What does "recursive cos" converge to?

Question

Found a few math geniuses on here and I figure I'd ask this question nagging at me for years.
When continuously taking the square root of a positive number x, i.e. if x[0] = p and x[n+1] = sqrt(x[n])
and if we take
lim (x[n])
x => infinity
We get 1.
What about the function x[0] = c, x[n+1] = cos(x[n])?
I know it converges to *something* but can that something be expressed in elementary functions? What is the limit of that recursive function as x approaches infinity, and please show your steps.

Answer 1

Repeatedly taking cosines, suppose we converge on the value z. Then the arccos(z), by the nature of the infinite length of the sequence, =z. Solving for this equation (Using Mathematica), we arrive at 0.739085 approximately. Does this have an expression in elementary functions? To be honest, I don't know, but my intuition tells me it is a transcendental number that does not have such an expression.

Steve

Answer 2

Your solution is intersection between y = cos(x) and y = x.
And it is 0.739085133.