[seqfan] Re: 2(x - 1)^x = x^x

[neil greubel]

The exact solution to x^x = 2*(1-x)^x is
x = 1 / (1 + W(ln2/2)/ln2) , where W(x) is the Lambert W-function.

Proof: By taking the ln of both sides of the equation
x lnx = ln2 + x ln(1-x)
leads to
- (ln2/x) = ln((1-x)/x)
exp(- ln2/x) = (1-x)/x
(1/2) = (1/x - 1) * exp( ln2 / x) * (1/2)
(ln2 / 2) = ln2 * (1/x - 1) * exp(ln2 / x - ln2)
=>
ln2 *(1/x - 1) = W(ln2 / 2)
1/x = 1 + W(ln2 / 2) / ln2
and finally
x = ln2 / (ln2 + W(ln2/2)) = 0.72289208118314779553.....

On Wed, Feb 28, 2018 at 11:17 PM, <rgwv at rgwv.com> wrote:

> 0.722892081183147795532365351219771674355877736768641720125436
> 5151991334348752469431309465384425833668792370029595892746086...
>
> -----Original Message-----
> From: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of WALTER
> KEHOWSKI
> Sent: Wednesday, 28 February, 2018 6:18 PM
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>; юрий
> герасимов <2stepan at rambler.ru>
> Subject: [seqfan] Re: 2(x - 1)^x = x^x
>
> The equation 2(1-x)^x = x^x has the zero x=0.7228920812. HTH.
>
> > On February 26, 2018 at 2:32 AM юрий герасимов <2stepan at rambler.ru>
> wrote:
> >
> >
> > Dear SeqFans.
> > Help find the first members of this sequence:
> > Decimal expansion of the real positive solution to 2(x - 1)^x = x^x.
> > Thanks. JSG.
> >
> > --
> > Seqfan Mailing list - http://list.seqfan.eu/
>
> --
> Seqfan Mailing list - http://list.seqfan.eu/
>
>
> --
> Seqfan Mailing list - http://list.seqfan.eu/
>