All posts by math

Solutions to the book Galois Theory by Ian Stewart

Chapter 1: Classical Algebra

Chapter 2: The Fundamental Theorem of Algebra

Chapter 3: Factorisation of Polynomials

Chapter 4: Field Extensions

Chapter 5: Simple Extensions

Chapter 6: The Degree of an Extension

Chapter 7: Ruler-and-Compass Constructions

Chapter 8: The Idea Behind Galois Theory

Chapter 9: Normality and Separability

Chapter 10: Counting Principles

Chapter 11: Field Automorphisms

Chapter 12: The Galois Correspondence

Chapter 13: A Worked Example

Chapter 14: Solubility and Simplicity

Chapter 15: Solution by Radicals

Chapter 16: Abstract Rings and Fields

Chapter 17: Abstract Field Extensions

Chapter 18: The General Polynomial Equation

Chapter 19: Finite Fields

Chapter 20: Regular Polygons

Chapter 21: Circle Division

Chapter 22: Calculating Galois Groups

Chapter 23: Algebraically Closed Fields

Chapter 24: Transcendental Numbers

Chapter 25: What Did Galois Do or Know?

Solutions to the book Complex Variables by M. Ya. Antimirov, A. A. Kolyshkin and Remi Vaillancourt

Chapter 1: Functions of a Complex Variable

Complex numbers

Continuity in the complex plane

Functions of a complex variable

Analytic functions

Elementary analytic functions

Chapter 2: Elementary Conformal Mappings

Geometric meaning of f'(z)

Basic problems and principles of conformal mappings

Linear mapping and inversion

Linear fractional transformations

Symmetry and linear fractional transformations

Mapping by z^n and w = z^(1/n)

Exponential and logarithmic mappings

Mapping by Joukowsky’s function

Mapping by trigonometric functions

Chapter 3: Complex Integration and Cauchy’s Theorem

Paths in the complex plane

Complex line integrals

Cauchy’s Theorem

Cauchy’s integral formula and applications

Goursat’s Theorem

Chapter 4: Taylor and Laurent Series

Infinite series

Integer power series

Taylor series

Laurent series

Chapter 5: Singular Points and the Residue Theorem

Singular points of analytic functions

The residue theorem

Chapter 6: Elementary Definite Integrals

Rational functions over (-Infinity,+Infinity)

Rational functions times sine or cosine

Rational functions times exponential functions

Rational functions times a power of x

Chapter 7: Intermediate Definite Integrals

Rational functions over (0, +Infinity)

Forms containing (ln x)^p in the numerator

Forms containing In g(x) or arctan g(x)

Forms containing In in the denominator

Forms containing P(n)(e^x)/Q(m)(e^x)

Poisson’s integral

Fresnel integrals

Chapter 8: Advanced Definite Integrals

Rational functions times trigonometric functions

Forms containing (x^2 – 2a sinx + a^2)-1

Forms containing (h sin a.x + x cos a.x)^-1

Forms containing Bessel functions

Chapter 9: Further Applications of the Theory of Residues

Counting zeros and poles of meromorphic functions

The argument principle

Rouche’s Theorem

Simple-pole expansion of meromorphic functions

Infinite product expansion of entire functions

Chapter 10: Series Summation by Residues

Type of series considered

Summation of S1

Summation of S2

Summation of S3 and S4

Series with neither even nor odd terms

Series involving real zeros of entire functions

Series involving complex zeros of entire functions

Solutions to the book Introduction to Analytic Number Theory by Tom M. Apostol

Chapter 1: The Fundamental Theorem of Arithmetic

Chapter 2: Arithmetical Functions and Dirichlet Multiplication

Chapter 3: Averages of Arithmetical Functions

Chapter 4: Some Elementary Theorems on the Distribution of Prime Numbers

Chapter 5: Congruences

Chapter 6: Finite Abelian Groups and Their Characters

Chapter 7: Dirichlet’s Theorem on Primes in Arithmetic Progressions

Chapter 8: Periodic Arithmetical Functions and Gauss Sums

Chapter 9: Quadratic Residues and the Quadratic Reciprocity Law

Chapter 10: Primitive Roots

Chapter 11: Dirichlet Series and Euler Products

Chapter 12: The Functions $\zeta(s)$ and $L(s,\chi)$

Chapter 13: Analytic Proof of the Prime Number Theorem

Chapter 14: Partitions

One Finite Product

In this post, I wish to find the value of some finite product. I begin by defining a function F_n(z) as follows and trying to find it’s roots

\displaystyle F_n(z)=\frac1{2i}\left((1+\frac{iz}n)^n-(1-\frac{iz}n)^n\right)= \displaystyle\frac{(-1)^m}{n^n} \displaystyle z^n+\cdots+ \displaystyle1 \displaystyle z

