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> 2. The infinite tower sqrt(2)^sqrt(2)^sqrt(2)^... has a finite value, which is 2. (It's the fixed point, sqrt(2)^x = x). Yet again, we have a simple (integer)

by codeflo 12y ago

> 2. The infinite tower sqrt(2)^sqrt(2)^sqrt(2)^... has a finite value, which is 2. (It's the fixed point, sqrt(2)^x = x). Yet again, we have a simple (integer) value coming out of something that seems infinitely complex. However, yet again, once you make the conceptual leap necessary to evaluate such "towers", the answer is trivial. Is it really that simple? sqrt(2)^4 = 4, so the answer should also 4 by the same argument. Which makes the whole expression not well-defined IMHO.

dragonwriter 12y ago
The expression is well-defined, the statement that the infinite tower is equal to the fixed point is just slightly oversimplified. That's only true for attracting fixed points, but its possible to have repelling fixed points, as well.
- codeflo 12y ago
  Fair enough, I'm starting to read up on this. Obviously, you can define sqrt(2)^sqrt(2)^... to mean the (unique?) attracting fixed point of x -> sqrt(2)^x. But that makes your explanation a bit circular (it is so because it's defined to be so). Probably that's the only useful definition, but that fact is certainly not trivial or self-evident.
  - dragonwriter 12y ago
    Well, the thing is that you can test whether a fixed point is an attracting fixed point; per Wikipedia (which I am relying on for the details of the test because I haven't used this for years, but the test looks correct), any fixed point x of a function f which has an open neighborhood where f is continuously differentiable and f'(x) < 1 is attracting.
- michaelochurch 12y ago
  Very good point. Thanks for stating this.
- Chinjut 12y ago
  But a function could just as well have multiple attracting fixed points (consider the cube root function, with attracting fixed points at both +1 and -1). Instead, we can look at it like this: If f is a continuous, order-preserving function from a closed interval to itself, and x <= f(x), then x <= f(x) <= f(f(x)) <= ..., with this sequence converging to the least fixed point >= x. In our case, we can think of f(x) = sqrt(2)^x as a continuous function from [-infinity, +infinity] to itself, and take our starting point to be x = sqrt(2) [or perhaps 1, or 0, or -infinity, each one step back from the previous]; thus, the sequence sqrt(2), sqrt(2)^sqrt(2), sqrt(2)^(sqrt(2)^sqrt(2)), ..., converges to the least fixed point of f, which will be 2 rather than 4.
tel 12y ago
There's no obvious, singular definition of "is" for something like that tower. You have to describe it via a generating process or an equation. If you ask for the values "x" such that "x = sqrt(2)^x" then you need to consider the chance there are multiple such "x"es. If you consider it the limit result of continually applying f(x) = sqrt(2)^x then you'll get the least fixed point.

by codeflo 12y ago

dragonwriter 12y ago

The expression is well-defined, the statement that the infinite tower is equal to the fixed point is just slightly oversimplified. That's only true for attracting fixed points, but its possible to have repelling fixed points, as well.

codeflo 12y ago
Fair enough, I'm starting to read up on this. Obviously, you can define sqrt(2)^sqrt(2)^... to mean the (unique?) attracting fixed point of x -> sqrt(2)^x. But that makes your explanation a bit circular (it is so because it's defined to be so). Probably that's the only useful definition, but that fact is certainly not trivial or self-evident.
- dragonwriter 12y ago
  Well, the thing is that you can test whether a fixed point is an attracting fixed point; per Wikipedia (which I am relying on for the details of the test because I haven't used this for years, but the test looks correct), any fixed point x of a function f which has an open neighborhood where f is continuously differentiable and f'(x) < 1 is attracting.
michaelochurch 12y ago
Very good point. Thanks for stating this.
Chinjut 12y ago
But a function could just as well have multiple attracting fixed points (consider the cube root function, with attracting fixed points at both +1 and -1). Instead, we can look at it like this: If f is a continuous, order-preserving function from a closed interval to itself, and x <= f(x), then x <= f(x) <= f(f(x)) <= ..., with this sequence converging to the least fixed point >= x. In our case, we can think of f(x) = sqrt(2)^x as a continuous function from [-infinity, +infinity] to itself, and take our starting point to be x = sqrt(2) [or perhaps 1, or 0, or -infinity, each one step back from the previous]; thus, the sequence sqrt(2), sqrt(2)^sqrt(2), sqrt(2)^(sqrt(2)^sqrt(2)), ..., converges to the least fixed point of f, which will be 2 rather than 4.

tel 12y ago

There's no obvious, singular definition of "is" for something like that tower. You have to describe it via a generating process or an equation. If you ask for the values "x" such that "x = sqrt(2)^x" then you need to consider the chance there are multiple such "x"es. If you consider it the limit result of continually applying f(x) = sqrt(2)^x then you'll get the least fixed point.