This article has been viewed 12,813 times. For example, if you have 10 digits to choose from for a combination lock with 6 numbers to enter, and you're allowed to repeat all the digits, you're looking to find the number of permutations with repetition. If you're seeing this message, it means we're having trouble loading external resources on our website. Discuss: answer with explanation. Mathematically we can approach this question as follows: $$P=\frac{n!}{n_1! To create this article, volunteer authors worked to edit and improve it over time. = n (n - 1) (n - 2) (n - 3) …. n_2! Solution: There are 4 letters in the word love and making making 3 letter words is similar to arranging these 3 letters and order is important since LOV and VOL are different words because of the order of the same letters L, O and V. Hence it is a permutation problem. Then, find 7! 21300: C. 24400: D. 210: View Answer. Yes. The basic building block to solve permutation and combination problems is to understand the use of recursion in a for loop. Some graphing calculators offer a button to help you solve permutations without repetition quickly. To create this article, volunteer authors worked to edit and improve it over time. 0! other thatn the given sequence in the argument of the function where c is in the last position. Important Permutation Formulas. Solution:You need two points to draw a line. We use cookies to make wikiHow great. Last Updated: October 14, 2020 A committee including 3 boys and 4 girls is to be formed from a group of 10 boys and 12 girls. This image may not be used by other entities without the express written consent of wikiHow, Inc. \n<\/p> \n<\/p><\/div>"}, https://www.mathsisfun.com/combinatorics/combinations-permutations.html, https://betterexplained.com/articles/easy-permutations-and-combinations/, https://www.learneroo.com/modules/10/nodes/62, https://en.wikipedia.org/wiki/Permutation#Permutations_with_repetition, https://www.vitutor.com/statistics/combinatorics/permutations_repetition.html, https://www.learneroo.com/modules/10/nodes/55, https://mathbits.com/MathBits/TISection/Algebra1/Probability.htm, https://www.mathsisfun.com/combinatorics/combinations-permutations-calculator.html, consider supporting our work with a contribution to wikiHow. If you really can’t stand to see another ad again, then please consider supporting our work with a contribution to wikiHow. At this point, we have to make the permutations of only one digit with the index 3 and it has only one permutation i.e., itself. P (n,r) represents the number of permutations of n items r at a time. A permutation is an arrangement of objects in which the order is important[1] This article has been viewed 12,813 times. This program will find all possible combinations of the given string and print them. You will more details about each type of problem in the problem definition section. In a certain country, the car number plate is formed by 4 digits from the digits 1, 2, 3, 4, 5, 6, 7, 8 and 9 followed by 3 letters from the alphabet. You'd then calculate 3,628,800/5,040. If you're using Google Calculator, click on the, If you have to solve by hand, remember that, for each. Find the order in lexicographical sorted order If we want to find all the permutations of a string in a lexicographically sorted order means all the elements are arranged in alphabetical order and if the first element is equal then sorting them based on the next elements and so on. Given a positive integer n and a string s consisting only of letters D or I, you have to find any permutation of first n positive integer that satisfy the given input string. Enter "7" for "Number of sample points in set ". wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. The order of arrangement of the object is very crucial. To solve this problem using the Combination and Permutation Calculator, do the following: Choose "Count permutations" as the analytical goal. In other words, if the set is already ordered, then the rearranging of its elements is called the process of permuting. In a certain state’s lottery, 48 balls numbered 1 through 48 are placed in a machine and six of them … letters in our case which is 6 \(n_1$$ is the number of objects of type 1, for example, the number of b which is 2 $$n_2$$ is the number of objects of type 2, for example, the number of a which is 1 Research source Iteration : : Iteration, in the context of computer programming, is a process wherein a set of instructions or structures are repeated in a sequence a specified number of times or until a condition is met. Mathematically we can approach this question as follows: $$P=\frac{n!}{n_1! We leave the 3 out and recursively find all permutations of the set {1, 2}. For example, consider finding all permutations of {1, 2, 3). The permutation problems are arrangement problems and the combination problems are selection problems. This image is not<\/b> licensed under the Creative Commons license applied to text content and some other images posted to the wikiHow website. in Physics and Engineering, Exercises de Mathematiques Utilisant les Applets, Trigonometry Tutorials and Problems for Self Tests, Elementary Statistics and Probability Tutorials and Problems, Free Practice for SAT, ACT and Compass Math tests, Arithmetic Sequences Problems with Solutions, High School Math (Grades 10, 11 and 12) - Free Questions and Problems With Answers, Mathematical Induction - Problems With Solutions, Geometric Sequences Problems with Solutions. You can use a simple mathematical formula to find the number of different possible ways to order the items. For example 7 and 4. In how many ways can you select a committee of 3 students out of 10 students? (3) (2) (1) Permutations of n items taken r at a time. Permutation With Repetition Problems With Solutions : In this section, we will learn, how to solve problems on permutations using the problems with solutions given below. The is below the up/down arrow, at the top left of the answer. You may also vote on the quality/helpfulness of an answer, with the up or down arrow, if you have a 15+ reputation. The problem is to select 2 points out of 3 to draw different lines. 25200: B. References. P (n,r) = n!/ (n - r)! This consists of 2 permutations: {1, 2} {2, 1] Now we insert the 3 into every position for these permutations. In mathematics, permutation relates to the act of arranging all the members of a set into some sequence or order. How many different committee can be formed from the group? Problem Statement. For instance, you might be selecting 3 representatives for student government for 3 different positions from a set of 10 students. In mathematics, a permutation of a set is, loosely speaking, an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its elements.The word "permutation" also refers to the act or process of changing the linear order of an ordered set. Plug your numbers in for n {\displaystyle n} and r {\displaystyle r}. ). }$$ Where: $$n$$ is the total number of object, i.e. Calculate the number of permutations of n elements taken r at the time. That number means that, if you're picking from 10 different students for 3 student government positions, where order matters and there is no repetition, there are 720 possibilities. wikiHow, Inc. is the copyright holder of this image under U.S. and international copyright laws. A. Let’s take an example to understand the problem, Input xyz Output xyz, xzy, yxz, yzx, zxy, zyx Explanation These are all permutations take in order. This kind of problem... 2. This method takes a list as an input and returns an object list of tuples that contain all … = 24 . }\) Where: $$n$$ is the total number of object, i.e. = 5*4*3*2*1 = 120. So out of that set of 4 horses you want to pick the subset of … By using our site, you agree to our. Solved Examples(Set 1) - Permutation and Combination. Is there an easy way in order to memorize the formula for this? letters in our case which is 6 $$n_1$$ is the number of objects of type 1, for example, the number of b which is 2 $$n_2$$ is the number of objects of type 2, for example, the number of a which is 1 Permutations is not an easy problem. Permutations and combinations are used to solve problems. A permutation is a reordered arrangement of elements or characters of a string. by doing (10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1), which gives you 3,628,800 as a result. Suppose we have three people named A, B, and C. We have already determined that they can be seated in a straight line in 3! / (4 - 3)! n_2! Permutation With Repetition Problems With Solutions - Practice questions. n_3!…n_k! n_3!…n_k! 1. Include your email address to get a message when this question is answered. Find Permutation. How many 3 digit numbers can we make using the digits 2, 3, 4, 5, and 6 without repetitions? % of people told us that this article helped them. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed? The first problem comes under the category of Circular Permutations, and the second under Permutations with Similar Elements. The C program to find permutation and combination solves 4 different types of problems. 1! The order is not important. Enter "3" for "Number of sample points in each permutation". Solve the equation to find the number of permutations. Think of it like this: subtract the total amount by the total items. means (n factorial). Permutations occur, in more or less prominent ways, in almost every area of mathematics. The number of words is given by 4 P 3 = 4! Now in this permutation (where elements are 2, 3 and 4), we need to make the permutations of 3 and 4 first. = 1 Let us take a look at some examples: Problem 1: Find the number of words, with or without meaning, that can be formed with the letters of the word ‘CHAIR’. Ending index of the string. A pemutation is a sequence containing each element from a finite set of n elements once, and only once. Calculating Permutations without Repetition 1. By signing up you are agreeing to receive emails according to our privacy policy. In the example, you should get 720. If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked. How many triangles can you make using 6 non collinear points on a plane? Second … Permutations . Intuition If there were no Kleene stars (the * wildcard character for regular expressions), the problem would be easier - we simply check from left to right if each character of the text matches the pattern. For those who haven’t seen a backtracking question before, there is no clear naive solution, and this poses a real threat for software engineers during… If you're working with combinatorics and probability, you may need to find the number of permutations possible for an ordered set of items. Start with an example problem where you'll need a number of permutations without repetition. (unlike combinations, which are groups of items where order doesn't matter[2] These methods are present in an itertools package. Hopefully, you have seen … n! To start off, you just need to know whether repetition is allowed in your problem or not, and then pick your method and formula accordingly. Calculating Permutations without Repetition, {"smallUrl":"https:\/\/www.wikihow.com\/images\/thumb\/b\/b0\/Calculate-Permutations-Step-1.jpg\/v4-460px-Calculate-Permutations-Step-1.jpg","bigUrl":"\/images\/thumb\/b\/b0\/Calculate-Permutations-Step-1.jpg\/aid11015428-v4-728px-Calculate-Permutations-Step-1.jpg","smallWidth":460,"smallHeight":345,"bigWidth":728,"bigHeight":546,"licensing":"