Let z=n\tan(\theta), then we have

\displaystyle{1+\frac{iz}n=1+i\tan\theta=1+i\frac{\sin\theta}{\cos\theta}=\frac{1}{\cos\theta}(\cos\theta+i\sin\theta)=\sec\theta e^{i\theta}}

and similarly 1-\frac{iz}n=\sec\theta e^{-i\theta}. So we have

\displaystyle F_n(n\tan\theta)=\frac1{2i}\sec^n\theta(e^{in\theta}-e^{-in\theta})=\sec^n\theta\sin n\theta

The sine function vanishes at integer multiples of \pi, so it follows that F_n(n\tan\theta)=0 where n\theta=k\pi for all integers k, that is, for \theta=k\pi/n for all k\in\mathbb Z. Thus, F_n(z_k)=0 for

\displaystyle z_k=n\tan\frac{k\pi}n=(2m+1)\tan\frac{k\pi}{2m+1}

where we recall that n=2m+1. Since \tan\theta is strictly increasing on the interval (-\pi/2,\pi/2), it follows that

\displaystyle{z_{-m}<z_{-m+1}<...<z_{-1}<z_0<z_1<...<z_{m-1}<z_m}

moreover, since tangent is an odd function, we have z_{-k}=-z_k for each k. In particular we have found 2m+1=n distinct roots of F_n(z), so as a consequence of the fundamental theorem of algebra, we can write

\displaystyle F_n(z)= \displaystyle\frac{(-1)^m}{n^n} \displaystyle{z\prod_{k=1}^m(z^2-z_k^2)=\frac{(-1)^m}{n^n}z\prod_{k=1}^m(z^2-n^2\tan^2\frac{k\pi}n)}

                                        \displaystyle=\frac{(-1)^m}{n^n}\prod_{k=1}^m(-n^2\tan^2\frac{k\pi}n)z\prod_{k=1}^m\left(1-\frac{z^2}{n^2\tan^2\frac{k\pi}{n}}\right)

                                        \displaystyle= \displaystyle\frac1 n\prod_{k=1}^m\tan^2\frac{k\pi}{n} \displaystyle z\prod_{k=1}^m\left(1-\frac{z^2}{n^2\tan^2\frac{k\pi}{n}}\right)

So we have \frac1{2m+1}\prod_{k=1}^m\tan^2\frac{k\pi}{2m+1}=1, and

\boxed{\displaystyle\prod_{k=1}^m\tan\frac{k\pi}{2m+1}=\sqrt{2m+1}}

More Elementary Proof for Euler’s Sine Expansion

Our goal in this post is to provide an elementary proof for the real version of Euler’s famous Sine expansion.

I first prove that for any n that is a power of 2 and for any x\in\mathbb R,

\displaystyle{\sin(x)=n p_n(x)\sin\left(\frac x n\right)\cos\left(\frac x n\right),~p_n(x)=\prod_{k=1}^{\frac n 2-1}\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)}~~~~~(1)

by using the following more familiar identity

\displaystyle\sin(x)=2\sin\frac x 2\sin\frac{\pi+x}2

we can easily arrive to the following identity that is valid for n equal to any power of 2,

                    \displaystyle\sin(x)=2^{n-1}\prod_{k=0}^{n-1}\sin\frac{k\pi+x}n

\displaystyle=2^{n-1}\sin\frac x n\sin\frac{\frac n 2\pi+x}n\prod_{k=1}^{\frac n 2-1}\sin\frac{k\pi+x}n\sin\frac{k\pi-x}n

  \displaystyle=2^{n-1}\sin\frac x n\cos\frac x n\prod_{k=1}^{\frac n 2-1}\left(\sin^2\frac{k\pi}n-\sin^2\frac x n\right)~~~~~~~~~~(2)

By considering what happens as x\to0 in the recent formula, we can obtain that for n a power of 2

\displaystyle n=2^{n-1}\prod_{k=1}^{\frac n 2-1}\sin^2\frac{k\pi}n~~~~~~~~~~(3)

and then I replace (3) in (2) and prove (1).

Now I choose m<\frac n 2-1 and break up p_n(x) as follows

\displaystyle p_n(x)=\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)\prod_{k=m+1}^{\frac n 2-1}\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)~~~~~~~~~~(4)

Since for 0<\theta<\frac{\pi}2, \sin(\theta)>\frac2{\pi}\theta, we have

\displaystyle\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}<\frac{x^2}{4k^2}

and if I choose m and n large enough such that \frac{x^2}{4m^2}<1, then

\displaystyle0<1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}<1,k=m+1,...,\frac n 2-1

which implies that

\displaystyle0<\prod_{k=m+1}^{\frac n 2-1}\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)<1

Therefore from (4) we have

\displaystyle p_n(x)\le\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)

Also we have

\displaystyle{\prod_{k=m+1}^{\frac n 2-1}\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)\ge1-\sum_{k=m+1}^{\frac n 2-1}\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\ge1-\frac{x^2}4\sum_{k=m+1}^\infty\frac1{k^2}=1-s_m}

and once again by (4), we get

\displaystyle p_n(x)\ge(1-s_m)\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)

