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Topic: Schwarz Theorem (Read 11727 times) 

knightfischer
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Schwarz Theorem
« on: Aug 22^{nd}, 2007, 6:10am » 
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In the Schwarz Theorem, it starts out with f(z) analytical in region z<=R, f(0)=0, then states f(z)/z is analytical in that region. I do not see how f(z)/z is analytical at z=0. Can anyone help me with this?


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mikedagr8
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A rich man is one who is content; not wealthy.
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Re: Schwarz Theorem
« Reply #1 on: Aug 22^{nd}, 2007, 6:21am » 
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You don't need to post the same topic in different areas of the forum. Someone will respond to you soon enough, As for me I'm of to bed, its's 11.20 here at night, so everywhere else it is probably early morning or evening. Night All.


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"It's not that I'm correct, it's that you're just not correct, and so; I am right."  M.P.E.



Grimbal
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Re: Schwarz Theorem
« Reply #2 on: Aug 22^{nd}, 2007, 6:57am » 
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Iceman, you can start one of your riddles now!


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Grimbal
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Re: Schwarz Theorem
« Reply #3 on: Aug 22^{nd}, 2007, 7:04am » 
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This is from distant memories, so I offer no warranty. I think that for z<>0, you can differentiate the function with the usual rule for f/g. For z=0, you can apply L'Hospital rule.


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knightfischer
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Re: Schwarz Theorem
« Reply #4 on: Aug 22^{nd}, 2007, 10:27am » 
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I'm not sure L'Hopital applies. Here is my reasoning, given f(z) analytic, and f(0)=0, to show g(z)=f(z)/z is analytic at z=0, we must show the derivative exists at z=0. That is, the lim (delta z>0) of {(g(0+delta z)g(0))/(delta z)} =lim (delta z >0) of {(f(delta z)/(delta z)  f(0)/0)/(delta z)} I can apply L'Hopital to first term in numerator, but there is no limit involved in second term, f(0)/0, so how can I apply L'Hopital to that? Confused!


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Aryabhatta
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Re: Schwarz Theorem
« Reply #5 on: Aug 22^{nd}, 2007, 1:13pm » 
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I am not sure what approach to analytic functions you are using, but using the power series approach we can easily see that if f(z) = Sum a_{i} z^{i} is analytic in disc D such that f(z) = 0 (i.e a_{0} = 0) Then H(z) = Sum a_{i+1} z^{i} is also analytic in disc D and H(z) = f(z)/z for all z =/= 0. and H(0) = f'(0).


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knightfischer
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Re: Schwarz Theorem
« Reply #6 on: Aug 23^{rd}, 2007, 4:13am » 
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Thanks. Your explanation is clear. The book has not yet discussed power series approach to analytic functions. I was trying to see how f(z)/z is analytic, using the definition of an anlytic function as one with a derivative at all points in a region. Using this derivative approach I could not see how it was analytic at z=0.


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Wiliam_smith
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Re: Schwarz Theorem
« Reply #7 on: Nov 19^{th}, 2009, 6:43am » 
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Simply because z < R . Field. The same equation we use in Search analytics. Creating a peak within a range of low value keywords. I work for (link removed by moderator)

« Last Edit: Nov 19^{th}, 2009, 7:03am by SMQ » 
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SMQ
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Re: Schwarz Theorem
« Reply #8 on: Nov 19^{th}, 2009, 7:04am » 
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If you're going to spam the forum by resurrecting twoyearold threads, at least try to make sense. SMQ

« Last Edit: Nov 19^{th}, 2009, 7:04am by SMQ » 
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SMQ



