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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!

Franklin Pezzuti Dyer
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    Any explanation for the downvote? It better not be because "the purpose of MSE is just for stuff you need help with"... I used the "recreational-mathematics" tag, which specifies math done "just for fun." – Franklin Pezzuti Dyer Apr 09 '18 at 23:51
  • another way of saying it: find an initial condition for the discrete-time dynamical system such that it arrives to a period-4 cycle. Hope to see the answer, recently I am studying dynamical systems and this will be interesting. – iadvd Apr 09 '18 at 23:57
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    @iadvd Ha ha, I've been flirting with dynamical systems for a while now but I've never really had time to get into it. And I'm afraid that using all of that vocab will scare off lots of potential answerers (though that is a more concise statement of the problem). XD – Franklin Pezzuti Dyer Apr 09 '18 at 23:58
  • I am studying one I found by trial-error... the shapes are very curious, just in case: https://math.stackexchange.com/questions/2626063/does-this-dynamical-system-show-an-absorbing-area-or-a-chaotic-area – iadvd Apr 10 '18 at 00:04
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    I didn't downvote, but I would venture that a possible reason to is because this seems to lack context. What's the difference between posting a question with no context aside from "This is just for fun, can you find the answer?" and someone posting a homework question with no context aside from "I don't know where to start, can you help me find the answer?" (This is a genuine question, not rhetorical). If the former is accepted then anyone who wants their homework done should just tag recreational mathematics and say they are doing it for fun before copy-pasting their question. – wgrenard Apr 10 '18 at 01:03

4 Answers4

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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!

B. Mehta
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  • Wow, this is a cool solution. I'm glad I didn't use the dynamical systems statement of the problem, b/c then I may not have become privy to this interesting new method! $\checkmark$ – Franklin Pezzuti Dyer Apr 10 '18 at 20:54
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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 $4$.

orangeskid
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    I'm slightly suspicious of this answer because I can't see where the $\sqrt{5}$ terms (in Robert Israel's computational answer) come from. – B. Mehta Apr 10 '18 at 01:34
  • @B. Mehta: take a look here http://mathworld.wolfram.com/TrigonometryAnglesPi15.html – orangeskid Apr 10 '18 at 02:27
  • Ah, your edit clears this up. Nice solution! – B. Mehta Apr 10 '18 at 02:44
  • @orangeskid Well done! This is very similar to what I did. – Franklin Pezzuti Dyer Apr 10 '18 at 20:52
  • @Frpzzd: Thank you! Yes indeed. It's a very nice problem! Even considering the simple map $x\mapsto x^d$ finding the number of cycles of a fixed length $n$ is interesting. As a curiosity, what would be a generalization of your problem. Something conjugate to a power map. What is the most general form? Interesting.... – orangeskid Apr 11 '18 at 00:48
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    @orangeskid Since you're interested, I know that one can find a nice formula for the nth iterate of the function $$f(x)=\frac{x^2}{2x+a}$$ for any number $a$. In fact, my discovery of this led me to come up with this question. :) – Franklin Pezzuti Dyer Apr 11 '18 at 22:56
  • @Frpzzd: I wonder how to find most general rational fractions $f$, $g$, so that $f(g(x)) = g(x^2)$. Does $g$ have to be a quotient of degree at most $1$ (that is, invertible)? – orangeskid Apr 14 '18 at 03:27
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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.

Robert Israel
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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.

Franklin Pezzuti Dyer
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