Question and Answers Forum
All Questions Topic List
AlgebraQuestion and Answers: Page 341
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 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 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 30571 Answers: 0 Comments: 0
$${decompose}\:{F}\left({x}\right)=\:\frac{\mathrm{1}}{\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} \right)^{{n}} }\:{on}\:{C}\left[{x}\right]. \\ $$
Question Number 30541 Answers: 0 Comments: 0
$$\mathrm{4}{n}\mathrm{568} \\ $$
Question Number 30540 Answers: 0 Comments: 0
$$\mathrm{Show}\:\mathrm{that}\:\:\begin{vmatrix}{\mathrm{bc}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{ca}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{ab}}\\{\:\mathrm{a}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{b}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{c}}\\{\:\mathrm{a}^{\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:} \:\mathrm{b}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{c}^{\mathrm{2}} }\end{vmatrix}\:\:=\:\:\left(\mathrm{b}\:−\:\mathrm{a}\right)\left(\mathrm{c}\:−\:\mathrm{a}\right)\left(\mathrm{c}\:−\:\mathrm{b}\right)\left(\mathrm{ab}\:+\:\mathrm{bc}\:+\:\mathrm{ac}\right) \\ $$
Question Number 30529 Answers: 0 Comments: 0
$${let}\:{f}\left({x}\right)={e}^{−{x}^{\mathrm{2}} } \:\:{prove}\:{that}\:{f}^{\left({n}\right)} \:{is}\:{at}\:{form} \\ $$$${f}^{\left({n}\right)} =\:{p}_{{n}\:} \:{e}^{−{x}^{\mathrm{2}} } \:\:{find}\:{relation}\:{between}\:{p}_{{n}} {and}\:{p}_{{n}+\mathrm{1}\:} . \\ $$$$\left.\mathrm{2}\right)\:{find}\:{p}_{\mathrm{0}} \:,{p}_{\mathrm{1}} ,\:{p}_{\mathrm{2}} ,{p}_{\mathrm{3}} \\ $$
Question Number 30523 Answers: 0 Comments: 0
$$\:\left(\alpha_{{k}} \right)_{\mathrm{0}\leqslant{k}\leqslant{n}−\mathrm{1}} {are}\:{roots}\:{of}\:\:{x}^{{n}} −\mathrm{1}\:\:{simplify} \\ $$$$\prod_{{n}} =\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\left({x}+\alpha_{{k}} {y}\right)\:. \\ $$
Question Number 30522 Answers: 1 Comments: 1
$${let}\:{p}\left({x}\right)=\:\left(\mathrm{1}+{ix}\right)^{{n}} \:−\left(\mathrm{1}−{ix}\right)^{{n}} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{the}\:{roots}\:{of}\:{p}\left({x}\right)\:{and}\:{factorize}\:{p}\left({x}\right) \\ $$$$\left.\right)\:{give}\:{p}\left({x}\right)\:{at}\:{form}\:{of}\:{arcs}. \\ $$$$ \\ $$
Question Number 30513 Answers: 0 Comments: 1
Question Number 30501 Answers: 0 Comments: 0
$${let}\:{put}\:{w}={e}^{{i}\frac{\mathrm{2}\pi}{{n}}} \:\:\:\:\:{find}\:\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\left({x}+{w}^{{k}} \right)^{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:\:\sum_{{k}=\mathrm{1}} ^{{n}} {n}\left({x}+{w}^{{k}} \right)^{{n}−\mathrm{1}} \:\:. \\ $$$$ \\ $$
Question Number 30456 Answers: 1 Comments: 0
$${proof}\:{that} \\ $$$$\frac{{a}^{\mathrm{2}} }{\left({a}−{b}\right)\left({a}−{c}\right)}+\frac{{b}^{\mathrm{2}} }{\left({b}−{c}\right)\left({b}−{a}\right)}+\frac{{c}^{\mathrm{2}} }{\left({c}−{a}\right)\left({c}−{b}\right)}=\:{a}+{b}+{c} \\ $$
