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AlgebraQuestion and Answers: Page 341

Question Number 30588    Answers: 0   Comments: 0

(n_k )_(1≤k≤n) is a family of integrs numbers let put p(x)=Σ_(k=1) ^n x^n_k and q(x)= Σ_(j=0) ^(n−1) x^j if n_k ≡k−1[n] prove that q divide p.

$$\left({n}_{{k}} \right)_{\mathrm{1}\leqslant{k}\leqslant{n}} \:{is}\:{a}\:{family}\:{of}\:{integrs}\:{numbers}\:{let}\:{put} \\ $$$${p}\left({x}\right)=\sum_{{k}=\mathrm{1}} ^{{n}} \:{x}^{{n}_{{k}} } \:\:\:{and}\:{q}\left({x}\right)=\:\sum_{{j}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{x}^{{j}} \: \\ $$$${if}\:{n}_{{k}} \equiv{k}−\mathrm{1}\left[{n}\right]\:{prove}\:{that}\:{q}\:{divide}\:{p}. \\ $$

Question Number 30587    Answers: 0   Comments: 0

find Σ_(k=0) ^(n−1) (−1)^k cos^n (((kπ)/n)).

$${find}\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\left(−\mathrm{1}\right)^{{k}} \:{cos}^{{n}} \left(\frac{{k}\pi}{{n}}\right). \\ $$

Question Number 30586    Answers: 0   Comments: 0

let p=1+x+x^2 +....+x^(2^(n+1) −1) and q= 1+x^2^n find α= (p/q) .

$${let}\:{p}=\mathrm{1}+{x}+{x}^{\mathrm{2}} \:+....+{x}^{\mathrm{2}^{{n}+\mathrm{1}} −\mathrm{1}} \:{and}\:\:{q}=\:\mathrm{1}+{x}^{\mathrm{2}^{{n}} } \\ $$$${find}\:\alpha=\:\frac{{p}}{{q}}\:. \\ $$

Question Number 30583    Answers: 0   Comments: 0

decompose F =(1/((x^2 −1)^n )) inside C[x].n from N.

$${decompose}\:{F}\:=\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{{n}} }\:{inside}\:{C}\left[{x}\right].{n}\:{from}\:{N}. \\ $$

Question Number 30582    Answers: 0   Comments: 0

x_1 , x_2 , x_(3 ) are roots of the polynomial x^3 −x+1 find the polynomial wich have for roots x_1 ^3 ,x_2 ^3 and x_3 ^3 .

$${x}_{\mathrm{1}} ,\:{x}_{\mathrm{2}} ,\:{x}_{\mathrm{3}\:} \:{are}\:{roots}\:{of}\:{the}\:{polynomial}\:{x}^{\mathrm{3}} \:−{x}+\mathrm{1}\:{find} \\ $$$${the}\:{polynomial}\:{wich}\:{have}\:{for}\:{roots}\:{x}_{\mathrm{1}} ^{\mathrm{3}} \:,{x}_{\mathrm{2}} ^{\mathrm{3}} \:{and}\:{x}_{\mathrm{3}} ^{\mathrm{3}} \:\:. \\ $$

Question Number 30581    Answers: 0   Comments: 0

decompose inside C[x] F= (1/((x+iy)^n )) .

$${decompose}\:{inside}\:{C}\left[{x}\right]\:{F}=\:\:\frac{\mathrm{1}}{\left({x}+{iy}\right)^{{n}} }\:. \\ $$

Question Number 30579    Answers: 0   Comments: 0

decompose inside C[x] F= ((x^n −1)/(x^(2n) −1)) .

$${decompose}\:{inside}\:{C}\left[{x}\right]\:{F}=\:\frac{{x}^{{n}} −\mathrm{1}}{{x}^{\mathrm{2}{n}} −\mathrm{1}}\:. \\ $$

Question Number 30578    Answers: 0   Comments: 0

decompose F(x)= (1/((x^2 +1)^n )) on C[x].with n fromN.

$${decompose}\:{F}\left({x}\right)=\:\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{{n}} }\:{on}\:{C}\left[{x}\right].{with}\:{n}\:{fromN}. \\ $$

Question Number 30576    Answers: 0   Comments: 1

let consider the equation x^3 +px +q find S= Σ_(i≠j) (x_i /x_j ) .

$${let}\:{consider}\:{the}\:{equation}\:{x}^{\mathrm{3}} \:+{px}\:+{q}\: \\ $$$${find}\:{S}=\:\sum_{{i}\neq{j}} \:\frac{{x}_{{i}} }{{x}_{{j}} }\:. \\ $$

Question Number 30571    Answers: 0   Comments: 0

decompose F(x)= (1/((1+x+x^2 )^n )) on C[x].

$${decompose}\:{F}\left({x}\right)=\:\frac{\mathrm{1}}{\left(\mathrm{1}+{x}+{x}^{\mathrm{2}} \right)^{{n}} }\:{on}\:{C}\left[{x}\right]. \\ $$

