Evaluating $lim_{epsilonto 0^{+}} frac{ _2F_1left( tfrac{1}{2} - nu, tfrac{1}{2} + nu; epsilon; y right)...












1












$begingroup$


For $nu in mathbb{C}$ and negative $y<0$ is there a way to compute the limit
$$
f(nu,y) equiv lim_{epsilon to 0^{+}} frac{ _2F_1left( tfrac{1}{2} - nu, tfrac{1}{2} + nu; epsilon; y right) }{Gamma(epsilon)}
$$
in terms of simpler special functions (ideally not hypergeometrics...)? So far, I thought about writing this in terms of the series
$$
f(nu,y) = sum_{n=0}^infty frac{Gamma(frac{1}{2} - nu + n )Gamma(frac{1}{2} + nu + n )}{(n-1)! n!} y^n
$$
I think that this series converges at least for $-1 < y < 0$, but I am also interested in values of $y leq -1$.



(EDIT: I know that $_2F_1(a,b;c;z)$ has poles at $c=0,-1,-2,ldots$, this makes it hard to learn anything about this limit using Mathematica)






hypergeometric-function







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    1












    $begingroup$


    For $nu in mathbb{C}$ and negative $y<0$ is there a way to compute the limit
    $$
    f(nu,y) equiv lim_{epsilon to 0^{+}} frac{ _2F_1left( tfrac{1}{2} - nu, tfrac{1}{2} + nu; epsilon; y right) }{Gamma(epsilon)}
    $$
    in terms of simpler special functions (ideally not hypergeometrics...)? So far, I thought about writing this in terms of the series
    $$
    f(nu,y) = sum_{n=0}^infty frac{Gamma(frac{1}{2} - nu + n )Gamma(frac{1}{2} + nu + n )}{(n-1)! n!} y^n
    $$
    I think that this series converges at least for $-1 < y < 0$, but I am also interested in values of $y leq -1$.



    (EDIT: I know that $_2F_1(a,b;c;z)$ has poles at $c=0,-1,-2,ldots$, this makes it hard to learn anything about this limit using Mathematica)






    hypergeometric-function







    share|cite|improve this question









    $endgroup$















      1












      1








      1





      $begingroup$


      For $nu in mathbb{C}$ and negative $y<0$ is there a way to compute the limit
      $$
      f(nu,y) equiv lim_{epsilon to 0^{+}} frac{ _2F_1left( tfrac{1}{2} - nu, tfrac{1}{2} + nu; epsilon; y right) }{Gamma(epsilon)}
      $$
      in terms of simpler special functions (ideally not hypergeometrics...)? So far, I thought about writing this in terms of the series
      $$
      f(nu,y) = sum_{n=0}^infty frac{Gamma(frac{1}{2} - nu + n )Gamma(frac{1}{2} + nu + n )}{(n-1)! n!} y^n
      $$
      I think that this series converges at least for $-1 < y < 0$, but I am also interested in values of $y leq -1$.



      (EDIT: I know that $_2F_1(a,b;c;z)$ has poles at $c=0,-1,-2,ldots$, this makes it hard to learn anything about this limit using Mathematica)






      hypergeometric-function







      share|cite|improve this question









      $endgroup$




      For $nu in mathbb{C}$ and negative $y<0$ is there a way to compute the limit
      $$
      f(nu,y) equiv lim_{epsilon to 0^{+}} frac{ _2F_1left( tfrac{1}{2} - nu, tfrac{1}{2} + nu; epsilon; y right) }{Gamma(epsilon)}
      $$
      in terms of simpler special functions (ideally not hypergeometrics...)? So far, I thought about writing this in terms of the series
      $$
      f(nu,y) = sum_{n=0}^infty frac{Gamma(frac{1}{2} - nu + n )Gamma(frac{1}{2} + nu + n )}{(n-1)! n!} y^n
      $$
      I think that this series converges at least for $-1 < y < 0$, but I am also interested in values of $y leq -1$.



      (EDIT: I know that $_2F_1(a,b;c;z)$ has poles at $c=0,-1,-2,ldots$, this makes it hard to learn anything about this limit using Mathematica)






      hypergeometric-function




      hypergeometric-function






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 30 at 16:20









      Greg.PaulGreg.Paul

      787921




      787921






















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          $begingroup$

          We may express
          begin{equation}
          _2F_1left( tfrac{1}{2} - nu, tfrac{1}{2} + nu; epsilon; y right) =frac{Gammaleft(epsilonright)}{Gammaleft(tfrac{1}{2} - nuright)Gamma%
          left(tfrac{1}{2} + nuright)}sum_{s=0}^{infty}frac{Gammaleft(tfrac{1}{2} - nu+sright)Gammaleft(tfrac{1}{2} + nu+s%
          right)}{Gammaleft(epsilon+sright)s!}y^{s}
          end{equation}
          and thus,
          begin{align}
          lim_{epsilon to 0^{+}} frac{ _2F_1left( tfrac{1}{2} - nu, tfrac{1}{2} + nu; epsilon; y right) }{Gamma(epsilon)}&=frac{1}{Gammaleft(tfrac{1}{2} - nuright)Gamma%
          left(tfrac{1}{2} + nuright)}sum_{s=1}^{infty}frac{Gammaleft(tfrac{1}{2} - nu+sright)Gammaleft(tfrac{1}{2} + nu+s%
          right)}{Gammaleft(sright)s!}y^{s}\
          &=frac{y}{Gammaleft(tfrac{1}{2} - nuright)Gamma%
          left(tfrac{1}{2} + nuright)}sum_{t=0}^{infty}frac{Gammaleft(tfrac{3}{2} - nu+tright)Gammaleft(tfrac{3}{2} + nu+s%
          right)}{Gammaleft(t+2right)t!}y^{t}\
          &=frac{yGammaleft(tfrac{3}{2} - nuright)Gamma%
          left(tfrac{3}{2} + nuright)}{Gamma(2)Gammaleft(tfrac{1}{2} - nuright)Gamma%
          left(tfrac{1}{2} + nuright)} ,_2F_1left(tfrac{3}{2} - nu,tfrac{3}{2} +nu;2;y right)\
          &=yleft( frac{1}{4}-nu^2 right),_2F_1left(tfrac{3}{2} - nu,tfrac{3}{2} +nu;2;y right)
          end{align}
          From this representation of the associated Legendre function
          begin{equation}
          _2F_1left(a,b;tfrac{1}{2}(a+b+1);zright)=left(-z(1-z)right)^{%
          frac{(1-a-b)}{4}},P^{(1-a-b)/2}_{(a-b-1)/2}left(1-2zright)
          end{equation}
          with $a=3/2+nu,b=3/2-nu$ we can express
          begin{equation}
          f(nu,y)=left( frac{1}{4}-nu^2 right)sqrt{frac{-y}{1-y}}P^{-1}_{nu-1/2}left( 1-2y right)
          end{equation}
          where $1-2y>0$.