\u00a9 2021 wikiHow, Inc. All rights reserved. The permutation is an arrangement of objects in a specific order. Combinations P(n) = n! A new solution can be accepted if a better one shows up. D means the next number is smaller, while I means the next number is greater. to 4, so 7x6x5 and then find the answer, and you’ll get the permutations. Example 6: How many lines can you draw using 3 non collinear (not in a single line) points A, B and C on a plane? If your question is solved, say thank you by accepting the solution that is best for your needs. Solution: ‘CHAIR’ contains 5 letters. would be (7 * 6 * 5 * 4 * 3 * 2 * 1), which would equal 5,040. Given a set of ‘n’ elements, find their Kth permutation. How many 4 digit numbers can we make using the digits 3, 6, 7 and 8 without repetitions? Each question has four choices out … Approach #1 Using Stack [Accepted] Let's revisit the important points of the given problem statement. Therefore, the number of words that can be formed with these 5 letters = 5! First import itertools package to implement the permutations method in python. Thanks to all authors for creating a page that has been read 12,813 times. FYI: Thoroughly answering questions is time-consuming. Research source It usually looks like, All tip submissions are carefully reviewed before being published. Question 1 : 8 women and 6 men are standing in a line. If we proceed as we did with permutations, we get the following pairs of points to draw lines.AB , ACBA , BCCA , CBThere is a problem: line AB is the same as line BA, same for lines AC and CA and BC and CB.The lines are: AB, BC and AC ; 3 lines only.So in fact we can draw 3 lines and not 6 and that's because in this problem the order of the points A, B and C is not important.This is a combination problem: combining 2 items out of 3 and is written as follows: eval(ez_write_tag([[580,400],'analyzemath_com-large-mobile-banner-1','ezslot_4',700,'0','0'])); Calculate the number of combinations of n elements taken r at the time. Graphs of Functions, Equations, and Algebra, The Applications of Mathematics Permutations of the same set differ just in the order of elements. Permutation and Combination Solve Problems Quickly. For example, you would calculate 10! How many number plates can be formed if neither the digits nor the letters are repeated. Explanation: Number of ways of selecting 3 consonants from 7 X If you have a calculator handy, find the factorial setting and use that to calculate the number of permutations. Learn How to Solve Permutation and Combination Question Quickly form PrepInsta. Solution: n-factorial gives the number of permutations of n items. Find all Permutations of the word baboon. Consider the following set of elements: 7! Recursive functions are very useful to solve many mathematical problems, such as calculating the factorial of a number, generating Fibonacci series, etc. For the first permutation we insert … Circular Permutations . When a star is present, we may need to check many different suffixes of the text and see if they match the rest of the pattern. X = 1. wikiHow is where trusted research and expert knowledge come together. Python provides a package to find permutations and combinations of the sequence. Find all Permutations of the word baboon. The list of problems is given below. Permutation Problem 1 Choose 3 horses from group of 4 horses In a race of 15 horses you beleive that you know the best 4 horses and that 3 of them will finish in the top spots: win, place and show (1st, 2nd and 3rd). 4. wikiHow is a “wiki,” similar to Wikipedia, which means that many of our articles are co-written by multiple authors. Output First Swap generates the other permutation, i.e. 3. Introductory permutation problems. def permute (a, l, r): if l = = r: print toString (a) else: for i in xrange (l,r + 1 ): a [l], a [i] = a [i], a [l] permute (a, l + 1, r) a [l], a [i] = a [i], a [l] # backtrack. Line AB is the same as line BA. or 6 ways. The permutation test is designed to determine whether the observed difference between the sample means is large enough to reject, at some significance level, the null hypothesis H that the data drawn from is from the same distribution as the data drawn from. How many 6 letter words can we make using the letters in the word LIBERTY without repetitions? The number of permutations on the set of n elements is given by n! In how many ways can you arrange 5 different books on a shelf? Question.How many three digit numbers can be formed using digits 2, 3, 4, 7, 9 so that the digits can be repeated. Permutations differ from combinations, which are selections of some members of a set regardless of … PERMUTATION WORD PROBLEMS WITH SOLUTIONS Problem 1 : A student appears in an objective test which contain 5 multiple choice questions. Given a string, we have to find all the permutations of that string. Answer: Option A. Problem Statement. In this regard, what is the importance of permutation and combination? Input. Permutation is the arrangement of all parts of an object, in all possible orders of arrangement. Permutations with repetition n 1 – # of the same elements of the first cathegory n 2 - # of the same elements of the second cathegory For example, string “abc” have six permutations [“abc”, “acb”, “bac”, “bca”, “cab”, “cba”]. We know ads can be annoying, but they’re what allow us to make all of wikiHow available for free. We will be given a single string input. No student can be used in more than one position (no repetition), but the order still matters, since the student government positions are not interchangeable (a permutation where the first student is President is different from a permutation where they're Vice President). Please help us continue to provide you with our trusted how-to guides and videos for free by whitelisting wikiHow on your ad blocker. And thus, permutation(2,3) will be called to do so. Similarly, permutation(3,3) will be called at the end. Really can ’ t stand to see another ad again, then consider... 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# Of all polygons inscribed in a given circle which one has the maximum sum of squares of side lengths?

My son presented me with an interesting problem:

Of all possible polygons inscribed in a circle of radius $R$, find the one that has the sum $S$ of squared side lengths maximized: $S=a_1^2+a_2^2+\dots+a_n^2$, with $a_i$ representing the length of the $i$-th side. The number of sides is not fixed, you should consider all triangles, quadrilaterals, pentagons...

It's not that complicated, at least in the beginning. It's easy to show that the optimal polygon (with $n>3$) cannot have obtuse ($>90^\circ$) angles. For example, if such an angle $A_{i-1}A_{i}A_{i+1}$ exists, by cosine theorem:

$$|A_{i-1}A_{i}|^2+|A_{i}A_{i+1}|^2<|A_{i-1}A_{i+1}|^2$$

So if you drop vertex $A_i$, you get a polygon with a bigger $S$. This quickly eliminates all polygons with $n>4$.

All candidate polygons with $n=4$ must be rectangles and if their sides are $a$ and $b$, the sum $S$ is $2a^2+2b^2=8R^2$. So with respect to $S$ all rectangles inscribed in the circle are equivalent. In fact, a right triangle with sides $a$, $b$ and $2R$ has the same $S$ as any inscribed rectangle.

But maybe there is an inscribed triangle with $S>8R^2$. I was able to show that for an inscribed triangle with sides $a,b,c$ and $b\ne c$, an isosceles triangle with all acute angles and base $a$ has better value of $S$. So the optimal triangle must be isosceles. Looking from all three sides, the only possible solution is the equilateral triangle and the sum $S$ in that case is $9R^2$.

However, to prove that fact I had to use trigonometry which is not so complicated (and I can present it here if you want so), but it leaves impression that there has to be some simpler explanation why the equilateral triangle is the best choice. My trigonometry proof takes a few lines of text, I want something more elegant.

