So, I've been fiddling with Kneser's tetration, and I have been trying to think of different ways to express it. Let's write \( \text{tet}_K \) for Kneser's tetration. Now depending on how it grows as \( \Re(z) \to \infty \) we have different manners of expressing it. But there is one way I can say for sure.

Let's take \( \varphi(z) = i\sqrt{z} \) which maps the upper half plane of \( \mathbb{C} \) to the upper left quadrant of \( \mathbb{C} \). As \( |z|\to\infty \) in the upper left quadrant \( \text{tet}_K(z) \to L \) so long as we stay away from the real line. This implies the function \( \text{tet}_K(i\sqrt{iz}) : \mathbb{C}_{\Re(z) > 0} \to \mathbb{C} \) and this function satisfies \( \text{tet}_K(i\sqrt{iz}) \to L \) as \( |z|\to\infty \) in \( \mathbb{C}_{\Re(z) > 0} \). And from this the Mellin transform is a viable option.

Therefore, if we write,

\(

a_n = \text{tet}_K(i\sqrt{i(n+1)})\\

\)

Then, for \( \Re(z) > 0 \),

\(

\Gamma(1-z)\text{tet}_K(i\sqrt{iz}) = \sum_{n=0}^\infty a_{n} \frac{(-1)^n}{n!(n+1-z)} + \int_1^\infty (\sum_{n=0}^\infty a_{n}\frac{(-x)^n}{n!})x^{-z}\,dx\\

\)

Or if you prefer, for \( \Im(z) > 0 \) and \( \Re(z) < 0 \),

\(

\Gamma(1-iz^2) \text{tet}_K(z) = \sum_{n=0}^\infty a_{n} \frac{(-1)^n}{n!(n+1-iz^2)} + \int_1^\infty (\sum_{n=0}^\infty a_{n}\frac{(-x)^n}{n!})x^{-iz^2}\,dx\\

\)

This form gives us a way of representing Kneser's tetration using only the data points \( \text{tet}_K(i\sqrt{i(n+1)}) \). What I would really like to know is how fast Kneser's tetration grows as \( \Re(z) \to \infty \) in the complex plane and how fast \( \frac{1}{\text{tet}_K(z)} \) grows as \( \Re(z) \to \infty \). Which would translate, how close does Kneser's tetration get to zero (which it must), and how fast does it do so. Ideally, I wonder if we can weaken this with a less intrusive function than \( i\sqrt{iz} \). The best option being, simply using the data points \( \frac{1}{\text{tet}_K(i(n+1))} \) to express tetration in the upper-half plane. This would require a careful analysis though.

I believe this may be helpful, as we'd only have to approximate \( a_n \) which for large \( n \) should look something like,

\(

L + e^{iL\sqrt{i(n+1)}}\\

\)

Which has fairly fast convergence. It may also make sense to take a double sequence \( a_{nk} \) where \( \lim_{k\to\infty} a_{kn} = a_n \) and each \( a_{nk} \) generates an exponential-like function which is Mellin Transformable as well. Think of using the partial sums of an exponential series which approximates \( \text{tet}_K(z) \) as \( |z|\to\infty \) in the upper left half quadrant; which will Mellin-transformable as well.

Anywho, I'll let you guys know if I can think of a better kind of representation up this alley.

Regards, James

EDIT:

For instance, if we write,

\(

\text{tet}_K(z) = \Psi^{-1}(e^{Lz}\theta(z))\\

\)

For a one-periodic function \( \theta \) and \( \Psi^{-1} \) the inverse Schroder function at a fixed point \( L \) of \( e^z \). Then,

\(

\text{tet}_K(z) = L + \sum_{j=1}^\infty \sum_{m=0}^\infty c_{mj} e^{Ljz + 2\pi i m z}\\

\)

If we truncate this series, then,

\(

F_k(z) = L + \sum_{j=1}^k \sum_{m=0}^k c_{mj} e^{Ljz + 2\pi i m z}\\

\)

Then calling \( F_k(i\sqrt{iz}) : \mathbb{C}_{\Re(z) > 0} \to \mathbb{C} \) and as \( |z|\to\infty \) in this half plane \( F_k(z) \to L \). So if we define,

\(

a_{nk} = L + \sum_{j=1}^k \sum_{m=0}^k c_{mj} e^{(iLj - 2\pi m)\sqrt{i(n+1)} }\\

\)

And we define,

\(

f_k(x) = \sum_{n=0}^\infty a_{nk} \frac{(-x)^n}{n!}\\

\)

Then, for \( 0 < \Re(z) < 1 \),

\(

F_k(i\sqrt{i(1-z)})\Gamma(z) = \int_0^\infty f_k(x)x^{z-1}\,dx\\

\)

Whereupon,

\(

\lim_{p\to\infty} f_k^{(p)}(x) = Le^{-x}\\

\)

And for large values of \( n > N \) we may be able to uncover an asymptotic relationship that is easier to derive than bruteforcing \( c_{mj} \) using a Riemann Mapping theorem.

