If f(*) is analytic everywhere in , then its Taylor expansion in any point is...
(1)
... where...
(2)
Now if is then the derivative of order n of f(*) vanishes everywhere and that means that f(*) is a polynomial of order k<n...
Kind regards
I have seen the following problem and I didn't manage to solve it, so I will appreciate any help.
Suppose f is an analytic function defined everywhere in and such that for each at least one coefficient in the expansion f(z)=\sum_{n=0}^\infty c_n(z-z_0)^n is equal to 0. Prove that f is a polynomial.
It is easy to prove that for each z_0 there exists n such that f^{(n)}(z_0)=0, which means that each z_0 is root for a derivative of f, but how do I conclude that f s a polynomial?
If f(*) is analytic everywhere in , then its Taylor expansion in any point is...
(1)
... where...
(2)
Now if is then the derivative of order n of f(*) vanishes everywhere and that means that f(*) is a polynomial of order k<n...
Kind regards
I was thinking something like this: the union of sets {f^(k)(z)=0} where k runs through natural numbers is uncountable from the hypothesis, that means that there must be a number N which satisfies f^(N)=0 for all z, which demonstrates that f is a polynomial. But I am not sure if there are some missing arguments.
About the set of the zeros of a holomorphic function, I suppose it must be finite? If a holomorphic function has countable many zeros, that means that it is the function constant=0?
Put . is a closed subset of and . Using Baire category's theorem, we can find such that has a nonempty interior. We can find a ball on which is identically on this ball. Now you can conclude, using the fact that the zeros of an holomorphic function are isolated.