Find the upper bound of $|frac{d^2}{dx^2}(e^{-x^2})|leq6$ in $xin[0,1]$












-1












$begingroup$



Show by finding the second derivative of $e^{-x^2}$ that for all $xin[0,1]$



$$|frac{d^2}{dx^2}(e^{-x^2})|leq6$$



(if you obtain a better bound, that is fine )




My Try



let, $f(x)=e^{-x^2}$



$f''(x)=2e^{-x^2}(2x^2-1)$



From the plot in wolfram alpha I could see that the bound is 2/e or 0.73. But how to solve this analytically?






calculus upper-lower-bounds







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What does the single $|$ mean in the title and the text of the question?
    $endgroup$
    – Christian Blatter
    Dec 15 '18 at 9:57










  • $begingroup$
    Sorry It should be modulus sign. I'll fix it.
    $endgroup$
    – emil
    Dec 15 '18 at 18:34
















-1












$begingroup$



Show by finding the second derivative of $e^{-x^2}$ that for all $xin[0,1]$



$$|frac{d^2}{dx^2}(e^{-x^2})|leq6$$



(if you obtain a better bound, that is fine )




My Try



let, $f(x)=e^{-x^2}$



$f''(x)=2e^{-x^2}(2x^2-1)$



From the plot in wolfram alpha I could see that the bound is 2/e or 0.73. But how to solve this analytically?






calculus upper-lower-bounds







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    What does the single $|$ mean in the title and the text of the question?
    $endgroup$
    – Christian Blatter
    Dec 15 '18 at 9:57










  • $begingroup$
    Sorry It should be modulus sign. I'll fix it.
    $endgroup$
    – emil
    Dec 15 '18 at 18:34














-1












-1








-1





$begingroup$



Show by finding the second derivative of $e^{-x^2}$ that for all $xin[0,1]$



$$|frac{d^2}{dx^2}(e^{-x^2})|leq6$$



(if you obtain a better bound, that is fine )




My Try



let, $f(x)=e^{-x^2}$



$f''(x)=2e^{-x^2}(2x^2-1)$



From the plot in wolfram alpha I could see that the bound is 2/e or 0.73. But how to solve this analytically?






calculus upper-lower-bounds







share|cite|improve this question











$endgroup$





Show by finding the second derivative of $e^{-x^2}$ that for all $xin[0,1]$



$$|frac{d^2}{dx^2}(e^{-x^2})|leq6$$



(if you obtain a better bound, that is fine )




My Try



let, $f(x)=e^{-x^2}$



$f''(x)=2e^{-x^2}(2x^2-1)$



From the plot in wolfram alpha I could see that the bound is 2/e or 0.73. But how to solve this analytically?






calculus upper-lower-bounds




calculus upper-lower-bounds






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 15 '18 at 18:35







emil

















asked Dec 15 '18 at 9:36









emilemil

465410




465410








  • 1




    $begingroup$
    What does the single $|$ mean in the title and the text of the question?
    $endgroup$
    – Christian Blatter
    Dec 15 '18 at 9:57










  • $begingroup$
    Sorry It should be modulus sign. I'll fix it.
    $endgroup$
    – emil
    Dec 15 '18 at 18:34














  • 1




    $begingroup$
    What does the single $|$ mean in the title and the text of the question?
    $endgroup$
    – Christian Blatter
    Dec 15 '18 at 9:57










  • $begingroup$
    Sorry It should be modulus sign. I'll fix it.
    $endgroup$
    – emil
    Dec 15 '18 at 18:34








1




1




$begingroup$
What does the single $|$ mean in the title and the text of the question?
$endgroup$
– Christian Blatter
Dec 15 '18 at 9:57




$begingroup$
What does the single $|$ mean in the title and the text of the question?
$endgroup$
– Christian Blatter
Dec 15 '18 at 9:57












$begingroup$
Sorry It should be modulus sign. I'll fix it.
$endgroup$
– emil
Dec 15 '18 at 18:34




$begingroup$
Sorry It should be modulus sign. I'll fix it.
$endgroup$
– emil
Dec 15 '18 at 18:34










2 Answers
2






active

oldest

votes


















0












$begingroup$

We want to find $max{(exp(-x^2))''}$ so we evaluate its stationary points. This occurs when $$(exp(-x^2))'''=(2(2x^2-1)exp(-x^2))'=4x(3-2x^2)exp(-x^2)=0$$ so $x=0,sqrt{3/2}$. At the former, $(exp(-x^2))''=-2$ and at the latter, $(exp(-x^2))''=4exp(-3/2)<1$ which is a much tighter bound than $6$.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    Just use the triangle inequality. The given bound holds for all $xinmathbb{R}$, not just $xin[0,1]$.



    You know
    $$
    f''(x)=2cdot[2x^2exp(-x^2)-exp(-x^2)]
    $$
    So
    begin{align*}
    lvert f''(x)rvert
    &leq 2cdotbigg[sup_{xinmathbb{R}}2x^2exp(-x^2)+sup_{xinmathbb{R}}exp(-x^2)bigg]\
    &= 2cdotleft[2sup_{xinmathbb{R}}frac{x^2}{exp(x^2)}+1right]\
    &leq 2cdotleft(2sup_{xinmathbb{R}}frac{x^2}{1+x^2}+1right)\
    &leq 2cdot(2cdot 1+1)=6.
    end{align*}



    For $xin[0,1]$, you can obtain the bound a little more quickly using $x^2leq 1$ and $exp(-x^2)leq 1$.






    share|cite|improve this answer









    $endgroup$














      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%2f3040318%2ffind-the-upper-bound-of-fracd2dx2e-x2-leq6-in-x-in0-1%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









      0












      $begingroup$

      We want to find $max{(exp(-x^2))''}$ so we evaluate its stationary points. This occurs when $$(exp(-x^2))'''=(2(2x^2-1)exp(-x^2))'=4x(3-2x^2)exp(-x^2)=0$$ so $x=0,sqrt{3/2}$. At the former, $(exp(-x^2))''=-2$ and at the latter, $(exp(-x^2))''=4exp(-3/2)<1$ which is a much tighter bound than $6$.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        We want to find $max{(exp(-x^2))''}$ so we evaluate its stationary points. This occurs when $$(exp(-x^2))'''=(2(2x^2-1)exp(-x^2))'=4x(3-2x^2)exp(-x^2)=0$$ so $x=0,sqrt{3/2}$. At the former, $(exp(-x^2))''=-2$ and at the latter, $(exp(-x^2))''=4exp(-3/2)<1$ which is a much tighter bound than $6$.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          We want to find $max{(exp(-x^2))''}$ so we evaluate its stationary points. This occurs when $$(exp(-x^2))'''=(2(2x^2-1)exp(-x^2))'=4x(3-2x^2)exp(-x^2)=0$$ so $x=0,sqrt{3/2}$. At the former, $(exp(-x^2))''=-2$ and at the latter, $(exp(-x^2))''=4exp(-3/2)<1$ which is a much tighter bound than $6$.






          share|cite|improve this answer









          $endgroup$



          We want to find $max{(exp(-x^2))''}$ so we evaluate its stationary points. This occurs when $$(exp(-x^2))'''=(2(2x^2-1)exp(-x^2))'=4x(3-2x^2)exp(-x^2)=0$$ so $x=0,sqrt{3/2}$. At the former, $(exp(-x^2))''=-2$ and at the latter, $(exp(-x^2))''=4exp(-3/2)<1$ which is a much tighter bound than $6$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 15 '18 at 10:31









          TheSimpliFireTheSimpliFire

          13.2k62464




          13.2k62464























              0












              $begingroup$

              Just use the triangle inequality. The given bound holds for all $xinmathbb{R}$, not just $xin[0,1]$.



              You know
              $$
              f''(x)=2cdot[2x^2exp(-x^2)-exp(-x^2)]
              $$
              So
              begin{align*}
              lvert f''(x)rvert
              &leq 2cdotbigg[sup_{xinmathbb{R}}2x^2exp(-x^2)+sup_{xinmathbb{R}}exp(-x^2)bigg]\
              &= 2cdotleft[2sup_{xinmathbb{R}}frac{x^2}{exp(x^2)}+1right]\
              &leq 2cdotleft(2sup_{xinmathbb{R}}frac{x^2}{1+x^2}+1right)\
              &leq 2cdot(2cdot 1+1)=6.
              end{align*}



              For $xin[0,1]$, you can obtain the bound a little more quickly using $x^2leq 1$ and $exp(-x^2)leq 1$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                Just use the triangle inequality. The given bound holds for all $xinmathbb{R}$, not just $xin[0,1]$.



                You know
                $$
                f''(x)=2cdot[2x^2exp(-x^2)-exp(-x^2)]
                $$
                So
                begin{align*}
                lvert f''(x)rvert
                &leq 2cdotbigg[sup_{xinmathbb{R}}2x^2exp(-x^2)+sup_{xinmathbb{R}}exp(-x^2)bigg]\
                &= 2cdotleft[2sup_{xinmathbb{R}}frac{x^2}{exp(x^2)}+1right]\
                &leq 2cdotleft(2sup_{xinmathbb{R}}frac{x^2}{1+x^2}+1right)\
                &leq 2cdot(2cdot 1+1)=6.
                end{align*}



                For $xin[0,1]$, you can obtain the bound a little more quickly using $x^2leq 1$ and $exp(-x^2)leq 1$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  Just use the triangle inequality. The given bound holds for all $xinmathbb{R}$, not just $xin[0,1]$.



                  You know
                  $$
                  f''(x)=2cdot[2x^2exp(-x^2)-exp(-x^2)]
                  $$
                  So
                  begin{align*}
                  lvert f''(x)rvert
                  &leq 2cdotbigg[sup_{xinmathbb{R}}2x^2exp(-x^2)+sup_{xinmathbb{R}}exp(-x^2)bigg]\
                  &= 2cdotleft[2sup_{xinmathbb{R}}frac{x^2}{exp(x^2)}+1right]\
                  &leq 2cdotleft(2sup_{xinmathbb{R}}frac{x^2}{1+x^2}+1right)\
                  &leq 2cdot(2cdot 1+1)=6.
                  end{align*}



                  For $xin[0,1]$, you can obtain the bound a little more quickly using $x^2leq 1$ and $exp(-x^2)leq 1$.






                  share|cite|improve this answer









                  $endgroup$



                  Just use the triangle inequality. The given bound holds for all $xinmathbb{R}$, not just $xin[0,1]$.



                  You know
                  $$
                  f''(x)=2cdot[2x^2exp(-x^2)-exp(-x^2)]
                  $$
                  So
                  begin{align*}
                  lvert f''(x)rvert
                  &leq 2cdotbigg[sup_{xinmathbb{R}}2x^2exp(-x^2)+sup_{xinmathbb{R}}exp(-x^2)bigg]\
                  &= 2cdotleft[2sup_{xinmathbb{R}}frac{x^2}{exp(x^2)}+1right]\
                  &leq 2cdotleft(2sup_{xinmathbb{R}}frac{x^2}{1+x^2}+1right)\
                  &leq 2cdot(2cdot 1+1)=6.
                  end{align*}



                  For $xin[0,1]$, you can obtain the bound a little more quickly using $x^2leq 1$ and $exp(-x^2)leq 1$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 16 '18 at 8:30









                  user10354138user10354138

                  7,5372925




                  7,5372925






























                      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%2f3040318%2ffind-the-upper-bound-of-fracd2dx2e-x2-leq6-in-x-in0-1%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

                      mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924

                      Image
                      0 I just uploaded a new plugin to my wordpress site and I got this warning: mysqli_query(): Empty query in /home/lucindabrummitt/public_html/blog/wp-includes/wp-db.php on line 1924 Usually if I have a problem, I can find answers on google but this one seems pretty particular and I'm not finding any info on it except for this forum. I'm not terribly techie but if someone can tell me how to read this and how to change it I can follow directions pretty well. Thanks All! PS If I don't have the proper tags here could someone add the appropriate tags? php mysql wordpress share | improve this question asked Nov 16 '18 at 20:41 ...
                      Read more

                      How to change which sound is reproduced for terminal bell?

                      Image
                      3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question ...
                      Read more

                      Can I use Tabulator js library in my java Spring + Thymeleaf project?

                      Image
                      0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"...
                      Read more