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   A linear algebra upper bound.
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   Author  Topic: A linear algebra upper bound.  (Read 7809 times)
acarchau
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A linear algebra upper bound.  
« on: Sep 14th, 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 14th, 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| > (R-C)/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 15th, 2009, 5:44pm »
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Nice argument. Thanks.
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