Bifurcation of tetration below E^-E GFR Member Posts: 174 Threads: 4 Joined: Aug 2007 05/01/2008, 05:36 PM (This post was last modified: 05/02/2008, 12:03 PM by GFR.) Concerning my "old" comment about "parametration" of the perimeter of the h/sup and h/inf zone, for 0 < b < e^(-e): GFR Wrote:andydude Wrote:About GFR's "yellow zone" (YZ) or Ivars' "phase transition region" I was very interested to see the formula $x = (y^{-y} -1)\frac{e^{-e}}{e^{1/e}-1}$. I think it is a very good approximation, but I think it is not accurate..... you are right! My formula is wrong, because it is only a very approximated .... approximation. The only way, I think, is to take into consideration [/b]log x = b^x. I shall come back to that asap.Actually, I should like to propose the following strategy and formulas: a) - Let us define the joint union of the two real branches of the Lambert Function (the product-logarithm) as: plog(z) = {W/-1 (z), W/0 (z)}, where W/-1 and W/0 are "Mathematica's" ProductLog[-1,z] and ProductLog[x], respectively. b) - As we know, a formula for h(b), the height of the "infinite towers", is given by h = plog(-ln b)/(-ln b), inverse of b = h^(1/h), the selfroot function. We also know that this expression is not sufficient to describe what happens at 0 < b < e^(-e), where a zone suddenly appears, indicating the asymptotic max and min values of an oscillating y = b[4]x (real part of a complex function), for b < e^(-e) [my interpretation]. It is also clear that these formulas were obtained taking into consideration y = b[3]y, implying y = b[4]oo, according to the DL. c) - Now, if we take into consideration: y = b[3]y = b[3](b[3]y), we can write: b[3]\y = y = b[3]y, i.e.: log[/b] y = y = b^y, i.e.: ln (y^y) = y*(ln b)*e^y*(ln b), i.e.: y*(ln b) = plog(ln (y^y)), i.e.: b = e^((plog(ln(y^y)))/y). I think that this formula is very accurate and, perhaps, the correct one, to be used instead of the simple selfroot. It goes without saying that we need to invert it for obtaining y = h = h(b). In fact, with plog(z) = W/0 (z), the standard Lambert function, we get the function shown in annex Fig. 1, covering the upper part of the "yellow zone" [b = b(y), with b vertical], for y < 1/e, and the remaining part described by the selfroot, for y > 1/e. Using the W/-1 branch of the Lambert Function, for y < 1/e, we get the middle b = b(y), also given by the selfroot b = y^(1/y). For y > 1/e, we obtain the second branch of the perimeter of the "yellow zone" (see Fig. 2). Fig. 3 shows the two plots obtained by b = e^((plog(ln(y^y)))/y), jointly superposed. We see again the continuous line described by the selfroot, together with the perimeter of the "yellow zone", with a maximum at y = 1/e, with b = e^(-e). The plot is extremely similar to what was obtained by Andydude. We also see that b = b(y) = b(h) has two values, for 0 < y < 1. Fig 4 shows the comparison between my old (insufficient) approximations and my new (I hope correct) formulas, for b = b(y) or b = b(h). Fig. 5, 6, 7 are plots of low iterated logs and exps (to the base b = 0.025 < e^(-e)), showing the crosspoints for even and odd iterations, as well as the fixpoints, whenever they exist [f(x)=x]. Now, what is needed is to "invert" this description (Fig3) for getting h = h(b). But tis, will probably come next time. Fig. 8 is a qualitative presentation of what I am expecting. May Day, May Day, please help me!!! GFR Attached Files   Fig. 1-8.pdf (Size: 46.69 KB / Downloads: 417) andydude Long Time Fellow Posts: 509 Threads: 44 Joined: Aug 2007 05/02/2008, 05:35 AM Wow, You found it! Aside from the proof you gave, I also did some tests just to be sure. Here is the Mathematica code I used: Code:In[1]:= f[k_, y_] := Exp[ProductLog[k, Log[y^y]]/y] In[2]:= FullSimplify@Table[Limit[D[f[0, y], {y, k}]/k!, y -> 1/E, Direction -> 1], {k, 0, 4}] Out[2]= {E^(-E), 0, -E^(3 - E)/6, E^(4 - E)/18, (E^(5 - E)*(-82 + 15*E))/1080} In[3]:= FullSimplify@Table[Limit[D[f[-1, y], {y, k}]/k!, y -> 1/E, Direction -> -1], {k, 0, 4}] Out[3]= {E^(-E), 0, -E^(3 - E)/6, E^(4 - E)/18, (E^(5 - E)*(-82 + 15*E))/1080}and as you can see, these coefficients are the same as those I derived on page 2 of this thread, so this is definitely the function! Good Job GFR! Andrew Robbins bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/02/2008, 04:22 PM GFR Wrote:It is also clear that these formulas were obtained taking into consideration y = b[3]y, implying y = b[4]oo Gianfranco, please stop repeating wrong results. I showed already here that this implication is wrong. The left side has multiple solutions for $y$ while there can only be one limit $y$ on the right side. Only one solution of the left side can be the actual limit of the right side. Then I rephrased your considerations into a more clear mathematical context: What you want to know - if I got this right - is $y_0=\lim_{n\to\infty} b[4](2n)$ and $y_1=\lim_{n\to\infty} b[4](2n+1)$. By exchanging $n$ with $n+1$ the limit must remain the same, so we have $y_0=\lim_{n\to\infty} b[4](2(n+1))=\lim_{n\to\infty} b[4](2n+2)=\lim_{n\to\infty} b^{b^{b[4](2n)}} = b^{b^{\lim_{n\to\infty} b[4](2n)}} = b^b^{y_0}$ and similar as above $y_1=b^{b^{y_1}}$. We have also the correspondence $y_1=b^{y_0}$ and $y_0=b^{y_1}$, as you can easily derive in the above manner. From there you can apply your formula $b=e^{\text{W}(y\ln(y))/y}$ and find that $y_0$ corresponds to the solution when taking the branch 0 of $W$ and $y_1$ corresponds to the solution when taking the branch -1 of W. (or vice versa?) For the actual computation of $y$ in dependence of $b$ it only crosses my mind that: $y=b^{b^y}$ is equivalent to $y^{(1/b)^y}=b$, so if we had a symbol for the inversion of $f_c(x)=x^{c^x}$ then we had the solution $y={f_{1/b}}^{-1}(b)$. $f_{1/b}$ is a strictly increasing function for $b\ge e^{-e}$, here the limit case $b=e^{-e}$:     but has 3 (?) solutions $f_{1/b}(y)=b$ for $0:     I would guess the biggest solution is $y_0$ and the smallest solution is $y_1$ (but what is then the middle solution?). GFR Member Posts: 174 Threads: 4 Joined: Aug 2007 05/02/2008, 10:59 PM bo198214 Wrote:GFR Wrote:It is also clear that these formulas were obtained taking into consideration y = b[3]y, implying y = b[4]ooGianfranco, please stop repeating wrong results.Dear Henryk! Finally, my formula for the perimeter of the yellow zone is correct, despite my annoying repetitions of ... wrong statements. Despite that, I am still thinking that my "implicartion" was not completely wrong. What somebody could argue about is that: yes, y = b[4]oo has two branches (two values) and only the lower one can be reached. Therefore, it would be "cleaner" to start from the lower "civilized rational" branch of y = b[4]oo, and stipulate that this implies that y = b[3]y. But, you see, this is too difficult for me. This has nothing to do with my IQ factor, but it is only question of space (where things have to be done) and time (for doing things) of my ordinary life. Unfortunately, as I said, I will not have much time any more to spend in these interesting debates. It was nice to meet you all. GFR bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 05/03/2008, 08:09 AM (This post was last modified: 05/03/2008, 08:26 AM by bo198214.) GFR Wrote:Finally, my formula for the perimeter of the yellow zone is correct,yup thats a nice formula and one wouldnt have guessed that $b^{b^x}=x$ is solvable for $b$ with help of the plog/Lambert W. Quote:What somebody could argue about is that: yes, y = b[4]oo has two branches (two values) and only the lower one can be reached. What you are probably referring to is $H_b(x)=\lim_{n\to\infty} \exp_b^{\circ n}(x)$ which is more general then just $b[4]\infty$, i.e. $b[4]\infty = H_b(1)$. However the behaviour of this function for $1< b\le e^{1/e}$ with real fixed points at $1 is as follows: $H_b(x)=\infty$ for $x>a_2$ $H_b(x)=a_2$ for $x=a_2$ $H_b(x)=a_1$ for $0 This is because $a_1$ is an attracting fixed point, while $a_2$ is a repelling fixed point. I.e. only if we give exactly $a_2$ to $H_b$ it stays at $a_2$ otherwise the limit is scared away from $a_2$ (either to $a_1$ or to $\infty$). For $e^{-e}\le b\le 1$ there is only one real fixed point $a\le 1$, which is attracting, for $b^x$. The behaviour hence is $H_b(x)=a$ for each $0. For $0 the limit $b[4]\infty=H_b(1)$ is not defined. However we have two accumulation points $y_1. They form a 2-cycle as the dynamics people call it. This is because $b^{y_1}=y_2$ and $b^{y_2}=y_1$. So here would be the right place to introduce multi numbers. One could say that $\exp_b^{\circ n}(1)$ converges to the set $\{y_1,y_2\}$. And if you consider convergence of filters in general topology they indeed converge to sets. The upper accumulation point $y_2$ can be described as $\lim_{n\to\infty} f_b^{\circ n}(1)$ where $f_b(x)=b^{b^x}$. When we have a look at the