Taylor Expansion around $0$, for $\frac{1}{\cosh(x)}$, easy method? I understand it is essentially a maclaurin series, however, is their an easy method of calculating this? Such as working out the Maclaurin for $\cosh(x)$ then inversing the series? (First few terms)
 A: $\frac{1}{\cosh x}$ is a meromorphic function with simple poles at $\frac{\pi i}{2}+\pi i\mathbb{Z}$. By computing the residues we have
$$\frac{1}{\cosh x}=\sum_{n\geq 0}\frac{4\pi(2n+1)(-1)^n}{4x^2+\pi^2(2n+1)^2} $$
and by expanding the terms of the last series as geometric series we have:
$$\begin{eqnarray*}\frac{1}{\cosh x}&=&\sum_{n\geq 0}(-1)^n\sum_{m\geq 0}(-1)^m\left(\frac{4}{\pi(2n+1)}\right)^{2m+1}x^{2m}\\&=&\sum_{m\geq 0}\frac{(-1)^m 4^m x^{2m}}{\pi^{2m+1}}\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^{2m+1}}\\&\stackrel{(*)}{=}&\sum_{m\geq 0}\frac{(-1)^m x^{2m}}{4\cdot(2m)! }\,E_{2m}\end{eqnarray*}$$
where $(*)$ is a consequence of the relation between $\zeta\left(2m+1,\frac{1}{4}\right)-\zeta\left(2m+1,\frac{3}{4}\right)$, Euler numbers and alternating permutations. Euler numbers can be computed from the recursion
$$
\color{#C00000}{\mathrm{E}_{2n}=-\sum_{k=0}^{n-1}\binom{2n}{2k}\mathrm{E}_{2k}}
$$
proved here by robjohn, which is just a consequence of $1=\frac{1}{\cosh x}\cosh(x)=\frac{1}{\cosh x}\sum_{m\geq 0}\frac{x^{2m}}{(2m)!}$.
A: If you fix an order of expansion, the simplest way is division by increasing powers of $1$ by the expansion of $\cosh x$ up to this order. At order $4$, you should obtain
$$\frac1{\cosh x}=1-\frac{x^2}2+\frac{5x^4}{24}+O(x^6).$$
A: you can consider for example that
$$\cosh x=1+\frac{x^2}{2}+\frac{x^4}{24}+o(x^6)$$
then
$$\frac{1}{\cosh x}=\left(1+\frac{x^2}{2}+\frac{x^4}{24}+o(x^6)\right)^{-1}$$
an then use the series for
$$(1+x)^a$$ with $$a=-1$$
A: Well, to find the Taylor expansion about $0$, we can just compute the derivatives.
$$f(0) = \frac{1}{\cosh(0)} = 1$$
$$f'(0) = -\sinh(0)\frac{1}{\cosh^2(0)} = 0$$
$$f^{(2)} = \sinh(0)\frac{2}{\cosh^3(0)} -\cosh(0)\frac{1}{\cosh^2(0)} = -1$$
And in general, we can split up $f'$ into $g(x)=-\sinh(x)$ and $h(x)=\frac{1}{\cosh^2(x)}$ and use the generalized product rule to obtain
$$f^{(n+1)}(0)=\sum_{i=1}^{n} \binom{n}{i} g^{(i)}(0)f^{(n-i)}(0) =\sum_{i=1}^{\lfloor n/2 \rfloor}-\binom{n}{2i}\cosh(0)f^{(n-2i)}(0)$$
Using $h^{(k)}(0) = (-1)^k\frac{(k+1)!}{\cosh^{k+2}(0)} = (-1)^k (k+1)!$ we have
$$f^{(n+1)}(0) = \sum_{i=1}^{\lfloor n/2 \rfloor}(-1)^{n-2i+1} \binom{n}{n-2i}(n-2i+1)!$$
So we have for our Taylor expansion
$$a_{n}\cdot n! = f^{(n)}(0) = \sum_{i=1}^{\lfloor (n-1)/2 \rfloor} (-1)^{n-2i}\frac{(n-1)!}{(2i)!}(n-2i)$$
And our Taylor expansion is thus
$$a_0 + a_1 + \frac{a_2}{2!} + \frac{a_3}{3!}...$$
There probably exists a simplified formula for $a_n$. I will come back to this later and see if I can find one. This should, however, let you calculate the first few terms of the expansion.
