Recreational math problem: $f(x)=\frac{x^2}{2x+101}$

Question

Here is a math problem (just for fun) for the residents of MSE to enjoy:

Let $f(z)$ be defined as $$f(z)=\frac{z^2}{2z+101}$$ Find a 4-cycle of $f$ - that is, find four distinct complex numbers $z_1,z_2,z_3,z_4$ such that $$f(z_1)=z_2$$ $$f(z_2)=z_3$$ $$f(z_3)=z_4$$ $$f(z_4)=z_1$$

The answer (my answer, at least) is a bit messy, so watch out! After a correct answer is posted, I will post my solution method.

Enjoy!

Answer 1

Here's a fairly clean solution to find one of the cycles:

First simplify the algebra dramatically by substituting $y = -\frac{2i}{101} (z + \frac{101}{2})$, so we can solve for the fixed points of

$$g(y)=\frac{y^2 - 1}{2y}$$

instead (Without the factor of $-i$, you get a plus sign on the numerator - I worked it out with the $-i$, so sticking with that for now. This also gives you factors of $i$ right at the end, which are avoided this way.)

Now, we can compute $$\begin{align}h(y) := (g \circ g)(y) &= {\left({y^2-1\over2y}\right)^2 - 1\over2\left({y^2-1\over2y}\right)} \\ &=\frac{(y^2 - 1)^2 - 4y^2}{4(y^2-1) y} \\&=\frac{y^4 - 6y^2 + 1}{4y^3 - 4y} \end{align}$$

noting that this is an odd function. Observe if $h(y) = -y$, then we have a fixed point of $h \circ h$. But pleasantly, $h(y)$ is odd, so solutions to $h(y) = -y$ should come in $\pm$ pairs, and since we have polynomials this indicates that they should be easy to solve. $$\begin{align}h(y) &= -y \\ y^4 - 6y^2 + 1 &= -4y^4 + 4y^2 \\ 5y^4 - 10y^2 + 1 &= 0\end{align}$$ But the roots of this are easy to find: $5y^4 - 10y^2 + 1 = 5(y^4 - 2y^2) + 1=5(y^2-1)^2-4$, so $y = \pm \sqrt{1 \pm \frac{2}{\sqrt{5}}}$. Now this is certainly a period $2$ point of $h$ since it is non-zero, so gives a genuine period $4$ point of $g$!

Transform back to give

$$z = -\frac{101}{2} \pm \frac{101i}{2} \sqrt{1 \pm \frac{2}{\sqrt{5}}}$$

a 4-cycle.

---

Kudos to OP for posting this without the dynamical systems context - that certainly would've put me off this interesting problem!

Answer 2

According to Maple, $$f(f(f(f(x)))) - x = -{\frac { \left( x+101 \right) \left( 3\,{x}^{2}+303\,x+10201 \right) \left( 5\,{x}^{4}+1010\,{x}^{3}+102010\,{x}^{2}+5151505\,x+ 104060401 \right) \left( {x}^{8}+404\,{x}^{7}+142814\,{x}^{6}+ 28848428\,{x}^{5}+4058355639\,{x}^{4}+378363618036\,{x}^{3}+ 22291923162621\,{x}^{2}+750494746474907\,x+10828567056280801 \right) x }{ \left( 2\,x+101 \right) \left( 2\,{x}^{2}+202\,x+10201 \right) \left( 2\,{x}^{4}+404\,{x}^{3}+61206\,{x}^{2}+4121204\,x+104060401 \right) \left( 2\,{x}^{8}+808\,{x}^{7}+285628\,{x}^{6}+57696856\,{x} ^{5}+7284228070\,{x}^{4}+588565628056\,{x}^{3}+29722564216828\,{x}^{2} +857708281685608\,x+10828567056280801 \right) }} $$

The factors $x$, $x+101$ and $3 x^2+303 x+10201$ of the numerator give solutions of $f(f(x))=x$, so for $x$ to be in a $5$-cycle we need $x$ to be a root of the irreducible quartic $$5x^4+1010x^3+102010x^2+5151505x+104060401$$ or the irreducible octic $${x}^{8}+404\,{x}^{7}+142814\,{x}^{6}+28848428\,{x}^{5}+4058355639\,{x} ^{4}+378363618036\,{x}^{3}+22291923162621\,{x}^{2}+750494746474907\,x+ 10828567056280801 $$ These turn out to be less fearsome if you substitute $x = - 101 (1+t)/2$, obtaining the quartic $$ 5\,{t}^{4}+10 t^2+1=0$$ and the octic $$ {t}^{8}+28 t^6 + 134 t^4 + 92 t^2 + 1 = 0 $$ as these are quadratic and quartic in $t^2$. The solutions end up as (for the quartic) $$ x = -\frac{101}{2} \pm \frac{101 i}{10} \sqrt{25 \pm 10 \sqrt{5}}$$ and (for the octic) $$ \eqalign{x &= -\frac{101}{2} \pm \frac{101 i}{2} \sqrt{7 + 2 \sqrt{5} + 2 \sqrt{15 + 6 \sqrt{5}}}\cr x &= -\frac{101}{2} \pm \frac{101 i}{2} \sqrt{7 - 2 \sqrt{5} + 2 \sqrt{15 - 6 \sqrt{5}}}\cr x &= -\frac{101}{2} \pm \frac{101 i}{2} \sqrt{7 + 2 \sqrt{5} - 2 \sqrt{15 + 6 \sqrt{5}}}\cr x &= -\frac{101}{2} \pm \frac{101 i}{2} \sqrt{7 - 2 \sqrt{5} - 2 \sqrt{15 - 6 \sqrt{5}}}\cr } $$

I'll leave it to you to see how these split up into four-cycles.

Answer 3

HINT:

Let's do some reductions of $f$:

$$f(101 x) = \frac{(101 x)^2}{2\cdot 101 x + 101}= 101\cdot \frac{x^2}{2x+1}$$ so we have $$f= s \circ g \circ s^{-1}$$ where $s(x) = 101 x$ and $g(x)=\frac{x^2}{2x+1}$. Further

$$g(x)=\frac{x^2}{2x+1}= \frac{x^2}{(x+1)^2 - x^2}= \frac{1}{(1+\frac{1}{x})^2 -1}= t^{-1}\circ h \circ t(x)$$ where $t(x) = \frac{1}{x}+1$ and $h(x)=x^2$. Finally $$f=s \circ t^{-1} \circ h \circ t \circ s^{-1}= u \circ h \circ u^{-1}$$ where $u(x)=s\circ t^{-1}(x)=\frac{101}{x-1}$. In other words, $f\circ u = u\circ h$, $$f(\frac{101}{x-1})=\frac{101}{x^2-1}$$

So $f(x)$ is conjugate to the function $h(x)=x^2$. One checks easily that $x$ starts a cycle of length $4$ for $h$ if and only if $u(x)$ starts a cycle of length $4$ for $f$.

To find cycles of length $4$ for $h$, one looks for solutions of $x^{16}=x$ that are not solutions of $x^4=x$. That means, $x$ satisfies $x^{15}=1$ but not $x^{3}=1$. There are $15-3=12$ such solutions, $$x= \exp \frac{2 k \pi i}{15}, \ \ k = 1,2,3,4,6,7,8,9,11,12,13,14$$ They break into $3$ cycles of length $4$.

Answer 4

Here is my solution, as promised. First, I restate the problem:

Let $f(z)$ be defined as $$f(z)=\frac{z^2}{2z+101}$$ Find a 4-cycle of $f$ - that is, find four distinct complex numbers $z_1,z_2,z_3,z_4$ such that $$f(z_1)=z_2$$ $$f(z_2)=z_3$$ $$f(z_3)=z_4$$ $$f(z_4)=z_1$$

Notice that $f$ is equal to $$f(z)=\frac{1}{101\bigg(\frac{1}{z}+\frac{1}{101}\bigg)^{2}-\frac{1}{101}}$$ from which it easily follows that $$f^{\circ n}(z)=\frac{1}{101\bigg(\frac{1}{z}+\frac{1}{101}\bigg)^{2^n}-\frac{1}{101}}$$ where $f^{\circ n}$ represents $f$ composed $n$ times. In order for $f$ to have a $4$-cycle, $f^{\circ n}(z_1)$ must be a periodic function of $n$ with period $4$. This occurs if and only if $$\bigg(\frac{1}{z_1}+\frac{1}{101}\bigg)^{2^n}$$ is also a periodic function of $n$. Now recall some properties of the complex roots of unity. If $\omega_k$ is defined as $$\omega_k=\cos\frac{2\pi}{k}+i\sin\frac{2\pi}{k}$$ then an elementary property is that $$\omega_k^p=\omega_k^{p\bmod k}$$ Now notice that $2^n\bmod 5$ is periodic with period $4$, with $$2^{4n}\bmod 5 =1$$ $$2^{4n+1}\bmod 5 =2$$ $$2^{4n+2}\bmod 5 =4$$ $$2^{4n+3}\bmod 5 =3$$ Thus, $f^{\circ n}(z_1)$ should be periodic if we let $$\frac{1}{z_1}+\frac{1}{101}=\omega_5$$ so that $$\frac{1}{z_2}+\frac{1}{101}=\omega_5^2$$ $$\frac{1}{z_3}+\frac{1}{101}=\omega_5^4$$ $$\frac{1}{z_4}+\frac{1}{101}=\omega_5^3$$ giving us the values $$z_1=\frac{101}{101\omega_5-1}$$ $$z_2=\frac{101}{101\omega_5^2-1}$$ $$z_3=\frac{101}{101\omega_5^4-1}$$ $$z_4=\frac{101}{101\omega_5^3-1}$$ ...which can be simplified, of course, but you guys have already done that part for me.