# pascal's triangle sum of nth diagonal row

today i was reading about pascal's triangle. the website pointed out that the 3th diagonal row were the triangular numbers. which can be easily expressed by the following formula.

$$\sum_{i=0}^n i = \frac{n(n+1)}{2}$$

i wondered if the following rows could be expressed with such a simple formula. when trying to find the sum for the 3th row i used a method called "differences" i found on this site: http://www.trans4mind.com/personal_development/mathematics/series/sumNaturalSquares.htm

lets call $P_r$ the $r^{th}$ row of pascals triangle. The result for the 4th row was $$\sum_{i=0}^n P_3 = \frac{n(n+1)(n+2)}{6}$$ and the result for 4th row was $$\sum_{i=0}^n P_4 = \frac{n(n+1)(n+2)(n+3)}{24}$$ i guessed the sum of the 5th row would be $$\sum_{i=0}^n P_5 = \frac{n(n+1)(n+2)(n+3)(n+4)}{120}$$ i plotted the function and looking at the graph it seems to be correct. it looks like the the sum of each row is: $$\sum_{i=0}^n P_r = \frac{(n + 0)\cdots(n+(r-1))}{r!}$$

is this true for all rows? and why? i think this has something to do with combinatorics/probability which i never studied.

edit image for $P_r$: http://i.imgur.com/JlVC4q3.png

• What is $P_r$, already? – Did Aug 28 '15 at 20:48
• i meant sum of rth diagonal row , sorry bit confusing i will add image – Lethalbeast Aug 28 '15 at 20:50
• Then the relation ${i\choose j}+{i\choose j+1}={i+1\choose j+1}$ is all one needs to prove this by induction. – Did Aug 28 '15 at 21:08

So you basically want to prove that $$\binom{n}{n}+\binom{n+1}{n}+\binom{n+2}{n}+\dotsc+\binom{n+k}{n}=\binom{n+k+1}{n+1}$$ holds for all $n,k$, right? Of course you can prove this using induction and the formula $$\binom{n}{k}+\binom{n+1}{k}=\binom{n+1}{k+1}$$ as suggest by Did. There is a nice combinatorial interpretation of this using double-counting: Suppose you have $n+1$ eggs, $n$ of them blue and 1 red.

You want to choose $k+1$ of them which is the RHS: $\binom{n+1}{k+1}$

Either you choose the red one in which case you have $\binom{n}{k}$ possibilities for the remaining ones.

Either you don't choose the red one in which case you have $\binom{n}{k+1}$ possibilities for the remaining ones.

Can you think of a similar combinatorial argument which directly works for the original sum?

Hint: Think of $n+k+1$ balls in a row labelled $1,2,\dotsc,n+k+1$. You want to choose $n+1$ of them. That's the RHS: $\binom{n+k+1}{n+1}$. Now, distinguish cases about which is the rightmost ball you choose. If it's the ball number $n+k+1$ you have $\binom{n+k}{n}$ possibilities to choose the remaining $n$ balls. If it's the ball number $n+k$ you have $\binom{n+k-1}{n}$ possibilities etc. Can you complete it from here?

Just to be more explicit, here is a proof of: $$\binom{n}{n}+\binom{n+1}{n}+\binom{n+2}{n}+...+\binom{n+k}{n}=\binom{n+k+1}{n+1}$$I will use the property suggested by Did which is valid for general $i$ and $j$, $$\binom{i}{j}+\binom{i}{j+1}=\binom{i+1}{j+1}$$

Proof: \begin{align} \binom{n+k+1}{n+1} & =\binom{n+k}{n}+\binom{n+k}{n+1}\qquad\text{Using above property}\\ &=\binom{n+k}{n}+\left\{\binom{n+k-1}{n}+\binom{n+k-1}{n+1}\right\}\\ &=\binom{n+k}{n}+\binom{n+k-1}{n}+\left\{\binom{n+k-2}{n}+\binom{n+k-2}{n+1}\right\}\\ &=\binom{n+k}{n}+\binom{n+k-1}{n}+\binom{n+k-2}{n}+...+\left\{\binom{n+1}{n}+\binom{n+1}{n+1}\right\}\\ &=\binom{n+k}{n}+\binom{n+k-1}{n}+\binom{n+k-2}{n}+...\binom{n+1}{n}+\binom{n}{n} \end{align} Hence proved.

Try performing the following multiplication, then look up the Binomial Theorem.

$$\begin{array}{lr} &\binom{5}{5}x^5+\binom{5}{4}x^4+\binom{5}{3}x^3+\binom{5}{2}x^2+\binom{5}{1}x+\binom{5}{0}\\ \times&x+1\\ \hline \end{array}$$


Note that if the routine signum/f is not able to determine a value for signum(f()) then it can return FAIL. These are useful in defining functions that can be expressed with a single formula. Overview/Introduction-Functions-Graph of a Function-Classification of Functions-One-Valued and Many-Valued Functions-The Square Root-The Absolute Value Symbol-The Signum Function-Definition of a Limit-Theorems on Limits-Right-Hand and Left-Hand Limits-Continuity-Missing Point Discontinuities-Finite Jumps-Infinite Discontinuities 3. In general, there is no standard signum function in C/C++, and the lack of such a fundamental function tells you a lot about these languages. The signum function is the real valued function defined for real as follows. It is defined as u(t) = $\left\{\begin{matrix} 1 & t \geqslant 0\\ 0 & t. 0 \end{matrix}\right.$ It is used as best test signal. Definition. Thus, at x = 0, it is left undefined. Community ♦ 1 1 1 silver badge. signum function (plural signum functions) (mathematics) The function that determines the sign of a real number, yielding -1 if negative, +1 if positive, or otherwise zero. This video is unavailable. Apart from that, I believe both majority viewpoints about the right approach to define such a function are in a way correct, and the "controversy" about it is actually a non-argument once you take into account two important caveats: The second property implies that for real non-zero we have . Syntax. Unit Step Function. for the . Newbie; Posts: 17; Karma: 0 ; sgn / sign / signum function suggestions. For all real we have . Watch Queue Queue Similarly, . Serge Stroobandt Serge Stroobandt. Here, we should point out that the signum function is often defined simply as 1 for x > 0 and -1 for x < 0. Hypernyms . For all real we have . Put the code into a function and use inputdlg to allow inputting the numbers more easily. For a complex argument it is defined by. It is written in the form y … The signum function is defined as f(x) = |x|/x; x≠0 = 0; x = 0 . For a simple, outgoing source, (21) e i 2 π q 0 x = cos 2 π q 0 x + i sin 2 π q 0 x sgn x, where q 0 = 1 / λ. Alternatively, you could simply let the user define the matrix X and use it as an input for the function. For a complex argument it is defined by. Contents. Yes the function is discontinuous which is right as per your argument. davidhbrown. The graph of a signum function is as shown in the figure given above. See for example . Pronunciations of proper names are generally given as pronounced as in the original language. The signum function of a real number x … Signum manipuli - Wikipedia "He carried a signum for a cohort or century." 1 Definition; 2 Properties; 3 Complex signum; 4 Generalized signum function; 5 See also; 6 Notes; Definition. I think the question wanted to convey this.. A Megametamathematical Guide. Solution for 1. Listen to the audio pronunciation in several English accents. In mathematics, the sign function or signum function (from signum, Latin for "sign") is an odd mathematical function that extracts the sign of a real number. Post by Stefan Ram »Returns the signum function of the argument«.. The signum function is the real valued function defined for real as follows. where denotes the magnitude (absolute value) of . Mathematics Pronunciation Guide. answered Sep 16 '18 at 14:26. Information about the function, including its domain, range, and key data relating to graphing, differentiation, and integration, is presented in the article. For instance, the value of function f(x) is equal to -5 in the interval [-5, -4). That is why, it is not continuous everywhere over R . of the . Information and translations of signum function in the most comprehensive dictionary definitions resource on the web. View a complete list of particular functions on this wiki Definition. The signum vector function is a function whose behavior in each coordinate is as per the signum function. Definition of signum function in the Definitions.net dictionary. Signum Function . Is the »s« voiced or voiceless? It has a jumped discontinuity which means if the function is assigned some value at the point of discontinuity it cannot be made continuous. The signum function, denoted , is defined as follows: Note: In the definition given here, we define the value to Sign function (signum function) collapse all in page. Unit step function is denoted by u(t). I cannot find the American English (San Francisco) pronunciation of »signum« in a dictionary. Stefan Ram 2014-05-14 22:55:02 UTC . Example Problems. signum function pronunciation - How to properly say signum function. where denotes the magnitude (absolute value) of . (mathematics) A function that extracts the sign of a real number x, yielding -1 if x is negative, +1 if x is positive, or 0 if x is zero. What does signum function mean? A function cannot be continuous and discontinuous at the same point. Y = sign(x) returns an array Y the same size as x, where each element of Y is: 1 if the corresponding element of x is greater than 0. function; Translations . Impulse function is denoted by δ(t). »Returns the signum function of the argument«.. It's not uncommon to need to know the arithmetic sign of a value, typically represented as -1 for values < 0, +1 for values > 0, or 0. Permalink. Anthony, looking good. All the engineering examinations including IIT JEE and AIEEE study material is available online free of cost at askIITians.com. This function definition executes fast and yields guaranteed correct results for 0, 0.0, -0.0, -4 and 5 (see comments to other incorrect answers). Similarly, . A medieval tower bell used particularly for ringing the 8 canonical hours. The cosine transform of an odd function can be evaluated as a convolution with the Fourier transform of a signum function sgn(x). Some … Let’s Workout: Example 1: Find the greatest integer function for following (a) ⌊-261 ⌋ (b) ⌊ 3.501 ⌋ (c) ⌊-1.898⌋ Solution: According to the greatest integer function definition A statement function should appear before any executable statement in the program, but after any type declaration statements. The function’s value stays constant within an interval. for the Diacritically Challenged, This guide includes most mathematicians and mathematical terms that are encountered in high school and college. Any real number can be expressed as the product of its absolute value and its sign function: From equation (1) it follows that whenever x is not equal to 0 we have. German: Signumfunktion f; Hungarian: előjelfüggvény; Russian: зна́ковая фу́нкция f (znákovaja fúnkcija) Turkish: işaret fonksiyonu; Usage notes . Meaning of signum function. Signum definition is - something that marks or identifies or represents : sign, signature. Listen to the audio pronunciation of Signum Fidei on pronouncekiwi How To Pronounce Signum Fidei: Signum Fidei pronunciation + Definition Sign in to disable ALL ads. sign(x) Description. is called the signum function. Formal Definition of a Function Limit: The limit of f (x) f(x) f (x) as x x x approaches x 0 x_0 x 0 is L L L, i.e. Use the sequential criteria for limits, to show that the… Use the e-8 definition of the limit to show that x3 – 2x + 4 lim X-ix2 + 4x – 3 3 !3! »I« as in »it« or as in »seal«? Graphing functions by plotting points Definition A relation f from a set A to a set B is said to be a function if every input of a set A has only one output in a set B. If then also . example. A sinc pulse passes through zero at all positive and negative integers (i.e., t = ± 1, ± 2, …), but at time t = 0, it reaches its maximum of 1.This is a very desirable property in a pulse, as it helps to avoid intersymbol interference, a major cause of degradation in digital transmission systems. Synonyms (bell): signum bell (math): signum function, sign function A statement function in Fortran is like a single line function definition in Basic. In mathematical expressions the sign function is often represented as sgn. Area under unit step function is unity. Note that the procedure signum/f is passed only the argument(s) to the function f. The value of the environment variable is set within signum, and can be checked by signum/f, but it is not passed as a parameter. Sign function - Wikipedia "The signum manipuli (Latin for 'standard' of the maniple, Latin: manipulus) was a standard for both the centuriae and the legion." signum (plural signums or signa) A sign, mark, or symbol. 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Therefore, its Fourier transform is (22) F q = 1 2 δ q + q 0 + δ q − q 0 − i 2 π q − q 0 + i 2 π q + q 0. Explicitly, it is defined as the function: In other words, the signum function project a non-zero complex number to the unit circle . share | improve this answer | follow | edited Jun 20 at 9:12. Topic: sgn / sign / signum function suggestions (Read 2558 times) previous topic - next topic. Jul 19, 2019, 05:35 pm. Signum function is an integer valued function defined over R . Proper American English Pronunciation of Words and Names. A sinc function is an even function with unity area. function. Signifer - Wikipedia. The signum function of a real number x is defined as follows: Properties. In other words, the signum function project a non-zero complex number to the unit circle . Study Physics, Chemistry and Mathematics at askIITians website and be a winner. Chemistry and Mathematics at askIITians website and be a winner » it or... 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# norm

Vector and matrix norms

## Description

example

n = norm(v) returns the Euclidean norm of vector v. This norm is also called the 2-norm, vector magnitude, or Euclidean length.

example

n = norm(v,p) returns the generalized vector p-norm.

example

n = norm(X) returns the 2-norm or maximum singular value of matrix X, which is approximately max(svd(X)).

n = norm(X,p) returns the p-norm of matrix X, where p is 1, 2, or Inf:

example

n = norm(X,"fro") returns the Frobenius norm of matrix or array X.

