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Question : Given a function f holomorphic in the unit disk D, and such that for every point z ∈ D, there exists an n ∈ N such that the nth derivative of f vanishes at z. Prove that f is a polynomial.

Intuitively I think the question is quite obvious, however , when I truly try to prove it formally , I am stuck. Does anyone have any idea about this question? Truly grateful to any help!

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    $\begingroup$ hint: if $f^{(n)}$ is not identically zero then it has at most countably many zeros in D $\endgroup$ – user8268 May 4 '17 at 18:35
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Let be $D_n = \{z\in D: f^{(n)}(z) = 0\}$. By hypothesis, $D = \bigcup_{n\in\Bbb N}D_n$. As $D$ is uncountable, Some $D_n$ is uncountable...

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    $\begingroup$ I changed $f_n$ to $f^{(n)}$ to make it clearer that it is the $n$th derivative. Hope you don't mind! :) $\endgroup$ – Cameron Williams May 4 '17 at 18:59
  • $\begingroup$ Thanks a lot for your answer!!! $\endgroup$ – fxy May 5 '17 at 0:41
  • $\begingroup$ But I still have some confusion. Say there exists some integer N , s.t DN is uncountable , how can we claim that f coincides with a polynomial in DN? $\endgroup$ – fxy May 5 '17 at 0:43
  • $\begingroup$ Since if we can prove the claim , then by identity theorem , we can say f has to be a polynomial. $\endgroup$ – fxy May 5 '17 at 0:44
  • $\begingroup$ @CameronWilliams, thanks for the correction. $\endgroup$ – Martín-Blas Pérez Pinilla May 5 '17 at 6:08

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