Why didn't they simplify $x^y=y^x$ to $x=y$?












8












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 7




    $begingroup$
    Why should it reduce to that? $x=4$ and $y=2$ has $x neq y$.
    $endgroup$
    – Randall
    2 days ago






  • 3




    $begingroup$
    No, $2^4=16=4^2$.
    $endgroup$
    – Randall
    2 days ago






  • 7




    $begingroup$
    I'm just confused why the solution "should" simplify to $x=y$ when there are solutions that do not satisfy $x = y$.
    $endgroup$
    – Randall
    2 days ago






  • 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
    2 days ago






  • 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
    2 days ago


















8












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 7




    $begingroup$
    Why should it reduce to that? $x=4$ and $y=2$ has $x neq y$.
    $endgroup$
    – Randall
    2 days ago






  • 3




    $begingroup$
    No, $2^4=16=4^2$.
    $endgroup$
    – Randall
    2 days ago






  • 7




    $begingroup$
    I'm just confused why the solution "should" simplify to $x=y$ when there are solutions that do not satisfy $x = y$.
    $endgroup$
    – Randall
    2 days ago






  • 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
    2 days ago






  • 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
    2 days ago
















8












8








8


2



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Daniel R

2,46832035




2,46832035










asked 2 days ago









user14554user14554

465




465








  • 7




    $begingroup$
    Why should it reduce to that? $x=4$ and $y=2$ has $x neq y$.
    $endgroup$
    – Randall
    2 days ago






  • 3




    $begingroup$
    No, $2^4=16=4^2$.
    $endgroup$
    – Randall
    2 days ago






  • 7




    $begingroup$
    I'm just confused why the solution "should" simplify to $x=y$ when there are solutions that do not satisfy $x = y$.
    $endgroup$
    – Randall
    2 days ago






  • 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
    2 days ago






  • 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
    2 days ago
















  • 7




    $begingroup$
    Why should it reduce to that? $x=4$ and $y=2$ has $x neq y$.
    $endgroup$
    – Randall
    2 days ago






  • 3




    $begingroup$
    No, $2^4=16=4^2$.
    $endgroup$
    – Randall
    2 days ago






  • 7




    $begingroup$
    I'm just confused why the solution "should" simplify to $x=y$ when there are solutions that do not satisfy $x = y$.
    $endgroup$
    – Randall
    2 days ago






  • 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
    2 days ago






  • 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
    2 days ago










7




7




$begingroup$
Why should it reduce to that? $x=4$ and $y=2$ has $x neq y$.
$endgroup$
– Randall
2 days ago




$begingroup$
Why should it reduce to that? $x=4$ and $y=2$ has $x neq y$.
$endgroup$
– Randall
2 days ago




3




3




$begingroup$
No, $2^4=16=4^2$.
$endgroup$
– Randall
2 days ago




$begingroup$
No, $2^4=16=4^2$.
$endgroup$
– Randall
2 days ago




7




7




$begingroup$
I'm just confused why the solution "should" simplify to $x=y$ when there are solutions that do not satisfy $x = y$.
$endgroup$
– Randall
2 days ago




$begingroup$
I'm just confused why the solution "should" simplify to $x=y$ when there are solutions that do not satisfy $x = y$.
$endgroup$
– Randall
2 days ago




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
2 days ago




$begingroup$
Anyway, to potentially answer your question, your algebra moves are invalid if $x$ is negative, and there are solutions with negative $x$.
$endgroup$
– Randall
2 days ago




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
2 days ago






$begingroup$
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
2 days ago












2 Answers
2






active

oldest

votes


















11












$begingroup$

The Lambert $W$ function is not single-valued for negative arguments.



enter image description here



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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
    2 days ago












  • $begingroup$
    +1 for the first parenthetical.
    $endgroup$
    – Randall
    2 days 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
    2 days ago








  • 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
    2 days ago








  • 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
    2 days ago



















1












$begingroup$

The solution is:



$$y = -frac{x Wleft(-frac{log (x)}{x}right)}{log (x)}$$



which has the following form:



enter image description here



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).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
    2 days 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
    2 days ago












  • $begingroup$
    Right, why isn't it $y=x$ all the way.
    $endgroup$
    – user14554
    2 days ago










  • $begingroup$
    @user14554 I see your question now.
    $endgroup$
    – Randall
    2 days ago










  • $begingroup$
    user14554 and Randall: There must be a branch cut in the Lambert W function.
    $endgroup$
    – David G. Stork
    2 days ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









11












$begingroup$

The Lambert $W$ function is not single-valued for negative arguments.



enter image description here



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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
    2 days ago












  • $begingroup$
    +1 for the first parenthetical.
    $endgroup$
    – Randall
    2 days 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
    2 days ago








