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Question Number 30600 Answers: 0 Comments: 0
$${let}\:{w}_{{k}} ={e}^{{i}\frac{\mathrm{2}{k}\pi}{{n}}} \:\:\:\:{find}\:{A}=\:\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left({a}\:+{bw}_{{k}} \:\right). \\ $$
Question Number 30599 Answers: 0 Comments: 1
$${decompose}\:{inside}\:{C}\left({x}\right)\:\:{F}=\:\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)\left({x}^{{n}} \:−\mathrm{1}\right)}\:. \\ $$
Question Number 30598 Answers: 0 Comments: 1
$${prove}\:{that}\:{it}\:{exist}\:{one}\:{polynomial}\:{p}/ \\ $$$${p}\left({cosx}\right)={cos}\left({nx}\right)\:{find}\:{the}\:{roots}\:{of}\:{p}\left({x}\right)\:. \\ $$
Question Number 30597 Answers: 0 Comments: 0
$${let}\:{p}\left({x}\right)=\left(\mathrm{1}+{x}\right)^{{m}} \:−{e}^{\mathrm{2}{imx}} \left(\mathrm{1}−{x}\right)^{{m}} \:{factorize}\:{p}\left({x}\right) \\ $$$${inside}\:{C}\left[{x}\right]. \\ $$
Question Number 30596 Answers: 0 Comments: 0
$${find}\:{all}\:{polynomial}\:{wich}\:{verify}\: \\ $$$${p}\left({x}^{\mathrm{2}} \right)\:+{p}\left({x}\right){p}\left({x}+\mathrm{1}\right)=\mathrm{0}. \\ $$
Question Number 30595 Answers: 0 Comments: 1
$${let}\:{f}\left({x}\right)=\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} \:−\mathrm{2}{cos}\alpha{x}+\mathrm{1}}\:\:{find}\:{f}^{\left({n}\right)} . \\ $$
Question Number 30594 Answers: 0 Comments: 0
$${let}\:{p}\left({x}\right)={x}^{\mathrm{3}} \:+\mathrm{1}\:{and}\:{q}\left({x}\right)={x}^{\mathrm{4}} \:+\mathrm{1}\:{prove}\:{that} \\ $$$${D}\left({p},{q}\right)=\mathrm{1}. \\ $$
Question Number 30593 Answers: 1 Comments: 0
$${factorize}\:{inside}\:{C}\left[{x}\right]\:{p}\left({x}\right)=\left(\mathrm{1}+{i}\frac{{x}}{{n}}\right)^{{n}} \:−\left(\mathrm{1}−{i}\frac{{x}}{{n}}\right)^{{n}} . \\ $$
Question Number 30592 Answers: 1 Comments: 0
$${let}\:{p}\left({x}\right)={x}^{\mathrm{2}{n}} \:−\mathrm{2}{cos}\alpha\:{x}^{{n}} \:+\mathrm{1} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{roots}\:{lf}\:{p}\left({x}\right) \\ $$$$\left.\mathrm{2}\right){factorize}\:{p}\left({x}\right)\:{inside}\:{C}\left[{x}\right] \\ $$$$\left.\mathrm{3}\right){factorize}\:{p}\left({x}\right)\:{inside}\:{R}\left[{x}\right]. \\ $$
Question Number 30590 Answers: 1 Comments: 0
$${decompose}\:{sur}\:{R}\left[{x}\right]\:\:{x}^{\mathrm{2}{n}+\mathrm{1}} \:−\mathrm{1}. \\ $$
Question Number 30589 Answers: 0 Comments: 0
$${let}\:{U}_{{n}} =\:\left\{{z}\in{C}\:/\:{z}^{{n}} =\mathrm{1}\right\}\:\:{find} \\ $$$${S}=\:\sum_{{z}\in{U}_{{n}} } \:\:\frac{{z}}{\left({x}−{z}\right)^{\mathrm{2}} }\:\:. \\ $$
Question Number 30588 Answers: 0 Comments: 0
$$\left({n}_{{k}} \right)_{\mathrm{1}\leqslant{k}\leqslant{n}} \:{is}\:{a}\:{family}\:{of}\:{integrs}\:{numbers}\:{let}\:{put} \\ $$$${p}\left({x}\right)=\sum_{{k}=\mathrm{1}} ^{{n}} \:{x}^{{n}_{{k}} } \:\:\:{and}\:{q}\left({x}\right)=\:\sum_{{j}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{x}^{{j}} \: \\ $$$${if}\:{n}_{{k}} \equiv{k}−\mathrm{1}\left[{n}\right]\:{prove}\:{that}\:{q}\:{divide}\:{p}. \\ $$
