0
$\begingroup$

How can I find the Maclaurin series for $f(x)=e^x$/$(1-x^2)$? I have tried expanding it out but I am having trouble with the algebra of it.

$\endgroup$
0
1
$\begingroup$

One may recall that, as $x \to 0$, $$ \begin{align} e^x&=\sum_{n=0}^{\infty}\frac{x^n}{n!}, \quad x \in \mathbb{C}, \tag1\\\\ \frac{1}{1-x^2}&=\sum_{n=0}^{\infty}x^{2n}=\frac1{2}\sum_{n=0}^{\infty}(1+(-1)^n)x^{n}, \quad |x|<1, \tag2 \end{align} $$ then using the Cauchy product we get

$$ \frac{e^x}{1-x^2}=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}\frac{(1+(-1)^{n-k})}{2k!}\right)x^{n}, \quad |x|<1. \tag3 $$

The series on the right hand side of $(3)$ starts as $$ \frac{e^x}{1-x^2}=1+x+\frac3{2}x^2+\frac7{6}x^3+\cdots $$

$\endgroup$
3
  • 1
    $\begingroup$ A small nitpick: (1) and (2) are not only when $x\to 0$ (the first holds for all $x\in\mathbb{C}$, the second when $\lvert x\rvert \leq 1$). $\endgroup$
    – Clement C.
    May 21 '15 at 13:59
  • $\begingroup$ @ClementC. It is not true that the second holds when $|x|\leq1$ :) $\endgroup$ May 21 '15 at 18:06
  • $\begingroup$ Sorry. $\lvert x\rvert < 1$, of course... nitpicking away! $\endgroup$
    – Clement C.
    May 21 '15 at 18:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.