Question Number 32933 by abdo imad last updated on 06/Apr/18 $$\Sigma\:{u}_{{n}} \:{is}\:{a}\:{convergent}\:{serie}\:{with}\:{positif}\:{terms} \\ $$$${find}\:{the}\:{nature}\:{of}\:\:\sum_{{n}\geqslant\mathrm{1}} \:\frac{\sqrt{{u}_{{n}} }}{{n}}\:\:{and}\:\:\:\sum_{{n}\geqslant{o}} \:\:\frac{{u}_{{n}} }{\mathrm{1}+{u}_{{n}} }\:\:. \\ $$ Terms of Service Privacy…
Question Number 32934 by abdo imad last updated on 06/Apr/18 $${let}\:{give}\:\mid{x}\mid<\mathrm{1}\:{prove}\:{that} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:{x}^{{n}} {cos}\left({n}\theta\right)=\:\frac{{xcos}\theta\:−{x}^{\mathrm{2}} }{\mathrm{1}−\mathrm{2}{xcos}\theta\:+{x}^{\mathrm{2}} } \\ $$ Terms of Service Privacy Policy…
Question Number 32932 by abdo imad last updated on 06/Apr/18 $${find}\:{the}\:{nature}\:{of}\:\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}\sum_{{k}=\mathrm{1}} ^{{n}} \:\frac{\mathrm{1}}{{k}}}\:. \\ $$ Commented by prof Abdo imad last updated on…
Question Number 98430 by mathmax by abdo last updated on 13/Jun/20 $$\mathrm{let}\:\mathrm{g}\left(\mathrm{x}\right)\:=\frac{\mathrm{2}}{\mathrm{cos}\left(\pi\mathrm{x}\right)}\:\:\mathrm{developp}\:\mathrm{g}\:\mathrm{at}\:\mathrm{fourier}\:\mathrm{serie} \\ $$ Answered by abdomathmax last updated on 13/Jun/20 $$\mathrm{we}\:\mathrm{have}\:\mathrm{g}\left(\mathrm{x}\right)\:=\frac{\mathrm{2}}{\mathrm{cos}\left(\pi\mathrm{x}\right)}\:=\frac{\mathrm{4}}{\mathrm{e}^{\mathrm{i}\pi\mathrm{x}\:} \:+\mathrm{e}^{−\mathrm{i}\pi\mathrm{x}} } \\…
Question Number 98429 by mathmax by abdo last updated on 13/Jun/20 $$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{cos}\left(\alpha\mathrm{x}\right)\:\mathrm{developp}\:\mathrm{f}\:\mathrm{at}\:\mathrm{fourier}\:\mathrm{serie} \\ $$ Answered by abdomathmax last updated on 15/Jun/20 $$\mathrm{f}\:\mathrm{is}\:\mathrm{even}\:\Rightarrow\mathrm{f}\left(\mathrm{x}\right)\:=\frac{\mathrm{a}_{\mathrm{0}} }{\mathrm{2}}\:+\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\mathrm{a}_{\mathrm{n}}…
Question Number 98424 by mathmax by abdo last updated on 13/Jun/20 $$\mathrm{calculste}\:\mathrm{A}_{\mathrm{n}} =\int_{−\frac{\mathrm{1}}{\mathrm{2}}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:\mathrm{x}^{\mathrm{n}} \sqrt{\frac{\mathrm{1}−\mathrm{x}}{\mathrm{1}+\mathrm{x}}}\mathrm{dx} \\ $$$$\mathrm{find}\:\mathrm{nature}\:\mathrm{of}\:\mathrm{the}\:\mathrm{serie}\:\Sigma\:\mathrm{A}_{\mathrm{n}} \\ $$ Answered by maths mind last…
Question Number 98425 by mathmax by abdo last updated on 13/Jun/20 $$\mathrm{let}\:\mathrm{f}\left(\mathrm{x}\right)\:=\mathrm{e}^{−\mathrm{x}} \:\mathrm{arctan}\left(\pi\mathrm{x}\right) \\ $$$$\left.\mathrm{1}\right)\:\mathrm{calculate}\:\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{x}\right)\mathrm{and}\:\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{2}\right)\:\mathrm{developp}\:\mathrm{f}\:\mathrm{at}\:\mathrm{integr}\:\mathrm{serie} \\ $$ Terms of Service Privacy…
Question Number 98310 by mathmax by abdo last updated on 12/Jun/20 $$\mathrm{prove}\:\mathrm{by}\:\mathrm{using}\:\mathrm{serie}\:\mathrm{that}\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{cos}\left(\mathrm{x}^{\mathrm{2}} \right)\mathrm{dx}\:=\int_{\mathrm{0}} ^{\infty} \mathrm{sin}\left(\mathrm{x}^{\mathrm{2}} \right)\mathrm{dx} \\ $$ Terms of Service Privacy Policy…
Question Number 32735 by caravan msup abdo. last updated on 01/Apr/18 $${let}\:{give}\:{A}_{{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{\mathrm{1}+{t}^{{n}} } \\ $$$$\left.\mathrm{1}\right)\:{find}\:{l}={lim}_{{n}\rightarrow\infty} \:{A}_{{n}} \\ $$$$\left.\mathrm{2}\right){give}\:{a}\:{equivalent}\:{of}\:{A}_{{n}} −{l} \\ $$$$\left.\mathrm{3}\right)\:{find}\:{a}\:{equivalent}\:{of}\:{A}_{{n}} \\…
Question Number 32734 by caravan msup abdo. last updated on 01/Apr/18 $$\left.\mathrm{1}\right)\:{a}\geqslant\mathrm{0}\:\:{calculate}\:\int_{\mathrm{0}} ^{{a}} \:\frac{{n}^{\mathrm{2}} \:−{x}^{\mathrm{2}} }{\left({n}^{\mathrm{2}} \:+{x}^{\mathrm{2}} \right)^{\mathrm{2}} }{dx}\:{with} \\ $$$${n}\:{integr} \\ $$$$\left.\mathrm{2}\right)\:{find}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{n}^{\mathrm{2}}…