# Prove that the critical value for the appearance of period 3-orbits in the tent map is $\frac{1+\sqrt5}{2}$

Given the tent map: $$T(x)=\mu(1-2|x-\frac{1}{2}|)=\mu\min\{x,1-x\}$$ Prove that the critical value of $\mu$ for the appearance of period 3-orbits in the tent map is $\frac{1+\sqrt5}{2}$.

I think you might have to solve it algebraically, for the tent map iterated three times, but this is a very complicated problem and I'm not sure how to solve it. Thanks

• it's the other way around, 3-orbits imply chaos (and that's a nice result because checking if there are 3-orbits is much easier than finding out if there is chaos). Hint: apply the intermediate value theorem to $h(x)=f^3(x)-x$ between well-chose points Commented May 10, 2017 at 14:10
• @Glougloubarbaki I don't think the IVT will help. Here is a graph of $f^3(x)-x$ over the unit interval. It only crosses the $x$-axis once and that point is a fixed point. The relevant points of intersection are points where the graph just touches the $x$-axis. Commented May 10, 2017 at 19:47

Yes, you do need to do a little algebra on this problem but the amount can be alleviated with a little geometric insight. Let's start with a look at the graphs of $T_{\mu}$ and $T_{\mu}^3$ for a typical value of $\mu$ with $\mu>1$ - specifically, $\mu=1.5$, which is a little less than $(1+\sqrt{5})/2$.
Note that the graph of $T_{\mu}^3$ consists of 8 segments. It's pretty easy to see why. The inverse image $T_{\mu}^{-1}(y)$ of any point $y\in[0,1]$ consists of at most two points; thus, $T^{-3}(y)$ can consist of 8 points and, hence, the 8 segments. Note also that three vertices of this piecewise linear graph are close to the line $y=x$. The period 3 cycle will appear when these simultaneously touch the line; this must happen simultaneously because those points together will form the 3 cycle. As a result, you simply need to solve the equation $$T_{\mu}^3\left(\frac{1}{2}\right) = \frac{1}{2}$$ for $\mu$. But $$T_{\mu}^3\left(\frac{1}{2}\right) = \left(1-\frac{\mu }{2}\right) \mu ^2$$ as you can see by first computing $T_{\mu}(1/2) =\mu/2$ and then computing $T_{\mu}(\mu/2)$ accounting for the fact that $\mu>1$ and finally computing $T_{\mu}$ of that result.
Finally, the solutions of $$\left(1-\frac{\mu }{2}\right) \mu^2 = \frac{1}{2}$$ are $\mu=1$ and $\mu=(1\pm\sqrt{5})/2$ and only one of these makes sense in your context.