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 \( \eta = e^{\frac{1}{e}} \) 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 \( \eta \) to construct hyper-operations base \( e \).

Start with the function \( \eta^z \) which has a neutral fixed point at \( e \). There is an attractive petal \( \mathcal{P} \) which includes the line \( (-\infty,e) \). We can define an Abel function on this petal \( \alpha : \mathcal{P} \to \mathbb{C} \) such that,

\(

\alpha(\eta^z) = \alpha(z) + 1\\

\)

Let, \( \mathcal{U} = \{|z-e| < \delta\} \cap \mathcal{P} \) be a neighborhood of the point \( e \) in the attracting petal \( \mathcal{P} \). The function \( \alpha \) is univalent on \( \mathcal{U} \) (for a small enough \( \delta \)) in this neighborhood. Further, as \( z \to e \) in this neighborhood \( \Re(\alpha(z)) \to \infty \)--which corresponds to iterating \( \eta^z \) which converts to right translations. This allows us to define a function,

\(

\eta^{\circ s}(z) : \mathbb{C}_{\Re(s) > 0} \times \mathcal{U} \to \mathcal{U}\\

\)

Which can be written,

\(

\eta^{\circ s}(z) = \alpha^{-1}(\alpha(z) + s)\\

\)

Upon which, as \( |s|\to\infty \) while \( 0 \le \arg(s) < \pi/2 \) we get that \( \eta^{\circ s}(z) \to e \); as this relates to right translations which eventually approach the fixed point \( e \) and approaches in the manner \( \mathcal{O}(\frac{1}{\sqrt{\Re(s)}}) \). What we want now, is to know what happens as \( \Im(s) \to \pm\infty \); upon which it can't grow faster than \( \mathcal{O}(\Im(s)) \) by nature of Abel functions on neutral fixed points (this is an argument from John Milnor and the manner of constructing \( \alpha \)), 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),

\(

|\eta^{\circ s}(z)| \le C e^{\tau|\Im(s)| + \rho|\Re(s)|}\\

\)

For, \( 0 < \tau < \pi/2 \) and \( C,\rho > 0 \)--which is again, a very weak bound. And now we enter the inverse mellin transform. Let \( 0 < \sigma < 1 \) then,

\(

f(x) = \frac{1}{2\pi i} \int_{\sigma - i\infty}^{\sigma + i\infty} \Gamma(s) \eta^{\circ 1-s}(z) x^{-s}\,ds\\

\)

The poles of this integrand in the left half plane are at \( s = -n \) and their residues are,

\(

\text{Res}(s=-n;\Gamma(s) \eta^{\circ 1-s}(z) x^{-s}) = \eta^{\circ n+1}(z) \frac{(-x)^n}{n!}\\

\)

Since we have the exponential bounds of \( \eta^{\circ s} \) and that,

\(

\Gamma(s) \sim \sqrt{2\pi}s^{s-\frac{1}{2}}e^{-s}\,\,\text{as}\,\,|s|\to\infty\,\,\text{while}\,\,|\arg(s)| < \pi\\

\)

By a standard exercise in contour integration, we get,

\(

f(x) = \sum_{n=0}^\infty \eta^{\circ n+1}(z) \frac{(-x)^n}{n!}\\

\)

And also, by Mellin's inversion theorem, we know that,

\(

\int_0^\infty |f(x)|x^{\sigma - 1}\,dx < \infty\\

\int_0^\infty f(x)x^{s-1}\,dx = \eta^{\circ 1-s}(z)\Gamma(s)\\

\)

Since \( f \) is entire, we can analyticially continue this expression by breaking the integral into \( \int_0^1 + \int_1^\infty \)--and we make the substitution \( 1-s \mapsto s \) so that,

\(

\Gamma(1-s) \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\\

\)

Which provides a formula for \( \eta^{\circ s} \) depending solely on the natural iterates \( \eta^{\circ n+1} \). Now since, there exists \( K \) such that \( \eta^{\circ K}(1) \in \mathcal{U} \), which implies the function \( \eta^{\circ s}(1) \) is holomorphic for \( \Re(s) > K \). And now we extend this function to its maximal domain \( \mathbb{C}/(-\infty,-2] \). To do this, we go by induction.

