# 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 $$x$$ $$\newcommand\vp{\varepsilon}$$ $$\begin{equation} F_n(x)=\text{vol}_n(I^n\cap H_{n;c^{(n)},x}) =\frac1{n!\prod_1^n c_i}\,\sum_{\vp\in\{0,1\}^n}(-1)^{|\vp|}\, \big(x-c^{(n)}\cdot\vp\big)_+^n, \end{equation}$$ 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 $$y$$.
• 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).