Maclaurin series for $\frac{x}{e^x+1}$

Question

I need to find the first two terms of the Maclaurin series of

$$ f (x) = \frac {x} {e ^ x + 1} $$

That's to say,

$$ \frac {x} {2} - \frac {x ^ 2} {4} + O (x ^ 4) $$

I'm assuming $ x $ is small so that the function can be approximated with these terms. Is there a way to calculate them faster than by applying Taylor's formula? Maybe using known series expansions such as $ e ^ x $ or $\frac {1} {1 + x} $?

Answer 1

If$$\frac x{e^x+1}=a_0+a_1x+a_2x^2+\cdots,$$then\begin{align}x&=\left(2+x+\frac{x^2}2+\cdots\right)\left(a_0+a_1x+a_2x^2+\cdots\right)\\&=2a_0+\left(2a_1+a_0\right)x+\left(2a_2+a_1+\frac{a_0}2\right)x^2+\cdots\end{align}So, solve the system$$\left\{\begin{array}{l}2a_0=0\\2a_1+a_0=1\\2a_2+a_1+\frac{a_0}2=0.\end{array}\right.$$

Answer 2

Using $$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+O\left(x^5\right)$$ $$\frac x {e^x+1}=\frac x {2+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}+O\left(x^5\right)}$$ Now, long division to get $$\frac x {e^x+1}=\frac{x}{2}-\frac{x^2}{4}+\frac{x^4}{48}+O\left(x^6\right)=\frac{x}{2}-\frac{x^2}{4}+O\left(x^4\right)$$