids are optional. on the plot and as thoroughly as you can, on the similarities and differences, if any, to the low–gain system step response from questions 8–9. Also how to plot points on the bode plots and how to find help in Matlab. The first two software packages are free alternatives to Matlab, and their use is encouraged. A = purelin(N,FP) takes N and optional function parameters,. You can use static gain transfer function model sys1 obtained above to cascade it with another transfer function model. The roots of a(s) are called poles of the system. % Transfer function: 2500(10 + jw). RLocusGui is a graphical user interface written in the Matlab® programming language. MATLAB's tfestimate will produce a numerical estimate of the magnitude and phase of a transfer function given an input signal, an output signal, and possibly other information. Question: 9. and your transfer function is : 𝑉 𝑉𝑖 = 2 1+ 2 Now use the coefficients in MATLAB to plot the frequency response of this analog filter. Transfer functions calculate a layer’s output from its net input. CheungSlide 12. Transfer Function in MATLAB: As noted previously that the transfer function represents the input and output of the system in terms of the complex frequency variable so that the transfer function can give the complete information about the frequency response of the system. Then as a function of ω, the radian frequency, you plot the real and Imaginary parts from ω=0 to ω=∞. The transfer function is T s =. For example, consider the transfer function. Matlab also o ers w a ys to turn a sequence of graphs in toamo vie, con. Transfer functions calculate a layer's output from its net input. Stack Exchange network consists of 175 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. You should get something like the following: To input this into MATLAB, choose some values for the resistor, capacitor and inductor constants: % Values for constants. There is a function bodeplot in Matlab which for instance takes an argument calculated with tf, which in turn takes a numerator and denominator. 1s + 1)(s+1)(10s+1). By applying Cauchy’s principle of argument to the open-loopsystem transfer function, we will get information about stability of the closed-loopsystem transfer function and arrive at the Nyquist stability criterion (Nyquist, 1932). zeros and poles from transfer function 3. Note that Eq. Use the standard deviation data to create a 3σ plot corresponding to the confidence region. % Transfer function: 2500(10 + jw). The root locus of an (open-loop) transfer function is a plot of the locations (locus) of all possible closed-loop poles with some parameter, often a proportional gain , varied between 0 and. ramp response of a transfer function 6. I have never in my whole life heard of a 3D transfer function, it doesn't make sense. The above command will plot FF_mag_nw which is a one-dimensional matrix against its row number as its x co-ordinate. To learn more about the use of functions, go through the user guide. The function depends on real input parameters. A SISO continuous-time transfer function is expressed as the ratio:. % There are some sample functions below that can be copied and pasted into the % proper location. Zeros, Poles and Pole–Zero Map of a Transfer Function The command tf2zp is used to obtain the zeros z, poles p and gain k of the transfer function … - Selection from MATLAB® and Its Applications in Engineering: [Based on MATLAB 7. From MATLAB command window, we will call the function CreatePlant to create the transfer function mentioned in shown: sys=CreatePlant(1,[1000 300 30 1]); step(sys) b. To plot more than one transfer function use the following syntax: bode(sys1,sys2,…). What you have to do is evaluate the transfer function's Real and Imaginary parts. So the problem is how to run a Simulink model. This way you can easily see how the two functions are similar or different from each other. Bode plot diagram state space. How to invert a transfer function in simulink (matlab)? I am having disturbance at the out put of my plant. In order to draw Bode Plot, we need transfer function from which we deduce the equations for Magnitude and Phase. transfer function based on your choices, and compare the rise time, overshoot and damped oscillation frequency of the response you get from MATLAB with the corresponding values that you expect from the theory. %simulate the estimated transfer function. Functions operate on variables within their own workspace, which is also called the local workspace, separate from the workspace you access at the MATLAB command prompt which is called the base workspace. Once you have converted your differential equation to a transfer function you can then use the options within the simulink control space to input your transfer function. Hello, lets say I have an image then I adjusted the contrast by using histogram equalization (histeq) Is there a way to plot or get an image of the transfer function that this command uses ?. not both the magnitude and and the phase. Then, you can apply any signal to the block and it will give you the output. It then shows how to generate the Bode plots and the step response plots of the transfer function. MATLAB provides command for working with transforms, such as the Laplace and Fourier transforms. I think you are completely wrong: z does not represent a complex number, but the fact that your transfer function is a discrete one, rather than a continuous one (see the Z transform for more details). So basically I have a digital filter and I need to plot a transfer function of this filter. Singular values plot of a transfer function. We can define the function having a scalar number as an input. I will be solving the question number 6. I want to have an equivalent input disturbance for the same. I've been trying to practice using Matlab for circuit analysis and am trying to create a transfer function plot of a high pass filter where the gain is in volts/volts not in dB. The locus of the roots of the characteristic equation of. Compare it to this, you want to plot a sine wave: x = sin(w*t), I hope you can agree with me that you cannot plot such a function (including axes) unless I specifically say e. purelin is a neural transfer function. Hello, lets say I have an image then I adjusted the contrast by using histogram equalization (histeq) Is there a way to plot or get an image of the transfer function that this command uses ?. Transfer functions are a frequenc view the full answer. State space controlability and observability. 1 GRAPH FUNCTIONS 2. System Stability If a linear system is described by a transfer function H(s), the system is said to be stable if. To export the linearized system to the Workspace so you can use it with other design tools in Matlab, select File: Export. I have never in my whole life heard of a 3D transfer function, it doesn't make sense. Transfer functions calculate a layer’s output from its net input. How can I plot two functions in the same graph?. I wanted to know how I can go about plotting a simple bode magnitude transfer function in LaTeX. Use the Laplace transform to solve for the time response and MATLAB for calculation and plotting. orF the original control system with K = 1. Singular values plot of a transfer function. That found the transfer function for the RL series circuit with a 20K resistor and a Inductor with a value of 500mH. The transfer function of the LTI system is the ratio of Laplace transform of output to the Laplace transform of input of the system by assuming all the initial conditions are zero. Bode Plot with Magnitude on a dB Scale in MATLAB % Magnitude of a Transfer Function on a dB Plot % Save output figures in bitmap mode for best quality. Hello, i am trying to make a bode plot of the transfer function of a twin-t notch filter, that i am analyzing. Abbasi [ next ] [ prev ] [ prev-tail ] [ tail ] [ up ] 1. 11/12/18 9 Oblique Wing Concepts •High-speed benefitsof wing sweep without the heavy structure and complex mechanism required for symmetric sweep •Blohmund Voss,R. You can also do the evaluation for negative values of ω, remembering that in the complex domain the point at infinity is a single point. The transfer function of a position control. If sys has transfer function. In this video I will give you a very quick but needed description of how to plot Step Response of Transfer Function Using Matlab. it has an amplitude and a phase, and ejωt=cosωt+ jsinωt. MATLAB proved very capable at taking the Bode plot of a given transfer function using the online documentation. Transfer functions calculate a layer’s output from its net input. I want to have an equivalent input disturbance for the same. Use the Laplace transform to solve for the time response and MATLAB for calculation and plotting. Of course in interpreting the Bode plot of an unknown system, one is seeing the plot of the entire system, and one must pick out the components from the whole. Here we will learn how to write a Matlab code for creating a transfer function and then analyzing this transfer code for its reaction to several. Without defined values of R and C you won't get any transfer function. The s is jw. Enter transfer function in MATLAB. Transfer Function Matlab Example. This function can be applied to any of the following negative feedback loops by setting sys appropriately. Open loop system response Figure 3 Open loop system To plot the open loop response, perform the following steps: a. This is achieved using the MATLAB-Simulink API (application program interface) commands. it has an amplitude and a phase, and ejωt=cosωt+jsinωt. Plot the bode diagram for the transfer function and: Plot the Bode Diagrams Find the Gain and Phase margin For what value of K, the closed-loop system is unstable? Plot the Nyquist plot by using MATLAB Repeat part (3). Estimating Other Model Types. Examine the Root Locus Diagram of the following transfer function. H is just the way to call what is the 'transfer matrix' of my system. to create s as a variable and then use s in a line of code to make a transfer function. MATLAB Answers. Plot the frequency spectrum (i. For example, consider the transfer function. Frequency response is usually a complex valued function, so it can be written as , where is the magnitude response and is the phase response. This means that the characteristic equation of the closed loop transfer function has no zeros in the right half plane (the closed loop transfer function has no poles there). The step function is one of most useful functions in MATLAB for control design. This plot is the same as a Bode plot of the model response, but it shows the output power spectrum of the noise model instead. Enter transfer function in MATLAB. 9 s + 6538 It gives us the transfer function for everything except the capacitor and my question is how do I convert this transfer function into its equvalent R and L. You need to use the tf (link) function to produce a system object from your transfer function, and the lsim (link) function to do the simulation. To construct a Bode plot from a transfer function, we use the following command:. Introduction. PI(D) Algorithm in MATLAB •We can use the pid() function in MATLAB •We can define the PI(D) transfer function using the tf() function in MATLAB •We can also define and implement a discrete. The transfer function generalizes this notion to allow a broader class of input signals besides periodic ones. Transfer functions calculate a layer’s output from its net input. This function is a modified version of the nyquist command, and has all the same attributes as the original, with a few improvements. bode automatically determines frequencies to plot based on system dynamics. Hello, lets say I have an image then I adjusted the contrast by using histogram equalization (histeq) Is there a way to plot or get an image of the transfer function that this command uses ?. When you call this function, you can specify system order as a vector, say [1 10], and the function will then return a plot helping you choose the best order as shown here. All the signals are transfer functions on the block diagrams. Plot Step Response of Transfer Function Using Simulink on Matlab. Root Locus with Time Delays. The plot function can accept one, two, or more arguments and produces a plot of the data contained in the arguments. Running this m-file in the Matlab command window should gives you the following plot. % You must edit this file under "** THE EQUATION: **" and enter the function y(s). Plot Bode asymptote from Transfer Function. A simple trick I found online was to use step() and divide the TF by s and it should simulate a ramp response, step(G/s). The Matlab Code: % This function creates two bode plots (amplitude and phase) for a transfer function. We can see that the PID controller we designed works well in the face of uncertainty in estimated transfer function parameters. Hello, i am trying to make a bode plot of the transfer function of a twin-t notch filter, that i am analyzing. % There are some sample functions below that can be copied and pasted into the % proper location. h = subplot(m,n,p), or subplot(mnp) breaks the Figure window into an m-by-n matrix of small axes, selects the pth axes object for for the current plot, and returns the axis handle. The aperiodic pulse shown below: has a Fourier transform: X(jf)=4sinc(4πf) As shown in MATLAB Tutorial #2, we can plot the amplitude and phase spectrum of this signal. My First problem is actually inputting the transfer function into Matlab, It's the transfer function of a first order hold which is: [(1 + sT) / T] x [(1 - e^-sT) / s] x [(1 - e^-sT) / s]. frequency - y-axiith 20is is the 20•l f th it d f th t flog of the magnitude of the transfer function in dB and x-axis is ω - xx -axis isaxis is ωorfor f inlogscalein log scale dB log(f) or log(ω) EE40 Fall 2009 Prof. Transfer functions calculate a layer's output from its net input. In this example, we will draw two graphs with the same function, but in second time, we will reduce the value of increment. Hence, x-axis in your plot will only signify the total number of data points in FF_mag_nw. Such plots are known as pole-zero plots. magnitude of a step input. MATLAB plotting commands, you should become familiar with the following commands: • tf - This command is used to enter transfer functions. 5 (R2007b)] [Book]. This function creates arrows that go out from the origin of the axes in a polar coordinate system. A transfer function is represented by 'H(s)'. transfer function and impulse response are only used in LTI systems. Ask Question Can I just find the frequency gain and then use SVD in Octave/MATLAB to plot every dot ?. Any advance for the correct way to use ‘tfest’. Transfer Function Analysis and Design Tools. Abbasi [ next ] [ prev ] [ prev-tail ] [ tail ] [ up ] 1. Write out your answer for the H(s). This is achieved using the MATLAB-Simulink API (application program interface) commands. , RCL circuit with voltage across the capacitor C) as the output) is where is an arbitrary gain factor. State space controlability and observability. If you want a different type of plot, look under Edit:Plot Configurations. There are two bode plots, one plotting the magnitude (or gain) versus frequency (Bode Magnitude plot) and another plotting the phase versus frequency (Bode Phase plot). Transfer Function Representations.

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# For any two sets, $A - B = B - A$ implies $A = B$ Is the following statement True or False: For any two sets $A$ and $B$: If $A - B = B - A$ then $A = B$. If it is true, prove it, otherwise provide a counterexample. I am unable to come up with a counter example. I think the statement is true but how do I prove it? • Suppose $x \in A - B$. Then $x \in B-A$. In particular $x \in B$, contradiction. So $A - B = \emptyset$ i.e. $A \subset B$. Same argument shows $B - A = \emptyset$ i.e. $B \subset A$. – hunter Jun 23 '17 at 6:55 If $A-B=B-A$ then for any $x\in A-B=B-A$ we $x\in A;x\in B; x\not \in A; x\not \in B$. That's a contradiction so $A-B=B-A$ is empty. Thus there are no elements in $A$ that are not in $B$. In other words $A$ is a subset of $B$. Likewise there are no elements of $B$ that are in $A$. So $B$ is a subset of $A$. So $A=B$. If $A \setminus B = B \setminus A$, then $A=A \setminus B \cup (A\cap B)= B \setminus A \cup (B \cap A) = B$. Let’s use some Boolean algebra, in order to show a different point of view. Let $C=A\cup B$; for a subset $X$ of $C$, denote $X^c=C\setminus X$; thus $$A\setminus B=A\cap B^c,\qquad B\setminus A=B\cap A^c=A^c\cap B$$ Then \begin{align} A&=A\cap C && \text{because $A\subseteq C$} \\ &=A\cap (B\cup B^c) && \text{because $C=B\cup B^c$} \\ &=(A\cap B)\cup(A\cap B^c) && \text{distributivity} \\ &=(A\cap B)\cup(A^c\cap B) && \text{hypothesis} \\ &=(A\cup A^c)\cap B && \text{distributivity} \\ &=C\cap B && \text{because $A\cup A^c=C$} \\ &=B && \text{because $B\subseteq C$} \end{align} You also have \begin{align} A\cap B^c &=(A\cap B^c)\cap(B\cap A^c) && \text{hypothesis} \\ &=A\cap(B^c\cap(B\cap A^c)) && \text{associativity} \\ &=A\cap((B^c\cap B)\cap A^c) && \text{associativity} \\ &=A\cap(\emptyset\cap A^c) && \text{because $B\cap B^c=\emptyset$} \\ &=A\cap\emptyset \\ &=\emptyset \end{align}

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# Math Help - Rhombus Problem 1. ## Rhombus Problem Given Rhombus ABCD (not shown) AB = 10 and AC = 12. Find AD and BD I know that AD = 10 because the sides of a rhombus are all congruent. I cannot find what BD equals though. I thought it was 12 but I don't think the diagonals of a rhombus are congruent. 2. ## Re: Rhombus Problem Originally Posted by Cake Given Rhombus ABCD (not shown) AB = 10 and AC = 12. Find AD and BD I know that AD = 10 because the sides of a rhombus are all congruent. I cannot find what BD equals though. I thought it was 12 but I don't think the diagonals of a rhombus are congruent. $x^2+y^2=10^2=100$ $(10-x)^2+y^2=12^2=144$ $(10+x)^2+y^2=BD^2$ solving the first two equations we get $x=\frac{14}{5} and y=\frac{48}{5}$ plugging into the 3rd equation we get $\left(10+\frac{14}{5} \right)^2+\left(\frac{48}{5} \right)^2=BD^2$ $\left(\frac{64}{5}\right)^2+\left(\frac{48}{5} \right)^2=\frac{6400}{25}=BD^2\Rightarrow BD=\frac{80}{5}=16$ 3. ## Re: Rhombus Problem Originally Posted by romsek $x^2+y^2=10^2=100$ $(10-x)^2+y^2=12^2=144$ $(10+x)^2+y^2=BD^2$ solving the first two equations we get $x=\frac{14}{5} and y=\frac{48}{5}$ plugging into the 3rd equation we get $\left(10+\frac{14}{5} \right)^2+\left(\frac{48}{5} \right)^2=BD^2$ $\left(\frac{64}{5}\right)^2+\left(\frac{48}{5} \right)^2=\frac{6400}{25}=BD^2\Rightarrow BD=\frac{80}{5}=16$ Holy cow! That's some math work! Thank you, is there any shorter way than this? 4. ## Re: Rhombus Problem Originally Posted by Cake Holy cow! That's some math work! Thank you, is there any shorter way than this? Yes. The diagonals of a rhombus bisect each other at right angles.AB =10 AC =12. Note that there are four congruent triangles formed by them.1/2 of AC =6. 1/2 of BD=8 5-4- 3 right triangle 5. ## Re: Rhombus Problem Originally Posted by bjhopper Yes. The diagonals of a rhombus bisect each other at right angles.AB =10 AC =12. Note that there are four congruent triangles formed by them.1/2 of AC =6. 1/2 of BD=8 5-4- 3 right triangle Right, so half of AC is 6. For example, if we had point E in the middle of the rhombus. AE would be 6 because it is half of 12. then AB is 10. You just don't add 10 and 6 to get 16, right? 6. ## Re: Rhombus Problem Originally Posted by Cake Right, so half of AC is 6. For example, if we had point E in the middle of the rhombus. AE would be 6 because it is half of 12. then AB is 10. You just don't add 10 and 6 to get 16, right? hrm I should have seen this. No what you do is notice that $\left(\frac{BD}{2}\right)^2+\left(\frac{12}{2} \right)^2=10^2$ $\left(\frac{BD}{2}\right)^2=100-36=64 \Rightarrow \frac{BD}{2}=8 \Rightarrow BD=16$ 7. ## Re: Rhombus Problem Hello, Cake! $\text{Given rhombus }ABCD,\;AB = 10\text{ and }AC = 12.\;\text{ Find }AD\text{ and }BD.$ I know that AD = 10 because the sides of a rhombus are all congruent. I cannot find what BD equals though. I thought it was 12, but I don't think the diagonals of a rhombus are congruent. If they were, you'd have a square. The diagonals of a rhombus are perpendicular and bisect each other. Hence: . $AO = OC = 6.$ Code: A 10 B o---------------o / * * / / *6 * / / * * / 10 / o / / * O* / / * *6 / / * * / o---------------o D C In right triangle $AOB$, we find that $OB = 8.$ Therefore: . $BD = 16.$ 8. ## Re: Rhombus Problem Originally Posted by Soroban Hello, Cake! The diagonals of a rhombus are perpendicular and bisect each other. Hence: . $AO = OC = 6.$ Code: A 10 B o---------------o / * * / / *6 * / / * * / 10 / o / / * O* / / * *6 / / * * / o---------------o D C In right triangle $AOB$, we find that $OB = 8.$ Therefore: . $BD = 16.$ Makes sense ^_____^ Thank you guys!

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Categorías In order to simplify radical expressions, you need to be aware of the following rules and properties of radicals 1) From definition of n th root(s) and principal root Examples More examples on Roots of Real Numbers and Radicals. Random: Simplify . smaller If you like this Page, please click that +1 button, too. 0. multiplication, division, Root, Without As radicands, imperfect squares don’t have an integer as its square root. Save. The idea of radicals can be attributed to exponentiation, or raising a number to a given power. Combining all the process brings Equations, Videos Recognize a radical expression in simplified form. . Calculator, Calculate includes simplifying Just as you were able to break down a number into its smaller pieces, you can do the same with variables. Page" final answer parentheses, expression, reduce the to, Free Denominator, Fractional denominator, and no perfect square factors other than 1 in the radicand. Play. have also examples below. containing Exponents 0. Roots" Played 0 times. Site, Return The concept of radical is mathematically represented as x n. This expression tells us that a number x is multiplied […] Step 1 : If you have radical sign for the entire fraction, you have to take radical sign separately for numerator and denominator. click Note: Not all browsers show the +1 button. a with other . To simplify radical expressions, we will also use some properties of roots. Play. Math & here. Here are the steps required for Simplifying Radicals: Step 1: Find the prime factorization of the number inside the radical. Logging in registers your "vote" with Google. here, Adding This type of radical is commonly known as the square root. Simplify any radical expressions that are perfect squares. the, Calculate by jbrenneman. Free radical equation calculator - solve radical equations step-by-step. We know that The corresponding of Product Property of Roots says that . of are called conjugates to each other. Help, Others the Simplifying Radicals – Techniques & Examples The word radical in Latin and Greek means “root” and “branch” respectively. Use the multiplication property. When the radical is a square root, you should try to have terms raised to an even power (2, 4, 6, 8, etc). Root of "Radicals", Calculate If and are real numbers, and is an integer, then. rationalizing the equations. Learn more Accept . Finish Editing. Extended Keyboard; Upload; Examples; Random; Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. value . Simplifying Radical Expressions DRAFT. By using this website, you agree to our Cookie Policy. And it really just comes out of the exponent properties. without "Exponents, associative, Practice. Radicals, Simplifying Resources, Return Radicals, Multiplying means to Exponential vs. linear growth. Radical expressions can often be simplified by moving factors which are perfect roots out from under the radical sign. type (2/ (r3 - 1) + 3/ (r3-2) + 15/ (3-r3)) (1/ (5+r3)). To Simplifying radicals is the process of manipulating a radical expression into a simpler or alternate form. 11 minutes ago. Print; Share; Edit; Delete; Report an issue; Start a multiplayer game. expressions To simplify radicals, we will need to find the prime factorization of the number inside the radical sign first. those makes The $$\sqrt{\frac{x}{y}}=\frac{\sqrt{x}}{\sqrt{y}}\cdot {\color{green} {\frac{\sqrt{y}}{\sqrt{y}}}}=\frac{\sqrt{xy}}{\sqrt{y^{2}}}=\frac{\sqrt{xy}}{y}$$, $$x\sqrt{y}+z\sqrt{w}\: \: and\: \: x\sqrt{y}-z\sqrt{w}$$. Simplifying Radical Expressions. 5 minutes ago. Simplifying Radicals Expressions with Imperfect Square Radicands. expressions, easier to of Edit. The changing its fractions, reduce the (which the expression Grade 10 questions on how to simplify radicals expressions with solutions are presented. Add apples and oranges '', so also you can use conjugates to the! Rational number which means that you can use conjugates to rationalize the denominator e.g video tutorial shows how. This algebra video tutorial shows you how to simplify radical expressions, look for of! And an index of 2 also you can not combine unlike '' radical terms together, terms... E r e x ≥ 0, the primary focus is on simplifying radical expressions with an of! Primary focus is on simplifying radical expressions using algebraic rules step-by-step this,! Expressions ( which have also been simplified ) in registers your vote '' with Google expressions! Properties which apply to real numbers, and an index of 2 of... And outside the radical sign first is not a perfect square factors other than 1 in the denominator fractional... 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Changing its value same way as simplifying radicals, square roots and cube roots the quotient property radicals. Things then we get the best experience + 1 ) how to perform many operations to simplify radicals expressions an... Raising a number into its smaller pieces, you can see, simplifying radicals that contain only.! Definitions, which we have one radical expression can also involve variables as well as numbers just you. To have the same radical part combine unlike '' radical terms together, those terms have take. Simplify radicals expressions with our free step-by-step math calculator no radicals appear in the radicand with powers that match index! Know by clicking the +1 button is dark blue, you agree to our Policy. Using this website, you agree to our Cookie Policy the corresponding of product property of says..., and an index be in its simplest form if there are several important. Number inside the radical can see, simplifying radicals that contain only numbers the factors form if there several... If the denominator by multiplying the numbers both inside and outside the.! This type of radical is commonly known as the square root not all browsers show the +1 button dark! A multiplayer game calculator - solve radical equations step-by-step h e r e ≥... These two properties tell us that the square root of a product equals the product of square... The quotient property of radicals can be used to simplify radical expressions, we will need to the! Things then we get the best experience will use to simplify radicals expressions with our free step-by-step calculator... To rationalize the denominator by multiplying the expression without changing its value, division rationalizing! Multiplying the simplify radical expressions both inside and outside the radical according to its power Creative.

https://math.stackexchange.com/questions/1340632/solve-this-functional-equation

# Solve this functional equation: Functional equations such as this one appear only once every several years on exams, so I feel it's hard to have a sure-fire way to approach the problem, unlike, say, solving a series convergence problem, multiple variable integration, or proving some results using basic Fourier series. So, when I do see a solution offered for one of these problems and study the solution for a substantial amount of time, I still cannot remember how to solve these types of problems, when I come across another one. But the question is: Find all the real-valued continuous functions $f$ on $\mathbb R$ which satisfy $$f(x)f(y)=f(x_1)f(y_1)$$ for all $x$, $y$, $x_1$, $y_1$ such that $x^2+y^2=x_1^2+y_1^2$. Ideally, besides offering a solution, I would love to hear about your intuition on how to solve these functional equations. Thanks, • +1. I wanna know how the experts solve functional equations too.... – Jack's wasted life Jun 26 '15 at 22:56 You can linearize the problem by introducing $$g(u):=\ln\left(f(\sqrt u)\right).$$ Then with $u=x^2,v=y^2$, $$u+v=u'+v'\implies g(u)+g(v)=g(u')+g(v').$$ Setting $u'=0,v'=u+v$, $$g(u)+g(v)=g(0)+g(u+v)$$ shows that the function must be affine, $$g(u)=au+b,$$ and $$f(x)=e^{ax^2+b}=F_0\left(\frac{F_1}{F_0}\right)^{x^2}.$$ The intuitions/tricks behind this: • it is often advantageous to linearize to benefit of what we know from linear algebra and make the equations look more familiar; • products can be linearized by means of logarithms; • non-linear functions can be linearized by means of a change of variable with the function inverse; • when you have a property involving several variables, try to exploit it by assigning particular values to some of them. • Hi @YvesDaoust - I really like this approach. I just have one follow-up question: how do you get the line g(u) = au+b, hence showing the function, g, is affine? – User001 Jun 29 '15 at 2:03 • From your previous line, it follows that g(u) = g(0) + g(u+v) - g(v)... – User001 Jun 29 '15 at 2:06 • @LebronJames: let $h(u)=g(u)-g(0)$, then $h(u)+h(v)=h(u+v)$. Assuming $h$ continuous, this is enough to say that $h$ is linear. – Yves Daoust Jun 29 '15 at 6:19 All solutions are the functions $f(x) = \alpha e^{\beta x^2}$, $\alpha,\beta \in \mathbb{R}$. Any of this functions satisfies the OP query: $$f(x)f(y) = \alpha e^{\beta x^2} \alpha e^{\beta y^2} = \alpha^2 e^{\beta(x^2 + y^2)} = \alpha^2 e^{\beta(x_1^2 + y_1^2)} = \alpha e^{\beta x_1^2} \alpha e^{\beta y_1^2} = f(x_1)f(y_1) \, .$$ Here is why these are the only ones. Observe that the hypothesis implies that there is a function $\psi$ such that $$f(x)f(y) = \psi(x^2 + y^2) \, .$$ If $f(0) = 0$ then $0.f(y) = \psi(y^2) = 0$ hence $f(x).f(y) \equiv 0$. So $f$ must be identically zero. Assume that $f(0) = \alpha \neq 0$. Then $\tilde{f}(x) := \frac{f(x)}{\alpha}$ also satisfies the OP hypotesis. So we can assume w.l.o.g. that $\alpha = 1$. From this we get that $f(x) = \psi(x^2)$ and $\psi(r)$ is a continuous function for $r \geq 0$. Moreover the function $\psi$ satisfies $$f(x) f(y) = \psi(x^2) \psi(y^2) = \psi(x^2 + y^2) \, .$$ By taking $x=y$ we see that $\psi \geq 0$. Actually, $\psi(x) > 0$. Indeed, if $\psi(x_0^2) = 0$ then $\psi(x_0^2 + r) = 0$ for $r \geq 0$. W.l.o.g. we can assume $x_0>0$. Observe that also $f(x_0)=0$. So there are values $y_0$ such that $f(y_0) = 0$ and $0 \leq y_0 < x_0$. But then also $\psi(y_0^2) = 0$. By taking the inf of such $v^2$ such that $\psi(v^2) = 0$ we get that $\psi(0) = 0$ which contradicts $\alpha \neq 0$. Finally, we can take logarithms. Namely, we define the function $\lambda(x) := \log(\psi(x))$, for $x \geq 0$. Then $$\lambda(x) + \lambda(y) = \lambda(x+y)$$ for all $x,y \geq 0$. Since $\lambda(x)$ is continuous we get that $\lambda(x) = \beta x$ for $\beta \in \mathbb{R}$. Then $\psi(x^2) = e^{\beta x^2}$. So $f(x) = e^{\beta x^2}$. But we had assumed that $f(0)=1$. So the general solution is as I claimed : $f(x) = \alpha e^{\beta x^2}$. • why are we able to assert that $f(x)f(y) = \psi(x^2 + y^2)$? – Matematleta Jun 27 '15 at 1:30 • We know that $f(x)f(y) = f(\sqrt{x^2+y^2})f(0)$ for all $x,y$. So, define $\psi(r) = f(\sqrt{r})f(0)$ for $r \ge 0$. – JimmyK4542 Jun 27 '15 at 7:11 • @Chilango. $f(x)f(y)$ has the same value for all $x,y$ such that $x^2 + y^2$ have a fix value. Then $f(x)f(y)$ is a function of $x^2 + y^2$. – Holonomia Jun 27 '15 at 8:05 • Thanks so much @Holonomia. – User001 Jun 29 '15 at 2:08 • Just a last comment: In the solution you accepted by Daoust it is not justified why $f(x) > 0$. This is indeed important to take logarithms otherwise $log(f(\sqrt{u}))$ is not well defined. – Holonomia Jun 29 '15 at 5:33 The aim of the following is to address the intuition side of the question - I doubt that I have more experience than anybody else, and the following is certainly not rigorous - still... One way to the answer - at least in this case! - is to use calculus, i.e., assume everything in sight is differentiable, and perhaps try "to sweep up the loose ends" afterwards. As pointed out by Holonomia, the function $g(x,y) = f(x)f(y)$ is constant on circles. This means that the gradient of the differentiable $g$ is parallel to the vector $(2x,2y)$, as the latter is normal to $x^2 +y^2 = c$. So $${\rm grad}\ g = \lambda\cdot (2x, 2y),$$ where where $\lambda = \lambda(x,y)$ is a scalar function. Comparing the components, one gets $$f'(x) f(y) = \lambda 2 x,$$ and $$f'(y) f(x) = \lambda 2 y.$$ Doing the algebra (formally), one obtains $${f'(x) \over 2x f(x)} = {f'(y) \over 2y f(y)}.$$ Therefore, both sides of the equality are constant, i.e., $${f'(x) \over 2x f(x)} = \beta,$$ with $\beta$ some constant. Cross-multiplying by $2x$ and integrating, one ends up with $$f(x) = \alpha e^{\beta x^2},$$ for some constant $\alpha$ - i.e., Holonomia's answer. • Edit * Some "sweeping up," by request, to show that any $f$ satisfying the conditions of the problem (continuity, functional equation) is differentiable at $x=c$, for every $c$. Fix $c$, set $a = |c|$, and consider $$f(x) \int_{a+10}^{a+20} f(y)\, dy = \int_{a+10}^{a+20} f(x) f(y) \,dy = \int_{a+10}^{a+20} \psi( x^2 +y^2) \, dy,$$ where $\psi$ is Holonomia's $\psi$. The integral multiplying $f(x)$ is not zero if $f$ is not identically zero (using a Holonomia-style argument and $f(\sqrt 2 x)f(0) = f(x)^2$, for instance, to conclude that the continuous $f$ is nowhere $0$ if not identically $0$). With the change of variables $y= \sqrt{r^2-x^2}$, the integral on the right becomes $$\int_{\sqrt {(a+10)^2 +x^2}}^{\sqrt {(a+20)^2 +x^2}} \psi ( r^2) {r\over \sqrt{r^2-x^2} }\, dr,$$ which is differentiable at (in a neighborhood of) $x=c$, because $r^2-x^2 \ge (a+10)^2 >0$, $\psi$ is continuous, and the limits of the integral are differentiable. Thus $f(x)$ is differentiable. • Hi @peterag, how do we then back out of the differentiability assumption? Really cool intuition... – User001 Jun 29 '15 at 1:44 • @LebronJames - actually, I just rolled back the edit to address your comment, as I messed something up - Tomorrow... – peter a g Jun 29 '15 at 3:47 • Ok, got it. Thanks @peterag! :-) – User001 Jun 29 '15 at 3:52 • @LebronJames see the 'edit'. However, as advertised in the first line of this answer, the point really was not 'rigor' - although the first version of this when you posed the question actually had a <problematic> version of this 'edit', but it seemed pointless given Holonomia's answer. Be that as it may, the argument in the edit is based on the standard proof that a measurable character on the reals is continuous (and differentiable). – peter a g Jun 29 '15 at 21:24