  • 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
    2 days ago








  • 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
    2 days ago
















11












$begingroup$

The Lambert $W$ function is not single-valued for negative arguments.



enter image description here



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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
    2 days ago












  • $begingroup$
    +1 for the first parenthetical.
    $endgroup$
    – Randall
    2 days 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
    2 days ago








  • 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
    2 days ago








  • 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
    2 days ago














11












11








11





$begingroup$

The Lambert $W$ function is not single-valued for negative arguments.



enter image description here



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.






share|cite|improve this answer











$endgroup$



The Lambert $W$ function is not single-valued for negative arguments.



enter image description here



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 days ago

























answered 2 days ago









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
    2 days ago












  • $begingroup$
    +1 for the first parenthetical.
    $endgroup$
    – Randall
    2 days 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
    2 days ago








  • 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
    2 days ago








  • 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
    2 days 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
    2 days ago












  • $begingroup$
    +1 for the first parenthetical.
    $endgroup$
    – Randall
    2 days 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
    2 days ago








  • 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
    2 days ago








  • 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
    2 days 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
2 days 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
2 days ago














$begingroup$
+1 for the first parenthetical.
$endgroup$
– Randall
2 days ago




$begingroup$
+1 for the first parenthetical.
$endgroup$
– Randall
2 days 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
2 days 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
2 days ago






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
2 days 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
2 days ago






1




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
2 days ago




$begingroup$
@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
2 days ago











1












$begingroup$

The solution is:



$$y = -frac{x Wleft(-frac{log (x)}{x}right)}{log (x)}$$



which has the following form:



enter image description here



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).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
    2 days 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
    2 days ago












  • $begingroup$
    Right, why isn't it $y=x$ all the way.
    $endgroup$
    – user14554
    2 days ago










  • $begingroup$
    @user14554 I see your question now.
    $endgroup$
    – Randall
    2 days ago










  • $begingroup$
    user14554 and Randall: There must be a branch cut in the Lambert W function.
    $endgroup$
    – David G. Stork
    2 days ago
















1












$begingroup$

The solution is:



$$y = -frac{x Wleft(-frac{log (x)}{x}right)}{log (x)}$$



which has the following form:



enter image description here



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).






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
    2 days 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
    2 days ago












  • $begingroup$
    Right, why isn't it $y=x$ all the way.
    $endgroup$
    – user14554
    2 days ago










  • $begingroup$
    @user14554 I see your question now.
    $endgroup$
    – Randall
    2 days ago










  • $begingroup$
    user14554 and Randall: There must be a branch cut in the Lambert W function.
    $endgroup$
    – David G. Stork
    2 days ago














1












1








1





$begingroup$

The solution is:



$$y = -frac{x Wleft(-frac{log (x)}{x}right)}{log (x)}$$



which has the following form:



enter image description here



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).






share|cite|improve this answer











$endgroup$



The solution is:



$$y = -frac{x Wleft(-frac{log (x)}{x}right)}{log (x)}$$



which has the following form:



enter image description here



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).







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 days ago

























answered 2 days ago









David G. StorkDavid G. Stork

10.4k31332




10.4k31332












  • $begingroup$
    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
    2 days 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
    2 days ago












  • $begingroup$
    Right, why isn't it $y=x$ all the way.
    $endgroup$
    – user14554
    2 days ago










  • $begingroup$
    @user14554 I see your question now.
    $endgroup$
    – Randall
    2 days ago










  • $begingroup$
    user14554 and Randall: There must be a branch cut in the Lambert W function.
    $endgroup$
    – David G. Stork
    2 days 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
    2 days 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
    2 days ago












  • $begingroup$
    Right, why isn't it $y=x$ all the way.
    $endgroup$
    – user14554
    2 days ago










  • $begingroup$
    @user14554 I see your question now.
    $endgroup$
    – Randall
    2 days ago










  • $begingroup$
    user14554 and Randall: There must be a branch cut in the Lambert W function.
    $endgroup$
    – David G. Stork
    2 days 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
2 days 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
2 days 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
2 days 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
2 days ago














$begingroup$
Right, why isn't it $y=x$ all the way.
$endgroup$
– user14554
2 days ago




$begingroup$
Right, why isn't it $y=x$ all the way.
$endgroup$
– user14554
2 days ago












$begingroup$
@user14554 I see your question now.
$endgroup$
– Randall
2 days ago




$begingroup$
@user14554 I see your question now.
$endgroup$
– Randall
2 days ago












$begingroup$
user14554 and Randall: There must be a branch cut in the Lambert W function.
$endgroup$
– David G. Stork
2 days ago




$begingroup$
user14554 and Randall: There must be a branch cut in the Lambert W function.
$endgroup$
– David G. Stork
2 days ago


















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