Experimental Mathematics

Introduction To L Functions (part 2)

2019-11-08T00:00:00+00:00

L_functions (Part II)

Introduction to L-Functions (Part II)

Introduction to L-Functions (Part II)

In the last post, we dived into the history and motivation and surrounding L-Functions in this post we continue with the exploration. We shall dedicate this post to make an in-depth study of the Riemann Zeta Function.
In the last post we ended up with the proof that $\zeta(s)$ converges for any $s \in \mathbb{R}$ and $s > 1$ . This is an extremely profound statement. This opens up the possibilities of asking questions about what values these sums evaluate to and how to determine that. For answering such questions we need some prerequisites.

Bernoulli Numbers

Bernoulli was well aware of the formulas for the partial sums of the powers of the integers, the first few being:
$\begin{aligned} \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \end{aligned}$

$\begin{aligned} \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \end{aligned}$

$\begin{aligned} \sum_{k=1}^{n} k^3 = \frac{n^2(n+1)^2}{4} \end{aligned}$

Given the parallel nature between increasing exponents in the summands and the exponents in the explicit formulas, it is natural to search for some generalisation. The result gave rise to an interesting sequence of numbers, called the Bernoulli Numbers; named by Euler in honour of his teacher. We have the general formula as we denote the $k^{th}$ Bernoulli Numbers as $B_k$ .

$\sum_{k=1}^{n-1} k^m = \frac{1}{m+1}\bigg[ \sum_{j=0}^{m}\binom{m+1}{j} n^{m+1-j}B_j \bigg] \tag{2.1}$

From here, we can produce the recursive relationship between the Bernoulli numbers by plugging $n = 1$ into equation $(2.1)$ , resulting in

$0 = \sum_{j=0}^m \binom{m+1}{j}B_j$

Using the recursive formula and $B_0 = 1$ we have a fairly straight forward, although not very speedy, method for computing the Bernoulli numbers. For reference, the first couple Bernoulli numbers are:

$B_{0}=1, \quad B_{1}=-\frac{1}{2}, \quad B_{2}=\frac{1}{6}, \quad B_{3}=0, \quad B_{4}=-\frac{1}{30}, \quad B_{5}=0$

We see that the third and fifth Bernoulli number is $0$ , and its also can be showed that $B_{2k+1} = 0$ for all $k \ge 1$ .

Fourier Series

The theory of Fourier Series is extremely important in mathematics as a fact I feel its crucial in our understanding nature as it pops up everywhere when we try to model something.
The whole idea behind Fourier series is to write a periodic function $f(x)$ as an infinite sum of sines and cosines:
$f(x)=\frac{a_{0}}{2}+\sum_{n=1}^{\infty}\left(a_{n} \cos (n x)+b_{n} \sin (n x)\right)$ where the coefficients of the sines and cosines are given by Euler’s formulas:
$\begin{aligned} a_{n} &=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos (n x) d x \tag{2.2}\\ b_{n} &=\frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin (n x) d x \end{aligned}$ Now, the function $f(x)$ can be represented as a Fourier series provided the following properties hold:

$f(x) = f(x+2 \pi)$
$f(x)$ is sectionally smooth.
At every point $x$ we have $f(x)=\frac{1}{2}\left(\lim _{a \rightarrow x^{+}} f(a)+\lim _{a \rightarrow x^{-}} f(a)\right)$

For our analysis of Riemann Zeta function we require a special product formula for the $\textbf{sinc}$ function. To start off our derivation we will find the Fourier series of $\cos(\mu x)$ on the interval $(−\pi, \pi)$ where $\mu$ is a real number that is not an integer. Note that all three properties are satisfied, so our quest for a Fourier series is feasible. Since $\cos(\mu x)$ is an even function we immediately know that our Fourier series will be a cosine series, thus $b_n = 0$ .

Now from $(2.2)$ we have
$\begin{aligned} a_{n} &=\frac{1}{\pi} \int_{-\pi}^{\pi} \cos (\mu x) \cos (n x) d x \\ &=\frac{1}{\pi} \int_{0}^{\pi} \cos ((\mu+n) x)+\cos ((\mu-n) x) d x \\ &=\frac{1}{\pi}\left[\frac{\sin ((\mu+n) \pi)}{\mu+n}+\frac{\sin ((\mu-n) \pi)}{\mu-n}\right] \\ &=\frac{1}{\pi}\left[\frac{(\mu-n) \sin ((\mu+n) \pi)+(\mu+n) \sin ((\mu-n) \pi)}{\mu^{2}-n^{2}}\right] \\ &=\frac{1}{\pi}\left[\frac{2 \mu \sin (\mu x) \cos (n x)}{\mu^{2}-n^{2}}\right]\\ &=\frac{2 \mu(-1)^{n}}{\pi\left(\mu^{2}-n^{2}\right)} \sin (\mu \pi) \end{aligned}$ We have used basic trigonometric identities and Now that we have found our coefficients, we have the Fourier series representation:
$\cos (\mu x)=\frac{\sin (\mu \pi)}{\mu \pi}+\frac{2 \mu \sin (\mu \pi)}{\pi} \sum_{n=1}^{\infty}(-1)^{n} \frac{\cos (n x)}{\mu^{2}-n^{2}}$
Since our function satisfied the property that the value at every point (specifically the endpoints $−\pi$ and $\pi$ ) was the average of the limits from both sides, we can plug $x=\pi$ into our Fourier expression without any trouble. After doing this and dividing by $\sin(\mu x)$ (which we can do since $\mu$ is not an integer) we get:
$\cot (\mu \pi)=\frac{1}{\mu \pi}+\frac{2 \mu}{\pi} \sum_{n=1}^{\infty} \frac{1}{\mu^{2}-n^{2}}$
Now rearranging the terms we have
$\cot (\mu \pi)-\frac{1}{\mu \pi}=\frac{2 \mu}{\pi} \sum_{n=1}^{\infty} \frac{1}{\mu^{2}-n^{2}}$
After multiplying by $\pi$ and integrating $\mu$ from $0$ to $x$ we have:
$\pi \int_{0}^{x}\left(\cot (\mu \pi)-\frac{1}{\mu \pi}\right) d \mu=\int_{0}^{x} \sum_{n=1}^{\infty} \frac{-2 \mu}{n^{2}-\mu^{2}} d \mu \tag{2.3}$
Rearranging the left side
$\begin{aligned} \pi \int_{0}^{x}\left(\cot (\mu \pi)-\frac{1}{\mu \pi}\right) d \mu &=\log \frac{\sin (\pi x)}{\pi x}-\lim _{a \rightarrow 0} \frac{\sin (\pi a)}{\pi a} \\ &=\log \frac{\sin (\pi x)}{\pi x} \end{aligned}$
On the right hand side of $(2.3)$ we can swap the sum and integral because we showed that equation $(2.3)$ was uniformly convergent. Therefore we end up with
$\begin{aligned} \int_{0}^{x} \sum_{n=1}^{\infty} \frac{-2 \mu}{n^{2}-\mu^{2}} d \mu &=\sum_{n=1}^{\infty} \int_{0}^{x} \frac{-2 \mu}{n^{2}-\mu^{2}} \\ &=\sum_{n=1}^{\infty} \log \left(1-\frac{x^{2}}{n^{2}}\right) \\ &=\lim _{\alpha \rightarrow \infty} \sum_{n=1}^{\alpha} \log \left(1-\frac{x^{2}}{n^{2}}\right) \\ &=\lim _{\alpha \rightarrow \infty} \log \prod_{n=1}^{\alpha}\left(1-\frac{x^{2}}{n^{2}}\right) \\ &=\log \lim _{\alpha \rightarrow \infty} \prod_{n=1}^{\alpha}\left(1-\frac{x^{2}}{n^{2}}\right) \\ &=\log \prod_{n=1}^{\infty}\left(1-\frac{x^{2}}{n^{2}}\right) \end{aligned}$
This gives us
$\log \frac{\sin (\pi x)}{\pi x}=\log \prod_{n=1}^{\infty}\left(1-\frac{x^{2}}{n^{2}}\right)$
After exponentiation both sides and multiplying the $\pi x$ over, we end up with the product formula for sine:
$\frac{\sin (\pi x)}{\pi x}= \prod_{n=1}^{\infty}\left(1-\frac{x^{2}}{n^{2}}\right) \tag{2.4}$

