(03/29/2015, 11:25 PM)JmsNxn Wrote: This one is a little more rough but it explains how continuum sums can be created.Did you mean this link? http://arxiv.org/abs/1503.06211 (I guess you pasted the wrong link)

http://arxiv.org/abs/1503.07555

Quote:I'm currently trying to make this work on half operators, I do believe thatThat is exatly what I was thinking you were hoping for!

produces a holomorphic function. This will interpolate the hyper operators and satisfy the recursive identity.

We hope that the sequence for a fixed pair of suitable values and where the uparrow is computed using your method.

The main problem is that if we believe that and ... how can we check the recursive identity for the sequence ? I mean... ackermann function is defined using recursion at the same time on two different variables (rank and exponent) so that the only recurrence that we have should be of this form

Let then

In other words, the sequence doens't occours on the right side of the recurrence... so where is the recurrence if is evaluated whithout using another element of the sequence (rememebr that x is fixed)?

Anyways I see that if we watch the problem from the point of view of creating a new auxiliary functions relative to another Operator (and its natural iterations) the problem seems to disappear: you outlined this at page 4 of your paper. The concept seems really mysterious an at the same time amazing... I'm asking to my self (mhh... asking to you) if there are ... "hidden simmetries" or something like a big picture behind it. In fact I can see alot of analogies:

An interesting pattern appears: For E=D is the Maclaurin series of f at zero. If then is the auxiliary function that you are using in your paper.

So if we define the "Superfunction operator" and then iteration of iteration should be reachable working on this auxiliary function

Since you say

Quote:Sadly it will only work for operators greater than or equal to multiplication. It fails for addition by simple means.

Let for a suitable fixed alpha and for beta=1 then so that

There seems to be some indexes that look weird/wrong (the sequence inside the summation should start with n=one?) but nothing that you can't fix.

Anyways I think we could use the antirecursion/subfunction operator as well:

And that is equivalent to using my tau-sequence or theta-sequence.

Maybe those sequences are more easy to handle... if the the rank of f is finite then after some n the tau sequence becomes t+1 (t-1 if the theta sequence) and also all the information about its tau-sequence is already inside f.

And there exist surely functions with a tau sequence that converges to t+1... because the successor is a fixed point of the subfunction operator.

Quote: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.I really wish you good luck!

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