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Jun
30
asked Is there a way to turn this summation into a matrix multiplication?
Jun
9
revised What are some results that shook the foundations of one or more fields of mathematics?
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Jun
8
awarded  Popular Question
Jun
8
comment What are some results that shook the foundations of one or more fields of mathematics?
I understand the question may have many possible answers, but I think it is sufficiently narrow that we can distinguish an answer from a non-answer in general, no?
Jun
8
revised What are some results that shook the foundations of one or more fields of mathematics?
added 688 characters in body
Jun
8
comment True or False. $K_{2n}$ ( complete graph with 2n vertices) has Euler circuit.
@BrianTung for n=1 it's K_2
Jun
8
comment True or False. $K_{2n}$ ( complete graph with 2n vertices) has Euler circuit.
n=1 is an obvious counterexample
Jun
8
awarded  Nice Question
Jun
8
revised What are some results that shook the foundations of one or more fields of mathematics?
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Jun
7
asked What are some results that shook the foundations of one or more fields of mathematics?
Jun
2
awarded  Popular Question
May
4
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Apr
28
awarded  Popular Question
Feb
19
revised Given a vector $x\in \mathbb R^n$, how can we find $z\in \mathbb Z^n$ which is closest to a scalar multiple of $x$?
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Feb
19
revised Given a vector $x\in \mathbb R^n$, how can we find $z\in \mathbb Z^n$ which is closest to a scalar multiple of $x$?
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Feb
18
revised Given a vector $x\in \mathbb R^n$, how can we find $z\in \mathbb Z^n$ which is closest to a scalar multiple of $x$?
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Feb
18
comment Given a vector $x\in \mathbb R^n$, how can we find $z\in \mathbb Z^n$ which is closest to a scalar multiple of $x$?
@ThomasAndrews yes that's right
Feb
18
asked Given a vector $x\in \mathbb R^n$, how can we find $z\in \mathbb Z^n$ which is closest to a scalar multiple of $x$?
Feb
10
comment Is closed unit disk in $R^2$ be a vector space ?
@RobertIsrael what if we consider the open unit disc?
Feb
9
comment What kind of algorithm might solve this type of optimization problem?
Very nice. I like to think I would have eventually made the observation about the sum of the terms, but thank you very much for being smarter than me.