Complete the last step in an analytic function proof

Question

let $ f$ be analytic function in segment A. and assume $ \in A $ and $ f(0)=0 $

prove that exists $ n \in N $ and analytic function $ g(x) $ in segment A such that:

$ f(x)=x^n\cdot g(x) $ for any $ x∈A$

So, I want to prove that $ g $ is analytic in any point in segment $ A $

so, i can just define $ g= \frac{1}{x^{n}}f\left(x\right) $ and since f analytic and $ \frac{1}{x^n} $ analytic, g will be analytic in any point $ f $ is analytic, except $ x=0 $

so, we know that $ f(0)=0 $ and we can define $ g\left(0\right)=a_{n_{0}} $ when $ n_0 $ is the minimal index such $ a_{n_{0}}\neq0 $ in the power series $ f\left(x\right)=\sum_{n=0}^{\infty}a_{n}x^{n} $ around $x_0=0 $ so I've proved that g is analytic in any point except $ 0 $. how can i prove that its also analytic in x=0 ? Thanks.

Answer 1

Since $f(0)=0,$ it means the constant term in the power series for $f$ about $0$ vanishes. Thus it is enough to take $n=1,$ although some higher value of $n$ may exist. Then $g(x)$ is also a power series, which defines an analytic function perforce.