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212

Part of 2007 Today's Calculation Of Integral

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Today's calculation of Integral 212

Source: Kyoto University entrance exam 1978

6/16/2007
For integers k (0≤k≤5)k\ (0\leq k\leq 5), positive numbers m, nm,\ n and real numbers a, ba,\ b, let f(k)=∫−ππ(sin⁡kx−asin⁡mx−bsin⁡nx)2 dxf(k)=\int_{-\pi}^{\pi}(\sin kx-a\sin mx-b\sin nx)^{2}\ dx, p(k)=5!k!(5−k)!(12)5, E=∑k=05p(k)f(k)p(k)=\frac{5!}{k!(5-k)!}\left(\frac{1}{2}\right)^{5}, \ E=\sum_{k=0}^{5}p(k)f(k). Find the values of m, n, a, bm,\ n,\ a,\ b for which EE is minimized.
calculusintegrationtrigonometrycalculus computations