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Fourier Integral

  1. Nov 7, 2005 #1
    Can someone please show me how to solve these three integrals, or proivde a link that does. I cant think of a way to solve it right now and I dont have any free time to do so. Thank you!

    [tex] \int cos(nx)cos(mx)dx [/tex]

    [tex] \int sin(nx)sin(mx)dx [/tex]

    [tex] \int sin(nx)cos(mx) dx [/tex]

    The thing that makes it tricky is the different coefficients of m and n in each trig term, so you cant make them cos^2(x) or sin^2(x) or use trig identities. (at least I cant see how). Anywho, thanks once more.

    Cyrus. :smile:
    Last edited: Nov 7, 2005
  2. jcsd
  3. Nov 7, 2005 #2


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    Trig identities.

    sin(a+ b)= sin(a)cos(b)+ cos(a)sin(b)
    sin(a- b)= sin(a)cos(-b)+ cos(a)sin(-b) but cos(-b)= cos(b) sin(-b)= -sin(b)
    = sin(a)cos(b) - cos(a)sin(b)

    Adding, sin(a+b)+ sin(a- b)= 2sin(a)cos(b).

    To integrate sin(nx)cos(mx) let a= nx, b= mx to write
    sin(nx)cos(mx)= (1/2)sin((m+n)x)+ (1/2)sin((m-n)x).

    cos(a+ b)= cos(a)cos(b)- sin(a)sin(b)
    cos(a- b)= cos(a)cos(b)+ sin(a)sin(b)

    Adding, cos(a+b)+ cos(a-b)= 2cos(a)cos(b)
    To integrate cos(nx)cos(mx), let a= nx, b= mx to write
    cos(nx)cos(mx)= (1/2)cos((m+n)x)+ (1/2)cos((m-n)x)

    Subtracting, cos(a+b)- cos(a-b)= 2sin(a)sin(b)
    To integrate sin(nx)sin(mx), let a= nx, b= mx to write
    sin(nx)sin(mx)= (1/2)cos((m+n)x)- (1/2)cos((m-n)x)
  4. Nov 7, 2005 #3
    Im not following how this is true, pluging in a value for b and none of them are comming out equal?

    Ahhh, I figure your just missing a comma inbetween there. You meant cos(-b) = cos(b) and sin(-b)= - sin(b).

    Thats a VERY clever trick to solve it. I would never have spotted it! Thanks!
    Last edited: Nov 7, 2005
  5. Nov 10, 2005 #4
    A follow up, can you show me how the series is reduced, its been a while since i touched a series, a long while.

    [tex] cos(n \pi) = (-1) ^n [/tex]

    My book says [tex] cos(x) = (-1)^n x^{2n} / (2n!) [/tex]

    what happened to the fractional part?

    The n is different for each equation, poor notation sorry.

    Thanks twice.

    Last edited: Nov 10, 2005
  6. Nov 11, 2005 #5


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    Isn't it more like

    [tex] cos(x) = \sum_{n=0}^{\infty}(-1)^n x^{2n} / (2n!) [/tex].

    Your question is strangely worded. What do you mean by reduced?
  7. Nov 11, 2005 #6
    oops your right.
    My question is how did they arrive at this:

    [tex] cos(n \pi) = (-1) ^n [/tex]

    oh, I see now, because its just positive of negative 1 depending on the coefficent. Crap. never mind........................I got it now. :-)
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