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Category: Relation and Functions

calculate-U-n-0-1-2-x-1-cos-n-x-dx-find-nature-of-U-n-

Question Number 108711 by mathmax by abdo last updated on 18/Aug/20 $$\mathrm{calculate}\:\mathrm{U}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{2}\left[\mathrm{x}\right]−\mathrm{1}} \mathrm{cos}\left(\mathrm{n}\left[\mathrm{x}\right]\right)\mathrm{dx} \\ $$$$\mathrm{find}\:\mathrm{nature}\:\mathrm{of}\:\Sigma\:\mathrm{U}_{\mathrm{n}} \\ $$ Answered by mathmax by abdo…

if-k-1-n-u-k-n-2-n-3-determine-lim-n-k-1-n-1-u-k-

Question Number 108705 by mathmax by abdo last updated on 18/Aug/20 $$\mathrm{if}\:\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{u}_{\mathrm{k}} =\mathrm{n}\left(\mathrm{2}^{\mathrm{n}} +\mathrm{3}\right)\:\:\mathrm{determine}\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{u}_{\mathrm{k}} } \\ $$ Answered by mathmax…

prove-that-n-1-ax-1-n-pi-a-n-lim-x-1-a-1-n-1-cotan-pix-n-1-

Question Number 108688 by abdomsup last updated on 18/Aug/20 $${prove}\:{that}\: \\ $$$$\sum_{{n}=−\infty} ^{\infty} \:\frac{\mathrm{1}}{\left({ax}+\mathrm{1}\right)^{{n}} } \\ $$$$=−\frac{\pi}{{a}^{{n}} }\:{lim}_{{x}\rightarrow−\frac{\mathrm{1}}{{a}}} \:\:\:\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\left\{{cotan}\left(\pi{x}\right)\right\}^{\left({n}−\mathrm{1}\right)} \\ $$ Commented by mathdave last…

let-u-n-1-i-lt-j-n-1-ij-1-find-a-equivalent-of-u-n-2-calculate-lim-n-u-n-

Question Number 43003 by abdo.msup.com last updated on 06/Sep/18 $${let}\:{u}_{{n}} =\:\sum_{\mathrm{1}\leqslant{i}<{j}\leqslant{n}} \:\:\frac{\mathrm{1}}{\:\sqrt{{ij}}} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{a}\:{equivalent}\:{of}\:{u}_{{n}} \\ $$$$\left.\mathrm{2}\right){calculate}\:{lim}_{{n}\rightarrow+\infty} \:{u}_{{n}} \\ $$ Answered by maxmathsup by imad last…

prove-that-2-2-2-2cos-pi-2-n-

Question Number 43000 by abdo.msup.com last updated on 06/Sep/18 $${prove}\:{that}\:\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+….+\sqrt{\mathrm{2}}}}\:\:=\mathrm{2}{cos}\left(\frac{\pi}{\mathrm{2}^{{n}} }\right) \\ $$ Answered by tanmay.chaudhury50@gmail.com last updated on 06/Sep/18 $${cos}\mathrm{2}\alpha=\mathrm{2}{cos}^{\mathrm{2}} \alpha−\mathrm{1} \\ $$$${cos}^{\mathrm{2}} \alpha=\frac{\mathrm{1}+{cos}\mathrm{2}\alpha}{\mathrm{2}}…