# Tetration Forum

Full Version: Could be tetration if this integral converges
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(05/07/2014, 03:25 AM)mike3 Wrote: [ -> ]$\beta(x) = \frac{1}{2 \pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \G(z)F(z)x^{-z}\,dx$

Yes that should read $dz$ My mistake ^_^

$\beta(x) = \frac{1}{2 \pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \G(z)F(z)x^{-z}\,dz$

As for your question on the tetration integral not working I hsould mention with $\lambda =1$ its no longer going to approximate tetration. I was just hoping we would still have some nice decay properties, but I guess not.
I'm really stumped on applying this to tetration at the moment. But I feel like theres got to be a way using fractional calculus, I'm just missing it.
(05/07/2014, 03:18 PM)JmsNxn Wrote: [ -> ]
(05/07/2014, 03:25 AM)mike3 Wrote: [ -> ]$\beta(x) = \frac{1}{2 \pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \G(z)F(z)x^{-z}\,dx$

Yes that should read $dz$ My mistake ^_^

$\beta(x) = \frac{1}{2 \pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} \G(z)F(z)x^{-z}\,dz$

As for your question on the tetration integral not working I hsould mention with $\lambda =1$ its no longer going to approximate tetration. I was just hoping we would still have some nice decay properties, but I guess not.
I'm really stumped on applying this to tetration at the moment. But I feel like theres got to be a way using fractional calculus, I'm just missing it.

I'm curious here: what kind of strategies are available for evaluating that integral for $\beta(x)$ analytically? Contour integration doesn't seem to work with $F(z) = \mathrm{tet}(z)$ due to the branch singularities in the left half-plane.
(05/03/2014, 01:19 AM)mike3 Wrote: [ -> ]JmsNxn,

I am a little suspicious of this method. In particular, I'm not sure the integral

$I = \int_{0}^{\infty} \vartheta(-x) x^{z-1} dx$, $0 < z < 1$

converges.

I totally agree. Although I added a minus sign to it, which I assume was a typo.

So lets think about $\vartheta(-x)$.

Since the Taylor coefficients of $\vartheta(-x)$ decay EXTREMELY FAST , I consider this as a function that is well approximated by a polynomial for a long time.
( many remainder theorems for Taylor series imply this )

This means the main behaviour of this $\vartheta(-x)$ is like $(-1)^n a_n x^n$ where n increases slowly with x.

This implies that $\vartheta(-x)$ is not bounded by a polynomial and also that $\vartheta(-x)$ = 0 infinitely often.

Therefore the integral diverges.

Even if we consider taking the limit of x going to +oo as the limit of the sequence x_i with $\vartheta(-x_i)$ = 0.

regards

tommy1729
If $F$ is holomorphic for $\Re(z) < b$ and $|F(z)| < C e^{\alpha |\im(z)| + \rho|\Re(z)|}$ then:

$\beta(x) = \sum_{n=0}^\infty \frac{(-x)^n}{n!}F(-n)$

When F has singularities we see we pull on a second balancing function $\psi$ such that:

$\beta(x) = \psi(x) + \sum_{n=0}^\infty \frac{(-x)^n}{n!}F(-n)$

For simple functions like $F(z) = 1/(z^2 - 1)$ then $\psi$ is very easy to calculate. For more complicated functions like tetration, which no doubt has essential singularities instead of poles, it becomes much more complicated. But the general result on essential singularities is just applying cauchy's residue formula around all the poles of $F$ on the function $\G(z)x^{-z}$

As in, if $F(z) = G(z) + \sum_{n=0}^\infty \frac{a_n}{(z-\alpha)^{n+1}$ where $G(z)$ is analytic in a neighbourhood of $\alpha$ then:

$\psi(x) = g(x) + \sum_{n=0}^\infty a_n [\frac{d^n}{dz^n} \G(z)x^{-z} ]_{z=\alpha}$

where $g(x)$ carries information about the other poles of $F$

I haven't looked into much of how these balancing functions behave. I'm more familiar with just working with entire $\beta$ and entire $F$. I wish I could help you more on this but I feel discouraged looking at tetration. I think my iteration method might be restricted to simpler functions that don't behave quite as eraddictly.

And on a different note. I've successfully shown that, if $\phi$ is holomorphic in the strip $0< a-1 < \Re(z) < b$ and satisfies the bounds $|\phi(z)| < Ce^{\alpha |\Im(z)|}$ for $0 \le \alpha < \pi/2$. Then for $a < \Re(z) < b$ and $\Re(s) > 0$ I can calculate:

$\bigtriangledown_z^{-s} \phi(z)$

where it satisfies the composition rule and interpolates the iterated continuum sum at natural values. Pcha!! Holomorphic in z and s. Pcha!

(05/11/2014, 04:26 PM)tommy1729 Wrote: [ -> ]I totally agree. Although I added a minus sign to it, which I assume was a typo.

So lets think about $\vartheta(-x)$.

Since the Taylor coefficients of $\vartheta(-x)$ decay EXTREMELY FAST , I consider this as a function that is well approximated by a polynomial for a long time.
( many remainder theorems for Taylor series imply this )

This means the main behaviour of this $\vartheta(-x)$ is like $(-1)^n a_n x^n$ where n increases slowly with x.

This implies that $\vartheta(-x)$ is not bounded by a polynomial and also that $\vartheta(-x)$ = 0 infinitely often.

Therefore the integral diverges.

