# Tetration Forum

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Consider iterations of exponentials of a fixed height. Also called growth.

For instance semi-exponentials.

In algebra the main thing is Sum and product.
Kinda.

When we consider asymptotics i call it pseudoalgebra.

So for semiexp we get the natural questions such as
The best fit ( given by the symbol = )

$Exp_2^{[0.5]}(x) Exp_3^{[0.5]}(x) Exp_5^{[0.5]}(x) = Exp_y^{[0.5]}(x)$

Where y is the value we seek and x > 1.

This is - for clarity - an asymptotic equation for bases ( 2,3,5,y).

It reminds me of base change and others.

How about these ?

How to find such identities ?

Regards

Tommy1729
@ means approximation.

Lemma

$Exp_a^[b] (x) @ Exp^[b] ( @ Ln(a) x)$

From there we get

$Exp_q^{[1/2]}(x) Exp_s^{[1/2]}(x) @ Exp_d^{[1/2]}(x)$

Where d is

$Exp(d) = @ Ln^{[1/2]}( Exp^{[1/2]}( Ln(q) x) Exp^{[1/2]} ( Ln(s) x) ) / x$

( notice this can be rewritten with 1 semi-exp and 2 semi-logs too )

But this is not the full story ofcourse.

We need proofs.

Perhaps consider other ways to handle the issue.

And a qualitative understanding of the formula for d such as d ~ (qs)^2 or the alike.

I wonder if you would have done it differently ?

Also a table would be nice.

Still alot of work to do.

Regards

Tommy1729
The master
Im afraid the strategy fails.

For exp_b^[a] <*> and a < 1 we get

<*> @ = exp^[a] ( T x )

Where T Goes to 1 as x grows and for a >= 1 , T Goes to ln(b).

Proof sketch

S commutes with exp.

S(T x) = ln S ( T b^x) / ln(B).

= S ln ( T b^x) / ln(B)

= S ( ln T + ln B x ) / ln B

Maybe.

Still thinking ...

Regards

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

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

--

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 :

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

--

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.

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

Although that might be hard to read.

Regards

Tommy1729
(10/05/2016, 12:21 PM)tommy1729 Wrote: [ -> ]....
So for semiexp we get the natural questions such as
The best fit ( given by the symbol = )

$Exp_2^{[0.5]}(x) Exp_3^{[0.5]}(x) Exp_5^{[0.5]}(x) = Exp_y^{[0.5]}(x)$

Mick's question on mathstack exchange is related to this post. In my answer, I considered $\exp_2^{0.5}(x)\;$ and $\exp_e^{0.5}(x)\;$. See math.stackexchange.com If one uses the analytic solution for the half iterates of base_2 and the half iterate of base_e (ignoring the conjectured nowhere analytic basechange type solutions), the fractional exponentials are not at all well ordered. If a<b, half the time $\exp_a^{0.5}(x)\;>\exp_b^{0.5}(x)\;$ as x grows super exponentially large.

There are more details in my post, but if g(x)=0, then $\;\exp_e^{0.5}(x)=\exp_2^{0.5}(x)\;\;$

$g(x) = \text{slog}_e(\text{sexp}_2(x+0.5))-\text{slog}_e(\text{sexp}_2(x))-0.5$

Consider the 2nd peak for base2 occurs near 5.668, where the half iterate base_2 is larger than the half iterate base_e.

x2=sloge(sexp2(5.668 +0.5))= 5.03973936018302
xe=sloge(sexp2(5.668 ))+0.5 =5.03945210684265

The question is how much larger is x2 than xe? We can take the logarithm twice of both numbers, and compare sexp_e(x2-2) vs sexp_e(xe-2), and they differ by about +418960.3! This very large difference can be compared to the difference in log(log(2^x)) vs log(log(e^x)) which is always log(log(2)) or -0.3665