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  "body": "\n## Statement\n\nLet $k \\ge 1$, and let $a_1, \\dots, a_k \u003e 0$ and $0 \u003c b_1, \\dots, b_k \u003c 1$ be real constants.\nLet $g \\colon [1, \\infty) \\to [0, \\infty)$ satisfy the polynomial growth condition: there are constants $c_1, c_2 \u003e 0$ such that $c_1 g(x) \\le g(u) \\le c_2 g(x)$ for all $x \\ge 1$, all $i$ and all $u \\in [b_i x, x]$.\nLet $h_1, \\dots, h_k$ be real functions with $\\abs{h_i(x)} = O(x / (\\log x)^2)$, in [[big O notation|0TGV80E6A4]].\nLet $x_0$ be large enough, and let $T \\colon [1, \\infty) \\to \\R$ be positive and bounded above and below by positive constants on $[1, x_0]$, and satisfy the recurrence\n$$T(x) = g(x) + \\sum_{i=1}^{k} a_i\\, T\\big(b_i x + h_i(x)\\big) \\qquad (x \u003e x_0).$$\nLet $p$ be the unique real number with $\\sum_{i=1}^{k} a_i b_i^p = 1$.\nThen\n$$T(x) = \\Theta\\left( x^p \\left( 1 + \\int_1^x \\frac{g(u)}{u^{p+1}}\\, du \\right) \\right),$$\nwhere $f(x) = \\Theta(F(x))$ means that $c F(x) \\le f(x) \\le C F(x)$ for some constants $0 \u003c c \\le C$ and all large $x$.\n\n## Remarks\n\nThe number $p$ exists and is unique because $s \\mapsto \\sum a_i b_i^s$ is continuous and strictly decreasing, with limits $\\infty$ and $0$.\nThe proof is by induction on $x$: one guesses the bound with suitable constants and checks that the recurrence preserves it.\nThe growth condition on $g$ is used to compare $g$ at $b_i x$ and at $x$.\nThe perturbations $h_i$ cover floors and ceilings, such as $T(\\lceil x/2 \\rceil)$, which appear in real algorithms.\nExample: for $T(x) = T(x/3) + T(2x/3) + x$, one has $p = 1$ since $1/3 + 2/3 = 1$, and $\\int_1^x u/u^2\\, du = \\log x$, so $T(x) = \\Theta(x \\log x)$.\nThe growth condition on $g$ is needed.\nFor $T(x) = T(x/2) + 2^x$, one has $p = 0$, and the formula would give the order $2^x / x$, since $\\int_1^x 2^u u^{-1}\\, du$ grows like $2^x/(x \\log 2)$; but $T(x) \\ge 2^x$.\nHere $g(x) = 2^x$ fails the growth condition, as $g(x)/g(x/2) = 2^{x/2}$ is unbounded.\n\n## Explanations\n\n### L1 human\n\nDivide-and-conquer algorithms split a problem of size $x$ into smaller pieces of sizes $b_i x$, solve them, and pay $g(x)$ to split and combine.\nTheir running time satisfies a recurrence like the one above.\nThe theorem gives the order of growth of the running time directly, without solving the recurrence.\nFor merge sort, $T(x) = 2 T(x/2) + x$ gives $p = 1$ and $T(x) = \\Theta(x \\log x)$.\n\n### L2 human\n\nThe theorem generalizes the master theorem of algorithm analysis, which handles only equal subproblem sizes $T(x) = a T(x/b) + g(x)$.\nHere the pieces may have different sizes $b_i x$, and small perturbations $h_i$ are allowed.\nTo use it, first solve $\\sum a_i b_i^p = 1$ for $p$, then compute the integral of $g(u)/u^{p+1}$.\nIf $g(x) = x^q$ with $q \u003c p$, the integral stays bounded and $T = \\Theta(x^p)$; if $q = p$, one gets $\\Theta(x^p \\log x)$; if $q \u003e p$, one gets $\\Theta(x^q)$.\nIn Mathlib a version is `AkraBazziRecurrence.isTheta_asympBound`.\n",
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