Evaluation of Zeta Function For Even Integers

Since we know that $\zeta(1)$ diverges the natural question is to ask for the value of $\zeta(2)$ . And there is amazing history related with zeta function evaluated at $2$ . In 1644 Pietro Mengoli proposed a problem to all mathematicians. Determine the value of the infinite sum
$\sum_{k=1}^{\infty}\frac{1}{k^2}$
This problem, which is known as the Basel problem, was eventually solved by Euler in 1735 after multiple renowned mathematicians had failed. Euler’s solution granted him instant fame in the mathematical community, but that was hardly a reason for Euler to stop there. Euler ingeniously derived and proved the general formula for which the Basel problem is a special case . The purpose of this section is to present this formula, which happens to be the value of Euler’s zeta series at the even integers.

We start with the product formula for $\sin(x)$ , which was produced in $(2.4)$
${\sin (x)}= x\prod_{k=1}^{\infty}\left(1-\frac{x^{2}}{\pi^2 k^{2}}\right)$
After taking the natural logarithm of both sides and substituting $x = −iu/2$ we have
$\log \left(\sin \left(-\frac{i u}{2}\right)\right)=\log \left(-\frac{i u}{2}\right)+\sum_{k=1}^{\infty} \log \left(1-\frac{\left(-\frac{i u}{2}\right)^{2}}{k^{2} \pi^{2}}\right)$ and after some simplification we get
$\log \left(\sin \left(-\frac{i u}{2}\right)\right)=\log \left(-\frac{i u}{2}\right)+\sum_{k=1}^{\infty} \log \left(1+\frac{u^{2}}{4 k^{2} \pi^{2}}\right)$
We can then differentiate both sides with respect to $u$ .
$\left(-\frac{i}{2}\right) \frac{\cos \left(-\frac{i u}{2}\right)}{\sin \left(-\frac{i u}{2}\right)}=\frac{1}{u}+\sum_{k=1}^{\infty} \frac{2 u}{4 k^{2} \pi^{2}+u^{2}} \tag{2.5}$ Focusing on the left hand side and using the following identities
$\cos (z)=\frac{1}{2}\left(e^{i z}+e^{-i z}\right) \quad \text { and } \quad \sin (z)=\frac{1}{2 i}\left(e^{i z}-e^{-i z}\right)$ gives

