I know how to find the Taylor expansion of both $\sinh x$ and $\cosh x$, but how would you find the Taylor expansion of $\tanh x$. It seems you can't just divide both the Taylor series of $\sinh x$ and $\cosh x$ so how would you do it?
Any suggestions? I saw it contains the Bernoulli series, what is that exactly?
Kind Regards
Best Answer
You may too use the method I used here for the expansion of $\tan$ :
Integrate repetitively $\ \tanh'(x)=1-\tanh(x)^2\ $ starting with $\,\tanh(x)\approx x$ :
\begin{align} \tanh(x)&\small{=}\ x+O\bigl(x^2\bigr)\\ &\small{=\int 1-\left(x+O\bigl(x^2\bigr)\right)^2\,dx}=x-\frac {x^3}3+O\bigl(x^4\bigr)\\ &\small{=\int 1-\left(x-\frac {x^3}3+O\bigl(x^4\bigr)\right)^2dx=\int 1-x^2+\frac {2x^4}3\,dx-O\bigl(x^6\bigr)}=x-\frac {x^3}3+\frac {2x^5}{15}-O\bigl(x^6\bigr)\\ &= \cdots\\ \end{align}
Every integration gives another coefficient of $\ \displaystyle\tanh(x)=\sum_{n\ge 0} a_n\ (-1)^n\,x^{2n+1}\ $ and we get simply : $$a_0=1,\; a_{n+1}=\frac 1{2n+3} \sum_{k=0}^n a_k\ a_{n-k},\ \text{for}\;n>0$$ i.e. the sequence (with alternating signs for $\tanh$) : $$(a_n)_{n\in\mathbb{N}}=\left(\frac 11,\frac 13, \frac 2{15}, \frac {17}{315}, \frac {62}{2835}, \frac{1382}{155925},\cdots\right)$$
We may probably deduce the recurrence relation of Bernoulli numbers in function of this one (or vice et versa) but I didn't try that yet.