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 some questions about sexp sheldonison Long Time Fellow Posts: 640 Threads: 22 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 »