Tangential Chaos

Dan Sitaru has posted the following problem and its solution at the CutTheKnotMath facebook page:

Solve in real numbers:

$\begin{cases} yx^{4}+4x^{3}+y=6x^{2}y+4x\\ zy^{4}+4y^{3}+z=6y^{2}z+4y\\ xz^{4}+4z^{3}+x=6z^{2}x+4z \end{cases}$

Solution

Let $x=\tan a,$ $\displaystyle a\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right).$ Then

$\displaystyle y=\frac{4x-4x^{3}}{x^{4}-6x^{2}+1}=\tan 4a.$

Thus, $\displaystyle z=\tan 16a$ and $\displaystyle x=\tan 64a,$ implying $\tan a=\tan 64a$ from which $\displaystyle a=\frac{k\pi}{63},$ with $k$ an integer. But, since $\displaystyle a\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right),$ $k=0,\pm 1,\ldots,\pm 31.$

It follows that

$(x,y,z)\in\left\{\displaystyle\left(\tan\frac{k\pi}{63},\tan\frac{4k\pi}{63},\tan\frac{16k\pi}{63}\right):\; k\in\{0,\pm 1,\ldots,\pm31\}\right\}.$

Remark

Let $\displaystyle f(x)=\frac{4x-4x^{3}}{x^{4}-6x^{2}+1}.$ Then the solution to the system $y=f(x),$ $z=f(y),$ $x=f(z)$ could be seen as having iterations on $f$ run into $3\mbox{-cycle}$ which, reminds (if only spuriously) of Sharkovskys theorem (see also Period Three Implies Chaos) means that the iteration on function $f$ have cycles of any length and are, in principle, chaotic. Dan's solution makes it obvious that the substitution $x=\tan a$ will solve $n\mbox{-cycles}$ for any $n=2,3,4,\ldots$ Moreover, the union of all such solutions is the countable set of numbers in the form $\displaystyle\frac{k\pi}{4^{n}-1},$ where $|k|\lt 4^n/2.$ Iterations that start with any other point will be chaotic.

Quite obviously the same can be said of a simpler function $\displaystyle f(x)=\frac{2x}{1-x^{2}},$ that, for example, could be converted to a system of three much simpler equations:

$\begin{cases} y-2x=x^{2}y\\ z-2y=y^{2}z\\ x-2z=z^{2}x. \end{cases}$

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