http://www.danaetobajas.com/furniture-manufacturers-auh/9e2609-standard-form-of-a-quadratic-function-examples

standard form of a quadratic function examples 23303 The functions above are examples of quadratic functions in standard quadratic form. How to Graph Quadratic Functions given in Vertex Form? The standard form of a quadratic function. Sometimes, a quadratic function is not written in its standard form, $$f(x)=ax^2+bx+c$$, and we may have to change it into the standard form. ax² + bx + c = 0. R1 cannot be negative, so R1 = 3 Ohms is the answer. We like the way it looks up there better. Algebra Examples. If the quadratic polynomial = 0, it forms a quadratic equation. can multiply all terms by 2R1(R1 + 3) and then simplify: Let us solve it using our Quadratic Equation Solver. The quadratic function f(x) = a(x − h)2 + k, not equal to zero, is said to be in standard quadratic form. Note: You can find exactly where the top point is! 1 R1 Quadratic Function The general form of a quadratic function is f ( x ) = a x 2 + b x + c . Yes, a Quadratic Equation. Here are some examples: Move all terms to the left side of the equation and simplify. The standard form of the quadratic function helps in sketching the graph of the quadratic function. Two resistors are in parallel, like in this diagram: The total resistance has been measured at 2 Ohms, and one of the resistors is known to be 3 ohms more than the other. Standard Form of a Quadratic Equation The general form of the quadratic equation is ax²+bx+c=0 which is always put equals to zero and here the value of x is always unknown, which has to be determined by applying the quadratic formula while … Graphing Quadratic Functions in Vertex Form The vertex form of a quadratic equation is y = a(x − h) 2 + k where a, h and k are real numbers and a is not equal to zero. Because the vertex appears in the standard form of the quadratic function, this form is also known as the vertex form of a quadratic function.. The standard form of a quadratic function presents the function in the form $f\left(x\right)=a{\left(x-h\right)}^{2}+k$ where $\left(h,\text{ }k\right)$ is the vertex. The standard form of quadratic equations looks like the one below:. The constants ‘a’, ‘b’ and ‘c’ are called the coefficients. Quadratic functions follow the standard form: f(x) = ax 2 + bx + c. If ax 2 is not present, the function will be linear and not quadratic. Because (0, 8) is point on the parabola 2 units to the left of the axis of symmetry, x  =  2, (4, 8) will be a point on the parabola 2 units to the right of the axis of symmetry. the standard form of a quadratic function from a graph or information about a graph (as we’ll see in the next lesson), the value of the leading coefficient will need to be found first, while the vertex will be given. This means that they are equations containing at least one term that is squared. shows the profit, a company earns for selling items at different prices. Solved Example on Quadratic Function Ques: Graph the quadratic function y = - (1/4)x 2.Indicate whether the parabola opens up or down. Graphing a Quadratic Function in Standard Form. Here are some examples of functions and their standard forms. Once we have three points associated with the quadratic function, we can sketch the parabola based on our knowledge of its general shape. And how many should you make? Standard Form of a Quadratic Equation. The quadratic equations refer to equations of the second degree. Quadratic Equation in "Standard Form": ax2 + bx + c = 0, Answer: x = −0.39 or 10.39 (to 2 decimal places). Therefore, the standard form of a quadratic equation can be written as: ax 2 + bx + c = 0 ; where x is an unknown variable, and a, b, c are constants with ‘a’ ≠ 0 (if a = 0, then it becomes a linear equation). The quadratic function given by is in standard form. Factorize x2 − x − 6 to get; (x + 2) (x − 3) < 0. Example. Confirm that the graph of the equation passes through the given three points. The standard form of a quadratic function is. When a quadratic function is in general form, then it is easy to sketch its graph by reflecting, shifting and stretching/shrinking the parabola y = x 2. This means that they are equations containing at least one term that is squared. \"x\" is the variable or unknown (we don't know it yet). (3,0) says that at 3 seconds the ball is at ground level. How to Graph Quadratic Functions given in Vertex Form? Any function of the type, y=ax2+bx+c,a≠0y=a{{x}^{2}}+bx+c,\text{ }a\ne 0 y = Let us look at some examples of a quadratic equation: Solving linear equations using elimination method, Solving linear equations using substitution method, Solving linear equations using cross multiplication method, Solving quadratic equations by quadratic formula, Solving quadratic equations by completing square, Nature of the roots of a quadratic equations, Sum and product of the roots of a quadratic equations, Complementary and supplementary worksheet, Complementary and supplementary word problems worksheet, Sum of the angles in a triangle is 180 degree worksheet, Special line segments in triangles worksheet, Proving trigonometric identities worksheet, Quadratic equations word problems worksheet, Distributive property of multiplication worksheet - I, Distributive property of multiplication worksheet - II, Writing and evaluating expressions worksheet, Nature of the roots of a quadratic equation worksheets, Determine if the relationship is proportional worksheet, Trigonometric ratios of some specific angles, Trigonometric ratios of some negative angles, Trigonometric ratios of 90 degree minus theta, Trigonometric ratios of 90 degree plus theta, Trigonometric ratios of 180 degree plus theta, Trigonometric ratios of 180 degree minus theta, Trigonometric ratios of 270 degree minus theta, Trigonometric ratios of 270 degree plus theta, Trigonometric ratios of angles greater than or equal to 360 degree, Trigonometric ratios of complementary angles, Trigonometric ratios of supplementary angles, Domain and range of trigonometric functions, Domain and range of inverse  trigonometric functions, Sum of the angle in a triangle is 180 degree, Different forms equations of straight lines, Word problems on direct variation and inverse variation, Complementary and supplementary angles word problems, Word problems on sum of the angles of a triangle is 180 degree, Domain and range of rational functions with holes, Converting repeating decimals in to fractions, Decimal representation of rational numbers, L.C.M method to solve time and work problems, Translating the word problems in to algebraic expressions, Remainder when 2 power 256 is divided by 17, Remainder when 17 power 23 is divided by 16, Sum of all three digit numbers divisible by 6, Sum of all three digit numbers divisible by 7, Sum of all three digit numbers divisible by 8, Sum of all three digit numbers formed using 1, 3, 4, Sum of all three four digit numbers formed with non zero digits, Sum of all three four digit numbers formed using 0, 1, 2, 3, Sum of all three four digit numbers formed using 1, 2, 5, 6, Equation of Line with a Point and Ratio of Intercept is Given, Graphing Linear Equations Using Intercepts Worksheet, Find x Intercept and y Intercept of a Line. Therefore, the standard form of a quadratic equation can be written as: ax 2 + bx + c = 0 ; where x is an unknown variable, and a, b, c are constants with ‘a’ ≠ 0 (if a = 0, then it becomes a linear equation). What are the values of the two resistors? Let us solve this one by Completing the Square. f(x) = a x 2+ b x + c If a > 0, the vertex is a minimum point and the minimum value of the quadratic function f is equal to k. This minimum value occurs at x = h. If a < 0, the vertex is a maximum point and the maximum value of the quadratic function f is equal to k. This maximum value occurs at x = h. The quadratic function f(x) = a x 2+ b x + c can be written in vertex form as follows: f(x) = a (x - h) 2+ k General and Standard Forms of Quadratic Functions The general form of a quadratic function presents the function in the form f (x)= ax2 +bx+c f (x) = a x 2 + b x + c where a a, b b, and c c are real numbers and a ≠0 a ≠ 0. If you have any feedback about our math content, please mail us : You can also visit the following web pages on different stuff in math. 1. y = x^{2} , y = 3x^{2} - 2x , y = 8x^{2} - 16x - 15 , y = 16x^{2} + 32x - 9 , y = 6x^{2} + 12x - 7 , y = \left ( x - 2 \right )^{2} . To find the roots of such equation, we use the formula, (root1,root2) = (-b ± √b 2-4ac)/2. Here, “a” is the coefficient of which is generally called as leading coefficient,“b” is the coefficient of “x” and the “c” is called as the constant term. Example 1. Find the roots of the equation as; (x + 2) … Using Vertex Form to Derive Standard Form. Quadratic functions in standard form: $$y=ax^2+bx+c$$ where $$x=-\frac{b}{2a}$$ is the value of $$x$$ in the vertex of the function. (Note: t is time in seconds). Examples of Quadratic Equations in Standard Form. In "Standard Form" it looks like: −5t 2 + 14t + 3 = 0. Quadratic equations pop up in many real world situations! The standard form of a quadratic equation: The standard form of a quadratic equation is given by It contains three terms with a decreasing power of “x”. Now we use our algebra skills to solve for "x". So, the selling price of $35 per item gives the maximum profit of$6,250. Substitute the value of h into the equation for x to find k, the y-coordinate of the vertex. But we want to know the maximum profit, don't we? y = a(x 2 - 2xh + h 2) + k. y = ax 2 - 2ahx + ah 2 + k The vertex of a quadratic function is (h, k), so to determine the x-coordinate of the vertex, solve b = -2ah for h. Because h is the x-coordinate of the vertex, we can use this value to find the y-value, k, of the vertex. Find the vertex of the quadratic function. Quadratic functions make a parabolic U-shape on a graph. Graph vertical compressions and stretches of quadratic functions. Answer: Boat's Speed = 10.39 km/h (to 2 decimal places), And so the upstream journey = 15 / (10.39−2) = 1.79 hours = 1 hour 47min, And the downstream journey = 15 / (10.39+2) = 1.21 hours = 1 hour 13min. The quadratic equations refer to equations of the second degree. Write the vertex form of a quadratic function. Quadratic Equations are useful in many other areas: For a parabolic mirror, a reflecting telescope or a satellite dish, the shape is defined by a quadratic equation. x2 − x − 6 < 0. Show Step-by-step Solutions Try the free Mathway calculator and problem solver below to practice various math topics. If a is negative, the parabola is flipped upside down. Write the vertex form of a quadratic function. And many questions involving time, distance and speed need quadratic equations. The vertex form of a quadratic equation is y = a (x − h) 2 + k where a, h and k are real numbers and a is not equal to zero. x = −0.39 makes no sense for this real world question, but x = 10.39 is just perfect! Area of steel after cutting out the 11 × 6 middle: The desired area of 28 is shown as a horizontal line. The standard form of quadratic equations looks like the one below:. the standard form of a quadratic function from a graph or information about a graph (as we’ll see in the next lesson), the value of the leading coefficient will need to be found first, while the vertex will be given. f (x)= a(x−h)2 +k f ( x) = a ( x − h) 2 + k. The Standard Form of a Quadratic Equation looks like this: 1. a, b and c are known values. Factoring Quadratic Functions. The "basic" parabola, y = x 2 , … Write the equation of a transformed quadratic function using the vertex form. It looks even better when we multiply all terms by −1: 5t 2 − 14t − 3 = 0. Examples of quadratic inequalities are: x 2 – 6x – 16 ≤ 0, 2x 2 – 11x + 12 > 0, x 2 + 4 > 0, x 2 – 3x + 2 ≤ 0 etc.. The squaring function f(x)=x2is a quadratic function whose graph follows. Quadratic functions are symmetric about a vertical axis of symmetry. Example : Graph the quadratic function : f(x) = x 2 - 4x + 8. ax² + bx + c = 0. Examples of Quadratic Equations in Standard Form. The frame will be cut out of a piece of steel, and to keep the weight down, the final area should be 28 cm2, The inside of the frame has to be 11 cm by 6 cm. Let us solve it using the Quadratic Formula: Where a, b and c are To find out if the table represents pairs of a quadratic function we should find out if the second difference of the y-values is constant. The following video shows how to use the method of Completing the Square to convert a quadratic function from standard form to vertex form. The x-axis shows the selling price and the y-axis shows the profit. The standard form of a quadratic function is y = ax 2 + bx + c. where a, b and c are real numbers, and a ≠ 0. + Step-by-Step Examples. Subtract from . Graph the equation y = x2 + 2. Here are some points: Here is a graph: Connecting the dots in a "U'' shape gives us. The general form to vertex form problem solver below to practice various math.. Second degree company earns for selling items at different prices price is $or. Graph quadratic functions in standard form graphing quadratic functions from general form to vertex form Exercise. Any other stuff in math, please use our google custom search here meters high the value h. 5T standard form of a quadratic function examples = 0, -3 for b, c are constants as! Move all terms by −1: 5t 2 − 14t − 5t 2 = 0, the of! Reaches the highest order of is 2 function the general form to vertex form quadratic.... Below to practice various math topics the values of,, and if gt... That at 3 seconds the ball reaches the highest order of is 2 equation when its coefficients are values... Multiply the coefficient of x 2, except we 've moved the whole up! Multiply all terms by −1: 5t 2 − 14t − 3 = 0 the... Transformed quadratic function helps in sketching the graph of a quadratic function presents the function to find the of! That they are equations containing at least one term that is squared math topics where the top is!, except we 've moved the whole picture up by 2 what the... An equal sign Solutions examples of quadratic equations looks like the one below: just perfect price and ball! H, the vertex and axis of symmetry of a quadratic equation looks like one! All terms by −1: 5t 2 − 14t − 3 ) <.. Meters after 1.4 seconds Graph-A ; opens down from the stuff given above, you... Equations containing at least one term that is given in vertex form or factored form we! Standard quadratic form +bx+c, where a, b and c = 8 profit, do know! 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Know the maximum profit, do n't know it yet ) if the quadratic function can called... Hit the ground when the height is zero: 3 + 14t +.. Expect sales to follow this Demand curve '': so... what is the best?. A is negative, so r1 = 3 Ohms is the variable or unknown ( we do n't we all!, c are real numbers, and a can not be 0 standard form the functions are. The dots in a U '' shape gives us just perfect and c = standard form of a quadratic function examples 13 high. Through the given three points Demand curve '': so standard form of a quadratic function examples what is the answer given above, you. You can expect sales to follow this Demand curve '': so... what is the answer ( ). Its parabola below to practice various math topics - 2axh + ah2 + k is a inequality! Negative time, distance and speed need quadratic equations refer to equations of the second degree dots in ... By is in standard form of a quadratic function is a polynomial function, since the highest of.

http://clay6.com/qa/55480/at-3-40-the-hour-hand-and-the-minute-hand-of-a-clock-form-an-angle-of

Comment Share Q) # At 3.40, the hour hand and the minute hand of a clock form an angle of ( A ) 120 ( B ) 135 ( C ) 130 ( D ) 125 Comment A) Comment A) 130 is correct answer or not Yes, that is correct Comment A) Solution : Angle traced by hour hand in 12 hrs $=360^{\circ}$ Angle traced by 1t in $\large\frac{11}{3}$$hrs = \bigg[\large\frac{360}{12} \times \frac{11}{3}\bigg] \qquad= 110^{\circ} Angle traced by minute hand in 60 mins =360^{\circ} Angle traced by it in 40 min =\bigg[\large\frac{360}{60}$$ \times 40 \bigg]$ $\quad= 240^{\circ}$ Required angle $=[240 -110]^{\circ}$ $\qquad= 130^{\circ}$

https://brilliant.org/discussions/thread/show-why-the-value-converges-to-pi/

# Show why the value converges to $\pi$ $a_0=1$ $a_{n+1}=a_n+\sin{(a_n)}$ Explain why the following occurs: $a_0=1$ $a_1=1+\sin{(1)}\approx 1.841470985$ $a_2=1+\sin{(1)}+\sin{(1+\sin{(1)})}\approx 2.805061709$ $a_3=1+\sin{(1)}+\sin{(1+\sin{(1)})}+\sin{(1+\sin{(1)}+\sin{(1+\sin{(1)})})}\approx 3.135276333$ $a_4\approx 3.141592612$ $a_5\approx 3.141592654\approx\pi$ Note by Jack Han 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. 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: Let's see here... The interesting thing I was talking about is the fact that the given series always converges into a value $a_n$ such that $sin(a_n)=0$ for different values of $a_0$. To put it in more exact terms, it always converges to a value of $a_n$ such that $cos(a_n)=-1\Rightarrow a_n=m\pi$, where $m$ is an odd integer. ($m$ can also be even, but that is a degenerate case where all terms are same) I'm going to use something here that I actually learned from gradient descent. If you don't know what it is, you can google it. But, the mathematics used below is an extremely tame form and is easy to understand with little knowledge of calculus. Consider the function $f\left( x \right) =\cos { (x) } \\ \Rightarrow \frac { df\left( x \right) }{ dx } =-\sin { (x) }$ Now see what happens when we take some arbitrary value of $x$ (say $x=1$)and then do the following repeatedly: $x:=x-\frac { df\left( x \right) }{ dx }=x+\sin { (x) }$ (":=" is the assignment operator ) In the above figure, we can see two points marked. One is red, which represents the first value of $x$($=1$). The other is brown, and is after one iteration of above step. We can see that when we do $x:=x+sin(x)$, what is actually happening is that $x$ is surfing along the slope of the curve $cos(x)$. We move the value of $x$ down the tangent. Change $x$ little by little, so that finally, after many iterations it moves closer and closer to the minima, i.e $x=\pi$. I know this is not a definitive proof of what happens... I'm sure you will realize the importance of this once you understand what is happening. In general, series defined as $a_n=a_{n-1}-\alpha\frac { df\left(a_{n-1} \right) }{ da_{n-1} }$ Will converge to the nearest value of $a$ (nearest to $a_0$) such that $f(a)$ is minimum, provided the value of $\alpha$ is not too large. - 4 years, 11 months ago Yes, that was awesome, that is basically the newton Rhapsody method of estimation of roots, doing the following iteration for any curve will eventually lead us to the nearest root, that is great, actually i think this is pretty much the solution +1 - 4 years, 11 months ago Good work! Newton's method for estimating roots. Pretty much seals the deal. Great solution +1. - 4 years, 11 months ago Good work. - 4 years, 11 months ago - 4 years, 11 months ago As usual, since the series converges.. this means that when $n\to\infty$ , $a_{n+1}=a_{n}$. But $a_{n+1}=a_{n}+sin(a_{n})$ $\Rightarrow sin(a_{n})=0$ Now how do we know that $a_{n}=\pi$? We know this since $a_0=1$ and the series is constantly increasing. Therefore, it converges onto the first value of $x>1$ such that $sin(x)=0$. - 4 years, 11 months ago Bro , But It is not always true $\lim _{ n\rightarrow \infty }{ ({ a }_{ n }) } =\lim _{ n\rightarrow \infty }{ { (a }_{ n+1 }) }$ . - 4 years, 11 months ago Why not? Do you have a counter-example? - 4 years, 11 months ago Which value converges to pi ?? - 4 years, 11 months ago If you call L the value of the limit you obtain sin(L)=0. Now L can be pi or zero but zero is impossible because of the initial condition. More precisely you can say that the value of the sequence is LOW bounded - 4 years, 11 months ago This is turning out to be very interesting... I want to know if we can find a general form for a function $f(x)$ such that the series $a_1,a_2,...$ defined by: $a_{n+1}=a_{n-1}+f(a_{n-1})$ Converges for a given value of $a_0$. Further, is it true that all of the values of such $a_n$ as $n\to \infty$ satisfy $f(a_n)=0$? - 4 years, 11 months ago I think the answer to this is going to be extremely interesting.... I have a feeling... Is anyone else thinking what I'm thinking? - 4 years, 11 months ago What are you thinking? - 4 years, 11 months ago

https://math.stackexchange.com/questions/2399770/in-how-many-ways-can-we-permute-the-digits-2-3-4-5-2-3-4-5-if-identical-digits

# In how many ways can we permute the digits $2,3,4,5,2,3,4,5$ if identical digits must not be adjacent? In how many ways can we permute the digits $2,3,4,5,2,3,4,5$ if identical digit must not be adjacent? I tried this by first taking total permutation as $\dfrac{8!}{2^4}$ Now $n_1$ as $22$ or $33$ or $44$ or $55$ occurs differently $N_1 = \left(^7C_1\times \dfrac{7!}{8}\right)$ And $n_2 = \left(^4C_1 \times 4!\right)$ Using the inclusion-exclusion principle I got: $\dfrac{8!}{16}-\left(^7C_1\times\dfrac{7!}{8}\right)+\left(^4C_1\times4!\right)$ This question is from combinatorics and helpful for RMO • I made some edits to help the layout and appearance - see this linked article for more help on formatting - but I am not clear how you derived your formulas. Also I interpretted IEP as inclusion-exclusion principle but I don't know what RMO means. Note that you need two extra spaces on the end of a line to produce a line break. Aug 20 '17 at 2:26 You need a few more inclusion-exclusion steps to complete this approach. Without constraints, you do indeed have $\dfrac {8!}{2^4} = 2520$ arrangements. Then there are $\dfrac {7!}{2^3} = 630$ cases where a $22$ is found in the arrangement, and similarly for the other digits. Then there are $\dfrac {6!}{2^2} = 180$ cases where both a $22$ and a $33$ are found, and similarly for other pairs, etc. So by inclusion-exclusion, we have to subtract the paired cases then add back the double-paired cases, then subtract off triple-paired again and finally add in the cases where all digits appear in pairs. $$\frac {8!}{2^4} - \binom 41\frac {7!}{2^3} + \binom 42\frac {6!}{2^2} - \binom 43\frac {5!}{2} + \binom 44\frac {4!}{1} \\[3ex] =2520 -4\cdot 630 +6\cdot 180-4\cdot60 + 24 = 864$$ [Sharp eyes might notice that $\frac {8!}{2^4} = \binom 41\frac {7!}{2^3}$, shortening the calculation process.] Here is a variation based upon generating functions of Smirnov words. These are words with no equal consecutive characters. (See example III.24 Smirnov words from Analytic Combinatorics by Philippe Flajolet and Robert Sedgewick for more information.) We encode the digits \begin{align*} 2,3,4,5 \qquad\text{as}\qquad a,b,c,d \end{align*} and look for Smirnov words of length $8$ built from $a,b,c,d$ with each letter occurring exactly twice. A generating function for the number of Smirnov words over a four letter alphabet $V=\{a,b,c,d\}$ is given by \begin{align*} \left(1-\frac{4z}{1+z}\right)^{-1} \end{align*} We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series $A(z)$. The number of all Smirnov words of length $8$ over a four letter alphabet is therefore \begin{align*} [z^8]\left(1-\frac{4z}{1+z}\right)^{-1} \end{align*} Since we want to count the number of words of length $8$ with each character in $V$ occurring twice, we keep track of each character. We obtain with some help of Wolfram Alpha \begin{align*} [a^2b^2c^2d^2]\left(1-\frac{a}{1+a}-\frac{b}{1+b}-\frac{c}{1+c}-\frac{d}{1+d}\right)^{-1}=\color{blue}{864} \end{align*} • (+1) This approach is so elegant and considerably malleable! I remember answering a similar question earlier this year using an exciting variation of this. I hope you don't mind me putting the link here, Markus, but it seemed appropriate as it relates more closely to your method than the others. Aug 21 '17 at 2:07 Here is another approach: Assume for the moment that the first appearance of the four digits is in increasing order. The places of their first appearance can be distributed in six ways, see the following figure. The places for the prospective second appearances have been marked by empty boxes, next to which is written the number of choices we have when filling them in. The last column shows the product of these numbers in each row. $$\matrix{ 2&3&4&5&\square_3&\square_3&\square_2&\square_1&&18\cr 2&3&4&\square_2&5&\square_2&\square_2&\square_1&&8\cr 2&3&4&\square_2&\square_2&5&\square_1&\square_1&&4\cr 2&3&\square_1&4&5&\square_2&\square_2&\square_1&&4\cr 2&3&\square_1&4&\square_1&5&\square_1&\square_1&&1\cr 2&3&\square_1&\square_1&4&5&\square_1&\square_1&&1\cr}$$ Summing the last column gives $36$. This has to be multiplied by $4!$ in order to compensate for the chosen order $2345$. It follows that there are $864$ admissible arrangements of the eight digits. • One of the charms of mathematics is that there is always another way.... Aug 20 '17 at 13:52

https://math.stackexchange.com/questions/1506871/with-4-rooks-on-a-4-times4-chessboard-such-that-no-rook-can-attack-another-wh

# With 4 rooks on a $4\times4$ chessboard such that no rook can attack another, what is the probability there are no rooks on the diagonal? Four rooks are randomly placed on a $4 \times 4$ chessboard. Suppose no rook can attack another. Under this condition, what is the probability that the leading diagonal of the chessboard has no rooks at all? Since no rook can attack another, we know that each row and each column contains exactly one rook each. Let $A_i$ be the event that row $i$ has its rook on the diagonal. Then $P\{A_i\} = \frac{1}{4}$ for each $i = 1,\dots,4$. We want to find the probability that the diagonal of the chessboard has no rooks at all, or equivalently that none of the rows have their rook on the diagonal. Therefore we have \begin{align} P\{A^c_1 \cap A^c_2 \cap A^c_3 \cap A^c_4\} & = P\{(A_1 \cup A_1 \cup A_3 \cup A_4)^c\} \\ & = 1 - P\{A_1 \cup A_1 \cup A_3 \cup A_4\} \\ & = 1 - (P\{A_1\} + P\{A_2\} + P\{A_3\} + P\{A_4\} - P\{A_1 \cap A_2\} - P\{A_1 \cap A_3\} - P\{A_1 \cap A_4\} - P\{A_2 \cap A_3\} - P\{A_2 \cap A_4\} - P\{A_3 \cap A_4\} + P\{A_1 \cap A_2 \cap A_3\} + P\{A_1 \cap A_2 \cap A_4\} + P\{A_1 \cap A_3 \cap A_4\} + P\{A_2 \cap A_3 \cap A_4\} - P\{A_1 \cap A_2 \cap A_3 \cap A_4\}) \\ & = 1 - (4 \cdot \frac{1}{4} - 6 \cdot \frac{1}{16} + 4 \cdot \frac{1}{64} - \frac{1}{256}) \\ & = 1 - \frac{175}{256} \\ & = \frac{81}{256} \end{align} using De Morgan's Law and the inclusion-exclusion principle. However, it seems that this is incorrect since if we consider the number of ways that we can place the rooks such that no rook can attack each other we have $\frac{(4!)^2}{4!} = 4! = 24$ [as per this answer for a similar problem] and so the answer should have denominator of 24. Having said that I don't see where my answer is wrong, so would someone be able to show me the correct solution? • No rooks on either diagonal, or just one specified diagonal? Oct 31 '15 at 20:47 • Look at those 24 rook configurations. Is $P(A_i)=1/4$ ? Oct 31 '15 at 20:51 • Just the leading diagonal, i.e. a4, b3, c2, d1 @BrianTung Oct 31 '15 at 20:51 • OK, thanks. I've given the answer to both interpretations, just in case. Oct 31 '15 at 20:59 Each non-attacking placement of the rooks defines a permutation $c_1c_2c_3c_4$ of $\{1,2,3,4\}$: $c_k$ is the number of the column containing the rook in row $k$. There are $4!=24$ such permutations, all equally likely. Those that have no rook on the main diagonal are derangements, and there are $9$ of them, so the desired probability is $\frac9{24}$. If you know the formula for the number of derangements of a set of $n$ objects, you can use it, but $4$ is small enough that it’s almost as easy just to list them: \begin{align*} &2143,2341,2413\\ &3142,3412,3421\\ &4123,4312,4321 \end{align*} Your answer assumes independence between rook placements. But given the condition that they cannot attack each other, their placements are clearly not independent; therefore, you cannot multiply individual probabilities to obtain joint probabilities. You are correct in observing that the total number of possible non-attacking arrangements is $4! = 24$. If the rooks cannot be on either diagonal, then there are two choices for the rook in the first file, two choices for the rook in the second file, and then the rooks in the third and fourth file have their placements determined by the first two. There are therefore $2 \times 2 = 4$ placements that avoid both diagonals. If you only need to avoid one diagonal (say, the black diagonal), we merely need the number of derangements of four objects. These can be grouped into two categories: those that involve two pairs swapping, of which there are $\binom{4}{2} \div 2 = 3$; and those that involve a cyclic permutation of all four, of which there are $3! = 6$; for a total of $9$ derangements. The number of allowable placements should be the number of derangements of $4$ items, which is $9$. And as you point out there are a total of $24$ possible non-attacking placements. You are misapplying Inclusion-Exclusion. The $A_i$ need to be the number of elements of the set satisfying condition $i$, not their probability. So: \begin{equation*} |A_i| = 6,\quad |A_i\cap A_j| = 2,\quad |A_i\cap A_j\cap A_k| = 1,\quad |A_1\cap A_2\cap A_3\cap A_4| = 1. \end{equation*} This gives for the probability that the placement is in none of the $A_i$ \begin{equation*} 24 - (4\cdot 6 + 6\cdot 2 - 4\cdot 1 + 1\cdot 1) = 9. \end{equation*} So the probability is $\frac{9}{24} = \frac{3}{8}$. As has been pointed out in another answer, this is just the number of derangements of a four-element set.

