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
\( f_a(f_a^{-1}(x) + \frac{1}{h}) \)
is \( \frac{1}{h} \) iterate of the "subfunction" of \( f_a \), which is uniquely given by \( f_a(f_a^{-1}(x) + 1) \). 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?

