Legendre Polynomials: proofs $int_{-1}^1P_n^2(x)dx=frac{2}{(2n+1)}$












2












$begingroup$


Does any one know, how to compute any of those two things?





  1. The relationship between Legendre polynomials and Shifted Legendre Polynomials.


  2. $displaystyleint_{-1}^1P_n^2(x)dx=dfrac{2}{(2n+1)}$ for $ngeq0$.





I tried to use Bonnet's equation:



$(2n-1)xP_{n-1}(x)=nP_n(x)+(n-1)P_{n-2}(x)$



but I couldn't move. Thanks :)



Edit: The second problem refers to regular Legendre Polynomials.






ordinary-differential-equations polynomials







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are you sure Item 2 is written correctly? I assume that it is referring to the regular Legendre polynomials (not the shifted ones).
    $endgroup$
    – Amzoti
    Mar 19 '13 at 14:58












  • $begingroup$
    Yes, the second one reffers to regular ones. I'm sorry :/
    $endgroup$
    – Adam
    Mar 19 '13 at 18:34










  • $begingroup$
    I also think it should be $2/(2n+1)$ for item $2$. Can you check that?
    $endgroup$
    – Amzoti
    Mar 19 '13 at 18:39












  • $begingroup$
    Yea, sorry :/ my fault
    $endgroup$
    – Adam
    Mar 19 '13 at 18:48
















2












$begingroup$


Does any one know, how to compute any of those two things?





  1. The relationship between Legendre polynomials and Shifted Legendre Polynomials.


  2. $displaystyleint_{-1}^1P_n^2(x)dx=dfrac{2}{(2n+1)}$ for $ngeq0$.





I tried to use Bonnet's equation:



$(2n-1)xP_{n-1}(x)=nP_n(x)+(n-1)P_{n-2}(x)$



but I couldn't move. Thanks :)



Edit: The second problem refers to regular Legendre Polynomials.






ordinary-differential-equations polynomials







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are you sure Item 2 is written correctly? I assume that it is referring to the regular Legendre polynomials (not the shifted ones).
    $endgroup$
    – Amzoti
    Mar 19 '13 at 14:58












  • $begingroup$
    Yes, the second one reffers to regular ones. I'm sorry :/
    $endgroup$
    – Adam
    Mar 19 '13 at 18:34










  • $begingroup$
    I also think it should be $2/(2n+1)$ for item $2$. Can you check that?
    $endgroup$
    – Amzoti
    Mar 19 '13 at 18:39












  • $begingroup$
    Yea, sorry :/ my fault
    $endgroup$
    – Adam
    Mar 19 '13 at 18:48














2












2








2





$begingroup$


Does any one know, how to compute any of those two things?





  1. The relationship between Legendre polynomials and Shifted Legendre Polynomials.


  2. $displaystyleint_{-1}^1P_n^2(x)dx=dfrac{2}{(2n+1)}$ for $ngeq0$.





I tried to use Bonnet's equation:



$(2n-1)xP_{n-1}(x)=nP_n(x)+(n-1)P_{n-2}(x)$



but I couldn't move. Thanks :)



Edit: The second problem refers to regular Legendre Polynomials.






ordinary-differential-equations polynomials







share|cite|improve this question











$endgroup$




Does any one know, how to compute any of those two things?





  1. The relationship between Legendre polynomials and Shifted Legendre Polynomials.


  2. $displaystyleint_{-1}^1P_n^2(x)dx=dfrac{2}{(2n+1)}$ for $ngeq0$.





