Can we refine this asymptotic for Laguerre polynomials?












4














I just found an interesting and useful limit for Laguerre polynomials:



$$lim_{n to infty} L_n left( frac{2r}{n+1/2} right)=J_0(2 sqrt{2r})$$



I'm using specifically this form of the argument because it's the one I'm working with in the application. Of course, we can set any fixed number instead of $1/2$ in the denominator.



I found this limit in a paper, which references G. Szego. Orthogonal Polynomials. Amer. Math. Soc. Colloq. Publ. 23, Amer. Math. Soc. Providence,
RI, 1975. Fourth Edition., , Theorem 8.1.3.



While the limit is useful for very large orders and smallish $r$, I would really like to know if there's a refinement that could be applied to derive an asymptotic expansion, which would depend on $n$.



Here's an illustration which shows that the limit is not that good for larger $r$ (though it does approximate the roots better than the magnitude):



enter image description here



Not sure how we could refine this asymptotic or how the original limit was derived (as I don't have the linked book).



One way is considering the differential equations for both functions.



There's also an interesting result from Gradshteyn-Ryzhik:



$$L_n(z)= frac{2}{n!} e^z int_0^infty e^{-t^2} t^{2n+1} J_0(2t sqrt{z}) dt$$



Which may or may not be related to the limit above.






limits asymptotics approximation bessel-functions orthogonal-polynomials







share|cite|improve this question



























    4














    I just found an interesting and useful limit for Laguerre polynomials:



    $$lim_{n to infty} L_n left( frac{2r}{n+1/2} right)=J_0(2 sqrt{2r})$$



    I'm using specifically this form of the argument because it's the one I'm working with in the application. Of course, we can set any fixed number instead of $1/2$ in the denominator.



    I found this limit in a paper, which references G. Szego. Orthogonal Polynomials. Amer. Math. Soc. Colloq. Publ. 23, Amer. Math. Soc. Providence,
    RI, 1975. Fourth Edition.    , , Theorem 8.1.3.



    While the limit is useful for very large orders and smallish $r$, I would really like to know if there's a refinement that could be applied to derive an asymptotic expansion, which would depend on $n$.



    Here's an illustration which shows that the limit is not that good for larger $r$ (though it does approximate the roots better than the magnitude):



    enter image description here



    Not sure how we could refine this asymptotic or how the original limit was derived (as I don't have the linked book).



    One way is considering the differential equations for both functions.



    There's also an interesting result from Gradshteyn-Ryzhik:



    $$L_n(z)= frac{2}{n!} e^z int_0^infty e^{-t^2} t^{2n+1} J_0(2t sqrt{z}) dt$$



    Which may or may not be related to the limit above.






    limits asymptotics approximation bessel-functions orthogonal-polynomials







    share|cite|improve this question

























      4












      4








      4


      2





      I just found an interesting and useful limit for Laguerre polynomials:



      $$lim_{n to infty} L_n left( frac{2r}{n+1/2} right)=J_0(2 sqrt{2r})$$



      I'm using specifically this form of the argument because it's the one I'm working with in the application. Of course, we can set any fixed number instead of $1/2$ in the denominator.



      I found this limit in a paper, which references G. Szego. Orthogonal Polynomials. Amer. Math. Soc. Colloq. Publ. 23, Amer. Math. Soc. Providence,
      RI, 1975. Fourth Edition.      , , Theorem 8.1.3.



      While the limit is useful for very large orders and smallish $r$, I would really like to know if there's a refinement that could be applied to derive an asymptotic expansion, which would depend on $n$.



      Here's an illustration which shows that the limit is not that good for larger $r$ (though it does approximate the roots better than the magnitude):



      enter image description here



      Not sure how we could refine this asymptotic or how the original limit was derived (as I don't have the linked book).



      One way is considering the differential equations for both functions.



      There's also an interesting result from Gradshteyn-Ryzhik:



      $$L_n(z)= frac{2}{n!} e^z int_0^infty e^{-t^2} t^{2n+1} J_0(2t sqrt{z}) dt$$



      Which may or may not be related to the limit above.






      limits asymptotics approximation bessel-functions orthogonal-polynomials







      share|cite|improve this question













      I just found an interesting and useful limit for Laguerre polynomials:



      $$lim_{n to infty} L_n left( frac{2r}{n+1/2} right)=J_0(2 sqrt{2r})$$



      I'm using specifically this form of the argument because it's the one I'm working with in the application. Of course, we can set any fixed number instead of $1/2$ in the denominator.



