Re:Have any long-suspected irrational numbers turned out to ...
by Chinjut on Tuesday September 16, 2014 @09:16PM
(#8327980)
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.
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.