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  "body": "\n## Definition\n\nLet $f \\colon [0, \\infty) \\to \\R$ satisfy $\\abs{f(t)} \\le C (1 + t)^m$ for some constants $C \\ge 0$, $m \\ge 0$ and all $t \\ge 0$.\nFor an integer $n \\ge 1$, the *Baskakov operator* $V_n$ maps $f$ to the function\n$$V_n(f)(x) = \\sum_{k=0}^{\\infty} \\binom{n + k - 1}{k} \\frac{x^k}{(1 + x)^{n + k}} \\, f\\left(\\frac{k}{n}\\right), \\qquad x \\ge 0,$$\nwhere $\\binom{n + k - 1}{k}$ is a [[binomial coefficient|092GEF38YD]].\n\n## Remarks\n\nWith $y = x/(1 + x)$, the weights are $\\binom{n + k - 1}{k} y^k (1 - y)^n$, the negative binomial probabilities, so they are nonnegative and sum to $1$.\nHence $V_n$ is a [[positive linear operator|AQDXWH4812]] with $V_n(1) = 1$.\n\nExample: the mean of these probabilities is $n y / (1 - y) = n x$, so $V_n(t)(x) = x$.\nTheir variance is $n x (1 + x)$, which gives $V_n(t^2)(x) = x^2 + \\frac{x(1 + x)}{n}$.\nBy Korovkin's theorem, $V_n f \\to f$ uniformly on each interval $[0, b]$ for every bounded continuous $f$.\n\nNon-example: $f(t) = e^{t}$ is not in the domain: the terms grow like $k^{n-1} (x/(1 + x))^k e^{k/n}$, which diverges when $\\frac{x}{1 + x} e^{1/n} \\ge 1$.\n\nBaskakov's general construction takes functions $\\varphi_n$ and sets $L_n(f)(x) = \\sum_{k \\ge 0} \\frac{(-x)^k \\varphi_n^{(k)}(x)}{k!} f(k/n)$.\nThe choice $\\varphi_n(x) = (1 + x)^{-n}$ gives $V_n$, $\\varphi_n(x) = e^{-nx}$ gives the Szász–Mirakyan operator, and $\\varphi_n(x) = (1 - x)^n$ gives the Bernstein polynomials on $[0, 1]$.\n",
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