+- Tetration Forum (https://tetrationforum.org)
+-- Forum: Tetration and Related Topics (https://tetrationforum.org/forumdisplay.php?fid=1)
+--- Forum: Mathematical and General Discussion (https://tetrationforum.org/forumdisplay.php?fid=3)
+--- Thread: Pseudoalgebra (/showthread.php?tid=1098)
Pseudoalgebra - tommy1729 - 10/05/2016
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
RE: Pseudoalgebra - tommy1729 - 10/08/2016
@ 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
RE: Pseudoalgebra - tommy1729 - 10/13/2016
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
RE: Pseudoalgebra - tommy1729 - 10/19/2016
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.
See also
http://math.stackexchange.com/questions/198757/a-tetration-limit-for-base-a-e1-e
Although that might be hard to read.
Regards
Tommy1729
RE: Pseudoalgebra - sheldonison - 10/23/2016
(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.
![[Image: halfsize2.gif]](http://sheltx.com/share_stuff/halfsize2.gif)
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