# Wonderful exponent tower property!

1. Jan 25, 2006

### 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?

2. Jan 25, 2006

### 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.

3. Jan 25, 2006

### 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.

4. Jan 25, 2006

### Hurkyl

Staff Emeritus
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.

5. Feb 18, 2006

### 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

Last edited: Feb 18, 2006