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  "body": "\n## Statement\n\nLet $f$ be a real function that is three times continuously differentiable on an open interval $I$.\nLet $r \\in I$ be a simple zero of $f$, that is, $f(r) = 0$ and $f'(r) \\ne 0$.\n*Halley's method* is the iteration\n$$x_{n+1} = x_n - \\frac{2 f(x_n) f'(x_n)}{2 f'(x_n)^2 - f(x_n) f''(x_n)}.$$\nThen there is $\\delta \u003e 0$ such that for every $x_0$ with $\\abs{x_0 - r} \u003c \\delta$ the iterates are defined and converge to $r$.\nThe [[convergence|YWZQC5E54B]] has order at least $3$:\n$$\\abs{x_{n+1} - r} \\le C \\abs{x_n - r}^3$$\nfor some constant $C$ and all $n$.\n\n## Remarks\n\nThe step is [[Newton's method|07EXVNF9VP]] applied to $g = f / \\sqrt{\\abs{f'}}$.\nAt a simple zero, $g$ has the zero $r$ and satisfies $g''(r) = 0$, and Newton's method on such a function converges cubically.\nThe same formula can be written $x_{n+1} = x_n - \\dfrac{f/f'}{1 - f f''/(2 f'^2)}$, evaluated at $x_n$.\nHere $f'$ is the [[derivative|0F1WR9C83K]] of $f$.\nFor $f(x) = x^2 - a$ with $a \u003e 0$, the method gives $x_{n+1} = x_n \\dfrac{x_n^2 + 3a}{3x_n^2 + a}$, which roughly triples the number of correct digits of $\\sqrt{a}$ per step near the root.\nAt a multiple zero the order of convergence drops to $1$.\n\n## Explanations\n\n### L1 human\n\nNewton's method replaces the graph by a tangent line.\nHalley's method also uses the second derivative, so it follows the bending of the graph.\nNear a simple root, each step roughly triples the number of correct digits, against doubling for Newton's method.\n\n### L2 human\n\nEach step costs one evaluation of $f$, $f'$ and $f''$.\nSo it pays off when the second derivative is cheap, as for polynomials.\nIt is the case $d = 2$ of the Householder family of methods, whose case $d = 1$ is Newton's method.\n",
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  "markdown": "https://tamnd.github.io/math2/node/RPME2B5MS2.md",
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  "name": "Halley's method",
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