As usual in analysis, this infinite sum can be split into two parts: a finite bulk and an infinite tail term, each of which is to be separately handled. There are finitely many <math>a_k</math> with modulus <math>< |z_k|/2</math> and infinitely many <math>a_k</math> with modulus <math>\geq |z_k|/2</math>. So we have to bound:<math display="block">\sum_{|a_k| < |z_k|/2} \ln |E_p(z/a_k)| + \sum_{|a_k| \geq |z_k|/2} \ln |E_p(z/a_k)| </math>These can be accomplished by the bounds (2), (3) on <math>|E_p|</math>.
As usual in analysis, this infinite sum can be split into two parts: a finite bulk and an infinite tail term, each of which is to be separately handled. There are finitely many <math>a_k</math> with modulus <math>< |z_k|/2</math> and infinitely many <math>a_k</math> with modulus <math>\geq |z_k|/2</math>. So we have to bound:<math display="block">\sum_{|a_k| < |z_k|/2} \ln |E_p(z/a_k)| + \sum_{|a_k| \geq |z_k|/2} \ln |E_p(z/a_k)| </math>These can be accomplished by the bounds (2), (3) on <math>|E_p|</math>.
=== <math>\rho \leq g + 1</math> ===
=== <math>\rho \leq g + 1</math><ref>{{Cite web |last=Garrett |first=Paul |date=September 15, 2019 |title=Weierstrass and Hadamard products |url=https://www-users.cse.umn.edu/~garrett/m/mfms/notes_2019-20/05c_Hadamard_products.pdf}}</ref> ===
This is Theorem 2.6 of Chapter 11 of <ref>{{Cite book |last=Conway |first=John B. |url=http://link.springer.com/10.1007/978-1-4612-6313-5 |title=Functions of One Complex Variable I |date=1978 |publisher=Springer New York |isbn=978-0-387-94234-6 |series=Graduate Texts in Mathematics |volume=11 |location=New York, NY |doi=10.1007/978-1-4612-6313-5}}</ref>.
== References ==
== References ==
Revision as of 20:21, 8 April 2023
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.
Formal statement
Define the Hadamard canonical factors
Entire functions of finite order have Hadamard's canonical representation[1]:
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 series
converges. 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 .
For example, , and are entire functions of genus .
Critical exponent
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 ([2]): If is an entire function with infinitely many roots, then
Note: These two equalities are purely about the limit behaviors of a real number sequence that diverges to infinity.
Then the goal is to show that is of order . This does not exactly work, however, due to bad behavior of near . Consequently, we need to pepper the complex plane with "forbidden disks", one around each , each with radius . Then since by the previous result on , we can pick an increasing sequence of radii that diverge to infinity, such that each circle avoids all these forbidden disks.
Thus, if we can prove a bound of form for all large[nb 2] that avoids these forbidden disks, then by the same application of Borel–Carathéodory theorem, for any , and so as we take , we obtain .
Since by the definition of , it remains to show that , that is, there exists some constant such that
for all large that avoids these forbidden disks.
As usual in analysis, this infinite sum can be split into two parts: a finite bulk and an infinite tail term, each of which is to be separately handled. There are finitely many with modulus and infinitely many with modulus . So we have to bound:
These can be accomplished by the bounds (2), (3) on .