# Tetration Forum

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Is there an entire function known that is close to sexp(z) near the positive real line ?

Such a function must exist.

Apart from interpolation I do not know in what way this should be constructed.

regards

tommy1729
This smells like an integral transform. And a proof with contour integration.

Exactly how is another thing.
Maybe something for James Nixon ? (user JmsNxn's name I assume)

regards

tommy1729
I got a function that is very close to tetration:

Let us take the function:

$\vartheta(w) = \sum_{n=0}^\infty \frac{w^n}{n!(^ne)}$

Then $\vartheta$ is exponential order zero.

If $\int_0^\infty |\vartheta(-w)|w^{\sigma-1}\,dw<\infty$ for $0 < \sigma < 1$

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

Which IS TETRATION and satisfies the recursion. However, I haven't been able to prove absolute convergence of that integral.

BUT! We can take:

$\phi(s) = \frac{1}{\Gamma(s)}\int_0^\infty e^{-\lambda w}\vartheta(-w)w^{s-1}\,dw$

for $0<\lambda<\epsilon$ and $\phi(s) \approx \frac{1}{(^{-s}e)}$ for small $\epsilon$ hopefully not too small to blow up.

Furthermore this can be made even better for $\forall s \in \mathbb{C}$

$\frac{1}{(^s e)} \approx \psi(s) = \frac{1}{\Gamma(-s)} (\sum_{n=0}^\infty \frac{(-1)^n}{n!(^n e)(n-s)} + \int_1^\infty e^{-\lambda w}\vartheta(-w)w^{-s-1}\,dw)$

These functions are so close to tetration it makes me want to punch a hole in the wall that I can' prove absolute convergence. -_- lol. It's Real to real and interpolates tetration too. The limit as $\lambda \to 0$ is tetration in both cases if it converges. but both functions are not the same $\psi \neq \phi$.
Im a bit puzzled by the " -s ".

e^[3] = e^(e^e)

But e^[-3] = ln(ln(ln(1))) = ln(ln(0)) = ??

So I do not know what the "-s" means.

However your last formula seems to end that problem so it seems

$\frac{1}{(^s e)} \approx \psi(s) = \frac{1}{\Gamma(-s)} (\sum_{n=0}^\infty \frac{(-1)^n}{n!(^n e)(n-s)} + \int_1^\infty e^{-\lambda w}\vartheta(-w)w^{-s-1}\,dw)$

Is the kind of formula we look for.

However, Im more concerned about wheither or not this is analytic !?
Convergeance is an intresting subject, but if it is analytic somewhere then by analytic continuation we can extend it.

But why should that be analytic ? Assuming it is analytic then its derivative must equal the term by term derivative. Can you show it - the term by term derivative formulation - converges ?

Actually I bet that this function is not analytic and I even bet sheldon agrees with me.

Intresting though. But I suspect you have mentioned this function before, not ?

regards

tommy1729
(04/29/2014, 09:25 PM)tommy1729 Wrote: [ -> ]Im a bit puzzled by the " -s ".

e^[3] = e^(e^e)

But e^[-3] = ln(ln(ln(1))) = ln(ln(0)) = ??

So I do not know what the "-s" means.

However your last formula seems to end that problem so it seems

$\frac{1}{(^s e)} \approx \psi(s) = \frac{1}{\Gamma(-s)} (\sum_{n=0}^\infty \frac{(-1)^n}{n!(^n e)(n-s)} + \int_1^\infty e^{-\lambda w}\vartheta(-w)w^{-s-1}\,dw)$

Is the kind of formula we look for.

However, Im more concerned about wheither or not this is analytic !?
Convergeance is an intresting subject, but if it is analytic somewhere then by analytic continuation we can extend it.

But why should that be analytic ? Assuming it is analytic then its derivative must equal the term by term derivative. Can you show it - the term by term derivative formulation - converges ?

Actually I bet that this function is not analytic and I even bet sheldon agrees with me.

