A non-recursive, explicit formula for the Fabius function

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?

• Is there a question, or is this just a way to provide a place to put your answer? – LSpice Jan 23 at 14:41
• @LSpice : Indeed, I may have to think of a better way to present the MathSE question. – Iosif Pinelis Jan 23 at 14:44
• @LSpice : Now I have an explicit question. – 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 $$x$$ $$\newcommand\vp{\varepsilon}$$ $$$$F_n(x)=\text{vol}_n(I^n\cap H_{n;c^{(n)},x}) =\frac1{n!\prod_1^n c_i}\,\sum_{\vp\in\{0,1\}^n}(-1)^{|\vp|}\, \big(x-c^{(n)}\cdot\vp\big)_+^n,$$$$ where $$\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$$ $$$$F_n(x)=\frac{2^{n(n+1)/2}}{n!}\,\sum_{y\in D_{n,x}}(-1)^{s(y)}\, \big(x-y\big)^n, \tag{2}$$$$ where $$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).
• 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. – Greg Zitelli Mar 5 at 23:38
• I don't know what you mean by "this formula terminates". Anyhow, the sum in (2) is over a finite set. – Iosif Pinelis Mar 6 at 3:14
• 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). – Greg Zitelli Mar 6 at 16:46
• @GregZitelli : This sequence is strictly increasing for each $x$ in $(0,1)$. – Iosif Pinelis Mar 6 at 20:34