To summarize, I have shown that

\displaystyle{(1-s_m)\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)\le p_n(x)\le\prod_{k=1}^m\left(1-\frac{\sin^2\frac x n}{\sin^2\frac{k\pi}n}\right)}~~~~~~~~~~(5)

Now taking n\to\infty in (5) I arrive to

\displaystyle(1-s_m)\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)\le\frac{\sin(x)}x\le\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)

After rearrangement and then taking absolute values, we get

\displaystyle\left|\frac{\sin(x)}x-\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)\right|\le|s_m|\left|\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)\right|

but since we have

\displaystyle{\left|\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)\right|\le\prod_{k=1}^m\left(1+\frac{x^2}{k^2\pi^2}\right)\le\prod_{k=1}^me^{\frac{x^2}{k^2\pi^2}}\le e^{\sum_{k=1}^\infty\frac{x^2}{k^2\pi^2}}=e^{\ell}}

I can write

\displaystyle\left|\frac{\sin(x)}x-\prod_{k=1}^m\left(1-\frac{x^2}{k^2\pi^2}\right)\right|\le|s_m|e^{\ell}

Finally since \lim_{m\to\infty}s_m=0, this inequality implies our goal, indeed

\boxed{\displaystyle\prod_{k=1}^{\infty}\left(1-\frac{x^2}{k^2\pi^2}\right)=\frac{\sin(x)}x}

One Finite Sum

In this post I want to find the value of the following finite sum,

\displaystyle\sum_{k=1}^{m-1}\frac{1}{\sin^2(\frac{k\pi}{m})}

I use the following amazing identity that you can see it’s proof in my previous posts

\displaystyle\frac{1}{\sin^2(x)}=\sum_{n\in\mathbb{Z}}\frac{1}{(x+n\pi)^2}

By this identity, I can write

                    \displaystyle\sum_{k=0}^{m-1}\frac{1}{\sin^2(\frac{x+k\pi}{m})}=\sum_{k=0}^{m-1}\sum_{n\in\mathbb{Z}}\frac{1}{(\frac{x+k\pi}{m}+n\pi)^2}

                                                           \displaystyle=\sum_{k=0}^{m-1}\sum_{n\in\mathbb{Z}}\frac{1}{\frac{(x+k\pi+mn\pi)^2}{m^2}}

                                                          \displaystyle=m^2\sum_{k=0}^{m-1}\sum_{n\in\mathbb{Z}}\frac{1}{(x+k\pi+mn\pi)^2}

                                                          \displaystyle=m^2\sum_{n\in\mathbb{Z}}\sum_{k=0}^{m-1}\frac{1}{(x+(k+mn)\pi)^2}=\frac{m^2}{\sin^2(x)}

and this follows that

\displaystyle\sum_{k=1}^{m-1}\frac{1}{\sin^2(\frac{x+k\pi}{m})}=\frac{m^2}{\sin^2(x)}-\frac{1}{\sin^2(\frac{x}{m})}

Hence,

\boxed{\displaystyle\sum_{k=1}^{m-1}\frac{1}{\sin^2(\frac{k\pi}{m})}=\lim_{x\to0}\frac{m^2}{\sin^2(x)}-\frac{1}{\sin^2(\frac{x}{m})}=\frac{m^2-1}{3}}

Euler Sum

Sums of the form \displaystyle E(p,q)=\sum_{n=1}^\infty\frac{H_n^{(p)}}{n^q} where \displaystyle H_n^{(p)}=\sum_{m=1}^n\frac{1}{m^p}, sometimes are called Euler sums. There are several ways to evaluate these  sums. Aabout 240 years ago, Leonhard Euler(1707-1783) in 1775 proved that for q\geq2,

\boxed{\displaystyle{\sum_{n=1}^\infty\frac{H_n}{n^q}=\frac{q+2}{2}\zeta(q+1)-\frac{1}{2}\sum_{r=1}^{q-2}\zeta(q-r)\zeta(r+1)}}

In the following you can see an elementary proof of this formula in the three steps:

Step 1:

\displaystyle\sum_{n=1}^\infty\frac{H_n}{n^q}=\sum_{n=1}^\infty\sum_{m=1}^n\frac{1}{mn^q}

                    \displaystyle=\sum_{m=1}^\infty\sum_{n=m}^\infty\frac{1}{mn^q}

                    \displaystyle=\zeta(q+1)+\sum_{m=1}^\infty\sum_{n=m+1}^\infty\frac{1}{mn^q}

                    \displaystyle=\zeta(q+1)+\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{1}{m(n+m)^q}

                    \displaystyle=\zeta(q+1)+\frac{1}{2}\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{1}{m(n+m)^q}+\frac{1}{n(n+m)^q}

