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There is at least one such function. It is the cdf of the equilibrium probability distribution for the chaotic sequence $x(n+1) = |\log x(n)|$ with $x(1) = 2$. Its graph (approximation) is pictured below. I am interested in a series expansion for the density $f(x)$, which is the derivative of $F(x)$.

enter image description here

Note

I expect that if you start with a different seed, say $x(1) =3$, you end up with the same distribution, unless you pick up one of the very rare seeds (called bad seed) that results in a different $F$. The set of bad seeds has Lebesgue measure zero, but it is infinite and even dense. My intuition is based on the following: consider instead $x(n+1) = bx(n) - \lfloor bx(n) \rfloor$. The equilibrium distribution is uniform on $[0, 1]$ this time (if $b$ is an integer larger than 1) unless you pick up a bad seed. All rational numbers are bad seeds. Tons of other numbers are bad seeds. But the vast majority are good seeds. A good seed is equivalent to a normal number: its digits in base $b$ are evenly distributed. No one knows if $\pi, e, \log 2$ or $\sqrt{2}$ is a good seed. More on this in my article on the theory of randomness or my book on organized chaos.

Similarly, in our context here, proving that $x(1) = 2$ is a good seed is a very hard problem, and possibly un-provable. Yet plenty of evidence makes you believe that it is a good seed. Some sequences such as $x(n+1) = b+x(n) - \lfloor b + x(n)\rfloor$ do not have bad seeds if $b$ is irrational. The logistic map $x(n+1) = 4x(n)\cdot (1-x(n))$ has plenty of bad seeds. In our example $x(1) = 0.567143...$ is a bad seed because $x(2) = x(1)$ and thus $x(n) = x(1)$ for all $n$.

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  • $\begingroup$ By the definition the function has to be odd and increasing. Neither of these properties holds for the graph. $\endgroup$ – user May 29 at 20:04
  • $\begingroup$ Still $F(0)=1$ in this... Shouldn't $F(0)=0$ instead? $\endgroup$ – DinosaurEgg May 29 at 21:20
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    $\begingroup$ The functional equation alone is not very informative. You can define $F(x)$ on $[0,1]$ in any way you want (just keeping $F(0)=0$), after which it extends in a unique way to the rest of the positive ray. $\endgroup$ – fedja May 30 at 3:08
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    $\begingroup$ @DinosaurEgg If we first ensure $F(\infty)$ is finite, we can scale. $\endgroup$ – J.G. May 30 at 6:30
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    $\begingroup$ "show (that's the difficult part, I haven't solved it) " Yeah, I "haven't solved" it either. Moreover, I do not even know how to proceed from $E_n\to 0$ to the convergence of $F_n$ themselves. OK, let me think :-) $\endgroup$ – fedja May 31 at 1:21
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If you take as a seed the function $F_0(x)=\exp(-\exp(-x))-\exp(-\exp(x))$ and plug it into the iteration

$$F_{n+1}(x) = F_n(\exp(x))-F_n(\exp(-x))$$

you can generate ever better approximations to the distribution. After only two iterations I got this result:

Cumulative distibution

Density

You can also apply the trick directly on the density.

See also my answer to this question.

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  • $\begingroup$ Are you sure that you'll get the same distribution? As I said, the fixed points are many. $\endgroup$ – fedja May 30 at 12:36
  • $\begingroup$ If you mean sure as in having a proof: no. But I'm pretty sure this does give the right distribution yes. $\endgroup$ – Raskolnikov May 30 at 12:39
  • $\begingroup$ Why? (I'm really curious: if you have a good answer to this question, we may discern some extra property of $F$ that, together with the functional equation will allow us to determine it uniquely) $\endgroup$ – fedja May 30 at 12:41
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    $\begingroup$ See my reply to the question I linked to in my answer. I explain how I came with the idea for using this specific seed and not another one. $\endgroup$ – Raskolnikov May 30 at 12:45
  • $\begingroup$ It sounds like my recursion is known as the logarithmic map. $\endgroup$ – Vincent Granville May 30 at 16:20

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