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







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  • 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

438410




438410








  • 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






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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









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

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      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

          12.7k62461




          12.7k62461























              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

                  7,4322925




                  7,4322925






























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