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Question Number 139474 by mathdanisur last updated on 27/Apr/21

$$let:{\mathrm{\Omega }}_{\mathit{n}}=\underset{0}{\stackrel{2\pi }{\int }}cos\left(x\right)\cdot cos\left(2x\right)\cdot ...\cdot cos\left(nx\right)dx$$$$forwhichintegersn,1⩽n⩽10,is{\mathrm{\Omega }}_{\mathit{n}}\ne 0?$$

Answered by mathmax by abdo last updated on 28/Apr/21

$${\mathrm{\Phi }}_{\mathrm{n}}={\int }_0^{\pi }\mathrm{cosx}.\cos \left(2\mathrm{x}\right)....\cos \left(\mathrm{nx}\right)\mathrm{dx}+{\int }_{\pi }^{2\pi }\mathrm{cosx}\cos \left(2\mathrm{x}\right)....\cos \left(\mathrm{nx}\right)\mathrm{dx}$$$$(\to \mathrm{x}=\pi +\mathrm{t})={\int }_0^{\pi }\mathrm{cosx}.\cos \left(2\mathrm{x}\right)...\cos \left(\mathrm{nx}\right)\mathrm{dx}+{\int }_0^{\pi }\cos (\pi +\mathrm{t}).\cos (\pi +2\mathrm{t})...\cos (\mathrm{n}\pi +\mathrm{nt})\mathrm{dt}$$$$={\int }_0^{\pi }\prod _{\mathrm{k}=1}^{\mathrm{n}}\cos \left(\mathrm{kx}\right)\mathrm{dx}+{\int }_0^{\pi }\prod _{\mathrm{k}=1}^{\mathrm{n}}\cos (\mathrm{k}\pi +\mathrm{kx})\mathrm{dx}$$$$={\int }_0^{\pi }\prod _{\mathrm{k}=1}^{\mathrm{n}}\cos \left(\mathrm{kx}\right)\mathrm{dx}+{\int }_0^{\pi }\prod _{\mathrm{k}=1}^{\mathrm{n}}{(-1)}^{\mathrm{k}}\cos \left(\mathrm{kx}\right)\mathrm{dx}$$$$={\int }_0^{\pi }\prod _{\mathrm{k}=1}^{\mathrm{n}}\cos \left(\mathrm{kx}\right)\mathrm{dx}+{(-1)}^{\frac{\mathrm{n}(\mathrm{n}+1)}{2}}{\int }_0^{\pi }\prod _{\mathrm{k}=1}^{\mathrm{n}}\cos \left(\mathrm{kx}\right)\mathrm{dx}$$$$=(1+{(-1)}^{\frac{\mathrm{n}(\mathrm{n}+1)}{2}}){\int }_0^{\pi }\prod _{\mathrm{k}=1}^{\mathrm{n}}\cos \left(\mathrm{kx}\right)\mathrm{dx}\mathrm{so}{\mathrm{\Phi }}_{\mathrm{n}}\ne 0\Rightarrow$$$$1+{(-1)}^{\frac{\mathrm{n}(\mathrm{n}+1)}{2}}\ne 0\mathrm{and}{\int }_0^{\pi }\cos \left(\mathrm{x}\right).\cos \left(2\mathrm{x}\right)...\cos \left(\mathrm{nx}\right)\mathrm{dx}\ne 0$$$$....\mathrm{be}\mathrm{continued}...$$

Commented by mathdanisur last updated on 30/Apr/21

$$thankssir$$

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