if f has a zero of order 2 at z0 and A=f''(z0) and B=f'''(z0) and h(z)=1/f(z), how does one find a formula for Res(h;z0) in terms of A and B? thanks!
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Originally Posted by morganfor if f has a zero of order 2 at z0 and A=f''(z0) and B=f'''(z0) and h(z)=1/f(z), how does one find a formula for Res(h;z0) in terms of A and B? thanks! Just expand them in terms of Laurent series: Therefore Multiply it out, take derivatives, equate coefficients and solve for I'd recommend you find "Basic Complex Analysis" by Marsden and Hoffman. It's a nice read on basic stuff and it goes over this derivation.
Last edited by mr fantastic; Jun 4th 2009 at 09:26 PM. Reason: Fixed a couple of subscripts
You could try the following: then using the product rule And then since the forms of both terms are indeterminate, apply L'Hospital's rule. But multiple applications will be required.
Thank you! Shawsend, I'm just confused about where the g(z) came from and what it is - do you mean f(z)?
Originally Posted by shawsend Just expand them in terms of Laurent series: Therefore Multiply it out, take derivatives, equate coefficients and solve for I'd recommend you find "Basic Complex Analysis" by Marsden and Hoffman. It's a nice read on basic stuff and it goes over this derivation. I'm sorry my conflicts with your . Mine was just general terms. You have: and: and therefore so that Equating coefficients: and: Solving for
Originally Posted by shawsend I'm sorry my conflicts with your . Mine was just general terms. You have: and: and therefore so that Equating coefficients: and: Solving for This might be a stupid question, but how do we know that ?
If has a zero of order 2 at , then can be written as: but that's the Taylor series which is unique so that: and so on. That means the first two terms of the Taylor series must be zero or and .
Originally Posted by shawsend If has a zero of order 2 at , then can be written as: but that's the Taylor series which is unique so that: and so on. That means the first two terms of the Taylor series must be zero or and . But basically what you're using is the fact that if is a zero of of order k, then , right? I don't why I didn't realize that before.
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