(non-)Linear Composition Algebra

Investigate a composition product representation of a set of equations.

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 \[\beta(0)= \frac{\pi}{6}\\ \gamma(1)=\frac{\pi}{3}\]

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

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 \[A=\{\cdot^2,0,0,\cdot^2\}\\ A^{-1}=\{ \pm\sqrt{\cdot},0,0,\pm \sqrt{\cdot}\}\]

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{bmat

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