Question Number 30541    Answers: 0   Comments: 0

4n568

$$\mathrm{4}{n}\mathrm{568} \\ $$

Question Number 30540    Answers: 0   Comments: 0

Show that determinant (((bc ca ab)),(( a b c)),(( a^(2 ) b^2 c^2 ))) = (b − a)(c − a)(c − b)(ab + bc + ac)

$$\mathrm{Show}\:\mathrm{that}\:\:\begin{vmatrix}{\mathrm{bc}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{ca}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{ab}}\\{\:\mathrm{a}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{b}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{c}}\\{\:\mathrm{a}^{\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:} \:\mathrm{b}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{c}^{\mathrm{2}} }\end{vmatrix}\:\:=\:\:\left(\mathrm{b}\:−\:\mathrm{a}\right)\left(\mathrm{c}\:−\:\mathrm{a}\right)\left(\mathrm{c}\:−\:\mathrm{b}\right)\left(\mathrm{ab}\:+\:\mathrm{bc}\:+\:\mathrm{ac}\right) \\ $$

Question Number 30529    Answers: 0   Comments: 0

let f(x)=e^(−x^2 ) prove that f^((n)) is at form f^((n)) = p_(n ) e^(−x^2 ) find relation between p_n and p_(n+1 ) . 2) find p_0 ,p_1 , p_2 ,p_3

$${let}\:{f}\left({x}\right)={e}^{−{x}^{\mathrm{2}} } \:\:{prove}\:{that}\:{f}^{\left({n}\right)} \:{is}\:{at}\:{form} \\ $$$${f}^{\left({n}\right)} =\:{p}_{{n}\:} \:{e}^{−{x}^{\mathrm{2}} } \:\:{find}\:{relation}\:{between}\:{p}_{{n}} {and}\:{p}_{{n}+\mathrm{1}\:} . \\ $$$$\left.\mathrm{2}\right)\:{find}\:{p}_{\mathrm{0}} \:,{p}_{\mathrm{1}} ,\:{p}_{\mathrm{2}} ,{p}_{\mathrm{3}} \\ $$

Question Number 30523    Answers: 0   Comments: 0

(α_k )_(0≤k≤n−1) are roots of x^n −1 simplify Π_n =Π_(k=0) ^(n−1) (x+α_k y) .

$$\:\left(\alpha_{{k}} \right)_{\mathrm{0}\leqslant{k}\leqslant{n}−\mathrm{1}} {are}\:{roots}\:{of}\:\:{x}^{{n}} −\mathrm{1}\:\:{simplify} \\ $$$$\prod_{{n}} =\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\left({x}+\alpha_{{k}} {y}\right)\:. \\ $$

Question Number 30522    Answers: 1   Comments: 1

let p(x)= (1+ix)^n −(1−ix)^n 1) find the roots of p(x) and factorize p(x) ) give p(x) at form of arcs.

$${let}\:{p}\left({x}\right)=\:\left(\mathrm{1}+{ix}\right)^{{n}} \:−\left(\mathrm{1}−{ix}\right)^{{n}} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{the}\:{roots}\:{of}\:{p}\left({x}\right)\:{and}\:{factorize}\:{p}\left({x}\right) \\ $$$$\left.\right)\:{give}\:{p}\left({x}\right)\:{at}\:{form}\:{of}\:{arcs}. \\ $$$$ \\ $$

Question Number 30513    Answers: 0   Comments: 1

Question Number 30501    Answers: 0   Comments: 0

let put w=e^(i((2π)/n)) find Σ_(k=1) ^n (x+w^k )^n 2) find Σ_(k=1) ^n n(x+w^k )^(n−1) .

$${let}\:{put}\:{w}={e}^{{i}\frac{\mathrm{2}\pi}{{n}}} \:\:\:\:\:{find}\:\:\sum_{{k}=\mathrm{1}} ^{{n}} \:\left({x}+{w}^{{k}} \right)^{{n}} \\ $$$$\left.\mathrm{2}\right)\:{find}\:\:\sum_{{k}=\mathrm{1}} ^{{n}} {n}\left({x}+{w}^{{k}} \right)^{{n}−\mathrm{1}} \:\:. \\ $$$$ \\ $$

Question Number 30456    Answers: 1   Comments: 0

proof that (a^2 /((a−b)(a−c)))+(b^2 /((b−c)(b−a)))+(c^2 /((c−a)(c−b)))= a+b+c

$${proof}\:{that} \\ $$$$\frac{{a}^{\mathrm{2}} }{\left({a}−{b}\right)\left({a}−{c}\right)}+\frac{{b}^{\mathrm{2}} }{\left({b}−{c}\right)\left({b}−{a}\right)}+\frac{{c}^{\mathrm{2}} }{\left({c}−{a}\right)\left({c}−{b}\right)}=\:{a}+{b}+{c} \\ $$

