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 complex iteration (complex "height") Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 03/23/2008, 11:24 PM (This post was last modified: 03/24/2008, 10:22 AM by Ivars.) Anyway, I made a graph cause it looks nice. The part from x= 0.9-1.1 (base) is actually going to -infinity left of base=1, + infinity right of base=1 , but i did not calculate the points since then the rest of graph becomes invisible. If the idea is right, there one can choose pairs (base<1) or triples (base>1) of real bases and heights giving 1 as a result of b[4]height. base[4]height=1     I changed the name of the graph as it was wrong- it is finite real tetrations h of base b=a^(1/a) such that b[4]h=1 . Ivars bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 03/24/2008, 11:16 PM Ivars Wrote:lower fixed point $a=\Omega=0.567143..$ we get the base by $b={1/e}=\Omega^{1/\Omega}$ if $z={1/\Omega}$ then Hm, perhaps I have to add that the formula given is designed for strictly increasing functions with an attracting fixed point. However if you set $b<1$ then the function $b^x$ is strictly decreasing. It was discussed somewhere on this forum already that it most likely can not have real-valued fractional iterations there (real valued only at integer iterations). I dont know how this formula behaves in this case (whether it converges), at least the fixed point is still attractive, so try your luck! Would be interesting whether this supports my conjecture of complex values in the range of $e^{-e}. Quote:${1/e}[4]{1/\Omega}=\lim_{n\to\infty} \log_{1/e}^{\circ n}(\Omega*2 -1*\exp_{1/e}^{\circ n}(1))$ At this point I do not know what to do with limits -how do You apply n to iterating logarithms, and at the same time to iteration of exponentiation of 1/e inside it? It just means that if you set $n$ high enough you get quite close to the "right" value. And we discussed already how $f^{\circ n}(x)$ is defined. For example $\exp_b^{\circ 3}(x)=b^{b^{b^x}}$. Quote: If we could move limit inside, than: $\lim_{n\to\infty} \exp_{1/e}^{\circ n}(1))=h({1/e)}=\Omega$ And $2*\Omega-\Omega=\Omega$ $\log_{1/e}(\Omega)= \Omega$ Yeah, but unfortunately its not possible to move the limit inside! Thats the strange beauty of this formula. Quote:If we take the same but: $z={-1/\Omega}$ And then I dont want to think about negative bases, perhaps first finish with bases $e^{-1}! Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 03/25/2008, 08:29 AM (This post was last modified: 03/25/2008, 08:32 AM by Ivars.) bo198214 Wrote:If we take the same but: $z={-1/\Omega}$ And then I dont want to think about negative bases, perhaps first finish with bases $e^{-1}! The base in this case is positive, it is $a=\Omega$, it is height $z={-1/\Omega}$ that is negative. For all a<1 this formula will lead to negative heights: $a^{(1/a)} [4] {1/ln(a)} = 1$ Ivars bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 03/25/2008, 09:22 AM Ivars Wrote:bo198214 Wrote:$z={-1/\Omega}$ The base in this case is positive, it is $a=\Omega$, it is height $z={-1/\Omega}$ that is negative. Oh, right, I wasnt sure what you meant by $z$, however now its clear that it is the right argument of tetration. bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 03/26/2008, 04:46 PM As it seems that Ivars is too fearful of complex numbers. I give here the plot (1/e)[4]t by regular iteration at the lower fixed point (1/e is indeed in the range of bases $e^{-e}\dots e^{1/e}$ where there a real fixed point exists) via the previously mentioned formula. As we already discussed in the thread Tetration below 1 the values for non-integer iterations are supposed to be complex. So I give here the curve of (1/e)[4]t, t=0...7 in the complex plane. We see that (1/e)[4]0=1 and of course (1/e)[4]1=1/e$\approx$0.36, (1/e)[4]2=(1/e)^(1/e)$\approx$0.69, etc.     Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 03/26/2008, 06:22 PM (This post was last modified: 03/26/2008, 08:30 PM by Ivars.) Congratulations! This opens whole lot of new worlds. bo198214 Wrote:As it seems that Ivars is too fearful of complex numbers. Not afraid, but difficult for me to work with before I learn enough software how to handle complex number iterations... I love them, at least imaginary unit So You mean Your formula works at least in the range $e^{-e}; e^{1/e}$ also below 1? That is wonderful achievement because it leads to infinity of very simple exactly definable points of negative real heights (when a<1, $a^{1/a}[4]{1/\ln(a)}=1$ which can also be utilized to check various real and complex heigths calculation approaches. For base ${1/e}$, ${1/e}[4]-{1/\Omega} =1$ - you can add this to your graph as well perhaps another spiral will emerge which should pass via this point? Your nice spiral graph (see-spirals are involved) anyway converges to my beloved $\Omega$, as it should as ${1/e}[4]\infty=\Omega=0.567143..$. Which just confirms its true. But what about base $e^{1/e}$ which is limit case? Ivars bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 03/26/2008, 08:54 PM Ivars Wrote:Not afraid, but difficult for me to work with before I learn enough software how to handle complex number iterationsYa right, probably excel can not compute with complex numbers ... Quote:So You mean Your formula works at least in the range $e^{-e}; e^{1/e}$ also below 1? That is wonderful achievement Yes, exactly, isnt it?! Bases below 1 were what *I* was afraid of! Quote:For base ${1/e}$, ${1/e}[4]-{1/\Omega} =1$ - you can add this to your graph as well perhaps another spiral will emerge which should pass via this point? You mean $\frac{1}{e}[4](-\frac{1}{\Omega})=1$? Thats not true. $b[4]t$ is only real for integer $t$ (considering $e^{-e}). Quote:Your nice spiral graph (see-spirals are involved) anyway converges to my beloved $\Omega$, as it should as ${1/e}[4]\infty=\Omega=0.567143..$. Exactly: $a^{1/a}[4]\infty=a$. Quote:But what about base $e^{1/e}$ which is limit case? There are also iteration formulas, I think you can find them in Ecalle's work. However if I remember right, its not one formula, but some intermediate steps required. Meanwhile however you can just consider it as a limit of the other case. Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 03/26/2008, 10:05 PM (This post was last modified: 03/26/2008, 10:13 PM by Ivars.) bo198214 Wrote:You mean $\frac{1}{e}[4](-\frac{1}{\Omega})=1$? Thats not true. $b[4]t$ is only real for integer $t$ (considering $e^{-e}). But .. You have not really considered negative heights, have You? May be its different then. $b[4]t$ is real at t=integers only for t(height ) >0 ; for t=0 it is real for all b and $b[4]0=1$; what about t<0? May be there the points t (when t<0) where $b[4]t$ are real (or at least $b[4]t =1$) and t are not negative integers? Just asking, the idea came from your formula anyway, setting $(1-\ln{(a)^z)} = 0$ , so that infinite iterations in both directions cancel out (log and exp). If formula works for bases in that range in general, why not for negative heights? Ivars bo198214 Administrator Posts: 1,389 Threads: 90 Joined: Aug 2007 03/26/2008, 10:19 PM Ivars Wrote:$\ln(\Omega)=-\Omega$ $\ln(\Omega)^{1/\Omega }=({1/\Omega })*\ln(\Omega)={-\Omega/\Omega }=-1$ here is an error $\ln(\Omega)^{1/\Omega}\neq \frac{1}{\Omega}\ln(\Omega)$. Ivars Long Time Fellow Posts: 366 Threads: 26 Joined: Oct 2007 03/26/2008, 10:30 PM bo198214 Wrote:here is an error $\ln(\Omega)^{1/\Omega}\neq \frac{1}{\Omega}\ln(\Omega)$. Thanks, that is probably even better...learning by doing...I was wandering can it be so simple .So back to textbook... Ivars « Next Oldest | Next Newest »

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