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Question Number 30601 by abdo imad last updated on 23/Feb/18

for 0<r≤1 and (θ,x)∈R^2 find S=Σ_(n=0) ^∞ r^n cos(nθ).

$${for}\:\mathrm{0}<{r}\leqslant\mathrm{1}\:{and}\:\left(\theta,{x}\right)\in{R}^{\mathrm{2}} \:\:{find} \\ $$ $${S}=\sum_{{n}=\mathrm{0}} ^{\infty} \:{r}^{{n}} {cos}\left({n}\theta\right). \\ $$

Commented byabdo imad last updated on 24/Feb/18

S=Re(Σ_(n=0) ^∞ r^n e^(inθ) )but Σ_(n=0) ^∞ r^n e^(inθ) = Σ_(n=0) ^∞ (re^(iθ) )^n =(1/(1−re^(iθ) )) = (1/(1−r cosθ −irsinθ))=((1−r cosθ +isinθ)/((1−r cosθ)^2 +r^2 sin^2 θ)) ⇒ S= ((1−r cosθ)/((1−rcosθ)^2 +r^2 sin^2 θ))=((1−r cosθ)/(1−2rcosθ +r^2 )) but we must study the case r=1.

$${S}={Re}\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:{r}^{{n}} \:{e}^{{in}\theta} \right){but} \\ $$ $$\sum_{{n}=\mathrm{0}} ^{\infty} \:{r}^{{n}} \:{e}^{{in}\theta} \:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \left({re}^{{i}\theta} \right)^{{n}} \:=\frac{\mathrm{1}}{\mathrm{1}−{re}^{{i}\theta} } \\ $$ $$=\:\:\:\frac{\mathrm{1}}{\mathrm{1}−{r}\:{cos}\theta\:−{irsin}\theta}=\frac{\mathrm{1}−{r}\:{cos}\theta\:\:+{isin}\theta}{\left(\mathrm{1}−{r}\:{cos}\theta\right)^{\mathrm{2}} \:+{r}^{\mathrm{2}} {sin}^{\mathrm{2}} \theta}\:\Rightarrow \\ $$ $${S}=\:\:\:\frac{\mathrm{1}−{r}\:{cos}\theta}{\left(\mathrm{1}−{rcos}\theta\right)^{\mathrm{2}} \:+{r}^{\mathrm{2}} \:{sin}^{\mathrm{2}} \theta}=\frac{\mathrm{1}−{r}\:{cos}\theta}{\mathrm{1}−\mathrm{2}{rcos}\theta\:+{r}^{\mathrm{2}} }\:\:{but}\:{we}\:{must} \\ $$ $${study}\:{the}\:{case}\:{r}=\mathrm{1}. \\ $$

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