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Question Number 67307 by aliesam last updated on 25/Aug/19

Commented by mathmax by abdo last updated on 25/Aug/19

S =Σ_(n=0) ^∞ ((cos(nθ))/2^n ) ⇒ S =Re(Σ_(n=0) ^∞ (e^(inθ) /2^n )) =Re(Σ_(n=0) ^∞ ((e^(iθ) /2))^n ) we have ∣(e^(iθ) /2)∣<1 ⇒Σ_(n=0) ^∞ ((e^(iθ) /2))^n =(1/(1−(e^(iθ) /2))) =(2/(2−e^(iθ) )) =(2/(2−cosθ −isinθ)) =((2(2−cosθ +isinθ))/((2−cosθ)^2 +sin^2 θ)) =((4−2cosθ +2i sinθ)/(4−4cosθ +1)) =((4−2cosθ +2i sinθ)/(5−4cosθ)) ⇒ S =((4−2cosθ)/(5−4cosθ))

$${S}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{cos}\left({n}\theta\right)}{\mathrm{2}^{{n}} }\:\Rightarrow\:{S}\:={Re}\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{e}^{{in}\theta} }{\mathrm{2}^{{n}} }\right)\:={Re}\left(\sum_{{n}=\mathrm{0}} ^{\infty} \left(\frac{{e}^{{i}\theta} }{\mathrm{2}}\right)^{{n}} \right) \\ $$$${we}\:{have}\:\mid\frac{{e}^{{i}\theta} }{\mathrm{2}}\mid<\mathrm{1}\:\Rightarrow\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(\frac{{e}^{{i}\theta} }{\mathrm{2}}\right)^{{n}} \:=\frac{\mathrm{1}}{\mathrm{1}−\frac{{e}^{{i}\theta} }{\mathrm{2}}}\:=\frac{\mathrm{2}}{\mathrm{2}−{e}^{{i}\theta} } \\ $$$$=\frac{\mathrm{2}}{\mathrm{2}−{cos}\theta\:−{isin}\theta}\:=\frac{\mathrm{2}\left(\mathrm{2}−{cos}\theta\:+{isin}\theta\right)}{\left(\mathrm{2}−{cos}\theta\right)^{\mathrm{2}} \:+{sin}^{\mathrm{2}} \theta}\:=\frac{\mathrm{4}−\mathrm{2}{cos}\theta\:+\mathrm{2}{i}\:{sin}\theta}{\mathrm{4}−\mathrm{4}{cos}\theta\:+\mathrm{1}} \\ $$$$=\frac{\mathrm{4}−\mathrm{2}{cos}\theta\:+\mathrm{2}{i}\:{sin}\theta}{\mathrm{5}−\mathrm{4}{cos}\theta}\:\Rightarrow\:{S}\:=\frac{\mathrm{4}−\mathrm{2}{cos}\theta}{\mathrm{5}−\mathrm{4}{cos}\theta} \\ $$

Answered by mind is power last updated on 25/Aug/19

1+(1/2)cos (θ)+(1/4)cos(2θ)+.......... s=1+(1/2)cos (θ)+(1/4)cos(2θ)+.......... s=lim Σ_(n=0) ^k ((cos(nθ))/2^n )=limΣ_(n=0) ^k Re(((e^(inθ) )/2^n ))=limRE(Σ_(n=0) ^k ((e^(iθ) /2))^n ) =lim RE(((1−((e^(iθ) /2))^(n+1) )/(1−(e^(iθ) /2))))=lim RE((((1−(e^(i(n+1)θ) /2^(n+1) ))(1−(e^(−iθ) /2)))/((1−(e^(iθ) /2))(1−(e^(−iθ) /2))))) =lim RE(((1−(e^(i(n+1)θ) /2^(n+1) )−(e^(−iθ) /2^ )+(e^(inθ) /2^(n+2) ))/(1−cos(θ)+(1/4))))=lim((1−((cos(n+1))/2^(n+1) )−((cos(−θ))/2^ )+((cos(nθ))/2^(n+2) ))/((5/4)−cos(θ))) lim ((cos(n+1))/2^(n+1) )=lim((cos(nθ))/2^(n+2) )=0 so we get ((1−((cos(θ))/2))/((5/4)−cos(θ)))=((4−2cos(θ))/(5−4cos(θ)))

