# Multiplicative Inverse of a Power Series

For a formal power series $$F(x) = \sum p_i x^i$$ a multiplicative inverse of $F$ exists iff $p_0 \neq 0$. The inverse $\sum q_i x^i$ satisfies the recursion $$q_0 =\frac{1}{p_0}\\ q_{n} = -\frac{1}{p_0}\sum_{0 \leq i < n}p_{n-i}q_{i}$$

What's the closed form of this recurrence? Writing out a bunch of terms hasn't yet revealed to me a pattern. It's helpful to assume $p_0 = 1$.

Write $F = 1 + F'$ where $F'$ has constant coefficient $0$. The inverse $F^{-1}$ satisfies $$(1 + F')F^{-1} = 1$$ Therefore $$F^{-1} = (1 + F')^{-1}$$ Now expand using the generalized binomial theorem, \begin{align*} F^{-1} & = \displaystyle\sum_{n \geq 0} \binom{-1}{n}(F')^n\\ &= \displaystyle\sum_{ n \geq 0} (-1)^n (F')^n \end{align*} Letting $F = \displaystyle\sum_{n \geq 0} a_nx^n$, $F' = \displaystyle\sum_{n \geq 1} a_{n}x^n$,

\begin{align*} F^{-1} &= \displaystyle\sum_{n \geq 0} (-1)^n \Big(\sum_{i \geq 1} a_{i}x^i\Big)^n\\ &= \displaystyle\sum_{n \geq 0} (-1)^n \Big(\sum_{\substack{ \beta_1, \beta_2,\dots,\\\sum_{i} \beta_i = n}} \binom{n}{\beta_1, \beta_2,\dots} \prod_{i \geq 1} (a_{i}x^i)^{\beta_i} \Big)\\ &= \displaystyle\sum_{n \geq 0}\sum_{\substack{\beta_1, \beta_2,\dots,\\\sum_{i} \beta_i = n}} \Big((-1)^n\binom{n}{\beta_1,\beta_2,\dots}\prod_{i \geq 1} a_{i}^{\beta_i}\Big) x^{\sum_i i\beta_i} \end{align*}

where the inner sum is over all natural number sequences $\langle \beta_i\rangle$, and $\binom{n}{\beta_1,\beta_2, \dots}$ is a multinomial coefficient. Grouping terms by exponent on $x$, we have the somewhat-closed form $$F^{-1} = \displaystyle\sum_{n \geq 0} \Bigg(\sum_{\substack{\beta_1, \beta_2, \dots\\\sum_{i}i\beta_i= n}} (-1)^{\sum_i \beta_i}\binom{\sum_i \beta_i}{\beta_1, \beta_2, \dots} \prod_{i \geq 1} a_i^{\beta_i}\Bigg) x^n$$

For the multiplicative inverse, applying Faà di Bruno's formula (Wikipedia: Bell polynomials - Faà di Bruno's formula) yields the following formula.

$$q_{i}=\frac{1}{i!}\sum_{k=0}^{i}(-1)^{k}k!p_{0}^{-(k+1)}B_{i,k}(1!p_{1},2!p_{2},...,(i-k+1)!p_{i-k+1})$$

$B_{i,k}$: Partial exponential Bell polynomial (Wikipedia: Bell polynomials - Exponential Bell polynomials)

This is written e.g. in Singh, M.: nth-order derivatives of certain inverses and the Bell polynomials. J. Phys. A 23 (1990) (12) 2307-2313.

$$q_{i}=\sum_{k=0}^{i}(-1)^{k}p_{0}^{-(k+1)}\hat{B}_{i,k}(p_{1},p_{2},...,p_{i-k+1})$$

$\hat{B}_{i,k}$: Partial ordinary Bell polynomial (Wikipedia: Bell polynomials - Ordinary Bell polynomials)

For certain formal power series, there is a general formula for $B_{i,k}$ and $\hat{B}_{i,k}$.

I wrote an article "On partial Bell polynomials for the higher derivatives of composed functions".

The formula given by Chris Jones seems to be the best available, and I just want to make a couple of remarks.

1) With $$F = 1 + \displaystyle\sum_{n \geq 1} a_nx^n$$, the formula \begin{align*} F^{-1} &= \displaystyle\sum_{n \geq 0} (-1)^n \Big(\sum_{i \geq 1} a_{i}x^i\Big)^n \end{align*}

is just a consequence of the geometric series expansion:

$$\frac1{1+w} = \sum_{n\geq 0} (-1)^n w^n,$$

with $$w=\sum_{i \geq 1} a_{i}x^i$$, ie, the "generalized binomial theorem" is not really needed.

2) The formula for $$F^{-1}$$ can be rewritten in terms of partitions of the natural number $$n$$.

Indeed, partitions of $$n$$ (into equal or different non-zero parts) are in bijection with sequences of non-negative integers $$(\beta_1,\cdots,\beta_n)$$, satisfying $$\sum_{i}i\beta_i= n.$$ This gives the interpretation of each $$\beta_i$$ as the number of parts of size equal to $$i$$ (it can, of course, be zero). Moreover $$\sum_{i}\beta_i$$ becomes now the total number of parts of the partition (usually called the length of the partition).

Then, the formula admits this shorter version: $$F^{-1} = \displaystyle\sum_{n \geq 0} \Bigg(\sum_{\beta=(\beta_i)\in P_n} (-1)^{|\beta|} |\beta| ! \prod_{i=1}^n \frac{a_i^{\beta_i}}{\beta_i !} \Bigg) x^n$$

where $$P_n$$ denotes the finite set of all partitions of $$n$$, and $$|\beta|$$ is the length of the partition $$\beta = (\beta_1,\cdots,\beta_n)\in P_n$$