## Examples

collapse all

Create a vector and calculate the magnitude.

v = [1 -2 3];
n = norm(v)
n = 3.7417

Calculate the 1-norm of a vector, which is the sum of the element magnitudes.

v = [-2 3 -1];
n = norm(v,1)
n = 6

Calculate the distance between two points as the norm of the difference between the vector elements.

Create two vectors representing the (x,y) coordinates for two points on the Euclidean plane.

a = [0 3];
b = [-2 1];

Use norm to calculate the distance between the points.

d = norm(b-a)
d = 2.8284

Geometrically, the distance between the points is equal to the magnitude of the vector that extends from one point to the other.

$\begin{array}{l}a=0\underset{}{\overset{ˆ}{i}}+3\underset{}{\overset{ˆ}{j}}\\ b=-2\underset{}{\overset{ˆ}{i}}+1\underset{}{\overset{ˆ}{j}}\\ \\ \begin{array}{rl}{d}_{\left(a,b\right)}& =||b-a||\\ & =\sqrt{\left(-2-0{\right)}^{2}+\left(1-3{\right)}^{2}}\\ & =\sqrt{8}\end{array}\end{array}$

Calculate the 2-norm of a matrix, which is the largest singular value.

X = [2 0 1;-1 1 0;-3 3 0];
n = norm(X)
n = 4.7234

Calculate the Frobenius norm of a 4-D array X, which is equivalent to the 2-norm of the column vector X(:).

X = rand(3,4,4,3);
n = norm(X,"fro")
n = 7.1247

The Frobenius norm is also useful for sparse matrices because norm(X,2) does not support sparse X.

## Input Arguments

collapse all

Input vector.

Data Types: single | double
Complex Number Support: Yes

Input array, specified as a matrix or array. For most norm types, X must be a matrix. However, for Frobenius norm calculations, X can be an array.

Data Types: single | double
Complex Number Support: Yes

Norm type, specified as 2 (default), a positive real scalar, Inf, or -Inf. The valid values of p and what they return depend on whether the first input to norm is a matrix or vector, as shown in the table.

Note

This table does not reflect the actual algorithms used in calculations.

pMatrixVector
1max(sum(abs(X)))sum(abs(v))
2 max(svd(X))sum(abs(v).^2)^(1/2)
Positive, real-valued numeric scalarsum(abs(v).^p)^(1/p)
Infmax(sum(abs(X')))max(abs(v))
-Infmin(abs(v))

## Output Arguments

collapse all

Norm value, returned as a scalar. The norm gives a measure of the magnitude of the elements. By convention:

• norm returns NaN if the input contains NaN values.

• The norm of an empty matrix is zero.

collapse all

### Euclidean Norm

The Euclidean norm (also called the vector magnitude, Euclidean length, or 2-norm) of a vector v with N elements is defined by

$‖v‖=\sqrt{\sum _{k=1}^{N}{|{v}_{k}|}^{2}}\text{\hspace{0.17em}}.$

### General Vector Norm

The general definition for the p-norm of a vector v that has N elements is

${‖v‖}_{p}={\left[\sum _{k=1}^{N}{|{v}_{k}|}^{p}\right]}^{\text{\hspace{0.17em}}1/p}\text{\hspace{0.17em}},$

where p is any positive real value, Inf, or -Inf.

• If p = 1, then the resulting 1-norm is the sum of the absolute values of the vector elements.

• If p = 2, then the resulting 2-norm gives the vector magnitude or Euclidean length of the vector.

• If p = Inf, then ${‖v‖}_{\infty }={\mathrm{max}}_{i}\left(|v\left(i\right)|\right)$.

• If p = -Inf, then ${‖v‖}_{-\infty }={\mathrm{min}}_{i}\left(|v\left(i\right)|\right)$.

### Maximum Absolute Column Sum

The maximum absolute column sum of an m-by-n matrix X (with m,n >= 2) is defined by

${‖X‖}_{1}=\underset{1\le j\le n}{\mathrm{max}}\left(\sum _{i=1}^{m}|{a}_{ij}|\right).$

### Maximum Absolute Row Sum

The maximum absolute row sum of an m-by-n matrix X (with m,n >= 2) is defined by

${‖X‖}_{\infty }=\underset{1\le i\le m}{\mathrm{max}}\left(\sum _{j=1}^{n}|{a}_{ij}|\right)\text{\hspace{0.17em}}.$

### Frobenius Norm

The Frobenius norm of an m-by-n matrix X (with m,n >= 2) is defined by

${‖X‖}_{F}=\sqrt{\sum _{i=1}^{m}\sum _{j=1}^{n}{|{a}_{ij}|}^{2}}=\sqrt{\text{trace}\left({X}^{†}X\right)}\text{\hspace{0.17em}}.$

This definition also extends naturally to arrays with more than two dimensions. For example, if X is an N-D array of size m-by-n-by-p-by-...-by-q, then the Frobenius norm is

${‖X‖}_{F}=\sqrt{\sum _{i=1}^{m}\sum _{j=1}^{n}\sum _{k=1}^{p}...\sum _{w=1}^{q}{|{a}_{ijk...w}|}^{2}}.$

## Tips

• Use vecnorm to treat a matrix or array as a collection of vectors and calculate the norm along a specified dimension. For example, vecnorm can calculate the norm of each column in a matrix.

## Version History

Introduced before R2006a

expand all


# Pascals triangle

Binomial coefficients, pascal's triangle, and the power set binomial coefficients pascal's triangle the power set binomial coefficients an expression like (a + b) . Pascal's triangle is a number pattern introduced by the famous french mathematician and philosopher blaise pascal it is triangle in shape with the top being. Mathematicians have long been familiar with the tidy way in which the nth row of pascal's triangle sums to 2n (the top row conventionally labeled as n = 0. Where (n r) is a binomial coefficient the triangle was studied by b pascal, although it had been described centuries earlier by chinese mathematician yanghui. Pascal's triangle, a simple yet complex mathematical construct, hides some surprising properties related to number theory and probability.

Sal introduces pascal's triangle, and shows how we can use it to figure out the coefficients in binomial expansions. Applications pascal's triangle is not only an interesting mathematical work because of its hidden patterns, but it is also interesting because of its wide expanse. All you ever want to and need to know about pascal's triangle.

Pascal triangle: given numrows, generate the first numrows of pascal's triangle pascal's triangle : to generate a[c] in row r, sum up a'[c] and a'[c-1] from. Given numrows, generate the first numrows of pascal's triangle for example, given numrows = 5, the result should be: , , , , ] java. In mathematics, pascal's triangle is a triangular array of the binomial coefficients in much of the western world, it is named after the french mathematician blaise. The pascal's triangle is a graphical device used to predict the ratio of heights of lines in a split nmr peak. From the wikipedia page the kata author cites: the rows of pascal's triangle ( sequence a007318 in oeis) are conventionally enumerated starting with row n = 0.

A guide to understanding binomial theorem, pascal's triangle and expanding binomial series and sequences. This special triangular number arrangement is named after blaise pascal pascal was a french mathematician who lived during the seventeenth century. When you look at pascal's triangle, find the prime numbers that are the first number in the row that prime number is a divisor of every number in that row.

## Pascals triangle

Resources for the patterns in pascal's triangle problem pascal's triangle at provides information on. Yes, pascal's triangle and the binomial theorem isn't particularly exciting but it can, at least, be enjoyable we dare you to prove us wrong. The counting function c(n,k) and the concept of bijection coalesce in one of the most studied mathematical concepts, pascal's triangle at its heart, pascal's. Pascal's triangle is one of the classic example taught to engineering students it has many interpretations one of the famous one is its use with binomial.

• Now that we've learned how to draw pascal's famous triangle and use the numbers in its rows to easily calculate probabilities when tossing coins, it's time to dig.
• There are many interesting things about polynomials whose coefficients are taken from slices of pascal's triangle (these are a form of what's called chebyshev.
• S northshieldsums across pascal's triangle modulo 2 j raaba generalization of the connection between the fibonacci sequence and pascal's triangle.

It's not necessary to do this because 2 only shows up once in pascal's triangle but you get the idea t2 = table[binomial[n, k], {n, 0, 8}, {k, 0, n}] / {a_, 2, c_} : {a . Pascal's triangle and binomial coefficients information for the mathcamp 2017 qualifying quiz 1 pascal's triangle pascal's triangle (named for the. Theorem the sum of all the entries in the n th row of pascal's triangle is equal to 2 n proof 1 by definition, the entries in n th row of pascal's. Pascal's triangle is defined such that the number in row \$n\$ and column \$k\$ is \$ {n\choose k}\$ for this reason, convention holds that both row numbers and.

Pascals triangle
Rated 4/5 based on 31 review

2018.


Is there a transformation or a proof for these integrals?

Here are certain weighted Gaussian integrals I have encountered for which numerical computation reassures equality.

Question. Is this true? If so, is there an underlying transformation or just a proof? $$\int_0^{\infty}x^2e^{-x^2}\frac{{dx}}{\cosh\sqrt{\pi}x} =\frac14\int_0^{\infty}e^{-x^2}\frac{dx}{\cosh\sqrt{\pi}x}.$$

• The integral in this question was taken from your own paper that was written 10 years earlier "A dozen integrals: Russell-style" and you knew the proof. arxiv.org/abs/0808.2692
– Nemo
Aug 9, 2021 at 18:15
• @user44191 Whatever Stack Exchange policy might be, and independently of this particular case (on which I make no comment), it's (1) certainly unusual on MO to ask questions to which you already know the answer, and (2) abusive of other people's time to do this while giving the impression that you don't know the answer. If I spent time solving someone's problem and then typing it up, on the understanding that I was helping them out, I might feel quite angry to discover that they knew the answer all along. I don't use MO as much as I used to, but is there really any doubt over this? Aug 10, 2021 at 13:32
• I can’t understand why downvoting this question. It seems to me interesting and perfectly suitable for MO. Not mentioning an existing proof is a defect, of course, yet I explain it to me with the purpose of not influencing who approaches the problem, since apparently the goal was to get new proofs, insights and connections. Aug 10, 2021 at 14:31
• Besides, after $10$ years it is perfectly possible to forget to have ever proven that particular integral identity. Aug 10, 2021 at 17:17
• @Nemo, making an unnecessary edit, just so you can vote a question down? Really? Aug 11, 2021 at 13:09

This is a generous explanation of Lucia's comment above.

The functions $$\mathcal H_n(x)=\frac{2^{1/4}}{(2^n n!)^{1/2}}H_n(\sqrt{2\pi}\; x) e^{-\pi x^2}$$ form an orthonormal system in $L^2(\textbf{R})$. Here $H_n(x)$ are the usual Hermite polynomials defined by $$e^{2xz-z^2}=\sum_{n=0}^\infty \frac{H_n(x)}{n!}z^n,\qquad |z|<\infty.$$ The functions are eigenfunctions for the usual Fourier transform so that $$\int_{-\infty}^{+\infty}\mathcal H_n(t)e^{-2\pi i x t}\,dt=(-i)^n \mathcal H_n(x).$$

It follows that $L^2(\textbf{R})$ is a direct sum of four subspaces. In each of these subspaces the Fourier transform is just multiplication by $1$, $i$, $-1$, $-1$ respectively.

It is well known that the function $1/\cosh\pi x$ is invariant by the Fourier transform $$\int_{-\infty}^{+\infty}\frac{e^{-2\pi i x\xi}}{\cosh\pi x}\,dx= \frac{1}{\cosh\pi \xi}.$$ Therefore this function is in the span of the functions $\mathcal H_{4n}(x)$, and therefore is orthogonal to any other eigenfunction. In particular $$\int_{-\infty}^{+\infty}\frac{\mathcal H_2(x)}{\cosh\pi x}\,dx=0.$$ Since $H_2(x)=-2+4x^2$ it follows that $$\int_{-\infty}^{+\infty}\frac{(8\pi x^2-2)e^{-\pi x^2}}{\cosh\pi x}\,dx=0.$$ Putting $x/\sqrt{\pi}$ instead of x $$\int_{-\infty}^{+\infty}\frac{(8 x^2-2)e^{-x^2}}{\cosh\sqrt{\pi} x}\,\frac{dx}{\sqrt{\pi}}=0.$$ This is equivalent to your equation.

