# Thread: using periodicity of a funtion

1. ## using periodicity of a funtion

Hello!
I got the following function:

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

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

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

Letīs consider $i > n$.

If we look at
$f_{i} (0.b_{1}b_{2}b_{3} \cdots)$ why do we have by the periodicity of $f_{i}$ for i > n
$f_{i} (0.b_{1}b_{2}b_{3} \cdots) = f_{i} (0.000....0b_{n+1}b_{n+2}b_{n+3} \cdots)$ ?

I guess that the period is subtracted exactly i times, but I can not show it.
Can anyone help please? Thanks.

2. ## Re: using periodicity of a funtion

For $n, we have $f\left(\sum_{k=1}^{+\infty}b_k2^{-k}\right) = f\left(\sum_{k=1}^nb_k2^{-k} +\sum_{k=n+1}^{+\infty}b_k2^{-k}\right)$ and you have to show that $\sum_{k=1}^nb_k2^{-k} =j2^{1-i}$ for an integer $j$.

3. ## Re: using periodicity of a funtion

Originally Posted by girdav
For $n, we have $f\left(\sum_{k=1}^{+\infty}b_k2^{-k}\right) = f\left(\sum_{k=1}^nb_k2^{-k} +\sum_{k=n+1}^{+\infty}b_k2^{-k}\right)$ and you have to show that $\sum_{k=1}^nb_k2^{-k} =j2^{1-i}$ for an integer $j$.
Your advice has been great!
What is left to show is not equivalent to show that
$2^{i-1} \cdot \sum_{k=1}^{n} b_{k} 2^{-k}$ is an integer, correct?. I guess I actually have to calculate that integer j?

Thanks.

4. ## Re: using periodicity of a funtion

You have to show that $2^{i-1} \cdot \sum_{k=1}^{n} b_{k} 2^{-k} = \sum_{k=1}^{n} b_{k} 2^{i-1-k}$ is an integer. Use the fact that $k\leq n

5. ## Re: using periodicity of a funtion

Originally Posted by girdav
You have to show that $2^{i-1} \cdot \sum_{k=1}^{n} b_{k} 2^{-k} = \sum_{k=1}^{n} b_{k} 2^{i-1-k}$ is an integer. Use the fact that $k\leq n
Yeah! That is what I did. I just wasnīt sure I had to show exactly that one cause I first thought I had to calculate a specific j.

I know that $i-k > 0$ and then $i-k-1 \geq 0$. Thus the sum will be an integer.