Maclaurin expansion for sech$(x)$

Question

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?

Answer 1

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$( x)=\dfrac2{e^x+e^{-x}}=\dfrac{2}{2+2\dfrac{x^2}2+2\dfrac{x^4}{4!}+\cdots}=\dfrac1{1+\dfrac{x^2}2+\dfrac{x^4}{4!}+\cdots}$

$=1-\left(\dfrac{x^2}2+\dfrac{x^4}{4!}+\cdots\right)+\left(\dfrac{x^2}{2}+\dfrac{x^4}{4!}+\cdots\right)^2\cdots$

$=1-\dfrac{x^2}2-\dfrac{x^4}{24}+\dfrac{x^4}{4}\cdots=1-\dfrac{x^2}2+\dfrac{5x^4}{24}\cdots$

Answer 2

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.

Answer 3

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 $x_0$: $$ f(x)=\sum_{k=0}^{+\infty}\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k $$

Answer 4

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: $$\begin{array}[t]{r|} 1\phantom{{}-\frac{x^2}2-\frac{x^4}{24}} \\ -1-\frac{x^2}2-\frac{x^4}{24}\\\hline -\frac{x^2}2-\frac{x^4}{24} \\ \frac{x^2}2+\frac{x^4}{4} \\ \hline \frac{5x^4}{24}\\ -\frac{5x^4}{24}\\\hline 0 \end{array} \begin{array}[t]{l} 1+\frac{x^2}2+\frac{x^4}{24} \\\hline \color{red}{1 -\frac{x^2}2+\frac{5x^4}{24}} \end{array}$$

Answer 5

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 $\zeta(4)=\frac{\pi^4}{90}$.