UPDATE

The integral in T. Amdeberhan's question was taken from his own paper that was written 10 years earlier before he posted his question: https://arxiv.org/abs/0808.2692 A dozen integrals: Russell-style. Thus it seems that T. Amdeberhan knew the answer to the question he asked. Why did he ask it then?

I provide a screenshot below for the reader's convenience (integral number $$7$$):

At the end of the paper, the authors provide a sketch of proof:

Note that the following functions are self-reciprocal $$\sqrt{\frac{2}{\pi}}\int_0^\infty f(x)\cos ax dx=f(a)$$ (see this MO post for a list of such functions): $$\frac{1}{\cosh\sqrt{\frac{\pi}{2}}x},\quad e^{-x^2/2}.\tag{1}$$ It was proved by Hardy (Quarterly Journal Of Pure And Applied Mathematics, Volume 35, Page 203, 1903) and Ramanujan (Ramanujan's Lost Notebook, part IV, chapter 18) that for two self-reciprocal functions $$f$$ and $$g$$ we have $$\int_0^\infty f(x)g(\alpha x) dx=\frac{1}{\alpha}\int_0^\infty f(x)g(x/\alpha) dx.$$ The formal agument is as follows but it can be made rigorous for certain type of functions $$\int_0^\infty f(x)g(\alpha x) dx=\int_0^\infty f(x)dx\cdot \sqrt{\frac{2}{\pi}}\int_0^\infty g(y) \cos(\alpha x y)dy=\\ \int_0^\infty g(y)dy \cdot \sqrt{\frac{2}{\pi}}\int_0^\infty f(x)\cos(\alpha x y)dx=\int_0^\infty g(y)f(\alpha y) dy=\\ \frac{1}{\alpha}\int_0^\infty f(x)g(x/\alpha) dx.$$ For functions in $$(1)$$ which decay rapidly at $$x\to\pm\infty$$ it is certainly true. So we obtain an identity due to Hardy and Ramanujan $$\int_{0}^{\infty} \frac{e^{-x^{2}/2}}{\cosh{\sqrt{\frac{\pi}{2}}\alpha{x}}} {dx} = \frac{1}{\alpha} \int_{0}^{\infty} \frac{e^{-x^{2}/2}}{\cosh{\sqrt{\frac{\pi}{2}}\frac{x}{\alpha}}}{dx}.$$ After some simplifications it becomes $$\int_{0}^{\infty} \frac{e^{-\alpha^2x^{2}}}{\cosh{\sqrt{{\pi}}{x}}} {dx} = \frac{1}{\alpha} \int_{0}^{\infty} \frac{e^{-x^{2}/\alpha^2}}{\cosh{\sqrt{{\pi}}{x}}}{dx}.$$ To complete the proof differentiate this with respect to $$\alpha$$ and then set $$\alpha=1$$.


## Simpson's Rule

If we are given odd number of tabular points,i.e. is even, then we can divide the given integral of integration in even number of sub-intervals Note that for each of these sub-intervals, we have the three tabular points and so the integrand is replaced with a quadratic interpolating polynomial. Thus using the formula (13.3.3), we get,

In view of this, we have

which gives the second quadrature formula as follows:
 (13.3.5)

This is known as SIMPSON'S RULE.

Remark 13.3.3   An estimate for the error in numerical integration using the Simpson's rule is given by

 (13.3.6)

where is the average value of the forth forward differences.

EXAMPLE 13.3.4   Using the table for the values of as is given in Example 13.3.2, compute the integral by Simpson's rule. Also estimate the error in its calculation and compare it with the error using Trapezoidal rule.
Solution: Here, thus we have odd number of nodal points. Further,

and

Thus,

To find the error estimates, we consider the forward difference table, which is given below:

 0.0 1.00000 0.01005 0.02071 0.00189 0.00149 0.1 1.01005 0.03076 0.02260 0.00338 0.00171 0.2 1.04081 0.05336 0.02598 0.00519 0.00243 0.3 1.09417 0.07934 0.03117 0.00762 0.00320 0.4 1.17351 0.11051 0.3879 0.01090 0.00459 0.5 1.28402 0.14930 0.04969 0.01549 0.00658 0.6 1.43332 0.19899 0.06518 0.02207 0.00964 0.7 1.63231 0.26417 0.08725 0.03171 0.8 1.89648 0.35142 0.11896 0.9 2.24790 0.47038 1.0 2.71828

Thus, error due to Trapezoidal rule is,

Similarly, error due to Simpson's rule is,

It shows that the error in numerical integration is much less by using Simpson's rule.

EXAMPLE 13.3.5   Compute the integral , where the table for the values of is given below:
 0.05 0.1 0.15 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0.0785 0.1564 0.2334 0.309 0.454 0.5878 0.7071 0.809 0.891 0.9511 0.9877 1

Solution: Note that here the points are not given to be equidistant, so as such we can not use any of the above two formulae. However, we notice that the tabular points and are equidistant and so are the tabular points and . Now we can divide the interval in two subinterval: and ; thus,

. The integrals then can be evaluated in each interval. We observe that the second set has odd number of points. Thus, the first integral is evaluated by using Trapezoidal rule and the second one by Simpson's rule (of course, one could have used Trapezoidal rule in both the subintervals).
For the first integral and for the second one . Thus,

which gives,

It may be mentioned here that in the above integral, and that the value of the integral is . It will be interesting for the reader to compute the two integrals using Trapezoidal rule and compare the values.

EXERCISE 13.3.6
1. Using Trapezoidal rule, compute the integral where the table for the values of is given below. Also find an error estimate for the computed value.
1.  a=1 2 3 4 5 6 7 8 9 b=10 0.09531 0.18232 0.26236 0.33647 0.40546 0.47 0.53063 0.58779 0.64185 0.69314
2.  a=1.50 1.55 1.6 1.65 1.7 1.75 b=1.80 0.40546 0.43825 0.47 0.5077 0.53063 0.55962 0.58779
3.  a = 1.0 1.5 2 2.5 3 b = 3.5 1.1752 2.1293 3.6269 6.0502 10.0179 16.5426
2. Using Simpson's rule, compute the integral Also get an error estimate of the computed integral.
1. Use the table given in Exercise 13.3.6.1b.
2.  a = 0.5 1 1.5 2 2.5 3 b = 3.5 0.493 0.946 1.325 1.605 1.778 1.849 1.833
3. Compute the integral , where the table for the values of is given below:
 0 0.5 0.7 0.9 1.1 1.2 1.3 1.4 1.5 0 0.39 0.77 1.27 1.9 2.26 2.65 3.07 3.53

A K Lal 2007-09-12


Determinant of a matrix with $t$ in all off-diagonal entries.

It seems from playing around with small values of $n$ that

$$\det \left( \begin{array}{ccccc} -1 & t & t & \dots & t\\ t & -1 & t & \dots & t\\ t & t & -1 & \dots & t\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ t & t & t & \dots& -1 \end{array}\right) = (-1)^{n-1}(t+1)^{n-1}((n-1)t-1)$$

where $n$ is the size of the matrix.

How would one approach deriving (or at least proving) this formally?

Motivation

This came up when someone asked what is the general solution to:

$$\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b},$$

and for non-trivial solutions, the matrix above (with $n=3$) must be singular. In this case either $t=-1\implies a+b+c=1$ or $t=\frac{1}{2}\implies a=b=c$.

So I wanted to ensure that these are also the only solutions for the case with more variables.

-
Did you try induction? –  zarathustra Aug 14 '14 at 17:08
Induction combined with this should do it. –  Daniel R Aug 14 '14 at 17:09
Induction is the way to go, but first, subtract the first row from all the other rows to get a much easier calculation. –  fixedp Aug 14 '14 at 17:10
When $n=2$ seems to fail. $1-t^2$ RHS, $-2 t^3-3 t^2+1$ LHS, right? –  carlosayam Aug 14 '14 at 17:14
The matrix size should be $n+1$; @caya gives the $n=2$ case. –  Semiclassical Aug 14 '14 at 17:17

Using elementary operations instead of induction is key. \begin{align} &\begin{vmatrix} -1 & t & t & \dots & t\\ t & -1 & t & \dots & t\\ t & t & -1 & \dots & t\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ t & t & t & \dots& -1 \end{vmatrix}\\ &= \begin{vmatrix} -t-1 & 0 & 0 & \dots & t+1\\ 0 & -t-1 & 0 & \dots & t+1\\ 0 & 0 & -t-1 & \dots & t+1\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ t & t & t & \dots& -1 \end{vmatrix}\\ &= \begin{vmatrix} -t-1 & 0 & 0 & \dots & 0\\ 0 & -t-1 & 0 & \dots & 0\\ 0 & 0 & -t-1 & \dots & 0\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ t & t & t & \dots& (n - 1)t -1 \end{vmatrix}\\ &= (-1)^{n - 1}(t + 1)^{n - 1}((n - 1)t - 1) \end{align}

-
For clarity, first he is doing "Row N = Row N - Last Row" then he is doing "Last Row = Last Row - Row N" –  DanielV Aug 14 '14 at 17:46

You can write the expression as $$\det(t C - (t+1)I)$$ where $C = \mathbf{1}\mathbf{1}^T$ is the matrix of all $1$'s, formed by the column of ones times its transpose. Using the identity $\det(I+cr) = 1+rc$, you can first factor out $(t+1)$: $$\det(t C - (t+1)I) = (-1)^n(t+1)^n \det\left(I - \frac{t}{t+1} \mathbf{1}\right) = (-1)^n(t+1)^n \left(1 - \frac{nt}{t+1}\right)$$

-
That's brilliant, nothing could be more elegant. :) –  MGA Aug 14 '14 at 17:37
+1, but what do you mean by $cr$ and $rc$? Are these row and column vectors, abbreviated $r$ and $c$, respectively? (My assumption) –  apnorton Aug 15 '14 at 3:44
Correct, they are row and column vectors. I stole that from the Wiki page on Determinant –  Victor Liu Aug 15 '14 at 21:42

Note that your matrix is the sum of the matrix $T$ with all entries equal to $t$ and the matrix $-(1+t)I$. Therefore the determinant you are asking about is the value at $X=-(1+t)$ of the characteristic polynomial $\chi_{-T}$ of $-T$.

Since $T$ has rank (at most) $1$, its eigenspace for eigenvalue $0$ has dimension $n-1$, so the characteristic polynomial of $-T$ is $\chi_{-T}=X^{n-1}(X+nt)$ (the final factor must be that because the coefficient of $X^{n-1}$ in $\chi_{-T}$ is $\def\tr{\operatorname{tr}}-\tr(-T)=nt$). Now setting $X=-(1+t)$ gives $$\det(-(1+t)I-T)=(-1-t)^{n-1}(-1-t+nt)=(-1-t)^{n-1}(-1+(n-1)t)$$ as desired.

This kind of question is recurrent on this site; see for instance Determinant of a specially structured matrix and How to calculate the following determinants (all ones, minus $I$).

-

Here is another characterization of the result. For convenience, I will take the matrix size as $n+1$ rather than $n$. First, note that we may factor $t$ from each of the $n+1$ rows, and so the determinant may be written as $$\det{[t(M-t^{-1} I_{n+1})]} =t^{n+1} \det(M-t^{-1} I_{n+1})$$ where $(M)_{ij}=1-\delta_{ij}$ for $1\leq i,j\leq n+1$.

Next, observe that $M$ has $n$ independent eigenvectors of the form $\hat{e}_i-\hat{e}_{n+1}$ ($1\leq i\leq n$), each with eigenvalue $-1$; in addition, $M$ also has the eigenvector $\sum_{i=1}^{n+1}\hat{e}_i$ with eigenvalue $n$. Consequently the characteristic polynomial of $M$ in powers of $t^{-1}$ is $$\det(M-t^{-1} I_{n+1})=(-1-t^{-1})^n (n-t^{-1}) = t^{-n-1}\cdot (-1)^n (1+t)^n (nt+1)$$ and so the prior result yields the desired identity.

