11

The Fabius function $F$ can be defined on $[0,1]$ by

  • $F(0)=0$
  • $F(1)=1$
  • on $[0,\frac{1}{2}]$ $F'(x)=2F(2x)$
  • on $[\frac{1}{2},1]$ $F'(x)=2F(2(1-x))$

It's a known example of a not analytic $C^\infty$ function.

The Fabius function can also be defined as the CDF of the random variable $X$ such that $$X=\sum_{i=1}^\infty 2^{-i}U_i$$ where $U_i$ are independent random variables uniform on $[0,1]$.

Is it possible to find a usual function $f$ such that $$\lim_{x\rightarrow 0}\frac{F(x)}{f(x)}=1$$

Some functions like $f(x)=e^{-\frac{1}{x^2}}$ could be good candidates, but I don't really know how to find such a function, or given a function $f$, how to compute the limit.

Varun Vejalla
  • 8,873
Xoff
  • 10,310
  • 2
    $\displaystyle f(x)=\frac{1}{\Gamma!\left(-\log_2!x\right),\sqrt{x^{\log_2!x+1}}}\int_0^1 \frac{1-F(t)}{t^{\log_2!x+1}},dt$ looks like a plausible candidate. It is real-analytic and we can show that the equality $f(x)=F(x)$ holds exactly on an infinite set of points in any neighborhood of $x=0$. Its graph suggests that the required limit holds, but I still have to fill some gaps in my proof. – Vladimir Reshetnikov Mar 29 '18 at 02:53
  • @VladimirReshetnikov If you are using F itself, it's not really useful, because you can just say $f(x)=F(x)$ or $f(x)=2\int F(2t)dt$. But your expressions are really complex and puzzling – Xoff Mar 31 '18 at 14:55
  • 1
    The question ask for a "usual" function asymptotic to $F(x)$. It is not really clear what "usual" means, so I interpreted it as "analytic". Although my example is defined in terms of an integral of $F(x)$, it is an analytic function (in some vicinity of $x=0$, but excluding that point itself; see https://math.stackexchange.com/questions/2495466/an-integral-involving-a-smooth-function). Of course, it would be much more interesting to find an elementary(-ish) asymptotic function, not defined in terms of $F(x)$; I tried to do that for quite some time, but I do not know if it is possible. – Vladimir Reshetnikov Mar 31 '18 at 17:31
  • 2
    Here is an alternative form, not involving the Fabius function at all (it is the same analytic function as in my first comment above):$\displaystyle f(x) = \left.\lim_{n\to\infty}\frac{(-1)^{n+1} , x^{\frac{a-1}2}}{2^{a,n+\binom n2} , \Gamma\left(a+n\right)}\sum_{k=0}^{2^{n-1}-1}t_k,\left(2k+1\right)^{a+n-1}\right|{a,=-\log_2!x, ,, t_0 = 1, ,, t_k = (-1)^k , t{\lfloor k/2 \rfloor}}$ – Vladimir Reshetnikov Apr 02 '18 at 19:41

1 Answers1

3

Based oh this answer, the asymptotic of the Fabius function for $x\to0^+$: $$\small F(x)\sim \frac{1}{2^{7/12} \sqrt{\pi \, x}}\, \exp \left(\frac{1}{\log 2} \left(\gamma _1+\frac{\gamma_0^2}{2}-\frac{\pi ^2}{12}-q-\frac{q^2}{2}\right)\right),\quad q=W_{-1}(-x \log2),$$ where $W_{-1}(x)$ is the non-principal real-valued branch of the Lambert W-function, and $\gamma_n$ are the Stieltjes constants (in particular, $\gamma_0$ is the Euler–Mascheroni constant $\gamma$).

Vladimir Reshetnikov
  • 47,122