Why didn't they simplify this?
$begingroup$
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
asked 2 hours ago
user14554user14554
234
$endgroup$
|
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
asked 2 hours ago
user14554user14554
234
$endgroup$
1
$begingroup$
Why should it reduce to that? $x=4$ and $y=2$ has $x neq y$.
$endgroup$
– Randall
2 hours ago
$begingroup$
$4=2^2$? Why wouldn't it? It makes sense that if x and y are the same, they would satisfy this equation. If it doesn't though, then that's what I want to know. According to the algebraic properties of the productlog function, it should simplify in the way I presented, so if it doesn't actually work that way, it begs the question "why not?".
$endgroup$
– user14554
2 hours ago
$begingroup$
No, $2^4=16=4^2$.
$endgroup$
– Randall
2 hours ago
$begingroup$
@Randall But then you still haven't explained why the algebra simplifies yet a counterexample remains. Why should it? Because that's what the algebra says it should do, according to the definition of the productlog which has the properties $W(ze^z)=z$. That property was the basis of deriving that solution in the first place.
$endgroup$
– user14554
2 hours ago
1
$begingroup$
As motivation, you might consider whether the relation $y = arcsin(sin(x))$ is "algebraically equivalent" to $y=x$. (What about when $y=0$ and $x=pi$?) The same issue is at play here: $v = W(u)$ means that $u e^u = v$, but there is not always a unique $u$ with that property, and depending on which one you choose, your claim may or may not be valid.
$endgroup$
– Nate Eldredge
1 hour ago
|
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
asked 2 hours ago
user14554user14554
234
$endgroup$
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
asked 2 hours ago
user14554user14554
234
asked 2 hours ago
user14554user14554
234
edited 1 hour ago
user14554
asked 2 hours ago
user14554user14554
234
asked 2 hours ago
user14554user14554
234
asked 2 hours ago
user14554user14554
234
234
1
$begingroup$
Why should it reduce to that? $x=4$ and $y=2$ has $x neq y$.
$endgroup$
– Randall
2 hours ago
$begingroup$
$4=2^2$? Why wouldn't it? It makes sense that if x and y are the same, they would satisfy this equation. If it doesn't though, then that's what I want to know. According to the algebraic properties of the productlog function, it should simplify in the way I presented, so if it doesn't actually work that way, it begs the question "why not?".
$endgroup$
– user14554
2 hours ago
$begingroup$
No, $2^4=16=4^2$.
$endgroup$
– Randall
2 hours ago
$begingroup$
@Randall But then you still haven't explained why the algebra simplifies yet a counterexample remains. Why should it? Because that's what the algebra says it should do, according to the definition of the productlog which has the properties $W(ze^z)=z$. That property was the basis of deriving that solution in the first place.
$endgroup$
– user14554
2 hours ago
1
$begingroup$
As motivation, you might consider whether the relation $y = arcsin(sin(x))$ is "algebraically equivalent" to $y=x$. (What about when $y=0$ and $x=pi$?) The same issue is at play here: $v = W(u)$ means that $u e^u = v$, but there is not always a unique $u$ with that property, and depending on which one you choose, your claim may or may not be valid.
$endgroup$
– Nate Eldredge
1 hour ago
|
show 7 more comments
1
$begingroup$
Why should it reduce to that? $x=4$ and $y=2$ has $x neq y$.
$endgroup$
– Randall
2 hours ago
$begingroup$
$4=2^2$? Why wouldn't it? It makes sense that if x and y are the same, they would satisfy this equation. If it doesn't though, then that's what I want to know. According to the algebraic properties of the productlog function, it should simplify in the way I presented, so if it doesn't actually work that way, it begs the question "why not?".
$endgroup$
– user14554
2 hours ago
$begingroup$
No, $2^4=16=4^2$.
$endgroup$
– Randall
2 hours ago
$begingroup$
@Randall But then you still haven't explained why the algebra simplifies yet a counterexample remains. Why should it? Because that's what the algebra says it should do, according to the definition of the productlog which has the properties $W(ze^z)=z$. That property was the basis of deriving that solution in the first place.
$endgroup$
– user14554
2 hours ago
1
$begingroup$
As motivation, you might consider whether the relation $y = arcsin(sin(x))$ is "algebraically equivalent" to $y=x$. (What about when $y=0$ and $x=pi$?) The same issue is at play here: $v = W(u)$ means that $u e^u = v$, but there is not always a unique $u$ with that property, and depending on which one you choose, your claim may or may not be valid.
$endgroup$
– Nate Eldredge
1 hour ago
1
1
$begingroup$
Why should it reduce to that? $x=4$ and $y=2$ has $x neq y$.
$endgroup$
– Randall
2 hours ago
$begingroup$
Why should it reduce to that? $x=4$ and $y=2$ has $x neq y$.
$endgroup$
– Randall
2 hours ago
$begingroup$
$4=2^2$? Why wouldn't it? It makes sense that if x and y are the same, they would satisfy this equation. If it doesn't though, then that's what I want to know. According to the algebraic properties of the productlog function, it should simplify in the way I presented, so if it doesn't actually work that way, it begs the question "why not?".
$endgroup$
– user14554
2 hours ago
$begingroup$
$4=2^2$? Why wouldn't it? It makes sense that if x and y are the same, they would satisfy this equation. If it doesn't though, then that's what I want to know. According to the algebraic properties of the productlog function, it should simplify in the way I presented, so if it doesn't actually work that way, it begs the question "why not?".
$endgroup$
– user14554
2 hours ago
$begingroup$
No, $2^4=16=4^2$.
$endgroup$
– Randall
2 hours ago
$begingroup$
No, $2^4=16=4^2$.
$endgroup$
– Randall
2 hours ago
$begingroup$
@Randall But then you still haven't explained why the algebra simplifies yet a counterexample remains. Why should it? Because that's what the algebra says it should do, according to the definition of the productlog which has the properties $W(ze^z)=z$. That property was the basis of deriving that solution in the first place.
$endgroup$
– user14554
2 hours ago
$begingroup$
@Randall But then you still haven't explained why the algebra simplifies yet a counterexample remains. Why should it? Because that's what the algebra says it should do, according to the definition of the productlog which has the properties $W(ze^z)=z$. That property was the basis of deriving that solution in the first place.
$endgroup$
– user14554
2 hours ago
1
1
$begingroup$
As motivation, you might consider whether the relation $y = arcsin(sin(x))$ is "algebraically equivalent" to $y=x$. (What about when $y=0$ and $x=pi$?) The same issue is at play here: $v = W(u)$ means that $u e^u = v$, but there is not always a unique $u$ with that property, and depending on which one you choose, your claim may or may not be valid.
$endgroup$
– Nate Eldredge
1 hour ago
$begingroup$
As motivation, you might consider whether the relation $y = arcsin(sin(x))$ is "algebraically equivalent" to $y=x$. (What about when $y=0$ and $x=pi$?) The same issue is at play here: $v = W(u)$ means that $u e^u = v$, but there is not always a unique $u$ with that property, and depending on which one you choose, your claim may or may not be valid.
$endgroup$
– Nate Eldredge
1 hour ago
|
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 1 hour ago
Eric TowersEric Towers
32.3k22267
$endgroup$
$begingroup$
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
1 hour ago
$begingroup$
+1 for the first parenthetical.
$endgroup$
– Randall
1 hour ago
$begingroup$
@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
1 hour ago
$begingroup$
Ok, one other problem then. Clearly $6^6 = 6^6$ and satisfies $x^y=y^x$ when $x=6, y=6$, right? Well, in your graph, the $x=y$ portion is only valid up to $x=e$, so where did the remaining $x=y$ solutions go? How do I get them back?
$endgroup$
– user14554
1 hour ago
$begingroup$
@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
1 hour ago
|
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 1 hour ago
David G. StorkDavid G. Stork
10.2k21332
$endgroup$
$begingroup$
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
1 hour ago
$begingroup$
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
1 hour ago
$begingroup$
Right, why isn't it $y=x$ all the way.
$endgroup$
– user14554
1 hour ago
$begingroup$
@user14554 I see your question now.
$endgroup$
– Randall
1 hour ago
$begingroup$
user14554 and Randall: There must be a branch cut in the Lambert W function.
$endgroup$
– David G. Stork
1 hour ago
|
show 1 more comment
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
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active
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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 1 hour ago
Eric TowersEric Towers
32.3k22267
$endgroup$
$begingroup$
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
1 hour ago
$begingroup$
+1 for the first parenthetical.
