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


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 $n$.

  • $\begingroup$ 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
  • 3
    $\begingroup$ 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
  • 1
    $\begingroup$ @DinosaurEgg If we first ensure $F(\infty)$ is finite, we can scale. $\endgroup$ – J.G. May 30 at 6:30
  • 1
    $\begingroup$ "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

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


You can also apply the trick directly on the density.

See also my answer to this question.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.