Is there an analytic function $f$ on $B(0,1)$ the open ball with radius 1 such that $f(1/n)=e^{-n}$ for $n=2,3,4,...$?
Solution 1:
Here's another way to do it. Assume that such an $f$ exists. Then we can write $$ f(z) = z^k g(z) $$ for some (positive due to the assumptions) integer $k$ and some holomorphic function $g$ on $B(0,1)$ with $g(0) \neq 0$. Plug in $z=1/n$: $$ e^{-n} = f(1/n) = \frac{1}{n^k} g(1/n) $$ i.e. $$ g(1/n) = n^k e^{-n}. $$ Let $n \to \infty$. This gives (by continuity of $g$) that $g(0)=0$, which is a contradiction.