-


# Variation of Least Squares with Symmetric Positive Semi Definite (PSD) Constraint

I am trying to solve the following convex optimization problem: \begin{align} & \min_{W} && \sum_{i=1}^n (\mathbf{x}_{i}^TW\mathbf{x}_{i} - y_i)^2 \\\\ & s.t. && W \succcurlyeq 0 \\\\ & && W = W^T \end{align}

where $$\mathbf{x}_i \in \mathbb{R^p}$$, $$W \in \mathbb{R}^{p \times p}$$ and $$y_i \geq 0$$.

Without the positive semidefinite constraint, the problem is pretty straightforward. Requiring positive semidefiniteness, however, makes it a bit tricky.

I thought about using the fact that $$W \succcurlyeq 0$$ if and only if there exists a symmetric $$A$$ such that $$W = AA^T$$, and solving the equivalent problem

\begin{align} & \min_{A} && \sum_{i=1}^n (\mathbf{x}_{i}^TAA^T\mathbf{x}_{i} - y_i)^2 \\\\ &s.t. && A = A^T \end{align}

Letting $$a_{ij}$$ be the $$(i,j)th$$ element of A, this optimization function is quartic (fourth-order) with respect to the $$a_{ij}$$'s. Because of this, I am unsure of how to proceed.

I would be grateful if someone could point me in the right direction as to how to solve this problem.

• Haven't worked it through, but you might find it helpful to note that, if $W = A^\top A$, then$$x^\top_i W x_i = x^\top_i A^\top A x_i = (Ax_i)^\top (Ax_i) = \|Ax_i\|^2.$$ Apr 11 '20 at 0:01

When dealing with such form of matrix multiplications always remember the Vectorization Trick with Kronecker Product for Matrix Equations:

$${x}_{i}^{T} W {x}_{i} - {y}_{i} \Rightarrow \left({x}_{i}^{T} \otimes {x}_{i}^{T} \right) \operatorname{Vec} \left( W \right) - \operatorname{Vec} \left( {y}_{i} \right) = \left({x}_{i}^{T} \otimes {x}_{i}^{T} \right) \operatorname{Vec} \left( W \right) - {y}_{i}$$

Since the problem is given by summing over $${x}_{i}$$ one could build the matrix:

$$X = \begin{bmatrix} {x}_{1}^{T} \otimes {x}_{1}^{T} \\ {x}_{2}^{T} \otimes {x}_{2}^{T} \\ \vdots \\ {x}_{n}^{T} \otimes {x}_{n}^{T} \end{bmatrix}$$

Then:

$$\arg \min_{W} \sum_{i = 1}^{n} {\left( {x}_{i}^{T} W {x}_{i} - {y}_{i} \right)}^{2} = \arg \min_{W} {\left\| X \operatorname{Vec} \left( W \right) - \boldsymbol{y} \right\|}_{2}^{2}$$

Where $$\boldsymbol{y}$$ is the column vector composed by $${y}_{i}$$.

Now the above has nice form of regular Least Squares. The handling of the constraint can be done using Projected Gradient Descent Method. The projection onto the set of Symmetric Matrices and Positive Semi Definite (PSD) Matrices cone are given by:

1. $$\operatorname{Proj}_{\mathcal{S}^{n}} \left( A \right) = \frac{1}{2} \left( A + {A}^{T} \right)$$. See Orthogonal Projection of a Matrix onto the Set of Symmetric Matrices.
2. $$\operatorname{Proj}_{\mathcal{S}_{+}^{n}} \left( A \right) = Q {\Lambda}_{+} {Q}^{T}$$ where $$A = Q \Lambda {Q}^{T}$$ is the eigen decomposition of $$A$$ and $${\Lambda}_{+}$$ means we zero any negative values in $$\Lambda$$. See Find the Matrix Projection of a Symmetric Matrix onto the set of Symmetric Positive Semi Definite (PSD) Matrices.

Since both the Symmetric Matrices Set and PSD Cone are Linear Sub Space hence even the greedy iterative projection on the set will yield an orthogonal projection on the intersection of the 2 sets. See Orthogonal Projection onto the Intersection of Convex Sets.

So, with all the tools above one could create his own solver using basic tools with no need for external libraries (Which might be slow or not scale).

I implemented the Projected Gradient Descent Method with the above projections in MATLAB. I compared results to CVX to validate the solution. This is the solution:

My implementation is vanilla Gradient Descent with constant Step Size and no acceleration. If you add those you'll see convergence which is order of magnitude faster (I guess few tens of iterations). Not bad for hand made solver.

The MATLAB Code is accessible in my StackExchange Mathematics Q3619669 GitHub Repository.

• This was a very clear and practical response. Thanks! Assuming I understand correctly, I have one question: what is the difference between iteratively projecting vs. projecting just once after finding the optimal solution to the unconstrained problem? Would they give different solutions? Apr 13 '20 at 15:43
• @tygaking, Please mark it as answer and +1. Regarding your question, you can't do that. Think of "Gradient Descent" as some kind of projection by itself. Hence what we do above is basically iterative projection onto iterative set, one for the gradient descent, one for each other.
– Royi
Apr 13 '20 at 15:45

Are you doing FGLS or something?

You could try substituting the constraint into the object. For the two by two case, for example, solve $$\sum_{i = 1}^N \left(x_i'\left[\array{ \array{w_{11} & w_{12}} \\ \array{w_{12}& w_{22} }} \right]x_i - y_i \right)^2$$ where $$w_{12} = w_{21}$$ now by construction. Then the matrix will be symmetric.

To ensure positive semi-definiteness, you can then use the standard principal minors test: $$w_{ii} \ge 0$$ for each $$i$$, $$w_{11} w_{22} - w_{12}^2 \ge 0$$, and so on, with the determinant of the upper left-hand minor weakly positive.

That at least subsumes positive semi-definiteness into concrete constraints and adjustments to the problem. That sounds like a nightmare to solve however, using Kuhn-Tucker. A simpler sufficient condition for semi-definiteness is the dominant diagonal condition, that $$w_{ii} \ge \sum_{j \neq i} |w_{ij}|$$ for each row $$i$$, which would be much more computationally tractable. Perhaps it could give you a good initial guess before you try relaxing it to the standard principal minors constraints.

• There is no need for this. While your approach might lead to problem formulated as Least Squares with Linear Equality and Inequality Constraints, it will also require some kind of iterative solver (As there is no closed form solution for LS with Inequality Constraints). Since we have the projection onto the Symmetric and PSD Matrices we can use it. See my answer. Still, nice idea of yours! Especially with the Diagonally Dominant Matrix. I might post a question and answer about this.
– Royi
Apr 12 '20 at 0:00

This is a convex optimization problem which can be easily formulated, and then numerically solved via a convex optimization tool, such as CVX, YALMIP, CVXPY, CVXR. This is a linear Semidefinite Programming problem (SDP), for which numerical solvers exist.

Here is the code for CVX.

Assume $$x_i$$ is the ith column of matrix $$X$$

cvx_begin
variable W(p,p) semidefinite % constrains W to be symmetric positive semidefinite
Objective = 0;
for i=1:n
Objective = Objective + square(X(:,i)'*W*X(:,i) - y(i))
end
minimize(Objective)
cvx_end

CVX will transform the problem into the form required by the solver, call the solver, and return the solution.

• I think the use of square_pos() here is wrong. It will lead to the wrong solution. You can use square() or if you want to be sure square_abs() or something like that.
– Royi
Apr 11 '20 at 23:35
• @Royi Right you are. Now fixed. Thanks for pointing it out. Do you want to take over for me as moderator at CVX Forum? Apr 11 '20 at 23:43
• Care to explain what you meant? Thanks...
– Royi
Apr 18 '20 at 22:43
• @Royi You can take over answering many of the questions there, and I can go into semi-retirement from that forum. Apr 18 '20 at 23:02
• I would answer you with a PM on CVX Forum. Yet the PM feature is disabled. Want to move it to an email?
– Royi
Apr 22 '20 at 15:08

This does seem to be a straight forward 'nonnegative' least squares though with the sdp constraint. n = 10; p = 5; X = zeros(n,p^2); for ii = 1:n x = randn(p,1); temp = xx'; X(ii,:) = temp(:)'; end y = randn(n,1); cvx_begin sdp variable W(p,p) semidefinite minimize(norm(XW(:)-y)) cvx_end


# Deduce that there exists a prime $p$ where $p$ divides $x^2 +2$ and $p≡3$ (mod 4)

I am revising for a number theory exam and have a question that I am struggling with, any help would be greatly appreciated.

First I am asked to show that for an odd number $x$, $x^2+2 ≡3$(mod 4).

I can do this part of the question, but next I am asked to deduce that there exists a prime $p$ where $p$ divides $x^2 +2$ and $p≡3$ (mod 4)

I am struggling to see how to attempt the second part and how the first part relates.

My thoughts so far are that I want to show $x^2≡-2$(mod p) ? And perhaps Fermat's Little Theorem could be of use here somehow?

Not sure if I'm barking up the wrong tree though.

Consider the prime factorization of $x^2+2$. Since $x$ is odd, $x^2+2$ is odd implying $2$ will not show up in the factorization. Now consider the primes that DO show up in the prime factorization. If they are all $1$ in modulo 4, then their product will also be one in modulo $4$. This is not true though, since you know that $x^2+2$ is $3$ modulo $4$. Therefore, there must be a prime that in the prime factorization of $x^2+2$ s.t. it is not $1$ modulo 4. Since primes other than 2 that are not $1$ modulo 4 are $3$ modulo $4$, this completes the reasoning.

• Fantastic! Thank you so much – Olivia77989 May 15 '13 at 20:53

Our number $x^2+2$ is odd, and greater than $1$, so it is a product of odd not necessarily distinct primes. If all these primes were congruent to $1$ modulo $4$, their product would be congruent to $1$ modulo $4$. But you have shown that $x^2+2\equiv 3\pmod{4}$.

• You are welcome. You probably already ran into the fact that every positive integer of the form $4k+3$ has a prime divisor of the form $4k+3$. That fact plays a key role in the standard proof that there are infinitely many primes of the form $4k+3$. – André Nicolas May 15 '13 at 20:59
• A useful insight! So would I be right in thinking that in order to prove there are infinitely any primes of the form $4k +3$ , I would split $4k +3$ into its prime divisors $p1,p2...pn$ noting that at least one of these divisors is of the form $4k +3$, then, following a similar approach to Euclid's proof, consider N=4(p1. p2.....pn)+3 ? Or what would we equate N to and why? – Olivia77989 May 15 '13 at 21:31
• Yes, suppose there are finitely many primes $p_1,\dots,p_s$ of the right shape. Consider $N=4p_1\cdots p_s-1$. This is of the shape $4k+3$, so has a prime divisor of that shape, which must be different from the $p_i$. I am avoiding $N=4p_1\cdots p_s+3$, since maybe $N$ could be a power of $3$ (there are other ways around that issue). – André Nicolas May 15 '13 at 21:38
• I see. Possibly a silly question, but how do we know that the prime divisor of N in the form $4k+3$ is different from the $p_i$ already in our list? – Olivia77989 May 15 '13 at 21:46
• With $N=4p_1\cdots p_s -1$, if it was in list, it would divide $4p_1\cdots p_s$. Since it divides $4p_1\cdots p_s-1$, it would divide the difference, which is $1$. Impossible! – André Nicolas May 15 '13 at 21:56

The question, as asked, has been answered. Nevertheless, I'll show how quadratic residues can help in problems like these, and show that moreover $x^2+2$ ($x$ odd) has a prime factor congruent to $3 \!\!\!\mod 8$:

Suppose that $x$ is odd. If $p$ divides $x^2+2$, then $p$ is odd and $x^2 \equiv -2 \!\!\mod p$. In particular, $-2$ is a square mod $p$, hence $$1=\left(\frac{-2}{p}\right)=\left(\frac{-1}{p}\right)\left(\frac{2}{p}\right)=(-1)^{(p^2-1)/2} (-1)^{(p-1)/2}=(-1)^{(p^2+p-2)/2}.$$ This implies $p \equiv 1,3 \!\!\mod \!8$. If each prime $p$ dividing $x^2+2$ were congruent to $1 \!\!\mod 8$, then $x^2+2$ would be congruent to $1\!\! \mod 8$. This is a contradiction, hence there exists some $p \mid x^2+2$ congruent to $3 \!\!\mod 8$. (In fact, there must be an odd number of such primes.)

Of course, any one of the primes is congruent to $3 \!\!\mod 4$, which gives your result.

