Calculating integral

**Tahoe** · June 29th 2011, 10:30 AM

Hello!

I am having some problems proving the following:
The function $f_{i}(x) = \int_{0}^{x} \left(-1 \right)^{\lfloor 2^{i} \cdot t \rfloor} \ dt$ has a period of $2^{1-i}$ where $x \in \left[0, 1\right]$ and $i \geq 1$ .

I got to prove that
$f_{i} \left(x + 2^{1-i} \right) = f_{i} \left(x \right)$

What I did was the following:
$f_{i} \left(x + 2^{1-i} \right) = \int_{0}^{x} \left(-1 \right)^{\lfloor 2^{i} \cdot t \rfloor} \ dt} + \int_{x}^{x + 2^{1-i}} \left(-1 \right)^{\lfloor 2^{i} \cdot t \rfloor} \ dt = f_{i} (x) + \int_{x}^{x + 2^{1-i}} \left(-1 \right)^{\lfloor 2^{i} \cdot t \rfloor} \ dt$

What is left to prove is that the integral
$\int_{x}^{x + 2^{1-i}} \left(-1 \right)^{\lfloor 2^{i} \cdot t \rfloor} \ dt$ takes the value 0.

How can I show the last step?

Thanks for your help! I appreciate it.

**chisigma** · June 29th 2011, 09:55 PM

Originally Posted by Tahoe

Hello!

I am having some problems proving the following:
The function $f_{i}(x) = \int_{0}^{x} \left(-1 \right)^{\lfloor 2^{i} \cdot t \rfloor} \ dt$ has a period of $2^{1-i}$ where $x \in \left[0, 1\right]$ and $i \geq 1$ .

I got to prove that
$f_{i} \left(x + 2^{1-i} \right) = f_{i} \left(x \right)$

What I did was the following:
$f_{i} \left(x + 2^{1-i} \right) = \int_{0}^{x} \left(-1 \right)^{\lfloor 2^{i} \cdot t \rfloor} \ dt} + \int_{x}^{x + 2^{1-i}} \left(-1 \right)^{\lfloor 2^{i} \cdot t \rfloor} \ dt = f_{i} (x) + \int_{x}^{x + 2^{1-i}} \left(-1 \right)^{\lfloor 2^{i} \cdot t \rfloor} \ dt$

What is left to prove is that the integral
$\int_{x}^{x + 2^{1-i}} \left(-1 \right)^{\lfloor 2^{i} \cdot t \rfloor} \ dt$ takes the value 0.

How can I show the last step?

Thanks for your help! I appreciate it.

If a function $f(t)$ is periodic of period $2 \tau$ , then it can be expanded in Fourier series as...

$f(t)= \frac{a_{0}}{2} + \sum_{n=1}^{\infty} (a_{n}\ \cos \frac{\pi\ n\ t}{\tau} + b_{n}\ \sin \frac{\pi\ n\ t}{\tau})$ (1)

... where...

$a_{n}= \frac{1}{\tau}\ \int_{0}^{2 \tau} f(t)\ \cos \frac{\pi\ n\ t}{\tau}\ dt$

$b_{n}= \frac{1}{\tau}\ \int_{0}^{2 \tau} f(t)\ \sin \frac{\pi\ n\ t}{\tau}\ dt$ (2)

If we consider the integral...

$F(t)= \int_{0}^{t} f(x)\ dx$ (3)

... it is easy to see that if $a_{0}=0$ then $F(t)$ is also periodic of period $2 \tau$ . Now $\varphi_{i} (t) = (-1)^{\lfloor 2^{i}\ t \rfloor}$ is periodic of period $2^{1-i}$ and for it is $a_{0}=0$ , so that also...

$f_{i} (t)= \int_{0}^{t} \varphi_{i} (x)\ dx$ (4)

... is periodic of period $2^{1-i}$ ...

Kind regards

$\chi$ $\sigma$

**Tahoe** · June 30th 2011, 02:28 AM

Originally Posted by chisigma

If a function $f(t)$ is periodic of period $2 \tau$ , then it can be expanded in Fourier series as...

$f(t)= \frac{a_{0}}{2} + \sum_{n=1}^{\infty} (a_{n}\ \cos \frac{\pi\ n\ t}{\tau} + b_{n}\ \sin \frac{\pi\ n\ t}{\tau})$ (1)

... where...

$a_{n}= \frac{1}{\tau}\ \int_{0}^{2 \tau} f(t)\ \cos \frac{\pi\ n\ t}{\tau}\ dt$

$b_{n}= \frac{1}{\tau}\ \int_{0}^{2 \tau} f(t)\ \sin \frac{\pi\ n\ t}{\tau}\ dt$ (2)

If we consider the integral...

$F(t)= \int_{0}^{t} f(x)\ dx$ (3)

... it is easy to see that if $a_{0}=0$ then $F(t)$ is also periodic of period $2 \tau$ . Now $\varphi_{i} (t) = (-1)^{\lfloor 2^{i}\ t \rfloor}$ is periodic of period $2^{1-i}$ and for it is $a_{0}=0$ , so that also...

$f_{i} (t)= \int_{0}^{t} \varphi_{i} (x)\ dx$ (4)

... is periodic of period $2^{1-i}$ ...

Kind regards

$\chi$ $\sigma$

Thanks for your answer!

Why it is easy to see that if $a_{0}=0$ then $F(t)$ is also periodic of period $2 \tau$ ?

I guess you meant
$f_{i} (x)= \int_{0}^{x} \varphi_{i} (t)\ dt$ instead of
$f_{i} (t)= \int_{0}^{t} \varphi_{i} (x)\ dx$ .

I can see that your approach seems to be a solution to the problem, but on the other hand there must also be a way to show the periodicity the way I approached it cause that seems to be the most obvious one.

Thanks for your help!

Regards

**Tahoe** · June 30th 2011, 02:56 PM

I think the idea using a Fourier series to prove the whole thing is a good one but it is not a solution which comes automatically to somebody´s mind.

Can anyone think of a solution starting the way I did in my first posting? There must be a way to prove that the integral left takes only the value 0. I just can´t see at the moment how to solve it though. Does anyone have an idea?

Thanks.

Kind regards

**Tahoe** · July 3rd 2011, 07:54 AM

Does anyone have any further ideas? Thank you.

**Opalg** · July 3rd 2011, 08:46 AM

Originally Posted by Tahoe

What is left to prove is that the integral
$\int_{x}^{x + 2^{1-i}} \left(-1 \right)^{\lfloor 2^{i} \cdot t \rfloor} \ dt$ takes the value 0.

How can I show the last step?

In any interval $x\leqslant t\leqslant x+2$ of length 2, the integer $\lfloor t\rfloor$ will be even for a total length of exactly half the length of the interval, and odd for the other half. It follows that in any interval $x\leqslant t\leqslant x+2^{1-i}$ of length $2^{1-i}$ , $\lfloor 2^it\rfloor$ will again be even for half the length of the interval, and odd for the other half. Therefore $(-1)^{\lfloor 2^it\rfloor}$ will be +1 over half the interval, and –1 over the other half, and so its integral over the whole interval will be 0.

**Tahoe** · July 3rd 2011, 09:10 AM

Hello Opalg!

Thanks for your post. Your solution to the problem is great. Once again thanks and I do appreciate the time you took to read my post!

Regards

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