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Jan
13
comment Groups with no nontrivial topology
@PedroTamaroff Thanks, these are the sort of things I really appreciate in trying to work out how to intuit the topologisation of groups.
Jan
13
comment Groups with no nontrivial topology
@NajibIdrissi Okay, edited.
Jan
13
revised Groups with no nontrivial topology
added 4 characters in body
Jan
12
asked Groups with no nontrivial topology
Jan
7
comment Which $n$th order differential equations have $n$ linearly independent solutions?
@user1537366 No, I didn't.
Dec
28
accepted Is there a general way to prove series and products are modular?
Dec
28
comment Is there a general way to prove series and products are modular?
@guest Thanks, edited.
Dec
28
revised Is there a general way to prove series and products are modular?
added 63 characters in body
Dec
28
revised Is there a general way to prove series and products are modular?
deleted 12 characters in body
Dec
28
awarded  Nice Question
Dec
27
accepted Motivation for constructing $F$ s.t. $\ker(\text{curl}) \subset \text{Im}(\text{grad})$, $\ker(\text{div}) \subset \text{Im}(\text{curl})$
Dec
25
comment Motivation for constructing $F$ s.t. $\ker(\text{curl}) \subset \text{Im}(\text{grad})$, $\ker(\text{div}) \subset \text{Im}(\text{curl})$
Thanks for the thorough answer, and sorry for not replying sooner. The answer has helped understand the theory more, but I still feel confused as to how the author came up with $F(x_1,x_2)$ and $G(x,t)$ in the first line of each proof, their use seems a little magical to me still.
Dec
24
comment Motivation for constructing $F$ s.t. $\ker(\text{curl}) \subset \text{Im}(\text{grad})$, $\ker(\text{div}) \subset \text{Im}(\text{curl})$
@Travis I don't.
Dec
24
comment Motivation for constructing $F$ s.t. $\ker(\text{curl}) \subset \text{Im}(\text{grad})$, $\ker(\text{div}) \subset \text{Im}(\text{curl})$
I apologise for the dreadful LaTeX. I am using a keyboard I am not used to using.
Dec
24
asked Motivation for constructing $F$ s.t. $\ker(\text{curl}) \subset \text{Im}(\text{grad})$, $\ker(\text{div}) \subset \text{Im}(\text{curl})$
Dec
19
awarded  Constituent
Dec
8
awarded  Caucus
Nov
29
comment How is half-contour integration possible?
@AlexR. I am. Apologies. I meant a contour that does not stop at any finite point (that is, goes off to infinity in both directions).
Nov
29
asked How is half-contour integration possible?
Nov
19
comment Is there a general way to prove series and products are modular?
@user45195 The mild moral qualm about reposting this question is stronger than my attachment to reputation.