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Question Number 62225 by maxmathsup by imad last updated on 17/Jun/19
let j =e^((i2π)/3) and P(x) =(1+jx)^n −(1−jx)^n 1) find P(x) at form of arctan 2) find the roots of P(x) 3)factorize inside C[x] the polynome P(x) 4) calculate ∫_0 ^1 P(x)dx
$${let}\:{j}\:={e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \:\:\:{and}\:{P}\left({x}\right)\:=\left(\mathrm{1}+{jx}\right)^{{n}} −\left(\mathrm{1}−{jx}\right)^{{n}} \\ $$$$\left.\mathrm{1}\right)\:{find}\:{P}\left({x}\right)\:{at}\:{form}\:{of}\:{arctan} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{roots}\:{of}\:{P}\left({x}\right) \\ $$$$\left.\mathrm{3}\right){factorize}\:{inside}\:{C}\left[{x}\right]\:\:{the}\:{polynome}\:{P}\left({x}\right) \\ $$$$\left.\mathrm{4}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{P}\left({x}\right){dx} \\ $$
Commented by mathmax by abdo last updated on 24/Jun/19
1) we have j =cos(((2π)/3))+isin(((2π)/3)) =−(1/2)+i((√3)/2) ⇒ P(x)=(1+(−(1/2)+i((√3)/2))x)^n −(1−(−(1/2) +i((√3)/2))x)^n =(1−(x/2) +i((√3)/2)x)^n −(1+(x/2)−i((√3)/2)x)^n =A^n (x)−B^n (x) ∣A(x)∣ =(√((1−(x/2))^2 +(3/4)x^2 )) =(√(((2−x)^2 +3x^2 )/4))=(1/2)(√((x−2)^2 +3x^2 )) we know that x+iy =(√(x^2 +y^2 )) e^(i arctan((y/x))) ⇒ A(x) =(1/2)(√((x−2)^2 +3x^2 )) e^(i arctan((((√3)x)/(2−x)))) ⇒A^n (x) =(1/2^n ){(x−2)^2 +3x^2 }^(n/2) e^(i n arctan((((√3)x)/(2−x)))) ∣B(x)∣ =(1/2)(√((x+2)^2 +3x^2 )) ⇒B(x) =(1/2)(√((x+2)^2 +3x^2 ))e^(i arctan(((−(√3)x)/(2+x)))) ⇒ B^n (x) =(1/2^n ){ (x+2)^2 +3x^2 }^(n/2) e^(−in arctan((((√3)x)/(2+x)))) ⇒ P(x) =(1/2^n ){ (x−2)^2 +3x^2 }^(n/2) e^(in arctan((((√3)x)/(2−x)))) −(1/2^n ){(x+2)^2 +3x^2 }^(n/2) e^(−in arctan((((√3)x)/(2+x)))) and we see that B(x) =A(−x).
