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  "body": "\n## Definition\n\nLet $f \\colon [0, \\infty) \\to \\R$ satisfy $\\abs{f(t)} \\le C e^{A t}$ for some constants $C, A \\ge 0$ and all $t \\ge 0$.\nFor an integer $n \\ge 1$, the *Szász–Mirakyan operator* $S_n$ maps $f$ to the function\n$$S_n(f)(x) = e^{-nx} \\sum_{k=0}^{\\infty} \\frac{(nx)^k}{k!} \\, f\\left(\\frac{k}{n}\\right), \\qquad x \\ge 0.$$\n\n## Remarks\n\nThe growth bound makes the series converge for all $x \\ge 0$.\nEach $S_n$ is a [[positive linear operator|AQDXWH4812]].\nThe weights $e^{-nx} (nx)^k / k!$ are the probabilities of the [[Poisson distribution|0XB9GQTN9G]] with mean $nx$.\n\nExample: if $N$ is Poisson with mean $nx$, then $E[N] = nx$ and $E[N^2] = nx + n^2 x^2$.\nSo $S_n(1) = 1$, $S_n(t)(x) = x$ and $S_n(t^2)(x) = x^2 + x/n$.\n\nNon-example: $f(t) = e^{t^2}$ is not in the domain, since $e^{k^2/n^2} (nx)^k / k!$ does not tend to $0$ as $k \\to \\infty$ when $x \u003e 0$.\n\nFrom the three moments and Korovkin's theorem, $S_n f \\to f$ uniformly on each interval $[0, b]$ for every bounded continuous $f$.\n",
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