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

Question

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!

Answer 1

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 $z=0$.

Answer 2

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 $|z|<r$.

Answer 3

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?