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Let $f(x)=\sum_{n=0}^{\infty}a_nx^n$ be a function on the real line with a positive(or infinite) radius of convergence about $x=0$, and $a_0\neq0$. Then is it true that the function $1/f(x)$ can also be Taylor expanded about the origin? The question is essentially whether the formal inverse series has a positive or infinite radius of convergence.

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What you call inverse is more commonly called reciprocal. –  lhf Jun 1 '11 at 17:07
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@lhf's comment is important because there is something else called the inverse -- i.e., the inverse function. –  Pete L. Clark Jun 1 '11 at 17:12
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I took the liberty of editing the title. –  Hans Lundmark Jun 1 '11 at 17:15
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Sorry for the ambiguity. I was thinking of ring of analytic functions at 0. It is a local ring with anything outside the maximal ideal having an inverse element in the ring. –  ashpool Jun 2 '11 at 3:41

3 Answers 3

up vote 8 down vote accepted

Yes. I think the following is a way to do this without complex analysis (although I think it would be the easiest way).

For simplicity, let's assume $a_0 = 1$ (otherwise you just divide by $a_0$). And take a radius $\rho > 0$ such that $\sum_{k=1}^{\infty} |a_n| |x|^n \le 1$ for all $|x| \le \rho$ (there exists one since $x \mapsto \sum_{k=1}^{\infty} |a_n| |x|^n$ is continuous, and zero at the origin). Let's denote $b_n$ for the coefficient of the formal inverse and show by recurrence that $|b_n| \le \rho^{-n}$ (which will prove it converges with radius $\ge \rho^{-1}$).

For $n = 0$ it is obvious since $b_0 = 1$. Then for $n \ge 1$, you have

$$b_n = - \sum_{k=1}^{n} a_k b_{n-k}$$

And

$$|b_n| \le \sum_{k=1}^{n} |a_k| \rho^{k-n}$$

So

$$|b_n| \rho^n \le \sum_{k=1}^{n} |a_k| \rho^{k} \le \sum_{k=1}^{\infty} |a_k| \rho^{k} \le 1$$

Which concludes.

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Just beat me to the punch. +1 –  Jesse Madnick Jun 1 '11 at 17:29

The answer is yes: the given function has a Taylor expansion about the origin with positive radius of convergence (note that most consider $\infty$ to be a positive radius of convergence).

Probably the easiest way to see this is to appeal to complex analysis: the function $f$ equally well defines a complex analytic function on a neighborhood of $0$. Recall that a function is complex analytic iff it is differentiable. Moreover, the usual calculus proof adapts to this context to show that if $f$ is complex differentiable and $f(0) \neq 0$, then $\frac{1}{f}$ is differentiable at $0$, thus complex analytic.

It is also possible to prove this result by direct power series manipulations, but I confess I don't remember the argument at the moment: I believe it's a little tricky. But you could consult a book on real analytic function theory, e.g. this one by Krantz and Parks. (Added: the argument is given on pages 7 and 8 of this text. I am about to sign off for a little while, but upon request I could paraphrase it here a little later on.)

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The proof in the book has an obvious error but I got the idea. But I have to say I like Joel Cohen's proof above better. Thanks, though. –  ashpool Jun 2 '11 at 9:45

ashpool: Yes, $1/f(x)$ is analytic, and in fact it is analytic in the largest disk centered at the origin where $f(x)$ is defined and non-zero. This disk exists, since $f(0)=a_0\ne0$.

One way to check this is to use that having a convergent power series expansion is equivalent to being complex differentiable. And one can check directly that $1/f$ is differentiable if $f$ is.


(Though I prefer the complex analytic approach, one can argue directly. The idea is simple, though I don't have time right now to give full details. Note that $f(x)=a_0(1+\epsilon(x))$ where $\epsilon(x)$ is analytic (so, continuous) and small when $x$ is small. Then $1/f(x)$ can be expanded as a geometric series in terms of $\epsilon(x)$, and you can expand $\epsilon$ and its powers term by term. The fact that it is small can be quantified to see that the coefficients resulting from this expansion give a series that converges in a small interval. In a sense, you are just comparing the formal series term by term with a numerical series that you know converges (Fix $0<\delta<1$ and pick $r$ so that $|\epsilon(x)|<\delta$ for all $x$ with $|x|<r$. Then work inside the interval of radius $r$). It is a bit messy but straightforward.)

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Thanks. Do you think there is a way to prove it without using complex analysis? –  ashpool Jun 1 '11 at 17:02
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@ashpool: I've added a small comment indicating an approach (what one would think works actually does). –  Andres Caicedo Jun 1 '11 at 17:18

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