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Where did I make a mistake in simplifying this?
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So I found general solution to $x^y=y^x$ for positive values on the Internet via Lambert $W$ function and it goes like this: $$y=frac-xcdot Wleft(frac-log(x)xright)log(x).$$ Now there is an identity for Lambert $W$ function $$Wleft(frac-log(x)xright)=log(x^-1).$$ Using this identity you can simplify the solution to $y=x$. Which certainly is not general solution, so I think I made a mistake somewhere here when used this identity, maybe it is something connected with the values for which the identity is true. So where is the mistake ?
lambert-w
edited 2 days ago
TheSimpliFire
9,20751751
asked 2 days ago
îріù ïрþш
979412
add a comment |Â
up vote
2
down vote
favorite
So I found general solution to $x^y=y^x$ for positive values on the Internet via Lambert $W$ function and it goes like this: $$y=frac-xcdot Wleft(frac-log(x)xright)log(x).$$ Now there is an identity for Lambert $W$ function $$Wleft(frac-log(x)xright)=log(x^-1).$$ Using this identity you can simplify the solution to $y=x$. Which certainly is not general solution, so I think I made a mistake somewhere here when used this identity, maybe it is something connected with the values for which the identity is true. So where is the mistake ?
lambert-w
edited 2 days ago
TheSimpliFire
9,20751751
asked 2 days ago
îріù ïрþш
979412
Lambert W function is defined as an inverse of function that is not injective, so there are actually many different Lambert W functions (branches), so you need to be careful which one are you using...
– Sil
2 days ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
So I found general solution to $x^y=y^x$ for positive values on the Internet via Lambert $W$ function and it goes like this: $$y=frac-xcdot Wleft(frac-log(x)xright)log(x).$$ Now there is an identity for Lambert $W$ function $$Wleft(frac-log(x)xright)=log(x^-1).$$ Using this identity you can simplify the solution to $y=x$. Which certainly is not general solution, so I think I made a mistake somewhere here when used this identity, maybe it is something connected with the values for which the identity is true. So where is the mistake ?
lambert-w
edited 2 days ago
TheSimpliFire
9,20751751
asked 2 days ago
îріù ïрþш
979412
So I found general solution to $x^y=y^x$ for positive values on the Internet via Lambert $W$ function and it goes like this: $$y=frac-xcdot Wleft(frac-log(x)xright)log(x).$$ Now there is an identity for Lambert $W$ function $$Wleft(frac-log(x)xright)=log(x^-1).$$ Using this identity you can simplify the solution to $y=x$. Which certainly is not general solution, so I think I made a mistake somewhere here when used this identity, maybe it is something connected with the values for which the identity is true. So where is the mistake ?
lambert-w
edited 2 days ago
TheSimpliFire
9,20751751
asked 2 days ago
îріù ïрþш
979412
edited 2 days ago
TheSimpliFire
9,20751751
edited 2 days ago
TheSimpliFire
9,20751751
edited 2 days ago
TheSimpliFire
9,20751751
9,20751751
asked 2 days ago
îріù ïрþш
979412
asked 2 days ago
îріù ïрþш
979412
asked 2 days ago
îріù ïрþш
979412
979412
Lambert W function is defined as an inverse of function that is not injective, so there are actually many different Lambert W functions (branches), so you need to be careful which one are you using...
– Sil
2 days ago
add a comment |Â
Lambert W function is defined as an inverse of function that is not injective, so there are actually many different Lambert W functions (branches), so you need to be careful which one are you using...
– Sil
2 days ago
Lambert W function is defined as an inverse of function that is not injective, so there are actually many different Lambert W functions (branches), so you need to be careful which one are you using...
– Sil
2 days ago
Lambert W function is defined as an inverse of function that is not injective, so there are actually many different Lambert W functions (branches), so you need to be careful which one are you using...
– Sil
2 days ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
Lambert $W$ function is defined as an inverse function to $f(x)=xe^x$, which is not injective on $mathbbR$. So actually there are multiple different Lambert $W$ functions (branches), or, you can view this as $W(x)$ being a multi-valued function.
