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!