Results 1 to 11 of 11

Math Help - Fourier Serries - Integration Help

  1. #1
    Member
    Joined
    Mar 2007
    Posts
    206
    Awards
    1

    Fourier Serries - Integration Help

    Ok, just attempted another Fourier serries for the half range sine extension of x(l-x) over the range 0<x<l. I havent finished the question yet but before i go any futher i would like to get someone else to have a look over my working if anyone has the time.

    The problem and my solution to finding the coeficient bn, are attached and was hopeing someone could take a look at my integration and make sure i didn't make a mistake somewhere. Also, is there a simple way to test to see if i have the right expression for bn?

    Thanks
    Elbarto
    Attached Thumbnails Attached Thumbnails Fourier Serries - Integration Help-question2.jpg   Fourier Serries - Integration Help-q2alljpg.jpg  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Here is how you should extend the function.

    The thin red line is the actual function.

    The dotted line is the extension to make it ODD.

    The bold red lines are just the periodical repeats.
    Attached Thumbnails Attached Thumbnails Fourier Serries - Integration Help-picture2.gif  
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Mar 2007
    Posts
    206
    Awards
    1
    Yes the sketch i did looks pretty much the same, tho i left my graph in terms of l and said the maximum occurs at y=l^2/4.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    \frac{a_0}{2}+\sum_{n=1}^{\infty} a_n \cos \frac{\pi n x}{L} + b_n \sin \frac{\pi n x}{L}
    ---

    You made the function odd, so this is a half-range sine Fourier expansion.

    We can ignore the constant term a_0 and the cosine terms a_n = 0 \mbox {for }n\geq 1.

    That means we simply need to find,
    b_n = \frac{1}{L} \int_{-L}^L f(x) \sin \frac{\pi n x}{L} dx = b_n = \frac{2}{L} \int_0^L f(x) \sin \frac{\pi n}{L}.

    We know that f(x) = x(L-x) on [0,L].

    b_n = \frac{2}{L} \int_0^L xL\sin \frac{\pi n x}{L} - x^2\sin \frac{\pi n x}{L}dx

    So now you need to boringly compute each summand.
    The first,

    \int_0^L xL\sin \frac{\pi n x}{L} dx = - \frac{L^2}{\pi n}\cos \pi n

    I did not do the second one. And I have to leave now.
    Last edited by ThePerfectHacker; May 27th 2007 at 06:41 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Mar 2007
    Posts
    206
    Awards
    1
    Theperfecthacker,
    My first integral was the same as the integral you found so I looked for a problem in working out the second part. Sure enough, I found a problem in evaluating the limits at the end of the problem. I fixed this and my answer came out as


    <br />
b_n=\frac{4L^2} {n^3\pi^3}[1-(-1)^n]<br />

    I graphed this in Mircosoft Excel for just one term in the serries and it looks pretty good. Tho i can see that the first term of the serries slightly over estimates the curve x(l-x) so i guess it makes sense that the next term should be subtracted but my serries for bn makes every odd term a possitive so can i expect the estimation of the curve x(l-x) to get worste as n approaches infinity?

    I am looking at part C and it looks alot like my series apart from the alternating integers.

    Thanks For the help.
    Elbarto
    Last edited by elbarto; May 26th 2007 at 09:49 PM.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    I check the first one again and it was correct.
    Let me just do the second one.

    \int_0^L x^2 \sin \frac{\pi n x}{L}dx

     -\frac{L}{\pi n}\cdot x^2 \cos \frac{\pi n x}{L} \big|_0^L + \frac{2L}{\pi n}\int_0^L x \cos \frac{\pi n x}{L}dx

    Look at second integral: \int_0^L x \cos \frac{\pi n x}{L} dx

    \frac{L}{\pi n}\cdot x\sin \frac{\pi n x}{L} \big|_0^L - \frac{L}{\pi n}\int_0^L \sin \frac{\pi n x}{L} dx = \frac{L}{\pi^2 n^2}\cos \frac{\pi n x}{L} \big|_0^L = (1-(-1)^n)\cdot \frac{L^2}{\pi^2 n^2}

    Return to line two, and evaluate:

    (-1)^{n+1} \cdot \frac{L^3}{\pi n} + \frac{2L}{\pi n} \left( (1-(-1)^n)\cdot \frac{L^2}{\pi^2 n^2} \right) = (-1)^{n+1} \cdot \frac{L^3}{\pi n} +(1-(-1)^n) \cdot \frac{2L^3}{\pi^3 n^3} =  \left\{ \begin{array}{c} -L^3/\pi n \mbox{ even } \\ L^3/\pi n + 4L^3/\pi^3n^3 \mbox{ odd} \end{array} \right\}

    So b_n terms are (do not forget the factor outside the integral):

     \frac{2(-1)^{n+1}L^2}{\pi n} - \frac{2}{L} \cdot c_n where c_n are desribed in that box.

