# 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 $$\beta = (\beta_1,\cdots,\beta_n)\in P_n$$

The Wronski's formula \eqref{f(t)g(t)=1=determ} below is a best answer to this question.

If $$a_0\ne0$$ and $$P(t)=a_0+a_1t+a_2t^2+\dotsm$$ is a formal series, then the coefficients of the reciprocal series $$\frac{1}{P(t)}=b_0+b_1t+b_2t^2+\dotsm$$ are given by $$$$\label{f(t)g(t)=1=determ}\tag{1} b_r=\frac{(-1)^r}{a_0^{r+1}} \begin{vmatrix} a_1 & a_0 & 0 & 0 & \dotsm & 0& 0 & 0\\ a_2 & a_1 & a_0 & 0 & \dotsm & 0 & 0& 0\\ a_3 & a_2 & a_1 & a_0 & \dotsm & 0 & 0 & 0\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots &\vdots & \vdots\\ a_{r-2} & a_{r-3} & a_{r-4} & a_{r-5} & \dotsm & a_1 & a_0 & 0\\ a_{r-1} & a_{r-2} & a_{r-3} & a_{r-4} & \dotsm & a_2 & a_1 & a_0\\ a_{r} & a_{r-1} & a_{r-2} & a_{r-3} & \dotsm & a_3 & a_2 & a_1 \end{vmatrix}, \quad r=1,2,\dotsc.$$$$

Wronski's formula \eqref{f(t)g(t)=1=determ} can be found on page 17 Theorem 1.3 in the book [1], on page 347 in the paper [2], in Lemma 2.4 of the paper [3], in Section 2 of the paper [4], and the paper [5].

References

1. P. Henrici, Applied and Computational Complex Analysis, Volume 1, Pure and Applied Mathematics, Wiley-Interscience [John Wiley & Sons], New York-London-Sydney, 1974.
2. A. Inselberg, On determinants of Toeplitz-Hessenberg matrices arising in power series, J. Math. Anal. Appl. 63 (1978), no. 2, 347--353; available online at https://doi.org/10.1016/0022-247X(78)90080-X.
3. Feng Qi and Robin J. Chapman, Two closed forms for the Bernoulli polynomials, Journal of Number Theory 159 (2016), 89--100; available online at https://doi.org/10.1016/j.jnt.2015.07.021.
4. H. Rutishauser, Eine Formel von Wronski und ihre Bedeutung fur den Quotienten-Differenzen-Algorithmus, Z. Angew. Math. Phys. 7 (1956), 164--169; available online at https://doi.org/10.1007/BF01600787. (German)
5. M. H. Wronski, Introduction a la Philosophie des Mathematiques: Et Technie de l'Algorithmie, Chez COURCIER, Imprimeur-Libraire pour les Matheooatiqtte, quai des Augustins, n°57, Paris, 1811; available online at https://gallica.bnf.fr/ark:/12148/bpt6k6225961k. (French)

I think that the solution can be derived from the formula $$$$\label{Sitnik-Bourbaki}\tag{1} \frac{\textrm{d}^k}{\textrm{d}z^k}\biggl(\frac{u}{v}\biggr) =\frac{(-1)^k}{v^{k+1}} \begin{vmatrix} u & v & 0 & \dotsm & 0\\ u' & v' & v & \dotsm & 0\\ u'' & v'' & 2v' & \dotsm & 0\\ \dotsm & \dotsm & \dotsm & \ddots & \dotsm\\ u^{(k-1)} & v^{(k-1)} & \binom{k-1}1v^{(k-2)} & \dots & v\\ u^{(k)} & v^{(k)} & \binom{k}1v^{(k-1)} & \dots & \binom{k}{k-1}v' \end{vmatrix}$$$$ where $$u=u(z)$$ and $$v=v(z)\ne0$$ are differentiable functions. For more information on the formula \eqref{Sitnik-Bourbaki}, please refer to another answer of mine at the site https://math.stackexchange.com/a/4261705/945479.