Euler - Riemann Zeta Function.

“Madam, I have just come from a country where people are hanged if
they talk.” ― Leonhard Euler

The Riemann Zeta function, denoted as $\zeta{(s)}$ is a function of a complex variable $s$ that analytically continues the sum of the Dirichlet Series, [1]
$$\LARGE \zeta{(s)}= \sum_{n=1}^{\infty}\frac{1}{n^{s}}$$
The entire blog will be divided into the following parts:

1. What is Analytic Continuation?
2. What is Dirichlet Series?
3. Recurrence relation between Bernoulli Numbers.
4. Relationship between Bernoulli, Riemann and Euler.

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1.

Before we talk about Analytic continuation, we will have to know what an Analytic function is.

There are multiple ways of defining the Analytic functions[2]. The one that I find most comfortable is given below:
$\underline{\text{Definition 1}}$: A function $f(z)$ is said to be analytic in a region $\mathcal{R}$ of the complex plane if $f(z)$ has a derivative at each point of $\mathcal{R}$ and if $f(z)$ is single valued.

For better understanding, I will extend an example,
$\underline{\text{Example 1}}$: Consider $f(z)$ = $2xy + i(x^{2}-y^{2})$, determine if the given function is analytic or not.
$\underline{\text{Solution}}$: We will use the Cauchy-Riemann equations.

For a given, $f(z) = f(x+iy) = u(x,y) + i v(x,y)$
Now we have the following definitions,
$$u_{x} = \frac{\partial{u}}{\partial{x}}, \\ u_{y} = \frac{\partial{u}}{\partial{y}}, \\ v_{x} = \frac{\partial{v}}{\partial{x}}, \\ v_{y} = \frac{\partial{v}}{\partial{y}}.$$
Now, according to the Cauchy-Riemann Equation, any $f(z)$ is an analytic function, iff,
$$u_{x}=v_{y}, \\ u_{y} = -v_{x}.$$

Now, for the given $f(z)$,
$$u(x,y) = 2xy, \\ v(x,y) = x^{2}-y^{2}.$$
Hence, the relations are,
$$u_{x} = \frac{\partial{(2xy)}}{\partial{x}} = 2y, \\ u_{y} = \frac{\partial{(2xy)}}{\partial{y}}=2x, \\ v_{x} = \frac{\partial{(x^{2}-y^{2})}}{\partial{x}}=2x, \\ v_{y} = \frac{\partial{(x^{2}-y^{2})}}{\partial{y}}=-2y.$$
It is clear that,
$$u_{x} \neq v_{y}, \\ u_{y} \neq -v_{x}.$$
Therefore, the given $f(z)$ is not an analytic function.

Analytic continuation is a pretty simple concept to understand really.
$\underline{\text{Definition 2}}$: Say $f(z)$ is an analytic function defined over a non-empty open subset $W$ of the complex plane $\mathcal{C}$. If $V$ is a larger open subset of $\mathcal{C}$, containing $W$, and $\mathcal{F}$ is an analytic function defined on $V$ such that,
$$\mathcal{F}(z) = f(z), \forall z \in V$$
In other words, Analytic continuation is the method of extending the domain of an analytic function. [3]

For some cool examples on Analytic continuation refer Virginia Tech’s paper here [4].

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2.

Dirichlet Series[5] is any series of general form,
$$\sum_{n=1}^{\infty}\frac{a_n}{n^{s}}$$
Now, with a slight modification,
$$a_n = 1 , \forall n \in \mathcal{R}$$
We get,
$$\sum_{n=1}^{\infty}\frac{1}{n^{s}}$$
For a function defined as,
$$\zeta{(s)} = \sum_{n=1}^{\infty}\frac{1}{n^{s}}$$
is the Riemann Zeta function.

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3.

