Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Verify the Rodrigues formula of the Legendre polynomials

  1. Jan 11, 2016 #1
    How does (6.79) satisfy (6.70)?

    After substitution, I get

    $$(1-w^2)\frac{d^{l+2}}{dw^{l+2}}(w^2-1)^l-2w\frac{d^{l+1}}{dw^{l+1}}(w^2-1)^l+l(l+1)\frac{d^{l}}{dw^{l}}(w^2-1)^l$$

    Using product rule in reverse on the first two terms,
    $$(1-w^2)\frac{d^{l+1}}{dw^{l+1}}(w^2-1)^l+l(l+1)\frac{d^{l}}{dw^{l}}(w^2-1)^l$$
    $$(1-w^2)\frac{d^{l+1}}{dw^{l+1}}(w^2-1)^l-2w\frac{d^{l}}{dw^{l}}(w^2-1)^l+2w\frac{d^{l}}{dw^{l}}(w^2-1)^l+l(l+1)\frac{d^{l}}{dw^{l}}(w^2-1)^l$$

    Using product rule in reverse on the first two terms,
    $$(1-w^2)\frac{d^{l}}{dw^{l}}(w^2-1)^l+2w\frac{d^{l}}{dw^{l}}(w^2-1)^l+l(l+1)\frac{d^{l}}{dw^{l}}(w^2-1)^l$$
    $$\big[(1+2w-w^2)+l(l+1)\big]\frac{d^l}{dw^l}(w^2-1)^l$$

    How do we prove it is equal to 0?

    Screen Shot 2016-01-12 at 12.38.59 am.png
    Screen Shot 2016-01-12 at 12.39.29 am.png
    Screen Shot 2016-01-12 at 12.40.08 am.png
    Screen Shot 2016-01-12 at 12.40.26 am.png
     
  2. jcsd
  3. Jan 11, 2016 #2

    blue_leaf77

    User Avatar
    Science Advisor
    Homework Helper

    I think it can be easily solved with the help of Leibniz rule for the differentiation of a product of two functions
    $$
    \frac{d^n}{dx^n}(f(x)g(x)) = \sum_{k=0}^n C(n,k) \frac{d^{n-k}f}{dx^{n-k}} \frac{d^k g}{dx^n}
    $$
    In the terms which contain differentiation ##l+1## or ##l+2## times, differentiate the function 1 or 2 times first.
     
  4. Jan 12, 2016 #3
    I could't solve it.

    Let ##y=(w^2-1)^l##.
    ##\frac{dy}{dw}=2l(w^2-1)^{l-1}w##
    ##\frac{d^2y}{dw^2}=2l(w^2-1)^{l-2}\big[2(l-1)w+w^2-1\big]##

    ##\begin{align}\frac{d^{l+1}y}{dw^{l+1}}&=\frac{d^{l}}{dw^{l}}2l(w^2-1)^{l-1}w\\&=2l\Big[w\frac{d^l}{dw^l}(w^2-1)^{l-1}+l\frac{d^{l-1}}{dw^{l-1}}(w^2-1)^{l-1}\Big]\end{align}##

    ##\begin{align}\frac{d^{l+2}y}{dw^{l+2}}&=\frac{d^{l}}{dw^{l}}2l(w^2-1)^{l-2}\big[2(l-1)w+w^2-1\big]\\&=4l(l-1)\Big[w\frac{d^l}{dw^l}(w^2-1)^{l-2}+l\frac{d^{l-1}}{dw^{l-1}}(w^2-1)^{l-2}\Big]+2l\Big[\frac{d^l}{dw^l}(w^2-1)^{l-1}\Big]\end{align}##

    After substituting (2) and (4) into the LHS of (6.70), it is not clear how the terms cancel out to give 0.
     
    Last edited: Jan 12, 2016
  5. Jan 12, 2016 #4
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Verify the Rodrigues formula of the Legendre polynomials
  1. Verify the equality (Replies: 25)

Loading...