Inverse of General Quintic by Determinants

I will work out how to potentially express the series reversion for a general quintic (and possibly the conditions that this fails) using the determinants of matrices. Let us define our quintic polynomial

\begin{equation} P(x)=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+a_{5}x^{5}\\ \end{equation}we can treat this is a series and do a series reversion about \(a_{0}\) giving

\begin{equation} P^{-1}(x)=\frac{x-a_{0}}{a_{1}}-\frac{a_{2}(x-a_{0})^{2}}{a_{1}^{3}}+\frac{(2a_{2}^{2}-a_{1}a_{3})(x-a_{0})^{3}}{a_{1}^{5}}+\frac{(-5a_{2}^{3}+5a_{1}a_{2}a_{3}-a_{1}^{2}a_{4})(x-a_{0})^{4}}{a_{1}^{7}}+\cdots\\ \end{equation}there is a clear pattern for the denominators and the \((x-a_{0})\) term. Can we express this sequence in the form

\begin{equation} P^{-1}(x)=\sum_{n=0}^{\infty}|M_{n}|\frac{(x-a_{0})^{n+1}}{a_{1}^{2n+1}}\\ \end{equation}where \(M_{n}\) is an \(n\times n\) matrix, and \(|\cdot|\) is the determinant. Then we have \(M_{0}\) as the empty matrix, define the determinant \(|M_{0}|=1\), then

\begin{equation} M_{1}=-[a_{2}]\\ M_{2}=-\begin{bmatrix}a_{1}&a_{2}\\ 2a_{2}&a_{3}\end{bmatrix}\\ M_{3}=-\begin{bmatrix}a_{1}&a_{2}&0\\ 0&a_{1}&5a_{2}\\ a_{2}&a_{3}&a_{4}\end{bmatrix}\\ M_{4}=-\begin{bmatrix}a_{1}&a_{2}&2a_{3}/3&0\\ 0&6a_{1}&14a_{2}&18a_{3}\\ 0&0&a_{1}/6&a_{2}\\ a_{2}&a_{3}&a_{4}&a_{5}\end{bmatrix}\\ M_{5}=-\begin{bmatrix}a_{1}&a_{2}&a_{3}&a_{4}&0\\ 0&7a_{1}&14a_{2}&7a_{3}&0\\ 0&0&a_{1}&3a_{2}&a_{3}\\ 0&0&0&a_{1}&a_{2}\\ a_{2}&a_{3}&a_{4}&a_{5}&0\end{bmatrix}\\ \end{equation}we could also swap rows and columns being careful about sign, there are likely other (potentially prettier) solutions. If we could work out a general pattern, then we would have a form for the inverse of a general quintic. However, it can be seen that if \(a_{1}\to 0\), the inverse function becomes ill defined.

We might quickly want to apply this method to lesser polynomials, for example the linear inverse

\begin{equation} P_{1}(x)=a_{0}+a_{1}x\\ \end{equation}then we have the inverse by series reversion

\begin{equation} P_{^{-1}}(x)=\frac{(x-a_{0})}{a_{1}}\\ \end{equation}we can note this is also the first term of the quintic expression above. For the quadratic we have

\begin{equation} P_{2}(x)=a_{0}+a_{1}x+a_{2}x^{2}\\ \end{equation}and by reversion the sequence

\begin{equation} P_{2}^{-1}(x)=\frac{(x-a_{0})}{a_{1}}-\frac{a_{2}(x-a_{0})^{2}}{a_{1}^{3}}+\frac{2a_{2}^{2}(x-a_{0})^{3}}{a_{1}^{5}}-\frac{5a_{2}^{3}(x-a_{0})^{4}}{a_{1}^{7}}+\cdots\\ \end{equation}where the coefficients \(1,1,2,5,\cdots\) are the Catalan numbers, A000108, which gives a form for this expression

\begin{equation} P_{2}^{-1}(x)=\sum_{n=0}^{\infty}\frac{(2n)!}{n!(n+1)!}\frac{(-1)^{n}a_{2}^{n}(x-a_{0})^{n+1}}{a_{1}^{2n+1}}=\frac{-a_{1}+\sqrt{a_{1}^{2}-4a_{0}a_{2}+4a_{2}x}}{2a_{2}}\\ \end{equation}which clearly resembles the quadratic formula, this however gives little opportunity to yield a determinant structure, it is clear that the \(a_{2}^{n}\) coefficients are arriving from the determinant. Next we may try the cubic equation

