Given the McLaurin series of $f(x)$, find the McLaurin series of $e^{f(x)}$ If I know the McLaurin series of
$$f(x)=\sum_{n=0}^\infty a_n x^n,$$
can I say something about the McLaurin series of
$$e^{f(x)}=\sum_{n=0}^\infty b_n x^n?$$
In other words, what is the relation between $a_n$ and $b_n$?
 A: Let $g(x) = e^{f(x)}=\sum_{n=0}^\infty b_n x^n$
By definition of McLaurin Series, $b_0=g(0)=e^{f(0)} = e^{a_0} $

Based on @zwim's comment, we use the fact that $g'(x)=f'(x)g(x)$ to write the following:
$\sum_{n=0}^{\infty} (n+1)b_{n+1}x^{n}  =  (\sum_{m=0}^{\infty} b_mx^{m})(\sum_{k=0}^{\infty} (k+1)a_{k+1}x^{k}) = $ t1*t2

Coefficient of $x^r$ on the LHS is $(r+1)b_{r+1}$
Coefficient of $x^r$ on the RHS is 
= $\sum_{k=0}^r $(Coefficient of $x^k$ in t1)* (Coefficient of $x^{r-k}$ in t2) 
= $\sum_{k=0}^r b_k*(r-k+1)*a_{r-k+1} $

That is, we have the relation for all $r$:  
$ b_{r+1} = (\sum_{k=0}^r b_k*(r-k+1)*a_{r-k+1})/((r+1)) $
We know $b_0$ and the above formula gives a way to compute rest of the $b_i$s using previous $b_i$s.
Its a recurrence relation. I am not sure of the explicit formula.
A: If $g(x)=e^{f(x)}$, then $a_n=\frac{f^{(n)}(0)}{n!}$ and $b_n=\frac{g^{(n)}(0)}{n!}$. So by Faà di Bruno's formula we have
\begin{align*}
b_n&=\frac{1}{n!}g^{(n)}(0)\\
&=\frac{1}{n!}\sum \frac{n!}{m_1!m_2!\dots m_n!}\exp^{(m_1+m_2+\dots+m_n)}(f(0))\prod_{j=1}^n (a_j)^{m_j}\\
&=e^{a_0}\sum \frac{\prod_{j=1}^n (a_j)^{m_h}}{m_1!m_2!\dots m_n!}
\end{align*}
where the sums in the second and third lines are taken over all tuples $(m_1,m_2,\dots,m_n)$ of nonnegative integers with $m_1+2m_2+3m_3+\dots+nm_n = n$.
