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  "body": "\n## Definition\n\nLet $f$ be an [[entire function|0419794R0T]].\nThe function $f$ is of *exponential type* if there are constants $C \u003e 0$ and $\\tau \\ge 0$ such that\n$$\\abs{f(z)} \\le C e^{\\tau \\abs{z}} \\quad \\text{for all } z \\in \\C.$$\nIn that case the *type* of $f$ is\n$$\\sigma = \\limsup_{r \\to \\infty} \\frac{\\log M(r)}{r}, \\qquad M(r) = \\max_{\\abs{z} = r} \\abs{f(z)},$$\nwith $\\log 0 = -\\infty$, and $f$ is said to be of exponential type $\\sigma$.\n\n## Remarks\n\nExample: $\\abs{\\sin z} \\le e^{\\abs{z}}$ for all $z$, and $M(r) = \\sinh r$ for $\\sin z$, so $\\sin z$ is of exponential type with type $1$.\nNon-example: $f(z) = e^{z^2}$ has $\\abs{f(x)} = e^{x^2}$ on the real line, and $e^{x^2} \\le C e^{\\tau \\abs{x}}$ fails for large $x$ for every $C$ and $\\tau$.\nIn terms of the [[order|SHF8Z7GMFX]], $f$ is of exponential type exactly when its order is less than $1$, or equal to $1$ with finite type.\nFor $f(z) = \\sum_n a_n z^n$, the type is $\\sigma = \\frac{1}{e} \\limsup_{n \\to \\infty} n \\abs{a_n}^{1/n}$; for $e^z$ this gives $1$.\nThe notion is also used for functions holomorphic in a sector or a half-plane, with the bound required only there.\n",
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