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AlgebraQuestion and Answers: Page 348

Question Number 27046 Answers: 0 Comments: 0

Considering y=x^3 +px+q If (dy/dx)∣_(x=α) =0 ⇒ α^2 =−(p/3) if ((d(y/x))/dx)∣_(x=β) =0 ⇒ β^( 3) =(q/2) roots of the cubic eq^n are: x=[−β^( 3) ±(√(β^( 6) −α^6 )) ]^(1/3) −[β^( 3) ±(√(β^( 6) −α^6 )) ]^(1/3) . Why such a connection? If equation is quadratic even_ y=ax^2 +bx+c (dy/dx)∣_(x=α) =0 ⇒ α=−(b/(2a)) ((d(y/x))/dx)∣_(x=β) =0 ⇒ β^( 2) =(c/a) roots of quadratic eq. are: x=𝛂±(√(𝛂^2 −𝛃^( 2) )) why such a connection ?

$${Considering}\:\boldsymbol{{y}}=\boldsymbol{{x}}^{\mathrm{3}} +\boldsymbol{{px}}+\boldsymbol{{q}} \\ $$$${If}\:\:\:\:\:\frac{{dy}}{{dx}}\mid_{{x}=\alpha} =\mathrm{0}\:\:\Rightarrow\:\:\alpha^{\mathrm{2}} =−\frac{{p}}{\mathrm{3}} \\ $$$${if}\:\:\:\frac{{d}\left({y}/{x}\right)}{{dx}}\mid_{{x}=\beta} =\mathrm{0}\:\:\:\Rightarrow\:\beta^{\:\mathrm{3}} =\frac{{q}}{\mathrm{2}} \\ $$$${roots}\:{of}\:{the}\:{cubic}\:\:{eq}^{{n}} \:{are}: \\ $$$$\:\:\:\:{x}=\left[−\beta^{\:\mathrm{3}} \pm\sqrt{\beta^{\:\mathrm{6}} −\alpha^{\mathrm{6}} }\:\right]^{\mathrm{1}/\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\left[\beta^{\:\mathrm{3}} \pm\sqrt{\beta^{\:\mathrm{6}} −\alpha^{\mathrm{6}} }\:\right]^{\mathrm{1}/\mathrm{3}} \:. \\ $$$$\:{Why}\:{such}\:{a}\:{connection}? \\ $$$${If}\:{equation}\:{is}\:{quadratic}\:{even\_} \\ $$$$\:\:\:\:\boldsymbol{{y}}=\boldsymbol{{ax}}^{\mathrm{2}} +\boldsymbol{{bx}}+\boldsymbol{{c}} \\ $$$$\frac{{dy}}{{dx}}\mid_{{x}=\alpha} =\mathrm{0}\:\:\:\Rightarrow\:\:\alpha=−\frac{{b}}{\mathrm{2}{a}} \\ $$$$\:\:\:\:\:\:\frac{{d}\left({y}/{x}\right)}{{dx}}\mid_{{x}=\beta} =\mathrm{0}\:\:\Rightarrow\:\beta^{\:\mathrm{2}} =\frac{{c}}{{a}} \\ $$$${roots}\:{of}\:{quadratic}\:{eq}.\:{are}: \\ $$$$\:\:\:\:{x}=\boldsymbol{\alpha}\pm\sqrt{\boldsymbol{\alpha}^{\mathrm{2}} −\boldsymbol{\beta}^{\:\mathrm{2}} }\: \\ $$$${why}\:{such}\:{a}\:{connection}\:?\: \\ $$

Question Number 27061 Answers: 2 Comments: 1

Question Number 27060 Answers: 1 Comments: 1

Question Number 26999 Answers: 1 Comments: 0

calculate Π_(k=1) ^n cos((a/2^k )) and0<a<π then find the value of lim_(n−>∝) Σ_(k=1) ^n ln(cos((a/2^k ))).

$${calculate}\:\prod_{{k}=\mathrm{1}} ^{{n}} {cos}\left(\frac{{a}}{\mathrm{2}^{{k}} }\right)\:\:{and}\mathrm{0}<{a}<\pi\:\:{then}\:{find}\:{the}\:{value}\:{of} \\ $$$${lim}_{{n}−>\propto} \:\sum_{{k}=\mathrm{1}} ^{{n}} {ln}\left({cos}\left(\frac{{a}}{\mathrm{2}^{{k}} }\right)\right). \\ $$

Question Number 26998 Answers: 0 Comments: 0

smlify X= Π_(p=2) ^n ((p^3 −1)/(p^3 +1)) by using 1,j,j^2 and j=e^((i2π)/3) .

$${smlify}\:{X}=\:\:\prod_{{p}=\mathrm{2}} ^{{n}} \frac{{p}^{\mathrm{3}} −\mathrm{1}}{{p}^{\mathrm{3}} \:+\mathrm{1}}\:{by}\:{using}\:\mathrm{1},{j},{j}^{\mathrm{2}} {and}\:{j}={e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} . \\ $$

Question Number 26997 Answers: 0 Comments: 3

let give ξ ∈C and ξ^n =1 (ξ is the n^(me) root of 1) simplify A= 1+ξ^p +ξ^(2p) +... +ξ^((n−1)p) and B= 1+2ξ +3ξ^2 +...+nξ^(n−1) .

$${let}\:{give}\:\xi\:\in\mathbb{C}\:{and}\:\xi^{{n}} =\mathrm{1}\:\left(\xi\:{is}\:{the}\:{n}^{{me}} \:{root}\:{of}\:\mathrm{1}\right) \\ $$$${simplify}\:\:{A}=\:\mathrm{1}+\xi^{{p}} +\xi^{\mathrm{2}{p}} +...\:+\xi^{\left({n}−\mathrm{1}\right){p}} \\ $$$${and}\:{B}=\:\mathrm{1}+\mathrm{2}\xi\:+\mathrm{3}\xi^{\mathrm{2}} +...+{n}\xi^{{n}−\mathrm{1}} . \\ $$

Question Number 26942 Answers: 1 Comments: 0

Question Number 27000 Answers: 0 Comments: 1

P is a polynomial havng n roots (x_i )_(1≤i≤n) with x_i ≠ x_j for i≠ j find the values of Σ_(k1) ^(k=n) (1/(x−x_k )) and Σ_(k=1) ^n (1/((x−x_k )^2 )) .

$${P}\:{is}\:{a}\:{polynomial}\:{havng}\:{n}\:{roots}\:\left({x}_{{i}} \right)_{\mathrm{1}\leqslant{i}\leqslant{n}} \:\:{with}\:{x}_{{i}} \neq\:{x}_{{j}} \:{for}\:{i}\neq\:{j} \\ $$$${find}\:{the}\:{values}\:{of}\:\sum_{{k}\mathrm{1}} ^{{k}={n}} \frac{\mathrm{1}}{{x}−{x}_{{k}} }\:\:{and}\:\:\sum_{{k}=\mathrm{1}} ^{{n}} \frac{\mathrm{1}}{\left({x}−{x}_{{k}} \right)^{\mathrm{2}} }\:. \\ $$

Question Number 26733 Answers: 1 Comments: 1