Just an idea: if you draw lines through the center of the circle perpendicular to the sides of a triangle and denote the pedal lengths with $h_a,h_b,h_c$, it's easy to show that in order to maximize $a^2+b^2+c^2$ you have to minimize $h_a^2+h_b^2+h_c^2$. And then - what?

EDIT: I want to present the part of the proof that I don't like. Take an arbitrary triangle $ABC$ with sides $a,b,c$ inscribed in a circle. Consider side $a$ fixed and play with angle $\gamma$ to get different values of $b,c$. I want to prove that isosceles triangle $BCA_1$ has bigger $S$ than any other triangle with one side equal to $a$.

$$b=2R\sin\frac{\pi-\alpha+\gamma}{2}=2R\cos\left(\frac\alpha2-\frac\gamma2\right)$$

$$c=2R\sin\frac{\pi-\alpha-\gamma}{2}=2R\cos\left(\frac\alpha2+\frac\gamma2\right)$$

$$b^2=4R^2\cos^2\left(\frac\alpha2-\frac\gamma2\right)=2R^2(1+\cos(\alpha-\gamma))$$

$$c^2=4R^2\cos^2\left(\frac\alpha2+\frac\gamma2\right)=2R^2(1+\cos(\alpha+\gamma))$$

$$b^2+c^2=4R^2+2R^2(\cos(\alpha-\gamma)+\cos(\alpha+\gamma))=4R^2(1+\cos\alpha\cos\gamma)$$

And this sum achieves maximum obviously for $\gamma=0$, or for $A\equiv A_1$. So for any given side $a$, $b$ and $c$ must be of equal. But you can look at the optimal triangle from sides $b$ and $c$ as well. The only triangle which has no better option is equilateral triangle.

EDIT 2: This “moving vertex” procedure can be repeated infinite number of times and the result is an equilateral triangle! Check excellent proof by Noah Schweber here.

• Note that your last sentence "The only triangle which has no better option is equilateral triangle." is insufficient to imply "The equilateral triangle is better than every other triangle.". See my answer for a sketch of how to do that part rigorously. – user21820 Sep 17 '18 at 17:45
• @Oldboy, I'm studying your topic and curiously I have a problem a little similar in here. The triangle's case was the most problematic for me too. – Na'omi Sep 18 '18 at 14:10

Yes, the maximal sum is the one of the equilateral triangle, that is $9R^2$.

Since Prove that in any triangle $ABC$, $\cos^2A+\cos^2B+\cos^2C\geq\frac{3}{4}$ then $$\sin^2 A+\sin^2 B+\sin^2 C=3-\cos^2 A-\cos^2 B-\cos^2 C\leq \frac{9}{4}$$ where $A$, $B$ and $C$ are non negative numbers such that $A+B+C=\pi$. Hence, for any inscribed triangle, the sum of the squares of the sides is $$(2R\sin A)^2+(2R\sin B)^2+(2R\sin C)^2\leq 9R^2.$$

• I have upvoted your answer, it's just that it's not much simpler compared with my solution. It uses a trigonometric fact about triangle that is not obvious. Anyway, if nothing better shows up in a day or two, I'll accept your answer as the best one. – Oldboy Sep 17 '18 at 9:27

This problem can be stated as

$$\max_{n}\sum_{k=1}^n\left(2r\sin\left(\frac{\theta_k}{2}\right)\right)^2$$

s. t.

$$\sum_{k=1}^n\theta_k = 2\pi$$

but

$$\sum_{k=1}^n\left(2r\sin\left(\frac{\theta_k}{2}\right)\right)^2\ge n\left(2^{2n}r^{2n}\prod_{k=1}^n\sin^2\left(\frac{\theta_k}{2}\right)\right)^{\frac 1n}$$

assuming $\theta_1=\cdots=\theta_n$ we have

$$\sum_{k=1}^n\left(2r\sin\left(\frac{\theta_k}{2}\right)\right)^2\ge n\left(2^{2n}r^{2n}\sin^{2n}\left(\frac{\pi}{n}\right)\right)^{\frac 1n} = n2^2r^2\sin^{2}\left(\frac{\pi}{n}\right)$$

Now calling

$$f(n) = n\sin^{2}\left(\frac{\pi}{n}\right)$$

we have clearly a maximum about $n = 3$ as can be depicted in the attached plot

• This is interesting, but as far as I can tell it assumes that the polygon is regular; and in some cases an irregular polygon can do better for fixed $n$. For example, for a regular hexagon, we have $S = 6 R^2$. But for a hexagon where four of the points are degenerate, the hexagon is "basically" a triangle with three sides of length zero; and if we pick the other three sides to have equal length, we have an equilateral triangle with $S = 9 R^2$. – Michael Seifert Sep 17 '18 at 14:50

Let $\theta_k$ be the successive angles subtended by the sides, but the last one. The sum of squares is given by

$$4\sum_{k=1}^n\sin^2\frac{\theta_k}2+4\sin^2\left(\pi-\frac12\sum_{k=1}^n\theta_k\right)$$ which has the same extrema as $$\sum_{k=1}^n\cos\theta_k-\cos\left(\sum_{k=1}^n\theta_k\right).$$

$$\sin\theta_k=\sin\left(\sum_{k=1}^n\theta_k\right).$$

This shows that all angles $\theta_k$ must be equal, and then

$$n\cos\theta-\cos n\theta$$ is minimized with $n\theta=2\pi$.

Finally,

$$n\cos\frac{2\pi} n-\cos\pi$$ is the smallest with $n=3$.

The objective function is continuous on the domain of interest (all triples of points on the circle), which is also compact. Therefore by the extreme value theorem it has a global maximum. That reduces the problem to the part you are interested in, namely proving that if the optimal triangle has sides $a,b,c$ then $b = c$. Firstly it has to be acute as you observed. Thus maximizing $b^2+c^2$ $= a^2+2bc·\cos(\angle A)$ is equivalent to maximizing $bc$, since fixing $B,C$ fixes $a$ and $\angle A$. Letting $x = \angle BAO$ and $y = \angle OAC$ we have $bc = 4R^2·\cos(x)\cos(y)$, and finally note $2 \cos(x)\cos(y)$ $= \cos(x+y) + \cos(x-y)$ $\le \cos(x+y) + 1$ $= \cos(\angle A) + 1$ with equality exactly when $x=y$.

• Part of the reason for the first part of my answer is that I have a small suspicion that you did not get that part right. One cannot automatically claim that every optimization problem has a global extremum. This has to be proven, and you cannot do so via an infinite iterative process unless you prove that it converges. One way around this is to use the extreme value theorem if the domain is compact, as here. – user21820 Sep 17 '18 at 17:43

\begin{align} AB^2 + BC^2 + CA^2 &= (\vec{OB} - \vec{OA})^2 + (\vec{OC} - \vec{OB})^2 + (\vec{OA} - \vec{OC})^2\\ &= 2(\vec{OA})^2 + 2(\vec{OB})^2 + 2(\vec{OC})^2 - 2 \times \vec{OA} \cdot \vec{OB} - 2 \times \vec{OB} \cdot \vec{OC} - 2 \times \vec{OC} \cdot \vec{OA} \\ &= 3(\vec{OA})^2 + 3(\vec{OB})^2 + 3(\vec{OC})^2 - (\vec{OA} + \vec{OB} + \vec{OC})^2 \\ &\leq 9R^2. \end{align}

• I would try to avoid $\times$, as it usually denotes cross product in this context. Also, this appears to be simply proving that the value for an equilateral triangle is $9R^2$, rather than, as the OP asks, proving that this is maximal. – Acccumulation Sep 17 '18 at 15:20
• @Acccumulation sorry, what exactly is unclear in proving that $AB^2+BC^2+CA^2 \leq 9R^2$ for an arbitrary inscribed triangle? – Roman Odaisky Sep 17 '18 at 15:23

We want to maximise $b^2+c^2$, which by the cosine rule is equal to $a^2+2bc\cos A$.

Angle $\angle BAC$ is fixed, so this means maximising $bc$.

