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Question Number 132496 by mathmax by abdo last updated on 14/Feb/21
$$\mathrm{calculateA}_{\mathrm{n}} =\:\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} }\mathrm{cos}\left(\mathrm{n}\right)\:\mathrm{andB}_{\mathrm{n}} =\:\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} }\mathrm{cos}\left(\frac{\mathrm{n}\pi}{\mathrm{3}}\right) \\ $$
Answered by mnjuly1970 last updated on 14/Feb/21
![by using dilogarithm function or fourier series... first method.. f(x)=x x∈[−π,π] f is artificial periodic function on [−π ,π] f(x)≈(a_0 /2)+Σ_(n=1) (a_n cos(nx)+b_n sin(nx)) a_n =(1/π)∫_(−π) ^( π) f(x)cos(nx)dx=0 b_n =(1/π)∫_(−π) ^( π) f(x)sin(nx)dx =(1/π)∫_(−π) ^( π) xsin(nx)dx=(2/π)∫_0 ^( π) xsin(nx)dx =(2/π){[−(x/n)cos(nx)]_(0 ) ^π +(1/n)∫_0 ^( π) cos(nx)dx 2(−1)^(n+1) (1/n) a_0 =(1/π)∫_(−π) ^( π) f(x)dx=0 x≈ Σ_(n=1) ^∞ ((2(−1)^(n+1) )/(n ))sin(nx) ∫xdx≈2Σ_(n=1) ^∞ ∫ (((−1)^(n+1) )/(n ))sin(nx)dx (x^2 /2)≈C+2Σ_(n=1) ^∞ (((−1)^n )/n^2 )cos(nx) x=π⇒ (π^2 /2)≈C+2Σ_(n=1) (1/n^2 ) C=(π^2 /2)−(π^2 /3)=(π^2 /6) (x^2 /2)=(π^2 /6)+2Σ_(n=1) ^∞ (((−1)^n )/n^2 )cos(nx) x=1 ⇒(1/2)−(π^2 /6)=2Σ_(n=1) ^∞ (((−1)^n )/n^2 ) Σ_(n=1) ^∞ (((−1)^n )/n^2 )=(1/4)−(π^2 /(12)) ...](Q132503.png)
$$\:{by}\:{using}\:{dilogarithm}\:{function} \\ $$$${or}\:{fourier}\:{series}... \\ $$$${first}\:\:{method}.. \\ $$$$\:\:\:{f}\left({x}\right)={x}\:\:\:\:\:\:\:\:\:{x}\in\left[−\pi,\pi\right] \\ $$$$\:\:\:\:{f}\:{is}\:{artificial}\:{periodic}\:{function} \\ $$$${on}\:\left[−\pi\:,\pi\right] \\ $$$$\:{f}\left({x}\right)\approx\frac{{a}_{\mathrm{0}} }{\mathrm{2}}+\underset{{n}=\mathrm{1}} {\sum}\left({a}_{{n}} {cos}\left({nx}\right)+{b}_{{n}} {sin}\left({nx}\right)\right) \\ $$$${a}_{{n}} =\frac{\mathrm{1}}{\pi}\int_{−\pi} ^{\:\pi} {f}\left({x}\right){cos}\left({nx}\right){dx}=\mathrm{0} \\ $$$${b}_{{n}} =\frac{\mathrm{1}}{\pi}\int_{−\pi} ^{\:\pi} {f}\left({x}\right){sin}\left({nx}\right){dx} \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{1}}{\pi}\int_{−\pi} ^{\:\pi} {xsin}\left({nx}\right){dx}=\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\:\pi} {xsin}\left({nx}\right){dx} \\ $$$$\:=\frac{\mathrm{2}}{\pi}\left\{\left[−\frac{{x}}{{n}}{cos}\left({nx}\right)\right]_{\mathrm{0}\:} ^{\pi} +\frac{\mathrm{1}}{{n}}\int_{\mathrm{0}} ^{\:\pi} {cos}\left({nx}\right){dx}\right. \\ $$$$\mathrm{2}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} \frac{\mathrm{1}}{{n}} \\ $$$${a}_{\mathrm{0}} =\frac{\mathrm{1}}{\pi}\int_{−\pi} ^{\:\pi} {f}\left({x}\right){dx}=\mathrm{0} \\ $$$$\:\:\:{x}\approx\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{2}\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}\:}{sin}\left({nx}\right)\:\: \\ $$$$\int{xdx}\approx\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int\:\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}\:}{sin}\left({nx}\right){dx} \\ $$$$\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\approx{C}+\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }{cos}\left({nx}\right) \\ $$$$\:\:{x}=\pi\Rightarrow\:\frac{\pi^{\mathrm{2}} }{\mathrm{2}}\approx{C}+\mathrm{2}\underset{{n}=\mathrm{1}} {\sum}\frac{\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$$$\:\:\:\:{C}=\frac{\pi^{\mathrm{2}} }{\mathrm{2}}−\frac{\pi^{\mathrm{2}} }{\mathrm{3}}=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$\:\:\:\frac{{x}^{\mathrm{2}} }{\mathrm{2}}=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}+\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }{cos}\left({nx}\right) \\ $$$$\:\:{x}=\mathrm{1}\:\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}=\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{4}}−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:\:... \\ $$
