I'm going to explain some things about my tetration and the auxiliary function I use to construct it. These are mostly things I either can prove but aren't so hard to prove, so I didn't include them; or things I can't quite prove yet, but I'm probably not far off the mark.

First of all, I'd like to clarify the definition of . If we call , then:

Now, , so this limit converges uniformly on compact subsets, in and for ; but it converges to a constant in . Tommy raised a good point in his thread, where this wasn't apparently obvious to him, where as it is to me. If we have an alternative solution to,

And for all ; then,

Which can be seen because,

And since is a normal sequence in this converges flatly.

***

As to the branch cuts; my solution probably (99% sure) has branch cuts other than the one at . This is a bit harder to identify, but I'll explain. Suppose there are no branch cuts and is holomorphic for . Then, firstly,

and,

Therefore for very large ; I'm pretty sure I can show that,

This implies that,

Is mellin transformable, which means,

And now we run into a whole lot of trouble. First of all, it's been discussed on this forum that this probably doesn't converge. I can actually prove it now--this cannot converge. If two holomorphic functions on , say , which satisfy these types of bounds, then:

If these function agree on the naturals they agree everywhere. Here's where we encounter problems. Huge problems. Suppose that,

Then surely and equal on the natural numbers, therefore,

But! IS NOT PERIODIC. Therefore this function must be injective. But it can't be injective. As per (I forget the name of the authors, but it can be found in Milnor's Complex Dynamics) the orbits of the exponential are dense. This means the sequence for any neighborhood is dense. This can't happen if is injective. CONTRADICTION.

Therefore CANNOT be holomorphic on . It has branch-cuts. Or rather, the nicer way of saying this is, has zeroes for .

Now these branch cuts work as you play your logarithms, which is; each one looks like ; for convenience we'll make the convention that the is holomorphic for in a neighborhood of the singularity. This way the branch cuts are a straight, horizontal line from the point of branching off to . From this we partition in the left-half plane into vertical sections. And each section limits to a fixed point of as .

I hope the reader can see this is not much different than what happens in Kneser's case. This is largely why I think this solution is Kneser's solution.

***

The next thing I'd like to say, or what I can think of; is a formula for the derivative of . Suppose we've isolated ourselves to one of these horizontal strips in which is holomorphic. Then, as where ; and it tends to this as . This is shown because functions (in this case ) when iterated tend to fixed points geometrically like the multiplier. There exists the formula in this strip,

For some . This implies our derivative is non-zero in this horizontal strip. Proving this theorem is just out of my grasp, because I've yet to understand the branch-cuts yet. But it's derived from the relationship,

And the geometric convergence of the above product, which when attached with , satisfies this functional equation... This is one of those things I'm pretty sure I can prove but I'm not 100%.

***

Anyone have any questions, I'm happy to answer them here. There's a lot of things I did not publish. I cut a lot of fat. I've always prescribed to the iceberg theory of writing; perhaps to my detriment...

...........I'll add more as I think of things and scrubble through my notes.

First of all, I'd like to clarify the definition of . If we call , then:

Now, , so this limit converges uniformly on compact subsets, in and for ; but it converges to a constant in . Tommy raised a good point in his thread, where this wasn't apparently obvious to him, where as it is to me. If we have an alternative solution to,

And for all ; then,

Which can be seen because,

And since is a normal sequence in this converges flatly.

***

As to the branch cuts; my solution probably (99% sure) has branch cuts other than the one at . This is a bit harder to identify, but I'll explain. Suppose there are no branch cuts and is holomorphic for . Then, firstly,

and,

Therefore for very large ; I'm pretty sure I can show that,

This implies that,

Is mellin transformable, which means,

And now we run into a whole lot of trouble. First of all, it's been discussed on this forum that this probably doesn't converge. I can actually prove it now--this cannot converge. If two holomorphic functions on , say , which satisfy these types of bounds, then:

If these function agree on the naturals they agree everywhere. Here's where we encounter problems. Huge problems. Suppose that,

Then surely and equal on the natural numbers, therefore,

But! IS NOT PERIODIC. Therefore this function must be injective. But it can't be injective. As per (I forget the name of the authors, but it can be found in Milnor's Complex Dynamics) the orbits of the exponential are dense. This means the sequence for any neighborhood is dense. This can't happen if is injective. CONTRADICTION.

Therefore CANNOT be holomorphic on . It has branch-cuts. Or rather, the nicer way of saying this is, has zeroes for .

Now these branch cuts work as you play your logarithms, which is; each one looks like ; for convenience we'll make the convention that the is holomorphic for in a neighborhood of the singularity. This way the branch cuts are a straight, horizontal line from the point of branching off to . From this we partition in the left-half plane into vertical sections. And each section limits to a fixed point of as .

I hope the reader can see this is not much different than what happens in Kneser's case. This is largely why I think this solution is Kneser's solution.

***

The next thing I'd like to say, or what I can think of; is a formula for the derivative of . Suppose we've isolated ourselves to one of these horizontal strips in which is holomorphic. Then, as where ; and it tends to this as . This is shown because functions (in this case ) when iterated tend to fixed points geometrically like the multiplier. There exists the formula in this strip,

For some . This implies our derivative is non-zero in this horizontal strip. Proving this theorem is just out of my grasp, because I've yet to understand the branch-cuts yet. But it's derived from the relationship,

And the geometric convergence of the above product, which when attached with , satisfies this functional equation... This is one of those things I'm pretty sure I can prove but I'm not 100%.

***

Anyone have any questions, I'm happy to answer them here. There's a lot of things I did not publish. I cut a lot of fat. I've always prescribed to the iceberg theory of writing; perhaps to my detriment...

...........I'll add more as I think of things and scrubble through my notes.