analysis

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analysis

apoorva
x,y belongs to euclidean space. To Prove,
Max{xk + yk:1<=K<=n} <= Max{xk:1<=K<n} + Max{xk:1<=K<=n}
Where xk represents x subscript k
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Re: analysis

Amit Goyal
Administrator
I am going to show:
Max{x(k):1 ≤ k ≤ n} + Max{y(k):1 ≤ k ≤ n} ≥ Max{x(k) +y(k): 1 ≤ k ≤ n}

By definition of Max(),
Max{x(k):1 ≤ k ≤ n} ≥ x(j) for all 1 ≤ j ≤ n  ---(1)
Max{y(k):1 ≤ k ≤ n} ≥ y(j) for all 1 ≤ j ≤ n  ---(2)
Adding (1) and (2), we get
Max{x(k):1 ≤ k ≤ n} + Max{y(k):1 ≤ k ≤ n} ≥ x(j) +y(j) for all 1 ≤ j ≤ n
This implies that:
Max{x(k):1 ≤ k ≤ n} + Max{y(k):1 ≤ k ≤ n} ≥ Max{x(j) +y(j): 1 ≤ j ≤ n} or equivalently, if you prefer,
Max{x(k):1 ≤ k ≤ n} + Max{y(k):1 ≤ k ≤ n} ≥ Max{x(k) +y(k): 1 ≤ k ≤ n}
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Re: analysis

Apoorva
thank you sir
i did till equation no 2 after that i was stuck...
now i feel i should have thought about that question little more, given the fact that ans was so simple
anyway thanks alot....
:)