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Math Help - Real Analysis stuff

  1. #1
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    Real Analysis stuff

    Hi guys,

    Still revising for exams! Looking at the following, any help would be appreciated!

    Consider the function f_i=PC(2PI), i=1,2
    f_1(x)=x f_2(x)=x^2 for 0<=x<2PI

    Show that f_1(x) ~ PI- 2 SUM [(sin(nx))/n] (n=1 to oo)
    ___________
    Given that the function is on (-PI,PI] and is 2PI periodic
    Calculate the Fourier Series

    xcosx~-(sinx)/2 + 2 SUM [((-1)^k)(k/((k^2)-1)]sin kx (k=2 to oo)
    ____________
    For the first I think I should be working from the fact that

    x~2 SUM [(-1)^(n-1).(Sin nx)/n for -PI<x<PI
    but I could be wrong?
    It gives almost the same but I'm not sure how or what the differences are.
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  2. #2
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    Quote Originally Posted by musicmental85 View Post
    Hi guys,

    Still revising for exams! Looking at the following, any help would be appreciated!

    Consider the function f_i=PC(2PI), i=1,2
    f_1(x)=x f_2(x)=x^2 for 0<=x<2PI

    Show that f_1(x) ~ PI- 2 SUM [(sin(nx))/n] (n=1 to oo)
    ___________
    Given that the function is on (-PI,PI] and is 2PI periodic
    Calculate the Fourier Series

    xcosx~-(sinx)/2 + 2 SUM [((-1)^k)(k/((k^2)-1)]sin kx (k=2 to oo)
    ____________
    For the first I think I should be working from the fact that

    x~2 SUM [(-1)^(n-1).(Sin nx)/n for -PI<x<PI
    but I could be wrong?
    It gives almost the same but I'm not sure how or what the differences are.
    Are you asking to find the Fourier series? If that is the case post what you have done. These problems that a long time to do.
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  3. #3
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    Joined
    Nov 2007
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    For the first part I can calculate x~2 SUM [(-1)^(n-1).(Sin nx)/n for -PI<x<PI by the following

    compute the fourier coefficients for the function gives a_n=0


    Then calculate . The is works out to be

    b_n= [2(-1)^(n-1)]/n

    whcih gives from the formula



    gives x~2 SUM [(-1)^(n-1).(Sin nx)/n

    I'm not sure how this works when you are calculating it with funtions?
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