03/17/2021, 11:15 PM
Hey, everyone.
I made a post last night, but it really wasn't what I wanted, so I deleted it today and decided to write it out better this time. I'll begin by constructing tetration base
and proving some properties about it, and then branch off into how I envision a base-change method looking. This is intended as a manner to use hyper-operations base 
to construct hyper-operations base 
.
Start with the function
which has a neutral fixed point at . There is an attractive petal 
which includes the line )
. We can define an Abel function on this petal 
such that,
 = \alpha(z) + 1\\<br />
)
Let,
be a neighborhood of the point in the attracting petal . The function 
is univalent on 
(for a small enough 
) in this neighborhood. Further, as 
in this neighborhood ) \to \infty)
--which corresponds to iterating which converts to right translations. This allows us to define a function,
 : \mathbb{C}_{\Re(s) > 0} \times \mathcal{U} \to \mathcal{U}\\<br />
)
Which can be written,
 = \alpha^{-1}(\alpha(z) + s)\\<br />
)
Upon which, as
while  < \pi/2)
we get that  \to e)
; as this relates to right translations which eventually approach the fixed point and approaches in the manner }}))
. What we want now, is to know what happens as  \to \pm\infty)
; upon which it can't grow faster than ))
by nature of Abel functions on neutral fixed points (this is an argument from John Milnor and the manner of constructing ), where at infinity it looks like the identity function through a substitution.
So now we are safe and have the following bounds (which are fairly weak from what we can say, but all we need),
| \le C e^{\tau|\Im(s)| + \rho|\Re(s)|}\\<br />
)
For,
and 
--which is again, a very weak bound. And now we enter the inverse mellin transform. Let 
then,
 = \frac{1}{2\pi i} \int_{\sigma - i\infty}^{\sigma + i\infty} \Gamma(s) \eta^{\circ 1-s}(z) x^{-s}\,ds\\<br />
)
The poles of this integrand in the left half plane are at
and their residues are,
 \eta^{\circ 1-s}(z) x^{-s}) = \eta^{\circ n+1}(z) \frac{(-x)^n}{n!}\\<br />
)
Since we have the exponential bounds of
and that,
 \sim \sqrt{2\pi}s^{s-\frac{1}{2}}e^{-s}\,\,\text{as}\,\,|s|\to\infty\,\,\text{while}\,\,|\arg(s)| < \pi\\ <br />
)
By a standard exercise in contour integration, we get,
 = \sum_{n=0}^\infty \eta^{\circ n+1}(z) \frac{(-x)^n}{n!}\\<br />
)
And also, by Mellin's inversion theorem, we know that,
|x^{\sigma - 1}\,dx < \infty\\<br />
\int_0^\infty f(x)x^{s-1}\,dx = \eta^{\circ 1-s}(z)\Gamma(s)\\<br />
)
Since
is entire, we can analyticially continue this expression by breaking the integral into 
--and we make the substitution 
so that,
 \eta^{\circ s}(z) = \sum_{n=0}^\infty \eta^{\circ n+1}(z)\frac{(-1)^n}{n!(n+1-s)} + \int_1^\infty (\sum_{n=0}^\infty\eta^{\circ n+1}(z) \frac{(-x)^n}{n!})x^{-s}\,dx\\<br />
)
Which provides a formula for depending solely on the natural iterates 
. Now since, there exists 
such that  \in \mathcal{U})
, which implies the function )
is holomorphic for  > K)
. And now we extend this function to its maximal domain 
. To do this, we go by induction.
Assume that is holomorphic for and  > 0)
. Assume that,
 = \eta^{\circ s_1}(1)\\<br />
)
Then certainly for all
,
Take
such that  = \eta^{\circ s_{01} + s + N})
is holomorphic for  > 0)
. Then, for all 
,
 = f_1(n)\\<br />
)
Now the function
is exponentially bounded as above. And therefore it's inverse mellin transform is,
 - f_1(n+1)) \frac{(-x)^n}{n!} = 0\\<br />
)
But, by the invertibility of the mellin transform, when we take the mellin transform we must get that it is zero while also equaling. And therefore,
This then implies that
for all 
. This implies that 
is a period of ; but has no period, therefore 
. Therefore, since  = 0)
there can be no other point where  = 0)
. Therefore the logarithm of is always defined for . A similar analysis works for  < 0)
. Therefore,
\,\,\text{is holomorphic for}\,\,\mathbb{C}/(-\infty,-2]\\<br />
)
Further, on the attractive petal we have that,
This implies a couple thing. First of all, if
then 
because 
and this is impossible by injectivity of tetration. We have a holomorphic slog function which is very well behaved, 
--which is equally as bijective.
When we use = e\log(z))
which is the functional inverse of , we know that is repelling on the petal and tends to infinity. This implies that  \to \infty)
while for | < \pi/2)
.
Now here is where we start for constructing a basechange function. Typically as I've seen attempts, we would use the unbounded branch of (the chi function, I believe we called it)--I want to use the bounded tetration to construct unbounded tetration.
Let's try to look for a function
such that,
\to(-\infty,\infty)\,\,\text{bijectively}\\<br />
\varphi: \mathcal{P} \to \mathbb{C}\\<br />
\varphi(\infty \pm it) = L^{\pm}\,\,\text{where}\,\,e^{L^{\pm}} = L^{\pm}\,\,\text{and}\,\,\overline{L^{\pm}} = L^{\mp}\\<br />
\varphi(0) = 0\\<br />
\overline{\varphi(z)} = \varphi(\overline{z})\\<br />
)
So we really only care about for  \ge 0, s\not \in (e,\infty))
. Now, it's obvious there are many functions which satisfy these properties; of which the goal I have in mind, is to sequentially approximate with 
in which,
} = \varphi(\eta^z)\\<br />
)
Now the existence of these functions are ABSOLUTELY guaranteed. The example I have in mind is to use Kneser's tetration,)
and simply take,
 = \text{tet}_{K}(\text{slog}_{\eta}(z))\\<br />
)
But, as the reader may guess, this is cheating. We use Kneser's solution to get, which is putting the cart before the horse really. But I'm wondering if we can use what we know about Kneser's tetration, to attempt to construct independently.
