# The Zeta Function: Finding a general formula Part 2

From the first part we got to

$\pi.s\frac{\cos \pi.s}{\sin \pi.s}=1+\displaystyle\sum_{k=1}^{\infty} \frac{1}{(1-\frac{s^2}{k^2})}.-\frac{2s^2}{k^2}$

Note that $\frac{\cos \pi.s}{\sin \pi.s}= cotan (\pi.s)$.

We can take the -2 up front and use the geometric series formula to obtain:

$\displaystyle\pi.s.cotan(\pi.s)=$

$=1-2.\displaystyle\sum_{k=1}^{\infty} [1+\frac{s^2}{k^2}+\frac{s^4}{k^4}...].\frac{s^2}{k^2}=1-2.\sum_{k=1}^{\infty} (\sum_{n=1}^{\infty} (\frac{s^2}{k^2})^n)=$

$=1-2.\displaystyle\sum_{n=1}^{\infty} \frac{s^{2n}}{1^{2n}}+\frac{s^{2n}}{2^{2n}}...=1-2.\sum_{n=1}^{\infty} \zeta (2n).s^{2n}$

Now let's try to find a different representation of $\pi.s.cotan(\pi.s)$

$\pi.s.cotan(\pi.s)=\pi.s.\frac{ \cos \pi.s}{ \sin \pi.s}=$

$=\displaystyle\pi.s.\frac{e^{i.\pi.s}+e^{-i.\pi.s}}{2}.\frac{2i}{e^{i.\pi.s}-e^{-i.\pi.s}}=$

$=\displaystyle\pi.s.i.\frac{e^{i.\pi.s}+e^{-i.\pi.s}}{e^{i.\pi.s}-e^{-i.\pi.s}}$.

Multiplying the numerator and denominator by $e^{i.\pi.s}$ we get

$\displaystyle \pi.s.cotan(\pi.s)=\pi.s.i.\frac{e^{2i.\pi.s}+1}{e^{2i.\pi.s}-1}=$

$=\displaystyle i.\pi.s + \frac{2.i.\pi.s}{e^{2i.\pi.s}-1}$

We now need to find an infinite series for $\frac{z}{e^z-1}$

Let's assume we have an infinite series representation of that

$\displaystyle\frac{z}{e^z-1}=\sum_{n=0}^{\infty}\frac{\beta_n}{n!}.z^n$

Then, multiplying both sides by $e^z-1$, using the Taylor expansion for $e^z$ and dividing by z, we get

$1=\displaystyle\sum_{n=0}^{\infty}\frac{\beta_n}{n!}.z^n.\sum_{n=0}^{\infty} \frac{z^{n}}{(n+1)!}$

Tune in next time for Part 3 Note by Bogdan Simeonov
5 years, 9 months ago

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