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I am just really starting to get into formal power series and understanding them. I'm particularly interested in looking at the coefficients generated by the inverse of a formal power series: $$\left(\sum_{n\ge 0}a_nx^n\right)^{-1}=\sum_{n\ge 0}b_nx^n$$

I first thought that my approach would be looking at $$\frac{1}{\sum_{n\ge 0}a_nx^n}$$

But I'm more thinking that since we know that a series is invertible in the ring if $a_0$ is invertible in the ring of coefficients. Thus, since if we assume it is, and since the unit series is $\{1,0,0,0,....\}$ then we have

$$\left(\sum_{n\ge 0}a_nx^n\right)\left(\sum_{n\ge 0}b_nx^n\right)=1$$

Thus we know that $a_0b_0=1$ and thus $b_0=\frac1{a_0}$. And for the remaining terms we are just looking at the convolution generated by the Cauchy Product and so

$$0=\sum_{j=0}^ka_jb_{k-j}$$ $$-a_0b_k=\sum_{j=1}^ka_jb_{k-j}$$ $$b_k=\frac{-1}{a_0}\sum_{j=1}^ka_jb_{k-j}$$ And thus we have a recursive definition.

Is there another approach that defines the numbers $b_k$ without recursive means? Are you forced to only recursive methods when operating on the ring of formal power series to calculate coefficents?

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    $\begingroup$ If there was an easy method, I think there'd be a nice equation for the coefficients of $\frac x{e^x-1}=1/(1+\frac1{2!}x+\frac1{3!}x^2+\dotsb)$. (The coefficients of $\frac x{e^x-1}$ are the Bernoulli numbers.) $\endgroup$ Commented May 3, 2015 at 15:25
  • $\begingroup$ that is interesting. $\endgroup$
    – Iceman
    Commented May 3, 2015 at 16:09
  • $\begingroup$ This question has answers here: math.stackexchange.com/questions/710252/… and here: mathoverflow.net/questions/53384/… . $\endgroup$
    – IV_
    Commented Dec 17, 2019 at 18:50

6 Answers 6

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just a couple of thoughts. the approach you indicate seems more useful and elegant than one based on, say $$ \frac1{1-xP} = 1 +xP+xP^2+\cdots $$ another method might be to use $$ \begin{align} D^1(f^{-1}) &= -f^{-2}f_1 \\ D^2(f^{-1}) &= -f^{-3}(ff_2-2f_1^2) \\ &\cdots \end{align} $$ to build a McLaurin expansion

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    $\begingroup$ in the first equation, shouldn't it read ...+x²P²+... (or ...+(xP)²+...) ? $\endgroup$
    – Max
    Commented Dec 16, 2022 at 15:16
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The followig Wronski's formula \eqref{f(t)g(t)=1=determ} 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 \begin{equation}\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. \end{equation}

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)
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As closed a form as I could get I posted here. It looks pretty ugly... but I'm not sure how much prettier it can get.

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You are looking for the reciprocal of a power series.

Applying Faà di Bruno's formula (Wikipedia: Bell polynomials - Faà di Bruno's formula) yields the following formula.

$$b_{n}=\frac{1}{n!}\sum_{k=0}^{n}(-1)^{k}k!a_{0}^{-(k+1)}B_{n,k}(1!a_{1},2!a_{2},...,(n-k+1)!a_{n-k+1})$$

$B_{n,k}$: 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.

$$b_{n}=\sum_{k=0}^{n}(-1)^{k}a_{0}^{-(k+1)}\hat{B}_{n,k}(a_{1},a_{2},...,a_{n-k+1})$$

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

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

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The question is very interesting, and Chris Jones gave a very nice answer (not ugly at all!) that can be interpreted in terms of partitions, as follows.

Without loss of generality, we can assume the zeroth coefficient to be $a_0=1$, so that we are inverting:

$$F = 1 + \displaystyle\sum_{n \geq 1} a_nx^n ,$$

and then \begin{align*} F^{-1} &= \displaystyle\sum_{n \geq 0} (-1)^n \Big(\sum_{i \geq 1} a_{i}x^i\Big)^n \end{align*}

(a consequence of the geometric series expansion: $\frac1{1+w} = \sum_{n\geq 0} (-1)^n w^n$ with $w=\sum_{i \geq 1} a_{i}x^i$.)

Next, use the multinomial theorem to expand the inner sum:

\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 $$F^{-1} = \sum_{n\geq 0} b_n x^n = \displaystyle\sum_{n \geq 0} \Bigg(\sum_{\substack{\beta_1, \beta_2, \dots\\\sum_{i}i\beta_i= n}} (-1)^{\sum_i \beta_i}\binom{\sum_i \beta_i}{\beta_1, \beta_2, \dots} \prod_{i \geq 1} a_i^{\beta_i}\Bigg) x^n $$

Finally, 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 for the coefficients $b_n$ of $F^{-1}$ is: $$b_n = \sum_{\beta=(\beta_i)\in P_n} (-1)^{|\beta|} |\beta| ! \prod_{i=1}^n \frac{a_i^{\beta_i}}{\beta_i !} $$

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

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Wronski's formula can be derived from the formula \begin{equation}\label{Sitnik-Bourbaki}\tag{2} \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}, \end{equation} where $u=u(z)$ and $v=v(z)\ne0$ are differentiable functions. For more details, please refer to Section 5 in the paper [1] below. So the formula \eqref{Sitnik-Bourbaki} is also an answer to this question.

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.

[1] Feng Qi, On signs of certain Toeplitz–Hessenberg determinants whose elements involve Bernoulli numbers, Contributions to Discrete Mathematics 18 (2023), no. 2, 48–59; available online at https://doi.org/10.55016/ojs/cdm.v18i2.73022.

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