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Question Number 48091 by peter frank last updated on 19/Nov/18

Commented by maxmathsup by imad last updated on 19/Nov/18

we have e^(−x) sinx =Im(e^(−x+ix) )=Im(e^((−1+i)x) ) but  e^((−1+i)x) =Σ_(n=0) ^∞  (((−1+i)^n x^n )/(n!))  but (−1+i)=(√2)(−(1/(√2)) +(i/(√2)))  =(√2)(cos(((3π)/4))+isin(((3π)/4))) =(√2)e^(i((3π)/4))  ⇒(−1+i)^n =((√2))^n  e^(i((3nπ)/4))   =((√2))^n { cos(((3nπ)/4))+isin(((3nπ)/4))} ⇒e^(−x) sin(x)=Σ_(n=0) ^∞ ((((√2))^n )/(n!)) sin(((3nπ)/4))x^n    x=(π/3) ⇒ e^(−(π/3))  sin((π/3))=Σ_(n=0) ^∞  ((((√2))^n )/(n!)) sin(((3nπ)/4))((π/3))^n  .

wehaveexsinx=Im(ex+ix)=Im(e(1+i)x)bute(1+i)x=n=0(1+i)nxnn!but(1+i)=2(12+i2)=2(cos(3π4)+isin(3π4))=2ei3π4(1+i)n=(2)nei3nπ4=(2)n{cos(3nπ4)+isin(3nπ4)}exsin(x)=n=0(2)nn!sin(3nπ4)xnx=π3eπ3sin(π3)=n=0(2)nn!sin(3nπ4)(π3)n.

Commented by peter frank last updated on 19/Nov/18

thanks

thanks

Answered by tanmay.chaudhury50@gmail.com last updated on 19/Nov/18

p=e^(−x) cosx  q=e^(−x) sinx  p+iq=e^(−x) (cosx+isinx)=e^(−x) .e^(ix) =e^(x(−1+i)) =e^(tx)   e^(tx) =1+tx+((t^2 x^2 )/(2!))+((t^3 x^3 )/(3!))+...  =1+(−1+i)x+(−1+i)^2 ×(x^2 /(2!))+(−1+i)^3 ×(x^3 /(3!)) other  terms not considered as to find coefficient of  upto x^3   =1+(−x+ix)+(−ix^2 )+(−2i)(−1+i)×(x^3 /(3!))  =1+(−x+ix)+(−ix^2 )+(2i+2)(x^3 /6)  =(1−x+(x^3 /3))+i(x−x^2 +(x^3 /3))  so e^(tx) =e^((−1+i)x) =(1−x+(x^3 /3))+i×(x/3)(x^2 −3x+3)  e^(−x) ×e^(ix) =(1−x+(x^3 /3))+i×(x/3)(x^2 −3x+3)  e^(−x) (cosx+isinx)=(1−x+(x^3 /3))+i×(x/3)(x^2 −3x+3)  hence  e^(−x) sinx=(x/3)(x^2 −3x+3)  proved  e^(−(π/3)) sin((π/3))=(π/9)((π^2 /9)−π+3)

p=excosxq=exsinxp+iq=ex(cosx+isinx)=ex.eix=ex(1+i)=etxetx=1+tx+t2x22!+t3x33!+...=1+(1+i)x+(1+i)2×x22!+(1+i)3×x33!othertermsnotconsideredastofindcoefficientofuptox3=1+(x+ix)+(ix2)+(2i)(1+i)×x33!=1+(x+ix)+(ix2)+(2i+2)x36=(1x+x33)+i(xx2+x33)soetx=e(1+i)x=(1x+x33)+i×x3(x23x+3)ex×eix=(1x+x33)+i×x3(x23x+3)ex(cosx+isinx)=(1x+x33)+i×x3(x23x+3)henceexsinx=x3(x23x+3)provedeπ3sin(π3)=π9(π29π+3)

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