$$\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\left(\theta\right)+\frac{\mathrm{1}}{\mathrm{4}}{cos}\left(\mathrm{2}\theta\right)+.......... \\ $$$${s}=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\left(\theta\right)+\frac{\mathrm{1}}{\mathrm{4}}{cos}\left(\mathrm{2}\theta\right)+.......... \\ $$$${s}={lim}\:\sum_{{n}=\mathrm{0}} ^{{k}} \frac{{cos}\left({n}\theta\right)}{\mathrm{2}^{{n}} }={lim}\sum_{{n}=\mathrm{0}} ^{{k}} {Re}\left(\frac{{e}^{{in}\theta} \:}{\mathrm{2}^{{n}} }\right)={limRE}\left(\sum_{{n}=\mathrm{0}} ^{{k}} \left(\frac{{e}^{{i}\theta} }{\mathrm{2}}\right)^{{n}} \right) \\ $$$$={lim}\:{RE}\left(\frac{\mathrm{1}−\left(\frac{{e}^{{i}\theta} }{\mathrm{2}}\right)^{{n}+\mathrm{1}} }{\mathrm{1}−\frac{{e}^{{i}\theta} }{\mathrm{2}}}\right)={lim}\:{RE}\left(\frac{\left(\mathrm{1}−\frac{{e}^{{i}\left({n}+\mathrm{1}\right)\theta} }{\mathrm{2}^{{n}+\mathrm{1}} }\right)\left(\mathrm{1}−\frac{{e}^{−{i}\theta} }{\mathrm{2}}\right)}{\left(\mathrm{1}−\frac{{e}^{{i}\theta} }{\mathrm{2}}\right)\left(\mathrm{1}−\frac{{e}^{−{i}\theta} }{\mathrm{2}}\right)}\right) \\ $$$$={lim}\:{RE}\left(\frac{\mathrm{1}−\frac{{e}^{{i}\left({n}+\mathrm{1}\right)\theta} }{\mathrm{2}^{{n}+\mathrm{1}} }−\frac{{e}^{−{i}\theta} }{\mathrm{2}^{} }+\frac{{e}^{{in}\theta} }{\mathrm{2}^{{n}+\mathrm{2}} }}{\mathrm{1}−{cos}\left(\theta\right)+\frac{\mathrm{1}}{\mathrm{4}}}\right)={lim}\frac{\mathrm{1}−\frac{{cos}\left({n}+\mathrm{1}\right)}{\mathrm{2}^{{n}+\mathrm{1}} }−\frac{{cos}\left(−\theta\right)}{\mathrm{2}^{} }+\frac{{cos}\left({n}\theta\right)}{\mathrm{2}^{{n}+\mathrm{2}} }}{\frac{\mathrm{5}}{\mathrm{4}}−{cos}\left(\theta\right)} \\ $$$${lim}\:\frac{{cos}\left({n}+\mathrm{1}\right)}{\mathrm{2}^{{n}+\mathrm{1}} }={lim}\frac{{cos}\left({n}\theta\right)}{\mathrm{2}^{{n}+\mathrm{2}} }=\mathrm{0} \\ $$$${so}\:{we}\:{get}\:\frac{\mathrm{1}−\frac{{cos}\left(\theta\right)}{\mathrm{2}}}{\frac{\mathrm{5}}{\mathrm{4}}−{cos}\left(\theta\right)}=\frac{\mathrm{4}−\mathrm{2}{cos}\left(\theta\right)}{\mathrm{5}−\mathrm{4}{cos}\left(\theta\right)} \\ $$$$ \\ $$

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