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Question-82519




Question Number 82519 by peter frank last updated on 22/Feb/20
Commented by mathmax by abdo last updated on 23/Feb/20
A_θ =Σ_(n=0) ^∞  (((−1)^n )/2^n )cos(nθ)  =Σ_(n=0) ^∞  (−(1/2))^n  cos(nθ)  Re(Σ_(n=0) ^∞  (−(1/2)e^(iθ) )^n )  but Σ_(n=0) ^∞ (−(1/2)e^(iθ) )^n  =(1/(1+(1/2)e^(iθ) ))  =(2/(2+cosθ +isinθ)) =((2(2+cosθ −isinθ))/((2+cosθ)^2  +sin^2 θ)) =((2(2+cosθ))/(4+4cosθ +1))−((isinθ)/(4+4cosθ +1))  ⇒A_θ =((4+2cosθ)/(5+4cosθ))
$${A}_{\theta} =\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{{n}} }{cos}\left({n}\theta\right)\:\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)^{{n}} \:{cos}\left({n}\theta\right) \\ $$$${Re}\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\frac{\mathrm{1}}{\mathrm{2}}{e}^{{i}\theta} \right)^{{n}} \right)\:\:{but}\:\sum_{{n}=\mathrm{0}} ^{\infty} \left(−\frac{\mathrm{1}}{\mathrm{2}}{e}^{{i}\theta} \right)^{{n}} \:=\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}{e}^{{i}\theta} } \\ $$$$=\frac{\mathrm{2}}{\mathrm{2}+{cos}\theta\:+{isin}\theta}\:=\frac{\mathrm{2}\left(\mathrm{2}+{cos}\theta\:−{isin}\theta\right)}{\left(\mathrm{2}+{cos}\theta\right)^{\mathrm{2}} \:+{sin}^{\mathrm{2}} \theta}\:=\frac{\mathrm{2}\left(\mathrm{2}+{cos}\theta\right)}{\mathrm{4}+\mathrm{4}{cos}\theta\:+\mathrm{1}}−\frac{{isin}\theta}{\mathrm{4}+\mathrm{4}{cos}\theta\:+\mathrm{1}} \\ $$$$\Rightarrow{A}_{\theta} =\frac{\mathrm{4}+\mathrm{2}{cos}\theta}{\mathrm{5}+\mathrm{4}{cos}\theta} \\ $$
Answered by mr W last updated on 22/Feb/20
z=(−(1/2))(cos θ+i sin θ)=−(e^(iθ) /2)  z^k =(−1)^k (1/2^k )(cos kθ+i sin kθ)=(−(e^(iθ) /2))^k   Σ_(k=0) ^∞ (−1)^k (1/2^k ) cos kθ+i Σ_(k=0) ^∞ (−1)^k (1/2^k ) sin kθ=Σ_(k=0) ^∞ (−(e^(iθ) /2))^k   Σ_(k=0) ^∞ (−1)^k (1/2^k ) cos kθ+i Σ_(k=0) ^∞ (−1)^k (1/2^k ) sin kθ=(1/(1+(e^(iθ) /2)))  Σ_(k=0) ^∞ (−1)^k (1/2^k ) cos kθ+i Σ_(k=0) ^∞ (−1)^k (1/2^k ) sin kθ=(2/(2+cos θ+i sin θ))  Σ_(k=0) ^∞ (−1)^k (1/2^k ) cos kθ+i Σ_(k=0) ^∞ (−1)^k (1/2^k ) sin kθ=((2(2+cos θ−i sin θ))/((2+cos θ)^2 +sin^2  θ))  Σ_(k=0) ^∞ (−1)^k (1/2^k ) cos kθ+i Σ_(k=0) ^∞ (−1)^k (1/2^k ) sin kθ=((2(2+cos θ)−i 2 sin θ(2+cos θ))/(5+4 cos θ))  ⇒Σ_(k=0) ^∞ (−1)^k (1/2^k ) cos kθ=((2(2+cos θ))/(5+4 cos θ))  ⇒Σ_(k=0) ^∞ (−1)^k (1/2^k ) sin kθ=((−2 sin θ(2+cos θ))/(5+4 cos θ))
