TITLE: Does there exist an analytic function s.t. $f\left(\frac{1}{n}\right)=2^{-n}.$
QUESTION [8 upvotes]: Prove that there does not exist an analytic function $f$  in an unit disc containing $0$ such that $$f\left(\frac{1}{n}\right)=2^{-n}.$$
I tried by using Identity theorem.
Suppose that $f$ is analytic in the unit disc.
Consider the function $g(z)=f(z)-2^{-\frac{1}{z}}$.
Then the zeros of the function $g(z)$ are $\{\frac{1}{n}:n\in \mathbb N\}$ which has a limit point $0$ in the disc. So, $g$ is identically zero. Then, $f(z)=2^{-1/z}$. But, I am unable to find a point such that we arrive at a contradiction.
Please help to find it OR any other technique to prove the question.

REPLY [6 votes]: Note that
$$f(z) = 2^{-1/z}$$
is not continuous at $z=0$. Indeed: For $(z_n)_{n \in \mathbb{N}} \subseteq (0,\infty)$ such that $z_n \to 0$, we have
$$f(z_n) \to 0.$$
On the other hand, for $w_n := -z_n \to 0$,
$$f(w_n) \to \infty.$$
Since an analytic function has to be continuous, this contradicts our assumption.