Assume that \( \eta^{\circ s}(1) \) is holomorphic for \( \Re(s) > K \) and \( \Im(s) > 0 \). Assume that,

\(

\eta^{\circ s_0}(1) = \eta^{\circ s_1}(1)\\

\)

Then certainly for all \( k \in \mathbb{N} \),

\(

\eta^{\circ s_0 + k} = \eta^{\circ s_1 + k}

\)

Take \( N \) such that \( f_{01}(s) = \eta^{\circ s_{01} + s + N} \) is holomorphic for \( \Re(s) > 0 \). Then, for all \( n \in \mathbb{N} \),

\(

f_0(n) = f_1(n)\\

\)

Now the function \( f_0 - f_1 \) is exponentially bounded as above. And therefore it's inverse mellin transform is,

\(

\sum_{n=0}^\infty (f_0(n+1) - f_1(n+1)) \frac{(-x)^n}{n!} = 0\\

\)

But, by the invertibility of the mellin transform, when we take the mellin transform we must get that it is zero while also equaling \( f_0 - f_1 \). And therefore,

\(

f_0 - f_1 = 0\\

f_0 = f_1\\

\)

This then implies that \( \eta^{\circ s_0 + s + N} = \eta^{\circ s_1 + s + N} \) for all \( s \). This implies that \( s_1 - s_0 \) is a period of \( \eta^{\circ s} \); but \( \eta^{\circ s} \) has no period, therefore \( s_0 = s_1 \). Therefore, since \( \eta^{\circ -1}(1) = 0 \) there can be no other point where \( \eta^{\circ s}(1) = 0 \). Therefore the logarithm of \( \eta^{\circ s} \) is always defined for \( \Im(s) > 0 \). A similar analysis works for \( \Im(s) < 0 \). Therefore,

\(

\eta^{\circ s}(1)\,\,\text{is holomorphic for}\,\,\mathbb{C}/(-\infty,-2]\\

\)

Further, on the attractive petal \( \mathcal{P} \) we have that,

\(

\eta^{\circ s} : \mathbb{C}/(-\infty,-2] \to \mathcal{P}\,\,\text{bijectively}\\

\)

This implies a couple thing. First of all, if \( z \in \mathcal{P} \) then \( z+ 2\pi i e \not \in \mathcal{P} \) because \( \eta^{z} = \eta^{z+2\pi i e} \) and this is impossible by injectivity of tetration. We have a holomorphic slog function which is very well behaved, \( \text{slog} : \mathcal{P} \to \mathbb{C}/(-\infty,-2] \)--which is equally as bijective.

When we use \( f(z) = e\log(z) \) which is the functional inverse of \( \eta^z \) , we know that \( f \) is repelling on the petal \( \mathcal{P} \) and tends to infinity. This implies that \( \eta^{\circ s}(z) \to \infty \) while \( |s|\to\infty \) for \( 0 < |\arg(-s)| < \pi/2 \).

Now here is where we start for constructing a basechange function. Typically as I've seen attempts, we would use the unbounded branch of \( \eta \) (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 \( \varphi \) such that,

\(

\varphi : (-\infty,e)\to(-\infty,\infty)\,\,\text{bijectively}\\

\varphi: \mathcal{P} \to \mathbb{C}\\

\varphi(\infty \pm it) = L^{\pm}\,\,\text{where}\,\,e^{L^{\pm}} = L^{\pm}\,\,\text{and}\,\,\overline{L^{\pm}} = L^{\mp}\\

\varphi(0) = 0\\

\overline{\varphi(z)} = \varphi(\overline{z})\\

\)

So we really only care about \( \varphi \) for \( \Im(s) \ge 0, s\not \in (e,\infty) \). Now, it's obvious there are many functions which satisfy these properties; of which the goal I have in mind, is to sequentially approximate \( \varphi \) with \( \varphi_n \) in which,