Question Number 30425 Answers: 1 Comments: 0
$${decompose}\:{inside}\:{R}\left[{x}\right]\: \\ $$$${F}\left({x}\right)=\:\:\:\frac{{x}^{\mathrm{2}{n}} }{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{{n}} }\:\:\:{with}\:{n}\:{from}\:{N}\:{and}\:{n}>\mathrm{0}. \\ $$
Question Number 30405 Answers: 1 Comments: 0
$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{13} \\ $$$${x}^{\mathrm{2}} −\mathrm{3}{xy}+{y}^{\mathrm{2}} =\mathrm{35} \\ $$$${find}\:{the}\:{value}\:{of}\:{x}\:{and}\:{y} \\ $$
Question Number 30401 Answers: 0 Comments: 0
$${is}\:{there}\:{exists}\:{a}\:{onto}\:{group}\:{homo}\:{from}\:{D}\mathrm{4}\:{to}\:{Z}\mathrm{4}? \\ $$
Question Number 30267 Answers: 0 Comments: 7
$${Can}\:{We}\:{expand}\:{the}\:{following} \\ $$$${expression}? \\ $$$$\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+\mathrm{2}{x}\right)\left(\mathrm{1}+\mathrm{3}{x}\right)......\left(\mathrm{1}+{nx}\right) \\ $$$${or}\:{is}\:{there}\:{any}\:{formula}\:{for}\:{this}? \\ $$
Question Number 30220 Answers: 0 Comments: 1
$${let}\:{p}\left({x}\right)=\:{x}^{\mathrm{3}} \:{px}\:+{q} \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:{p}\:{have}\:{double}\:{roots}\Leftrightarrow\:\mathrm{4}{p}^{\mathrm{3}} \:+\mathrm{27}{q}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left.\mathrm{3}\right)\:{let}\:{suppose}\:{p}\:{have}\:\mathrm{3}\:{real}\:{roots}\:{differnts}\:{prove}\:{that} \\ $$$$\mathrm{4}{p}^{\mathrm{3}} \:+\mathrm{27}{q}^{\mathrm{2}} \:<\mathrm{0}. \\ $$
Question Number 30218 Answers: 0 Comments: 0
$${prove}\:{that}\:{D}\left({x}^{\mathrm{5}} −\mathrm{1},{x}^{\mathrm{2}} +{x}+\mathrm{1}\right)=\mathrm{1}. \\ $$
Question Number 30119 Answers: 2 Comments: 1
$$\mathrm{Let}\:\:\mathrm{N}=\:\mathrm{2}^{\mathrm{1224}} \:−\mathrm{1}. \\ $$$$\mathrm{S}=\:\mathrm{2}^{\mathrm{153}} +\mathrm{2}^{\mathrm{77}} +\mathrm{1}. \\ $$$$\mathrm{T}=\:\mathrm{2}^{\mathrm{408}} −\mathrm{2}^{\mathrm{204}} +\mathrm{1}. \\ $$$$\mathrm{then}\:\mathrm{which}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{statment}\:\mathrm{is} \\ $$$$\:\mathrm{correct}? \\ $$$$\left.\mathrm{a}\right)\:\mathrm{S}\:\mathrm{and}\:\mathrm{T}\:\mathrm{both}\:\mathrm{divide}\:\mathrm{N}. \\ $$$$\left.\mathrm{b}\right)\:\mathrm{only}\:\mathrm{S}\:\mathrm{divides}\:\mathrm{N}. \\ $$$$\left.\mathrm{c}\right)\:\mathrm{only}\:\mathrm{T}\:\mathrm{divides}\:\mathrm{N}. \\ $$$$\left.\mathrm{d}\right)\:\mathrm{Neither}\:\mathrm{S}\:\mathrm{nor}\:\mathrm{T}\:\mathrm{divides}\:\mathrm{N}. \\ $$
Question Number 30092 Answers: 1 Comments: 0
$${If}\:{x}+\frac{\mathrm{1}}{{x}}=\mathrm{3}\:{find}\:{x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} } \\ $$
Pg 336 Pg 337 Pg 338 Pg 339 Pg 340 Pg 341 Pg 342 Pg 343 Pg 344 Pg 345
Terms of Service
Privacy Policy
Contact: info@tinkutara.com