https://tringham.net/h06rim/5a3ef8-how-to-graph-a-horizontal-stretch

This video explains to graph graph horizontal and vertical stretches and compressions in the A point on the object gets further away from the vertical axis on the image. J. JonathanEyoon. x). 1. This problem has been solved! Embedded content, if any, are copyrights of their respective owners. Horizontal And Vertical Graph Stretches And Compressions (Part 1) The general formula is given as well as a few concrete examples. This graph has a vertical asymptote at $$x=–2$$ and has been vertically reflected. Retain the y-intercepts’ position. Write the expressions for g(x) and h(x) in terms of f(x) given the following conditions: a. Images/mathematical drawings are created with GeoGebra. 8. This video reviews function transformation including stretches, compressions, shifts left, shifts right, The function, g(x), is obtained by horizontally stretching f(x) = 16x2 by a scale factor of 2. When in its original state, it has a certain interior. Apply the transformations to graph g(x). Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. A horizontal stretch or shrink by a factor of 1/kmeans that the point (x, y) on the graph of f(x) is transformed to the point (x/k, y) on the graph of g(x). In this video we discuss the effects on the parent function when: There are different types of math transformation, one of which is the type y = f(bx). This is called a horizontal stretch. ... k ----- 'k' is a horizontal stretch or compression, which means it will effect all the x-values of the coordinates of a parent function. transformation by using tables to transform the original elementary function. So I don't want to change any scales or values or limits. Related Pages Translation Of 2 Units Left IV. Horizontally stretched by a scale factor of 1/3. See the answer. horizontal/vertical stretch? Stretching a Graph Vertically or Horizontally : Suppose f is a function and c > 0. From this, we can see that q(x) is the result of p(x) being stretched horizontally by a scale factor of 1/4 and translated one unit downward. The graph of $$y = f(0.5x)$$ has a stretch factor of 2 from the vertical axis parallel to the horizontal axis. But do not divide outside of the parenthesis, it remains close to the X. Use the graph of f(x) shown below to guide you. We welcome your feedback, comments and questions about this site or page. Horizontal and vertical translations, as well as reflections, are called rigid transformations because the shape of the basic graph is left unchanged, or rigid. Use the graph of f(x) shown below to guide you. When one stretches the rubber band, the interior gets bigger or the edges get farther apart. The function g(x) is the result of f(x) being stretched horizontally by a factor of 1/4. The new x-coordinate of the point will be, 1. You da real mvps! Yes, it's contrary to believe that a stretch should divide a factor, and a compression would multiply. Cosine of x would be the same as these, but shifted πb/2 to the left. If you're seeing this message, it means we're having trouble loading external resources on our website. We can also stretch and shrink the graph of a function. This time, instead of moving the vertex of the graph, we will strech or compress the graph. More Pre-Calculus Lessons. What are the transformations done on f(x) so that it results in g(x) = 3√(x/2)? This video provides two examples of how to express a horizontal stretch or compression using function notation.Site: http://mathispower4u.com The general formula is given as well as a few concrete examples. This shifted the graph down 1 unit. When f (x) is stretched horizontally to f (ax), multiply the x-coordinates by a. The image below shows the graph of f(x). :) https://www.patreon.com/patrickjmt !! 2f (x) is stretched in the y direction by a factor of 2, and f (x) is shrunk in the y direction by a factor of 2 (or stretched by a factor of ). 4. more examples, solutions and explanations. 5. I just didn’t know how to animate that with my program. This video talks about reflections around the X axis and Y axis. Viewed 28k times 15. The simplest way to consider this is that for every x you want to put into your equation, you must modify x before actually doing the substitution. Vertical Stretch and Vertical Compression y = af(x), a > 1, will stretch the graph f(x) vertically by a factor of a. y = af(x), 0 < a < 1, will stretch the graph f(x) vertically by a factor of a. Horizontal Stretch and Horizontal Compression y = f(bx), b > 1, will compress the graph f(x) horizontally. When using transformations to graph a function in the fewest steps, you can apply a and k together, and then c and d together. by horizontally stretching f(x) by a factor of 1/k. (Part 3). Hence, we’ve just shown how g(x) can be graphed using the parent function of absolute value functions, f(x) = |x|. Try the free Mathway calculator and Translation means moving an object without rotation, and can be described as “sliding”. 0=square root of x - … Though both of the given examples result in stretches of the graph of y = sin(x), they are stretches of a certain sort. To easily graph this, you have to stretch the graph to infinity, ripping the space-time continuum until it flips back around upside down. Notice that the coefficient needed for a horizontal stretch or compression is the reciprocal of the stretch or compression. A point $\,(a,b)\,$ on the graph of $\,y=f(x)\,$ moves to a point $\,(k\,a,b)\,$ on the graph of [beautiful math coming... please be patient] $\,y=f(\frac{x}{k})\,$. Ask Question Asked 7 years ago. Try the given examples, or type in your own Teams. transformations include vertical shifts, horizontal shifts, and reflections. 7. Vertical stretch on a graph will pull the original graph outward by a given scale factor. In all seriousness, you flip your graph upside down. We carefully make a 90° angle around the third peg, so that one side is vertical and the other is horizontal. Lastly, let’s translate the graph one unit downward. Please submit your feedback or enquiries via our Feedback page. This type of The function, f(x), passes through the point (10, 8). Meaning, n(x) is the result of m(x) being vertically stretched by a scale factor of 3 and horizontally stretched by a scale factor of 1/4. problem and check your answer with the step-by-step explanations. If g(x) is the result of f(x) being horizontally stretched by a scale factor of 3, construct its table of values and retain the current output values. y = c f(x), vertical stretch, factor of c; y = (1/c)f(x), compress vertically, factor of c; y = f(cx), compress horizontally, factor of c; y = f(x/c), stretch horizontally, factor of c; y = - f(x), reflect at x-axis Replacing x with x n results in a horizontal stretch by a factor of n . and reflections across the x and y axes. Q&A for Work. Substituting $$(–1,1)$$, This video explains to graph graph horizontal and vertical translation in the form af(b(x-c))+d. This video discusses the horizontal stretching and compressing of graphs. problem solver below to practice various math topics. I want a simple x,y plot created with matplotlib stretched physically in x-direction. Scroll down the page for So to stretch the graph horizontally by a scale factor of 4, we need a coefficient of $\frac{1}{4}$ in our function: $f\left(\frac{1}{4}x\right)$. Vertically stretched by a scale factor of 2. Thanks to all of you who support me on Patreon. b. Horizontal Stretching and Compression of Graphs This applet helps you explore the changes that occur to the graph of a function when its independent variable x is multiplied by a positive constant a (horizontal stretching or compression). Jul 2007 290 3. When a base function is multiplied by a certain factor, we can immediately be able to graph the new function by applying the vertical stretch. Horizontal Stretch and Shrink. To stretch vertically do you multiply the y-values of the parent function, by the number your stretching it by? Copyright © 2005, 2020 - OnlineMathLearning.com. It looks at how a and b affect the graph of f(x). If f(x) is horizontally stretched by a scale factor of 5, what would be the new x-coordinate of the point? This means that the input values must be four times larger to produce the same result, requiring the input to be larger, causing the horizontal stretching. It might be simpler to think of a stretch or a compression in terms of a rubber band. The graphs below summarize the key features of the resulting graphs of vertical stretches and compressions of logarithmic functions. The table of values for f(x) is shown below. These lessons with videos and examples help Pre-Calculus students learn about horizontal and vertical Observe the functions shown below. The function, g(x), is obtained by horizontally stretching f(x) = 16x, Horizontal Stretch – Properties, Graph, & Examples, Since the y-coordinates will remain the same, the, We can only horizontally stretch a graph by a factor of. Horizontal Stretch/Compression Replacing x with n x results in a horizontal compression by a factor of n . physically stretch plot in horizontal direction in python. When f(x) is stretched horizontally to f(ax). We know so far that the equation will have form: $$f(x)=−a\log(x+2)+k$$ It appears the graph passes through the points $$(–1,1)$$ and $$(2,–1)$$. We can only horizontally stretch a graph by a factor of 1/a when the input value is also increased by a. To perform a horizontal compression or stretch on a graph, instead of solving your equation for f(x), you solve it for f(c*x) for stretching or f(x/c) for compressing, where c is the stretch factor. g(x) = f(kx), can be sketched by horizontally shrinking f(x) by a factor of 1/kif k > 1. or. Let’s go ahead and express g(x) in terms of f(x). 2. Translation Of 2 Units Right O I And IV O II And III Oll And IV I And III. Take a look at the following graph. form af(b(x-c))+d. Lastly, let’s observe the translations done on p(x). In describing transformations of graphs, some textbooks use the formal term “translate”, while others use an informal term like “shift”.Our first question comes from 1998:These examples represent the three main transformations: translation (shifting), reflection (flipping), and dilation (stretching). Functions that are multiplied by a real number other than 1, depending on the real number, appear to be stretched vertically or stretched horizontally. How To: Given a logarithmic function Of the form $f\left(x\right)=a{\mathrm{log}}_{b}\left(x\right)$, $a>0$, graph the Stretch … Define functions g and h by g (x) = c f (x) and h (x) = f (cx). Graph h(x) using the fact that it is the result of f(x) being stretched horizontally by a factor of 1/3. This means that the translations on f(x) to obtain g(x) are: Let’s slowly apply these transformations on f(x) starting with horizontally stretching f(x). Im in algebra one and we need to know how to change a parent function's graph by stretching it vertically/horizontally. Expert Answer . This transformation type is formally called horizontal scaling (stretching/shrinking). Horizontal Stretch By A Factor Of 3 II. if 0 < k< 1. Parent Functions And Their Graphs Other important 6. The intention is to get a result were it is easier for me to detect features in the signal. You make horizontal changes by adding a number to or subtracting a number from the input variable x, or by multiplying x by some number. We do not know yet the vertical shift or the vertical stretch. Subtracting from x makes the function go right. Question: How Is The Graph Y =3(x - 2)2 Related To The Graph Of Y = 1. You start with y=square root of (x-1) it becomes 0<=x-1. The following table gives a summary of the Transformation Rules for Graphs. Apply the transformations to graph g(x). \$1 per month helps!! then 1 <=x. Show transcribed image text . Active 2 years ago. math transformation is a horizontal compression when b is greater than one. PLEASE give an easy way to stretch! Which of the following is the correct expression for g(x)? We can graph this math What is the relationship between f(x) and g(x)? Sal graphs y=-2.5*cos(1/3*x) by considering it as a vertical stretch and reflection, and a horizontal stretch, of y=cos(x). Now we stretch one part of the rubber band straight up from the left peg and around a third peg to make the sides of a right triangle as shown in Figure $$\PageIndex{2}$$. y = c f(x), vertical stretch, factor of c, y = (1/c)f(x), compress vertically, factor of c, y = f(cx), compress horizontally, factor of c, y = f(x/c), stretch horizontally, factor of c. Stretching a graph involves introducing a coefficient into the function, whether that coefficient fronts the equation as in y = 3 sin(x) or is acted upon by the trigonometric function, as in y = sin(3x). Describe the transformations done on the following functions shown below. 3. Horizontal Stretches/Compressions - multiply the x value directly. The resulting function will have the same range but may have a different domain. To stretch or shrink the graph in the y direction, multiply or divide the output by a constant. Let’s now stretch the resulting graph vertically by a scale factor of 2. All horizontal transformations, except reflection, work the opposite way you’d expect: Adding to x makes the function go left. It looks at how c and d affect the graph of f(x). What are the transformations done on f(x) so that it results to g(x) = 2|x/3| – 1? In general, the graph of $$y = f(ax)$$ has a stretch value of $$\frac{1}{a}$$ from the vertical axis parallel to the horizontal axis. if we say we stretched it by 1/4, that means it only increased by 1/4 of its original length as opposed to 4 times its original length . A horizontal stretch can be applied to a function by multiplying its input values by a scale factor, Let’s go ahead and take a look at how f(x) = x, Remember that when we horizontally stretch a function by, When we stretch a graph horizontally, we multiply the base function’s x-coordinate by the given scale factor’s denominator, Hence, we have (6, 4) → (2 ∙ 6, 4). You use the graph and solve it as you would for any function using small values first, then you have y=square root of x - 1, the domain 0<=x. When we horizontally stretch g(x) by a scale factor of 1/3, we obtain h(x). Transformations Of Trigonometric Graphs Then. graph stretches and compressions. the graph will be stretched horizontally so that its horizontal length on any finite interval will be 4 times what it was originally, stretching by a factor of 4 is the way we would describe that. For a horizontal stretch of 2, x 2 would become (x/2) 2. Vertical Stretch By A Factor Of 3 III. To stretch a function f(x) vertically, we have to multiply the entire function by a constant greater than 1. This type of non-rigid transformation is called a The first example creates a vertical stretch, the second a horizontal stretch. where p is the horizontal stretch factor, (h, k) is the coordinates of the vertex. Graphs Of Functions Make sure to include the new critical points for g(x). Solver below to guide you stretches and compressions ( Part 1 ) general... To animate that with my program transformations of Trigonometric Graphs More Pre-Calculus how to graph a horizontal stretch divide a factor 2. X/2 ) 2 Related to the left shifted πb/2 to the left factor, ( h k! ( Part 1 ) the general formula is given as well as a concrete... On our website step-by-step explanations = 3√ ( x/2 ) 2 Related to the graph one unit downward start! General formula is given as well as a few concrete examples want change! 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Shift or the vertical shift or the edges get farther apart examples, and! Interior gets bigger or the vertical axis on the following functions shown below original elementary.. Support me on Patreon only horizontally stretch g ( x ) would become ( x/2?! The translations done on f ( x ) by a factor, and a compression would multiply will! Reciprocal of the stretch or shrink the graph, we have to multiply the y-values of graph... Value is also increased by a given scale factor of n reciprocal of the point (,! That a stretch should divide a factor of 2, x 2 become. That a stretch should divide a factor of 5, what would be the new x-coordinate the... To change any scales or values or limits through the point in (. Done on f ( x ) is stretched horizontally to f ( ax ), multiply y-values. 10, 8 ) been vertically reflected ( ax ) 3√ ( )... 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# If I flip $1$ of $3$ modified coins $3$ times, what's the probability that I will get tails? We have $3$ modified coins: $M_1$ which has tails on the both sides, $M_2$ which has heads on the both sides and $M_3$ which is a fair coin. We extract a coin from the urn and we flip it $3$ times. 1. What is the probability that if I flip the coin $3$ times I will get all tails? 2. If I got all tails at all $3$ flips what is the probability that the extracted coin is $M_3$? My attempt: 1. I have tried this way: There is a $\frac{1}{3}$ chance to get $M_1$ or $M_2$ or $M_3$. If we get $M_1$ the probability to get tails is $1$, for $M_2$ is $0$ and for $M_3$ is $\frac{1}{2}$. Then the probability to get tails at one flip is $$\frac{1}{3}\cdot 1 + \frac{1}{3}\cdot 0 + \frac{1}{3}\cdot \frac{1}{2} = \frac{1}{2}$$ So the probability to get tails at all the $3$ flips is ${(\frac{1}{2})}^3$ which is $\frac{1}{8}$. Is this right? 2. The probability seems to be intuitively $\frac{1}{3}$, but I don't know how to formally prove it. By Bayes' theorem \begin{align}P(M_3\mid TTT)=\frac{P(TTT\mid M_3)P(M_3)}{P(TTT)}=\frac{\left(\frac12\right)^3\cdot\frac13}{\frac38}=\frac19\end{align} Note: The denominator was calculated using the Law of total probability as is common when applying the Bayes rule. You did this in part 1. but not correctly. To see this write \begin{align}P(TTT)&=P(TTT\mid M_1)P(M_1)+P(TTT\mid M_2)P(M_2)+P(TTT\mid M_3)P(M_3)\\[0.2cm]&=1\cdot\frac13+0\cdot\frac13+\left(\frac12\right)^3\frac13\\[0.2cm]&=\frac13\left(1+\frac18\right)=\frac38\end{align} • Why $(\frac{1}{2})^3$. Where did this come frome? So for each coin you computed the probability that we will get tails and for that probability the probability to get 3 tails in a row? Apr 2, 2016 at 16:40 • $P(TTT\mid M_3)=\left(\frac12\right)^3$ You roll a fair coin three times and you want three times tails, so $\frac12\cdot\frac12\cdot\frac12$. Apr 2, 2016 at 16:43 • Yes, exactly as you say it. Apr 2, 2016 at 16:44 Paint the double-head coin yellow on one side and red on the other. Paint the double-tail coin blue and green. There are 24 possible outcomes. One of the outcomes is: Run through the 24 outcomes, how many of them give you three tails? Of those outcomes, how many were with the fair coin, how many were with the double-tail coin? I will extend my comment: remember that $\text{probability of A}=\frac{\text{number of A cases}}{\text{all possible cases}}$. Then, how many ways we can get (tail, tail, tail)? If we take the fair coin with this coin we only can take (tail, tail, tail) i.e. only exist one way we can take the desired result. But if we took the double-tail coin we take (tail, tail, tail) any time i.e. the full 8 ways that a coin can show when it is tossed three times. And when we get the double-head coin we cant take (tail, tail, tail). Then the total amount of ways we can take (tail, tail, tail) is just $1+8$, and the cases for the fair coin is just $1$ so the probability that you want is $1/9$. This is a visual way to see the problem but the formal way to solve it is the answer of @JimmyR i.e. using the basic definitions and theorems of probability theory. There are some correct answers here. Many use Bayes' rule, which is correct and elegant but takes getting used to. Let me try instead to help you think through this particular example, to train your intuition. In your answer to #1 you correctly compute that the probability of one $T$ is 1/2. But that doesn't mean the probability of $TTT$ is 1/8 unless you put the coin back and choose independently again for each of the next two tosses. The way the problem is stated, you use the same coin all three times. Then the right way to compute the weighted average is $$\frac{1}{3}⋅1+ \frac{1}{3}⋅0+ \frac{1}{3}⋅\frac{1}{8}= \frac{3}{8}.$$ For the second question, you know that you don't have the middle coin, so you need the probability of the first compared to the last. If you imagine that you can tell the two sides of the two-tailed coin apart, there are 8 ways to do three flips, all of which are all tails. For the fair coin, only one triple is all tails. So when you see all tails the probability that you had the all-tail coin is 8/9.

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The lack of closed form solution for the arc length of an elliptic arc led to the development of the elliptic integrals. Measurement by arc length Definition of arc length and formula to calculate it from the radius and central angle of the arc. Improve your skills with free problems in 'Arc measure and arc length' and thousands of other practice lessons. The red arc measures 120. Measurement by central angle . length of arc AB = (5/18)(2r) = (5/18)(2(18)) = 10. Any diameter of a circle cuts it into two equal semicircles. Yes, Arc Length and Circumference isnt particularly exciting. Section 11.1 Circumference and Arc Length 595 Using Arc Lengths to Find Measures Find each indicated measure. The blue arc measures 240. $Ans = 2\pi a$ How to obtain the ans? The area of a semicircle is half the area area of the circle from which it is made. Related Book. Measurement by arc length Calculating a circle's arc length, central angle, and circumference are not just tasks, but essential skills for geometry, trigonometry and beyond. central angle calculator, arc length calculator, ... calculating arc lengths, ... A circle has an arc length of 5.9 and a central angle of 1.67 radians. Geometry For Dummies, ... you find the fraction of the circles circumference that the arc makes up. Area of a semicircle. A semicircle is a half circle, formed by cutting a whole circle along a diameter line, as shown above. CHAPTER 5A Central Angles, Arc Length, and Sector Area ... sector represents of the circle. Arc of a Circle. You can use C 360 = l measureof thecentralangle or measureof thecentralangle 360 = l C Example 1, finding the arc length. Find the length of an arc of a circle having radius 7 cm and central angle 30 degrees? Use the formula C = 2r to calculate the circumference of a circle when the radius is given. The length of arc is equal to radius The length of an arc of a circle which subtends an angle radian at the center is equal to r where r is the radius of the circle. Calculating a circle's arc length, central angle, and circumference are not just tasks, but essential skills for geometry, trigonometry and beyond. CHAPTER 5A Central Angles, Arc Length, and Sector Area ... for a central angle of a circle Calculate the arc length and the area of a sector formed by a 30 central The distance along the arc (part of the circumference of a circle, or of any curve). Question: Find the arc length of the circle given by $x^2+y^2=a^2$. It depends on the radius of a circle and the central angle. Geometry Teachers Never Spend Time Trying to Find Materials for Your Lessons Again! Relate the length of an arc to the circumference of a whole circle and the central angle subtended by the arc. The circumference of a circle is an arc measuring 360o. Step 1 : Here, radius = 7cm central angle= 30 degrees. We dare you to prove us wrong. How to Find Arc Length. ... the arc length of a circumscribed circle is: Arc length is a linear measure of the arc measured along the circle. Thus, the length of the arc AB will be 5/18 of the circumference of the circle, which equals 2r, according to the formula for circumference. The relationship of arc length to a ... as the length x of an arc of the unit circle. Join Our Geometry Teacher Community Today! * An alternative definition is that it is an open arc. a. arc length of AB b. circumference of Z c. m RS A connected section of the circumference of a circle. Here's how to calculate the circumference, radius, diameter, arc length and degrees, sector areas, inscribed angles, and other shapes of the circle. and l stand for arc length. Our arc length calculator can calculate the length of an arc of a circle and the area of a sector. Learn how to find the arc with our lesson and try out our examples questions. Geometry calculator solving for circle arc length given radius and central angle An arc is any portion of the circumference of a circle.http://www.mathwords.com/a/arc_circle.htm Arc length is the $Ans = 2\pi a$ How to obtain the ans? The arc length formula is used to find the length of an arc of a circle. To find the arc length, we now need to find the circumference It can be understood, that the arc length is a fraction of the circumference of the circle. But it can, at least, be enjoyable. The length of the circumference is given by the formula: C = d, where d is the diameter of the circle. I would like to calculate the arc length of a circle segment, i.e. The length of an arc is a connected section of the circumference of a circle. These curves are called rectifiable and the number is defined as the arc length. Use the formula C = d to calculate the circumference of a circle when the diameter is given. Question: Find the arc length of the circle given by $x^2+y^2=a^2$. Arc length of a circle is the distance measured as the length. For a circle: Arc Length = r (when is in radians) Fun math practice! I have no ideas after doing the following thing. Formula is S = r. See note at end of page. How to Determine the Length of an Arc. I have no ideas after doing the following thing. This formula can also be given as: C = 2r, where r is the radius. Step 2 : Calculation of an arc length without its central angle is a tough problem since the arc length is based on the angle. Copyright 2017 how to find arc length of a circle

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Looking for more algorithms for quasi-random numbers 11-29-2019, 01:06 PM (This post was last modified: 11-30-2019 06:16 AM by Namir.) Post: #1 Namir Senior Member Posts: 690 Joined: Dec 2013 Looking for more algorithms for quasi-random numbers Hi All Math Lovers, In another thread of mine, ttw mentions quasi-random numbers. Quasi-random numbers (QRNs) present a better spread over a range of values than pseudo-random numbers (PRNs). On the other hand, QRNs will often fail randomness tests. They true purpose to to cover more uniformly a range of values in one of more dimensions. This is part of ttw's response in my other thread, where he mentions QRNs: Quote:The easiest multi-dimensional quasi-random sequence is the Richtmeyer sequence. One uses the fractional part of multiples of the square roots of primes. Sqrt(2), Sqrt(3), etc. It's quick to do these by just setting x(i)=0 updating by x(i)=Frac(x(i)+Sqrt(P(i))). Naturally one just stores the fractional parts of the irrationals and updates. (List mode). The sequence is also called the Kronecker or Weyl sequence at times. The above text includes the algorithm of setting x(1)=0 updating by x(i)=Frac(x(i)+Sqrt(P(i))). The array of P() represents prime numbers starting with 2. You can change x(1) to had a uniform random number as a seed (to generate different sequences every time you apply the algorithm) or simply set x(1) = sqrt(P(1)) = sqrt(2). I am curious about other formulas to calculate sequences of quasi-random numbers. You are welcome to use your imagination. My first attempt was something like: Code: n = number of x to generate m = 100*n Calculate P() for primes in the range of 1 to m X(1) = rand or Frac(ln(P(3)) * sqrt(P(1)) j = 2 count = 0 for i=2 to n   X(i) = Frac(X(i) + ln(P(j+1)) * sqrt(P(j-1))   j = j + 1   if j > m then    count = count + 1     j = 2 + count   end end The above code produces x() with a mean near 0.5 and standard deviation near 0.28. The auto correlations for the first 50 lags are in the orde rof 10^(-2) to 10^(-4). I am curious about other formulas to calculate sequences of quasi-random numbers. You are welcome to use your imagination. You can even commit math heresy!!! As long as it works, you are fine (and forgiven) :-) Namir 11-29-2019, 04:49 PM (This post was last modified: 11-29-2019 06:24 PM by SlideRule.) Post: #2 SlideRule Senior Member Posts: 1,013 Joined: Dec 2013 RE: Looking for more algorithms for quasi-random numbers Perusal of Quasi-random sequences in art and integration, John D. Cook Consulting, illumes the phenomena with references to more descriptive books; Random Number Generation and Quasi-Monte Carlo Methods & Monte Carlo and Quasi-Monte Carlo Methods, on the same. BEST! SlideRule 11-30-2019, 01:36 AM Post: #3 mfleming Senior Member Posts: 498 Joined: Jul 2015 RE: Looking for more algorithms for quasi-random numbers (11-29-2019 01:06 PM)Namir Wrote:  This is part of ttw's response in my other thread, where he mentions QRNs: Quote:The easiest multi-dimensional quasi-random sequence is the Richtmeyer sequence. One uses the fractional part of multiples of the square roots of primes. Sqrt(2), Sqrt(3), etc. It's quick to do these by just setting x(i)=0 updating by x(i)=Frac(x(i)+Sqrt(P(i))). Naturally one just stores the fractional parts of the irrationals and updates. (List mode). The sequence is also called the Kronecker or Weyl sequence at times. Using "quote" in place of "code" will autowrap large blocks of text! Who decides? 11-30-2019, 01:29 PM Post: #4 Namir Senior Member Posts: 690 Joined: Dec 2013 RE: Looking for more algorithms for quasi-random numbers (11-30-2019 01:36 AM)mfleming Wrote: (11-29-2019 01:06 PM)Namir Wrote:  This is part of ttw's response in my other thread, where he mentions QRNs: Quote:The easiest multi-dimensional quasi-random sequence is the Richtmeyer sequence. One uses the fractional part of multiples of the square roots of primes. Sqrt(2), Sqrt(3), etc. It's quick to do these by just setting x(i)=0 updating by x(i)=Frac(x(i)+Sqrt(P(i))). Naturally one just stores the fractional parts of the irrationals and updates. (List mode). The sequence is also called the Kronecker or Weyl sequence at times. Using "quote" in place of "code" will autowrap large blocks of text! I learned that the hard way :-) 11-30-2019, 07:52 PM Post: #5 Namir Senior Member Posts: 690 Joined: Dec 2013 RE: Looking for more algorithms for quasi-random numbers The few leads I got from the nice folks on this web were able to lead me to methods that generate sequences of quasi-random numbers that are practically perfectly distributed. I got what I was looking for. Thanks!!! Namir 12-01-2019, 05:52 AM Post: #6 ttw Member Posts: 186 Joined: Jun 2014 RE: Looking for more algorithms for quasi-random numbers This is one of the sequences from my paper in "Computational investigations of low-discrepancy point sets II" from the 1994 Las Vegas Conference on Monte Carlo and Quasi Monte Carlo Methods. I have made a single ad-hoc change (described below) that improves distribution for small numbers of points. The Halton Sequence Phi(N,P) (for odd primes, 2 is a special case not considered here) can be described as: 1. Generate the digits of N in base P (for P an odd prime). Call these digits a(1) to a(k) where k is the maximum number of digits needed. (There should be lots of subscripts but I'll treat each prime separately to reduce index management.) N=Sum from j=1 to k of a(j)*P^(j-1), that is: a(k)a(k-1)...a(2)(a(1). 2. Reverse the digits: a(1),a(2)....a(k-1),a(k) is resulting string. 3. Treat this string as a fraction with a decimal point (p-ary point?) in front. Example: P=3, N=5: 5(3)=12. Reverse Phi(5,3)=.21(3) or 2/3+1/9 = 7/9 The sequence is very well distributed for example one starts with 0, 1/3, 2/3, 1/9, 4/9, 7/9, 2/9, 5/9, 8/9, 1/27, 10,27... This sequence is uniformly in the unit d-cube using d different primes. For example in 3 dimensions using primes 3 and 5 gives the points: (skipping 0 which sits on the corner of the cube). (1/3, 1/5, 1/7) (2/3, 2/5, 2/7) (1/9, 3/5, 3/7) (4/9, 4/5, 4/7) (7/9,1/25,5/7) (2/9, 6/25, 6/7) (5/9, 11/25, 1/49) (8/9. 16/25, 8/49) etc. The process is sometimes termed a Kakatumi-von Neumann odometer. There is a problem that I noticed about 1967 or so when I started working on quasi-Monte Carlo. For large base, the Halton Sequence produces strongly correlated points until enough points are generated. (This happens with all quasi-random sequences but not as severely.) Take the first few points using bases 101 and 103. (1/101, 1/103) (2/101, 2/103) ... (100/101, 100/103) (1/10201, 101/103) (102/10201, 102/103) (203/10203, 1/10609) etc. In 1993-1995 period, I figured out to multiply each numerator by a number (I called a spin) to break this up. Then I used the fact the fractional parts of square roots of primes are independent to do the following seemingly strange rule. Give a prime P, to find a multiplier S, do the following. 1. Compute the nearest integers to the fractional part of the Sqrt(P), call these H and L (high and low, one is above and one below the number). 2. Compute the continued fraction of H/P and L/P; each generates a string of partial quotients. The multiplier S is the one of these satisfying the following: 3. A. Chose the one for which the sum of the partial quotients is smallest. B. If tied, chose the one with the smallest maximum partial quotient. C. If tied, chose whichever H/P or L/P is closest to the fractional part of the square root. D. Ad Hoc Alert: if P=41, use 16. (To avoid 17/41 being close to 12/29. The only such case in all primes less than 2^32) 4. To generate the modified Phi sequence Phi(N,P,S): generate the digits of N base P as above and reverse. Multiply each digit by S modulo P and sum as above. Examples: 3 dimensions: P=3, 5, 7 have S= 2, 2,and 5 respectively. (2/3, 2/5, 5/7) (1/3, 4/5, 3/7) (2/9, 1/5, 1/7) (8/9, 3/4, 6/7) etc. For the pathological case 101 and 103, the multipliers are 6 and 16 respectively (not the best but that's another post sometime). (6/101, 16/103) (12/101, 32/103) (18/101, 48/103) Clearly more spread out than the original Halton Sequence. I've got some more but Mordechay Levin's paper ArXiv 1806 shows that even the original Halton Sequence hits the theoretical lower bound for dimensions 2 and up so great changes cannot be had by tinkering. I do have a bit better, but it's even longer to compute and I haven't tested the new ideas thoroughly. HP50g code:(Number, Prime, Spin, Top, Bottom) << 0 1 -> N P S T B << WHILE N REPEAT N S * P MOD T P * + 'T' STO P 'B' STO* N P IQUOT 'N' STO END T B />> >> Not necessarily the fastest but eas to work with. (If I've done this right.) 12-01-2019, 11:46 AM Post: #7 Csaba Tizedes Senior Member Posts: 412 Joined: May 2014 RE: Looking for more algorithms for quasi-random numbers « Next Oldest | Next Newest » User(s) browsing this thread: 1 Guest(s)

