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n-Involutory Rational Functions

2017 April 17

Find a function with the property $f^3(x)=x$. Find a function with the property $f^4(x)=x$. Find a function with the property $f^6(x)=x$.

Lately I've been fascinated with a different class of functions - involutory functions, or functions that invert themselves. These functions have the property $f^2(x)=x$. A few of these functions include: $$f_1(x)=-x$$ $$f_2(x)=1-x$$ $$f_3(x)=\frac{1}{x}$$ $$f_4(x)=-\frac{1}{x}$$ $$f_5(x)=\frac{x+1}{x-1}$$

In fact, there are infinitely many such functions, because if $f$ is in the form $$f(x)=(g\circ h\circ g^{-1})(x)$$

then $f$ is an involution whenever $h$ is. A very special type of function to consider is a function of the type $$f(x)=\frac{ax+b}{cx+d}$$

which does some interesting things when iterated. Notice what happens when we compose two rational functions of the form $$f(x)=\frac{ax+b}{cx+d}$$ $$g(x)=\frac{ex+f}{gx+h}$$ $$(f\circ g)(x)=\frac{a\frac{ex+f}{gx+h}+b}{c\frac{ex+f}{gx+h}+d}$$ $$(f\circ g)(x)=\frac{a(ex+f)+b(gx+h)}{c(ex+f)+d(gx+h)}$$ $$(f\circ g)(x)=\frac{(ae+bg)x+(af+bh)}{(ce+dg)x+(cf+dh)}$$

This may not seem remarkable at first, but notice what happens when we multiply two matrices: $$\begin{bmatrix}a & b\\c & d\end{bmatrix}\begin{bmatrix}e & f\\g & h\end{bmatrix}$$ $$\begin{bmatrix}ae+bg & af+bh\\ce+dg & cf+dh\end{bmatrix}$$

Which is analogous to what we got for our iterated rational function. Therefore, if we map the rational function $$f(x)=\frac{ax+b}{cx+d}$$

onto the matrix $$\begin{bmatrix}a & b\\c & d\end{bmatrix}$$

so that we map the coefficients of $x$ on the numerator and denominator to the leftmost entries and the constants to the rightmost entries, then $f^n(x)$ maps onto the matrix $$\begin{bmatrix}a & b\\c & d\end{bmatrix}^n$$

Notice then that if such a function is involutory, then the matrix that it maps onto has the property that $$\begin{bmatrix}a & b\\c & d\end{bmatrix}^2=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$$

which is the $2$ x $2$ identity matrix. Interestingly, this holds even for fractional and negative iterates of $f$. Using this result, we can find a couple very interesting types of rational functions. For example, functions of the form $$f(x)=\frac{ax-a^2}{x}$$

are 3-involutory, meaning that $f^3(x)=x$. The function $$f(x)=\frac{ax-a^2}{x+a}$$

is 4-involutory. We can even find a function that is $6-involutory$ by finding the halfth iterate of our 3-involutory function. To find the halfth iterate or functional square root of a rational function $$f(x)=\frac{ax+b}{cx+d}$$

we map it onto a matrix $$\begin{bmatrix}a & b\\c & d\end{bmatrix}$$

and use the matrix square root formula: $$\sqrt{\begin{bmatrix}a & b\\c & d\end{bmatrix}}=\frac{1}{t}\begin{bmatrix}a+s & b\\c & d+s\end{bmatrix}$$

where $s$ is the positive or negative root of the determinant. We don't have to worry about what $t$ is, because when we revert this back to a rational function, it will cancel out. So the square root of our rational function is $$f(x)=\frac{(a\pm\sqrt{ad-bc})x+b}{cx+(d\pm\sqrt{ad-bc})}$$

and using this, we can find the functional square root of our 3-involutory function: $$f(x)=\frac{(a\pm\sqrt{a^2})x+b}{cx+(d\pm\sqrt{a^2})}$$ $$f(x)=\frac{(a\pm a)x-a^2}{x\pm a}$$

and so now we know that functions of the form $$f(x)=\frac{2ax-a^2}{x+a}$$ $$f(x)=\frac{a^2}{a-x}$$

