08/13/2007, 09:33 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.Dont tell nonsense. Log has a singularity at 0, thatswhy it is not entire (thatswhy its radius of convergence is 1, if developed at 1).

Thats the whole thing about entire functions, they have no singularities and hence their radius of convergence is infinity.

Analytic continuation is necessary only if there are complex singularities.

exp,sin,cos are entire functions, their series converge for arbitrary large arguments.

Quote: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.I was talking about , do we talk about the same?

Quote: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

sin converges for arbitrary x, however it does not satisfy Walker's condition because the coefficients are partly negative.

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

Yes it does also not converge for , but we seem anyway not to talk about the same thing ...

Quote: 0, 1.0, 1.500000000, 1.250000000, 1.500000000, 1.187500000, 1.609375000, 1.046875000, 1.714843750, 1.175781250, 0.8930664062, 3.502685547, -3.853393555, 10.37207032, -8.21649169

But of course it could be that it stabilizes for greater n, however the computation is quite resource using, so I can not see it above 15.