      I found this limit in a paper, which references G. Szego. Orthogonal Polynomials. Amer. Math. Soc. Colloq. Publ. 23, Amer. Math. Soc. Providence,
      RI, 1975. Fourth Edition.      , , Theorem 8.1.3.



      While the limit is useful for very large orders and smallish $r$, I would really like to know if there's a refinement that could be applied to derive an asymptotic expansion, which would depend on $n$.



      Here's an illustration which shows that the limit is not that good for larger $r$ (though it does approximate the roots better than the magnitude):



      enter image description here



      Not sure how we could refine this asymptotic or how the original limit was derived (as I don't have the linked book).



      One way is considering the differential equations for both functions.



      There's also an interesting result from Gradshteyn-Ryzhik:



      $$L_n(z)= frac{2}{n!} e^z int_0^infty e^{-t^2} t^{2n+1} J_0(2t sqrt{z}) dt$$



      Which may or may not be related to the limit above.






      limits asymptotics approximation bessel-functions orthogonal-polynomials




      limits asymptotics approximation bessel-functions orthogonal-polynomials






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 26 at 23:05









      Yuriy S

      15.7k433117




      15.7k433117






















          1 Answer
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          From the given representation, we can express
          begin{equation}
          L_n(frac{2r}{n+1/2})= frac{2}{n!} e^{frac{2r}{n+1/2}} int_0^infty e^{-t^2} t^{2n+1} J_0left(2t sqrt{frac{2r}{n+1/2}}right) ,dt
          end{equation}
          changing $ t=sqrt{uleft( n+1/2 right)}$,
          begin{equation}
          L_n(frac{2r}{n+1/2})= frac{(n+1/2)^{n+1}}{n!} e^{frac{2r}{n+1/2}}int_0^infty e^{-uleft( n+1/2 right)}u^{n+1/2}J_0left( 2sqrt{2ru} right)frac{du}{sqrt{u}}
          end{equation}
          or
          begin{equation}
          L_n(frac{2r}{n+1/2})= frac{(n+1/2)^{n+1}}{n!} e^{frac{2r}{n+1/2}}int_0^infty e^{-left( n+1/2 right)left( u-ln u right)}J_0left( 2sqrt{2ru} right)frac{du}{sqrt{u}}
          end{equation}
          The argument of the exponential is minimum at $u=1$. We can use the Laplace method to derive an asymptotic approximation for the integral. Near $u=1$,
          $u-ln(u)sim 1+(u-1)^2/2$ and $u^{-1/2}J_0left( 2sqrt{2ru}right)sim J_0left( 2sqrt{2r}right)$, then
          begin{equation}
          L_n(frac{2r}{n+1/2})sim sqrt{2pi}frac{(n+1/2)^{n+1/2}}{n!} e^{frac{2r}{n+1/2}}
          e^{-(n+1/2)}J_0left( 2sqrt{2r}right)
          end{equation}
          Now, plugging the Stirling approximation for $n!$ and an expansion for the exponential term, we obtain the expected result $ L_n(frac{2r}{n+1/2})sim J_0left( 2sqrt{2r}right)$. To improve the approximation, the above formula and/or higher orders in the Laplace expansion can be used.






          share|cite|improve this answer























          • Thank you! I will wait a little while before accepting the answer, but this is exactly what I wanted
            – Yuriy S
            Nov 30 at 1:02











