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for-0-lt-r-1-and-x-R-2-find-S-n-0-r-n-cos-n-




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 by abdo 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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