11/06/2010, 04:57 AM

(11/06/2010, 01:30 AM)mike3 Wrote: Well this approach looks to be of no use. Note that

is iterate of the "subfunction" of , which is uniquely given by . Thus h iterates of it yields that subfunction, regardless of the specific superfunction used (i.e. a 1-cyclic transform of it will give the same subfunction, hence the same "iterative derivative".). Essentially, there is apparently no difference between this and the "discrete iterative derivative".

Yes. I have noticed that. What about using the x instead of 1 as in the current post?