The area of the triangle is $\frac12bc\sin A$, and $\sin A$ is fixed, so this means maximising the area of the triangle.

The area of the triangle is $\frac12$base$\times$height $= \frac12a\times$ height, so this means maximising the height.

And the height is maximum when $\triangle BA_1C$ is isosceles.

Suppose we have three unit vectors a, b, and c. This will define a triangle with side lengths (a-b), (a-c), and (b-c), so the sum of squares will be (a-b)^2+(a-c)^2+(b-c)^2. Taking the derivative with respect to a, we get 2a'(a-b)+2a'(a-c)=2a'(2a-(b+c)). Because a is constrained to be on the unit circle, a' is perpendicular to a, hence a'a=0. So the derivative simplifies to -2a'(b+c). Thus, the derivative is zero if a' is perpendicular to b+c, which is equivalent to a being parallel to b+c, which happens when the angle between a and b is equal to the angle between a and c. Applying the same argument to the derivatives with respect to b and c shows that all the angles must be equal.


# A definite integral $\int_0^\infty\frac{2-\cos x}{\left(1+x^4\right)\,\left(5-4\cos x\right)}dx$

I need to find a value of this definite integral: $$\int_0^\infty\frac{2-\cos x}{\left(1+x^4\right)\,\left(5-4\cos x\right)}dx.$$ Its numeric value is approximately $0.7875720991394284$, and lookups in Inverse Symbolic Calculator Plus and WolframAlpha did not return a plausible closed-form candidate.

Do you have any ideas how I can approach this problem?

• Looks like you already have a value there. Do you have a particular reason to think it has a closed form? – Henning Makholm Sep 28 '13 at 23:55
• @HenningMakholm One should always believe there is a closed form. It is just that some functions and constants are not yet well studied, understood and named :) – Vladimir Reshetnikov Sep 29 '13 at 0:06
• @hmedan.mnsh Just to clarify: Mathematica does not return a closed form solution, but can give a numerical approximation using NIntegrate. – Vladimir Reshetnikov Sep 29 '13 at 0:31
• @HenningMakholm It seems to me the phrase "you already have a value" is somewhat objectionable. A numerical integration in this case converges quite slowly, and I doubt that one can get more than $25$ correct digits in a reasonable time (and it would be a very difficult to make sure they all are indeed correct). On the other hand, having a closed form, I can compute $10^5$ digits in about a second. And I can be pretty sure all these digits are correct, because numerical algorithms for elementary functions have been thoroughly designed and polished for years, and verified for correctness. – Vladimir Reshetnikov Sep 29 '13 at 18:11
• @StevenStadnicki I do not need $10^5$ digits, I used this number just to demonstrate the speed of calculations. But I can easily imagine that someone wants to get $30$ digits that are certainly correct. It is trivial task when one has an elementary closed form, but it might be quite difficult if one has to resort to numerical integration. Also, closed forms sometimes have a useful property to partially cancel each other or otherwise simplify when several factors or terms are combined into a single expression, while combining multiple approximate numeric values results in a precision loss. – Vladimir Reshetnikov Oct 1 '13 at 18:15

Yes, there is an elementary closed form for this integral: $$\int_0^\infty\frac{2-\cos x}{\left(1+x^4\right)\,\left(5-4\cos x\right)}dx=\frac{\pi}{2\,\sqrt2}\cdot\exp\left(\frac1{\sqrt2}\right)\cdot\frac{\sin\left(\frac1{\sqrt2}\right)-\cos\left(\frac1{\sqrt2}\right)+2\,\exp\left(\frac1{\sqrt2}\right)}{1-4\,\exp\left(\frac1{\sqrt2}\right)\cos\left(\frac1{\sqrt2}\right)+4\,\exp\left(\sqrt2\right)}\tag1$$

Proof:

Let us denote the integral in question as $$\mathcal{I}=\int_0^\infty\frac{2-\cos x}{\left(x^4+1\right)\,\left(5-4\cos x\right)}dx\tag2$$ Note that the trigonometric part of the integrand is a periodic function and can be expanded to a Fourier series with particularly simple coefficients: $$\frac{2-\cos x}{5-4\cos x}=\sum_{n=0}^\infty\frac{\cos(n\,x)}{2^{n+1}}\tag3$$ (this can be easily checked by expressing cosines via exponents of an imaginary argument).

Now we can integrate it term-wise: $$\mathcal{I}=\sum_{n=0}^\infty\left(\frac1{2^{n+1}}\int_0^\infty\frac{\cos(n\,x)}{x^4+1}dx\right)=\sum_{n=0}^\infty\left(\frac1{2^{n+1}}\cdot\frac{\pi}{2\,\sqrt2}\cdot\exp\left(-\frac{n}{\sqrt2}\right)\cdot\left(\sin\left(\frac{n}{\sqrt2}\right)+\cos\left(\frac{n}{\sqrt2}\right)\right)\right)\tag4$$ (for the integral, see DLMF 1.14, vii, Table 1.14.2, $4^{th}$ row).

Trig functions in the last sum can again be expressed via exponents of an imaginary argument, and then the sum is easily evaluated. Converting exponents back to trig functions and getting rid of complex numbers, we get the final result $(1)$.

• Clever expansion! – Ron Gordon Sep 29 '13 at 17:51
• Very nice answer! – Start wearing purple Sep 29 '13 at 17:53

There is an alternate way to compute this integral w/o a summation over $n$ in the middle steps.

Notice

$$\frac{2-\cos z}{5 - 4\cos z} = \frac12 \left[\frac{(2-e^{iz})+(2-e^{-iz})}{(2-e^{iz})(2-e^{-iz})}\right] = \frac12\left[\frac{1}{2-e^{iz}} + \frac{1}{2-e^{-iz}}\right]$$ and $\displaystyle\;\frac{1}{1+z^4}$ is an even function, we have

$$\mathcal{I} \stackrel{def}{=} \int_0^\infty \frac{2-\cos x}{(1+x^4)(5-4\cos x)} dx = \frac12\int_{-\infty}^\infty \frac{1}{(1+z^4)(2 - e^{iz})}dz$$

Since $\displaystyle\;\frac{1}{2-e^{iz}}$ is entire on the upper half plane $\Im z \ge 0$ and $\displaystyle\;\left|\frac{1}{2-e^{iz}}\right| \le \frac{1}{2-1} = 1$ there, we can complete the contour in the upper half-plane and

$$\mathcal{I} = \lim_{R\to\infty}\frac12 \oint_{C_R} \frac{1}{(1+z^4)(2 - e^{iz})}dz \quad\text{ where }\quad C_R = [-R, R ] \cup \big\{\; Re^{i\theta} : \theta \in [0,\pi]\big\}$$

$\displaystyle\;\frac{1}{1+z^4}$ has $4$ poles $\omega_k = e^{\frac{(2k+1)i}{4\pi}}, k = 0..3$ over $\mathbb{C}$. Two of them $\omega_0 = \frac{1+i}{\sqrt{2}}$ and $\omega_1 = \frac{-1+i}{\sqrt{2}}$ belongs to the upper half-plane. Since

$$\frac{1}{1+z^4} = \sum_{k=0}^3 \frac{1}{(z-\omega_k) 4\omega_k^3} = -\frac14 \sum_{k=0}^3\frac{\omega_k}{z-\omega_k}$$