We can ignore quadratic residues in your particular problem because there are only two odd residues mod $4$. In more difficult questions, looking at quadratic residues can give more information.


# Why are the last two digits of a perfect square never both odd?

Earlier today, I took a test with a question related to the last two digits of perfect squares.

I wrote out all of these digits pairs up to $20^2$.

I noticed an interesting property, and when I got home I wrote a script to test it. Sure enough, my program failed before it was able to find a square where the last two digits are both odd.

Why is this?

Is this always true, or is the rule broken at incredibly large values?

• do you know about modular arithmetic ? that might be a starting place. – user451844 Aug 19 '17 at 0:51
• The last two digit of a number is the number modulus $100$ Talking about squares the last two digits are cyclical. They repeat every 50 squares from $0$ to $49$ or from $10^9 + 2017$ to $10^9 + 2017 + 49$ they are always the following $00,\;01,\;04,\;09,\;16,\;25,\;36,\;49,\;64,\;81,\;00,\;21,\;44,\;69,\;96,\;25,\;56,\;89,\;24,\;61,\;00,\;41,\;84,\;29,\;76,\;25,\;76,\;29,\;84,\;41,\;00,\;61,\;24,\;89,\;56,\;25,\;96,\;69,\;44,\;21,\;00,\;81,\;64,\;49,\;36,\;25,\;16,\;09,\;04,\;01$ and there is never a combination of two odd digits. – Raffaele Aug 19 '17 at 14:39
• Not only, but if a number does not end with one of the following pair of digits it cannot be a perfect square $00,\; 01,\; 04,\; 09,\; 16,\; 21,\; 24,\; 25,\; 29,\; 36,\; 41,\; 44,\; \\ 49,\; 56,\; 61,\; 64,\; 69,\; 76,\; 81,\; 84,\; 89,\; 96$ – Raffaele Aug 19 '17 at 14:39
• You can also note that the last two digits of squares from 0 to 25 are the same as from 50 to 25 so it is both cyclical and symetrical :) – Rafalon Aug 20 '17 at 9:21
• @Raffaele Sadly if you'd made that into an answer it probably would have earned you a decent chunk of rep and might have been the accepted answer. – Pharap Aug 20 '17 at 16:24

Taking the last two digits of a number is equivalent to taking the number $\bmod 100$. You can write a large number as $100a+10b+c$ where $b$ and $c$ are the last two digits and $a$ is everything else. Then $(100a+10b+c)^2=10000a^2+2000ab+200ac+100b^2+20bc+c^2$. The first four terms all have a factor $100$ and cannot contribute to the last two digits of the square. The term $20bc$ can only contribute an even number to the tens place, so cannot change the result. To have the last digit of the square odd we must have $c$ odd. We then only have to look at the squares of the odd digits to see if we can find one that squares to two odd digits. If we check the five of them, none do and we are done.

• Ah, it's so obvious in hindsight. Although, I suppose most problems like this are. Thanks! – Pavel Aug 19 '17 at 1:00

Others have commented on the trial method. Just to note that $3^2$ in base $8$ is $11_8$ which has two odd digits. This is an example to show that the observation here is not a trivial one.

But we can also note that $(2m+1)^2=8\cdot \frac {m(m+1)}2+1=8n+1$ so an odd square leaves remainder $1$ when divided by $8$.

The final odd digits of squares can be $1,5,9$ so odd squares are $10p+4r+1$ with $r=0,1,2$. $10p+4r$ must be divisible by $8$ and hence by $4$, so $p$ must be even.

In the spirit of experimentation, the last two digits of the squares of numbers obtained by adding the column header to the row header:

$$\begin {array}{c|ccc} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9\\ \hline 0 & 00 & 01 & 04 & 09 & 16 & 25 & 36 & 49 & 64 & 81\\ 10 & 00 & 21 & 44 & 69 & 96 & 25 & 56 & 89 & 24 & 61\\ 20 & 00 & 41 & 84 & 29 & 76 & 25 & 76 & 29 & 84 & 41\\ 30 & 00 & 61 & 24 & 89 & 56 & 25 & 96 & 69 & 44 & 21\\ 40 & 00 & 81 & 64 & 49 & 36 & 25 & 16 & 09 & 04 & 01\\ 50 & 00 & 01 & 04 & 09 & 16 & 25 & 36 & 49 & 64 & 81\\ 60 & 00 & 21 & 44 & 69 & 96 & 25 & 56 & 89 & 24 & 61\\ 70 & 00 & 41 & 84 & 29 & 76 & 25 & 76 & 29 & 84 & 41\\ 80 & 00 & 61 & 24 & 89 & 56 & 25 & 96 & 69 & 44 & 21\\ 90 & 00 & 81 & 64 & 49 & 36 & 25 & 16 & 09 & 04 & 01\\ 100 & 00 & 01 & 04 & 09 & 16 & 25 & 36 & 49 & 64 & 81\\ 110 & 00 & 21 & 44 & 69 & 96 & 25 & 56 & 89 & 24 & 61\\ 120 & 00 & 41 & 84 & 29 & 76 & 25 & 76 & 29 & 84 & 41\\ \end{array}$$

The patterns are clear, after which the search for a reason for such patterns is well given by the answer of @RossMillikan - you can see that the parity of both final digits of the square is entirely dependent on the final digit of the number that you square.

As a hint, consider what determines the last two digits of a multiplication. Do you remember doing multiplication by hand? If you have square a ten digit number, do all the digits matter when considering just the last two digits of the answer? You will realize that you can put a bound on the number of squares you need to check before you can prove the assertion you are making for all n

• In fact more than enough values of n has been checked (within $20^2$) to see that this result is true. Question is, how does one know enough values have been checked? – Jihoon Kang Aug 19 '17 at 1:02
• That's right - a lazy bound would be checking with a computer all two digit squares, as you know that in a 3 digit number, the hundreds digit will not impact on the final two digits in the multiplication (the same for larger numbers). Obviously you can get sharper than that, as the other answer showed. I was just pointing out that you can very easily come up with a lazy bound by thinking about what happens when you multiply numbers. – Franz Aug 19 '17 at 1:23

This is just another version of Ross Millikan's answer.

Let $N \equiv 10x+n \pmod{100}$ where n is an odd digit.

\begin{align} (10x + 1)^2 \equiv 10(2x+0)+1 \pmod{100} \\ (10x + 3)^2 \equiv 10(6x+0)+9 \pmod{100} \\ (10x + 5)^2 \equiv 10(0x+2)+5 \pmod{100} \\ (10x + 7)^2 \equiv 10(4x+4)+9 \pmod{100} \\ (10x + 9)^2 \equiv 10(8x+8)+1 \pmod{100} \\ \end{align}

A simple explanation.

1. Squaring means multiplication, multiplication means repeatative additions.

2. Now if you add even no.s for odd no. of times or odd no.s for even no. of times you will always get an even no.

Hence, square of all the even no.s are even, means the last digit is always even.

1. If you add odd no.s for odd no. of times you will always get an odd no.

Coming to the squares of odd no.s whose results are >= 2 digits. Starting from 5^2 = 25, break it as 5+5+5+5+5, we have a group with even no. of 5 and one extra 5. According to my point no. 2 the even group will always give you a even no. i.e. 20, means the last digit is always even. Addition of another 5 with 20 makes it 25, 2 is even.

Taking 7^2, 7+7+7+7+7+7+7, group of six 7's = 42 plus another 7 = 49.

Now consider 9^2, 9+9+9+9+9+9+9+9+9, group of eight 9's = 72 plus another 9 = 81, (72+9 gets a carry of 1 making the 2nd last digit even)

35^2 = group of twenty four 34's (1190) plus 35 = 1225, carry comes.

In short just check the last digit of no. that you can think of in the no. co-ordinate (Real and Imaginary) it will always be b/w 0-9 so the basic principle (point 2 and 3) will never change. Either the last digit will be an even or the 2nd last digit will become even with a carry. So the 1 digit sq can come odd, 1 and 9, as there is no carry. I have kept it as an exception in point 3.

BTW many, including the author may not like my lengthy explanation as mine is not a mathematical one, full of tough formulae. Sorry for that. I'm not from mathematical background and never like maths.

• "35^2 = group of twenty four 34's (1190) plus 35 = 1225, carry comes."? One of the burdens of being a mathematician is to present your arguments and then deal with any criticism that follows. It's got nothing to do with you. Whoever you are, if you publish a turkey, someones going to roast it. – steven gregory Aug 20 '17 at 23:13
• That's a typo. That will be odd+odd = even. I saw that error but didn't edit it. – NewBee Aug 23 '17 at 14:26

Let b be last digit of odd perfect square a,then b can be 1,9 or 5. For b=1,9; $a^2-b$ is divisible by 4, $(a^2-b)/10$ is even. For b=5 ;a always ends in 25.


# A curious property of an acute triangle

Many years back in high school I happened to stumble upon the following property that seems to hold for any acute triangle:

$CD$ and $BE$ are altitudes, the arcs are semicircles with diameters $AB$ and $AC$ respectively.

The property is that $AF = AG$

Proof:

Let $H$ be the midpoint of $AB$ (and the centre of the respective semicircle) $$AG^2 = AD^2 + GD^2 = \left(AC\cdot \cos\angle A\right)^2 + GD^2$$ Since $HG = AH = \frac{AB}{2}$ is the radius of the semicircle $$GD^2 = HG^2 - HD^2 = \left(\frac{AB}{2}\right)^2 - \left(\frac{AB}{2} - AC\cdot\cos\angle A\right)^2 = \\ = AB\cdot AC\cdot \cos\angle A - \left(AC\cdot\cos\angle A\right)^2$$

which gives $$AG^2 = AB\cdot AC\cdot \cos\angle A$$

Analogously ($I$ is the midpoint of $AC$) $$AF^2 = AE^2 + FE^2 = \left(AB\cdot \cos\angle A\right)^2 + FE^2$$ $$FE^2 = FI^2 - EI^2 = \left(\frac{AC}{2}\right)^2 - \left(AB\cdot \cos\angle A - \frac{AC}{2}\right)^2 = \\ = AC\cdot AB\cdot \cos\angle A - \left(AB\cdot \cos\angle A\right)^2$$ which finally gives $$AF^2 = AC\cdot AB\cdot \cos\angle A$$

Questions:

1. Is this a known property?
2. Is there a better more elegant proof?
• You can say something more: The four points where the (full) circles with diameters $\overline{AB}$ and $\overline{AC}$ meet the altitudes from $B$ and $C$, lie on a common circle about $A$.
– Blue
Apr 3 '18 at 16:49
• @Blue, that's a very, very good one! Thank you. Apr 3 '18 at 21:54

HINT:

$$AF^2 = AE \cdot AC\\ AG^2 = AD \cdot AB \\ AE \cdot AC = AD \cdot AB$$

• what a beautiful proof! Apr 3 '18 at 16:42
• Brilliant indeed. Although not so trivial (to me at least). Thanks a lot. Apr 3 '18 at 21:41
• @ageorge: My pleasure! A nice problem indeed! Apr 3 '18 at 22:32

This is a possibly similar proof, but here goes anyway...

Let the triangle be labelled in the usual way, so $AC=b$ and $AB=c$, and let angle $FAC=\theta$

Then $$AF=b\cos\theta$$ $$\implies FE=AF\sin\theta=b\sin\theta\cos\theta$$ $$\implies EC=FE\tan\theta=b\sin^2\theta$$ But $$EC=b-c\cos A=b\sin^2\theta$$ $$\implies b\cos^2\theta=c\cos A$$ $$\implies AF=\sqrt{bc \cos A}$$

To get $AG$ we only need to exchange $b$ and $c$, so the result follows.


# Math Help - Trouble Simplifying/Factorising

1. ## Trouble Simplifying/Factorising

I have a rather straightforward Physics problem to solve, only I can't for the life of me simplify the answer to what it needs to be. What am I doing wrong?

$
v_1=\frac{m}{m+M}v_o
$

$\frac{1}{2}mv_o^2=\frac{1}{2}mv_1^2+\frac{1}{2}Mv_ 1^2+\frac{1}{2}kd^2$

The answer is in the form $d=$ and it's a simplification of the information I provided here in the post.

I get as far as $d=\sqrt{\frac{1}{k}(mv_o^2-(m+M)v_1^2)}$

The book plugs in $v_1$

and simplifies to $d=\sqrt{\frac{mM}{k(m+M)}}\times{v_o}$

How?

