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Suppose i have a formal power series

$$f(t)=e^t-1-t=\sum_{k=2}^\infty \frac{t^k}{k!}$$

How would I go about finding its compositional inverse? Trying to find the inverse of $f(t)=e^t-1-t$ i think is impossible to solve for $t$ w/o using like the Lambert W-function. However, is there a way to find it using power series? Perhaps there is a way to extract the coefficients then...

AveryJessup
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2 Answers2

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A compositional inverse $g(t)$ of the function $f(t)$ fulfilling $$f(g(t))=t=g(f(t))$$does not exist in the ring of formal power series.

We find in section 5.4 The Lagrange Inversion Formula of Enumerative Combinatorics 2 by R.P. Stanley:

  • The set $xK[[x]]$ of all formal power series $a_1x+a_2x^2+\cdots$ with zero constant term over a field $K$ forms a monoid under the operation of functional composition. The identity element of this monoid is the power series $x$.

    Recall ... that if $f(x)=a_1x+a_2x^2+\cdots\in K[[x]]$, then we call a power series $g(x)$ a compositional inverse of $f$ if $f(g(x))=g(f(x))=x$, in which case we write $g(x)=f^{\langle-1\rangle}(x)$. The following simple proposition explains when $f(x)$ has a compositional inverse.

  • Proposition 5.4.1 A power series $f(x)=a_1x+a_2x^2+\cdots\in K[[x]]$ has a compositional inverse $f^{\langle-1\rangle}$ if and only if $a_1\ne 0$, in which case $f^{\langle -1\rangle}(x)$ is unique. Moreover, if $g(x)=b_1x+b_2x^2+\cdots$ satisfies either $f(g(x))=x$ or $g(f(x))=x$, then $g(x)=f^{\langle-1\rangle}(x)$.

    Proof. Assume that $g(x)=b_1x+b_2x^2+b_3x^3+\cdots$ satisfies $f(g(x))=x$. We then have

    \begin{align*} \color{blue}{a_1(b_1x}+b_2x^2+b_3x^3+\cdots) +a_2(b_1x+b_2x^2+\cdots)^2+a_3(b_1x+\cdots)^3+\cdots\color{blue}{=x} \end{align*} Equating coefficients on both sides yields the infinite system of equations \begin{align*} \color{blue}{a_1b_1}&\color{blue}{=1}\\ a_1b_2+a_2b_1^2&=0\\ a_1b_3+2a_2b_1b_2+a_3b_1^3&=0\\ &\vdots\\ \end{align*} We can solve the first equation (uniquely) for $b_1$ if and only if $a_1\ne 0$.

Since $a_1=[t^1]f(t)=[t^1](e^t-1-t)=0$ we conclude $f(t)$ has no compositional inverse in the ring of formal power series.

Markus Scheuer
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  • Great answer! Moreover you can weaken the hp, i.e the compositional inverse is defined even if $b_0 \neq 0$, in such case $f$ has to be a polynomial, so you essentially have two cases to deal with, but it's the same idea. – jacopoburelli Aug 22 '23 at 08:16
  • @jacopoburelli: Many thanks. Here the author focus on the monoid $xK[[x]]$ as he uses this setting also in the following parts. – Markus Scheuer Aug 22 '23 at 19:23
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Given $f(x):=e^x-1-x,$ let $\;g(x) := x - 1/6 x^2 + 1/36 x^3 - 1/270 x^4 + 1/4320 x^5 + O(x^6).\;$ The function $\;\sqrt{2f(x)}=x + 1/6 x^2 + 1/36 x^3 + 1/270 x^4 + 1/2592 x^5 + O(x^6)\;$ is its inverse. Thus $h(x):=g(\sqrt{2x}) = \sqrt{2x} - 1/3x + 1/18\sqrt{2x}x -2/135x^2 + O(x^{5/2})\;$ is the Puiseux series inverse of $f(x).\;$ Solve for the coefficients of $\;g(x),\;$ and hence $h(x),\;$ one at a time using either one of the equations $\;f(g(x))=x^2/2\;$ or $\;g(\sqrt{2f(x)})=x.$

Somos
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