Question Number 30587 Answers: 0 Comments: 0
$${find}\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\left(−\mathrm{1}\right)^{{k}} \:{cos}^{{n}} \left(\frac{{k}\pi}{{n}}\right). \\ $$
Question Number 30586 Answers: 0 Comments: 0
$${let}\:{p}=\mathrm{1}+{x}+{x}^{\mathrm{2}} \:+....+{x}^{\mathrm{2}^{{n}+\mathrm{1}} −\mathrm{1}} \:{and}\:\:{q}=\:\mathrm{1}+{x}^{\mathrm{2}^{{n}} } \\ $$$${find}\:\alpha=\:\frac{{p}}{{q}}\:. \\ $$
Question Number 30585 Answers: 0 Comments: 0
$${find}\:\:{F}_{{n}} \left({x}\right)=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{x}^{{n}} }{{e}^{{x}+{n}} \:+\mathrm{1}}{dx}\:. \\ $$
Question Number 30584 Answers: 0 Comments: 0
$${find}\:\:{I}=\:\int_{−\infty} ^{+\infty} \:\:\:\:\:\frac{{e}^{−{x}^{\mathrm{2}} } }{{a}^{\mathrm{2}} \:+\left({v}−{x}\right)^{\mathrm{2}} }{dx}. \\ $$
Question Number 30583 Answers: 0 Comments: 0
$${decompose}\:{F}\:=\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{{n}} }\:{inside}\:{C}\left[{x}\right].{n}\:{from}\:{N}. \\ $$
Question Number 30582 Answers: 0 Comments: 0
$${x}_{\mathrm{1}} ,\:{x}_{\mathrm{2}} ,\:{x}_{\mathrm{3}\:} \:{are}\:{roots}\:{of}\:{the}\:{polynomial}\:{x}^{\mathrm{3}} \:−{x}+\mathrm{1}\:{find} \\ $$$${the}\:{polynomial}\:{wich}\:{have}\:{for}\:{roots}\:{x}_{\mathrm{1}} ^{\mathrm{3}} \:,{x}_{\mathrm{2}} ^{\mathrm{3}} \:{and}\:{x}_{\mathrm{3}} ^{\mathrm{3}} \:\:. \\ $$
Question Number 30581 Answers: 0 Comments: 0
$${decompose}\:{inside}\:{C}\left[{x}\right]\:{F}=\:\:\frac{\mathrm{1}}{\left({x}+{iy}\right)^{{n}} }\:. \\ $$
Question Number 30580 Answers: 0 Comments: 1
$${decompose}\:{inside}\:{C}\left[{x}\right]\:{F}=\:\frac{{x}^{{n}} }{{x}^{{m}} \:+\mathrm{1}}\:{with}\:{m}\geqslant{n}+\mathrm{2} \\ $$$${then}\:{find}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{{n}} }{{x}^{{m}} \:+\mathrm{1}}{dx}. \\ $$
Question Number 30579 Answers: 0 Comments: 0
$${decompose}\:{inside}\:{C}\left[{x}\right]\:{F}=\:\frac{{x}^{{n}} −\mathrm{1}}{{x}^{\mathrm{2}{n}} −\mathrm{1}}\:. \\ $$
Question Number 30578 Answers: 0 Comments: 0
$${decompose}\:{F}\left({x}\right)=\:\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{{n}} }\:{on}\:{C}\left[{x}\right].{with}\:{n}\:{fromN}. \\ $$
Question Number 30576 Answers: 0 Comments: 1
$${let}\:{consider}\:{the}\:{equation}\:{x}^{\mathrm{3}} \:+{px}\:+{q}\: \\ $$$${find}\:{S}=\:\sum_{{i}\neq{j}} \:\frac{{x}_{{i}} }{{x}_{{j}} }\:. \\ $$
Question Number 30575 Answers: 0 Comments: 0
$${find}\:\int\int_{{D}} \:\left({x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \right){dxdy}\:\:{with} \\ $$$${D}=\left\{\left({x},{y}\right)/\:{x}\leqslant\mathrm{1}\:{and}\:{x}^{\mathrm{2}} \leqslant{y}\leqslant\mathrm{2}\:\right\}. \\ $$
Question Number 30574 Answers: 0 Comments: 0
$${find}\:\int\int_{\left[\mathrm{1},{e}\right]^{\mathrm{2}} } \:\:\:{ln}\left({xy}\right){dxdy}. \\ $$
Question Number 30573 Answers: 0 Comments: 0
$${find}\:\int\int_{\left[\mathrm{0},\mathrm{1}\right]×\left[\mathrm{0},\mathrm{1}\right]} \:\:\:\:\frac{{x}^{\mathrm{2}} }{\mathrm{1}+{y}^{\mathrm{2}} }{dxdy}. \\ $$
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