2. Originally Posted by dkaksl
$
v_1=\frac{m}{m+M}v_o
$

$\frac{1}{2}mv_o^2=\frac{1}{2}mv_1^2+\frac{1}{2}Mv_ 1^2+\frac{1}{2}kd^2$
Changing your v1 to v and v0 to w, then your 2 equations are:

v = mw / (m + M) [1]

mw^2 = mv^2 + Mv^2 + kd^2 [2]

Simplifying:
m + M = mw / v [1]
mw^2 - kd^2 = v^2(m + M) [2]

Substitute [1] in [2] ; OK?

3. Very useful, easy-to-understand solution. Thanks a lot.

4. Hello again.

Just got to checking the textbook and unfortunately the question was to define $d$ in the terms $m, M, k$ and $v_o$.

Cancelling out $(m + M)$ got me an answer with $v_1$.

Edit:

$kd^2=mv_o^2-(m+M)v_1^2$

if $v_1=\frac{mv_o}{m+M}$ then what is $v_1^2$?

$\frac{mv_o}{m+M}\times\frac{mv_o}{m+M}$ which is

$\frac{m^2v_o^2}{(m+M)^2}$

$kd^2=mv_o^2-(m+M)\times\frac{m^2v_o^2}{(m+M)^2}$

$kd^2=mv_o^2-\frac{m^2v_o^2}{(m+M)}$

$kd^2=\frac{mv_o^2(m+M)}{(m+M)}-\frac{m^2v_o^2}{(m+M)}$

$kd^2=\frac{mv_o^2(m+M)-m^2v_o^2}{(m+M)}$

$kd^2=\frac{m^2v_o^2+mMv_o^2-m^2v_o^2}{(m+M)}=\frac{mM}{(m+M)}\times{v_o^2}$

$d=\sqrt{\frac{1}{k}\times{\frac{mM}{(m+M)}}}\times {v_o}$

Last edit: Thanks a lot. I see where I made a mistake in expanding. Learned a lot today. Merry Christmas.

5. Originally Posted by dkaksl
Just got to checking the textbook and unfortunately the question was to define $d$ in the terms $m, M, k$ and $v_o$.
Ahhh; well then, we need to get rid of your v1 (my v); the 2 equations:

v = mw / (m + M) [1]

mw^2 = mv^2 + Mv^2 + kd^2 [2]

square [1]: v^2 = m^2w^2 / (m + M)^2
rearrange [2]: v^2 = (mw^2 - kd^2) / (m + M)

m^2w^2 / (m + M)^2 = (mw^2 - kd^2) / (m + M)

m^2w^2 / (m + M) = mw^2 - kd^2

m^2w^2 = mw^2(m + M) - kd^2(m + M)

kd^2(m + M) = mw^2(m + M) - m^2w^2

kd^2(m + M) = mw^2(m + M - m)

kd^2(m + M) = mMw^2

d^2 = mMw^2 / (k(m + M))

d = wSQRT[mM / (k(m + M))] ........Merry Xmas!


# What is $\lim\limits_{n\to\infty} \frac{n^d}{ {n+d \choose d} }$?

What is $\lim\limits_{n\to\infty} \frac{n^d}{ {n+d \choose d} }$ in terms of $d$? Does the limit exist? Is there a simple upper bound interms of $d$?

-
I think the limit sould exist and is a function of $d$. Because the denominator is $\sim n^d$ when $n\rightarrow \infty$ – Seyhmus Güngören Sep 20 '12 at 11:59

Write $$\frac{n^d}{(n+d)!}d!n!=d!\prod_{j=1}^d\frac{n}{n+j}$$ to see that the expected limit is $d!$.

-
You mean "... is $d!$" instead of $n!$. – martini Sep 20 '12 at 12:00
I think there is a little mistake here. – Seyhmus Güngören Sep 20 '12 at 12:02
The denominator in the product should be $n+j$. – celtschk Sep 20 '12 at 12:15
Thanks guys, I've corrected. – Davide Giraudo Sep 20 '12 at 12:26

$$n^d\frac{n!\,d\,!}{(n+d)!}=d\,!\,(\frac{n}{n+1})(\frac{n}{n+2})\ldots(\frac{n}{n+d})$$

Note that the R.H.S. contains a finite number of terms ($=d$) each tending to 1.

Hence the limit is $d\,!$

To find an upper bound you can simply do this

$d\,!\,(\frac{n}{n+1})(\frac{n}{n+2})\ldots(\frac{n}{n+d})<d\,!\,(\frac{n}{n})(\frac{n}{n})\ldots(\frac{n}{n})$

$d\,!\,(\frac{n}{n+1})(\frac{n}{n+2})\ldots(\frac{n}{n+d})<d\,!$

So your upper bound is $d\,!$

Another Method (Just to show that it the sequence is convergent)

Let us treat the given term $n^d\frac{n!\,d\,!}{(n+d)!}$ as term of sequces $\{x_n\}$.

Since we have shown that the sequence is bounded, it will suffice to show that the sequence is monotonically increasing.

We will now show that it is monotonically increasing.

$$\frac{n}{n+r}-\frac{m}{m+r}=\frac{r(n-m)}{(n+r)(m+r)}>0 \quad\forall\; n>m>0\quad \& \quad\forall r>0$$

Let $n>m$

Then $n!>m!$ and each term $\frac{n}{n+r}>\frac{m}{m+r} \quad \forall r \in \{1, 2, \ldots ,d\}$, we have proved this above.

Hecne $x_n \gt m_m$ (as $a>a' \quad \& \quad b\gt b'\implies a.b>a'.b' \quad \forall a,a',b,b'>0$ )

If you want to see how a monotonically increasing sequence which is upper bounded is convergent go to this link.

-

You can use the Stirling approximation $$n! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n\,,$$

after writing your expression in the form

$$d!\frac{n^d n!}{(n+d)!}$$

$$d!\frac{n^d n!}{(n+d)!} \sim d! \frac{n^d \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n }{\sqrt{2 \pi (n+d)} \left(\frac{n+d}{e}\right)^{n+d}}= d!\, {\rm e}^d \left(\frac{n}{n+d}\right)^{n} \left(\frac{n}{n+d}\right)^{d} \left(\frac{n}{n+d}\right)^{\frac{1}{2}}$$

$$\rightarrow d! \,{\rm e}^d \, \,{\rm e}^{-d}\, 1.1 = d!\,, \quad n \rightarrow \infty$$

-


# Does an n by n Hermitian matrix always has n independent eigenvectors?

I am learning the MIT ocw 18.06 Linear Algebra, and I have learnt: an arbitrary $n×n$ matrix A with n independent eigenvectors can be written as $A=SΛS^{-1}$, and then for the Hermitian matrices, because the eigenvectors can be chosen orthonormal, it can be written as $A=QΛQ^H$ further.

I wonder does every $n×n$ Hermitian matrix has n independent eigenvectors and why? Thank you!

P.S. MIT 18.06 Linear Algebra Lecture 25: Symmetric Matrices and Positive Definiteness. You may wish to start from 4:20. From the course, I think the spectral theorem comes from diagonalizable matrix, it's just a special case, it's just the case eigenvectors are orthonormal. The eigenvectors of Hermitian matrices can be chosen orthnormal, but is every Hermitian matrix diagonalizable? If it is, why?

• You mean independent eigenvectors, not independent eigenvalues. – uniquesolution Feb 8 '18 at 12:04
• Oh, thanks, I have edited it. – Thomas Feb 9 '18 at 2:28
• It's on the Wikipedia page: en.wikipedia.org/wiki/Hermitian_matrix. – Robert Wolfe Feb 9 '18 at 4:25
• @Robert Thank you for the link, I had read the page though, and there wasn't an answer to my question. – Thomas Feb 9 '18 at 10:34
• Look here for a simple proof. – user Feb 11 '18 at 20:15

This is a theorem with a name: it is called the Spectral Theorem for Hermitian (or self-adjoint) matrices. As pointed out by Jose' Carlos Santos, it is a special case of the Spectral Theorem for normal matrices, which is just a little bit harder to prove.

Actually we can prove the spectral theorem for Hermitian matrices right here in a few lines.

We are going to have to think about linear operators rather than matrices. If $T$ is a linear operator on a finite dimensional complex inner product space $V$, its adjoint $T^*$ is another linear operator determined by $\langle T v, w\rangle = \langle v, T^* w \rangle$ for all $v, w \in V$. (Note this is a basis-free description.) $T$ is called Hermitian or self-adjoint if $T = T^*$.

Let $B$ be an $n$-by-$n$ complex matrix and $B^*$ the conjugate transpose matrix. Let $T_B$ and $T_{B^*}$ be the corresponding linear operators. Then $(T_B)^* = T_{B^*}$, so a a matrix is Hermitian if and only if the corresponding linear operator is Hermitian.

Let $A$ be a Hermitian linear operator on a complex inner product space $V$ of dimension $n$. We need to consider $A$--invariant subspaces of $V$, that is linear subspaces $W$ such that $A W \subseteq W$. We should think about such a subspace as on an equal footing as our original space $V$. In particular, any such subspace is itself an inner product space, $A_{|W} : W \to W$ is a linear operator on $W$, and $A_{|W}$ is also Hermitian. If $\dim W \ge 1$, $A_{|W}$ has an least one eigenvector $w \in W$ -- because any linear operator at all acting on a (non-zero) finite dimensional complex vector space has at least one eigenvector.

The basic phenomenon is this: Let $W$ be any invariant subspace for $A$. Then $W^\perp$ is also invariant under $A$. The reason is that if $w \in W$ and $x \in W^\perp$, then $$\langle w, A x\rangle = \langle A^* w , x \rangle = \langle A w, x \rangle = 0,$$ because $Aw \in W$ and $x \in W^\perp$. Thus $A x \in W^\perp$.

Write $V = V_1$. Take one eigenvector $v_1$ for $A$ in $V_1$. Then $\mathbb C v_1$ is $A$--invariant. Hence $V_2 = (\mathbb C v_1)^\perp$ is also $A$ invariant. Now just apply the same argument to $V_2$: the restriction of $A$ to $V_2$ has an eigenvector $v_2$ and the perpendicular complement $V_3$ to $\mathbb C v_2$ in $V_2$ is $A$--invariant. Continuing in this way, one gets a sequence of mutually orthogonal eigenvectors and a decreasing sequence of invariant subpsaces, $V = V_1 \supset V_2 \supset V_3 \dots$ such that $V_k$ has dimension $n - k + 1$. The process will only stop when we get to $V_n$ which has dimension 1.

• You really hit the nail on the head as I'm afraid my question is too unusual to be understand. May I repeat you answer in my imprecise words? – Thomas Feb 9 '18 at 12:42
• Regard A as an operator on V, its invariant subspaces contains at least one eigenvector, their orthogonal complement are invariant subspaces too because A is a Hermitian matrix, we start from $\mathbb C^n$, every time we take a eigenvector from the space, it on a line which is a invariant subspace, and the space left which like a plane perpendicular to the line is also a invariant subspace, and we can take n times (i.e. n orthogonal/independent eigenvectors). Do I understand it correctly? – Thomas Feb 9 '18 at 12:47
• I still have some trouble with this, what if I take an eigenvector from the last space? As we know "Let $W$ be any invariant subspace for $A$. Then $W^⊥$ is also invariant under $A$", I take the eigenvector out, then there will be only zero vector left, and I think it really orthogonal to that eigenvector we take, is zero vector an invariant subspace? It may be, but "invariant space has at least one eigenvector" would be false. This "Let $W$ be any invariant subspace for $A$. Then $W^⊥$ is also invariant under $A$" makes me confused. – Thomas Feb 9 '18 at 13:43
• When you have reached $V_n$, you already have your collection of $n$ mutually orthogonal eigenvectors. You can't go further and you don't need to go further. I corrected the answer at one point to say that a linear operator on a non-zero finite dimensional complex vector space has an eigenvector. – fredgoodman Feb 9 '18 at 14:19
• It's important in the second paragraph that "acting on" means "mapping it into itself". – Ian Feb 9 '18 at 15:26

Yes, it has. That is due to the spectral theorem: every normal $n\times n$ matrix is diagonalizable. And Hermitian matrices are normal.

