Menu Close

calculate-n-0-1-n-2-1-and-n-0-1-n-n-2-1-

Tinku Tara 0 Comments
Pin




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))
$${calculate}\:\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}^{\mathrm{2}\:} +\mathrm{1}}\:\:{and}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} \:+\mathrm{1}} \\ $$
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)))
$${let}\:{named}\:\:{f}\left({a}\right)=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{a}^{\mathrm{2}} }\:\:{and}\:\:{g}\left({a}\right)=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} +{a}^{\mathrm{2}} }\:\: \\ $$$${f}\left({a}\right)=\frac{\mathrm{1}}{\mathrm{2}}\underset{−\infty} {\overset{\infty} {\sum}}\:\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} +{a}^{\mathrm{2}} }\:=−\frac{\mathrm{1}}{\mathrm{2}}\left[{Res}\left({A}\left({z}\right),{ia}\right)+{Res}\left({A}\left({z}\right),−{ia}\right)\right]\:\:\:{with}\:\:{A}\left({z}\right)=\frac{\pi{cot}\left(\pi{z}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}\:\:\:\:\left({that}\:{formula}\:{comes}\:{from}\:{the}\:{Residu}\:{theorem}…\right. \\ $$$${Res}\left({A}\left({z}\right),{ia}\right)=\frac{\pi{cot}\left({i}\pi{a}\right)}{\mathrm{2}{ia}}=\:−\frac{\pi{coth}\left(\pi{a}\right)}{\mathrm{2}{a}}\:\:\:{cause}\:\:{tanx}=−{ith}\left({ix}\right)\Rightarrow{cot}\left({x}\right)={icoth}\left({ix}\right) \\ $$$${Res}\left({A}\left({z}\right),−{ia}\right)=\frac{\pi{cot}\left(−{i}\pi{a}\right)}{−\mathrm{2}{ia}}\:=−\frac{\pi{coth}\left(\pi{a}\right)}{\mathrm{2}{a}}\: \\ $$$${So}\:\:{f}\left({a}\right)=\frac{\pi}{\mathrm{2}{a}}\:{coth}\left(\pi{a}\right) \\ $$$${g}\left({a}\right)=\frac{\mathrm{1}}{\mathrm{2}}\underset{−\infty} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} +{a}^{\mathrm{2}} }\:=−\frac{\mathrm{1}}{\mathrm{2}}\:\left[{Res}\left({B},{ia}\right)+{Res}\left({B},−{ia}\right)\right]\:\:\:\:\:\:{with}\:{B}\left({z}\right)=\:\:\frac{\pi{csc}\left(\pi{z}\right)}{{z}^{\mathrm{2}} +\mathrm{1}}=\frac{\pi}{\left({z}^{\mathrm{2}} +\mathrm{1}\right){sin}\left(\pi{z}\right)}\: \\ $$$${Res}\left({B},{ia}\right)=\frac{\pi}{\mathrm{2}{iasin}\left({i}\pi{a}\right)}\:=\:\frac{\pi}{\mathrm{2}{ia}\left(\frac{{e}^{−\pi{a}} −{e}^{\pi{a}} }{\mathrm{2}{i}}\right)}=\frac{\pi}{−\mathrm{2}{ash}\left(\pi{a}\right)} \\ $$$${Res}\left({B},−{ia}\right)=\frac{\pi}{−\mathrm{2}{iasin}\left(−{i}\pi{a}\right)}=\frac{\pi}{−\mathrm{2}{ash}\left(\pi{a}\right)}\: \\ $$$${So}\:\:{g}\left({a}\right)=\frac{\pi}{\mathrm{2}{ash}\left(\pi{a}\right)}\:\: \\ $$$$ \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 28/Aug/19
thank you sir.
$${thank}\:{you}\:{sir}. \\ $$
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) ))
$${we}\:{have}\:{proved}\:{tbat}\:\:{e}^{−\mid{x}\mid} \:=\frac{\mathrm{1}−{e}^{−\pi} }{\pi}\:+\frac{\mathrm{2}}{\pi}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(\mathrm{1}−\left(−\mathrm{1}\right)^{{n}} {e}^{−\pi} \right)}{\mathrm{1}+{n}^{\mathrm{2}} }{cos}\left({nx}\right) \\ $$$$\left({developpement}\:{at}\:{fourier}\:{serie}\right) \\ $$$${x}=\mathrm{0}\:\Rightarrow\mathrm{1}\:=\frac{\mathrm{1}−{e}^{−\pi} }{\pi}\:+\frac{\mathrm{2}}{\pi}\:\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\mathrm{1}}{{n}^{\mathrm{2}} \:+\mathrm{1}}\:−\frac{\mathrm{2}}{\pi}{e}^{−\pi} \sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$\Rightarrow\pi\:=\mathrm{1}−{e}^{−\pi} \:+\mathrm{2}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} \:+\mathrm{1}}\:−\mathrm{2}\:{e}^{−\pi} \:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${x}=\pi\:\Rightarrow{e}^{−\pi} \:=\frac{\mathrm{1}−{e}^{−\pi} }{\pi}\:+\frac{\mathrm{2}}{\pi}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(\mathrm{1}−\left(−\mathrm{1}\right)^{{n}} {e}^{−\pi} \right)}{{n}^{\mathrm{2}} \:+\mathrm{1}}\left(−\mathrm{1}\right)^{{n}} \:\Rightarrow \\ $$$${e}^{−\pi} \:=\frac{\mathrm{1}−{e}^{−\pi} }{\pi}\:+\frac{\mathrm{2}}{\pi}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} \:+\mathrm{1}}\:−\frac{\mathrm{2}}{\pi}\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{{e}^{−\pi} }{{n}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$$\pi\:{e}^{−\pi} \:=\mathrm{1}−{e}^{−\pi} \:+\mathrm{2}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} \:+\mathrm{1}}\:−\mathrm{2}\:{e}^{−\pi} \:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${let}\:\:{x}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} \:+\mathrm{1}}\:{and}\:{y}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} \:+\mathrm{1}}\:{we}\:{get} \\ $$$$\pi\:=\mathrm{1}−{e}^{−\pi} \:+\mathrm{2}{x}−\mathrm{2}{e}^{−\pi} {y}\:\:\:{and}\:\:\pi{e}^{−\pi} =\mathrm{1}−{e}^{−\pi} +\mathrm{2}{y}\:−\mathrm{2}{e}^{−\pi} {x}\:\Rightarrow \\ $$$$\begin{cases}{\mathrm{2}{x}−\mathrm{2}{e}^{−\pi} {y}\:=\pi−\mathrm{1}+{e}^{−\pi} }\\{−\mathrm{2}{e}^{−\pi} {x}\:+\mathrm{2}{y}\:=\pi{e}^{−\pi} −\mathrm{1}\:+{e}^{−\pi} }\end{cases} \\ $$$$\Delta\:=\begin{vmatrix}{\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2}{e}^{−\pi} }\\{−\mathrm{2}{e}^{−\pi} \:\:\:\:\:\:\:\:\mathrm{2}}\end{vmatrix}=\mathrm{4}−\mathrm{4}\:{e}^{−\mathrm{2}\pi} \:\neq\mathrm{0}\:\Rightarrow \\ $$$$\Delta_{{x}} =\begin{vmatrix}{\pi−\mathrm{1}+{e}^{−\pi} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{2}{e}^{−\pi} }\\{\pi{e}^{−\pi} −\mathrm{1}\:+{e}^{−\pi} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}}\end{vmatrix}=\mathrm{2}\pi−\mathrm{2}+\mathrm{2}{e}^{−\pi} \:+\mathrm{2}{e}^{−\pi} \left(\pi{e}^{−\pi} −\mathrm{1}+{e}^{−\pi} \right) \\ $$$$=\mathrm{2}\pi\:−\mathrm{2}\:+\mathrm{2}\:{e}^{−\pi} \:+\mathrm{2}\pi\:{e}^{−\mathrm{2}\pi} −\mathrm{2}{e}^{−\pi} \:+\mathrm{2}\:{e}^{−\mathrm{2}\pi} \\ $$$$=\mathrm{2}\pi\:−\mathrm{2}\:+\mathrm{4}\pi\:{e}^{−\mathrm{2}\pi} \\ $$$$\Delta_{{y}} =\begin{vmatrix}{\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\pi−\mathrm{1}+{e}^{−\pi} }\\{−\mathrm{2}{e}^{−\pi} \:\:\:\:\:\:\:\:\pi{e}^{−\pi} −\mathrm{1}+{e}^{−\pi} }\end{vmatrix}=\mathrm{2}\pi{e}^{−\pi} −\mathrm{2}+\mathrm{2}{e}^{−\pi} \\ $$$$+\mathrm{2}{e}^{−\pi} \left(\pi−\mathrm{1}\:+{e}^{−\pi} \right)=\mathrm{2}\pi\:{e}^{−\pi} −\mathrm{2}\:+\mathrm{2}\:{e}^{−\pi} \:\:+\mathrm{2}\pi\:{e}^{−\pi} −\mathrm{2}{e}^{−\pi} \:+\mathrm{2}{e}^{−\mathrm{2}\pi} \\ $$$$=\mathrm{4}\pi\:{e}^{−\pi} −\mathrm{2}\:+\mathrm{2}{e}^{−\mathrm{2}\pi} \:\:\Rightarrow \\ $$$${x}\:=\frac{\Delta_{{x}} }{\Delta}\:=\frac{\mathrm{2}\pi−\mathrm{2}\:+\mathrm{4}\pi\:{e}^{−\mathrm{2}\pi} }{\mathrm{4}−\mathrm{4}{e}^{−\mathrm{2}\pi} }\:=\frac{\pi−\mathrm{1}+\mathrm{2}\pi\:{e}^{−\mathrm{2}\pi} }{\mathrm{2}−\mathrm{2}\:{e}^{−\mathrm{2}\pi} }\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${y}\:=\frac{\Delta_{{y}} }{\Delta}\:=\frac{\mathrm{4}\pi\:{e}^{−\pi} −\mathrm{2}\:+\mathrm{2}{e}^{−\mathrm{2}\pi} }{\mathrm{4}−\mathrm{4}{e}^{−\mathrm{2}\pi} }\:=\frac{\mathrm{2}\pi\:{e}^{−\pi} −\mathrm{1}\:+{e}^{−\mathrm{2}\pi} }{\mathrm{2}−\mathrm{2}{e}^{−\mathrm{2}\pi} }\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} +\mathrm{1}}\:=\mathrm{1}+{x}\:=\mathrm{1}+\frac{\pi−\mathrm{1}+\mathrm{2}\pi\:{e}^{−\mathrm{2}\pi} }{\mathrm{2}−\mathrm{2}{e}^{−\mathrm{2}\pi} }\: \\ $$$$\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} +\mathrm{1}}\:=\mathrm{1}+{y}\:=\mathrm{1}+\frac{\mathrm{2}\pi{e}^{−\pi} −\mathrm{1}\:+{e}^{−\mathrm{2}\pi} }{\mathrm{2}−\mathrm{2}{e}^{−\mathrm{2}\pi} } \\ $$$$ \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *