TITLE: There does not exist an entire function which satisfies $f({1\over n})={1\over 2^n}$?
QUESTION [7 upvotes]: There does not exist an entire function which satisfies $f({1\over n})={1\over 2^n}$, what I tried is if possible then define $g(z)=f(z)-{1\over 2^{1\over z}}$ Then $g({1\over n})=0$ and so $g(z)$ is entire and its $0$ set has limit point in it and so $f(z)={1\over 2^{1\over z}}$ which is not analytic at $0$?
Please help!
Edit: OOps! the way I defined $g(z)$ that is not entire! could any one give me hint?

REPLY [11 votes]: There is no  $f$, analytic in a neighborhood $U$ of $0$, with $f\bigl({1\over n}\bigr)=2^{-n}$ for all $n\geq1$, let alone an entire function with this property.
Proof. Such an $f$ would be not identically zero. By general principles about analytic functions there would exist an $r\geq0$ and a function $g$, analytic in $U$, with
$$f(z)=z^r \>g(z)\quad(z\in U);\qquad g(0)=:a\ne0\ .$$
It follows that
$$f\bigl({\textstyle{1\over n}}\bigr)\ 2^n={2^n\over n^r}\> g\bigl({\textstyle{1\over n}}\bigr)\to\infty\qquad(n\to\infty)\ ,$$
contradicting our basic assumption about $f$.
(Ayman Hourieh beat me by 9 minutes.)