# The mysterious numbers $\frac{13}{20}$ and $20$?

Let $$g(x) = x^6 - 30 x$$

Let $$h(x) = x^6$$

Let $$f(x) = x^2 - 2$$

Let $$r$$ be a reduced fraction $$0 < \frac{p}{q} < 2$$ with integers $$p,q > 1$$

Let $$f_{n+1}(x) = f(f_n(x)) = f_n(f(x)) , f_0(x) = x$$.

Now consider for n going to infinity the following ' averages ' :

$$a(r,v) = \lim n^{-1} \sum_{i=1}^n f_n(r)^{2v+1}$$

For $$v > -1$$ an integer.

$$b(r) = \lim n^{-1} \sum_{i=1}^n g(f_n(r))$$

$$c(r) = \lim n^{-1} \sum_{i=1}^n h(f_n(r))$$

Now Independant of our choices of $$r,v$$ we get

$$a(r,v) = 0$$ (property a)

However it appears that for most $$v$$ the choice $$r = 13/20$$ converges as one of the fastest !?

That is a bit mysterious to me.

From property a , it is easy to see that $$b(r) = c(r)$$. However ... It seems that usually $$b(r)$$ converges Faster than $$c(r)$$.

That is the reason d'être of b. Another mystery.

The third mystery is that again for $$b,c$$ that value $$r = 13/20$$ is one of the fastest.

The Fourth mystery is that apparently for all $$r$$ ;

$$b(r) = c(r) = 20$$

I have been told this related to the logistic map, but I do not know how.

Since I can only ask one question at a time the main question is this

Is it true that

$$b(13/20) = 20$$ V And if so, how to prove it?

Update.

Gerry Myerson's hint that

$$\alpha=(x\pm\sqrt{x^2-4})/2$$

$$f_n(x)=\alpha^{2^n}+\alpha^{-2^n}$$

Together with $$(c + 1/c)^6 = * + 20$$

Proves where they number $$20$$ came from.

The average of even Powers of iterations of $$f$$ are the central binomial coefficients. Nice.

So that solved Mystery 4.

Below is however another mystery :

Mystery 5 :

It appears that for $$t = 13/20$$

$$\lim \sum^n \frac{ \frac{1}{t} + \frac{1}{f(t)} + ... \frac{1}{f_n(t)} }{n} = \frac{-1}{2}$$

• Also posted at MSE math.stackexchange.com/questions/3335399/… – mick Aug 28 '19 at 11:18
• $f_n(x)=\alpha^{2^n}+\alpha^{-2^n}$ where $\alpha=(x\pm\sqrt{x^2-4})/2$. – Gerry Myerson Aug 28 '19 at 13:10
• The map you're looking at is conjugate to the map $x\mapsto 4x(1-x)$ on $[0,1]$ which is well-studied. This in turn is conjugate to the "tent map" $x\mapsto 1-|1-2x|$. You can write down an invariant measure for that map. Your mysteries should be resolvable using Birkhoff's theorem. – Anthony Quas Aug 28 '19 at 22:06
• See the important edit ! – mick Aug 28 '19 at 22:52