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{j}\:={cos}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)+{isin}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\:=−\frac{\mathrm{1}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\Rightarrow \\ $$$${P}\left({x}\right)=\left(\mathrm{1}+\left(−\frac{\mathrm{1}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right){x}\right)^{{n}} \:−\left(\mathrm{1}−\left(−\frac{\mathrm{1}}{\mathrm{2}}\:+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right){x}\right)^{{n}} \\ $$$$=\left(\mathrm{1}−\frac{{x}}{\mathrm{2}}\:+{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{x}\right)^{{n}} \:−\left(\mathrm{1}+\frac{{x}}{\mathrm{2}}−{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{x}\right)^{{n}} \:={A}^{{n}} \left({x}\right)−{B}^{{n}} \left({x}\right) \\ $$$$\mid{A}\left({x}\right)\mid\:=\sqrt{\left(\mathrm{1}−\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}{x}^{\mathrm{2}} }\:=\sqrt{\frac{\left(\mathrm{2}−{x}\right)^{\mathrm{2}} \:+\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{4}}}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\left({x}−\mathrm{2}\right)^{\mathrm{2}} \:+\mathrm{3}{x}^{\mathrm{2}} } \\ $$$${we}\:{know}\:{that}\:{x}+{iy}\:=\sqrt{{x}^{\mathrm{2}} \:+{y}^{\mathrm{2}} \:}\:{e}^{{i}\:{arctan}\left(\frac{{y}}{{x}}\right)} \:\Rightarrow \\ $$$${A}\left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\left({x}−\mathrm{2}\right)^{\mathrm{2}} \:+\mathrm{3}{x}^{\mathrm{2}} }\:{e}^{{i}\:{arctan}\left(\frac{\sqrt{\mathrm{3}}{x}}{\mathrm{2}−{x}}\right)} \:\Rightarrow{A}^{{n}} \left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\left\{\left({x}−\mathrm{2}\right)^{\mathrm{2}} \:+\mathrm{3}{x}^{\mathrm{2}} \right\}^{\frac{{n}}{\mathrm{2}}} \:{e}^{{i}\:{n}\:{arctan}\left(\frac{\sqrt{\mathrm{3}}{x}}{\mathrm{2}−{x}}\right)} \\ $$$$\mid{B}\left({x}\right)\mid\:=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\left({x}+\mathrm{2}\right)^{\mathrm{2}} \:+\mathrm{3}{x}^{\mathrm{2}} \:}\:\Rightarrow{B}\left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\left({x}+\mathrm{2}\right)^{\mathrm{2}} \:+\mathrm{3}{x}^{\mathrm{2}} \:}{e}^{{i}\:{arctan}\left(\frac{−\sqrt{\mathrm{3}}{x}}{\mathrm{2}+{x}}\right)} \:\Rightarrow \\ $$$${B}^{{n}} \left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\left\{\:\left({x}+\mathrm{2}\right)^{\mathrm{2}} \:+\mathrm{3}{x}^{\mathrm{2}} \right\}^{\frac{{n}}{\mathrm{2}}} \:{e}^{−{in}\:{arctan}\left(\frac{\sqrt{\mathrm{3}}{x}}{\mathrm{2}+{x}}\right)} \:\Rightarrow \\ $$$${P}\left({x}\right)\:=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\left\{\:\left({x}−\mathrm{2}\right)^{\mathrm{2}} \:+\mathrm{3}{x}^{\mathrm{2}} \right\}^{\frac{{n}}{\mathrm{2}}} \:{e}^{{in}\:{arctan}\left(\frac{\sqrt{\mathrm{3}}{x}}{\mathrm{2}−{x}}\right)} −\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\left\{\left({x}+\mathrm{2}\right)^{\mathrm{2}} \:+\mathrm{3}{x}^{\mathrm{2}} \right\}^{\frac{{n}}{\mathrm{2}}} \:{e}^{−{in}\:{arctan}\left(\frac{\sqrt{\mathrm{3}}{x}}{\mathrm{2}+{x}}\right)} \\ $$$${and}\:{we}\:{see}\:{that}\:\:{B}\left({x}\right)\:={A}\left(−{x}\right). \\ $$
Commented by mathmax by abdo last updated on 24/Jun/19
2) P(x)=0 ⇔(1+jx)^n =(1−jx)^n ⇒(((1+jx)/(1−jx)))^n =1 ⇒Z^n =1 with Z =((1+jx)/(1−jx)) the roots of Z^n =1 are Z_k =e^(i((2kπ)/n)) with k∈[[0,n−1]] Z =((1+jx)/(1−jx)) ⇒Z−jZx =1+jx ⇒Z−1 =jx(1+Z) ⇒x =((Z−1)/(j(1+Z))) so the roots of P(x)=0 are x_k =−(1/j) ((1−Zk)/(1+Z_k )) =−(1/j) ((1−cos(((2kπ)/n))−i sin(((2kπ)/n)))/(1+cos(((2kπ)/n))+i sin(((2kπ)/n)))) =−(1/j)((2sin^2 (((kπ)/n))−2i sin(((kπ)/n))cos(((kπ)/n)))/(2cos^2 (((kπ)/n))+2i cos(((kπ)/n))sin(((kπ)/n)))) =−(1/j) ((−i sin(((kπ)/n)) e^(i((kπ)/n)) )/(cos(((kπ)/n))e^((ikπ)/n) )) =(i/j)tan(((kπ)/n)) ⇒ x_k =(i/j) tan(((kπ)/n)) and k ∈[[0,n−1]].
