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a) It is defined as the limit at a of the function existing (i.e. the limit is the same as you approach from all sides, but the function is defined as something else when it is actually at a. (sorry I don't feel like inserting the latex right now).
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The proof given here strikes me as rather strange. In particular, it's necessary to show that the Taylor series won't have a first order term. And what is the motivation for multiplying by (z-a)^2 rather than by (z-a)? Ahlfors,
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Yes, this phenomenon is unique to complex analysis. If the function is bounded at the singularity, that's enough to guarantee not only its continuity, but even infinite differentiability. I clarified this in the article.
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I believe that your statement of
Reimann's theorem is incorrect. While I can't seem to find a good description of the theorem online anywhere I am pretty sure that:
170:=1 we have a function that is obviously bounded at 0, but this is not a removable singularity. No redefinition of the Heaviside function will make this continuous.
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Perhaps this definition of
Reimann's theorem makes more sense in the complex analysis sense, but if so you should be more specific. --anon
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220:(on page 124 in the 3rd edition, 1979) gives a neat proof using the Cauchy integral formula instead.
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on
Knowledge. If you would like to participate, please visit the project page, where you can join
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I agree that the proof is not very clear. I guess that the motivation for multiplying by (
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