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Topic: A linear algebra upper bound. (Read 7849 times) 

acarchau
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A linear algebra upper bound.
« on: Sep 14^{th}, 2009, 9:29pm » 
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Let X and Y be linearly independent vectors in R^2. Let the lattice U be the set of all vectors of the form: mX+nY, where m and n are integers. Choose an appropriately small and positive d, and let W(d) be the non empty set { v in R^2 : v + u > d for all u in U}. For any v in W(d) let g(v) = sup_{ u in U} (  u  /  u + v). Then is sup_{v in W(d)} g(v) < infinity?


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Eigenray
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Re: A linear algebra upper bound.
« Reply #1 on: Sep 14^{th}, 2009, 10:41pm » 
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Each g(v) is finite but rather than sup_{v in W(d)} g(v) being finite, we have in fact g(v) as v . Indeed, there exists a constant C such that for all v, there exists u in U with u+v<C. For v > R, pick such a u; then C > u+v v  u > R  u, so u > R  C, and g(v) u/u+v > (RC)/C, which goes to infinity as R does.


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acarchau
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Re: A linear algebra upper bound.
« Reply #2 on: Sep 15^{th}, 2009, 5:44pm » 
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Nice argument. Thanks.


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