some questions about sexp  Printable Version + Tetration Forum (https://math.eretrandre.org/tetrationforum) + Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) + Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) + Thread: some questions about sexp (/showthread.php?tid=466) Pages:
1
2

RE: some questions about sexp  sheldonison  07/02/2010 (06/28/2010, 11:18 PM)tommy1729 Wrote: (question 3)I think all of the other fixed points involve nonprimary 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) !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 RE: some questions about sexp  tommy1729  07/05/2010 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 RE: some questions about sexp  sheldonison  07/06/2010 (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. > 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 nth 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 halfiterate 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 