x↑↑x = -1

[tommy1729]

\( \alpha_1(\text{tet}_b(z)) = z + \theta_1(z) \)

\( \alpha_2(\text{tet}_b(z)) = z + \theta_2(z) \)

SO

\( \alpha_1(\text{tet}_b(z)) - \alpha_1(\text{tet}_b(z-1)) = 1 \)

And

\( \alpha_2(\text{tet}_b(z)) - \alpha_2(\text{tet}_b(z-1)) = 1 \)

Thus

\( \alpha_1(\text{tet}_b(z)) - \alpha_1(\text{tet}_b(z-1)) - \alpha_2(\text{tet}_b(z)) + \alpha_2(\text{tet}_b(z-1)) = 0 \)

Let \( G(z) = \G(z) = \alpha_1(z) - \alpha_2(z) \)

Then we get

\( \G(\text{tet}_b(z)) - \G(\text{tet}_b(z-1))= 0 \)

or

\( \G(\text{tet}_b(z)) = \G(\text{tet}_b(z-1)) \)

This implies :

\( \text{tet}_b(z) = G^{[-1]}( \G(\text{tet}_b(z-1)) ) \)

Hence because G^[-1](G(z)) = id(z) is absurd we get ( by branches )

one of the following potential conclusions.

1) G^[-1](G(z)) =/= b^z

And this implies that \( \text{tet}_b(z) \) is a superfunction of 2 functions !!??

2) G^[-1](G(z)) = b^z or G^[-1](G(z)) = b^z + 2pi i / ln(b)

Very unlikely.

Since both potential conclusions are very very likely wrong , this implies

\( \alpha_1(\text{tet}_b(z)) = z + \theta_1(z) \)

\( \alpha_2(\text{tet}_b(z)) = z + \theta_2(z) \)

has no solution.

And this idea can easily be generalized to other functions then b^z , such as entire functions with 2 repelling fixpoints.

Hence my pessimistic attitude.


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But then what did mike and sheldon compute !??

Another fake function ???

Hmmm.

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Lets take another look :
\( \G(\text{tet}_b(z)) - \G(\text{tet}_b(z-1))= 0 \)

The only hope seems G(z) = p(inv tet_b(z)).

where inv tet_b is the inverse function of tet_b and p is a 1-periodic function.

Or said differently : \( \alpha_1(\text{tet}_b(z)) - \alpha_2(\text{tet}_b(z)) = p(z) \)

But this seems to bring us back at the beginning : \( p(z) = \theta_1(z) - \theta_2(z) \)

Lets analyze further : in the pessimistic case we wrote

G(tet(z)) = G(tet(z-1))

the key is that if G(z) = p(tet^[-1](z))

We get p(z) = p(z-1).

THIS IS IMPORTANT because we got the hidden paradox :

p(z) = p(z-1)

=> z = p^[-1](p(z-1))

=> z = z - 1.

So the step

\( \G(\text{tet}_b(z)) = \G(\text{tet}_b(z-1)) \)

This implies :

\( \text{tet}_b(z) = G^{[-1]}( \G(\text{tet}_b(z-1)) ) \)

Might not be as valid as it might seem.

Hmm.

regards

tommy1729