Lasse's answer expanded: Let $\psi$ be the map of the exterior of the unit disk onto the exterior of the Mandelbrot set, with Laurent series $$ \psi(w) = w + \sum_{n=0}^\infty b_n w^{-n} = w - \frac{1}{2} + \frac{1}{8} w^{-1} - \frac{1}{4} w^{-2} + \frac{15}{128} w^{-3} + 0 w^{-4} -\frac{47}{1024} w^{-5} + \dots $$ Then of course the boundary of the Mandelbrot set is the image of the unit circle under this map. I think, howeverHowever, this depends on the (not yet proved?proved) local connectedness of that boundary. Here, for the coefficients $b_n$ there is no known closed form, but they can be computed recursively. Of course we put $w = e^{i\theta}$ and then this is a Fourier series.
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Let $\psi$ be the map of the exterior of the unit disk onto the exterior of the Mandelbrot set, with Laurent series $$ \psi(w) = w + \sum_{n=0}^\infty b_n w^{-n} = w - \frac{1}{2} + \frac{1}{8} w^{-1} - \frac{1}{4} w^{-2} + \frac{15}{128} w^{-3} + 0 w^{-4} -\frac{47}{1024} w^{-5} + \dots $$ Then of course the boundary of the Mandelbrot set is the image of the unit circle under this map. I think, however, this depends on the (not yet proved?) local connectedness of that boundary. Here, for the coefficients $b_n$ there is no known closed form, but they can be computed recursively. |
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