Co-efficient of $x^n$ in Rodrigues formula for Legendre's polynomials












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I want to show that the co-efficient of $x^n$ in the function
$$ f(x) = (x^2-1)^n $$ when taken the derivative n times is $$ frac{(2n)!}{n!} $$



To simplify the expression in order to apply Leibnitz formula, I factorize the function
$$ f(x) = (x^2-1)^n = (x-1)^n(x+1)^n $$
Now the co-efficient of each term in the Leibnitz formula is a co-efficient of $x^n$, therefore $$ frac{d^n}{dx^n}(x-1)^n(x+1)^n = sum_{k=0}^nfrac{n!n!n!}{(n-k)!k!(n-k)!k!} $$



I am unable to prove this equal to $ frac{(2n)!}{n!} $.



For odd k, the number of terms are even and the formula can be slightly simplified to $$ frac{d^n}{dx^n}(x-1)^n(x+1)^n = 2sum_{k=0}^frac{n-1}{2}frac{n!n!n!}{(n-k)!k!(n-k)!k!} $$ However this doesn't help me prove the identity.






ordinary-differential-equations derivatives







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    1












    $begingroup$


    I want to show that the co-efficient of $x^n$ in the function
    $$ f(x) = (x^2-1)^n $$ when taken the derivative n times is $$ frac{(2n)!}{n!} $$



    To simplify the expression in order to apply Leibnitz formula, I factorize the function
    $$ f(x) = (x^2-1)^n = (x-1)^n(x+1)^n $$
    Now the co-efficient of each term in the Leibnitz formula is a co-efficient of $x^n$, therefore $$ frac{d^n}{dx^n}(x-1)^n(x+1)^n = sum_{k=0}^nfrac{n!n!n!}{(n-k)!k!(n-k)!k!} $$



    I am unable to prove this equal to $ frac{(2n)!}{n!} $.



    For odd k, the number of terms are even and the formula can be slightly simplified to $$ frac{d^n}{dx^n}(x-1)^n(x+1)^n = 2sum_{k=0}^frac{n-1}{2}frac{n!n!n!}{(n-k)!k!(n-k)!k!} $$ However this doesn't help me prove the identity.






    ordinary-differential-equations derivatives







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      I want to show that the co-efficient of $x^n$ in the function
      $$ f(x) = (x^2-1)^n $$ when taken the derivative n times is $$ frac{(2n)!}{n!} $$



      To simplify the expression in order to apply Leibnitz formula, I factorize the function
      $$ f(x) = (x^2-1)^n = (x-1)^n(x+1)^n $$
      Now the co-efficient of each term in the Leibnitz formula is a co-efficient of $x^n$, therefore $$ frac{d^n}{dx^n}(x-1)^n(x+1)^n = sum_{k=0}^nfrac{n!n!n!}{(n-k)!k!(n-k)!k!} $$



      I am unable to prove this equal to $ frac{(2n)!}{n!} $.



      For odd k, the number of terms are even and the formula can be slightly simplified to $$ frac{d^n}{dx^n}(x-1)^n(x+1)^n = 2sum_{k=0}^frac{n-1}{2}frac{n!n!n!}{(n-k)!k!(n-k)!k!} $$ However this doesn't help me prove the identity.






      ordinary-differential-equations derivatives







      share|cite|improve this question









      $endgroup$




      I want to show that the co-efficient of $x^n$ in the function
      $$ f(x) = (x^2-1)^n $$ when taken the derivative n times is $$ frac{(2n)!}{n!} $$



      To simplify the expression in order to apply Leibnitz formula, I factorize the function
      $$ f(x) = (x^2-1)^n = (x-1)^n(x+1)^n $$
      Now the co-efficient of each term in the Leibnitz formula is a co-efficient of $x^n$, therefore $$ frac{d^n}{dx^n}(x-1)^n(x+1)^n = sum_{k=0}^nfrac{n!n!n!}{(n-k)!k!(n-k)!k!} $$



      I am unable to prove this equal to $ frac{(2n)!}{n!} $.



      For odd k, the number of terms are even and the formula can be slightly simplified to $$ frac{d^n}{dx^n}(x-1)^n(x+1)^n = 2sum_{k=0}^frac{n-1}{2}frac{n!n!n!}{(n-k)!k!(n-k)!k!} $$ However this doesn't help me prove the identity.






      ordinary-differential-equations derivatives




      ordinary-differential-equations derivatives






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      asked Jan 7 at 15:10









      SSBSSB

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          You need the co-efficient of $x^{2n}$ in $(x^2-1)^n$, because differentiating $x^{2n}$ $n$-times gives you $x^n$. Using the binomial theorem, the coefficient is $1$. On differentiating $n$-times $x^{2n}$, you get $$2n(2n-1)...(2n-(n-1))=frac{(2n)!}{n!}$$






          share|cite|improve this answer









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          • $begingroup$
            Thanks a lot. SO simple :)
            $endgroup$
            – SSB
            Jan 7 at 15:44












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          $begingroup$

          You need the co-efficient of $x^{2n}$ in $(x^2-1)^n$, because differentiating $x^{2n}$ $n$-times gives you $x^n$. Using the binomial theorem, the coefficient is $1$. On differentiating $n$-times $x^{2n}$, you get $$2n(2n-1)...(2n-(n-1))=frac{(2n)!}{n!}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks a lot. SO simple :)
            $endgroup$
            – SSB
            Jan 7 at 15:44
















          1












          $begingroup$

          You need the co-efficient of $x^{2n}$ in $(x^2-1)^n$, because differentiating $x^{2n}$ $n$-times gives you $x^n$. Using the binomial theorem, the coefficient is $1$. On differentiating $n$-times $x^{2n}$, you get $$2n(2n-1)...(2n-(n-1))=frac{(2n)!}{n!}$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Thanks a lot. SO simple :)
            $endgroup$
            – SSB
            Jan 7 at 15:44














          1












          1








          1





          $begingroup$

          You need the co-efficient of $x^{2n}$ in $(x^2-1)^n$, because differentiating $x^{2n}$ $n$-times gives you $x^n$. Using the binomial theorem, the coefficient is $1$. On differentiating $n$-times $x^{2n}$, you get $$2n(2n-1)...(2n-(n-1))=frac{(2n)!}{n!}$$






          share|cite|improve this answer









          $endgroup$



          You need the co-efficient of $x^{2n}$ in $(x^2-1)^n$, because differentiating $x^{2n}$ $n$-times gives you $x^n$. Using the binomial theorem, the coefficient is $1$. On differentiating $n$-times $x^{2n}$, you get $$2n(2n-1)...(2n-(n-1))=frac{(2n)!}{n!}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 7 at 15:19









          Shubham JohriShubham Johri

          5,558818




          5,558818












          • $begingroup$
            Thanks a lot. SO simple :)
            $endgroup$
            – SSB
            Jan 7 at 15:44


















          • $begingroup$
            Thanks a lot. SO simple :)
            $endgroup$
            – SSB
            Jan 7 at 15:44
















          $begingroup$
          Thanks a lot. SO simple :)
          $endgroup$
          – SSB
          Jan 7 at 15:44




          $begingroup$
          Thanks a lot. SO simple :)
          $endgroup$
          – SSB
          Jan 7 at 15:44


















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