08/13/2007, 09:16 PM

jaydfox Wrote: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).

Wait, now I'm really mixed up. You posted a different power series from the one we were looking at earlier. The previous one matched my calculations. What's this new one? Is this the series for the iterating function itself? If so, I stand by my first post about trying x=3.

~ Jay Daniel Fox