Re: Solve y^x = x^y ?

From: David W. Cantrell (DWCantrell_at_sigmaxi.org)
Date: 11/13/04 

Date: 13 Nov 2004 22:23:54 GMT

Ioannis <morpheus@olympus.mons> wrote:
> David W. Cantrell wrote:
> > Ioannis <morpheus@olympus.mons> wrote:
> >
> >>David W. Cantrell wrote:
> >>[snip]
> >>
> >>>>I *think* that when BOTH x and y are > 0, the only solutions seem to
> >>>>me to be the ones exhibited by the Lemma above, so it was a
> >>>>misunderstanding of the term "all".
> >>>
> >>>Does the lemma cover the trivial solutions: x = y ?
> >>
> >>Obviously not.
> >
> >
> > I wasn't being flippant when I asked that, BTW. The trivial solutions
> > when both x and y are positive can be written in terms of the Lambert W
> > function, as I did in my article.
> >
> > When 0 < x <= e, y = LambertW(0, -log(x)/x)/(-log(x)/x)
> >
> > simplifies to y = x, as does
> >
> > y = LambertW(-1, -log(x)/x)/(-log(x)/x) when x >= e.
> >
> > So, not remembering your lemma well when I asked the question above,
> > I wasn't 100% certain that it didn't also yield those trivial
> > solutions.
>
> The only trivial case I see in my Lemma is the case w=e^(1/e). In this
> case,
>
> x=W(0,-log(w))/(-log(w)),
> y=W(-1,-log(w))/(-log(w))
>
> give the same value, since W(-1,-1/e)=W(-1/e), so
>
> x=y=e.
>
> Also, your expressions for x and y seem to be somewhat different than
> mine. How do you get the "-log(x)/x" factor?

It's different, of course, because I had solved for y explicitly as a
function of x. (In so doing, the quantity -log(x)/x arises naturally.)
You, OTOH, had expressed x and y in terms of a parameter.

> I checked your expression with x=e, i.e. "LambertW(0,
> -log(x)/x)/(-log(x)/x)" and it gives e^(1/e). Mine gives e.

Huh? Mine also gives y = e when x = e.

> Of course for the trivial cases (where x=y) it doesn't matter, since in
> this case, I say e^e=e^e and you say e^(1/e)^(e^(1/e))=e^(1/e)^(e^(1/e)),
>
> but for non-trivial cases it matters. I.e. if your expressions evaluate
> to something different from that which my expressions evaluate, then you
> are missing all the solutions in my Lemma.

AFAIK, I'm not missing any real solutions.

David

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