EDIT2:

As I remember people aren't so used to the mellin transform/fractional calculus approach as I am, this also generates a uniqueness condition. Let \( h(z) \) be the function such that \( i\sqrt{ih(z)} = i \sqrt{iz} + 1 \), and assume there exists a function,

\(

G(z) : \mathbb{C}_{\Re(z) > 0} \to \mathbb{C}\\

\)

Such that,

\(

|G(z)| \le C e^{\tau|\Im(z)| + \rho|\Re(z)|}\,\,\text{for}\,\,0 < \tau <\pi/2\,\,C,\rho > 0\\

G(n) = \text{tet}_K(i\sqrt{in})\\

\)

Then necessarily \( e^{G(z)} = G(h(z)) \) and \( G = \text{tet}_K(i\sqrt{iz}) \). We can also derive a uniqueness condition by ignoring the data points and requiring only that \( G \) be bounded exponentially (as in the above), tends to the same fixed point, satisfies \( e^{G(z)} = G(h(z)) \), and only requiring that \( G(b_n) = \text{tet}_K(i\sqrt{ib_n}) \) for some sequence \( b_n \to \infty \). Though this is a tad more complicated to derive.

EDIT3:

All in all, these transformations are intended as methods of understanding tetration through equations of the form,

\(

\vartheta_\lambda(w) = \sum_{n=0}^\infty e^{\lambda\sqrt{n+1}}\frac{w^n}{n!}\\

\frac{d^z}{dw^z} \vartheta_\lambda(w) = \sum_{n=0}^\infty e^{\lambda\sqrt{n+z+1}}\frac{w^n}{n!}\,\,\text{where}\,\,\\

\frac{d^z}{dw^z} \vartheta_\lambda(w)|_{w=0} = e^{\lambda \sqrt{z+1}}\\

\)

For \( \lambda \in \mathbb{C} \). And summing over \( \lambda_n \) with coefficients \( c_n \in \mathbb{C} \) in an effort to approximating \( G(h(z)) = e^{G(z)} \) with \( \frac{d^z}{dw^z} \sum_n c_n \vartheta_{\lambda_n}(w) |_{w=0} = G(z) \). Whereby Kneser's solution, we know such sequences \( c_n,\lambda_n \) exist.

Let's take \( \varphi(z) = i\sqrt{z} \) which maps the upper half plane of \( \mathbb{C} \) to the upper left quadrant of \( \mathbb{C} \). As \( |z|\to\infty \) in the upper left quadrant \( \text{tet}_K(z) \to L \) so long as we stay away from the real line. This implies the function \( \text{tet}_K(i\sqrt{iz}) : \mathbb{C}_{\Re(z) > 0} \to \mathbb{C} \) and this function satisfies \( \text{tet}_K(i\sqrt{iz}) \to L \) as \( |z|\to\infty \) in \( \mathbb{C}_{\Re(z) > 0} \). And from this the Mellin transform is a viable option.

Therefore, if we write,

\(

a_n = \text{tet}_K(i\sqrt{i(n+1)})\\

\)

Then, for \( \Re(z) > 0 \),

\(

\Gamma(1-z)\text{tet}_K(i\sqrt{iz}) = \sum_{n=0}^\infty a_{n} \frac{(-1)^n}{n!(n+1-z)} + \int_1^\infty (\sum_{n=0}^\infty a_{n}\frac{(-x)^n}{n!})x^{-z}\,dx\\

\)

Or if you prefer, for \( \Im(z) > 0 \) and \( \Re(z) < 0 \),

\(

\Gamma(1-iz^2) \text{tet}_K(z) = \sum_{n=0}^\infty a_{n} \frac{(-1)^n}{n!(n+1-iz^2)} + \int_1^\infty (\sum_{n=0}^\infty a_{n}\frac{(-x)^n}{n!})x^{-iz^2}\,dx\\

\)

This form gives us a way of representing Kneser's tetration using only the data points \( \text{tet}_K(i\sqrt{i(n+1)}) \). What I would really like to know is how fast Kneser's tetration grows as \( \Re(z) \to \infty \) in the complex plane and how fast \( \frac{1}{\text{tet}_K(z)} \) grows as \( \Re(z) \to \infty \). Which would translate, how close does Kneser's tetration get to zero (which it must), and how fast does it do so. Ideally, I wonder if we can weaken this with a less intrusive function than \( i\sqrt{iz} \). The best option being, simply using the data points \( \frac{1}{\text{tet}_K(i(n+1))} \) to express tetration in the upper-half plane. This would require a careful analysis though.

