Soo, fix 0 &#x2264;<!-- ≤ --> &#x03F5;<!-- ϵ --> &#x2264;<!-- ≤ --> 1 . Given &

rjawbreakerca 2022-07-02 Answered
Soo, fix 0 ϵ 1. Given λ 1 , x i 0, I know that i = 1 n x i λ n. I also know that i = 1 n x i 2 λ 2 n + ϵ. I am trying to prove a multiplicative error on each x i , mainly something along the lines of
| x i λ | f ( ϵ ) λ
Where f ( ϵ ) is some function of ϵ, say f ( ϵ ) = 2 ϵ. Is there any inequality that would bound that distance?
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Answers (1)

Zachery Conway
Answered 2022-07-03 Author has 7 answers
Here is the solution which I literally thought of two minutes after typing the question. Let x i = λ + δ i for some δ i . Then we have that
i ( x i 2 ) = i ( λ + δ i ) 2 = λ 2 n + 2 λ i δ i + i δ i 2 < λ 2 n + ϵ
But we have that
i x i = i λ + δ i = λ n + i δ i > λ n
Which implies
i δ i > 0
Combining with first inequality
λ 2 n + 0 + i δ i 2 < λ 2 n + 2 λ i δ i + i δ i 2 < λ 2 n + ϵ
Or simply
i δ i 2 < ϵ
Or
| δ i | ϵ ϵ λ
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