1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Binomial expansion comparison with legendre polynomial expansion

  1. Apr 15, 2012 #1
    Hi,

    I've been working on this question which asks to show that

    [itex]{{P}_{n}}(x)=\frac{1}{{{2}^{n}}n!}\frac{{{d}^{n}}}{d{{x}^{n}}}{{\left( {{x}^{2}}-1 \right)}^{n}}[/itex]

    So first taking the n derivatives of the binomial expansions of (x2-1)n

    [itex]{{({{x}^{2}}-1)}^{n}}=\sum\limits_{k=0}^{n}{{{(-1)}^{k}}\frac{n!}{k!(n-k)!}{{x}^{2n-2k}}}[/itex]

    [itex]\frac{{{d}^{n}}}{d{{x}^{n}}}...=\sum\limits_{k=0}^{n}{{{(-1)}^{k}}\frac{n!}{k!(n-k)!}(2n-2k)(2n-2k-1)...(2n-2k-n+1){{x}^{2n-2k}}}[/itex]
    [itex]=n!\sum\limits_{k=0}^{n}{{{(-1)}^{k}}\frac{(2n-2k)!}{k!(n-k)!(n-2k)!}{{x}^{2n-2k}}}[/itex]


    and comparing it with


    [itex]{{P}_{n}}(x)=\sum\limits_{m=0}^{M}{{{(-1)}^{m}}\frac{(2n-2m)!}{{{2}^{n}}m!(n-m)!(n-2m)!}{{x}^{n-2m}}},\,\,\,\,M=\frac{n}{2},\frac{n-1}{2}[/itex]

    [itex]=\frac{1}{{{2}^{n}}}\sum\limits_{m=0}^{\frac{n}{2}}{{{(-1)}^{m}}\frac{(2n-2m)!}{m!(n-m)!(n-2m)!}{{x}^{n-2m}}}[/itex]

    I'm having trouble with the final part,

    It's clear that there's a factor of 1/n!2n difference between them but also

    the Pn(x) series has m=0...n/2, and also xn , where as the n'th derivative series has k=0...n and x2n.

    How can you rewrite one in terms of the other so they both have the same sum limits?

    I've tried setting k=2s in the n'th derivative series and a bunch of other similar changes, but non will change the n'th powers of x.

    The reason I noticed this was because the last terms of the series arn't the same,

    the first series has last term, (-n)! on the bottom, means 1/infinity right?

    [itex]n!{{(-1)}^{n}}\frac{0!}{n!0!(-n)!}[/itex]

    and the second

    [itex]\frac{1}{{{2}^{n}}}{{(-1)}^{\frac{n}{2}}}\frac{n!}{n!(\frac{n}{2})!0!}{{x}^{0}}[/itex]

    Have I made a mistake early on or is there a clever way to combine the two series?

    Thanks,

    Linda
     
  2. jcsd
  3. Apr 16, 2012 #2

    I like Serena

    User Avatar
    Homework Helper

    Hi linda300! :smile:

    The factor 1/n!2n is already present in your first equation, so that is not a difference.

    When you took the n-th derivative, you didn't lower the power of x by n, so you should have xn-2k instead of x2n-2k

    Finally, when you take the derivative of x0 you should get zero, and not a negative power of x.
    So you should leave out the first n/2 terms, since they are all zero.
     
  4. Apr 16, 2012 #3
    Thanks heaps for taking the time to find my silly mistakes!

    =D
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Binomial expansion comparison with legendre polynomial expansion
  1. Binomial Expansion (Replies: 2)

  2. Binomial Expansion (Replies: 11)

  3. Binomial expansion (Replies: 10)

Loading...