# Abstract

Investigate a composition product representation of a set of equations.

# Main

Consider the composition product of a matrix of functions on a vector of variables as follows $\begin{bmatrix} f_{11} & f_{12} \\ f_{21} & f_{22} \end{bmatrix} \odot \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} =\begin{bmatrix} c_1 \\ c_2 \end{bmatrix}$ Which is the statement that $f_{11}(x_1)+f_{12}(x_2) = c_1 \\ f_{21}(x_1)+f_{22}(x_2) = c_2$

and example might be $\begin{bmatrix} \sin & \cos \\ \cos & -\sin \end{bmatrix}\odot\begin{bmatrix} x \\ y \end{bmatrix} =\begin{bmatrix} 1 \\ 0 \end{bmatrix}$

Then $\sin(x)+\cos(y)=1\\ \cos(x)-\sin(y)=0$ this has solutions $x=\frac{1}{6}(12\pi n+\pi),\;y=\frac{1}{3}(6\pi m + \pi), \; n,m \in \mathbb{Z}\\ x=\frac{1}{6}(12\pi n+5\pi),\;y=\frac{1}{3}(6\pi m - \pi), \; n,m \in \mathbb{Z}$

It would be lovely to have the concept of an inverse such that for $A \odot v = c\\ v= A^{-1} \odot c$

Although there clearly exist two families of vectors which “transform” using $$A$$ to get to $$c$$ given two solutions. However, picking periodic functions may be the cause of this. There are a few potential answers but one of them is $\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \mathrm{acos} & \beta \\ \gamma & \mathrm{asin} \end{bmatrix} \odot \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ where we require that $\beta(0)= \frac{\pi}{6}\\ \gamma(1)=\frac{\pi}{3}$

# An Example With Less Solutions

$\begin{bmatrix} \cdot & \cdot^2 \\ \cdot^2 & \cdot \end{bmatrix} \odot \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$

where $$\cdot$$ represents the function $$x$$, and $$\cdot^2$$ represents the function $$x^2$$. with approximate solutions (which have exact forms but they are too horrible to write here) $x\approx -1.9263,\; y\approx -1.71064\\ x\approx -0.809265, \; y\approx 1.34509$

another $\begin{bmatrix} \cdot & \cdot \\ \cdot & \cdot^2 \end{bmatrix} \odot \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} \phi^2 \\ \phi^2 + 1 \end{bmatrix}$ then one solution is $x=1 \\ y=\phi$ the other is $x=1+\sqrt{5}\\ y=-\frac{1}{\phi}$ this gives us four equations for the elements of the inverse matrix$\alpha(\phi^2)+\beta(\phi^2 + 1) = 1 \\ \gamma(\phi^2)+\delta(\phi^2 + 1) = \phi\\ \alpha(\phi^2)+\beta(\phi^2 + 1) = 1 + \sqrt{5} \\ \gamma(\phi^2)+\delta(\phi^2 + 1) = -\frac{1}{\phi}$ which must mean different functions are used for each set of solutions. We can do simpler “uncoupled” systems $A=\{\cdot^2,0,0,\cdot^2\}\\ A^{-1}=\{ \pm\sqrt{\cdot},0,0,\pm \sqrt{\cdot}\}$

# Coupled System

Problems arise when systems are coupled. For example $\begin{bmatrix} \cdot^2 & 1 \\ 0 & \cdot^2 \end{bmatrix}\odot \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} a \\ b \end{bmatrix}$ expressing $x^2+y=a\\ y^2=b\\$ we can clearly solve for the inverse, but find $y=\pm\sqrt{b}\\ x=\pm\sqrt{a\mp\sqrt{b}}$

In this sense we can’t express the inverse as hoped. But can instead write $\begin{bmatrix} x \\ y \end{bmatrix} =\begin{bmatrix} \pm\sqrt{\cdot} & 0 \\ 0 & \pm\sqrt{\cdot} \end{bmatrix} \odot \begin{bmatrix} a\mp\sqrt{b} \\ b \end{bmatrix}$ we see the inverse matrix is the same as the uncoupled version above. But we now require an additional matrix transformation to make the equation complete, $\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix}\pm\sqrt{\cdot} & 0 \\ 0 & \pm\sqrt{\cdot}\end{bmatrix} \odot\left( \begin{bmatrix}1 & \mp\sqrt{\cdot} \\ 0 & 1 \end{bmatrix} \odot\begin{bmatrix}a \\ b\end{bmatrix}\right)$ Note the parenthesis must be present as there is a non-associativity present

# Larger Sets of Equations

We can represent more complicated expressions such as $\begin{bmatrix} w\\x\\y\\z \end{bmatrix} =\begin{bmatrix} \sqrt{a-\sqrt{b-\sqrt{c-\sqrt{d}}}}\\ \sqrt{b-\sqrt{c-\sqrt{d}}}\\ \sqrt{c-\sqrt{d}}\\ \sqrt{d} \end{bmatrix}$ we may write a chain of transformations (compositions) $\begin{bmatrix} w\\x\\y\\z \end{bmatrix} = \begin{bmatrix} \sqrt{\cdot} & 0 & 0 & 0 \\ 0 & \sqrt{\cdot} & 0 & 0 \\ 0 & 0 & \sqrt{\cdot} & 0 \\ 0 & 0 & 0 & \sqrt{\cdot} \\ \end{bmatrix} \odot \left( \begin{bmatrix} 1 & -\sqrt{\cdot} & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \odot \left( \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & -\sqrt{\cdot} & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \odot \left( \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & -\sqrt{\cdot} \\ 0 & 0 & 0 & 1 \\ \end{bmatrix} \odot \begin{bmatrix} a\\b\\c\\d \end{bmatrix} \right)\right)\right)$

this example is the solution to the set of equations $\begin{bmatrix} \cdot^2 & 1 & 0 & 0 \\ 0 & \cdot^2 & 1 & 0 \\ 0 & 0 & \cdot^2 & 1 \\ 0 & 0 & 0 & \cdot^2 \\ \end{bmatrix} \odot \begin{bmatrix} w\\x\\y\\z \end{bmatrix} = \begin{bmatrix} a\\b\\c\\d \end{bmatrix}$ noting that one transformation matrix is required per additional $$1$$ in the initial statement of the equations.