There does not exist an entire function which satisfies $f({1\over n})={1\over 2^n}$?

Solution 1:

Hint: If $f(0) = 0$ and $f$ is not identically zero, we can find another entire function $g$ so that $f(z) = z^k g(z)$ for some positive integer $k$ and $g(0) \ne 0$.

Solution 2:

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.)

Solution 3:

Hints:

$$(1)\;\;\;\;\;f\left(\frac1n\right)\xrightarrow[n\to\infty]{}0$$

$$(2)\;\;\;\;\;\text{The zeros of entire functions are isolated.}$$