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 Iterating at fixed points of b^x bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 09/12/2007, 08:13 PM (This post was last modified: 09/12/2007, 08:19 PM by bo198214.) Daniel Wrote:bo198214 Wrote:I dont know whether I am the first one who realizes that regularly iterating $\exp$ at a complex fixed point yields real coefficients! Moreover they do not depend on the chosen fixed point! I don't think that is true, if I understand you.Yes I withdrawed this assertion (based on an error in my computation) in the preprevious post. However I numerically verified (hopefully this time without error) that the regular iterations at both *real* fixed points coincide for $b. Quote:Just consider the term $\ln(a)^n$ in $\;^{n}b = a + \ln(a)^n \; (1-a) + \ldots$. Öhm, a bit more explanative? Quote:As a side note, consider any two fixed points $a_j,a_k$ for the same $b$, then the fixed point commute under exponentiation ${a_j}^{a_k}= {a_k}^{a_j}$. Yes, this is clear as $a_1^{a_2}=b^{a_1a_2}$ which is independent on the order of $a_1$ and $a_2$. Daniel Fellow Posts: 51 Threads: 20 Joined: Aug 2007 09/12/2007, 09:50 PM bo198214 Wrote:Daniel Wrote:Just consider the term $\ln(a)^n$ in $\;^{n}b = a + \ln(a)^n \; (1-a) + \ldots$.Öhm, a bit more explanative?This is just a version of my definition for extending tetration from http://tetration.org/tetration_net/tetra...omplex.htm . The Taylor series of $\;^{n}b$ taken at $a$ has the zero term at the fixed point of course and then $D f^n(a)= f'(a)^n = \ln(a)^n$. bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 09/12/2007, 10:00 PM (This post was last modified: 09/12/2007, 10:01 PM by bo198214.) Daniel Wrote:bo198214 Wrote:Daniel Wrote:Just consider the term $\ln(a)^n$ in $\;^{n}b = a + \ln(a)^n \; (1-a) + \ldots$.Öhm, a bit more explanative?This is just a version of my definition for extending tetration from http://tetration.org/tetration_net/tetra...omplex.htm . The Taylor series of $\;^{n}b$ taken at $a$ has the zero term at the fixed point of course and then $D f^n(a)= f'(a)^n = \ln(a)^n$. This is surely true, but I dont see the connection to showing that the regular iteration at fixed point $a$ is non-real for real arguments $x$. Natural numbered iterations of $b^x$ at any fixed point of course yield real values for real arguments $x$, just $\exp_b^{\circ n}(x)$. Which is no more true for fractional iterations. bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 10/02/2007, 06:39 PM Here is a little animated gif that shows how the fixed points in the complex plane change in dependency on the base $b$. For $\eta=e^{1/e}$ the animation shows b going through $\eta+0.2 \dots \eta-0.1$. One can clearly see, how the conjugated primary complex fixed points transform into two real fixed points after $b$ goes below $\eta$.     bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 10/02/2007, 07:57 PM Here is a zoom of the primary fixed point showing its trajectory for $b=\eta+10\dots \eta+0.05$. You see that it heads to $e$ for $b\to\eta$.     There is also a mysterious base where the primary fixed point is exactly $i$. This is the case for $i=b^i$ which is clearly satisfied for $b=e^{\frac{\pi}{2}}\approx 4.8104$. Now a zoom of the secondary fixed point showing its trajectory for $b=11\dots 1.05$.     jaydfox Long Time Fellow Posts: 440 Threads: 31 Joined: Aug 2007 10/03/2007, 03:58 AM bo198214 Wrote:Here is a little animated gif that shows how the fixed points in the complex plane change in dependency on the base $b$. For $\eta=e^{1/e}$ the animation shows b going through $\eta+0.2 \dots \eta-0.1$. Random, off-topic question, and feel free to move this, but what do you use for animating gifs? I have yet to find a decent gif animating program that isn't free. ~ Jay Daniel Fox bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 10/03/2007, 09:03 AM jaydfox Wrote:Random, off-topic question, and feel