Definitions,[6]
$$s_{0}(n)=n,\\ s_{1}(n) = \frac{n(n-1)}{2},\\ s_{2}(n) = \frac{n(n-1)(2n-1)}{6}...$$
Therefore, $s_{p}(n)$ is given as,
$$s_{p}(n) = \frac{1}{p+1}n^{p+1}-\frac{1}{2}n^{p}+\frac{p}{12}n^{(p-1)}+0 \times n^{(p-2)}+...$$
More generally,
$$s_{p}(n)=\sum_{k=0}^{p} \frac{B_{k}}{k!} \frac{p!}{(p+1-k)!}n^{(p+1-k)}$$
for $n=1$,
$$0=\sum_{k=0}^{p} \frac{B_{k}}{k!} \frac{p!}{(p+1-k)!}$$
or equivalently,
$$B_p=-\frac{1}{p+1}\sum_{k=0}^{p-1} \binom{p+1}{k}$$
The above relation is symbolically written as,
$$(B+1)^{(p+1)}-B^{(p+1)}=0$$
On expansion, all $k$-th powers of $B$, $B^{k}$ must be written as $B_k$ and treated as Bernoulli Numbers.
The expansion is done on the basis of Binomial Theorem, which statest that,
$$\forall a,b \in \mathcal{Z} \ \text{and} \ n \in \mathcal{N}\\ \large (a+b)^{n}= \sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^{k}$$
Therefore, for $n=5$,
$$(a+b)^{5} = \sum_{k=0}^{5}\binom{5}{k}a^{5-k}b^{k}\\ \implies(a+b)^{5} = \binom{5}{0}a^{5-0}b^{0}+\binom{5}{1}a^{5-1}b^{1}+\binom{5}{2}a^{5-2}b^{2}+\\ \binom{5}{3}a^{5-3}b^{3}+\binom{5}{4}a^{5-4}b^{4}++\binom{5}{5}a^{5-5}b^{5}\\ \implies(a+b)^{5} = 1 \times a^{5}b^{0}+5\times a^{5-1}b^{1}+10\times a^{5-2}b^{2}+\\ 10\times a^{5-3}b^{3}+5\times a^{5-4}b^{4}+1\times a^{5-5}b^{5}\\ $$
Therefore, for $(B+1)^{(p+1)}-B^{(p+1)}=0$, after expanding for $p=4$ we have,
$$(B+1)^{(4+1)}-B^{(4+1)}=0 \\ \implies (B+1)^{5}-B^{5}=0 \\ \implies 1\times B_5+5\times B_4+10 \times B_3 + 10 \times B_2 + \\ 5 \times B_1+1\times B_0 =0\\ \implies 1\times 0+5\times B_4+10 \times 0+ 10 \times (\frac{1}{6}) + \\ 5 \times (\frac{-1}{2})+1\times 1 =0\\ \implies 5\times B_4+10 \times (\frac{1}{6}) + 5 \times (\frac{-1}{2})+1 =0\\ \implies B_4=-\frac{1}{30}\\ $$

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4.

Euler found a formula which easily defined the **even-numbered zeta**functions as follows[7]:
$$\zeta(2n)=(-1)^{(n+1)}\frac{{B_{2n}}{(2\pi)^{2n}}}{2(2n)!}$$
Interestingly, $B_{2n+1}=0 \ \ \forall n \geq 1$, so there is no counter-part for $\zeta(2n+1)$ and yet a value of $\zeta(3)$ exists. Well, that is beyond my scope of this blog.
Cheers!

Bernoulli Numbers - Explanation

Augusta Ada King-Noel, Countess of Lovelace (10 December 1815 – 27 November 1852) was an English mathematician and writer, chiefly known for her work on Charles Babbage’s proposed mechanical general-purpose computer, the Analytical Engine. [1].
She is widely regarded as the first computer programmer. She wrote an algorithm to calculate the Bernoulli numbers, for more please visit the previous blog here.
So, I thought to study and analyze Bernoulli numbers. $\mathcal Crazy \ math \ ahead!$
$B_0 = 1; B_1= \pm \frac{1}{2}; B_2 = \frac{1}{6}; B_3 = 0; B_4=-\frac{1}{30}; B_5 = 0...$

Approach 1 :

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Let us begin,
Definition: The Bernoulli numbers are defined as the co-efficients of the power series of the expansion of $\large \frac{t}{e^{t} -1}$. For $m \geq 0$, we define $B_m$ so that,
$$\large \frac{t}{e^{t} -1} = \sum_{m=0}^{\infty}\frac{B_m}{m!}t^{m}-(1)$$


