How can we solve for $w$ : $2we^{w^2}-2e^w=0$












0












$begingroup$


This was the first dervative obtained with respect to $w$ from equation: $e^x+e^y+e^{w^2}-2e^w-x-y$.



We had to find the maxima and minima for which the first derivative is equal to zero and got this above equation $2we^{w^2}-2e^w=0$ which i couldnot solve to get the value of $w$.






ordinary-differential-equations







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  • 3




    $begingroup$
    The obvious solution is $w = 1$. With some CAS you see that there is no other solution.
    $endgroup$
    – Alpha001
    Dec 24 '18 at 16:10






  • 1




    $begingroup$
    @Alpha001 What is 'CAS'?
    $endgroup$
    – shwetha
    Dec 24 '18 at 16:18










  • $begingroup$
    @shwetha CAS = Computer Algebra System. For example WolframAlpha.
    $endgroup$
    – mrtaurho
    Dec 24 '18 at 16:20










  • $begingroup$
    @mrtauro Thank you!
    $endgroup$
    – shwetha
    Dec 24 '18 at 16:24
















0












$begingroup$


This was the first dervative obtained with respect to $w$ from equation: $e^x+e^y+e^{w^2}-2e^w-x-y$.



We had to find the maxima and minima for which the first derivative is equal to zero and got this above equation $2we^{w^2}-2e^w=0$ which i couldnot solve to get the value of $w$.






ordinary-differential-equations







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    The obvious solution is $w = 1$. With some CAS you see that there is no other solution.
    $endgroup$
    – Alpha001
    Dec 24 '18 at 16:10






  • 1




    $begingroup$
    @Alpha001 What is 'CAS'?
    $endgroup$
    – shwetha
    Dec 24 '18 at 16:18










  • $begingroup$
    @shwetha CAS = Computer Algebra System. For example WolframAlpha.
    $endgroup$
    – mrtaurho
    Dec 24 '18 at 16:20










  • $begingroup$
    @mrtauro Thank you!
    $endgroup$
    – shwetha
    Dec 24 '18 at 16:24














0












0








0


1



$begingroup$


This was the first dervative obtained with respect to $w$ from equation: $e^x+e^y+e^{w^2}-2e^w-x-y$.



We had to find the maxima and minima for which the first derivative is equal to zero and got this above equation $2we^{w^2}-2e^w=0$ which i couldnot solve to get the value of $w$.






ordinary-differential-equations







share|cite|improve this question











$endgroup$




This was the first dervative obtained with respect to $w$ from equation: $e^x+e^y+e^{w^2}-2e^w-x-y$.



We had to find the maxima and minima for which the first derivative is equal to zero and got this above equation $2we^{w^2}-2e^w=0$ which i couldnot solve to get the value of $w$.






ordinary-differential-equations




ordinary-differential-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 24 '18 at 16:06









mrtaurho

5,75551540




5,75551540










asked Dec 24 '18 at 16:05









Pritee SubediPritee Subedi

31




31








  • 3




    $begingroup$
    The obvious solution is $w = 1$. With some CAS you see that there is no other solution.
    $endgroup$
    – Alpha001
    Dec 24 '18 at 16:10






  • 1




    $begingroup$
    @Alpha001 What is 'CAS'?
    $endgroup$
    – shwetha
    Dec 24 '18 at 16:18










  • $begingroup$
    @shwetha CAS = Computer Algebra System. For example WolframAlpha.
    $endgroup$
    – mrtaurho
    Dec 24 '18 at 16:20










  • $begingroup$
    @mrtauro Thank you!
    $endgroup$
    – shwetha
    Dec 24 '18 at 16:24














  • 3




    $begingroup$
    The obvious solution is $w = 1$. With some CAS you see that there is no other solution.
    $endgroup$
    – Alpha001
    Dec 24 '18 at 16:10






  • 1




    $begingroup$
    @Alpha001 What is 'CAS'?
    $endgroup$
    – shwetha
    Dec 24 '18 at 16:18










  • $begingroup$
    @shwetha CAS = Computer Algebra System. For example WolframAlpha.
    $endgroup$
    – mrtaurho
    Dec 24 '18 at 16:20










