Thread Rating:
  • 0 Vote(s) - 0 Average
  • 1
  • 2
  • 3
  • 4
  • 5
Rational operators (a {t} b); a,b > e solved
(06/08/2011, 08:32 PM)sheldonison Wrote: So we have a definition for t=0..2 (addition, multiplication, exponentiation), for all values of a. Nice!

So, now, I'm going to define the function f(b), which returns the base which has the fixed point of "b".
, since e is the fixed point of b=
, since 2 is the lower fixed point of b=sqrt(2)
, since 3 is the upper fixed point of this base
, since 4 is the upper fixed point of b=sqrt(2)
, since 5 is the upper fixed point of this base
I don't know how to calculate the base from the fixed point, but that's the function we need, and we would like the function to be analytic! This also explains why the approximation of using base eta pretty well, since the base we're going to use isn't going to be much smaller than eta, as b gets bigger or smaller than e.

Now, we use this new function in place of eta, in James's equation. Here, f=f(b).

- Sheldon

Woah, I wonder what consequences this will have on the algebra.

I guess

which I guess isn't too drastic.

But ,the question is, of course, does the following still hold :

, so no it doesn't. That's not good, we want operators to be recursive.

And I'm unsure if the inverse is still well defined, so I think we lose:

where S(q) is the identity function. We may even lose the identity function altogether, this is really bad.

We also lose:

These are all too many valuable qualities that are lost when redefining semi-operators the way that you do. Sure it's analytic over , but it loses all its traits which make it an operator in the first place. I'm going to have to stick with the original definition of that isn't fully analytic.

However, I am willing to concede the idea of changing from base eta to base root 2.

That is to say if we define:

This will give the time honoured result, and aesthetic necessity in my point of view, of:
for all

I like this also because it makes and potentially analytic over since 2 and 4 are fix points.

I also propose writing

Sheldon's analytic function is then:

that's still very pretty though, that isn't piecewise over and potentially analytic.

Messages In This Thread
RE: Rational operators (a {t} b); a,b > e solved - by JmsNxn - 06/08/2011, 11:47 PM

Possibly Related Threads...
Thread Author Replies Views Last Post
  Thoughts on hyper-operations of rational but non-integer orders? VSO 2 2,210 09/09/2019, 10:38 PM
Last Post: tommy1729
  Hyper operators in computability theory JmsNxn 5 6,900 02/15/2017, 10:07 PM
Last Post: MphLee
  Recursive formula generating bounded hyper-operators JmsNxn 0 2,417 01/17/2017, 05:10 AM
Last Post: JmsNxn
  holomorphic binary operators over naturals; generalized hyper operators JmsNxn 15 22,595 08/22/2016, 12:19 AM
Last Post: JmsNxn
  The bounded analytic semiHyper-operators JmsNxn 2 5,201 05/27/2016, 04:03 AM
Last Post: JmsNxn
  Bounded Analytic Hyper operators JmsNxn 25 30,016 04/01/2015, 06:09 PM
Last Post: MphLee
  Incredible reduction for Hyper operators JmsNxn 0 3,066 02/13/2014, 06:20 PM
Last Post: JmsNxn
  interpolating the hyper operators JmsNxn 3 6,975 06/07/2013, 09:03 PM
Last Post: JmsNxn
  Number theory and hyper operators JmsNxn 7 10,627 05/29/2013, 09:24 PM
Last Post: MphLee
  Number theoretic formula for hyper operators (-oo, 2] at prime numbers JmsNxn 2 5,459 07/17/2012, 02:12 AM
Last Post: JmsNxn

Users browsing this thread: 1 Guest(s)