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Recently, I made the following question Integrating the Fabius function $q(x)=\int_0^x F(u)\ du$ trying to make a function that would behave like a continuous smooth version of a Random walk by integrating the smooth Fabius function, which lobes alternates from positive to negative values following the Thue–Morse sequence, so it could be imagined as a fixed realizations of a random variable taking two values (as is done when summing the binary $\pm1$ random variables to have a random walk).

But for my surprise, as shown by @MartinR answer:

The Fabius function satisfies the “functional differential equation” $$ F'(x) = 2 F(2x) \, , $$ therefore is $$ \int_0^x F(u) \, du = \int_0^{x/2} 2 F(2t) \, dt = \int_0^{x/2} F'(t)\, dt = F(x/2) \, . $$

So we have again another scaled Fabius function, meaning that the sequence of lobes never accumulates such as the function lobes detaches from the x-axis.

I believe that this proves the following:

Let define the sequence $S(n)$ as: $$S(n) = \sum_{k=0}^n (-1)^{\text{ThueMorse}[k]}$$ then $S(n)$ only can take the values $\{-1,\ 0,\ 1\}$ for all integer $n\geq 0$... $S(n)$ will never accumulates or add up to something different, even when Wolfram-Alpha says something against this,

Is this $|S(n)| \leq 1, \forall n\geq 0$ conjecture true?

Joako
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Yes, because if we split the Thue-Morse sequence into pairs, they all will be either $01$ or $10$, so for any even $n$ we have $S(n)=0$, and for odd $n$ $S(n)$ differs by $1$.

WolframAlpha is right when it says that the series diverges: it doesn't mean that it converges to $\infty$, it only means that the limit does not exist.

colt_browning
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  • Thanks for answering. I didn't noticed before that the TM series never have groups of contiguous ones or zeros bigger than $2$. – Joako Oct 04 '23 at 00:31