Is $h:xto expleft[-frac{x^2}{2}right]left(expleft[frac{x^4}{24n}right]-1right)$ increasing?












3












$begingroup$


I struggling to show this function ins increasing on $mathbb{R}^+$



$$h:xto expleft[-dfrac{x^2}{2}right]left(expleft[dfrac{x^4}{24n}right]-1right) qquad nin mathbb{N}^*$$



With the derivative



I've found



$$h'(x)=xe^{-frac{x^2}{2}}left[left(dfrac{x^2}{6n}-1right)e^{frac{x^4}{24n}}+1right]$$



$h'(x)ge 0iff left(dfrac{x^2}{6n}-1right)e^{frac{x^4}{24n}}+1ge 0$



and now I'm blocked...






real-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is the parameter $n$?
    $endgroup$
    – gimusi
    Dec 2 '18 at 12:37










  • $begingroup$
    $$nin mathbb{N}^*$$
    $endgroup$
    – Stu
    Dec 2 '18 at 12:44
















3












$begingroup$


I struggling to show this function ins increasing on $mathbb{R}^+$



$$h:xto expleft[-dfrac{x^2}{2}right]left(expleft[dfrac{x^4}{24n}right]-1right) qquad nin mathbb{N}^*$$



With the derivative



I've found



$$h'(x)=xe^{-frac{x^2}{2}}left[left(dfrac{x^2}{6n}-1right)e^{frac{x^4}{24n}}+1right]$$



$h'(x)ge 0iff left(dfrac{x^2}{6n}-1right)e^{frac{x^4}{24n}}+1ge 0$



and now I'm blocked...






real-analysis







share|cite|improve this question











$endgroup$












  • $begingroup$
    What is the parameter $n$?
    $endgroup$
    – gimusi
    Dec 2 '18 at 12:37










  • $begingroup$
    $$nin mathbb{N}^*$$
    $endgroup$
    – Stu
    Dec 2 '18 at 12:44














3












3








3





$begingroup$


I struggling to show this function ins increasing on $mathbb{R}^+$



$$h:xto expleft[-dfrac{x^2}{2}right]left(expleft[dfrac{x^4}{24n}right]-1right) qquad nin mathbb{N}^*$$



With the derivative



I've found



$$h'(x)=xe^{-frac{x^2}{2}}left[left(dfrac{x^2}{6n}-1right)e^{frac{x^4}{24n}}+1right]$$



$h'(x)ge 0iff left(dfrac{x^2}{6n}-1right)e^{frac{x^4}{24n}}+1ge 0$



and now I'm blocked...






real-analysis







share|cite|improve this question











$endgroup$




I struggling to show this function ins increasing on $mathbb{R}^+$



$$h:xto expleft[-dfrac{x^2}{2}right]left(expleft[dfrac{x^4}{24n}right]-1right) qquad nin mathbb{N}^*$$



With the derivative



I've found



$$h'(x)=xe^{-frac{x^2}{2}}left[left(dfrac{x^2}{6n}-1right)e^{frac{x^4}{24n}}+1right]$$



$h'(x)ge 0iff left(dfrac{x^2}{6n}-1right)e^{frac{x^4}{24n}}+1ge 0$



and now I'm blocked...






real-analysis




real-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 12:43







Stu

















asked Dec 2 '18 at 12:32









StuStu

1,1761413




1,1761413












  • $begingroup$
    What is the parameter $n$?
    $endgroup$
    – gimusi
    Dec 2 '18 at 12:37










  • $begingroup$
    $$nin mathbb{N}^*$$
    $endgroup$
    – Stu
    Dec 2 '18 at 12:44


















  • $begingroup$
    What is the parameter $n$?
    $endgroup$
    – gimusi
    Dec 2 '18 at 12:37










  • $begingroup$
    $$nin mathbb{N}^*$$
    $endgroup$
    – Stu
    Dec 2 '18 at 12:44
















$begingroup$
What is the parameter $n$?
$endgroup$
– gimusi
Dec 2 '18 at 12:37




$begingroup$
What is the parameter $n$?
$endgroup$
– gimusi
Dec 2 '18 at 12:37












$begingroup$
$$nin mathbb{N}^*$$
$endgroup$
– Stu
Dec 2 '18 at 12:44




$begingroup$
$$nin mathbb{N}^*$$
$endgroup$
– Stu
Dec 2 '18 at 12:44










1 Answer
1






active

oldest

votes


















1












$begingroup$

We have that



$$h'(x)=-xe^{-frac{x^2}{2}}left(e^{frac{x^4}{24n}}-1right)+frac{x^3}{6n}e^{-frac{x^2}{2}}e^{frac{x^4}{24n}}
=-xe^{-frac{x^2}{2}}left(left(1-frac{x^2}{6n}right)e^{frac{x^4}{24n}}-1right)$$



then we need to prove that



$$left(frac{x^2}{6n}-1right)e^{frac{x^4}{24n}}+1 ge 0 $$



and for $frac{x^2}{6n}-1<0 iff x<sqrt{6n}$ it is equivalent to



$$left(1-frac{x^2}{6n}right)e^{frac{x^4}{24n}}-1le 0 $$



and by $y=frac{x^2}{6n}in (0,1)$



$$(1-y)e^{frac32ny^2}-1le 0 $$



which doesn't hold for $n=2$ indeed



$$f(y)=(1-y)e^{3y^2}-1 implies f'(y)=e^{3y^2}(-6y^2+6y-1)=0 implies y_{1,2}=frac12pmfrac{sqrt 3}6$$



and $f(y_1)>0$.