I tried to use Bonnet's equation:



$(2n-1)xP_{n-1}(x)=nP_n(x)+(n-1)P_{n-2}(x)$



but I couldn't move. Thanks :)



Edit: The second problem refers to regular Legendre Polynomials.






ordinary-differential-equations polynomials




ordinary-differential-equations polynomials






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 7 '18 at 15:30









Nosrati

26.5k62354




26.5k62354










asked Mar 19 '13 at 6:47









AdamAdam

385113




385113












  • $begingroup$
    Are you sure Item 2 is written correctly? I assume that it is referring to the regular Legendre polynomials (not the shifted ones).
    $endgroup$
    – Amzoti
    Mar 19 '13 at 14:58












  • $begingroup$
    Yes, the second one reffers to regular ones. I'm sorry :/
    $endgroup$
    – Adam
    Mar 19 '13 at 18:34










  • $begingroup$
    I also think it should be $2/(2n+1)$ for item $2$. Can you check that?
    $endgroup$
    – Amzoti
    Mar 19 '13 at 18:39












  • $begingroup$
    Yea, sorry :/ my fault
    $endgroup$
    – Adam
    Mar 19 '13 at 18:48


















  • $begingroup$
    Are you sure Item 2 is written correctly? I assume that it is referring to the regular Legendre polynomials (not the shifted ones).
    $endgroup$
    – Amzoti
    Mar 19 '13 at 14:58












  • $begingroup$
    Yes, the second one reffers to regular ones. I'm sorry :/
    $endgroup$
    – Adam
    Mar 19 '13 at 18:34










  • $begingroup$
    I also think it should be $2/(2n+1)$ for item $2$. Can you check that?
    $endgroup$
    – Amzoti
    Mar 19 '13 at 18:39












  • $begingroup$
    Yea, sorry :/ my fault
    $endgroup$
    – Adam
    Mar 19 '13 at 18:48
















$begingroup$
Are you sure Item 2 is written correctly? I assume that it is referring to the regular Legendre polynomials (not the shifted ones).
$endgroup$
– Amzoti
Mar 19 '13 at 14:58






$begingroup$
Are you sure Item 2 is written correctly? I assume that it is referring to the regular Legendre polynomials (not the shifted ones).
$endgroup$
– Amzoti
Mar 19 '13 at 14:58














$begingroup$
Yes, the second one reffers to regular ones. I'm sorry :/
$endgroup$
– Adam
Mar 19 '13 at 18:34




$begingroup$
Yes, the second one reffers to regular ones. I'm sorry :/
$endgroup$
– Adam
Mar 19 '13 at 18:34












$begingroup$
I also think it should be $2/(2n+1)$ for item $2$. Can you check that?
$endgroup$
– Amzoti
Mar 19 '13 at 18:39






$begingroup$
I also think it should be $2/(2n+1)$ for item $2$. Can you check that?
$endgroup$
– Amzoti
Mar 19 '13 at 18:39














$begingroup$
Yea, sorry :/ my fault
$endgroup$
– Adam
Mar 19 '13 at 18:48




$begingroup$
Yea, sorry :/ my fault
$endgroup$
– Adam
Mar 19 '13 at 18:48










2 Answers
2






active

oldest

votes


















3












$begingroup$

I will answer your question about determining the value of $int_{-1}^1 P_n(x)^2dx$, using Rodrigues' formula
$$P_n(x) = frac{1}{2^nn!}[(x^2-1)^n]^{(n)}$$



$newcommand{partial}[1]{left[#1right]}$
$newcommand{bracket}[1]{left(#1right)}$
begin{equation}
begin{split}
I&=int_{-1}^1 P_n(x)^2dx
\ &=frac{1}{2^{2n}n!^2}int_{-1}^1 [(x^2-1)^n]^{(n)} cdot [(x^2-1)^n]^{(n)}dx
\ &=frac{1}{2^{2n}n!^2} bracket{partial{[(x^2-1)^n]^{(n)}cdot [(x^2-1)^n]^{(n-1)}}_{-1}^{+1}-int_{-1}^1 [(x^2-1)^n]^{(n+1)}cdot [(x^2-1)^n]^{(n-1)}dx}
\ &=frac{1}{2^{2n}n!^2} bracket{0-int_{-1}^1 [(x^2-1)^n]^{(n+1)}cdot [(x^2-1)^n]^{(n-1)}dx}
\ &=...
\ &=frac{(-1)^n}{2^{2n}n!^2} int_{-1}^1 [(x^2-1)^n]^{(2n)}cdot [(x^2-1)^n]^{(0)}dx
\ &=frac{(-1)^n}{2^{2n}n!^2} cdot (2n)! cdot int_{-1}^1 (x^2-1)^ndx
end{split}
end{equation}