Specifically we have two branches (well in complex numbers there are actually infinitely many, but that is another story), usually denoted $W_0(x)$ defined for $(-1/e,infty)$ (also called principal branch, corresponding to the inverse on $(-1,infty)$), and $W_-1(x)$ defined for $(-1/e,0)$ (corresponding to the inverse on $(-infty,-1)$).
Now both branches satisfy the first equation you came to, so far no problem there. Problem becomes with the identity $Wleft(frac-log xxright)=log(x^-1)$, which is true for each of the branches only with certain restrictions. Specifically, $$W_0left(frac-log xxright)=log(x^-1)$$ for $x in (0,e]$, and similarly $$W_-1left(frac-log xxright)=log(x^-1)$$ for $x in (e,infty)$.
How does that apply to your problem? Well it just means that by applying the identity above, you have only considered subset of solutions. To get all solutions, you also need to add $y=frac-xlog xcdot W_0left(frac-log xxright)$ for $x>e$, and similarly $y=frac-xlog xcdot W_-1left(frac-log xxright)$ for $x leq e$. And indeed, this covers some previously missing cases, for example with $x=4$ we obtain $y=frac-4log4 cdot W_0left(frac-log 44right)=2$.
answered 2 days ago
Sil
5,06821342
Why can't we look at the principal branch for all positive numbers ? Because $-log(x)/x>=-1/e$ for every $x$, so can't we look at principal branch for every $x$ ?
– Ã®Ñ€Ñ–ù ïрþш
2 days ago
I mean why can't we look at $W_0(-log(x)/x)$ for all positive $x$.
– Ã®Ñ€Ñ–ù ïрþш
2 days ago
@îріùïрþш You can use $W_0$ for $x in (-1/e, infty)$, but the identity is not satisfied for $x>e$. For example $W_0(frac-log 1010)approx -.3157508629$, but $log(1/10)approx -2.302585093$. On the other hand $W_-1(frac-log 1010)approx -2.302585093$ ...
– Sil
2 days ago
So for $x>e$ or $x<=e$, because in the end of the post you seem to write opposite.
– Ã®Ñ€Ñ–ù ïрþш
2 days ago
@îріùïрþш Yes because at the end of post we are considering remaining solutions, which are not handled by the identity, hence negated inequality.
– Sil
2 days ago
 |Â
show 1 more comment
up vote
1
down vote
Hint.
Attached a plot for $x^y = y^x$ (blue) with the superimposed plot for $y = -x Wleft(-fracln xxright)/ln x$ (dashed red)
answered 2 days ago
Cesareo
5,4812412
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Lambert $W$ function is defined as an inverse function to $f(x)=xe^x$, which is not injective on $mathbbR$. So actually there are multiple different Lambert $W$ functions (branches), or, you can view this as $W(x)$ being a multi-valued function.
Specifically we have two branches (well in complex numbers there are actually infinitely many, but that is another story), usually denoted $W_0(x)$ defined for $(-1/e,infty)$ (also called principal branch, corresponding to the inverse on $(-1,infty)$), and $W_-1(x)$ defined for $(-1/e,0)$ (corresponding to the inverse on $(-infty,-1)$).
Now both branches satisfy the first equation you came to, so far no problem there. Problem becomes with the identity $Wleft(frac-log xxright)=log(x^-1)$, which is true for each of the branches only with certain restrictions. Specifically, $$W_0left(frac-log xxright)=log(x^-1)$$ for $x in (0,e]$, and similarly $$W_-1left(frac-log xxright)=log(x^-1)$$ for $x in (e,infty)$.
How does that apply to your problem? Well it just means that by applying the identity above, you have only considered subset of solutions. To get all solutions, you also need to add $y=frac-xlog xcdot W_0left(frac-log xxright)$ for $x>e$, and similarly $y=frac-xlog xcdot W_-1left(frac-log xxright)$ for $x leq e$. And indeed, this covers some previously missing cases, for example with $x=4$ we obtain $y=frac-4log4 cdot W_0left(frac-log 44right)=2$.
answered 2 days ago
Sil
5,06821342
Why can't we look at the principal branch for all positive numbers ? Because $-log(x)/x>=-1/e$ for every $x$, so can't we look at principal branch for every $x$ ?
