Thread Rating:
  • 0 Vote(s) - 0 Average
  • 1
  • 2
  • 3
  • 4
  • 5
extension of the Ackermann function to operators less than addition
Well the result is surprisingly simple, and is derived from a single defining axiom. I feel if there are any qualms with the proof it comes from the axiom itself.

we assume for now that all variables can be extended to the complex domain

and thus, the Ackermann series of operators are defined by the single property:

starting the Ackermann function with addition at 1 gives us:

therefore, if we plug in from our defining axiom (I) and try to solve for when sigma is zero we get:

with this, we try to solve for the more general:

which by (II) gives

this was derived from only a single defining axiom and no other assumptions were made. But that's not it, we can continue this sequence and try to solve for when sigma is negative one.

Again, we'll start with the defining axiom:

of course, by (III) we have the simple formula:

which of course, gives the fantastic equation:

now, by induction, we can make the complete argument:

Or basically, we have the result that every operator less than addition which is equal to an integer is the equivalent to successorship.


Now if we add the additional property that sigma plus one be an iteration count of sigma, we run into many problems. And I will shortly argue here:

The second axiom which need not be in play is the axiom of iteration given as

with this axiom we instantly have a contradiction with the first axiom when we extend operators less than addition:

by (III)


therefore if we allow the axiom of iteration we cannot extend the Ackermann function to operators less than addition.

Furthermore, if we allow the axiom of iteration we also have the cheap result:

this causes the function

to no longer possibly be analytic.

we can however accept a modified axiom of iteration which is written as:

this allows for different identities at non-integer values of sigma, however, this still disallows sigma to be extended to values less than addition.

So, I leave it open ended, hoping someone else has some comment. Do we use axioms (I) and (IV), or (I) and (V), or, as I prefer, (I) alone?
furthermore, if we define the identity function as:

since S(1) = 0, and S(2) = 1. If we want S(x) to be analytic and continuous (which we do), by the mean value theorem there is an operator who's identity is 0.5. Let's call this q.

doing some simple manipulations and we get

this is by the first axiom.

Given that. We can let a = 2, and get the contradictory result:

if we want:

we arrive at a contradiction or a result that is definitely not desired.



this would imply our function at q has the following behaviour (if we simply reapply the first axiom recusively):

this result can be extended to any value of sigma:


therefore we have three options

1. the identity function is not analytic


3. or in general is not a smooth monotonic increasing function unless sigma is an integer.

If we let 3 then we also get the result that operators extended less than addition are impossible. In this we imply

but we already know

The first option, that the identity function be not analytic isn't desirable at all. Furthermore, it's more than it just not be analytic, but it implies the only results possible are 0 and 1. And that it only takes 0 at addition and everywhere else we have 1.

In conclusion, if we want to have the Ackermann function extended to the complex domain everywhere and be monotonically increasing everywhere we're going to have to lose the aesthetic property:

I realize now we have to create a second axiom in order that the series of operators become the true Ackermann function.

Consider the possibility that:

we still have the result that

the only difference is that

and in return we get

So in order to get the true Ackermann function we must make the second assertion:

This is actually the equivalent to the iteration axiom:

though only true for integer values of sigma greater than or equal to two.

Therefore we define the Ackermann function from axioms:

such that

Does anyone see any modifications necessary?

Possibly Related Threads…
Thread Author Replies Views Last Post
  The modified Bennet Operators, and their Abel functions JmsNxn 6 243 07/22/2022, 12:55 AM
Last Post: JmsNxn
  The \(\varphi\) method of semi operators, the first half of my research JmsNxn 13 681 07/17/2022, 05:42 AM
Last Post: JmsNxn
  The bounded analytic semiHyper-operators JmsNxn 4 7,735 06/29/2022, 11:46 PM
Last Post: JmsNxn
  Between addition and product ( pic ) tommy1729 9 8,520 06/25/2022, 09:34 PM
Last Post: tommy1729
  A fundamental flaw of an operator who's super operator is addition JmsNxn 6 13,113 06/16/2022, 10:33 PM
Last Post: MphLee
  Holomorphic semi operators, using the beta method JmsNxn 71 5,330 06/13/2022, 08:33 PM
Last Post: JmsNxn
  Hyper operators in computability theory JmsNxn 5 10,832 02/15/2017, 10:07 PM
Last Post: MphLee
  Recursive formula generating bounded hyper-operators JmsNxn 0 3,710 01/17/2017, 05:10 AM
Last Post: JmsNxn
  Rational operators (a {t} b); a,b > e solved JmsNxn 30 75,374 09/02/2016, 02:11 AM
Last Post: tommy1729
  holomorphic binary operators over naturals; generalized hyper operators JmsNxn 15 31,170 08/22/2016, 12:19 AM
Last Post: JmsNxn

Users browsing this thread: 1 Guest(s)