How to show $frac{Gamma((n-1)/2)}{Gamma(n/2)} approx frac{sqrt{2}}{sqrt{n-2}}$












3















Show $frac{Gamma((n-1)/2)}{Gamma(n/2)} approx frac{sqrt{2}}{sqrt{n-2}}$




Try



Using the facts:





  • $(1 + alpha/m)^m = e^alpha ( 1+ r_m)$, where $lim_{m to infty} sqrt{m}r_m = 0$


  • $Gamma(n+1) = n^{n + 1/2} e^{-n} sqrt{2 pi} (1 + r_n)$, where $lim_{n to infty} sqrt{n}r_n = 0$


Note :



$$
begin{aligned}
frac{Gamma((n-1)/2)}{Gamma(n/2)} &= frac{left(frac{n-3}{2}right)^{(n-2)/2} e^{-(n-3)/2}(sqrt{2pi}(1 + r_{1n})}{left(frac{n-2}{2}right)^{(n-1)/2} e^{-(n-2)/2}(sqrt{2pi}(1 + r_{2n})} \
&=left( left(frac{n-3}{n-2}right)^{(n-2)/2} sqrt{e} frac{1+r_{1n}}{1 + r_{2n}} right) times frac{sqrt{2}}{sqrt{n-2}}
end{aligned}
$$



But I'm stuck at how I should proceed to eliminate the $left( left(frac{n-3}{n-2}right)^{(n-2)/2} sqrt{e} frac{1+r_{1n}}{1 + r_{2n}} right)$ term.






approximation gamma-function







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  • 2




    What is your definition of $Gamma$? According to it, proving Gautschi's inequality might be more or less trivial.
    – Jack D'Aurizio
    Nov 21 '18 at 2:21






  • 2




    what is your definition of $approx$??
    – Masacroso
    Nov 21 '18 at 2:39
















3















Show $frac{Gamma((n-1)/2)}{Gamma(n/2)} approx frac{sqrt{2}}{sqrt{n-2}}$




Try



Using the facts:





  • $(1 + alpha/m)^m = e^alpha ( 1+ r_m)$, where $lim_{m to infty} sqrt{m}r_m = 0$


  • $Gamma(n+1) = n^{n + 1/2} e^{-n} sqrt{2 pi} (1 + r_n)$, where $lim_{n to infty} sqrt{n}r_n = 0$


Note :



$$
begin{aligned}
frac{Gamma((n-1)/2)}{Gamma(n/2)} &= frac{left(frac{n-3}{2}right)^{(n-2)/2} e^{-(n-3)/2}(sqrt{2pi}(1 + r_{1n})}{left(frac{n-2}{2}right)^{(n-1)/2} e^{-(n-2)/2}(sqrt{2pi}(1 + r_{2n})} \
&=left( left(frac{n-3}{n-2}right)^{(n-2)/2} sqrt{e} frac{1+r_{1n}}{1 + r_{2n}} right) times frac{sqrt{2}}{sqrt{n-2}}
end{aligned}
$$



But I'm stuck at how I should proceed to eliminate the $left( left(frac{n-3}{n-2}right)^{(n-2)/2} sqrt{e} frac{1+r_{1n}}{1 + r_{2n}} right)$ term.






approximation gamma-function







share|cite|improve this question


















  • 2




    What is your definition of $Gamma$? According to it, proving Gautschi's inequality might be more or less trivial.
    – Jack D'Aurizio
    Nov 21 '18 at 2:21






  • 2




    what is your definition of $approx$??
    – Masacroso
    Nov 21 '18 at 2:39














3












3








3


1






Show $frac{Gamma((n-1)/2)}{Gamma(n/2)} approx frac{sqrt{2}}{sqrt{n-2}}$




Try



Using the facts:





  • $(1 + alpha/m)^m = e^alpha ( 1+ r_m)$, where $lim_{m to infty} sqrt{m}r_m = 0$


  • $Gamma(n+1) = n^{n + 1/2} e^{-n} sqrt{2 pi} (1 + r_n)$, where $lim_{n to infty} sqrt{n}r_n = 0$


Note :



$$
begin{aligned}
frac{Gamma((n-1)/2)}{Gamma(n/2)} &= frac{left(frac{n-3}{2}right)^{(n-2)/2} e^{-(n-3)/2}(sqrt{2pi}(1 + r_{1n})}{left(frac{n-2}{2}right)^{(n-1)/2} e^{-(n-2)/2}(sqrt{2pi}(1 + r_{2n})} \
&=left( left(frac{n-3}{n-2}right)^{(n-2)/2} sqrt{e} frac{1+r_{1n}}{1 + r_{2n}} right) times frac{sqrt{2}}{sqrt{n-2}}
end{aligned}
$$



But I'm stuck at how I should proceed to eliminate the $left( left(frac{n-3}{n-2}right)^{(n-2)/2} sqrt{e} frac{1+r_{1n}}{1 + r_{2n}} right)$ term.






approximation gamma-function







share|cite|improve this question














Show $frac{Gamma((n-1)/2)}{Gamma(n/2)} approx frac{sqrt{2}}{sqrt{n-2}}$




Try



Using the facts:





  • $(1 + alpha/m)^m = e^alpha ( 1+ r_m)$, where $lim_{m to infty} sqrt{m}r_m = 0$


  • $Gamma(n+1) = n^{n + 1/2} e^{-n} sqrt{2 pi} (1 + r_n)$, where $lim_{n to infty} sqrt{n}r_n = 0$


Note :



$$
begin{aligned}
frac{Gamma((n-1)/2)}{Gamma(n/2)} &= frac{left(frac{n-3}{2}right)^{(n-2)/2} e^{-(n-3)/2}(sqrt{2pi}(1 + r_{1n})}{left(frac{n-2}{2}right)^{(n-1)/2} e^{-(n-2)/2}(sqrt{2pi}(1 + r_{2n})} \
&=left( left(frac{n-3}{n-2}right)^{(n-2)/2} sqrt{e} frac{1+r_{1n}}{1 + r_{2n}} right) times frac{sqrt{2}}{sqrt{n-2}}
end{aligned}
$$



But I'm stuck at how I should proceed to eliminate the $left( left(frac{n-3}{n-2}right)^{(n-2)/2} sqrt{e} frac{1+r_{1n}}{1 + r_{2n}} right)$ term.






approximation gamma-function




approximation gamma-function






share|cite|improve this question













share|cite|improve this question











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asked Nov 21 '18 at 2:20









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  • 2




    What is your definition of $Gamma$? According to it, proving Gautschi's inequality might be more or less trivial.
    – Jack D'Aurizio
    Nov 21 '18 at 2:21






  • 2




    what is your definition of $approx$??
    – Masacroso
    Nov 21 '18 at 2:39














  • 2




    What is your definition of $Gamma$? According to it, proving Gautschi's inequality might be more or less trivial.
    – Jack D'Aurizio
    Nov 21 '18 at 2:21






  • 2




    what is your definition of $approx$??
    – Masacroso
    Nov 21 '18 at 2:39








2




2




What is your definition of $Gamma$? According to it, proving Gautschi's inequality might be more or less trivial.
– Jack D'Aurizio
Nov 21 '18 at 2:21




What is your definition of $Gamma$? According to it, proving Gautschi's inequality might be more or less trivial.
– Jack D'Aurizio
Nov 21 '18 at 2:21




2




2




what is your definition of $approx$??
– Masacroso
Nov 21 '18 at 2:39




what is your definition of $approx$??
– Masacroso
Nov 21 '18 at 2:39










1 Answer
1






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oldest

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6














A possible approach:



$$ frac{Gammaleft(tfrac{n-1}{2}right)}{Gammaleft(tfrac{n}{2}right)}=tfrac{1}{sqrt{pi}},Bleft(tfrac{n-1}{2},tfrac{1}{2}right)=frac{1}{sqrt{pi}}int_{0}^{1}x^{frac{n-3}{2}}(1-x)^{-1/2},dx=frac{2}{sqrt{pi}}int_{0}^{1}frac{x^{n-2}}{sqrt{1-x^2}},dx $$
gives
$$ frac{Gammaleft(tfrac{n-1}{2}right)}{Gammaleft(tfrac{n}{2}right)}= frac{2}{sqrt{pi}}int_{0}^{pi/2}left(costhetaright)^{n-2},dthetasimfrac{2}{sqrt{pi}}int_{0}^{+infty}expleft[-(n-2)frac{theta^2}{2}right]dtheta=sqrt{frac{2}{n-2}}. $$
Notice that the integral representation (as a moment) instantly gives that the LHS is a log-convex function. The asymptotic equivalence $sim$ can be seen as an instance of Laplace/Hayman's method.






share|cite|improve this answer























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    1 Answer
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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6














    A possible approach:



    $$ frac{Gammaleft(tfrac{n-1}{2}right)}{Gammaleft(tfrac{n}{2}right)}=tfrac{1}{sqrt{pi}},Bleft(tfrac{n-1}{2},tfrac{1}{2}right)=frac{1}{sqrt{pi}}int_{0}^{1}x^{frac{n-3}{2}}(1-x)^{-1/2},dx=frac{2}{sqrt{pi}}int_{0}^{1}frac{x^{n-2}}{sqrt{1-x^2}},dx $$
    gives
    $$ frac{Gammaleft(tfrac{n-1}{2}right)}{Gammaleft(tfrac{n}{2}right)}= frac{2}{sqrt{pi}}int_{0}^{pi/2}left(costhetaright)^{n-2},dthetasimfrac{2}{sqrt{pi}}int_{0}^{+infty}expleft[-(n-2)frac{theta^2}{2}right]dtheta=sqrt{frac{2}{n-2}}. $$
    Notice that the integral representation (as a moment) instantly gives that the LHS is a log-convex function. The asymptotic equivalence $sim$ can be seen as an instance of Laplace/Hayman's method.






    share|cite|improve this answer




























      6














      A possible approach:



      $$ frac{Gammaleft(tfrac{n-1}{2}right)}{Gammaleft(tfrac{n}{2}right)}=tfrac{1}{sqrt{pi}},Bleft(tfrac{n-1}{2},tfrac{1}{2}right)=frac{1}{sqrt{pi}}int_{0}^{1}x^{frac{n-3}{2}}(1-x)^{-1/2},dx=frac{2}{sqrt{pi}}int_{0}^{1}frac{x^{n-2}}{sqrt{1-x^2}},dx $$
      gives
      $$ frac{Gammaleft(tfrac{n-1}{2}right)}{Gammaleft(tfrac{n}{2}right)}= frac{2}{sqrt{pi}}int_{0}^{pi/2}left(costhetaright)^{n-2},dthetasimfrac{2}{sqrt{pi}}int_{0}^{+infty}expleft[-(n-2)frac{theta^2}{2}right]dtheta=sqrt{frac{2}{n-2}}. $$
      Notice that the integral representation (as a moment) instantly gives that the LHS is a log-convex function. The asymptotic equivalence $sim$ can be seen as an instance of Laplace/Hayman's method.






      share|cite|improve this answer


























        6












        6








        6






        A possible approach:



        $$ frac{Gammaleft(tfrac{n-1}{2}right)}{Gammaleft(tfrac{n}{2}right)}=tfrac{1}{sqrt{pi}},Bleft(tfrac{n-1}{2},tfrac{1}{2}right)=frac{1}{sqrt{pi}}int_{0}^{1}x^{frac{n-3}{2}}(1-x)^{-1/2},dx=frac{2}{sqrt{pi}}int_{0}^{1}frac{x^{n-2}}{sqrt{1-x^2}},dx $$
        gives
        $$ frac{Gammaleft(tfrac{n-1}{2}right)}{Gammaleft(tfrac{n}{2}right)}= frac{2}{sqrt{pi}}int_{0}^{pi/2}left(costhetaright)^{n-2},dthetasimfrac{2}{sqrt{pi}}int_{0}^{+infty}expleft[-(n-2)frac{theta^2}{2}right]dtheta=sqrt{frac{2}{n-2}}. $$
        Notice that the integral representation (as a moment) instantly gives that the LHS is a log-convex function. The asymptotic equivalence $sim$ can be seen as an instance of Laplace/Hayman's method.






        share|cite|improve this answer














        A possible approach:



        $$ frac{Gammaleft(tfrac{n-1}{2}right)}{Gammaleft(tfrac{n}{2}right)}=tfrac{1}{sqrt{pi}},Bleft(tfrac{n-1}{2},tfrac{1}{2}right)=frac{1}{sqrt{pi}}int_{0}^{1}x^{frac{n-3}{2}}(1-x)^{-1/2},dx=frac{2}{sqrt{pi}}int_{0}^{1}frac{x^{n-2}}{sqrt{1-x^2}},dx $$
        gives
        $$ frac{Gammaleft(tfrac{n-1}{2}right)}{Gammaleft(tfrac{n}{2}right)}= frac{2}{sqrt{pi}}int_{0}^{pi/2}left(costhetaright)^{n-2},dthetasimfrac{2}{sqrt{pi}}int_{0}^{+infty}expleft[-(n-2)frac{theta^2}{2}right]dtheta=sqrt{frac{2}{n-2}}. $$
        Notice that the integral representation (as a moment) instantly gives that the LHS is a log-convex function. The asymptotic equivalence $sim$ can be seen as an instance of Laplace/Hayman's method.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 21 '18 at 2:57

























        answered Nov 21 '18 at 2:27









        Jack D'Aurizio

        287k33280657




        287k33280657






























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