Hey everyone. Sorry I've taken so long to comment. Been very busy working on the super factorial I think I've found a way to solve it.

This does work on the continuum sum Tommy. It takes a lot of extra fiddling but look here:

http://arxiv.org/abs/1503.07555

This one is a little more rough but it explains how continuum sums can be created.

I made a few modifications (had some typos and etc) but heres the paper in full force:

http://arxiv.org/abs/1503.07555

I'm currently trying to make this work on half operators, I do believe that

produces a holomorphic function. This will interpolate the hyper operators and satisfy the recursive identity. Sadly it will only work for operators greater than or equal to multiplication. It fails for addition by simple means.

Thanks for the replies, all your questions have been things I'm interested in and investigating. This will probably work on the super root and more complicated patterns. The goal is to encourage using only integer valued iterations to produce complex iterations.

Any questions I'll be glad to answer. This is really a simple idea that has been pushed to its extreme. If this works on bases greater than eta Id be blown away. I think we can modify it somehow to produce this, but it would require understanding iteration when the multiplier is complex and sends to complex domains rather than when it is only real. This is something I've been working on for a while and I have a lot more built up knowledge than just what's in this paper.

If you have qualms or something look at the arxiv paper, it is slightly touched up, again, I fixed some typos and a few brazen quick techniques that weren't right.

you're quite right, see my second paper on the indefinite sum. It's a little rough but true. I actually develop an analytic formula involving a triple integral that justifies the expression you put there. I do believe this will work on many many recursive relationships.

Again, thanks to everyone for reading this. If you want your mind blown away just wait for how I apply this to I think it's almost completely solved. I just need to prove a few lemmas. This method will generalize to many more operators. Including the super iterations. (iterations of iteration.)

This does work on the continuum sum Tommy. It takes a lot of extra fiddling but look here:

http://arxiv.org/abs/1503.07555

This one is a little more rough but it explains how continuum sums can be created.

I made a few modifications (had some typos and etc) but heres the paper in full force:

http://arxiv.org/abs/1503.07555

I'm currently trying to make this work on half operators, I do believe that

produces a holomorphic function. This will interpolate the hyper operators and satisfy the recursive identity. Sadly it will only work for operators greater than or equal to multiplication. It fails for addition by simple means.

Thanks for the replies, all your questions have been things I'm interested in and investigating. This will probably work on the super root and more complicated patterns. The goal is to encourage using only integer valued iterations to produce complex iterations.

Any questions I'll be glad to answer. This is really a simple idea that has been pushed to its extreme. If this works on bases greater than eta Id be blown away. I think we can modify it somehow to produce this, but it would require understanding iteration when the multiplier is complex and sends to complex domains rather than when it is only real. This is something I've been working on for a while and I have a lot more built up knowledge than just what's in this paper.

If you have qualms or something look at the arxiv paper, it is slightly touched up, again, I fixed some typos and a few brazen quick techniques that weren't right.

(03/29/2015, 03:35 AM)fivexthethird Wrote:(03/28/2015, 08:59 PM)tommy1729 Wrote: I repeat : Does this method agree with koenigs ?What is koenigs? Are you talking about the regular iteration? If so, in the case of JmsNxn's tetration, yes, any two tetrations for bases are equal if they have the same period. See proposition 10 and corollary 8 in this paper

I'm petty sure that this can also be proved using the techniques in JmsNxn's paper

As for the super root thing... it's not even clear if it even results in a tetration, since the super root has no nice recurrence relation we can exploit to prove it.

Quote:Im wondering if this method also simplifies towards the continuum sum.With this we can easily define a continuum sum:

( that might have affect on methods based on continuum sums that are in an " unfinished state " )

That this works can be easily verified with the newton series identity. Using that, it can also be seen that it extends faulhaber's formula (i.e. it maps polynomials to polynomials)

Now the continuum sum formula is

for all positive integers z. If the LHS satisfies the bounds for Ramanujan's master theorem to apply, then this equality applies for all z in the right half-plane and is just a restatement of that fact.

I'm almost certain that the equality applies to JmsNxn's tetration, but I have no idea how to prove that.

It would also imply that the iterational form probably converges to it, too.

you're quite right, see my second paper on the indefinite sum. It's a little rough but true. I actually develop an analytic formula involving a triple integral that justifies the expression you put there. I do believe this will work on many many recursive relationships.

Again, thanks to everyone for reading this. If you want your mind blown away just wait for how I apply this to I think it's almost completely solved. I just need to prove a few lemmas. This method will generalize to many more operators. Including the super iterations. (iterations of iteration.)