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tommy1729 · 05/04/2014, 01:25 PM

Surely you are joking ?

sheldonison · 05/04/2014, 03:28 PM

(05/04/2014, 11:50 AM)mike3 Wrote: WOW, this is easier than I thought...

All we do is take that $a_n = {^n} e$ and $c_n = {^n} e$ (there seems to be a mistake as the post as written seems to suggest we should take $c_n = -{^n} e$ and that doesn't seem to work), and $z_n = n$ (the indexes at which we are to interpolate), which gives

$F(z) = \sum_{n=0}^{\infty} e^{({^n} e) (z - n)} ({^n} e) \frac{(-1)^n \sin(\pi z)}{\pi (z - n)}$

as an entire interpolant for the tetrational sequence.

What are the values for F (0,1,2,3)?

JmsNxn · (This post was last modified: 05/04/2014, 08:36 PM by JmsNxn.)

(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:

http://en.wikipedia.org/wiki/Weyl_differintegral

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:
http://mathoverflow.net/questions/2944/w...anns-funct

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 $\rho \ge 0$

Then

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

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

JmsNxn · 05/04/2014, 07:36 PM

I figure it's probably a good idea to post my paper now. I feel it's 90% there and it certainly is legible and will clarify what I am talking about much more. I can apply more powerful techniques than what's in the paper but nonetheless it still portrays what I am trying to do.

James Nixon's paper on fractional calculus.
$.pdf$ On a representation of the Weyl differintegral.pdf (Size: 343.78 KB / Downloads: 365)

MphLee · 05/04/2014, 07:44 PM

(05/04/2014, 07:36 PM)JmsNxn Wrote: I figure it's probably a good idea to post my paper now. I feel it's 90% there and it certainly is legible and will clarify what I am talking about much more. I can apply more powerful techniques than what's in the paper but nonetheless it still portrays what I am trying to do.

James Nixon's paper on fractional calculus.

I'll read it now $Big Grin$ I hope it will answer some of the question i've pm-ed you $Big Grin$

JmsNxn · (This post was last modified: 05/04/2014, 09:07 PM by JmsNxn.)

As a note on a similar technique you are applying Mike but trying to keep the vibe much more fractional calculus'y (since it's what I am familiar with) We will try the following function:

$0 < \lambda$ We want $\lambda$ fairly small

$\beta(w) = \sum_{n=0}^\infty \frac{w^n}{n!(^n e)}$

We know that $|\beta(w)| < C e^{\kappa |w|}$ for $\kappa > 0$ because $\frac{1}{(^n e)} < C_\kappa \kappa^n$

So that $\Re(z) > 0$ .

$F(-z) = \frac{1}{\Gamma(z)}\int_0^\infty e^{-\lambda x}\beta(-x)x^{z-1} \,dx$

This function should be smaller then tetration at natural values.

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

We would get the entire expression for $F(z)$ by Lemma 3 of my paper:

$F(z) = \frac{1}{\Gamma(-z)}(\sum_{n=0}^\infty F(n)\frac{(-1)^n}{n!(n-z)} + \int_1^\infty e^{-\lambda x} \beta(-x)x^{z-1}\,dx)$

Now F(z) will be susceptible to alot of the techniques I have in my belt involving fractional calculus. This Idea just popped into my head but I'm thinking working with a function like this will pull down the imaginary behaviour and pull down the real behaviour.

We also note that

$e^{F(z) }\approx F(z+1)$ . Which again will be more obvious if you look at the paper, but it basically follows because:

$F(n) \approx (^n e)$

mike3 · (This post was last modified: 05/04/2014, 10:49 PM by mike3.)

(05/04/2014, 07:36 PM)JmsNxn Wrote: I figure it's probably a good idea to post my paper now. I feel it's 90% there and it certainly is legible and will clarify what I am talking about much more. I can apply more powerful techniques than what's in the paper but nonetheless it still portrays what I am trying to do.

James Nixon's paper on fractional calculus.

Amazing. I only saw the first bit of the paper, I'm really excited right now because it seems there's a simple integral-transform representation for the "Weyl differintegral". This is really really really really really interesting to me, because of ... dum dum dum ... the continuum sum!

