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Question Number 68206 by turbo msup by abdo last updated on 07/Sep/19

find S(θ)=Σ_(n=0) ^∞ ((sin^3 (nθ))/(n!))

$${find}\:{S}\left(\theta\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{sin}^{\mathrm{3}} \left({n}\theta\right)}{{n}!} \\ $$

Answered by Smail last updated on 07/Sep/19

sin^2 (nθ)=(3/4)sin(nθ)−(1/4)sin(3nθ) Thus, S(θ)=(1/4)Σ_(n=0) ^∞ ((3sin(nθ)−sin(3nθ))/(n!)) =(3/4)Σ_(n=0) ^∞ ((sin(nθ))/(n!))−(1/4)Σ_(n=0) ^∞ ((sin(3nθ))/(n!)) =(3/4)Im(Σ_(n=0) ^∞ (e^(inθ) /(n!)))−(1/4)Im(Σ_(n=0) ^∞ (e^(3inθ) /(n!))) =(3/4)Im(e^e^(iθ) )−(1/4)Im(e^e^(3iθ) ) Im(e^e^(iθ) )=Im(e^(cosθ) e^(isinθ) )=e^(cosθ) sin(sinθ) Im(e^e^(3iθ) )=e^(cos3θ) sin(sin(3θ)) S(θ)=(1/4)(3e^(cosθ) sin(sinθ)−e^(cos3θ) sin(sin(3θ)))

$${sin}^{\mathrm{2}} \left({n}\theta\right)=\frac{\mathrm{3}}{\mathrm{4}}{sin}\left({n}\theta\right)−\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{3}{n}\theta\right) \\ $$$${Thus}, \\ $$$${S}\left(\theta\right)=\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{3}{sin}\left({n}\theta\right)−{sin}\left(\mathrm{3}{n}\theta\right)}{{n}!} \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{sin}\left({n}\theta\right)}{{n}!}−\frac{\mathrm{1}}{\mathrm{4}}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{sin}\left(\mathrm{3}{n}\theta\right)}{{n}!} \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}{Im}\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{e}^{{in}\theta} }{{n}!}\right)−\frac{\mathrm{1}}{\mathrm{4}}{Im}\left(\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{e}^{\mathrm{3}{in}\theta} }{{n}!}\right) \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}{Im}\left({e}^{{e}^{{i}\theta} } \right)−\frac{\mathrm{1}}{\mathrm{4}}{Im}\left({e}^{{e}^{\mathrm{3}{i}\theta} } \right) \\ $$$${Im}\left({e}^{{e}^{{i}\theta} } \right)={Im}\left({e}^{{cos}\theta} {e}^{{isin}\theta} \right)={e}^{{cos}\theta} {sin}\left({sin}\theta\right) \\ $$$${Im}\left({e}^{{e}^{\mathrm{3}{i}\theta} } \right)={e}^{{cos}\mathrm{3}\theta} {sin}\left({sin}\left(\mathrm{3}\theta\right)\right) \\ $$$${S}\left(\theta\right)=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{3}{e}^{{cos}\theta} {sin}\left({sin}\theta\right)−{e}^{{cos}\mathrm{3}\theta} {sin}\left({sin}\left(\mathrm{3}\theta\right)\right)\right) \\ $$

Commented by mathmax by abdo last updated on 07/Sep/19

thank you sir.

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

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