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I have to show, whether there is a holomorphic funtion $f$ defined in an open neighborhood of zero, such that: $$ f\left(\frac{1}{n}\right)=(-1)^n \frac{1}{n^3}$$ for all positive integer $n$.

My idea was to apply the identity theorem for holomorphic funtions. How can I do that? Maybe I must consider the subsequences $ \frac{1}{2k}, \frac{1}{2k+1}$. Can somebody help me?

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  • $\begingroup$ "f in zero"? What do you mean by that? $\endgroup$ – Lord Shark the Unknown Dec 9 '18 at 17:59
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    $\begingroup$ I don't know for sure what you want. This is my guess. Please change it if it is not what you wanted. $\endgroup$ – Batominovski Dec 9 '18 at 23:49
  • $\begingroup$ Thank you, Your guess was right:) $\endgroup$ – Steven33 Dec 10 '18 at 7:30
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While I don't know clearly what your question is, I assume you want to find all entire function $f$ such that $f(1/n)=(-1)^n/n^3$ for all positive integers $n$. I want to point out that $f$ does not exist, and your idea of considering $1/(2k)$ and $1/(2k+1)$ is a good idea.

Note that $f(z)=z^3$ for all $z$ of the form $1/(2k)$ where $k$ is a positive integer. Since the set of $1/(2k)$ has an accumulation point (namely, $0$) in $\Bbb{C}$, so $f(z)=z^3$ must hold for all $z\in \Bbb{C}$, but then the condition says that $f\big(1/(2k+1)\big)$ is $-1/(2k+1)^3$, not $1/(2k+1)^3$. This is a contradiction.

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  • $\begingroup$ Thank you for your answer:) I dont understand how you get $ f(z)=z^3$ When you consider $1/2k$, then you get $ f(1/2k)= 1/(2k)^3 = \frac{1}{8 z^3 } $? $\endgroup$ – Steven33 Dec 9 '18 at 18:53
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    $\begingroup$ If $z=1/(2k)$, then $1/(2k)^3=z^3$. $\endgroup$ – user614671 Dec 9 '18 at 21:34
  • $\begingroup$ Ok, you put $z=\frac{1}{2k} \Rightarrow g(z)=z^3 $, so $f(z)=g(z) \forall z=1/2k \Rightarrow f=g$ , but $ f(1/2k+1)= .-1/(2k+1)^3 \ne 1/(2k+1)^3 =g(1/2k+1) $ $\endgroup$ – Steven33 Dec 10 '18 at 7:26
  • $\begingroup$ Ok I think I understand that now:) $\endgroup$ – Steven33 Dec 10 '18 at 7:29
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    $\begingroup$ I think something is wrong with your last example. If $f(1/n)=\frac{1}{n^2-1}$, then $$f(1/n)=\frac{(1/n)^2}{1-(1/n)^2}.$$ Therefore, if $z=1/n$, then $f(z)=\frac{1}{1-z^2}$. $\endgroup$ – user614671 Dec 10 '18 at 11:55

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