In mathematics, and particularly in the field of complex analysis, the Hadamard factorization theorem asserts that every entire function with finite order can be represented as a product involving its zeroes and an exponential of a polynomial. It is named for Jacques Hadamard. The theorem may be viewed as an extension of the fundamental theorem of algebra, which asserts that every polynomial may be factored into linear factors, one for each root. It is closely related to Weierstrass factorization theorem, which does not restrict to entire functions with finite orders. Define the Hadamard canonical factors Entire functions of finite order have Hadamard's canonical representation:where are those roots of that are not zero (), is the order of the zero of at (the case being taken to mean ), a polynomial (whose degree we shall call ), and is the smallest non-negative integer such that the seriesconverges. The non-negative integer is called the genus of the entire function . In this notation,In other words: If the order is not an integer, then is the integer part of . If the order is a positive integer, then there are two possibilities: or . Furthermore, Jensen's inequality implies that its roots are distributed sparsely, with critical exponent . For example, , and are entire functions of genus . Define the critical exponent of the roots of as the following:where is the number of roots with modulus . In other words, we have an asymptotic bound on the growth behavior of the number of roots of the function:It's clear that . Theorem: If is an entire function with infinitely many roots, thenNote: These two equalities are purely about the limit behaviors of a real number sequence that diverges to infinity. It does not involve complex analysis. Proposition: , by Jensen's formula. Since is also an entire function with the same order and genus, we can wlog assume . If has only finitely many roots, then with the function of order . Thus by an application of the Borel–Carathéodory theorem, is a polynomial of degree , and so we have .