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Generating function for Legendre polynomials

  1. May 8, 2010 #1
    1. The problem statement, all variables and given/known data

    Using binomial expansion, prove that

    [tex]
    \frac{1}{\sqrt{1 - 2 x u + u^2}} = \sum_{k} P_k(x) u^k.
    [/tex]

    2. Relevant equations

    [tex]
    \frac{1}{\sqrt{1 + v}} = \sum_{k} (-1)^k \frac{(2k)!}{2^{2k} (k!)^2} v^k
    [/tex]

    3. The attempt at a solution

    I simply inserted [itex]v = u^2 - 2 x u[/itex], then expanded the [itex]v^k[/tex] to obtain the double sum

    [tex]
    \sum_{k} (-1)^k \frac{(2k)!}{2^{2k} (k!)^2} \sum_{n \leq k} \left( \begin{array}{c} k \\ n \end{array} \right) (-2 x)^n u^{2 k - n}.
    [/tex]

    Now I need to turn this into a single sum by collecting like powers of [itex]u[/itex], which is what I'm stuck at. I don't see how to go about that.
     
  2. jcsd
  3. May 9, 2010 #2

    vela

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    You might find this relationship helpful. It's just a rearrangement of terms in a series.

    [tex]\sum_{p=0}^\infty\sum_{q=0}^p\,a_{q,p-q} = \sum_{r=0}^\infty\sum_{s=0}^{[r/2]}\,a_{s,r-2s}[/tex]

    where [r/2] = r/2 if r is even or (r-1)/2 if r is odd.
     
  4. May 9, 2010 #3
    I suppose you mean to apply this using [itex]p = 2 k[/itex], so that my powers of [itex]u[/itex] become [itex]2 k - 2 n[/itex] instead of [itex]2 k - n[/itex]? I'm not sure how that simplifies things. I'm still not sure what kind of expressions I'm going to get in front of my [itex]u[/itex]'s.
     
  5. May 9, 2010 #4

    vela

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    I think you have to have p=k and q=n from comparing your expression to the LHS of the relationship.

    What definition do you have for the Legendre polynomials? I'm wondering what you should be shooting for.
     
  6. May 9, 2010 #5

    vela

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    You can factor out uk from the second summation and swap the exponents on the (-2x) and u to get

    [tex]\sum_{k=0}^\infty (-1)^k \frac{(2k)!}{2^{2k} (k!)^2} u^k \sum_{n=0}^k \begin{pmatrix}k \\ n \end{pmatrix} (-1)^{k-n} (2x)^{k-n} u^n = \sum_{k=0}^\infty \sum_{n=0}^k (-1)^{n} \frac{(2k)!}{2^{2k} (k!)^2} u^k \begin{pmatrix}k \\ n \end{pmatrix} (2x)^{k-n} u^n[/tex]

    After you switch to the new indices, us will be multiplied by a polynomial with just even or odd powers of x, just like the Legendre polynomials.
     
  7. May 9, 2010 #6
    Though I hadn't given it a lot of thought yet, I expected to have to use the Rodriguez formula. I now see where you are going, though, since I just discovered the expression

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

    So, after using the relation you suggested, I get the sum

    [tex]
    \sum_k \sum_{n=0}^{[k/2]} (-1)^k \frac{(2k-2n)!}{2^{2k-n} (k-n)! n! (k-2n)!} x^n u^{2k-3n}.
    [/tex]

    This is starting to look like what I need, but not quite yet.

    As for your second post, I'm not sure what you're doing. First of all, your expression isn't equal to mine, since you have [itex]u^{k+n}[/itex], while I have [itex]u^{2k-n}[/itex]. Is my expression incorrect? Second, what do you mean by 'swap the exponents'?
     
  8. May 9, 2010 #7

    vela

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    Both expressions are correct. It's just how you applied the binomial theorem. You wrote

    [tex](x+y)^n = \sum_{k=0}^n \begin{pmatrix}n \\ k\end{pmatrix} x^k y^{n-k}[/tex]

    while I'm suggesting you instead write

    [tex](x+y)^n = \sum_{k=0}^n \begin{pmatrix}n \\ k\end{pmatrix} x^{n-k} y^k[/tex]

    That's what I meant by "swap the exponents." It's just to make the exponent of u look more like what you're shooting for.
     
  9. May 9, 2010 #8
    Ah, I see what you mean now. I hadn't actually used this form of the binomial expansion, I was using the special case [itex](1+x)^k[/itex] and was factoring out to get that form, so the symmetry got a little lost. Using the second form of the expansion indeed made things a lot neater and I've got my answer now. Thanks a lot!
     
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