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Question Number 36923 Answers: 0 Comments: 1
$${for}\:{t}\geqslant\mathrm{0}\:{and}\:\:{f}\left({t}\right)=\:\frac{{t}}{\sqrt{\mathrm{1}+{t}}}\:\:{let} \\ $$$${S}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}} \:{f}\left(\frac{{k}}{{n}^{\mathrm{2}} }\right)\:\:{study}\:{the}\:{convergence}\:{of}\:{S}_{{n}} \:\:. \\ $$
Question Number 36922 Answers: 0 Comments: 1
$$\left({u}_{{n}} \right){is}\:{a}\:{sequence}\:{and}\:{lim}_{{n}\rightarrow+\infty} {u}_{{n}} ={l}\:{let} \\ $$$${v}_{{n}} =\:\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\:\sum_{{k}=\mathrm{0}} ^{{n}} {C}_{{n}} ^{{k}} \:{u}_{{k}} \:\:{prove}\:{that}\:{v}_{{n}} \:\rightarrow{l}\left({n}\rightarrow+\infty\:\right) \\ $$
Question Number 36921 Answers: 0 Comments: 0
$${study}\:{the}\:{convergence}\:{of}\:\:{u}_{\mathrm{1}} ={ln}\left(\mathrm{2}\right)\:{and}\:{u}_{{n}} =\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} {ln}\left(\mathrm{2}−{u}_{{k}} \right). \\ $$
Question Number 36920 Answers: 0 Comments: 1
$${let}\:\alpha\:{from}\:{R}\:{and}\:\:{u}_{{n}} \:−\mathrm{2}{cos}\left(\alpha\right){u}_{{n}−\mathrm{1}} \:+{u}_{{n}−\mathrm{2}} =\mathrm{0}\:\:\:{withn}\geqslant\mathrm{2} \\ $$$${find}\:{u}_{{n}} \:{and}\:{study}\:{its}\:{convrgence}. \\ $$
Question Number 36919 Answers: 0 Comments: 1
$${calculate}\:{f}\left(\alpha\right)=\:\int_{−\infty} ^{+\infty} \:\left(\mathrm{1}+\alpha{i}\right)^{−{x}^{\mathrm{2}} } {dx}\:. \\ $$
Question Number 36918 Answers: 0 Comments: 0
$${calculate}\:\int_{\mathrm{0}} ^{+\infty} \:\left(\mathrm{1}−{i}\right)^{−{x}^{\mathrm{2}} } {dx}\: \\ $$
Question Number 36917 Answers: 0 Comments: 1
$${calculate}\:\int_{\mathrm{0}} ^{+\infty} \left(\mathrm{1}+{i}\right)^{−{x}^{\mathrm{2}} } {dx} \\ $$
Question Number 36916 Answers: 0 Comments: 1
$${let}\:{z}={r}\:{e}^{{i}\theta} \:\:\:\:{fins}\:{f}\left({z}\right)\:=\:\int_{−\infty} ^{+\infty} \:\:{z}^{−{x}^{\mathrm{2}} } {dx} \\ $$
Question Number 36915 Answers: 0 Comments: 1
$${let}\:{z}\:={a}+{ib}\:\:\:{find}\:\:{f}\left({z}\right)\:=\:\int_{−\infty} ^{+\infty} \:{z}^{−{x}^{\mathrm{2}} } {dx} \\ $$
Question Number 36912 Answers: 0 Comments: 1
$${let}\:\:\langle{p},{q}\rangle=\:\int_{−\mathrm{1}} ^{\mathrm{1}} {p}\left({x}\right){q}\left({x}\right){dx}\:\:{with}\:{p}\:{and}\:{q}\:{are} \\ $$$${two}\:{polynoms}\:{fromR}\left[{x}\right] \\ $$$$\left.\mathrm{1}\right){let}\:{p}\left({x}\right)={x}^{{n}} \:\:\:{calculate}\:\langle{p},{p}\rangle \\ $$$$\left.\mathrm{2}\right){let}\:{p}\left({x}\right)=\mathrm{1}+{x}+{x}^{\mathrm{2}} \:+....+{x}^{{n}} \\ $$$${find}\:\langle{p},{p}\rangle. \\ $$
Question Number 36911 Answers: 0 Comments: 1
$${p}\:{is}\:{a}\:{polynome}\:{having}\:{nroots}\:{simples} \\ $$$${x}_{{i}} \:\left(\mathrm{1}\leqslant{x}_{{i}} \leqslant{n}\:\right)\:{with}\:{x}_{{i}} ^{\mathrm{2}} \:\neq\mathrm{1}\:\:{calculste} \\ $$$$\sum_{{k}=\mathrm{1}} ^{{n}} \:\:\frac{\mathrm{1}}{\mathrm{1}−{x}_{{k}} }\:. \\ $$
Question Number 36910 Answers: 0 Comments: 0
$$\left.\mathrm{1}\right)\:{decompose}\:{inside}\:{R}\left({x}\right)\:{the}\:{fraction} \\ $$$${F}\left({x}\right)=\:\:\frac{\mathrm{1}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left(\mathrm{1}−{x}^{\mathrm{3}} \right)} \\ $$$$\left.\mathrm{2}\right)\:{find}\:\int\:{F}\left({x}\right){dx}\:. \\ $$