In particular, I am really interested in your Definition 3. How does that expression come from the original Weyl differintegral which was defined for Fourier series? I.e. how do you generalize the Fourier series definition to that definition?

JmsNxn · (This post was last modified: 05/05/2014, 12:32 AM by JmsNxn.)

(05/04/2014, 10:42 PM)mike3 Wrote: Amazing. I only saw the first bit of the paper, I'm really excited right now because it seems there's a simple integral-transform representation for the "Weyl differintegral". This is really really really really really interesting to me, because of ... dum dum dum ... the continuum sum!

In particular, I am really interested in your Definition 3. How does that expression come from the original Weyl differintegral which was defined for Fourier series? I.e. how do you generalize the Fourier series definition to that definition?

Well Weyl defined the operator on periodic functions slightly differently but it's equivalent. I did not come up with definition 3. It's been long standing. It's also called the exponential differintegral, The Riemann liouville operator with lower limit negative infinity. Etc etc..

I am lacking proofs on convergence for the mellin transform of lots of interesting forms of functions. However, I have made progress on continuum sums. That is actually my next paper. Essentially I've shown that if $\phi$ is holomorphic in the strip $0<a < \Re(s) < b$ and:

$|\phi(s)|< C e^{\alpha |\Im(s)|}$ for $0 \le \alpha < \pi/2$

i can generate an indefinite sum of $\phi$ on that same strip> Im currently trying to show i can take as many indefinite sums as i want.

I can't explain now but its all involving the weyl differintegral.

mike3 · (This post was last modified: 05/05/2014, 01:44 AM by mike3.)

(05/04/2014, 07:27 PM)JmsNxn Wrote: 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 $|\frac{F(-z)}{(^{-z} e)}|$

However, $F(n) = {^{n+1}} e$ would make this condition hold. So if that doesn't work, then there must be additional conditions beyond that.

Also, if it's not supposed to interpolate tetration, then what about just interpolating a sequence which grows as fast or faster than tetration but is not tetration? E.g. use the interpolation method I mentioned but with a different sequence, e.g. perhaps not tetration itself but something close to it, so that the function in the absolute value still behaves like an exponential or otherwise decays to 0. What about $F(n)$ as I gave but with a small perturbation term like $F(n) + 2^{-n}$ , or something? Then it no longer interpolates tetration but the terms in the series will asymptotically approach those of the exponential function, so it should have similar asymptotic behavior, I'd think, in particular the decay to 0.

Also, with regards to the "bounded" approach, the given $F(n)$ that I mentioned seemed like it might be a good candidate for that too -- in particular, it demonstrates strong growth as $\Re(z) \rightarrow +\infty$ , regardless of the value of $\Im(z)$ , so it might be able to cancel out tetration's (or reciprocal tetration's) growth. However, it has zeros in the right half-plane, which would mean dividing by it would produce nasty poles, i.e. the quotient would fail to be holomorphic in the right half-plane and be only meromorphic. So the question then becomes as to whether there's a function with such growth with no zeros in the right half-plane.

(05/04/2014, 07:27 PM)JmsNxn Wrote:
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 $\rho \ge 0$

Then

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

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

Hmm.

mike3 · 05/05/2014, 01:00 AM

(05/04/2014, 03:28 PM)sheldonison Wrote:
(05/04/2014, 11:50 AM)mike3 Wrote: WOW, this is easier than I thought...

All we do is take that $a_n = {^n} e$ and $c_n = {^n} e$ (there seems to be a mistake as the post as written seems to suggest we should take $c_n = -{^n} e$ and that doesn't seem to work), and $z_n = n$ (the indexes at which we are to interpolate), which gives

$F(z) = \sum_{n=0}^{\infty} e^{({^n} e) (z - n)} ({^n} e) \frac{(-1)^n \sin(\pi z)}{\pi (z - n)}$

as an entire interpolant for the tetrational sequence.
What are the values for F (0,1,2,3)?

$^0 e$ , $^1 e$ , $^2 e$ , $^3 e$ , ... .

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