# Existence of a holomorphic function $f$ on the unit disc, such that $f(1/n) =\frac{(-1)^{n}}{n^2}$ for any integer $n>1$

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.

• 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... Oct 15, 2018 at 11:24
• 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. Oct 15, 2018 at 12:33
• 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. Oct 15, 2018 at 14:17

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 $$n$$.

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.

• Ah! Thanks! I see the grave folly in my approach Oct 15, 2018 at 14:32

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

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