Even if we consider taking the limit of x going to +oo as the limit of the sequence x_i with $\vartheta(-x_i)$ = 0.

regards

tommy1729

yes yes, I'm quite aware it diverges. That's only another trick we need to come up with to handle that. I have a few but I need to look deeper into the laplace transform.
(05/11/2014, 04:30 PM)JmsNxn Wrote: [ -> ]
(05/11/2014, 04:26 PM)tommy1729 Wrote: [ -> ]I totally agree. Although I added a minus sign to it, which I assume was a typo.

So lets think about $\vartheta(-x)$.

Since the Taylor coefficients of $\vartheta(-x)$ decay EXTREMELY FAST , I consider this as a function that is well approximated by a polynomial for a long time.
( many remainder theorems for Taylor series imply this )

This means the main behaviour of this $\vartheta(-x)$ is like $(-1)^n a_n x^n$ where n increases slowly with x.

This implies that $\vartheta(-x)$ is not bounded by a polynomial and also that $\vartheta(-x)$ = 0 infinitely often.

Therefore the integral diverges.

Even if we consider taking the limit of x going to +oo as the limit of the sequence x_i with $\vartheta(-x_i)$ = 0.

regards

tommy1729

yes yes, I'm quite aware it diverges. That's only another trick we need to come up with to handle that. I have a few but I need to look deeper into the laplace transform.

I assumed you were aware of it. But some readers might not have been convinced.
With that in the back of my mind, I felt the neccessity to reply.

Its pretty hard to combine the properties of convergeance and the functional equation with fractional calculus and integral transforms ... or so it seems.

Maybe a bit of topic but finding an approximation to $\exp^{[1/2]}(x)$ in terms of an integral transform seems to be closer to a solution due to the recent work and talk of myself and sheldon.

$\exp^{[1/2]}(x) = \theta(x) \int_{\0}^{\infty} \((z (\2sinh^{[-1]}(z) - 1) )!)^{-1} x^{z}\,dz$

where $\theta(x)$ is bounded by a constant above and a constant below , and x > 27.

(And the factorial is computed with the gamma function ofcourse )

with approximation I mean that they have the same " growth rate ".

regards

tommy1729
(05/11/2014, 04:29 PM)JmsNxn Wrote: [ -> ]If $F$ is holomorphic for $\Re(z) < b$ and $|F(z)| < C e^{\alpha |\im(z)| + \rho|\Re(z)|}$ then:

$\beta(x) = \sum_{n=0}^\infty \frac{(-x)^n}{n!}F(-n)$

When F has singularities we see we pull on a second balancing function $\psi$ such that:

$\beta(x) = \psi(x) + \sum_{n=0}^\infty \frac{(-x)^n}{n!}F(-n)$

For simple functions like $F(z) = 1/(z^2 - 1)$ then $\psi$ is very easy to calculate. For more complicated functions like tetration, which no doubt has essential singularities instead of poles, it becomes much more complicated. But the general result on essential singularities is just applying cauchy's residue formula around all the poles of $F$ on the function $\G(z)x^{-z}$

Well, I guess then these formulas aren't of much use for continuum-summing tetration, since it is most definitely not bounded with the bound

$|F(z)| < C e^{\alpha |\Im(z)| + \rho |\Re(z)|}$

. In fact it is unbounded on the right half-plane (where it behaves chaotically) and has branch point singularities (which are neither poles nor essential singularities) on the left half-plane (these are logarithmic, double-logarithmic, triple-logarithmic, etc. in that order at $z = -2$, $z = -3$, $z = -4$, ...).

I'm curious: how did you get that first formula? Is it possible to get a similar formula for

$\mathrm{semi}\beta_U(x) = \frac{1}{2\pi i} \int_{\sigma}^{\sigma + i\infty} \Gamma(z) F(z) x^{-z} dz$

and

$\mathrm{semi}\beta_L(x) = \frac{1}{2\pi i} \int_{\sigma - i\infty}^{\sigma} \Gamma(z) F(z) x^{-z} dz$

and $F$ satisfying the given bound? As then I might have something, perhaps. I'll have to see, though.
(05/11/2014, 11:26 PM)mike3 Wrote: [ -> ]I'm curious: how did you get that first formula? Is it possible to get a similar formula for

I'm indisposed at the moment but the formula is derived in the paper I posted. Its a very brief proof and follows from cauchy's residue theorem and a meromorphic representation of the Gamma function. I could write some of it out, but it wouldn't be completely formal and might not leave you convinced .
(05/12/2014, 01:44 AM)JmsNxn Wrote: [ -> ]
(05/11/2014, 11:26 PM)mike3 Wrote: [ -> ]I'm curious: how did you get that first formula? Is it possible to get a similar formula for

I'm indisposed at the moment but the formula is derived in the paper I posted. Its a very brief proof and follows from cauchy's residue theorem and a meromorphic representation of the Gamma function. I could write some of it out, but it wouldn't be completely formal and might not leave you convinced .

Contour integration, right?
(05/12/2014, 02:15 AM)mike3 Wrote: [ -> ]
(05/12/2014, 01:44 AM)JmsNxn Wrote: [ -> ]
(05/11/2014, 11:26 PM)mike3 Wrote: [ -> ]I'm curious: how did you get that first formula? Is it possible to get a similar formula for

I'm indisposed at the moment but the formula is derived in the paper I posted. Its a very brief proof and follows from cauchy's residue theorem and a meromorphic representation of the Gamma function. I could write some of it out, but it wouldn't be completely formal and might not leave you convinced .

Contour integration, right?

Yeah. It's a very nifty contour integral. The gamma function is just so beautiful, I wish I could just kiss it.
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