# Why is $\frac{1}{1-x}\circ\frac{1}{1-x}\circ\frac{1}{1-x}=x$ (graphically)?

Of course, I can evaluate this composition of functions algebraically which is a fairly elementary exercise. But I'm curious as to whether one can find a (simple) graphical or visual argument (likely based on symmetry, perhaps due to representing this as a composition of the functions $$1/x$$ and $$1-x$$). To this extent, I've had a look at the following cobweb plot:

Which represents the composition. But this provides me with no geometric clarity or insight, perhaps because I'm not familiar enough with the geometric properties of hyperbolas? Anyways, if anyone has such an insight or ideas then I'd be grateful.

• What has this got to do with group theory? Nov 16, 2021 at 11:44
• Are you willing/able to accept an argument involving complex numbers? Nov 16, 2021 at 11:56
• If you rearrange the function in the form $\frac{ax+b}{cx+d}$, the solution to the fractional iteration formula on Wikipedia would involve complex values. It is your task to graph it graphically.
– Max
Nov 16, 2021 at 11:59
• @Shaun Perhaps that was not the best tag, but this problem arose after I considered the group of functions arising from the composition of $A=1/x$ and $B=1-x$ and decided it would be nice if I had deeper intuition for why $ABA=BAB$ (the product there is composition). (I don't know much about group theory I was just messing around with the concept in my head) Nov 16, 2021 at 12:15
• @PolymorphismPrince: Both $1/x$ and $1-x$ are “fractional linear transformations” which map the set $\{ 1, 0, \infty \}$ onto itself, therefore the generated group is isomorphic to the permutation group of three elements. Every element of that group has period $1$, $2$, or $3$. – A similar argument had been given here: math.stackexchange.com/a/3713614/42969. Nov 16, 2021 at 13:16

## 2 Answers

There is a geometric explanation, but it's not a precalculus explanation. And the explanation does involve hyperbolic stuff, but not hyperbolas per se. Instead, the explanation involves the understanding that the formula $$\frac{1}{1-x}$$ represents an isometry of the upper half plane model of the hyperbolic plane $$\mathbb H^2$$.

If we rewrite your formula with a complex coordinate as $$\frac{1}{-z+1}$$ then we can see that it is a special case of a fractional linear transformation $$f(z) = \frac{az+b}{cz+d}, \quad a,b,c,d \in \mathbb R, \quad ad-bc=1$$ The upper half plane model of hyperbolic geometry is $$\mathbb H^2 = \{z = x + iy \mid y > 0\}$$ equipped with the Riemannian metric $$\frac{dx^2+dy^2}{y^2}$$. Every fractional linear transformation is an isometry of $$\mathbb H^2$$ in that metric. And here's where some group theory comes in: the fractional linear transformations are exactly the group of all orientation preserving isometries of $$\mathbb H^2$$ (leaving out orientation reversing things like reflections and glide reflections).

Back to your example $$f(z) = \frac{1}{-z+1}$$. If you solve the fixed point equation $$f(z)=z$$ you'll obtain $$z^2-z+1=0$$ which gives $$z = \frac{1}{2} + \frac{\sqrt{3}}{2} i \in \mathbb H^2$$. In other words, your map fixes the point $$z = \frac{1}{2} + \frac{\sqrt{3}}{2} i$$.

If you compute $$\frac{df}{dz}(\frac{1}{2} + \frac{\sqrt{3}}{2} i)$$ you'll get $$-\frac{1}{2} \pm \frac{\sqrt{3}}{2} i$$ (I'm not sure about the $$+$$ or $$-$$ sign) which is a cube root of unity. In other words, your map rotates around the point $$z = \frac{1}{2} + \frac{\sqrt{3}}{2} i$$ by an angle $$\pm 2 \pi / 3$$.

It follows that $$f^3$$ fixes the point $$\frac{1}{2} \pm \frac{\sqrt{3}}{2} i$$ and, by the chain rule, has derivative $$1$$ at that point. In other words, $$f^3$$ rotates by angle $$0$$ around that point.

And in the hyperbolic plane (just as in the Euclidean plane), any isometry that fixes a point and rotates by angle $$0$$ at that point is the identity map.

• Great answer! Do the fractional iterates of $f$ just rotate around the same point or is it more complex?
– Max
Nov 16, 2021 at 12:34
• Another (similar) interpretation is that $f$ is conjugate to a rotation of the complex plane, compare math.stackexchange.com/a/3713614/42969. Nov 16, 2021 at 13:02
• If by "fractional" iterates you mean solutions of equations like $g^n=f$, then yes, that's exactly right @Max. Nov 16, 2021 at 15:36
• That conjugation interpretation is interesting. One point is that the conjugating map takes the upper half plane model of $\mathbb H^2$ to the Poincare disc model of $\mathbb H^2$, and then the conjugated map restricts to a rotation of the Poincare disc model around the origin. But this won't work for fractional linear transformations that represent translations of $\mathbb H^2$. @MartinR Nov 16, 2021 at 15:39

We cube the matrix of your Mobius transformation;

$$\left( \begin{array}{cc} a&b \\ c&d \end{array} \right)$$ means mapping $$\frac{ax+b}{cx+d}$$ where we usually write $$z$$ instead of $$x$$

$$\left( \begin{array}{rr} 0&1 \\ -1&1 \end{array} \right) \left( \begin{array}{rr} 0&1 \\ -1&1 \end{array} \right) \left( \begin{array}{rr} 0&1 \\ -1&1 \end{array} \right) = \left( \begin{array}{rr} -1&0 \\ 0&-1 \end{array} \right)$$

The corresponding Mobius transformation is $$\frac{-x}{-1} = x$$