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.
