TITLE: Expand: $e^{\sin z}$ at $z = 0$
QUESTION [7 upvotes]: I need hints to find the Laurent expansion of $\displaystyle e^{\sin z}$ at $z = 0$ in a simpler way. I am getting double series which I can't simplify.
ADDED:: Can it be simpler that the below? I also need to find the radius of convergence.
$$\Large e^{\sin z} = e^{\sum_{k = 0}^\infty \frac{(-1)^nz^{2n+1}}{(2n+1)!}} = \sum_{n = 0}^\infty \frac{\small{B(1, -1/3!, \dots , (-1)^n/(2n+1)!)} z^n}{n!}$$

REPLY [6 votes]: Note that $\frac{\mathrm{d}}{\mathrm{d}x}e^{\sin(x)}=\cos(x)e^{\sin(x)}$. Equating the coefficients for the series on both sides, that is
$$
\sum_{k=0}^\infty(k+1)a_{k+1}x^k=\sum_{k=0}^\infty\frac{(-1)^k}{(2k)!}x^{2k} \sum_{k=0}^\infty a_kx^k
$$
we get the following recursion for the coefficients:
$$
a_{n+1}=\frac1{n+1}\sum_{k=0}^{\lfloor n/2\rfloor}\frac{(-1)^k}{(2k)!}a_{n-2k}
$$
Since $a_0=1$, we get
$$
e^{\sin(x)}=1+x+\frac12x^2+0x^3-\frac18x^4-\frac1{15}x^5-\frac1{240}x^6+\frac1{90}x^7+\dots
$$
Not working from the sequence of coefficients, but just knowing that the radii of convergence of both $\sin(x)$ and $e^x$ are infinite, the radius of convergence for the composition will also be infinite.

Generalization on the Same Idea
To get a result similar to the Exponential formula cited by Michael Hardy, we can use the preceding idea to get a recursion for general $f$. Assuming that $f(0)=0$, and $\displaystyle f(x)=\sum_{k=1}^\infty f_kx^k$
we get the following recursion for $\displaystyle e^{f(x)}=\sum_{n=0}^\infty a_nx^n$:
$$
a_0=1\quad\text{and}\quad a_{n+1}=\frac1{n+1}\sum_{k=0}^n(k+1)f_{k+1}a_{n-k}
$$