Hence,

\boxed{\displaystyle{\sum_{n=1}^\infty\frac{H_n}{n^q}=\zeta(q+1)+\frac{1}{2}\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{1}{m(n+m)^q}+\frac{1}{n(n+m)^q}}}

Step 2:

\displaystyle H_n=\sum_{m=1}^n\frac{1}{m}=\sum_{i=0}^{n-1}\sum_{m=1}^\infty\frac{1}{m+i}-\frac{1}{m+i+1}

                              \displaystyle=\sum_{m=1}^\infty\sum_{i=0}^{n-1}\frac{1}{m+i}-\frac{1}{m+i+1}

                              \displaystyle=\sum_{m=1}^\infty\frac{1}{m}-\frac{1}{n+m}

Now by recent formula that is also an alternate definition of the Harmonic numbers, we can write:

\boxed{\displaystyle\sum_{n=1}^\infty\frac{H_n}{n^q}=\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{1}{mn^q}-\frac{1}{(n+m)n^q}}

Step 3:

\displaystyle\sum_{n=1}^\infty\frac{H_n}{n^q} \displaystyle=2\sum_{n=1}^\infty\frac{H_n}{n^q} \displaystyle-\sum_{n=1}^\infty\frac{H_n}{n^q}

\displaystyle{=2\zeta(q+1)+\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{1}{m(n+m)^q}+\frac{1}{n(n+m)^q}-\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{1}{mn^q}-\frac{1}{(n+m)n^q}}

\displaystyle=2\zeta(q+1)+\sum_{n=1}^\infty\sum_{m=1}^\infty\frac{1}{m(n+m)^{q-1}n}-\frac{1}{(n+m)n^{q-1}m}

\displaystyle=2\zeta(q+1)+\sum_{n=1}^\infty\sum_{m=n+1}^\infty\frac{1}{nm^{q-1}(m-n)}-\frac{1}{mn^{q-1}(m-n)}

\displaystyle=2\zeta(q+1)-\frac{1}{2}\sum_{\substack{m,n=1\\m\ne n}}^\infty\frac{1}{mn^{q-1}(m-n)}-\frac{1}{nm^{q-1}(m-n)}

\displaystyle=2\zeta(q+1)-\frac{1}{2}\sum_{\substack{m,n=1\\m\ne n}}^\infty\frac{m^{q-2}-n^{q-2}}{m^{q-1}n^{q-1}(m-n)}

\displaystyle=2\zeta(q+1)-\frac{1}{2}\sum_{\substack{m,n=1\\m\ne n}}^\infty\sum_{r=1}^{q-2}\frac{1}{m^{q-r}n^{r+1}}

\displaystyle=2\zeta(q+1)-\frac{1}{2}\left(\sum_{m,n=1}^\infty\sum_{r=1}^{q-2}\frac{1}{m^{q-r}n^{r+1}}-\sum_{\substack{m,n=1\\m=n}}^\infty\sum_{r=1}^{q-2}\frac{1}{m^{q-r}n^{r+1}}\right)

\displaystyle=2\zeta(q+1)+\frac{q-2}{2}\zeta(q+1)-\frac{1}{2}\sum_{r=1}^{q-2}\sum_{m=1}^\infty\sum_{n=1}^\infty\frac{1}{m^{q-r}n^{r+1}}

\displaystyle{=\frac{q+2}{2}\zeta(q+1)-\frac{1}{2}\sum_{r=1}^{q-2}\zeta(q-r)\zeta(r+1)}.

Another Result By the Flajolet-Vardi Theorem

By using the Flajolet-Vardi theorem we can find the value of the another amazing convergent series. Indeed, following series

\displaystyle\sum_{n=2}^{\infty}\frac{\zeta(n)}{k^n}

I first recall the Flajolet-Vardi theorem that you can find it’s proof in my second post:

Flajolet-Vardi Theorem:

If \displaystyle f\left(z \right)=\sum_{n=2}^{\infty}a_{n}z^n and \displaystyle\sum_{n=2}^{\infty}|a_n| converges then‎,

\displaystyle\sum_{n=1}^{\infty}f\left(\frac{1}{n}\right)=\sum_{n=2}^{\infty}a_n\zeta\left(n\right).

This theorem shows that \displaystyle\sum_{n=2}^{\infty}\frac{\zeta(n)}{k^n}=\sum_{n=1}^{\infty}\frac{1}{kn(kn-1)}, because if we let f(z)=\displaystyle\sum_{n=2}^{\infty}\frac{z^n}{k^n}, then f(z)=\frac{z^2}{k(k-z)} and by this theorem

\displaystyle{\sum_{n=2}^\infty\frac{\zeta(n)}{k^n}=\sum_{n=1}^\infty f\left(\frac{1}{n}\right)=\sum_{n=1}^\infty\frac{1}{kn(kn-1)}}

Therefore now we must find the value of \displaystyle\sum_{n=1}^\infty\frac{1}{kn(kn-1)}. we use the Taylor expansion of \log(1-x) and the fact that the sum \displaystyle\sum_{\alpha^k=1}\alpha^n is k if k divides n and 0 otherwise.

