formal power series derivative multiplicative inverse

Question

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.

Answer 1

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 $g$.

Answer 2

It is unclear. But if you have a formal power series $f(x) = \sum_{n=0}^\infty c_n x^n$ then $\frac{1}{f(x)}$ is a formal power series iff $c_0 \ne 0$.

In that case, using the "geometric series" $$\frac{1}{f(x)} = \frac{\frac{1}{c_0}}{1-(1-\frac{f(x)}{c_0})} = \frac{1}{c_0}\sum_{k=0}^\infty \left(1-\frac{f(x)}{c_0}\right)^k$$ $$ = c_0+\sum_{k=1}^\infty \sum_{n=k}^\infty x^n \sum_{n=m_1+\ldots+m_k, m_j > 0}\prod_{j=1}^k (-1)^k \frac{c_{m_j}}{c_0}= \sum_{n=0}^\infty a_n x^n $$ where $$a_n= \sum_{k=1}^n\sum_{n=m_1+\ldots+m_k, m_j > 0}\prod_{j=1}^k (-1)^k \frac{c_{m_j}}{c_0}$$ and $\sum_{n=m_1+\ldots+m_k, m_j > 0}$ means summing over all the partitions of $n$ in $k$ terms, where the order counts and each term is $> 0$.