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

2,27231028




2,27231028










asked Jan 8 at 12:03









MinaThumaMinaThuma

457




457












  • $begingroup$
    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

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









$endgroup$





















    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











    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3066090%2fshow-that-lambdade-d-frac-pi-fracd2-gamma-fracd21-a%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    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

        16.6k43391




        16.6k43391























            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

            3,15217




            3,15217












            • $begingroup$
              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


















            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3066090%2fshow-that-lambdade-d-frac-pi-fracd2-gamma-fracd21-a%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            1300-talet

            Image
            Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
            Read more

            1300-talet

            Image
            Den här artikeln handlar om hundraårsperioden 1300–1399. För decenniet 1300–1309, se 1300-talet (decennium). .mw-parser-output .infobox{border:1px solid #aaa;background-color:#f9f9f9;color:black;margin:.5em 0 .5em 1em;padding:.2em;float:right;clear:right;width:22em;text-align:left;font-size:88%;line-height:1.6em}.mw-parser-output .infobox td,.mw-parser-output .infobox th{vertical-align:top;padding:0 .2em}.mw-parser-output .infobox caption{font-size:larger}.mw-parser-output .infobox.bordered{border-collapse:collapse}.mw-parser-output .infobox.bordered td,.mw-parser-output .infobox.bordered th{border:1px solid #aaa}.mw-parser-output .infobox.bordered .borderless td,.mw-parser-output .infobox.bordered .borderless th{border:0}.mw-parser-output .infobox-showbutton .mw-collapsible-text{color:inherit}.mw-parser-output .infobox.bordered .mergedtoprow td,.mw-parser-output .infobox.bordered .mergedtoprow th{border:0;border-top:1px solid #aaa;border-right:1px solid #aaa}.mw-parser-output ...
            Read more

            Display a custom attribute below product name in the front-end Magento 1.9.3.8

            Image
            0 I need to display a custom attribute bellow the product name and I don't know how can I call this attribute. I am editing the file /app/design/frontend/templatemela/MAG100219_1/template/catalog/product/view.phtml But how can I call this attribute in php? I tried calling a string echo with the name I put in the attribute, but it didn't work. Code is as below: <div class="page-title product-title"> <h1><?php echo $_helper->productAttribute($_product, $_product->getName(), 'name') ?></h1> </div> <div class="codigoinmetro"> <h2><?php echo $_helper->productAttribute($_product, $_product->getName(), 'registroinmetro') ?></h2> </div> php attrib...
            Read more