There are some \log‘s of complex numbers. Those numbers have always non-negative real part, for the Argument we take the angle between \displaystyle\frac{-\pi}{2} and \displaystyle\frac{\pi}{2}, so that it fits with the power series for \log(1-x).

\displaystyle{\sum_{n=1}^\infty \frac{x^{kn}}{kn}=\frac{-\log(1-x^k)}{k}=\frac{-1}{k}\sum_{\alpha^k=1}\log(1-\alpha x)}

\displaystyle\sum_{\alpha^k=1}\sum_{m=1}^\infty\frac{\alpha(\alpha x)^m}{m}=k\sum_{n=1}^{\infty}\frac{x^{kn-1}}{(kn-1)}

but also

\displaystyle{\sum_{\alpha^k=1}\sum_{m=1}^\infty\frac{\alpha(\alpha x)^m}{m}=-\sum_{\alpha^k=1}\alpha\log(1-\alpha x)}.

We thus have

\displaystyle{\sum_{n=1}^\infty\frac{x^{kn}}{kn(kn-1)}=\sum_n x^{kn}\left(\frac{1}{kn-1}-\frac{1}{kn}\right)}

\displaystyle{=\frac{1}{k}\sum_{\alpha^k=1}(1-x\alpha)\log(1-\alpha x)}.

We have to take the limit x\to 1. The \alpha=1 term disappears, so we get

\displaystyle{\sum_{n=1}^\infty\frac{1}{kn(kn-1)}=\frac{1}{k}\sum_{\alpha^k=1,\alpha\neq1}(1-\alpha)\log(1-\alpha),\alpha=e^{\frac{2\pi im}{k}}}

\displaystyle{=\frac{1}{k}\sum_{m=1}^{k-1}\left(1-\cos\frac{2\pi m}{k}-i\sin\frac{2\pi m}{k}\right)\left(\log\left(2\sin\frac{\pi m}{k}\right)+\pi i\left(\frac{m}{k}-\frac{1}{2}\right)\right)}

\displaystyle{=\frac{1}{k}\sum_{m=1}^{k-1}\left[\left(1-\cos\frac{2\pi m}{k}\right)\log\left(2\sin\frac{\pi m}{k}\right)+\frac{(2m-k)\pi}{2k}\sin\frac{2\pi m}{k}\right]}.

Examples:

For k=2 we have

\displaystyle{\sum_{n=2}^\infty\frac{\zeta(n)}{2^n}=\sum_{n=1}^\infty\frac{1}{2n(2n-1)}=\frac{1}{2}\left[(1-\cos\pi)\log\left(2\sin\frac{\pi}{2}\right)\right]=\log(2)}.

and also for k=3,

\displaystyle\sum_{n=2}^\infty\frac{\zeta(n)}{3^n}=\sum_{n=1}^\infty\frac{1}{3n(3n-1)}

\displaystyle=\frac{1}{3}\left[\left(1-\cos\frac{2\pi}{3}\right)\log\left(2\sin\frac{\pi}{3}\right)-\frac{\pi}{6}\sin\frac{2\pi}{3}+\left(1-\cos\frac{4\pi}{3}\right)\log\left(2\sin\frac{2\pi}{3}\right)+\frac{\pi}{6}\sin\frac{4\pi}{3}\right]

\displaystyle=\frac{1}{2}\log(3)-\frac{\pi}{6\sqrt{3}}.

A Beautiful Convergent Series II

In this post I want to generalize the given formula at my second post as follows

\boxed{\displaystyle\sum_{n=1}^\infty\frac{(m-1)^n-1}{m^n}\zeta(n+1)=\pi\cot\frac{\pi}{m}}

To prove this , I first prove following useful identity

\displaystyle\frac{1}{\sin^2x}=\sum_{n\in\mathbb{Z}}\frac{1}{(x+n\pi)^2}

PROOF: (Trigonometric Method) Note that

\displaystyle{\frac{1}{\sin^2x}=\frac{1}{4\sin^2\frac{x}{2}\cos^2\frac{x}{2}}=\frac{1}{4}\left(\frac{1}{\sin^2\frac{x}{2}}+\frac{1}{\cos^2\frac{x}{2}}\right)=\frac{1}{4}\left(\frac{1}{\sin^2\frac{x}{2}}+\frac{1}{\sin^2\frac{\pi+x}{2}}\right)}

\displaystyle=\frac{1}{4^2}\left(\frac{1}{\sin^2\frac{x}{2^2}}+\frac{1}{\sin^2\frac{2\pi+x}{2^2}}+\frac{1}{\sin^2\frac{\pi+x}{2^2}}+\frac{1}{\sin^2\frac{3\pi+x}{2^2}}\right)

