Menu Close

calculateA-n-n-1-x-n-cos-n-and-B-n-n-1-x-n-sin-n-

Tinku Tara 0 Comments
Pin




Question Number 53781 by maxmathsup by imad last updated on 25/Jan/19
calculateA_n =Σ_(n=1) ^∞ x^n cos(nθ) and B_n =Σ_(n=1) ^∞ x^n sin(nθ)
$${calculateA}_{{n}} =\sum_{{n}=\mathrm{1}} ^{\infty} \:{x}^{{n}} {cos}\left({n}\theta\right)\:\:{and}\:{B}_{{n}} =\sum_{{n}=\mathrm{1}} ^{\infty} \:{x}^{{n}} {sin}\left({n}\theta\right) \\ $$
Commented by maxmathsup by imad last updated on 26/Jan/19
we have A_n +i B_n =Σ_(n=1) ^∞ x^n e^(inθ) =Σ_(n=0) ^∞ (x e^(iθ) )^n −1 so if ∣x∣<1 we get A_n +iB_n =(1/(1−xe^(iθ) )) −1 =(1/(1−xcosθ −ix sinθ)) −1 = ((1−xcosθ +ix sinθ)/((1−xcosθ)^2 +x^2 sin^2 θ)) −1 =((1−xcosθ)/(1−2xcosθ +x^2 )) −1 +i((xsinθ)/(x^2 −2xcosθ +1)) =((1−xcosθ −1 +2xcosθ−x^2 )/(x^2 −2xcosθ +1)) +i ((xsinθ)/(x^2 −2x cosθ +1)) ⇒ A_n =((−x^2 +x cosθ +1)/(x^2 −2xcosθ +1)) and B_n =((xsinθ)/(x^2 −2x cosθ +1))
$${we}\:{have}\:{A}_{{n}} +{i}\:{B}_{{n}} =\sum_{{n}=\mathrm{1}} ^{\infty} \:{x}^{{n}} \:{e}^{{in}\theta} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \left({x}\:{e}^{{i}\theta} \right)^{{n}} \:−\mathrm{1} \\ $$$$\:{so}\:{if}\:\mid{x}\mid<\mathrm{1}\:\:\:{we}\:{get}\:{A}_{{n}} \:+{iB}_{{n}} =\frac{\mathrm{1}}{\mathrm{1}−{xe}^{{i}\theta} }\:−\mathrm{1}\:=\frac{\mathrm{1}}{\mathrm{1}−{xcos}\theta\:−{ix}\:{sin}\theta}\:−\mathrm{1} \\ $$$$=\:\frac{\mathrm{1}−{xcos}\theta\:+{ix}\:{sin}\theta}{\left(\mathrm{1}−{xcos}\theta\right)^{\mathrm{2}} \:+{x}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta}\:−\mathrm{1} \\ $$$$=\frac{\mathrm{1}−{xcos}\theta}{\mathrm{1}−\mathrm{2}{xcos}\theta\:+{x}^{\mathrm{2}} }\:−\mathrm{1}\:+{i}\frac{{xsin}\theta}{{x}^{\mathrm{2}} \:−\mathrm{2}{xcos}\theta\:+\mathrm{1}} \\ $$$$=\frac{\mathrm{1}−{xcos}\theta\:−\mathrm{1}\:+\mathrm{2}{xcos}\theta−{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} \:−\mathrm{2}{xcos}\theta\:+\mathrm{1}}\:+{i}\:\frac{{xsin}\theta}{{x}^{\mathrm{2}} −\mathrm{2}{x}\:{cos}\theta\:+\mathrm{1}}\:\Rightarrow \\ $$$${A}_{{n}} =\frac{−{x}^{\mathrm{2}} \:+{x}\:{cos}\theta\:+\mathrm{1}}{{x}^{\mathrm{2}} \:−\mathrm{2}{xcos}\theta\:+\mathrm{1}}\:\:{and}\:{B}_{{n}} =\frac{{xsin}\theta}{{x}^{\mathrm{2}} \:−\mathrm{2}{x}\:{cos}\theta\:+\mathrm{1}} \\ $$$$ \\ $$
Answered by Smail last updated on 26/Jan/19
A_n +iB_n =Σ_(n=1) ^∞ x^n e^(inθ) =Σ_(n=0) ^∞ (xe^(iθ) )^n −1=(1/(1−xe^(iθ) ))−1=((xe^(iθ) )/(1−xe^(iθ) )) =((xe^(iθ) )/(1−xcosθ−xisinθ)) =((xe^(iθ) (1−xe^(−iθ) ))/((1−xe^(iθ) )(1−xe^(−iθ) )))=((x(e^(iθ) −x))/(1−x(e^(iθ) +e^(−iθ) )+x^2 )) =((x(cosθ−x+isinθ))/(1+x^2 −2xcosθ)) A_n +iB_n =((x(cosθ−x))/(1+x^2 −2xcosθ))+i((xsinθ)/(1+x^2 −2xcosθ)) A_n =((x(cosθ−x))/((x−cosθ)^2 +sin^2 θ)) B_n =((xsinθ)/((x−cosθ)^2 +sin^2 θ))
$${A}_{{n}} +{iB}_{{n}} =\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{x}^{{n}} {e}^{{in}\theta} \\ $$$$=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left({xe}^{{i}\theta} \right)^{{n}} −\mathrm{1}=\frac{\mathrm{1}}{\mathrm{1}−{xe}^{{i}\theta} }−\mathrm{1}=\frac{{xe}^{{i}\theta} }{\mathrm{1}−{xe}^{{i}\theta} } \\ $$$$=\frac{{xe}^{{i}\theta} }{\mathrm{1}−{xcos}\theta−{xisin}\theta} \\ $$$$=\frac{{xe}^{{i}\theta} \left(\mathrm{1}−{xe}^{−{i}\theta} \right)}{\left(\mathrm{1}−{xe}^{{i}\theta} \right)\left(\mathrm{1}−{xe}^{−{i}\theta} \right)}=\frac{{x}\left({e}^{{i}\theta} −{x}\right)}{\mathrm{1}−{x}\left({e}^{{i}\theta} +{e}^{−{i}\theta} \right)+{x}^{\mathrm{2}} } \\ $$$$=\frac{{x}\left({cos}\theta−{x}+{isin}\theta\right)}{\mathrm{1}+{x}^{\mathrm{2}} −\mathrm{2}{xcos}\theta} \\ $$$${A}_{{n}} +{iB}_{{n}} =\frac{{x}\left({cos}\theta−{x}\right)}{\mathrm{1}+{x}^{\mathrm{2}} −\mathrm{2}{xcos}\theta}+{i}\frac{{xsin}\theta}{\mathrm{1}+{x}^{\mathrm{2}} −\mathrm{2}{xcos}\theta} \\ $$$${A}_{{n}} =\frac{{x}\left({cos}\theta−{x}\right)}{\left({x}−{cos}\theta\right)^{\mathrm{2}} +{sin}^{\mathrm{2}} \theta} \\ $$$${B}_{{n}} =\frac{{xsin}\theta}{\left({x}−{cos}\theta\right)^{\mathrm{2}} +{sin}^{\mathrm{2}} \theta} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *