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visits member for 1 year, 2 months
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Aug
16
asked Why are normal subgroups called “Normal”?
Aug
13
accepted Proof Essential Singularities are Isolated
Aug
13
comment Proof Essential Singularities are Isolated
Haha I was just reading all of your responses to other questions on this subject. I was defining them as the second definition, so I see then.
Aug
13
asked Proof Essential Singularities are Isolated
Aug
12
revised Translating to English: $\forall x ¬(\forall y P(x, y)) \rightarrow \forall x \forall y ¬P(x, y)$
Added in the quantifier...
Aug
12
suggested suggested edit on Translating to English: $\forall x ¬(\forall y P(x, y)) \rightarrow \forall x \forall y ¬P(x, y)$
Aug
12
asked Laurent Series for $\frac{1}{e^z-1}$-Radius
Aug
11
accepted Proof Bounded Implies a Limit Exists
Aug
11
accepted Identity with Dirac delta function: $\delta (x^2-a^2) = \frac{1}{2|a|}[\delta(x-a)+\delta(x+a)]$
Aug
11
comment Residue of $\frac{z}{Log{z}}$ at $z=1$?
I wasn't aware I was allowed to do that...
Aug
11
asked Residue of $\frac{z}{Log{z}}$ at $z=1$?
Aug
6
comment Proof Bounded Implies a Limit Exists
I suppose I could work with the fact that the maximum must occur on the boundary. Then if there was any periodicity I could just shrink the radius until that was the maximum/minimum. If it wasn't on the boundary it would be constant. I see.
Aug
6
revised Proof Bounded Implies a Limit Exists
deleted 74 characters in body
Aug
6
asked Proof Bounded Implies a Limit Exists
Aug
4
comment Proof that Radius of Convergence Extend to Nearest Singularity
Did you say why it can't be smaller...?
Aug
4
asked Proof that Radius of Convergence Extend to Nearest Singularity
Jul
29
comment Uniform convergence on compact sets allows switching the limit and the integral.
It was with regard to the last line.
Jul
29
comment Uniform convergence on compact sets allows switching the limit and the integral.
Ugh I didn't want to whip out the LaTeX. Given a sequence of analytic functions $\{f_{n}\}_{n}$ on a domain D, suppose $\sum_{n} f_{n}$ converges at least normally on D. Then $F = \sum_{n} f_{n}$ is analytic on D and $F^{(k)} = \sum_{n} f^{(k)}_{n}.$ Given a path $\Gamma \subset D$, then $\displaystyle\int_{\Gamma} \sum_{n} f_{n} \, dz = \sum_{n} \displaystyle\int_{\Gamma} f_{n} \, dz.$
Jul
29
comment Uniform convergence on compact sets allows switching the limit and the integral.
Both would be nice- But I'd like the theorem especially.
Jul
29
asked Uniform convergence on compact sets allows switching the limit and the integral.