Why didn't they simplify $x^y=y^x$ to $x=y$?
$begingroup$
Solving $x^y = y^x$ analytically in terms of the Lambert $W$ function
This "solution" for $x^y=y^x$ should simplify to $y=x$, but for some reason no pointed that out in the OP.
According to the stack exchange, the answer is $y= frac{-xW(-frac{ln(x)}{x})}{ln(x)}$. However, the term $frac{-ln(x)}{x}$ itself can be rewritten as
$$frac{-ln(x)}{x}=-ln(x)e^{-ln(x)}$$
Therefore, the productlog of that expression should simplify as follows,
$y= frac{-xW(-frac{ln(x)}{x})}{ln(x)}, $ $y= frac{-xW(-ln(x)e^{-ln(x)})}{ln(x)}, $ $y=frac{-x(-ln(x))}{ln(x)}=x$
Did this simplification just slip past everyone or is there something wrong about my algebra?
algebra-precalculus
edited Jan 14 at 15:13
Daniel R
2,46832035
asked Jan 14 at 3:51
user14554user14554
445
$endgroup$
|
show 7 more comments
$begingroup$
Solving $x^y = y^x$ analytically in terms of the Lambert $W$ function
This "solution" for $x^y=y^x$ should simplify to $y=x$, but for some reason no pointed that out in the OP.
According to the stack exchange, the answer is $y= frac{-xW(-frac{ln(x)}{x})}{ln(x)}$. However, the term $frac{-ln(x)}{x}$ itself can be rewritten as
$$frac{-ln(x)}{x}=-ln(x)e^{-ln(x)}$$
Therefore, the productlog of that expression should simplify as follows,
$y= frac{-xW(-frac{ln(x)}{x})}{ln(x)}, $ $y= frac{-xW(-ln(x)e^{-ln(x)})}{ln(x)}, $ $y=frac{-x(-ln(x))}{ln(x)}=x$
Did this simplification just slip past everyone or is there something wrong about my algebra?
algebra-precalculus
edited Jan 14 at 15:13
Daniel R
2,46832035
asked Jan 14 at 3:51
user14554user14554
445
$endgroup$
7
$begingroup$
Why should it reduce to that? $x=4$ and $y=2$ has $x neq y$.
$endgroup$
– Randall
Jan 14 at 3:54
4
$begingroup$
No, $2^4=16=4^2$.
$endgroup$
– Randall
Jan 14 at 3:56
8
$begingroup$
I'm just confused why the solution "should" simplify to $x=y$ when there are solutions that do not satisfy $x = y$.
$endgroup$
– Randall
Jan 14 at 3:58
1
$begingroup$
Anyway, to potentially answer your question, your algebra moves are invalid if $x$ is negative, and there are solutions with negative $x$.
$endgroup$
– Randall
Jan 14 at 3:59
1
$begingroup$
But if the solution is algebraically equivalent to $y=x$, so why does the original representation contain any more solutions than $y=x$? There is definitely something more complicated being left out here.
$endgroup$
– user14554
Jan 14 at 4:07
|
show 7 more comments
$begingroup$
Solving $x^y = y^x$ analytically in terms of the Lambert $W$ function
This "solution" for $x^y=y^x$ should simplify to $y=x$, but for some reason no pointed that out in the OP.
According to the stack exchange, the answer is $y= frac{-xW(-frac{ln(x)}{x})}{ln(x)}$. However, the term $frac{-ln(x)}{x}$ itself can be rewritten as
$$frac{-ln(x)}{x}=-ln(x)e^{-ln(x)}$$
Therefore, the productlog of that expression should simplify as follows,
$y= frac{-xW(-frac{ln(x)}{x})}{ln(x)}, $ $y= frac{-xW(-ln(x)e^{-ln(x)})}{ln(x)}, $ $y=frac{-x(-ln(x))}{ln(x)}=x$
Did this simplification just slip past everyone or is there something wrong about my algebra?
algebra-precalculus
edited Jan 14 at 15:13
Daniel R
2,46832035
asked Jan 14 at 3:51
user14554user14554
445
$endgroup$
Solving $x^y = y^x$ analytically in terms of the Lambert $W$ function
This "solution" for $x^y=y^x$ should simplify to $y=x$, but for some reason no pointed that out in the OP.
