# Taylor series for $\tanh(z)$

Find the Taylor series of $\tanh(z)$ around $z_0=0$.

$z$ is a complex variable.

I can use all the basic series as facts like the $\cosh$ and $\sinh$ series. I know how to calculate the series with the derivatives but I want a closed form like the others have. Wikipedia says something about Bernoulli numbers; I don't know a thing about that and there was no mention of those numbers. So I kinda need to find a formula using only the series of the other basic functions. I am not good using properties of series so don't make big steps as to what you might think it is trivial.

• You can use $\sinh z=-i\sin(iz)$ and $\cosh z=\cos(iz)$. And use the series of $\tan$.
– user228113
Jun 5 '16 at 12:05
• yea ok and then how can i break the series after substituting .to get to Known series.?
– Jam
Jun 5 '16 at 12:08
• Well, how bad of me. I forgot that the series of $\tan$ is not a good one!
– user228113
Jun 5 '16 at 12:21
• $tanh(z)=-itan(iz)$ so the series is the same as $tan(z)$ only instead of z is $iz$ and multiplied by $-i$ ??
– Jam
Jun 5 '16 at 12:41
• Yes, that was the thing I suggested but, well: it's as hard as finding the taylor series of $\tan$. Which is as hard as calculating the Bernoulli numbers.
– user228113
Jun 5 '16 at 12:44

You may exploit $\tanh x=\frac{d}{dz}\log\cosh z$ together with: $$\cosh z=\prod_{n\geq 0}\left(1+\frac{4z^2}{(2n+1)^2\pi^2}\right)\tag{1}$$ that give: $$\begin{eqnarray*}\tanh z = \sum_{n\geq 0}\frac{8z}{4z^2+(2n+1)^2\pi^2}&=&\sum_{n\geq 0}\frac{8z}{(2n+1)^2\pi^2}\sum_{m\geq 0}(-1)^m\frac{4^m z^{2m}}{(2n+1)^{2m}\pi^{2m}}\\&=&\sum_{m\geq 0}(-1)^m 2^{2m+3} z^{2m+1}\sum_{n\geq 0}\frac{1}{\pi^{2m+2}(2n+1)^{2m+2}}\\&=&\sum_{m\geq 0}(-1)^m 2^{2m+3} z^{2m+1}(1-2^{-2m-2})\frac{\zeta(2m+2)}{\pi^{2m+2}}\\&=&\sum_{m\geq 0}\color{red}{(-1)^m(2^{2m+3}-2)\frac{\zeta(2m+2)}{\pi^{2m+2}}}\,z^{2m+1}.\tag{2}\end{eqnarray*}$$ Accidentally, that gives a nice result about the values of $\zeta(2m)$. Since $f(z)=\tanh(z)$ fullfills $$f'(z)+f(z)^2 = 1,\tag{3}$$ we have: $$(2m-1)!\cdot (2^{2m+1}-2)\frac{\zeta(2m)}{\pi^{2m}}\in\color{red}{\mathbb{Z}}\tag{4}$$ due to $f^{(n)}(0)\in\mathbb{Z}$, that follows from $(3)$.