Question Number 30425    Answers: 1   Comments: 0

decompose inside R[x] F(x)= (x^(2n) /((x^2 +1)^n )) with n from N and n>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^2 +y^2 =13 x^2 −3xy+y^2 =35 find the value of x and y

$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{13} \\ $$$${x}^{\mathrm{2}} −\mathrm{3}{xy}+{y}^{\mathrm{2}} =\mathrm{35} \\ $$$${find}\:{the}\:{value}\:{of}\:{x}\:{and}\:{y} \\ $$

Question Number 30401    Answers: 0   Comments: 0

is there exists a onto group homo from D4 to Z4?

$${is}\:{there}\:{exists}\:{a}\:{onto}\:{group}\:{homo}\:{from}\:{D}\mathrm{4}\:{to}\:{Z}\mathrm{4}? \\ $$

Question Number 30267    Answers: 0   Comments: 7

Can We expand the following expression? (1+x)(1+2x)(1+3x)......(1+nx) or is there any formula for this?

$${Can}\:{We}\:{expand}\:{the}\:{following} \\ $$$${expression}? \\ $$$$\left(\mathrm{1}+{x}\right)\left(\mathrm{1}+\mathrm{2}{x}\right)\left(\mathrm{1}+\mathrm{3}{x}\right)......\left(\mathrm{1}+{nx}\right) \\ $$$${or}\:{is}\:{there}\:{any}\:{formula}\:{for}\:{this}? \\ $$

Question Number 30220    Answers: 0   Comments: 1

let p(x)= x^3 px +q 1) prove that p have double roots⇔ 4p^3 +27q^2 =0 3) let suppose p have 3 real roots differnts prove that 4p^3 +27q^2 <0.

$${let}\:{p}\left({x}\right)=\:{x}^{\mathrm{3}} \:{px}\:+{q} \\ $$$$\left.\mathrm{1}\right)\:{prove}\:{that}\:{p}\:{have}\:{double}\:{roots}\Leftrightarrow\:\mathrm{4}{p}^{\mathrm{3}} \:+\mathrm{27}{q}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left.\mathrm{3}\right)\:{let}\:{suppose}\:{p}\:{have}\:\mathrm{3}\:{real}\:{roots}\:{differnts}\:{prove}\:{that} \\ $$$$\mathrm{4}{p}^{\mathrm{3}} \:+\mathrm{27}{q}^{\mathrm{2}} \:<\mathrm{0}. \\ $$

Question Number 30218    Answers: 0   Comments: 0

prove that D(x^5 −1,x^2 +x+1)=1.

$${prove}\:{that}\:{D}\left({x}^{\mathrm{5}} −\mathrm{1},{x}^{\mathrm{2}} +{x}+\mathrm{1}\right)=\mathrm{1}. \\ $$

Question Number 30119    Answers: 2   Comments: 1

Let N= 2^(1224) −1. S= 2^(153) +2^(77) +1. T= 2^(408) −2^(204) +1. then which of the following statment is correct? a) S and T both divide N. b) only S divides N. c) only T divides N. d) Neither S nor T divides N.

$$\mathrm{Let}\:\:\mathrm{N}=\:\mathrm{2}^{\mathrm{1224}} \:−\mathrm{1}. \\ $$$$\mathrm{S}=\:\mathrm{2}^{\mathrm{153}} +\mathrm{2}^{\mathrm{77}} +\mathrm{1}. \\ $$$$\mathrm{T}=\:\mathrm{2}^{\mathrm{408}} −\mathrm{2}^{\mathrm{204}} +\mathrm{1}. \\ $$$$\mathrm{then}\:\mathrm{which}\:\mathrm{of}\:\mathrm{the}\:\mathrm{following}\:\mathrm{statment}\:\mathrm{is} \\ $$$$\:\mathrm{correct}? \\ $$$$\left.\mathrm{a}\right)\:\mathrm{S}\:\mathrm{and}\:\mathrm{T}\:\mathrm{both}\:\mathrm{divide}\:\mathrm{N}. \\ $$$$\left.\mathrm{b}\right)\:\mathrm{only}\:\mathrm{S}\:\mathrm{divides}\:\mathrm{N}. \\ $$$$\left.\mathrm{c}\right)\:\mathrm{only}\:\mathrm{T}\:\mathrm{divides}\:\mathrm{N}. \\ $$$$\left.\mathrm{d}\right)\:\mathrm{Neither}\:\mathrm{S}\:\mathrm{nor}\:\mathrm{T}\:\mathrm{divides}\:\mathrm{N}. \\ $$

Question Number 30092    Answers: 1   Comments: 0

If x+(1/x)=3 find x^5 +(1/x^5 )

$${If}\:{x}+\frac{\mathrm{1}}{{x}}=\mathrm{3}\:{find}\:{x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} } \\ $$

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