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1-2-3-3-1-2-3-3-1-6-3-3-1-6-3-3-1-10-3-3-1-10-3-3-

Question Number 130449 by Dwaipayan Shikari last updated on 25/Jan/21 $$\frac{\mathrm{1}}{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{3}} }+\frac{\mathrm{1}}{\left(\mathrm{6}−\sqrt{\mathrm{3}}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\left(\mathrm{6}+\sqrt{\mathrm{3}}\right)^{\mathrm{3}} }+\frac{\mathrm{1}}{\left(\mathrm{10}−\sqrt{\mathrm{3}}\right)^{\mathrm{3}} }−\frac{\mathrm{1}}{\left(\mathrm{10}+\sqrt{\mathrm{3}}\right)^{\mathrm{3}} }+.. \\ $$ Terms of Service Privacy Policy Contact:…

pi-e-sin-1-pi-2-2e-2-sin-2-pi-3-3e-3-sin-3-pi-4-4e-4-sin-4-

Question Number 130415 by Dwaipayan Shikari last updated on 25/Jan/21 $$\frac{\pi}{{e}}{sin}\left(\mathrm{1}\right)−\frac{\pi^{\mathrm{2}} }{\mathrm{2}{e}^{\mathrm{2}} }{sin}\left(\mathrm{2}\right)+\frac{\pi^{\mathrm{3}} }{\mathrm{3}{e}^{\mathrm{3}} }{sin}\left(\mathrm{3}\right)−\frac{\pi^{\mathrm{4}} }{\mathrm{4}{e}^{\mathrm{4}} }{sin}\left(\mathrm{4}\right)+… \\ $$ Commented by Dwaipayan Shikari last updated…

Question-130399

Question Number 130399 by gowsalya last updated on 25/Jan/21 Answered by Dwaipayan Shikari last updated on 25/Jan/21 $${f}\left({x}\right)=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}{a}_{{n}} {x}^{{n}} ={a}_{\mathrm{0}} +{a}_{\mathrm{1}} {x}+{a}_{\mathrm{2}} {x}^{\mathrm{2}}…

Question-130369

Question Number 130369 by gowsalya last updated on 24/Jan/21 Answered by Olaf last updated on 24/Jan/21 $$\Omega\:=\:\int_{\mid{z}\mid=\mathrm{1}} {xdz} \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mathrm{cos}\theta{d}\left(\mathrm{cos}\theta+{i}\mathrm{sin}\theta\right) \\ $$$$\Omega\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi}…

Question-130370

Question Number 130370 by gowsalya last updated on 24/Jan/21 Answered by TheSupreme last updated on 24/Jan/21 $${z}={e}^{{i}\theta} \\ $$$$\mid{dz}\mid=\mid{ie}^{{i}\theta} {d}\theta\mid={d}\theta \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mid{e}^{{i}\theta} −\mathrm{1}\mid{d}\theta…