The residue of the integrand at $\omega_k$ are $\displaystyle\;-\frac14 \frac{\omega_k}{2 - e^{i\omega_k}}$ for $k = 0, 1$. This leads to

\begin{align} \mathcal{I} &= \frac12\left[ -\frac{2\pi i}{4}\left(\frac{\omega_0}{2 - e^{i\omega_0}} + \frac{\omega_1}{2 - e^{i\omega_1}} \right)\right]\\ &= \frac{-\pi i}{4\sqrt{2}}\left( \frac{1+i}{2 - e^{-1/\sqrt{2}} e^{i/\sqrt{2}}} + \frac{-1+i}{2 - e^{-1/\sqrt{2}} e^{-i/\sqrt{2}}}\right)\\ &= \frac{-\pi i}{4\sqrt{2}} \left[ \frac{ (1+i)(2 - e^{-1/\sqrt{2}} e^{-i/\sqrt{2}}) + (-1+i)(2 -e^{-1/\sqrt{2}} e^{ i/\sqrt{2}}) }{ 4 - 4 e^{-1/\sqrt{2}}\cos\left(\frac{1}{\sqrt{2}}\right) + e^{-\sqrt{2}} } \right]\\ &= \frac{\pi}{2\sqrt{2}} \left[ \frac{2 - e^{-1/\sqrt{2}}\left( \cos\left(\frac{1}{\sqrt{2}}\right) - \sin\left(\frac{1}{\sqrt{2}}\right) \right) }{ 4 - 4 e^{-1/\sqrt{2}}\cos\left(\frac{1}{\sqrt{2}}\right) + e^{-\sqrt{2}} } \right]\\ &= \frac{\pi e^{1/\sqrt{2}} }{2\sqrt{2}} \left[ \frac{2 e^{1/\sqrt{2}} - \left( \cos\left(\frac{1}{\sqrt{2}}\right) - \sin\left(\frac{1}{\sqrt{2}}\right) \right) }{ 4 e^{\sqrt{2}} - 4 e^{1/\sqrt{2}}\cos\left(\frac{1}{\sqrt{2}}\right) + 1 } \right] \end{align} Reproducing what Vladimir get in his answer.


# Show that you can distribute powers to commuting elements

## Homework Statement

If ##a## and ##b## are commuting elements of ##G##, prove that ##(ab)^n = a^nb^n## for all ##n \in \mathbb{Z}##.

## The Attempt at a Solution

We prove two lemmas:
1) If ##a## and ##b## commute, then so do their inverses: ##ab=ba \implies (ab)^{-1} = (ba)^{-1} \implies b^{-1}a^{-1} = a^{-1}b^{-1}##.

2) If ##a## and ##b## commute, then ##b^n a = ab^n##: Base case is trivial. Suppose for some ##k## we have ##b^ka = ab^k##. Then ##b^{k+1}a = bb^ka = bab^k = abb^k = ab^{k+1}##.

Now to the actual result.

Clearly ##(ab)^0 = a^0b^0##. So first we prove the result for the positive integers, by induction. The base case is trivial. Suppose for some ##k \in \mathbb{Z}^+## we have ##(ab)^k = a^kb^k##. Then ##(ab)^{k+1} = (ab)^k(ab) = a^kb^kab = a^kab^kb = a^{k+1}b^{k+1}##.

Now we prove the result for negative integers. ##(ab)^{-n} = (b^{-1}a^{-1})^n = (a^{-1}b^{-1})^n = (a^{-1})^n(b^{-1})^n = a^{-n}b^{-n}##.

Does this argument work? Were the lemmas really necessary or could I have assumed they held since their proofs are trivial?

mfb
Mentor
since their proofs are trivial?
So is the whole statement you have to show. Better to show it explicitly.

fresh_42
Mentor
There is another argument, why you will not need to prove the negative ones explicitly: If you prove the statement for all ##n \in \mathbb{N}_0## and for all ##a,b \in G##, then you have also proven it for inverse elements. This is in words what you have written.

To do it by induction is a very formal way to prove it. In cases like the above, some dots will equally be acceptable, although from a logical point of view certainly not sufficient. But with the dots, every reader knows how the induction goes.

Mr Davis 97
There is another argument, why you will not need to prove the negative ones explicitly: If you prove the statement for all ##n \in \mathbb{N}_0## and for all ##a,b \in G##, then you have also proven it for inverse elements. This is in words what you have written.

To do it by induction is a very formal way to prove it. In cases like the above, some dots will equally be acceptable, although from a logical point of view certainly not sufficient. But with the dots, every reader knows how the induction goes.
But isn't it the case that I am not proving for all ##a,b \in G##, rather just in the case ##a,b## commute?

fresh_42
Mentor
But isn't it the case that I am not proving for all ##a,b \in G##, rather just in the case ##a,b## commute?
Yes, sure. But that doesn't change by taking the inverses: they commutate if and only if ##a## and ##b## do.

Mr Davis 97
StoneTemplePython


# transpose matrix definition

A new matrix is obtained the following way: each [i, j] element of the new matrix gets the value of the [j, i] element of the original one. Transpose of a matrix is the interchanging of rows and columns. In practical terms, the matrix transpose is usually thought of as either (a) flipping along the diagonal entries or (b) “switching” the rows for columns. Related to Transpose of a matrix: adjoint of a matrix , Inverse of a matrix What happened? Noun . So we now get that C transpose is equal to D. Or you could say that C is equal to D transpose. The number $$4$$ was in the first row and the second column and it ended up in the second row and first column. In other words if A= [aij], then At ji = aij. The matrix B is called the transpose of A. This is the definition of a transpose. The way the concept was presented to me was that an orthogonal matrix has orthonormal columns. The transpose of a matrix is defined as a matrix formed my interchanging all rows with their corresponding column and vice versa of previous matrix. For permissions beyond the … See more. Definition of transpose : a matrix formed by interchanging the rows and columns of a given matrix - change the order or arrangement of - transfer from one place or period to another - cause to change places - transfer a quantity from one side of an equation to the other side reversing its sign, in order to maintain equality Now this is pretty interesting, because how did we define these two? Find the transpose of that matrix. For example if you transpose a 'n' x 'm' size matrix you'll get a new one of 'm' x … synonym : reverse . Definition. The transpose of an m × n matrix A is the n × m matrix A T whose rows are the columns of A. In other words, transpose of A[][] is obtained by changing A[i][j] to A[j][i]. Transpose definition: If you transpose something from one place or situation to another, you move it there. So if X is a 3x2 matrix, X' will be a 2x3 matrix. Dictionary ! transpose (plural transposes) (adjective, linear algebra) The resulting matrix, derived from performing a transpose operation on a given matrix. Solution: It is an order of 2*3. This is a transpose which is written and A superscript T, and the way you compute the transpose of a matrix is as follows. Let A be a nonsingular matrix. If there are two Matrix with dimension A (2 x 3 ) and B (3 x 2). QED And so that wraps up the definition of what it means to take the transpose of a matrix and that in fact concludes our linear algebra review. Meaning of Transpose. Here are a couple of ways to accomplish this in Python. The conjugate transpose of a matrix is the matrix defined by where denotes transposition and the over-line denotes complex conjugation. All content on this website, including dictionary, thesaurus, literature, geography, and other reference data is for informational purposes only. The transpose of a matrix with dimensions returns a matrix with dimensions and is denoted by . The definition of the transpose is as follows. en.wiktionary.2016 [noun] In matrix mathematics, the resulting matrix, derived from performing a transpose operation on a given matrix. The first thing to know is that you can separate rows by semi-colons (;) and that you define rows by just placing elements next to one another. Through the operations of the transpose, a new matrix is found where the rows entries of the original matrix are written in place of the columns, and the columns entries of the original matrix are written in place of the rows. Do the transpose of matrix. Definition. Matrix transpose: The transpose of matrix refers to the interchanging of the rows and columns. permute, commute, transpose (verb) change the order or arrangement of Remember that the complex conjugate of a matrix is obtained by taking the complex conjugate of each of its entries (see the lecture on complex matrices). In other words, the ij entry of A T is a ji. Initialize a 2D array to work as matrix. The first column became the first row and the second column became the second row. "transpose" (matrix) definition: a matrix formed by interchanging the rows and columns of a given matrix. We said that our matrix C is equal to the matrix product A and B. The transpose of a matrix by Duane Q. Nykamp is licensed under a Creative Commons Attribution-Noncommercial-ShareAlike 4.0 License. (Problems and Solutions in Linear Algebra. ) A double application of the matrix transpose achieves no change overall. | Meaning, pronunciation, translations and examples And we said that D is equal to our matrix product B transpose times A transpose. At t = A; 2. In order to state the transpose property, we need to define the transpose of a matrix. It is denoted as X'. Then At, the transpose of A, is the matrix obtained by interchanging the rows and columns of A. The transpose will also be of dimension (2x2). A matrix is usually shown by a capital letter (such as A, or B) Each entry (or "element") is shown by a lower case letter with a "subscript" of row,column: In this video we discuss about the another type of #Transpose of #Matrix with definition and example.If you have a question about then comment here . Dictionary Thesaurus Examples Sentences Quotes ... A matrix obtained by interchanging the rows and columns of a given matrix. Disclaimer. This will be the left hand side of (AB)⊺=B⊺A⊺ Solving for right hand side, if I take transpose of A and B then the dimension of resultant matrix … By, writing another matrix B from A by writing rows of A as columns of B. Do the transpose of matrix. And that's it. Stack Exchange Network. Consequently At is n m. Here are some properties: 1. We can find its transpose by swapping the column and row elements as follows. Definition of Transpose in the Definitions.net dictionary. We put a "T" in the top right-hand corner to mean transpose: Notation. Adjacency Matrix Definition. ... (0.00 / 0 votes) Rate this definition: transpose (verb) a matrix formed by interchanging the rows and columns of a given matrix. Let’s start by defining matrices. To "transpose" a matrix, swap the rows and columns. transpose: To reverse or transfer the order or place of; interchange. Ask Question Asked 4 years, 3 months ago. Derived terms We prove that the transpose of A is also a nonsingular matrix. The algorithm of matrix transpose is pretty simple. Consider the following example-Problem approach. Thus the $$3\times 2$$ matrix became a $$2\times 3$$ matrix. Therefore it occurred to me that the definition in the book of Weinberg is not consistent with that in the book of Tung: in one of them the symbol ${\Lambda_\mu}^\nu$ is defined as the inverse of the Lorentz transformation of contravariant vectors, while in the other case, the same symbol is defined as the transpose of the original matrix. What does Transpose mean? TRANSPOSE OF A MATRIX DEFINITION. Store values in it. We have: . The element at ith row and jth column in X will be placed at jth row and ith column in X'. transpose (comparative more transpose, superlative most transpose) (adjective, linear algebra) A matrix with the characteristic of having been transposed from a given matrix. Synonyms: tr. Menu. Definition. For a matrix defined as = , the transpose matrix is defined as = . For example, consider a matrix . (The transpose of a matrix) Let Abe an m nmatrix. Find transpose by using logic. Transpose definition, to change the relative position, order, or sequence of; cause to change places; interchange: to transpose the third and fourth letters of a word. Example 1: Consider the matrix . Learn more. Recommended: Please solve it on “ PRACTICE ” first, before moving on to the solution. Dimension also changes to the opposite. (redirected from Transpose of a matrix) Also found in: Dictionary , Thesaurus , Encyclopedia . Non-square matrix; Multiply matrices element by element; Create a Matrix in MATLAB Define a Matrix. ... Why is the inverse of an orthogonal matrix equal to its transpose? Example 2: Consider the matrix . Or is it a definition? Transpose of a matrix is obtained by changing rows to columns and columns to rows. [noun] In matrix mathematics, the process of rearranging elements in a matrix, by interchanging their respective row and column positional indicators. 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# Proving the existence and number of *real* roots for $x^3 - 3x + 2$

