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AlgebraQuestion and Answers: Page 341

Question Number 31035 Answers: 0 Comments: 0

1) solve x^5 =1 2) if x_i are roots of this equation prove that Σ_i x_i =0 3)prove that cos(((2π)/5)) +cos(((4π)/5)) =−(1/2) 4)calculate cos(((4π)/5)) interms cos(((2π)/5)) then find its values.

$$\left.\mathrm{1}\right)\:{solve}\:{x}^{\mathrm{5}} =\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:{if}\:{x}_{{i}} \:{are}\:{roots}\:{of}\:{this}\:{equation}\:{prove}\:{that}\:\sum_{{i}} \:{x}_{{i}} =\mathrm{0} \\ $$$$\left.\mathrm{3}\right){prove}\:{that}\:{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)\:+{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right)\:=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left.\mathrm{4}\right){calculate}\:{cos}\left(\frac{\mathrm{4}\pi}{\mathrm{5}}\right)\:{interms}\:{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)\:{then}\:{find}\:{its} \\ $$$${values}. \\ $$

Question Number 31034 Answers: 0 Comments: 0

1) solve inside C z^4 =6 2)sove inside C (((z+i)/(z−i)))^3 +(((z+i)/(z−i)))^2 +(((z+i)/(z−i))) +1=0

$$\left.\mathrm{1}\right)\:{solve}\:{inside}\:{C}\:\:{z}^{\mathrm{4}} =\mathrm{6} \\ $$$$\left.\mathrm{2}\right){sove}\:{inside}\:{C}\:\left(\frac{{z}+{i}}{{z}−{i}}\right)^{\mathrm{3}} \:+\left(\frac{{z}+{i}}{{z}−{i}}\right)^{\mathrm{2}} \:+\left(\frac{{z}+{i}}{{z}−{i}}\right)\:+\mathrm{1}=\mathrm{0} \\ $$

Question Number 31033 Answers: 0 Comments: 0

factorize p(x)=(1+ix)^n −e^(inθ) wth θ∈R.

$${factorize}\:{p}\left({x}\right)=\left(\mathrm{1}+{ix}\right)^{{n}} \:−{e}^{{in}\theta} \:\:\:{wth}\:\theta\in{R}. \\ $$

Question Number 31032 Answers: 0 Comments: 0

let give x_0 =0 ,y_0 =1 and {_(y_n =x_(n−1) +y_(n−1) ) ^(x_n =x_(n−1) −y_(n−1) ) for n≥1 let z_n =x_n +i y_n ∀n∈N 1)calculate z_0 ,z_1 and z_2 2)prove that ∀n∈N^ ,n≥1 z_n =(1+i)z_(n−1) find z_n then find the expression of x_n and y_n 3)let put S_n =z_0 +z_1 +....z_n s_n =x_0 +x_1 +...+x_n s_n ^′ =y_0 +y_1 +...+y_n find those sum interms of n.

$${let}\:{give}\:{x}_{\mathrm{0}} =\mathrm{0}\:,{y}_{\mathrm{0}} =\mathrm{1}\:{and}\:\left\{_{{y}_{{n}} ={x}_{{n}−\mathrm{1}} \:+{y}_{{n}−\mathrm{1}} } ^{{x}_{{n}} ={x}_{{n}−\mathrm{1}} \:−{y}_{{n}−\mathrm{1}} } \:\:\:\:\:\:{for}\:{n}\geqslant\mathrm{1}\:{let}\right. \\ $$$${z}_{{n}} ={x}_{{n}} \:+{i}\:{y}_{{n}} \:\:\:\:\:\forall{n}\in{N} \\ $$$$\left.\mathrm{1}\right){calculate}\:{z}_{\mathrm{0}} \:,{z}_{\mathrm{1}} \:{and}\:{z}_{\mathrm{2}} \\ $$$$\left.\mathrm{2}\right){prove}\:{that}\:\forall{n}\in{N}^{} ,{n}\geqslant\mathrm{1}\:\:{z}_{{n}} =\left(\mathrm{1}+{i}\right){z}_{{n}−\mathrm{1}} \:{find}\:{z}_{{n}} {then} \\ $$$${find}\:{the}\:{expression}\:{of}\:{x}_{{n}} \:{and}\:{y}_{{n}} \\ $$$$\left.\mathrm{3}\right){let}\:{put}\:{S}_{{n}} ={z}_{\mathrm{0}} \:+{z}_{\mathrm{1}} \:+....{z}_{{n}} \\ $$$${s}_{{n}} ={x}_{\mathrm{0}} \:+{x}_{\mathrm{1}} \:+...+{x}_{{n}} \\ $$$${s}_{{n}} ^{'} ={y}_{\mathrm{0}} \:+{y}_{\mathrm{1}} \:+...+{y}_{{n}} \:\:{find}\:{those}\:{sum}\:{interms}\:{of}\:{n}. \\ $$$$ \\ $$

Question Number 31031 Answers: 0 Comments: 0

1) solve inside C z^(12) =1 and give the solution at form r e^(iθ) 2)calculate 1+u +u^2 +... +u^n then find the solution of z∈C z^8 +z^4 +1=0

$$\left.\mathrm{1}\right)\:{solve}\:{inside}\:{C}\:\:{z}^{\mathrm{12}} =\mathrm{1}\:{and}\:{give}\:{the}\:{solution}\:{at}\:{form} \\ $$$${r}\:{e}^{{i}\theta} \\ $$$$\left.\mathrm{2}\right){calculate}\:\mathrm{1}+{u}\:+{u}^{\mathrm{2}} \:+...\:+{u}^{{n}} \:{then}\:{find}\:{the}\:{solution} \\ $$$${of}\:{z}\in{C}\:\:\:\:\:\:\:{z}^{\mathrm{8}} \:+{z}^{\mathrm{4}} \:+\mathrm{1}=\mathrm{0} \\ $$

Question Number 31023 Answers: 0 Comments: 0

Question Number 31018 Answers: 0 Comments: 1

Question Number 31003 Answers: 1 Comments: 0

Number of positive integers x for which f(x)=x^3 −8x^2 +20x−13 is a prime number are ?

$${Number}\:{of}\:{positive}\:{integers}\:{x}\:{for} \\ $$$${which}\:{f}\left({x}\right)={x}^{\mathrm{3}} −\mathrm{8}{x}^{\mathrm{2}} +\mathrm{20}{x}−\mathrm{13}\:{is}\:\:{a}\: \\ $$$${prime}\:{number}\:{are}\:? \\ $$

Question Number 30990 Answers: 1 Comments: 0

Question Number 30958 Answers: 1 Comments: 4

Question Number 30956 Answers: 1 Comments: 0

Question Number 30916 Answers: 1 Comments: 1

Question Number 30860 Answers: 1 Comments: 0

S= 3(1!)−4(2!)+5(3!)−6(4!)+.... .....−(2008)(2006!)+2007! Find value of S.

$${S}=\:\mathrm{3}\left(\mathrm{1}!\right)−\mathrm{4}\left(\mathrm{2}!\right)+\mathrm{5}\left(\mathrm{3}!\right)−\mathrm{6}\left(\mathrm{4}!\right)+.... \\ $$$$\:\:\:\:.....−\left(\mathrm{2008}\right)\left(\mathrm{2006}!\right)+\mathrm{2007}! \\ $$$${Find}\:{value}\:{of}\:{S}. \\ $$

Question Number 30849 Answers: 0 Comments: 5

x^7 +x^6 +x^5 +x^4 +x^3 +x^2 +x+1=0 Σ_(k=1) ^7 [ℜ(x_k )]^2 = ? x_k = k^( th) root of the equation ℜ(x_k ) = real part of the root

$${x}^{\mathrm{7}} +{x}^{\mathrm{6}} +{x}^{\mathrm{5}} +{x}^{\mathrm{4}} +{x}^{\mathrm{3}} +{x}^{\mathrm{2}} +{x}+\mathrm{1}=\mathrm{0} \\ $$$$\: \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{7}} {\sum}}\left[\Re\left({x}_{{k}} \right)\right]^{\mathrm{2}} \:=\:? \\ $$$${x}_{{k}} \:=\:{k}^{\:\mathrm{th}} \:\mathrm{root}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\Re\left({x}_{{k}} \right)\:=\:\mathrm{real}\:\mathrm{part}\:\mathrm{of}\:\mathrm{the}\:\mathrm{root} \\ $$

Question Number 30745 Answers: 0 Comments: 0

let U_n ={z∈C/z^n =1} simlify A_n = Σ_(α∈U_n ) (x+α)^n and B_n =Σ_(α∈ U_n ) (x−α)^n .

$${let}\:\:{U}_{{n}} =\left\{{z}\in{C}/{z}^{{n}} =\mathrm{1}\right\}\:\:{simlify} \\ $$$${A}_{{n}} =\:\sum_{\alpha\in{U}_{{n}} } \:\left({x}+\alpha\right)^{{n}} \:{and}\:{B}_{{n}} =\sum_{\alpha\in\:{U}_{{n}} } \:\:\left({x}−\alpha\right)^{{n}} . \\ $$

