Maclaurin expansion for sech$(x)$ I am a bit unsure where I have gone wrong in working this out.
Sech$(x)=2/(e^x+e^{-x}).$
Maclaurin expansions:
$e^x = 1+ x +  x^2/2+ x^3/6 + x^4/24;\; e^{-x} = 1- x + x^2/2 - x^3/6 -  x^4/24;$
so sech$(x)= (1+x^2/2+x^4/24)^{-1}.\;$  (I think this is where I have gone wrong.)
The actual answer is $1-x^2/2+ 5x^4/24$ (first 3 terms).
How would I work this out?
 A: From your expansions of $e^x$ and $e^{-x}$, we have $e^x+e^{-x}=2+2\dfrac{x^2}2+2\dfrac{x^4}{4!}+\cdots$,
so  sech$( x)=\dfrac2{e^x+e^{-x}}=\dfrac{2}{2+2\dfrac{x^2}2+2\dfrac{x^4}{4!}+\cdots}=\dfrac1{1+\dfrac{x^2}2+\dfrac{x^4}{4!}+\cdots}$
$=1-\left(\dfrac{x^2}2+\dfrac{x^4}{4!}+\cdots\right)+\left(\dfrac{x^2}{2}+\dfrac{x^4}{4!}+\cdots\right)^2\cdots$
$=1-\dfrac{x^2}2-\dfrac{x^4}{24}+\dfrac{x^4}{4}\cdots=1-\dfrac{x^2}2+\dfrac{5x^4}{24}\cdots$
A: Since $\operatorname{sech}x$ is even, write $\operatorname{sech}x=\sum_{j=0}a_j(x^2)^j$ so$$1=\sum_{j,\,k\ge0}\frac{a_j}{(2k)!}(x^2)^{j+k}\implies\delta_{n0}=\sum_{j=0}^n\frac{a_j}{(2n-2j)!}.$$Now get the first few $a_j$ by solving simultaneous equations.
A: You are wrong exactly where you thought to be. The reason is that the Taylor expansion of a function $f$ takes into account the derivatives of the function itself, hence writing what you did, means that the derivative of $\frac2{e^x+e^{-x}}$ is $\left[(e^x)'+(e^{-x})'\right]^{-1}$ which is clearly false.
The most straightforward way to compute the Taylor expansion near a point $x_0$ is to compute the derivatives of that function at $x_0$:
$$
f(x)=\sum_{k=0}^{+\infty}\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k
$$
A: The simplest way t get it is to write $\;\operatorname{sech} x=\frac1{\cosh x}$ and to use the MacLaurin expansion  of $\cosh x$ at order $4$:
$$\cosh x=1+\frac{x^2}2+\frac{x^4}{24}+o(x^4)$$
You obtain the expansion of its reciprocal  dividing  $1$ by the MacLaurin expansion of $\cosh x$ along increasing powers, up to order 4, truncating every term at order $4$ in this process:
$$\begin{array}[t]{r|}
1\phantom{{}-\frac{x^2}2-\frac{x^4}{24}} \\
-1-\frac{x^2}2-\frac{x^4}{24}\\\hline
-\frac{x^2}2-\frac{x^4}{24} \\ 
\frac{x^2}2+\frac{x^4}{4} \\ \hline
\frac{5x^4}{24}\\ -\frac{5x^4}{24}\\\hline 0
\end{array}
\begin{array}[t]{l}
1+\frac{x^2}2+\frac{x^4}{24} \\\hline
\color{red}{1 -\frac{x^2}2+\frac{5x^4}{24}}
\end{array}$$
A: Since in a neighbourhood of the origin $\cosh(x)=1+\frac{x^2}{2}+\frac{x^4}{24}+O(x^6)$ we have $\operatorname{sech}(x)=1-\frac{x^2}{2}+Ax^4+O(x^6)$
and since
$$ \left(1+\frac{x^2}{2}+\frac{x^4}{24}\right)\left(1-\frac{x^2}{2}+Ax^4\right)=1+\left(A-\frac{5}{24}\right)x^4+O(x^6) $$
we must have $A=\frac{5}{24}$. This unique non-trivial coefficient can also be found by considering that
$$\cosh(x)=\prod_{n\geq 0}\left(1+\frac{4x^2}{\pi^2(2n+1)^2}\right)\tag{1} $$
$$\operatorname{sech}(x)=\prod_{n\geq 0}\sum_{m\geq 0}\frac{(-1)^m 4^m x^{2m}}{\pi^{2m}(2n+1)^{2m}}\tag{2} $$
$$\operatorname{sech}(\sqrt{x})=\prod_{n\geq 0}\sum_{m\geq 0}\frac{(-1)^m 4^m x^{m}}{\pi^{2m}(2n+1)^{2m}}\tag{3} $$
$$ [x^2]\operatorname{sech}(\sqrt{x})=\frac{16}{\pi^4}\sum_{a>b\geq 0}\frac{1}{(2a+1)^2(2b+1)^2}+\frac{16}{\pi^4}\sum_{n\geq 0}\frac{1}{(2n+1)^4}\tag{4} $$
and the RHS of $(4)$ just depends on the values of $\zeta(2)=\frac{\pi^2}{6}$ and $\zeta(4)=\frac{\pi^4}{90}$.
