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Math Help - prove a certain function is a polynomial

  1. #1
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    prove a certain function is a polynomial

    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 \mathbb{C} and such that for each z_0\in\mathbb{C} 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?
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: prove a certain function is a polynomial

    If f(*) is analytic everywhere in \mathbb{C} , then its Taylor expansion in any point z=z_{0} is...

    f(z)= \sum_{n=0}^{\infty} a_{n}\ (z-z_{0})^{n} (1)

    ... where...

    a_{n}= \frac{1}{n!}\ \frac{d^{n}}{d z^{n}} f(z)_{z=z_{0}} (2)

    Now if \vorall z_{0} is a_{n}=0 then the derivative of order n of f(*) vanishes everywhere and that means that f(*) is a polynomial of order k<n...

    Kind regards

    \chi \sigma
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  3. #3
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    Re: prove a certain function is a polynomial

    you can solve it using 2 hints:

    -whats the union of sets {f^(k)(z)=0} where k runs through natural numbers?
    -what are the properties of the zero set of a holomorphic function?
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  4. #4
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    Re: prove a certain function is a polynomial

    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?
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  5. #5
    Super Member girdav's Avatar
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    Re: prove a certain function is a polynomial

    Put F_n:=\{z\in\mathbb C, f^{(n)}(z)=0\}. F_n is a closed subset of \mathbb C and \bigcup_{n\in\mathbb N}F_n=\mathbb C. Using Baire category's theorem, we can find n_0 such that F_{n_0} has a nonempty interior. We can find a ball on which f^{(n_0)} is identically 0 on this ball. Now you can conclude, using the fact that the zeros of an holomorphic function are isolated.
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