Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
2  
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
add comment

2 Answers 2

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 $z$.

share|improve this answer
    
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
add comment

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.