# Tetration Forum

You're currently viewing a stripped down version of our content. View the full version with proper formatting.
Well I've been having suspicions for quite a while that the Ackermann function cannot be analytic. I was having trouble visualizing it and then one step to the answer came to me through the identity function. It's actually incredibly simple the formulas involved. It's a proof by contradiction.

First, we start off by defining what we mean by the "Ackermann function", so as not to cause any confusion.

where:

and if

or

is the identity function

we set:

so that

and we make the final two requirements:

so that we have as multiplication and as exponentiation so on and so forth.

if

or is the "logarithmic" inverse operator of order

must be analytic across

therefore
is analytic across at all values of b if and only if our identity function () is analytic.

We will work to prove that cannot be analytic by at all values of b.

So we come to the weaker conclusion than the Ackermann function cannot be analytic across all values, to the "logarithmic" Ackermann function cannot be analytic across all values.

or the series

cannot be analytic if it holds the property:

Firstly, we make the argument that cannot be periodic if we want it to be analytic.

Or simply put if:

this is instantly invalid for and and an analytic periodic function must be periodic for all values, however, we do have it to be true for integer values of greater than 1.

Now we write,

since and since , must oscillate similarly to a periodic function only it is not periodic since if it was periodic and analytic therefore we designate it cannot be analytic and periodic.

So let us write what I call the "pseudo period" as

evidently is multivalued and not analytic and does not always return values.

for example does not necessarily exist, because it is fully possible that
;

however, it is evident that must exist for some values of in order that be analytic.

it is trivial that

Now we must derive a few lemmas

consider

by the first axiom it's easy to deduce:

continuing this process again we get:

this is how we get the famed identity

Now we move on to the "logarithmic" inverse function and defining a new function I call Q.

it's obvious that

which is a simple exercise in super operators

and

before continuing we need to acknowledge that

which is again a simple exercise.

now we have to make some odd observations that may get confusing

from the "logarithmic" inverse law

this implies:

which in turn gives

since

by the identity

but we also know by Q's definition

so that

we know that from earlier

so that we get

but we also know, by the definition of the "logarithmic" inverse:

so therefore

which since

and

we get the beautiful result:

which is the core identity of our proof

since is multivalued, we find that

for some value

if

then

which in turn applies

which iterated gives

or that is periodic. However we already noted that if is periodic then cannot be analytic since by definition of an analytic function that is periodic it must be periodic over every value in the domain and

Therefore

is not analytic by for all values of b

and

is not analytic by

Q.E.D.

This puts me one step closer to proving non analycity across for some value of a and b in the Ackermann function

Personally I think that non-analycity is intuitively implied since is not primitive recursive, but this is yet to be rigorously decided. To disprove universal analycity, it suffices to show that for some fixed , is not analytic across .

I understand the notation in this proof maybe a bit confusing, but when working with three variables I think that's inevitable.