          share|cite|improve this answer









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            $begingroup$

            We may express
            begin{equation}
            _2F_1left( tfrac{1}{2} - nu, tfrac{1}{2} + nu; epsilon; y right) =frac{Gammaleft(epsilonright)}{Gammaleft(tfrac{1}{2} - nuright)Gamma%
            left(tfrac{1}{2} + nuright)}sum_{s=0}^{infty}frac{Gammaleft(tfrac{1}{2} - nu+sright)Gammaleft(tfrac{1}{2} + nu+s%
            right)}{Gammaleft(epsilon+sright)s!}y^{s}
            end{equation}
            and thus,
            begin{align}
            lim_{epsilon to 0^{+}} frac{ _2F_1left( tfrac{1}{2} - nu, tfrac{1}{2} + nu; epsilon; y right) }{Gamma(epsilon)}&=frac{1}{Gammaleft(tfrac{1}{2} - nuright)Gamma%
            left(tfrac{1}{2} + nuright)}sum_{s=1}^{infty}frac{Gammaleft(tfrac{1}{2} - nu+sright)Gammaleft(tfrac{1}{2} + nu+s%
            right)}{Gammaleft(sright)s!}y^{s}\
            &=frac{y}{Gammaleft(tfrac{1}{2} - nuright)Gamma%
            left(tfrac{1}{2} + nuright)}sum_{t=0}^{infty}frac{Gammaleft(tfrac{3}{2} - nu+tright)Gammaleft(tfrac{3}{2} + nu+s%
            right)}{Gammaleft(t+2right)t!}y^{t}\
            &=frac{yGammaleft(tfrac{3}{2} - nuright)Gamma%
            left(tfrac{3}{2} + nuright)}{Gamma(2)Gammaleft(tfrac{1}{2} - nuright)Gamma%
            left(tfrac{1}{2} + nuright)} ,_2F_1left(tfrac{3}{2} - nu,tfrac{3}{2} +nu;2;y right)\
            &=yleft( frac{1}{4}-nu^2 right),_2F_1left(tfrac{3}{2} - nu,tfrac{3}{2} +nu;2;y right)
            end{align}
            From this representation of the associated Legendre function
            begin{equation}
            _2F_1left(a,b;tfrac{1}{2}(a+b+1);zright)=left(-z(1-z)right)^{%
            frac{(1-a-b)}{4}},P^{(1-a-b)/2}_{(a-b-1)/2}left(1-2zright)
            end{equation}
            with $a=3/2+nu,b=3/2-nu$ we can express
            begin{equation}
            f(nu,y)=left( frac{1}{4}-nu^2 right)sqrt{frac{-y}{1-y}}P^{-1}_{nu-1/2}left( 1-2y right)
            end{equation}
            where $1-2y>0$.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              We may express
              begin{equation}
              _2F_1left( tfrac{1}{2} - nu, tfrac{1}{2} + nu; epsilon; y right) =frac{Gammaleft(epsilonright)}{Gammaleft(tfrac{1}{2} - nuright)Gamma%
              left(tfrac{1}{2} + nuright)}sum_{s=0}^{infty}frac{Gammaleft(tfrac{1}{2} - nu+sright)Gammaleft(tfrac{1}{2} + nu+s%
              right)}{Gammaleft(epsilon+sright)s!}y^{s}
              end{equation}
              and thus,
              begin{align}
              lim_{epsilon to 0^{+}} frac{ _2F_1left( tfrac{1}{2} - nu, tfrac{1}{2} + nu; epsilon; y right) }{Gamma(epsilon)}&=frac{1}{Gammaleft(tfrac{1}{2} - nuright)Gamma%
              left(tfrac{1}{2} + nuright)}sum_{s=1}^{infty}frac{Gammaleft(tfrac{1}{2} - nu+sright)Gammaleft(tfrac{1}{2} + nu+s%
              right)}{Gammaleft(sright)s!}y^{s}\
              &=frac{y}{Gammaleft(tfrac{1}{2} - nuright)Gamma%
              left(tfrac{1}{2} + nuright)}sum_{t=0}^{infty}frac{Gammaleft(tfrac{3}{2} - nu+tright)Gammaleft(tfrac{3}{2} + nu+s%
              right)}{Gammaleft(t+2right)t!}y^{t}\
              &=frac{yGammaleft(tfrac{3}{2} - nuright)Gamma%
              left(tfrac{3}{2} + nuright)}{Gamma(2)Gammaleft(tfrac{1}{2} - nuright)Gamma%
              left(tfrac{1}{2} + nuright)} ,_2F_1left(tfrac{3}{2} - nu,tfrac{3}{2} +nu;2;y right)\
              &=yleft( frac{1}{4}-nu^2 right),_2F_1left(tfrac{3}{2} - nu,tfrac{3}{2} +nu;2;y right)
              end{align}
              From this representation of the associated Legendre function
              begin{equation}
              _2F_1left(a,b;tfrac{1}{2}(a+b+1);zright)=left(-z(1-z)right)^{%
              frac{(1-a-b)}{4}},P^{(1-a-b)/2}_{(a-b-1)/2}left(1-2zright)
              end{equation}
              with $a=3/2+nu,b=3/2-nu$ we can express
              begin{equation}
              f(nu,y)=left( frac{1}{4}-nu^2 right)sqrt{frac{-y}{1-y}}P^{-1}_{nu-1/2}left( 1-2y right)
              end{equation}
              where $1-2y>0$.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                We may express
                begin{equation}
                _2F_1left( tfrac{1}{2} - nu, tfrac{1}{2} + nu; epsilon; y right) =frac{Gammaleft(epsilonright)}{Gammaleft(tfrac{1}{2} - nuright)Gamma%
                left(tfrac{1}{2} + nuright)}sum_{s=0}^{infty}frac{Gammaleft(tfrac{1}{2} - nu+sright)Gammaleft(tfrac{1}{2} + nu+s%
                right)}{Gammaleft(epsilon+sright)s!}y^{s}
                end{equation}
                and thus,
                begin{align}
                lim_{epsilon to 0^{+}} frac{ _2F_1left( tfrac{1}{2} - nu, tfrac{1}{2} + nu; epsilon; y right) }{Gamma(epsilon)}&=frac{1}{Gammaleft(tfrac{1}{2} - nuright)Gamma%
                left(tfrac{1}{2} + nuright)}sum_{s=1}^{infty}frac{Gammaleft(tfrac{1}{2} - nu+sright)Gammaleft(tfrac{1}{2} + nu+s%
                right)}{Gammaleft(sright)s!}y^{s}\
                &=frac{y}{Gammaleft(tfrac{1}{2} - nuright)Gamma%
                left(tfrac{1}{2} + nuright)}sum_{t=0}^{infty}frac{Gammaleft(tfrac{3}{2} - nu+tright)Gammaleft(tfrac{3}{2} + nu+s%
                right)}{Gammaleft(t+2right)t!}y^{t}\
                &=frac{yGammaleft(tfrac{3}{2} - nuright)Gamma%
                left(tfrac{3}{2} + nuright)}{Gamma(2)Gammaleft(tfrac{1}{2} - nuright)Gamma%
                left(tfrac{1}{2} + nuright)} ,_2F_1left(tfrac{3}{2} - nu,tfrac{3}{2} +nu;2;y right)\
                &=yleft( frac{1}{4}-nu^2 right),_2F_1left(tfrac{3}{2} - nu,tfrac{3}{2} +nu;2;y right)
                end{align}
                From this representation of the associated Legendre function
                begin{equation}
                _2F_1left(a,b;tfrac{1}{2}(a+b+1);zright)=left(-z(1-z)right)^{%
                frac{(1-a-b)}{4}},P^{(1-a-b)/2}_{(a-b-1)/2}left(1-2zright)
                end{equation}
                with $a=3/2+nu,b=3/2-nu$ we can express
                begin{equation}
                f(nu,y)=left( frac{1}{4}-nu^2 right)sqrt{frac{-y}{1-y}}P^{-1}_{nu-1/2}left( 1-2y right)
                end{equation}
                where $1-2y>0$.