Let us apply some mathematical rigor into it and see what happens.
We have,
$$\large \frac{t}{e^{t} -1} = \sum_{m=0}^{\infty}\frac{B_m}{m!}t^{m}\\ \implies\large (e^{t} -1) \frac{t}{e^{t} -1} = (e^{t} -1)\sum_{m=0}^{\infty}\frac{B_m}{m!}t^{m} \\ \implies\large {t}= (e^{t} -1)\sum_{m=0}^{\infty}\frac{B_m}{m!}t^{m} -(2)$$
Now, we know the McLaurin Series is,
$f(t) = f(0) + f'(0) + \frac{f""(0)}{2!}t^{2} + \frac{f^{(3)}(0)}{3!}t^{3}+...+\frac{f^{(n)} }{n!}t^{n}$
Let, $f(t) = e^{t}$, to find its McLaurin coefficients we must evaluate $\large(\frac{d^{k}}{dt^{k}}f(t))|_{t=0}$ for all $k$ = $0, 1, 2, ...$

By substitution, the McLaurin Series expansion of $(e^{t})$ is,
$$\large \sum_{k=0}^{\infty}\frac{f^{k}(0)}{k!}(t-0)^{k} = \frac{x^{0}}{0!} + \frac{x^{1}}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ...\\ \implies \large e^{t} = \frac{x^{0}}{0!} + \frac{x^{1}}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ...\\ \implies \large e^{t} = 1 + \frac{x^{1}}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ...\\ \implies \large (e^{t} -1) = \frac{x^{1}}{1!} + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + ...\\ \implies \large (e^{t} -1) = \sum_{k=1}^{\infty}\frac{t^{k}}{k!} - (3)$$

Using $(3)$ in $(2)$, we get,
$$\implies\large {t}= \sum_{m=0}^{\infty}\frac{B_m}{m!}t^{m} .\sum_{k=1}^{\infty}\frac{t^{k}}{k!} \\ \implies\large {t}= \sum_{m=0}^{\infty}\frac{B_m}{m!}t^{m} .\sum_{k=1}^{\infty}\frac{t^{k}}{k!} \\ \implies\large \frac{t}{t}= \frac{1}{t}\sum_{m=0}^{\infty}\frac{B_m}{m!}t^{m} .\sum_{k=1}^{\infty}\frac{t^{k}}{k!} \\ \implies\large \frac{t}{t}= \sum_{m=0}^{\infty}\frac{B_m}{m!}t^{m} .\sum_{k=1}^{\infty}\frac{t^{k-1}}{k!} \\ \implies\large 1= \sum_{m=0}^{\infty}\frac{B_m}{m!}t^{m} .\sum_{k=1}^{\infty}\frac{t^{k}}{(k+1)!} - (4)$$
Let’s see which coefficient $a_n$ has the $t^n$ in the expanded right-hand part of this equation for some $n > 0$ (we expect this coefficient to be a zero). $t^n$ may appear if $t^m$ from the first sum multiplies the $t^{n-m}$ from the second one for some $0\le m\le n$. Thus,
$$\large a_n = \sum_{m=0}^{n}\frac{B_{n-k}}{m!(n-m+1)!}\\ \large = \sum_{m=0}^{n} {n+1\choose m} \frac{B_{m}}{(n+1)!}\\ \large = \sum_{m=0}^{n-1} {n+1\choose m} \frac{B_{m}}{(n+1)!} + \\\binom{n+1}{n}\frac{B_n}{(n+1)!} = 0$$
for any $n > 0$. Thus
$$\large (n+1)B_n = -\sum_{m=0}^{n-1} {n+1\choose m} \frac{B_{m}}{(n+1)!}$$
Now for $n = 0$ we have from $(1)$
$$\large \frac{B_0}{0!}\frac{1}{1!} = 1$$
and thus
$$B_0 = 1$$

Approach 2 :