Now



begin{equation}
begin{split}
int_{-1}^1 (x^2-1)^ndx&=int_{-1}^1 (x+1)^n(x-1)^ndx
\ &=bracket{partial{frac{(x+1)^{n+1}}{n+1}(x-1)^n}_{-1}^{+1} - int_{-1}^1 frac{(x+1)^{n+1}}{n+1}cdot n(x-1)^{n-1}dx}
\ &=bracket{0 - int_{-1}^1 frac{(x+1)^{n+1}}{n+1}cdot n(x-1)^{n-1}dx}
\ &=...
\ &=(-1)^nint_{-1}^1 frac{n!cdot (x+1)^{2n}}{(2n)!}cdot n!(x-1)^0dx
\ &=(-1)^nfrac{n!^2}{(2n)!}int_{-1}^1 (x+1)^{2n}dx
\ &=(-1)^nfrac{n!^2}{(2n)!}partial{frac{(x+1)^{2n+1}}{2n+1}}_{-1}^{+1}
\ &=(-1)^nfrac{n!^2}{(2n)!}cdot frac{2^{2n+1}}{2n+1}
end{split}
end{equation}



So finally we get our desired value:



begin{equation}
begin{split}
I&=int_{-1}^1 P_n(x)^2dx
\ &=frac{(-1)^n}{2^{2n}n!^2} cdot (2n)! cdot (-1)^nfrac{n!^2}{(2n)!}cdot frac{2^{2n+1}}{2n+1}
\ &=frac{2}{2n+1}
end{split}
end{equation}



Q.E.D.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Hints:



    Part 1:



    See Shifted Legendre Polynomials.



    I am not exactly sure what you intend to do for part 1., since it is not clear from your question. Maybe you can clarify.



    Look at the DLMF and what do you notice about the Legnedre versus SHifted Legendre. So, if you can prove one of them, do you see an approach to deriving the other?



    Part 2:



    Try evaluating the integral using Rodrigues' Formula.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Thanks for your hints, what I'm asking in first question is, that there is a relation between them. I know about it. And you posted a wikipedia article talking about this. But I need a proof of this. And that's my problem. Thanks:)
      $endgroup$
      – Adam
      Mar 19 '13 at 20:03










    • $begingroup$
      See update - no more comments for a while
      $endgroup$
      – Amzoti
      Mar 19 '13 at 20:22










    • $begingroup$
      In the part 2 I'm supposed (forced) to use Bonnet's equation.
      $endgroup$
      – Adam
      Mar 19 '13 at 20:43










    • $begingroup$
      Then, you might be interested in the answers here: math.stackexchange.com/questions/205932/…
      $endgroup$
      – Amzoti
      Mar 19 '13 at 21:07










    • $begingroup$
      How did this go unvoted until now? +1
      $endgroup$
      – amWhy
      Apr 17 '13 at 0:54











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    I will answer your question about determining the value of $int_{-1}^1 P_n(x)^2dx$, using Rodrigues' formula
    $$P_n(x) = frac{1}{2^nn!}[(x^2-1)^n]^{(n)}$$



