Talk:Removable singularity

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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:

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).

b) In the example of the Heaviside functin where f(x)=0 x<0 and f(x)=1 x>=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.

Perhaps this definition of Reimann's theorem makes more sense in the complex analysis sense, but if so you should be more specific. --anon

I believe Riemann's theorem applies only to holomorphic functions. Perhaps a different wording could make this clearer. —Caesura^(t) 22:07, 8 December 2005 (UTC)Reply

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. Oleg Alexandrov (talk) 01:02, 9 December 2005 (UTC)Reply

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, Complex Analysis (on page 124 in the 3rd edition, 1979) gives a neat proof using the Cauchy integral formula instead. Jowa fan (talk) 05:49, 25 January 2010 (UTC)Reply

I agree that the proof is not very clear. I guess that the motivation for multiplying by (z − a)² is to ensure that h'(a) exists (namely, it is 0), but this should be stated explicitly in the proof. The first two terms of the Taylor series of h then vanish since a₀ = h(a) = 0 and a₁ = h'(a) = 0. — Emil J. 11:40, 25 January 2010 (UTC)Reply