Pseudoalgebra

[tommy1729]

So originally i tried to work from " the inside " like \( Exp_b^{[1/2]} ( g(b,x) ) \) but from " the outside " like \( h ( Exp_b^{[1/2}] (x) ) \) we got already the following result.

( i Will omit x sometimes , since it Goes to oo )

For
\( 0 < t < 1 , 0 < e/b < 1 \)

\( Exp_b^[t] = Exp^[t] ^ z \)

Now z > 1 must be true.

\( z = ln Exp_b^[t] / ln Exp^[t] \)

Simplify

\( z = ln(b) Exp_b^{[t-1]} /Exp^{[t-1]} \)

Since z > 1 and \( Exp_b^{[t-1]} / Exp^{[t-1]} < 1 \) we get

\( 1 < z < ln(b) \)

and

\( 1/ln(b) < Exp_b^{[t-1]} /Exp^{[t-1]} < 1 \).

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Notice for integer n > 0 we get by the above and induction

\( Exp_b^{[t-1-n]} / Exp^{[t-1-n]} \) ~~ \( 1/ln(b) \)

I assume it holds for n = 0 , that would imply that powers dominate bases for subexponential tetration.

In other words

Conjecture for p > 1 :

\( ( Exp^{[t]} )^p > Exp_b^{[t]} \)

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However we need much better understanding and approximations.

We are not close to answering

semiexp_q * semiexp_s ~ semiexp_d ^ R

For a given pair (q,s) and a desired best fit (d,R).

I considered the base change but without succes. The approximation slog - slog_b ~~ constant is insufficient.

See also

http://math.stackexchange.com/questions/...ase-a-e1-e

Although that might be hard to read.


Regards

Tommy1729