According to the stack exchange, the answer is $y= frac{-xW(-frac{ln(x)}{x})}{ln(x)}$. However, the term $frac{-ln(x)}{x}$ itself can be rewritten as
$$frac{-ln(x)}{x}=-ln(x)e^{-ln(x)}$$
Therefore, the productlog of that expression should simplify as follows,
$y= frac{-xW(-frac{ln(x)}{x})}{ln(x)}, $ $y= frac{-xW(-ln(x)e^{-ln(x)})}{ln(x)}, $ $y=frac{-x(-ln(x))}{ln(x)}=x$
Did this simplification just slip past everyone or is there something wrong about my algebra?
algebra-precalculus
algebra-precalculus
edited Jan 14 at 15:13
Daniel R
2,46832035
asked Jan 14 at 3:51
user14554user14554
445
edited Jan 14 at 15:13
Daniel R
2,46832035
asked Jan 14 at 3:51
user14554user14554
445
edited Jan 14 at 15:13
Daniel R
2,46832035
edited Jan 14 at 15:13
Daniel R
2,46832035
edited Jan 14 at 15:13
Daniel R
2,46832035
2,46832035
asked Jan 14 at 3:51
user14554user14554
445
asked Jan 14 at 3:51
user14554user14554
445
asked Jan 14 at 3:51
user14554user14554
445
445
7
$begingroup$
Why should it reduce to that? $x=4$ and $y=2$ has $x neq y$.
$endgroup$
– Randall
Jan 14 at 3:54
4
$begingroup$
No, $2^4=16=4^2$.
$endgroup$
– Randall
Jan 14 at 3:56
8
$begingroup$
I'm just confused why the solution "should" simplify to $x=y$ when there are solutions that do not satisfy $x = y$.
$endgroup$
– Randall
Jan 14 at 3:58
1
$begingroup$
Anyway, to potentially answer your question, your algebra moves are invalid if $x$ is negative, and there are solutions with negative $x$.
$endgroup$
– Randall
Jan 14 at 3:59
1
$begingroup$
But if the solution is algebraically equivalent to $y=x$, so why does the original representation contain any more solutions than $y=x$? There is definitely something more complicated being left out here.
$endgroup$
– user14554
Jan 14 at 4:07
|
show 7 more comments
7
$begingroup$
Why should it reduce to that? $x=4$ and $y=2$ has $x neq y$.
$endgroup$
– Randall
Jan 14 at 3:54
4
$begingroup$
No, $2^4=16=4^2$.
$endgroup$
– Randall
Jan 14 at 3:56
8
$begingroup$
I'm just confused why the solution "should" simplify to $x=y$ when there are solutions that do not satisfy $x = y$.
$endgroup$
– Randall
Jan 14 at 3:58
1
$begingroup$
Anyway, to potentially answer your question, your algebra moves are invalid if $x$ is negative, and there are solutions with negative $x$.
$endgroup$
– Randall
Jan 14 at 3:59
1
$begingroup$
But if the solution is algebraically equivalent to $y=x$, so why does the original representation contain any more solutions than $y=x$? There is definitely something more complicated being left out here.
$endgroup$
– user14554
Jan 14 at 4:07
7
7
$begingroup$
Why should it reduce to that? $x=4$ and $y=2$ has $x neq y$.
$endgroup$
– Randall
Jan 14 at 3:54
$begingroup$
Why should it reduce to that? $x=4$ and $y=2$ has $x neq y$.
$endgroup$
– Randall
Jan 14 at 3:54
4
4
$begingroup$
No, $2^4=16=4^2$.
$endgroup$
– Randall
Jan 14 at 3:56
$begingroup$
No, $2^4=16=4^2$.
$endgroup$
– Randall
Jan 14 at 3:56
8
8
$begingroup$
I'm just confused why the solution "should" simplify to $x=y$ when there are solutions that do not satisfy $x = y$.
$endgroup$
– Randall
Jan 14 at 3:58
$begingroup$
I'm just confused why the solution "should" simplify to $x=y$ when there are solutions that do not satisfy $x = y$.
