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Please help integration using rodrigues formula.

  1. Jun 5, 2010 #1
    The book gave the integration of a function with the legendre polynomial formula:

    [tex]\int_{-1}^1 f(x)P_n(x)dx = \frac{1}{2^n n!}\int_{-1}^{1} f^n(x)(x^2-1)^n dx[/tex]

    It just said the formula can be obtained by repeat using Rodrigues formula and integral by parts but did not go into detail. I want to work out the steps and I got stuck. This is what I have:

    Using Rodrigues:

    [tex]\int_{-1}^{1} f(x)P_n(x)dx = \frac{1}{2^n n!}\int_{-1}^{1} f(x)\frac{d^n}{dx^n}[(x^2-1)^n] dx[/tex]

    After the first integral by parts:

    [tex]\frac{1}{2^n n!}\int_{-1}^{1} f(x)\frac{d^n}{dx^n}[(x^2-1)^n] dx = \frac{1}{2^n n!}[f(x)\frac{d^{n-1}}{dx^{n-1}}[(x^2-1)^n]]_{-1}^1 \;-\; \frac{1}{2^n n!}\int_{-1}^{1} f^{(1)}(x)\frac{d^{n-1}}{dx^{n-1}}[(x^2-1)^n] dx[/tex]

    In order for this to continue to the next integration by parts, the following has to be true:

    [tex]\frac{1}{2^n n!}[f(x)\frac{d^{n-1}}{dx^{n-1}}[(x^2-1)^n]]_{-1}^1 = 0[/tex]

    [tex]\Rightarrow\; [\frac{d^{n-1}}{dx^{n-1}}[(x^2-1)^n]]_{-1}^1 = 0[/tex]

    I don't know whether my assumption is correct. If so, I still don't know how it is equal to zero. can anyone give me some guidance.
     
    Last edited: Jun 5, 2010
  2. jcsd
  3. Jun 5, 2010 #2

    LCKurtz

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    I haven't checked you work, but assuming it is correct, this step:

    [tex]
    \Rightarrow\; [\frac{d^{n-1}}{dx^{n-1}}[(x^2-1)^n]]_{-1}^1 = 0
    [/tex]

    can be written as

    [tex]
    [\frac{d^{n-1}}{dx^{n-1}}[(x+1)^n(x-1)^n]]_{-1}^1 = 0
    [/tex]

    Using the product rule differentiating n-1 times I think you will find at least a factor of (x-1)(x+1) in every term of the expansion. That would give you the answer 0.
     
  4. Jun 6, 2010 #3
    Thanks for you help.
     
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