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Question Number 36741 by prof Abdo imad last updated on 04/Jun/18

calculate S(x)=Σ_(n=0) ^∞ ((sin(nx))/(n!))

$${calculate}\:{S}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{sin}\left({nx}\right)}{{n}!} \\ $$

Commented by abdo.msup.com last updated on 05/Jun/18

S(x)=Im(Σ_(n=0) ^∞ (e^(inx) /(n!))) but Σ_(n=0) ^∞ (e^(inx) /(n!)) =Σ_(n=0) ^∞ (((e^(ix) )^n )/(n!)) =e^e^(ix) =e^(cosx +isinx) =e^(cosx) {cos(sinx)+isin(sinx)} ⇒ S(x)= e^(cosx) sin(sinx) .

$${S}\left({x}\right)={Im}\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{e}^{{inx}} }{{n}!}\right)\:{but} \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{e}^{{inx}} }{{n}!}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left({e}^{{ix}} \right)^{{n}} }{{n}!}\:={e}^{{e}^{{ix}} } \\ $$$$={e}^{{cosx}\:+{isinx}} ={e}^{{cosx}} \left\{{cos}\left({sinx}\right)+{isin}\left({sinx}\right)\right\} \\ $$$$\Rightarrow\:{S}\left({x}\right)=\:{e}^{{cosx}} \:{sin}\left({sinx}\right)\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 05/Jun/18

p=((cosnx)/(n!)) q=((sinnx)/(n!)) p+iq=(e^(inx) /(n!))=(((e^(ix) )^n )/(n!)) T_n =(((e^(ix) )^n )/(n!))= T_1 =(((e^(ix) )^1 )/(1!)) T_2 =(((e^(ix) )^2 )/(2!)) T_3 =(((e^(ix) )^3 )/(3!)) ... ... T_n =(((e^(ix) )^n )/(n!)) ... ... T_1 +T_2 +...+T_n +... we know that e^x =(x/(1!))+(x^2 /(2!))+.... soT_1 +T_2 +T_3 +... (e^((e^(ix) )) /1)=e^(cosx+isinx) =e^(cosx) .e^(isinx) =e^(cosx) {cos(sinx)+isin(sinx)} e^(cosx) cos(sinx)+ie^(cosx) sin(sinx) so required ans is e^(cosx) sin(sinx)

$${p}=\frac{{cosnx}}{{n}!} \\ $$$${q}=\frac{{sinnx}}{{n}!} \\ $$$${p}+{iq}=\frac{{e}^{{inx}} }{{n}!}=\frac{\left({e}^{{ix}} \right)^{{n}} }{{n}!} \\ $$$${T}_{{n}} =\frac{\left({e}^{{ix}} \right)^{{n}} }{{n}!}= \\ $$$${T}_{\mathrm{1}} =\frac{\left({e}^{{ix}} \right)^{\mathrm{1}} }{\mathrm{1}!} \\ $$$${T}_{\mathrm{2}} =\frac{\left({e}^{{ix}} \right)^{\mathrm{2}} }{\mathrm{2}!} \\ $$$${T}_{\mathrm{3}} =\frac{\left({e}^{{ix}} \right)^{\mathrm{3}} }{\mathrm{3}!} \\ $$$$... \\ $$$$... \\ $$$${T}_{{n}} =\frac{\left({e}^{{ix}} \right)^{{n}} }{{n}!} \\ $$$$... \\ $$$$... \\ $$$${T}_{\mathrm{1}} +{T}_{\mathrm{2}} +...+{T}_{{n}} +... \\ $$$${we}\:{know}\:{that}\:{e}^{{x}} =\frac{{x}}{\mathrm{1}!}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}!}+.... \\ $$$${soT}_{\mathrm{1}} +{T}_{\mathrm{2}} +{T}_{\mathrm{3}} +... \\ $$$$\frac{{e}^{\left({e}^{{ix}} \right)} }{\mathrm{1}}={e}^{{cosx}+{isinx}} ={e}^{{cosx}} .{e}^{{isinx}} \\ $$$$={e}^{{cosx}} \left\{{cos}\left({sinx}\right)+{isin}\left({sinx}\right)\right\} \\ $$$$ \\ $$$${e}^{{cosx}} {cos}\left({sinx}\right)+{ie}^{{cosx}} {sin}\left({sinx}\right) \\ $$$${so}\:{required}\:{ans}\:{is}\: \\ $$$${e}^{{cosx}} {sin}\left({sinx}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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