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  "body": "\n## Statement\n\nLet $f$ be an [[entire function|0419794R0T]] that is not identically zero and has finite [[order|SHF8Z7GMFX]] $\\rho$.\nLet $p$ be the integer with $p \\le \\rho \u003c p + 1$.\nLet $m \\ge 0$ be the order of the [[zero|0CEGEASY49]] of $f$ at $0$.\nLet $(a_n)$ be the nonzero zeros of $f$, repeated according to multiplicity.\nLet $E_p(z) = (1 - z) \\exp(z + z^2/2 + \\cdots + z^p/p)$, with $E_0(z) = 1 - z$.\nThen $\\sum_n \\abs{a_n}^{-(p+1)} \u003c \\infty$.\nMoreover, there is a polynomial $g$ of degree at most $p$ such that\n$$f(z) = z^m e^{g(z)} \\prod_{n} E_p\\!\\left(\\frac{z}{a_n}\\right)$$\nfor all $z \\in \\C$, where the product converges uniformly on compact sets.\n\n## Remarks\n\nThe proof first bounds the number $n(r)$ of zeros in the disc $\\abs{z} \\le r$ by [[Jensen's formula|497K99BQ4Y]]: $n(r) = O(r^{\\rho + \\varepsilon})$ for every $\\varepsilon \u003e 0$.\nThis gives the convergence of $\\sum_n \\abs{a_n}^{-(p+1)}$, so the [[Weierstrass product|ACSF58HYDH]] with the fixed index $p$ converges.\nThe quotient of $f$ by $z^m$ times the product is $e^{g}$ with $g$ entire.\nLower bounds for the product away from its zeros show that $g$ is a polynomial of degree at most $\\rho$.\nFor example, $\\sin(\\pi z)$ has order $1$, so $p = 1$ and $\\sin(\\pi z) = \\pi z \\prod_{n \\ne 0} (1 - z/n) e^{z/n}$.\nThe finite order hypothesis is necessary: $f(z) = e^{e^z}$ has no zeros and infinite order, and $f = e^{g}$ with $g(z) = e^z$, which is not a polynomial.\n\n## Explanations\n\n### L1 human\n\nThe theorem refines the [[Weierstrass factorization theorem|11VC51YHQJ]] for entire functions that do not grow too fast.\nFor such a function, all convergence factors can be taken of the same small degree.\nThe leftover zero-free factor is the exponential of a polynomial, not of an arbitrary entire function.\nSo the zeros determine the function up to a few constants.\n\n### L2 human\n\nA useful consequence: an entire function of finite order that is not an integer has infinitely many zeros.\nIndeed, with finitely many zeros the factorization gives a polynomial times $e^{g}$ with $g$ a polynomial, whose order is an integer.\nAnother consequence is that an entire function of finite order without zeros is $e^{g}$ with $g$ a polynomial.\nThe theorem is a basic tool for the Riemann zeta function, through the product formula of the function $\\xi$, and for the [[Laguerre–Pólya class|X9SRCEENV3]].\n",
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