Solution
= Solution
The removable definition gives the entire <series>
$$
f(z)=\sum_{n=0}^{\infty}(-1)^n\frac{z^{2n}}{(2n+1)!}.
$$
Therefore
$$
f^{(k)}(0)=\begin{cases}0,&k\text{ odd},\\(-1)^{k/2}/(k+1),&k\text{ even}.
\end{cases}
$$
= Solution
The removable definition gives the entire <series>
$$
f(z)=\sum_{n=0}^{\infty}(-1)^n\frac{z^{2n}}{(2n+1)!}.
$$
Therefore
$$
f^{(k)}(0)=\begin{cases}0,&k\text{ odd},\\(-1)^{k/2}/(k+1),&k\text{ even}.
\end{cases}
$$