$endgroup$
– Randall
Jan 14 at 3:58
1
1
$begingroup$
Anyway, to potentially answer your question, your algebra moves are invalid if $x$ is negative, and there are solutions with negative $x$.
$endgroup$
– Randall
Jan 14 at 3:59
$begingroup$
Anyway, to potentially answer your question, your algebra moves are invalid if $x$ is negative, and there are solutions with negative $x$.
$endgroup$
– Randall
Jan 14 at 3:59
1
1
$begingroup$
But if the solution is algebraically equivalent to $y=x$, so why does the original representation contain any more solutions than $y=x$? There is definitely something more complicated being left out here.
$endgroup$
– user14554
Jan 14 at 4:07
$begingroup$
But if the solution is algebraically equivalent to $y=x$, so why does the original representation contain any more solutions than $y=x$? There is definitely something more complicated being left out here.
$endgroup$
– user14554
Jan 14 at 4:07
|
show 7 more comments
2 Answers
2
active
oldest
votes
$begingroup$
The Lambert $W$ function is not single-valued for negative arguments.
Using your "simplification" forces use of the lower branch, $W leq -1$ when you assume $W^{-1}(-ln x)$ only equals $-ln (x) mathrm{e}^{- ln x}$. (The same thing happens when you assume the only square root of $3^2$ is $3$ or the only arcsine of $1$ is $-3pi/2$.) You get two values from $W^{-1}(-ln x)$ having the same algebraic form, but one has $0 < x leq mathrm{e}$ and one has $x > mathrm{e}$. ("$3^2$" and "$(-3)^2$" have the same algebraic form, "$x^2$", but one has $x>0$ and one has $x < 0$.)
This is indicated explicitly in the identities at the Lambert $W$ function article on the English Wikipedia.
Edit: Got myself turned around with too many minus signs. I originally claimed the $x=y$ solutions were on $W geq -1$, but this is backwards. It is corrected above.
answered Jan 14 at 4:16
Eric TowersEric Towers
32.4k22268
$endgroup$
$begingroup$
What is confusing is how $W(z)e^{W(z)}=z$ always simplifies no matter which branch you use, but $W(ze^{z})$ does not.
$endgroup$
– user14554
Jan 14 at 4:18
$begingroup$
+1 for the first parenthetical.
$endgroup$
– Randall
Jan 14 at 4:19
$begingroup$
@user14554 : This is the usual problem with inverse functions. $sqrt{9} = 3$, but "the things which square to $9$" is ${-3,3}$. This is always lurking around when you are solving equations.
$endgroup$
– Eric Towers
Jan 14 at 4:19
1
$begingroup$
@user14554 : When $x = 6$, $W_{-1}$ gives $y = 6$ and $W_0$ gives $y = 1.624dots$. You get them back the same way you do with any other function whose domain must be restricted to obtain the inverse function: you use a full set of inverses whose ranges cover the entire domain of the unrestricted function.
$endgroup$
– Eric Towers
Jan 14 at 4:27
1
$begingroup$
@user14554 : I disagree. Every time you apply a $W^{-1}$, you get a contribution from $W_0$ and another from $W_{-1}$. You are, of course, free to incorrectly ignore solutions. I, on the other hand, will continue to find that $(x^2 - 3)^2-1=0$ has four real roots.
$endgroup$
– Eric Towers
Jan 14 at 5:08
|
show 12 more comments
$begingroup$
The solution is:
$$y = -frac{x Wleft(-frac{log (x)}{x}right)}{log (x)}$$
which has the following form:
Clearly there are solutions other than $x = y$. Indeed, we see that for $y=2$ we can have $x=2$ or $x=4$ (intersection between blue and red dashed line).
answered Jan 14 at 4:08
David G. StorkDavid G. Stork
10.7k31332
$endgroup$
$begingroup$
So it has something to do with the multiple branches of the log and productlog then. For $W_{0}(x)$ it simplifies, but when it changes to $W_{-1}(x)$ it doesn't.
$endgroup$
– user14554
Jan 14 at 4:09
$begingroup$
I think OP's question is why isn't the blue line simply $y=x$? It is tantalizing that it is $y=x$ for a while and then there's a sudden change.
$endgroup$
– Randall
Jan 14 at 4:09
$begingroup$
Right, why isn't it $y=x$ all the way.
$endgroup$
– user14554
Jan 14 at 4:09
$begingroup$
@user14554 I see your question now.
$endgroup$
– Randall
Jan 14 at 4:10
$begingroup$
user14554 and Randall: There must be a branch cut in the Lambert W function.
$endgroup$
– David G. Stork
Jan 14 at 4:11
|
show 1 more comment
Your Answer
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2 Answers
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active
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2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
The Lambert $W$ function is not single-valued for negative arguments.
Using your "simplification" forces use of the lower branch, $W leq -1$ when you assume $W^{-1}(-ln x)$ only equals $-ln (x) mathrm{e}^{- ln x}$. (The same thing happens when you assume the only square root of $3^2$ is $3$ or the only arcsine of $1$ is $-3pi/2$.) You get two values from $W^{-1}(-ln x)$ having the same algebraic form, but one has $0 < x leq mathrm{e}$ and one has $x > mathrm{e}$. ("$3^2$" and "$(-3)^2$" have the same algebraic form, "$x^2$", but one has $x>0$ and one has $x < 0$.)
This is indicated explicitly in the identities at the Lambert $W$ function article on the English Wikipedia.
Edit: Got myself turned around with too many minus signs. I originally claimed the $x=y$ solutions were on $W geq -1$, but this is backwards. It is corrected above.
answered Jan 14 at 4:16
Eric TowersEric Towers
32.4k22268
$endgroup$
$begingroup$
What is confusing is how $W(z)e^{W(z)}=z$ always simplifies no matter which branch you use, but $W(ze^{z})$ does not.
$endgroup$
– user14554
Jan 14 at 4:18
$begingroup$
+1 for the first parenthetical.
$endgroup$
– Randall
Jan 14 at 4:19
$begingroup$
@user14554 : This is the usual problem with inverse functions. $sqrt{9} = 3$, but "the things which square to $9$" is ${-3,3}$. This is always lurking around when you are solving equations.
$endgroup$
– Eric Towers
Jan 14 at 4:19
1
$begingroup$
@user14554 : When $x = 6$, $W_{-1}$ gives $y = 6$ and $W_0$ gives $y = 1.624dots$. You get them back the same way you do with any other function whose domain must be restricted to obtain the inverse function: you use a full set of inverses whose ranges cover the entire domain of the unrestricted function.
$endgroup$
– Eric Towers
Jan 14 at 4:27
1
$begingroup$
@user14554 : I disagree. Every time you apply a $W^{-1}$, you get a contribution from $W_0$ and another from $W_{-1}$. You are, of course, free to incorrectly ignore solutions. I, on the other hand, will continue to find that $(x^2 - 3)^2-1=0$ has four real roots.
$endgroup$
– Eric Towers
Jan 14 at 5:08
|
show 12 more comments
$begingroup$
The Lambert $W$ function is not single-valued for negative arguments.
Using your "simplification" forces use of the lower branch, $W leq -1$ when you assume $W^{-1}(-ln x)$ only equals $-ln (x) mathrm{e}^{- ln x}$. (The same thing happens when you assume the only square root of $3^2$ is $3$ or the only arcsine of $1$ is $-3pi/2$.) You get two values from $W^{-1}(-ln x)$ having the same algebraic form, but one has $0 < x leq mathrm{e}$ and one has $x > mathrm{e}$. ("$3^2$" and "$(-3)^2$" have the same algebraic form, "$x^2$", but one has $x>0$ and one has $x < 0$.)
This is indicated explicitly in the identities at the Lambert $W$ function article on the English Wikipedia.
Edit: Got myself turned around with too many minus signs. I originally claimed the $x=y$ solutions were on $W geq -1$, but this is backwards. It is corrected above.
answered Jan 14 at 4:16
Eric TowersEric Towers
32.4k22268
$endgroup$
$begingroup$
What is confusing is how $W(z)e^{W(z)}=z$ always simplifies no matter which branch you use, but $W(ze^{z})$ does not.
$endgroup$
– user14554
Jan 14 at 4:18
$begingroup$
+1 for the first parenthetical.
$endgroup$
– Randall
Jan 14 at 4:19
$begingroup$
@user14554 : This is the usual problem with inverse functions. $sqrt{9} = 3$, but "the things which square to $9$" is ${-3,3}$. This is always lurking around when you are solving equations.
$endgroup$
– Eric Towers
Jan 14 at 4:19
1
$begingroup$
@user14554 : When $x = 6$, $W_{-1}$ gives $y = 6$ and $W_0$ gives $y = 1.624dots$. You get them back the same way you do with any other function whose domain must be restricted to obtain the inverse function: you use a full set of inverses whose ranges cover the entire domain of the unrestricted function.
$endgroup$
– Eric Towers
Jan 14 at 4:27
1
$begingroup$
@user14554 : I disagree. Every time you apply a $W^{-1}$, you get a contribution from $W_0$ and another from $W_{-1}$. You are, of course, free to incorrectly ignore solutions. I, on the other hand, will continue to find that $(x^2 - 3)^2-1=0$ has four real roots.
$endgroup$
– Eric Towers
Jan 14 at 5:08
|
show 12 more comments
$begingroup$
The Lambert $W$ function is not single-valued for negative arguments.
Using your "simplification" forces use of the lower branch, $W leq -1$ when you assume $W^{-1}(-ln x)$ only equals $-ln (x) mathrm{e}^{- ln x}$. (The same thing happens when you assume the only square root of $3^2$ is $3$ or the only arcsine of $1$ is $-3pi/2$.) You get two values from $W^{-1}(-ln x)$ having the same algebraic form, but one has $0 < x leq mathrm{e}$ and one has $x > mathrm{e}$. ("$3^2$" and "$(-3)^2$" have the same algebraic form, "$x^2$", but one has $x>0$ and one has $x < 0$.)
This is indicated explicitly in the identities at the Lambert $W$ function article on the English Wikipedia.
Edit: Got myself turned around with too many minus signs. I originally claimed the $x=y$ solutions were on $W geq -1$, but this is backwards. It is corrected above.
answered Jan 14 at 4:16
Eric TowersEric Towers
32.4k22268
$endgroup$
The Lambert $W$ function is not single-valued for negative arguments.
Using your "simplification" forces use of the lower branch, $W leq -1$ when you assume $W^{-1}(-ln x)$ only equals $-ln (x) mathrm{e}^{- ln x}$. (The same thing happens when you assume the only square root of $3^2$ is $3$ or the only arcsine of $1$ is $-3pi/2$.) You get two values from $W^{-1}(-ln x)$ having the same algebraic form, but one has $0 < x leq mathrm{e}$ and one has $x > mathrm{e}$. ("$3^2$" and "$(-3)^2$" have the same algebraic form, "$x^2$", but one has $x>0$ and one has $x < 0$.)
This is indicated explicitly in the identities at the Lambert $W$ function article on the English Wikipedia.
Edit: Got myself turned around with too many minus signs. I originally claimed the $x=y$ solutions were on $W geq -1$, but this is backwards. It is corrected above.
answered Jan 14 at 4:16
Eric TowersEric Towers
32.4k22268
edited Jan 14 at 4:30
answered Jan 14 at 4:16
Eric TowersEric Towers
32.4k22268
answered Jan 14 at 4:16
Eric TowersEric Towers
32.4k22268
answered Jan 14 at 4:16
Eric TowersEric Towers
32.4k22268
32.4k22268
$begingroup$
What is confusing is how $W(z)e^{W(z)}=z$ always simplifies no matter which branch you use, but $W(ze^{z})$ does not.
$endgroup$
– user14554
Jan 14 at 4:18
$begingroup$
+1 for the first parenthetical.
$endgroup$
– Randall
Jan 14 at 4:19
$begingroup$
@user14554 : This is the usual problem with inverse functions. $sqrt{9} = 3$, but "the things which square to $9$" is ${-3,3}$. This is always lurking around when you are solving equations.
$endgroup$
– Eric Towers
Jan 14 at 4:19
1
$begingroup$
@user14554 : When $x = 6$, $W_{-1}$ gives $y = 6$ and $W_0$ gives $y = 1.624dots$. You get them back the same way you do with any other function whose domain must be restricted to obtain the inverse function: you use a full set of inverses whose ranges cover the entire domain of the unrestricted function.
$endgroup$
– Eric Towers
Jan 14 at 4:27
1
$begingroup$
@user14554 : I disagree. Every time you apply a $W^{-1}$, you get a contribution from $W_0$ and another from $W_{-1}$. You are, of course, free to incorrectly ignore solutions. I, on the other hand, will continue to find that $(x^2 - 3)^2-1=0$ has four real roots.
$endgroup$
– Eric Towers
Jan 14 at 5:08
|
show 12 more comments
$begingroup$
What is confusing is how $W(z)e^{W(z)}=z$ always simplifies no matter which branch you use, but $W(ze^{z})$ does not.
$endgroup$
– user14554
Jan 14 at 4:18
$begingroup$
+1 for the first parenthetical.
$endgroup$
– Randall
Jan 14 at 4:19
$begingroup$
@user14554 : This is the usual problem with inverse functions. $sqrt{9} = 3$, but "the things which square to $9$" is ${-3,3}$. This is always lurking around when you are solving equations.
$endgroup$
– Eric Towers
Jan 14 at 4:19
1
$begingroup$
@user14554 : When $x = 6$, $W_{-1}$ gives $y = 6$ and $W_0$ gives $y = 1.624dots$. You get them back the same way you do with any other function whose domain must be restricted to obtain the inverse function: you use a full set of inverses whose ranges cover the entire domain of the unrestricted function.
$endgroup$
– Eric Towers
Jan 14 at 4:27
1
$begingroup$
@user14554 : I disagree. Every time you apply a $W^{-1}$, you get a contribution from $W_0$ and another from $W_{-1}$. You are, of course, free to incorrectly ignore solutions. I, on the other hand, will continue to find that $(x^2 - 3)^2-1=0$ has four real roots.
$endgroup$
– Eric Towers
Jan 14 at 5:08
$begingroup$
What is confusing is how $W(z)e^{W(z)}=z$ always simplifies no matter which branch you use, but $W(ze^{z})$ does not.
$endgroup$
– user14554
Jan 14 at 4:18
$begingroup$
What is confusing is how $W(z)e^{W(z)}=z$ always simplifies no matter which branch you use, but $W(ze^{z})$ does not.
$endgroup$
– user14554
Jan 14 at 4:18
$begingroup$
+1 for the first parenthetical.
$endgroup$
– Randall
Jan 14 at 4:19
$begingroup$
+1 for the first parenthetical.
$endgroup$
– Randall
Jan 14 at 4:19
$begingroup$
@user14554 : This is the usual problem with inverse functions. $sqrt{9} = 3$, but "the things which square to $9$" is ${-3,3}$. This is always lurking around when you are solving equations.
$endgroup$
– Eric Towers
Jan 14 at 4:19
$begingroup$
@user14554 : This is the usual problem with inverse functions. $sqrt{9} = 3$, but "the things which square to $9$" is ${-3,3}$. This is always lurking around when you are solving equations.
$endgroup$
– Eric Towers
Jan 14 at 4:19
1
1
$begingroup$
@user14554 : When $x = 6$, $W_{-1}$ gives $y = 6$ and $W_0$ gives $y = 1.624dots$. You get them back the same way you do with any other function whose domain must be restricted to obtain the inverse function: you use a full set of inverses whose ranges cover the entire domain of the unrestricted function.
$endgroup$
– Eric Towers
Jan 14 at 4:27
$begingroup$
@user14554 : When $x = 6$, $W_{-1}$ gives $y = 6$ and $W_0$ gives $y = 1.624dots$. You get them back the same way you do with any other function whose domain must be restricted to obtain the inverse function: you use a full set of inverses whose ranges cover the entire domain of the unrestricted function.