free to move this, but what do you use for animating gifs? I have yet to find a decent gif animating program that isn't free. This was an export from a Maple animation to gif. I dont know about others but perhaps first search the documentation of sage whether there is a similar feature. GFR Member Posts: 174 Threads: 4 Joined: Aug 2007 10/03/2007, 11:06 AM bo198214 Wrote:There is also a mysterious base where the primary fixed point is exactly $i$. This is the case for $i=b^i$ which is clearly satisfied for $b=e^{\frac{\pi}{2}}\approx 4.8104$. The fixed points of functional equation x = b^x define, at the same time, the "heights" of the corresponding infinite towers: x = b#oo. We have seen that, for bases b < e^(1/e), there are real x solutions. For b > e^(1/e), our discussion (see Gottfried, for example) seems to bring us to agree on complex "fixed points" satisfying x = b^x, for x real >0. This means that we may have infinite towers with complex values (heights). In particular, there is the mysterious fact shown by Henryk: x = b^x, satisfied by x=i, if b = e^(Pi/2). Actually, it is well known that: e^(Pi/2) = i. This means that an infinite tower x = b#oo, with b = 4.810477381.., may have an imaginary "height" of x = i = +/- sqrt(-1). This is also corroborated by the recent plots shown by Gottfried and by a plot I got using the "Mathematica" product-logarithm operators [order 0 and -1]. By the way, very beautiful animated plot, Henryk! Are then we authorized to write: (4.810477381..)#oo = {-i, +i, +oo} ? Can we also see other complex "heights" (branches) ? In other words, for b > Eta, together with a set of complex "heights" solutions, shall we always be authorized (obliged?) to also admit at least one infinite tower with infinite height? Sorry to be so annoying and ... flat. GFR bo198214 Administrator Posts: 1,386 Threads: 90 Joined: Aug 2007 10/03/2007, 11:25 AM GFR Wrote:The fixed points of functional equation x = b^x define, at the same time, the "heights" of the corresponding infinite towers: x = b#oo. We have seen that, for bases b < e^(1/e), there are real x solutions. For b > e^(1/e), our discussion (see Gottfried, for example) seems to bring us to agree on complex "fixed points" satisfying x = b^x, for x real >0. This means that we may have infinite towers with complex values (heights). Though metaphysically this thought perhaps makes sense, we have to admit that by mathematical consideration $\lim_{n\to\infty} {^nb}=\infty$ for $b>\eta$. Quote:Actually, it is well known that: e^(Pi/2) = i. Hey guys, dont forget the $i$ in the exponent! $e^{\frac{\pi}{2}i}=i$. Quote:This means that an infinite tower x = b#oo, with b = 4.810477381.., may have an imaginary "height" of x = i = +/- sqrt(-1). There can at most be one limit however there are many fixed points. Quote:Are then we authorized to write: (4.810477381..)#oo = {-i, +i, +oo} ? Can we also see other complex "heights" (branches) ? In other words, for b > Eta, together with a set of complex "heights" solutions, shall we always be authorized (obliged?) to also admit at least one infinite tower with infinite height? No, no, no. GFR Member Posts: 174 Threads: 4 Joined: Aug 2007 10/03/2007, 11:03 PM (This post was last modified: 10/03/2007, 11:07 PM by GFR.) Sorry, it can happen, I meant: e^(Pi*i/2). I confused the two i's in the sequential notation. It may happen in ... the best families. Concerning your => (No, no, no.) remember that i = e^[(Pi/2)i] = [e^(Pi/2]^i, i. e.: i = k ^ i = k ^ (k^i) = k ^ (k ^ (k^i)) = .... = k#oo, i.e. : k#oo = i. Methaphysically speaking, we could say that an infinite tower (k#oo) with base b = 4.810477381... can be equal to "i". Concerning the infinite limit values of the infinite towers (b#oo), for b > Eta, I think we are in agreement. But I don't insist, just to avoid to say any other irrelevant nonsense. (; ->) GFR « Next Oldest | Next Newest »

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