Repeatedly applying \displaystyle\frac{1}{\sin^2x}=\frac{1}{4}\left(\frac{1}{\sin^2\frac{x}{2}}+\frac{1}{\sin^2\frac{\pi+x}{2}}\right), we arrive at the following formula:

\displaystyle\frac{1}{\sin^2x}=\frac{1}{4^k}\sum_{n=0}^{2^k-1}\frac{1}{\sin^2\frac{x+n\pi}{2^k}}

but

\displaystyle{\frac{1}{4^k}\sum_{n=0}^{2^k-1}\frac{1}{\sin^2\frac{x+n\pi}{2^k}}=\frac{1}{4^k}\sum_{n=-2^{k-1}}^{2^{k-1}-1}\frac{1}{\sin^2\frac{x+n\pi}{2^k}}=\lim_{k\to\infty}\sum_{n=-k}^k\frac{1}{(x+n\pi)^2}=\sum_{n\in\mathbb{Z}}\frac{1}{(x+n\pi)^2}}

and now note that

\displaystyle\sum_{n\in\mathbb{Z}}\frac{1}{x+n}=\pi\cot\pi x

Because,

\displaystyle{\cot x=-\int\frac{1}{\sin^2x}dx=-\int\sum_{n\in\mathbb{Z}}\frac{1}{(x+n\pi)^2}dx=\sum_{n\in\mathbb{Z}}\frac{1}{x+n\pi}}

By using this recent identity we can write

\displaystyle\sum_{n=1}^\infty\frac{(m-1)^n-1}{m^n}\zeta(n+1) \displaystyle=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{(m-1)^n-1}{m^n}\frac1{k^{n+1}}

                                                                         \displaystyle=\sum_{k=1}^\infty\frac1k\sum_{n=1}^\infty\left(\frac{(m-1)^n}{m^nk^n}-\frac1{m^nk^n}\right)

                                                                         \displaystyle=\sum_{k=1}^\infty\frac1k\left(\frac{\frac{m-1}{mk}}{1-\frac{m-1}{mk}}-\frac{\frac1{mk}}{1-\frac1{mk}}\right)

                                                                         \displaystyle=\sum_{k=1}^\infty\left(\frac1{\frac1m-k}+\frac1{\frac1m+k-1}\right)

                                                                         \displaystyle=\sum_{k\in\mathbb{Z}}\frac1{\frac1m+k} \displaystyle=\pi\cot\frac{\pi}{m}.

The Basel Problem, Double Integral Method II

One another way that can evaluate \zeta(2) by double integral is to write \zeta(2) as follows

\displaystyle\zeta(2)=\frac{4}{3}\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}

This is true, since we have

\displaystyle{\zeta(2)=\sum_{n=1}^{\infty}\frac{1}{n^2}=\sum_{n=1}^{\infty}\frac{1}{(2n)^2}+\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}=\frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{n^2}+\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}}.

Now integrating by parts yields

\displaystyle{\int_0^1(-x^{2n}\ln(x))dx=\left[-\frac{x^{2n+1}}{2n+1}\ln(x)\right]_0^1+\int_0^1\frac{x^{2n}}{2n+1}dx=\frac{1}{(2n+1)^2}}

and by using the monotone convergence theorem, I can write

\displaystyle{\zeta(2)=\frac{4}{3}\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}=\frac{4}{3}\sum_{n=0}^{\infty}\int_0^1(-x^{2n}\ln(x))dx=\frac{4}{3}\int_0^1\frac{\ln(x)}{x^2-1}dx}

Hence,

\displaystyle\zeta(2)=\frac{4}{3}\int_0^1\frac{\ln(x)}{x^2-1}dx

            \displaystyle=\frac{4}{3}\times\frac{1}{2}\left(\int_0^1\frac{\ln(u)}{u^2-1}du+\int_0^1\frac{\ln(v)}{v^2-1}dv\right)

            \displaystyle=\frac{2}{3}\left(\int_0^1\frac{\ln(u)}{u^2-1}du+\int_1^{\infty}\frac{\ln(v)}{v^2-1}dv\right)

            \displaystyle=\frac{2}{3}\left(\int_0^1\frac{\ln(u)}{u^2-1}du+\int_0^{\infty}\frac{\ln(u)}{u^2-1}du-\int_0^{1}\frac{\ln(u)}{u^2-1}du\right)

            \displaystyle=\frac{2}{3}\int_0^{\infty}\frac{\ln(u)}{u^2-1}du=\frac{2}{3}\times\frac{1}{2}\int_0^{\infty}\frac{\ln(u^2)}{u^2-1}du

            \displaystyle=\frac{1}{3}\int_0^{\infty}\frac{1}{1-u^2}\ln(\frac{1}{u^2})du=\frac{1}{3}\int_0^{\infty}\frac{1}{1-u^2}\left[\ln\left(\frac{1+v}{1+u^2v}\right)\right]_{v=0}^{v=\infty}du

