The Fabius function $F\colon\mathbb R\to[-1,1]$ may be defined as the unique solution of the functional integral equation $F(x)=\int_0^{2x}F(t)\,dt$ for all real $x$ such that $F(1)=1$.

The recent MO post provided a link to the MathSE question, asking to confirm a conjectured "non-recursive, self-contained formula for the Fabius function". The MO post has been overall negatively received and may get closed. I think the mentioned MathSE question may be of interest.

On this page, whereas the mentioned conjectured formula will not be confirmed so far, a simpler non-recursive, explicit formula for the Fabius function will be offered, which is expressed in terms similar to, but simpler than, those in the conjectured formula.

So, a question yet remains: Can one use the simpler formula below to confirm the conjecture on MathSE? Or maybe one could do that otherwise?

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    $\begingroup$ Is there a question, or is this just a way to provide a place to put your answer? $\endgroup$ – LSpice Jan 23 at 14:41
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    $\begingroup$ @LSpice : Indeed, I may have to think of a better way to present the MathSE question. $\endgroup$ – Iosif Pinelis Jan 23 at 14:44
  • $\begingroup$ @LSpice : Now I have an explicit question. $\endgroup$ – Iosif Pinelis Jan 23 at 14:50

As noted in the linked Wikipedia article on the Fabius function, on the interval $I:=[0,1]$ the Fabius function coincides with the cumulative distribution function (cdf) of $$\sum_{j=1}^\infty 2^{-j}U_j,$$ where the $U_j$'s are independent random variables uniformly distributed on $I$. So, for each $x\in I$ $$F(x)=\lim_{n\to\infty} F_n(x),\tag{1}$$ where $F_n$ is the cdf of $\sum_{j=1}^n 2^{-j}U_j$.

Next (see e.g. formula (2.2)), for any real $x$ $\newcommand\vp{\varepsilon}$ \begin{equation} F_n(x)=\text{vol}_n(I^n\cap H_{n;c^{(n)},x}) =\frac1{n!\prod_1^n c_i}\,\sum_{\vp\in\{0,1\}^n}(-1)^{|\vp|}\, \big(x-c^{(n)}\cdot\vp\big)_+^n, \end{equation} where $\text{vol}_n$ is the Lebesgue measure on $\mathbb R^n$, $H_{n;b,x}:=\{v\in\mathbb R^n\colon b\cdot v\le x\}$, $c^{(n)}:=(c_1,\dots,c_n)$, $c_j:=2^{-j}$, $|\vp|:=\vp_1+\dots+\vp_n$, $\cdot$ denotes the dot product, and $t_+^n:=\max(0,t)^n$. So, for $x\in I$ \begin{equation} F_n(x)=\frac{2^{n(n+1)/2}}{n!}\,\sum_{y\in D_{n,x}}(-1)^{s(y)}\, \big(x-y\big)^n, \tag{2} \end{equation} where $D_{n,x}$ is the set of all dyadic numbers in $[0,x]$ of the form $m2^{-n}$ for integers $m$, and $s(y)$ is the sum of the binary digits of $y$.

Formulas (1) and (2) provide the answer to the question.

Some of the differences between (1)--(2) and the formula conjectured in the linked MathSE post are as follows:

  • In the MathSE post, the main conjectured formula for $F(x)$ is stated only for dyadic numbers $x$, and then extended to all values of $x$ by the continuity of $F$.
  • The expression in that post for $F(x)$ for dyadic $x$ contains a double summation and a number of $q$-Pochhammer symbols (I have had no experience with those symbols).
  • $\begingroup$ I'm having trouble seeing how this formula terminates at a finite sum for dyadic x, or perhaps it doesn't. This seemed to be one of the goals of the original MathSE post. $\endgroup$ – Greg Zitelli Mar 5 at 23:38
  • $\begingroup$ I don't know what you mean by "this formula terminates". Anyhow, the sum in (2) is over a finite set. $\endgroup$ – Iosif Pinelis Mar 6 at 3:14
  • $\begingroup$ I mean that this approach takes in a value x in I and produces a sequence of numbers which converge to F(x). However, if x is a dyadic rational then F(x) is known to be rational, and it is not clear if the sequence of values that this method produces will return that value exactly after a finite number of steps. I think this would be a desirable quality of a "solution" to the problem of finding a fast approximation to F(x). $\endgroup$ – Greg Zitelli Mar 6 at 16:46
  • $\begingroup$ @GregZitelli : This sequence is strictly increasing for each $x$ in $(0,1)$. $\endgroup$ – Iosif Pinelis Mar 6 at 20:34

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