Prove relation between coefficients of power series 
If $f(x)=\sum_{m = 0}^{\infty}c_mx^m$ has positive radius of convergence and $e^{f(x)}=\sum_{m = 0}^{\infty}d_mx^m$, show that $$nd_n=\sum_{i=1}^nic_id_{n-i}$$
  James Stewart Calculus

 
My answer below, others welcome.
 A: If $f(x)=\sum_{m = 0}^{\infty}c_mx^m$ has positive radius of convergence and $e^{f(x)}=\sum_{m = 0}^{\infty}d_mx^m$, we have:
$$
\frac{d}{dx}e^{f(x)} = \frac{d}{dx}\sum_{m = 0}^{\infty}d_mx^m.
$$
Using the chain rule on the left of this equation and termwise differentiation on the right, this gives:
$$
\left(\frac{d}{dx}f(x)\right)e^{f(x)} = \sum_{m = 1}^{\infty}md_mx^{m-1}.
$$
Differentiating the power series for $f(x)$ termwise and using the equation for $e^{f(x)}$, the above is:
$$
\left(\sum_{m = 1}^{\infty}mc_mx^{m-1}\right)\left(\sum_{m = 0}^{\infty}d_mx^m \right)= \sum_{m = 1}^{\infty}md_mx^{m-1}.
$$
Multiplying out the product on the left gives:
$$
\sum_{m = 1}^{\infty}\left(\sum_{i = 1}^{m}ic_id_{m-i}\right)x^{m-1}= \sum_{m = 1}^{\infty}md_mx^{m-1}.
$$
Comparing the coefficients of $x^{n+1}$ gives the desired result:
$$nd_n=\sum_{i=1}^nic_id_{n-i}.$$
[Motivation: the appearance of $nd_n$ in the goal, strongly suggests that the power series $\sum d_nx^n$ has been differentiated. So see what happens if you differentiate both sides of the equation involving that power series.]
A: The sequence $c_n$ represents the taylor expansion of $f(x)$, so $c_n=\frac{f'^{(n)}(0)}{n!}$ where $f'^{(n)}$ is the nth derivative of $f$. Likewise, $d_n=\frac{g'^{(n)}(0)}{n!}$, where $g(x)=e^{f(x)}$. All $d_n=\frac{e^{f(0)}}{n!}k_n(0)$ for the value $k_n(0)$, where $k_n(0)=p(0)$ and $p(x)=\frac{g(x)'^{(n)}}{e^{f(x)}}=\frac{(e^{f(x)})'^{(n)}}{e^{f(x)}}$. Letting $k_n(x)=p(x)$ satisifies all the values for $k_n(0)$ and provides the useful identity
$$k_{n+1}(x)=f'(x)k_n+k_n'$$
Proof:$$k_n(x)=p(x)=\frac{(e^{f(x)})'^{(n)}}{e^{f(x)}} \implies
\\
e^{f(x)}k_n(x)=(e^{f(x)})'^{(n)}\implies
\\
f'(x)e^{f(x)}k_n(x)+e^{f(x)}k_n'(x)=(e^{f(x)})'^{(n+1)} \implies
\\
\frac{(e^{f(x)})'^{(n+1)}}{e^{f(x)}}=f'(x)k_n+k_n' \implies
\\
k_{n+1}(x)=f'(x)k_n+k_n'
$$
So $$nd_n=\sum_{i=1}^nic_id_{n-i} \\ \Longleftrightarrow n \frac{e^{f(0)}}{n!}k_n(0)=\sum_{i=1}^ni\frac{f'^{(i)}(0)}{i!}\frac{e^{f(0)}}{(n-i)!}k_{n-i}(0)\\ \Longleftrightarrow k_n(0)=\sum_{i=1}^n\left(\begin{array}{c}n-1\\ \Longleftrightarrow i-1\end{array}\right)f'^{(i)}(0)k_{n-i}(0)\\ \Longleftrightarrow k_{n+1}(0)=\sum_{i=1}^{n+1}\left(\begin{array}{c}n\\ i-1\end{array}\right)f'^{(i)}(0)k_{n+1-i}(0)$$ 
I will prove the last one and have proven the first. But first, I will prove by induction a helper theorem: $$k_n(x)'=\sum_{i=1}^{n}\left(\begin{array}{c}n\\ i\end{array}\right)f'^{(i+1)}(x)k_{n-i}(x)$$ Base case: $$k_1'(x)=\left(\begin{array}{c}1\\ 1\end{array}\right)f''(x)k_0(x)$$ is true, as $k_0(x)=\frac{e^{f(x)}}{e^{f(x)}}=1$ and $k_1(x)=\frac{f'(x)e^{f(x)}}{e^{f(x)}}=f'(x)$.
