08/13/2007, 09:13 PM

jaydfox Wrote:The logarithm is "entire" for reals > 0 (not quite "entire", but you know what I mean), but it's standard power series diverges outside the range (0, 2]. Analytic extension is used outside that range. So this function might only converge for x<=1.Sorry, I got mixed up. I was thinking of the iterating function itself. I was thinking of solving for the 5th iterate. You meant solving the half-iterate for x=5. Big difference.

Besides, you do know what the formula equals at x=5, don't you? Well, for starters, at x=4, it's approximately 5.03481484682034616908489989276E+41. Bear in mind, this function is equal to the second iterated logarithm (base e) of my cheta function, shifted by a constant in the x direction. So it has tetrational growth.

It's like trying to solve sin(x) using the power series, and then saying it appear divergent because you tried to solve the power series for x=100.

So no worries. Just make sure it converges for x=3, which should equal 96.0223655650268799109865292599.

Nevertheless, I would try various x values and see what the radius of convergence really is. It might be 1. But a radius of convergence of 1 is still far more than enough allow analytic extension.

To be a valid test, you'll need far more than the first 10 terms, though. (And to be a proof, you need a formula for all of them, etc., which is what I'm working on, in the general case).

~ Jay Daniel Fox