$\begin{aligned}\left(-\frac{i}{2}\right) \frac{\cos \left(-\frac{i u}{2}\right)}{\sin \left(-\frac{i u}{2}\right)} &=\left(-\frac{i}{2}\right) \frac{\left(\frac{e^{-u / 2}+e^{u / 2}}{2}\right)}{\left(\frac{-e^{-u / 2}+e^{a / 2}}{2 i}\right)} \\ &=\frac{1}{2} \frac{\left(e^{-u / 2}+e^{u / 2}\right)}{\left(-e^{-u / 2}+e^{u / 2}\right)} \\ &=\frac{1}{2} \frac{\left(e^{-u / 2}\right)}{\left(e^{-u / 2}\right)} \frac{\left(e^{u}-1+2\right)}{\left(-1+e^{u}\right)} \\ &=\frac{1}{2}+\frac{1}{e^{u}-1} \end{aligned}$
Then, after substituting this back into equation $(2.5)$ , multiplying by u and moving a term over we end up
$\frac{u}{e^{u}-1}+\frac{u}{2}-1=\sum_{k=1}^{\infty} \frac{2 u^{2}}{4 k^{2} \pi^{2}+u^{2}} \tag{2.6}$
Now for the homestretch, focusing on the left hand side we
$\begin{aligned} \frac{u}{e^{u}-1}+\frac{u}{2}-1 &=\sum_{k=0}^{\infty} \frac{B_{k} u^{k}}{k !}+\frac{u}{2}-1 \\ &=1-\frac{u}{2}+\sum_{k=2}^{\infty} \frac{B_{k} u^{k}}{k !}+\frac{u}{2}-1 \\ &=\sum_{k=2}^{\infty} \frac{B_{k} u^{k}}{k !} \\ &=\sum_{k=1}^{\infty} \frac{B_{2 k} u^{2 k}}{(2 k) !} \tag{2.7}\end{aligned}$
In the last step we were able to relabel the indices because all the odd terms were zero since $B_k = 0$ for odd $k > 1$ . The right hand side of equation $(2.6)$ becomes
$\begin{aligned} \sum_{k=1}^{\infty} \frac{2 u^{2}}{4 k^{2} \pi^{2}+u^{2}} &=\sum_{k=1}^{\infty} \frac{2 u^{2}}{(2 k \pi)^{2}}\left(\frac{1}{1+\left(\frac{u}{2 \pi}\right)^{2}}\right) \\ &=\sum_{k=1}^{\infty} \frac{2 u^{2}}{(2 k \pi)^{2}} \sum_{j=0}^{\infty}(-1)^{j}\left(\frac{u}{2 \pi k}\right)^{2 j} \\ &=\sum_{k=1}^{\infty} \sum_{k=1}^{\infty} 2(-1)^{j-1}\left(\frac{u}{2 \pi k}\right)^{2 j+2} \\ &=\sum_{j=1}^{\infty} \sum_{k=1}^{\infty} \frac{2(-1)^{j-1}}{(2 \pi)^{2 j}} \sum_{k=1}^{\infty}\left(\frac{1}{k^{2 j}}\right) u^{2 j} \\ &=\sum_{j=1}^{\infty} \frac{2(-1)^{j-1}}{(2 \pi)^{j-1}} \zeta(2 j) u^{2 j} \tag{2.8} \\ &=\sum_{j=1}^{\infty} \frac{2(-1)^{j-1}}{(2 \pi)^{2 j}} \zeta(2 j) u^{i j} \end{aligned}$
Finally, changing the index from j to k and substituting equations $(2.7)$ and $(2.8)$ back into equation $(2.6)$ yields:
$\sum_{k=1}^{\infty} \frac{B_{2 k}}{(2 k) !} u^{2 k}=\sum_{k=1}^{\infty} \frac{2(-1)^{k-1}}{(2 \pi)^{2 k}} \zeta(2 k) u^{2 k}$
Finally, we equate the corresponding coefficients in the sums and rearrange to arrive at our goal
$\boxed{ {\frac{(-1)^{k-1} B_{2 k}(2 \pi)^{2 k}}{2(2 k) !}=\zeta(2 k)}}$
This is a rather profound formula and it generators some of the most surprising and beautiful relations in all of mathematics. And hence for $k=1$ we have $\zeta(2) = \frac{B_2.(2 \pi)^2}{2(2!)} = \frac{\pi^2}{6}$
In its full glory , to appreciate the shear beauty of the result we see it in its fullest form :
$\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+ \cdots = \frac{\pi^2}{6}$

Introduction To L Functions ( Part 1)

2019-09-27T00:00:00+00:00

L_functions (Part I)

Introduction to L-Functions (Part I)
- Motivation
- Riemann Zeta Function

Introduction to L-Functions (Part I)

We dive into the history and motivation and surrounding L-Functions in this post. We begin with exploring the prime numbers.