Usually the constructive manner of creating this function is to use iterated logs. This won't work in this case because we are using the bounded branch of's iteration. And largely, by this point, I believe the iterated log method has become sort of hopeless to produce holomorphy. Instead, we want to look at the behaviour of as 
for | < \pi/2)
where  \to L^{\pm})
a complex fixed point, and use this to approximate way off in the left half plane. I am not certain how to do this, but I'm experimenting with certain processes.
For example,
 = \exp^{\circ n}(L^{\pm} + f(\text{slog}_\eta(z) - n)}\\<br />
)
satisfies,
} = \varphi_{n+1}(\eta^z)\\<br />
)
For a function. If we were able to choose this properly so that, convergence is guaranteed and the properties of are satisfied for each , we may have an approach as solving these equations. Finding a function will be the difficult part here. The alternative way I think we might be able to do this; rather than dealing directly with iterated exponentials, we look at the Inverse Schroder function of 
.
 : \mathbb{C} \to \mathbb{C}/\{0\}\\<br />
)
Where,
 = \Psi^{-1}(e^{L\text{slog}_\eta(z) + \theta(\text{slog}_{\eta}(z))})\\<br />
)
Where
is a 
-periodic function. Let's let this be holomorphic for  > 0)
. This function will satisfy the identity, ) = \theta(\text{slog}_\eta(z)))
and,
} = e^{\Psi^{-1}(e^{L\text{slog}_\eta(z) + \theta(\text{slog}_{\eta}(z))})}\\<br />
=\Psi^{-1}(e^{L\text{slog}_\eta(z) +L+ \theta(\text{slog}_{\eta}(z))})\\<br />
= \Psi^{-1}(e^{L(\text{slog}_\eta(z)+1) + \theta(\text{slog}_{\eta}(z))})\\<br />
=\Psi^{-1}(e^{L\text{slog}_\eta(\eta^z) + \theta(\text{slog}_{\eta}(\eta^z))})\\<br />
=\varphi(\eta^z)\\<br />
)
This function will pretty much satisfy all our requirements of except for the real-valued requirement, unless we choose properly. Where, spoiler alert, Kneser's choice is the right choice. Now, we have restricted our attention to the function  = \theta(\text{slog}(z)))
; but now we are asking for a function 
which is invariant under the operator  \mapsto f(\eta^z))
. Which is  = h(z))
for . And I wonder if we can find a suitable function 
by using a sequence of functions 
.
Now, the main reason I chose to write all this out, is a question about hyper-operators. Does there exist a tetration base
such that,
 : \mathbb{C}/(-\infty,-2] \to \mathbb{C}\\<br />
)
Such that,
 = x_0\\<br />
\eta_1'(x_0) = 1\\<br />
\eta_1^{\circ n}(1) \to x_0\,\,\text{as}\,\,n\to\infty\\<br />
)
If there is, we can construct a pentation function which satisfies pretty much all the above properties that satisfy--except we change the domains of holomorphy to suit pentation. But the mellin transform identity will still hold, and therefore so will the injectivity (which is crucial), non-periodicity will occur, so the inverse function will be unbounded; when using periodic functions the inverse function will be restricted to a horizontal strip the length of periodicity (why we didn't use 
and we used ).
All in all, I'm not quite too sure if this is doable. But upon which we'd have the function,
 = \varphi(\eta^{\circ s}(1))<br />
)
Any comments, suggested reading, questions regarding my construction, would be greatly appreciated,
Regards, James
I made a post last night, but it really wasn't what I wanted, so I deleted it today and decided to write it out better this time. I'll begin by constructing tetration base
Start with the function
Let,
Which can be written,
Upon which, as
So now we are safe and have the following bounds (which are fairly weak from what we can say, but all we need),
For,
The poles of this integrand in the left half plane are at
Since we have the exponential bounds of
By a standard exercise in contour integration, we get,
And also, by Mellin's inversion theorem, we know that,
Since
Which provides a formula for
Assume that
Then certainly for all
Take
Now the function
But, by the invertibility of the mellin transform, when we take the mellin transform we must get that it is zero while also equaling
This then implies that
Further, on the attractive petal
This implies a couple thing. First of all, if
When we use
Now here is where we start for constructing a basechange function. Typically as I've seen attempts, we would use the unbounded branch of
Let's try to look for a function
So we really only care about
Now the existence of these functions are ABSOLUTELY guaranteed. The example I have in mind is to use Kneser's tetration,
But, as the reader may guess, this is cheating. We use Kneser's solution to get
Usually the constructive manner of creating this function is to use iterated logs. This won't work in this case because we are using the bounded branch of
For example,
satisfies,
For a function
Where,
Where
This function will pretty much satisfy all our requirements of
Now, the main reason I chose to write all this out, is a question about hyper-operators. Does there exist a tetration base
Such that,
If there is, we can construct a pentation function which satisfies pretty much all the above properties that
All in all, I'm not quite too sure if this is doable. But upon which we'd have the function,
Any comments, suggested reading, questions regarding my construction, would be greatly appreciated,
Regards, James
