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bonanova
Daedalian Member

 Posted: Sun Jun 01, 2008 6:57 am    Post subject: 1 A friend pointed this one out to me. Interested in any comments. Consider the equation which involves an infinite string of exponentiation x to the x to the x to the x ...: [1] x x... = 2. Noting that the exponent of the first x equals the whole expression gives x 2 = 2, or x = sqrt(2). Then consider the equation [2] x x... = 4. Now x 4 = 4, or x = 4throot(4) = sqrt(2). Thus the expression sqrt(2) sqrt(2)... is multivalued, or perhaps indeterminate. Why?_________________ Vidi, vici, veni.
ralphmerridew
Daedalian Member

 Posted: Sun Jun 01, 2008 1:45 pm    Post subject: 2 You've gotten ahead of yourself. What you showed is that if there exists an x for which the power converges to 2 or 4, then that x must be sqrt(2). You did not show that a tower of sqrt(2) converges to 2 or 4, or even that it converges at all.
bonanova
Daedalian Member

Posted: Fri Jun 06, 2008 5:42 am    Post subject: 3

 ralphmerridew wrote: You did not show that a tower of sqrt(2) converges to 2 or 4, or even that it converges at all.

The infinite power would be the limit of the sequence sqrt(2), sqrt(2) sqrt(2) , ... for simplicity write it as a 1 , a 2 , a 3 , ...

Since a i+2 = sqrt(2) a 1+1 > sqrt(2) a i = a i+1 . then 1<a i <a i+1 for each i>=1.
If the topmost sqrt(2) in any a i is replaced by the larger value 2, a i itself becomes 2.
So the sequence is increasing and bounded above, thus has a limit and the expression converges.

I've partially exposed the puzzle at this point, but let's ask: what's with equation [2] in the OP?
_________________
Vidi, vici, veni.
Swepsie
Icarian Member

 Posted: Fri Jun 06, 2008 9:19 am    Post subject: 4 X 2 = 2 has 2 solutions. Playing around in Excel, shows that sqrt(2) = 1.41421... is indeed a solution that works. X 4 = 4 has 4 solutions. 1.41421..., -1.41421..., i * 1.41421... and i * -1.41421... One of the last 3 should be the solution for [2]._________________Antidisestablishmentarialism ?
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