# Taylor expansion of the multiplicative inverse of a function

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?

• (1) Are those exponents supposed to be $1$, $2$, and $3$? If so, you don't need parentheses. (2) There is no Taylor expansion for $f(x)$ if $f(0)=0$. (3) If $f(0)\not=0$, then you can get the Taylor expansion for $\displaystyle{1\over f(x)}$ by doing long division. Apr 26, 2016 at 7:02
• No, they are not supposed to be $1,2$ and $3$. They are just some coefficients that appear from the expansion of $f(x)$, and I'm interested in seeing how those coefficients can be used to write the expansion for $1/f(x)$ Apr 26, 2016 at 7:04
• So what you meant was $f(x)=a_0 + a_1 x + a_2 x^2 + \cdots$ ? Apr 26, 2016 at 7:04
• Yes, that is a much better way to write it. Thanks. Apr 26, 2016 at 7:05
• As for an answer to your problem, check out the second answer to mathoverflow.net/questions/53384/… . Apr 26, 2016 at 7:21

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 $$0<\rho so that \begin{align} \sum^\infty_{n=1}|a_n|\rho^n<|a_0|\tag{3}\label{three} \end{align} Then, $$|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)$$.