Question Number 101500 by mathmax by abdo last updated on 03/Jul/20 $$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{cosx}\:.\mathrm{cos}\left(\mathrm{2x}\right).\mathrm{cos}\left(\mathrm{3x}\right) \\ $$$$\left.\mathrm{1}\right)\mathrm{calculate}\:\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{x}\right)\:\mathrm{and}\:\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right)\mathrm{developp}\:\mathrm{f}\:\mathrm{at}\:\:\mathrm{integr}\:\mathrm{serie} \\ $$$$\mathrm{3}.\:\mathrm{calculate}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{f}\left(\mathrm{x}\right)\mathrm{dx} \\ $$ Commented…
Question Number 167003 by cortano1 last updated on 04/Mar/22 $$\:\:\:\:\:\underset{\mathrm{n}=\mathrm{2}} {\overset{\mathrm{n}=\infty} {\sum}}\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{n}} \mathrm{cos}\:\left(\mathrm{180}°\mathrm{n}\right)=\:? \\ $$ Answered by greogoury55 last updated on 04/Mar/22 $$\:\mathrm{cos}\:\left(\mathrm{180}°{n}\right)=\left(−\mathrm{1}\right)^{{n}} \\ $$$$\:\underset{{n}=\mathrm{2}}…
Question Number 35873 by Tinkutara last updated on 25/May/18 Answered by tanmay.chaudhury50@gmail.com last updated on 28/May/18 $${let}\:{sin}^{−\mathrm{1}} {x}=\theta \\ $$$${sin}\theta={x} \\ $$$${tan}\theta=\frac{{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:\:\:{so}\:\theta={tan}^{−\mathrm{1}} \left(\frac{{x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\right)…
Question Number 35833 by abdo mathsup 649 cc last updated on 24/May/18 $${fond}\:{lim}_{{n}\rightarrow+\infty} \:\:{n}^{{a}} \:\:\left\{{ln}\left(\mathrm{1}+{e}^{−{n}} \right)\:−{e}^{−{n}} \right\}\:{with}\:{a}>\mathrm{0} \\ $$ Commented by prof Abdo imad last…
Question Number 101342 by mathmax by abdo last updated on 02/Jul/20 $$\mathrm{find}\:\:\mathrm{U}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{x}^{\mathrm{2n}} −\mathrm{1}}{\mathrm{lnx}}\mathrm{dx}\:\:\mathrm{with}\:\mathrm{n}\:\mathrm{integr}\:\mathrm{natural}\:\mathrm{and}\:\mathrm{n}\geqslant\mathrm{2} \\ $$$$\mathrm{find}\:\mathrm{nature}\:\mathrm{of}\:\mathrm{the}\:\mathrm{serie}\:\Sigma\:\mathrm{U}_{\mathrm{n}} \\ $$ Answered by mathmax by abdo…
Question Number 35769 by Tinkutara last updated on 23/May/18 Answered by ajfour last updated on 23/May/18 $${let}\:\:{f}\left({x}\right)=\frac{\mathrm{2}{x}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$\frac{{df}\left({x}\right)}{{dx}}=\frac{\mathrm{2}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)−\mathrm{4}{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\:\frac{\mathrm{2}\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}\right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{2}}…
Question Number 35766 by abdo mathsup 649 cc last updated on 23/May/18 $${let}\:{f}\left({x}\right)={arctan}\left(\mathrm{1}+{x}\right)\:\:{and}\:\:{f}_{{n}} \left({x}\right)=\:{e}^{−{nx}} \\ $$$$\left.\mathrm{1}\right)\:{define}\:{fof}_{{n}} \left({x}\right)\:{and}\:{f}_{{n}} {of}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{study}\:{the}\:{nature}\:{of}\:{the}\:{series}\:\sum_{{n}=\mathrm{0}} ^{+\infty} {fof}_{{n}} \left({x}\right) \\ $$$${and}\:\sum_{{n}=\mathrm{0}}…
Question Number 101273 by mathmax by abdo last updated on 01/Jul/20 $$\mathrm{caoculate}\:\mathrm{lim}_{\mathrm{x}\rightarrow\mathrm{0}} \:\:\:\frac{\mathrm{sh}\left(\mathrm{sin}\left(\mathrm{2x}\right)\right)−\mathrm{sin}\left(\mathrm{sh}\left(\mathrm{2x}\right)\right)}{\mathrm{x}^{\mathrm{3}} } \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 35690 by prof Abdo imad last updated on 22/May/18 $${let}\:{B}_{{n}} \:=\sum_{{k}=\mathrm{1}} ^{{n}} \:\:{sin}\left(\frac{{k}\pi}{{n}}\right)\:{sin}\left(\frac{{k}}{{n}^{\mathrm{2}} }\right) \\ $$$${find}\:{lim}_{{n}\rightarrow+\infty} \:\:{B}_{{n}} \\ $$ Terms of Service Privacy…
Question Number 35688 by prof Abdo imad last updated on 22/May/18 $${let}\:{A}_{{n}} \:=\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\:\frac{\mathrm{1}}{{k}+{n}}{ln}\left(\mathrm{1}+\frac{{k}}{{n}}\right) \\ $$$${calculate}\:{lim}_{{n}\rightarrow+\infty} {A}_{{n}} \\ $$ Commented by prof Abdo imad…