Hello!

I got the following function:

$\displaystyle f_{i} (x) = \int_{0}^{x} (-1)^{\lfloor{t \cdot 2^{i}}\rfloor} \ dt$

The function has a periodicity of $\displaystyle 2^{1-i}$.

Let $\displaystyle x_{0}=0.b_{1}b_{2}b_{3} \cdots$ be the binary expansion of a number between 0 and 1.

Letīs consider $\displaystyle i > n$.

If we look at

$\displaystyle f_{i} (0.b_{1}b_{2}b_{3} \cdots)$ why do we have by the periodicity of $\displaystyle f_{i}$ for i > n

$\displaystyle f_{i} (0.b_{1}b_{2}b_{3} \cdots) = f_{i} (0.000....0b_{n+1}b_{n+2}b_{n+3} \cdots) $ ?

I guess that the period is subtracted exactly i times, but I can not show it.

Can anyone help please? Thanks.