$egingroup$ If everyone cares, it seems that the two components intersect in ~ the suggest $(e,e)$. $endgroup$
David ns was talking around the two components on the graph of $x^y=y^x$. Are you talking about the two curves or the two components of the first curve? and I used Wolfram. $endgroup$
Let $y=inter-base.netcalRx$, $inter-base.netcalR$ a variable. Then,$$x^inter-base.netcalRx=(inter-base.netcalRx)^x\implies inter-base.netcalRxln x=xln (inter-base.netcalRx)=xln inter-base.netcalR+xln x\implies (inter-base.netcalR-1)ln x=ln inter-base.netcalR$$We can select $inter-base.netcalR=1$, or, if $inter-base.netcalR>1$,$$x=expleft(dfracln inter-base.netcalRr-1 ight)=inter-base.netcalR^1/(inter-base.netcalR-1)$$I.e.,$$y=inter-base.netcalR^inter-base.netcalR/(inter-base.netcalR-1)$$The non-straight graph is the graph because that $y$ because that which $inter-base.netcalR eq 1$.
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Since $inter-base.netcalRx eq dfrac1x$, girlfriend don"t obtain the very same graph.
edited might 8 "14 at 1:57
answered may 8 "14 at 1:45
$egingroup$ Isn't that just a line? $endgroup$
might 8 "14 at 1:48
NicholasStull $xinter-base.netcalR=inter-base.netcalR^inter-base.netcalR/(inter-base.netcalR-1)$. $endgroup$
might 8 "14 in ~ 1:52
NotNotLogical The graph i m sorry the OP is talking about is an implicit plot, and the graph is offered by the over equation. $endgroup$
might 8 "14 at 1:53
SanathDevalapurkar so that's a single point? Or a line? I'm not clear top top what that is. $endgroup$
may 8 "14 at 1:54
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Here is the way I an initial saw it,many year ago:
If $x^y = y^x$with $x > 0$ and $y > 0$,let $r = y/x$,so $y = rx$.We will derive a parameterizationfor $x$ and $y$in terms of $r$.
Then$x^rx = (rx)^x$or,taking the $x^th$ root,$x^r = rx$or$x^r-1 = r$or,if $r
e 1$,$x = r^1/(r-1)$and$y = rx = r^1+1/(r-1)= r^r/(r-1)$.
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Letting $r$ walk throughthe reals $> 1$gives allude $(x, y)$ with$x^y = y^x$.
Another method to look at thisis to think about the curve$v = u^1/u$.For each $v$ such that$1