The Zeta Function: Finding a general formula Part 2

From the first part we got to

π.scosπ.ssinπ.s=1+k=11(1s2k2).2s2k2\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 cosπ.ssinπ.s=cotan(π.s)\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:

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

=12.k=1[1+s2k2+s4k4...].s2k2=12.k=1(n=1(s2k2)n)==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)=

=12.n=1s2n12n+s2n22n...=12.n=1ζ(2n).s2n=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 π.s.cotan(π.s)\pi.s.cotan(\pi.s)

π.s.cotan(π.s)=π.s.cosπ.ssinπ.s=\pi.s.cotan(\pi.s)=\pi.s.\frac{ \cos \pi.s}{ \sin \pi.s}=

=π.s.ei.π.s+ei.π.s2.2iei.π.sei.π.s==\displaystyle\pi.s.\frac{e^{i.\pi.s}+e^{-i.\pi.s}}{2}.\frac{2i}{e^{i.\pi.s}-e^{-i.\pi.s}}=

=π.s.i.ei.π.s+ei.π.sei.π.sei.π.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 ei.π.se^{i.\pi.s} we get

π.s.cotan(π.s)=π.s.i.e2i.π.s+1e2i.π.s1=\displaystyle \pi.s.cotan(\pi.s)=\pi.s.i.\frac{e^{2i.\pi.s}+1}{e^{2i.\pi.s}-1}=

=i.π.s+2.i.π.se2i.π.s1=\displaystyle i.\pi.s + \frac{2.i.\pi.s}{e^{2i.\pi.s}-1}

We now need to find an infinite series for zez1\frac{z}{e^z-1}

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

zez1=n=0βnn!.zn\displaystyle\frac{z}{e^z-1}=\sum_{n=0}^{\infty}\frac{\beta_n}{n!}.z^n

Then, multiplying both sides by ez1e^z-1, using the Taylor expansion for eze^z and dividing by z, we get

1=n=0βnn!.zn.n=0zn(n+1)!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, 6 months ago

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