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04/03/2009, 06:12 PM
(This post was last modified: 04/04/2009, 02:27 PM by nuninho1980.)
lastest news :
I calculated (w/ 410 digits) by slog (by 200 dimensions of matrix) during 80~90 seconds per a time  my cpu c2de6600 running vista x64sp1.
^^^ is pentation.
now the fixed point more smooth
1.6353244967 ^^^ oo ~= 3.0885549441 (200 dimensions of matrix)
b ^^^ oo = x => slog_b (y) = x, y=x, if b<=1.6353244967 and b=/=1.6353244968 then x<y and x>y.
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04/04/2009, 02:21 PM
(This post was last modified: 04/04/2009, 02:21 PM by nuninho1980.)
you have warning to this next page you read my message #11.
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nuninho1980 Wrote:now the fixed point more smooth
1.6353244967 ^^^ oo ~= 3.0885549441 (200 dimensions of matrix)
b ^^^ oo = x => slog_b (y) = x, y=x, if b<=1.6353244967 and b=/=1.6353244968 then x<y and x>y.
Oh, you mean we have an upper fixed point of the tetrational for and the fixed point can then be computed by or . Ya, interesting. I dont know whether we even have a thread on the forum that dealt with the topic of the fixed point of tetrationals.
Of course there maybe always the dependency of the values from the chosen method of tetration.
Quote:"tetration and slog" original by Andrew Robbins is as smoother as "new regular slog"
So how big is the difference between both methods, with respect to the computed fixed point?
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04/07/2009, 01:35 AM
(This post was last modified: 04/07/2009, 02:00 AM by nuninho1980.)
bo198214 Wrote:Oh, you mean we have an upper fixed point of the tetrational for and the fixed point can then be computed by or . Ya, interesting. I dont know whether we even have a thread on the forum that dealt with the topic of the fixed point of tetrationals.
Of course there maybe always the dependency of the values from the chosen method of tetration. slog_b (x) = x <=> b ^^ x = x => b ^^^ oo = x
yeah!
to remember:
one  1^oo = 1
Euler  (e^(1/e)) ^^ oo = e
now new fixed point
(1.63532...) ^^^ oo ~= (3.08855...) what is this new result? it's a SuperEuler?
bo198214 Wrote:So how big is the difference between both methods, with respect to the computed fixed point?
to remember  you follow "regular slog"  http://en.wikipedia.org/wiki/Talk:Tetrat...on_methods lol
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04/14/2009, 10:12 PM
(This post was last modified: 04/15/2009, 10:57 AM by gent999.)
Hi, guys.
Let d := e^(1/e).
Facts:
1. a(n) := d + 1/n, Lim n>inf a(n)[4]n = inf.
2. b(n) := d + 1/n^2, Lim n>inf b(n)[4]n = e exists.
Questions:
1. Can be defined real function f(n,x) that c(n,x) := d + f(n,x), and Lim n>inf c(n,x)[4]n = l(x) exists, and e < l < inf?
2. For which x function l(x) can be defined?
3. How "good" functions f(n,x) and l(x) can be relative to x?
4. How function l(x) relates to tetration definition given by D.Kuznetsov in collaboration with you?
5. If l(x) can be defined, can be it used to give yet another definition of tetration for real positive heights (analogy with classics)?
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Excellent questions, however can you prove your "facts"?
Or are they the outcome of numerical experiments?
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04/15/2009, 12:18 AM
(This post was last modified: 04/15/2009, 10:51 AM by gent999.)
bo198214 Wrote:Excellent questions, however can you prove your "facts"?
Or are they the outcome of numerical experiments?
I do not have objective to give formal proof for these statements. But I’ve checked statements by using Taylor formula and induction process to calculate first four coefficients and residue in resulting series. This check is not very hard exercise, if you interested, please, verify. Then I have made numerical check. So there is chance to choose corresponding function.. may be it is possible to choose quite simple function, may be not...
My questions are what I am interesting, no more. I have made posting because I like this forum and article in Wikipedia, it is very interesting, thanx.
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04/15/2009, 01:35 PM
(This post was last modified: 04/15/2009, 01:38 PM by bo198214.)
I just wanted to know whether these are wellknown facts (and only i didnt hear of them yet) or whether it are observations on the research front.
Because I think only these two facts are already noticable alone.
And it would desirable to have proper proofs for them.
A more thorough investigation of these cases,
could perhaps also provide ideas for your other questions.
Did you experiment with them? Can you plot a graph in dependence of ? Or is the limit for all ?
And what about other fixed points, e.g. ?
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04/15/2009, 04:41 PM
(This post was last modified: 04/15/2009, 06:25 PM by gent999.)
bo198214 Wrote:I just wanted to know whether these are wellknown facts (and only i didnt hear of them yet) or whether it are observations on the research front.
Because I think only these two facts are already noticable alone.
And it would desirable to have proper proofs for them.
A more thorough investigation of these cases,
could perhaps also provide ideas for your other questions.
Did you experiment with them? Can you plot a graph in dependence of ? Or is the limit for all ?
And what about other fixed points, e.g. ?
These calculations were made several years ago but currently my notes in another country. I can try to restore them, if it is interestring. Thereat I considered function: c(n,x) = d + x ^(1+s), s>0 and wrote Taylor expansion for n = 1, 2, 3. Comparing expansions I wrote recurrent expressions for first coefficients and using induction proved them for all n. Thus, constant part was d[4]n, linear part was reduced (due to selection of d) and to investigate the question, quadratic part (quite bulky) and order of residue was sufficient. The conclusion was that for any s>0 the sequence converges to e. Numeric experiment was fulfilled in Mathematica, for s=1 convergence confirmed. It is interesting, that the numeric process has two stages: first, when resulting value grows (the more, the less is s), and second, where it is converges. I’ve tried several 0<s<1, but, in most cases, calculation was overflowed on first stage.
Fixed points. I do not made any accurate analytical investigation of possible sequences which are not converge to d[4]inf, 0<=d<e^(1/e). Case n^s, 0<s leads to expected value d[n]inf.
Interesting to look also at sequence (d+1/ (i*(ni+1)))[4]ni=1..n.