I believe this may be helpful, as we'd only have to approximate \( a_n \) which for large \( n \) should look something like,

\(

L + e^{iL\sqrt{i(n+1)}}\\

\)

Which has fairly fast convergence. It may also make sense to take a double sequence \( a_{nk} \) where \( \lim_{k\to\infty} a_{kn} = a_n \) and each \( a_{nk} \) generates an exponential-like function which is Mellin Transformable as well. Think of using the partial sums of an exponential series which approximates \( \text{tet}_K(z) \) as \( |z|\to\infty \) in the upper left half quadrant; which will Mellin-transformable as well.

Anywho, I'll let you guys know if I can think of a better kind of representation up this alley.

Regards, James

EDIT:

For instance, if we write,

\(

\text{tet}_K(z) = \Psi^{-1}(e^{Lz}\theta(z))\\

\)

For a one-periodic function \( \theta \) and \( \Psi^{-1} \) the inverse Schroder function at a fixed point \( L \) of \( e^z \). Then,

\(

\text{tet}_K(z) = L + \sum_{j=1}^\infty \sum_{m=0}^\infty c_{mj} e^{Ljz + 2\pi i m z}\\

\)

If we truncate this series, then,

\(

F_k(z) = L + \sum_{j=1}^k \sum_{m=0}^k c_{mj} e^{Ljz + 2\pi i m z}\\

\)

Then calling \( F_k(i\sqrt{iz}) : \mathbb{C}_{\Re(z) > 0} \to \mathbb{C} \) and as \( |z|\to\infty \) in this half plane \( F_k(z) \to L \). So if we define,

\(

a_{nk} = L + \sum_{j=1}^k \sum_{m=0}^k c_{mj} e^{(iLj - 2\pi m)\sqrt{i(n+1)} }\\

\)

And we define,

\(

f_k(x) = \sum_{n=0}^\infty a_{nk} \frac{(-x)^n}{n!}\\

\)

Then, for \( 0 < \Re(z) < 1 \),

\(

F_k(i\sqrt{i(1-z)})\Gamma(z) = \int_0^\infty f_k(x)x^{z-1}\,dx\\

\)

Whereupon,

\(

\lim_{p\to\infty} f_k^{(p)}(x) = Le^{-x}\\

\)

And for large values of \( n > N \) we may be able to uncover an asymptotic relationship that is easier to derive than bruteforcing \( c_{mj} \) using a Riemann Mapping theorem.

EDIT2:

As I remember people aren't so used to the mellin transform/fractional calculus approach as I am, this also generates a uniqueness condition. Let \( h(z) \) be the function such that \( i\sqrt{ih(z)} = i \sqrt{iz} + 1 \), and assume there exists a function,

\(

G(z) : \mathbb{C}_{\Re(z) > 0} \to \mathbb{C}\\

\)

Such that,

\(

|G(z)| \le C e^{\tau|\Im(z)| + \rho|\Re(z)|}\,\,\text{for}\,\,0 < \tau <\pi/2\,\,C,\rho > 0\\

G(n) = \text{tet}_K(i\sqrt{in})\\

\)

Then necessarily \( e^{G(z)} = G(h(z)) \) and \( G = \text{tet}_K(i\sqrt{iz}) \). We can also derive a uniqueness condition by ignoring the data points and requiring only that \( G \) be bounded exponentially (as in the above), tends to the same fixed point, satisfies \( e^{G(z)} = G(h(z)) \), and only requiring that \( G(b_n) = \text{tet}_K(i\sqrt{ib_n}) \) for some sequence \( b_n \to \infty \). Though this is a tad more complicated to derive.

EDIT3:

All in all, these transformations are intended as methods of understanding tetration through equations of the form,

\(

\vartheta_\lambda(w) = \sum_{n=0}^\infty e^{\lambda\sqrt{n+1}}\frac{w^n}{n!}\\

\frac{d^z}{dw^z} \vartheta_\lambda(w) = \sum_{n=0}^\infty e^{\lambda\sqrt{n+z+1}}\frac{w^n}{n!}\,\,\text{where}\,\,\\

\frac{d^z}{dw^z} \vartheta_\lambda(w)|_{w=0} = e^{\lambda \sqrt{z+1}}\\

\)

For \( \lambda \in \mathbb{C} \). And summing over \( \lambda_n \) with coefficients \( c_n \in \mathbb{C} \) in an effort to approximating \( G(h(z)) = e^{G(z)} \) with \( \frac{d^z}{dw^z} \sum_n c_n \vartheta_{\lambda_n}(w) |_{w=0} = G(z) \). Whereby Kneser's solution, we know such sequences \( c_n,\lambda_n \) exist.