TITLE: formal power series derivative multiplicative inverse
QUESTION [0 upvotes]: Suppose $f(x)*g(x) = 1 $. Why does the product of their formal power series expansions (finding the coefficients by Taylor Series expansion) also equal $1$ ?
  Any help is appreciated.

REPLY [0 votes]: Let's restate what I believe to be the question:

Suppose that $f(x)$ and $g(x)$ are smooth functions on some interval $I \ni 0$ and that $f(x) g(x) = 1$ for $x \in I$.  Are the formal power series
  $$A := f(0) + f'(0) x + \frac{f''(0)}{2} x^2 + \cdots \in \mathbb{R}[[x]]$$
  and
  $$B := g(0) + g'(0) x + \frac{g''(0)}{2} x^2 + \cdots \in \mathbb{R}[[x]]$$
  inverses?

For $A$ and $B$ to be inverses means that their product,
\begin{align*}
A B &= [f(0) g(0)] + [f(0) g'(0) + f'(0) g'(0)] x + \cdots \\
 &= \sum_{n = 0}^\infty \left[\sum_{k = 0}^n \frac{1}{k!} f^{(k)}(0) \frac{1}{(n-k)!} g^{(n-k)}(0)\right] x^n
\end{align*}
is equal to one.
Starting from 
$$f(x) g(x) = 1$$
we can take $n$-th derivatives on each side, giving
$$\sum_{k=0}^n \binom{n}{k} f^{(k)}(x) g^{(n-k)}(x) = 0 \qquad (n \geq 1)$$
Plugging in $x = 0$ we get
$$\sum_{k=0}^n \binom{n}{k} f^{(k)}(0) g^{(n-k)}(0) = 0 \qquad (n \geq 1)$$
Plugging this back into the original expression we see that $A B = 1$ since
$$\binom{n}{k} = \frac{n!}{k! (n-k)!}$$
Note that we made no assumption that the power series for $f$ and $g$ actually converge to $f$ and $g$.
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