Problem: Determine whether there exists an entire function $f$ such that $f'(\frac{1}{n}) = \frac{1}{n}f(\frac{1}{n})$ for every positive integer $n$. If there does exist such an $f$, is it unique?
Attempt at solution: it seems that there does exist such a function $f$, namely $f(z) = 0, \forall z \in \mathbb{C}$. However, I am having a hard time seeing if this is actually the only entire function that satisfies the property. Any help is appreciated!
If $f$ is entire, so is $g(z)=f'(z)-zf(z)$. The condition implies that $g(1/n)=0$ for all $n$. Since $1/n$ accumulates at $z=0$, we must have $g(z)=0$ for all $z$. Therefore $f$ satisfies the differential equation $$f'(z)=zf(z)$$
We can have $f\equiv 0$. If not, we can use separation of variables
$$(\ln(f(z))'=z$$
Which gives $f(z)=\exp(z^2/2+C)$