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Question Number 67559 by Abdo msup. last updated on 28/Aug/19
calculate  Σ_(n=0) ^∞   (1/(n^(2 ) +1))  and Σ_(n=0) ^∞  (((−1)^n )/(n^2  +1))
calculaten=01n2+1andn=0(1)nn2+1
Commented by ~ À ® @ 237 ~ last updated on 28/Aug/19
let named  f(a)=Σ_(n=0) ^∞  (1/(n^2 +a^2 ))  and  g(a)=Σ_(n=0) ^∞   (((−1)^n )/(n^2 +a^2 ))    f(a)=(1/2)Σ_(−∞) ^∞   (1/(n^2 +a^2 )) =−(1/2)[Res(A(z),ia)+Res(A(z),−ia)]   with  A(z)=((πcot(πz))/(z^2 +1))    (that formula comes from the Residu theorem...  Res(A(z),ia)=((πcot(iπa))/(2ia))= −((πcoth(πa))/(2a))   cause  tanx=−ith(ix)⇒cot(x)=icoth(ix)  Res(A(z),−ia)=((πcot(−iπa))/(−2ia)) =−((πcoth(πa))/(2a))   So  f(a)=(π/(2a)) coth(πa)  g(a)=(1/2)Σ_(−∞) ^∞  (((−1)^n )/(n^2 +a^2 )) =−(1/2) [Res(B,ia)+Res(B,−ia)]      with B(z)=  ((πcsc(πz))/(z^2 +1))=(π/((z^2 +1)sin(πz)))   Res(B,ia)=(π/(2iasin(iπa))) = (π/(2ia(((e^(−πa) −e^(πa) )/(2i)))))=(π/(−2ash(πa)))  Res(B,−ia)=(π/(−2iasin(−iπa)))=(π/(−2ash(πa)))   So  g(a)=(π/(2ash(πa)))
letnamedf(a)=n=01n2+a2andg(a)=n=0(1)nn2+a2f(a)=121n2+a2=12[Res(A(z),ia)+Res(A(z),ia)]withA(z)=πcot(πz)z2+1(thatformulacomesfromtheResidutheoremRes(A(z),ia)=πcot(iπa)2ia=πcoth(πa)2acausetanx=ith(ix)cot(x)=icoth(ix)Res(A(z),ia)=πcot(iπa)2ia=πcoth(πa)2aSof(a)=π2acoth(πa)g(a)=12(1)nn2+a2=12[Res(B,ia)+Res(B,ia)]withB(z)=πcsc(πz)z2+1=π(z2+1)sin(πz)Res(B,ia)=π2iasin(iπa)=π2ia(eπaeπa2i)=π2ash(πa)Res(B,ia)=π2iasin(iπa)=π2ash(πa)Sog(a)=π2ash(πa)
Commented by mathmax by abdo last updated on 28/Aug/19
thank you sir.
thankyousir.
Commented by mathmax by abdo last updated on 29/Aug/19
we have proved tbat  e^(−∣x∣)  =((1−e^(−π) )/π) +(2/π)Σ_(n=1) ^∞  (((1−(−1)^n e^(−π) ))/(1+n^2 ))cos(nx)  (developpement at fourier serie)  x=0 ⇒1 =((1−e^(−π) )/π) +(2/π) Σ_(n=1) ^∞ (1/(n^2  +1)) −(2/π)e^(−π) Σ_(n=1) ^∞    (((−1)^n )/(n^2  +1))  ⇒π =1−e^(−π)  +2 Σ_(n=1) ^∞  (1/(n^2  +1)) −2 e^(−π)  Σ_(n=1) ^∞  (((−1)^n )/(n^2  +1))  x=π ⇒e^(−π)  =((1−e^(−π) )/π) +(2/π) Σ_(n=1) ^∞  (((1−(−1)^n e^(−π) ))/(n^2  +1))(−1)^n  ⇒  e^(−π)  =((1−e^(−π) )/π) +(2/π)Σ_(n=1) ^∞  (((−1)^n )/(n^2  +1)) −(2/π)Σ_(n=1) ^∞   (e^(−π) /(n^2  +1)) ⇒  π e^(−π)  =1−e^(−π)  +2 Σ_(n=1) ^∞  (((−1)^n )/(n^2  +1)) −2 e^(−π)  Σ_(n=1) ^∞  (1/(n^2  +1))  let  x =Σ_(n=1) ^∞  (1/(n^2  +1)) and y =Σ_(n=1) ^∞  (((−1)^n )/(n^2  +1)) we get  π =1−e^(−π)  +2x−2e^(−π) y   and  πe^(−π) =1−e^(−π) +2y −2e^(−π) x ⇒   { ((2x−2e^(−π) y =π−1+e^(−π) )),((−2e^(−π) x +2y =πe^(−π) −1 +e^(−π) )) :}  Δ = determinant (((2           −2e^(−π) )),((−2e^(−π)         2)))=4−4 e^(−2π)  ≠0 ⇒  Δ_x = determinant (((π−1+e^(−π)                  −2e^(−π) )),((πe^(−π) −1 +e^(−π)                2)))=2π−2+2e^(−π)  +2e^(−π) (πe^(−π) −1+e^(−π) )  =2π −2 +2 e^(−π)  +2π e^(−2π) −2e^(−π)  +2 e^(−2π)   =2π −2 +4π e^(−2π)   Δ_y = determinant (((2                    π−1+e^(−π) )),((−2e^(−π)         πe^(−π) −1+e^(−π) )))=2πe^(−π) −2+2e^(−π)   +2e^(−π) (π−1 +e^(−π) )=2π e^(−π) −2 +2 e^(−π)   +2π e^(−π) −2e^(−π)  +2e^(−2π)   =4π e^(−π) −2 +2e^(−2π)   ⇒  x =(Δ_x /Δ) =((2π−2 +4π e^(−2π) )/(4−4e^(−2π) )) =((π−1+2π e^(−2π) )/(2−2 e^(−2π) )) =Σ_(n=1) ^∞  (1/(n^2  +1))  y =(Δ_y /Δ) =((4π e^(−π) −2 +2e^(−2π) )/(4−4e^(−2π) )) =((2π e^(−π) −1 +e^(−2π) )/(2−2e^(−2π) )) =Σ_(n=1) ^∞  (((−1)^n )/(n^2  +1))  Σ_(n=0) ^∞   (1/(n^2 +1)) =1+x =1+((π−1+2π e^(−2π) )/(2−2e^(−2π) ))   Σ_(n=0) ^∞    (((−1)^n )/(n^2 +1)) =1+y =1+((2πe^(−π) −1 +e^(−2π) )/(2−2e^(−2π) ))
wehaveprovedtbatex=1eππ+2πn=1(1(1)neπ)1+n2cos(nx)(developpementatfourierserie)x=01=1eππ+2πn=11n2+12πeπn=1(1)nn2+1π=1eπ+2n=11n2+12eπn=1(1)nn2+1x=πeπ=1eππ+2πn=1(1(1)neπ)n2+1(1)neπ=1eππ+2πn=1(1)nn2+12πn=1eπn2+1πeπ=1eπ+2n=1(1)nn2+12eπn=11n2+1letx=n=11n2+1andy=n=1(1)nn2+1wegetπ=1eπ+2x2eπyandπeπ=1eπ+2y2eπx{2x2eπy=π1+eπ2eπx+2y=πeπ1+eπΔ=|22eπ2eπ2|=44e2π0Δx=|π1+eπ2eππeπ1+eπ2|=2π2+2eπ+2eπ(πeπ1+eπ)=2π2+2eπ+2πe2π2eπ+2e2π=2π2+4πe2πΔy=|2π1+eπ2eππeπ1+eπ|=2πeπ2+2eπ+2eπ(π1+eπ)=2πeπ2+2eπ+2πeπ2eπ+2e2π=4πeπ2+2e2πx=ΔxΔ=2π2+4πe2π44e2π=π1+2πe2π22e2π=n=11n2+1y=ΔyΔ=4πeπ2+2e2π44e2π=2πeπ1+e2π22e2π=n=1(1)nn2+1n=01n2+1=1+x=1+π1+2πe2π22e2πn=0(1)nn2+1=1+y=1+2πeπ1+e2π22e2π

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