Could be tetration if this integral converges

[mike3]

\( F(n) = \sum_{j=0}^n \frac{n!(-\lambda)^{n-j}}{j!(n-j)!(^j e)} \)

So is this \( F \) supposed to approximate tetration if \( \lambda \) is small? As if so, then it doesn't seem to be working for me. If I take \( \lambda = 0.01 \) and the integral upper bound at 2000, I get \( F(1.5) \) as ~443444.33873479713260158296678612894384. Clearly, that can't be right -- it should be between \( e \) and \( e^e \) (if this is supposed to reproduce the Kneser tetrational then it should be ~5.1880309584291901006085359610758671512). It gets worse the smaller you make \( \lambda \) -- 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'll note firstly that

\( F(n) = \sum_{j=0}^n \frac{n!\lambda^{n-j}}{j!(n-j)!(^j e)} \)

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 \( \lambda = 0 \) diverges this doesn't happen. Maybe if you try \( \lambda = 1 \) I've done more research into this form of the operator so perhaps we can work with this one.

\( F(n) = \sum_{j=0}^n \frac{(-1)^{n-j} n!}{j!(n-j)!(^j e)} \)

Well I tried this \( F \) and the integral also didn't seem to converge. Trying \( \lambda = 1 \) yields a finite value but the recurrence \( F(z+1) = \exp(F(z)) \) 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 \( f(z) \) is periodic with period \( 2\pi \), we have a Fourier series

\( f(z) = \sum_{n=-\infty}^{\infty} a_n e^{inz) \).

We assume \( a_0 = 0 \). Then,

\( D^s f(z) = \sum_{n=-\infty}^{\infty} a_n (in)^s e^{inz} \).

This is the Weyl differintegral. Note that taking \( s = -1 \) yields the integral (though we have to drop the term at \( n = 0 \)).

Similarly, for the finite difference operator,

\( \Delta^s f(z) = \sum_{n=-\infty}^{\infty} a_n (e^{in} - 1)^s e^{inz} \).

Note that taking \( s = -1 \) yields the continuum sum (though, again, we have to drop the term at \( n = 0 \)).

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?