# A neutral orbit of period 2

#1

Show that f(x)=x^2-5/4 has a neutral orbit of period 2.

#2

Please explain if this is the right way to go about it.
So f(x) = x^2 - 5/4, to find orbits of period two, we look at f^2(x) = (x^2 - 5/4)^2 - 5/4.
Expanding, we get F(x) =x^4 - \frac{5}{2}x^2 + \frac{5}{16}.
So, F'(x) = 4 x^3 - 5 x.
Solving F(x) = x, we get four potential solutions. The only ones where |F'(x)|=1 are x=- \frac{1}{2}-\frac{1}{\sqrt{2}} and x=-\frac{1}{2}+\frac{1}{\sqrt{2}}. Thus, these are our only neutral period 2 orbits.

#3

@MikhailTal I think you’re really quite close but missing one subtle point and making it bit harder than it needs to be. Certainly, the points in any orbit of period two must, indeed, be solutions of f^2(x)=x and you’ve correctly found those two points to be

x = \frac{1}{\sqrt{2}} \pm\frac{1}{2}.

I think it’s incorrect, though, to say that those are two “neutral period 2 orbits”. There’s just one period two orbit and it contains both of those points. That is,

\frac{1}{\sqrt{2}} + \frac{1}{2} \to\frac{1}{\sqrt{2}} - \frac{1}{2} \to \frac{1}{\sqrt{2}} +\frac{1}{2}.

The other two solutions to f^2(x)=x are actually fixed points of f, as you can check by plugging them into f.

Once you know that an orbit has period two, you can indeed check that it is neutral by plugging them into F' to see that the result has absolute value 1. Alternatively, you could use the fact that if x_0 and x_1 are the points in an orbit of period two, then the multiplier is

f'(x_0)\times f'(x_1) = 4\times x_0 \times x_1.

I guess I think that’s a little simpler.