Question Number 36909 Answers: 0 Comments: 0
$${let}\:{p}\left({x}\right)={x}^{\mathrm{3}} \:−\mathrm{2}{x}^{\mathrm{2}} \:−\mathrm{1}\:{and}\:\alpha\:{is}\:{root}\:{of}\:{p}\left({x}\right) \\ $$$${prove}\:{that}\:\alpha\notin\:{Q}\:. \\ $$
Question Number 36908 Answers: 0 Comments: 0
$${calculate}\:{S}_{{n}} =\:\sum_{{p}=\mathrm{1}} ^{{n}} \:\:\frac{{p}}{\mathrm{1}+{p}\:+{p}^{\mathrm{2}} } \\ $$$$\left.\mathrm{2}\right)\:{find}\:{lim}_{{n}\rightarrow+} \:{S}_{{n}} \:\:. \\ $$
Question Number 36907 Answers: 0 Comments: 0
$${let}\:\:{f}\left({x}\right)=\:\:\frac{\mathrm{1}}{{cosx}}\:\:{find}\:{f}^{\left({n}\right)} \left({x}\right) \\ $$
Question Number 36969 Answers: 1 Comments: 0
$$\left[\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{2}.\mathrm{2}^{\mathrm{3}} .\mathrm{2}^{\mathrm{5}} .....\mathrm{2}^{\mathrm{n}−\mathrm{1}} .\mathrm{3}^{\mathrm{2}} .\mathrm{3}^{\mathrm{4}} .....\mathrm{3}^{\mathrm{n}} \right)^{\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} +\mathrm{1}}} \right]^{\mathrm{4}} =? \\ $$
Question Number 36905 Answers: 0 Comments: 0
$${p}\:{is}\:{apolynom}\:{with}\:{n}\:{roots}\:{differents} \\ $$$${let}\:{Q}\:=\:{p}^{\mathrm{2}} \:+{p}^{'} \:\:\:\:{let}\:\alpha\:{the}\:{number}\:{of}\:{roots}\:{of} \\ $$$${Q}\:{prove}\:{that}\:\:\:{n}−\mathrm{1}\leqslant\alpha\leqslant{n}+\mathrm{1}\:. \\ $$
Question Number 36904 Answers: 0 Comments: 1
$$\left.\mathrm{1}\right){decompose}\:{inside}\:{C}\left[{x}\right] \\ $$$${p}\left({x}\right)={x}^{\mathrm{2}{n}} \:−\mathrm{2}\left({cos}\alpha\right){x}^{{n}} \:+\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:{decopose}\:{p}\left({x}\right){inside}\:{R}\left[{x}\right] \\ $$
Question Number 36903 Answers: 0 Comments: 0
$${prove}\:{that}\:\:\mathrm{2}^{{n}+\mathrm{1}} \:{divide}\:\left[\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}{n}+\mathrm{1}} \right]\: \\ $$$$\left[{x}\right]\:{mean}\:{integr}\:{part}\:{of}\:{x} \\ $$
Question Number 36892 Answers: 1 Comments: 1
$$\mathrm{2}.\:\int\left[\sqrt{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)/\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\right]{dx}=? \\ $$
Question Number 36886 Answers: 0 Comments: 1
Question Number 36884 Answers: 0 Comments: 0
$$\mathrm{1}.\:\mathrm{What}\:\mathrm{will}\:\mathrm{be}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{cueved}\:\mathrm{line}\:\mathrm{which}\:\mathrm{is}\:\mathrm{made}\:\mathrm{by}\:\mathrm{a} \\ $$$$\mathrm{fixed}\:\mathrm{point}\:\mathrm{in}\:\mathrm{the}\:\mathrm{boundary}\:\mathrm{of}\:\mathrm{a} \\ $$$$\mathrm{moving}\:\mathrm{circular}\:\mathrm{object}\:\mathrm{in}\:\mathrm{respect}\:\mathrm{to} \\ $$$$\mathrm{an}\:\mathrm{another}\:\mathrm{fiexd}\:\mathrm{point}\:\mathrm{on}\:\mathrm{the}\:\mathrm{way}\:\mathrm{of} \\ $$$$\mathrm{moving}? \\ $$
Question Number 36880 Answers: 1 Comments: 3
Question Number 36877 Answers: 0 Comments: 0
Question Number 36876 Answers: 0 Comments: 0
Question Number 36874 Answers: 0 Comments: 0
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