# Every power series expansion for an entire function converges everywhere

I would like to show that every power series expansion for an entire function converges everywhere.

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This is Taylor's theorem. – Antonio Vargas Mar 27 '12 at 21:09
What is your definition of an entire function? – Aryabhata Mar 27 '12 at 21:13

Using Cauchy's Theorem and integration by parts yields \begin{align} \left|\frac{f^{(n)}(w)}{n!}\right| &=\left|\frac{1}{2\pi i}\oint\frac{f^{(n)}(w+z)}{n!\,z}\mathrm{d}z\right|\\ &=\left|\frac{1}{2\pi i}\oint\frac{f(w+z)}{z^{n+1}}\mathrm{d}z\right|\\ &\le\frac{1}{r^n}\max_{B(w,r)} |f|\tag{1} \end{align} where the integration is around the circle $z=r\,e^{it}$ for $t$ from $0$ to $2\pi$.

Estimate $(1)$, called Cauchy's Estimates, says that the radius of convergence of the Taylor series for $f$ is at least $r$. Since $f$ is entire, we can set $r$ as large as we want.

Therefore, the Taylor series for $f$ at $w$ converges for all $z$.

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It might be worth pointing out that the family of inequalities you prove is often called Cauchy's estimates (on the Taylor coefficients). – t.b. Mar 28 '12 at 3:08
@tb: Thanks! I dredged this up from somewhere, but I don't think I ever knew what they were called. – robjohn Mar 28 '12 at 4:19

Maybe something is wrong with this answer, but it seems to be pretty simple.

First, we know that the power series of an analytic function is unique. So if a function is entire (analytic in the whole complex plane), then its power series is unique on the whole plane, and by definition is convergent.

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We know that the power series around any point is unique, but what does it mean to say "its power series is unique on the whole plane"? – Antonio Vargas Mar 27 '12 at 21:13
Yeah, that was pretty sloppy, sorry. I meant as you said: that the power series expansion of the function about some point in the plane is unique, and by entirety converges at every point in the complex plane. – Keaton Mar 27 '12 at 21:20
If we're taking the definition of entire to be "all power series converge everywhere to the function" then your answer is a circular detour back to this definition. Otherwise you're begging the question. – Antonio Vargas Mar 27 '12 at 21:22