Question Number 73486 by abdomathmax last updated on 13/Nov/19
$${let}\:{P}\left({x}\right)=\left(\mathrm{1}+{ix}\right)^{{n}} −\left(\mathrm{1}−{ix}\right)^{{n}} \:{with}\:{n}\:{integr} \\ $$$${decompose}\:{the}\:{Fraction}\:{F}\:\left({x}\right)=\frac{\mathrm{1}}{{P}\left({x}\right)} \\ $$
Commented by abdomathmax last updated on 17/Nov/19
![P(x)=0 ⇔(((1−ix)^n )/((1+ix)^n )) =1 ⇔(((1+ix)/(1−ix)))^n =1 let z =((1+ix)/(1−ix)) (e) ⇒z^n =1 ⇒z_k =e^((i2kπ)/n) k∈[[0,n−1]] we have z−izx=1+ix ⇒ z−1=ix+izx =i(z+1)x ⇒x=(1/i)((z−1)/(1+z)) =−i((z−1)/(z+1)) =i((1−z)/(1+z)) so the roots of P are x_k =i((1−z_k )/(1+z_k )) =i((1−e^(i((2kπ)/n)) )/(1+e^(i((2kπ)/n)) )) =i((1−cos(((2kπ)/n))−isin(((2kπ)/n)))/(1+cos(((2kπ)/n))+isin(((2kπ)/n)))) =i((2sin^2 (((kπ)/n))−2i sin(((kπ)/n))cos(((kπ)/n)))/(2cos^2 (((kπ)/n))+2isin(((kπ)/n))cos(((kπ)/n)))) =i((−isin(((kπ)/n))e^((ikπ)/n) )/(cos(((kπ)/n))e^((ikπ)/n) )) =tan(((kπ)/n)) ⇒x_k =tan(((kπ)/n)) k ∈[[0,n−1]] ⇒P(x)=a Π_(k=0) ^(n−1) (x−tan(((kπ)/n))} a? we have P(x)=(1+ix)^n −(1−ix)^n =Σ_(k=0) ^n C_n ^k (ix)^k −Σ_(k=0) ^n C_n ^k (−ix)^k =Σ_(k=0) ^n C_n ^k { (i)^k −(−i)^k }x^k =Σ_(p=0) ^([((n−1)/2)]) 2i(−1)^p C_n ^(2p+1) x^(2p+1) ⇒a =2i(−1)^([((n−1)/2)]) C_n ^(2[((n−1)/2)]+1) ⇒ P(x)=2i(−1)^([((n−1)/2)]) C_n ^(2[((n−1)/2)]+1) Π_(k=0) ^(n−1) (x−tan(((kπ)/n))) F(x)=(1/(P(x)))=(1/(aΠ_(k=0) ^(n−1) (x−x_k ))) =Σ_(k=0) ^(n−1) (λ_k /(x−x_k )) λ_k =(1/(P^′ (x_k ))) but we have P(x)=(1+ix)^n −(1−ix)^n ⇒P^( ′) (x)=ni(1+ix)^(n−1) +ni(1−ix)^(n−1) =ni{ (1+ix)^(n−1) −(1−ix)^(n−1) } =ni (2i)Im( (1+ix)^(n−1) } 1+ix =(√(1+x^2 )) e^(i arctanx) ⇒(1+ix)^(n−1) =(1+x^2 )^((n−1)/2) e^(i(n−1)arctan(x)) ⇒Im((1+ix)^(n−1) ) =(1+x^2 )^((n−1)/2) sin((n−1)arctan(x))⇒ P^′ (x)=−2n (1+x^2 )^((n−1)/2) sin((n−1)arctan(x)) ⇒ P^′ (x_k )=−2n(1+ tan^2 (((kπ)/n)))^((n−1)/2) sin((n−1)(((kπ)/n))) =−2n((1/(cos^2 (((kπ)/n)))))^((n−1)/2) sin{kπ−((kπ)/n)} =((−2n)/((cos(((kπ)/n)))^(n−1) )){−(−1)^k sin(((kπ)/n)) =((2n(−1)^k sin(((kπ)/n)))/({cos(((kπ)/n))}^(n−1) )) =(1/λ_k )](https://www.tinkutara.com/question/Q73975.png)
$${P}\left({x}\right)=\mathrm{0}\:\Leftrightarrow\frac{\left(\mathrm{1}−{ix}\right)^{{n}} }{\left(\mathrm{1}+{ix}\right)^{{n}} }\:=\mathrm{1}\:\Leftrightarrow\left(\frac{\mathrm{1}+{ix}}{\mathrm{1}−{ix}}\right)^{{n}} =\mathrm{1} \\ $$$${let}\:{z}\:=\frac{\mathrm{1}+{ix}}{\mathrm{1}−{ix}}\:\:\:\left({e}\right)\:\Rightarrow{z}^{{n}} =\mathrm{1}\:\Rightarrow{z}_{{k}} ={e}^{\frac{{i}\mathrm{2}{k}\pi}{{n}}} \\ $$$${k}\in\left[\left[\mathrm{0},{n}−\mathrm{1}\right]\right]\:\:\:\:{we}\:{have}\:{z}−{izx}=\mathrm{1}+{ix}\:\Rightarrow \\ $$$${z}−\mathrm{1}={ix}+{izx}\:={i}\left({z}+\mathrm{1}\right){x}\:\Rightarrow{x}=\frac{\mathrm{1}}{{i}}\frac{{z}−\mathrm{1}}{\mathrm{1}+{z}} \\ $$$$=−{i}\frac{{z}−\mathrm{1}}{{z}+\mathrm{1}}\:={i}\frac{\mathrm{1}−{z}}{\mathrm{1}+{z}}\:\:{so}\:{the}\:{roots}\:{of}\:{P}\:{are} \\ $$$${x}_{{k}} ={i}\frac{\mathrm{1}−{z}_{{k}} }{\mathrm{1}+{z}_{{k}} }\:={i}\frac{\mathrm{1}−{e}^{{i}\frac{\mathrm{2}{k}\pi}{{n}}} }{\mathrm{1}+{e}^{{i}\frac{\mathrm{2}{k}\pi}{{n}}} }\:={i}\frac{\mathrm{1}−{cos}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)−{isin}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)}{\mathrm{1}+{cos}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)+{isin}\left(\frac{\mathrm{2}{k}\pi}{{n}}\right)} \\ $$$$={i}\frac{\mathrm{2}{sin}^{\mathrm{2}} \left(\frac{{k}\pi}{{n}}\right)−\mathrm{2}{i}\:{sin}\left(\frac{{k}\pi}{{n}}\right){cos}\left(\frac{{k}\pi}{{n}}\right)}{\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{{k}\pi}{{n}}\right)+\mathrm{2}{isin}\left(\frac{{k}\pi}{{n}}\right){cos}\left(\frac{{k}\pi}{{n}}\right)} \\ $$$$={i}\frac{−{isin}\left(\frac{{k}\pi}{{n}}\right){e}^{\frac{{ik}\pi}{{n}}} }{{cos}\left(\frac{{k}\pi}{{n}}\right){e}^{\frac{{ik}\pi}{{n}}} }\:={tan}\left(\frac{{k}\pi}{{n}}\right)\:\Rightarrow{x}_{{k}} ={tan}\left(\frac{{k}\pi}{{n}}\right) \\ $$$${k}\:\in\left[\left[\mathrm{0},{n}−\mathrm{1}\right]\right]\:\Rightarrow{P}\left({x}\right)={a}\:\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left({x}−{tan}\left(\frac{{k}\pi}{{n}}\right)\right\} \\ $$$${a}?\:\:\:\:{we}\:{have}\:{P}\left({x}\right)=\left(\mathrm{1}+{ix}\right)^{{n}} −\left(\mathrm{1}−{ix}\right)^{{n}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \left({ix}\right)^{{k}} −\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \left(−{ix}\right)^{{k}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} {C}_{{n}} ^{{k}} \left\{\:\left({i}\right)^{{k}} −\left(−{i}\right)^{{k}} \right\}{x}^{{k}} \\ $$$$=\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \mathrm{2}{i}\left(−\mathrm{1}\right)^{{p}} \:{C}_{{n}} ^{\mathrm{2}{p}+\mathrm{1}} \:{x}^{\mathrm{2}{p}+\mathrm{1}} \\ $$$$\Rightarrow{a}\:=\mathrm{2}{i}\left(−\mathrm{1}\right)^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\:{C}_{{n}} ^{\mathrm{2}\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]+\mathrm{1}} \:\Rightarrow \\ $$$${P}\left({x}\right)=\mathrm{2}{i}\left(−\mathrm{1}\right)^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:{C}_{{n}} ^{\mathrm{2}\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]+\mathrm{1}} \prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left({x}−{tan}\left(\frac{{k}\pi}{{n}}\right)\right) \\ $$$${F}\left({x}\right)=\frac{\mathrm{1}}{{P}\left({x}\right)}=\frac{\mathrm{1}}{{a}\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left({x}−{x}_{{k}} \right)}\:=\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \frac{\lambda_{{k}} }{{x}−{x}_{{k}} } \\ $$$$\lambda_{{k}} =\frac{\mathrm{1}}{{P}\:^{'} \left({x}_{{k}} \right)}\:\:{but}\:{we}\:{have}\:\:{P}\left({x}\right)=\left(\mathrm{1}+{ix}\right)^{{n}} −\left(\mathrm{1}−{ix}\right)^{{n}} \\ $$$$\Rightarrow{P}^{\:'} \left({x}\right)={ni}\left(\mathrm{1}+{ix}\right)^{{n}−\mathrm{1}} +{ni}\left(\mathrm{1}−{ix}\right)^{{n}−\mathrm{1}} \\ $$$$={ni}\left\{\:\left(\mathrm{1}+{ix}\right)^{{n}−\mathrm{1}} −\left(\mathrm{1}−{ix}\right)^{{n}−\mathrm{1}} \right\} \\ $$$$={ni}\:\left(\mathrm{2}{i}\right){Im}\left(\:\left(\mathrm{1}+{ix}\right)^{{n}−\mathrm{1}} \right\} \\ $$$$\mathrm{1}+{ix}\:=\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:{e}^{{i}\:{arctanx}} \:\Rightarrow\left(\mathrm{1}+{ix}\right)^{{n}−\mathrm{1}} =\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\frac{{n}−\mathrm{1}}{\mathrm{2}}} {e}^{{i}\left({n}−\mathrm{1}\right){arctan}\left({x}\right)} \\ $$$$\Rightarrow{Im}\left(\left(\mathrm{1}+{ix}\right)^{{n}−\mathrm{1}} \right)\:=\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\frac{{n}−\mathrm{1}}{\mathrm{2}}} {sin}\left(\left({n}−\mathrm{1}\right){arctan}\left({x}\right)\right)\Rightarrow \\ $$$${P}\:^{'} \left({x}\right)=−\mathrm{2}{n}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\frac{{n}−\mathrm{1}}{\mathrm{2}}} \:{sin}\left(\left({n}−\mathrm{1}\right){arctan}\left({x}\right)\right)\:\Rightarrow \\ $$$${P}\:^{'} \left({x}_{{k}} \right)=−\mathrm{2}{n}\left(\mathrm{1}+\:{tan}^{\mathrm{2}} \left(\frac{{k}\pi}{{n}}\right)\right)^{\frac{{n}−\mathrm{1}}{\mathrm{2}}} {sin}\left(\left({n}−\mathrm{1}\right)\left(\frac{{k}\pi}{{n}}\right)\right) \\ $$$$=−\mathrm{2}{n}\left(\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \left(\frac{{k}\pi}{{n}}\right)}\right)^{\frac{{n}−\mathrm{1}}{\mathrm{2}}} {sin}\left\{{k}\pi−\frac{{k}\pi}{{n}}\right\} \\ $$$$=\frac{−\mathrm{2}{n}}{\left({cos}\left(\frac{{k}\pi}{{n}}\right)\right)^{{n}−\mathrm{1}} }\left\{−\left(−\mathrm{1}\right)^{{k}} \:{sin}\left(\frac{{k}\pi}{{n}}\right)\right. \\ $$$$=\frac{\mathrm{2}{n}\left(−\mathrm{1}\right)^{{k}} \:{sin}\left(\frac{{k}\pi}{{n}}\right)}{\left\{{cos}\left(\frac{{k}\pi}{{n}}\right)\right\}^{{n}−\mathrm{1}} }\:=\frac{\mathrm{1}}{\lambda_{{k}} } \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 17/Nov/19
$${forgive}\:{P}\:^{'} \left({x}\right)=−\mathrm{2}{n}\:{Re}\left(\left(\mathrm{1}+{ix}\right)^{{n}−\mathrm{1}} \right) \\ $$$$=−\mathrm{2}{n}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\frac{{n}−\mathrm{1}}{\mathrm{2}}} \:{cos}\left\{\left({n}−\mathrm{1}\right){arctanx}\right\}\:\Rightarrow \\ $$$${P}\:^{'} \left({x}_{{k}} \right)=−\mathrm{2}{n}\left(\mathrm{1}+{x}_{{k}} ^{\mathrm{2}} \right)^{\frac{{n}−\mathrm{1}}{\mathrm{2}}} \:{cos}\left\{\left({n}−\mathrm{1}\right){arctan}\left({x}_{{k}} \right)\right\} \\ $$$$=−\mathrm{2}{n}\:\left(\mathrm{1}+{tan}^{\mathrm{2}} \left(\frac{{k}\pi}{{n}}\right)\right)^{\frac{{n}−\mathrm{1}}{\mathrm{2}}} \:{cos}\left\{\left({n}−\mathrm{1}\right)\frac{{k}\pi}{{n}}\right\} \\ $$$$=−\mathrm{2}{n}\:\left(\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \left(\frac{{k}\pi}{{n}}\right)}\right)^{\frac{{n}−\mathrm{1}}{\mathrm{2}}} \:{cos}\left\{{k}\pi−\frac{{k}\pi}{{n}}\right\} \\ $$$$=−\frac{\mathrm{2}{n}}{\left({cos}\left(\frac{{k}\pi}{{n}}\right)\right)^{{n}−\mathrm{1}} }\left(−\mathrm{1}\right)^{{k}} {cos}\left(\frac{{k}\pi}{{n}}\right)\:=\frac{\mathrm{2}{n}\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} }{\left({cos}\left(\frac{{k}\pi}{{n}}\right)\right)^{{n}−\mathrm{2}} }\:=\frac{\mathrm{1}}{\lambda_{{k}} }\:\Rightarrow \\ $$$$\lambda_{{k}} =\frac{\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} \left({cos}\left(\frac{{k}\pi}{{n}}\right)\right)^{{n}−\mathrm{2}} }{\mathrm{2}{n}}\:\Rightarrow \\ $$$${P}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{2}{n}}\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\:\:\frac{\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} \left({cos}\left(\frac{{k}\pi}{{n}}\right)\right)^{{n}−\mathrm{2}} }{{x}−{tan}\left(\frac{{k}\pi}{{n}}\right)}\:\:\:\:\left(\:{n}\:>\mathrm{1}\right) \\ $$