$endgroup$
– Randall
1 hour ago
$begingroup$
@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
1 hour ago
$begingroup$
Ok, one other problem then. Clearly $6^6 = 6^6$ and satisfies $x^y=y^x$ when $x=6, y=6$, right? Well, in your graph, the $x=y$ portion is only valid up to $x=e$, so where did the remaining $x=y$ solutions go? How do I get them back?
$endgroup$
– user14554
1 hour ago
$begingroup$
@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
1 hour ago
|
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 1 hour ago
Eric TowersEric Towers
32.3k22267
$endgroup$
$begingroup$
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
1 hour ago
$begingroup$
+1 for the first parenthetical.
$endgroup$
– Randall
1 hour ago
$begingroup$
@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
1 hour ago
$begingroup$
Ok, one other problem then. Clearly $6^6 = 6^6$ and satisfies $x^y=y^x$ when $x=6, y=6$, right? Well, in your graph, the $x=y$ portion is only valid up to $x=e$, so where did the remaining $x=y$ solutions go? How do I get them back?
$endgroup$
– user14554
1 hour ago
$begingroup$
@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
1 hour ago
|
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 1 hour ago
Eric TowersEric Towers
32.3k22267
$endgroup$
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 1 hour ago
Eric TowersEric Towers
32.3k22267
edited 1 hour ago
answered 1 hour ago
Eric TowersEric Towers
32.3k22267
answered 1 hour ago
Eric TowersEric Towers
32.3k22267
answered 1 hour ago
Eric TowersEric Towers
32.3k22267
32.3k22267
$begingroup$
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
1 hour ago
$begingroup$
+1 for the first parenthetical.
$endgroup$
– Randall
1 hour ago
$begingroup$
@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
1 hour ago
$begingroup$
Ok, one other problem then. Clearly $6^6 = 6^6$ and satisfies $x^y=y^x$ when $x=6, y=6$, right? Well, in your graph, the $x=y$ portion is only valid up to $x=e$, so where did the remaining $x=y$ solutions go? How do I get them back?
$endgroup$
– user14554
1 hour ago
$begingroup$
@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
1 hour ago
|
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
1 hour ago
$begingroup$
+1 for the first parenthetical.
$endgroup$
– Randall
1 hour ago
$begingroup$
@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
1 hour ago
$begingroup$
Ok, one other problem then. Clearly $6^6 = 6^6$ and satisfies $x^y=y^x$ when $x=6, y=6$, right? Well, in your graph, the $x=y$ portion is only valid up to $x=e$, so where did the remaining $x=y$ solutions go? How do I get them back?
$endgroup$
– user14554
1 hour ago
$begingroup$
@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
1 hour ago
$begingroup$
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
1 hour ago
$begingroup$
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
1 hour ago
$begingroup$
+1 for the first parenthetical.
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– Randall
1 hour ago
$begingroup$
+1 for the first parenthetical.
$endgroup$
– Randall
1 hour ago
$begingroup$
@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
1 hour ago
$begingroup$
@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
1 hour ago
$begingroup$
Ok, one other problem then. Clearly $6^6 = 6^6$ and satisfies $x^y=y^x$ when $x=6, y=6$, right? Well, in your graph, the $x=y$ portion is only valid up to $x=e$, so where did the remaining $x=y$ solutions go? How do I get them back?
$endgroup$
– user14554
1 hour ago
$begingroup$
Ok, one other problem then. Clearly $6^6 = 6^6$ and satisfies $x^y=y^x$ when $x=6, y=6$, right? Well, in your graph, the $x=y$ portion is only valid up to $x=e$, so where did the remaining $x=y$ solutions go? How do I get them back?
$endgroup$
– user14554
1 hour ago
$begingroup$
@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
1 hour ago
$begingroup$
@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
1 hour ago
|
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 1 hour ago
David G. StorkDavid G. Stork
10.2k21332
$endgroup$
$begingroup$
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
1 hour ago
$begingroup$
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
1 hour ago
$begingroup$
Right, why isn't it $y=x$ all the way.
$endgroup$
– user14554
1 hour ago
$begingroup$
@user14554 I see your question now.
