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"Does there exist an analytic function $f:D\to\mathbb{C}$ such that $f(1/n)=f(-1/n)=1/n^3$?"

This is one of the past qualifying exam problems that I am working on and I found that $f(0)=0$, $f^{(n)}(0)=0,n=1,2$, $f^{(3)}(0)=1$ using the definition of derivative of a function. I am trying to use a Taylor expansion at z=0 since f is analytic in $D=\{z\in \mathbb{C}||z|=1\}$. However I do not know how to use $f(1/n)=f(-1/n)=1/n^3$ to prove or disprove the existence of such function $f$.

Any help would be appreciated.

Thank you in advance.

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@Andres Caicedo: We know that $f$ is not a constant function but I still do not understand how to use the fact that zero being isolated to disprove the existence of such function $f$. Would you give me more clues? –  Yeonjoo Yoo Dec 30 '12 at 21:06

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No. Use identity theorem to show that if $g(z)=z^3$ and $h(z)=-z^3$ then $f=g=h↯.$

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@ Pambos: I understood your idea that if you can find a nbhd of $\frac{1}{n}$ s.t. $f=g$ and a nbhd of $\frac{-1}{n}$ s.t. $f=h$ then by the identity theorem that $f=g=h$ in $D$ which is a contradiction. However, how do you guarantee that such neighborhoods exist? –  Yeonjoo Yoo Dec 30 '12 at 21:01
I supposed that $D=\{z\in\mathbb C: |Z|<1\}$. What is $D$? –  P.. Dec 31 '12 at 8:43
D is D={z∈C:|Z|<1}. –  Yeonjoo Yoo Jan 1 '13 at 15:54
Read the section "An improvement" here. –  P.. Jan 1 '13 at 18:56

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