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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 $f'(r) \\ne 0$.\nLet $(x_k)$ be the iterates of the [[secant method|9ZC177PVPC]] from starting points close enough to $r$ that $x_k \\to r$, and assume $x_k \\ne r$ for all $k$.\nLet $e_k = x_k - r$.\n\n1. The convergence is superlinear: $\\lim_{k \\to \\infty} \\abs{e_{k+1}}/\\abs{e_k} = 0$.\n2. If moreover $f''(r) \\ne 0$, the convergence has order $\\varphi = (1 + \\sqrt{5})/2$, the [[golden ratio|0SC8T7548D]]:\n$$\\lim_{k \\to \\infty} \\frac{\\abs{e_{k+1}}}{\\abs{e_k}^{\\varphi}} = \\abs{\\frac{f''(r)}{2 f'(r)}}^{\\varphi - 1}.$$\n\n## Remarks\n\nThe secant method satisfies $e_{k+1} = C_k e_k e_{k-1}$ with $C_k \\to C = f''(r)/(2f'(r))$.\nClause 1 follows at once: $\\abs{e_{k+1}}/\\abs{e_k} = \\abs{C_k} \\abs{e_{k-1}} \\to 0$.\nFor clause 2, the logarithms $y_k = \\log \\abs{e_k}$ satisfy approximately $y_{k+1} = y_k + y_{k-1} + \\log \\abs{C}$.\nThis is a Fibonacci-type recursion, whose solutions grow like $\\varphi^k$.\nA careful version of this argument gives the limit; the exponent $\\varphi - 1 = 1/\\varphi$ comes from $\\varphi^2 = \\varphi + 1$.\nExample: for $f(x) = x^2 - 2$ from $x_0 = 1$, $x_1 = 2$, the errors are about $4.1 \\times 10^{-1}$, $5.9 \\times 10^{-1}$, $8.1 \\times 10^{-2}$, $1.4 \\times 10^{-2}$, $4.2 \\times 10^{-4}$, $2.1 \\times 10^{-6}$, $3.2 \\times 10^{-10}$.\nOnce the errors are small, the number of correct digits grows by a factor of about $1.6$ per step.\nAt a zero with $f'(r) = 0$ the conclusion fails: for $f(x) = x^2$ the convergence is only linear.\n\n## Explanations\n\n### L1 human\n\nNear a simple zero the secant method improves very fast.\nAt each step the number of correct digits is multiplied by about $1.618$, the golden ratio.\nThat is less than the doubling of Newton's method, but each secant step needs only one new function value and no derivative.\n\n### L2 human\n\nIn terms of [[rate of convergence|YWZQC5E54B]], the secant method has order $\\varphi \\approx 1.618$ and Newton's method has order $2$.\nOne Newton step costs a value of $f$ and a value of $f'$.\nIf these cost about the same, two secant steps cost as much as one Newton step, and together they have order $\\varphi^2 \\approx 2.618 \u003e 2$.\nThis efficiency is why secant-type steps appear in practical solvers such as Brent's method.\nThe appearance of the golden ratio comes from the recursion $e_{k+1} \\approx C e_k e_{k-1}$, whose exponents follow the Fibonacci numbers.\n",
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