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Question Number 45518 by maxmathsup by imad last updated on 14/Oct/18
calculate Σ_(n=1) ^∞  ((cos(nθ))/n^2 )  and Σ_(n=1) ^∞   ((sin(nθ))/n^2 )
calculaten=1cos(nθ)n2andn=1sin(nθ)n2
Commented by maxmathsup by imad last updated on 23/Oct/18
we have proved that Σ_(n=1) ^∞  ((cos(nx))/n) =ln(sin((x/2))) ⇒  ∫_0 ^x (Σ_(n=1) ^∞  ((cos(nt))/n))dt =∫_0 ^x ln(sin((t/2)))dt ⇒  Σ_(n=1) ^∞  (1/n)[(1/n)sin(nt)]_0 ^x = ∫_0 ^x ln(sin((t/2)))dt⇒  Σ_(n=1) ^∞  ((sin(nx))/n^2 ) = ∫_0 ^x ln(sin((t/2)))dt=_((t/2)=u )    2 ∫_0 ^(x/2) ln(sin(u))du also we have  Σ_(n=1) ^∞  ((sin(nx))/n) =((π−x)/2) ⇒ ∫_0 ^x (Σ_(n=1) ^∞  ((sin(nt))/n))dt =∫_0 ^x  ((π−t)/2)dt ⇒  Σ_(n=1) ^∞  (1/n) ∫_0 ^x  sin(nt)dt =((πx)/2) −(1/2) (x^2 /2) ⇒  Σ_(n=1) ^∞  (1/n)[−(1/n)cos(nt)]_0 ^x  =((πx)/2) −(x^2 /4) ⇒  Σ_(n=1) ^∞  (1/n^2 ){1−cos(nx)} =((πx)/2) −(x^2 /4) ⇒  Σ_(n=1) ^∞ ((cos(nx))/n^2 ) =(π^2 /6) −((πx)/2) +(x^2 /4) .
wehaveprovedthatn=1cos(nx)n=ln(sin(x2))0x(n=1cos(nt)n)dt=0xln(sin(t2))dtn=11n[1nsin(nt)]0x=0xln(sin(t2))dtn=1sin(nx)n2=0xln(sin(t2))dt=t2=u20x2ln(sin(u))dualsowehaven=1sin(nx)n=πx20x(n=1sin(nt)n)dt=0xπt2dtn=11n0xsin(nt)dt=πx212x22n=11n[1ncos(nt)]0x=πx2x24n=11n2{1cos(nx)}=πx2x24n=1cos(nx)n2=π26πx2+x24.
Answered by tanmay.chaudhury50@gmail.com last updated on 14/Oct/18
P_n +iQ_n =((cosnθ)/n^2 )+i((sinnθ)/n^2 )=(e^(inθ) /n^2 )  T_n =P+iQ_n =(e^(inθ) /n^2 )  S_n =Σ_(n=1) ^∞ T_n   S_n =(e^(iθ) /1^2 )+(e^(i2θ) /2^2 )+(e^(i3θ) /3^2 )+...∞  S_n =(x/1^2 )+(x^2 /2^2 )+(x^3 /3^2 )+...∞  wait plz...
Pn+iQn=cosnθn2+isinnθn2=einθn2Tn=P+iQn=einθn2Sn=n=1TnSn=eiθ12+ei2θ22+ei3θ32+Sn=x12+x222+x332+waitplz
Answered by Meritguide1234 last updated on 15/Oct/18
Commented by maxmathsup by imad last updated on 21/Oct/18
your answer is not correct sir because ∣((sinθ)/n^2 )∣≤(1/n^2 ) and Σ (1/n^2 ) converges...
youranswerisnotcorrectsirbecausesinθn2∣⩽1n2andΣ1n2converges

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