\(

e^{\varphi(z)} = \varphi(\eta^z)\\

\)

Now the existence of these functions are ABSOLUTELY guaranteed. The example I have in mind is to use Kneser's tetration, \( \text{tet}_{K}(z) \) and simply take,

\(

\varphi(z) = \text{tet}_{K}(\text{slog}_{\eta}(z))\\

\)

But, as the reader may guess, this is cheating. We use Kneser's solution to get \( \varphi \), 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 \( \varphi \) 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 \( \eta \)'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 \( \varphi \) as \( |z| \to \infty \) for \( 0 < |\arg(-z)| < \pi/2 \) where \( \varphi(z) \to L^{\pm} \) 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,

\(

\varphi_n(z) = \exp^{\circ n}(L^{\pm} + f(\text{slog}_\eta(z) - n)}\\

\)

satisfies,

\(

e^{\varphi_n(z)} = \varphi_{n+1}(\eta^z)\\

\)

For a function \( f \). If we were able to choose this \( f \) properly so that, convergence is guaranteed and the properties of \( \varphi \) are satisfied for each \( \varphi_n \), we may have an approach as solving these equations. Finding a function \( f \) 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 \( e^z \).

\(

\Psi^{-1}(z) : \mathbb{C} \to \mathbb{C}/\{0\}\\

\)

Where,

\(

\varphi(z) = \Psi^{-1}(e^{L\text{slog}_\eta(z) + \theta(\text{slog}_{\eta}(z))})\\

\)

Where \( \theta \) is a \( 1 \)-periodic function. Let's let this be holomorphic for \( \Im(z) > 0 \). This function will satisfy the identity, \( \theta(\text{slog}_\eta(\eta^z)) = \theta(\text{slog}_\eta(z)) \) and,

\(

e^{\varphi(z)} = e^{\Psi^{-1}(e^{L\text{slog}_\eta(z) + \theta(\text{slog}_{\eta}(z))})}\\

=\Psi^{-1}(e^{L\text{slog}_\eta(z) +L+ \theta(\text{slog}_{\eta}(z))})\\

= \Psi^{-1}(e^{L(\text{slog}_\eta(z)+1) + \theta(\text{slog}_{\eta}(z))})\\

=\Psi^{-1}(e^{L\text{slog}_\eta(\eta^z) + \theta(\text{slog}_{\eta}(\eta^z))})\\

=\varphi(\eta^z)\\

\)

This function will pretty much satisfy all our requirements of \( \varphi \) except for the real-valued requirement, unless we choose \( \theta \) properly. Where, spoiler alert, Kneser's choice is the right choice. Now, we have restricted our attention to the function \( h(z) = \theta(\text{slog}(z)) \); but now we are asking for a function \( h \) which is invariant under the operator \( \mathcal{C}_\eta : f(z) \mapsto f(\eta^z) \). Which is \( h(\eta^z) = h(z) \) for \( \Im(z) > 0 \). And I wonder if we can find a suitable function \( F \) by using a sequence of functions \( h_n \).

Now, the main reason I chose to write all this out, is a question about hyper-operators. Does there exist a tetration base \( \eta_1 \) such that,

\(

\eta_1(z) : \mathbb{C}/(-\infty,-2] \to \mathbb{C}\\

\)

Such that,

\(

\eta_1(x_0) = x_0\\

\eta_1'(x_0) = 1\\

\eta_1^{\circ n}(1) \to x_0\,\,\text{as}\,\,n\to\infty\\

\)

If there is, we can construct a pentation function which satisfies pretty much all the above properties that \( \eta^{\circ s} \) 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 \( \sqrt{2} \) and we used \( \eta \)).

All in all, I'm not quite too sure if this is doable. But upon which we'd have the function,

\(

\text{tet}(s) = \varphi(\eta^{\circ s}(1))

\)

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 \( \eta = e^{\frac{1}{e}} \) 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 \( \eta \) to construct hyper-operations base \( e \).