    Now consider even and odd terms:
    \mbox{Even}: \frac{-2L^2}{\pi n} + \frac{2}{L} \cdot \frac{L^3}{\pi n} = 0
    \mbox{Odd}: \frac{2L^2}{\pi n} - \frac{2}{L} \cdot \frac{L^3}{\pi n} - \frac{4L^3}{\pi^3 n^3} = - \frac{4L^3}{\pi^3 n^3}

    So the terms of this sequence are only odd and they are:
     - \frac{4L^3}{\pi^3 n^3}
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    The function periodically extended as a half-range sine expansion has the following sum:

    f(x) = -\frac{4L^3}{1^3\pi ^3} \sin \frac{\pi x}{L} -  \frac{4L^3 }{3^3\pi ^3} \sin \frac{3\pi x}{L}  - \frac{4L^3}{5^3\pi^3} \sin \frac{5\pi x}{L} - ... =  -\frac{4L^3}{   \pi^3} \sum_{n=1}^{\infty} \frac{\sin\frac{(2n-1) \pi x}{L}}{(2n-1)^3}

    If L=2 we have,

    f(x) = - \frac{32}{\pi^3} \sum_{n=1}^{\infty} \frac{\sin \frac{(2n-1)\pi x}{2}}{(2n-1)^3}

    Now evaluate at x=1:

    -1=f(1) = - \frac{32}{\pi^3} \left( 1 - \frac{1}{3^3} + \frac{1}{5^3} - \frac{1}{7^3} + ... \right)

    Thus,

    1 - \frac{1}{3^3} + \frac{1}{5^3} - \frac{1}{7^3} + ... = \frac{\pi^3}{32}
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Mar 2007
    Posts
    206
    Awards
    1
    Thank you for your replies. I am just haveing a bit of trouble understanding your proceedure. when i have a look at your working, you get this.

    \frac{L}{\pi n}\cdot x\sin \frac{\pi n x}{L} \big|_0^L - \frac{L}{\pi n}\int_0^L \sin \frac{\pi n x}{L} dx = \frac{L}{\pi^2 n^2}\cos \frac{\pi n x}{L} \big|_0^L = (1-(-1)^n)\cdot \frac{L^2}{\pi^2 n^2}<br />

    but when i attempt to solve the equation i get

    \frac{L}{\pi^2 n^2}\cos \frac{\pi n x}{L} \big|_0^L = ((-1)^n-1)\cdot \frac{L}{\pi^2 n^2}<br />


    Then after multiplying through by the constants out the front i get

    \ <br />
((-1)^n-1)\cdot \frac{L}{\pi^2 n^2}\cdot -\frac{L}{\pi n}= (1-(-1)^n)\cdot \frac{L^2}{\pi^3 n^3}<br />

    what step did you take from here to get the integral in that form??

    Thank you
    Elbarto
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by elbarto View Post
    \frac{L}{\pi n}\cdot x\sin \frac{\pi n x}{L} \big|_0^L - \frac{L}{\pi n}\int_0^L \sin \frac{\pi n x}{L} dx = \frac{L}{\pi^2 n^2}\cos \frac{\pi n x}{L} \big|_0^L = (1-(-1)^n)\cdot \frac{L^2}{\pi^2 n^2}<br />
    Look at the first term,
    \frac{L}{\pi n}x\sin \frac{\pi n x}{L} \big|_0^L
    Evaluate,
    \frac{L}{\pi n}\cdot L \cdot\sin \frac{\pi n L}{L} - 0
    But,
    \frac{L}{\pi n}\cdot L \cdot\sin \frac \pi n = 0
    Because \sin \pi n =0.

    Look at the second term,
    -\frac{L}{\pi n}\int_0^L \sin \frac{\pi n x}{L} dx = \frac{L}{\pi n} \frac{L}{\pi n} \cos \frac{\pi n x}{L}\big|_0^L = \frac{L^2}{\pi^2n^2} (-1)^n - \frac{L^2}{\pi^2 n^2}
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Mar 2007
    Posts
    206
    Awards
    1
    Once again thank you for your reply. I still couldn't find the same solution so i used my integral tables rather then useing integration by parts. This is what i keep getting

    <br />
b_n=-\frac{4L^2} {n^3\pi^3}[(-1)^n-1]<br />

    Is there a possibility that you missed a negative sign in your integrals? i graphed my solution and it appears to work.


    I have attached a copy of my spread sheet if you could have a look at it?

    Regards
    Elbarto
    Attached Files Attached Files
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    I do not have Excel on the computer I am using. Nor have I ever used excel in my life. If you can do:
    n=10,20,50,100
    And post them as screenshots I would love to see that.
    I think the animation of how a Fourier Series converges to the function is beautiful to watch.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Numerical Integration of Fourier Transform.
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: November 21st 2009, 01:22 AM
  2. representing function as power serries
    Posted in the Calculus Forum
    Replies: 2
    Last Post: August 3rd 2009, 01:03 PM
  3. alternating serries test
    Posted in the Calculus Forum
    Replies: 3
    Last Post: August 1st 2009, 10:55 PM
  4. what values of p make this serries convergent
    Posted in the Calculus Forum
    Replies: 1
    Last Post: July 31st 2009, 11:41 AM
  5. converging serries
    Posted in the Calculus Forum
    Replies: 6
    Last Post: July 27th 2009, 12:40 AM

Search Tags


/mathhelpforum @mathhelpforum