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  "body": "\n## Definition\n\nLet $f \\colon [0, \\infty) \\to \\C$ be a [[function|0XHDTE5YMZ]] and let $a \\in \\R$.\nThe function $f$ is *of exponential order $a$* if there are constants $M \u003e 0$ and $T \\ge 0$ such that\n$$\\abs{f(t)} \\le M e^{at} \\quad \\text{for all } t \\ge T,$$\nwhere $e^{at}$ is the [[exponential function|0RMVS7RXZX]].\nThe function $f$ is *of exponential order* if it is of exponential order $a$ for some real $a$.\n\n## Remarks\n\nExample: $f(t) = t^n$ with $n \\in \\N$ is of exponential order $1$, since $e^t \\ge t^n / n!$ for $t \\ge 0$ by the power series of $e^t$; so $\\abs{f(t)} \\le n!\\, e^t$ with $T = 0$.\nIn fact $t^n$ is of exponential order $a$ for every $a \u003e 0$.\nNon-example: $f(t) = e^{t^2}$ is not of exponential order, since $e^{t^2}/e^{at} = e^{t(t - a)} \\to \\infty$ for every real $a$.\nA bounded function, such as $\\sin t$, is of exponential order $0$.\nSome books require the bound for all $t \\ge 0$ ($T = 0$); for functions that are bounded on bounded intervals the two versions agree, after enlarging $M$.\n",
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