using periodicity of a funtion

**Tahoe** · July 8th 2011, 01:43 PM

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.

Let´s 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.

**girdav** · July 8th 2011, 01:52 PM

For $n<i$ , 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 $j$ .

**Tahoe** · July 8th 2011, 02:56 PM

Originally Posted by girdav

For $n<i$ , 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.

**girdav** · July 8th 2011, 03:03 PM

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 $k\leq n<i$

**Tahoe** · July 8th 2011, 03:19 PM

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 $k\leq n<i$

Yeah! That is what I did. I just wasn´t 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.

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