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          From the given representation, we can express
          begin{equation}
          L_n(frac{2r}{n+1/2})= frac{2}{n!} e^{frac{2r}{n+1/2}} int_0^infty e^{-t^2} t^{2n+1} J_0left(2t sqrt{frac{2r}{n+1/2}}right) ,dt
          end{equation}
          changing $ t=sqrt{uleft( n+1/2 right)}$,
          begin{equation}
          L_n(frac{2r}{n+1/2})= frac{(n+1/2)^{n+1}}{n!} e^{frac{2r}{n+1/2}}int_0^infty e^{-uleft( n+1/2 right)}u^{n+1/2}J_0left( 2sqrt{2ru} right)frac{du}{sqrt{u}}
          end{equation}
          or
          begin{equation}
          L_n(frac{2r}{n+1/2})= frac{(n+1/2)^{n+1}}{n!} e^{frac{2r}{n+1/2}}int_0^infty e^{-left( n+1/2 right)left( u-ln u right)}J_0left( 2sqrt{2ru} right)frac{du}{sqrt{u}}
          end{equation}
          The argument of the exponential is minimum at $u=1$. We can use the Laplace method to derive an asymptotic approximation for the integral. Near $u=1$,
          $u-ln(u)sim 1+(u-1)^2/2$ and $u^{-1/2}J_0left( 2sqrt{2ru}right)sim J_0left( 2sqrt{2r}right)$, then
          begin{equation}
          L_n(frac{2r}{n+1/2})sim sqrt{2pi}frac{(n+1/2)^{n+1/2}}{n!} e^{frac{2r}{n+1/2}}
          e^{-(n+1/2)}J_0left( 2sqrt{2r}right)
          end{equation}
          Now, plugging the Stirling approximation for $n!$ and an expansion for the exponential term, we obtain the expected result $ L_n(frac{2r}{n+1/2})sim J_0left( 2sqrt{2r}right)$. To improve the approximation, the above formula and/or higher orders in the Laplace expansion can be used.






          share|cite|improve this answer























          • Thank you! I will wait a little while before accepting the answer, but this is exactly what I wanted
            – Yuriy S
            Nov 30 at 1:02
















          2














          From the given representation, we can express
          begin{equation}
          L_n(frac{2r}{n+1/2})= frac{2}{n!} e^{frac{2r}{n+1/2}} int_0^infty e^{-t^2} t^{2n+1} J_0left(2t sqrt{frac{2r}{n+1/2}}right) ,dt
          end{equation}
          changing $ t=sqrt{uleft( n+1/2 right)}$,
          begin{equation}
          L_n(frac{2r}{n+1/2})= frac{(n+1/2)^{n+1}}{n!} e^{frac{2r}{n+1/2}}int_0^infty e^{-uleft( n+1/2 right)}u^{n+1/2}J_0left( 2sqrt{2ru} right)frac{du}{sqrt{u}}
          end{equation}
          or
          begin{equation}
          L_n(frac{2r}{n+1/2})= frac{(n+1/2)^{n+1}}{n!} e^{frac{2r}{n+1/2}}int_0^infty e^{-left( n+1/2 right)left( u-ln u right)}J_0left( 2sqrt{2ru} right)frac{du}{sqrt{u}}
          end{equation}
          The argument of the exponential is minimum at $u=1$. We can use the Laplace method to derive an asymptotic approximation for the integral. Near $u=1$,
          $u-ln(u)sim 1+(u-1)^2/2$ and $u^{-1/2}J_0left( 2sqrt{2ru}right)sim J_0left( 2sqrt{2r}right)$, then
          begin{equation}
          L_n(frac{2r}{n+1/2})sim sqrt{2pi}frac{(n+1/2)^{n+1/2}}{n!} e^{frac{2r}{n+1/2}}
          e^{-(n+1/2)}J_0left( 2sqrt{2r}right)
          end{equation}
          Now, plugging the Stirling approximation for $n!$ and an expansion for the exponential term, we obtain the expected result $ L_n(frac{2r}{n+1/2})sim J_0left( 2sqrt{2r}right)$. To improve the approximation, the above formula and/or higher orders in the Laplace expansion can be used.






          share|cite|improve this answer























          • Thank you! I will wait a little while before accepting the answer, but this is exactly what I wanted
            – Yuriy S
            Nov 30 at 1:02














