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Good grief ...., bo! Thanks a lot! Why ... one less ... ? Twice its decimal mirror is fiftyfour. The difference is the double of the cube of three, symbol of the dialectics (thesisantithesissynthesis), or three times (again...!) the sublime perfect number (1 + 2 + 3 = 1 x 2 x 3), which is the third triangular one. Waaawww! However, perhaps you didn't pay enough attention to the fact that I am still a "Junior Member"! See you soon ... .
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GFR Wrote:Good grief ...., bo! Thanks a lot! Why ... one less ... ?
Oh my mistake! You are one year younger than the number I described, you see the first impression is the most permanent.
Ok, second try:
It is the maximum number of spheres that can touch another sphere in a lattice packing in 6 dimensions
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09/02/2007, 05:38 PM
(This post was last modified: 09/03/2007, 02:05 PM by GFR.)
Oh, I see! My fault, sorry about that. Yes, in six dimensions ... it is really better. I simply thought it was the decimal inverse of result of 3#2 (3tower2) and, at the same time, the result of: (2^3) x (3^2) [the brackets are not necessary].
GFR (;<)==== . Amended on 070903 ... forget it !!!
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09/03/2007, 01:16 PM
(This post was last modified: 09/03/2007, 02:05 PM by Gottfried.)
GFR Wrote:Therefore, as I promised to Enryk, I submit to your attention the attached pdf notes, hoping that you will not be annoyed and that find some ideas to be developed. Please see also what should, in my opinion, happen in the range 0 < b < e^(e). It is a simulation but, just tell me what you think. In fact, I agree that that interval corresponds to the negative bases, for exponentiation. The appearence of oscillations, for b<1 must be admitted, if we accept to go out of the "reality". I think to understand a little bit better what Euler meant by saying that the "infinite towers" don't converge
for b < e ^(e). Perhaps, they just oscillate > oo.
Please also investigate, in the 1 < b < e^(1/e) interval, what could be the role of the second (upper) "unreachable" asymptote. The notation of the limits of b in my eigendecomposition concept is very intriguing.
Assume b=h^(1/h), or h=h(b) , where h() is the function described by Ioannis Galidakis, then the log of admissible h, hl=log(h) is 1< hl <1.
The eigenvalues of the operator for tetration are the consecutive powers of hl, so the diagonal contains a convergent sequence 1,hl,hl^2,hl^3,... if hl is in the admissible limit, and a divergent sequence if hl is outside. There is a degeneration if hl is exactly 1 or 1.
So the tetration behaves properly if we begin by selecting 1<hl<1, use then h and s when computed from this. Thus we are bound for e^e < s < e^e^1.
If we assume an s outside of this bounds, we need solutions of h(s) for this assumption. The eigenvectors can be determined formally anyway (seemingly with some interesting exceptions, I'll be looking at this next days), but the needed complex solutions for h(s) determine the "extravagant" behave because of
a) the divergence of the sequence of eigenvalues,
b) the occurence of divergent series, when the operator [*1] is actually computed by matrixmultiplication of its components.
Since the evaluation of the matrixmultiplication of the eigensystemcomponents implies a powerseries in hl, I think, the radius of convergence is best described by the complex unit disk (except of its bounding circle) for hl. I'll check this with my formulae and the messages here in the forum so far.
(Consequences of the analytical solution for the eigenvectors, as far as this is not in error)
Gottfried
[*1] edit: "eigensystem" exchanged by the better choice "operator"
Gottfried Helms, Kassel
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Gottfried Wrote:Assume b=h^(1/h), or h=h(b) , where h() is the function described by Ioannis Galidakis, then the log of admissible h, hl=log(h) is 1< hl <1.
The eigenvalues of the operator for tetration are the consecutive powers of hl, so the diagonal contains a convergent sequence 1,hl,hl^2,hl^3,... if hl is in the admissible limit, and a divergent sequence if hl is outside.
Gottfried, this a wonderful assist. Before I could ask what the matrix operator method would answer for this case, you nearly gave the answer. So let me conclude.