• From my understanding, the factorization $A=QΛQ^H$ (i.e., the spectral theorem) is the special form of $A=SΛS^{-1}$ when A is a Hermitian matrix, and $A=SΛS^{-1}$ exists under the situation that there is a full set of independent eigenvectors. I wonder, if it is, why a Hermitian matrix is always diagonalizable? – Thomas Feb 9 '18 at 2:58
• @Thomas Because the spectral theorem says so. – José Carlos Santos Feb 9 '18 at 6:30
• From the course (I add the link into the question now), I think the spectral theorem comes from diagonalizable matrix, it's just a special case, it's just the case eigenvectors are orthonormal. The eigenvectors of Hermitian matrices can be chosen orthnormal, but is every Hermitian matrix diagonalizable? If it is, why? – Thomas Feb 9 '18 at 10:57
• @Thomas Every Heritian matrix is normal and every normal matrix is diagonalizable. – José Carlos Santos Feb 9 '18 at 11:08
• I had not learnt the concept of normal matrix from the course. I search for it now, and I realize that all matrices can be written as $A=QTQ^H$ with triangular T, so $A^HA=AA^H$ would be $T^HT=TT^H$, for a 2 by 2 matrix, it's easy to prove the upper right corner entry of such T is 0, extend it to size n, and we prove that T is diagonal. Is that a right poof? Thank you for your patience. – Thomas Feb 9 '18 at 11:58


# Approaching modular arithmetic problems

I'm a little stumbled on two questions.

How do I approach a problem like $x*41 \equiv 1 \pmod{99}$.

And given $2$ modulo, $7x+9y \equiv 0 \pmod{31}$ and $2x−5y \equiv 2 \pmod{31}$ (solve for $x$ only)?

When I solve for $x$ for the latter, I got a fraction as the answer and I'm not sure if I can have a fraction as an answer? I'm not sure how to approach the first problem either.

• What fraction did you get as answer? – Git Gud Sep 22 '13 at 22:20
• The first problem involves inverses. What is the inverse of 41 mod 99? The second problem is a set of linear equations in $x$ and $y$ modulo 31, which means the two can be added/subtracted/etc. as if they were a set of normal equations looking for a linear solution. – abiessu Sep 22 '13 at 22:25

Finding the solution to $$x \times 41 \equiv 1 \pmod {99}$$ is equivalent to asking for the multiplicative inverse of $41$ modulo $99$. Since $\gcd(99,41)=1$, we know $41$ actually has an inverse, and it can be found using the Extended Euclidean Algorithm:

\begin{align*} 99-2 \times 41 &= 17 \\ 41-2 \times 17 &= 7 \\ 17-2 \times 7 &= 3 \\ 7-2\times 3 &= 1 &=\gcd(99,41). \\ \end{align*} Going back, we see that \begin{align*} 1 &= 7-2\times 3 \\ &= 7-2\times (17-2 \times 7) \\ &= 5 \times 7-2\times 17 \\ &= 5 \times (41-2 \times 17)-2\times 17 \\ &= -12 \times 17+5 \times 41 \\ &= -12 \times (99-2 \times 41)+5 \times 41 \\ &= 29 \times 41-12 \times 99 \\ \end{align*} Hence $29 \times 41 \equiv 1 \pmod {99}$ and thus $x=29$.

In the second case, we have $$7x+9y \equiv 0 \pmod {31}$$ and $$2x-5y\equiv 2 \pmod {31}.$$ Here we want to take $7x+9y=0 \pmod {31}$ and rearrange it to get $x \equiv ?? \pmod {31}$, then substitute it into the other equation and solve for $y$. This requires finding the multiplicative inverse of $7$ modulo $31$ (which we can do as above). It turns out $7 \times 9 \equiv 1 \pmod {31}$. Hence \begin{align*} & 7x+9y=0 \pmod {31} \\ \iff & 7x \equiv -9y \pmod {31} \\ \iff & x \equiv -9y \times 9 \pmod {31} \\ \iff & x \equiv 12y \pmod {31}. \end{align*} We then substitute this into the equation $2x-5y\equiv 2 \pmod {31}$, which implies $$2 \times 12y-5y \equiv 2 \pmod {31}$$ or equivalently $$19y \equiv 2 \pmod {31}.$$ Yet again, we find a multiplicative inverse, this time of $19$ modulo $31$, which turns out to be $18$. So \begin{align*} & 19y \equiv 2 \pmod {31} \\ \iff & y \equiv 2 \times 18 \pmod {31} \\ \iff & y \equiv 5 \pmod {31}. \end{align*} Hence $$x \equiv 12y \equiv 29 \pmod {31}.$$ Thus we have the solution $(x,y)=(29,5)$.

• Are you sure the answer to the first part is x=20? – Mufasa Sep 22 '13 at 22:35
• Thanks for point that out; it was an arithmetic error (actually, I found the inverse mod 91 by mistake) and it should be fixed now. – Rebecca J. Stones Sep 22 '13 at 22:47
• You're welcome. BTW: I am learning a lot from your posts - so THANK YOU! :) – Mufasa Sep 22 '13 at 22:49

For the first one, you could approach it as follows:
$41x=1$ mod $99$
$140x=1$ mod $99$ (because 41 mod 99 = (41+99) mod 99)
$140x=100$ mod $99$ (because 1 mod 99 = (1+99) mod 99)
$7x=5$ mod $99$ (divide both sides by 20)
$7x=203$ mod $99$ (because 5 mod 99 = (5+99+99) mod 99)
$x=29$ mod $99$ (divide both sides by 7)

For the first one, you have to find a multiplicative inverse for $41$ mod $99$. Use Euclid's algorithm to find the solutions of $41\cdot x + 99 \cdot y = 1$ to find $x$.

Are you familiar with abstract algebra? If yes, do you know that $\mathbb{Z}_{31}$ is a field because $31$ is a prime number and you can use tools of linear algebra because it works overy any field?

You have the following system of equations in $\mathbb{Z}_{31}$:

$$7x+9y=0$$ $$2x-5y=2$$

You can use Gaussian elimination in $\mathbb{Z}_{31}$ to find $x$ and $y$. But you have to be careful that your coefficients are in $\mathbb{Z}_{31}$ not in $\mathbb{R}$.

EDIT(suggested by Git Gud):

Notice that if you get a fraction like $\displaystyle \frac{a}{b}$ then you should think of it as $a \cdot b^{-1}$ and then find $b^{-1}$ in $\mathbb{Z}_{31}$. As previously said, $\mathbb{Z}_{31}$ is a field, so talking about fractions in it makes sense. The notation $\frac{a}{b}$ is actually nothing but $a.b^{-1}$ where $.$ denotes the multiplication in the field and $b^{-1}$ denotes the multiplicative inverse of $b$ in the field.

• @GitGud: OK. Done. – user66733 Sep 22 '13 at 22:32


Properties of the Absolute Value

1. Jul 25, 2008

roam

Just wanted to say hi before I start my post!

As you may know there is a property of the absolute value that states; for $$a, b \in R$$;

$$|ab| = |a||b|$$

Well, my friend asked me if I knew a proof for this... but I don't know...
How can we prove this statement/property? I know there is a proof for the triangle inequality but for this I really don't know but I'm curious.

I'd appreciate it if anyone could show me any kind of proof or send me some links etc. Thanks!

2. Jul 25, 2008

HallsofIvy

Staff Emeritus
You prove that |ab|= |a||b| the same way you prove any such elementary statement: use the definitions.

The simplest definition of |x| (there are several equivalent definitions) is that |x|= x if x is positive or 0, -x if x is negative.

Now break it into "cases":

case 1: x and y are both positive: |x|= x and |y|= y. xy is also positive so |xy|= xy= |x||y|.

case 2: x is positive while y is negative: |x|= x and |y|= -y. xy is negative so |xy|= -xy= x(-y)= |x||y|.

case 3: x is negative while y is positive: |x|= -x and |y|= y. xy is negative so |xy|= -xy= (-x)y= |x||y|.

case 4: x and y are both negative: |x|= -x and |y|= -y. xy is positive so |xy|= xy= (-x)(-y)= |x||y|.

case 5: x= 0 and y is positive: |x|= 0 and |y|= y. xy= 0 so |xy|= 0= 0(y)= |x||y|.

case 6: x= 0 and y is negative: |x|= 0 and |y|= -y. xy= 0 so |xy|= 0= (0)(-y)= |x||y|.

case 7: x is positive and y is 0: |x|= x and |y|= 0. xy= 0 so |xy|= 0= x(0)= |x||y|.

case 8: x is negative and y is 0: |x|= -x and |y|= 0. xy= 0 so |xy|= 0 = (-x)(0)= |x||y|.

case 9: both x and y are 0: |x|= 0 and |y|= 0. xy= 0 so |xy|= 0= (0)(0)= |x||y|.

There are simpler ways to prove that but I thought this would be conceptually clearest.

3. Jul 25, 2008

sagardipak

First, we have to understand that the absolute value is a function defined by:

$$|x| = \begin{cases} x & \text{if } x\geq 0 \\ -x & \text{if } x<0 \end{cases}$$

So,

$$|ab| = \begin{cases} ab & \text{if } ab\geq 0 \\ -ab & \text{if } ab<0 \end{cases}$$

Now, let's see what |a||b| is:

$$|a||b| = \begin{cases} ab & \text{if } a\geq 0 \wedge b\geq0 \\ (-a)(-b) & \text{if } a\leq 0 \wedge b\leq0 \\ (-a)b & \text{if } a> 0 \wedge b<0 \\ a(-b) & \text{if } a<0 \wedge b<0 \end{cases} \Leftrightarrow$$

$$|a||b| = \begin{cases} ab & \text{if } a\geq 0 \wedge b\geq0 \\ ab & \text{if } a\leq 0 \wedge b\leq0 \\ -ab & \text{if } a> 0 \wedge b<0 \\ -ab & \text{if } a<0 \wedge b<0 \end{cases} \Leftrightarrow$$

Notice that you have ab if a and b have the same sign and that you use -ab otherwise.

Now, if a and b have the same sign, $$ab\geq0$$. If they have opposite signs (and are different than zero), $$ab<0$$.

Using this,

$$|a||b| = \begin{cases} ab & \text{if } ab \geq 0 \\ -ab & \text{if } ab<0 \end{cases} = |ab|$$

Quod erat demonstrandum :tongue2:

Last edited: Jul 25, 2008
4. Jul 28, 2008

quark1005

let a = mcis(kpi) where m => 0, k is integer
and b = ncis(hpi) where n => 0, h integer
(clearly a,b are real)

|a||b|= mn
|ab|=|mcis(kpi)*ncis(hpi)| = |mncis[pi(k+h)]| = mn

as required

5. Jul 28, 2008

HallsofIvy

Staff Emeritus
This would make more sense if you had said that "mcis(hpi)" is
$$m (cos(h\pi)+ i sin(h\pi))$$
That is much more an "engineering notation" than mathematics.

If you really want to go to complex numbers, why not
if
$$x= r_xe^{i\theta_x}$$
and
$$y= r_ye^{i\theta_y}$$, then
$$|xy|= |r_xe^{i\theta_x}r_ye^{i\theta_y}|= |(r_xr_y)e^{i(\theta_x+\theta_y)}|$$

But for any $z= re^{i\theta}$, |z|= r, so
$$|xy|= r_x r_y= |x||y|$$

6. Jan 8, 2011

Vincent Mazzo

$$\Huge |ab|=\sqrt{(ab)^2}=\sqrt{a^2b^2}=\sqrt{a^2}\sqrt{b^2}=|a||b|$$

and

$$\large |a+b|=\sqrt{(a+b)^2}=\sqrt{a^2+2ab+b^2}\leq \sqrt{a^2+|2ab|+b^2}=\sqrt{(|a|+|b|)^2}=||a|+|b||$$

where I utilize the following fact:

|2ab|=|2(ab)|=|2||ab|=2|a||b|

-----
À bientôt
?;-D


# Even functions and inflection points

Can an even function have an inflection point? If yes, then give an example while if not then give a proof. If needed you may assume that $f$ is two times differentiable .

Intution says that even functions don't have inflection points, but I cannot settle down an appropriate proof.

On the contrary for an odd function this is possible. For instance $f(x)=x^3$ has an inflection point at $x=0$.

• Does it have to be at $x=0$? – mickep Sep 15 '15 at 18:22
• Thank you all for your answers. – Tolaso Sep 15 '15 at 18:58

$\cos\colon\mathbb{R}\to\mathbb{R}$ is even and it has infinitely many inflection points (at its zeros, which are found at $\{n\pi\}_{n\in\mathbb{Z}}$).
Consider $$f(x)=\max\bigl((x-1)^3,-(x+1)^3\bigr).$$ Or, for a nicer example, consider $$g(x)=(x-1)^3(x+1)^3.$$
the function $y=x^4-4x^2$ has at least one inflection point WA plot


+0

# Geometry question

+1
125
11

In triangle PQR, PQ=15 and QR=25 The circumcircle of triangle PQR is constructed. The tangent to the circumcircle at Q is drawn, and the line through P parallel to this tangent intersects QR at S. Find RS.