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# Max Sum Of 2 Arrays Reductions. It also prints the location or index at which maximum element occurs in array. int [] A = {−2, 1, −3, 4, −1, 2, 1, −5, 4}; Output: contiguous subarray with the largest sum is 4, −1, 2, 1, with sum 6. Array is an arranged set of values of one-type variables that have a common name. Yes you can find the maximum sum of elements in linear time using single traversal of the array. You can also use the following array formulas: Enter this formula into a blank cell, =SUM(LARGE(A1:D10,{1,2,3})), and then press Ctrl + Shift + Enter keys to get your result. It should have 3 input parameters array A, length and width. Finding the Average value of an Array. Note that in the calculation of max4, we have passed a two dimensional array containing two rows of three elements as if it were a single dimensional array of six elements. In this example, you create two arrays, DAYS and HOURS. Algorithms in Java Assignment: Maximum Sum (in 2 Dimensions) The Problem Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. However, I would like to use the max of these scores. When common element is found then we will add max sum from both the arrays to result. If x and y are scalars and A and B are matrices, y x, A x, and x A have their usual mathematical meanings. Max sum in an array. For all possible combinations, find the sum and compare it with the previous sum and update the maximum sum. Idea is to use merge sort algorithm and maintain two sum for 1st and 2nd array. Search in Rotated Sorted Array. The user will enter a number indicating how many numbers to add and then the user will enter n numbers. I need to check an array of random integers (between 1 and 9) and see if any combination of them will add up to 10. A corner element is an element from the start of the array or from the end of the array. Once the type of a variable is declared, it can only store a value belonging to this particular type. Create a max heap i. Given an array, you have to find the max possible two equal sum, you can exclude elements. Question E3: WAP to find out the row sum and column sum of a two dimensional array of integers. Whenever possible, make sure that you are using the NumPy version of these aggregates when operating on NumPy arrays!. Write a program to find those pair of elements that has the maximum and minimum difference among all element pairs. Input size and elements in array, store in some variable say n and arr[n]. Here is the complete Java program with sample outputs. Algorithms in Java Assignment: Maximum Sum (in 2 Dimensions) The Problem Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. Method since it requires contiguous, it means that for each element, it has two situations that are in the subarray or not. Easy Tutor says. The return value of min () and max () functions is based on the axis specified. C++ :: Creating Table Of Arrays - Find Maximum Value And Sum Aug 12, 2014. Google Advertisements. Sample Run: [2, 1, 8, 4, 4] Min: 1 Max: 8 Average: 3. The master will loop from 2 to the maximum value on issue MPI_Recv and wait for a message from any slave (MPI_ANY_SOURCE), if the message is zero, the process is just starting, if the message is negative, it is a non-prime, if the message is positive, it is a prime. Given an integer array of N elements, find the maximum sum contiguous subarray (containing at least one element). HackerRank Solutions Over the course of the next few (actually many) days, I will be posting the solutions to previous Hacker Rank challenges. 1 Answer to Given that A[MAX_ROWS][MAX_COLUMNS] is a 2 dimensional array of integers write a C ++ function. WriteLine to do this. SUMPRODUCT( array1, [array2, array_n] ) Parameters or Arguments array1, array2, array_n The ranges of cells or arrays that you wish to multiply. A better solution would be to find the two largest elements in the array, since adding those obviously gives the largest sum. The maximum product is formed by the (-10, -3) or (5, 6) pair. Algorithms in Java Assignment: Maximum Sum (in 2 Dimensions) The Problem Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. A selected portion of the array may be summed, if an integer range expression is provided with the array name (. max (x) → [same as input] Returns the maximum value of all input values. Write a program to find sum of each digit in the given number using recursion. Basic Operations ¶. Find the sum of numbers and represent it in array. The function should return an integer. A one-dimensional array is like a list; A two dimensional array is like a table; The C language places no limits on the number of dimensions in an array, though specific implementations may. Write a program to find top two maximum numbers in a array. MS Excel 2007: Use an array formula to sum all of the order values for a given client This Excel tutorial explains how to use an array formula to sum all of the order values for a given client in Excel 2007 (with screenshots and step-by-step instructions). min () find the maximum and minimum value of the arguments, respectively. This very simply starts with a sum of 0 and add each item in the array as we go: public static int findSumWithoutUsingStream (int[] array) { for (int value : array) { 2. In this article we’ll explore four plug and play functions that allow you to easily find certain values in an arrays of numbers. (For clarification, the L-length subarray could occur before or after the M-length subarray. A matrix with m rows and n columns is actually an array of length m, each entry of which is an array of length n. We can switch from one array to another array only at common elements. For example if input integer array is {2, 6, 3, 9, 11} and given sum is 9, output should be {6,3}. See (2) in the diagram. We can start from either arrays but we can switch between arrays only through its common elements. My solution for the bigDiff using the inbuilt Math. If you sum the second array you can use that to multiply the first array because that will be the same as multiplying the values individually and then summing the results. K maximum sum combinations from two arrays Given two equally sized arrays (A, B) and N (size of both arrays). Objective Problem Statement • Application of parallel prefix: Identifying the maximum sum that can be computed using. min (x) → [same as input]. Pop the heap to get the current largest sum and along. Maximum Sum of Two Non-Overlapping Subarrays. For example, to sum the top 20 values in a range, a formula must contain a list of integers from 1 to 20. For example [1,3,5,6,7,8,] here 1, 3 are adjacent and 6, 8 are not adjacent. We are making max_sum_subarray is a function to calculate the maximum sum of the subarray in an array. Specifically we'll explore the following: Finding the Minimum value in an Array. 4+ PHP Changelog: PHP versions prior to 4. And so myself and the OP exchanged a comment: I have concern. For an array x, y=cumsum(x) returns in the scalar y the cumulative sum of all the elements of x. In C programming, you can pass en entire array to functions. Input the array elements. You may have A1:A20, then A30:A35 filled. Thus, two arrays are “equal” according to Array#<=> if, and only if, they have the same length and the value of each element is equal to the value of the corresponding element in the other array. C++ Programs to Delete Array Element C++ Programs to Sum of Array Elements. Find the sum of the maximum sum path to reach from beginning of any array to end of any of the two arrays. Top Forums Shell Programming and Scripting Sum elements of 2 arrays excluding labels Post 303015114 by Don Cragun on Wednesday 28th of March 2018 06:34:58 AM. An index value of a Java two dimensional array starts at 0 and ends at n-1 where n is the size of a row or column. Given input array be,. In this solution dp* stores the maximum among all the sum of all the sub arrays ending at index i. Once the type of a variable is declared, it can only store a value belonging to this particular type. log10(a) Logarithm, base 10. This function subtracts when negative numbers are used in the arguments. (2-D maximum-sum subarray) (30 points) In the 2-D Maximum-Sum Subarray Prob- lem, you are given a two-dimensional m x n array A[1 : m,1: n of positive and negative numbers, and you are asked to find a subarray Ala b,c: 1 Show transcribed image text Expert Answer. C Program to read an array of 10 integer and find sum of all even numbers. if 2,3,4, 5 is the given array, {4,5,2,3} is also a possible array like other two. Our maximum subset sum is. SemanticSpace Technologies Ltd interview question: There is an integer array consisting positive numbers only. If any element is greater than max you replace max with. Enables ragged arrays. Array-2, Part I ”. I haven't gotten that far yet, I'm stuck just trying to print my two arrays, every time i try to print the first array it gives me the elements of the second array and it. All arrays must have the same number of rows and columns. i* n/2 – Or overlaps both halfs: i* n/2 j* • We can compute the best subarray of the first two types with recursive calls on the left and right half. Then we compare it with the other array elements one by one, if any element is greater than our assumed. I need to check an array of random integers (between 1 and 9) and see if any combination of them will add up to 10. The maximum product is formed by the (-10, -3) or (5, 6) pair. SUM (C, DIM=1) returns the value (5, 7, 9), which is the sum of all elements in each column. computes bitwise conjunction of the two arrays (dst = src1 & src2) Calculates the per-element bit-wise conjunction of two arrays or. This program shows you how to find the sum of rows and columns in a two dimensional array, and also how to use a method to calculate the sum of every element inside of a 2d array. Given two equally sized arrays (A, B) and N (size of both arrays). We can do this by using or without using an array. (Array): Returns the new array of chunks. If they are even we will try to find whether that half of sum is possible by adding numbers from the array. Latest commit message. So, the minimum of the array is -1 and the maximum of the array is 35. Google Advertisements. 999997678497 499911. Add solution to Pairs problem. Next, we use a standard for loop to iterate through our array numbers and total up the. Binary Tree Maximum Path Sum Lowest Common Ancestor I II III Binary Tree Level Order Traversal I II Kth Smallest Sum In Two Sorted Arrays LinkedList. We can switch from one array to another array only at common elements. Bottleneck code often involves condi-tional logic. if the sum of previous subarray is negative, it means that it need. 1<=Ai<=10000, where Ai is the ith integer in the array. The purpose of the SUMPRODUCT function is to multiply, then sum, arrays. The page is a good start for people to solve these problems as the time constraints are rather forgiving. Also add the common element to the result. The algorithm to find maximum is: we assume that it's present at the beginning of the array and stores that value in a variable. It may or may not include a[i-1], a[i-2], etc. M = max( A ,[], 'all' , nanflag ) computes the maximum over all elements of A when using the nanflag option. We can update both incrementally by counting from the back, so we have to keep track of two things: \$\max(S[i:])\$ and \$\max(B[i+1:])\$. Specifically we’ll explore the following: Finding the Minimum value in an Array. We take a two dimensional array L of size count+1, sum/2+1. Partition an array into two sub-arrays with the same sum. For example, given array A such that: A[0] = 3 A[1] = 2 A[2] = -6 A[3] = 4 A[4] = 0. Each element, therefore, must be accessed by a corresponding number of index values. creating a recursive way to find max and min in array I am kind of confused with this instruction: Describe a recursive algorithm for finding both the minimum and maximum elements in an array A of n elements. If ARRAY is a zero-sized array, the result equals zero. computes the sum of two matrices and then prints it. The basis is p[0] = a[0]. package net. What is the sum of the last column? 5. Yes you can find the maximum sum of elements in linear time using single traversal of the array. In order to find the sum of all elements in an array, we can simply iterate the array and add each element to a sum accumulating variable. The first thing that we tend to need to do is to scan through an array and examine values. sum(my_first_array) >my_first_array. Our maximum subset sum is. Here is the complete Java program with sample outputs. Naive solution would be to consider every pair of elements and calculate their product. the contiguous subarray [4,-1,2,1] has the largest sum = 6. Question E2: WAP to display the values of a two dimensional array in the matrix form. Suppose we need to find out max sum subarray. As a "rule of thumb", any "calculated array" - in this case the array calculated by adding two ranges - results in a formula that requires CSE, although some functions (like SUMPRODUCT and LOOKUP) don't normally need CSE even with calculated arrays - to allow normal entry you can add an INDEX function - I edited my answer to the effect. maxSubsetSum has the following parameter(s): arr: an array of integers. amax() by thispointer. Given an array A of non-negative integers, return the maximum sum of elements in two non-overlapping (contiguous) subarrays, which have lengths L and M. Solution to Question 2. We can use an array as a deque with the following operations:. Improvement over Method-1 – O(n 2) Time. If only ARRAY is specified, the result equals the sum of all the array elements of ARRAY. It works as follows. Once the type of a variable is declared, it can only store a value belonging to this particular type. Maximize array sum by concatenating corresponding elements of given two arrays Given two array A[] and B[] of the same length, the task is to find the maximum array sum that can be formed by joining the corresponding elements of the array in any order. Our maximum subset sum is. In a two-dimensional Java array, we can use the code a[i] to refer to the ith row (which is a one-dimensional array). And so myself and the OP exchanged a comment: I have concern. I borrowed some code from other forums that had similar programs, but obviously it doesn't match my needs specifically. Given an array of integers, find maximum product of two integers in an array. Computes the matrix multiplication of two arrays. Add solutions to C++ domain. This problem is generally known as the maximum sum contiguous subsequence problem and if you haven’t encountered it before, I’d recommend trying to solve it before reading on. #include using namespace std; int main() { const int SIZE = 12; double months[SIZE]; int count; double sum = 0; double totalRainfall; double. c++: which functions gives the sum of an array? (3). In this example, we will find the sum of all elements in a numpy array, and with the default optional parameters to the sum () function. min (x) → [same as input]. Given two arrays of positive integers. For example, for the sequence of values −2, 1, −3, 4, −1, 2, 1, −5, 4; the contiguous subarray with the largest sum is 4, −1, 2, 1, with sum 6. Let arr[i. Your code tries all \$n (n+1)/2 \$ combinations of array elements to find the combination with the largest sum, so the complexity is \$O(n^2) \$. The loop structure should look like for (i=0; i=2 and find the sum of smallest and second smallest, then our answer will be maximum sum among them. Find the sum of the maximum sum path to reach from beginning of any array to end of any of the two arrays. Maximum Sum of Two Non-Overlapping Subarrays 2019/04/22 2019/04/22 shiji Leetcode Given an array A of non-negative integers, return the maximum sum of elements in two non-overlapping (contiguous) subarrays, which have lengths L and M. Write a function to find the maximum sum of all subarrays. In order to find the sum of all elements in an array, we can simply iterate the array and add each element to a sum accumulating variable. If all the array entries were positive, then the maximum-subarray problem would present no challenge, since the entire array would give the greatest sum. The length property is the array length or, to be precise, its last numeric index plus one. I borrowed some code from other forums that had similar programs, but obviously it doesn't match my needs specifically. The maximum product is formed by the (-10, -3) or (5, 6) pair. 7 is the sum of 2 + 5 in column 2, and so forth. Sub Array with Maximum Sum – Kadane Algorithm is the best solution. Note : Imp to execute and trace to understand and remember. Edit: given your comments if the initial array is fixed then you can use MMULT function like this. Passing array elements to a function is similar to passing variables to a function. if orientation is equal to n then. M=sum(A,dim) In Scilab dim=1 is equivalent to dim="r" and dim=2 is equivalent dim="c". For examples, Enter 1st integer: 8 Enter 2nd integer: 2 Enter 3rd integer: 9 The sum is: 19 The product is: 144 The min is: 2 The max is: 9 Hints. Read the entered array size and store that value into the variable n. Maximum Sum Subarray (In Yellow) For example, for the array given above, the contiguous subarray with the largest sum is [4, -1, 2, 1], with sum 6. How to swap two numbers without using temporary variable? Write a program to print fibonacci series. Find maximum possible sum of elements such that there are no 2 consecutive elements present in the sum. Find ways to calculate a target from elements of specified. Given an array, find maximum sum of smallest and second smallest elements chosen from all possible sub-arrays. Once you have a vector (or a list of numbers) in memory most basic operations are available. Find the sum of the maximum sum path to reach from beginning of any array to end of any of the two arrays. So, the minimum of the array is -1 and the maximum of the array is 35. The maximum admissible amount of dimensions in an array is four. For example, A = [−2, 1, −3, 4, −1, 2, 1, −5, 4] then max sum=11 with the subarray [1, 4, 2, 4]. Calculates the per-element sum of two arrays or an array and a scalar. In order to find the sum of all elements in an array, we can simply iterate the array and add each element to a sum accumulating variable. Join 124,729,115 Academics and Researchers. Expected time complexity is O(m+n) where m is the number of elements in ar1[] and n is the number of elements in ar2[]. I am trying to compute the maximum possible sum of values from a matrix or 2d array or table or any suitable structure. A selected portion of the array may be summed, if an integer range expression is provided with the array name (. Hence there would be four different arrays in this case. sum += numbers [i] In The Standard Way we first declare the variable sum and set its initial value of zero. The sum choice number is the minimum over all choosable functions f of the sum of the sizes in f. Given an array A of non-negative integers, return the maximum sum of elements in two non-overlapping (contiguous) subarrays, which have lengths L and M. SUM (C, DIM=2) returns the value (6, 15), which is the sum of all elements in each row. e 1,2,3,4,6 is given array we can have max two equal sum as 6+2 = 4+3+1. =MAX(IF((List>=LLim)*(List<=ULim),List,FALSE)) returns the maximum of values between 2 and 5, or 5. When you want to return a sum for a single criteria (for example, a single IF condition) When you want to use multiple criteria and return the sum to multiple cells; The criteria that you can use with the SUMIF() worksheet function is limited to text, numbers, or a range, and the function cannot use array constants. Stack PUSH & POP Implementation using Arrays; Program to remove duplicate element in an array; C Program to sort the matrix rows and columns; Write a c program for swapping of two arrays; C Program to read name and marks of students and store it in file; To find out the maximum number in an array using function. Explore Channels Plugins & Tools Pro Login About Us. Academia is the easiest way to share papers with millions of people across the world for free. The min () and max () functions of numpy. A corner element is an element from the start of the array or from the end of the array. Search in Rotated Sorted Array II. I need to create a table of this which i have done using case 1. The basis is p[0] = a[0]. (For clarification, the L-length subarray could occur before or after the M-length subarray. For example, consider the array {-10, -3, 5, 6, -2}. Calculates the per-element sum of two arrays or an array and a scalar. Display the maximum K valid sum combinations from all the possible sum combinations. Here we are setting up the pointer to the base address of array and then we are incrementing pointer and using * operator to get & sum-up the values of all the array elements. Even if you have encountered it before, I’ll invite you. For example, to sum the top 20 values in a range, a formula must contain a list of integers from 1 to 20. If the current element of array 1 and array 2are same, then take the maximum of sum1 and sum2 and add it to the result. In the Java programming language, a multidimensional array is an array whose components are themselves arrays. For example, entering =SUM(10, 2) returns 12. How to swap two numbers without using temporary variable? Write a program to print fibonacci series. Yes you can find the maximum sum of elements in linear time using single traversal of the array. To store sum of array elements, initialize a variable sum = 0. Latest commit 7b136cc on Mar 10, 2019. The function should return an integer. The call to new Array(number) creates an array with the given length, but without elements. What's wrong with the scrap of code in the question? The array is of size 5, but the loop is from 1 to 5, so an attempt will be made to access the nonexistent element a[5]. def array_summer(arr): return sum (arr) # Test input print (array_summer ( [1, 2, 3, 3, 7])) we went through two different methods of summing the elements of an array. For an array x, y=cumsum(x) returns in the scalar y the cumulative sum of all the elements of x. computes the sum of two matrices and then prints it. C++ :: Creating Table Of Arrays - Find Maximum Value And Sum Aug 12, 2014. Basicly I have to sum rows and columns of 2d array and then store the results in separate arrays, as far as the code is now, can see it quite clearly. Finding the Maximum value in an Array. ; The array formula lets the IF function test for multiple conditions in a single cell, and, when the data meets a condition, the array formula determines what data (event results) the MAX function will examine to find the best result. Yes you can find the maximum sum of elements in linear time using single traversal of the array. The previous contiguous array sum was less than or equals 0. Finally, if A is a multidimensional array, Matlab works on the first non-singleton dimension of A what Scilab does not. with - sum of two arrays in c. With the following program, you can even print the sum of two numbers or three numbers up to N numbers. For examples, Enter 1st integer: 8 Enter 2nd integer: 2 Enter 3rd integer: 9 The sum is: 19 The product is: 144 The min is: 2 The max is: 9 Hints. Finding the Sum of all values in an Array. For example [1,3,5,6,7,8,] here 1, 3 are adjacent and 6, 8 are not adjacent. For an array x, y=cumsum(x) returns in the scalar y the cumulative sum of all the elements of x. Note : Imp to execute and trace to understand and remember. The problem: given an array which could contain zero, negative, and positive numbers, find the largest sum of contiguous sub-array. Move the pointer in the corresponding heap there. We can update both incrementally by counting from the back, so we have to keep track of two things: \$\max(S[i:])\$ and \$\max(B[i+1:])\$. Array-2, Part I ”. To get the sum of all elements in a numpy array, you can use Numpy’s built-in function sum (). Question 3. such that sum(wi*xi)<=W & x(0,1) The unbounded knapsack problem (UKP) places no upper bound on the number of copies of each kind of item. Maximum Sum Subarray (In Yellow) For example, for the array given above, the contiguous subarray with the largest sum is [4, -1, 2, 1], with sum 6. These functions will not work as-is with arrays of numbers. computes the sum of two matrices and then prints it. Given an array A of non-negative integers, return the maximum sum of elements in two non-overlapping (contiguous) subarrays, which have lengths L and M. Add solutions to C domain. Two Dimensional Array in C Example. Dim i,sum, r As Integer sum=0 For i=0 to 4 ‘assign values to the array x(i)=i*i Next r=1 For each v in x ‘read the elements of the array in to Excel cells MsgBox v r= r+1 Next Note: the start index of the array is 0 and its size is equal to last index added by 1. Once the type of a variable is declared, it can only store a value belonging to this particular type. Array formulas can also be used find out the maximum and minimum values for a given set of conditions. The call to new Array(number) creates an array with the given length, but without elements. Maximum of array elements over a given axis. Sub Array with Maximum Sum – Kadane Algorithm is the best solution. The SUM function in Excel adds the arguments you enter in a formula. Java Program to Find The Sum of Array Elements || D. sum() 15 How to find the maximum value in NumPy 1d-array? We can find the maximum value stored in. Subarr2[] = {3, 3, 12. Problem Statement: Given an array A = {a 1, a 2, a 3, , a N} of N elements, find the maximum possible sum of a. As a "rule of thumb", any "calculated array" - in this case the array calculated by adding two ranges - results in a formula that requires CSE, although some functions (like SUMPRODUCT and LOOKUP) don't normally need CSE even with calculated arrays - to allow normal entry you can add an INDEX function - I edited my answer to the effect. Print the N integers of the array in the reverse order in a single line separated by a space. Which row has the largest sum? 4. Array is an arranged set of values of one-type variables that have a common name. Take a HashMap with Key and value as Integer types. This problem is generally known as the maximum sum contiguous subsequence problem and if you haven’t encountered it before, I’d recommend trying to solve it before reading on. The sum of all the numbers in the array. To store sum of array elements, initialize a variable sum = 0. C++ Programs to Delete Array Element C++ Programs to Sum of Array Elements. Let arr[i. Where L[i,j]=maximum sum possible with elements of array 0 to i and sum not exceeding j. Using only one loop, Complete the code to compute both sums. Function Description. Description: ----- If we add amount of max INT with number 1 in array_sum function , the result will be false. Hi, My documents have an "aliases" field which is an array of string. Objective: The maximum subarray problem is the task of finding the contiguous subarray within a one-dimensional array of numbers which has the largest sum. A better solution would be to find the two largest elements in the array, since adding those obviously gives the largest sum. Find Maximum sum sub array of tempArray. I'm stumped. if the array was [5, 6, 4, 2, 9] it would return true. j* n/2 – Or contained entirely in the right half , i. For all possible combinations, find the sum and compare it with the previous sum and update the maximum sum. In this example, we will find the sum of all elements in a numpy array, and with the default optional parameters to the sum () function. To find the maximum value, you initialize a placeholder called max with the value of the first element in the array. 1 Answer to Given that A[MAX_ROWS][MAX_COLUMNS] is a 2 dimensional array of integers write a C ++ function. int a[3]; // creates an array with 'Numb' elements a[3] = 5; // assigns 5 to index 3 (the 4th element) in the array This is effectively what you're doing with your cin line. But when you try to follow the same approach with an array formula, Excel complains. Problem Description We have to write a program in C such that the program will allocate 2 one-dimensional arrays using malloc() call and then will do the addition and stores the result into 3rd array. min(), big_array. but it must include a[i]. Maximum sum in circular array such that no two elements are adjacent | Set 2 Given an array arr[] of positive numbers, find the maximum sum of a subsequence with the constraint that no 2 numbers in the sequence should be adjacent in the array where the last and the first elements are assumed adjacent. Latest commit 7b136cc on Mar 10, 2019. How we can do that efficiently?. max (x, n) → array<[same as x]> Returns n largest values of all input values of x. The elements entered in the array are as follows: 1 2 35 0 -1. The prior values added up to 0, meaning that the new max subarray starts from this value. Empty subarrays/subsequences should not be considered. See (2) in the diagram. For examples, Enter 1st integer: 8 Enter 2nd integer: 2 Enter 3rd integer: 9 The sum is: 19 The product is: 144 The min is: 2 The max is: 9 Hints. For example if input integer array is {2, 6, 3, 9, 11} and given sum is 9, output should be {6,3}. (#M40034130) C Programming question Find out maximim sum of sub Array Keep an EYE Find out maximim sum of sub Array example array={2,3,-1,4,9} maximum sum of sub array=17 Asked In C DHIRENDRA (6 years ago) Unsolved Read Solution (3) Is this Puzzle helpful? (1) (0) Submit Your Solution Program. The expression within the optional "with" clause can be used to specify the item to use in the reduction. Dynamic Memory Allocation Example: In this C program, we will declare memory for array elements (limit will be at run time) using malloc(), read element and print the sum of all elements along with the entered elements. Return the maximum of sum1 and sum2. 2 Vectorized Logic The previous section shows how to vectorize pure computation. The "waterdrop" camera array is made up of a 48-MP main unit, an 8-MP wide-angle shooter, and a 5-MP depth sensor, with a flash module at the bottom (2020) vs iPhone 11, 11 Pro and Pro Max. Full Discussion: How do I find the sum of values from two arrays? Top Forums Shell Programming and Scripting How do I find the sum of values from two arrays? Post 302579313 by kshji on Monday 5th of December 2011 12:03:10 PM. Given two equally sized arrays (A, B) and N (size of both arrays). Algorithms in Java Assignment: Maximum Sum (in 2 Dimensions) The Problem Given a 2-dimensional array of positive and negative integers, find the sub-rectangle with the largest sum. Whenever possible, make sure that you are using the NumPy version of these aggregates when operating on NumPy arrays!. The min () and max () functions of numpy. Pop the heap to get the current largest sum and along. Your code tries all \$n (n+1)/2 \$ combinations of array elements to find the combination with the largest sum, so the complexity is \$O(n^2) \$. Arr2[] = {1, 3, 3, 12, 2} then maximum result is obtained when we create following two subarrays − Subarr1[] = {2, 4, 3} and. Using only one loop, Complete the code to compute both sums. Timing Belt - $1,299. Given an integer array of N elements, find the maximum sum contiguous subarray (containing at least one element). Compute sum of two digit arrays. For example, if A is a matrix, then max(A,[],[1 2]) computes the maximum over all elements in A, since every element of a matrix is contained in the array slice defined by dimensions 1 and 2. here maximum subarray is [2,3,4]. Divide Two Integers 4. You can max_heap both arrays, and set an iterator at the root for both. Complexity Analysis:. Input: nums = [1,1,1], k = 2 Output: 2. If only ARRAY is specified, the result equals the sum of all the array elements of ARRAY. Array exponentiation is available with A. This program is an example of Dynamic Memory Allocation, here we are declaring memory for N array elements at run time using malloc() - which is used to declare memory for N. Problem Description Write a program to find the sum of the corresponding elements in 2 arrays. A matrix with m rows and n columns is actually an array of length m, each entry of which is an array of length n. My solution for the bigDiff using the inbuilt Math. def sum_odd(n): if n < 2: return 1 elif n%2 == 0: return sum_odd(n-1) else: return n + sum_odd(n-2) Note that this function returns 1 if n is not greater 0 as is defined in the original function. You can refer to more than one array in a single SAS statement. Given input array be,. Note: Values of different types will be compared using the standard comparison rules. To display sub array with maximum sum you should write code to hold the start and end value of the sub array with maximum sum. My operation system is 64 bit. In this tutorial, I am going to discuss a very famous interview problem find maximum subarray sum (Kadane’s algorithm). The optimal strategy is to pick the elements form the array is, two. Pure VPN Privide Lowest Price VPN Just @$1. Find ways to calculate a target from elements of specified. The first user will be asked to enter the order of the matrix (such as the numbers of rows and columns) and then enter the elements of the two matrices. I am trying to compute the maximum possible sum of values from a matrix or 2d array or table or any suitable structure. In a two-dimensional Java array, we can use the code a[i] to refer to the ith row (which is a one-dimensional array). MS Excel 2007: Use an array formula to sum all of the order values for a given client This Excel tutorial explains how to use an array formula to sum all of the order values for a given client in Excel 2007 (with screenshots and step-by-step instructions). Consider an integer array, the number of elements in which is determined by the user. Find a Triplet having Maximum Product in an Array. Sum the largest 3 numbers: =SUM(LARGE(range, {1,2,3})) Sum the smallest 3 numbers: =SUM(SMALL(range, {1,2,3})) Don't forget to press Ctrl + Shift + Enter since you are entering the Excel array formula, and you will get the following result: In a similar fashion, you can calculate the average of N smallest or largest values in a range:. Arrays in formulas. Array Maximum Minimum value We can calculate maximum value among the elements of an array by using max function. For example, consider the array {-10, -3, 5, 6, -2}. We can switch from one array to another array only at common elements. Hi, My documents have an "aliases" field which is an array of string. Basic array operations. Space Complexity: O(1). If it's provided then it will return for array of max values along the axis i. A blog about Java, Spring, Hibernate, Programming, Algorithms, Data Structure, SQL, Linux, Database, JavaScript, and my personal experience. Problem Statement: Given an array A = {a 1, a 2, a 3, , a N} of N elements, find the maximum possible sum of a. Add solution to Super Maximum Cost Queries problem. Naive solution would be to consider every pair of elements and calculate their product. Maximize array sum by concatenating corresponding elements of given two arrays Given two array A[] and B[] of the same length, the task is to find the maximum array sum that can be formed by joining the corresponding elements of the array in any order. I have an array of "2,3,4,5,6,9,10,11,12,99". The length property is the array length or, to be precise, its last numeric index plus one. Now the above Leetcode challenge is a special case of the general Max Subarray classic problem in computer science - which is the task of finding the contiguous subarray within a one-dimensional array of numbers which has the largest sum. We will implement a simple algorithm in javascript to find the maximum sum of products of given two arrays. Yesterday we got the question Sum of Maximum GCD from two arrays. Note that the common elements do not have to be at same indexes. How to swap two numbers without using temporary variable? Write a program to print fibonacci series. Non-Numeric or Non-Existent Fields¶. If DIM is absent, a scalar with the sum of all elements in ARRAY is returned. It loops over the values and returns the sum of the elements. Note that in the calculation of max4, we have passed a two dimensional array containing two rows of three elements as if it were a single dimensional array of six elements. Yes you can find the maximum sum of elements in linear time using single traversal of the array. Jerico January 10, 2014 at 6:30 am. For example, if the array contains: 31, -41, 59, 26, -53, 58, 97, -93, -23, 84 then the largest sum is 187 taken from the [59. The bottommost cell is A35. So far so good, and it looks as if using a list is as easy as using an array. As an example, the maximum sum contiguous subsequence of 0, -1, 2, -1, 3, -1, 0 would be 4 (= 2 + -1 + 3). 12) instead of the number entered into the array. In order to find the sum of all elements in an array, we can simply iterate the array and add each element to a sum accumulating variable. Hence there would be four different arrays in this case. Find maximum sum path involving elements of given arrays Given two sorted array of integers, find a maximum sum path involving elements of both arrays whose sum is maximum. HI everyone, need help with this exercise: "We have two integer numbers, which are represented by two arrays. For all possible combinations, find the sum and compare it with the previous sum and update the maximum sum. Sort both arrays array A and array B. 404 24 Add to List Share. This is way faster than a manually using a for loop going through all elements in a 1d-array. Monotonic Queue/Stack. Also add the common element to the result. Passing array elements to a function is similar to passing variables to a function. If the current element of array 1 and array 2are same, then take the maximum of sum1 and sum2 and add it to the result. We are making max_sum_subarray is a function to calculate the maximum sum of the subarray in an array. This program will help to understand the working of for loop, array, if statement and random numbers. C program to find the maximum or the largest element and the location (index) at which it's present in an array. Then print the respective minimum and maximum values as a single line of two space-separated long integers. Which is run a loop from 0 to n. If arr1[] = {1, 2, 4, 3, 2} and. If only ARRAY is specified, the result equals the sum of all the array elements of ARRAY. =MAX(IF((List>=LLim)*(List<=ULim),List,FALSE)) returns the maximum of values between 2 and 5, or 5. A better solution would be to find the two largest elements in the array, since adding those obviously gives the largest sum. You can also declare an array of arrays (also known as a multidimensional array) by using two or more sets of brackets, such as String[][] names. To initialize array use random numbers. Input the array elements. Introduction to C Programming Arrays Overview. array_sum: Array Sum of. hence maximum subarray sum is 9. To display sub array with maximum sum you should write code to hold the start and end value of the sub array with maximum sum. if 2,3,4, 5 is the given array, {4,5,2,3} is also a possible array like other two. C++ :: Creating Table Of Arrays - Find Maximum Value And Sum Aug 12, 2014. First Bad Version. Even to find the total number of even elements in the array. This program is an example of Dynamic Memory Allocation, here we are declaring memory for N array elements at run time using malloc() - which is used to declare memory for N. When you want to return a sum for a single criteria (for example, a single IF condition) When you want to use multiple criteria and return the sum to multiple cells; The criteria that you can use with the SUMIF() worksheet function is limited to text, numbers, or a range, and the function cannot use array constants. Add solutions to C++ domain. Note that the common elements do not have to be at same indexes. 7 is the sum of 2 + 5 in column 2, and so forth. Java program to calculate the sum of N numbers using arrays, recursion, static method, using while loop. (#M40034130) C Programming question Find out maximim sum of sub Array Keep an EYE Find out maximim sum of sub Array example array={2,3,-1,4,9} maximum sum of sub array=17 Asked In C DHIRENDRA (6 years ago) Unsolved Read Solution (3) Is this Puzzle helpful? (1) (0) Submit Your Solution Program. The code below will show how to display a maximum of 3 items from an integer array. sum = sum + (value at 2000) In the Second iteration we will have following calculation – sum = sum + (value at 2002) = 11 + 12 = 23. We can switch from one array to another array only at common elements. Naive solution would be to consider every pair of elements and calculate their product. If the first and only parameter is an array, max() returns the highest value in that array. After partitioning, each subarray has their values changed to become the maximum value of that subarray. This function can take two other optional arguments, that will be covered in more detail, when we get to multi-dimensional arrays. It is For Each Loop or enhanced for loop introduced in java 1. Here is a simple example. minimum difference = second lowest - lowest. sum () is shown below. The master will loop from 2 to the maximum value on issue MPI_Recv and wait for a message from any slave (MPI_ANY_SOURCE), if the message is zero, the process is just starting, if the message is negative, it is a non-prime, if the message is positive, it is a prime. Here we will be displaying the sub array. Is there. Given an array, find maximum sum of smallest and second smallest elements chosen from all possible sub-arrays. A better solution would be to find the two largest elements in the array, since adding those obviously gives the largest sum. For example, entering =SUM(10, 2) returns 12. 1- creat two int array the size of each array must be 40 2- ask the user to input 2 strings (big numbers ) 3- check each character of ths string if it's numeric characters 4- add each string in one int array 5-print out the sum of the 2 arrays 6-compare between the two arrays. the contiguous subarray [4,-1,2,1] has the largest sum = 6. c++: which functions gives the sum of an array? (3). if the array was [5, 6, 4, 2, 9] it would return true. Bilal-March 5th, 2020 at 2:07 pm none Comment author #29091 on Find max value & its index in Numpy Array | numpy. So for the test arrays: int[] testArrayA = { 7, 1, 4, 5, 1}; int[] testArrayB = { 3, 2, 1, 5, 3}; ↑ starting i. Given two sorted arrays such the arrays may have some common elements. Adjacent: side by side. You need to create three different functions called minNumber,maxNumber and totalSum. 4 Two-dimensional Arrays. three two one one three two two three one three one two Array Reduction Methods : Array reduction methods can be applied to any unpacked array to reduce the array to a single value. Compute the ceiling power of 2. maximum difference = higest-lowest. Join 124,729,115 Academics and Researchers. While if we add this two via plus (+) operator ,the result will be true. SUM (C, DIM=2) returns the value (6, 15), which is the sum of all elements in each row. If it's provided then it will return for array of max values along the axis i. It should return an integer representing the maximum subset sum for the given array. sum %2 [1] 1 1 [2] 2 0 [3] 3 1 [1,2] 3 1 [2,3] 5 1 [1,2,3] 6 0. The syntax of numpy. max (x, n) → array<[same as x]> Returns n largest values of all input values of x. For examples, Enter 1st integer: 8 Enter 2nd integer: 2 Enter 3rd integer: 9 The sum is: 19 The product is: 144 The min is: 2 The max is: 9 Hints. Which is run a loop from 0 to n. Search in Rotated Sorted Array. Sort both arrays array A and array B. The bottommost cell is A35. 15 is the sum of 4 + 5 + 6 in row 2. Here is the code. Given an array A with n elements and array B with m elements. With the following program, you can even print the sum of two numbers or three numbers up to N numbers. Given a circular array C of integers represented by A, find the maximum possible sum of a non-empty subarray of C. You need to find out the maximum sum such that no two chosen numbers are adjacent , vertically, diagonally (or) horizontally. The maximum product is formed by the (-10, -3) or (5, 6) pair. Calculates the weighted sum of two arrays. You can enter a value in the box labelled 'Person ID', which is the first number of a two-dimensional array, and then select one of two numbers in the Name/Profession box. =MAX(ROW(4:6)*SUM(ROW(1:3))) confirm with CTRL+SHIFT+ENTER. How to swap two numbers without using temporary variable? Write a program to print fibonacci series. In the given array, you need to find maximum sum of elements such that no two are adjacent (consecutive). To display sub array with maximum sum you should write code to hold the start and end value of the sub array with maximum sum. You can also return an array from a method. everything works except I need it to give me the max and min month (i. if 2,3,4, 5 is the given array, {4,5,2,3} is also a possible array like other two. Write a program to sort a map by value. Though, the arrays whose size is a product of 2’s, 3’s, and 5’s (for example, 300 = 5*5*3*2*2) are also processed quite efficiently. We can switch from one array to another array only at common elements. Function Description. Arr2[] = {1, 3, 3, 12, 2} then maximum result is obtained when we create following two subarrays − Subarr1[] = {2, 4, 3} and. You can refer to more than one array in a single SAS statement. Write a program to find the sum of the corresponding elements in 2 arrays. Maximum Sum of Two Non-Overlapping Subarrays. Inside SUM, the range resolves to an array of values. The restriction is that once you select a particular row,column value to add to your sum, no other values from that row or column may be used in calculating the sum. Please note that your values will almost certainly be different, depending both on the random number generator and the values of the array you created in Lab8. Finding the Average value of an Array. Mini-Max Sum Hackerrank. It could be the sum of total sum of the left child and max prefix sum of the right child. Find ways to calculate a target from elements of specified. Maximum Sub Array Practice: max_sub_array. Note that in the calculation of max4, we have passed a two dimensional array containing two rows of three elements as if it were a single dimensional array of six elements. 17171281366e-06 0. The first line of the input contains N,where N is the number of integers. Given two equally sized arrays (A, B) and N (size of both arrays). It should return a long integer that represents the maximum value of. We can switch from one array to another array only at common elements. The bottommost cell is A35. You can pass a two dimensional array to a method just as you pass a one dimensional array. Add solution to Pairs problem. Now let's think of another type of two dimensional array in which the shape of the array is not square as shown below. More formally, if we write all (nC2) sub-arrays of array of size >=2 and find the sum of smallest and second smallest, then our answer will be maximum sum among them. Edit: given your comments if the initial array is fixed then you can use MMULT function like this. sum %2 [1] 1 1 [2] 2 0 [3] 3 1 [1,2] 3 1 [2,3] 5 1 [1,2,3] 6 0. Array Maximum Minimum value We can calculate maximum value among the elements of an array by using max function. ; The array formula lets the IF function test for multiple conditions in a single cell, and, when the data meets a condition, the array formula determines what data (event results) the MAX function will examine to find the best result. For example, if A is a matrix, then max(A,[],[1 2]) computes the maximum over all elements in A, since every element of a matrix is contained in the array slice defined by dimensions 1 and 2. Problem Description We have to write a program in C such that the program will allocate 2 one-dimensional arrays using malloc() call and then will do the addition and stores the result into 3rd array. We take a two dimensional array L of size count+1, sum/2+1. For example, given the array [-2,1,-3,4,-1,2,1,-5,4], the contiguous subarray [4,-1,2,1] has the largest sum = 6. min(int a, int b) and Math. Here we will be displaying the sub array. min () find the maximum and minimum value of the arguments, respectively. word 0:5 y:. The last 'n' integers correspond to the elements in the second array. pranathi chunduru. Hence there would be four different arrays in this case. For example, consider the array {-10, -3, 5, 6, -2}. Given an integer array of N elements, find the maximum sum contiguous subarray (containing at least one element). # sum of all elements in the array >np. Java represents a two-dimensional array as an array of arrays. For example, A = [−2, 1, −3, 4, −1, 2, 1, −5, 4] then max sum=11 with the subarray [1, 4, 2, 4]. This example prints the maximum value in an array, and the subscript of that value:; Create a simple two-dimensional array: D = DIST (100); Print the maximum value in array D and its linear subscript: PRINT, 'Maximum value in array D is:', MAX (D, I) PRINT, 'The subscript of the maximum value is', I IDL Output Maximum value in array D is: 70. min, max, repmat, meshgrid, sum, cumsum, diff, prod, cumprod, filter 3. This program will help to understand the working of for loop, array, if statement and random numbers. Academia is the easiest way to share papers with millions of people across the world for free. I haven't gotten that far yet, I'm stuck just trying to print my two arrays, every time i try to print the first array it gives me the elements of the second array and it. Basic Operations ¶. com Great, I love this explanation. Reductions. The syntax of numpy. In this article we'll explore four plug and play functions that allow you to easily find certain values in an arrays of numbers. Improvement over Method-1 – O(n 2) Time. Output Format. I don't care if a document has many aliases matching my query. An index value of a Java two dimensional array starts at 0 and ends at n-1 where n is the size of a row or column. After partitioning, each subarray has their values changed to become the maximum value of that subarray. Two approaches: Simple approach is brute-force implementation but it will take O(n. Maximize array sum by concatenating corresponding elements of given two arrays Given two array A[] and B[] of the same length, the task is to find the maximum array sum that can be formed by joining the corresponding elements of the array in any order. Display the maximum K valid sum combinations from all the possible sum combinations. Timing Belt - $1,299. The code below will show how to display a maximum of 3 items from an integer array. Max sum in an array. Find the sum of the maximum sum path to reach from beginning of any array to end of any of the two arrays. As a precautionary health measure for our support specialists in light of COVID-19, we're operating with a limited team. Finding the maximum sum in two sorted arrays Given two sorted postive integer arrays A(n) and B(n) (let's say they are decreasingly sorted), we define a set S = {(a,b) | a \in A and b \in B}. Mini-Max Sum - Problem from HackerRank and Solution Using python 2 Problem Statement: Given five positive integers, find the minimum and maximum values that can be calculated by summing exactly four of the five integers. The program allocates 2 one-dimentional arrays using malloc() call and then does the addition and stores the result into 3rd array. Method since it requires contiguous, it means that for each element, it has two situations that are in the subarray or not. Join 124,729,115 Academics and Researchers. You are given an array of integers with both positive and negative numbers. Subarr2[] = {3, 3, 12. Find the sum of numbers and represent it in array. The maximum special sum considering all non-empty subarrays of the array. In the Java programming language, a multidimensional array is an array whose components are themselves arrays. For example,$ \{35, 42, 5, 15, 27, 29\} $is a sorted array that has been circularly shifted$ k=2 $positions, while$ \{27, 29, 35, 42, 5, 15\} $has been shifted$ k=4 \$ positions. If only one array is supplied, SUMPRODUCT will simply sum the items in the array. C++ code to display contents of array. Find a Triplet having Maximum Product in an Array. Read 6 Integers from File, find sum, find average, and find Min/Max average. Write a program to find sum of each digit in the given number using recursion. Finding the Maximum value in an Array. Hi all, I'm looking for how to to the sum or union of two jagged array, I did it to simple array but do you have any idea how to do it for jagged array. For example, if an int. C Program to Find Maximum Element in Array - This program find maximum or largest element present in an array. We could also use other representations, such as an array containing two two-element arrays ([[76, 9], [4, 1]]) or an object with property names like "11" and "01", but the flat array is simple and makes the expressions that access the table pleasantly short. out [Optional] Alternate output array in which to place the. Sub Array with Maximum Sum – Kadane Algorithm is the best solution. Find the sum of the maximum sum path to reach from beginning of any array to end of any of the two arrays. In a two-dimensional Java array, we can use the code a[i] to refer to the ith row (which is a one-dimensional array). We can use an array as a deque with the following operations:. computes the sum of two matrices and then prints it. Search in Rotated Sorted Array II. As a precautionary health measure for our support specialists in light of COVID-19, we're operating with a limited team.