            \displaystyle=\frac{1}{3}\int_0^{\infty}\frac{1}{1-u^2}\left(\int_0^{\infty}\left(\frac{1}{1+v}-\frac{u^2}{1+u^2v}\right)dv\right)du

            \displaystyle{=\frac{1}{3}\int_0^{\infty}\int_0^{\infty}\frac{1}{(1+v)(1+u^2v)}dv\,du=\frac{1}{3}\int_0^{\infty}\frac{1}{1+v}\int_0^{\infty}\frac{1}{1+u^2v}du\,dv}

            \displaystyle{=\frac{1}{3}\int_0^{\infty}\left(\frac{1}{1+v}\left[\frac{\arctan(\sqrt{v}u)}{\sqrt{v}}\right]_{u=0}^{u=\infty}\right)dv=\frac{1}{3}\times\frac{\pi}{2}\int_0^{\infty}\frac{1}{\sqrt{v}(1+v)}dv}

            \displaystyle{=\frac{\pi}{3}\int_0^{\infty}\frac{1}{1+w^2}dw=\frac{\pi}{3}\left[\arctan(w)\right]_0^{\infty}=\frac{\pi}{3}\times\frac{\pi}{2}=} \displaystyle{\frac{\pi^2}{6}}.

The Basel Problem, Double Integral Method I

Solutions to the book Introduction to Analytic Number Theory by Tom M. Apostol

Solutions to the book Complex Variables by M. Ya. Antimirov, A. A. Kolyshkin and Remi Vaillancourt

The Italian mathematician Pietro Mengoli (1625–1686), in his 1650 book Novae quadraturae arithmeticae, seu de additione fractionum, posed the following question: What’s the value of the sum

\zeta(2)=\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}

Here’s what Mengoli said:

Having concluded with satisfaction my consideration of those arrangements of fractions,
I shall move on to those other arrangements that have the unit as numerator, and square
numbers as denominators. The work devoted to this consideration has bore some fruit — the question itself still awaiting solution — but it [the work] requires the support of a richer
mind, in order to lead to the evaluation of the precise sum of the arrangement [of fractions]
that I have set myself as a task. 

First time, Leonhard Euler (1707-1783) in 1735 proved that above series converges to \displaystyle\frac{\pi^2}{6}.

In page 122 of the book “Topics in Number Theory”(1956), by William J. LeVeque there is an exercise for evaluating the following integral in two ways.

\displaystyle\int_0^1\!\!\!\int_0^1\frac1{1-xy}\,dy\,dx

First way is to write the integrand as a geometric series,

\displaystyle\int_0^1\!\!\!\int_0^1\frac1{1-xy}\,dy\,dx=\int_0^1\!\!\!\int_0^1\left(\sum_{n=1}^\infty(xy)^{n-1}\right)\,dy\,dx=\sum_{n=1}^\infty\frac1{n^2}

and the second way by use of a suitable change of variables (y:=u-v,x:=u+v) which is also published by Tom M. Apostol in 1983 in Mathematical Intelligencer.

apostol2013

Note that

\displaystyle{\frac{1}{n^2}}=\displaystyle\int_0^1\displaystyle\int_0^1x^{n-1}y^{n-1}dx\,dy

and by the monotone convergence theorem we get

\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^2}=\displaystyle\int_0^1\displaystyle\int_0^1\left(\sum_{n=1}^{\infty}(xy)^{n-1}\right)dx\,dy

=\displaystyle\int_0^1\displaystyle\int_0^1\frac{1}{1-xy}dx\,dy

We change variables in this by putting (u,v)=\left(\frac{x+y}{2},\frac{y-x}{2}\right), so that (x,y)=(u-v,u+v). Hence

\zeta(2) =2\displaystyle\iint_S\frac{du\,dv}{1-u^2+v^2}

where S is the square with vertices (0,0), \left(\frac{1}{2},\frac{-1}{2}\right), (1,0) and (\frac{1}{2},\frac{1}{2}).

Exploiting the symmetry of the square we get

\zeta(2) =4\displaystyle\int_0^{\frac{1}{2}}\displaystyle\int_0^u\frac{dv\,du}{1-u^2+v^2}+4\displaystyle\int_{\frac{1}{2}}^1\displaystyle\int_0^{1-u}\frac{dv\,du}{1-u^2+v^2}

               =4\displaystyle\int_0^{\frac{1}{2}}\frac{\tan^{-1}\left(\frac{u}{\sqrt{1-u^2}}\right)}{\sqrt{1-u^2}}du+4\displaystyle\int_{\frac{1}{2}}^1\frac{\tan^{-1}\left(\frac{1-u}{\sqrt{1-u^2}}\right)}{\sqrt{1-u^2}}du

Now \tan^{-1}\left(\frac{u}{\sqrt{1-u^2}}\right)=\sin^{-1}(u), and if \alpha=\tan^{-1}\left(\frac{1-u}{\sqrt{1-u^2}}\right) then \tan^2(\alpha)=\frac{1-u}{1+u} and \sec^2(\alpha)=\frac{2}{1+u}.