\begin{equation} P_{3}(x)=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}\\ \end{equation}with reversion

\begin{equation} P_{3}^{-1}(x)=\frac{(x-a_{0})}{a_{1}}-\frac{a_{2}(x-a_{0})^{2}}{a_{1}^{3}}+\frac{(2a_{2}^{2}-a_{1}a_{3})(x-a_{0})^{3}}{a_{1}^{5}}+\frac{(-5a_{2}^{3}+5a_{1}a_{2}a_{3})(x-a_{0})^{4}}{a_{1}^{7}}+\cdots\\ \end{equation}this is sharing three terms with the quintic equation reversion, it also gives rise to the use of the determinants. The coefficient of \(a_{2}^{n}\) in each term appears to be the Catalan numbers as in \(P_{2}^{-1}(x)\). We can try to derive the series of matricies \({}_{3}M_{n}\) which describe the cubic pattern,

\begin{equation} {}_{3}M_{1}=-[a_{2}]\\ _{3}M_{2}=-\begin{bmatrix}a_{1}&2a_{2}\\ a_{2}&a_{3}\end{bmatrix}\\ _{3}M_{3}=-\begin{bmatrix}a_{1}&a_{2}&0\\ 0&a_{1}&5a_{2}\\ a_{2}&a_{3}&0\end{bmatrix}\\ \end{equation}which are the same, if not similar. We now note that there seems to always be a coeffient that is the current Catalan number in the matrix, and it is on \(a_{2}\). We might be better off assiging these coefficients to the lower left corner which is consistently \(a_{2}\),

\begin{equation} {}_{3}M_{1}=-[a_{2}]\\ _{3}M_{2}=-\begin{bmatrix}a_{1}&a_{2}\\ 2a_{2}&a_{3}\end{bmatrix}\\ _{3}M_{3}=-\begin{bmatrix}a_{1}&a_{2}&0\\ 0&a_{1}&a_{2}\\ 5a_{2}&a_{3}&0\end{bmatrix}\\ _{3}M_{4}=-\begin{bmatrix}3a_{1}&a_{2}&2a_{3}/7&0\\ 0&a_{1}&a_{2}&a_{3}\\ 0&0&a_{1}&a_{2}\\ 14a_{2}&a_{3}&0&0\end{bmatrix}\\ _{3}M_{5}=-\begin{bmatrix}7a_{1}&-a_{2}&-37a_{3}/126&0&0\\ 0&3a_{1}&a_{2}&2a_{3}/7&0\\ 0&0&a_{1}&a_{2}&a_{3}\\ 0&0&0&a_{1}&a_{2}\\ 42a_{2}&a_{3}&0&0&0\end{bmatrix}\\ \end{equation}We could also attempt this from the infinite limit, and then set the corresponding coefficients to \(0\) afterwards.

\begin{equation} P(x)=a_{0}+a_{1}x+a_{2}x^{2}+\cdots=\sum_{k=0}^{\infty}a_{k}x^{k}\\ \end{equation}then we can describe the inverse term by

\begin{equation} P^{-1}(x)=\sum_{n=1}^{\infty}\frac{1}{na_{1}^{n}}\left(\sum_{s,t,u,\cdots}(-1)^{s+t+u+\cdots}\frac{n(n+1)\cdots(n-1+s+t+u\cdots)}{s!t!u!\cdots}\left(\frac{a_{2}}{a_{1}}\right)^{s}\left(\frac{a_{3}}{a_{1}}\right)^{t}\cdots\right)(x-a_{0})^{n}\\ \end{equation}where the sum is over all combinations such that, \(s+2t+3u+\cdots=n-1\) (all positive integers). [Morse Feschbach 1953] For example, when \(n=1\), the only combination is for \(s=t=u=\cdots=0\). When \(n=2\), then then only combination is \(s=1\), for \(n=3\) we may have either, \(s=2,t=0\) or \(s=0,t=1\), giving two terms etc. The goal at this stage would be to develop a regular family of \(n\times n\) matrices that can express these rules through their determinants. We may find that the matrices are made of the sum or product of matrices expressing each term.