Taylor expansion of the multiplicative inverse of a function

Question

The Taylor expansion of a function is given by $$ f(x) = \sum^\infty_{n=1}a_n(z-a)^n $$ for $z\in B(a;r)$. Suppose $a_0\neq0$. From this, we can establish that $1/f$ is analytic in a neighborhood of $a$. Is there a way to obtain a nicer form of the Taylor expansion of $1/f(z)$ and the radius of convergence?

Answer 1

Suppose that $f(z)=\sum^\infty_{n=0}a_n (z-a)^n$ and $f(a)=a_0\neq0$. Furthermore, suppose $R>0$ is is the radius of convergence of $f$. Then $g(z)=1/f(z)$ is analytic in a neighborhood $B(a;\rho)\subset B(a;R)$. Suppose $g(z)=\sum^\infty_{n=0}b_n(z-a)^n$ on $B(a;\rho)$. Since $1=f(z)g(z)$, the Cauchy product formula yields $$\begin{align} a_0b_0=1\tag{1}\label{one} \end{align}$$ and $$\begin{align} 0=\sum^n_{k=0}a_kb_{n-k}=a_0b_n+a_1b_{n-1}+\ldots + a_nb_0\tag{2}\label{two} \end{align}$$ for $n\geq 1$. From \eqref{one} and \eqref{two} we obtained expressions for $b_n$, $n\in\mathbb{Z}_+$, recursively to get $$ \begin{align} b_0&=\frac{1}{a_0}\tag{1'}\label{onep}\\ b_n&=-\frac{1}{a_0}\big(a_1b_{n-1}+\ldots + a_nb_0\big)\tag{2'}\label{twop} \end{align} $$

All one can say about the radius of convergence of the power series $g(z)=\sum_{n\geq0}b_n(z-a)^n$ is that it is positive. Some Cauchy-type estimates for the $b_n$'s can be obtained as follows. Choose $0<\rho<R$ so that $$\begin{align} \sum^\infty_{n=1}|a_n|\rho^n<|a_0|\tag{3}\label{three} \end{align}$$ Then, $|b_n|\leq |b_0|\rho^{-n}$ for all $n\in\mathbb{Z}_+$: for $n=0$ this holds trivially; assuming this hols for $0\leq j\leq n-1$, we get from \eqref{twop} that $$\begin{align} |b_n|&\leq |b_0|\big(|a_1||b_0|\rho^{-(n-1)}+\ldots +|a_n||b_0|\big)\\ &=\rho^{-n}|b_0|^2\big(|a_1|\rho+\ldots+|a_n|\rho^n\big)\\ &\leq \rho^{-n}|b_0|^2\sum^\infty_{k=1}|a_k|\rho^k\leq |b_0|\rho^{-n} \end{align} $$ Consequently, the power series for $g$ converges in $B(a;\rho)$.