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 1 hour 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 1 hour ago
user14554user14554
234
$endgroup$
1
$begingroup$
Why should it reduce to that? $x=4$ and $y=2$ has $x neq y$.
$endgroup$
– Randall
1 hour 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
1 hour ago
$begingroup$
No, $2^4=16=4^2$.
$endgroup$
– Randall
1 hour 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
1 hour 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 1 hour 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 1 hour ago
user14554user14554
234
asked 1 hour ago
user14554user14554
234
edited 1 hour ago
user14554
asked 1 hour ago
user14554user14554
234
asked 1 hour ago
user14554user14554
234
asked 1 hour ago
user14554user14554
234
234
1
$begingroup$
Why should it reduce to that? $x=4$ and $y=2$ has $x neq y$.
$endgroup$
– Randall
1 hour 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
1 hour ago
$begingroup$
No, $2^4=16=4^2$.
$endgroup$
– Randall
1 hour 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
1 hour 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
1 hour 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
1 hour ago
$begingroup$
No, $2^4=16=4^2$.
$endgroup$
– Randall
1 hour 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
1 hour 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
1 hour ago
$begingroup$
Why should it reduce to that? $x=4$ and $y=2$ has $x neq y$.
$endgroup$
– Randall
1 hour 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
1 hour 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
1 hour ago
$begingroup$
No, $2^4=16=4^2$.
$endgroup$
– Randall
1 hour ago
$begingroup$
No, $2^4=16=4^2$.
$endgroup$
– Randall
1 hour 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
1 hour 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
1 hour 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
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3072828%2fwhy-didnt-they-simplify-this%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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 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.
$endgroup$
– 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
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3072828%2fwhy-didnt-they-simplify-this%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
Why should it reduce to that? $x=4$ and $y=2$ has $x neq y$.
$endgroup$
– Randall
1 hour 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
1 hour ago
$begingroup$
No, $2^4=16=4^2$.
$endgroup$
– Randall
1 hour 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
1 hour 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