I need to find how many real roots this polynomial has and prove there existence. I was wondering if my logic and thought process was correct.

Determine the number of real roots and prove it for $x^3 - 3x + 2$

First, note that $f'(x) = 3x^2 - 3$ and so

$f'(x) > 0$ for $x \in (-\infty, -1) \cup (1, \infty)$ and since $f'$ is strictly increasing on those intervals, there can be at most one root in each of them.

$f'(x) < 0$ for $x \in (-1,1)$ and since $f'$ is strictly decreasing on this interval it can have at most one root.

Now examine $f(-3) = -16$ and $f(-1) = 4$. By the Intermediate Value Theorem (IVT) $f(c) = 0$ for some $c \in (-3, 1)$ and so $f$ has a root on the interval $(-\infty, 1)$.

Again examine $f(-1) = 4$ and $f(1) = 0$. We cannot say anything about $f$ having a root on the interval $(-1, 1)$.

Likewise examine $f(1) = 0$ and $f(3) = 16$. Again, we cannot say anything about $f$ having a root on $(1, \infty)$.

However, $f(1) = 1 - 3 + 2 = 0$ is clearly a root. And by factorizing the polynomial we get $f(x) = (x+2)(x-1)^2$. Indeed, $1$ is a root with a multiplicity of two.

Hence, $f(x)$ has two real roots.

Also, do we say two real roots (because of the multiplicity), or three real roots, or do we say two distinct real roots?

While I realize factoring the polynomial gives me the answer I believe the purpose of the question was to do the former analysis, which when the polynomial isn't easily factorized, can provide a lot of insight. That is why I did it all

• I assume you mean real roots, because it has 3 complex roots. – ÍgjøgnumMeg Apr 4 '17 at 13:51
• "First, note that $f′(x)=3x-3$ and so", you mean $f′(x)=3x^{\color{red}{2}}-3$...? – StackTD Apr 4 '17 at 13:52
• Haha yes! Let me make these adjustments. – student_t Apr 4 '17 at 13:52

However, $f(1) = 1 - 3 + 2 = 0$ is clearly a root. And by factorizing the polynomial we get $f(x) = (x+2)(x-1)^2$. Indeed, $1$ is a root with a multiplicity of two.

All the work you did before this becomes unnecessary; after factoring, the roots (and hence the number of roots) are clear - right?

Addition after some comments: when you are asked about the number of roots (real or not), it is usually meant to count the number of distinct (i.e. different) roots. Your equation has two (real) roots, one of which has multiplicity 2 but that doesn't change the fact that there are only two real numbers where the polynomial becomes 0.

• Yes, but I think the purpose of the exercise was to do the former analysis. I only did that to show the multiplicity of the root. But yes in general all my work before would have been a waste haha! – student_t Apr 4 '17 at 13:55

Since we have $$x^3-3x+2=(x-1)^2(x+2),$$ we have three real roots $1,1,-2$. Here we count with multiplicities (which is standard for many results in geometry and other areas).

• I would say there are two real roots, one of which has multiplicity two. – gandalf61 Apr 4 '17 at 13:56
• Yes, but I am suppose to do the little analysis before for the question I believe. Of course factoring would be much faster. – student_t Apr 4 '17 at 13:57
• @danny Yes, this may be the case. But I think, it does not matter so much what you are supposed to do or think, but what you yourself think is the best way. – Dietrich Burde Apr 4 '17 at 13:59
• I agree with gandalf61. In the context of the fundamental theorem of algebra, we often say an $n$th-order polynomial "has $n$ complex roots" but this is an abbreviation where we mean to count the multiplicities. That doesn't change the fact that $x^3$ only has one (distinct) root. – StackTD Apr 4 '17 at 13:59
• @danny see comment above; I would say yours has two (distinct) roots, we usually omit 'distinct' and mean different roots when we're talking about the number of roots. – StackTD Apr 4 '17 at 14:00


# Integral of $\sin x \cdot \cos x$ [duplicate]

I've found 3 different solutions of this integral. Where did I make mistakes? In case there is no errors, could you explain why the results are different?

$\int \sin x \cos x dx$

1) via subsitution $u = \sin x$ $u = \sin x; du = \cos x dx \Rightarrow \int udu = \frac12 u^2 \Rightarrow \int \sin x \cos x dx = \frac12 \sin^2 x$

2) via subsitution $u = \cos x$ $u = \cos x; du = -\sin x dx \Rightarrow -\int udu = -\frac12 u^2 \Rightarrow \int \sin x \cos x dx = -\frac12 \cos^2 x = -\frac12 (1 - \sin^2 x) = -\frac12 + \frac12 \sin^2 x$

3) using $\sin 2x = 2 \sin x \cos x$

$\int \sin x \cos x dx = \frac12 \int \sin 2x = \frac12 (- \frac12 \cos 2x) = - \frac14 \cos 2x = - \frac14 (1 - 2 \sin^2 x) = - \frac14 + \frac12 \sin^2 x$

So, we have: $$\frac12 \sin^2 x \neq -\frac12 + \frac12 \sin^2 x \neq - \frac14 + \frac12 \sin^2 x$$

• Ahh! I always give this problem in my first year calculus course. – Jyrki Lahtonen May 4 '13 at 14:40
• $+C$... ${}{}{}$ – David Mitra May 4 '13 at 14:40
• $*$ is usually used for convolution in this context. I removed it. – Ayman Hourieh May 4 '13 at 14:42
• @AymanHourieh: Didn't I? – Inceptio May 4 '13 at 14:44
• @Inceptio Check the edit history. I edited the body; you edited the title. – Ayman Hourieh May 4 '13 at 14:49

Antiderivatives are only unique up to adding a constant ('of integration'). If you were to stick limits in your integrals then you'd always get the same number.