Motivation

Mathematics evolved to understand the structure and properties of numbers. Of-course it has come a long way from just that. But basically number theory tries to understand the structure of whole numbers ( ie. $n \in \mathbb{Z}$ ) .The building blocks of these whole numbers are prime numbers. To recall these numbers are numbers which are only divisible by $1$ and themselves. The prime numbers can be denoted by the set $\mathbb{P}$ and it is not hard to verify that $\mathbb{P} \in \{2,3,5,7,11,\cdots\}$ . Now the question which Euclid tackled was whether this set is infinite.

Lets say that there finitely many prime numbers $p_1,p_2,p_3,\cdots,p_n$ . From this we can define another number $S=1 +\prod_{i=1}^{n}p_i$ Now we see that $p_1$ does not divide $S$ , nor does $p_2$ , nor $p_3$ or any $p_i$ because of the extra added factor of $1$ . Hence this means that $S$ is a prime itself or there exists some other prime $p'$ which divides $S$ . Hence given a prime list we were able to find a prime which was out of that list. So if we do this indefinitely, we see that we can have infinitely many primes.

This was one of the most important realization made about prime numbers since antiquity. But this theorem lead to some really interesting questions like.

Are there infinitely many primes in the form of $5n+3$ , $18n+9$ or any such arithmetic progressions?
Or even harder questions like whether there are infinite primes in the form $n^2+1$ or $2^p - 1$ .
Are there infinite primes which occur in certain groups, like the existence of infinite pairs $(p,p+2)$ or $(p,p+4)$ or $(p,2p-1)$ , these types of prime pairs are called Twin Primes, Cousin Primes, Sophie Germain Primes respectively.

These questions are naturally occurring but the proof technique due to Euclid gives us totally no clue on even approaching these problems. Hence it was realized that to tackle these problems newer and deeper techniques needed to be proposed. This inspired Euler and other mathematicians who came centuries later.

Riemann Zeta Function

One of the most important tool which became central to developing these new techniques became the Riemann Zeta Function, its seems unrelated and very simple to state. But has far reaching consequences on the theory of numbers.
The most basic version of the zeta function was studied by Euler. For $s \in \mathbb{R}$ , define
$\zeta(s) = \sum_{n=1}^{\infty}\frac{1}{n^s} \tag{1.1}$
This function $\zeta(s)$ is the zeta function. Now Euler was able to find the connection of this product with the prime numbers as
$\zeta(s) = \sum_{n=1}^{\infty}\frac{1}{n^s} = \prod_{p \in \mathbb{P}} \left( \frac{1}{1-{p^{-s}}} \right)$
The products of these types are called Euler products. Lets look at the proof of the Euler Product.