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– Eric Towers
Jan 14 at 4:27
1
1
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@user14554 : I disagree. Every time you apply a $W^{-1}$, you get a contribution from $W_0$ and another from $W_{-1}$. You are, of course, free to incorrectly ignore solutions. I, on the other hand, will continue to find that $(x^2 - 3)^2-1=0$ has four real roots.
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– Eric Towers
Jan 14 at 5:08
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@user14554 : I disagree. Every time you apply a $W^{-1}$, you get a contribution from $W_0$ and another from $W_{-1}$. You are, of course, free to incorrectly ignore solutions. I, on the other hand, will continue to find that $(x^2 - 3)^2-1=0$ has four real roots.
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– Eric Towers
Jan 14 at 5:08
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show 12 more comments
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The solution is:
$$y = -frac{x Wleft(-frac{log (x)}{x}right)}{log (x)}$$
which has the following form:
Clearly there are solutions other than $x = y$. Indeed, we see that for $y=2$ we can have $x=2$ or $x=4$ (intersection between blue and red dashed line).
answered Jan 14 at 4:08
David G. StorkDavid G. Stork
10.7k31332
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So it has something to do with the multiple branches of the log and productlog then. For $W_{0}(x)$ it simplifies, but when it changes to $W_{-1}(x)$ it doesn't.
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– user14554
Jan 14 at 4:09
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I think OP's question is why isn't the blue line simply $y=x$? It is tantalizing that it is $y=x$ for a while and then there's a sudden change.
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– Randall
Jan 14 at 4:09
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Right, why isn't it $y=x$ all the way.
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– user14554
Jan 14 at 4:09
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@user14554 I see your question now.
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– Randall
Jan 14 at 4:10
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user14554 and Randall: There must be a branch cut in the Lambert W function.
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– David G. Stork
Jan 14 at 4:11
|
show 1 more comment
$begingroup$
The solution is:
$$y = -frac{x Wleft(-frac{log (x)}{x}right)}{log (x)}$$
which has the following form:
Clearly there are solutions other than $x = y$. Indeed, we see that for $y=2$ we can have $x=2$ or $x=4$ (intersection between blue and red dashed line).
answered Jan 14 at 4:08
David G. StorkDavid G. Stork
10.7k31332
$endgroup$
$begingroup$
So it has something to do with the multiple branches of the log and productlog then. For $W_{0}(x)$ it simplifies, but when it changes to $W_{-1}(x)$ it doesn't.
$endgroup$
– user14554
Jan 14 at 4:09
$begingroup$
I think OP's question is why isn't the blue line simply $y=x$? It is tantalizing that it is $y=x$ for a while and then there's a sudden change.
$endgroup$
– Randall
Jan 14 at 4:09
$begingroup$
Right, why isn't it $y=x$ all the way.
$endgroup$
– user14554
Jan 14 at 4:09
$begingroup$
@user14554 I see your question now.
$endgroup$
– Randall
Jan 14 at 4:10
$begingroup$
user14554 and Randall: There must be a branch cut in the Lambert W function.
$endgroup$
– David G. Stork
Jan 14 at 4:11
|
show 1 more comment
$begingroup$
The solution is:
$$y = -frac{x Wleft(-frac{log (x)}{x}right)}{log (x)}$$
which has the following form:
Clearly there are solutions other than $x = y$. Indeed, we see that for $y=2$ we can have $x=2$ or $x=4$ (intersection between blue and red dashed line).
answered Jan 14 at 4:08
David G. StorkDavid G. Stork
10.7k31332
$endgroup$
The solution is:
$$y = -frac{x Wleft(-frac{log (x)}{x}right)}{log (x)}$$
which has the following form:
Clearly there are solutions other than $x = y$. Indeed, we see that for $y=2$ we can have $x=2$ or $x=4$ (intersection between blue and red dashed line).
answered Jan 14 at 4:08
David G. StorkDavid G. Stork
10.7k31332
edited Jan 14 at 4:09
answered Jan 14 at 4:08
David G. StorkDavid G. Stork
10.7k31332
answered Jan 14 at 4:08
David G. StorkDavid G. Stork
10.7k31332
answered Jan 14 at 4:08
David G. StorkDavid G. Stork
10.7k31332
10.7k31332
$begingroup$
So it has something to do with the multiple branches of the log and productlog then. For $W_{0}(x)$ it simplifies, but when it changes to $W_{-1}(x)$ it doesn't.
