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immanuel78
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About the pole  
« on: Dec 9th, 2006, 5:05am »
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Let f be analytic in G={z:0<|z-a|<r} except that there is a sequence of poles {a_n} in G with a_n -> a.  
Show that for any w in C there is a sequence {z_n} in G with a=lim z_n and w=lim f(z_n)
« Last Edit: Dec 9th, 2006, 5:06am by immanuel78 » IP Logged
Icarus
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Re: About the pole  
« Reply #1 on: Dec 9th, 2006, 7:00am »
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If there is a sequence of poles converging to a, then a is an essential singularity of f. By Weierstrass, the image of any neighborhood of an essential singularity is dense in C. Hence such a sequence must exist.
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immanuel78
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Re: About the pole  
« Reply #2 on: Dec 9th, 2006, 7:29am »
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But z=a is not an isolated singularity of f.
Hence z=a is not an essential singularity of f.
« Last Edit: Dec 9th, 2006, 7:30am by immanuel78 » IP Logged
Icarus
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Re: About the pole  
« Reply #3 on: Dec 9th, 2006, 12:57pm »
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Consider g(z) = 1/(f(z) - w).  In many ways, poles are not really a failure of the function to be analytic.
« Last Edit: Dec 9th, 2006, 1:27pm by Icarus » IP Logged

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I wonder: Which is larger
When their digits are reversed? " - Anonymous
immanuel78
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Re: About the pole  
« Reply #4 on: Dec 11th, 2006, 10:50pm »
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Icarus, may you tell me the way in detail?
I think this problem by usuing your hint but I can't solve it perfectly.
But I can understand this problem more than before because of your hint.
 
« Last Edit: Dec 11th, 2006, 10:52pm by immanuel78 » IP Logged
Icarus
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Re: About the pole  
« Reply #5 on: Dec 12th, 2006, 4:15pm »
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A pole is really just a place where an analytic function takes on the value . Around poles, the function is as well-behaved as it is everywhere else. In particular, if you take the inverse, then you get a function that is analytic at that point (there is no such thing as a "removable singularity" - if you ever encounter one, it just means someone has been lazy and not got around to removing it yet).
 
If f(b) = w for some b in G, then we could just take zn to be any sequence converging to b. So assume that f does not take on the value w. Then g(z) = 1/(f(z) - w) is analytic everywhere in G. the point a is also an essential singularity of g (If it was a pole or removable, then f(z) would have had a worst at pole at a.) Therefore by Weierstrass's theorem, g(G) is dense in C. In particular, for every n, there must be zn in G such that |g(zn)| > n. From this it follows that f(zn) --> w.
 
 
(We differ on the definition of essential singularity, by the way - my definition is that any place where analycity fails that is not removable or a pole is an essential singularity, regardless of whether it is isolated. However, Weierstrass's theorem does require an isolated essential singularity - though it can be extended to allow poles by exactly the means I've used here.)
« Last Edit: Dec 12th, 2006, 4:16pm by Icarus » IP Logged

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And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
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