          2












          2








          2






          From the given representation, we can express
          begin{equation}
          L_n(frac{2r}{n+1/2})= frac{2}{n!} e^{frac{2r}{n+1/2}} int_0^infty e^{-t^2} t^{2n+1} J_0left(2t sqrt{frac{2r}{n+1/2}}right) ,dt
          end{equation}
          changing $ t=sqrt{uleft( n+1/2 right)}$,
          begin{equation}
          L_n(frac{2r}{n+1/2})= frac{(n+1/2)^{n+1}}{n!} e^{frac{2r}{n+1/2}}int_0^infty e^{-uleft( n+1/2 right)}u^{n+1/2}J_0left( 2sqrt{2ru} right)frac{du}{sqrt{u}}
          end{equation}
          or
          begin{equation}
          L_n(frac{2r}{n+1/2})= frac{(n+1/2)^{n+1}}{n!} e^{frac{2r}{n+1/2}}int_0^infty e^{-left( n+1/2 right)left( u-ln u right)}J_0left( 2sqrt{2ru} right)frac{du}{sqrt{u}}
          end{equation}
          The argument of the exponential is minimum at $u=1$. We can use the Laplace method to derive an asymptotic approximation for the integral. Near $u=1$,
          $u-ln(u)sim 1+(u-1)^2/2$ and $u^{-1/2}J_0left( 2sqrt{2ru}right)sim J_0left( 2sqrt{2r}right)$, then
          begin{equation}
          L_n(frac{2r}{n+1/2})sim sqrt{2pi}frac{(n+1/2)^{n+1/2}}{n!} e^{frac{2r}{n+1/2}}
          e^{-(n+1/2)}J_0left( 2sqrt{2r}right)
          end{equation}
          Now, plugging the Stirling approximation for $n!$ and an expansion for the exponential term, we obtain the expected result $ L_n(frac{2r}{n+1/2})sim J_0left( 2sqrt{2r}right)$. To improve the approximation, the above formula and/or higher orders in the Laplace expansion can be used.






          share|cite|improve this answer














          From the given representation, we can express
          begin{equation}
          L_n(frac{2r}{n+1/2})= frac{2}{n!} e^{frac{2r}{n+1/2}} int_0^infty e^{-t^2} t^{2n+1} J_0left(2t sqrt{frac{2r}{n+1/2}}right) ,dt
          end{equation}
          changing $ t=sqrt{uleft( n+1/2 right)}$,
          begin{equation}
          L_n(frac{2r}{n+1/2})= frac{(n+1/2)^{n+1}}{n!} e^{frac{2r}{n+1/2}}int_0^infty e^{-uleft( n+1/2 right)}u^{n+1/2}J_0left( 2sqrt{2ru} right)frac{du}{sqrt{u}}
          end{equation}
          or
          begin{equation}
          L_n(frac{2r}{n+1/2})= frac{(n+1/2)^{n+1}}{n!} e^{frac{2r}{n+1/2}}int_0^infty e^{-left( n+1/2 right)left( u-ln u right)}J_0left( 2sqrt{2ru} right)frac{du}{sqrt{u}}
          end{equation}
          The argument of the exponential is minimum at $u=1$. We can use the Laplace method to derive an asymptotic approximation for the integral. Near $u=1$,
          $u-ln(u)sim 1+(u-1)^2/2$ and $u^{-1/2}J_0left( 2sqrt{2ru}right)sim J_0left( 2sqrt{2r}right)$, then
          begin{equation}
          L_n(frac{2r}{n+1/2})sim sqrt{2pi}frac{(n+1/2)^{n+1/2}}{n!} e^{frac{2r}{n+1/2}}
          e^{-(n+1/2)}J_0left( 2sqrt{2r}right)
          end{equation}
          Now, plugging the Stirling approximation for $n!$ and an expansion for the exponential term, we obtain the expected result $ L_n(frac{2r}{n+1/2})sim J_0left( 2sqrt{2r}right)$. To improve the approximation, the above formula and/or higher orders in the Laplace expansion can be used.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 29 at 23:07

























          answered Nov 29 at 22:47









          Paul Enta

          4,11811029




          4,11811029












          • Thank you! I will wait a little while before accepting the answer, but this is exactly what I wanted
            – Yuriy S
            Nov 30 at 1:02


















          • Thank you! I will wait a little while before accepting the answer, but this is exactly what I wanted
            – Yuriy S
            Nov 30 at 1:02
















          Thank you! I will wait a little while before accepting the answer, but this is exactly what I wanted
          – Yuriy S
          Nov 30 at 1:02




          Thank you! I will wait a little while before accepting the answer, but this is exactly what I wanted
          – Yuriy S
          Nov 30 at 1:02


















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