TITLE: How to compute the values of this function ? ( Fabius function )
QUESTION [14 upvotes]: How to compute the values of this function ? ( Fabius function )
It is said not to be analytic but $C^\infty$ everywhere.
But I do not even know how to compute its values. Im confused.
Here is the link : http://www.math.osu.edu/~edgar.2/selfdiff/

REPLY [7 votes]: The Fabius function assumes rational values at dyadic rational arguments. We will give an explicit formula for those values. Because dyadic rationals form a dense subset of the reals, and the Fabius function is continuous, its value at any real point can be found as a limit of its values on a sequence of dyadic rationals converging to that point.
The following is based on the book Rvachev V. L., Rvachev V. A., "Non-classical methods of the approximation theory in boundary value problems", Naukova Dumka, Kiev (1979), pp. 116-126$^{[1]}$. The book is in Russian, and as far as I know, no English translation exists. But, fortunately, it mostly consists of math formulae.
Let's define $t_n$ by the recurrence
$$t_0 = 1, \quad t_n = (-1)^n \, t_{\lfloor n/2\rfloor}\tag1$$
It is easy to see that $|t_n|=1$, and the signs follow the same pattern as  the Thue–Morse sequence. The same sequence is given by a non-recurrent, but rather clumsy formula
$$t_n = 1-\frac83 \, \sin^2\left(\frac\pi3 \cdot \sum_{m=1}^n \left[2 + (-1)^{\binom n m}\right]\right)\tag2$$
Let's define $c_n$ by the recurrence
$$c_0 = 1, \quad c_n = \frac1{(4^n - 1)(2n+1)} \, \sum_{m\ge1} \binom{2n+1}{2m+1} \, c_{n-m},\tag3$$
so we have
$$c_1 = \frac19, \quad c_2 = \frac{19}{675}, \quad c_3 = \frac{583}{59535}, \quad c_4 = \frac{132809}{32531625}, \quad \text{etc.}\tag4$$
Note that only finite number of terms in the sum are non-zero. I am not aware of any non-recurrent formula for this sequence, but it would be very nice to find one.
The values of the Fabius function at dyadic rationals are given by
$$F\!\left(\tfrac{s}{2^n}\right) = \frac1{n! \, 2^{\binom{n+1}2}} \, \sum_{m\ge0}\binom n {2m} \, c_m \sum_{1 \le r \le s}(2r-1)^{n-2m} \, t_{s-r}.\tag5$$
Again, note that only finite number of terms in each sum are non-zero.