– Ã®Ñ€Ñ–ù ïрþш
2 days ago
I mean why can't we look at $W_0(-log(x)/x)$ for all positive $x$.
– Ã®Ñ€Ñ–ù ïрþш
2 days ago
@îріùïрþш You can use $W_0$ for $x in (-1/e, infty)$, but the identity is not satisfied for $x>e$. For example $W_0(frac-log 1010)approx -.3157508629$, but $log(1/10)approx -2.302585093$. On the other hand $W_-1(frac-log 1010)approx -2.302585093$ ...
– Sil
2 days ago
So for $x>e$ or $x<=e$, because in the end of the post you seem to write opposite.
– Ã®Ñ€Ñ–ù ïрþш
2 days ago
@îріùïрþш Yes because at the end of post we are considering remaining solutions, which are not handled by the identity, hence negated inequality.
– Sil
2 days ago
 |Â
show 1 more comment
up vote
2
down vote
accepted
Lambert $W$ function is defined as an inverse function to $f(x)=xe^x$, which is not injective on $mathbbR$. So actually there are multiple different Lambert $W$ functions (branches), or, you can view this as $W(x)$ being a multi-valued function.
Specifically we have two branches (well in complex numbers there are actually infinitely many, but that is another story), usually denoted $W_0(x)$ defined for $(-1/e,infty)$ (also called principal branch, corresponding to the inverse on $(-1,infty)$), and $W_-1(x)$ defined for $(-1/e,0)$ (corresponding to the inverse on $(-infty,-1)$).
Now both branches satisfy the first equation you came to, so far no problem there. Problem becomes with the identity $Wleft(frac-log xxright)=log(x^-1)$, which is true for each of the branches only with certain restrictions. Specifically, $$W_0left(frac-log xxright)=log(x^-1)$$ for $x in (0,e]$, and similarly $$W_-1left(frac-log xxright)=log(x^-1)$$ for $x in (e,infty)$.
How does that apply to your problem? Well it just means that by applying the identity above, you have only considered subset of solutions. To get all solutions, you also need to add $y=frac-xlog xcdot W_0left(frac-log xxright)$ for $x>e$, and similarly $y=frac-xlog xcdot W_-1left(frac-log xxright)$ for $x leq e$. And indeed, this covers some previously missing cases, for example with $x=4$ we obtain $y=frac-4log4 cdot W_0left(frac-log 44right)=2$.
answered 2 days ago
Sil
5,06821342
Why can't we look at the principal branch for all positive numbers ? Because $-log(x)/x>=-1/e$ for every $x$, so can't we look at principal branch for every $x$ ?
– Ã®Ñ€Ñ–ù ïрþш
2 days ago
I mean why can't we look at $W_0(-log(x)/x)$ for all positive $x$.
– Ã®Ñ€Ñ–ù ïрþш
2 days ago
@îріùïрþш You can use $W_0$ for $x in (-1/e, infty)$, but the identity is not satisfied for $x>e$. For example $W_0(frac-log 1010)approx -.3157508629$, but $log(1/10)approx -2.302585093$. On the other hand $W_-1(frac-log 1010)approx -2.302585093$ ...
– Sil
2 days ago
So for $x>e$ or $x<=e$, because in the end of the post you seem to write opposite.
– Ã®Ñ€Ñ–ù ïрþш
2 days ago
@îріùïрþш Yes because at the end of post we are considering remaining solutions, which are not handled by the identity, hence negated inequality.
– Sil
2 days ago
 |Â
show 1 more comment
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Lambert $W$ function is defined as an inverse function to $f(x)=xe^x$, which is not injective on $mathbbR$. So actually there are multiple different Lambert $W$ functions (branches), or, you can view this as $W(x)$ being a multi-valued function.
Specifically we have two branches (well in complex numbers there are actually infinitely many, but that is another story), usually denoted $W_0(x)$ defined for $(-1/e,infty)$ (also called principal branch, corresponding to the inverse on $(-1,infty)$), and $W_-1(x)$ defined for $(-1/e,0)$ (corresponding to the inverse on $(-infty,-1)$).
