13
$\begingroup$

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/

$\endgroup$
1

2 Answers 2

9
$\begingroup$

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:

enter image description here

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

enter image description here

$\endgroup$
2
  • 4
    $\begingroup$ I re-posted the Mathematica code from the screenshot in plain text format here. $\endgroup$ Commented Feb 12, 2017 at 1:31
  • $\begingroup$ I upvoted this answer. Hesitated to accept. $\endgroup$
    – mick
    Commented Nov 12, 2018 at 10:05
6
+450
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ An efficient algorithm to compute values of the Fabius function (for all arguments $0\le x\le1$) implemented in the Wolfram Language can be found here. It gives exact rational values for dyadic rational arguments and arbitrary precision values for other arguments. $\endgroup$ Commented Feb 14, 2017 at 18:18
  • $\begingroup$ A paper explaining some background for this algorithm: J. Arias de Reyna, Arithmetic of the Fabius Function. $\endgroup$ Commented Jun 14, 2018 at 5:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .