A question on the approximation of $ln(x!)$ for small x












0












$begingroup$


In a question of Arfken and Weber's Mathematical Methods for Physicists,



For small values of $x$, $ln(x!)=-gamma x+sumlimits_{n=2}^{infty}(-1)^nfrac{zeta(n)}{n}x^n$, where the symbols used have their usual meanings.



Now I need to prove that this can also be written as
$$ln(x!)=frac{1}{2}lnleft(frac{pi x}{sinpi x}right)-gamma x-sumlimits_{n=1}^{infty}frac{zeta(2n+1)}{2n+1}x^{2n+1}$$



I can only do this.



$$sumlimits_{n=2}^{infty}(-1)^nfrac{zeta(n)}{n}x^n=\
left(frac{zeta(2)}{2}x^2+frac{zeta(4)}{4}x^4+frac{zeta(6)}{6}x^6+...right)-left(frac{zeta(3)}{3}x^3+frac{zeta(5)}{5}x^5+frac{zeta(7)}{7}x^7+...right)
=\sumlimits_{n=1}^{infty}frac{zeta(2n)}{2n}x^{2n}-sumlimits_{n=1}^{infty}frac{zeta(2n+1)}{2n+1}x^{2n+1}$$



Putting this in the original equation and comparing it with what needs to be proven, it can be concluded that



$$frac{1}{2}lnleft(frac{pi x}{sinpi x}right)=sumlimits_{n=1}^{infty}frac{zeta(2n)}{2n}x^{2n}$$



I don't know how to prove this.



$frac{pi x}{sinpi x}=xGamma(x)Gamma(1-x)$ I think this has to do something with it since gamma function and zeta function are related to each other through an integral.






taylor-expansion factorial riemann-zeta







share|cite|improve this question











$endgroup$












  • $begingroup$
    Using $zeta(2n)=(-1)^{n+1}frac{B_{2n}(2pi)^{2n}}{2(2n)!} $ and $sum_{n=0}^infty frac{B_n}{n!} x^n = frac{x}{e^x-1}$ you obtain $ f(x)=sum_{n=1}^{infty}frac{zeta(2n)}{2n}x^{2n}=frac12 (frac{2pi ix}{e^{2pi ix}-1}+ frac{-2pi ix}{e^{-2pi ix}-1})$ and you can compare $f(x)/x$ with the derivative of $frac{1}{2}lnleft(frac{pi x}{sinpi x}right)$
    $endgroup$
    – reuns
    Jan 6 at 21:15










  • $begingroup$
    Using $sumlimits_{n=1}^{infty}frac{B_{2n}}{(2n)!}x^{2n}=frac{x}{e^x-1}+frac{x}{2}-1$, I got $sumlimits_{n=1}^{infty}(-1)^nfrac{B_{2n}}{(2n)!}(2pi x)^{2n}=frac{1}{2}[frac{2piiota x}{e^{2piiota x}-1}+frac{-2piiota x}{e^{-2piiota x}-1}]-1$. When zeta function is expressed in terms of Bernoulli numbers, $f(x)=frac{-1}{2}sumlimits_{n=1}^{infty}(-1)^nfrac{B_{2n}}{(2n)! (2n)}(2pi x)^{2n}$. $2n$ appears in the denomiantor. How to proceed?
    $endgroup$
    – Asit Srivastava
    Jan 6 at 21:56












  • $begingroup$
    Differentiate both side of $frac{1}{2}lnleft(frac{pi x}{sinpi x}right)=sumlimits_{n=1}^{infty}frac{zeta(2n)}{2n}x^{2n}$
    $endgroup$
    – reuns
    Jan 6 at 22:19










  • $begingroup$
    Thanks. It worked.
    $endgroup$
    – Asit Srivastava
    Jan 6 at 22:51
















0












$begingroup$


In a question of Arfken and Weber's Mathematical Methods for Physicists,



For small values of $x$, $ln(x!)=-gamma x+sumlimits_{n=2}^{infty}(-1)^nfrac{zeta(n)}{n}x^n$, where the symbols used have their usual meanings.



Now I need to prove that this can also be written as
$$ln(x!)=frac{1}{2}lnleft(frac{pi x}{sinpi x}right)-gamma x-sumlimits_{n=1}^{infty}frac{zeta(2n+1)}{2n+1}x^{2n+1}$$



I can only do this.



$$sumlimits_{n=2}^{infty}(-1)^nfrac{zeta(n)}{n}x^n=\
left(frac{zeta(2)}{2}x^2+frac{zeta(4)}{4}x^4+frac{zeta(6)}{6}x^6+...right)-left(frac{zeta(3)}{3}x^3+frac{zeta(5)}{5}x^5+frac{zeta(7)}{7}x^7+...right)
=\sumlimits_{n=1}^{infty}frac{zeta(2n)}{2n}x^{2n}-sumlimits_{n=1}^{infty}frac{zeta(2n+1)}{2n+1}x^{2n+1}$$



Putting this in the original equation and comparing it with what needs to be proven, it can be concluded that



$$frac{1}{2}lnleft(frac{pi x}{sinpi x}right)=sumlimits_{n=1}^{infty}frac{zeta(2n)}{2n}x^{2n}$$



I don't know how to prove this.



$frac{pi x}{sinpi x}=xGamma(x)Gamma(1-x)$ I think this has to do something with it since gamma function and zeta function are related to each other through an integral.






taylor-expansion factorial riemann-zeta







share|cite|improve this question











$endgroup$












  • $begingroup$
    Using $zeta(2n)=(-1)^{n+1}frac{B_{2n}(2pi)^{2n}}{2(2n)!} $ and $sum_{n=0}^infty frac{B_n}{n!} x^n = frac{x}{e^x-1}$ you obtain $ f(x)=sum_{n=1}^{infty}frac{zeta(2n)}{2n}x^{2n}=frac12 (frac{2pi ix}{e^{2pi ix}-1}+ frac{-2pi ix}{e^{-2pi ix}-1})$ and you can compare $f(x)/x$ with the derivative of $frac{1}{2}lnleft(frac{pi x}{sinpi x}right)$
    $endgroup$
    – reuns
    Jan 6 at 21:15










  • $begingroup$
    Using $sumlimits_{n=1}^{infty}frac{B_{2n}}{(2n)!}x^{2n}=frac{x}{e^x-1}+frac{x}{2}-1$, I got $sumlimits_{n=1}^{infty}(-1)^nfrac{B_{2n}}{(2n)!}(2pi x)^{2n}=frac{1}{2}[frac{2piiota x}{e^{2piiota x}-1}+frac{-2piiota x}{e^{-2piiota x}-1}]-1$. When zeta function is expressed in terms of Bernoulli numbers, $f(x)=frac{-1}{2}sumlimits_{n=1}^{infty}(-1)^nfrac{B_{2n}}{(2n)! (2n)}(2pi x)^{2n}$. $2n$ appears in the denomiantor. How to proceed?
    $endgroup$
    – Asit Srivastava
    Jan 6 at 21:56












  • $begingroup$
    Differentiate both side of $frac{1}{2}lnleft(frac{pi x}{sinpi x}right)=sumlimits_{n=1}^{infty}frac{zeta(2n)}{2n}x^{2n}$
    $endgroup$
    – reuns
    Jan 6 at 22:19










  • $begingroup$
    Thanks. It worked.
    $endgroup$
    – Asit Srivastava
    Jan 6 at 22:51














0












0








0





$begingroup$


In a question of Arfken and Weber's Mathematical Methods for Physicists,



For small values of $x$, $ln(x!)=-gamma x+sumlimits_{n=2}^{infty}(-1)^nfrac{zeta(n)}{n}x^n$, where the symbols used have their usual meanings.



Now I need to prove that this can also be written as
$$ln(x!)=frac{1}{2}lnleft(frac{pi x}{sinpi x}right)-gamma x-sumlimits_{n=1}^{infty}frac{zeta(2n+1)}{2n+1}x^{2n+1}$$



I can only do this.



$$sumlimits_{n=2}^{infty}(-1)^nfrac{zeta(n)}{n}x^n=\
left(frac{zeta(2)}{2}x^2+frac{zeta(4)}{4}x^4+frac{zeta(6)}{6}x^6+...right)-left(frac{zeta(3)}{3}x^3+frac{zeta(5)}{5}x^5+frac{zeta(7)}{7}x^7+...right)
=\sumlimits_{n=1}^{infty}frac{zeta(2n)}{2n}x^{2n}-sumlimits_{n=1}^{infty}frac{zeta(2n+1)}{2n+1}x^{2n+1}$$



Putting this in the original equation and comparing it with what needs to be proven, it can be concluded that



$$frac{1}{2}lnleft(frac{pi x}{sinpi x}right)=sumlimits_{n=1}^{infty}frac{zeta(2n)}{2n}x^{2n}$$



I don't know how to prove this.



$frac{pi x}{sinpi x}=xGamma(x)Gamma(1-x)$ I think this has to do something with it since gamma function and zeta function are related to each other through an integral.






taylor-expansion factorial riemann-zeta







share|cite|improve this question











$endgroup$




In a question of Arfken and Weber's Mathematical Methods for Physicists,



For small values of $x$, $ln(x!)=-gamma x+sumlimits_{n=2}^{infty}(-1)^nfrac{zeta(n)}{n}x^n$, where the symbols used have their usual meanings.



Now I need to prove that this can also be written as
$$ln(x!)=frac{1}{2}lnleft(frac{pi x}{sinpi x}right)-gamma x-sumlimits_{n=1}^{infty}frac{zeta(2n+1)}{2n+1}x^{2n+1}$$



I can only do this.



$$sumlimits_{n=2}^{infty}(-1)^nfrac{zeta(n)}{n}x^n=\
left(frac{zeta(2)}{2}x^2+frac{zeta(4)}{4}x^4+frac{zeta(6)}{6}x^6+...right)-left(frac{zeta(3)}{3}x^3+frac{zeta(5)}{5}x^5+frac{zeta(7)}{7}x^7+...right)
=\sumlimits_{n=1}^{infty}frac{zeta(2n)}{2n}x^{2n}-sumlimits_{n=1}^{infty}frac{zeta(2n+1)}{2n+1}x^{2n+1}$$



Putting this in the original equation and comparing it with what needs to be proven, it can be concluded that



$$frac{1}{2}lnleft(frac{pi x}{sinpi x}right)=sumlimits_{n=1}^{infty}frac{zeta(2n)}{2n}x^{2n}$$



I don't know how to prove this.



$frac{pi x}{sinpi x}=xGamma(x)Gamma(1-x)$ I think this has to do something with it since gamma function and zeta function are related to each other through an integral.






taylor-expansion factorial riemann-zeta




taylor-expansion factorial riemann-zeta






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 21:05









rtybase

11.5k31534




11.5k31534










asked Jan 6 at 20:54









Asit SrivastavaAsit Srivastava

257




257












  • $begingroup$
    Using $zeta(2n)=(-1)^{n+1}frac{B_{2n}(2pi)^{2n}}{2(2n)!} $ and $sum_{n=0}^infty frac{B_n}{n!} x^n = frac{x}{e^x-1}$ you obtain $ f(x)=sum_{n=1}^{infty}frac{zeta(2n)}{2n}x^{2n}=frac12 (frac{2pi ix}{e^{2pi ix}-1}+ frac{-2pi ix}{e^{-2pi ix}-1})$ and you can compare $f(x)/x$ with the derivative of $frac{1}{2}lnleft(frac{pi x}{sinpi x}right)$
    $endgroup$
    – reuns
    Jan 6 at 21:15










  • $begingroup$
    Using $sumlimits_{n=1}^{infty}frac{B_{2n}}{(2n)!}x^{2n}=frac{x}{e^x-1}+frac{x}{2}-1$, I got $sumlimits_{n=1}^{infty}(-1)^nfrac{B_{2n}}{(2n)!}(2pi x)^{2n}=frac{1}{2}[frac{2piiota x}{e^{2piiota x}-1}+frac{-2piiota x}{e^{-2piiota x}-1}]-1$. When zeta function is expressed in terms of Bernoulli numbers, $f(x)=frac{-1}{2}sumlimits_{n=1}^{infty}(-1)^nfrac{B_{2n}}{(2n)! (2n)}(2pi x)^{2n}$. $2n$ appears in the denomiantor. How to proceed?
    $endgroup$
    – Asit Srivastava
    Jan 6 at 21:56












  • $begingroup$
    Differentiate both side of $frac{1}{2}lnleft(frac{pi x}{sinpi x}right)=sumlimits_{n=1}^{infty}frac{zeta(2n)}{2n}x^{2n}$
    $endgroup$
    – reuns
    Jan 6 at 22:19










  • $begingroup$
    Thanks. It worked.
    $endgroup$
    – Asit Srivastava
    Jan 6 at 22:51


















  • $begingroup$
    Using $zeta(2n)=(-1)^{n+1}frac{B_{2n}(2pi)^{2n}}{2(2n)!} $ and $sum_{n=0}^infty frac{B_n}{n!} x^n = frac{x}{e^x-1}$ you obtain $ f(x)=sum_{n=1}^{infty}frac{zeta(2n)}{2n}x^{2n}=frac12 (frac{2pi ix}{e^{2pi ix}-1}+ frac{-2pi ix}{e^{-2pi ix}-1})$ and you can compare $f(x)/x$ with the derivative of $frac{1}{2}lnleft(frac{pi x}{sinpi x}right)$
    $endgroup$
    – reuns
    Jan 6 at 21:15










  • $begingroup$
    Using $sumlimits_{n=1}^{infty}frac{B_{2n}}{(2n)!}x^{2n}=frac{x}{e^x-1}+frac{x}{2}-1$, I got $sumlimits_{n=1}^{infty}(-1)^nfrac{B_{2n}}{(2n)!}(2pi x)^{2n}=frac{1}{2}[frac{2piiota x}{e^{2piiota x}-1}+frac{-2piiota x}{e^{-2piiota x}-1}]-1$. When zeta function is expressed in terms of Bernoulli numbers, $f(x)=frac{-1}{2}sumlimits_{n=1}^{infty}(-1)^nfrac{B_{2n}}{(2n)! (2n)}(2pi x)^{2n}$. $2n$ appears in the denomiantor. How to proceed?
    $endgroup$
    – Asit Srivastava
    Jan 6 at 21:56












  • $begingroup$
    Differentiate both side of $frac{1}{2}lnleft(frac{pi x}{sinpi x}right)=sumlimits_{n=1}^{infty}frac{zeta(2n)}{2n}x^{2n}$
    $endgroup$
    – reuns
    Jan 6 at 22:19










  • $begingroup$
    Thanks. It worked.
    $endgroup$
    – Asit Srivastava
    Jan 6 at 22:51
















$begingroup$
Using $zeta(2n)=(-1)^{n+1}frac{B_{2n}(2pi)^{2n}}{2(2n)!} $ and $sum_{n=0}^infty frac{B_n}{n!} x^n = frac{x}{e^x-1}$ you obtain $ f(x)=sum_{n=1}^{infty}frac{zeta(2n)}{2n}x^{2n}=frac12 (frac{2pi ix}{e^{2pi ix}-1}+ frac{-2pi ix}{e^{-2pi ix}-1})$ and you can compare $f(x)/x$ with the derivative of $frac{1}{2}lnleft(frac{pi x}{sinpi x}right)$
$endgroup$
– reuns
Jan 6 at 21:15




$begingroup$
Using $zeta(2n)=(-1)^{n+1}frac{B_{2n}(2pi)^{2n}}{2(2n)!} $ and $sum_{n=0}^infty frac{B_n}{n!} x^n = frac{x}{e^x-1}$ you obtain $ f(x)=sum_{n=1}^{infty}frac{zeta(2n)}{2n}x^{2n}=frac12 (frac{2pi ix}{e^{2pi ix}-1}+ frac{-2pi ix}{e^{-2pi ix}-1})$ and you can compare $f(x)/x$ with the derivative of $frac{1}{2}lnleft(frac{pi x}{sinpi x}right)$
$endgroup$
– reuns
Jan 6 at 21:15












$begingroup$
Using $sumlimits_{n=1}^{infty}frac{B_{2n}}{(2n)!}x^{2n}=frac{x}{e^x-1}+frac{x}{2}-1$, I got $sumlimits_{n=1}^{infty}(-1)^nfrac{B_{2n}}{(2n)!}(2pi x)^{2n}=frac{1}{2}[frac{2piiota x}{e^{2piiota x}-1}+frac{-2piiota x}{e^{-2piiota x}-1}]-1$. When zeta function is expressed in terms of Bernoulli numbers, $f(x)=frac{-1}{2}sumlimits_{n=1}^{infty}(-1)^nfrac{B_{2n}}{(2n)! (2n)}(2pi x)^{2n}$. $2n$ appears in the denomiantor. How to proceed?
$endgroup$
– Asit Srivastava
Jan 6 at 21:56






$begingroup$
Using $sumlimits_{n=1}^{infty}frac{B_{2n}}{(2n)!}x^{2n}=frac{x}{e^x-1}+frac{x}{2}-1$, I got $sumlimits_{n=1}^{infty}(-1)^nfrac{B_{2n}}{(2n)!}(2pi x)^{2n}=frac{1}{2}[frac{2piiota x}{e^{2piiota x}-1}+frac{-2piiota x}{e^{-2piiota x}-1}]-1$. When zeta function is expressed in terms of Bernoulli numbers, $f(x)=frac{-1}{2}sumlimits_{n=1}^{infty}(-1)^nfrac{B_{2n}}{(2n)! (2n)}(2pi x)^{2n}$. $2n$ appears in the denomiantor. How to proceed?
$endgroup$
– Asit Srivastava
Jan 6 at 21:56














$begingroup$
Differentiate both side of $frac{1}{2}lnleft(frac{pi x}{sinpi x}right)=sumlimits_{n=1}^{infty}frac{zeta(2n)}{2n}x^{2n}$
$endgroup$
– reuns
Jan 6 at 22:19




$begingroup$
Differentiate both side of $frac{1}{2}lnleft(frac{pi x}{sinpi x}right)=sumlimits_{n=1}^{infty}frac{zeta(2n)}{2n}x^{2n}$
$endgroup$
– reuns
Jan 6 at 22:19












$begingroup$
Thanks. It worked.
$endgroup$
– Asit Srivastava
Jan 6 at 22:51




$begingroup$
Thanks. It worked.
$endgroup$
– Asit Srivastava
Jan 6 at 22:51










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