12/11/2009, 10:35 PM
(12/11/2009, 08:16 AM)Gottfried Wrote: If it does not go to 0 but to a fixpoint: can we subtract that constant/fixpoint and compensate for that by introducing + zeta(0)*fixpoint ? (In some fiddling in other examples weeks ago such a compensation seemed to be meaningful, don't have it at hand at the moment, though)
Gottfried
Yes, which was the idea in the original post. Here, though, we're talking about bases like \( b = 0.04 \), for which you drew a graph of its regular iteration. I presume the tetration (that is, with \( F(0) = 1 \)) would look something like that, so \( F(x + n) \) (n = 0, 1, 2, 3, ...) would converge to a 2-cycle as \( n \rightarrow \infty \) for most \( x \), but diverge for a few others. For summing on the x-values where there is a 2-cycle, we compensate by subtracting a square wave and then adding its continuum sum as acquired via the Fourier series.
