Question-34117

Question Number 34117 by tanmay.chaudhury50@gmail.com last updated on 30/Apr/18

Commented by abdo imad last updated on 01/May/18

let put f(x)=((sinθ)/(x^2 −2xcosθ +1)) poles of f(x)? x^2 −2xcosθ +1=0 Δ^′ =cos^2 θ −1 =−sin^2 θ =(isinθ)^2 x_1 =cosθ +isinθ =e^(iθ) and x_2 =e^(−iθ) ⇒ f(x) = ((sinθ)/((x−e^(iθ) )(x−e^(−iθ) ))) =sinθ( (1/(x−e^(iθ) )) −(1/(x −e^(−iθ) ))).(1/(2isinθ)) =(1/(2i))( ((−1)/(e^(iθ) −x)) +(1/(e^(−iθ) −x))) =(1/(2i))( ((−e^(−iθ) )/(1−xe^(−iθ) )) + (e^(iθ) /(1−x e^(iθ) ))) =(1/(2i))( e^(iθ) Σ_(n=0) ^∞ x^n e^(inθ) −e^(−iθ) Σ_(n=0) ^∞ x^n e^(−inθ) ) =(1/(2i))( Σ_(n=0) ^∞ e^(i(n+1)θ) x^n −Σ_(n=0) ^∞ e^(−i(n+1)θ) x^n ) =(1/(2i))Σ_(n=0) ^∞ ( e^(i(n+1)θ) −e^(−i(n+1)θ) )x^n =Σ_(n=0) ^∞ sin((n+1)θ) x^n =sinθ +sin(2θ)x^2 +sin(3θ)x^3 +....

$${let}\:{put}\:{f}\left({x}\right)=\frac{{sin}\theta}{{x}^{\mathrm{2}} \:−\mathrm{2}{xcos}\theta\:+\mathrm{1}}\:\:{poles}\:{of}\:{f}\left({x}\right)? \\ $$$${x}^{\mathrm{2}} −\mathrm{2}{xcos}\theta\:+\mathrm{1}=\mathrm{0}\: \\ $$$$\Delta^{'} ={cos}^{\mathrm{2}} \theta\:−\mathrm{1}\:=−{sin}^{\mathrm{2}} \theta\:=\left({isin}\theta\right)^{\mathrm{2}} \\ $$$${x}_{\mathrm{1}} ={cos}\theta\:+{isin}\theta\:={e}^{{i}\theta} \:\:{and}\:{x}_{\mathrm{2}} ={e}^{−{i}\theta} \:\Rightarrow \\ $$$${f}\left({x}\right)\:=\:\frac{{sin}\theta}{\left({x}−{e}^{{i}\theta} \right)\left({x}−{e}^{−{i}\theta} \right)}\:={sin}\theta\left(\:\frac{\mathrm{1}}{{x}−{e}^{{i}\theta} }\:−\frac{\mathrm{1}}{{x}\:−{e}^{−{i}\theta} }\right).\frac{\mathrm{1}}{\mathrm{2}{isin}\theta} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\left(\:\:\frac{−\mathrm{1}}{{e}^{{i}\theta} \:−{x}}\:+\frac{\mathrm{1}}{{e}^{−{i}\theta} \:−{x}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\left(\:\frac{−{e}^{−{i}\theta} }{\mathrm{1}−{xe}^{−{i}\theta} }\:\:+\:\frac{{e}^{{i}\theta} }{\mathrm{1}−{x}\:{e}^{{i}\theta} }\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\left(\:\:{e}^{{i}\theta} \sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{{n}} \:{e}^{{in}\theta} \:\:−{e}^{−{i}\theta} \:\sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{{n}} \:{e}^{−{in}\theta} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\left(\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{{i}\left({n}+\mathrm{1}\right)\theta} \:{x}^{{n}} \:\:−\sum_{{n}=\mathrm{0}} ^{\infty} \:{e}^{−{i}\left({n}+\mathrm{1}\right)\theta} {x}^{{n}} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(\:{e}^{{i}\left({n}+\mathrm{1}\right)\theta} \:−{e}^{−{i}\left({n}+\mathrm{1}\right)\theta} \right){x}^{{n}} \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:{sin}\left(\left({n}+\mathrm{1}\right)\theta\right)\:{x}^{{n}} \\ $$$$={sin}\theta\:+{sin}\left(\mathrm{2}\theta\right){x}^{\mathrm{2}} \:\:+{sin}\left(\mathrm{3}\theta\right){x}^{\mathrm{3}} \:+…. \\ $$$$ \\ $$

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