$${z}=\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\mathrm{cos}\:\theta+{i}\:\mathrm{sin}\:\theta\right)=−\frac{{e}^{{i}\theta} }{\mathrm{2}} \\ $$$${z}^{{k}} =\left(−\mathrm{1}\right)^{{k}} \frac{\mathrm{1}}{\mathrm{2}^{{k}} }\left(\mathrm{cos}\:{k}\theta+{i}\:\mathrm{sin}\:{k}\theta\right)=\left(−\frac{{e}^{{i}\theta} }{\mathrm{2}}\right)^{{k}} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \frac{\mathrm{1}}{\mathrm{2}^{{k}} }\:\mathrm{cos}\:{k}\theta+{i}\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \frac{\mathrm{1}}{\mathrm{2}^{{k}} }\:\mathrm{sin}\:{k}\theta=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\frac{{e}^{{i}\theta} }{\mathrm{2}}\right)^{{k}} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \frac{\mathrm{1}}{\mathrm{2}^{{k}} }\:\mathrm{cos}\:{k}\theta+{i}\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \frac{\mathrm{1}}{\mathrm{2}^{{k}} }\:\mathrm{sin}\:{k}\theta=\frac{\mathrm{1}}{\mathrm{1}+\frac{{e}^{{i}\theta} }{\mathrm{2}}} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \frac{\mathrm{1}}{\mathrm{2}^{{k}} }\:\mathrm{cos}\:{k}\theta+{i}\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \frac{\mathrm{1}}{\mathrm{2}^{{k}} }\:\mathrm{sin}\:{k}\theta=\frac{\mathrm{2}}{\mathrm{2}+\mathrm{cos}\:\theta+{i}\:\mathrm{sin}\:\theta} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \frac{\mathrm{1}}{\mathrm{2}^{{k}} }\:\mathrm{cos}\:{k}\theta+{i}\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \frac{\mathrm{1}}{\mathrm{2}^{{k}} }\:\mathrm{sin}\:{k}\theta=\frac{\mathrm{2}\left(\mathrm{2}+\mathrm{cos}\:\theta−{i}\:\mathrm{sin}\:\theta\right)}{\left(\mathrm{2}+\mathrm{cos}\:\theta\right)^{\mathrm{2}} +\mathrm{sin}^{\mathrm{2}} \:\theta} \\ $$$$\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \frac{\mathrm{1}}{\mathrm{2}^{{k}} }\:\mathrm{cos}\:{k}\theta+{i}\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \frac{\mathrm{1}}{\mathrm{2}^{{k}} }\:\mathrm{sin}\:{k}\theta=\frac{\mathrm{2}\left(\mathrm{2}+\mathrm{cos}\:\theta\right)−{i}\:\mathrm{2}\:\mathrm{sin}\:\theta\left(\mathrm{2}+\mathrm{cos}\:\theta\right)}{\mathrm{5}+\mathrm{4}\:\mathrm{cos}\:\theta} \\ $$$$\Rightarrow\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \frac{\mathrm{1}}{\mathrm{2}^{{k}} }\:\mathrm{cos}\:{k}\theta=\frac{\mathrm{2}\left(\mathrm{2}+\mathrm{cos}\:\theta\right)}{\mathrm{5}+\mathrm{4}\:\mathrm{cos}\:\theta} \\ $$$$\Rightarrow\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}} \frac{\mathrm{1}}{\mathrm{2}^{{k}} }\:\mathrm{sin}\:{k}\theta=\frac{−\mathrm{2}\:\mathrm{sin}\:\theta\left(\mathrm{2}+\mathrm{cos}\:\theta\right)}{\mathrm{5}+\mathrm{4}\:\mathrm{cos}\:\theta} \\ $$
Commented by peter frank last updated on 22/Feb/20
thank you very much
$${thank}\:{you}\:{very}\:{much} \\ $$

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