wu :: forums - Print Page


    
      
        wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        general >> complex analysis >> Finite Genus
        
(Message started by: Jason on Feb 2^nd, 2004, 8:36pm)

Title: Finite Genus
Post by Jason on Feb 2^nd, 2004, 8:36pm

Hello people. Help me please to solve this problem.

Prove that finite genus implies finite order.

Title: Re: Finite Genus
Post by towr on Feb 3^rd, 2004, 1:45am

any context? What sort of genus and order are we talking about?

Title: Re: Finite Genus
Post by Jason on Feb 3^rd, 2004, 6:51pm

Entire function f is of finite order if there are numbers a,r>0 such that |f(z)|<=exp(|z|^a)

Title: Re: Finite Genus
Post by Icarus on Feb 3^rd, 2004, 7:21pm

Presumably, given the forum this post is in, he is talking about the order & genus of entire functions.

The order of an entire function f(z) is defined as limsup_{r[to][subinfty]} ln(ln(sup_|z|=r|f(z)|)) / ln(r).

The genus of an entire function is the smallest integer for which the series [sum]_n |a_n|^-1-h converges, where {a_n} is the sequence of zeros of the function (with each zero having as many entries as its multiplicity). An entire function f of finite genus may be expressed as an infinite product:

f(z) = e^g(z)[prod](1-z/a_n)e^q(z/a_n),

where g(z) is a polynomial of degree [le] h, and q(z) = z + z²/2 + ... + z^h/h.

The definition of order for an entire function in my reference (Complex Analysis, Lars Ahlfors; Third Edition, 1979; page 207) comes right before the following theorem:

The genus h and order [lambda] of an entire function satisfy the double inequality h [le] [lambda] [le] h + 1.

With this theorem, the problem is fairly trivial!

Assuming that you must prove the result without assuming this theorem, it will be harder. The proof of the result in my reference took 3 pages of some messy looking approximations. However, the far weaker result you want may not be bad. All you need to show is that any such product expansion as shown above has finite order.

Title: Re: Finite Genus
Post by Icarus on Feb 3^rd, 2004, 7:27pm

I see that while I was researching my post Jason has filled in part of the definition. But the definition you gave is obviously incomplete, since there is no "r" in the formula.

Before attempting a proof, perhaps you should provide exactly the definitions you are using for both "finite genus" and "finite order". Otherwise we may start from the wrong place.