share|cite|improve this answer











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    1 Answer
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    1 Answer
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    active

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    1












    $begingroup$

    We have that



    $$h'(x)=-xe^{-frac{x^2}{2}}left(e^{frac{x^4}{24n}}-1right)+frac{x^3}{6n}e^{-frac{x^2}{2}}e^{frac{x^4}{24n}}
    =-xe^{-frac{x^2}{2}}left(left(1-frac{x^2}{6n}right)e^{frac{x^4}{24n}}-1right)$$



    then we need to prove that



    $$left(frac{x^2}{6n}-1right)e^{frac{x^4}{24n}}+1 ge 0 $$



    and for $frac{x^2}{6n}-1<0 iff x<sqrt{6n}$ it is equivalent to



    $$left(1-frac{x^2}{6n}right)e^{frac{x^4}{24n}}-1le 0 $$



    and by $y=frac{x^2}{6n}in (0,1)$



    $$(1-y)e^{frac32ny^2}-1le 0 $$



    which doesn't hold for $n=2$ indeed



    $$f(y)=(1-y)e^{3y^2}-1 implies f'(y)=e^{3y^2}(-6y^2+6y-1)=0 implies y_{1,2}=frac12pmfrac{sqrt 3}6$$



    and $f(y_1)>0$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      We have that



      $$h'(x)=-xe^{-frac{x^2}{2}}left(e^{frac{x^4}{24n}}-1right)+frac{x^3}{6n}e^{-frac{x^2}{2}}e^{frac{x^4}{24n}}
      =-xe^{-frac{x^2}{2}}left(left(1-frac{x^2}{6n}right)e^{frac{x^4}{24n}}-1right)$$



      then we need to prove that



      $$left(frac{x^2}{6n}-1right)e^{frac{x^4}{24n}}+1 ge 0 $$



      and for $frac{x^2}{6n}-1<0 iff x<sqrt{6n}$ it is equivalent to



      $$left(1-frac{x^2}{6n}right)e^{frac{x^4}{24n}}-1le 0 $$



      and by $y=frac{x^2}{6n}in (0,1)$



      $$(1-y)e^{frac32ny^2}-1le 0 $$



      which doesn't hold for $n=2$ indeed



      $$f(y)=(1-y)e^{3y^2}-1 implies f'(y)=e^{3y^2}(-6y^2+6y-1)=0 implies y_{1,2}=frac12pmfrac{sqrt 3}6$$



      and $f(y_1)>0$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        We have that



        $$h'(x)=-xe^{-frac{x^2}{2}}left(e^{frac{x^4}{24n}}-1right)+frac{x^3}{6n}e^{-frac{x^2}{2}}e^{frac{x^4}{24n}}
        =-xe^{-frac{x^2}{2}}left(left(1-frac{x^2}{6n}right)e^{frac{x^4}{24n}}-1right)$$



        then we need to prove that



        $$left(frac{x^2}{6n}-1right)e^{frac{x^4}{24n}}+1 ge 0 $$



        and for $frac{x^2}{6n}-1<0 iff x<sqrt{6n}$ it is equivalent to



        $$left(1-frac{x^2}{6n}right)e^{frac{x^4}{24n}}-1le 0 $$



        and by $y=frac{x^2}{6n}in (0,1)$



        $$(1-y)e^{frac32ny^2}-1le 0 $$



        which doesn't hold for $n=2$ indeed



        $$f(y)=(1-y)e^{3y^2}-1 implies f'(y)=e^{3y^2}(-6y^2+6y-1)=0 implies y_{1,2}=frac12pmfrac{sqrt 3}6$$



        and $f(y_1)>0$.






        share|cite|improve this answer











        $endgroup$



        We have that



        $$h'(x)=-xe^{-frac{x^2}{2}}left(e^{frac{x^4}{24n}}-1right)+frac{x^3}{6n}e^{-frac{x^2}{2}}e^{frac{x^4}{24n}}
        =-xe^{-frac{x^2}{2}}left(left(1-frac{x^2}{6n}right)e^{frac{x^4}{24n}}-1right)$$



        then we need to prove that



        $$left(frac{x^2}{6n}-1right)e^{frac{x^4}{24n}}+1 ge 0 $$



        and for $frac{x^2}{6n}-1<0 iff x<sqrt{6n}$ it is equivalent to



        $$left(1-frac{x^2}{6n}right)e^{frac{x^4}{24n}}-1le 0 $$



        and by $y=frac{x^2}{6n}in (0,1)$



        $$(1-y)e^{frac32ny^2}-1le 0 $$



        which doesn't hold for $n=2$ indeed



        $$f(y)=(1-y)e^{3y^2}-1 implies f'(y)=e^{3y^2}(-6y^2+6y-1)=0 implies y_{1,2}=frac12pmfrac{sqrt 3}6$$



        and $f(y_1)>0$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 2 '18 at 13:26

























        answered Dec 2 '18 at 13:16









        gimusigimusi

        92.9k94494




        92.9k94494






























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