  • $begingroup$
    @mrtauro Thank you!
    $endgroup$
    – shwetha
    Dec 24 '18 at 16:24








3




3




$begingroup$
The obvious solution is $w = 1$. With some CAS you see that there is no other solution.
$endgroup$
– Alpha001
Dec 24 '18 at 16:10




$begingroup$
The obvious solution is $w = 1$. With some CAS you see that there is no other solution.
$endgroup$
– Alpha001
Dec 24 '18 at 16:10




1




1




$begingroup$
@Alpha001 What is 'CAS'?
$endgroup$
– shwetha
Dec 24 '18 at 16:18




$begingroup$
@Alpha001 What is 'CAS'?
$endgroup$
– shwetha
Dec 24 '18 at 16:18












$begingroup$
@shwetha CAS = Computer Algebra System. For example WolframAlpha.
$endgroup$
– mrtaurho
Dec 24 '18 at 16:20




$begingroup$
@shwetha CAS = Computer Algebra System. For example WolframAlpha.
$endgroup$
– mrtaurho
Dec 24 '18 at 16:20












$begingroup$
@mrtauro Thank you!
$endgroup$
– shwetha
Dec 24 '18 at 16:24




$begingroup$
@mrtauro Thank you!
$endgroup$
– shwetha
Dec 24 '18 at 16:24










1 Answer
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4












$begingroup$

Maybe it is even possible to show that this equation only has got one real root without invoking a CAS. Therefore lets rewrite the equation in the following way



$$begin{align}
2we^{w^2}-2e^w&=0\
we^{w^2}&=e^w\
we^{w^2-w}&=1\
we^{w(1-w)}&=1
end{align}$$



From the last line we can directly conclude that $w=1$ satisfies the equation. Howsoever assuming only positive $w$ we may take the logarithm of both sides to further get



$$begin{align}
lnleft(we^{w(1-w)}right)&=ln(1)\
ln(w)+w(w-1)&=0\
-w^2+w&=log(w)\
-left(w-frac12right)^2+frac14&=log(w)
end{align}$$



The LHS describes as parabola with vertex at $(1/2,1/4)$ and roots at $w_1=0$ and $w_2=1$. Due the minus sign infront we can further conclude that the vertex is a maximum. Basically we are done now. The logarithm intersects with the $w$-axis at $w=1$ and tends to negative infinity as $w$ goes to $0$. But hence the parabola has got a root at $w=0$ they cannot intersect on the left hand side of $w=1$ whereas on the right hand side of $w=1$ the logarithm is strictly increasing and the parabola stricly decreasing.

So the only possible real root is $w=1$ due the above argumentation with regard to the graphs of both functions.






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    1 Answer
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    1 Answer
    1






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    oldest

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    active

    oldest

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    4












    $begingroup$

    Maybe it is even possible to show that this equation only has got one real root without invoking a CAS. Therefore lets rewrite the equation in the following way



    $$begin{align}
    2we^{w^2}-2e^w&=0\
    we^{w^2}&=e^w\
    we^{w^2-w}&=1\
    we^{w(1-w)}&=1
    end{align}$$



    From the last line we can directly conclude that $w=1$ satisfies the equation. Howsoever assuming only positive $w$ we may take the logarithm of both sides to further get



    $$begin{align}
    lnleft(we^{w(1-w)}right)&=ln(1)\
    ln(w)+w(w-1)&=0\
    -w^2+w&=log(w)\
    -left(w-frac12right)^2+frac14&=log(w)
    end{align}$$



    The LHS describes as parabola with vertex at $(1/2,1/4)$ and roots at $w_1=0$ and $w_2=1$. Due the minus sign infront we can further conclude that the vertex is a maximum. Basically we are done now. The logarithm intersects with the $w$-axis at $w=1$ and tends to negative infinity as $w$ goes to $0$. But hence the parabola has got a root at $w=0$ they cannot intersect on the left hand side of $w=1$ whereas on the right hand side of $w=1$ the logarithm is strictly increasing and the parabola stricly decreasing.