• Oh, yes. I checked it with 0 and \pi/2 and those 'strange' fractions deducted each other. Thank you – Jimch May 4 '13 at 14:46

Note: You are calculating indefinite integral and constants can be anything(they may differ). In fact the general solution to that would be just $C+\dfrac{\sin^2 x}{2}$

$$\frac{d\{f(x)+c\}}{dx}=f'(x)$$ for any arbitrary constant $c$

$$\implies \int f'(x)dx=f(x)+d$$ for any arbitrary constant $d$

So, in indefinite integral we can get answers which differ by some constant

A primitive is unique up to a constant


# How to show that $\sum_{k=1}^n k(n+1-k)=\binom{n+2}3$?

While thinking about another question I found out that this equality might be useful there: $$n\cdot 1 + (n-1)\cdot 2 + \dots + 2\cdot (n-1) + 1\cdot n = \frac{n(n+1)(n+2)}6$$ To rewrite it in a more compact way: $$\sum_{k=1}^n k(n+1-k)=\frac{n(n+1)(n+2)}6.$$

This equality is relatively easy to prove: $$\sum_{k=1}^n k(n+1-k)= (n+1)\sum_{k=1}^n k - \sum_{k=1}^n k^2 = (n+1) \frac{n(n+1)}2 - \frac{n(n+1)(2n+1)}6 = n(n+1) \left(\frac{n+1}2-\frac{2n+1}6\right) = n(n+1)\frac{3(n+1)-(2n+1)}6 = \frac{n(n+1)(n+2)}6.$$ (We only used the known formulas for the sum of the first $n$ squares and the sum of the first $n$ numbers.)

Are there some other nice proofs of this equality? (Induction, combinatorial arguments, visual proofs, ...)

EDIT: Now I found another question which asks about the same identity: Combinatorial interpretation of a sum identity: $\sum_{k=1}^n(k-1)(n-k)=\binom{n}{3}$ (I have tried to search before posting. But the answers posted here so far gave me some new ideas for good keywords to search which lead me to finding that question.) The questions are, in my opinion, not exact duplicates since the other question asks specifically about combinatorial proofs and my question does not have that restriction. But I agree that this is a very minor distinction. In any case, if you think that one of them should be closed as a duplicate, then you can vote to close. I will refrain from voting to close/reopen on this question. (If one of the two questions is voted to be a duplicate of the other one, they probably cannot be merged, since the summation variables are off by one.)

• Generating functions? Coefficient of $x^{n+1}$ in $\left(\sum\limits_{n=1}^{\infty} nx^{n}\right)^2$ – sciona Jan 21 '15 at 14:33
• Here is another post about the same formula. – Martin Sleziak Oct 2 '15 at 17:31

Let us choose three numbers from $\{0,1,2,\ldots, n+1\}$, beginning with the middle one, which has to be some $k\in \{1,\ldots,n\}$. We then have $k$ choices for the smallest and $n+1-k$ choices for the largest of the three. It follows that $${n+2\choose3}=\sum_{k=1}^n k(n+1-k)\ .$$

• Straight to the point! +1 – user2345215 Jan 21 '15 at 15:02

I think of this as the "twelve days of Christmas equality", because if $n = 12$ then we get $1 \times 12 + 2 \times 11 + \cdots + 12 \times 1 = {14 \choose 3}$ and both sides represent the number of gifts which are given in the song "The Twelve Days of Christmas". (This happens to be 364, one less than the number of days in a year.)

Here's a combinatorial proof of that equality, which I have previously written about at my blog. As I wrote there, let's try to prove the identity $\sum_{j=2}^{n+1} (j-1)(n+2-j) = {n+2 \choose 3}$, which differs from your equality by a change of index. To do this, we count subsets of the set $\{ 1, 2, \ldots, n+2 \}$ of size 3. We can write each such subset as $\{ x, y, z \}$ where we require $x < y < z.$ Then we’ll count these subsets according to the difference $z - x$. To construct such a set with $z - x = j$ we must:

• choose $x$. $x$ must be between $1$ and $n+2-j$ inclusive, so there are $n+2-j$ possible choices.
• choose $y$. $y$ must be between $x$ and $z=x+j$ exclusive, so there are $j-1$ possible choices.

At this point $z$ is fixed. So there are $(j-1)(n+2-j)$ ways to choose a $3$-set with $z - x = j$; summing over the possible values of $j$ gives the desired identity.

• I guess more-or-less the same argument could be reformulated like this. Let us start by choosing $y$. We only can choose $2, 3, \dots, n+1$. To make the notation the same as in your sum, let us denote $j=y$. If $y$ is chosen, we can choose any of elements on the left for $x$, which gives $(j-1)$ possibilities. We can choose any of the elements on the right for $z$, which gives $(n+2-j)$ possibilities. Together we have $\sum_{j=2}^{n+1} (j-1)(n+2-j)$. BTW +1 for the nice combinatorial argument and thanks for the link to your blog. – Martin Sleziak Jan 21 '15 at 14:46

A convolution is always a good way. Since: $$\sum_{k\geq 0} k z^k = \frac{z}{(1-z)^2}, \tag{1}$$ we have: $$\begin{eqnarray*}\sum_{k=1}^{n}k(n+1-k) = \sum_{k=0}^{n+1}k(n+1-k) &=& [z^{n+1}]\frac{z^2}{(1-z)^4}\\&=&[z^n]\frac{z}{(1-z)^4}\tag{2}\end{eqnarray*}$$ and the claim follows from: $$\sum_{k\geq 0}\binom{k+2}{3}z^k = \frac{z}{(1-z)^4}.\tag{3}$$

• I suppose the notation $[z^k]f(z)$ means: "the coefficient of $z^k$ in the power series $f(z)$", right? – Martin Sleziak Jan 21 '15 at 14:49
• @MartinSleziak: yes, it is the standard notation in analytic combinatorics. I am just borrowing it from Wilf or Flajolet. – Jack D'Aurizio Jan 21 '15 at 14:52

Induction is pretty straighforward: \begin{align*}\sum_{k=1}^{n+1}&k(n+2-k)-\sum_{k=1}^n k(n+1-k)=(n+1)+\sum_{k=1}^n k\cdot 1=\sum_{k=1}^{n+1} k\\&=\frac{(n+1)(n+2)}2=\frac{(n+1)(n+2)(n+3)}6-\frac{n(n+1)(n+2)}6\end{align*}

Here is another solution. $$n\cdot 1+(n-1)\cdot2+\ldots1\cdot n=\\ (n+(n-1)+\ldots +1)+((n-1)+(n-2)+\ldots+1)+\ldots+1=\\ \sum_{i=1}^ni+\sum_{i=1}^{n-1}i+\ldots+1=\\ \frac{(n+1)n}{2}+\frac{n(n-1)}{2}+\ldots +\frac{2\cdot 1}{2}.$$ Now, this sum is equal $$\frac{(n+2)(n+1)n}{6}$$ by a straightforward induction.

• Your equality can be seen also as $\binom{n+1}2+\binom{n}2+\dots+\binom22=\binom{n+2}3$, which can be a bit generalized to what some people call hockey-stick identity. See, for example, article on AoPS Wiki or this post (and some of the posts linked to it). – Martin Sleziak Jan 22 '15 at 8:13
• True, thank you for pointinig it out. – Ofir Schnabel Jan 22 '15 at 8:14


Probability with Poker cards

You deal 5 cards from a well-shuffled deck of playing cards. What is the probability that the 5th card is the queen of spades?

Just from analysis, P(5th queen spade) = (51*50*49*48*1)/(52*51*50*49*48) = 1/52

However why wont this method of logic thinking incorrect?

