Question Number 204754 by York12 last updated on 27/Feb/24
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Commented by York12 last updated on 27/Feb/24
Commented by Rasheed.Sindhi last updated on 27/Feb/24
$${sir},\:{if}\:{you}\:{can}\:{upload}\:{an}\:{image} \\ $$$${as}\:{a}\:{comment},\:{you}\:{can}\:{do}\:{so} \\ $$$${as}\:{a}\:{question}\:{also}! \\ $$
Commented by York12 last updated on 27/Feb/24
$$\mathrm{Rasheed}\:\mathrm{sindhai}\:\mathrm{yes}\:\mathrm{sir}\:\mathrm{I}\:\mathrm{know}\:\mathrm{but}\:\mathrm{my}\:\mathrm{phone} \\ $$$$\mathrm{Ram}\:\mathrm{is}\:\mathrm{pretty}\:\mathrm{bad} \\ $$
Answered by mr W last updated on 27/Feb/24
![t_k =x^k cos (kθ) s_k =x^k sin (kθ) t_k +is_k =x^k [cos (kθ)+i sin (kθ)]=x^k e^(kθi) =(xe^(θi) )^k T_n =Σ_(k=0) ^n t_k S_n =Σ_(k=0) ^n s_k T_n +iS_n =Σ_(k=0) ^n (t_k +is_k ) =Σ_(k=0) ^n (xe^(θi) )^k =(((xe^(θi) )^(n+1) −1)/(xe^(θi) −1)) =((x^(n+1) [cos (n+1)θ+i sin (n+1)θ]−1)/(x(cos θ+i sin θ)−1)) =(([x^(n+1) cos (n+1)θ−1]+i x^(n+1) sin (n+1)θ)/((xcos θ−1)+i xsin θ)) =(({[x^(n+1) cos (n+1)θ−1]+i x^(n+1) sin (n+1)θ}{(xcos θ−1)−i xsin θ})/((xcos θ−1)^2 +x^2 sin^2 θ)) =(({1+x^(n+2) cos (nθ)−x^(n+1) cos (n+1)θ−x cos θ}+i{x^(n+2) sin (nθ)−x^(n+1) sin (n+1)θ+x sin θ})/(x^2 +1−2x cos θ)) ⇒S_n =((x^(n+2) sin (nθ)−x^(n+1) sin (n+1)θ+x sin θ)/(x^2 +1−2x cos θ)) ⇒lim_(n→∞) S_n =((x sin θ)/(x^2 +1−2x cos θ))](https://www.tinkutara.com/question/Q204785.png)
$${t}_{{k}} ={x}^{{k}} \mathrm{cos}\:\left({k}\theta\right) \\ $$$${s}_{{k}} ={x}^{{k}} \mathrm{sin}\:\left({k}\theta\right) \\ $$$${t}_{{k}} +{is}_{{k}} ={x}^{{k}} \left[\mathrm{cos}\:\left({k}\theta\right)+{i}\:\mathrm{sin}\:\left({k}\theta\right)\right]={x}^{{k}} {e}^{{k}\theta{i}} =\left({xe}^{\theta{i}} \right)^{{k}} \\ $$$${T}_{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{t}_{{k}} \\ $$$${S}_{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{s}_{{k}} \\ $$$${T}_{{n}} +{iS}_{{n}} =\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left({t}_{{k}} +{is}_{{k}} \right) \\ $$$$\:\:\:\:=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\left({xe}^{\theta{i}} \right)^{{k}} =\frac{\left({xe}^{\theta{i}} \right)^{{n}+\mathrm{1}} −\mathrm{1}}{{xe}^{\theta{i}} −\mathrm{1}} \\ $$$$\:\:\:\:=\frac{{x}^{{n}+\mathrm{1}} \left[\mathrm{cos}\:\left({n}+\mathrm{1}\right)\theta+{i}\:\mathrm{sin}\:\left({n}+\mathrm{1}\right)\theta\right]−\mathrm{1}}{{x}\left(\mathrm{cos}\:\theta+{i}\:\mathrm{sin}\:\theta\right)−\mathrm{1}} \\ $$$$\:\:\:\:=\frac{\left[{x}^{{n}+\mathrm{1}} \mathrm{cos}\:\left({n}+\mathrm{1}\right)\theta−\mathrm{1}\right]+{i}\:{x}^{{n}+\mathrm{1}} \mathrm{sin}\:\left({n}+\mathrm{1}\right)\theta}{\left({x}\mathrm{cos}\:\theta−\mathrm{1}\right)+{i}\:{x}\mathrm{sin}\:\theta} \\ $$$$\:\:\:\:=\frac{\left\{\left[{x}^{{n}+\mathrm{1}} \mathrm{cos}\:\left({n}+\mathrm{1}\right)\theta−\mathrm{1}\right]+{i}\:{x}^{{n}+\mathrm{1}} \mathrm{sin}\:\left({n}+\mathrm{1}\right)\theta\right\}\left\{\left({x}\mathrm{cos}\:\theta−\mathrm{1}\right)−{i}\:{x}\mathrm{sin}\:\theta\right\}}{\left({x}\mathrm{cos}\:\theta−\mathrm{1}\right)^{\mathrm{2}} +{x}^{\mathrm{2}} \:\mathrm{sin}^{\mathrm{2}} \:\theta} \\ $$$$\:\:\:\:=\frac{\left\{\mathrm{1}+{x}^{{n}+\mathrm{2}} \mathrm{cos}\:\left({n}\theta\right)−{x}^{{n}+\mathrm{1}} \mathrm{cos}\:\left({n}+\mathrm{1}\right)\theta−{x}\:\mathrm{cos}\:\theta\right\}+{i}\left\{{x}^{{n}+\mathrm{2}} \mathrm{sin}\:\left({n}\theta\right)−{x}^{{n}+\mathrm{1}} \mathrm{sin}\:\left({n}+\mathrm{1}\right)\theta+{x}\:\mathrm{sin}\:\theta\right\}}{{x}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}{x}\:\mathrm{cos}\:\theta} \\ $$$$\Rightarrow{S}_{{n}} =\frac{{x}^{{n}+\mathrm{2}} \mathrm{sin}\:\left({n}\theta\right)−{x}^{{n}+\mathrm{1}} \mathrm{sin}\:\left({n}+\mathrm{1}\right)\theta+{x}\:\mathrm{sin}\:\theta}{{x}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}{x}\:\mathrm{cos}\:\theta} \\ $$$$\Rightarrow\underset{{n}\rightarrow\infty} {\mathrm{lim}}{S}_{{n}} =\frac{{x}\:\mathrm{sin}\:\theta}{{x}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}{x}\:\mathrm{cos}\:\theta} \\ $$
Commented by York12 last updated on 27/Feb/24
$$\mathrm{Perfect} \\ $$
Commented by York12 last updated on 27/Feb/24
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir} \\ $$