Now both branches satisfy the first equation you came to, so far no problem there. Problem becomes with the identity $Wleft(frac-log xxright)=log(x^-1)$, which is true for each of the branches only with certain restrictions. Specifically, $$W_0left(frac-log xxright)=log(x^-1)$$ for $x in (0,e]$, and similarly $$W_-1left(frac-log xxright)=log(x^-1)$$ for $x in (e,infty)$.
How does that apply to your problem? Well it just means that by applying the identity above, you have only considered subset of solutions. To get all solutions, you also need to add $y=frac-xlog xcdot W_0left(frac-log xxright)$ for $x>e$, and similarly $y=frac-xlog xcdot W_-1left(frac-log xxright)$ for $x leq e$. And indeed, this covers some previously missing cases, for example with $x=4$ we obtain $y=frac-4log4 cdot W_0left(frac-log 44right)=2$.
answered 2 days ago
Sil
5,06821342
Lambert $W$ function is defined as an inverse function to $f(x)=xe^x$, which is not injective on $mathbbR$. So actually there are multiple different Lambert $W$ functions (branches), or, you can view this as $W(x)$ being a multi-valued function.
Specifically we have two branches (well in complex numbers there are actually infinitely many, but that is another story), usually denoted $W_0(x)$ defined for $(-1/e,infty)$ (also called principal branch, corresponding to the inverse on $(-1,infty)$), and $W_-1(x)$ defined for $(-1/e,0)$ (corresponding to the inverse on $(-infty,-1)$).
Now both branches satisfy the first equation you came to, so far no problem there. Problem becomes with the identity $Wleft(frac-log xxright)=log(x^-1)$, which is true for each of the branches only with certain restrictions. Specifically, $$W_0left(frac-log xxright)=log(x^-1)$$ for $x in (0,e]$, and similarly $$W_-1left(frac-log xxright)=log(x^-1)$$ for $x in (e,infty)$.
How does that apply to your problem? Well it just means that by applying the identity above, you have only considered subset of solutions. To get all solutions, you also need to add $y=frac-xlog xcdot W_0left(frac-log xxright)$ for $x>e$, and similarly $y=frac-xlog xcdot W_-1left(frac-log xxright)$ for $x leq e$. And indeed, this covers some previously missing cases, for example with $x=4$ we obtain $y=frac-4log4 cdot W_0left(frac-log 44right)=2$.
answered 2 days ago
Sil
5,06821342
edited 2 days ago
answered 2 days ago
Sil
5,06821342
answered 2 days ago
Sil
5,06821342
answered 2 days ago
Sil
5,06821342
5,06821342
Why can't we look at the principal branch for all positive numbers ? Because $-log(x)/x>=-1/e$ for every $x$, so can't we look at principal branch for every $x$ ?
– Ã®Ñ€Ñ–ù ïрþш
2 days ago
I mean why can't we look at $W_0(-log(x)/x)$ for all positive $x$.
– Ã®Ñ€Ñ–ù ïрþш
2 days ago
@îріùïрþш You can use $W_0$ for $x in (-1/e, infty)$, but the identity is not satisfied for $x>e$. For example $W_0(frac-log 1010)approx -.3157508629$, but $log(1/10)approx -2.302585093$. On the other hand $W_-1(frac-log 1010)approx -2.302585093$ ...
– Sil
2 days ago
So for $x>e$ or $x<=e$, because in the end of the post you seem to write opposite.
– Ã®Ñ€Ñ–ù ïрþш
2 days ago
@îріùïрþш Yes because at the end of post we are considering remaining solutions, which are not handled by the identity, hence negated inequality.
– Sil
2 days ago
 |Â
show 1 more comment
Why can't we look at the principal branch for all positive numbers ? Because $-log(x)/x>=-1/e$ for every $x$, so can't we look at principal branch for every $x$ ?
– Ã®Ñ€Ñ–ù ïрþш
2 days ago
I mean why can't we look at $W_0(-log(x)/x)$ for all positive $x$.