Start with the function \( \eta^z \) which has a neutral fixed point at \( e \). There is an attractive petal \( \mathcal{P} \) which includes the line \( (-\infty,e) \). We can define an Abel function on this petal \( \alpha : \mathcal{P} \to \mathbb{C} \) such that,

\(

\alpha(\eta^z) = \alpha(z) + 1\\

\)

Let, \( \mathcal{U} = \{|z-e| < \delta\} \cap \mathcal{P} \) be a neighborhood of the point \( e \) in the attracting petal \( \mathcal{P} \). The function \( \alpha \) is univalent on \( \mathcal{U} \) (for a small enough \( \delta \)) in this neighborhood. Further, as \( z \to e \) in this neighborhood \( \Re(\alpha(z)) \to \infty \)--which corresponds to iterating \( \eta^z \) which converts to right translations. This allows us to define a function,

\(

\eta^{\circ s}(z) : \mathbb{C}_{\Re(s) > 0} \times \mathcal{U} \to \mathcal{U}\\

\)

Which can be written,

\(

\eta^{\circ s}(z) = \alpha^{-1}(\alpha(z) + s)\\

\)

Upon which, as \( |s|\to\infty \) while \( 0 \le \arg(s) < \pi/2 \) we get that \( \eta^{\circ s}(z) \to e \); as this relates to right translations which eventually approach the fixed point \( e \) and approaches in the manner \( \mathcal{O}(\frac{1}{\sqrt{\Re(s)}}) \). What we want now, is to know what happens as \( \Im(s) \to \pm\infty \); upon which it can't grow faster than \( \mathcal{O}(\Im(s)) \) by nature of Abel functions on neutral fixed points (this is an argument from John Milnor and the manner of constructing \( \alpha \)), 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),

\(

|\eta^{\circ s}(z)| \le C e^{\tau|\Im(s)| + \rho|\Re(s)|}\\

\)

For, \( 0 < \tau < \pi/2 \) and \( C,\rho > 0 \)--which is again, a very weak bound. And now we enter the inverse mellin transform. Let \( 0 < \sigma < 1 \) then,

\(

f(x) = \frac{1}{2\pi i} \int_{\sigma - i\infty}^{\sigma + i\infty} \Gamma(s) \eta^{\circ 1-s}(z) x^{-s}\,ds\\

\)

The poles of this integrand in the left half plane are at \( s = -n \) and their residues are,

\(

\text{Res}(s=-n;\Gamma(s) \eta^{\circ 1-s}(z) x^{-s}) = \eta^{\circ n+1}(z) \frac{(-x)^n}{n!}\\

\)

Since we have the exponential bounds of \( \eta^{\circ s} \) and that,

\(

\Gamma(s) \sim \sqrt{2\pi}s^{s-\frac{1}{2}}e^{-s}\,\,\text{as}\,\,|s|\to\infty\,\,\text{while}\,\,|\arg(s)| < \pi\\

\)

By a standard exercise in contour integration, we get,

\(

f(x) = \sum_{n=0}^\infty \eta^{\circ n+1}(z) \frac{(-x)^n}{n!}\\

\)

And also, by Mellin's inversion theorem, we know that,

\(

\int_0^\infty |f(x)|x^{\sigma - 1}\,dx < \infty\\

\int_0^\infty f(x)x^{s-1}\,dx = \eta^{\circ 1-s}(z)\Gamma(s)\\

\)

Since \( f \) is entire, we can analyticially continue this expression by breaking the integral into \( \int_0^1 + \int_1^\infty \)--and we make the substitution \( 1-s \mapsto s \) so that,

\(

\Gamma(1-s) \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\\

\)

Which provides a formula for \( \eta^{\circ s} \) depending solely on the natural iterates \( \eta^{\circ n+1} \). Now since, there exists \( K \) such that \( \eta^{\circ K}(1) \in \mathcal{U} \), which implies the function \( \eta^{\circ s}(1) \) is holomorphic for \( \Re(s) > K \). And now we extend this function to its maximal domain \( \mathbb{C}/(-\infty,-2] \). To do this, we go by induction.