    $newcommand{partial}[1]{left[#1right]}$
    $newcommand{bracket}[1]{left(#1right)}$
    begin{equation}
    begin{split}
    I&=int_{-1}^1 P_n(x)^2dx
    \ &=frac{1}{2^{2n}n!^2}int_{-1}^1 [(x^2-1)^n]^{(n)} cdot [(x^2-1)^n]^{(n)}dx
    \ &=frac{1}{2^{2n}n!^2} bracket{partial{[(x^2-1)^n]^{(n)}cdot [(x^2-1)^n]^{(n-1)}}_{-1}^{+1}-int_{-1}^1 [(x^2-1)^n]^{(n+1)}cdot [(x^2-1)^n]^{(n-1)}dx}
    \ &=frac{1}{2^{2n}n!^2} bracket{0-int_{-1}^1 [(x^2-1)^n]^{(n+1)}cdot [(x^2-1)^n]^{(n-1)}dx}
    \ &=...
    \ &=frac{(-1)^n}{2^{2n}n!^2} int_{-1}^1 [(x^2-1)^n]^{(2n)}cdot [(x^2-1)^n]^{(0)}dx
    \ &=frac{(-1)^n}{2^{2n}n!^2} cdot (2n)! cdot int_{-1}^1 (x^2-1)^ndx
    end{split}
    end{equation}



    Now



    begin{equation}
    begin{split}
    int_{-1}^1 (x^2-1)^ndx&=int_{-1}^1 (x+1)^n(x-1)^ndx
    \ &=bracket{partial{frac{(x+1)^{n+1}}{n+1}(x-1)^n}_{-1}^{+1} - int_{-1}^1 frac{(x+1)^{n+1}}{n+1}cdot n(x-1)^{n-1}dx}
    \ &=bracket{0 - int_{-1}^1 frac{(x+1)^{n+1}}{n+1}cdot n(x-1)^{n-1}dx}
    \ &=...
    \ &=(-1)^nint_{-1}^1 frac{n!cdot (x+1)^{2n}}{(2n)!}cdot n!(x-1)^0dx
    \ &=(-1)^nfrac{n!^2}{(2n)!}int_{-1}^1 (x+1)^{2n}dx
    \ &=(-1)^nfrac{n!^2}{(2n)!}partial{frac{(x+1)^{2n+1}}{2n+1}}_{-1}^{+1}
    \ &=(-1)^nfrac{n!^2}{(2n)!}cdot frac{2^{2n+1}}{2n+1}
    end{split}
    end{equation}



    So finally we get our desired value:



    begin{equation}
    begin{split}
    I&=int_{-1}^1 P_n(x)^2dx
    \ &=frac{(-1)^n}{2^{2n}n!^2} cdot (2n)! cdot (-1)^nfrac{n!^2}{(2n)!}cdot frac{2^{2n+1}}{2n+1}
    \ &=frac{2}{2n+1}
    end{split}
    end{equation}



    Q.E.D.






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      I will answer your question about determining the value of $int_{-1}^1 P_n(x)^2dx$, using Rodrigues' formula
      $$P_n(x) = frac{1}{2^nn!}[(x^2-1)^n]^{(n)}$$



      $newcommand{partial}[1]{left[#1right]}$
      $newcommand{bracket}[1]{left(#1right)}$
      begin{equation}
      begin{split}
      I&=int_{-1}^1 P_n(x)^2dx
      \ &=frac{1}{2^{2n}n!^2}int_{-1}^1 [(x^2-1)^n]^{(n)} cdot [(x^2-1)^n]^{(n)}dx
      \ &=frac{1}{2^{2n}n!^2} bracket{partial{[(x^2-1)^n]^{(n)}cdot [(x^2-1)^n]^{(n-1)}}_{-1}^{+1}-int_{-1}^1 [(x^2-1)^n]^{(n+1)}cdot [(x^2-1)^n]^{(n-1)}dx}
      \ &=frac{1}{2^{2n}n!^2} bracket{0-int_{-1}^1 [(x^2-1)^n]^{(n+1)}cdot [(x^2-1)^n]^{(n-1)}dx}
      \ &=...
      \ &=frac{(-1)^n}{2^{2n}n!^2} int_{-1}^1 [(x^2-1)^n]^{(2n)}cdot [(x^2-1)^n]^{(0)}dx
      \ &=frac{(-1)^n}{2^{2n}n!^2} cdot (2n)! cdot int_{-1}^1 (x^2-1)^ndx
      end{split}
      end{equation}



