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Question Number 54700 by peter frank last updated on 09/Feb/19

Find the sum to infinity 1−(1/(2 ))cos θ+(1/4)cos 2θ−(1/8)cos 3θ+..........

$$\mathrm{F}{i}\mathrm{nd}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{to}\:\mathrm{infinity} \\ $$$$\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}\:}\mathrm{cos}\:\theta+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{cos}\:\mathrm{2}\theta−\frac{\mathrm{1}}{\mathrm{8}}\mathrm{cos}\:\mathrm{3}\theta+.......... \\ $$$$ \\ $$

Commented by maxmathsup by imad last updated on 09/Feb/19

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

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

Answered by tanmay.chaudhury50@gmail.com last updated on 09/Feb/19

p=1−((cosθ)/2)+((cos2θ)/4)−((cos3θ)/8)+... q=−((sinθ)/2)+((sin2θ)/4)−((sin3θ)/8)+... p+iq=1−(1/2)(cosθ+isinθ)+(1/4)(cos2θ+isin2θ)−(1/8)(cos3θ+isin3θ)+... p+iq=1−(e^(iθ) /2)+(e^(i2θ) /2^2 )−(e^(i3θ) /2^3 )+... p+iq=1−a+a^2 −a^3 +... p+iq=(1/(1+a))=(1/(1+(e^(iθ) /2)))=(2/(2+e^(iθ) ))=(2/(2+cosθ+isinθ)) p+iq=((2(2+cosθ−isinθ))/((2+cosθ)^2 +(sinθ)^2 )) p+iq=((4+2cosθ)/(4+4cosθ+1))+((−2isinθ)/(5+4cosθ)) so required ans is=((4+2cosθ)/(5+4cosθ)) 2

$${p}=\mathrm{1}−\frac{{cos}\theta}{\mathrm{2}}+\frac{{cos}\mathrm{2}\theta}{\mathrm{4}}−\frac{{cos}\mathrm{3}\theta}{\mathrm{8}}+... \\ $$$${q}=−\frac{{sin}\theta}{\mathrm{2}}+\frac{{sin}\mathrm{2}\theta}{\mathrm{4}}−\frac{{sin}\mathrm{3}\theta}{\mathrm{8}}+... \\ $$$${p}+{iq}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\left({cos}\theta+{isin}\theta\right)+\frac{\mathrm{1}}{\mathrm{4}}\left({cos}\mathrm{2}\theta+{isin}\mathrm{2}\theta\right)−\frac{\mathrm{1}}{\mathrm{8}}\left({cos}\mathrm{3}\theta+{isin}\mathrm{3}\theta\right)+... \\ $$$${p}+{iq}=\mathrm{1}−\frac{{e}^{{i}\theta} }{\mathrm{2}}+\frac{{e}^{{i}\mathrm{2}\theta} }{\mathrm{2}^{\mathrm{2}} }−\frac{{e}^{{i}\mathrm{3}\theta} }{\mathrm{2}^{\mathrm{3}} }+... \\ $$$${p}+{iq}=\mathrm{1}−{a}+{a}^{\mathrm{2}} −{a}^{\mathrm{3}} +... \\ $$$${p}+{iq}=\frac{\mathrm{1}}{\mathrm{1}+{a}}=\frac{\mathrm{1}}{\mathrm{1}+\frac{{e}^{{i}\theta} }{\mathrm{2}}}=\frac{\mathrm{2}}{\mathrm{2}+{e}^{{i}\theta} }=\frac{\mathrm{2}}{\mathrm{2}+{cos}\theta+{isin}\theta} \\ $$$${p}+{iq}=\frac{\mathrm{2}\left(\mathrm{2}+{cos}\theta−{isin}\theta\right)}{\left(\mathrm{2}+{cos}\theta\right)^{\mathrm{2}} +\left({sin}\theta\right)^{\mathrm{2}} } \\ $$$${p}+{iq}=\frac{\mathrm{4}+\mathrm{2}{cos}\theta}{\mathrm{4}+\mathrm{4}{cos}\theta+\mathrm{1}}+\frac{−\mathrm{2}{isin}\theta}{\mathrm{5}+\mathrm{4}{cos}\theta} \\ $$$${so}\:{required}\:{ans}\:{is}=\frac{\mathrm{4}+\mathrm{2}{cos}\theta}{\mathrm{5}+\mathrm{4}{cos}\theta} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\mathrm{2} \\ $$

Commented by peter frank last updated on 09/Feb/19

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Commented by peter frank last updated on 09/Feb/19

thank you

$${thank}\:{you} \\ $$

Commented by peter frank last updated on 09/Feb/19

sorry sir ellaborate second line

$${sorry}\:{sir}\:{ellaborate}\: \\ $$$${second}\:{line} \\ $$$$ \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 09/Feb/19

p=the sum of series of cosθ q=sum of series of sinθ the method is find p+iq and convert it to series of e^(iθ) then separate real snd imaginary part real part→p imaginary part→q

$${p}={the}\:{sum}\:{of}\:{series}\:{of}\:{cos}\theta \\ $$$${q}={sum}\:{of}\:{series}\:{of}\:{sin}\theta \\ $$$${the}\:{method}\:{is}\:{find}\:{p}+{iq}\:{and}\:{convert}\:{it}\:{to}\:{series}\:{of}\:{e}^{{i}\theta} \\ $$$${then}\:{separate}\:{real}\:{snd}\:{imaginary}\:{part} \\ $$$${real}\:{part}\rightarrow{p}\:\:\:{imaginary}\:{part}\rightarrow{q} \\ $$

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