# Inverse of General Quintic by Determinants

## Main

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

$$P(x)=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+a_{5}x^{5}\\$$

we can treat this is a series and do a series reversion about $$a_{0}$$ giving

$$P^{-1}(x)=\frac{x-a_{0}}{a_{1}}-\frac{a_{2}(x-a_{0})^{2}}{a_{1}^{3}}+\frac{(2a_{2}^{2}-a_{1}a_{3})(x-a_{0})^{3}}{a_{1}^{5}}+\frac{(-5a_{2}^{3}+5a_{1}a_{2}a_{3}-a_{1}^{2}a_{4})(x-a_{0})^{4}}{a_{1}^{7}}+\cdots\\$$

there is a clear pattern for the denominators and the $$(x-a_{0})$$ term. Can we express this sequence in the form

$$P^{-1}(x)=\sum_{n=0}^{\infty}|M_{n}|\frac{(x-a_{0})^{n+1}}{a_{1}^{2n+1}}\\$$

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

$$M_{1}=-[a_{2}]\\ M_{2}=-\begin{bmatrix}a_{1}&a_{2}\\ 2a_{2}&a_{3}\end{bmatrix}\\ M_{3}=-\begin{bmatrix}a_{1}&a_{2}&0\\ 0&a_{1}&5a_{2}\\ a_{2}&a_{3}&a_{4}\end{bmatrix}\\ M_{4}=-\begin{bmatrix}a_{1}&a_{2}&2a_{3}/3&0\\ 0&6a_{1}&14a_{2}&18a_{3}\\ 0&0&a_{1}/6&a_{2}\\ a_{2}&a_{3}&a_{4}&a_{5}\end{bmatrix}\\ M_{5}=-\begin{bmatrix}a_{1}&a_{2}&a_{3}&a_{4}&0\\ 0&7a_{1}&14a_{2}&7a_{3}&0\\ 0&0&a_{1}&3a_{2}&a_{3}\\ 0&0&0&a_{1}&a_{2}\\ a_{2}&a_{3}&a_{4}&a_{5}&0\end{bmatrix}\\$$

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.

## Inverse of Lesser Polynomials

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

$$P_{1}(x)=a_{0}+a_{1}x\\$$

then we have the inverse by series reversion

$$P_{^{-1}}(x)=\frac{(x-a_{0})}{a_{1}}\\$$

we can note this is also the first term of the quintic expression above. For the quadratic we have

$$P_{2}(x)=a_{0}+a_{1}x+a_{2}x^{2}\\$$

and by reversion the sequence

$$P_{2}^{-1}(x)=\frac{(x-a_{0})}{a_{1}}-\frac{a_{2}(x-a_{0})^{2}}{a_{1}^{3}}+\frac{2a_{2}^{2}(x-a_{0})^{3}}{a_{1}^{5}}-\frac{5a_{2}^{3}(x-a_{0})^{4}}{a_{1}^{7}}+\cdots\\$$

where the coefficients $$1,1,2,5,\cdots$$ are the Catalan numbers, A000108, which gives a form for this expression

$$P_{2}^{-1}(x)=\sum_{n=0}^{\infty}\frac{(2n)!}{n!(n+1)!}\frac{(-1)^{n}a_{2}^{n}(x-a_{0})^{n+1}}{a_{1}^{2n+1}}=\frac{-a_{1}+\sqrt{a_{1}^{2}-4a_{0}a_{2}+4a_{2}x}}{2a_{2}}\\$$

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

$$P_{3}(x)=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}\\$$

with reversion

$$P_{3}^{-1}(x)=\frac{(x-a_{0})}{a_{1}}-\frac{a_{2}(x-a_{0})^{2}}{a_{1}^{3}}+\frac{(2a_{2}^{2}-a_{1}a_{3})(x-a_{0})^{3}}{a_{1}^{5}}+\frac{(-5a_{2}^{3}+5a_{1}a_{2}a_{3})(x-a_{0})^{4}}{a_{1}^{7}}+\cdots\\$$

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,

$${}_{3}M_{1}=-[a_{2}]\\ _{3}M_{2}=-\begin{bmatrix}a_{1}&2a_{2}\\ a_{2}&a_{3}\end{bmatrix}\\ _{3}M_{3}=-\begin{bmatrix}a_{1}&a_{2}&0\\ 0&a_{1}&5a_{2}\\ a_{2}&a_{3}&0\end{bmatrix}\\$$

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 $$a_{2}$$,

$${}_{3}M_{1}=-[a_{2}]\\ _{3}M_{2}=-\begin{bmatrix}a_{1}&a_{2}\\ 2a_{2}&a_{3}\end{bmatrix}\\ _{3}M_{3}=-\begin{bmatrix}a_{1}&a_{2}&0\\ 0&a_{1}&a_{2}\\ 5a_{2}&a_{3}&0\end{bmatrix}\\ _{3}M_{4}=-\begin{bmatrix}3a_{1}&a_{2}&2a_{3}/7&0\\ 0&a_{1}&a_{2}&a_{3}\\ 0&0&a_{1}&a_{2}\\ 14a_{2}&a_{3}&0&0\end{bmatrix}\\ _{3}M_{5}=-\begin{bmatrix}7a_{1}&-a_{2}&-37a_{3}/126&0&0\\ 0&3a_{1}&a_{2}&2a_{3}/7&0\\ 0&0&a_{1}&a_{2}&a_{3}\\ 0&0&0&a_{1}&a_{2}\\ 42a_{2}&a_{3}&0&0&0\end{bmatrix}\\$$

## General Series

We could also attempt this from the infinite limit, and then set the corresponding coefficients to $$0$$ afterwards.

$$P(x)=a_{0}+a_{1}x+a_{2}x^{2}+\cdots=\sum_{k=0}^{\infty}a_{k}x^{k}\\$$

then we can describe the inverse term by

$$P^{-1}(x)=\sum_{n=1}^{\infty}\frac{1}{na_{1}^{n}}\left(\sum_{s,t,u,\cdots}(-1)^{s+t+u+\cdots}\frac{n(n+1)\cdots(n-1+s+t+u\cdots)}{s!t!u!\cdots}\left(\frac{a_{2}}{a_{1}}\right)^{s}\left(\frac{a_{3}}{a_{1}}\right)^{t}\cdots\right)(x-a_{0})^{n}\\$$

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.