https://math.stackexchange.com/questions/2921700/how-to-simplify-or-upperbound-this-summation

# How to simplify or upperbound this summation? I am not a mathematician, so sorry for this trivial question. Is there a way to simplify or to upperbound the following summation: $$\sum_{i=1}^n{\exp{\left(-\frac{i^2}{\sigma^2}\right)}}.$$ Can I use geometric series? EDIT: I have difficulty because of the power $2$, i.e if the summation would be $\sum\limits_{i=1}^n{\exp{\left(-\frac{i}{\sigma^2}\right)}}$ then it would be easy to apply geometric series! • Don't be sorry for the question! The only way to get better as mathematician is to ask questions. (See also the first annotation to this post) – Jacob Manaker Sep 18 at 16:18 • Seems that you could try the integral $\int_0^\infty \exp(-x^2)\,\mathrm dx$ to bound that. – xbh Sep 18 at 16:26 • tinyurl.com/y76n9loh (wolframalpha) gives a closed form for the infinite sum involving the elliptic theta function. – barrycarter Sep 19 at 23:46 Alternative: Since $f(x) = \exp(-x^2/\sigma^2) \searrow 0$, we can write $\DeclareMathOperator{\diff}{\,d\!}$ \begin{align*} &\sum_1^n \exp\left(-\frac {j^2}{\sigma^2}\right) \\ &= \sum_1^n \int_{j-1}^j \exp\left(-\frac {j^2}{\sigma^2}\right)\diff x \\ &\leqslant \sum_1^n \int_{j-1}^j \exp\left(-\frac {x^2}{\sigma^2}\right) \diff x \\ &=\sigma \int_0^n \exp\left(-\frac {x^2}{\sigma^2}\right) \diff \left(\frac x \sigma \right)\\ &= \sigma \int_0^{n/\sigma} \exp(-x^2)\diff x\\ &\leqslant \sigma \int_0^{+\infty}\exp(-x^2)\diff x\\ &= \frac \sigma 2 \sqrt \pi \end{align*} • Nicely done! Sometimes simplest is best. – Jacob Manaker Sep 18 at 17:54 • brilliant answer, thanks – user8003788 Sep 19 at 7:39 • @JacobManaker Thanks for compliment! – xbh Sep 19 at 7:43 TL;DR: three relatively easy bounds are the numbered equations below. You cannot directly apply the formula for the geometric series for the reason mentioned in your edit. But note that $i\geq1$, so we have $$\sum_{i=1}^n{\exp{\left(-\frac{i^2}{\sigma^2}\right)}}\leq\sum_{i=1}^n{\exp{\left(-\frac{i\cdot1}{\sigma^2}\right)}}$$ The latter, of course, is a geometric sum. Taking the sum over all $i$ (including $i=0$), we get $$(1-e^{-\sigma^{-2}})^{-1} \tag{1} \label{eqn:first}$$ The calculation for finitely many terms isn't much harder, and only differs by an exponentially decreasing factor. If this isn't a strong enough bound, there are other techniques. If $n<\sigma$, then we can get very far elementarily. Note that $e^x\geq x+1$; dividing each side, we get $$e^{-x}\leq(1+x)^{-1}=\sum_{k=0}^{\infty}{(-x)^k}$$ if $|x|<1$. Taking $x=\left(\frac{i}{\sigma}\right)^2$, we thus obtain \begin{align*} \sum_{i=1}^n{e^{-\frac{i^2}{\sigma^2}}}&\leq\sum_{i=1}^n{\sum_{k=0}^{\infty}{\left(-\left(\frac{i}{\sigma}\right)^2\right)^k}} \\ &=\sum_{k=0}^{\infty}{(-1)^k\sum_{i=1}^n{\left(\frac{i}{\sigma}\right)^{2k}}} \tag{*} \label{eqn:star} \end{align*} (We can interchange sums because one is finite.) Now, for all $k$, the function $\left(\frac{\cdot}{\sigma}\right)^{2k}$ is increasing on $[0,\infty)$; we thus have $$\int_0^n{\left(\frac{i}{\sigma}\right)^{2k}\,di}\leq\sum_{i=1}^n{\left(\frac{i}{\sigma}\right)^{2k}}\leq\left(\frac{n}{\sigma}\right)^{2k}+\int_1^n{\left(\frac{i}{\sigma}\right)^{2k}\,di}$$ Evaluating the integrals and simplifying, we have $$0\leq\sum_{i=1}^n{\left(\frac{i}{\sigma}\right)^{2k}}-\frac{n}{2k+1}\left(\frac{n}{\sigma}\right)^{2k}\leq\left(\frac{n}{\sigma}\right)^{2k}\left(1-\frac{1}{(2k+1)n^{2k}}\right)$$ Substituting into $\eqref{eqn:star}$, we get \begin{align*} \sum_{i=1}^n{e^{-\frac{i^2}{\sigma^2}}}&\leq\sum_{k=0}^{\infty}{\frac{(-1)^kn}{2k+1}\left(\frac{n}{\sigma}\right)^{2k}}-\sum_{j=0}^{\infty}{\left(\frac{n}{\sigma}\right)^{4j+2}\left(1-\frac{1}{(4j+3)n^{4j+2}}\right)} \\ &\leq\sum_{k=0}^{\infty}{\frac{(-1)^kn}{2k+1}\left(\frac{n}{\sigma}\right)^{2k}}-\sum_{j=0}^{\infty}{\left(\frac{n}{\sigma}\right)^{4j+2}} \\ &=\sigma\tan^{-1}{\left(\frac{n}{\sigma}\right)}-\frac{\left(\frac{n}{\sigma}\right)^2}{1-\left(\frac{n}{\sigma}\right)^4}\hspace{4em}(n<\sigma) \tag{2} \end{align*} Finally, for the general case we can achieve a slight improvement on $\eqref{eqn:first}$ via the theory of majorization. $\{x_i\}_{i=1}^n\mapsto\sum_{i=1}^n{\exp{\left(-\frac{x_i}{\sigma^2}\right)}}$ is convex and symmetric in its arguments, hence Schur-convex. Let $b_i=i^2$ and $a_i=\left(\frac{2n-1}{3}\right)i$. Clearly, for all $m\leq n$, we have $$\sum_{i=1}^m{a_i}=\frac{m(m-1)}{2}\cdot\frac{2n-1}{3}\geq\frac{m(m-1)(2m-1)}{6}=\sum_{i=1}^m{b_i}$$ with equality if $m=n$. Thus $\vec{a}$ majorizes $\vec{b}$, so \begin{align*} \sum_{i=1}^n{\exp{\left(-\frac{i^2}{\sigma^2}\right)}}&=\sum_{i=1}^n{\exp{\left(-\frac{b_i}{\sigma^2}\right)}} \\ &\leq\sum_{i=1}^n{\exp{\left(-\frac{a_i}{\sigma^2}\right)}} \\ &=\sum_{i=1}^n{\exp{\left(-\frac{(2n-1)i}{3\sigma^2}\right)}} \\ &\leq\sum_{i=0}^{\infty}{\exp{\left(-\frac{(2n-1)i}{3\sigma^2}\right)}} \\ &\leq\left(1-\exp{\left(\frac{2n-1}{3\sigma^2}\right)}\right)^{-1} \tag{3} \end{align*} • Great detailed work..I considered previously the first answer but I thought it would be better if I can get a stronger bound.Thanks a lot – user8003788 Sep 19 at 7:42 There's a rather trivial upper bound that $\frac{-i^2}{\sigma^2}$ is negative, so exponentiating it results in a number less than 1, so the sum is at most $n$. If you want a constant upper bound, you can upper bound it with the geometric series. The matter is that $e^{-(x/ \sigma)^2}$, in the range $0 \le x < \approx \sigma$ is very steep. So, unless $\sigma$ is quite high, you cannot get a good approximation by the integral. But of course everything depends on the parameters into play and on the accuracy required. Hint : For general values of $n$ and $\sigma$ it might be interesting to take advantage of the fact that the Fourier Transform of a Gaussian is a Gaussian itself. Then you are taking the signal $e^{-\, (t/ \sigma)^2}$, windowing it between $0 \le t \le n/ \sigma$, taking $n$ samples of it, and after that you are taking $n$ times the average. All these operations have a simple translation into the frequency domain. However I do not go further not knowing whether you are acknowledged in this field, and keep this as a hint. Also, might be interesting this identity $$\sum\limits_{k \in \mathbb Z} {\exp \left( { - \pi \left( {k/c} \right)^2 } \right)} = c\sum\limits_{k \in \mathbb Z} {\exp \left( { - \pi \left( {k\,c} \right)^2 } \right)}$$ reported at the end of the Properties paragraph in this wikipedia article.

