There does not exist an entire function which satisfies $f({1\over n})={1\over 2^n}$? 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?
 A: Hints:
$$(1)\;\;\;\;\;f\left(\frac1n\right)\xrightarrow[n\to\infty]{}0$$
$$(2)\;\;\;\;\;\text{The zeros of entire functions are isolated.}$$
A: 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.)
A: 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 $g(0) \ne 0$.
A: Since $f$ is supposed to be entire, it is continuous at $0$.  Let $n \to \infty$. Then we must have $f(0) = 0$.  Now let $n \to -\infty$. What is $f(0)$ now?
A: Another proof.
Lemma. Let $w\in\mathbb{C}$. If $u(z)$ is a holomorphic function in an open subset $G\subset \mathbb{C}$ that is not a constant function $w$, then elements in $u^{-1}(w)$ are isolated.
[Proof of the Lemma]: Without loss of generality, we may assume $w=0$. Suppose $u(z_0)=0$. Let the order of $z_0$ be $m$ (must be finite, otherwise $u\equiv 0$), then $u(z)=(z-z_0)^m h(z)$, where $h(z)$ is holomorphic in $G$ and $h(z_0)\ne0$. Then there is a neighbourhood $U\subset G$ of $z_0$ such that $h(z)\ne 0$ (thus $u(z)\ne0$) in $U\backslash \{z_0\}$. So $z_0$ is an isolated point of $u^{-1}(0)$. The proof of the Lemma is complete.
Suppose there is a holomorphic function $f$ in the open unit disk such that $f(\frac{1}{n})=\frac{1}{2^n}$ for $n=2,3,\ldots$. Then $2^{n}f(\frac{1}{n})=1$ for $n=2,3,\ldots$.
Note that $g(z)=2^{\frac{1}{z}}f(z)=\exp(\frac{1}{z}\log 2) f(z)$ is holomorphic in $D\backslash \{0\}$ where $D$ is the open unit disk, and $g(\frac1n)=1,n=2,3,\ldots$.
If $g$ is constant $1$ in $D\backslash\{0\}$, then $f(z)=2^{-\frac{1}{z}}=\exp(-\frac1z \log 2)$, by any textbook about functions of complex variable, $\exp(-\frac1z)$ has a nonremovable singularity at $z=0$, so is $f(z)$. So $f$ cannot be extended to a holomorphic function in $D$, a contradiction.
If $g$ is not constant. Note that $g^{-1}(1)$ has a limit point, a contradiction with the Lemma.
The proof is complete.
