August 12th, 2018, 09:54 PM  #1 
Senior Member Joined: Jan 2012 Posts: 123 Thanks: 2  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. 
August 12th, 2018, 10:25 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 2,200 Thanks: 1155 
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. 
August 13th, 2018, 09:32 AM  #3 
Senior Member Joined: Jan 2012 Posts: 123 Thanks: 2 
Thx.


Tags 
inverse, series, sin, sum 
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