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August 12th, 2018, 08:54 PM   #1
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Series sum of inverse sin

Hi,

How can we evaluate $\displaystyle \sum_{i=1}^{2n}x_{i}$

given, $\displaystyle \sum_{i=1}^{2n}\sin^{-1} (x_{i})=n\pi $

Thx.
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August 12th, 2018, 09:25 PM   #2
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I see nothing that keeps us from keeping $x_i$ constant.

$n \pi = \displaystyle \sum \limits_{i=1}^{2n}\sin^{-1}(x_i) =

2n \sin^{-1}(x)$

$\dfrac{\pi}{2} = \sin^{-1}(x)$

$x = 1$

and thus

$\displaystyle \sum \limits_{i=1}^{2n}~x_i = 2n(1) = 2n$

We note that $x_i \leq 1$ and so $n\pi$ is the maximum value the 2nd sum can attain and it attains it only when $x_i = 1,~\forall i$

Thus the value of the first sum is unique.
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August 13th, 2018, 08:32 AM   #3
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Thx.
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