About the Bessel function












0














I was looking at this Bessel function, we got these two sums



$$sum_{n=0}^{infty}left(-frac{1}{x^2}right)^nfrac{{4n choose 2n}{2n choose n}}{[(2n-1)!!]^2}=cosleft(frac{4}{x}right)J_0left(frac{4}{x}right)tag1$$



$$sum_{n=0}^{infty}left(-frac{1}{x^2}right)^nfrac{H_{n,k}}{(n!)^2}=zeta(k)J_0left(frac{2}{x}right)tag2$$



Where $H_{n,k}$ and $zeta(k)$



Just out of curiosity.



How does one show that the sums are correct?






sequences-and-series







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    0














    I was looking at this Bessel function, we got these two sums



    $$sum_{n=0}^{infty}left(-frac{1}{x^2}right)^nfrac{{4n choose 2n}{2n choose n}}{[(2n-1)!!]^2}=cosleft(frac{4}{x}right)J_0left(frac{4}{x}right)tag1$$



    $$sum_{n=0}^{infty}left(-frac{1}{x^2}right)^nfrac{H_{n,k}}{(n!)^2}=zeta(k)J_0left(frac{2}{x}right)tag2$$



    Where $H_{n,k}$ and $zeta(k)$



    Just out of curiosity.



    How does one show that the sums are correct?






    sequences-and-series







    share|cite|improve this question

























      0












      0








      0







      I was looking at this Bessel function, we got these two sums



      $$sum_{n=0}^{infty}left(-frac{1}{x^2}right)^nfrac{{4n choose 2n}{2n choose n}}{[(2n-1)!!]^2}=cosleft(frac{4}{x}right)J_0left(frac{4}{x}right)tag1$$



      $$sum_{n=0}^{infty}left(-frac{1}{x^2}right)^nfrac{H_{n,k}}{(n!)^2}=zeta(k)J_0left(frac{2}{x}right)tag2$$



      Where $H_{n,k}$ and $zeta(k)$



      Just out of curiosity.



      How does one show that the sums are correct?






      sequences-and-series







      share|cite|improve this question













      I was looking at this Bessel function, we got these two sums



      $$sum_{n=0}^{infty}left(-frac{1}{x^2}right)^nfrac{{4n choose 2n}{2n choose n}}{[(2n-1)!!]^2}=cosleft(frac{4}{x}right)J_0left(frac{4}{x}right)tag1$$



      $$sum_{n=0}^{infty}left(-frac{1}{x^2}right)^nfrac{H_{n,k}}{(n!)^2}=zeta(k)J_0left(frac{2}{x}right)tag2$$



      Where $H_{n,k}$ and $zeta(k)$



      Just out of curiosity.



      How does one show that the sums are correct?






      sequences-and-series




      sequences-and-series






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 4 at 9:41









      user583851user583851

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      4117






















          1 Answer
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          About $(2)$: we may replace $x$ by $frac{1}{x}$, then consider the Laplace transform of both sides.
          The problem boils down to showing



          $$sum_{ngeq 0}binom{2n}{n}frac{(-1)^nH_{n}^{(k)}}{s^{2n+1}} = frac{zeta(k)}{sqrt{4+s^2}} $$
          where
          $$ H_n^{(k)}=sum_{m=1}^{n}frac{1}{m^k}=zeta(k)-int_{0}^{+infty}frac{z^{k-1}e^{-nz}}{(k-1)!}cdotfrac{dz}{e^z-1}$$
          allows to state the previous (conjectural) identity as
          $$ frac{zeta(k)}{sqrt{4+s^2}}-frac{1}{(k-1)!}color{blue}{int_{0}^{+infty}frac{z^{k-1}}{(e^z-1)sqrt{s^2+4 e^{-z}}},dz} = frac{zeta(k)}{sqrt{4+s^2}} $$
          or in the form: the $color{blue}{text{blue}}$ integral is constantly zero for any $s$. This is not true, since the blue integral is a decreasing function with respect to the parameter $s$. In particular $(2)$ is not correct.



          $(1)$ is correct: it is enough to consider the Cauchy product between the Maclaurin series of $cos(4x)$ and $J_0(4x)$.






          share|cite|improve this answer





















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            1 Answer
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            active

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            1 Answer
            1






            active

            oldest

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            active

            oldest

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            active

            oldest

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            2














            About $(2)$: we may replace $x$ by $frac{1}{x}$, then consider the Laplace transform of both sides.
            The problem boils down to showing



            $$sum_{ngeq 0}binom{2n}{n}frac{(-1)^nH_{n}^{(k)}}{s^{2n+1}} = frac{zeta(k)}{sqrt{4+s^2}} $$
            where
            $$ H_n^{(k)}=sum_{m=1}^{n}frac{1}{m^k}=zeta(k)-int_{0}^{+infty}frac{z^{k-1}e^{-nz}}{(k-1)!}cdotfrac{dz}{e^z-1}$$
            allows to state the previous (conjectural) identity as
            $$ frac{zeta(k)}{sqrt{4+s^2}}-frac{1}{(k-1)!}color{blue}{int_{0}^{+infty}frac{z^{k-1}}{(e^z-1)sqrt{s^2+4 e^{-z}}},dz} = frac{zeta(k)}{sqrt{4+s^2}} $$
            or in the form: the $color{blue}{text{blue}}$ integral is constantly zero for any $s$. This is not true, since the blue integral is a decreasing function with respect to the parameter $s$. In particular $(2)$ is not correct.



