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Showing an analytic function is identically zero

  • Thread starter Dunkle
  • Start date
  • #1
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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 [tex](r_{n})[/tex] is a sequence of distinct numbers such that [tex]f(r_{n}) = 0[/tex] 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



[tex]f(x) = \sum a_{n}x^{n}[/tex]

[tex]f(x_{n})=0[/tex]

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 [tex](r_n) = 1/n[/tex], then the zeros of f bunch up near 0. It is clear to me that f(0)=0 (since f is continuous), so [tex]a_{0}=0[/tex]. It is also clear to me that f '(0)=0, and so [tex]a_{1}=0[/tex]. It seems like the way to go is to show (using induction) that each [tex]a_{n}[/tex] is 0 by showing that all derivatives at 0 are equal to 0.

Thoughts?
 

Answers and Replies

  • #2
1,838
7
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.
 
  • #3
56
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I am having problems with the induction for the derivatives. I already showed f '(0)=0. Assuming

[tex]f^{(k)}(0)=0[/tex],

we need to show that

[tex]f^{(k+1)}(0)=0[/tex].

We compute

[tex]f^{(k+1)}(0)=\frac{f^{(k)}(x)-f^{(k)}(0)}{x-0}=\frac{f^{(k)}(x)}{x}[/tex]

But I don't see why this is 0. Any suggestions?
 
  • #4
1,838
7
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:

[tex]\frac{f^{\left k\right)}\left(r_{n}\right)}{r_{n}}=0[/tex]

So, the limit is clearly zero.
 

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