Finding holomorphic functions satisfying certain conditions Determine all the holomorphic functions $f$ from the open unit disk $D=\{z:|z|<1\}$ to whole complex plane which satisfies $f''(\frac{1}{n})+f(\frac{1}{n})=0$ for all $n=2,3,4,\ldots$
At first when i saw the differential equation, all i thought is the sine and cosine function, but the trouble is i dont know how should i use the given unit disk and the "$\frac{1}{n}$" in the parentheses.
Anyone can guide me or give some hints for me?
 A: Here's a hint: let $g = f'' + f$. What then do you know about $g$? (Think identity theorem.)
A: Some people are mentioning the "identity theorem", which is reported in this Wikipedia article to say that if two holomorphic functions defined on some open subset $D$ of the plane agree with each other on some non-empty open subset of $D$, then the agree on $D$.  But here what you're given is not agreement on some non-empty open subset of $D$, but satisfaction of a differential equation on a convergent sequence of points in $D$.
You know $f$ is holomorphic in $D$; hence according to the proof given in this article, we have
$$
f(z) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} z^n \tag 1
$$
for all $z$ in $D$ (generally for all $z$ closer to $0$ than the nearest point to which $f$ cannot be analytically continued, and there are no such ill-behaved points within the disk $D$).
Power series can be differentiated term by term within the interior of their disk of convergence.  So if you know $(1)$ and if you can show that $f''(0)+f(0)=0$, then all of the coefficients of the power series will be determined by $f(0)$ and $f'(0)$.  And then you have a series that you know converges to $f(0)\cos z + f'(0)\sin z$.
So how do you show that if $f''+f=0$ at $1/n$ for $n=2,3,4,\ldots$ then $f''+f=0$ at $0$?  Here I'd think about the smooth nature of functions defined by convergent power series.
