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Question Number 34116 by tanmay.chaudhury50@gmail.com last updated on 30/Apr/18

Commented by abdo mathsup 649 cc last updated on 02/May/18

let put f(x)= ((1−x^2 )/(x^2 −2xcosθ +1)) the poles of f are x_1 = ((1−x^2 )/((x −e^(iθ) )(x−e^(−iθ) ))) =(1−x^2 )( (1/(x −e^(iθ) )) −(1/(x −e^(−iθ) )))(1/(2isinθ)) =((1−x^2 )/(2isinθ))(−(1/(e^(iθ) −x)) +(1/(e^(−iθ) −x))) =((1−x^2 )/(2isinθ))( (e^(iθ) /(1−x e^(iθ) )) −(e^(−iθ) /(1−x e^(−iθ) ))) = ((1−x^2 )/(2isinθ))( e^(iθ) Σ_(n=0) ^∞ x^n e^(inθ) −e^(−iθ) Σ_(n=0) ^∞ x^n e^(−inθ) ) = ((1−x^2 )/(2i sinθ)) ( Σ_(n=0) ^∞ x^n e^(i(n+1)θ) −Σ_(n=0) ^∞ x^n e^(−i(n+1)θ) ) =((1−x^2 )/(2isinθ)) ( Σ_(n=0) ^∞ x^n (2isin(n+1)θ)) =((1−x^2 )/(sinθ)) Σ_(n=0) ^∞ (sin(n+1)θ)x^n = (1/(sinθ)) Σ_(n=0) ^∞ sin(n+1)θ .x^n −(1/(sinθ)) Σ_(n=0) ^∞ sin(n+1)θ.x^(n+2) =....

$${let}\:{put}\:{f}\left({x}\right)=\:\frac{\mathrm{1}−{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} \:−\mathrm{2}{xcos}\theta\:+\mathrm{1}}\:{the}\:{poles}\:{of}\:{f}\:{are} \\ $$$${x}_{\mathrm{1}} =\:\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\left({x}\:−{e}^{{i}\theta} \right)\left({x}−{e}^{−{i}\theta} \right)} \\ $$$$=\left(\mathrm{1}−{x}^{\mathrm{2}} \right)\left(\:\:\frac{\mathrm{1}}{{x}\:−{e}^{{i}\theta} }\:−\frac{\mathrm{1}}{{x}\:−{e}^{−{i}\theta} }\right)\frac{\mathrm{1}}{\mathrm{2}{isin}\theta} \\ $$$$=\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{2}{isin}\theta}\left(−\frac{\mathrm{1}}{{e}^{{i}\theta} −{x}}\:\:+\frac{\mathrm{1}}{{e}^{−{i}\theta} −{x}}\right) \\ $$$$=\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{2}{isin}\theta}\left(\:\:\:\:\frac{{e}^{{i}\theta} }{\mathrm{1}−{x}\:{e}^{{i}\theta} }\:\:−\frac{{e}^{−{i}\theta} }{\mathrm{1}−{x}\:{e}^{−{i}\theta} }\right) \\ $$$$=\:\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{2}{isin}\theta}\left(\:{e}^{{i}\theta} \:\sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{{n}} \:{e}^{{in}\theta} \:−{e}^{−{i}\theta} \:\sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{{n}} \:{e}^{−{in}\theta} \right) \\ $$$$=\:\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{2}{i}\:{sin}\theta}\:\left(\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:{x}^{{n}} \:{e}^{{i}\left({n}+\mathrm{1}\right)\theta} \:\:−\sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{{n}} \:{e}^{−{i}\left({n}+\mathrm{1}\right)\theta} \right) \\ $$$$=\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{2}{isin}\theta}\:\left(\:\:\sum_{{n}=\mathrm{0}} ^{\infty} {x}^{{n}} \left(\mathrm{2}{isin}\left({n}+\mathrm{1}\right)\theta\right)\right) \\ $$$$=\frac{\mathrm{1}−{x}^{\mathrm{2}} }{{sin}\theta}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\left({sin}\left({n}+\mathrm{1}\right)\theta\right){x}^{{n}} \\ $$$$=\:\frac{\mathrm{1}}{{sin}\theta}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{sin}\left({n}+\mathrm{1}\right)\theta\:.{x}^{{n}} \:−\frac{\mathrm{1}}{{sin}\theta}\:\sum_{{n}=\mathrm{0}} ^{\infty} {sin}\left({n}+\mathrm{1}\right)\theta.{x}^{{n}+\mathrm{2}} \\ $$$$=.... \\ $$$$ \\ $$

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