# Analytical function

talolard

## Homework Statement

I had this question on a test. No one I talked to had a clue how to solve it, and our professor will not be publishing solutions. Any insight as to how to aproach this would be great because I am totally stumped.

Let $$f \in C^\infty[-1,1]$$ such that for all $$j \in N |f^{(j)}|<M$$. given that $$f(1/k)=0 \forall k \in N$$ prove that $$f=0$$ in $$[-1,1]$$

## The Attempt at a Solution

The only idea I've had is that I need to use the taylor series. But I have no idea what to do with it.
Thanks
Tal

latentcorpse
hmm. can only offer comments really as im not much of an analyst. but if its zero at every rational number in the interval then by the density of the rationals, it would need to be zero everywhere else since the derivatives are bounded which prevents things such as discontinuities etc.

talolard
I thought of that, but it's npot the case. we know nothing about, say, F(5/7)

Staff Emeritus
Homework Helper
Can you make use of the theorem that says analytic functions have at least one singular point or are constant?

talolard
Never heard that one, so I guess not.

Count Iblis
Hints:

1) What is f(0) ?

2) What are the derivatives of f at zero?

Homework Helper
By the way, you title this "analytic functions" but refer $$\displaystyle f\in C^\infty [-1, 1]$$. That is NOT the set of "analytic functions on [-1, 1]. The fact that a function has all derivatives continuous does NOT imply that it is analytic. For example, the function
$$f(x)= \begin{array}{c}0 if x= 0 \\ e^{-\frac{1}{x}} if x\ne 0$$
is in $$C^\infty[-1, 1]$$ but is not analytic on that set.

talolard
@countbliss
I asume f(0) is 0 but I don't know how to prove it. I thought that since K can go to infity values of 1/k that are arbitrarily close to 0 are zero implies that f(0) is 0. But I'm not sure that really proves it and I don't have an idea as to how to prove it.

Halls ofIvy, duly noted. Thanks

Staff Emeritus
That's a general property of continuous functions. A function is continuous if and only if given any convergent sequence $$x_n \to x$$, you get $$f(x_n) \to f(x)$$