      Now



      begin{equation}
      begin{split}
      int_{-1}^1 (x^2-1)^ndx&=int_{-1}^1 (x+1)^n(x-1)^ndx
      \ &=bracket{partial{frac{(x+1)^{n+1}}{n+1}(x-1)^n}_{-1}^{+1} - int_{-1}^1 frac{(x+1)^{n+1}}{n+1}cdot n(x-1)^{n-1}dx}
      \ &=bracket{0 - int_{-1}^1 frac{(x+1)^{n+1}}{n+1}cdot n(x-1)^{n-1}dx}
      \ &=...
      \ &=(-1)^nint_{-1}^1 frac{n!cdot (x+1)^{2n}}{(2n)!}cdot n!(x-1)^0dx
      \ &=(-1)^nfrac{n!^2}{(2n)!}int_{-1}^1 (x+1)^{2n}dx
      \ &=(-1)^nfrac{n!^2}{(2n)!}partial{frac{(x+1)^{2n+1}}{2n+1}}_{-1}^{+1}
      \ &=(-1)^nfrac{n!^2}{(2n)!}cdot frac{2^{2n+1}}{2n+1}
      end{split}
      end{equation}



      So finally we get our desired value:



      begin{equation}
      begin{split}
      I&=int_{-1}^1 P_n(x)^2dx
      \ &=frac{(-1)^n}{2^{2n}n!^2} cdot (2n)! cdot (-1)^nfrac{n!^2}{(2n)!}cdot frac{2^{2n+1}}{2n+1}
      \ &=frac{2}{2n+1}
      end{split}
      end{equation}



      Q.E.D.






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        I will answer your question about determining the value of $int_{-1}^1 P_n(x)^2dx$, using Rodrigues' formula
        $$P_n(x) = frac{1}{2^nn!}[(x^2-1)^n]^{(n)}$$



        $newcommand{partial}[1]{left[#1right]}$
        $newcommand{bracket}[1]{left(#1right)}$
        begin{equation}
        begin{split}
        I&=int_{-1}^1 P_n(x)^2dx
        \ &=frac{1}{2^{2n}n!^2}int_{-1}^1 [(x^2-1)^n]^{(n)} cdot [(x^2-1)^n]^{(n)}dx
        \ &=frac{1}{2^{2n}n!^2} bracket{partial{[(x^2-1)^n]^{(n)}cdot [(x^2-1)^n]^{(n-1)}}_{-1}^{+1}-int_{-1}^1 [(x^2-1)^n]^{(n+1)}cdot [(x^2-1)^n]^{(n-1)}dx}
        \ &=frac{1}{2^{2n}n!^2} bracket{0-int_{-1}^1 [(x^2-1)^n]^{(n+1)}cdot [(x^2-1)^n]^{(n-1)}dx}
        \ &=...
        \ &=frac{(-1)^n}{2^{2n}n!^2} int_{-1}^1 [(x^2-1)^n]^{(2n)}cdot [(x^2-1)^n]^{(0)}dx
        \ &=frac{(-1)^n}{2^{2n}n!^2} cdot (2n)! cdot int_{-1}^1 (x^2-1)^ndx
        end{split}
        end{equation}