– Ã®Ñ€Ñ–ù ïрþш
2 days ago
@îріùïрþш You can use $W_0$ for $x in (-1/e, infty)$, but the identity is not satisfied for $x>e$. For example $W_0(frac-log 1010)approx -.3157508629$, but $log(1/10)approx -2.302585093$. On the other hand $W_-1(frac-log 1010)approx -2.302585093$ ...
– Sil
2 days ago
So for $x>e$ or $x<=e$, because in the end of the post you seem to write opposite.
– Ã®Ñ€Ñ–ù ïрþш
2 days ago
@îріùïрþш Yes because at the end of post we are considering remaining solutions, which are not handled by the identity, hence negated inequality.
– Sil
2 days ago
Why can't we look at the principal branch for all positive numbers ? Because $-log(x)/x>=-1/e$ for every $x$, so can't we look at principal branch for every $x$ ?
– Ã®Ñ€Ñ–ù ïрþш
2 days ago
Why can't we look at the principal branch for all positive numbers ? Because $-log(x)/x>=-1/e$ for every $x$, so can't we look at principal branch for every $x$ ?
– Ã®Ñ€Ñ–ù ïрþш
2 days ago
I mean why can't we look at $W_0(-log(x)/x)$ for all positive $x$.
– Ã®Ñ€Ñ–ù ïрþш
2 days ago
I mean why can't we look at $W_0(-log(x)/x)$ for all positive $x$.
– Ã®Ñ€Ñ–ù ïрþш
2 days ago
@îріùïрþш You can use $W_0$ for $x in (-1/e, infty)$, but the identity is not satisfied for $x>e$. For example $W_0(frac-log 1010)approx -.3157508629$, but $log(1/10)approx -2.302585093$. On the other hand $W_-1(frac-log 1010)approx -2.302585093$ ...
– Sil
2 days ago
@îріùïрþш You can use $W_0$ for $x in (-1/e, infty)$, but the identity is not satisfied for $x>e$. For example $W_0(frac-log 1010)approx -.3157508629$, but $log(1/10)approx -2.302585093$. On the other hand $W_-1(frac-log 1010)approx -2.302585093$ ...
– Sil
2 days ago
So for $x>e$ or $x<=e$, because in the end of the post you seem to write opposite.
– Ã®Ñ€Ñ–ù ïрþш
2 days ago
So for $x>e$ or $x<=e$, because in the end of the post you seem to write opposite.
– Ã®Ñ€Ñ–ù ïрþш
2 days ago
@îріùïрþш Yes because at the end of post we are considering remaining solutions, which are not handled by the identity, hence negated inequality.
– Sil
2 days ago
@îріùïрþш Yes because at the end of post we are considering remaining solutions, which are not handled by the identity, hence negated inequality.
– Sil
2 days ago
 |Â
show 1 more comment
up vote
1
down vote
Hint.
Attached a plot for $x^y = y^x$ (blue) with the superimposed plot for $y = -x Wleft(-fracln xxright)/ln x$ (dashed red)
answered 2 days ago
Cesareo
5,4812412
add a comment |Â
up vote
1
down vote
Hint.
Attached a plot for $x^y = y^x$ (blue) with the superimposed plot for $y = -x Wleft(-fracln xxright)/ln x$ (dashed red)
answered 2 days ago
Cesareo
5,4812412
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint.
Attached a plot for $x^y = y^x$ (blue) with the superimposed plot for $y = -x Wleft(-fracln xxright)/ln x$ (dashed red)
answered 2 days ago
Cesareo
5,4812412
Hint.
Attached a plot for $x^y = y^x$ (blue) with the superimposed plot for $y = -x Wleft(-fracln xxright)/ln x$ (dashed red)
answered 2 days ago
Cesareo
5,4812412
answered 2 days ago
Cesareo
5,4812412
answered 2 days ago
Cesareo
5,4812412
answered 2 days ago
Cesareo
5,4812412
5,4812412
add a comment |Â
add a comment |Â
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Lambert W function is defined as an inverse of function that is not injective, so there are actually many different Lambert W functions (branches), so you need to be careful which one are you using...
– Sil
2 days ago