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1-find-n-1-cos-nx-n-and-n-1-sin-nx-n-2-calculate-n-1-1-n-cos-2npi-3-and-n-1-1-n-sin-2npi-3-

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Question Number 45968 by maxmathsup by imad last updated on 19/Oct/18
1)find Σ_(n=1) ^∞ ((cos(nx))/n) and Σ_(n=1) ^∞ ((sin(nx))/n) 2) calculate Σ_(n=1) ^∞ (1/n)cos(((2nπ)/3)) and Σ_(n=1) ^∞ (1/n)sin(((2nπ)/3))
1)find∑n=1∞cos(nx)nand∑n=1∞sin(nx)n2)calculate∑n=1∞1ncos(2nπ3)and∑n=1∞1nsin(2nπ3)
Commented by maxmathsup by imad last updated on 23/Oct/18
1) let f(x)=Σ_(n=1) ^∞ ((cos(nx))/n) and g(x)=Σ_(n=1) ^∞ ((sin(nx))/n) we have f(x)+ig(x)=Σ_(n=1) ^∞ (e^(inx) /n) =Σ_(n=1) ^∞ (((e^(ix) )^n )/n) =−ln(1−e^(ix) ) =−ln(1−cosx −isinx)=−ln(2 sin^2 ((x/2))−2isin((x/2))cos((x/2))) =−ln(−i sin((x/2))( e^(i(x/2)) ))=−ln(−i) +ln(sin((x/2)))−i(x/2) =−ln(e^(−i(π/2)) )−i(x/2) +ln(sin((x/2)))=((iπ)/2)−i(x/2) +ln(sin((x/2))) =i((π−x)/2) +ln(sin((x/2))) ⇒Σ_(n=1) ^∞ ((cos(nx))/n) =ln(sin((x/2))) and Σ_(n=1) ^∞ ((sin(nx))/n) =((π−x)/2) .
1)letf(x)=∑n=1∞cos(nx)nandg(x)=∑n=1∞sin(nx)nwehavef(x)+ig(x)=∑n=1∞einxn=∑n=1∞(eix)nn=−ln(1−eix)=−ln(1−cosx−isinx)=−ln(2sin2(x2)−2isin(x2)cos(x2))=−ln(−isin(x2)(eix2))=−ln(−i)+ln(sin(x2))−ix2=−ln(e−iπ2)−ix2+ln(sin(x2))=iπ2−ix2+ln(sin(x2))=iπ−x2+ln(sin(x2))⇒∑n=1∞cos(nx)n=ln(sin(x2))and∑n=1∞sin(nx)n=π−x2.
Commented by maxmathsup by imad last updated on 23/Oct/18
2)we have proved that Σ_(n=1) ^∞ ((cos(nx))/n)=ln(sin((x/2)) for x=((2π)/3) ⇒ Σ_(n=1) ^∞ ((cos(((2nπ)/3)))/n) =ln(sin((π/3)))=ln(((√3)/2))=(1/2)ln(3)−ln(2) also we have Σ_(n=1) ^∞ ((sin(nx))/n) =((π−x)/2) ⇒Σ_(n=1) ^∞ ((sin(((2nπ)/3)))/n) =((π−((2π)/3))/2) =(π/2)−(π/3) =(π/6) .
2)wehaveprovedthat∑n=1∞cos(nx)n=ln(sin(x2)forx=2π3⇒∑n=1∞cos(2nπ3)n=ln(sin(π3))=ln(32)=12ln(3)−ln(2)alsowehave∑n=1∞sin(nx)n=π−x2⇒∑n=1∞sin(2nπ3)n=π−2π32=π2−π3=π6.
Answered by Smail last updated on 22/Oct/18
p(x)=Σ_(n=1) ^∞ (e^(inx) /n) ln(1−x)=−Σ_(n=1) ^∞ (x^n /n) with ∣x∣<1 So p(x)=−(−Σ_(n=1) ^∞ (((e^(ix) )^n )/n))=−ln(1−e^(ix) ) p(x)=a+ib=−ln(1−e^(ix) ) Σ_(n=1) ^∞ (e^(inx) /n)=Σ_(n=1) ^∞ ((cos(nx))/n)+iΣ_(n=1) ^∞ ((sin(nx))/n) e^(a+ib) =(1/(1−cos(x)−isin(x))) e^a cos(b)+ie^a sin(b)=((1−cos(x)+isin(x))/(2(1−cos(x)))) e^a cos(b)=(1/2) and e^a sin(b)=((sin(x))/(2(1−cos(x)))) tan(b)=((sin(x))/(1−cos(x))) b=tan^(−1) (((sin(x))/(1−cos(x)))) which is b=Σ_(n=1) ^∞ ((sin(nx))/n)=tan^(−1) (((sinx)/(1−cosx))) and e^a cos(b)=(1/2) e^a =(1/(2cos(b)))=(1/2)(√(1+tan^2 (b))) e^a =(1/2)(√(1+((sin^2 x)/((1−cosx)^2 ))))=(1/2)(√((2(1−cos(x)))/((1−cosx)^2 ))) =((√2)/2)(√(1/(1−cosx))) So a=ln((1/( (√2))))+ln((1/( (√(1−cosx))))) a=Σ_(n=1) ^∞ ((cos(nx))/n)=−(1/2)(ln(2)+ln(1−cosx))
p(x)=∑∞n=1einxnln(1−x)=−∑∞n=1xnnwith∣x∣<1Sop(x)=−(−∑∞n=1(eix)nn)=−ln(1−eix)p(x)=a+ib=−ln(1−eix)∑∞n=1einxn=∑∞n=1cos(nx)n+i∑∞n=1sin(nx)nea+ib=11−cos(x)−isin(x)eacos(b)+ieasin(b)=1−cos(x)+isin(x)2(1−cos(x))eacos(b)=12andeasin(b)=sin(x)2(1−cos(x))tan(b)=sin(x)1−cos(x)b=tan−1(sin(x)1−cos(x))whichisb=∑∞n=1sin(nx)n=tan−1(sinx1−cosx)andeacos(b)=12ea=12cos(b)=121+tan2(b)ea=121+sin2x(1−cosx)2=122(1−cos(x))(1−cosx)2=2211−cosxSoa=ln(12)+ln(11−cosx)a=∑∞n=1cos(nx)n=−12(ln(2)+ln(1−cosx))
Commented by Smail last updated on 22/Oct/18
a(x)=−(1/2)(ln(2)+ln(1−cosx)) a(((2π)/3))=Σ_(n=1) ^∞ ((cos(2nπ/3))/n)=−(1/2)(ln(2)+ln(1−cos(2π/3))) =−(1/2)(ln(2)+ln(1+(1/2)))=−((ln3)/2) b(x)=tan^(−1) (((sinx)/(1−cosx))) b(2π/3)=tan^(−1) (((sin(2π/3))/(1−cos(2π/3)))) =tan^(−1) (((√3)/3))=(π/6)+kπ so Σ_(n=1) ^∞ ((sin(2nπ/3))/n)=(π/6)
a(x)=−12(ln(2)+ln(1−cosx))a(2π3)=∑∞n=1cos(2nπ/3)n=−12(ln(2)+ln(1−cos(2π/3)))=−12(ln(2)+ln(1+12))=−ln32b(x)=tan−1(sinx1−cosx)b(2π/3)=tan−1(sin(2π/3)1−cos(2π/3))=tan−1(33)=π6+kπso∑∞n=1sin(2nπ/3)n=π6

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