How can we solve for $w$ : $2we^{w^2}-2e^w=0$
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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
edited Dec 24 '18 at 16:06
mrtaurho
6,22571641
asked Dec 24 '18 at 16:05
Pritee SubediPritee Subedi
31
$endgroup$
add a comment |
$begingroup$
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
edited Dec 24 '18 at 16:06
mrtaurho
6,22571641
asked Dec 24 '18 at 16:05
Pritee SubediPritee Subedi
31
$endgroup$
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
add a comment |
$begingroup$
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
edited Dec 24 '18 at 16:06
mrtaurho
6,22571641
asked Dec 24 '18 at 16:05
Pritee SubediPritee Subedi
31
$endgroup$
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
edited Dec 24 '18 at 16:06
mrtaurho
6,22571641
asked Dec 24 '18 at 16:05
Pritee SubediPritee Subedi
31
edited Dec 24 '18 at 16:06
mrtaurho
6,22571641
asked Dec 24 '18 at 16:05
Pritee SubediPritee Subedi
31
edited Dec 24 '18 at 16:06
mrtaurho
6,22571641
edited Dec 24 '18 at 16:06
mrtaurho
6,22571641
edited Dec 24 '18 at 16:06
mrtaurho
6,22571641
6,22571641
asked Dec 24 '18 at 16:05
Pritee SubediPritee Subedi
31
asked Dec 24 '18 at 16:05
Pritee SubediPritee Subedi
31
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Dec 24 '18 at 16:37
mrtaurhomrtaurho
6,22571641
$endgroup$
add a comment |
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1 Answer
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$begingroup$
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.
answered Dec 24 '18 at 16:37
mrtaurhomrtaurho
6,22571641
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 24 '18 at 16:37
mrtaurhomrtaurho
6,22571641
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 24 '18 at 16:37
mrtaurhomrtaurho
6,22571641
$endgroup$
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.
answered Dec 24 '18 at 16:37
mrtaurhomrtaurho
6,22571641
answered Dec 24 '18 at 16:37
mrtaurhomrtaurho
6,22571641
answered Dec 24 '18 at 16:37
mrtaurhomrtaurho
6,22571641
answered Dec 24 '18 at 16:37
mrtaurhomrtaurho
6,22571641
6,22571641
add a comment |
add a comment |
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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