I saw the other answers to this question from a year ago but I'm still quite confused. I have tried using identity theorem and several of its corollaries, factoring the nth derivative of f, and more, but every time I assert something I realize that it relies only on one point z having a derivative vanish, rather than all points in $\mathbb{C}$ having some derivative vanish. Any insight into the start of this problem, or a theorem I might use?
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1Would you care to share a link to the question? – Apr 04 '16 at 01:41
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http://math.stackexchange.com/questions/943952/prove-that-f-is-a-polynomial-if-one-of-the-coefficients-in-its-taylor-expansio – mathiest Apr 04 '16 at 01:43
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Does $n$ depend on $z$? – lhf Apr 04 '16 at 01:48
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Yes, n depends on z. – mathiest Apr 04 '16 at 02:01
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Hint: Show that since $D=\{z: |z|\leq 1\}$ is uncountable, there must be an $n$ such that $f^{(n)}(z)=0$ for infinitely many $z\in D$.
Thomas Andrews
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2Would this mean that ${ z \in \mathbb{C} : f^{(n)}(z) = 0 }$ has a limit point in $\mathbb{C}$, so $f^{(n)} \equiv 0$ in $\mathbb{C}$ ? – mathiest Apr 04 '16 at 02:13
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Hint: Let $E_n = \{z\in \mathbb C : f^{(n)}(z) = 0\}.$ Then $\mathbb C = \cup_n E_n.$ Hence some $E_n$ is uncountable.
zhw.
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We can establish a relation $$ zR j \iff f^{(j)}(z) = 0 $$ This relation is left-total, so $\bigcup_{j \geq 0} R^{-1}[\{j\}] = \mathbb{C}$. Since $\mathbb{C}$ is uncountable, one of the $R^{-1}[\{j\}]$ is uncountable, so $f^{(j)}$ has uncountably many zeroes.
Since in $\mathbb{R}^n$, any uncountable set has a limit point, we can use An improvement to derive that $f^{(j)} = 0$.
Hence $f$ is a polynomial.
Henricus V.
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