Re: How to show x=ln(x) has no solution ?
- From: "Ioannis" <morpheus@xxxxxxxxxxxx>
- Date: Thu, 14 Jul 2005 21:52:20 +0300
Ο "Ronald Bruck" <bruck@xxxxxxxxxxxxxxxxx> έγραψε στο μήνυμα
news:140720051138517716%bruck@xxxxxxxxxxxxxxxxxxxx
>
> In article <1121250101.650172.312790@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
> Herb <fuzzykyh@xxxxxxxx> wrote:
>
> > Clearly, it has no solution for x<=1.
> > but
> > I don't know how to show it also has no solution for x>1.
> > If it has k(>1) as a solution,
> > Then, Is there any contradiction ?
> >
> > Thanks for any help...
> >
>
> Others have shown you the canonical way to think about this.
>
> I'll add just one thought: it DOES have solutions for complex z.
For complex z the equation is equivalent to e^z=z, which can be solved via
the Lambert W function:
z*e^{-z}=1 <=>
-z*e^{-z}=-1 <=>
-z=W(k,-1), k in Z
z=-W(k,-1), k in Z.
where k denotes the complex branch of W chosen.
So in effect there are infinitely many complex solutions.
> --Ron Bruck
--
I. N. Galidakis
http://users.forthnet.gr/ath/jgal/
Eventually, _everything_ is understandable
.
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