$$\left.\mathrm{2}\right)\:{P}\left({x}\right)=\mathrm{0}\:\Leftrightarrow\left(\mathrm{1}+{jx}\right)^{{n}} \:=\left(\mathrm{1}−{jx}\right)^{{n}} \:\Rightarrow\left(\frac{\mathrm{1}+{jx}}{\mathrm{1}−{jx}}\right)^{{n}} \:=\mathrm{1}\:\Rightarrow{Z}^{{n}} \:=\mathrm{1}\:\:{with}\:{Z}\:=\frac{\mathrm{1}+{jx}}{\mathrm{1}−{jx}} \\ $$$$\:{the}\:{roots}\:{of}\:{Z}^{{n}} \:=\mathrm{1}\:{are}\:{Z}_{{k}} ={e}^{{i}\frac{\mathrm{2}{k}\pi}{{n}}} \:\:{with}\:{k}\in\left[\left[\mathrm{0},{n}−\mathrm{1}\right]\right] \\ $$$${Z}\:=\frac{\mathrm{1}+{jx}}{\mathrm{1}−{jx}}\:\Rightarrow{Z}−{jZx}\:=\mathrm{1}+{jx}\:\Rightarrow{Z}−\mathrm{1}\:={jx}\left(\mathrm{1}+{Z}\right)\:\Rightarrow{x}\:=\frac{{Z}−\mathrm{1}}{{j}\left(\mathrm{1}+{Z}\right)}\:{so}\:{the}\:{roots} \\ $$$${of}\:{P}\left({x}\right)=\mathrm{0}\:{are}\:{x}_{{k}} =−\frac{\mathrm{1}}{{j}}\:\frac{\mathrm{1}−{Zk}}{\mathrm{1}+{Z}_{{k}} } \\ $$$$=−\frac{\mathrm{1}}{{j}}\:\frac{\mathrm{1}−{cos}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)−{i}\:{sin}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)}{\mathrm{1}+{cos}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)+{i}\:{sin}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)}\:=−\frac{\mathrm{1}}{{j}}\frac{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{k}\pi}{{n}}\right)−\mathrm{2}{i}\:{sin}\left(\frac{{k}\pi}{{n}}\right){cos}\left(\frac{{k}\pi}{{n}}\right)}{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{{k}\pi}{{n}}\right)+\mathrm{2}{i}\:{cos}\left(\frac{{k}\pi}{{n}}\right){sin}\left(\frac{{k}\pi}{{n}}\right)} \\ $$$$=−\frac{\mathrm{1}}{{j}}\:\frac{−{i}\:{sin}\left(\frac{{k}\pi}{{n}}\right)\:{e}^{{i}\frac{{k}\pi}{{n}}} }{{cos}\left(\frac{{k}\pi}{{n}}\right){e}^{\frac{{ik}\pi}{{n}}} }\:=\frac{{i}}{{j}}{tan}\left(\frac{{k}\pi}{{n}}\right)\:\:\:\Rightarrow\:{x}_{{k}} =\frac{{i}}{{j}}\:{tan}\left(\frac{{k}\pi}{{n}}\right)\:{and}\:{k}\:\in\left[\left[\mathrm{0},{n}−\mathrm{1}\right]\right]. \\ $$
Commented by mathmax by abdo last updated on 24/Jun/19
3) P(x) =λ Π_(k=0) ^(n−1) (x−(i/j)tan(((kπ)/n))) with λ is the dominent coefficientof P(x).