                share|cite|improve this answer









                $endgroup$



                We may express
                begin{equation}
                _2F_1left( tfrac{1}{2} - nu, tfrac{1}{2} + nu; epsilon; y right) =frac{Gammaleft(epsilonright)}{Gammaleft(tfrac{1}{2} - nuright)Gamma%
                left(tfrac{1}{2} + nuright)}sum_{s=0}^{infty}frac{Gammaleft(tfrac{1}{2} - nu+sright)Gammaleft(tfrac{1}{2} + nu+s%
                right)}{Gammaleft(epsilon+sright)s!}y^{s}
                end{equation}
                and thus,
                begin{align}
                lim_{epsilon to 0^{+}} frac{ _2F_1left( tfrac{1}{2} - nu, tfrac{1}{2} + nu; epsilon; y right) }{Gamma(epsilon)}&=frac{1}{Gammaleft(tfrac{1}{2} - nuright)Gamma%
                left(tfrac{1}{2} + nuright)}sum_{s=1}^{infty}frac{Gammaleft(tfrac{1}{2} - nu+sright)Gammaleft(tfrac{1}{2} + nu+s%
                right)}{Gammaleft(sright)s!}y^{s}\
                &=frac{y}{Gammaleft(tfrac{1}{2} - nuright)Gamma%
                left(tfrac{1}{2} + nuright)}sum_{t=0}^{infty}frac{Gammaleft(tfrac{3}{2} - nu+tright)Gammaleft(tfrac{3}{2} + nu+s%
                right)}{Gammaleft(t+2right)t!}y^{t}\
                &=frac{yGammaleft(tfrac{3}{2} - nuright)Gamma%
                left(tfrac{3}{2} + nuright)}{Gamma(2)Gammaleft(tfrac{1}{2} - nuright)Gamma%
                left(tfrac{1}{2} + nuright)} ,_2F_1left(tfrac{3}{2} - nu,tfrac{3}{2} +nu;2;y right)\
                &=yleft( frac{1}{4}-nu^2 right),_2F_1left(tfrac{3}{2} - nu,tfrac{3}{2} +nu;2;y right)
                end{align}
                From this representation of the associated Legendre function
                begin{equation}
                _2F_1left(a,b;tfrac{1}{2}(a+b+1);zright)=left(-z(1-z)right)^{%
                frac{(1-a-b)}{4}},P^{(1-a-b)/2}_{(a-b-1)/2}left(1-2zright)
                end{equation}
                with $a=3/2+nu,b=3/2-nu$ we can express
                begin{equation}
                f(nu,y)=left( frac{1}{4}-nu^2 right)sqrt{frac{-y}{1-y}}P^{-1}_{nu-1/2}left( 1-2y right)
                end{equation}
                where $1-2y>0$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 30 at 18:33









                Paul EntaPaul Enta

                5,38611434




                5,38611434






























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