Wonderful exponent tower property

[meemoe_uk]

Hi dudes, don't be put off by the clumsy notation here.

I was wondering about these particular exponent towers and this curious property of theirs...

Let p be a positive integer. Then the exponent tower, composed of p+1 parts each of value p^(1/p), equals p.

e.g. for p=2.
tower part = 2^(1/2)
(2^(1/2))^ ((2^(1/2))^(2^(1/2)))=2
bah, this looks clumsy, but it's concise written by hand, i.e. a 3 part exponent tower.

Anyway, I heard that it's difficult to prove any particular case for p, let alone the general case. I had a go myself for case p=2. I set x equals exponent tower and tried to show x=2, but I got nowhere.

Can anyone post the easiest proof for case p=2?, or any other case?

 

[StatusX]

Well, that isn't true as you've written it, since the only number you can raise \sqrt{2} to to get 2 is 2 (did I mention 2?). However, if you write it like this:

((\sqrt{2})^{\sqrt{2}})^{\sqrt{2}} = (\sqrt{2})^{\sqrt{2}\cdot\sqrt{2}} = (\sqrt{2})^2=2

Then it is true, and it is clear how this extends to the general case.

 

[meemoe_uk]

Doops, well that's another of reality's amazing mysterys unweaved. Feels like the time I discovered santa claus didn't really exist.

 

[Hurkyl]

And FYI, a power tower looks like this:

a^{a^{a^\ldots}}

and not

((a^a)^a)^\ldots

A quick google search reveals that people seem to use exponent tower to mean the same thing... so it sounds like whoever posed the problem to you has their terms wrong.

 

[arildno]

However, the power tower SEQUENCE of square roots of 2 does converge to 2.

Here's a proof:
1. The sequence is bounded above by 2. This is seen in that each member of the sequence must be less than the number where the last square root of 2 is replaced by a 2. That number is easily sen to be 2.

2. The sequence is increasing, by 1., it must therefore converge to some number x.

3. x must satisfy the equation:
x=(\sqrt{2})^{x}
This equation has two solutions; x=2 and x=4
Since 2 is the lesser upper bound, the tower converges to 2, rather than to 4