# Taylor expansion of rational function

I want to taylor expand the following function (in the complex plane):

$$(1-\frac{z^2}{3!}+\frac{z^4}{5!})^{-1}$$

I am told that the result is:

$$1+\frac{z^2}{3!}+\frac{14}{6!}z^4+O(z^6)$$

I do not see how to obtain this. The only thing I have tried is using the taylor expansion of $(1-z)^{-1}$. But using this I can only obtain the second term of the expansion written above.

## 1 Answer

HINT:

Recall from Taylor's Theoremthat $\frac{1}{1-x}=1+x+x^2+O(x^3)$

Now, set $x= z^2/3!-z^4/5!$ and don't forget the second order term $(z^2/3!-z^4/5!)^2$.

• Thank you. But isn't this only true for $|z|<1$? – john melon Apr 3 '17 at 20:23
• OK. So the expression above is only valid in a domain that satisfies $\left|\frac{z^2}{3!}-\frac{z^4}{5!}\right|<1$? – john melon Apr 3 '17 at 20:26
• No. The expansion is not a full series. Note from Taylor's Theorem that $\frac{1}{1-x}=1+x+x^2+o(x^2)$where we are using the little $o$ notation. – Mark Viola Apr 3 '17 at 20:26
• What do you mean by full series? – john melon Apr 3 '17 at 20:28
• We are truncating a series, not writing one. Taylor's Theorem is not Taylor's Series. If we write $\frac{1}{1-x}=\sum_{n=0}^\infty x^n$, then the series converges for $|x|<1$ and diverges elsewhere. But we can write $\frac{1}{1-x}=1+x+x^2+O(x^3)$ without restricting $x$. – Mark Viola Apr 3 '17 at 20:29