Franklin Pezzuti Dyer’s Blog

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
\[f^n(x)=x\]