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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 $

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

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    $\begingroup$ Also posted at MSE math.stackexchange.com/questions/3335399/… $\endgroup$ – mick Aug 28 '19 at 11:18
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    $\begingroup$ $f_n(x)=\alpha^{2^n}+\alpha^{-2^n}$ where $\alpha=(x\pm\sqrt{x^2-4})/2$. $\endgroup$ – Gerry Myerson Aug 28 '19 at 13:10
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    $\begingroup$ 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. $\endgroup$ – Anthony Quas Aug 28 '19 at 22:06
  • $\begingroup$ See the important edit ! $\endgroup$ – mick Aug 28 '19 at 22:52

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