$$\displaystyle\lim\limits_{n\to \infty} \int_0^1 \int_0^1...\int_0^1 \cos^2\left(\frac{\pi}{2n}(x_1+x_2+...x_n)\right)dx_1 dx_2...dx_n$$
I don't know how to begin.
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3Please show some effort. Stated like this it looks like home work you want us to do for you. – mickep Jun 03 '16 at 16:16
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I try, $u=x_1+x_2+x_3+....+x_n$ but ......... – Micheal Brain Hurts Jun 03 '16 at 16:17
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2http://math.stackexchange.com/questions/728173/putnam-exam-integral?rq=1 – student forever Jun 03 '16 at 16:23
1 Answers
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Better to exploit a probabilistic approach. Let $X_1,\ldots X_n$ be independent random variables, uniformly distributed over $[0,1]$. Our integral computes: $$ \mathbb{E}\left[\cos^2\left(\frac{\pi}{2} M_n\right)\right] $$ where: $$ M_n = \frac{X_1+\ldots +X_n}{n}. $$ Since $M_n\to \frac{1}{2}$ by the law of large numbers, the given limit equals: $$ \cos^2\frac{\pi}{4} = \color{red}{\frac{1}{2}}. $$
Jack D'Aurizio
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