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awarded  Caucus
Dec
12
comment What is the name of this letter?
This is question fits better on tex.stackexchange.com.
Nov
21
accepted Minimum dimension to hold $N$ points with given distances?
Nov
21
revised How to determine if a vector belongs to the conical hull of a set of vectors?
edited title
Nov
10
comment How to determine if a vector belongs to the conical hull of a set of vectors?
@p.s. Can you give more details please?
Nov
7
comment How to determine if a vector belongs to the conical hull of a set of vectors?
@John $0\le \alpha_i < \infty$. Yes, the infinite cone.
Nov
7
asked How to determine if a vector belongs to the conical hull of a set of vectors?
Oct
31
asked Minimum dimension to hold $N$ points with given distances?
Oct
20
comment Is the set of all invertible $n \times n$ matrices a vector space?
@JacquesUng You should accept this answer.
Oct
17
comment The Wicked Integral
I think it is very unintuitive that this is independent of $n$. It would be nice if someone offered an intuitive explanation of that fact.
Oct
9
comment Creative way to solve a linear system
Try Gaussain elimination.
Oct
3
revised Evaluating a multiplicity factor
edited tags
Oct
3
revised Evaluating a multiplicity factor
added 23 characters in body
Oct
3
comment Evaluating a multiplicity factor
I couldn't come up with a more specific title. I'm open to title suggestions.
Oct
3
asked Evaluating a multiplicity factor
Sep
18
comment 1/1000 chance of a reaction. If you do the action 1000 times, whats the new chance the reaction occurs?
@Dancrumb Agree.
Sep
18
suggested rejected edit on 1/1000 chance of a reaction. If you do the action 1000 times, whats the new chance the reaction occurs?
Sep
18
comment 1/1000 chance of a reaction. If you do the action 1000 times, whats the new chance the reaction occurs?
@Dancrumb's comment has a lot of votes, it's confusing. I think the answer should be edited to clarify that no such assumption is made.
Aug
29
revised Prove that if $\frac{x+y}{3a-b}=\frac{y+z}{3b-c}=\frac{z+x}{3c-a}$ then $\frac{x+y+z}{a+b+c}=\frac{ax+by+cz}{a^2+b^2+c^2}$
Original title suggests that the equality holds always.
Aug
29
suggested approved edit on Prove that if $\frac{x+y}{3a-b}=\frac{y+z}{3b-c}=\frac{z+x}{3c-a}$ then $\frac{x+y+z}{a+b+c}=\frac{ax+by+cz}{a^2+b^2+c^2}$