$endgroup$
– user14554
Jan 14 at 4:09
$begingroup$
I think OP's question is why isn't the blue line simply $y=x$? It is tantalizing that it is $y=x$ for a while and then there's a sudden change.
$endgroup$
– Randall
Jan 14 at 4:09
$begingroup$
Right, why isn't it $y=x$ all the way.
$endgroup$
– user14554
Jan 14 at 4:09
$begingroup$
@user14554 I see your question now.
$endgroup$
– Randall
Jan 14 at 4:10
$begingroup$
user14554 and Randall: There must be a branch cut in the Lambert W function.
$endgroup$
– David G. Stork
Jan 14 at 4:11
|
show 1 more comment
$begingroup$
So it has something to do with the multiple branches of the log and productlog then. For $W_{0}(x)$ it simplifies, but when it changes to $W_{-1}(x)$ it doesn't.
$endgroup$
– user14554
Jan 14 at 4:09
$begingroup$
I think OP's question is why isn't the blue line simply $y=x$? It is tantalizing that it is $y=x$ for a while and then there's a sudden change.
$endgroup$
– Randall
Jan 14 at 4:09
$begingroup$
Right, why isn't it $y=x$ all the way.
$endgroup$
– user14554
Jan 14 at 4:09
$begingroup$
@user14554 I see your question now.
$endgroup$
– Randall
Jan 14 at 4:10
$begingroup$
user14554 and Randall: There must be a branch cut in the Lambert W function.
$endgroup$
– David G. Stork
Jan 14 at 4:11
$begingroup$
So it has something to do with the multiple branches of the log and productlog then. For $W_{0}(x)$ it simplifies, but when it changes to $W_{-1}(x)$ it doesn't.
$endgroup$
– user14554
Jan 14 at 4:09
$begingroup$
So it has something to do with the multiple branches of the log and productlog then. For $W_{0}(x)$ it simplifies, but when it changes to $W_{-1}(x)$ it doesn't.
$endgroup$
– user14554
Jan 14 at 4:09
$begingroup$
I think OP's question is why isn't the blue line simply $y=x$? It is tantalizing that it is $y=x$ for a while and then there's a sudden change.
$endgroup$
– Randall
Jan 14 at 4:09
$begingroup$
I think OP's question is why isn't the blue line simply $y=x$? It is tantalizing that it is $y=x$ for a while and then there's a sudden change.
$endgroup$
– Randall
Jan 14 at 4:09
$begingroup$
Right, why isn't it $y=x$ all the way.
$endgroup$
– user14554
Jan 14 at 4:09
$begingroup$
Right, why isn't it $y=x$ all the way.
$endgroup$
– user14554
Jan 14 at 4:09
$begingroup$
@user14554 I see your question now.
$endgroup$
– Randall
Jan 14 at 4:10
$begingroup$
@user14554 I see your question now.
$endgroup$
– Randall
Jan 14 at 4:10
$begingroup$
user14554 and Randall: There must be a branch cut in the Lambert W function.
$endgroup$
– David G. Stork
Jan 14 at 4:11
$begingroup$
user14554 and Randall: There must be a branch cut in the Lambert W function.
$endgroup$
– David G. Stork
Jan 14 at 4:11
|
show 1 more comment
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7
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Why should it reduce to that? $x=4$ and $y=2$ has $x neq y$.
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– Randall
Jan 14 at 3:54
4
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No, $2^4=16=4^2$.
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– Randall
Jan 14 at 3:56
8
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I'm just confused why the solution "should" simplify to $x=y$ when there are solutions that do not satisfy $x = y$.
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– Randall
Jan 14 at 3:58
1
$begingroup$
Anyway, to potentially answer your question, your algebra moves are invalid if $x$ is negative, and there are solutions with negative $x$.
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– Randall
Jan 14 at 3:59
1
$begingroup$
But if the solution is algebraically equivalent to $y=x$, so why does the original representation contain any more solutions than $y=x$? There is definitely something more complicated being left out here.
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– user14554
Jan 14 at 4:07