Showing an analytic function is identically zero

[Dunkle]

Homework Statement

This is a claim from a Wikipedia page about analytic functions (http://en.wikipedia.org/wiki/Analytic_function), and I can't seem to prove it.

If (r_{n}) is a sequence of distinct numbers such that f(r_{n}) = 0 for all n and this sequence converges to a point r in the domain of D, then f is identically zero on the connected component of D containing r.

Homework Equations

f(x) = \sum a_{n}x^{n}

f(x_{n})=0

The Attempt at a Solution

I'm trying to prove the case when r=0. If f is analytic, then it can be represented by some power series. If you look at some sequence (r_n) = 1/n, then the zeros of f bunch up near 0. It is clear to me that f(0)=0 (since f is continuous), so a_{0}=0. It is also clear to me that f '(0)=0, and so a_{1}=0. It seems like the way to go is to show (using induction) that each a_{n} is 0 by showing that all derivatives at 0 are equal to 0.

Thoughts?

 

[Count Iblis]

That's correct. It follows from this that two analytic functions that have the same values in the r_n are identitical. Analytic continuation is thus unique.

 

[Dunkle]

I am having problems with the induction for the derivatives. I already showed f '(0)=0. Assuming

f^{(k)}(0)=0,

we need to show that

f^{(k+1)}(0)=0.

We compute

f^{(k+1)}(0)=\frac{f^{(k)}(x)-f^{(k)}(0)}{x-0}=\frac{f^{(k)}(x)}{x}

But I don't see why this is 0. Any suggestions?

 

[Count Iblis]

You need to take limit of x to zero. You can compute that limit by taking for x the sequence of the rn that tend to zero and computing the lmit for n to infinity. For all n you have that:

\frac{f^{\left k\right)}\left(r_{n}\right)}{r_{n}}=0

So, the limit is clearly zero.