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 some questions about sexp sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 07/02/2010, 03:52 PM (This post was last modified: 07/02/2010, 05:40 PM by sheldonison.) (06/28/2010, 11:18 PM)tommy1729 Wrote: (question 3) what happens to limit cycles and n-ary fixpoints ?? sure we can set the fixpoints exp(L) = L at oo i but how about the fixpoints of exp(exp(.. q)) = q and limit cycles of the exp iterations ... e.g. let e^q1 = q2 , e^q2 = q1 , if we want half-iterates , 1/3 iterates and sqrt(2) iterates to have the same fixpoints , this is a problem , not ? perhaps we can 'hide' L at +/- oo i ( like kouznetsov ) and 'hide' the other points at - oo ?? (question 4)I think all of the other fixed points involve non-primary branch points of the logarithm. Those other fixed points show up via their primary natural logarithm branches. In Knesser's solution, the primary fixed point via its primary branch is at +/- i*infinity, but it does show up via its non primary branches. For example, the secondary fixed point of "e" is ~2.062+i7.589, but it's primary logarithm is ~2.062+i1.305. The inverse superfunction of (2.062+i1.305) is well defined. tommy1729 Wrote:i meant sqrt(e) ! because my method works for bases > sqrt(e) , not for all bases > e^(1/e).Tommy is referring to his proposed solution based on iterates of e^(kx)+e^(-kx). The equation has a fixed point of zero. The fixed point is diverging for k>0.5, and for k>0.5 the superfunction grows superexponentially. K=0.5 corresponds to the sqrt(e). Then you iterate logarithms for base e^k (analogous to the base change). However, the resulting equation may only be well defined for real numbers. - Sheldon tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 07/05/2010, 09:59 PM more about those cycles and attractors : you can find g that can't be written as f°f°...°f for any n > 1. Namely, suppose g has exactly one fixed point p_0, and exactly one p_1 <> p_0 such that g(p_1) = p_0, and at least one p_2 such that g(p_2) = p_1. If g = f°f°...°f, it's easy to see that f(p_0) = p_0, f(p_1) = p_0, and (for some p_3) f(p_3) = p_1. But then f(f(p_3)) = p_0 which would imply g(p_3) = p_0, contradiction. A suitable g is g(x) = x + 2 - 2 exp(x). hope that helps regards tommy1729 sheldonison Long Time Fellow Posts: 683 Threads: 24 Joined: Oct 2008 07/06/2010, 01:45 PM (This post was last modified: 07/07/2010, 03:44 AM by sheldonison.) (07/05/2010, 09:59 PM)tommy1729 Wrote: you can find g that can't be written as f°f°...°f for any n > 1. Namely, suppose g has exactly one fixed point p_0, and exactly one p_1 <> p_0 such that g(p_1) = p_0,....and g = f°f°...°f, it's easy to see that .... f(p_1) = p_0 > f(p_1) = p_0 I don't understand your equation, but it isn't true if g=e^x. However, that could be because e^x has an infinite number of p_1 alternatives such that e^p_1=p_0, the fixed point of base "e". For exp(x), p_1=p_0 + i2pi*n (07/05/2010, 09:59 PM)tommy1729 Wrote: A suitable g is g(x) = x + 2 - 2 exp(x).g(x) has a fixed point at zero, (the other fixed points are 0+i2pi*n). The slope of g(x) at x=0 is -1. This is a strange oscillatory fixed point; actually it appears to be an attracting fixed point for real values, and a repelling fixed point for complex values. I wouldn't know how to expand g(x) into a superfunction .... unless there is another more well behaved fixed point. But if there was another fixed point, that would contradict Tommy's assumption, "suppose g has exactly one fixed point p_0". Because the slope of the fixed point of zero is -1, there isn't a well defined superfunction, and because there isn't a well defined superfunction, then there isn't a well defined half iterate (or n-th iterate). The "p_1" value for g(x) is approximately -1.5936. g(-1.5936)~=0. But, again, we can't calculate the half iterate of p_1. Nor can we calculate the half-iterate of p_0. added stuff I discovered upon looking a little closer; 2nd update for clarity Now for the surprise; the function g(x) does half another fixed point -- actually a fixed slope -- of sorts, and that can be used to define an unexpected alternative definition of the half iterate of p_1=-1.5936. Consider the sequence p(0), p(-1), p(-2), p(-3), p(-4), p(-5), p(-6) .... p(-n), where p(0)=0, and g(p(-1))=p(0)=0, and p(n+1)=g(p(n)). Here's how the sequence looks, accurate to four decimal places. Notice the pattern? For large enough negative values of n, p(n+1)=p(n)+2, because the exponential term becomes insignificant. p(-9 )=-17.5261 p(-8 )=-15.5261 p(-7 )=-13.5261 p(-6 )=-11.5261 p(-5 )=-9.5261 p(-4 )=-7.5263 p(-3 )=-5.5274 p(-2 )=-3.5353 p(-1 )=-1.5936 p( 0 )=0.0000 p( 1 )=0.0000 p( n )=0.0000 This sequence can be extended to half iterates, with results as follows, and can also be used to generate a superfunction (the limit definition is obvious, but I can post it if others are interested). With this definition of the superfunction, the half iterate of p(-1)=p(-0.5)=0.7071. Then there are an infinite number of half iterates of zero depending on the path, and beginning with p(0.5)=0.3068. p(1.5), p(2.5), p(3.5) ... are all alternative half iterates of zero. p(-9.5)=-18.5261 p(-8.5)=-16.5261 p(-7.5)=-14.5261 p(-6.5)=-12.5261 p(-5.5)=-10.5261 p(-4.5)=-8.5262 p(-3.5)=-6.5266 p(-2.5)=-4.5295 p(-1.5)=-2.5511 p(-0.5)=-0.7071 p(0.5)=0.3068 p(1.5)=-0.4113 p(2.5)=0.2631 p(3.5)=-0.3388 p(4.5)=0.2360 p(5.5)=-0.2963 p(6.5)=0.2166 p(7.5)=-0.2670 p(8.5)=0.2017 p(9.5)=-0.2452 - Sheldon « Next Oldest | Next Newest »

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