Assuming the theorem is true for case $n$, we have 
$$k_n'(x)=\sum_{i=1}^{n}\left(\begin{array}{c}n\\ i\end{array}\right)f'^{(i+1)}(x)k_{n-i}(x) 
\\
 =\sum_{i=1}^{n}\left(\begin{array}{c}n\\ i\end{array}\right)f'^{(i+1)}(x)\left(\frac{k_{n-i+1}(x)-k_{n-i}'(x)}{f'(x)}\right)$$ 
Because $k_{p+1}(x)=k_p'(x)+f'(x)k_p(x) \implies k_p(x)=\frac{k_{p+1}(x)-k_{p}'(x)}{f'(x)}$. So 
$$f'(x)k_n'(x)
=
 \sum_{i=1}^{n}\left(\begin{array}{c}n\\ i\end{array}\right)f'^{(i+1)}(x)\left(k_{n-i+1}(x)-k_{n-i}'(x)\right)
=
\\
 \sum_{i=1}^{n}\left(\begin{array}{c}n\\ i\end{array}\right)f'^{(i+1)}(x)k_{n-i+1}(x)- \sum_{i=1}^{n}\left(\begin{array}{c}n\\ i\end{array}\right)f'^{(i+1)}(x)k_{n-i}'(x)
=
\\
 \sum_{i=1}^{n}\left(\left(\begin{array}{c}n+1\\ i\end{array}\right)-\left(\begin{array}{c}n\\ i-1\end{array}\right)\right)f'^{(i+1)}(x)k_{n-i+1}(x)- \sum_{i=1}^{n}\left(\begin{array}{c}n\\ i\end{array}\right)f'^{(i+1)}(x)k_{n-i}'(x)
= 
\\
\sum_{i=1}^{n}\left(\begin{array}{c}n+1\\ i\end{array}\right)f'^{(i+1)}(x)k_{n-i+1}(x)-\sum_{i=1}^{n}\left(\begin{array}{c}n\\ i-1\end{array}\right)f'^{(i+1)}(x)k_{n-i+1}(x)- \sum_{i=1}^{n}\left(\begin{array}{c}n\\ i\end{array}\right)f'^{(i+1)}(x)k_{n-i}'(x)
\\ 
\implies
\sum_{i=1}^{n}\left(\begin{array}{c}n+1\\ i\end{array}\right)f'^{(i+1)}(x)k_{n-i+1}(x)=
\\
 \sum_{i=1}^{n}\left(\begin{array}{c}n\\ i\end{array}\right)f'^{(i+1)}(x)k_{n-i}'(x)+\sum_{i=1}^{n}\left(\begin{array}{c}n\\ i-1\end{array}\right)f'^{(i+1)}(x)k_{n-i+1}(x)+f'(x)k_n'
 \implies \\
\sum_{i=1}^{n+1}\left(\begin{array}{c}n+1\\ i\end{array}\right)f'^{(i+1)}(x)k_{n+1-i}(x)=
\\
\sum_{i=1}^{n}\left(\begin{array}{c}n\\ i\end{array}\right)f'^{(i+1)}(x)k_{n-i}'(x)+
\sum_{i=1}^{n}\left(\begin{array}{c}n\\ i-1\end{array}\right)f'^{(i+1)}(x)k_{n-i+1}(x)+f'(x)k_n'(x)+f'^{(n+2)}(x)k_0(x)
=
\\
\sum_{i=1}^{n}\left(\begin{array}{c}n\\ i\end{array}\right)f'^{(i+1)}(x)k_{n-i}'(x)+
\sum_{i=0}^{n-1}\left(\begin{array}{c}n\\ i\end{array}\right)f'^{(i+2)}(x)k_{n-i}(x)+f'(x)k_n'(x)+f'^{(n+2)}(x)k_0(x)
=
\\
\sum_{i=1}^{n}\left(\begin{array}{c}n\\ i\end{array}\right)f'^{(i+1)}(x)k_{n-i}'(x)+
\sum_{i=1}^{n}\left(\begin{array}{c}n\\ i\end{array}\right)f'^{(i+2)}(x)k_{n-i}(x)+f'(x)k_n'(x)+f''(x)k_n(x)
$$
$$ $$
Now look at the above. It's the derivative of $\left(\sum_{i=1}^{n}\left(\begin{array}{c}n\\ i\end{array}\right)f'^{(i+1)}(x)k_{n-i}(x)\right)+f'(x)k_n(x) = k_n'(x)+f'(x)k_n(x)$ by the induction hypothesis. But $k_n'(x)+f'(x)k_n(x)=k_{n+1}(x)$, so the expression must equal $k_{n+1}'(x)$ and the theorem is proven.
Now $$k_n'(x)=\sum_{i=1}^{n}\left(\begin{array}{c}n\\ i\end{array}\right)f'^{(i+1)}(x)k_{n-i}(x)\\ \implies k_{n+1}(x)=k_n'(x)+f'(x)k_n(x)=\sum_{i=1}^{n}\left(\begin{array}{c}n\\ i\end{array}\right)f'^{(i+1)}(x)k_{n-i}(x)+f'(x)k_n(x)\\=\sum_{i=2}^{n+1}\left(\begin{array}{c}n\\ i-1\end{array}\right)f'^{(i)}(x)k_{n+1-i}(x)+f'(x)k_n(x)\\=\sum_{i=1}^{n+1}\left(\begin{array}{c}n\\ i-1\end{array}\right)f'^{(i)}(x)k_{n+1-i}(x)$$
QED.