If then we have the negative Eigenvalues in the power derivation matrix A of . Now we compute . It has the Eigenvalues . Take for example then we see that has also nonreal Eigenvalues and hence has also nonreal entries. Supposed had only real coefficients then would have only real coefficients. So it is clear that must have some nonreal coefficients and is a nonreal function.
However perhaps it could be that is real or generally that is real, which I dont believe. Can someone just compute it?
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09/03/2007, 03:58 PM
(This post was last modified: 09/03/2007, 05:09 PM by jaydfox.)
I was thinking about (was it Daniel?)'s idea about using the complex fixed points to compute continuous iteration of bases greater than eta, which in general would give complex solutions. For multiple fixed points (not including conjugates), the question came up as to whether the solutions would be the same. And in general, we would expect real solutions anyway, so why should we expect these complex solutions to also be correct in some way?
On the other hand, we have these bases less than 1, where continuous iteration only seems to make sense with complex outputs at nonintegers.
Finally, thinking back to Henryk's analogy with exponentiation of negative bases, I made a realization. Consider exponentiation as fractional multiplication. Now consider that there are multiple complex numbers of the form that are equal to b. In the normal course of our mathematical day, we would treat these numbers as the same. But what if we momentarily considered them as different?
Then the fractional iterate of multiplication for and the fractional iterate of would be different. The first would give real values, and indeed would be what we think of as exponentiation in base b. The second would be a complex spiral. There would be multiple such spirals, with pairs spiraling in opposite directions to represent the conjugates. All would pass through the same point for a particular integer input.
For each "positive" b, there is an infinite family of values which are equal to b. Only one of them has 0 for the imaginary part (at k=0), but we get spoiled by the fact that there is a "primary" branch of exponentiation that gives us real values for real exponents. For "negative" b, there is an infinite family of values , none of which has 0 for the imaginary part. Being spoiled by the our experience with bases b = for k=0, we naively ask ourselves questions like "Why isn't there a realvalued function for negative b?" Such as silly question! (Yes, I'm being somewhat sarcastic.)
We should expect multiple exponentiation functions, as separate branches of some master relation (is "relation" the right word?). For positive b, only one of these branches returns real values for real exponents. For negative b, the best we can hope for is to pick the branch or its conjugate. Note that in these cases, we get alternating positive and negative outputs at the integers.
For complex b, the best we can hope for is to pick the branch with the smallest (absolute value) imaginary part in its logarithm. But there would always be those other branches, equally valid in some sense.
Getting back to tetration, we would expect a realvalued function for b>1, but there would likely be additional complex "branches", some of which may coincide with continuous iteration from one of the complex fixed points. For bases less than 1, we would only expect complex outputs, but there would be multiple branches (in addition to conjugates). Note that we get alternating upper and lower outputs, much as we got alternating positive and negative values for exponentiation of negative bases.
The analogy seems pretty good, but now we must dig deeper to understand the nature of it.
Edit: Heh, forgot to close all my tex tags.
edit 2: updated to include i's, and escaped the ln functions for good measure. See below evidence of my omission.
~ Jay Daniel Fox
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09/03/2007, 04:07 PM
(This post was last modified: 09/03/2007, 04:15 PM by bo198214.)
Before continue reading, the usual formula is:
there is an in the exponent.
Did you left it out by intention or by mistake?
Edit: corrected my own mistake
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Yes, left out the i by mistake. It was a quick reply, I'm about to leave for the beach. Sorry.
~ Jay Daniel Fox
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Gottfried Wrote:Quote:However perhaps it could be that is real or generally that is real, which I dont believe. Can someone just compute it?
Hmm, I don't know, whether I understand you correctly. If h=1 then all eigenvalues except the first are zero ( = [1,0,0,....]) and the result is always the same, independent of any power of log(h) since the "height" y of the tower occurs only as exponent of the eigenvalues....
Did I misread something obvious?
As said is not 1 but .
, is the power derivation matrix of (I think this is the transpose of your matrix ), and is (though we can also apply the powerseries directly to ). Hence the power series has as coefficients the first row of (think transposed in your notation).
And now tetration is defined as . We set not .
