Could be tetration if this integral converges
(05/05/2014, 02:11 AM)mike3 Wrote:
(05/04/2014, 09:06 PM)JmsNxn Wrote: \( F(n) = \sum_{j=0}^n \frac{n!(-\lambda)^{n-j}}{j!(n-j)!(^j e)} \)

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.

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

I do have a nice result that will work for these operators that is slightly off topic but is related to continuum sums. If \( \beta = \sum_{n=0}^\infty \frac{x^n}{n!(^n e)} \) is as before and \( z \in \mathbb{C} \):

\( \frac{d^{-z}}{dx^{-z}}|_{x=0} e^{x}\beta(-x)= F(-z) = \frac{1}{\G(z)}(\sum_{n=0}^\infty F(n) \frac{1}{n!(n+z)} + \int_1^\infty e^{-x}\beta(x)x^{z-1}\,dx) \)

Quite fantabulously if \( \bigtriangledown F(-z) = F(1-z) - F(-z) \) we get the really interesting result:

\( \bigtriangledown^n F(-z) = \frac{d^{-z}}{dx^{-z}}|_{x=0} e^x \beta^{(n)}(x) \)

And of course:

\( [\bigtriangledown^n F(-z)]_{z=0} = \frac{1}{(^n e)} \)

I'm a little fuzzy on the following but if we use a fractional iteration of \( \bigtriangledown \) we can find a very nice interpolant of \( (^n e) \). Since it's all using fractional calculus I have an impressionistic vision of how a similar proof of recursion would go (picking a certain fractional iteration of \( \bigtriangledown \) I have in my mind and then using a similar contour integral technique).

I'm really intrigued by this idea but I think I have a more general result we need. I wonder if there exists a theorem in complex analysis on the following.

If \( \phi(z) \) is holomorphic on \( \Re(z) > -b \) does there exist some holomorphic function \( \pi(z)\neq 0 \) holomorphic on \( \Re(z) > -b \) such that:

\( |\pi(z)\phi(z)|, |\pi(z)\phi(z+1)| < C e^{\alpha |\Im(z)| + \rho|\Re(z)|} \) for \( 0 \le \alpha < \pi/2 \) and \( \rho \ge 0 \)

such that \( \pi \) satisfies some conditions I'm not sure of yet. It cannot interpolate \( \phi(n) \), it cannot interpolate the inverse and it cannot be a fair amount of obvious easy functions.

Messages In This Thread
RE: Could be tetration if this integral converges - by JmsNxn - 05/05/2014, 04:27 PM

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