I recently submitted the solution (proof) to a problem posed in the April 2009 edition of Math Magazine. It is embedded below as a PDF. I hope you enjoy it, though it is a bit lengthy. (There's a button on the top right of the document window to make it full screen, and then a zoom button so the font is big enough to read.)
Game Show Problem

On the Power Tower

I recently looked into the properties of infinite power towers, or an infinite tower of powers. I first asked if the nested power tower with the square root of 2,

n^(n^(n^(n^(..., n=2^(1/2),

converges or not.

So, I simply asked Mathematica, and it turns out that the power tower of square roots of 2 converges to 2.

To prove this, we can set x equal to the tower and solve for it by some nice manipulation. (We can only do this when the number actually converges, but in this case I know it does so we're alright.)

x=n^n^n^n..., n=2^(1/2)

x=n^x

This equation, we can now see, has two solutions: x=2 and x=4. (Because Sqrt[2]^2=2 and Sqrt[2]^4=4.) However, we know that the power tower converges to 2- so why do we get the extraneous solution x=4 when solving analytically?

As it turns out, the power tower converges for any n in the interval (0,2^1/2]. As we've seen, for n=2^1/2, the equation has two solutions. However, it is interesting to note that for n on (1,2^1/2], the equation has two solutions, whereas for n on (0,1], it has exactly one solution. We can see this by looking at the function f(x)=n^x-x defined for n on (0,2^1/2].

In terms of the stated problem, f(x)'s zeroes tell us what the power tower of n converges to. I will now show that on the intervals stated, (0,1] and (1, 2^1/2], the equation has one and two solutions, respectively.

f(x)=n^x-x

f'(x)=ln[n]n^x-1

f'(x)=0 when

1=ln[n]n^x

n^x=1/ln[n]

x ln[n]=-ln[ln[n]]

x=-ln[ln[n]]/ln[n]

f(x) decreases on (-Infinity, -ln[ln[n]]/ln[n])
f(x) increases on (-ln[ln[n]]/ln[n],Infinity)

This is the single critical point of our function, and we can now see why we get two solutions for 1
Why the second zero is trivial, I have no idea. But I do like this problem, as well as some related ones. In writing this post I noticed that a similar "nested" infinite square root runs into a similar, but more obvious problem in the algebra.

Take x=Sqrt[n+Sqrt[n+Sqrt[n+..., out to infinity.

This transforms, just as above, into

x^2 = n+x

x^2-x-n=0

Using the quadratic formula, we obtain

x=(1+-Sqrt[1+4n])/2

Taking n=2 as an example,

x=2 or -1.

Mathematica agrees that the more logical solution is the correct one- naturally, 2 is the answer. Still, it is strange that the -1 shows up in our analytical approach.

Paralleling my other analysis, this function has two solutions most of the time, namely for n>0, and is useless when n<0 because we have no business assuming the nested square root's convergence in that case. (Clearly Sqrt[-.5+Sqrt[-.5+... doesn't converge to any real number, so setting the expression equal to x is against the law.)

There is one more of these that runs into the same problem: the continued fraction [n,n,n,n...] However, I see no need to share the specifics of this one as it comes out to a nearly identical quadratic problem. I leave it to the reader to investigate the case further.

Finally, I'd like to add a quick fun fact about the nested square root and the continued fraction. If we take the value of n=1, in BOTH cases we get the golden ratio, (i.e. Sqrt[1+Sqrt[1+...=1+1/(1+1/(1+1/...=(1+Sqrt[5])/2). You can check it yourself.

I discovered a new summation formula regarding the consecutive powers of a given base. I noticed, to start, that any given power of 2 could be written as a sum involving the previous powers of two- for example, 4-1 = 2+1, or 32-1 = 16+8+4+2+1, generalized in the formula Sum(i=0, n-1) {2ⁱ } = 2ⁿ - 1. My idea was to expand this formula to include the powers of any base.

So, I first asked- is it even possible to write every power of 3 as a sum of the terms preceding it? Instinct said yes, and with a little guesswork I noticed a pattern: 3-1 = 2(1), 9-1 = 2(1+3), 27-1 = 2(1+3+9), and so on. This led to the general formula, Sum(i=0, n-1) {3ⁱ} = (3ⁿ - 1)/2.]

Before I state the general formula for xⁿ, which you might be able to guess, I would like to note that there is a somewhat non-traditional proof of this kind of formula that I used to verify my results. I'll give an example using 3.

Our goal is to prove that 1+2(1+3+9+...+3^n-1) = 3ⁿ

Multiplying out and adding the 1 to the first term of the sum,

3+2(3)+2(9)+...+2(3^n-1) = 3ⁿ

Adding the first two terms of the LHS,

9+2(9)+2(27)+...+2(3^n-1) = 3ⁿ

Adding the first two terms of the LHS,

27+2(27)+...+2(3^n-1) = 3ⁿ

And so on. This domino process by which each preceeding sum adds with the next term to create a power of three can be quantified by this relationship: 3^k-1 + 2(3^k-1) = 3^k. This series of summations eventually ends with 3ⁿ.

A similar proof can be can be used to show that the formula for 2^n works, or for any consequent base. Here's the general formula: 1+(x-1)*Sum(i=0, n-1) {xⁱ} = xⁿ. With this formula, we can now see what the main hinge of the given proof really is- it is that the first "domino" in the series comes from 1+(x-1)(1) = x, which is exactly what is needed.

After creating this formula and remembering one of my older finds on the same topic, I had the desire to somehow combine the two finds and generate some cool new relationship. (The other formula came up with the same kind of result, a formula for xⁿ, though in that case the formula was a sum of polynomial values instead of exponential values.) I'll have to go back and see if there's any way to reconcile the two.