S.O.S. Mathematics CyberBoard

Your Resource for mathematics help on the web!
It is currently Fri, 24 May 2013 19:16:01 UTC

All times are UTC [ DST ]




Post new topic Reply to topic  [ 5 posts ] 
Author Message
 Post subject: singularities
PostPosted: Wed, 24 Oct 2007 14:56:41 UTC 
Offline
Senior Member

Joined: Thu, 19 Aug 2004 10:11:02 UTC
Posts: 72
I was looking at the function \frac{\sin{z}}{z \cdot \cos{z}}.

I see that this has a removable singularity at z = 0 by looking at the power series expansions (centered at 0) of the functions.

For the poles of the function, I know that \cos{z} = 0 \Leftrightarrow z = \frac{\pi}{2} + k\pi where k is an integer.

So if I want to look at the power series expansions centered at \frac{\pi}{2} + k\pi, is there a shortcut to change the original power series?

Apart from this I also know that the cos function has a zero of order 1. Does this mean that the order of the pole will also be 1?


Top
 Profile  
 
 Post subject:
PostPosted: Thu, 25 Oct 2007 20:23:42 UTC 
Offline
Senior Member

Joined: Thu, 19 Aug 2004 10:11:02 UTC
Posts: 72
Given f(z) = \frac{1}{1 - \exp(z)} I see that the denominator has a zero of degree 2 at z = 0. Hence the function has a pole of degree 2 at z = 0. How Should I go about finding the power series expansion so that I can find the singular part of the function? That is, the first two terms of its Laurent series expansion.

Any suggestions?


Top
 Profile  
 
 Post subject:
PostPosted: Fri, 26 Oct 2007 20:25:59 UTC 
Offline
Member of the 'S.O.S. Math' Hall of Fame

Joined: Mon, 25 Jul 2005 13:17:07 UTC
Posts: 1017
Location: PA, USA
"the cos function has a zero of order 1. Does this mean that the order of the pole will also be 1?"

Yes. If you have not encountered this lemma, it's probably in the next section. In one text I have, it appears right before Picard's Theorem. You'll find it.

"zero of degree 2"

Are you sure?


Top
 Profile  
 
 Post subject:
PostPosted: Fri, 26 Oct 2007 23:23:45 UTC 
Offline
Senior Member

Joined: Thu, 19 Aug 2004 10:11:02 UTC
Posts: 72
1 - \exp{z} = -z - \frac{z^2}{2} - \ldots
So as you point out, it is in fact a pole of degree one.

How do I go about calculating the singular part of the this function?

f(z) = \frac{1}{1 - \exp{z}}


Top
 Profile  
 
 Post subject:
PostPosted: Sat, 27 Oct 2007 02:40:52 UTC 
Offline
Member of the 'S.O.S. Math' Hall of Fame

Joined: Mon, 25 Jul 2005 13:17:07 UTC
Posts: 1017
Location: PA, USA
I am sorely tempted to notice that it looks an awful lot like a simple geometric series expansion, but that seems to overlook the criteria for doing that sort of thing.

Why not just calculate it directly? The long division algorithm certainly is tedious, but you can get couple of terms quickly enough.

Of course, if I actually remembered anything about this, I would be more help. You're about at my limit on this one.


Top
 Profile  
 
Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 5 posts ] 

All times are UTC [ DST ]


Who is online

Users browsing this forum: No registered users


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum

Search for:
Jump to:  
Contact Us | S.O.S. Mathematics Homepage
Privacy Statement | Search the "old" CyberBoard

users online during the last hour
Powered by phpBB © 2001, 2005-2011 phpBB Group.
Copyright © 1999-2013 MathMedics, LLC. All rights reserved.
Math Medics, LLC. - P.O. Box 12395 - El Paso TX 79913 - USA