x^(x^(x^(x^x^...)))....)))) = 2 - sci.math

Message from discussion x^(x^(x^(x^x^...)))....)))) = 2

Denis Feldmann

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More options May 16, 1:39 pm

Newsgroups: sci.math

From: Denis Feldmann <feldmann.denis.asuppri...@neuf.fr>

Date: Fri, 16 May 2008 19:39:35 +0200

Local: Fri, May 16 2008 1:39 pm

Subject: Re: x^(x^(x^(x^x^...)))....)))) = 2

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David W. Cantrell a écrit :

> Sujit <sujit.gu...@gmail.com> wrote:
>> Hello all,

>> What is solution to
>> x^(x^(x^(x^x^...)))....)))) = 2?
>> ( That is x raised to x raised to x ..infinite times... = 2)

>> One immediate thing comes to mind is:
>> it is same as x^(the whole thing which is 2) =2
>> so x = sqrt 2

> That is correct.

>> but if we consider,
>> x^(x^(x^(x^x^...)))....)))) = 4
>> still we get (if apply same logic)
>> x=sqrt 2
>> but then x^(x^(x^(x^x^...)))....)))) is 2 or 4, if x=sqrt 2?

> It's only 2.

> I think there is a result due to Euler that the equation
> x^(x^(x^(x^x^...)))....)))) = c has a real solution iff 1/e <= c <= e.

> David

Euler, I dont know, but it is reasonably easy to prove. Except it is
first *essential* to define x^(x^(x^(x^x^...)))....)))) as the limit of
the sequence u_0=x, u_1=x^x,..., u_(n+1)=x^(u_n) ; then classical tests
of convergence of this sequence (here using the mean value theorem) give
the result above

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