For a function $f:C\to C$ and $n \geq 1$, let $f^{(n)}$ denote the nth derivative of f and $f^{0}=f$. Let f be entire function such that for some $n \geq 1$, $f^{(n)}(\frac {1}{k}) = 0 \forall k=1,2,3 ...$. Show that f is polynomial
Can you tell me is this following correct? Given $\forall n \geq N$, we have $f^{(n)}(\frac {1}{k}) = 0$
As $f^{(n)}$ exists as $f$is analytic through out. So $f^{(n)}$ is continuous at z=0, So $f^{(n)}(0)=0 \forall n \geq N$.
$f(z)=\sum_{n=0}^{\infty} a_nz^n \implies f(z) = a_0+a_1z+....a_Nz^N$ --> polynomial of degree N
Pls correct me if am going wrong.
Your proof is not correct. The hypothesis only says 'for some $n$' and you are assuming that the hypothesis is 'for all $n \geq N$. Where did $N$ come form? Here is the correct proof: $f^{(n)}$ has zeros at the points $1,1/2,1/3,....$ so its zeros have have a limit point. Of course, $f^{(n)}$ is analytic. Hence $f^{(n)}\equiv 0$. Now you can use your power series argument to conclude that $f$ is a polynomial of degree at most $n$.