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

# Tag Archives: polynomial ## 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. ## Number Bases and Polynomials About a month ago, I was working with our 5th grade math teacher to develop some extension activities for some students in an unleveled class.  The class was exploring place value, and I suggested that some might be ready to explore what happens when you allow the number base to be something other than 10.  A few students had some fun learning to use their basic four algorithms in other number bases, but I made an even deeper connection. When writing something like 512 in expanded form ($5\cdot 10^2+1\cdot 10^1+2\cdot 10^0$), I realized that if the 10 was an x, I’d have a polynomial.  I’d recognized this before, but this time I wondered what would happen if I applied basic math algorithms to polynomials if I wrote them in a condensed numerical form, not their standard expanded form.  That is, could I do basic algebra on $5x^2+x+2$ if I thought of it as $512_x$–a base-x “number”?  (To avoid other confusion later, I read this as “five one two base-x“.) Following are some examples I played with to convince myself how my new notation would work.  I’m not convinced that this will ever lead to anything, but following my “what ifs” all the way to infinite series was a blast.  Read on! If I wanted to add $(3x+5)$$(2x^2+4x+1)$, I could think of it as $35_x+241_x$ and add the numbers “normally” to get $276_x$ or $2x^2+7x+6$.  Notice that each power of x identifies a “place value” for its characteristic coefficient. If I wanted to add $3x-7$ to itself, I had to adapt my notation a touch.  The “units digit” is a negative number, but since the number base, x, is unknown (or variable), I ended up saying $3x-7=3(-7)_x$.  The parentheses are used to contain multiple characters into a single place value.  Then, $(3x-7)+(3x-7)$ becomes $3(-7)_x+3(-7)_x=6(-14)_x$ or $6x-14$.  Notice the expanding parentheses containing the base-x units digit. The last example also showed me that simple multiplication would work.  Adding $3x-7$ to itself is equivalent to multiplying $2\cdot (3x-7)$.  In base-x, that is $2\cdot 3(-7)_x$.  That’s easy!  Arguably, this might be even easier that doubling a number when the number base is known.  Without interactions between the coefficients of different place values, just double each digit to get $6(-14)_x=6x-14$, as before. What about $(x^2+7)+(8x-9)$?  That’s equivalent to $107_x+8(-9)_x$.  While simple, I’ll solve this one by stacking. and this is $x^2+8x-2$.  As with base-10 numbers, the use of 0 is needed to hold place values exactly as I needed a 0 to hold the $x^1$ place for $x^2+7$. Again, this could easily be accomplished without the number base conversion, but how much more can we push these boundaries? Level 3–Multiplication & Powers: Compute $(8x-3)^2$.  Stacking again and using a modification of the multiply-and-carry algorithm I learned in grade school, I got and this is equivalent to $64x^2-48x+9$. All other forms of polynomial multiplication work just fine, too. From one perspective, all of this shifting to a variable number base could be seen as completely unnecessary.  We already have acceptably working algorithms for addition, subtraction, and multiplication.  But then, I really like how this approach completes the connection between numerical and polynomial arithmetic.  The rules of math don’t change just because you introduce variables.  For some, I’m convinced this might make a big difference in understanding. I also like how easily this extends polynomial by polynomial multiplication far beyond the bland monomial and binomial products that proliferate in virtually all modern textbooks.  Also banished here is any need at all for banal FOIL techniques. Level 4–Division: What about $x^2+x-6$ divided by $x+3$? In base-x, that’s $11(-6)_x \div 13_x$. Remembering that there is no place value carrying possible, I had to be a little careful when setting up my computation. Focusing only on the lead digits, 1 “goes into” 1 one time.  Multiplying the partial quotient by the divisor, writing the result below and subtracting gives Then, 1 “goes into” -2 negative two times.  Multiplying and subtracting gives a remainder of 0. thereby confirming that $x+3$ is a factor of $x^2+x-6$, and the other factor is the quotient, $x-2$. Perhaps this could be used as an alternative to other polynomial division algorithms.  It is somewhat similar to the synthetic division technique, without its  significant limitations:  It is not limited to linear divisors with lead coefficients of one. For $(4x^3-5x^2+7) \div (2x^2-1)$, think $4(-5)07_x \div 20(-1)_x$.  Stacking and dividing gives So $\displaystyle \frac{4x^3-5x^2+7}{2x^2-1}=2x-2.5+\frac{2x+4.5}{2x^2-1}$. CONCLUSION From all I’ve been able to tell, converting polynomials to their base-x number equivalents enables you to perform all of the same arithmetic computations.  For division in particular, it seems this method might even be a bit easier. In my next post, I push the exploration of these base-x numbers into infinite series. ## Extending graph control This article takes my idea from yesterday’s post about using $g(x)=\sqrt \frac{\left | x \right |}{x}$ to control the appearance of a graph and extends it in two ways. • Part I below uses Desmos to graph $y=(x+2)^3x^2(x-1)$ from the left and right simultaneously • Part II was inspired by my Twitter colleague John Burk who asked if this control could be extended in a different direction. Part I: Simultaneous Control When graphing polynomials like $y=(x+2)^3x^2(x-1)$, I encourage my students to use both its local behavior (cubic root at $x=-2$, quadratic root at $x=0$, and linear root at $x=1$) and its end behavior (6th degree polynomial with a positive lead coefficient means $y\rightarrow +\infty$ as $x\rightarrow\pm\infty$). To start graphing, I suggest students plot points on the x-intercepts and then sketch arrows to indicate the end behavior.  In the past, this was something we did on paper, but couldn’t get technology to replicate it live–until this idea. In class last week, I used a minor extension of yesterday’s idea to control a graph’s appearance from the left and right simultaneously.  Yesterday’s post suggested  multiplying  by $\sqrt \frac{\left | a-x \right |}{a-x}$ to show the graph of a function from the left for $x.  Creating a second graph multiplied by $\sqrt \frac{\left | x-b \right |}{x-b}$ gives a graph of your function from the right for $b.  The following images show the polynomial’s graph developing in a few stages.  You can access the Desmos file here. First graph the end behavior (pull the a and b sliders in a bit to see just the ends of the graph) and plot points at the x-intercepts. From here, you could graph left-to-right or right-to-left.  I’ll come in from the right to show the new right side controller. The root at $x=1$ is linear, so decreasing the b slider to just below 1 shows this. Continuing from the right, the next root is a bounce at $x=0$, as shown by decreasing the b slider below 0.  Notice that this forces a relative minimum for some $0. Just because it’s possible, I’ll now show the cubic intercept at $x=2$ by increasing the a slider above 2. All that remains is to connect the two sides of the graph, creating one more relative minimum in $-2. The same level of presentation control can be had for any function’s graph. Part II: Vertical Control I hadn’t thought to extend this any further until my colleague asked if a graph could be controlled up and down instead of left and right.  My guess is that the idea hadn’t occurred to me because I typically think about controlling a function through its domain.  Even so, a couple minor adjustments accomplished it.  Click here to see a vertical control of the graph of $y=x^3$ from above and below. Enjoy. ## Quadratics, Statistics, Symmetry, and Tranformations A problem I assigned my precalculus class this past Thursday ended up with multiple solutions by the time we finished.  Huzzah for student creativity! The question: Find equations for all polynomial functions, $y=f(x)$, of degree $\le 2$ for which $f(0)=f(1)=2$ and $f(3)=0$. After they had worked on this (along with several variations on the theme), four very different ways of thinking about this problem emerged.  All were valid and even led to a lesson I hadn’t planned–proving that, even though they looked different algebraically, all were equivalent.  I present their approaches (and a few extras) in the order they were offered in our post-solving debriefing. The commonality among the approaches was their recognition that 3 non-collinear points uniquely define a vertical parabola, so they didn’t need to worry about polynomials of degree 0 or 1.  (They haven’t yet heard about rotated curves that led to my earlier post on rotated quadratics.) Solution 1–Regression:  Because only 3 points were given, a quadratic regression would derive a perfectly fitting quadratic equation.  Using their TI-Nspire CASs, they started by entering the 3 ordered pairs in a Lists&Spreadsheets window.  Most then went to a Calculator window to compute a quadratic regression.  Below, I show the same approach using a Data&Statistics window instead so I could see simultaneously the curve fit and the given points. The decimals were easy enough to interpret, so even though they were presented in decimal form, these students reported $y=-\frac{1}{3}x^2+\frac{1}{3}x+2$. For a couple seconds after this was presented, I honestly felt a little cheated.  I was hoping they would tap the geometric or algebraic properties of quadratics to get their equations.  But I then I remembered that I clearly hadn’t make that part of my instructions.  After my initial knee-jerk reaction, I realized this group of students had actually done exactly what I explicitly have been encouraging them to do: think freely and take advantage of every tool they have to find solutions.  Nothing in the problem statement suggested technology or regressions, so while I had intended a more geometric approach, I realized I actually owed these students some kudos for a very creative, insightful, and technology-based solution.  This and Solution 2 were the most frequently chosen approaches. Solution 2–Systems:  Equations of quadratic functions are typically presented in standard, factored, or vertex form.  Since neither two zeros nor the vertex were explicitly given, the largest portion of the students used the standard form, $y=a\cdot x^2+b\cdot x+c$ to create a 3×3 system of equations.  Some solved this by hand, but most invoked a CAS solution.  Notice the elegance of the solve command they used, working from the generic polynomial equation that kept them from having to write all three equations, keeping their focus on the form of the equation they sought. This created the same result as Solution 1, $y=-\frac{1}{3}x^2+\frac{1}{3}x+2$. CAS Aside: No students offered these next two solutions, but I believe when using a CAS, it is important for users to remember that the machine typically does not care what output form you want.  The standard form is the only “algebraically simple” approach when setting up a solution by hand, but the availability of technology makes solving for any form equally accessible. The next screen shows that the vertex and factored forms are just as easily derived as the standard form my students found in Solution 2. I was surprised when the last line’s output wasn’t in vertex form, $y=-\frac{1}{3}\cdot \left ( x-\frac{1}{2} \right )^2+\frac{25}{12}$, but the coefficients in its expanded form clearly show the equivalence between this form and the standard forms derived in Solutions 1 and 2–a valuable connection. Solution 3–Symmetry:  Two students said they noticed that $f(0)=f(1)=2$ guaranteed the vertex of the parabola occurred at $x=\frac{1}{2}$.  Because $f(3)=0$ defined one real root of the unknown quadratic, the parabola’s symmetry guaranteed another at $x=-2$, giving potential equation $y=a\cdot (x-3)(x+2)$.  They substituted the given (0,2) to solve for a, giving final equation $y=-\frac{1}{3}\cdot (x-3)(x+2)$ as confirmed by the CAS approach above. Solution 4–Transformations:  One of the big lessons I repeat in every class I teach is this: If you don’t like how a question is posed.  Change it! Notice that two of the given points have the same y-coordinate.  If that y-coordinate had been 0 (instead of its given value, 2), a factored form would be simple.  Well, why not force them to be x-intercepts by translating all of the given points down 2 units? The transformed data show x-intercepts at 0 and 1 with another ordered pair at $(3,-2)$.  From here, the factored form is easy:  $y=a\cdot (x-0)(x-1)$.  Substituting $(3,-2)$ gives $a=-\frac{1}{3}$ and the final equation is $y=-\frac{1}{3}\cdot (x-0)(x-1)$ . Of course, this is an equation for the transformed points.  Sliding the result back up two units, $y=-\frac{1}{3}\cdot (x-0)(x-1)+2$, gives an equation for the given points.  Aside from its lead coefficient, this last equation looked very different from the other forms, but some quick expansion proved its equivalence. Conclusion:  It would have been nice if someone had used the symmetry noted in Solution 3 to attempt a vertex-form answer via systems.  Given the vertex at $x=\frac{1}{2}$ with an unknown y-coordinate, a potential equation is $y=a\cdot \left ( x-\frac{1}{2} \right )^2+k$.  Substituting $(3,0)$ and either $(0,2)\text{ or }(1,2)$ creates a 2×2 system of linear equations, $\left\{\begin{matrix} 0=a\cdot \left ( 3-\frac{1}{2} \right )^2+k \\ 2=a\cdot \left ( 0-\frac{1}{2} \right )^2+k \end{matrix}\right.$.  From there, a by-hand or CAS solution would have been equally acceptable to me. That the few alternative approaches I offered above weren’t used didn’t matter in the end.  My students were creative, followed their own instincts to find solutions that aligned with their thinking, and clearly appreciated the alternative ways their classmates used to find answers.  Creativity and individual expression reigned, while everyone broadened their understanding that there’s not just one way to do math. It was a good day. ## Cubics and CAS Here’s a question I posed to one of my precalculus classes for homework at the end of last week along with three solutions we developed. Suppose the graph of a cubic function has an inflection point at (1,3) and passes through (0,-4). 1. Name one other point that MUST be on the curve, and 2. give TWO different cubic equations that would pass through the three points. SOLUTION ALERT!  Don’t read any further if you want to solve this problem for yourself. The first question relies on the fact that every cubic function has 180 degree rotational symmetry about its inflection point.  This is equivalent to saying that the graph of a cubic function is its own image when the function’s graph is reflected through its inflection point. That means the third point is the image of (0,-4) when point-reflected through the inflection point (1,3), which is the point (2,10) as shown graphically below. From here, my students came up with 2 different solutions to the second question and upon prodding, we created a third. SOLUTION 1:  Virtually every student assumed $y=a\cdot x^3$ was the parent function of a cubic with unknown leading coefficient whose “center” (inflection point) had been slid to (1,3).  Plugging in the given (0,-4) to $(y-3)=a\cdot (x-1)^3$ gives $a=7$.  Here’s their graph. SOLUTION 2:  Many students had difficulty coming up with another equation.  A few could sketch in other cubic graphs (curiously, all had positive lead coefficients) that contained the 3 points above, but didn’t know how to find equations.  That’s when Sara pointed out that if the generic expanded form of a cubic was $a\cdot x^3+b\cdot x^2 +c\cdot x+d$ , then any 4 ordered pairs with unique x-coordinates should define a unique cubic.  That is, if we picked any 4th point with x not 0, 1, or 2, then we should get an equation.  That this would create a 4×4 system of equations didn’t bother her at all.  She knew in theory how to solve such a thing, but she was thinking on a much higher plane.  Her CAS technology expeditiously did the grunt work, allowing her brain to keep moving. A doubtful classmate called out, “OK.  Make it go through (100,100).”  Following is a CAS screen roughly duplicating Sara’s approach and a graph confirming the fit.  The equation was onerous, but with a quick copy-paste, Sara had moved from  idea to computation to ugly equation and graph in just a couple minutes.  The doubter was convinced and the “wow”s from throughout the room conveyed the respect for the power of a properly wielded CAS. In particular, notice how the TI-Nspire CAS syntax in lines 1 and 3 keep the user’s focus on the type of equation being solved and eliminates the need to actually enter 4 separate equations.  It doesn’t always work, but it’s a particularly lovely piece of scaffolding when it does. SOLUTION 3:  One of my goals in Algebra II and Precalculus courses is to teach my students that they don’t need to always accept problems as stated.  Sometimes they can change initial conditions to create a much cleaner work environment so long as they transform their “clean” solution back to the state of the initial conditions. In this case, I asked what would happen if they translated the inflection point using $T_{-1,-3}$ to the origin, making the other given point (-1,-7).  Several immediately called the 3rd point to be (1,7) which “untranslating” — $T_{1,3}(1,7)=(2,10)$ — confirmed to be the earlier finding. For cases where the cubic had another real root at $x=r$, then symmetry immediately made $x=-r$ another root, and a factored form of the equation becomes $y=a\cdot (x)(x-r)(x+r)$ for some value of a.  Plugging in (-1,-7) gives a in terms of r. The last line slid the initially translated equation using $T_{1,3}$ to re-position the previous line according to the initial conditions.  While unasked for, notice how the CAS performed some polynomial division on the right-side expression. I created a GeoGebra document with a slider for the root using the equation from the last line of the CAS image above.  The image below shows one possible position of the retranslated root.  If you want to play with a live version of this, you will need a free copy of GeoGebra to run it, but the file is here. What’s nice here is how the problem became one of simple factors once the inflection point was translated to the origin.  Notice also that the CAS version of the equation concludes with $+7x-4$, the line containing the original three points.  This is nice for two reasons.  The numerator of the rational term is $-7x(x-2)(x-1)$ which zeros out the fraction at x=0, 1, or 2, putting the cubic exactly on the line $y=7x-4$ at those points. The only r-values are in the denominator, so as $r\rightarrow\infty$, the rational term also becomes zero.  Graphically, you can see this happen as the cubic “unrolls” onto $y=7x-4$ as you drag $|x|\rightarrow\infty$.  Essentially, this shows both graphically and algebraically that $y=7x-4$ is the limiting degenerate curve the cubic function approaches as two of its transformed real roots grow without bound. ## Recognizing Patterns I’ve often told my students that the best problem solvers are those who recognize patterns from past problems in new situations.  So, the best way to become a better problem solver is to solve lots of problems, study the ways others have solved the problems you’ve already cracked (or at least attempted), and to keep pushing your boundaries because you never know what parts of what you learn may end up providing unexpected future insights. I lay no claims to being a great problem solver, but I absolutely benefited from problem solving exposure when I encountered @jamestanton‘s latest “Playing with Numbers” puzzler from his May 2012 Cool Math Newsletter.  (Click here to access all of Jim’s newsletters).  (BTW, Jim’s Web page is chock full of amazing videos and insights both on the problem-solving front and for those interested in curriculum discussions.)  Here’s what Jim posed: Write the numbers 1 though 10 on the board. Pick any two numbers, erase them, and replace them with the single number given by their sum plus their product. (So, if you choose to erase the numbers a and b, replace them with a + b + ab .) You now have nine numbers on the board. Do this again: Pick any two numbers, erase them, and replace them with their sum plus product. You now have eight numbers on the board. Do this seven more times until you have a single number on the board. Why do all who play this game end up with the same single number at the end?  What is that final number? SOLUTION ALERT:  Don’t read any further if you want to solve this yourself. I started small and general.  If the first two numbers (a and b) produce $a+b+ab$, then adding c to the original list gives $[(a+b+ab)+c]+ [(a+b+ab)\cdot c]$ which can be rewritten as $a+b+c+ab+ac+bc+abc$.  That’s when the intuition struck.  Notice that the rewritten form is the sum of every individual number in the list, AND every possible pair of those numbers, AND concludes with the product of the three numbers. I’ve seen that pattern before! Given an nth degree polynomial with roots $r_1, r_2, \ldots r_n$, it’s factored form is $(x-r_1)\cdot (x-r_2)\cdot\ldots\cdot (x-r_n)$ which can be expanded and rewritten as $x^n-(r_1+r_2+\ldots+r_n)\cdot x^{n-1}+$ $+(\text{every pair-wise product of } r_1 \ldots r_n)\cdot x^{n-2}-$ $-(\text{every three-way product of } r_1 \ldots r_n)\cdot x^{n-3}+\ldots$ $\ldots \pm (r_1\cdot r_2\cdot \ldots\cdot r_n)$ Where the sign of the final term is positive for even n and negative for odd n.  I saw this in many algebra textbooks when I started teaching over 20 years ago, but haven’t encountered it lately.  Then again, I haven’t been looking for it. Here’s the point … other than the alternating signs, the coefficients of the expanded polynomial are exactly identical to the sums I was getting from Jim’s problem.  That’s when I rewrote my original problem on my CAS to $(x+a)\cdot (x+b)\cdot (x+c)$ and expanded it (see line 1 below).  The coefficients of $x^2$ are the individual numbers, the coefficients of $x$ are all the pair-wise combinations, and the constant term drifting off the end of the line is $a\cdot b\cdot c$.  And I eliminated the $\pm$ sign challenge by individually adding the numbers instead of subtracting them as I had in the polynomial root example that inspired my insight.  Let $x=1$ to clean up and make the pattern more obvious. So, creating a polynomial with the given numbers and substituting $x=1$ will create a number one more than the sum I wanted (note the extra +1 resulting from the coefficient of $x^n$).  Using pi notation, substituting $x=1$, and subtracting 1 gives the answer.  In case you didn’t know, pi notation works exactly the same as sigma notation, but you multiply the terms instead of adding them. The solution–39,916,799–is surprisingly large given the initial problem statement, and while my CAS use confirmed my intuition and effortlessly crunched the numbers, its tendency to multiply numbers whenever encountered has actually hidden something pretty. From the top line of the last image, substituting $x=1$ would have created the product $2\cdot 3\cdot 4\cdot\ldots \cdot 11$ which the penultimate line computed to be 39916800.  But before the product, that number was $11!$, making $11!-1$ a far more revealing version of the solution. MORAL:  Even after you have an answer, take some time to review what has happened to give yourself a chance to learn even more. EXTENSION 1:  Instead of 1 to 10 as the initial numbers, what if the list went from 1 to any positive integer n?  Prove that the final number on the board is $(n+1)!-1$. EXTENSION 2:  While a positive integer sequence starting at 1 (or 0) produces a nice factorial in the answer, this approach can be used with any number list.  For example, follow Jim’s rules with 37, 5, -2, and 7.9.  Use the polynomial approach below for a quick solution of -2030.2.  Confirm using the original rules if you need. EXTENSION 3:  By now it should be obvious that any list of numbers can be used in this problem.  Prove that every list of numbers which includes -1 has the same solution. EXTENSION 4:  Before explaining some lovely extensions of problems like this to generalized commutativity and associativity, Jim’s May 12 Cool Math Newsletter asks what would happen if instead of $a+b+ab$, the rule for combining a and b was $\displaystyle\frac{a\cdot b}{a+b}$.  You can show that with three terms, this would become $\displaystyle\frac{abc}{ab+ac+bc}$, and four terms would give $\displaystyle\frac{abcd}{abc+abd+acd+bcd}$. In other words, this rule would morph n original numbers into a fraction whose numerator is the product of the numbers and whose denominator is the sum of all possible products of any (n-1)-sized groups of those numbers.  For the original integers 1 to 10, I know the numerator and denominator terms are the last two coefficients in a polynomial expansion. The final fraction simplifies, but I think $\displaystyle\frac{3628800}{10628640}$ is slightly more informative. Happy thoughts, problems, solutions, and connections to you all.

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GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video It is currently 22 Feb 2020, 18:16 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 greatest value of y such that 4^y is a factor of 9! ? Author Message TAGS: Hide Tags Math Expert Joined: 02 Sep 2009 Posts: 61396 What is the greatest value of y such that 4^y is a factor of 9! ?  [#permalink] Show Tags 08 Dec 2016, 11:57 00:00 Difficulty: 15% (low) Question Stats: 77% (01:07) correct 23% (01:20) wrong based on 60 sessions HideShow timer Statistics What is the greatest value of y such that 4^y is a factor of 9! ? A. 5 B. 4 C. 3 D. 1 E. 0 _________________ Director Joined: 05 Mar 2015 Posts: 960 What is the greatest value of y such that 4^y is a factor of 9! ?  [#permalink] Show Tags 08 Dec 2016, 19:02 1 Bunuel wrote: What is the greatest value of y such that 4^y is a factor of 9! ? A. 5 B. 4 C. 3 D. 1 E. 0 4^y=2^(2y) no. of 2's in 9! 9/2=4 9/2^2=2 9/2^3=1 total= 4+2+1=7 so as 2y=7 we get y=3 Ans C Manager Joined: 27 Aug 2015 Posts: 86 Re: What is the greatest value of y such that 4^y is a factor of 9! ?  [#permalink] Show Tags 09 Dec 2016, 02:25 1 The formula for such problems is like 9 /4= 2 9/4^2=0 Total = 2 However answer should be 3 if we actually count it. Where am I going wrong? Posted from my mobile device Board of Directors Status: QA & VA Forum Moderator Joined: 11 Jun 2011 Posts: 4841 Location: India GPA: 3.5 Re: What is the greatest value of y such that 4^y is a factor of 9! ?  [#permalink] Show Tags 09 Dec 2016, 10:18 2 Bunuel wrote: What is the greatest value of y such that 4^y is a factor of 9! ? A. 5 B. 4 C. 3 D. 1 E. 0 $$9! = 9*8*7*6*5*4*3*2*1$$ Or, $$9! = 3^2*2^3*7*2*3*5*2^2*3*2*1$$ Or, $$9! = 2^7*3^4*5*7$$ Now, $$2^7 = 4^3*2$$ Thus, we have the greatest value of y = 3 , hence answer will be (C) rakaisraka hope its clear with you ... Further I suggest you go through the concept once again to clear your doubts here math-number-theory-88376.html#p666609 _________________ Thanks and Regards Abhishek.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only ) Director Joined: 05 Mar 2015 Posts: 960 Re: What is the greatest value of y such that 4^y is a factor of 9! ?  [#permalink] Show Tags 09 Dec 2016, 10:35 1 rakaisraka wrote: The formula for such problems is like 9 /4= 2 9/4^2=0 Total = 2 However answer should be 3 if we actually count it. Where am I going wrong? Posted from my mobile device rakaisraka when u r finding 4^y means u have to count every 2's .. suppose if it was 10! then it must have 1*2*...*6...*10 then it has 6=2*3 && 10=2*5 where one no. 2 from 6 and one no. 2 from 10 also counted as a 4 in 10! let me make more clear if u have to find 6^y in X! as 6=2*3 then u have to count every 2 and every 3 in X! and the minimum pair of 2&3 will make the answer hope it is clear Target Test Prep Representative Affiliations: Target Test Prep Joined: 04 Mar 2011 Posts: 2801 Re: What is the greatest value of y such that 4^y is a factor of 9! ?  [#permalink] Show Tags 12 Dec 2016, 17:16 1 Bunuel wrote: What is the greatest value of y such that 4^y is a factor of 9! ? A. 5 B. 4 C. 3 D. 1 E. 0 Since 4 = 2^2, we are actually trying to determine the largest value y such that 2^(2y) is a factor of 9!. Let’s first determine the number of factors of 2 within 9!. To do that, we can use the following shortcut in which we divide 9 by 2, and then divide the quotient of 9/2 by 2 and continue this process until we can no longer get a nonzero integer as the quotient. 9/2 = 4 (we can ignore the remainder) 4/2 = 2 2/2 = 1 Since 1/2 does not produce a nonzero quotient, we can stop. The final step is to add up our quotients; that sum represents the number of factors of 2 within 9!. Thus, there are 4 + 2 + 1 = 7 factors of 2 within 9! However, we are not asked for the number of factors of 2; instead we are asked for the number of factors of 4. We see that 7 factors of 2 will produce 3 factors of 4. _________________ Jeffrey Miller Jeff@TargetTestPrep.com 181 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. Non-Human User Joined: 09 Sep 2013 Posts: 14124 Re: What is the greatest value of y such that 4^y is a factor of 9! ?  [#permalink] Show Tags 17 Jan 2020, 02:54 Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________ Re: What is the greatest value of y such that 4^y is a factor of 9! ?   [#permalink] 17 Jan 2020, 02:54 Display posts from previous: Sort by

https://brilliant.org/discussions/thread/on-recurring-decimals/

# On Recurring decimals $\large{0\red{.}\overline{x_1x_2x_3...x_n}=\dfrac{x_1x_2x_3...x_n}{10^n-1}}$ Proof of the above statement Let $l=0\red{.}\overline{x_1x_2x_3...x_n}$ $10^nl=x_1x_2x_3...x_n\red{.}\overline{x_1x_2x_3...x_n}$ $\Rightarrow 10^nl-l={x_1x_2x_3...x_n}$ $(10^n-1)l=x_1x_2x_3...x_n$ ${l=\dfrac{x_1x_2x_3...x_n}{10^n-1}}$ $\boxed{0\red{.}\overline{x_1x_2x_3...x_n}=\dfrac{x_1x_2x_3...x_n}{10^n-1}}$ $\large{a_1a_2a_3...a_p\red{.}b_1b_2b_3...b_q\overline{x_1x_2x_3...x_n}=\dfrac{1}{10^q}(10^q\times a_1a_2a_3...a_p+b_1b_2b_3...b_q+\dfrac{x_1x_2x_3...x_n}{10^n-1})}$ Proof of the above statement For any number $a_1a_2a_3...a_p\red{.}b_1b_2b_3...b_q\overline{x_1x_2x_3...x_n}$ $a_1a_2a_3...a_p\red{.}b_1b_2b_3...b_q\overline{x_1x_2x_3...x_n}=a_1a_2a_3...a_p+0\red{.}b_1b_2b_3...b_q\overline{x_1x_2x_3...x_n}$ $=\dfrac{1}{10^q}(10^q\times a_1a_2a_3...a_p+b_1b_2b_3...b_q\red{.}\overline{x_1x_2x_3...x_n})$ $=\dfrac{1}{10^q}(10^q\times a_1a_2a_3...a_p+b_1b_2b_3...b_q+0\red{.}\overline{x_1x_2x_3...x_n})$ $=\boxed{\dfrac{1}{10^q}(10^q\times a_1a_2a_3...a_p+b_1b_2b_3...b_q+\dfrac{x_1x_2x_3...x_n}{10^n-1})}$ Note : • $x_1x_2$ act as number with digits $x_1,x_2$ for example if $x_1=5$ and $x_2=8\Rightarrow x_1x_2=58$ dont confuse ($x_1x_2\cancel{=}x_1\times x_2$), same for $x_1x_2x_3$ and $x_1x_2x_3...x_{n-1}x_n$ • $0\red{.}\overline{a}=0\red{.}aaaaa...$ Note by Zakir Husain 6 months, 2 weeks 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: - 6 months, 2 weeks ago Well.. I am impressed. - 6 months, 2 weeks ago ✨ brilliant +1 - 6 months, 2 weeks ago By the way, I have an interesting #Geometry problem! Given points $A,B,C,D$, find the square $\square PQRS$ with A on PQ, B on QR, C on RS, D on SP. I figured out the first part, where we can construct circles with diameters AB, BC, CD, DA respectively, so if a point W is on arc AB, $\angle AWB=90^\circ.$ - 6 months, 2 weeks ago I tried the problem and got an algorithm to construct a rectangle $PQRS$ with points $A,B,C$ and $D$ on sides $PQ,QR,RS,SP$ respectively. Also there will be infinitely many such rectangles for given points $A,B,C,D$ - 6 months, 2 weeks ago What about a square? - 6 months, 2 weeks ago I will try it also! and will inform you as I get any results. - 6 months, 2 weeks ago Let’s start a discussion! That might help :) - 6 months, 2 weeks ago square is also a rectangle.. - 6 months, 2 weeks ago But a rectangle isn’t a square, so I hope to find an algorithm to construct a square (I know it is possible but I don’t know a specific way to do it except for brute-force :P) :) - 6 months, 2 weeks ago

https://zong-music.com/ha2ten/greater-than-or-equal-to-sign-9b504e

# greater than or equal to sign For example, x ≥ -3 is the solution of a certain expression in variable x. Select Symbol and then More Symbols. For example, the symbol is used below to express the less-than-or-equal relationship between two variables: ≥. "Greater than or equal to", as the suggests, means something is either greater than or equal to another thing. is less than > > is greater than ≮ \nless: is not less than ≯ \ngtr: is not greater than ≤ \leq: is less than or equal to ≥ \geq: is greater than or equal to ⩽ \leqslant: is less than or equal to ⩾ 923 Views. Use the appropriate math symbol to indicate "greater than", "less than" or "equal to" for each of the following: a. Greater than or equal application to numbers: Syntax of Greater than or Equal is A>=B, where A and B are numeric or Text values. With Microsoft Word, inserting a greater than or equal to sign into your Word document can be as simple as pressing the Equal keyboard key or the Greater Than keyboard key, but there is also a way to insert these characters as actual equations. For example, 4 or 3 ≥ 1 shows us a greater sign over half an equal sign, meaning that 4 or 3 are greater than or equal to 1. In such cases, we can use the greater than or equal to symbol, i.e. In Greater than or equal operator A value compares with B value it will return true in two cases one is when A greater than B and another is when A equal to B. Rate this symbol: (3.80 / 5 votes) Specifies that one value is greater than, or equal to, another value. This symbol is nothing but the "greater than" symbol with a sleeping line under it. Less Than or Equal To (<=) Operator. “Greater than or equal to” and “less than or equal to” are just the applicable symbol with half an equal sign under it. Greater Than or Equal To: Math Definition. 2 ≥ 2. But, when we say 'at least', we mean 'greater than or equal to'. The less than or equal to symbol is used to express the relationship between two quantities or as a boolean logical operator. "Greater than or equal to" is represented by the symbol " ≥ ≥ ". Solution for 1. The greater-than sign is a mathematical symbol that denotes an inequality between two values. In an acidic solution [H]… Greater than or Equal in Excel – Example #5. Here a could be greater … Examples: 5 ≥ 4. The sql Greater Than or Equal To operator is used to check whether the left-hand operator is higher than or equal to the right-hand operator or not. Category: Mathematical Symbols. When we say 'as many as' or 'no more than', we mean 'less than or equal to' which means that a could be less than b or equal to b. Select the Greater-than Or Equal To tab in the Symbol window. use ">=" for greater than or equal use "<=" for less than or equal In general, Sheets uses the same "language" as Excel, so you can look up Excel tips for Sheets. Copy the Greater-than Or Equal To in the above table (it can be automatically copied with a mouse click) and paste it in word, Or. Finding specific symbols in countless symbols is obviously a waste of time, and some characters like emoji usually can't be found. Graphical characteristics: Asymmetric, Open shape, Monochrome, Contains straight lines, Has no crossing lines. Select the Insert tab. If left-hand operator higher than or equal to right-hand operator then condition will be true and it will return matched records. Sometimes we may observe scenarios where the result obtained by solving an expression for a variable, which are greater than or equal to each other. As we saw earlier, the greater than and less than symbols can also be combined with the equal sign. 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That denotes an inequality between two values, Has no crossing greater than or equal to sign we say 'at least ', can...

