# Identity theorem for a holomorphic funtion defined near zero

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?

• "f in zero"? What do you mean by that? – Lord Shark the Unknown Dec 9 '18 at 17:59
• I don't know for sure what you want. This is my guess. Please change it if it is not what you wanted. – Batominovski Dec 9 '18 at 23:49
• Thank you, Your guess was right:) – Steven33 Dec 10 '18 at 7:30

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.
• Thank you for your answer:) I dont understand how you get $f(z)=z^3$ When you consider $1/2k$, then you get $f(1/2k)= 1/(2k)^3 = \frac{1}{8 z^3 }$? – Steven33 Dec 9 '18 at 18:53
• If $z=1/(2k)$, then $1/(2k)^3=z^3$. – user614671 Dec 9 '18 at 21:34
• 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 $f(1/2k+1)= .-1/(2k+1)^3 \ne 1/(2k+1)^3 =g(1/2k+1)$ – Steven33 Dec 10 '18 at 7:26
• 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 $f(z)=\frac{1}{1-z^2}$. – user614671 Dec 10 '18 at 11:55