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  "body": "\n## Statement\n\nLet $d \\ge 1$ be an integer.\nLet $f$ be a real function that is $d + 1$ 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*Householder's method of order $d + 1$* is the iteration\n$$x_{n+1} = x_n + d \\, \\frac{(1/f)^{(d-1)}(x_n)}{(1/f)^{(d)}(x_n)},$$\nwhere $(1/f)^{(k)}$ is the $k$-th [[derivative|0F1WR9C83K]] of the function $1/f$.\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|YWZQC5E54B]] to $r$ with order $d + 1$.\n\n## Remarks\n\nFor $d = 1$ the formula gives $x_{n+1} = x_n - f(x_n)/f'(x_n)$, which is [[Newton's method|07EXVNF9VP]].\nFor $d = 2$ it gives [[Halley's method|RPME2B5MS2]].\nThe idea of the proof: near $r$, the function $1/f$ has a simple pole at $r$, so it equals $c/(x - r)$ plus a function that is smoother at $r$.\nFor the pole part alone, the ratio of consecutive derivatives gives exactly $-(x - r)/d$, so the step lands on $r$.\nThe remainder adds an error of size $O(\\abs{x - r}^{d+1})$.\nThe hypothesis $f'(r) \\ne 0$ is needed: at a multiple zero the pole of $1/f$ has higher order and the convergence becomes linear.\n\n## Explanations\n\n### L1 human\n\nHouseholder's methods form a ladder of root-finding methods.\nThe rung $d$ uses derivatives of $f$ up to order $d$, and near a simple root it multiplies the number of correct digits by about $d + 1$ at each step.\nThe first rung is Newton's method and the second is Halley's method.\n\n### L2 human\n\nHigher order does not always mean less work, because each step needs more derivatives.\nIn practice $d = 1$ or $d = 2$ is used, unless the derivatives come almost for free.\nThe formula is a convenient way to see all these methods as one family built on the reciprocal $1/f$.\n",
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