P(5th queen spade) = (51Cr4) / (52Cr5) = 5/52. Reasoning: choose any first 4 cards and last card is queen spade, divide by all possible choice

• Isn't that fifth card as likely to be the queen of spades as the three of clubs? – Lord Shark the Unknown Sep 14 '18 at 6:46
• But what if the three of clubs was selected before the 5th card? Is it the order of when cards are drawn does not matter? – userName Sep 14 '18 at 6:49
• You are drawing four cards and then a fifth card. You don't care about order of the first four cards, but pay special attention to the fifth. Count ways to do so in both numerator and denominator (that is: favoured event and total space). – Graham Kemp Sep 14 '18 at 8:28
• @graham So since first four cards order does not matter, the probability of that cancel each other in numerator and denominator is that what it is? – userName Sep 14 '18 at 8:39

Your (51Cr4) / (52Cr5), or as I prefer to write it $\dfrac{51 \choose 4}{52\choose 5}$, is the probability that the first five cards contain the Queen of Spades.

If that happens, there then is a $1$ in $5$ chance that the Queen of Spades is the fifth of these five cards,

making the result to the original question $\dfrac5{52}\times \dfrac{1}{5} = \dfrac{1}{52}$

• Oh okay this makes sense... So does it mean that if I use the formula P(number of sample points in an event) / P(number of sample points in sample space) it will always be like the explanation that you have given? – userName Sep 14 '18 at 6:55
• @userName - it depends on the precise question, especially on whether each element of the sample space is equally likely – Henry Sep 14 '18 at 7:06
• What about this question? Find probability that the 5th card is the queen of spades, given that the first 4 cards are hearts? I consider P(first 4 cards hearts ^ last card queen of spade) / P(first 4 cards hearts) = (13Cr4 * 1 / 5) / (13Cr4 x 39Cr1 x 1 / 5) why is this reasoning incorrect? [multiplied by 1/5 due to the order, like in the previous example] correct ans: 1/48 – userName Sep 14 '18 at 7:16
• @userName After you have chosen four particular hearts, there would be $52-4=48$ cards remaining. So you could answer that question with $\dfrac{{13 \choose 4}{1 \choose 1}}{ {13 \choose 4}{48 \choose 1}} =\dfrac{1}{48}$ – Henry Sep 14 '18 at 7:35
• Alright thanks Henry. But just like to clarify why do I not need to multiply by 1/5 in this situation? Isn't P(first 4 cards hearts and last card queen spade) also mean that we have to be careful and consider that the queen spade is the last card? – userName Sep 14 '18 at 8:11

Since you are asking for specifically the fifth card to be $\spadesuit Q$, this is a problem in which order is important.

The number of ways to select $5$ different cards with order important is $P(52,5)$. The number of ways in which the fifth card is $\spadesuit Q$ is $P(51,4)$. So the probability is $$\frac{P(51,4)}{P(52,5)}=\frac{51\times50\times49\times48}{52\times51\times50\times49\times48}=\frac1{52}\ .$$ Your mistake was to use $C$s instead of $P$s.

• Thanks David. Can I check how can the use of Permutation solve this qn other qn? Find probability that the 5th card is the queen of spades, given that the first 4 cards are hearts? If I use that formula I got stuck when (13Pr4) / (13Pr4 x 39Cr1) which gives 1/39... The correct ans is 1/48 instead – userName Sep 14 '18 at 7:27
• Is it suppose to be (13Pr4) / (13Pr4 x 48Cr1) since I only select 4 cards and now left with 48 cards to choose from? – userName Sep 14 '18 at 7:30
• You seem to have got an answer from someone else so I trust that is OK. But can I point out that on this site it is preferred that new questions should not be asked in comments. This is because it makes it harder for other people to find the question later. It would be better if you asked a new question. You can always cross-reference your present question if relevant. Thanks. – David Sep 17 '18 at 0:09

$\def\cbinom#1#2{{^{#1}\mathsf C_{#2}}}\frac{\cbinom{51}4\cbinom 11}{\cbinom{52}5}$ is the probability for selecting the queen of spades and four from the fifty-one other cards when selecting any five from all fifty two cards. It does not consider the fifth place as a seperate group from the first four.

You require the probability for selecting any four from the fifty-one non-queen of spaces then the queen of spades when selecting any four from fifty-two cards then any one from the fourty-eight other card.

That is $\frac{\cbinom {51}4\cbinom 11}{\cbinom {52}4\cbinom{48}{1}}$, which equals $1/52$.


1. ## Algebra

Q1= show that if x is real, 2(x^2) + 6x + 9 is always positive.

Q2= Solve the simultaneous equations for x,y > 0

2log y = log 2 + log x

2^y = 4^x

cheers

2. for Q1 check the discriminant of the quadratic $b^2 -4ac$

for Q2 using the first equation you should be able to get $\log y^2 = \log 2x \Rightarrow y^2 = 2x$ using the laws of logs.

log both sides on the second equation and it should be easy form there.

3. Originally Posted by sparky69er
Q1= show that if x is real, 2(x^2) + 6x + 9 is always positive.

Q2= Solve the simultaneous equations for x,y > 0

2log y = log 2 + log x

2^y = 4^x

cheers
An alternative "looking" way to show Q1 would be completing squares and examining the expression.
$
2(x^2 + 3x + \frac92) = 2([x^2 + 2.\frac32.x + (\frac32)^2] + \frac92 - (\frac32)^2) = 2(x + \frac32)^2 + \frac94$

Now since $(x + \frac32)^2$ is always greater than or equal to zero, for all real x, and since $\frac94$ is positive , its sum is strictly positive.

4. Originally Posted by bobak
for Q1 check the discriminant of the quadratic $b^2 -4ac$

for Q2 using the first equation you should be able to get $\log y^2 = \log 2x \Rightarrow y^2 = 2x$ using the laws of logs.

log both sides on the second equation and it should be easy form there.
Hi, for Q1 i have done as you said and got -36 but what do you mean by the discriminant????

5. Hello, sparky69er!

1) Show that if $x$ is real, $2x^2 + 6x + 9$ is always positive.

We have: . $x^2 + x^2 + 6x + 9 \:=\:x^2 + (x+3)^2$

The sum of two squares is always nonnegative.

When $x = 0$, the polynomial has a minimum value of 9.

2) Solve the simultaneous equations for x, y > 0

. . $\begin{array}{cc}2\log y \:=\: \log 2 + \log x & {\color{blue}[1]} \\ 2^y \: =\: 4^x & {\color{blue}[2]} \end{array}$

Equation [1] is: . $2\log y \:=\:\log(2x)\;\;{\color{blue}[3]}$

From Equation [2]: . $2^y \:=\:(2^2)^x \:=\:2^{2x}\quad\Rightarrow\quad y \:=\:2x\;\;{\color{blue}[4]}$

Substitute into [3]: . $2\log(2x) \:=\:\log(2x)\quad\Rightarrow\quad\log(2x) \:=\:0$

. . Hence: . $2x \:=\:1\quad\Rightarrow\quad \boxed{x \:=\:\frac{1}{2}}$

Substitute into [4]: . $y \:=\:2\left(\frac{1}{2}\right) \quad\Rightarrow\quad\boxed{ y \:=\:1}$

6. Originally Posted by sparky69er
Hi, for Q1 i have done as you said and got -36 but what do you mean by the discriminant????
Given a quadratic polynomial equation of the form
$ax^2 + bx + c = 0$,
we have solutions
$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

The "discriminant" is defined as $D = b^2 - 4ac$. We have three different classes of solution based on the value of the discriminant:
$D > 0 \implies$ two real, unequal roots.

$D = 0 \implies$ one real root. (Or two equal real roots, however you wish to look at it.)

$D < 0 \implies$ two complex conjugate roots.

In the case of D < 0 the curve $y = ax^2 + bx + c$ never crosses the x axis. If a > 0 then this would imply that the quadratic is always positive.

-Dan

7. Originally Posted by Soroban
Hello, sparky69er!

We have: . $x^2 + x^2 + 6x + 9 \:=\:x^2 + (x+3)^2$

The sum of two squares is always nonnegative.
Smart, really smart solution