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  "body": "\n## Example\n\nLet $p(z) = z^3 - 1$, with roots $1$, $\\omega = e^{2\\pi i/3}$ and $\\omega^2$.\n[[Newton's method|07EXVNF9VP]] for $p$ iterates the [[rational function|0XDK9CGYKJ]]\n$$N(z) = z - \\frac{p(z)}{p'(z)} = \\frac{2z^3 + 1}{3z^2}.$$\nThe *Newton fractal* of $p$ is the [[Julia set|5M4S7P04GR]] $J(N)$.\nIt is usually drawn with each point of the [[Fatou set|C0N0XGC6GK]] coloured by the root its orbit tends to.\n\nCheck of the main features:\n\n1. Each root $r$ is a superattracting fixed point: $N(r) = r$ since $p(r) = 0$, and $N'(z) = p(z) p''(z) / p'(z)^2$ vanishes at $r$.\n2. So each root has an open basin of attraction, contained in the Fatou set.\n3. The map $N$ commutes with $z \\mapsto \\omega z$, so the picture has threefold rotational symmetry.\n4. The Julia set is the common boundary of the three basins: every point of $J(N)$ lies in the closure of each basin.\n\n## Remarks\n\nNewton's method can therefore fail in a complicated way: arbitrarily close to the boundary of one basin there are starting points that converge to each of the other roots.\nFor a quadratic polynomial with two distinct roots, the Newton map is conjugate to $z \\mapsto z^2$.\nThen the Julia set is the perpendicular bisector of the two roots, a result going back to Cayley.\n\n## Explanations\n\n### L1 human\n\nNewton's method guesses a root of a polynomial and improves the guess again and again.\nColour each starting point in the plane by the root that the guesses settle on.\nFor $z^3 - 1$ there are three colours, and the borders between them form an endlessly detailed fractal.\nNear any border point all three colours appear, so a tiny change of the starting guess can change which root is found.\n",
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