(08/30/2010, 11:14 PM)tommy1729 Wrote: i.e. z<> z^z => z^z =/= (z + z^^2)/2.
I explained in an earlier post my misuse of the z^^h-notation. (Oh, that was in the other thread only?, Then - sorry)
Ok. To make ascii-writing short I use it there (and now here) in that meaning:
using a base b, z^^h is b^b^...^b^z where h is the number of b,
in case of continuous: z^^h = \( \exp_b^{oh}(z) \)
I think that "misuse" can be justified since we usually talk about a fixed chosen base ("b") and discuss the change of parameter z as iteration start, h as iteration "height" and z^^h as iteration result. Reading some "z^z" above I think there is a misunderstanding there.
Gottfried
Gottfried Helms, Kassel