Proof of Euler Product Formula

$\zeta(s)=1+\frac{1}{2^{s}}+\frac{1}{3^{s}}+\frac{1}{4^{s}}+\frac{1}{5^{s}}+\ldots$
Now we can multiply both sides by $\frac{1}{2^s}$ and get
${\frac{1}{2^{s}} \zeta(s)=\frac{1}{2^{s}}+\frac{1}{4^{s}}+\frac{1}{6^{s}}+\frac{1}{8^{s}}+\frac{1}{10^{s}}+\ldots}$
Subtracting the second equation from the first we remove all elements that have a factor of 2:
$\left(1-\frac{1}{2^{s}}\right) \zeta(s)=1+\frac{1}{3^{s}}+\frac{1}{5^{s}}+\frac{1}{7^{s}}+\frac{1}{9^{s}}+\frac{1}{11^{s}}+\frac{1}{13^{s}}+\ldots$
Repeating for the next term:
$\frac{1}{3^{s}}\left(1-\frac{1}{2^{s}}\right) \zeta(s)=\frac{1}{3^{s}}+\frac{1}{9^{s}}+\frac{1}{15^{s}}+\frac{1}{21^{s}}+\frac{1}{27^{s}}+\frac{1}{33^{s}}+\ldots$
Subtracting again we get:
$\left(1-\frac{1}{3^{s}}\right)\left(1-\frac{1}{2^{s}}\right) \zeta(s)=1+\frac{1}{5^{s}}+\frac{1}{7^{s}}+\frac{1}{11^{s}}+\frac{1}{13^{s}}+\frac{1}{17^{s}}+\dots$
where all elements having a factor of 3 or 2 (or both) are removed. We see that we are removing the numbers having prime factors $2$ and $3$ and we can do similar multiplication with $\frac{1}{p^s}$ and subtraction for all the primes $p$ and hence nothing will remain in the R.H.S. Hence we see that:
$\ldots\left(1-\frac{1}{11^{s}}\right)\left(1-\frac{1}{7^{s}}\right)\left(1-\frac{1}{5^{s}}\right)\left(1-\frac{1}{3^{s}}\right)\left(1-\frac{1}{2^{s}}\right) \zeta(s)=1$
Which means
$\zeta(s)=\frac{1}{\left(1-\frac{1}{2^{s}}\right)\left(1-\frac{1}{3^{s}}\right)\left(1-\frac{1}{5^{s}}\right)\left(1-\frac{1}{7^{s}}\right)\left(1-\frac{1}{11^{s}}\right)\cdots}$
Which can be written in compact form as
$\zeta(s) = \prod_{p \in \mathbb{P}} \left( \frac{1}{1-{p^{-s}}} \right) \tag{1.2}$

This is an extremely important equality and would be useful in proving many important results about prime numbers and in understanding L Functions in general.

Understanding the Function at $s=1$ .

The simplest instance of zeta function can be when we look at $s=1$ , we see
$\zeta(1) = 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots$
We make the following key observation. This sereis has a special name its called The Harmonic Series.
$= {1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}+\cdots}$
${ \geq 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{16}+\cdots}$
Each term of the harmonic series is greater than or equal to the corresponding term of the second series, and therefore the sum of the harmonic series must be greater than or equal to the sum of the second series. However, the sum of the second series is infinite:
$=1+\left(\frac{1}{2}\right)+\left(\frac{1}{4}+\frac{1}{4}\right)+\left(\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}\right)+\left(\frac{1}{16}+\cdots+\frac{1}{16}\right)+\cdots$
When we evaluate each bracket we get
$1 + \frac{1}{2}+ \frac{1}{2}+ \frac{1}{2}+ \frac{1}{2}+\cdots = \infty$ .
Which means $\zeta(s) = \infty$ .

This statement implies a very important fact which follows from $(1.2)$ .
$\lim_{s \to 1 } \zeta(s) =\lim_{s \to 1 } \prod_{p \in \mathbb{P}} \left( \frac{1}{1-{p^{-s}}} \right) = \infty \tag{1.3}$ .

$\prod_{p \in \mathbb{P}} \left( \frac{p}{p-1} \right) \to \infty$
This implies that the product of primes grows unboundely as $s \to 1$ which imples that there are infinite primes. This proof is much more complex than that due to Euclid but as we shall see ,this proof is far more generalizable and lead to many important insights.

Exploring Other values of $s$

Before we explore other values of $s$ we need to first recall Integral test for convergence which is the following statement:

Consider an $N$ and a non-negative function $f$ defined on the unbounded $[N, ∞)$ , on which it is monotone decreasing. Then the infinite series : $\sum_{n=N}^{\infty}f(n)$ converges to a real number if and only if the improper integral $\int_{n=N}^{\infty}f(n) dn$ is finite. If the integral diverges, then the series diverges as well.

Hence for any $s \in \mathbb{R}$ and $s > 1$ , we consider $f(n) = \frac{1}{n^s}$ and hence
$\int_{1}^{\infty}f(n) dn = \int_{1}^{\infty}\frac{1}{n^s}dn = \left(\frac{n^{s+1}}{s+1}\right)\Big|_{1}^{\infty} = 1$
So we see that $\zeta(s)$ converges for any $s \in \mathbb{R}$ and $s > 1$ .