Fourier series

Discussion in 'Math' started by jagjit Sehra, Mar 17, 2008.

  1. jagjit Sehra

    Thread Starter Member

    Feb 29, 2008
    25
    0
    Hi Folks,
    Got a question regarding how to calculate the Fourier coefficents a0, an and bn.

    I have a problem with the orthogonal trig identities which are used to identify the particular co-efficent. I know it works and I can just take that and accept, however I would like to see a proof of it. The identitiy is as follows:
    I dont now how to show the integral sign so I will use (int instead) sorry!

    2*pi
    int cos mx. cos nx dx = 0 ( if m does not equal n)
    0 = 1 ( if m does equal n(therefore identifying term))

    Identity used

    cos mx. cos nx = = 1/2 (cos(m+n)x + cos(m-n)x)

    substuting this identity into the integral above.

    2*pi
    1/2 int [(cos(m+n)x + cos(m-n)x)]dx
    0

    = | sin(m+n)x sin(m-n)x | 2*pi
    | ---------- + ---------- |
    | (m+n) (m-n) |0

    I finally get to my question
    According to the orthogonal feature of this, the result will be 1 only when m=n
    all other values of m and n will result in the reslut bieng 0.

    The thing I cant understand is taht in each part of the integral the sin term is been multiplied by a multiple of 2*pi or 0, surly any multiple of that result in a zero output. So how is it possible to get a result of 1 when m equals n.

    I know I am wrong but I would appreciate it if someone will tell me where I am going wrong. I have tried doing it on my calculator and it always works out to be 0.

    Jag.
     
  2. Mark44

    Well-Known Member

    Nov 26, 2007
    626
    1
    Jagjit,
    I'll represent the integral you started with in this way:
    int[cos(mx)cos(nx)dx;0;2*pi]

    If m = n, the integral simplifies to
    int[(cos(mx)^2)dx; 0; 2*pi], which comes out for me as pi, not 1 as you stated.

    If m <> n, the original integral becomes
    1/2 * int[ (cos(m+n)x +cos(m-n)x)dx; 0; 2*pi]

    which is 1/2 * [ sin(m+n)x/(m+n) + sin(m-n)x/(m-n), evaluated at 2*pi and 0.
    Both sine terms are zero for integer multiples of 2pi and for x=0, so that integral's value is zero.

    In summary, the integral you're working with is nonzero only if m = n. This shows that the functions cos(mx) and cos(nx) are orthogonal when m and n are different.
    Mark
     
  3. jagjit Sehra

    Thread Starter Member

    Feb 29, 2008
    25
    0
    Cheers Mark,
    I can see where I went wrong.
    Jag.
     
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