$endgroup$
– Randall
1 hour ago
$begingroup$
user14554 and Randall: There must be a branch cut in the Lambert W function.
$endgroup$
– David G. Stork
1 hour ago
|
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 1 hour ago
David G. StorkDavid G. Stork
10.2k21332
$endgroup$
$begingroup$
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
1 hour ago
$begingroup$
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
1 hour ago
$begingroup$
Right, why isn't it $y=x$ all the way.
$endgroup$
– user14554
1 hour ago
$begingroup$
@user14554 I see your question now.
$endgroup$
– Randall
1 hour ago
$begingroup$
user14554 and Randall: There must be a branch cut in the Lambert W function.
$endgroup$
– David G. Stork
1 hour ago
|
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 1 hour ago
David G. StorkDavid G. Stork
10.2k21332
$endgroup$
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 1 hour ago
David G. StorkDavid G. Stork
10.2k21332
edited 1 hour ago
answered 1 hour ago
David G. StorkDavid G. Stork
10.2k21332
answered 1 hour ago
David G. StorkDavid G. Stork
10.2k21332
answered 1 hour ago
David G. StorkDavid G. Stork
10.2k21332
10.2k21332
$begingroup$
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
1 hour ago
$begingroup$
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
1 hour ago
$begingroup$
Right, why isn't it $y=x$ all the way.
$endgroup$
– user14554
1 hour ago
$begingroup$
@user14554 I see your question now.
$endgroup$
– Randall
1 hour ago
$begingroup$
user14554 and Randall: There must be a branch cut in the Lambert W function.
$endgroup$
– David G. Stork
1 hour ago
|
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
1 hour ago
$begingroup$
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
1 hour ago
$begingroup$
Right, why isn't it $y=x$ all the way.
$endgroup$
– user14554
1 hour ago
$begingroup$
@user14554 I see your question now.
$endgroup$
– Randall
1 hour ago
$begingroup$
user14554 and Randall: There must be a branch cut in the Lambert W function.
$endgroup$
– David G. Stork
1 hour ago
$begingroup$
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
1 hour ago
$begingroup$
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
1 hour ago
$begingroup$
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
1 hour ago
$begingroup$
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
1 hour ago
$begingroup$
Right, why isn't it $y=x$ all the way.
$endgroup$
– user14554
1 hour ago
$begingroup$
Right, why isn't it $y=x$ all the way.
$endgroup$
– user14554
1 hour ago
$begingroup$
@user14554 I see your question now.
$endgroup$
– Randall
1 hour ago
$begingroup$
@user14554 I see your question now.
$endgroup$
– Randall
1 hour ago
$begingroup$
user14554 and Randall: There must be a branch cut in the Lambert W function.
$endgroup$
– David G. Stork
1 hour ago
$begingroup$
user14554 and Randall: There must be a branch cut in the Lambert W function.
$endgroup$
– David G. Stork
1 hour ago
|
show 1 more comment
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1
$begingroup$
Why should it reduce to that? $x=4$ and $y=2$ has $x neq y$.
$endgroup$
– Randall
2 hours ago
$begingroup$
$4=2^2$? Why wouldn't it? It makes sense that if x and y are the same, they would satisfy this equation. If it doesn't though, then that's what I want to know. According to the algebraic properties of the productlog function, it should simplify in the way I presented, so if it doesn't actually work that way, it begs the question "why not?".
$endgroup$
– user14554
2 hours ago
$begingroup$
No, $2^4=16=4^2$.
$endgroup$
– Randall
2 hours ago
$begingroup$
@Randall But then you still haven't explained why the algebra simplifies yet a counterexample remains. Why should it? Because that's what the algebra says it should do, according to the definition of the productlog which has the properties $W(ze^z)=z$. That property was the basis of deriving that solution in the first place.
$endgroup$
– user14554
2 hours ago
1
$begingroup$
As motivation, you might consider whether the relation $y = arcsin(sin(x))$ is "algebraically equivalent" to $y=x$. (What about when $y=0$ and $x=pi$?) The same issue is at play here: $v = W(u)$ means that $u e^u = v$, but there is not always a unique $u$ with that property, and depending on which one you choose, your claim may or may not be valid.
$endgroup$
– Nate Eldredge
1 hour ago