Asymptotics
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From: RSimmons on 13 Apr 2010 01:52 Let a(n) be a real sequence. If sum_k=1..n [a(k)]^2 = O( 1/(n^p . log^q n)), n -> +infinity (p,q >= 0) then a(n) = O( 1/n^(p+1)/2 . log^q/2 n), n -> +infinity?
From: Robert Israel on 13 Apr 2010 11:36
> Let a(n) be a real sequence. If sum_k=1..n [a(k)]^2 = O( 1/(n^p . log^q > n)), n -> +infinity (p,q >= 0) then > a(n) = O( 1/n^(p+1)/2 . log^q/2 n), n -> +infinity? Are you sure this is what you meant to ask? sum_{k=1}^n a(k)^2 is nonnegative and nondecreasing, so it can't be O(1/(n^p log(n)^q) except in the case p=q=0 with the sum converging. And even then you can't make any conclusion a(n) = O(f(n)) with f(n) -> 0 because you could e.g. take a(n_k) = 2^(-k), a(n) = 0 otherwise, with n_k a sequence increasing fast enough that f(n_k) < 3^(-k). -- Robert Israel israel(a)math.MyUniversitysInitials.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada |