# $\frac{1}{f(x)}$ analytic at 0 when $f(x)$ is a power series with positive convergence radius

Assume $y=f(x)$ is a function represented by a power series with convergence radius $>0$; that is, $y=f(x)=\sum_{n=0}^{\infty} a_nx^n$ with $a_n\in \mathbb{C}$. When $|a_0|\neq 0$, does the function $\frac{1}{f(x)}$ analytic at point 0?

I've tried to slove the questio by writing $\frac{1}{f(x)}=\sum_{n=0}^{\infty} b_nx^n$, and determine $b_n$ in terms of $a_n$. Then I got $b_0=\frac{1}{a_0}, b_1a_0+b_0a_1=0\Rightarrow b_1=-\frac{a_1}{a_0}^2$... But I'm wondering if those $b_i$'s can have a simpler form rather than giving inexplicitly by recurrences. Also are there better approaches to this problem? Thanks!

Take $\;z\;$ pretty close to zero (say, $\;|z|<\epsilon\;$ for some $\;\epsilon>0\;$_ , then

$$f(z)=a_0+a_1z+\ldots\implies\frac1{f(z)}=\frac1{a_0+a_1z+\ldots}=$$

$$\frac1{a_0}\frac1{1+\frac{a_1}{a_0}z+\ldots}=\frac1{a_0}\left(1-\frac{a_1}{a_0}z+\frac{a_1^2}{a_0^2}z^2-\ldots\right)$$

and you get a power series for $\;\cfrac1f\;$ around zero$\;\;\implies\;\cfrac1f\;$ is analytic at $\;z=0\;$

Can you see now how to choose $\;\epsilon>0\;$ and use the development of a geometric series?

• Can I ask how you show that $$\frac{1}{1 + \frac{a_{1}}{a_{0}} z + \cdots} = 1 - \frac{a_{1}}{a_{0}} z + \cdots$$ ? I distinctly remember seeing it on math.stackexchange before, but I can't for the life of me remember where. Also (+1). Nov 18 '16 at 11:35
• do you mean that using geometric series to write $\frac{1}{1+\frac{a_1}{a_0}z+\cdots }$ to $1+\sum_{i=1}^{\infty} (-(\frac{a_1}{a_0}z+\cdots))^i$? but what is this to do with $\epsilon$? Is it $\epsilon$ is small enough to make $f(z)\neq 0$? Nov 18 '16 at 11:51

Similar to the answer of DonAntonio.

Since $f(0) = a_0\ne0$, we can write $$frac{1}{f(z)}= \frac{1}{a_0 - (a_0-f(z))} =\frac{1}{a_0}\frac{1}{1-(a_0-f(z))/a_0}$$ Now recall the geometric expansion: $$\frac{1}{1-x}=\sum_{n\ge0}x^n$$ valid for $|x|<1$. Hence we have, $$\frac{1}{f(z)}= \frac{1}{a_0}\sum_{n\ge0}\left(\frac{a_0-f(z)}{a_0}\right)^n$$ provided we choose $z$ such that $|a_0-f(z)|<|a_0|$, which is clearly possible. Moreover, we can apply the Weirstass M-test (do you see how) in order to ensure the convergence is uniform close to $z=0$.

Let $R$ be the convergence radius. We have $f(0)=a_0 \ne 0$. Since $f$ is continuous at $0$, there is $r$ such that $0<r \le R$ and $f(z) \ne 0$ for all $z$ with $|z|<r$.

Hence $g:=1/f$ is analytic for $|z|<r$.

• I think OP would also appreciate it if you've mentioned how you jump to the last conclusion just from nonvanishing of $f$. Nov 18 '16 at 11:23
• Thanks. Can you explain in detail for the last step as Wojowu mentioned? Nov 18 '16 at 11:26
• It should be clear that $g$ is complex differentiable, therefore g is represented by a power series around $0$
– Fred
Nov 18 '16 at 11:28