It follows that u=2\cos^2(\alpha)-1=\cos(2\alpha) and so \alpha=\frac{1}{2}\cos^{-1}(u)=\frac{\pi}{4}-\frac{\sin^{-1}(u)}{2}. Hence

\zeta(2) =4\displaystyle\int_0^{\frac{1}{2}}\frac{\sin^{-1}(u)}{\sqrt{1-u^2}}du+4\displaystyle\int_{\frac{1}{2}}^1\frac{\frac{\pi}{4}-\frac{\sin^{-1}(u)}{2}}{\sqrt{1-u^2}}du

                         \displaystyle=\left[2(\sin^{-1}(u))^2\right]_0^{\frac{1}{2}}+\left[\pi\sin^{-1}(u)-(\sin^{-1}(u))^2\right]_{\frac{1}{2}}^1

\displaystyle=\frac{\pi^2}{18}+\frac{\pi^2}{2}-\frac{\pi^2}{4}-\frac{\pi^2}{6}+\frac{\pi^2}{36}= \displaystyle\frac{\pi^2}{6}.

Flajolet-Vardi Theorem and Calculating Infinite Sums

In this post I want to find the value of the sum

\displaystyle\sum_{n=1}^{\infty}\frac{3^n-1}{4^n} \zeta \left(n+1 \right)

Note that for s>1, \zeta(s)=\displaystyle\sum_{n=1}^{\infty}\frac{1}{n^s}.

I use following formula that is called Gregory-Leibniz-Madhava’s series

\displaystyle\sum_{n=1}^{\infty}\frac{1}{4n-3}-\frac{1}{4n-1}=\frac{\pi}{4}

If we define f function as follows:

\displaystyle{f\left(z\right)=\frac{z}{4-3z}-\frac{z}{4-z}}

then

\displaystyle{f(z)=\frac{\frac{z}{4}}{1-\frac{3z}{4}}-\frac{\frac{z}{4}}{1-\frac{z}{4}}=\frac{z}{4}\sum_{n=0}^{\infty}\left(\frac{3z}{4}\right)^n-\left(\frac{z}{4}\right)^n=\frac{z}{4}\sum_{n=1}^{\infty}\left(\frac{3z}{4}\right)^n-\left(\frac{z}{4}\right)^n}

\displaystyle{=\frac{z}{4}\sum_{n=2}^{\infty}\left(\frac{3z}{4}\right)^{n-1}-\left(\frac{z}{4}\right)^{n-1}=\frac{1}{4}\sum_{n=2}^{\infty}\left[\left(\frac{3}{4}\right)^{n-1}-\left(\frac{1}{4}\right)^{n-1}\right]z^n=\frac{1}{4}\sum_{n=2}^{\infty}\frac{3^{n-1}-1}{4^{n-1}}z^n}.

Now I use this theorem

THEOREM (FlajoletVardi): If f\left(z \right)=\displaystyle\sum_{n=2}^{\infty}a_{n}z^n and \displaystyle\sum_{n=2}^{\infty}|a_n| converges then,

\displaystyle\sum_{n=1}^{\infty}f\left(\frac{1}{n}\right)=\displaystyle\sum_{n=2}^{\infty}a_n\zeta\left(n\right).

PROOF:

Because \displaystyle\sum_{m=2}^{\infty}|a_m|<\infty,

\displaystyle\sum_{n=1}^{\infty}\displaystyle\sum_{m=2}^{\infty}|a_m|\frac{1}{n^m}\leq\displaystyle\sum_{n=1}^{\infty}\displaystyle\sum_{m=2}^{\infty}|a_m|\frac{1}{n^2}<\infty

Hence, by Cauchy’s double series theorem, we can switch the order of summation:

\displaystyle{\sum_{n=1}^{\infty}f\left(\frac{1}{n}\right)=\sum_{n=1}^{\infty}\sum_{m=2}^{\infty}a_m\frac{1}{n^m}=\sum_{m=2}^{\infty}a_m\sum_{n=1}^{\infty}\frac{1}{n^m}=\sum_{n=2}^{\infty}a_n\zeta(n)}

This theorem implies that

\displaystyle{\frac{\pi}{4}=\sum_{n=1}^{\infty}f\left(\frac{1}{n}\right)=\frac{1}{4}\sum_{n=2}^{\infty}\frac{3^{n-1}-1}{4^{n-1}}\zeta(n)=\frac{1}{4}\sum_{n=1}^{\infty}\frac{3^{n}-1}{4^{n}}\zeta(n+1)}

and

\boxed{\displaystyle\sum_{n=1}^{\infty}\frac{3^n-1}{4^n} \zeta \left(n+1 \right)=\pi}