    So the only possible real root is $w=1$ due the above argumentation with regard to the graphs of both functions.






    share|cite|improve this answer









    $endgroup$


















      4












      $begingroup$

      Maybe it is even possible to show that this equation only has got one real root without invoking a CAS. Therefore lets rewrite the equation in the following way



      $$begin{align}
      2we^{w^2}-2e^w&=0\
      we^{w^2}&=e^w\
      we^{w^2-w}&=1\
      we^{w(1-w)}&=1
      end{align}$$



      From the last line we can directly conclude that $w=1$ satisfies the equation. Howsoever assuming only positive $w$ we may take the logarithm of both sides to further get



      $$begin{align}
      lnleft(we^{w(1-w)}right)&=ln(1)\
      ln(w)+w(w-1)&=0\
      -w^2+w&=log(w)\
      -left(w-frac12right)^2+frac14&=log(w)
      end{align}$$



      The LHS describes as parabola with vertex at $(1/2,1/4)$ and roots at $w_1=0$ and $w_2=1$. Due the minus sign infront we can further conclude that the vertex is a maximum. Basically we are done now. The logarithm intersects with the $w$-axis at $w=1$ and tends to negative infinity as $w$ goes to $0$. But hence the parabola has got a root at $w=0$ they cannot intersect on the left hand side of $w=1$ whereas on the right hand side of $w=1$ the logarithm is strictly increasing and the parabola stricly decreasing.

      So the only possible real root is $w=1$ due the above argumentation with regard to the graphs of both functions.






      share|cite|improve this answer









      $endgroup$
















        4












        4








        4





        $begingroup$

        Maybe it is even possible to show that this equation only has got one real root without invoking a CAS. Therefore lets rewrite the equation in the following way



        $$begin{align}
        2we^{w^2}-2e^w&=0\
        we^{w^2}&=e^w\
        we^{w^2-w}&=1\
        we^{w(1-w)}&=1
        end{align}$$



        From the last line we can directly conclude that $w=1$ satisfies the equation. Howsoever assuming only positive $w$ we may take the logarithm of both sides to further get



        $$begin{align}
        lnleft(we^{w(1-w)}right)&=ln(1)\
        ln(w)+w(w-1)&=0\
        -w^2+w&=log(w)\
        -left(w-frac12right)^2+frac14&=log(w)
        end{align}$$



        The LHS describes as parabola with vertex at $(1/2,1/4)$ and roots at $w_1=0$ and $w_2=1$. Due the minus sign infront we can further conclude that the vertex is a maximum. Basically we are done now. The logarithm intersects with the $w$-axis at $w=1$ and tends to negative infinity as $w$ goes to $0$. But hence the parabola has got a root at $w=0$ they cannot intersect on the left hand side of $w=1$ whereas on the right hand side of $w=1$ the logarithm is strictly increasing and the parabola stricly decreasing.

        So the only possible real root is $w=1$ due the above argumentation with regard to the graphs of both functions.






        share|cite|improve this answer









        $endgroup$



        Maybe it is even possible to show that this equation only has got one real root without invoking a CAS. Therefore lets rewrite the equation in the following way



        $$begin{align}
        2we^{w^2}-2e^w&=0\
        we^{w^2}&=e^w\
        we^{w^2-w}&=1\
        we^{w(1-w)}&=1
        end{align}$$



        From the last line we can directly conclude that $w=1$ satisfies the equation. Howsoever assuming only positive $w$ we may take the logarithm of both sides to further get



        $$begin{align}
        lnleft(we^{w(1-w)}right)&=ln(1)\
        ln(w)+w(w-1)&=0\
        -w^2+w&=log(w)\
        -left(w-frac12right)^2+frac14&=log(w)
        end{align}$$



        The LHS describes as parabola with vertex at $(1/2,1/4)$ and roots at $w_1=0$ and $w_2=1$. Due the minus sign infront we can further conclude that the vertex is a maximum. Basically we are done now. The logarithm intersects with the $w$-axis at $w=1$ and tends to negative infinity as $w$ goes to $0$. But hence the parabola has got a root at $w=0$ they cannot intersect on the left hand side of $w=1$ whereas on the right hand side of $w=1$ the logarithm is strictly increasing and the parabola stricly decreasing.

        So the only possible real root is $w=1$ due the above argumentation with regard to the graphs of both functions.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 24 '18 at 16:37









        mrtaurhomrtaurho

        5,75551540




        5,75551540






























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