Could be tetration if this integral converges
(05/04/2014, 02:18 AM)mike3 Wrote: So you mean "the inverse Mellin transform of the tetrational", right? Or do you mean of the reciprocal? Also, what do you mean by "Weyl differintegrable"? According to here:

that is something that only functions with a Fourier series, i.e. periodic, can have. Tetration is not periodic (although it does have a pair of "pseudo periods").

Sorry I should've mentioned that Weyl differintegrability is a different condition in the context I'm referring to. Weyl originally applied the operator on periodic functions but the general form of it applies on more functions than this.

\( \int_0^\infty |f(-t)|t^{\sigma - 1} \,dt < \infty \)

for \( a < \sigma < b \) impliess \( f \) is Weyl differintegrable on \( a < \Re(z) < b \).

(05/04/2014, 02:18 AM)mike3 Wrote:
(05/03/2014, 06:12 PM)JmsNxn Wrote: However after seeing what you just posted I have to draw the same conclusion as you. \( \vartheta \) has no hope of converging in a mellin transform. Which is what I pretty much figured. BUT! we're not out of the woods yet.

IF we can find some entire function \( F(z) \) such that for \( 0 < \sigma < 1 \):

\( \int_0^\infty |\sum_{n=0}^\infty F(n)\frac{(-w)^n}{n!(^ne)}|w^{\sigma - 1} < \infty \)

we are back in the game

OR IF we can find some entire function \( F(z) \) such that for \( |\frac{F(-z)}{(^{-z} e)} |<C e^{\alpha|\Im(z)| + \rho|\Re(z)|} \) for \( 0 \le \alpha < \pi/2 \) and \( 0 \le \rho \)

we are back in the game.

By back in the game I mean I think I can provide an analytic expression for tetration. I'm just finishing the paper I'm working on at the moment and it contains a fair amount of what I'm talking about a lot more rigorously. I'll attach it once I know it's in it's final form. It shows what I am talking about more c learly when I am using fractional calculus on recursion.

Hmm. Given the nasty, chaotic behavior of tetration I've mentioned, it would seem the second kind of function would be more difficult than the first.

Actually, I think it might be possible to get a function of the first type (for the series). If we could find an entire function \( F(z) \) such that \( F(n) =\ ^n e \) when \( n \in \mathbb{N} \), then we should be in luck, for then your sum will just be \( e^{-w} \) and your integral \( \Gamma(\sigma) \). Such a function need not be a tetration extension, for it need not satisfy the functional equation for tetration, merely interpolate the values at the natural numbers.

However, it seems you can get a different \( \vartheta \) for every \( F \), indeed, I believe, with judicious choice of the \( F \), you can make \( \vartheta \) anything you want, indeed, any function which decays to 0 and is analytic. So it would seem that any tetration extension constructed with this method would be highly non-unique, unless I'm missing something. Does the final tetration result not depend on the choice of \( F \)?

According to this:

there is a method to construct an entire interpolant of any increasing sequence, which would include \( {^n} e \). So this should provide (many!) suitable candidates \( F \) that will reduce your integral to the \( \Gamma \) function.

I see what you're trying to do by making \( F(n) = (^n e) \) but that won't work. If we try it this way we will get the following formula:

\( \frac{1}{(^{-s} e)} = \frac{1}{F(-s)\Gamma(s)} \int_0^\infty e^{-w}w^{s-1}\,dw = \frac{1}{F(-s)} \)

So that all we are doing is recovering \( F \) which doesn't satisfy the recursion. I'm not sure how we can create an \( F \) that satisfies the right conditions but it is necessary that it does not interpolate tetration. I realize I forgot to say we need the extra condition:

\( \int_0^\infty |\sum_{n=0}^\infty F(n)\frac{(-w)^n}{n!(^{n+1} e)}| w^{\sigma - 1} < \infty \) for \( 0 < \sigma < 1 \)

This ensures recursion. It also expresses that we need a more complicated choice for \( F(n) \). I really do believe our best hope is the second condition on the exponential boundedness of \( |\frac{F(-z)}{(^{-z} e)}| \)

Quote:Does the final tetration result not depend on the choice of \( F \)?

YES it does not depend in every situation I've come across. By this I mean:

if \( \beta(w) = \sum_{n=0} \phi(n) \frac{w^n}{n!} \) and \( \alpha(w) = \sum_{n=0} \psi(n) \frac{w^n}{n!} \) and \( \gamma(w) = \sum_{n=0}^\infty \phi(n) \psi(n)\frac{w^n}{n!} \) \( |\phi(z)|,|\psi(z)|, |\phi(z)\psi(z)| < C e^{\alpha |\Im(z)| + \rho |\Re(z)|} \)
for \( 0 \le \alpha < \pi/2 \) and \( \rho \ge 0 \)


\( \frac{d^z}{dw^z}|_{w=0} \beta(w) = \phi(z) \)
\( \frac{d^z}{dw^z}|_{w=0} \alpha(w) = \psi(z) \)
\( \frac{d^z}{dw^z}|_{w=0} \gamma(w) = \phi(z)\psi(z) \)

So there are some strong results on uniqueness and it preserves a fair amount of data.


Messages In This Thread
RE: Could be tetration if this integral converges - by JmsNxn - 05/04/2014, 07:27 PM

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