General Case

It is known that the superfunction of any rational linear function can be found. As I computed it by myself before finding it elsewhere, I will showcase my method here. It does not contain any big novelty but it showcases common methods to find superfunctions by hand.

Let

\[ p(x)=\frac{ax+b}{x+c}. \]

This describes all ratios of linear functions, while ignoring linear functions, which are simpler to treat. Let's define the \(j\)th composition of this function as

\[ p_j(x)=\frac{a_jx+b_j}{k_jx+c_j} \]

This form of function can be guessed from doing some composition. We could get rid of \(k_j\) but it would complicate the computation. We know that

\[ p_{j+1}(x)=p(p_j(x))=\frac{a_jax+b_jx+a_jb+b_jc}{k_jax+c_jx+bk_j+cc_j} \]

Meaning that

\[ \left\{ \begin{array}{ll} a_{j+1}=a_ja+b_j\\ b_{j+1}=a_jb+b_jc\\ c_{j+1}=bk_j+cc_j\\ k_{j+1}=k_ja+c_j \end{array} \right. \]

Which is equivalent to two systems of equations:

\[ \begin{pmatrix}a_{j+1}\\b_{j+1}\end{pmatrix} =\begin{pmatrix}a&1\\b&c\end{pmatrix} \begin{pmatrix}a_{j}\\b_{j}\end{pmatrix} \]

\[ \begin{pmatrix}c_{j+1}\\k_{j+1}\end{pmatrix} =\begin{pmatrix}c&b\\1&a\end{pmatrix} \begin{pmatrix}c_{j}\\k_{j}\end{pmatrix} \]

with initial conditions

\[ \begin{pmatrix}c_{0}\\k_{0}\end{pmatrix}= \begin{pmatrix}a_{0}\\b_{0}\end{pmatrix}= \begin{pmatrix}1\\0\end{pmatrix} \]

as \(p_0(x)=x\), the identity function. We now want to find

\[ M_1^n=\begin{pmatrix}a&1\\b&c\end{pmatrix}^n \]

\[ M_2^n=\begin{pmatrix}c&b\\1&a\end{pmatrix}^n \]

by diagonalisation, this is equivalent to

\[ M_{1/2}^n=S_{1/2}J_{1/2}^nS_{1/2}^{-1} \]

with \(J_{1/2}\) containing the eigenvalues of \(M_{1/2}\) and \(S_{1/2}\) containing its eigenvectors. We are in fact only interested to the first column of \(M_{1/2}^n\) as it will be multiplied by \(\begin{pmatrix} 1 \\ 0 \end{pmatrix}\).

The result is

\[ \left\{ \begin{array}{ll} a_{j}=\frac{c-a+\sqrt{\rho}}{2\sqrt{\rho}}\left(\frac{a+c-\sqrt{\rho}}{2}\right)^j+\frac{a-c+\sqrt{\rho}}{2\sqrt{\rho}}\left(\frac{a+c+\sqrt{\rho}}{2}\right)^j\\ b_{j}=\frac{b}{\sqrt{\rho}}\left(\left(\frac{a+c+\sqrt{\rho}}{2}\right)^j-\left(\frac{a+c-\sqrt{\rho}}{2}\right)^j\right)\\ c_{j}=\frac{a-c+\sqrt{\rho}}{2\sqrt{\rho}}\left(\frac{a+c-\sqrt{\rho}}{2}\right)^j+\frac{c-a+\sqrt{\rho}}{2\sqrt{\rho}}\left(\frac{a+c+\sqrt{\rho}}{2}\right)^j\\ k_{j}=\frac{1}{\sqrt{\rho}}\left(\left(\frac{a+c+\sqrt{\rho}}{2}\right)^j-\left(\frac{a+c-\sqrt{\rho}}{2}\right)^j\right)\\ \end{array} \right. \]

with

\[ \rho=(a-c)^2+4b \]

Let now

\[ p_j(x)=\frac{A_jx+B_j}{x+C_j} \]

which leads to

\[ \left\{ \begin{array}{ll} A_j=\frac{a_j}{k_j}=\frac{(c-a+\sqrt{\rho})(a+c-\sqrt{\rho})^j+(a-c+\sqrt{\rho})(a+c+\sqrt{\rho})^j}{2((a+c+\sqrt{\rho})^j-(a+c-\sqrt{\rho})^j)}\\ B_j=\frac{b_j}{k_j}=b\\ C_j=\frac{c_j}{k_j}=\frac{(a-c+\sqrt{\rho})(a+c-\sqrt{\rho})^j+(c-a+\sqrt{\rho})(a+c+\sqrt{\rho})^j}{2((a+c+\sqrt{\rho})^j-(a+c-\sqrt{\rho})^j)}\\ \end{array} \right. \]