How to compute the values of this function ? ( Fabius function ) 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/
 A: For the Fabius function we have $$\tag{1}F(0)=0\quad F(1)=1$$ $$F'(x)=\begin{cases} 
      2F(2x) & 0\leq x\leq 1/2 \\
      2F(2(1-x)) & 1/2\leq x\leq1
   \end{cases}\tag{2}
$$
Now for $h\in[0, 2^{-i}]$ consider the functions
$$P_i(h)=F(2^{-i}+h) - (-1)^i F(2^{-i}-h)$$
We have $$P_1'(h)=F'(1/2 + h) - F'(1/2 - h) \overset{(2)}{=} 0$$
hence $P_1(h)$ is a constant, which equals $1$ because of $(1)$.
For $i\geq 2$ using the first case of $(2)$ yields
$$P_i'(h)=2F(2^{1-i}+2h) - 2(-1)^{i-1} F(2^{1-i}-2h) = 2P_{i-1}(2h)$$
That is, $P_{i}(h)$ is just the integral of $2P_{i-1}(2h)$, i.e. a degree $i-1$ polynomial with coefficients that are simple to find, except for the additive constant from the integration. For the even $i$s, the constant can be found by solving $P_i(0) = 0$, and for the uneven $i$s, one needs to solve $P_i(0) = 2F(2^{-i})$.
This means that for $i < n$, $P_{i}(h)$ is easily found when knowing $F(2^{-i})$ for $i \in \{1, 3,...,N\}$
Knowing $P_i(h)$ allows us to solve $F(2^{1-i})$ solely in terms of $F(1)$ and $F(2^{-j})$ for $j\in\{1,3,...,i-2\}$. Doing so for the even $i\leq N$ turns out to make up a lower triangular linear system of equations with $U =2\lfloor N/2 \rfloor$ unknowns and rank $U-1$, because the first equation is $F(1/2) = F(1/2)$. However, the value of $F(1/2)$ is obviously $1/2$, by means of which forward substitution solves the rest.
Now that you know $P_i(h)$ for however many $i$s you'd like, you can approximate the value of $F(z)$ for any $z\in[0,1]$ by reflecting the $z$-value about the relevant symmetry-axes among $x=1/2$, $x=1/4$, $x=1/8$, etc. while solving the relevant $P_i(h)$-equation in terms of another Fabius function value which converge super fast towards 0, if you reflect your $z$ towards $x=0$.
On runtime the needed operations increases quadratically w.r.t. amount of reflections. Using the method for $z=2/3$ causes the need of a reflection about all of the axes because it's binary representation is $0.101010101$... That makes it a good choice to see the significant figures relative to the amount of reflections, which looks very much like a quadratic function as well:

In case anyone is interested, I created the method in Mathematica:

A: 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.