graph of $b^{b^x}-x$ we see that there are 3 fixed points $a_1 of $f_b(x)=b^{b^x}$:     The middle one is repelling, the lower and the upper are attracting. So we can say that for $0 $f_b^{\infty}(x)=a_3$ for $x>a_2$ $f_b^{\infty}(x)=a_2$ for $x=a_2$ $f_b^{\infty}(x)=a_1$ for $0 So we can complete our case distinction regarding the behaviour of $H_b$ with: For $0 where $a_1 are the 3 real fixed points of $b^{b^x}$: $H_b(x)=\{a_1,a_3\}$ for $x\neq a_2$ $H_b(x)=a_2$ for $x=a_2$ The middle fixed point $a_2$ is the fixed point of $b^x$ which is however now repelling for the chosen $b$. and of course for $b>e^{1/e}$ always $H_b(x)=\infty$. Quote:Therefore, it would be "cleaner" to start from the lower "civilized rational" branch of y = b[4]oo As we just have seen there are no "branches" but one value for $b[4]\infty=H_b(1)$ for $e^{-e}\le b\le e^{1/e}$. GFR Member Posts: 174 Threads: 4 Joined: Aug 2007 05/03/2008, 01:13 PM Well ... ! Ok. Thank you very much. GFR Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 06/16/2008, 05:08 PM (This post was last modified: 06/17/2008, 07:30 AM by Ivars.) For infinite tetration of z $z: $h(z)= 1/x; 1/y$ such that $z=(1/x)^y = (1/y)^x$ e.g if x=2, y =4, $z= (1/2)^4=(1/4)^2=1/16$ and $h(z)=1/2; 1/4$ e.g if x=1.78381425177, y =4.89536795553, $z= (1/1.78381425177)^{4.89536795553}=(1/4.89536795553)^{1.78381425177}=1/17$ and $h(z)=h(1/17)= 1/1.78381425177; 1/4.89536795553$ Oops I did not notice Henryk has already written the same in previous post. Quote:One could say that converges to the set . And if you consider convergence of filters in general topology they indeed converge to sets. I also like the idea that function (filter above) can converge to a multinumber set. This could be applicable also to the strange 3_logarithm I introduced in another thread Generalization of logarithms with property 3_log(z=x^y) = 3_log(x)*3_log(y) , which basically maps ALL values of Real z=x^y to set {0,1}, so for all z it converges to set {0,1}, useful or not. In case h(b), b< e^-e 2 number set moves with b, according to rule above and more exactly stated by Henryk in previous post. In limit b->0, this set becomes {0,1}. In case of 3-logarithm, it does not move at all at least for real x, and is {0,1}. so 3_log(z) = lim z->0 h(z) = {0,1} if z is real and h(z) stays purely real. Ivars Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 10/28/2008, 10:43 PM (This post was last modified: 11/19/2008, 02:46 PM by Ivars.) Long time, no see. I have come to (from physical considerations) an unproved conclusion that GFR was right in a sense of multivaluedness of W function below $-1/e$ : Simultaneous considerations of branches W1 and W-1 of the Lambert function (and may be other symmetric) , and , related values of particular value of h( e^pi/2) = +-I are in fact , spinors or , as Cartan states it (free quote): Pair of 2 complex numbers that connect 3 complex coordinates so that it leads to an isotropic (Fundamental quadratic form=0) vector in some (my addition) coordinate space. for example, values +I and -I satisfy these equations in 3D orthogonal coordinate system if x1=0: x1^2+x2^2+x3^2 = 0 x1=0 x2= -2I x3= -2 $\eta0=+I$, $\eta1=-I$ These coordinates can also be projective. In this case, the solution of an Absolute conic in plane ( 3 projective coordinates characterize point in plane, x1=0 means that point is somewhere at infinity (ideal point, vanishing point) ( usually it is x3=0, but that does not matter how we assign them) will be a spinor. In case +I, -I this Absolute conic will be a circle- in fact, all circles in that projective plane as they all cross in these imaginary points. So the actual transformation $h(I^{1/I})$ performs is not from a real to complex number, but from real to spinor in projective space( "restricted complex number pair with certain properties" ) Ivars mrob27 Junior Fellow Posts: 3 Threads: 0 Joined: Feb 2008 03/14/2011, 04:14 AM My pages moved, and that link died. It is now at: mrob.com/pub/math/numbers-4.html#la15_1542 I have also reorganized my numbers pages to have more permanent links (the chapter divisions won't keep changing) - Robert (01/14/2008, 08:01 PM)andydude Wrote: By the way, I like your itemization of the "remarkable numbers" related to e. If you look at Robert Munafo's http://home.earthlink.net/~mrob/pub/math....html#la15 he has a bit about e^e, just to let you know. Andrew Robbins « Next Oldest | Next Newest »