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Dec 23, 2009 at 13:44 comment added Mariano Suárez-Álvarez The $E$ should be a $D$, of course... (comments should be editable!)
Dec 23, 2009 at 13:43 comment added Mariano Suárez-Álvarez Hmm. Why doesn't $\sqrt{D-1}$ have a Taylor series at $0$? As long as you are willing to consider complex numbers (and why wouldn't you?!), $\sqrt{D-1}=i\sqrt{1-E}$ :)
Dec 23, 2009 at 5:48 history edited Qiaochu Yuan CC BY-SA 2.5
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Dec 23, 2009 at 5:36 history edited Qiaochu Yuan CC BY-SA 2.5
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Dec 23, 2009 at 5:30 history answered Qiaochu Yuan CC BY-SA 2.5