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Question Number 62570 by aliesam last updated on 23/Jun/19

Commented by mathmax by abdo last updated on 23/Jun/19

we have S(x) =Σ_(n=1) ^∞ ((sin(nx))/2^(n−1) ) =2 Σ_(n=1) ^∞ ((sin(nx))/2^n ) =2 Im(Σ_(n=1) ^∞ (e^(inx) /2^n )) Σ_(n=1) ^∞ (e^(inx) /2^n ) =Σ_(n=1) ^∞ ((e^(ix) /2))^n =Σ_(n=0) ^∞ ((e^(ix) /2))^n −1 =(1/(1−(e^(ix) /2))) −1 (∣(e^(ix) /2)∣=(1/2)<1) =(2/(2−e^(ix) )) −1 =(2/(2−cosx −isinx)) −1 =((2(2−cosx +isinx))/((2−cosx)^2 +sin^2 x)) −1 =((2(2−cosx)−{ (2−cosx)^2 +sin^2 x} +2i sinx)/((2−cosx)^2 +sin^2 x)) ⇒ S(x) =((4sinx)/((2−cosx)^2 +sin^2 x)) =((4sinx)/(4−4cosx +1)) =((4sinx)/(5−4cosx)) x =(π/4) ⇒S((π/4)) =((4sin((π/4)))/(5−4cos((π/4)))) =((4/(√2))/(5−(4/(√2)))) =(4/(5(√2)−4)) .

$${we}\:{have}\:{S}\left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{sin}\left({nx}\right)}{\mathrm{2}^{{n}−\mathrm{1}} }\:=\mathrm{2}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{sin}\left({nx}\right)}{\mathrm{2}^{{n}} }\:=\mathrm{2}\:{Im}\left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{e}^{{inx}} }{\mathrm{2}^{{n}} }\right)\:\:\:\: \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{e}^{{inx}} }{\mathrm{2}^{{n}} }\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\left(\frac{{e}^{{ix}} }{\mathrm{2}}\right)^{{n}} \:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(\frac{{e}^{{ix}} }{\mathrm{2}}\right)^{{n}} \:−\mathrm{1}\:=\frac{\mathrm{1}}{\mathrm{1}−\frac{{e}^{{ix}} }{\mathrm{2}}}\:−\mathrm{1}\:\:\:\:\left(\mid\frac{{e}^{{ix}} }{\mathrm{2}}\mid=\frac{\mathrm{1}}{\mathrm{2}}<\mathrm{1}\right) \\ $$$$=\frac{\mathrm{2}}{\mathrm{2}−{e}^{{ix}} }\:−\mathrm{1}\:=\frac{\mathrm{2}}{\mathrm{2}−{cosx}\:−{isinx}}\:−\mathrm{1}\:=\frac{\mathrm{2}\left(\mathrm{2}−{cosx}\:+{isinx}\right)}{\left(\mathrm{2}−{cosx}\right)^{\mathrm{2}} \:+{sin}^{\mathrm{2}} {x}}\:−\mathrm{1} \\ $$$$=\frac{\mathrm{2}\left(\mathrm{2}−{cosx}\right)−\left\{\:\left(\mathrm{2}−{cosx}\right)^{\mathrm{2}} +{sin}^{\mathrm{2}} {x}\right\}\:+\mathrm{2}{i}\:{sinx}}{\left(\mathrm{2}−{cosx}\right)^{\mathrm{2}} \:+{sin}^{\mathrm{2}} {x}}\:\Rightarrow \\ $$$${S}\left({x}\right)\:=\frac{\mathrm{4}{sinx}}{\left(\mathrm{2}−{cosx}\right)^{\mathrm{2}} \:+{sin}^{\mathrm{2}} {x}}\:=\frac{\mathrm{4}{sinx}}{\mathrm{4}−\mathrm{4}{cosx}\:+\mathrm{1}}\:=\frac{\mathrm{4}{sinx}}{\mathrm{5}−\mathrm{4}{cosx}} \\ $$$${x}\:=\frac{\pi}{\mathrm{4}}\:\Rightarrow{S}\left(\frac{\pi}{\mathrm{4}}\right)\:=\frac{\mathrm{4}{sin}\left(\frac{\pi}{\mathrm{4}}\right)}{\mathrm{5}−\mathrm{4}{cos}\left(\frac{\pi}{\mathrm{4}}\right)}\:=\frac{\frac{\mathrm{4}}{\sqrt{\mathrm{2}}}}{\mathrm{5}−\frac{\mathrm{4}}{\sqrt{\mathrm{2}}}}\:=\frac{\mathrm{4}}{\mathrm{5}\sqrt{\mathrm{2}}−\mathrm{4}}\:. \\ $$

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