            $(1)$ is correct: it is enough to consider the Cauchy product between the Maclaurin series of $cos(4x)$ and $J_0(4x)$.






            share|cite|improve this answer


























              2














              About $(2)$: we may replace $x$ by $frac{1}{x}$, then consider the Laplace transform of both sides.
              The problem boils down to showing



              $$sum_{ngeq 0}binom{2n}{n}frac{(-1)^nH_{n}^{(k)}}{s^{2n+1}} = frac{zeta(k)}{sqrt{4+s^2}} $$
              where
              $$ H_n^{(k)}=sum_{m=1}^{n}frac{1}{m^k}=zeta(k)-int_{0}^{+infty}frac{z^{k-1}e^{-nz}}{(k-1)!}cdotfrac{dz}{e^z-1}$$
              allows to state the previous (conjectural) identity as
              $$ frac{zeta(k)}{sqrt{4+s^2}}-frac{1}{(k-1)!}color{blue}{int_{0}^{+infty}frac{z^{k-1}}{(e^z-1)sqrt{s^2+4 e^{-z}}},dz} = frac{zeta(k)}{sqrt{4+s^2}} $$
              or in the form: the $color{blue}{text{blue}}$ integral is constantly zero for any $s$. This is not true, since the blue integral is a decreasing function with respect to the parameter $s$. In particular $(2)$ is not correct.



              $(1)$ is correct: it is enough to consider the Cauchy product between the Maclaurin series of $cos(4x)$ and $J_0(4x)$.






              share|cite|improve this answer
























                2












                2








                2






                About $(2)$: we may replace $x$ by $frac{1}{x}$, then consider the Laplace transform of both sides.
                The problem boils down to showing



                $$sum_{ngeq 0}binom{2n}{n}frac{(-1)^nH_{n}^{(k)}}{s^{2n+1}} = frac{zeta(k)}{sqrt{4+s^2}} $$
                where
                $$ H_n^{(k)}=sum_{m=1}^{n}frac{1}{m^k}=zeta(k)-int_{0}^{+infty}frac{z^{k-1}e^{-nz}}{(k-1)!}cdotfrac{dz}{e^z-1}$$
                allows to state the previous (conjectural) identity as
                $$ frac{zeta(k)}{sqrt{4+s^2}}-frac{1}{(k-1)!}color{blue}{int_{0}^{+infty}frac{z^{k-1}}{(e^z-1)sqrt{s^2+4 e^{-z}}},dz} = frac{zeta(k)}{sqrt{4+s^2}} $$
                or in the form: the $color{blue}{text{blue}}$ integral is constantly zero for any $s$. This is not true, since the blue integral is a decreasing function with respect to the parameter $s$. In particular $(2)$ is not correct.



                $(1)$ is correct: it is enough to consider the Cauchy product between the Maclaurin series of $cos(4x)$ and $J_0(4x)$.






                share|cite|improve this answer












                About $(2)$: we may replace $x$ by $frac{1}{x}$, then consider the Laplace transform of both sides.
                The problem boils down to showing



                $$sum_{ngeq 0}binom{2n}{n}frac{(-1)^nH_{n}^{(k)}}{s^{2n+1}} = frac{zeta(k)}{sqrt{4+s^2}} $$
                where
                $$ H_n^{(k)}=sum_{m=1}^{n}frac{1}{m^k}=zeta(k)-int_{0}^{+infty}frac{z^{k-1}e^{-nz}}{(k-1)!}cdotfrac{dz}{e^z-1}$$
                allows to state the previous (conjectural) identity as
                $$ frac{zeta(k)}{sqrt{4+s^2}}-frac{1}{(k-1)!}color{blue}{int_{0}^{+infty}frac{z^{k-1}}{(e^z-1)sqrt{s^2+4 e^{-z}}},dz} = frac{zeta(k)}{sqrt{4+s^2}} $$
                or in the form: the $color{blue}{text{blue}}$ integral is constantly zero for any $s$. This is not true, since the blue integral is a decreasing function with respect to the parameter $s$. In particular $(2)$ is not correct.



                $(1)$ is correct: it is enough to consider the Cauchy product between the Maclaurin series of $cos(4x)$ and $J_0(4x)$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 4 at 17:47









                Jack D'AurizioJack D'Aurizio

                287k33280658




                287k33280658






























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