Show that $lambda^{d}(E_{d})=frac{pi^{frac{d}{2}}}{Gamma(frac{d}{2}+1)}$ and determine behavior for $d to...












2












$begingroup$


Let $d in mathbb N$ and $E_{d}:={x in mathbb R^{d}:|x|leq 1}$



Prove that $$ lambda^{d}(E_{d})=frac{pi^{frac{d}{2}}}{Gamma(frac{d}{2}+1)} $$



and determine $lambda^{d}(E_{d})$ as $d to infty$



I struggle with d-dimensional volume, so I will try the behavior for $d to infty$



Note:



$$frac{pi^{frac{d}{2}}}{Gamma(frac{d}{2}+1)}=frac{pi^{frac{d}{2}}}{int_{0}^{infty}x^{frac{d}{2}}e^{-x}dx}$$



Looking particularly at:



$int_{0}^{infty}x^{frac{d}{2}}e^{-x}dx$ it looks like partial integration, but I wouldn't as $d in mathbb N$. I would use substitution, namely $y = x^{frac{d}{2}}Rightarrow frac{2}{d}dy=x^{frac{d}{2}-1}dx$. But this is a dead end, as $x$ does not disappear



Any ideas?






real-analysis integration measure-theory lebesgue-integral







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are you looking for the derivation of the volume or the limiting behavior?
    $endgroup$
    – user1337
    Jan 8 at 12:12










  • $begingroup$
    Rather the deviation as $d$ gets larger
    $endgroup$
    – MinaThuma
    Jan 8 at 12:21












  • $begingroup$
    I think you are looking for volume of n-ball.
    $endgroup$
    – StubbornAtom
    Jan 8 at 12:59










  • $begingroup$
    Also see math.stackexchange.com/q/67039/321264.
    $endgroup$
    – StubbornAtom
    Jan 8 at 13:05
















2












$begingroup$


Let $d in mathbb N$ and $E_{d}:={x in mathbb R^{d}:|x|leq 1}$



Prove that $$ lambda^{d}(E_{d})=frac{pi^{frac{d}{2}}}{Gamma(frac{d}{2}+1)} $$



and determine $lambda^{d}(E_{d})$ as $d to infty$



I struggle with d-dimensional volume, so I will try the behavior for $d to infty$



Note:



$$frac{pi^{frac{d}{2}}}{Gamma(frac{d}{2}+1)}=frac{pi^{frac{d}{2}}}{int_{0}^{infty}x^{frac{d}{2}}e^{-x}dx}$$



Looking particularly at:



$int_{0}^{infty}x^{frac{d}{2}}e^{-x}dx$ it looks like partial integration, but I wouldn't as $d in mathbb N$. I would use substitution, namely $y = x^{frac{d}{2}}Rightarrow frac{2}{d}dy=x^{frac{d}{2}-1}dx$. But this is a dead end, as $x$ does not disappear



Any ideas?






real-analysis integration measure-theory lebesgue-integral







share|cite|improve this question











$endgroup$












  • $begingroup$
    Are you looking for the derivation of the volume or the limiting behavior?
    $endgroup$
    – user1337
    Jan 8 at 12:12










  • $begingroup$
    Rather the deviation as $d$ gets larger
    $endgroup$
    – MinaThuma
    Jan 8 at 12:21












  • $begingroup$
    I think you are looking for volume of n-ball.
    $endgroup$
    – StubbornAtom
    Jan 8 at 12:59










  • $begingroup$
    Also see math.stackexchange.com/q/67039/321264.
    $endgroup$
    – StubbornAtom
    Jan 8 at 13:05














2












2








2





$begingroup$


Let $d in mathbb N$ and $E_{d}:={x in mathbb R^{d}:|x|leq 1}$



Prove that $$ lambda^{d}(E_{d})=frac{pi^{frac{d}{2}}}{Gamma(frac{d}{2}+1)} $$



and determine $lambda^{d}(E_{d})$ as $d to infty$



I struggle with d-dimensional volume, so I will try the behavior for $d to infty$



Note:



$$frac{pi^{frac{d}{2}}}{Gamma(frac{d}{2}+1)}=frac{pi^{frac{d}{2}}}{int_{0}^{infty}x^{frac{d}{2}}e^{-x}dx}$$



Looking particularly at:



$int_{0}^{infty}x^{frac{d}{2}}e^{-x}dx$ it looks like partial integration, but I wouldn't as $d in mathbb N$. I would use substitution, namely $y = x^{frac{d}{2}}Rightarrow frac{2}{d}dy=x^{frac{d}{2}-1}dx$. But this is a dead end, as $x$ does not disappear



Any ideas?






real-analysis integration measure-theory lebesgue-integral







share|cite|improve this question











$endgroup$




Let $d in mathbb N$ and $E_{d}:={x in mathbb R^{d}:|x|leq 1}$



Prove that $$ lambda^{d}(E_{d})=frac{pi^{frac{d}{2}}}{Gamma(frac{d}{2}+1)} $$



and determine $lambda^{d}(E_{d})$ as $d to infty$



I struggle with d-dimensional volume, so I will try the behavior for $d to infty$



Note:



$$frac{pi^{frac{d}{2}}}{Gamma(frac{d}{2}+1)}=frac{pi^{frac{d}{2}}}{int_{0}^{infty}x^{frac{d}{2}}e^{-x}dx}$$



Looking particularly at:



$int_{0}^{infty}x^{frac{d}{2}}e^{-x}dx$ it looks like partial integration, but I wouldn't as $d in mathbb N$. I would use substitution, namely $y = x^{frac{d}{2}}Rightarrow frac{2}{d}dy=x^{frac{d}{2}-1}dx$. But this is a dead end, as $x$ does not disappear



Any ideas?






real-analysis integration measure-theory lebesgue-integral




real-analysis integration measure-theory lebesgue-integral






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 8 at 12:22









Larry

2,54531131




2,54531131










asked Jan 8 at 12:03









MinaThumaMinaThuma

2709




2709












  • $begingroup$
    Are you looking for the derivation of the volume or the limiting behavior?
    $endgroup$
    – user1337
    Jan 8 at 12:12










  • $begingroup$
    Rather the deviation as $d$ gets larger
    $endgroup$
    – MinaThuma
    Jan 8 at 12:21












  • $begingroup$
    I think you are looking for volume of n-ball.
    $endgroup$
    – StubbornAtom
    Jan 8 at 12:59










  • $begingroup$
    Also see math.stackexchange.com/q/67039/321264.
    $endgroup$
    – StubbornAtom
    Jan 8 at 13:05


















  • $begingroup$
    Are you looking for the derivation of the volume or the limiting behavior?
    $endgroup$
    – user1337
    Jan 8 at 12:12










  • $begingroup$
    Rather the deviation as $d$ gets larger
    $endgroup$
    – MinaThuma
    Jan 8 at 12:21












  • $begingroup$
    I think you are looking for volume of n-ball.
    $endgroup$
    – StubbornAtom
    Jan 8 at 12:59










  • $begingroup$
    Also see math.stackexchange.com/q/67039/321264.
    $endgroup$
    – StubbornAtom
    Jan 8 at 13:05
















$begingroup$
Are you looking for the derivation of the volume or the limiting behavior?
$endgroup$
– user1337
Jan 8 at 12:12




$begingroup$
Are you looking for the derivation of the volume or the limiting behavior?
$endgroup$
– user1337
Jan 8 at 12:12












$begingroup$
Rather the deviation as $d$ gets larger
$endgroup$
– MinaThuma
Jan 8 at 12:21






$begingroup$
Rather the deviation as $d$ gets larger
$endgroup$
– MinaThuma
Jan 8 at 12:21














$begingroup$
I think you are looking for volume of n-ball.
$endgroup$
– StubbornAtom
Jan 8 at 12:59




$begingroup$
I think you are looking for volume of n-ball.
$endgroup$
– StubbornAtom
Jan 8 at 12:59












$begingroup$
Also see math.stackexchange.com/q/67039/321264.
$endgroup$
– StubbornAtom
Jan 8 at 13:05




$begingroup$
Also see math.stackexchange.com/q/67039/321264.
$endgroup$
– StubbornAtom
Jan 8 at 13:05










2 Answers
2






active

oldest

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2












$begingroup$

Let $t=d/2$. Observe that
$$0 leq frac{pi^t}{Gamma(t+1)} leq frac{pi^{lfloor t rfloor+1}}{Gamma(lfloor t rfloor+1)} $$
and recall that the series $$sum_{n=0}^infty frac{pi^n}{n!} $$ converges (to $e^pi$). Convergent series have their terms approaching zero, and the squeeze law shows that the same is true for $lambda^d (E_d)$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Hint for the derivation: compute $int_{mathbb{R}^d}{e^{-|x|^2}}$ by separating integrals and by polar coordinates.



    For the asymptotic behavior: use the functional equation of $Gamma$ to have a non-integral formula for $d=2p$ and $d=2p-1$. Then let $p$ go to infinity in both formulas and use Stirling’s theorem.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Why would I use the integral $int_{mathbb R^{d}}e^{-|x|^2}$ ?
      $endgroup$
      – MinaThuma
      Jan 8 at 16:42










    • $begingroup$
      It is the shortest way I know to derive the “volume” of the unit sphere thus the volume of the unit ball.
      $endgroup$
      – Mindlack
      Jan 8 at 17:13












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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

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    active

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    active

    oldest

    votes









    2












    $begingroup$

    Let $t=d/2$. Observe that
    $$0 leq frac{pi^t}{Gamma(t+1)} leq frac{pi^{lfloor t rfloor+1}}{Gamma(lfloor t rfloor+1)} $$
    and recall that the series $$sum_{n=0}^infty frac{pi^n}{n!} $$ converges (to $e^pi$). Convergent series have their terms approaching zero, and the squeeze law shows that the same is true for $lambda^d (E_d)$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Let $t=d/2$. Observe that
      $$0 leq frac{pi^t}{Gamma(t+1)} leq frac{pi^{lfloor t rfloor+1}}{Gamma(lfloor t rfloor+1)} $$
      and recall that the series $$sum_{n=0}^infty frac{pi^n}{n!} $$ converges (to $e^pi$). Convergent series have their terms approaching zero, and the squeeze law shows that the same is true for $lambda^d (E_d)$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Let $t=d/2$. Observe that
        $$0 leq frac{pi^t}{Gamma(t+1)} leq frac{pi^{lfloor t rfloor+1}}{Gamma(lfloor t rfloor+1)} $$
        and recall that the series $$sum_{n=0}^infty frac{pi^n}{n!} $$ converges (to $e^pi$). Convergent series have their terms approaching zero, and the squeeze law shows that the same is true for $lambda^d (E_d)$.






        share|cite|improve this answer









        $endgroup$



        Let $t=d/2$. Observe that
        $$0 leq frac{pi^t}{Gamma(t+1)} leq frac{pi^{lfloor t rfloor+1}}{Gamma(lfloor t rfloor+1)} $$
        and recall that the series $$sum_{n=0}^infty frac{pi^n}{n!} $$ converges (to $e^pi$). Convergent series have their terms approaching zero, and the squeeze law shows that the same is true for $lambda^d (E_d)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 8 at 12:30









        user1337user1337

        16.9k43594




        16.9k43594























            1












            $begingroup$

            Hint for the derivation: compute $int_{mathbb{R}^d}{e^{-|x|^2}}$ by separating integrals and by polar coordinates.



            For the asymptotic behavior: use the functional equation of $Gamma$ to have a non-integral formula for $d=2p$ and $d=2p-1$. Then let $p$ go to infinity in both formulas and use Stirling’s theorem.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Why would I use the integral $int_{mathbb R^{d}}e^{-|x|^2}$ ?
              $endgroup$
              – MinaThuma
              Jan 8 at 16:42










            • $begingroup$
              It is the shortest way I know to derive the “volume” of the unit sphere thus the volume of the unit ball.
              $endgroup$
              – Mindlack
              Jan 8 at 17:13
















            1












            $begingroup$

            Hint for the derivation: compute $int_{mathbb{R}^d}{e^{-|x|^2}}$ by separating integrals and by polar coordinates.



            For the asymptotic behavior: use the functional equation of $Gamma$ to have a non-integral formula for $d=2p$ and $d=2p-1$. Then let $p$ go to infinity in both formulas and use Stirling’s theorem.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Why would I use the integral $int_{mathbb R^{d}}e^{-|x|^2}$ ?
              $endgroup$
              – MinaThuma
              Jan 8 at 16:42










            • $begingroup$
              It is the shortest way I know to derive the “volume” of the unit sphere thus the volume of the unit ball.
              $endgroup$
              – Mindlack
              Jan 8 at 17:13














            1












            1








            1





            $begingroup$

            Hint for the derivation: compute $int_{mathbb{R}^d}{e^{-|x|^2}}$ by separating integrals and by polar coordinates.



            For the asymptotic behavior: use the functional equation of $Gamma$ to have a non-integral formula for $d=2p$ and $d=2p-1$. Then let $p$ go to infinity in both formulas and use Stirling’s theorem.






            share|cite|improve this answer











            $endgroup$



            Hint for the derivation: compute $int_{mathbb{R}^d}{e^{-|x|^2}}$ by separating integrals and by polar coordinates.



            For the asymptotic behavior: use the functional equation of $Gamma$ to have a non-integral formula for $d=2p$ and $d=2p-1$. Then let $p$ go to infinity in both formulas and use Stirling’s theorem.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 8 at 12:34

























            answered Jan 8 at 12:23









            MindlackMindlack

            4,910211




            4,910211












            • $begingroup$
              Why would I use the integral $int_{mathbb R^{d}}e^{-|x|^2}$ ?
              $endgroup$
              – MinaThuma
              Jan 8 at 16:42










            • $begingroup$
              It is the shortest way I know to derive the “volume” of the unit sphere thus the volume of the unit ball.
              $endgroup$
              – Mindlack
              Jan 8 at 17:13


















            • $begingroup$
              Why would I use the integral $int_{mathbb R^{d}}e^{-|x|^2}$ ?
              $endgroup$
              – MinaThuma
              Jan 8 at 16:42










            • $begingroup$
              It is the shortest way I know to derive the “volume” of the unit sphere thus the volume of the unit ball.
              $endgroup$
              – Mindlack
              Jan 8 at 17:13
















            $begingroup$
            Why would I use the integral $int_{mathbb R^{d}}e^{-|x|^2}$ ?
            $endgroup$
            – MinaThuma
            Jan 8 at 16:42




            $begingroup$
            Why would I use the integral $int_{mathbb R^{d}}e^{-|x|^2}$ ?
            $endgroup$
            – MinaThuma
            Jan 8 at 16:42












            $begingroup$
            It is the shortest way I know to derive the “volume” of the unit sphere thus the volume of the unit ball.
            $endgroup$
            – Mindlack
            Jan 8 at 17:13




            $begingroup$
            It is the shortest way I know to derive the “volume” of the unit sphere thus the volume of the unit ball.
            $endgroup$
            – Mindlack
            Jan 8 at 17:13


















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