are 6-involutory. We could go even further and find a 12-involutory function or a 24-involutory function, but... nah. There's one more interesting property about these types of functions. If $$f(x)=\frac{ax+b}{cx+d}$$

and $$f^n(x)=\frac{ex+f}{gx+h}$$

then if $$g(x)=\frac{ax+c}{bx+d}$$

it must be true that $$g^n(x)=\frac{ex+g}{fx+h}$$

Why is this true? Well, try mapping the functions $f$ and $g$ onto matrices $F$ and $G$. Then it is clear that $F$ is the transpose of $G$, or that $F^T=G$. There is a property of matrices stating that for any matrices $A$ and $B$, $$(AB)^T=B^TA^T$$

so it follows from this that for any matrix $A$, $$(A^{n})^T=(A^{T})^n$$

Since we are likening exponentiation of matrices to the iteration of functions, this means that the "transpose" of $f$ iterated $n$ times is equal to the the nth iterate of its transpose, proving our original statement. Furthermore, if a function of this type is n-involutory, then so is its "transpose", telling us that in addition to our above 6-involutory functions, we also have the functions $$f(x)=\frac{2ax+1}{a-a^2x}$$ $$f(x)=\frac{-1}{a^2x+a}$$

And this theorem can be applied to all types of n-involutory functions of this type.

Edit: 2017 June 3

Okay, there is a way to find an n-involutory rational function for any positive integer $n$. Such a function is given by $$f(x)=\frac{x\cos\zeta-\sin\zeta}{x\sin\zeta-\cos\zeta}$$

where $$\zeta=\frac{2\pi}{n}$$

To prove this, we will have to employ the trigonometric sum angle formulas for the sine and cosine: $$\sin(\theta+\phi)=\sin\theta\cos\phi+\cos\theta\sin\phi$$ $$\cos(\theta+\phi)=\cos\theta\cos\phi-\sin\theta\sin\phi$$

This can be derived by, once again, using matrices in place of actual rational functions. Suppose we are multiplying the matrices $$\begin{bmatrix}\cos a\zeta & -\sin a\zeta\\\sin a\zeta & \cos a\zeta\end{bmatrix}\begin{bmatrix}\cos b\zeta & -\sin b\zeta\\\sin b\zeta & \cos b\zeta\end{bmatrix}$$

When we carry out the multiplication, we get $$\begin{bmatrix}\cos a\zeta\cos b\zeta-\sin a\zeta\sin b\zeta & -\sin a\zeta\cos b\zeta-\cos a\zeta\sin b\zeta\\\sin a\zeta\cos b\zeta+\cos a\zeta\sin b\zeta & \cos a\zeta\cos b\zeta-\sin a\zeta\sin b\zeta\end{bmatrix}$$

Look! These are the sum angle formulas, and this can be simplified to $$\begin{bmatrix}\cos (a+b)\zeta & -\sin (a+b)\zeta\\\sin (a+b)\zeta & \cos (a+b)\zeta\end{bmatrix}$$

Now that we know this, we can say that $$\begin{bmatrix}\cos \zeta & -\sin \zeta\\\sin \zeta & \cos \zeta\end{bmatrix}^n=\begin{bmatrix}\cos n\zeta & -\sin n\zeta\\\sin n\zeta & \cos n\zeta\end{bmatrix}$$

And, if $\zeta=\frac{2\pi}{n}$, $$\begin{bmatrix}\cos \zeta & -\sin \zeta\\\sin \zeta & \cos \zeta\end{bmatrix}^n=\begin{bmatrix}\cos 2\pi & -\sin 2\pi\\\sin 2\pi & \cos 2\pi\end{bmatrix}$$ $$\begin{bmatrix}\cos \zeta & -\sin \zeta\\\sin \zeta & \cos \zeta\end{bmatrix}^n=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$$

Thus it is proven. Because of the relationship previously established between matrices and rational functions, if $$f(x)=\frac{x\cos\zeta-\sin\zeta}{x\sin\zeta-\cos\zeta}$$ and $$\zeta=\frac{2\pi}{n}$$ then $$f^n(x)=x$$

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