# Which positive continuous functions satisfy $F(x) = F(e^x)-F(e^{-x})$ for $x\geq 0$?

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 $$F(x)$$.

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 $$n$$.

• By the definition the function has to be odd and increasing. Neither of these properties holds for the graph. – user May 29 at 20:04
• Still $F(0)=1$ in this... Shouldn't $F(0)=0$ instead? – DinosaurEgg May 29 at 21:20
• 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. – fedja May 30 at 3:08
• @DinosaurEgg If we first ensure $F(\infty)$ is finite, we can scale. – J.G. May 30 at 6:30
• "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 :-) – 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:

You can also apply the trick directly on the density.

• 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) – fedja May 30 at 12:41