Assume that \( \eta^{\circ s}(1) \) is holomorphic for \( \Re(s) > K \) and \( \Im(s) > 0 \). Assume that,

\(

\eta^{\circ s_0}(1) = \eta^{\circ s_1}(1)\\

\)

Then certainly for all \( k \in \mathbb{N} \),

\(

\eta^{\circ s_0 + k} = \eta^{\circ s_1 + k}

\)

Take \( N \) such that \( f_{01}(s) = \eta^{\circ s_{01} + s + N} \) is holomorphic for \( \Re(s) > 0 \). Then, for all \( n \in \mathbb{N} \),

\(

f_0(n) = f_1(n)\\

\)

Now the function \( f_0 - f_1 \) is exponentially bounded as above. And therefore it's inverse mellin transform is,

\(

\sum_{n=0}^\infty (f_0(n+1) - f_1(n+1)) \frac{(-x)^n}{n!} = 0\\

\)

But, by the invertibility of the mellin transform, when we take the mellin transform we must get that it is zero while also equaling \( f_0 - f_1 \). And therefore,

\(

f_0 - f_1 = 0\\

f_0 = f_1\\

\)

This then implies that \( \eta^{\circ s_0 + s + N} = \eta^{\circ s_1 + s + N} \) for all \( s \). This implies that \( s_1 - s_0 \) is a period of \( \eta^{\circ s} \); but \( \eta^{\circ s} \) has no period, therefore \( s_0 = s_1 \). Therefore, since \( \eta^{\circ -1}(1) = 0 \) there can be no other point where \( \eta^{\circ s}(1) = 0 \). Therefore the logarithm of \( \eta^{\circ s} \) is always defined for \( \Im(s) > 0 \). A similar analysis works for \( \Im(s) < 0 \). Therefore,

\(

\eta^{\circ s}(1)\,\,\text{is holomorphic for}\,\,\mathbb{C}/(-\infty,-2]\\

\)

Further, on the attractive petal \( \mathcal{P} \) we have that,

\(

\eta^{\circ s} : \mathbb{C}/(-\infty,-2] \to \mathcal{P}\,\,\text{bijectively}\\

\)

This implies a couple thing. First of all, if \( z \in \mathcal{P} \) then \( z+ 2\pi i e \not \in \mathcal{P} \) because \( \eta^{z} = \eta^{z+2\pi i e} \) and this is impossible by injectivity of tetration. We have a holomorphic slog function which is very well behaved, \( \text{slog} : \mathcal{P} \to \mathbb{C}/(-\infty,-2] \)--which is equally as bijective.

When we use \( f(z) = e\log(z) \) which is the functional inverse of \( \eta^z \) , we know that \( f \) is repelling on the petal \( \mathcal{P} \) and tends to infinity. This implies that \( \eta^{\circ s}(z) \to \infty \) while \( |s|\to\infty \) for \( 0 < |\arg(-s)| < \pi/2 \).

Now here is where we start for constructing a basechange function. Typically as I've seen attempts, we would use the unbounded branch of \( \eta \) (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 \( \varphi \) such that,

\(

\varphi : (-\infty,e)\to(-\infty,\infty)\,\,\text{bijectively}\\

\varphi: \mathcal{P} \to \mathbb{C}\\

\varphi(\infty \pm it) = L^{\pm}\,\,\text{where}\,\,e^{L^{\pm}} = L^{\pm}\,\,\text{and}\,\,\overline{L^{\pm}} = L^{\mp}\\

\varphi(0) = 0\\

\overline{\varphi(z)} = \varphi(\overline{z})\\

\)

So we really only care about \( \varphi \) for \( \Im(s) \ge 0, s\not \in (e,\infty) \). Now, it's obvious there are many functions which satisfy these properties; of which the goal I have in mind, is to sequentially approximate \( \varphi \) with \( \varphi_n \) in which,