Jan 16, 2020

#1
-1

RS = 12.

Jan 17, 2020
#2
+107414
+2

THERE WAS AN ERROR SO I DELETED THE PICS.

Jan 17, 2020
edited by Melody  Jan 17, 2020
edited by Melody  Jan 17, 2020
#3
+28449
+2

Deleted as for some reason I used entirely the wrong lengths of triangle legs!

Alan  Jan 17, 2020
edited by Alan  Jan 17, 2020
edited by Alan  Jan 17, 2020
#5
+28449
+1

I now think Guest #4 below is correct.  There is a unique answer.  Because that answer is independent of the shape of the triangle it is legitimate to choose an easy one (a right-angled one) on which to do the calculation:

RS is indeed 16.

Alan  Jan 17, 2020
#4
+1

I think that the answer is unique, and I think that it is 16.

Using Guest's diagram, the triangles QPR and QSP are similar.

The angle between QP and the tangent, call it alpha, is equal to the angle QPS and also the angle QRP.

Similarly the angle between QR and the tangent is equal to the angle QSP and also the angle QPR.

So, QP/QS = QR/QP,

QS = QP.QP/QR =225/25 = 9, so SR = 16.

That fits the approximate values that Melody found by drawing.

Alan seems to chosen the particular case where angle PQR is a right angle.

Using the angles I called alpha earlier, from the triangle PQR, tan(alpha) = 15/25 = 3/5.

From the triangle PQS, tan(alpha) = QS/15, so QS = 45/5 = 9, as above.

Jan 17, 2020
#6
+107414
+1

Yes you are right guest.

I fixed an error in mine and now it always stays at 16.

I wss see if I can work out how to give a proper link to what i have created.

https://www.geogebra.org/classic/ekztcz7y

Melody  Jan 17, 2020
#7
+107414
0

Thanks Guest,

I am just trying to make sense of your answer.

Why is   $$\angle QRP = \alpha$$        ?

Melody  Jan 17, 2020
#9
+1

Love the geogebra Melody.

In answer to your question, it's a standard angle property of a circle.

Take a diameter from Q across the circle to T on the other side.

Angle QPT will be right angle,since it's based on the diameter, and then angle PTQ will equal the angle between PQ and the tangent, ( subtract from 90 twice).

Then, all angles based on the same arc, PQ, are equal etc.

Guest Jan 17, 2020
#11
+107414
0

Thanks guest for the complement as well as for the explanation.

I am all clear now.

Thanks Heureka and Alan as well

Melody  Jan 18, 2020
#8
+24031
+2

In triangle PQR, PQ=15 and QR=25 The circumcircle of triangle PQR is constructed.

The tangent to the circumcircle at Q is drawn, and the line through P parallel to this tangent intersects QR at S.

Find RS.

$$\text{Let C is the center of the circle } \\ \text{Let PS=TS=h } \\ \text{Let RS=x } \\ \text{Let QS=25-x }$$

$$\begin{array}{|lrcll|} \hline 1. & \left(25-x\right)^2+h^2 &=& 15^2 \qquad \text{or}\qquad \mathbf{ h^2 = 15^2-\left(25-x\right)^2} \\\\ 2. & x(25-x) &=& h * h \qquad \text{Intersecting chords theorem} \\ & x(25-x) &=& h^2 \\ & x(25-x) &=& 15^2-\left(25-x\right)^2 \\ & 25x-x^2 &=& 15^2-( 25^2-50x+x^2) \\ & 25x-x^2 &=& 15^2-25^2+50x-x^2 \\ & 25x &=& 15^2-25^2+50x \\ & 25^2-15^2 &=& 25x \\ & 25x &=& 25^2-15^2 \\ & 25x &=& 400 \\ & \mathbf{x} &=& \mathbf{16} \\ \hline \end{array}$$

Jan 17, 2020
#10
+24031
+1

In triangle PQR, PQ=15 and QR=25 The circumcircle of triangle PQR is constructed.
The tangent to the circumcircle at Q is drawn, and the line through P parallel to this tangent intersects QR at S.
Find RS

General solution:

$$\text{Let Center is the center of the circle }\\ \text{Let P-Center =T-Center=v }\\ \text{Let S-Center =w }\\ \text{Let T-Center =w+k=v\qquad k=v-w }\\ \text{Let ST = k =v-w }\\ \text{Let PS = v+w }\\ \text{Let RS = x }\\ \text{Let QS = 25-x }\\ \text{Let Q-Center = u  }$$

Intersecting chords theorem:

$$\begin{array}{|rcll|} \hline RS*QS &=& ST*PS \\ x*(25-x) &=& (v-w)*(v+w) \\ \mathbf{x*(25-x)} &=& \mathbf{v^2-w^2} & (1) \\ \hline \end{array}$$

Pythagoras:

$$\begin{array}{|lrcll|} \hline & v^2+u^2 &=& 15^2 & (2) \\ & w^2+u^2 &=& (25-x)^2 & (3) \\ \hline (2)-(3): & v^2 -w^2 &=& 15^2-(25-x)^2 \quad | \quad \mathbf{x*(25-x)=v^2-w^2} \\ & x*(25-x) &=& 15^2-(25-x)^2 \\ &25x-x^2 &=& 15^2 -(25^2-50x+x^2) \\ &25x-x^2 &=& 15^2 -25^2+50x-x^2 \\ &25x &=& 15^2 -25^2+50x \\ &25^2-15^2 &=& 25x \\ & 25x &=& 25^2-15^2 \\ & 25x &=& 400 \quad |\quad :25 \\ & \mathbf{x} &=& \mathbf{16} \\ \hline \end{array}$$

Jan 17, 2020


+0

# 3.00 x 10^8 / 6.50 x 10^-7 the result is 4.61 x 10^14 my question is how 10^8 / 10^-7 = 10^14? thank u

0
488
2

3.00 x 10^8 / 6.50 x 10^-7

4.61 x 10^14

my question is how 10^8 / 10^-7 = 10^14?

thank u

Guest Nov 21, 2014

#1
+92775
+5

3.00 x 10^8 / 6.50 x 10^-7

4.61 x 10^14

my question is how 10^8 / 10^-7 = 10^14?      IT DOES NOT!

$$\\(3\times 10^8)/(6.5\times 10^{-7})\\\\ =(3/6.5)\times 10^8/10^{-7}\\\\ =0.4615\times 10^{8--7}\\\\ =0.4615\times 10^{15}\\\\$$

BUT this answer is to be in scientific notation so there mus be just one non-zero number in front of the point

$$\\=4.4615\times 10^{-1} \times 10^{15}\\\\ =4.4615\times 10^{14}\qquad \mbox{Correct to 4 significant figures}\\\\$$

So does that all make sense now?

Melody  Nov 21, 2014
#1
+92775
+5

3.00 x 10^8 / 6.50 x 10^-7

4.61 x 10^14

my question is how 10^8 / 10^-7 = 10^14?      IT DOES NOT!

$$\\(3\times 10^8)/(6.5\times 10^{-7})\\\\ =(3/6.5)\times 10^8/10^{-7}\\\\ =0.4615\times 10^{8--7}\\\\ =0.4615\times 10^{15}\\\\$$

BUT this answer is to be in scientific notation so there mus be just one non-zero number in front of the point

$$\\=4.4615\times 10^{-1} \times 10^{15}\\\\ =4.4615\times 10^{14}\qquad \mbox{Correct to 4 significant figures}\\\\$$

So does that all make sense now?

Melody  Nov 21, 2014
#2
+92775
0

That was a good question anon

Melody  Nov 21, 2014


# Can a set containing 0 be purely imaginary?

A purely imaginary number is one which contains no non-zero real component.

If I had a sequence of numbers, say $\{0+20i, 0-i, 0+0i\}$, could I call this purely imaginary?

My issue here is that because $0+0i$ belongs to multiple sets, not just purely imaginary, is there not a valid case to say that the sequence isn't purely imaginary?

-
I think it would simplify your question a bit to just ask "Is $\textit{0}$ purely imaginary?" – curious Nov 9 '14 at 1:16
But my question is why would I consider only one classification 0+0i and ignore the others – chris Nov 9 '14 at 1:19

0 is both purely real and purely imaginary. The given set is purely imaginary. That's not a contradiction since "purely real" and "purely imaginary" are not fully incompatible. Somewhat similarly baffling is that "all members of X are even integers" and "all members of X are odd integers" is not a contradiction. It just means that X is an empty set.

-
$\ldots$ and $0$ is unique in being both purely real and purely imaginary. – Thumbnail Nov 9 '14 at 10:06

A complex number is said to be purely imaginary if it's real part is zero. Zero is purely imaginary, as it's real part is zero.

-
This is really important in fields like ordinary differential equations, where you look at having no purely imaginary eigenvalues for it to be hyperbolic – Alan Nov 9 '14 at 1:23
from my understanding zero can also be considered real? – chris Nov 9 '14 at 1:30
Yes, a complex number is real if it's imaginary part is zero. So zero is also real. – Seth Nov 9 '14 at 1:33
so then my question is why can I consider only the definition that suits my needs? – chris Nov 9 '14 at 1:38
By definition, $0$ is purely imaginary. The fact that $0$ has other properties (it is real; it is nonnegative; it is rational; it is an integer; it is algebraic; it is divisible by every prime number) does not mean you can’t use the property you need. Similarly, is $\{-2,4\}$ a set of even numbers? Yes. The number $-2$ is not only even, but it’s also negative. The fact that it’s negative doesn’t mean you can’t use the fact that it’s even. Some sets defined by properties overlap. – Steve Kass Nov 9 '14 at 1:53


# How can I find the speed and the angle of a water droplet in a rainy day as seen by a person running?

The problem is as follows:

A person is running following a constant speed of $$4.5\,\frac{m}{s}$$ over a flat (horizontal) track on a rainy day. The water droplets fall vertically with a measured speed of $$6\,\frac{m}{s}$$. Find the speed in $$\frac{m}{s}$$ of the water droplet as seen by the person running. Find the angle to the vertical should his umbrella be inclined to get wet the less possible. (You may use the relationship of $$37^{\circ}-53^{\circ}-90^{\circ}$$ for the $$3-4-5$$ right triangle ).

The alternatives given on my book are as follows:

$$\begin{array}{ll} 1.7.5\,\frac{m}{s};\,37^{\circ}\\ 2.7.5\,\frac{m}{s};\,53^{\circ}\\ 3.10\,\frac{m}{s};\,37^{\circ}\\ 4.10\,\frac{m}{s};\,53^{\circ}\\ 5.12.5\,\frac{m}{s};\,37^{\circ}\\ \end{array}$$

On this problem I'm really very lost. What sort of equation should I use to get the vectors or the angles and most importantly to get the relative speed which is what is being asked.

I assume that to find the relative speed can be obtaining by subtracting the speed from which the water droplet is falling to what the person is running. But He is in this case running horizontally. How can I subtract these?

Can somebody give me some help here?

Supposedly the answer is the first option or $$1$$. But I have no idea how to get there.

• The $3-4-5$ triangle is a hint as to the nature of the result. He is traveling horizontally, the droplets travel vertically, the resulting relative speed must combine these quantities in some manner. Since the two lines of travel are perpendicular, it makes sense to think about the hypotenuse between them... Nov 4 '19 at 3:43
• @abiessu Perhaps can you develop an answer so I could understand what you mean by combining the speeds?. Can you show me the equation should I use?. Had I not been given the alternatives, what should I have done with this question? Nov 4 '19 at 4:27

• water droplet speed as seen from a stationary point on earth is $\vect{v_d}=(0,-6)$. the speed of the runner as seen from the same point is $\vect{v_r}=(4.5,0)$. however, from the perspective of the runner, the water droplet both falls down with the speed $\vect{v_d}=(0,-6)$ and it moves towards him with the horizontal speed $\vect{v_h}=(-4.5,0)$. wherefore, for the runner the water moves with a resultant speed $\vect{v_r}=\vect{v_d}+\vect{v_h}=(0,-6)+(-4.5,0)=(-4.5,-6)$. this vector has a magnidue of 7.5 and a direction of $37^0$ West from vertical. Nov 4 '19 at 6:49