        Now



        begin{equation}
        begin{split}
        int_{-1}^1 (x^2-1)^ndx&=int_{-1}^1 (x+1)^n(x-1)^ndx
        \ &=bracket{partial{frac{(x+1)^{n+1}}{n+1}(x-1)^n}_{-1}^{+1} - int_{-1}^1 frac{(x+1)^{n+1}}{n+1}cdot n(x-1)^{n-1}dx}
        \ &=bracket{0 - int_{-1}^1 frac{(x+1)^{n+1}}{n+1}cdot n(x-1)^{n-1}dx}
        \ &=...
        \ &=(-1)^nint_{-1}^1 frac{n!cdot (x+1)^{2n}}{(2n)!}cdot n!(x-1)^0dx
        \ &=(-1)^nfrac{n!^2}{(2n)!}int_{-1}^1 (x+1)^{2n}dx
        \ &=(-1)^nfrac{n!^2}{(2n)!}partial{frac{(x+1)^{2n+1}}{2n+1}}_{-1}^{+1}
        \ &=(-1)^nfrac{n!^2}{(2n)!}cdot frac{2^{2n+1}}{2n+1}
        end{split}
        end{equation}



        So finally we get our desired value:



        begin{equation}
        begin{split}
        I&=int_{-1}^1 P_n(x)^2dx
        \ &=frac{(-1)^n}{2^{2n}n!^2} cdot (2n)! cdot (-1)^nfrac{n!^2}{(2n)!}cdot frac{2^{2n+1}}{2n+1}
        \ &=frac{2}{2n+1}
        end{split}
        end{equation}



        Q.E.D.






        share|cite|improve this answer











        $endgroup$



        I will answer your question about determining the value of $int_{-1}^1 P_n(x)^2dx$, using Rodrigues' formula
        $$P_n(x) = frac{1}{2^nn!}[(x^2-1)^n]^{(n)}$$



        $newcommand{partial}[1]{left[#1right]}$
        $newcommand{bracket}[1]{left(#1right)}$
        begin{equation}
        begin{split}
        I&=int_{-1}^1 P_n(x)^2dx
        \ &=frac{1}{2^{2n}n!^2}int_{-1}^1 [(x^2-1)^n]^{(n)} cdot [(x^2-1)^n]^{(n)}dx
        \ &=frac{1}{2^{2n}n!^2} bracket{partial{[(x^2-1)^n]^{(n)}cdot [(x^2-1)^n]^{(n-1)}}_{-1}^{+1}-int_{-1}^1 [(x^2-1)^n]^{(n+1)}cdot [(x^2-1)^n]^{(n-1)}dx}
        \ &=frac{1}{2^{2n}n!^2} bracket{0-int_{-1}^1 [(x^2-1)^n]^{(n+1)}cdot [(x^2-1)^n]^{(n-1)}dx}
        \ &=...
        \ &=frac{(-1)^n}{2^{2n}n!^2} int_{-1}^1 [(x^2-1)^n]^{(2n)}cdot [(x^2-1)^n]^{(0)}dx
        \ &=frac{(-1)^n}{2^{2n}n!^2} cdot (2n)! cdot int_{-1}^1 (x^2-1)^ndx
        end{split}
        end{equation}



        Now



        begin{equation}
        begin{split}
        int_{-1}^1 (x^2-1)^ndx&=int_{-1}^1 (x+1)^n(x-1)^ndx
        \ &=bracket{partial{frac{(x+1)^{n+1}}{n+1}(x-1)^n}_{-1}^{+1} - int_{-1}^1 frac{(x+1)^{n+1}}{n+1}cdot n(x-1)^{n-1}dx}
        \ &=bracket{0 - int_{-1}^1 frac{(x+1)^{n+1}}{n+1}cdot n(x-1)^{n-1}dx}
        \ &=...
        \ &=(-1)^nint_{-1}^1 frac{n!cdot (x+1)^{2n}}{(2n)!}cdot n!(x-1)^0dx
        \ &=(-1)^nfrac{n!^2}{(2n)!}int_{-1}^1 (x+1)^{2n}dx
        \ &=(-1)^nfrac{n!^2}{(2n)!}partial{frac{(x+1)^{2n+1}}{2n+1}}_{-1}^{+1}
        \ &=(-1)^nfrac{n!^2}{(2n)!}cdot frac{2^{2n+1}}{2n+1}
        end{split}
        end{equation}



