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Question Number 139826 by qaz last updated on 01/May/21

Σ_(n=0) ^∞ ((sin [(n−1)x])/4^(n+1) )=?

$$\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{sin}\:\left[\left({n}−\mathrm{1}\right){x}\right]}{\mathrm{4}^{{n}+\mathrm{1}} }=? \\ $$

Answered by mnjuly1970 last updated on 01/May/21

Ω:=Σ_(n=0) ^∞ ((sin((n−1)x))/4^(n+1) )=(1/4)ImΣ_(n=0) ^∞ (e^(i(n−1)x) /4^n ) :=(1/4)Im{(e^(ix) )Σ_(n=0) ^∞ (e^(inx) /4^n )} := (1/4)Im{(e^(ix) )Σ_(n=0) ^∞ ((e^(ix) /4))^n } :=(1/4) Im{(e^(ix) )((1/(1−(e^(ix) /4))))} :=Im((cos(x)+isin(x)).((1/(4−cos(x)−isin(x))))) :=Im{(cos(x)+isin(x)}.Im{((4−cos(x)+isin(x))/(16−8cos(x)+1))} := (sin(x)).(((sin(x))/(17−8cos(x))))=((sin^2 (x))/(17−8cos(x))) ...

$$\:\:\:\:\:\Omega:=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{sin}\left(\left({n}−\mathrm{1}\right){x}\right)}{\mathrm{4}^{{n}+\mathrm{1}} }=\frac{\mathrm{1}}{\mathrm{4}}{Im}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{e}^{{i}\left({n}−\mathrm{1}\right){x}} }{\mathrm{4}^{{n}} } \\ $$$$\:\:\:\:\:\:\::=\frac{\mathrm{1}}{\mathrm{4}}{Im}\left\{\left({e}^{{ix}} \right)\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{{e}^{{inx}} }{\mathrm{4}^{{n}} }\right\} \\ $$$$\:\:\:\:\:\:\::=\:\frac{\mathrm{1}}{\mathrm{4}}{Im}\left\{\left({e}^{{ix}} \right)\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{{e}^{{ix}} }{\mathrm{4}}\right)^{{n}} \right\} \\ $$$$\:\:\:\:\:\:\:\::=\frac{\mathrm{1}}{\mathrm{4}}\:{Im}\left\{\left({e}^{{ix}} \right)\left(\frac{\mathrm{1}}{\mathrm{1}−\frac{{e}^{{ix}} }{\mathrm{4}}}\right)\right\} \\ $$$$\:\:\:\:\:\:\::={Im}\left(\left({cos}\left({x}\right)+{isin}\left({x}\right)\right).\left(\frac{\mathrm{1}}{\mathrm{4}−{cos}\left({x}\right)−{isin}\left({x}\right)}\right)\right) \\ $$$$\:\:\:\:\:\:\:\::={Im}\left\{\left({cos}\left({x}\right)+{isin}\left({x}\right)\right\}.{Im}\left\{\frac{\mathrm{4}−{cos}\left({x}\right)+{isin}\left({x}\right)}{\mathrm{16}−\mathrm{8}{cos}\left({x}\right)+\mathrm{1}}\right\}\right. \\ $$$$\:\:\:\:\:\:\::=\:\left({sin}\left({x}\right)\right).\left(\frac{{sin}\left({x}\right)}{\mathrm{17}−\mathrm{8}{cos}\left({x}\right)}\right)=\frac{{sin}^{\mathrm{2}} \left({x}\right)}{\mathrm{17}−\mathrm{8}{cos}\left({x}\right)}\:... \\ $$

Answered by Dwaipayan Shikari last updated on 01/May/21

Σ_(n=1) ^∞ ((sin(nx)cos2x−cos(nx)sin(2x))/(4^n )) =(1/(2i))cos2xΣ_(n=1) ^∞ ((e^(ix) /4))^n −((e^(−ix) /4))^n −(1/2)sin(2x)Σ_(n=1) ^∞ ((e^(ix) /4))^n +((e^(−ix) /4))^n =((cos(2x))/(2i))((1/(1−(e^(ix) /4)))−(1/(1−(e^(−ix) /4))))−((sin(2x))/2)((1/(1−(e^(ix) /4)))+(1/(1−(e^(−ix) /4)))) =((cos(2x))/(2i))((4/(4−e^(ix) ))−(4/(4−e^(−ix) )))−((sin(2x))/2)((4/(4−e^(ix) ))+(4/(4−e^(−ix) ))) =((4cos(2x))/(2i))(((2isin(x))/(16−2cosx+1)))−((sin(2x))/2)(((8−2cosx)/(16−2cosx+1))) =((4cos(2x)sin(x)−4sin(2x)+sin(2x)cos(x))/(17−2cosx))...

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{{sin}\left({nx}\right){cos}\mathrm{2}{x}−{cos}\left({nx}\right){sin}\left(\mathrm{2}{x}\right)}{\mathrm{4}^{{n}} \:} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}{cos}\mathrm{2}{x}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{{e}^{{ix}} }{\mathrm{4}}\right)^{{n}} −\left(\frac{{e}^{−{ix}} }{\mathrm{4}}\right)^{{n}} −\frac{\mathrm{1}}{\mathrm{2}}{sin}\left(\mathrm{2}{x}\right)\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{{e}^{{ix}} }{\mathrm{4}}\right)^{{n}} +\left(\frac{{e}^{−{ix}} }{\mathrm{4}}\right)^{{n}} \\ $$$$=\frac{{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}{i}}\left(\frac{\mathrm{1}}{\mathrm{1}−\frac{{e}^{{ix}} }{\mathrm{4}}}−\frac{\mathrm{1}}{\mathrm{1}−\frac{{e}^{−{ix}} }{\mathrm{4}}}\right)−\frac{{sin}\left(\mathrm{2}{x}\right)}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{1}−\frac{{e}^{{ix}} }{\mathrm{4}}}+\frac{\mathrm{1}}{\mathrm{1}−\frac{{e}^{−{ix}} }{\mathrm{4}}}\right) \\ $$$$=\frac{{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}{i}}\left(\frac{\mathrm{4}}{\mathrm{4}−{e}^{{ix}} }−\frac{\mathrm{4}}{\mathrm{4}−{e}^{−{ix}} }\right)−\frac{{sin}\left(\mathrm{2}{x}\right)}{\mathrm{2}}\left(\frac{\mathrm{4}}{\mathrm{4}−{e}^{{ix}} }+\frac{\mathrm{4}}{\mathrm{4}−{e}^{−{ix}} }\right) \\ $$$$=\frac{\mathrm{4}{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}{i}}\left(\frac{\mathrm{2}{isin}\left({x}\right)}{\mathrm{16}−\mathrm{2}{cosx}+\mathrm{1}}\right)−\frac{{sin}\left(\mathrm{2}{x}\right)}{\mathrm{2}}\left(\frac{\mathrm{8}−\mathrm{2}{cosx}}{\mathrm{16}−\mathrm{2}{cosx}+\mathrm{1}}\right) \\ $$$$=\frac{\mathrm{4}{cos}\left(\mathrm{2}{x}\right){sin}\left({x}\right)−\mathrm{4}{sin}\left(\mathrm{2}{x}\right)+{sin}\left(\mathrm{2}{x}\right){cos}\left({x}\right)}{\mathrm{17}−\mathrm{2}{cosx}}... \\ $$

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