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1d
comment $f$ meromorphic on $\mathbb{\hat{C}}$ $\implies$ $f$ has a finite number of poles
Also, more directly (this is just the topology on $\hat{\mathbb{C}}$): we say $f$ has a pole at $\infty$ if $f(\frac 1z)$ has a pole at $0$. For this singularity to be isolated, say there is some neighborhood of radius $\frac 1\delta$ on which there are no other poles. Then $f(z)$ has no poles for $|z|>\delta$, and since the ball of radius $\delta$ is compact, there are finitely many poles in there.
Oct
25
comment Relation I found: $ (\sum_{r=1}^{\infty}\frac{z(r)}{r})\times \int_0^\infty f(x) dx = \lim_{h \rightarrow 0} \sum_{i=0}^{n} f(k_ih)h$
Like, what if $J(x^r)=x^r/r!$? Do you know if even they can or can't happen? If it is, then $\sum_{r=1}^\infty J(x^r)=e^x$, and you can't write that as $\sum_{i=1}^n x^{k_i}$, so your definition doesn't make sense.
Oct
25
awarded  Commentator
Oct
25
comment Relation I found: $ (\sum_{r=1}^{\infty}\frac{z(r)}{r})\times \int_0^\infty f(x) dx = \lim_{h \rightarrow 0} \sum_{i=0}^{n} f(k_ih)h$
Your definition for $J(x)$ only makes sense for $|x|<1$, right? How else can you guarantee the sum is convergent? And even with that assumption, how can you guarantee that the latter sum converges? Lastly how can you guarantee that latter sum is the sum of (finitely many) powers of $x$?
Oct
12
comment Subrepresentations of finite dimensional semisimple representations of an algebra
Yep :) I take it you're not since it seems like you already know this stuff?
Oct
9
comment Subrepresentations of finite dimensional semisimple representations of an algebra
Thanks, and sorry I didn't find that when I tried searching for related posts. I wonder if Kasper is in the same class I'm in...
Oct
8
asked Subrepresentations of finite dimensional semisimple representations of an algebra
Sep
5
answered Simple question on symmetric tensors
Aug
5
comment Toral sub algebra
Sorry, what is the contradiction here exactly? If the Lie algebra is abelian then the toral algebra is as well, so it's nilpotent.
Jul
22
answered Where's the problem with a false “proof”: $\;1^0 = 1^2 \overset{?}\implies 0 = 2$
Jul
11
comment Does $\operatorname{Hom}(M,T)\cong\operatorname{Hom}(N, T)$ for all $A$-modules $T$ mean $M\cong N$?
Also for context, in my case I was trying to prove that $\left(\oplus_i M_i\right)\otimes N\cong\oplus_i\left(M_i\otimes N\right)$ categorically, so I took $\operatorname{Hom}(\left(\oplus_i M_i\right)\otimes N, P)$ for arbitrary $P$ and used the adjunction of tensor product and Hom. So in my case at least, I'm pretty confidant the isomorphism is natural.
Jul
10
accepted Distributional differential equation, somehow related to compact support distributions
Jul
10
comment Does $\operatorname{Hom}(M,T)\cong\operatorname{Hom}(N, T)$ for all $A$-modules $T$ mean $M\cong N$?
Really nice counterexample, thanks for the thoroughness
Jul
10
awarded  Scholar
Jul
10
accepted Does $\operatorname{Hom}(M,T)\cong\operatorname{Hom}(N, T)$ for all $A$-modules $T$ mean $M\cong N$?
Jul
10
comment Does $\operatorname{Hom}(M,T)\cong\operatorname{Hom}(N, T)$ for all $A$-modules $T$ mean $M\cong N$?
@MartinBrandenburg I do assume $A$ is commutative. Also, I don't think naturality in $T$ is implied in general, correct me if I'm wrong, but your comment made me realize that it may be true in my case.
Jul
10
asked Does $\operatorname{Hom}(M,T)\cong\operatorname{Hom}(N, T)$ for all $A$-modules $T$ mean $M\cong N$?
Jun
16
answered Maximal ideal in the ring of polynomials over $\mathbb Z$
Jun
16
suggested suggested edit on Using the Bolzano's theorem to prove that exists only one solution in the interval.
Jun
7
comment Is intersection of connected subgroups connected?
Right, as long as $\mathcal{A}'$ has the finite intersection property, the theorem goes through! It seems reasonable that if the answer is 'yes', this is a good place to use the group property.