(05/05/2014, 11:45 PM)mike3 Wrote:(05/05/2014, 04:27 PM)JmsNxn Wrote:(05/05/2014, 02:11 AM)mike3 Wrote:I'll note firstly that(05/04/2014, 09:06 PM)JmsNxn Wrote:

So is this supposed to approximate tetration if is small? As if so, then it doesn't seem to be working for me. If I take and the integral upper bound at 2000, I get as ~443444.33873479713260158296678612894384. Clearly, that can't be right -- it should be between and (if this is supposed to reproduce the Kneser tetrational then it should be ~5.1880309584291901006085359610758671512). It gets worse the smaller you make -- i.e. it doesn't seem to converge. Also, picking values to put in that are near-natural numbers doesn't seem to work, either.

I accidentally added an extra negative. But that doesn't really affect convergence. I understand whats happening.

Hmm. That makes sense now that I think about it. I was hoping you could take lambda small but not too small and it wouldn't diverge too fast but because obviously diverges this doesn't happen. Maybe if you try I've done more research into this form of the operator so perhaps we can work with this one.

Well I tried this and the integral also didn't seem to converge. Trying yields a finite value but the recurrence does not appear to hold, nor are the values close to those of the Kneser tetrational.

I noticed your discussion after this point about the continuum sum thing and you mentioned about fractional iteration of the difference operator. This is why I was curious as to how the integral definition for the Weyl differintegral related to its definition for periodic functions. In particular, if is periodic with period , we have a Fourier series

.

We assume . Then,

.

This is the Weyl differintegral. Note that taking yields the integral (though we have to drop the term at ).

Similarly, for the finite difference operator,

.

Note that taking yields the continuum sum (though, again, we have to drop the term at ).

Now, if the first expression for the differintegral can be generalized to certain non-periodic holomorphic functions via an integral transform, can that also be done for the second? Is there a method to derive the integral transform from the given definition in the first case? If so, can it be generalized to the second?

If I understand your question correctly the answer is yes to both questions. I'll write it out. Using the operators from my paper we can completely represent the iterated difference:

then:

I'm working on writing this all up. So far all I have is a bunch of notes and papers compiled together unorganized.

Now for the first question, to work on these periodic functions define:

And then we can generate the differintegral using taylor series. now if

then