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X^x^x^x^x^x^...
By Anonymous on Tuesday, March 7, 2000 - 04:01 pm:

Take the sequence (x, x^x, x^x^x, x^x^x^x, ....) ad infinitum.
For which values of x does this sequence converge.
Well, the answer is e-e < x < e1/e

The second inequality can be found by the following:

Firstly, let u=x^x^x^x^x^....
i.e. assume the sequence does converge.

Then x and u must satisfy:
xu=u (*)

we can now find the maximal value of x by setting dx/du=0

(*) can be written as u ln(x) = ln(u)

differentiating with respect to u gives:
ln(x)+ u/x (dx/du)=1/u

but ln(x) = ln(u)/u
so
u/x (dx/du)=1/u(1-ln(u))

setting dx/du=0 gives
1-ln(u)=0 => u=e

but x=u1/u
so x(max)=e1/e

Of course, this is not a proof that x^x^x^x^x^... really converges for e1/e, but it shows that it certainly doesn't for larger values.
Does anyone know, how to prove the first inequality, i.e that the sequence doesn't converge for x<e-e?

By Dan Goodman (Dfmg2) on Wednesday, March 22, 2000 - 07:36 pm:

I just found an interesting web page which deals with this question very thoroughly, here it is:

http://users.forthnet.gr/ath/jgal/math/exponents.html

By Peter Conlon (P2714) on Tuesday, July 11, 2000 - 07:38 pm:

Also, look at "on x to the x" at this site:

http://clem.mscd.edu/~talmanl/
(Currently item 5 under Fun Stuff.)