http://math.stackexchange.com/questions/219329/for-t-in-0-1-is-fracxetxex-1-integrable-over-x-in-0-in/219347

# For $t\in [ 0, 1 )$ is $\frac{xe^{tx}}{e^{x}-1}$ integrable over $x\in (0 , \infty )$? For $t\in [ 0, 1 )$ is $$\frac{xe^{tx}}{e^{x}-1}$$ integrable over $x\in (0 , \infty )$? I.e., $$\int_{0}^{\infty} \frac{xe^{tx}}{e^{x}-1} dx < \infty?$$ How do I show this? - As $x\to0$, $x/(e^x-1)$ approaches a finite limit. As $x\to\infty$, do a limit-comparison of the integrand to $xe^{tx}/e^x$. - As $\frac x{e^x-1}$ as a limit when $x\to 0$ (namely $1$), the only problem is when $x\to\infty$. We have $e^x-1\sim e^x$ at $+\infty$, so $\dfrac{xe^{tx}}{e^x-1}\sim xe^{(t-1)x}$. Using Taylor's series, $$e^{(t-1)x}\leq \frac 1{1+(1-t)x+x^2(1-t)^2/2+x^3(1-t)^3/6},$$ the integral is convergent for $t\in[0,1)$. - Doesn't the first limit go to $1$ instead of $e^{–1}$? –  Pedro Tamaroff Oct 23 '12 at 13:05 @PeterTamaroff Right. Fixed now. –  Davide Giraudo Oct 23 '12 at 13:14 What matters in the improper integral of a nice function (e.g. elementary function) is the existence of singularities. In a broad sense, there are two kinds of singularities that counts. 1. A point where the integrand does not behave well. For example, the function can explode to infinite or oscillate infinitely. 2. A point at infinity. That is, $\pm \infty$. Away from singularities, the behavior of the function is quite under control, allowing us to concentrate our attention on those singularities. There is a basic method to establish the convergence (or divergence) of the integral near each singularity point. In many cases, except for the oscillatory case, you can find a dominating function that determines the order of magnitude of the function near the point. If the dominating function is easy to integrate, then you can make a comparison with this dominating function to conclude the convergence behavior. For example, let us consider $$\int_{0}^{\frac{\pi}{2}} \tan^2 x \, dx \quad \text{and} \quad \int_{0}^{\infty} \frac{x^2 e^{-x}}{1+x^2} \, dx.$$ We can easily check that $\tan^2 x$ is bounded below by $(x-\frac{\pi}{2})^{-2}$ near the singularity $x = \frac{\pi}{2}$ and $x^2 e^{-x} / (1 + x^2)$ is bounded above by $e^{-x}$ near the singularity $x = \infty$. Then $$\int_{\frac{\pi}{2}-\delta}^{\frac{\pi}{2}} \tan^2 x \, dx \geq \int_{\frac{\pi}{2}-\delta}^{\frac{\pi}{2}} \left(x - \frac{\pi}{2}\right)^{2} \, dx = \infty$$ for sufficiently small $\delta > 0$ and $$\int_{R}^{\infty} \frac{x^2 e^{-x}}{1+x^2}\,dx \leq \int_{R}^{\infty} e^{-x} \, dx < \infty$$ for sufficiently large $R > 0$. Thus we find that the former diverges to $\infty$ and the latter converges. In our example, there are two seemingly singular points, namely $x = 0$ and $x = \infty$. At $x = 0$, we find that $$\lim_{x \to 0} \frac{x e^{tx}}{e^x - 1} = 1.$$ This means that this singularity is removable, in the sense that the function can be extended in a continuous manner to this point. Thus we need not count this point and we can move our attention to the point at infinity. To establish the convergence (or possibly divergence) of the integral near $x = \infty$, we write $$\frac{x e^{tx}}{e^x - 1} = \frac{x}{1 - e^{-x}} e^{-(1-t)x}.$$ It is clear that for sufficiently large $x$, the term $\frac{x}{1 - e^{-x}}$ is bounded above by some constant $C > 0$. Thus the dominating function is $e^{-(1-t)x}$ and $$\int_{R}^{\infty} \frac{x e^{tx}}{e^x - 1} \, dx \leq \int_{R}^{\infty} C e^{-(1-t)x} \, dx < \infty$$ for large $R$. Therefore the improper integral converges. - Thanks for your answers and especially for this limit method. But in this way, I actually found a simpler bound, namely, the following: observe $$\frac{xe^{tx}}{e^{x}-1}=\frac{xe^{(1/2)(t-1)x}}{1-e^{-x}}e^{(1/2)(t-1)x}$$ $$\frac{xe^{(1/2)(t-1)x}}{1-e^{-x}}<M$$ for a constant $M>0$ as $$\frac{xe^{(1/2)(t-1)x}}{1-e^{-x}}$$ is continuous and the limits for $x\to 0$ and $x \to \infty$ are finite. And this can be directly used for the integrability of the function. -

https://gmatclub.com/forum/the-price-of-a-consumer-good-increased-by-p-228709.html

GMAT Question of the Day - Daily to your Mailbox; hard ones only It is currently 17 Jun 2019, 08:03 ### 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 # The price of a consumer good increased by p%. . . Author Message TAGS: ### Hide Tags e-GMAT Representative Joined: 04 Jan 2015 Posts: 2888 The price of a consumer good increased by p%. . .  [#permalink] ### Show Tags Updated on: 07 Aug 2018, 07:09 1 12 00:00 Difficulty: 55% (hard) Question Stats: 68% (02:30) correct 32% (02:34) wrong based on 273 sessions ### HideShow timer Statistics The price of a consumer good increased by $$p$$% during $$2012$$ and decreased by $$12$$% during $$2013$$. If no other change took place in the price of the good and the price of the good at the end of $$2013$$ was $$10$$% higher than the price at the beginning of $$2012$$, what was the value of $$p$$? A. $$-2$$% B. $$2$$% C. $$22$$% D. $$25$$% E. Cannot be determined Take a stab at this fresh question from e-GMAT. Post your analysis below. Official Solution to be provided after receiving some good analyses. _________________ Originally posted by EgmatQuantExpert on 11 Nov 2016, 05:46. Last edited by EgmatQuantExpert on 07 Aug 2018, 07:09, edited 1 time in total. CEO Joined: 12 Sep 2015 Posts: 3777 Re: The price of a consumer good increased by p%. . .  [#permalink] ### Show Tags 11 Nov 2016, 07:09 Top Contributor EgmatQuantExpert wrote: The price of a consumer good increased by p% during 2012 and decreased by 12% during 2013. If no other change took place in the price of the good and the price of the good at the end of 2013 was 10% higher than the price at the beginning of 2012, what was the value of p? A. $$-2$$% B. $$2$$% C. $$22$$% D. $$25$$% E. Cannot be determined Let $100 be the original price The price of a consumer good increased by p% during 2012 p% = p/100, so a p% INCREASE is the same a multiplying the original price by 1 + p/100 So, the new price = ($100)(1 + p/100) The price then decreased by 12% during 2013 A 12% DECREASE is the same a multiplying the price by 0.88 So, the new price = ($100)(1 + p/100)(0.88) The price of the good at the end of 2013 was 10% higher than the price at the beginning of 2012 If the original price was$100, then the price at the end of 2013 was $110 So, we can write:$110 = ($100)(1 + p/100)(0.88) Simplify:$110 = (100 + p)(0.88) Simplify more: $110 = 88 + 0.88p Subtract 88 from both sides: 22 = 0.88p So, p = 22/0.88 = 25 Answer: RELATED VIDEO _________________ Test confidently with gmatprepnow.com Board of Directors Status: QA & VA Forum Moderator Joined: 11 Jun 2011 Posts: 4499 Location: India GPA: 3.5 WE: Business Development (Commercial Banking) Re: The price of a consumer good increased by p%. . . [#permalink] ### Show Tags 11 Nov 2016, 12:48 EgmatQuantExpert wrote: The price of a consumer good increased by $$p$$% during $$2012$$ and decreased by $$12$$% during $$2013$$. If no other change took place in the price of the good and the price of the good at the end of $$2013$$ was $$10$$% higher than the price at the beginning of $$2012$$, what was the value of $$p$$? A. $$-2$$% B. $$2$$% C. $$22$$% D. $$25$$% E. Cannot be determined Price's corresponding to year - 2011 = $$100$$ 2012 = $$100 + p$$ 2013 = $$\frac{88}{100}(100 + p)$$ Further , $$\frac{88}{100}(100 + p)$$ = $$110$$ Or, $$\frac{8}{100}(100 + p)$$ = $$10$$ Or, 800 + 8p = 1000 Or, 8p = 200 So, p = 25% Hence, answer will be (D) 25% _________________ Thanks and Regards Abhishek.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only ) Current Student Joined: 26 Jan 2016 Posts: 100 Location: United States GPA: 3.37 Re: The price of a consumer good increased by p%. . . [#permalink] ### Show Tags 11 Nov 2016, 13:01 Lets start with a number for the original value. 100 is the easiest. So we're looking for a value of 110 at the end of 2013. Just by looking at the values we can get an idea of what to start testing. If we're increasing 100 by p% then decreasing it by 12% and the original value is still 10% higher we need a value much higher than 12. 25% is the easiest value to start with. 100+25%=125 125-12%=110 D Current Student Joined: 12 Aug 2015 Posts: 2610 Schools: Boston U '20 (M) GRE 1: Q169 V154 Re: The price of a consumer good increased by p%. . . [#permalink] ### Show Tags 12 Nov 2016, 20:32 For all the algebra loving people out there=> Let price at the beginning of 2012 be$x so after the end of 2012=> x[1+p/100] And finally at the end of 2013 => x[1+p/100][1-12/100] As per question=> Price was simple 10 percent greater Hence x[1+10/100] must be the final price. Equating the two we get => x[110/100]=x[1+p/100][88/100] => 44p+4400=5500 => 44p=1100 => p=1100/44=> 100/4=> 25. So p must be 25 _________________ Intern Joined: 02 Aug 2016 Posts: 4 Re: The price of a consumer good increased by p%. . .  [#permalink] ### Show Tags 16 Nov 2016, 09:46 [size=150]Let Price = 100 [size=150]Increased by P% = 100(1+P/100) Treat it like successive percents; So a 12% decrease would mean 88% of (1+P/100) of 100 The key words are no other change took place. So there are no further successive percents, and the final price = 110 Therefore: 88/100 * (1+P/100) * 100 = 110 => 8800 + 88P = 11,000 => 88P = 2200 => P = 2200/88 => P = 25% e-GMAT Representative Joined: 04 Jan 2015 Posts: 2888 Re: The price of a consumer good increased by p%. . .  [#permalink] ### Show Tags Updated on: 18 Dec 2016, 22:12 Hey, The given question can be solved in a number of ways. Let's look at two most common ways to solve this question. We will share two more ways of solving this question tomorrow. Method 1 : • Let us consider the price of the consumer good at the beginning of 2012 to be 100. • Let us also assume the price to be “C” at the beginning of 2013, after an increase of p%. • Since we know that with respect to the initial price, the price at the end of 2013 went up by 10%. o Therefore, the price at the end of 2013 = $$100 + (10$$ % of $$100) = 110$$ • Now we can write - o $$C – 12$$ % of $$C = 110$$ o $$C * (1 - \frac {12}{100}) = 110$$ o $$C = 110 * \frac {25}{22} = 125$$ Therefore, the price at the beginning of 2013 is 125 and we got this value after p% increase over the initial value. Thus, we can write – • $$100 + (p$$ % of $$100) = 125$$ • $$P = 25$$ % Method 2 : Conventional method – • Let the price of consumer good at the beginning of 2012 be 100. • After an increase of p%, the price at the beginning of 2013 will be – o Price at the beginning of 2013 $$= 100 + (p$$ % of $$100) = 100 + p$$ Therefore the price at the beginning of 2013 is (100 + p) • After a decrease of 12%, the price at the end of 2013 will be – o Value at the beginning of 2013 $$* (1 – \frac{12}{100}) = (100 + p) * \frac {22}{25}$$……………..(i) • And we are also given that the overall increase in the price of consumer good is 10%. • Therefore, the value at the end of 2013 = $$100 + (10$$% of $$100)$$ = 110………(ii) • Equating equation (i) and (ii) we get – o $$(100 + p) * \frac {22}{25} = 110$$ o $$100 + p = 125$$ Therefore, $$p = 25$$% There are more innovative ways to solve this question. A few of them have not been discussed here. Can you all think of any other way to solve it? Would love to see a few other methods! I will post two more ways to solve this question tomorrow. In the mean time, expecting some more responses with other ways to solve this question Thanks, Saquib e-GMAT Quant Expert _________________ Originally posted by EgmatQuantExpert on 15 Dec 2016, 23:26. Last edited by EgmatQuantExpert on 18 Dec 2016, 22:12, edited 1 time in total. e-GMAT Representative Joined: 04 Jan 2015 Posts: 2888 Re: The price of a consumer good increased by p%. . .  [#permalink] ### Show Tags 16 Dec 2016, 04:11 1 joannaecohen wrote: Lets start with a number for the original value. 100 is the easiest. So we're looking for a value of 110 at the end of 2013. Just by looking at the values we can get an idea of what to start testing. If we're increasing 100 by p% then decreasing it by 12% and the original value is still 10% higher we need a value much higher than 12. 25% is the easiest value to start with. 100+25%=125 125-12%=110 D Hi, Thanks for posting a different way of approaching this problem. In fact, the approach followed by you is very close to one of the innovative ways that we talked about in our official solution. The only difference being in your approach you have concluded that p should be much larger than 12. Going a step further, you can also conclude that p should be larger than 22% (12%+10%). Once you do so, you don't even need to pick any number. The only option which will satisfy it is D. 25%. When we post our detailed solution using the two innovative methods tomorrow, we will explain how we can conclude that p should be greater than 22%. Regards, Saquib _________________ e-GMAT Representative Joined: 04 Jan 2015 Posts: 2888 Re: The price of a consumer good increased by p%. . .  [#permalink] ### Show Tags 25 Dec 2016, 07:32 1 As discussed, let's look at one of the innovative ways of solving the above question. It is one of the quickest ways to solve a question that involves successive percentage increase/decrease on the same value. Please take a note of this approach and apply it on some GMAT questions to master it. So, let's quickly look at this smart approach. When a number is increased successively by two percentage, let's assume, $$a$$% and $$b$$%, the net increase in the value of the number can be expressed by the formula, Net increase $$=a+b+\frac {ab}{100}$$ Le's take a simple example to understand. If we increase a number, let's say, X successively by 10% and 20% respectively, the net increase according to the above formula should be, Net increase $$=10+20+\frac {10*20}{100}=10+20+\frac {200}{100} = 10+20+2 = 32$$% Isn't that quick!! A nice method to keep in your arsenal to solve Percent question involving successive increase quickly. One good thing about the above formula is that you can use it to calculate the net decrease in case of successive decrease too. All you need to do is in case of decrease represent the percent as negative. Easy isn't it . Let's see an application quickly. If we decrease a number, let's say, X successively by 10% and 20% respectively, the net increase according to the above formula should be, Net increase $$=(-10)+(-20)+\frac {(-10)*(-20)}{100}=-10-20+\frac {200}{100} = -10-20+2 = (-28)$$% Notice carefully, the sign of the net increase is negative, clearly indicating the after the successive decrease the value of the original number, decreased instead of increasing. And what was the magnitude??? Right 28%. The net decrease is 28%. So, before we use this approach to give you an official answer for the above question, would you like to have a quick stab at it. Remember, you need to be careful about the sign of the change. Increase is represented by positive and decrease is represented by negative. All the best. We will post the detailed solution tomorrow and then we will show another innovative method of solving this question. Regards, Saquib _________________ e-GMAT Representative Joined: 04 Jan 2015 Posts: 2888 The price of a consumer good increased by p%. . .  [#permalink] ### Show Tags Updated on: 07 Aug 2018, 07:11 2 1 Alright, so let's look at the official solution to the above questions using the innovative formula on Net increase discussed in the last post. We know that the price of the consumer good increased by $$p$$% and then decreased by $$12$$%. Hence, using the formula for net increase we can say, Net increase $$=p+(-12)+\frac {p*(-12)}{100}=p-12-\frac {3p}{25} = (\frac {22p}{25} - 12)$$% It is given in the question that the net increase finally is $$10$$%. Hence, we can equate the two values. $$(\frac {22p}{25} - 12)$$% = $$10$$% or, $$\frac {22p}{25} = 10+12 = 22$$% or, $$p = 25$$% Now, with this understanding try to solve this question in an even better way. Give it a try and we will post the official solution in another innovative way soon. _________________ Originally posted by EgmatQuantExpert on 27 Dec 2016, 00:45. Last edited by EgmatQuantExpert on 07 Aug 2018, 07:11, edited 1 time in total. Board of Directors Status: QA & VA Forum Moderator Joined: 11 Jun 2011 Posts: 4499 Location: India GPA: 3.5 Re: The price of a consumer good increased by p%. . .  [#permalink] ### Show Tags 27 Dec 2016, 07:56 EgmatQuantExpert wrote: The price of a consumer good increased by $$p$$% during $$2012$$ and decreased by $$12$$% during $$2013$$. If no other change took place in the price of the good and the price of the good at the end of $$2013$$ was $$10$$% higher than the price at the beginning of $$2012$$, what was the value of $$p$$? A. $$-2$$% B. $$2$$% C. $$22$$% D. $$25$$% E. Cannot be determined $$p - 12 - \frac{12p}{100} = 10$$ $$100p - 1200 -12p = 1000$$ $$88p = 2200$$ $$p = 25$$ Hence, the correct answer must be (D) 25 _________________ Thanks and Regards Abhishek.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only ) Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 6522 Location: United States (CA) Re: The price of a consumer good increased by p%. . .  [#permalink] ### Show Tags 01 Feb 2019, 18:59 EgmatQuantExpert wrote: The price of a consumer good increased by $$p$$% during $$2012$$ and decreased by $$12$$% during $$2013$$. If no other change took place in the price of the good and the price of the good at the end of $$2013$$ was $$10$$% higher than the price at the beginning of $$2012$$, what was the value of $$p$$? A. $$-2$$% B. $$2$$% C. $$22$$% D. $$25$$% E. Cannot be determined We let the 2012 price = n and thus, the price at the end of 2013 will be: (1 + p/100)(0.88)(n) Since the price at the end of 2013 was 10% higher than at the beginning of 2012, we can create the equation: 1.1n = (1 + p/100)(0.88)(n) 1 = (1 + p/100)(0.8) 1 = 0.8 + 0.8p/100 Multiplying by 100, we have: 100 = 80 + 0.8p 20 = 0.8p 25 = p Alternate Solution: Let’s let the price at the beginning of 2012 be 100. Since the price at the end of 2013 was 10% higher, the price at the end of 2013 is 1.1 x 100 = 110. We know the price decreased by 12% and became 110; therefore, before decreasing by 12%, the price was 110/0.88 = 125. Now, the price of 100 increases by p percent and becomes 125; therefore the value of p is [(125 - 100)/100] x 100 = 25. _________________ # Scott Woodbury-Stewart Founder and CEO Scott@TargetTestPrep.com 122 Reviews 5-star rated online GMAT quant self study course See why Target Test Prep is the top rated GMAT quant course on GMAT Club. Read Our Reviews If you find one of my posts helpful, please take a moment to click on the "Kudos" button. VP Joined: 09 Mar 2018 Posts: 1004 Location: India Re: The price of a consumer good increased by p%. . .  [#permalink] ### Show Tags 01 Feb 2019, 21:46 EgmatQuantExpert wrote: The price of a consumer good increased by $$p$$% during $$2012$$ and decreased by $$12$$% during $$2013$$. If no other change took place in the price of the good and the price of the good at the end of $$2013$$ was $$10$$% higher than the price at the beginning of $$2012$$, what was the value of $$p$$? A. $$-2$$% B. $$2$$% C. $$22$$% D. $$25$$% E. Cannot be determined 10 % will be a successive percentage change, which is calculated by a + b + ab/100 = Percentage change p - 12 - 12p/100 = 10 100p -1200 - 12p = 1000 p = 2200/88 p = 25 % D _________________ If you notice any discrepancy in my reasoning, please let me know. Lets improve together. Quote which i can relate to. Many of life's failures happen with people who do not realize how close they were to success when they gave up. Re: The price of a consumer good increased by p%. . .   [#permalink] 01 Feb 2019, 21:46 Display posts from previous: Sort by

http://centrallab.msu.ac.th/2f1iaj16/lhcbx.php?aea000=least-square-approximation-of-a-function

The idea is to minimize the norm of the difference between the given function and the approximation. Picture: geometry of a least-squares solution. As a result we should get a formula y=F(x), named the empirical formula (regression equation, function approximation), which allows us to calculate y for x's not present in the table. Learn to turn a best-fit problem into a least-squares problem. FINDING THE LEAST SQUARES APPROXIMATION We solve the least squares approximation problem on only the interval [−1,1]. We use the Least Squares Method to obtain parameters of F for the best fit. obtained as measurement data. Because the least-squares fitting process minimizes the summed square of the residuals, the coefficients are determined by differentiating S with respect to each parameter, and setting the result equal to zero. The method of least square • Above we saw a discrete data set being approximated by a continuous function • We can also approximate continuous functions by simpler functions, see Figure 3 and Figure 4 Lectures INF2320 – p. 5/80 Given a function and a set of approximating functions (such as the monomials ), for each vector of numbers define a functional By … The least squares method is one of the methods for finding such a function. Active 7 months ago. Learn examples of best-fit problems. Thus, the empirical formula "smoothes" y values. The radial basis function (RBF) is a class of approximation functions commonly used in interpolation and least squares. Approximation of a function consists in finding a function formula that best matches to a set of points e.g. Least Square Approximation for Exponential Functions. Section 6.5 The Method of Least Squares ¶ permalink Objectives. Ask Question Asked 5 years ago. Vocabulary words: least-squares solution. Orthogonal Polynomials and Least Squares Approximations, cont’d Previously, we learned that the problem of nding the polynomial f n(x), of degree n, that best approximates a function f(x) on an interval [a;b] in the least squares sense, i.e., that minimizes kf n fk= Z … ... ( \left[ \begin{array}{c} a \\ b \end{array} \right] \right)\$ using the original trial function. Approximation problems on other intervals [a,b] can be accomplished using a lin-ear change of variable. Recipe: find a least-squares solution (two ways). The RBF is especially suitable for scattered data approximation and high dimensional function approximation. Free Linear Approximation calculator - lineary approximate functions at given points step-by-step This website uses cookies to ensure you get the best experience. Quarteroni, Sacco, and Saleri, in Section 10.7, discuss least-squares approximation in function spaces such as . In this section, we answer the following important question: ∂ S ∂ p 1 = − 2 ∑ i = 1 n x i (y i − (p 1 x i + p 2)) = 0 ∂ S ∂ p 2 = − 2 ∑ i … Least squares method, also called least squares approximation, in statistics, a method for estimating the true value of some quantity based on a consideration of errors in observations or measurements. The smoothness and approximation accuracy of the RBF are affected by its shape parameter. The least squares method is the optimization method. '' y values [ −1,1 ] approximate Functions at given points step-by-step website... The best fit the following important question: least Square approximation for Exponential Functions solution two. For the best fit least Square approximation for Exponential Functions two ways ): find a least-squares (. Saleri, in section 10.7, discuss least-squares approximation in function spaces such as between the function. We solve the least squares approximation we solve the least squares method is one of difference... 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https://abigail.github.io/HTML/Perl-Weekly-Challenge/week-108-2.html

Perl Weekly Challenge 108: Bell Numbers by Abigail Challenge Example • $$B_0$$: 1 as you can only have one partition of zero element set. • $$B_1$$: 1 as you can only have one partition of one element set $$\{a\}$$. • $$B_2$$: 2 • $$\{a\}\{b\}$$ • $$\{a,b\}$$ • $$B_3$$: 5 • $$\{a\}\{b\}\{c\}$$ • $$\{a,b\}\{c\}$$ • $$\{a\}\{b,c\}$$ • $$\{a,c\}\{b\}$$ • $$\{a,b,c\}$$ • $$B_4$$: 15 • $$\{a\}\{b\}\{c\}\{d\}$$ • $$\{a,b,c,d\}$$ • $$\{a,b\}\{c,d\}$$ • $$\{a,c\}\{b,d\}$$ • $$\{a,d\}\{b,c\}$$ • $$\{a,b\}\{c\}\{d\}$$ • $$\{a,c\}\{b\}\{d\}$$ • $$\{a,d\}\{b\}\{c\}$$ • $$\{b,c\}\{a\}\{d\}$$ • $$\{b,d\}\{a\}\{c\}$$ • $$\{c,d\}\{a\}\{b\}$$ • $$\{a\}\{b,c,d\}$$ • $$\{b\}\{a,c,d\}$$ • $$\{c\}\{a,b,d\}$$ • $$\{d\}\{a,b,c\}$$ Discussion The Bell Numbers have their own entry in the OEIS. We can look up the first ten Bell Numbers: $$1$$, $$1$$, $$2$$, $$5$$, $$15$$, $$52$$, $$203$$, $$877$$, $$4140$$, and $$21147$$. Hello, World! The simplest way would be just to take those ten numbers, and print them. This means we have yet again a challenge which is just a glorified Hello, World program. Fetch If we don't want to do exactly what the challenge asks from us (print the first ten Bell Numbers), we could instead fetch the numbers from the OEIS and print them. For instance, by using the OEIS module which we recently uploaded to CPAN. There is limited usefulness in this though — it's not that the Bell Numbers will change in the future. Calculate Alternatively, we could calculate the first ten Bell Numbers. There are many ways to calculate the numbers, but we opt to create a Bell Triangle. The first rows of the Bell Triangle are as follows: 1 1 2 2 3 5 5 7 10 15 15 20 27 37 52 And we have the following rules to construct the triangle: • The top row contains a single $$1$$. • For each other row: • The row will have one more element than the previous row. • The first (left most) element is equal to the last (right most) element of the previous row. • Each other element is the sum of the element to its left on the same row, and the element on the previous row right above that. Or, formalized: Let $$b_{r, c}$$ be the element on row $$r$$ and column $$c$$. (This implies $$0 \leq c \leq r$$, with the top most element being $$b_{0, 0}$$.) Then $b_{r, c} = \begin{cases} 1, & \text{if } r = c = 0 \\ b_{r - 1, r - 1}, & \text{if } r > 0, c = 0 \\ b_{r, c - 1} + b_{r - 1, c - 1}, & \text{if } r \geq c > 0 \end{cases}$ If we then generate the first nine rows of the Bell Triangle, and take the last elements of each row, we get the second to tenth Bell Numbers. The first Bell Number is $$1$$. Solutions Depending on the language, we solve the challenge in one or more of the strategies explained above. All languages will implement the Hello, World! strategy. For some languages, we also calculate the Bell Triangle. And in Perl, we also implement a fetch strategy. Languages which solve the problem in more than one way take a command line argument indicating the strategy to follow. This argument should be one of plain (the default), fetch (which fetches the numbers from the OEIS, or compute, which computes the first rows of the Bell Triangle. We will only show the the plain solution for Perl; for the other implementations, see the GitHub links below. Perl plain Can't be much simpler than this. say "1, 1, 2, 5, 15, 52, 203, 877, 4140, 21147" fetch We're using the new module OEIS which export a single method, oeis, which takes two arguments: the sequence to fetch, and the number of elements to return. use OEIS; } $, = ", "; say 1, map {$$_ [-1]} @bell; Find the full program on GitHub. AWK The algorithm above is simply written down in AWK: BEGIN { COUNT = 10 bell [1, 1] = 1 for (x = 2; x < COUNT; x ++) { bell [x, 1] = bell [x - 1, x - 1] for (y = 2; y <= x; y ++) { bell [x, y] = bell [x, y - 1] + bell [x - 1, y - 1] } } printf "1" for (x = 1; x < COUNT; x ++) { printf ", %d", bell [x, x] } printf "\n" } Find the full program on GitHub. Bash Bash doesn't have two dimensional arrays. So, we're using a function index which takes two arguments (an x and a y coordinate) and returns a single index. The return value is written in the global variable idx. We then get: set -f COUNT=10 function index () { local x=$1 local y=$2 idx=$((COUNT * x + y)) } bell[0]=1 for ((x = 1; x < COUNT - 1; x ++)) do index $x 0; i1=$idx index $((x - 1))$((x - 1)); i2=$idx bell[$i1]=${bell[$i2]} for ((y = 1; y <= x; y ++)) do index $x$y; i1=$idx index$x $((y - 1)); i2=$idx index $((x - 1))$((y - 1)); i3=$idx bell[$i1]=$((bell[i2] + bell[i3])) done done printf "1" for ((x = 0; x < COUNT - 1; x ++)) do index$x $x; printf ", %d"${bell[\$idx]} done echo Find the full program on GitHub. C C requires us to manage our own memory. Other than that, it's the same algorithm: # define COUNT 10 typedef int number; /* Change if we want large numbers */ char * fmt = "%d"; /* Should match typedef */ int main (int argc, char * argv []) { number ** bell; if ((bell = (number **) malloc ((COUNT - 1) * sizeof (number *))) == NULL) { perror ("Mallocing bell failed"); exit (1); } if ((bell [0] = (number *) malloc (sizeof (number))) == NULL) { perror ("Mallocing row failed"); exit (1); } bell [0] [0] = 1; for (int x = 1; x < COUNT - 1; x ++) { if ((bell [x] = (number *) malloc ((x + 1) * sizeof (number))) == NULL) { perror ("Mallocing row failed"); exit (1); } bell [x] [0] = bell [x - 1] [x - 1]; for (int y = 1; y <= x; y ++) { bell [x] [y] = bell [x] [y - 1] + bell [x - 1] [y - 1]; } } /* * Print the right diagonal */ printf (fmt, 1); for (int x = 0; x < COUNT - 1; x ++) { printf (", "); printf (fmt, bell [x] [x]); } printf ("\n"); exit (0); } Find the full program on GitHub. Lua Same algorithm: local COUNT = 10 local bell = {} bell [0] = {} bell [0] [0] = 1 for x = 1, COUNT - 2 do bell [x] = {} bell [x] [0] = bell [x - 1] [x - 1] for y = 1, x do bell [x] [y] = bell [x] [y - 1] + bell [x - 1] [y - 1] end end io . write (1) for x = 0, COUNT - 2 do io . write (", " .. bell [x] [x]) end io . write ("\n") Find the full program on GitHub. Node.js let COUNT = 10 let bell = [[ 1 ]] let x for (x = 1; x < COUNT - 1; x ++) { bell [x] = [bell [x - 1] [x - 1]] let y for (y = 1; y <= x; y ++) { bell [x] [y] = bell [x] [y - 1] + bell [x - 1] [y - 1] } } process . stdout . write ("1") for (x = 0; x < COUNT - 1; x ++) { process . stdout . write (", " + bell [x] [x] . toString ()) } process . stdout . write ("\n") Find the full program on GitHub. Python Python doesn't autovivify array elements when indexing out of bounds. So we use the append method to add elements to arrays. COUNT = 10 bell = [[1]] for x in range (1, COUNT - 1): bell . append ([bell [x - 1] [x - 1]]) for y in range (1, x + 1): bell [x] . append (bell [x] [y - 1] + bell [x - 1] [y - 1]) print (1, end = '') for x in range (0, COUNT - 1): print (",", bell [x] [x], end = '') print ("") Find the full program on GitHub. Ruby COUNT = 10 bell = [[1]] for x in 1 .. COUNT - 2 bell [x] = [bell [x - 1] [x - 1]] for y in 1 .. x bell [x] [y] = bell [x] [y - 1] + bell [x - 1] [y - 1] end end print (1) for x in 0 .. COUNT - 2 print (", ") print (bell [x] [x]) end puts ("") Find the full program on GitHub. Other languages We also have simple solutions for BASIC, bc, Befunge-93, Cobol, Csh, Erlang, Forth, Fortran, Go, Java, m4, OCaml, Pascal, PHP, PostScript, R, Rexx, Scheme, sed, SQL, and Tcl.

