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Math Help - Fourier Series

  1. #1
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    Fourier Series

    Hey, I have some questions from my Differential class that I am really confused on. I still don't completely understand Fourier, but I really don't understand this. I don't want the answers, just someone to explain how to go about doing these. Thanks!

    Problem 1


    Let f(t)=t^2 for 0<t<2 and suppose f(t+2)=f(t) for all t  \epsilon R-2Z. By 2Z, we denote even integers. Define f(2k)=1 for k \epsilon Z. Calculate the Fourier Series for f.

    Problem 2


    Show, by the result of the last problem, \sum_{n=1}^\infty \frac{\sin{nt}}{n}=\frac{\pi-t}{2}. Derive a series which converges to \pi by setting t to a particular value and doing a little algebra. How many terms in this series would you have to sum in order to obtain the digits 3.14 in \pi with certainty?
    Last edited by shumaterm; April 20th 2013 at 04:53 PM.
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  2. #2
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    Re: Fourier Series

    Hey shumaterm.

    For problem 1, calculate a_n and b_n using the standard formula's and show us what you got as an answer.
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  3. #3
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    Re: Fourier Series

    would a_0 be 1 and a_n be \frac{4(\pi n \sin{\pi n} + \cos{\pi n}-1)}{\pi^2n^2}? I know the general forumla for t\epsilon(-\pi,\pi), but I'm not sure what happens when the interval changes to a different period... Thank you for your help. I really do appreciate it.
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  4. #4
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    Re: Fourier Series

    Oops. also, in problem 1, f(t)=t, not f(t)=t^2.
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  5. #5
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    Re: Fourier Series

    Its going to be a lot better if you setup the integral.

    Hint: The Integrals will involve sin(pi*n*t)*f(t)dt and cos(pi*n*t)*f(t)dt as well as f(t)dt. (You will have to normalize them as well).
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  6. #6
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    Re: Fourier Series

    So would a_0=\int_0^2 f(t)dt, a_n=\int_0^2 f(t)cos(\frac{\pi n t}{2})dt, and b_n=\int_0^2 f(t)sin(\frac{\pi n t}{2})dt? I don't know where these really originate from or how to normalize the function, my professor didn't really explain it. Sorry for my ignorance haha.
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  7. #7
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    Re: Fourier Series

    I suggest you take a look at this page:

    Fourier series - Wikipedia, the free encyclopedia

    Basically the idea is that you take the original fourier series (which is defined over [-pi,pi)) and you scale them for the general interval.

    This gives the integration with respect to cos(2*pi*n*t/tau) and sin(2*pi*n*t/tau) where tau is the length of the interval f(t + tau) = f(t).

    To normalize the integrals you need to calculate Integral over Main Branch cos^2(2*pi*n*t/tau)dt and Integral over main branch sin^2(2*pi*n*t/tau)dt and take the square roots of these and divide all your a_n's and b_n's by this square root value.

    It has the same meaning as if you had a normal vector v and normalize by calculating v / ||v|| where ||v|| is the length of the vector.

    The normalization for a_0 will be sqrt(2) since Integral (0 to 2) 1dt = 2 and taking square root is SQRT(2).

    Once you calculate the normalized values, you can then plug them into the fourier series expansion.
    Thanks from shumaterm
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  8. #8
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    Re: Fourier Series

    Ok. I think I understand the scaling. Normalization is still a little unclear to me. Why do we choose \cos^2{\frac{2 \pi n t}{\tau}}dt to integrate? Is this just finding the " ||v||"? Which you find by the length? I think it's making more sense.


    Why would you find a_0 by doing \int_0^2 1dt and not \int_0^2 t dt? This is where the first part doesn't make sense to me. In my mind, I would find \int_0^2 t dt and then divide by the length, since the function of t goes from 0 to 2, the length of the vector would be 2\sqrt{2}. That would make a_0=\sqrt{2}?


    In the same way, to find a_n I would do \int_0^2 t \cos{\pi n t} which would give me \frac{2\sin{2 \pi n}}{\pi n}+\frac{\cos{2 \pi n}}{\pi^2 n^2}-\frac{1}{\pi^2 n^2}, and then I'm not sure. haha
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  9. #9
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    Re: Fourier Series

    Yes its basically the idea of finding the length and dividing by it: just like in a normal vector in R^n (or a similar space).

    The integral will be t^2 instead of t by the way since f(t) = t^2 not t.
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  10. #10
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    Re: Fourier Series

    Ok, I got some help from the tutoring center yesterday. I think I understand it now. Would you mind to check and see if I'm doing this correctly? Also, I meant for f(t)=t, not t^2. My apologies.

    a_0=\frac{1}{2}\int_{-2}^2 f(t)dt=\frac{1}{2}\int_{-2}^0 (t+2)dt + \frac{1}{2}\int_0^2 tdt=2

    a_n=\frac{1}{2}\int_{-2}^2 f(t)cos{\frac{\pi n t}{2}}dt=\frac{1}{2}\int_{-2}^0(t+2)cos{\frac{\pi n t}{2}}dt+\frac{1}{2}\int_0^2 t cos{\frac{\pi n t}{2}}dt which after some calculations we find a_n=0

    b_n=\frac{2}{\pi n}[(-1)^n-1] which was found by the same method as we found a_n

    Then we get f = 2+\sum_{n=1}^\infty\frac{2}{\pi n}[(-1)^n-1]sin{\frac{\pi n t}{2}}

    Does this seem somewhat correct?

    Thank you so much for your help.
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  11. #11
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    Re: Fourier Series

    The answer for a0 is not right since you have to normalize it. This means dividing by SQRT(2) which would leave a0 = 2/SQRT(2) = SQRT(2).

    I have no idea what you are doing with the limits though: the principle branch is [0,2) but you are calculating limits for [-2,2). If the limits are really [-2,2) as opposed to [0,2) then you have to use different normalization constants as well as different scaling factors within the sine and cosine terms.
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  12. #12
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    Re: Fourier Series

    Gotcha. I went and talked to my professor yesterday. I have it figured out now. Thank you so much for your help!
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