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
for Kneser's tetration. Now depending on how it grows as
we have different manners of expressing it. But there is one way I can say for sure.
Let's take
which maps the upper half plane of
to the upper left quadrant of
. As
in the upper left quadrant
so long as we stay away from the real line. This implies the function
and this function satisfies
as
in
. And from this the Mellin transform is a viable option.
Therefore, if we write,
})\\<br />
)
Then, for
,
\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\\ <br />
)
Or if you prefer, for
and
,
 \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\\<br />
)
This form gives us a way of representing Kneser's tetration using only the data points
. What I would really like to know is how fast Kneser's tetration grows as
in the complex plane and how fast
grows as
. 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
. The best option being, simply using the data points
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
which for large
should look something like,
}}\\<br />
)
Which has fairly fast convergence. It may also make sense to take a double sequence
where
and each
generates an exponential-like function which is Mellin Transformable as well. Think of using the partial sums of an exponential series which approximates
as
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,
 = \Psi^{-1}(e^{Lz}\theta(z))\\<br />
)
For a one-periodic function
and
the inverse Schroder function at a fixed point
of
. Then,
 = L + \sum_{j=1}^\infty \sum_{m=0}^\infty c_{mj} e^{Ljz + 2\pi i m z}\\<br />
)
If we truncate this series, then,
 = L + \sum_{j=1}^k \sum_{m=0}^k c_{mj} e^{Ljz + 2\pi i m z}\\<br />
)
Then calling
and as
in this half plane
. So if we define,
\sqrt{i(n+1)} }\\<br />
)
And we define,
 = \sum_{n=0}^\infty a_{nk} \frac{(-x)^n}{n!}\\<br />
)
Then, for
,
})\Gamma(z) = \int_0^\infty f_k(x)x^{z-1}\,dx\\<br />
)
Whereupon,
}(x) = Le^{-x}\\<br />
)
And for large values of
we may be able to uncover an asymptotic relationship that is easier to derive than bruteforcing
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
be the function such that
, and assume there exists a function,
 : \mathbb{C}_{\Re(z) > 0} \to \mathbb{C}\\<br />
)
Such that,
| \le C e^{\tau|\Im(z)| + \rho|\Re(z)|}\,\,\text{for}\,\,0 < \tau <\pi/2\,\,C,\rho > 0\\<br />
G(n) = \text{tet}_K(i\sqrt{in})\\<br />
)
Then necessarily
and
. We can also derive a uniqueness condition by ignoring the data points and requiring only that
be bounded exponentially (as in the above), tends to the same fixed point, satisfies
, and only requiring that
for some sequence
. 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,
 = \sum_{n=0}^\infty e^{\lambda\sqrt{n+1}}\frac{w^n}{n!}\\<br />
\frac{d^z}{dw^z} \vartheta_\lambda(w) = \sum_{n=0}^\infty e^{\lambda\sqrt{n+z+1}}\frac{w^n}{n!}\,\,\text{where}\,\,\\<br />
\frac{d^z}{dw^z} \vartheta_\lambda(w)|_{w=0} = e^{\lambda \sqrt{z+1}}\\<br />
)
For
. And summing over
with coefficients
in an effort to approximating
with
. Whereby Kneser's solution, we know such sequences
exist.
Let's take
Therefore, if we write,
Then, for
Or if you prefer, for
This form gives us a way of representing Kneser's tetration using only the data points
I believe this may be helpful, as we'd only have to approximate
Which has fairly fast convergence. It may also make sense to take a double sequence
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,
For a one-periodic function
If we truncate this series, then,
Then calling
And we define,
Then, for
Whereupon,
And for large values of
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
Such that,
Then necessarily
EDIT3:
All in all, these transformations are intended as methods of understanding tetration through equations of the form,
For