Let $f(z)=\exp(z^{2})$. Determine a power series of $f$, i.e the coefficients $a_{k}$ such that $f(z)=\sum_{k=0}^{\infty}a_{k}z^{k}$ for all $z\in \mathbb{C}$, where $a_{k}=f^{(k)}(0)/k!$.
First we have $$f^{(1)}(z)=2zf(z)\\f^{(2)}(z)=2f(z)+2zf^{(1)}(z)\\f^{(3)}(z)=4f^{(1)}(z)+2zf^{(2)}(z)\\ f^{(4)}(z)=6f^{(2)}(z)+2zf^{(3)}(z)\\ f^{(n)}(z)=2(n-1)f^{(n-2)}(z)+2zf^{(n-1)}(z)$$ for all $n\geq 5$. I have observed that $f^{(k)}(0)=0$ for all $k\in 2\mathbb{N}_{0}+1$. Thus we get $f(0)=1$, $f^{(2)}(0)=2$, $f^{(4)}(0)=12$, $f^{(6)}(0)=120$, $f^{(8)}(0)=1680$ etc. How do I then determine $a_{k}$?
If you have $$e^z = \sum_{n\ge 0} {z^n\over n!},$$ then you have $$e^{z^2} = \sum_{n\ge 0} {z^{2n}\over n!}.$$ These series both converge globally.