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41# 分形几何


应该不止迭代80次吧,分次迭代?
接下去看能否做出清晰的边界。
43# 榕坚


迭代次数就是80次,没有用分次迭代,因为迭代次数不需要太大,用分次迭代反而慢了.接着放大可以通过增大色差得到清晰的边界.后面两图足以说明此变换在此中心邻域内的收敛速度相当快.另外从图形上大家也能看到以前作N集的很熟悉的感觉吧.那种由et着色呈显的等势层.
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主要信息都在图上.baillut=10^-25.
25# 柳烟


迭代公式就是标题给的数学变换.
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This Mandelbrot set is constructed from two conjugate critical points, and if we let z1 and z2 be the images of these by (1 - z^2)/(z - z^2cos(z)), we have that if c is real and numerical large and belonging to the Mandelbrot set, then z1 + c and z2 + c belong to different Fatou domains for (1 - z^2)/(z - z^2cos(z)) + c, therefore 1/cos(z1 + c) + c and 1/cos(z2 + c) + c belong to different Fatou domains, and if we add a multiple of 2π to c (so that the result still is numerically large), then these two numbers will belong to different Fatou domains. Therefore we can say: it is possible that the Mandelbrot set extends towards infinity in the horizontal direction. It seems really to be the case, as we can see if we draw the Mandelbrot set and zoom out.
The two following pictures show sections of the Julia set for c = -1. The point z = -1 is a fixed point for this iteration, and as the derivative of the function in this point is numerically larger than 1, z = -1 is repelling and belongs therefore to the Julia set. If x is real and numerically very large, the iteration of x is near 1/cos(x) - 1. Therefore, if x* is a real point of the Julia set such that x* - (-1) > 1, we can find a real point of large numerical value that iterates into this point of the Julia set, and this indicates that the Julia set extends towards infinity in the horizontal direction. And as cos(z) grows onentially in the vertical direction, the iteration of a point z of large numerical y-value is very near -1, therefore we can find points of arbitrary large y-values that iterate into -1, so that the Julia set extends towards infinity in the vertical direction.
才放到10^11就混沌了,可惜。看上去与牛顿非常象的。

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项链里找M集就象大海捞针一样的感觉:

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