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Does there exist a holomorphic function $f$ on the unit disc, such that $$f\left(\frac{1}{n}\right)=\frac{(-1)^{n}}{n^2}$$ for any integer $n>1$?

Now, a function like $$g(z) = z^2(-1)^{1/z}$$ would be a holomorphic function that would agree with $f$, but clearly $g$ is not holomorphic at $z=0$. So, can I say that no such holomorphic function exists? Is this correct? If it is, then I'm having trouble formalising the argument. Any kind of assistance will be nice.

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    $\begingroup$ Hint: if $f$ is holomorphic on the unit disc, then you can take the limit of $f\left(z\right)$ as $z \rightarrow 0$ along any path and it should exist, but consider taking it along the real axis... $\endgroup$ Oct 15, 2018 at 11:24
  • $\begingroup$ If each interval of the $\left(\frac{1}{n+1},\frac{1}{n}\right)$ kind contains a zero of $f$ then $x=0$ is a point of accumulation of zeroes. On the other hand the zeroes of a non-zero holomorphic function are isolated. $\endgroup$ Oct 15, 2018 at 12:33
  • $\begingroup$ No, your argument is certainly not correct - saying a certain function would work but it doesn't work doesn't show that there is no such function. $\endgroup$ Oct 15, 2018 at 14:17

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Let $g(z)=f(z)-z^{2}$. Then $g(\frac 1 {2n})=0$ for all $n$. This implies that $g \equiv 0$ in $\{z: |z|<1\}$. Hence $f(z)={z^{2}}$ in $\{z: |z|<1\}$. But this contradicts the given equation for odd $n$.

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The existing answers and comments give correct proofs that there is no such $f$, but don't seem to address the question you ask, regarding whether your proof is correct. It's far from correct. Consider the following theorem, which uses the same reasoning:

Theorem. There does not exist a function $f$ holomorphic in the unit disk sch that $f(1/n)=1/n$ for $n=1,2,\dots$.

Proof: The function $f(z)=|z|$ would work, since $|1/n|=1/n$, but $f$ is not holomorphic.

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  • $\begingroup$ Ah! Thanks! I see the grave folly in my approach $\endgroup$ Oct 15, 2018 at 14:32
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Suppose that such a function $f$ exists. Then $f(\frac{1}{2n})=\frac{1}{4n^2}$ for all natural $n$. The identity theorem shows that we must have $f(z)=z^2$. But then we get that

$$ -\frac{1}{(2n+1)^2}=f(\frac{1}{2n+1})=\frac{1}{(2n+1)^2},$$

a contradiction !

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  • $\begingroup$ Yes, but I think all the conditions are met $\endgroup$ Oct 15, 2018 at 11:30

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