Showing an analytic function is identically zero

In summary, the statement is that if a sequence of distinct numbers (r_n) converges to a point r in the domain of an analytic function f, and f(r_n) = 0 for all n, then f is identically zero on the connected component of the domain containing r. This follows from the fact that two analytic functions with the same values on a sequence of points are identical, and the derivative of an analytic function at zero can be computed by taking the limit of x to zero using the sequence of points (r_n).
  • #1
56
0

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?
 
Physics news on Phys.org
  • #2
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
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
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.
 

1. What does it mean for a function to be analytically zero?

An analytic function being identically zero means that the function has a value of zero at every point in its domain. This means that the graph of the function will lie completely on the x-axis.

2. How can you prove that an analytic function is identically zero?

To prove that an analytic function is identically zero, you can use the identity theorem which states that if two analytic functions have the same values at all points in a connected domain, then they are equal. So, if the given function is equal to the zero function at all points in its domain, it is identically zero.

3. What are the conditions for an analytic function to be identically zero?

The conditions for an analytic function to be identically zero are that it must be defined on a connected domain and have a value of zero at every point in that domain.

4. Can an analytic function be identically zero on a non-connected domain?

No, an analytic function cannot be identically zero on a non-connected domain. This is because the identity theorem only holds for functions defined on connected domains.

5. Are there any other methods for proving that an analytic function is identically zero?

Yes, there are other methods for proving that an analytic function is identically zero such as using the Cauchy-Riemann equations or the maximum modulus principle. However, the identity theorem is the most commonly used method for this proof.

Suggested for: Showing an analytic function is identically zero

Replies
6
Views
533
Replies
3
Views
492
Replies
16
Views
969
Replies
26
Views
269
Replies
7
Views
881
Replies
9
Views
652
Replies
10
Views
846
Back
Top