        So finally we get our desired value:



        begin{equation}
        begin{split}
        I&=int_{-1}^1 P_n(x)^2dx
        \ &=frac{(-1)^n}{2^{2n}n!^2} cdot (2n)! cdot (-1)^nfrac{n!^2}{(2n)!}cdot frac{2^{2n+1}}{2n+1}
        \ &=frac{2}{2n+1}
        end{split}
        end{equation}



        Q.E.D.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Apr 22 '14 at 23:04

























        answered Apr 22 '14 at 22:58









        Maestro13Maestro13

        1,051624




        1,051624























            1












            $begingroup$

            Hints:



            Part 1:



            See Shifted Legendre Polynomials.



            I am not exactly sure what you intend to do for part 1., since it is not clear from your question. Maybe you can clarify.



            Look at the DLMF and what do you notice about the Legnedre versus SHifted Legendre. So, if you can prove one of them, do you see an approach to deriving the other?



            Part 2:



            Try evaluating the integral using Rodrigues' Formula.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thanks for your hints, what I'm asking in first question is, that there is a relation between them. I know about it. And you posted a wikipedia article talking about this. But I need a proof of this. And that's my problem. Thanks:)
              $endgroup$
              – Adam
              Mar 19 '13 at 20:03










            • $begingroup$
              See update - no more comments for a while
              $endgroup$
              – Amzoti
              Mar 19 '13 at 20:22










            • $begingroup$
              In the part 2 I'm supposed (forced) to use Bonnet's equation.
              $endgroup$
              – Adam
              Mar 19 '13 at 20:43










            • $begingroup$
              Then, you might be interested in the answers here: math.stackexchange.com/questions/205932/…
              $endgroup$
              – Amzoti
              Mar 19 '13 at 21:07










            • $begingroup$
              How did this go unvoted until now? +1
              $endgroup$
              – amWhy
              Apr 17 '13 at 0:54
















            1












            $begingroup$

            Hints:



            Part 1:



            See Shifted Legendre Polynomials.



            I am not exactly sure what you intend to do for part 1., since it is not clear from your question. Maybe you can clarify.



            Look at the DLMF and what do you notice about the Legnedre versus SHifted Legendre. So, if you can prove one of them, do you see an approach to deriving the other?



            Part 2:



            Try evaluating the integral using Rodrigues' Formula.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thanks for your hints, what I'm asking in first question is, that there is a relation between them. I know about it. And you posted a wikipedia article talking about this. But I need a proof of this. And that's my problem. Thanks:)
              $endgroup$
              – Adam
              Mar 19 '13 at 20:03










            • $begingroup$
              See update - no more comments for a while
              $endgroup$
              – Amzoti
              Mar 19 '13 at 20:22










            • $begingroup$
              In the part 2 I'm supposed (forced) to use Bonnet's equation.
              $endgroup$
              – Adam
              Mar 19 '13 at 20:43










            • $begingroup$
              Then, you might be interested in the answers here: math.stackexchange.com/questions/205932/…
              $endgroup$
              – Amzoti
              Mar 19 '13 at 21:07










            • $begingroup$
              How did this go unvoted until now? +1
              $endgroup$
              – amWhy
              Apr 17 '13 at 0:54














            1












            1








            1





            $begingroup$

            Hints:



            Part 1:



            See Shifted Legendre Polynomials.



            I am not exactly sure what you intend to do for part 1., since it is not clear from your question. Maybe you can clarify.



            Look at the DLMF and what do you notice about the Legnedre versus SHifted Legendre. So, if you can prove one of them, do you see an approach to deriving the other?



            Part 2:



            Try evaluating the integral using Rodrigues' Formula.






            share|cite|improve this answer











            $endgroup$



            Hints:



            Part 1:



            See Shifted Legendre Polynomials.



            I am not exactly sure what you intend to do for part 1., since it is not clear from your question. Maybe you can clarify.



            Look at the DLMF and what do you notice about the Legnedre versus SHifted Legendre. So, if you can prove one of them, do you see an approach to deriving the other?