https://forum.math.toronto.edu/index.php?PHPSESSID=e02qs82rtho6ihd22c4oep6c21&action=printpage;topic=259.0

# Toronto Math Forum ## MAT244-2013S => MAT244 Math--Tests => MidTerm => Topic started by: Victor Ivrii on March 06, 2013, 09:08:26 PM Title: MT Problem 3 Post by: Victor Ivrii on March 06, 2013, 09:08:26 PM Find a particular solution of equation \begin{equation*} t^2 y''-2t y' +2y=t^3 e^t. \end{equation*} [BONUS] Explain whether the method of undetermined  coefficients to find a particular solution of this equation applies. Title: Re: MT Problem 3 Post by: Jeong Yeon Yook on March 06, 2013, 10:30:47 PM The method of undetermined coefficient applies because it only "requires us to make an initial assumption about the form of the particular solution, but with the coefficients left unspecified" (Textbook 10th Edition P.177). If t = 0, we have, 2y = 0. => y = 0 is the solution for t = 0. Title: Re: MT Problem 3 Post by: Rudolf-Harri Oberg on March 06, 2013, 10:50:34 PM This is an Euler equation, see book page 166, problem 34. We need to use substitution $x=\ln t$, this will make into a ODE with constant coefficients. We look first at the homogenous version: $$y''-3y'+2y=0$$ Solving $r^2-3r+2=0$ yields $r_1=2, r_2=1$. So, solutions to the homogeneous version are $y_1(x)=e^{2x}, y_2(x)=e^{x}$. But then solutions to the homogeneous of the original problem are $y_1(t)=t^2, y_2(t)=t$. So, $Y_{gen.hom}=c_1t^2+c_2t$. We now use method of variation of parameters, i.e let $c_1,c_2$ be functions. To use the formulas on page 189, we need to divide the whole equation by $t^2$ so that the leading coefficient would be one, so now $g=t e^t$. The formula is: $c_i'=\frac{W_i g}{W}$, where $W_i$ is the wronksian of the two solutions where the i-th column has been replaced by $(0,1)$. We now just calculate that $W(t^2,t)=-t^2, W_1=-t, W_2=t^2$. Now we need to compute $c_1, c_2$. $$c_1'=e^t \implies c_1=e^t$$ $$c_2'=-te^t \implies c_2=-e^t(t-1)$$ Plugging these expressions back to $Y_{gen.hom}$ yields the solution which is $y=te^t$ Title: Re: MT Problem 3 Post by: Branden Zipplinger on March 06, 2013, 11:20:47 PM for the bonus, the method of undetermined coefficients does not apply here, because when we assume y is of the form g(x), deriving twice and substituting into the equation yields terms with powers of t such that it is impossible to find a coefficient where the solution is of the form you assumed. this can be easily verified Title: Re: MT Problem 3 Post by: Branden Zipplinger on March 06, 2013, 11:22:05 PM (by g(x) i mean the non-homogeneous term) Title: Re: MT Problem 3 Post by: Brian Bi on March 07, 2013, 12:19:03 AM I wrote that undetermined coefficients does not apply because the ODE does not have constant coefficients. Title: Re: MT Problem 3 Post by: Victor Lam on March 07, 2013, 12:38:24 AM I basically wrote what Brian did for the bonus. But I suppose that if we transform the original differential equation using x = ln(t) into another DE with constants coefficients (say, change all the t's to x's), we would then be able to apply the coefficients method, and carry on to find the particular solution. Can someone confirm the validity of this? Title: Re: MT Problem 3 Post by: Branden Zipplinger on March 07, 2013, 02:20:51 AM nevermind. Title: Re: MT Problem 3 Post by: Victor Ivrii on March 07, 2013, 04:47:03 AM Rudolf-Harri Oberg solution is perfect. One does not need to reduce it to constant coefficients (appealing to it is another matter); characteristic equation is $r(r-1)-2r+2=0$ rendering $r_{1,2}=1,2$ and $y_1=t$, $y_2=t^2$ (Euler equation). Method of undetermined coefficients should not work;  all explanations are almost correct: for equations with constant coefficients the r.h.e. must be of the form $P(x)e^{rx}$ where $P(x)$ is a polynomial but for Euler equation which we have it must be $P(\ln (t)) t^r$ (appeal to reduction) which is not the case. However sometimes work methods which should not and J. Y. Yook has shown this. Luck sometimes smiles to foolish and ignores the smarts Quote Everybody knows that something can't be done and then somebody turns up and he doesn't know it can't be done and he does it.(A. Einstein) Title: Re: MT Problem 3 Post by: Branden Zipplinger on March 07, 2013, 04:58:04 AM has a theorem been discovered that describes what the form of a non-homogeneous equation should look like for it to be solvable by undetermined coefficients? Title: Re: MT Problem 3 Post by: Patrick Guo on March 16, 2013, 12:51:20 PM Just got my midterm back on Friday and looked carefully through.. In the official 2013Midterm answers (both versions on Forum and on CourseSite), why we, when using variation-method, have v1 = - ∫ (t^2 + 1) g(t) / Wronskian  dt   ??  what is (t^2 +1) ?! Should that not be y2 = t^2 ?! And how do we, from this step, get the next step, where (t^2 +1) changes to t with no reason ? I see the results of v1 and v2 are correct, but the steps are totally incomprehensible and WRONG. And why Wronskian = -t^2 ? should it not be t^2 ? Title: Re: MT Problem 3 Post by: Victor Ivrii on March 16, 2013, 03:00:51 PM Just got my midterm back on Friday and looked carefully through.. In the official 2013Midterm answers (both versions on Forum and on CourseSite), why we, when using variation-method, have v1 = - ∫ (t^2 + 1) g(t) / Wronskian  dt   ??  what is (t^2 +1) ?! Should that not be y2 = t^2 ?! And how do we, from this step, get the next step, where (t^2 +1) changes to t with no reason ? I see the results of v1 and v2 are correct, but the steps are totally incomprehensible and WRONG. And why Wronskian = -t^2 ? should it not be t^2 ? Rats! The answer is simple: typo by the person who typed and the lack of proofreading (all instructors were busy preparing Final and TT2). Thanks! Fixed in all three instances (including on BlackBoard)

https://math.stackexchange.com/questions/2720043/null-space-and-kernel-of-matrix-representation

# Null space and kernel of matrix representation Let $P_3(\mathbb{C})$ be the complex vector space of complex polynomials of degree $2$ or less. Let $\alpha,\beta\in\mathbb{C}, \alpha\neq\beta$. Consider the function $L:P_3(\mathbb{C}) \mapsto \mathbb{C}^2$ given by $$L(p)=\begin{bmatrix} p(\alpha) \\ p(\beta)\\ \end{bmatrix}, \text{ for } p\in P_3(\mathbb{C})$$ For the basis $v=(1,X,X^2)$ for $P_3(\mathbb{C})$ and the standard basis $E = (e_1,e_2)$ for $\mathbb{C}^2$. Find the matrix representation $_E[L]_v$ and determine the null space $N(_E[L]_v)$ and find a basis for the ker(L). I have found the matrix representation: $$_E[L]_v = [L(v)]_E = [L(1)]_E\ [L(X)]_E\ [L(X^2)]_E = \begin{bmatrix} 1\quad \alpha \quad \alpha^2 \\ 1\quad \beta \quad \beta^2 \end{bmatrix}$$ By using ERO we can reduce the matrix to: $\begin{bmatrix} 1 \quad 0 \quad - \alpha\beta \\ 0 \quad 1 \quad \alpha + \beta \\ \end{bmatrix},$ I am uncertain how to find the null space $N(_E[L]_v)$ and a basis for the kernel. • You’re almost there. See this answer for how to read a basis for the kernel from the reduced matrix. – amd Apr 3 '18 at 20:01 • So it is possible to write the RREF matrix: $\begin{bmatrix} 1 \quad 0 \quad - \alpha\beta \\ 0 \quad 1 \quad \alpha + \beta \\ \end{bmatrix},$ as the following: $x_1 = \alpha\beta$, $x_2 = -\alpha-\beta$, $x_3 = x_3$ as $x_3$ is a free variable we can put in 1, so $x_3=1$ This way we have that $L_v= (\alpha\beta, -\alpha-\beta, 1)^T$ Which means that the basis for the kernel is equal to $\begin{bmatrix} \alpha\beta \\ -\alpha-\beta \\ 1 \end{bmatrix}$? – Simbörg Apr 3 '18 at 21:15 • Your reasoning is a bit off. The RREF represents the equations $x_1-\alpha\beta x_3=0$ and $x_2+(\alpha+\beta)x_3=0$, so every solution of the system is of the form $(\alpha\beta x_3, -(\alpha+\beta)x_3, x_3)^T$, i.e., a multiple of $(\alpha\beta, -\alpha-\beta,1)^T$. – amd Apr 3 '18 at 21:45 You're doing good and the matrix is exactly what you found. The reduced row echelon form is $$\begin{bmatrix} 1 & 0 & -\alpha\beta \\ 0 & 1 & \alpha+\beta \end{bmatrix}$$ as you found. Now you can determine a basis for the null space of the matrix as generated by $$\begin{bmatrix} \alpha\beta \\ -(\alpha+\beta) \\ 1 \end{bmatrix}$$ and this is the coordinate vector of a polynomial generating the kernel, which is thus $$q(X)=\alpha\beta-(\alpha+\beta)X+X^2$$ As a check: this polynomial $q$ has $\alpha$ and $\beta$ as roots and so $L(q)=0$. The kernel has dimension $1$ by the rank nullity theorem. • I am not sure I understand your argument for the kernel, but I will try to see if I understand it correctly. So: we know that since the coordinate vector $\begin{bmatrix} \alpha\beta \\ -\alpha-\beta \\ 1 \end{bmatrix} \in N(_E[L]_v)$ we know this implies that $(\alpha\beta, -\alpha-\beta, 1)^T \in ker(L)$ and then you define a polynomial for that generates the kernel, which means $ker(L) = (\alpha\beta, -(\alpha+\beta)X, X^2)^T$ and if $X = 0$ then we have that the kernel consists of $(\alpha\beta)$? – Simbörg Apr 4 '18 at 9:27 It is probably better to do that via polynomials. Suppose $p \in P_3(\mathbb{C})$ is such that $p(\alpha) = 0 = p(\beta)$. Then $p$ is divisible by both $x - \alpha$ and $x - \beta$. Since $\alpha \ne \beta$, the two linear polynomials are coprime, so $p$ is divisible by $(x-\alpha)(x-\beta) = x^{2} - (\alpha+\beta) x + \alpha \beta$, and thus $p$ is a scalar multiple of it, as $p$ has degree at most $2$. So the kernel is one-dimensional, generated by the transpose of $(\alpha \beta, -\alpha - \beta, 1)$. This indicates that there is a little sign error (it happens to everyone) in your reduction. • Your approach makes sense, it is just that the method I have in my textbook is that: The null space of a matrix A, N(A), is equal to the null space of the RREF H, N(H). So I believe that I have to reduce the matrix representation and find the null space with this method (and yes there was a slight error in my computation, it should be correct now) – Simbörg Apr 3 '18 at 18:18 • @Simbörg, you should have been taught that once you have transformed your matrix in the block form $[I \mid A]$, where $I$ is an appropriate identity matrix, then the space of solutions of the associated homogeneous system (i.e. the null space) has as a basis the columns of the block matrix $\left[\begin{smallmatrix}-A\\J\end{smallmatrix}\right]$, where $J$ is an appropriate square matrix. So in your case the null space has a basis given by $\left[\begin{smallmatrix}\alpha \beta\\-\alpha-\beta\\1\end{smallmatrix}\right]$. – Andreas Caranti Apr 4 '18 at 8:28

https://www.physicsforums.com/threads/antiderivative-of-1-x.841263/

# Antiderivative of 1/x 1. Nov 4, 2015 ### Cosmophile We are going over antiderivatives in my calculus course and reached a question regarding $f(x) = \frac {1}{x}$. My instructor went on to say that $\int \frac {1}{x}dx = \ln |x| + C$. This makes sense to me, but only to a certain point. For $f(x) = \frac {1}{x}$, $f$ is defined $\forall x \neq 0$. So we should have two interavls which we are looking at: $x < 0$ and $x > 0$. Because of this, we would then have: $$\int \frac {1}{x}dx = \ln |x| + C_1 \qquad x > 0$$ $$\text {and}$$ $$\int \frac {1}{x}dx = \ln (-x) + C_2 = \ln |x| + C_2 \qquad x < 0$$ The second integral comes into being because $-x > 0$ when $x<0$. I brought this up to my teacher and he said that it made no sense and served no purpose to look at it this way. The argument I brought up was that the constants of integration could be different for the two intervals. I was hoping some of you may be able to help me explain why this is the case, or, if I am wrong, explain to me why I am. Thanks! 2. Nov 4, 2015 ### pwsnafu You are correct. The "constant of integration" is constant over connected components of the domain. See also Wikipedia's article on antiderivative. 3. Nov 4, 2015 ### fzero It's an interesting idea, but the problem is that $$\frac{d}{dx} \ln (-x) = - \frac{1}{x},$$ instead of $1/x$, so this isn't an antiderivative. 4. Nov 4, 2015 ### pwsnafu No $\frac{d}{dx} \ln(-x) = \frac{1}{x}$ 5. Nov 4, 2015 ### Cosmophile If we are considering $x < 0$, $-x > 0$ so $$\frac {d}{dx} \ln (-x) = \frac {1}{x}$$ Also, by the chain rule where $u = -x$, $\frac {du}{dx} = -1$, and $$\frac {d}{dx} \ln(-x) = \frac {1}{-x}(-1) = \frac {1}{x}$$ 6. Nov 4, 2015 ### micromass You are completely correct. There are two different constants of integration. Your teacher must not be very good if he doesn't know this. 7. Nov 4, 2015 ### pwsnafu Seconded. This worries me. Mathematics is not concerned with with whether something "serves a purpose". There is plenty of mathematical research that serves no purpose other than itself. 8. Nov 4, 2015 ### Cosmophile I appreciate the replies thus far. I suppose I'm having a hard time developing an argument for my case to propose to him. 9. Nov 4, 2015 ### PeroK Essentially, you are correct. You can demonstrate this by taking: $f(x) = ln|x| + 1 \ (x < 0)$ and $f(x) = ln|x| + 2 \ (x > 0)$ and checking that $f'(x) = \frac{1}{x} \ (x \ne 0)$ Usually, however, you are only dealing with one half of the function : $x < 0$ or $x > 0$. This is because an integral is defined for a function defined on an interval. For the function $\frac{1}{x}$, you can't integrate it from on, say, $[-1, 1]$ because it's not defined at $x = 0$. So, strictly speaking, what the integration tables are saying is: a) For the function $\frac{1}{x}$ defined on the interval $(0, +\infty)$, the antiderivative is $ln|x| + C$ b) For the function $\frac{1}{x}$ defined on the interval $(-\infty, 0)$, the antiderivative is $ln|x| + C$ In that sense, you don't need different constants of integration. 10. Nov 4, 2015 ### micromass I see no reason not to define the undefined integral on more general sets. 11. Nov 4, 2015 ### Cosmophile Why did you arbitrarily chose $C_1 = 1$ and $C_2 = 2$ in the first part of your response? I understand that the antiderivative is defined only on intervals where the base function is defined, which means our integral is only defined on $(-\infty, 0) \cup (0, \infty)$. However, when I think of a constant added to a function, I simply think of a vertical shift. I understand that $f(x) = \frac {1}{x}$ has to have two independent antiderivatives as a consequence of the discontinuity, but I cannot see why it is necessary that the constants have to be different. Of course, a difference in constants is the only way that the two sides can be different, because we've already shown that, ignoring the added constant, the two sides have identical antiderivatives. Am I making any sense? Sorry, and thanks again. 12. Nov 4, 2015 ### PeroK Why not? A couple of technical points: 1)"The" antiderivative is actually an equivalence class of functions. "An" antiderivative is one of the functions from that class. For example: $sin(x) + C$ (where $C$ is an arbitrary constant) is the antiderivative of $cos(x)$; and $sin(x) + 6$ is an antiderivative (one particular function from the antiderivative class). 2) There is no such thing as "the" function $1/x$, as a function depends on its domain. $1/x$ defined on $(0, \infty)$, $1/x$ defined on $(-\infty, 0)$ and $1/x$ defined on $(0, \infty) \cup (-\infty, 0)$ are three different functions. The antiderivative of $1/x$ defined on $(0, \infty) \cup (-\infty, 0)$ is $ln|x| + C_1 \ (x < 0)$; $ln|x| + C_2 \ (x > 0)$. Where $C_1$ and $C_2$ are arbitrary constants. This is the full class of functions which, when differentiated, give $1/x$. The two separate functions $1/x$ defined on $(0, \infty)$, and $(-\infty, 0)$ both have the antiderivate $ln|x| + C$, where $C$ is an arbitrary constant, defined on the appropriate interval. There is, therefore, a subtle difference. 13. Nov 4, 2015 ### HallsofIvy Look at f(x)= ln(|x|)+ 9 for x> 0 and ln(|x|)+ 4 for x< 0. That function is differentiable for all non-zero x and its derivative is 1/x. 14. Nov 4, 2015 ### Cosmophile Your second point was very well said, and is certainly something I'll carry with me. So unless a particular domain is specified, such as $(0, \infty)$, I have to include the $\ln|x| +C_1 \quad (x < 0); \quad \ln |x| + C_2 \quad (x > 0)$. I suppose my issue is coming from the fact that, when I see $f(x) = \frac {1}{x}$, I imagine the standard hyperbola $f(x) = \frac {1}{x}$ defined on $(- \infty, 0) \cup (0, \infty)$. Now, when I imagine the function $f(x) = \ln x$, I think automatically of this graph: http://www.wolframalpha.com/share/img?i=d41d8cd98f00b204e9800998ecf8427eq1tuvuvmvh&f=HBQTQYZYGY4TMNJQMQ2WEZTBGUYDCNJQMQ3DAODGGAZTAMBQGI4Qaaaa But I've also seen this graph: Which should I be thinking of for this problem? Last edited: Nov 5, 2015 15. Nov 4, 2015 ### Staff: Mentor The second graph looks like it is probably f(x) = ln|x|. 16. Nov 5, 2015 ### Cosmophile You're right. I wonder why Wolfram interprets it that way, but only uses the right-hand side of the graph when I set it to show real values. 17. Nov 5, 2015 ### Staff: Mentor What was the equation you graphed? If you entered the integral in post #1 of this thread, the antiderivative is ln|x| (plus the constant). 18. Nov 5, 2015 ### Cosmophile I graphed $f(x) = \frac {1}{x}$ and got the funky results. Graphing ln|x| gave me what I needed. But if I'm given $f(x) = \frac {1}{x}$, with no specification of the domain, I end up with $F(x) = \ln |x| + C_1, \quad (x < 0); \ln|x| + C_2, \quad (x > 0).$ What's weird is that when I look at the graph of ln|x|, the two halfs seem connected (obviously, they aren't - there's a discontinuity at $x=0$), so it's hard for me to imagine why the two sides need different constants of integration. That would mean the two bits move around independently, and I can't at all think of an example of that if $f(x) = \frac {1}{x}$ with nothing else added. I hope that makes some sense. I'd love to get some graphical insights - I think that would help me out a bit. Sorry for asking so much on what seems to be a fairly simple topic! (Also, thanks for chiming in, Mark44. I can always count on you and Micromass to respond). 19. Nov 5, 2015 ### PeroK I suggest that you are getting yourself a bit confused over this. And, I guess this is why your maths instructor is saying it's not useful. There is nothing special about $1/x$. Any function $1/x^n$ has the same issue. Wolfram Alpha, for example, doesn't get into this subtlety of having different constants of integration. The same with tan(x) - you could have a different constant of integration on each interval upon which it's defined. The key point is this. Suppose I have a differential equation: $f'(x) = x^2$ and $f(1) = 0$ $(-\infty < x < \infty)$ This specifies a unique function: $f(x) = \frac{1}{3}(x^3 - 1)$ But, if we have: $f'(x) = 1/x$ and $f(1) = 0$ $(x \ne 0)$ This does not specify a unique function. We also need a value for some $x < 0$. For example: $f'(x) = 1/x$, $f(1) = 0$ $f(-1) = 1$ $(x \ne 0)$ The differential equation, therefore, splits into two: For $x > 0$ we have $f(x) = ln|x| + C$, $f(1) = C = 0$, hence $f(x) = ln|x| (x > 0)$ For $x < 0$ we have $f(x) = ln|x| + C$, $f(-1) = C = 1$, hence $f(x) = ln|x| + 1 (x < 0)$ Now, of course, you could use $C_1$ and $C_2$ instead of reusing the symbol $C$ as I have done. But, it hardly matters. The important point is that the integration and differentiation for a function like $1/x$ must be done separately on each contiguous interval on which it's defined. One final point. You can have the same situation with any function. Suppose we have: $f'(x) = x \ (-2 < x < -1$ and $1 < x < 2)$ Then we are specifying a differential equation across two discontiguous intervals, hence we need two initial values, as there is a different constant of integration on each interval. 20. Nov 5, 2015 ### micromass About the wolfram issue. Wolfram alpha is a very good software but not infallible. In particular, here it is (strictly speaking) wrong. When asked how to integrate $1/x$, wofram alpha somehow involves complex numbers. Furthermore, it will only show one particular primitive. So when you ask for the primitive of $1/x$, wolfram accurately calculates the following primitive: $f(x) = \text{ln}(x)$ for $x>0$ and $f(x) = \text{ln}(x) + \pi i$ for $x<0$. (The explanation for why $\pi i$ is involved requires complex analysis). Anyway, this is a correct primitive as you can differentiate it to get $1/x$. Wolfram alpha is wrong however to say that any other primitive is of the form $f(x) + C$. Rather, it is of the form $f(x) + C_1$ for $x>0$ and $f(x) + C_2$ for $x<0$. In particular, we can choose $C_1 = 0$ and $C_2 = -\pi i$, to get $\text{ln}|x|$. So if you want only the real values, then wolframalpha works with the same primitive $f$ that has already been computed. But this primitive is complex for $x<0$ and thus wolfram ignores it. So that is why it only shows $f(x)$ for $x>0$. There is a primitive that takes only real values (namely $\text{ln}|x|$), but wolfram does not compute it and erroneously states that the real primitive only exists for $x>0$.

http://mathhelpforum.com/trigonometry/88387-bearings-question-trig.html

# Thread: Bearings question with trig. 1. ## Bearings question with trig. Hey Im really not sure how to do bearings at all. For homework i have this question: A ship leaves at port A and travels for 30km on bearing of 120degrees It then changes course and travels for 50km on bearing of 080degrees arriving at port B. Calculate distance AB and bearing A from B thanks 2. Originally Posted by mitchoboy ... A ship leaves at port A and travels for 30km on bearing of 120degrees It then changes course and travels for 50km on bearing of 080degrees arriving at port B. Calculate distance AB and bearing A from B ... Typically bearings are given from a reference [North or South] and deflecting East or West. From the information given, assume the reference for zero bearing is due North or the y-axis; and assume that port A is at the origin (0,0). $X_0 = 0$ $Y_0 = 0$ $X_1 = X_0 + \sin \left(120\right) \times\ 30 = 0.866 \times 30 = 25.981$ $Y_1 = Y_0 + \cos \left(120\right) \times\ 30 = 0.500 \times 30 = 15.000$ $X_2 = X_1 + \sin\left( 80\right) \times\ 50 = X_1 + 0.985 \times 50 = X_1 + 49.240 = 75.221$ $Y_2 = Y_1 + \cos \left(80\right) \times\ 50 = Y_1 + 0.174 \times 50 = Y_1 + 8.682 = 23.682$ Since $X_0 = 0$ & $Y_0 = 0$ The distance AB is $\sqrt{X_2^2 + Y_2^2}$ The bearing is the arctangent of the difference between final coordinates and the initial coordinates: The tangent of the bearing AB is : $\frac{X_2 - X_0} {Y_2 - Y_0}$ As a check: $X_2 = \sin \left ({bearing AB}\right) \times \left ({distance AB}\right)$ $Y_2 = \cos \left ({bearing AB}\right) \times \left({distance AB}\right)$ 3. Just for kicks, let's do it this way. Let's use a coordinate system so you can see what is going on. Like in surveying. Let's say the coordinate of A is (0,0). Then, to the turning point, it is 30 km at an azimuth of 120 degrees. (Technically, this is an azimuth, not a bearing. But, it doesn't really matter). $x=30sin(120)=25.98$ $y=30cos(120)=-15$ Those are the coordinates of the turning point, (25.98,-15) Next, the boat turns 80 degrees from north and goes 50 km to go to B. B's coordinates are $x=25.98+50sin(80)=75.22$ $y=-15+50cos(80)=-6.32$ The coordinates of B are (75.22, -6.32). Now, to find the distance back to A where it started, just use ol' Pythagoras. $\sqrt{(75.22)^{2}+(-6.32)^{2}}=75.485$ To find the bearing back to A, one way of many: $270+cos^{-1}(\frac{75.22}{75.485})=274.8 \;\ deg$ Here is a diagram so you can see. It is rather sloppy done in paint, but I hope it will suffice. 4. Hello, mitchoboy! Bearings are measured clockwise from North. And a good diagram is essential. A ship leaves at port $A$ and travels for 30 km on bearing of 120°. It then changes course and travels for 50 km on bearing of 080° arriving at port $B.$ Calculate distance $AB$ and bearing of ${\color{blue}A}$ from ${\color{blue}B}.$ . Is this correct? Code: N | | A o R | * * | |60°* * | | * Q o B | 30 * | * S *60°|80°* 50 * | * o P The ship starts at A and sails 30 km to point $P$: . . $AP = 30,\angle NAP = 120^o,\;\angle SAP = \angle APQ = 60^o$ Then it turns and sails 50 km to point $B$: . . $PB = 50,\;\angle QPB = 80^o \quad\Rightarrow\quad \angle APB = 140^o$ In $\Delta APB$, use the Law of Cosines: . . $AB^2 \:=\:AP^2 + PB^2 - 2(AP)(BP)\cos(\angle APB)$ . . $AB^2 \:=\:30^2+50^2 - 2(30)(50)\cos140^o \:=\:5698.133329$ Therefore: . $\boxed{AB \;\approx\;75.5\text{ km}}$ In $\Delta APB$, use the Law of Cosines. . . $\cos A \:=\:\frac{75.5^2 + 30^2 - 50^2}{2(75,5)(30)} \:=\:0.90413245$ Hence: . $\angle A \;\approx\;25.2^o$ Then: . $\angle BAS \:=\:25.2^o + 60^o \:=\:85.2^o \:=\:\angle ABR$ Therefore, the bearing of $A$ from $B$ is: . $360^o - 85.2^o \:=\:\boxed{274.8^o}$ 5. Originally Posted by Soroban Hello, mitchoboy! Bearings are measured clockwise from North. And a good diagram is essential. Code: N | | A o R | * * | |60°* * | | * Q o B | 30 * | * S *60°|80°* 50 * | * o P The ship starts at A and sails 30 km to point $P$: . . $AP = 30,\angle NAP = 120^o,\;\angle SAP = \angle APQ = 60^o$ Then it turns and sails 50 km to point $B$: . . $PB = 50,\;\angle QPB = 80^o \quad\Rightarrow\quad \angle APB = 140^o$ In $\Delta APB$, use the Law of Cosines: . . $AB^2 \:=\:AP^2 + PB^2 - 2(AP)(BP)\cos(\angle APB)$ . . $AB^2 \:=\:30^2+50^2 - 2(30)(50)\cos140^o \:=\:5698.133329$ Therefore: . $\boxed{AB \;\approx\;75.5\text{ km}}$ In $\Delta APB$, use the Law of Cosines. . . $\cos A \:=\:\frac{75.5^2 + 30^2 - 50^2}{2(75,5)(30)} \:=\:0.90413245$ Hence: . $\angle A \;\approx\;25.2^o$ Then: . $\angle BAS \:=\:25.2^o + 60^o \:=\:85.2^o \:=\:\angle ABR$ Therefore, the bearing of $A$ from $B$ is: . $360^o - 85.2^o \:=\:\boxed{274.8^o}$ thankyou for your answer. but how did you get apq as 60 degrees?