# Power series centered at $0$ which converge to $\sinh$

Determine all power series centered at $0$ (i.e. equal to $\sum_{n=0}^\infty a_n x^n$)which converge to the hyperbolic sine $\sinh: \mathbb{C} \to \mathbb{C}, z \mapsto \frac{\sin(iz)}{i}$.

My idea: Let $P(x)=\sum_{n=0}^\infty a_n x^n$ be a power series as claimed. Since the formula $P^{(k)}(x)=\sum_{n=k}^{\infty}k!\cdot a_n z^{n-k} \binom nk$ and $sinh(x)=P(x)$ locally at $0$, we see that $a_n=\frac{sinh^{(n)}(0)}{n!}$. Hence, $P(x)=\sum_{n=0}^\infty \frac{sinh^{(n)}(0)}{n!} x^n$ for any such power series. Is this proof valid?

You need to state that holomorphic power series are unique, and simplify using $$\sinh^{(n)}(0) = \begin{cases} 0 &\text{ if } n \equiv 0 \mod{2} \\ 1 &\text{ if } n \equiv 1 \mod{2} \end{cases}$$
• Does not uniqueness follow from the formula of the coefficient (i.e. $a_n=\frac{P^{(n)}(x)}{n!}$, where $P(x)=\sum_{n=0}^\infty a_n (x-x_0)^n$), which I have already derived? – bjn Mar 17 '16 at 18:08
• @bjn The point is if another power series converge to $\sinh$ then their coefficient must agree. – Henricus V. Mar 17 '16 at 18:08
• Yes, Can I not argue as above: if $Q(x)=\sum_{n=0}^{\infty} b_n x^n$ is another power series centered at $0$ and converges to the function $\sinh(x)$. Then by differentiating $Q$ n-times at $0$, we see that $\frac{\sinh^{(n)}(0)}{n!}=\frac{Q^{(n)} (0)}{n!}=b_n$. Hence the coefficient are completely determined by the condition that the power series centered at $0$ converges to $\sinh$ and in particular also the power series itself. – bjn Mar 17 '16 at 21:19
• @bjn Yes you can do that. As long as you show $P = Q$, your argument is valid. – Henricus V. Mar 22 '16 at 21:08