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Let $f$ be an entire function. Suppose, for each $a\in \mathbb{R}$, there exists at least one coefficient $c_{n}$ in $f(z)=\sum c_{n}(z-a)^{n},$ which is zero. Then prove that there exist $k\geq 0$ such that $f^{(n)}(0)= 0$ for all $n\geq k.$ Where $f^{(n)}(0)$ means its n th derivative at zero.

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closed as off-topic by Daniel W. Farlow, anomaly, Alice Ryhl, Christopher, user98602 May 12 '15 at 18:47

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  • $\begingroup$ The given condition just says that at each reals either function value is zero or its derivative is zero at that point. Then how to prove that after some stage function derivative is zero at zero? $\endgroup$ – neelkanth May 12 '15 at 16:07
  • $\begingroup$ Please do not simply dump your homework problems here. Instead, please indicate what you're stuck on, what you've tried so far, and the context of the question. $\endgroup$ – anomaly May 12 '15 at 16:34
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Hint: for $n=0,1,2,\cdots,$  let $E_n= \{a \in \mathbb {R}: f^{(n)}(a) = 0\}.$ Argue that one of the $E_n$'s is uncountable and proceed. (There's a proof using Baire here, but we don't need to use that.)

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  • $\begingroup$ @YOGESH please spend more than just a few minutes thinking about this. $\endgroup$ – zhw. May 12 '15 at 16:50
  • $\begingroup$ yes uncountable because of set of reals is uncountable.. $\endgroup$ – neelkanth May 12 '15 at 16:56
  • $\begingroup$ ok its uncountable...now i make two case firstly is bounded set then the problem is solved....second one its unbouded uncountable then how to proceeds?? $\endgroup$ – neelkanth May 12 '15 at 17:38
  • $\begingroup$ ok ok i got the point..... now by using identity theorem.....function will be a polynomial...... am i correct ...please tell am i correct.?? $\endgroup$ – neelkanth May 12 '15 at 17:45

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