Bifurcation of tetration below E^-E andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 01/15/2008, 10:29 PM Ivars Wrote:Quote: A complete analysis of such questions considering also the complex roots involves the T function as shown by Hayes in ND Hayes "The roots of the equation $x = c \exp^n(x)$ and the cycles of the substitution $x \rightarrow ce^x$ I think the tree function is T(x) = -W(-x). I like T because it allows H to be defined $h(x) = \frac{T(\ln x)}{\ln x}$ very simply. Andrew Robbins andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 01/16/2008, 04:47 AM Well, this is approximately what I was thinking. If $f(x) = b^{b^x}$, then we are trying to solve $f(x_0) = x_0$ and $f(x_1) = x_1$ where $x_0 \ne x_1$right? OK, so I started with a series expansion, and this is easier if you deal with a single exponential rather than an iterated exponential, so I found the series expansions: $ \begin{tabular}{rl} b^x & = x_1 + x_1\ln(b)(x-x_0) + \frac{1}{2}x_1\ln(b)^2(x-x_0)^2 + \cdots \\ \log_b(x) & = x_1 + \frac{(x-x_0)}{x_0\ln(b)} - \frac{(x-x_0)^2}{2(x_0^2\ln(b))} + \cdots \end{tabular}$ and with these expansions at hand, we notice that if $x_0 = b^{b^{x_0}}$, then $\log_b(x_0) = b^{x_0} = x_1$, so these two expansions should be the same (this is not true, but for some reason I was thinking it was... It is true for $x=x_0$ but not for all x). If these two expansions are the same, then the first two coefficients should be equal, which means $ \begin{tabular}{rl} x_1\ln(b) & = \frac{1}{x_0\ln(b)} \\ x_1\ln(b)x_0\ln(b) & = 1 \\ \ln(b^{x_1})\ln(b^{x_0}) & = 1 \\ \ln(x_0)\ln(b^{x_0}) & = 1 \\ \ln(x_0)\ln(x_1) & = 1 \end{tabular}$ Solving for $b$ in the second-to-last equation gives $b = \exp\left(\frac{1}{x_0 \ln(x_0)}\right)$, which I suppose is what I was thinking would be a general function, but I don't think it works that way. Also, solving for $x_0$ in the second-to-last equation gives $x_0 = \frac{1}{\ln(b) W\left(\frac{1}{\ln(b)}\right)}$. Andrew Robbins Gottfried Ultimate Fellow Posts: 763 Threads: 118 Joined: Aug 2007 01/16/2008, 09:09 AM Just a proposal - it might be useful to plot the graphs not with x-axis = base b, but, assuming b=t^(1/t), u=log(t), with x-axis =u. It rescales the plot a bit, so the behaviour may be better visible by such a graph. (your opinion?) Example:     Gottfried Gottfried Helms, Kassel Gottfried Ultimate Fellow Posts: 763 Threads: 118 Joined: Aug 2007 01/16/2008, 10:03 AM (This post was last modified: 01/16/2008, 09:48 PM by Gottfried.) andydude Wrote:Well, this is approximately what I was thinking. If $f(x) = b^{b^x}$, then we are trying to solve $f(x_0) = x_0$ and $f(x_1) = x_1$ where $x_0 \ne x_1$right? OK, so I started with a series expansion, and this is easier if you deal with a single exponential rather than an iterated exponential, so I found the series expansions:Hi Andrew - I actually do not understand the job, that x0 and x1 shall do. But it gives a mental association to me, which may be helpful for your considerations, too. From the eigensystem-analysis of decremented iterated exponentiation $U^{(h)}_t(x),$ where $U^{(1)}_t(x) = t^x -1,$ and $u=\log(t)$) I get a bivariate polynomial in u and u^h, where h indicates the "height" (see below). Since on the other hand, I found, that the iterated exponentiation $T^{(h)}_b(x),$ (with x=1 means our usual tetration) can be converted into $U^{(h)}_t(x),$ by scaling and shifting of x we may use these polynomials also for T-tetration. We have, using b=t^(1/t), (t complex for b>eta) the substitution x'=x/t-1, and inverse substitution x"=(x+1)*t, $T^{(h)}_b(x) = (U^{(h)}_t(x')) ''$ so we may apply the polynomials to x' and use the result y' by applying the inverse substitution y" Since we have denominators of the powerseries for U() , which generate singularities for all complex u, where abs(u)=1 (which means for instance, b=e^(e^-1) = eta,u=1, or b=(e^-1)^e = beta , u=-1) these cases come clearly out as special cases. However, by the construction of the polynomials, for integer heights h the denominators can be cancelled by the same factors in the numerators, and we have a powerseries with 0/0-terms, or, when cancelled, surely an equivalent solution to the logarithm-approach (but I didn't check this so far). Here is the powerseries for $U^{(h)}_t(x),$ (sorry , when editing the formula I overlooked one occurence of f(), which should be rewritten as U())     which I assume is the "ultimate" description for the tetration-problem (assuming continuation of tetration via powerseries-expansion) from where other considerations about the behaviour of tetration can be derived (at least for real u, which means for bases b 0e^1/e of so that his complexplane is situated in this construction perpendicularly x but behind x> e^1/e? With all the branches. Then we would be able to connect somehow that region above e^1/e on that complex plane with the region x(0, 1) in other complex plane I proposed via region in the middle-by moving this plane as a crossection of h(x) in 3 dimensions. May be this would not work without involving superroots as well. Would that not allow to understand the phenomena of bifurcation as such better? And phase transitions? Ivars andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 01/17/2008, 11:34 PM (This post was last modified: 01/18/2008, 09:25 AM by andydude.) @Ivars: First of all those are not my definitions, they are Knoebel's. His paper, called "Exponentials Reiterated" is an amazing exposition of the infinitely iterated exponential. It covers almost all aspects of it, and mentions many aspects of tetration as well. @Gottfried: Your graph confuses me, why is the 1-tick for b=... and the 1-tick for f=... not aligned? and why do you have X's for points, do you have any options for boxes? diamonds? circles? no dots? That would make the graph much nicer to look at. But anyway, I got the general idea of the graph, I just think it could use some work. I'm not sure what the benefit of such a change-of-variables would be, unless it would simplify the series for b^b^x. @Gottfried: As for your series expansion, I really, really like your presentation of the series for (b^x-1), it shows patterns in the coefficients, where I never saw patterns before Good job. About the role of x1 and x2, the role is as follows. The whole point of $b^{b^x} = x$ is to parameterize the two branches of odd-H, and even-H as described by Knoebel. Individually, both x0 and x1 satisfy this equation, but when viewed together, they satisfy: $b^{x_0} = x_1$ and $b^{x_1} = x_0$, as this represents the fact that exponentials oscilate between two values when $0 < b < e^{-e}$. The values x0 and x1 are the two values that the iterated exponential oscilates between. That is their role. Anyways, I found a way to construct a power series for this function that expresses the coefficients. Lets call it $b(x)$ since its a function that gives b as an output, and it takes x as an input, such that $x = b(x)^{b(x)^x}$ for all x in [0, 1]. Starting with the definition of this function, and differentiating: $ \begin{tabular}{rl} b^{b^x} & = x \\ (b^{b^x}) b^x (b \ln(b)^2 + (1 + x \ln(b)) b')/b & = 1 \\ (b^{b^x}) b^x (b \ln(b)^2 + (1 + x \ln(b)) b') & = b \\ x b^x (b \ln(b)^2 + (1 + x \ln(b)) b') & = b \\ x b^x b \ln(b)^2 + x b^x (1 + x \ln(b)) b' & = b \\ x b^x b \ln(b) \ln(b) + x b^x (1 + x \ln(b)) b' & = b \\ x b \ln(b^{b^x}) \ln(b) + x b^x (1 + x \ln(b)) b' & = b \\ x b \ln(x) \ln(b) + x b^x (1 + x \ln(b)) b' & = b \\ x b \ln(x) \ln(b) + x b^x b' + x^2 b^x b' \ln(b) & = b \\ x b \ln(x) \ln(b) + x b^x b' + x^2 b' \ln(b^{b^x}) & = b \\ x b \ln(x) \ln(b) + x b^x b' + x^2 b' \ln(x) & = b \\ x b^x b' + x^2 b' \ln(x) & = b - x b \ln(x) \ln(b)\\ x b^x b' + x^2 b' \ln(x) & = b - x b \ln(x) \ln(b)\\ x(b^x + x\ln(x))b' & = b(1 - x \ln(x) \ln(b)) \\ b'(x) & = \frac{b(1 - x \ln(x) \ln(b))}{x(b^x + x\ln(x))} \end{tabular}$ And we now have a differential equation, or in other words, a closed form for the derivative. Many cases in which we have a closed form for the derivative, we can find a series expansion around (almost) any point. So for example, we know the closed form of the derivative of the Lambert W function, which is one way its coefficients can be found (this is the hard way). Anyways, to get back to my point. Using the second-to-last equation (which allows either side to be 0, the last equation prevents the denominator from being 0), We can start with an "unknown" series expansion for $b(x)$ and solve for its coefficients to satisfy this differential equation. Doing this at $x=\frac{1}{e}$, we get (factoring out $e^{-e}$ to simplify things): $ \begin{tabular}{rl} \frac{b(x)}{e^{-e}} & = 1 \\ & +\ \frac{1}{6} e^3 (-1) (x-\frac{1}{e})^2 \\ & +\ \frac{1}{18} e^4 (x-\frac{1}{e})^3 \\ & +\ \frac{1}{1080} e^5 (-82 + 15e)(x-\frac{1}{e})^4 \\ & +\ \frac{1}{1620} e^6 (79 - 15e)(x-\frac{1}{e})^5 \\ & +\ \frac{1}{45360} e^7 (-2236 + 644e - 35e^2)(x-\frac{1}{e})^6 \\ & +\ \frac{1}{226800} e^8 (9004 - 2800e + 175e^2)(x-\frac{1}{e})^7 \\ & +\ \cdots \end{tabular}$ With this series expansion, we can then test to see if we have a "correct" function, by testing its derivatives, rather than continually guessing. Maybe all we have to do to find the right function is look it up on OEIS, maybe its already there... Andrew Robbins PS. I love finding new series GFR Member Posts: 174 Threads: 4 Joined: Aug 2007 01/17/2008, 11:56 PM (This post was last modified: 01/18/2008, 12:28 AM by GFR.) Hi IVARS! A small preliminary comment (during the weekend I shall supply more) to your interesting "vision": Ivars Wrote:............. When I was speaking in previous post about the missing dimension, I was thinking about rotating plane (or rotation in plane) perpendicular the graphs of GFR and parallel to y axis. Then I have not even a guess, but just a vision of these 3 definitions of h(x) forming a double helix (odd/even) with a thread (x^1/x) in the midlle in region e^-e + oo, seem to be describable as "permanent oscillations" around an asypmptotic axis, defined by: y = plog(-log/b x)(-log/b x). Therefore, I agree with you (IVARS) thet we need an extra dimension, but I think that it should be represented by the imaginary axis itself. In this case, it will be the "fourth dimension" of the diagram. I also agree with your "spirally-like vision", but I guess that the (deformed) spirals (for any fixed b) should be ... unique. What we "see" in the yx plane is (I guess) the "real projection" of the spiral, at any fixed b. On the contrary, what we see in the "traditional" yb diagram (the upper and lower h's) is (always in my guessing dream) the projection of the spiral on the yb plane, for x -> *oo. The upper and lower y, in the "Forbidden Zone for Integer Superexponents" should show the upper and lower maximum elongations around the center of the spiral, for any constamt b and for x -> +oo. Sorry for the ... approximated description and thank you for your kind attention. GFR andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 01/18/2008, 05:50 AM Ivars, I think I know what you might be talking about, or at least, I have something that may be described in similar words. This is a 3D plot of the function $b^{b^x} - x$ and a 3D plot of "zero". The interesting thing here, is that the intersection of the two is what we are trying to parameterize. You can see the intersection of the two in this plot:     What is also interesting about this plot is that is shows (in some sense), why the outer "legs" should be attracting periodic points, and the middle is a repeling fixed point. It shows this because the slope/derivative is in the opposite direction fot the middle than it is for the other two branches/legs. Andrew Robbins « Next Oldest | Next Newest »