Question Number 30744 Answers: 0 Comments: 0

let p(x)= (x−1)^n −x^n +1 with n integr find n in ordre that p(x) have a double root.

$${let}\:{p}\left({x}\right)=\:\left({x}−\mathrm{1}\right)^{{n}} \:−{x}^{{n}} \:+\mathrm{1}\:\:{with}\:{n}\:{integr}\:{find}\:{n} \\ $$$${in}\:{ordre}\:{that}\:{p}\left({x}\right)\:{have}\:{a}\:{double}\:{root}. \\ $$

Question Number 30743 Answers: 0 Comments: 1

decompose inside R[x] p(x)=x^(2n+1) −1 then find Π_(k=1) ^n sin( ((kπ)/(2n+1))) .

$${decompose}\:{inside}\:{R}\left[{x}\right]\:\:{p}\left({x}\right)={x}^{\mathrm{2}{n}+\mathrm{1}} \:−\mathrm{1}\:{then}\:{find} \\ $$$$\prod_{{k}=\mathrm{1}} ^{{n}} \:{sin}\left(\:\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)\:. \\ $$

Question Number 30742 Answers: 0 Comments: 0

prove that ∀p ∈N it exist one polynomial Q_(2p) / sin(2p+1)θ=sin^(2p+1) θ Q_(2p) (cotanθ) and degQ_(2p) =2p 2) prove that Π_(k=1) ^p tan(((kπ)/(2p+1)))=(√(2p+1)) .

$${prove}\:{that}\:\forall{p}\:\in{N}\:\:{it}\:{exist}\:{one}\:{polynomial}\:{Q}_{\mathrm{2}{p}} \:/ \\ $$$${sin}\left(\mathrm{2}{p}+\mathrm{1}\right)\theta={sin}^{\mathrm{2}{p}+\mathrm{1}} \theta\:{Q}_{\mathrm{2}{p}} \:\left({cotan}\theta\right)\:{and}\:{degQ}_{\mathrm{2}{p}} =\mathrm{2}{p} \\ $$$$\left.\mathrm{2}\right)\:{prove}\:{that}\:\prod_{{k}=\mathrm{1}} ^{{p}} \:{tan}\left(\frac{{k}\pi}{\mathrm{2}{p}+\mathrm{1}}\right)=\sqrt{\mathrm{2}{p}+\mathrm{1}}\:. \\ $$$$ \\ $$

Question Number 30739 Answers: 0 Comments: 0

let (u_n ) / u_1 =1−i and ∀p∈{2,3,...n} u_p =u_(p−1) j with j=e^(i((2π)/3)) 1)verify that u_1 +u_2 +u_3 =0 2)prove that ∀p∈ {4,5,...,n} u_p =u_(p−3) 3)find the value of S_n =Σ_(i=1) ^n u_i 4)calculate α_n = Σ_(p=0) ^(n−1) cos(−(π/4) +((2pπ)/3)) and β_n = Σ_(p=0) ^(n−1) sin(−(π/4) +((2pπ)/3)).

$${let}\:\left({u}_{{n}} \right)\:/\:{u}_{\mathrm{1}} =\mathrm{1}−{i}\:{and}\:\:\forall{p}\in\left\{\mathrm{2},\mathrm{3},...{n}\right\}\:{u}_{{p}} ={u}_{{p}−\mathrm{1}} {j}\:{with} \\ $$$${j}={e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \\ $$$$\left.\mathrm{1}\right){verify}\:{that}\:{u}_{\mathrm{1}} \:+{u}_{\mathrm{2}} \:+{u}_{\mathrm{3}} =\mathrm{0} \\ $$$$\left.\mathrm{2}\right){prove}\:{that}\:\forall{p}\in\:\left\{\mathrm{4},\mathrm{5},...,{n}\right\}\:\:{u}_{{p}} ={u}_{{p}−\mathrm{3}} \\ $$$$\left.\mathrm{3}\right){find}\:{the}\:{value}\:{of}\:{S}_{{n}} \:=\sum_{{i}=\mathrm{1}} ^{{n}} \:{u}_{{i}} \\ $$$$\left.\mathrm{4}\right){calculate}\:\:\alpha_{{n}} =\:\sum_{{p}=\mathrm{0}} ^{{n}−\mathrm{1}} {cos}\left(−\frac{\pi}{\mathrm{4}}\:+\frac{\mathrm{2}{p}\pi}{\mathrm{3}}\right)\:{and} \\ $$$$\beta_{{n}} =\:\sum_{{p}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{sin}\left(−\frac{\pi}{\mathrm{4}}\:+\frac{\mathrm{2}{p}\pi}{\mathrm{3}}\right). \\ $$

Question Number 30627 Answers: 1 Comments: 0