For an integer \(p\ge0\), let \(E_p\) be the canonical factor \[ E_0(z)=1-z,\qquad E_p(z)=(1-z)\exp\Big(z+\frac{z^2}{2}+\cdots+\frac{z^p}{p}\Big)\quad(p\ge1). \] Let \(f\) be an entire function of order \(\rho<\infty\), where \[ \rho\coloneqq\inf\ts{\sigma\ge0\st\text{there is }R>0\text{ with }\abs{f(z)}\le e^{\abs z^\sigma}\text{ for all }\abs z>R}, \] and let \(p\coloneqq\lfloor\rho\rfloor\). Suppose \(f\) is not identically zero, has a zero of order \(m\ge0\) at \(0\), and let \((z_k)_{k\ge1}\) be its nonzero zeros, repeated with multiplicity. Then there is a polynomial \(g\) of degree at most \(p\) such that \[ f(z)=z^me^{g(z)}\prod_{k\ge1}E_p\Big(\frac{z}{z_k}\Big), \] where the product converges uniformly on compact subsets of \(\CC\).
Hadamard factorization
T-2WJ4U
An entire function of finite order $\rho$ is $z^me^{g(z)}\prod_k E_p(z/z_k)$ over its nonzero zeros $z_k$, with $p=\lfloor\rho\rfloor$ and $\deg g\le p$.Suggest a slogan