Tetration below 1 GFR Member Posts: 174 Threads: 4 Joined: Aug 2007 09/02/2007, 01:30 PM Good grief ...., bo! Thanks a lot! Why ... one less ... ? Twice its decimal mirror is fifty-four. The difference is the double of the cube of three, symbol of the dialectics (thesis-antithesis-synthesis), 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 ... . bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 09/02/2007, 01:40 PM 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 GFR Member Posts: 174 Threads: 4 Joined: Aug 2007 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 (3-tower-2) and, at the same time, the result of: (2^3) x (3^2) [the brackets are not necessary]. GFR (;-<)====| . Amended on 07-09-03 ... forget it !!! Gottfried Ultimate Fellow Posts: 757 Threads: 116 Joined: Aug 2007 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 eigen-decomposition 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 -11, 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 bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 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: $b^x=e^{x(\ln(b)+2\pi i k)}$ there is an $i$ in the exponent. Did you left it out by intention or by mistake? Edit: corrected my own mistake jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 09/03/2007, 04:36 PM Yes, left out the i by mistake. It was a quick reply, I'm about to leave for the beach. Sorry. ~ Jay Daniel Fox Gottfried Ultimate Fellow Posts: 757 Threads: 116 Joined: Aug 2007 09/03/2007, 06:54 PM bo198214 Wrote:So let me conclude. If $-1<\log(h)<0$ then we have the negative Eigenvalues $\log(h)^{2n+1}$ in the power derivation matrix A of $f(x)=b^x$. Now we compute $A^t$. It has the Eigenvalues $(\log(h)^n)^t$. Take for example $t=\frac{1}{2}$ then we see that $A^t$ has also non-real Eigenvalues and hence has also non-real entries. Supposed $f^{\circ \frac{1}{2}}$ had only real coefficients then $A^{\frac{1}{2}$ would have only real coefficients. So it is clear that $f^{\circ \frac{1}{2}$ must have some non-real coefficients and is a non-real function. However perhaps it could be that ${}^{\frac{1}{2}}b=f^{\circ \frac{1}{2}}(1)$ is real or generally that ${}^tb=f^{\circ t}(1)$ 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? Gottfried Gottfried Helms, Kassel bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 09/03/2007, 07:25 PM Gottfried Wrote:Quote:However perhaps it could be that ${}^{\frac{1}{2}}b=f^{\circ \frac{1}{2}}(1)$ is real or generally that ${}^tb=f^{\circ t}(1)$ 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 $h$ is not 1 but $e^{-1}. $f(x)=b^x$, $A$ is the power derivation matrix of $f$ (I think this is the transpose of your matrix $B_b$), and $A^t$ is $\exp(t\cdot \log A)$ (though we can also apply the powerseries $(1+x)^t$ directly to $A$). Hence the power series $f^{\circ t}(x)$ has as coefficients the first row of $A^t$ (think transposed in your notation). And now tetration is defined as ${}^tb=f^{\circ t}(1)$. We set $x=1$ not $h$. « Next Oldest | Next Newest »