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  "body": "\n## Statement\n\nLet $I$ be an open interval, let $f \\colon I \\to \\R$ be [[twice continuously differentiable|0VW90BBSJ4]], and let $r \\in I$ with $f(r) = 0$ and $f'(r) \\ne 0$.\nFor $\\delta \u003e 0$ with $J = [r - \\delta, r + \\delta] \\subseteq I$ and $f' \\ne 0$ on $J$, let\n$$M = \\frac{\\max_{x \\in J} \\abs{f''(x)}}{2 \\min_{x \\in J} \\abs{f'(x)}}.$$\nIf $M \\delta \u003c 1$, then for every $x_0 \\in J$ the iterates of [[Newton's method|07EXVNF9VP]], $x_{k+1} = x_k - f(x_k)/f'(x_k)$, are defined, lie in $J$, converge to $r$, and satisfy\n$$\\abs{x_{k+1} - r} \\le M \\abs{x_k - r}^2 \\quad \\text{for all } k \\ge 0.$$\nSuch a $\\delta$ always exists.\nIf moreover $x_k \\ne r$ for all $k$, then\n$$\\lim_{k \\to \\infty} \\frac{x_{k+1} - r}{(x_k - r)^2} = \\frac{f''(r)}{2 f'(r)}.$$\n\n## Remarks\n\n[[Taylor's theorem with the Lagrange remainder|0H5ZAZCX2T]] at $x_k$ gives $0 = f(r) = f(x_k) + f'(x_k)(r - x_k) + \\tfrac12 f''(\\xi_k)(r - x_k)^2$ for some $\\xi_k$ between $x_k$ and $r$.\nDividing by $f'(x_k)$ and using the definition of $x_{k+1}$ gives $x_{k+1} - r = \\frac{f''(\\xi_k)}{2 f'(x_k)} (x_k - r)^2$, so $\\abs{x_{k+1} - r} \\le M \\abs{x_k - r}^2 \\le M\\delta \\abs{x_k - r}$; this keeps the iterates in $J$ and makes them converge, and the limit follows from $\\xi_k, x_k \\to r$.\nSince $f'(r) \\ne 0$ and $f'$ and $f''$ are continuous, $M$ stays bounded as $\\delta \\to 0$, so a small $\\delta$ gives $M\\delta \u003c 1$.\nExample: for $f(x) = x^2 - 2$ from $x_0 = 1$, the errors are about $4.1 \\times 10^{-1}$, $8.6 \\times 10^{-2}$, $2.5 \\times 10^{-3}$, $2.1 \\times 10^{-6}$, $1.6 \\times 10^{-12}$; each error is about $0.35$ times the square of the previous one, and $f''(r)/(2f'(r)) = 1/(2\\sqrt{2}) \\approx 0.35$.\nThe hypothesis $f'(r) \\ne 0$ is needed: for $f(x) = x^2$ the iteration is $x_{k+1} = x_k/2$, which converges only linearly.\nDifferentiability at $r$ is needed as well: for $f(x) = x^{1/3}$ the iteration is $x_{k+1} = -2x_k$, which diverges from every $x_0 \\ne 0$.\n\n## Explanations\n\n### L1 human\n\nNear a simple zero, Newton's method roughly doubles the number of correct digits at each step.\nIf a guess is right to $3$ digits, the next one is right to about $6$, then $12$.\nThis holds only once the guess is close enough and the graph crosses the axis at a nonzero slope.\n\n### L2 human\n\nThis is the basic local convergence theorem for [[Newton's method|07EXVNF9VP]]: order $2$ in the sense of [[rate of convergence|YWZQC5E54B]], with asymptotic constant $f''(r)/(2f'(r))$.\nThe same proof works for systems $F(x) = 0$ in $\\R^n$ with an invertible Jacobian at the zero, where it is a special case of the Newton–Kantorovich theory.\nAt a zero of multiplicity $m \\ge 2$ the convergence becomes linear with ratio $(m - 1)/m$, and the modified step $x_{k+1} = x_k - m f(x_k)/f'(x_k)$ restores quadratic convergence.\nGlobal convergence is not guaranteed, which is why practical solvers combine Newton steps with bisection or line searches.\n",
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