            Part 2:



            Try evaluating the integral using Rodrigues' Formula.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Mar 19 '13 at 20:22

























            answered Mar 19 '13 at 19:47









            AmzotiAmzoti

            51k125397




            51k125397












            • $begingroup$
              Thanks for your hints, what I'm asking in first question is, that there is a relation between them. I know about it. And you posted a wikipedia article talking about this. But I need a proof of this. And that's my problem. Thanks:)
              $endgroup$
              – Adam
              Mar 19 '13 at 20:03










            • $begingroup$
              See update - no more comments for a while
              $endgroup$
              – Amzoti
              Mar 19 '13 at 20:22










            • $begingroup$
              In the part 2 I'm supposed (forced) to use Bonnet's equation.
              $endgroup$
              – Adam
              Mar 19 '13 at 20:43










            • $begingroup$
              Then, you might be interested in the answers here: math.stackexchange.com/questions/205932/…
              $endgroup$
              – Amzoti
              Mar 19 '13 at 21:07










            • $begingroup$
              How did this go unvoted until now? +1
              $endgroup$
              – amWhy
              Apr 17 '13 at 0:54


















            • $begingroup$
              Thanks for your hints, what I'm asking in first question is, that there is a relation between them. I know about it. And you posted a wikipedia article talking about this. But I need a proof of this. And that's my problem. Thanks:)
              $endgroup$
              – Adam
              Mar 19 '13 at 20:03










            • $begingroup$
              See update - no more comments for a while
              $endgroup$
              – Amzoti
              Mar 19 '13 at 20:22










            • $begingroup$
              In the part 2 I'm supposed (forced) to use Bonnet's equation.
              $endgroup$
              – Adam
              Mar 19 '13 at 20:43










            • $begingroup$
              Then, you might be interested in the answers here: math.stackexchange.com/questions/205932/…
              $endgroup$
              – Amzoti
              Mar 19 '13 at 21:07










            • $begingroup$
              How did this go unvoted until now? +1
              $endgroup$
              – amWhy
              Apr 17 '13 at 0:54
















            $begingroup$
            Thanks for your hints, what I'm asking in first question is, that there is a relation between them. I know about it. And you posted a wikipedia article talking about this. But I need a proof of this. And that's my problem. Thanks:)
            $endgroup$
            – Adam
            Mar 19 '13 at 20:03




            $begingroup$
            Thanks for your hints, what I'm asking in first question is, that there is a relation between them. I know about it. And you posted a wikipedia article talking about this. But I need a proof of this. And that's my problem. Thanks:)
            $endgroup$
            – Adam
            Mar 19 '13 at 20:03












            $begingroup$
            See update - no more comments for a while
            $endgroup$
            – Amzoti
            Mar 19 '13 at 20:22




            $begingroup$
            See update - no more comments for a while
            $endgroup$
            – Amzoti
            Mar 19 '13 at 20:22












            $begingroup$
            In the part 2 I'm supposed (forced) to use Bonnet's equation.
            $endgroup$
            – Adam
            Mar 19 '13 at 20:43




            $begingroup$
            In the part 2 I'm supposed (forced) to use Bonnet's equation.
            $endgroup$
            – Adam
            Mar 19 '13 at 20:43












            $begingroup$
            Then, you might be interested in the answers here: math.stackexchange.com/questions/205932/…
            $endgroup$
            – Amzoti
            Mar 19 '13 at 21:07




            $begingroup$
            Then, you might be interested in the answers here: math.stackexchange.com/questions/205932/…
            $endgroup$
            – Amzoti
            Mar 19 '13 at 21:07












            $begingroup$
            How did this go unvoted until now? +1
            $endgroup$
            – amWhy
            Apr 17 '13 at 0:54




            $begingroup$
            How did this go unvoted until now? +1
            $endgroup$
            – amWhy
            Apr 17 '13 at 0:54


















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