Understanding the Proof of Theorem 5.1 in a Complex Analysis Paper

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The discussion focuses on the proof of Theorem 5.1 from the paper available at http://arxiv.org/PS_cache/math/pdf/0506/0506319.pdf. Ben clarifies that multiplying equation (22) by (s-1) and differentiating at s=1 is not applicable due to the conditions of Corollary 5.2. Instead, he correctly identifies the use of the limit definition for derivatives, specifically using the expression ∂f(1)/∂s = lim(s→1) (f(s)-f(1))/(s-1) to compute the derivative. The discussion concludes with the confirmation that equation (41) provides the necessary derivative at s=1.

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benorin
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The paper I am reading is http://arxiv.org/PS_cache/math/pdf/0506/0506319.pdf ).

I am trying to follow the proof of Theorem 5.1 [a.k.a. eqn. (53)] on pg. 16. Multiplying (22) by (s-1) and differentiating w.r.t. s at s=1 wouldn't be the same as Corrollary 5.2 since s=1, right? So instead use the limit definition, e.g. using

[tex]\frac{\partial f(1)}{\partial s} = \lim_{s\rightarrow 1} \frac{f(s)-f(1)}{s-1}[/tex]

to compute the derivative, is this the right direction? How does (41) come into play here?

Thanks in advance for getting sucked far enough into my problem to post a reply

-Ben
 
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I haven't read this very thoroughly, but that looks correct on why cor. 5.2 doesn't apply. (22) was only valid when s was not 1.

The limit is the right idea for the derivative. When you multiply by (s-1) in (22), you get an expression for [tex]f(s)=(s-1)\Phi(1,s,u)[/tex] that's valid when s is not 1. At s=1, you define f(1) to be 1 as [tex]\Phi(1,s,u)[/tex] has a simple pole there with residue 1. (41) is

[tex]\lim_{s\rightarrow 1}\frac{(s-1)\Phi(1,s,u)-1}{s-1}[/tex]

which is the derivative of this f(s) at s=1.
 
Thank you shmoe! That makes it vividly clear to me.

-Ben
 

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