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 [2014] The angle fractal. tommy1729 Ultimate Fellow Posts: 1,507 Threads: 358 Joined: Feb 2009 10/10/2014, 11:51 PM If an initial value x converges to a fixpoint at an angle y , we can give that value pair(x,y) a corresponding color depending on the angle y. We can make a plot that would then be the analogue of a fractal , the coloring based on the values y corresponding to the x's. --- That is the main idea ... Some comments : the derivative at the fixpoints x_0 needs to be positive real. SO : Arg ( f ' (x_0) ) = 0 0 < Abs ( f ' (x_0) ) < 1 Also the case is simplest when f(z) only has 2 conjugate fixpoints. *** Therefore I seek : Taking - without loss of generality ? - the fixpoints equal to +/- i . g(z) = real entire = ?? f(z) := exp(g(z)) (z^2+1) + z Arg ( f ' (i) ) = 0 0 < Abs ( f ' (i) ) < 1 And also f(z) ~ exp(z) z^A for some small real A. Many solutions g(z) must exist , but which is best ? And how do the " new fractals " look like ? --- Also can the angle be given by an integral ? Contour integral ? z_0 = initial value z_n = f ( z_(n-1) ) lim n -> +oo [z_n - z_(n-1)] / [ | z_n - z_(n-1) | ] = contour integral ( z_0 ) ??? regards tommy1729 tommy1729 Ultimate Fellow Posts: 1,507 Threads: 358 Joined: Feb 2009 10/19/2014, 03:15 PM (This post was last modified: 10/19/2014, 03:22 PM by tommy1729.) So f(z) is of the form f(z) = exp(z + fake_ln( (- e^(-z) z^3 - e^(-z) z + 1 ) / (z^2+1) )) + z , such that the derivative at both the fixpoints is a real Q. If Q lies between 0 and 1 that can give a nice " angle fractal ". However the case Q > 1 is also very intresting and gives an analogue of sexp for the superfunction of f(z). This should be worth an investigation ! regards tommy1729 « Next Oldest | Next Newest »

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