$$\left.\mathrm{3}\right)\:{P}\left({x}\right)\:=\lambda\:\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left({x}−\frac{{i}}{{j}}{tan}\left(\frac{{k}\pi}{{n}}\right)\right)\:\:\:{with}\:\lambda\:{is}\:{the}\:{dominent}\:{coefficientof}\:{P}\left({x}\right). \\ $$
Commented by mathmax by abdo last updated on 24/Jun/19
4) we have P(x) =Σ_(k=0) ^n C_n ^k j^k x^k −Σ_(k=0) ^n C_n ^k (−j)^k x^k =Σ_(k=0) ^n C_n ^k (j^k −(−j)^k )x^k but j^k −(−j)^k =e^(i((2kπ)/3)) −(−e^(i((2π)/3)) )^k ={1−(−1)^k } e^(i((2kπ)/3)) =0 if k=2p and =2 e^(i((2(2p+1)π)/3)) if k =2p+1 ⇒ P(x) =Σ_(p=0) ^([((n−1)/2)]) 2 C_n ^(2p+1) e^(i((2(2p+1)π)/3)) x^(2p+1) ⇒ ∫_0 ^1 P(x)dx =2Σ_(p=0) ^([((n−1)/2)]) C_n ^(2p+1) e^(i((4p+2)/3)π) (1/(2p+2)) =Σ_(p=0) ^([((n−1)/2)]) C_n ^(2p+1) (e^(i((4p+2)/3)π) /(p+1)) .
$$\left.\mathrm{4}\right)\:{we}\:{have}\:{P}\left({x}\right)\:=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} {j}^{{k}} \:{x}^{{k}} \:\:−\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left(−{j}\right)^{{k}} \:{x}^{{k}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left({j}^{{k}} \:−\left(−{j}\right)^{{k}} \right){x}^{{k}} \:\:\:\:{but}\:\:{j}^{{k}} \:−\left(−{j}\right)^{{k}} \:={e}^{{i}\frac{\mathrm{2}{k}\pi}{\mathrm{3}}} \:−\left(−{e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \right)^{{k}} \\ $$$$=\left\{\mathrm{1}−\left(−\mathrm{1}\right)^{{k}} \right\}\:{e}^{{i}\frac{\mathrm{2}{k}\pi}{\mathrm{3}}} \:=\mathrm{0}\:{if}\:\:{k}=\mathrm{2}{p}\:\:{and}\:\:=\mathrm{2}\:{e}^{{i}\frac{\mathrm{2}\left(\mathrm{2}{p}+\mathrm{1}\right)\pi}{\mathrm{3}}} \:{if}\:{k}\:=\mathrm{2}{p}+\mathrm{1}\:\Rightarrow \\ $$$${P}\left({x}\right)\:=\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\mathrm{2}\:{C}_{{n}} ^{\mathrm{2}{p}+\mathrm{1}} \:{e}^{{i}\frac{\mathrm{2}\left(\mathrm{2}{p}+\mathrm{1}\right)\pi}{\mathrm{3}}} \:\:{x}^{\mathrm{2}{p}+\mathrm{1}} \:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:{P}\left({x}\right){dx}\:=\mathrm{2}\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\:\:{C}_{{n}} ^{\mathrm{2}{p}+\mathrm{1}} \:\:\:\:{e}^{{i}\frac{\mathrm{4}{p}+\mathrm{2}}{\mathrm{3}}\pi} \:\frac{\mathrm{1}}{\mathrm{2}{p}+\mathrm{2}} \\ $$$$=\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\:{C}_{{n}} ^{\mathrm{2}{p}+\mathrm{1}} \:\:\:\frac{{e}^{{i}\frac{\mathrm{4}{p}+\mathrm{2}}{\mathrm{3}}\pi} }{{p}+\mathrm{1}}\:. \\ $$

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