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We know,
$$\large \sum_{i=1}^{n} (i) = 1 + 2 + 3 + ... = \frac{n(n+1)}{2},\\ \large \sum_{i=1}^{n} (i^{2}) = 1^{2} + 2^{2} + 3^{2} + ... = \frac{n(n+1)(2n+1)}{6},\\ \sum_{i=1}^{n} (i^{3}) = 1^{3} + 2^{3} + 3^{3} + ...=\frac{n^{2}(n+1)^{2}}{4},\\ \sum_{i=1}^{n} (i^{4}) = 1^{4} + 2^{4} + 3^{4} + ...=\frac{n(n+1)(2n+1)(3n^{2}+3n-1}{30}.$$
and so on…

Then, how about,
$$\large \sum_{i=1}^{\infty}(i^{p})=?$$
Mathematicians have always been fascinated with such classic general formulae. So was JohannFaulhaber.
Let, $$S_p(n) = \sum_{i=1}^{\infty}(i^{p}) \\ \forall p \geq 0$$
Define the following exponential generating function with (initially) indeterminate $z$
$G(z,n) = \sum_{p=0}^{\infty } S_{p}(n) \frac {1}{p!}z^{p}$
We find
$$G(z,n)=\sum _{p=0}^{\infty }\sum _{k=1}^{n}\frac {1}{p!}((kz)^{p} )\\ =\sum _{k=1}^{n}e^{kz} \\ =e^{z}.\frac {1-e^{nz}}{1-e^{z}} \\ ={\frac {1-e^{nz}}{e^{-z}-1}}. \\ $$

$$G(z,n) =\sum_{p=0}^{\infty} \sum_{k=1}^{n} \frac{1}{p!}(kz)^p\\ =\sum_{k=1}^{n}e^{kz}=e^{z}.\frac{1-e^{nz}}{1-e^{z}},\\ = \frac{1-e^{nz}}{e^{-z}-1}.$$
This is an entire function in $z$ so that $z$ can be taken to be any complex number.
We next recall the exponential generating function for the Bernoulli polynomials $B_{j}(x) B_j(x)$
$${\frac {ze^{zx}}{e^{z}-1}}=\sum _{j=0}^{\infty }B_{j}(x)\frac{z^{j}}{j!}$$
where $B_{j}=B_{j}(0)$ denotes the Bernoulli number (with the convention $B_{1}=\frac {1}{2}; B_{1}=-\frac{1}{2}$). We obtain the Faulhaber formula by expanding the generating function as follows:
$$G(z,n) = \sum_{j=0}^{\infty}B_j \frac{(-z)^{j-1}}{j!}(-\sum_{l=1}^{\infty}\frac{(nz)^l}{l!}\\ $$
Solving it is again very complex, so, finally we get,
$$\large \sum_{k=1}^{n}i^{p}=\frac{1}{p+1}\sum_{j=0}^{p}(-1)^{j}\binom{p+1}{j}B_jn^{p+1-j}$$
Note that, $B_j = 0$, $\forall$ odd $j>1$; that is why Faulhaber defines $B_{1} = \frac{1}{2}$.

Verification:

Let us consider the following values,
$$p = 2 \\ $$
$$L.H.S = \large \sum_{k=1}^{n}i^{p} \\ = 1^{2} + 2^{2} + 3^{2} +...+n^{2}= \frac{n(n+1)(2n+1)}{6}.$$
$$R.H.S = \frac{1}{2+1}\sum_{j=0}^{2}(-1)^{j}\binom{2+1}{j}B_jn^{2+1-j}\\ =\frac{1}{2+1}[(-1)^{0}\binom{3}{0}B_0n^{3-0}+(-1)^{1}\binom{3}{1}B_1n^{3-1}+\\(-1)^{2}\binom{3}{2}B_2n^{3-2}]\\ =\frac{1}{3}[1 \cdot 1\cdot B_0 \cdot n^{3}+(-1) \cdot 3\cdot B_1 \cdot n^{2}+\\1 \cdot 3 \cdot B_2 \cdot n^{1}]\\ $$
We put, $B_0=1$ , $B_1=\frac{1}{2}$ and $B_2=\frac{1}{6}$ and see that $LHS$ = $RHS$.

Hence, Proved.