- Thread starter
- #1

#### eddybob123

##### Active member

- Aug 18, 2013

- 76

$$\begin{align*}\cos(2t_1)=4\cos(t_1)\cos(t_2)\\

\cos(2t_2)=4\cos(t_2)\cos(t_3)\\

\cos(2t_3)=4\cos(t_3)\cos(t_4)\\

\cos(2t_4)=4\cos(t_4)\cos(t_1)

\end{align*}$$

- Thread starter eddybob123
- Start date

- Thread starter
- #1

- Aug 18, 2013

- 76

$$\begin{align*}\cos(2t_1)=4\cos(t_1)\cos(t_2)\\

\cos(2t_2)=4\cos(t_2)\cos(t_3)\\

\cos(2t_3)=4\cos(t_3)\cos(t_4)\\

\cos(2t_4)=4\cos(t_4)\cos(t_1)

\end{align*}$$

- Moderator
- #2

- Feb 7, 2012

- 2,799

Nobody else has tried this one, so I'll throw in my thoughts. First, I used the relation $\cos2\theta = 2\cos^2\theta-1$ to write the equations as $\cos t_{n+1} = \dfrac{2\cos^2t_n - 1}{4\cos t_n}\ (n=1,2,3,4)$ (where $t_5$ is interpreted to mean $t_1$). Next, let $x_n = \cos t_n$. Then $x_{n+1} = f(x_n)$, where $f(x) = \dfrac{2x^2-1}{4x}$. So $x_1$ is a periodic point of this mapping, with period $4$. There is also the requirement that $|x_n| \leqslant 1$ for each $n$.

$$\begin{align*}\cos(2t_1)=4\cos(t_1)\cos(t_2)\\

\cos(2t_2)=4\cos(t_2)\cos(t_3)\\

\cos(2t_3)=4\cos(t_3)\cos(t_4)\\

\cos(2t_4)=4\cos(t_4)\cos(t_1)

\end{align*}$$

So we want to find $x$ such that $f(f(f(f(x)))) = x$. The composition of four repetitions of the map $f$ seems impossibly complicated, and the best I can do is to form the composition of two repetitions of $f$. In fact, $$f(f(x)) = \frac{2\left(\frac{2x^2-1}{4x}\right)^2-1}{\frac{2x^2-1}{x}},$$ which simplifies to $\dfrac{4x^4 - 12x^2 + 1}{8x(2x^2-1)}.$ if you put that equal to $x$, you get the solutions $x = \pm\sqrt{1/6}$.

Now let $\theta = \arccos\sqrt{1/6}$. Then you can get a solution to the original problem by taking $\{t_1,t_2,t_3,t_4\} = \{\theta, \pi-\theta, \pi+\theta, 2\pi-\theta\}$ (as an unordered set). The trick there is to expand the orbit of period two in the $x$-map to an orbit of period four in the $t$-space by using the fact that there are two numbers in the interval $[0,2\pi]$ taking any given value of the cosine function. In that way, we get four distinct values for the $t_n$ although there are only two distinct values of the $x$'s.

That is the only solution that I can find to this problem, and my guess is that it is unique.