Formal power series, the Chain Rule and the Product Rule.

Definitons

Let $$\mathbb{C}[[x]] := \left\{ \sum_{n\geq 0} a_n x^n : a_n \in \mathbb{C} \right\}$$ be the set of formal power series of $x$.

Exercise

i)

If $F_1(x)$ and $F_2(x)$ are power series in $\mathbf{C}[[x]]$, and $F_3(x)=F_1(x)F_2(x)$ is their product, then their formal derivatives satisfy the usual ‘derivative of the product’ formula
$$
F_3′(x)=F_1(x)F_2′(x)+F_1′(x)F_2(x).
$$

I tried making a proof using just the substitution and product rules, but I can’t seem to come to an answer. Thanks in advance.

Solutions Collecting From Web of "Formal power series, the Chain Rule and the Product Rule."

It’s good practice working with summations.

Let $F_1(x)=\sum_{n\ge 0}a_nx^n$, $F_2(x)=\sum_{n\ge 0}b_nx^n$, and $F_3(x)=F_1(x)F_2(x)=\sum_{n\ge 0}c_nx^n$, where we know that

$$c_n=\sum_{k=0}^na_kb_{n-k}\;.$$

Now

$$\begin{align*}
F_1(x)F_2′(x)&=\sum_{n\ge 0}a_nx^n\sum_{n\ge 0}nb_nx^{n-1}\\
&=\sum_{n\ge 0}a_nx^n\sum_{n\ge 0}(n+1)b_{n+1}x^n\\
&=\sum_{n\ge 0}\sum_{k=0}^na_k(n-k+1)b_{n-k+1}x^n
\end{align*}$$

and

$$\begin{align*}
F_1′(x)F_2(x)&=\sum_{n\ge 0}na_nx^{n-1}\sum_{n\ge 0}b_nx^n\\
&=\sum_{n\ge 0}(n+1)a_{n+1}x^n\sum_{n\ge 0}b_nx^n\\
&=\sum_{n\ge 0}\sum_{k=0}^n(k+1)a_{k+1}b_{n-k}x^n\;,
\end{align*}$$

so the coefficient of $x^n$ in $F_1(x)F_2′(x)+F_1′(x)F_2(x)$ is

$$\begin{align*}
&\sum_{k=0}^n(n-k+1)a_kb_{n-k+1}+\sum_{k=0}^n(k+1)a_{k+1}b_{n-k}\\\\
&\quad=(n+1)a_0b_{n+1}+\sum_{k=1}^n(n-k+1)a_kb_{n-k+1}+\sum_{k=0}^{n-1}(k+1)a_{k+1}b_{n-k}+(n+1)a_{n+1}b_0\\\\
&\quad=(n+1)a_0b_{n+1}+\sum_{k=1}^n(n-k+1)a_kb_{n-k+1}+\sum_{k=1}^nka_kb_{n-k+1}+(n+1)a_{n+1}b_0\\\\
&\quad=(n+1)a_0b_{n+1}+\sum_{k=1}^n(n+1)a_kb_{n-k+1}+(n+1)a_{n+1}b_0\\\\
&\quad=(n+1)\sum_{k=0}^{n+1}a_kb_{n+1-k}\\\\
&\quad=(n+1)c_{n+1}\;,
\end{align*}$$

which is of course the coefficient of $x^n$ in $$F_3′(x)=\sum_{n\ge 0}(n+1)c_{n+1}x^n\;.$$

We want to show that
$$
F_3′(x) = F_1(x)F_2′(x) + F_1(x)F_2′(x)
$$
In other words, we want to show that the coefficients of $x^n$ on either side are the same. However, by the Cauchy Product, the coefficient of $x^n$ on the LHS and on the RHS is only dependent on finitely many terms in the series describing $F_1(x)$ and $F_2(x)$.

So we may assume without loss of generality that $F_1(x), F_2(x)$ and $F_3(x)$ are polynomials, where we know it is true.