\(

e^{\varphi(z)} = \varphi(\eta^z)\\

\)

Now the existence of these functions are ABSOLUTELY guaranteed. The example I have in mind is to use Kneser's tetration, \( \text{tet}_{K}(z) \) and simply take,

\(

\varphi(z) = \text{tet}_{K}(\text{slog}_{\eta}(z))\\

\)

But, as the reader may guess, this is cheating. We use Kneser's solution to get \( \varphi \), 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 \( \varphi \) 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 \( \eta \)'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 \( \varphi \) as \( |z| \to \infty \) for \( 0 < |\arg(-z)| < \pi/2 \) where \( \varphi(z) \to L^{\pm} \) 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,

\(

\varphi_n(z) = \exp^{\circ n}(L^{\pm} + f(\text{slog}_\eta(z) - n)}\\

\)

satisfies,

\(

e^{\varphi_n(z)} = \varphi_{n+1}(\eta^z)\\

\)

For a function \( f \). If we were able to choose this \( f \) properly so that, convergence is guaranteed and the properties of \( \varphi \) are satisfied for each \( \varphi_n \), we may have an approach as solving these equations. Finding a function \( f \) 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 \( e^z \).

\(

\Psi^{-1}(z) : \mathbb{C} \to \mathbb{C}/\{0\}\\

\)

Where,

\(

\varphi(z) = \Psi^{-1}(e^{L\text{slog}_\eta(z) + \theta(\text{slog}_{\eta}(z))})\\

\)

Where \( \theta \) is a \( 1 \)-periodic function. Let's let this be holomorphic for \( \Im(z) > 0 \). This function will satisfy the identity, \( \theta(\text{slog}_\eta(\eta^z)) = \theta(\text{slog}_\eta(z)) \) and,

\(

e^{\varphi(z)} = e^{\Psi^{-1}(e^{L\text{slog}_\eta(z) + \theta(\text{slog}_{\eta}(z))})}\\

=\Psi^{-1}(e^{L\text{slog}_\eta(z) +L+ \theta(\text{slog}_{\eta}(z))})\\

= \Psi^{-1}(e^{L(\text{slog}_\eta(z)+1) + \theta(\text{slog}_{\eta}(z))})\\

=\Psi^{-1}(e^{L\text{slog}_\eta(\eta^z) + \theta(\text{slog}_{\eta}(\eta^z))})\\

=\varphi(\eta^z)\\

\)

This function will pretty much satisfy all our requirements of \( \varphi \) except for the real-valued requirement, unless we choose \( \theta \) properly. Where, spoiler alert, Kneser's choice is the right choice. Now, we have restricted our attention to the function \( h(z) = \theta(\text{slog}(z)) \); but now we are asking for a function \( h \) which is invariant under the operator \( \mathcal{C}_\eta : f(z) \mapsto f(\eta^z) \). Which is \( h(\eta^z) = h(z) \) for \( \Im(z) > 0 \). And I wonder if we can find a suitable function \( F \) by using a sequence of functions \( h_n \).

Now, the main reason I chose to write all this out, is a question about hyper-operators. Does there exist a tetration base \( \eta_1 \) such that,

\(

\eta_1(z) : \mathbb{C}/(-\infty,-2] \to \mathbb{C}\\

\)

Such that,

\(

\eta_1(x_0) = x_0\\

\eta_1'(x_0) = 1\\

\eta_1^{\circ n}(1) \to x_0\,\,\text{as}\,\,n\to\infty\\

\)

If there is, we can construct a pentation function which satisfies pretty much all the above properties that \( \eta^{\circ s} \) 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 \( \sqrt{2} \) and we used \( \eta \)).

All in all, I'm not quite too sure if this is doable. But upon which we'd have the function,

\(

\text{tet}(s) = \varphi(\eta^{\circ s}(1))

\)

Any comments, suggested reading, questions regarding my construction, would be greatly appreciated,

Regards, James