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Question Number 54700 by peter frank last updated on 09/Feb/19
Find the sum to infinity  1−(1/(2 ))cos θ+(1/4)cos 2θ−(1/8)cos 3θ+..........
Findthesumtoinfinity112cosθ+14cos2θ18cos3θ+.
Commented by maxmathsup by imad last updated on 09/Feb/19
we have S =Σ_(n=0) ^∞  (((−1)^n )/2^n )cos(nθ) ⇒S=Re(Σ_(n=0) ^∞  (((−1)^n )/2^n ) e^(inθ) ) but  Σ_(n=0) ^∞   (((−1)^n )/2^n ) e^(inθ)  =Σ_(n=0) ^∞ (−(1/2) e^(iθ) )^n   ( we have ∣−(1/2) e^(iθ) ∣<1 )   =(1/(1+(1/2)e^(iθ) )) =(2/(2 +cosθ +isinθ)) =((2(2+cosθ−isinθ))/((2+cosθ)^2  +sin^2 θ)) ⇒  S =((2(2+cosθ))/(4+4cosθ +1)) =((4 +2cosθ)/(5 +4cosθ))
wehaveS=n=0(1)n2ncos(nθ)S=Re(n=0(1)n2neinθ)butn=0(1)n2neinθ=n=0(12eiθ)n(wehave12eiθ∣<1)=11+12eiθ=22+cosθ+isinθ=2(2+cosθisinθ)(2+cosθ)2+sin2θS=2(2+cosθ)4+4cosθ+1=4+2cosθ5+4cosθ
Answered by tanmay.chaudhury50@gmail.com last updated on 09/Feb/19
p=1−((cosθ)/2)+((cos2θ)/4)−((cos3θ)/8)+...  q=−((sinθ)/2)+((sin2θ)/4)−((sin3θ)/8)+...  p+iq=1−(1/2)(cosθ+isinθ)+(1/4)(cos2θ+isin2θ)−(1/8)(cos3θ+isin3θ)+...  p+iq=1−(e^(iθ) /2)+(e^(i2θ) /2^2 )−(e^(i3θ) /2^3 )+...  p+iq=1−a+a^2 −a^3 +...  p+iq=(1/(1+a))=(1/(1+(e^(iθ) /2)))=(2/(2+e^(iθ) ))=(2/(2+cosθ+isinθ))  p+iq=((2(2+cosθ−isinθ))/((2+cosθ)^2 +(sinθ)^2 ))  p+iq=((4+2cosθ)/(4+4cosθ+1))+((−2isinθ)/(5+4cosθ))  so required ans is=((4+2cosθ)/(5+4cosθ))                      2
p=1cosθ2+cos2θ4cos3θ8+q=sinθ2+sin2θ4sin3θ8+p+iq=112(cosθ+isinθ)+14(cos2θ+isin2θ)18(cos3θ+isin3θ)+p+iq=1eiθ2+ei2θ22ei3θ23+p+iq=1a+a2a3+p+iq=11+a=11+eiθ2=22+eiθ=22+cosθ+isinθp+iq=2(2+cosθisinθ)(2+cosθ)2+(sinθ)2p+iq=4+2cosθ4+4cosθ+1+2isinθ5+4cosθsorequiredansis=4+2cosθ5+4cosθ2
Commented by peter frank last updated on 09/Feb/19
thank you sir
thankyousir
Commented by peter frank last updated on 09/Feb/19
thank you
thankyou
Commented by peter frank last updated on 09/Feb/19
sorry sir ellaborate   second line
sorrysirellaboratesecondline
Commented by tanmay.chaudhury50@gmail.com last updated on 09/Feb/19
p=the sum of series of cosθ  q=sum of series of sinθ  the method is find p+iq and convert it to series of e^(iθ)   then separate real snd imaginary part  real part→p   imaginary part→q
p=thesumofseriesofcosθq=sumofseriesofsinθthemethodisfindp+iqandconvertittoseriesofeiθthenseparaterealsndimaginarypartrealpartpimaginarypartq

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