Intresting though. But I suspect you have mentioned this function before, not ?

regards

tommy1729

very easy to see this is analytic. No idea what you're talking about. lol. Mellin transform of an entire function is analytic. Multiply it by the inverse gamma which is entire and voila, analytic and entire.
I'd explain why but it's literally an excercise in analysis 101. These are all uniform limits of analytic functions, so it's neessarily analytic.
(04/26/2014, 12:23 PM)tommy1729 Wrote: [ -> ]Is there an entire function known that is close to sexp(z) near the positive real line ?

Such a function must exist.

Apart from interpolation I do not know in what way this should be constructed.

regards

tommy1729
Tommy,
There are many examples of entire superexponentials, such as iterating

f(z)=b^z, where b=exp(1/e), jaydfox calls this "cheta(z)", fixpoint=e
f(z)=exp(z)-1, this turns out to be cheta(z)/e-1, fixpoint=0
f(z)=2sinh(z), fixpoint=0
f(z)=sinh(z), fixpoint=0
f(z)=b^z, from the upper fixed point for bases less than 1<b<exp(1/e)

All of them grow super-exponentially at the real axis, and they are entire, but they don't behave like sexp(z) near z=0.
- Sheldon
(04/29/2014, 11:21 PM)JmsNxn Wrote: [ -> ]
(04/29/2014, 09:25 PM)tommy1729 Wrote: [ -> ]Im a bit puzzled by the " -s ".

e^[3] = e^(e^e)

But e^[-3] = ln(ln(ln(1))) = ln(ln(0)) = ??

So I do not know what the "-s" means.

However your last formula seems to end that problem so it seems

$\frac{1}{(^s e)} \approx \psi(s) = \frac{1}{\Gamma(-s)} (\sum_{n=0}^\infty \frac{(-1)^n}{n!(^n e)(n-s)} + \int_1^\infty e^{-\lambda w}\vartheta(-w)w^{-s-1}\,dw)$

Is the kind of formula we look for.

However, Im more concerned about wheither or not this is analytic !?
Convergeance is an intresting subject, but if it is analytic somewhere then by analytic continuation we can extend it.

But why should that be analytic ? Assuming it is analytic then its derivative must equal the term by term derivative. Can you show it - the term by term derivative formulation - converges ?

Actually I bet that this function is not analytic and I even bet sheldon agrees with me.

Intresting though. But I suspect you have mentioned this function before, not ?

regards

tommy1729

very easy to see this is analytic. No idea what you're talking about. lol. Mellin transform of an entire function is analytic. Multiply it by the inverse gamma which is entire and voila, analytic and entire.
I'd explain why but it's literally an excercise in analysis 101. These are all uniform limits of analytic functions, so it's neessarily analytic.

So it is analytic.
To be entire it needs to be analytic everywhere , what requires that it converges everywhere.

regards

tommy1729
(04/30/2014, 10:56 AM)sheldonison Wrote: [ -> ]
(04/26/2014, 12:23 PM)tommy1729 Wrote: [ -> ]Is there an entire function known that is close to sexp(z) near the positive real line ?

Such a function must exist.

Apart from interpolation I do not know in what way this should be constructed.

regards

tommy1729
Tommy,
There are many examples of entire superexponentials, such as iterating

f(z)=b^z, where b=exp(1/e), jaydfox calls this "cheta(z)", fixpoint=e
f(z)=exp(z)-1, this turns out to be cheta(z)/e-1, fixpoint=0
f(z)=2sinh(z), fixpoint=0
f(z)=sinh(z), fixpoint=0
f(z)=b^z, from the upper fixed point for bases less than 1<b<exp(1/e)

All of them grow super-exponentially at the real axis, and they are entire, but they don't behave like sexp(z) near z=0.
- Sheldon

Right. Sorry I should have remembered.
(04/30/2014, 12:23 PM)tommy1729 Wrote: [ -> ]Your right, sorry.
So it is analytic.
To be entire it needs to be analytic everywhere , what requires that it converges everywhere.

regards

tommy1729
It is entire. It converges everywhere. Should've said that.