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Math Help - Calculating integral

  1. #1
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    Calculating integral

    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.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Calculating integral

    Quote Originally Posted by Tahoe View Post
    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
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  3. #3
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    Re: Calculating integral

    Quote Originally Posted by chisigma View Post
    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
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  4. #4
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    Re: Calculating integral

    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
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  5. #5
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    Re: Calculating integral

    Does anyone have any further ideas? Thank you.
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  6. #6
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    Opalg's Avatar
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    Re: Calculating integral

    Quote Originally Posted by Tahoe View Post
    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.
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  7. #7
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    Re: Calculating integral

    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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