Commented by mathmax by abdo last updated on 14/Feb/21
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by mnjuly1970 last updated on 15/Feb/21
$${grateful}... \\ $$
Answered by mnjuly1970 last updated on 14/Feb/21
$$\:\:\:{second}\:{method} \\ $$$$\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\left(\frac{{e}^{{in}} +{e}^{−{in}} }{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−{e}^{{i}} \right)^{{n}} }{{n}^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−{e}^{−{i}} \right)^{{n}} }{{n}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({li}_{\mathrm{2}} \left(−{e}^{{i}} \right)+{li}_{\mathrm{2}} \left(−\frac{\mathrm{1}}{{e}^{{i}} }\right)\right) \\ $$$$\underset{\frac{−\pi^{\mathrm{2}} }{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left({z}\right)} {\overset{{li}_{\mathrm{2}} \left(−{z}\right)+{li}_{\mathrm{2}} \left(−\frac{\mathrm{1}}{{z}}\right)} {=}}\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{−\pi^{\mathrm{2}} }{\mathrm{6}}−\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\mathrm{2}} \left({e}^{{i}} \right)\right) \\ $$$$\:=\frac{−\pi^{\mathrm{2}} }{\mathrm{12}}−\frac{\mathrm{1}}{\mathrm{4}}\left({i}^{\mathrm{2}} \right)=\frac{−\pi^{\mathrm{2}} }{\mathrm{12}}+\frac{\mathrm{1}}{\mathrm{4}}\:\:.... \\ $$
Answered by mathmax by abdo last updated on 14/Feb/21
![let ϕ(x)=x ,2π periodic odd ⇒ϕ(x)=Σ_(n=1) ^∞ a_n sin(nx) a_n =(1/π)∫_(−π) ^π x sin(nx)dx =(2/π)∫_0 ^π xsin(nx)dx ⇒ (π/2)a_n =[−(x/n)cos(nx)]_0 ^π +∫_0 ^π (1/n)cos(nx)dx =−(π/n)(−1)^n +(1/n^2 )[sin(nx)]_0 ^π =−(π/n)(−1)^n ⇒a_n =(2/π).(−(π/n))(−1)^n =−(2/n)(−1)^n ⇒ x=−2Σ_(n=1) ^∞ (((−1)^n )/n)sin(nx) ⇒(x^2 /2) =2Σ_(n=1) ^∞ (((−1)^n )/n^2 )cos(nx) +C x=0 ⇒0 =2 Σ_(n=1) ^∞ (((−1)^n )/n^2 ) +c =2.(2^(1−2) −1)ξ(2)+C =2(−(1/2))(π^2 /6) +C =−(π^2 /6) +C ⇒C =(π^2 /6) ⇒ (x^2 /2) =2Σ_(n=1) ^∞ (((−1)^n )/n^2 )cos(nx)+(π^2 /6) ⇒2Σ(....)=(x^2 /2)−(π^2 /6) ⇒ Σ_(n=1) ^∞ (((−1)^n )/n^2 )cos(nx) =(x^2 /4)−(π^2 /(12)) x=1 ⇒Σ_(n=1) ^∞ (((−1)^n )/n^2 )cos(n) =(1/4)−(π^2 /(12)) x=(π/3) ⇒Σ_(n=1) ^∞ (((−1)^n )/n^2 )cos(n(π/3))=(π^2 /(36))−(π^2 /(12)) =(π^2 /(36))−((3π^2 )/(36)) =−(π^2 /(18))](Q132524.png)
$$\mathrm{let}\:\varphi\left(\mathrm{x}\right)=\mathrm{x}\:\:,\mathrm{2}\pi\:\mathrm{periodic}\:\mathrm{odd}\:\Rightarrow\varphi\left(\mathrm{x}\right)=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\mathrm{a}_{\mathrm{n}} \mathrm{sin}\left(\mathrm{nx}\right) \\ $$$$\mathrm{a}_{\mathrm{n}} =\frac{\mathrm{1}}{\pi}\int_{−\pi} ^{\pi} \:\mathrm{x}\:\mathrm{sin}\left(\mathrm{nx}\right)\mathrm{dx}\:=\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\pi} \:\mathrm{xsin}\left(\mathrm{nx}\right)\mathrm{dx}\:\Rightarrow \\ $$$$\frac{\pi}{\mathrm{2}}\mathrm{a}_{\mathrm{n}} =\left[−\frac{\mathrm{x}}{\mathrm{n}}\mathrm{cos}\left(\mathrm{nx}\right)\right]_{\mathrm{0}} ^{\pi} +\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}}{\mathrm{n}}\mathrm{cos}\left(\mathrm{nx}\right)\mathrm{dx}\:=−\frac{\pi}{\mathrm{n}}\left(−\mathrm{1}\right)^{\mathrm{n}} \:+\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\left[\mathrm{sin}\left(\mathrm{nx}\right)\right]_{\mathrm{0}} ^{\pi} \\ $$$$=−\frac{\pi}{\mathrm{n}}\left(−\mathrm{1}\right)^{\mathrm{n}} \:\Rightarrow\mathrm{a}_{\mathrm{n}} =\frac{\mathrm{2}}{\pi}.\left(−\frac{\pi}{\mathrm{n}}\right)\left(−\mathrm{1}\right)^{\mathrm{n}} \:=−\frac{\mathrm{2}}{\mathrm{n}}\left(−\mathrm{1}\right)^{\mathrm{n}} \:\Rightarrow \\ $$$$\mathrm{x}=−\mathrm{2}\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}}\mathrm{sin}\left(\mathrm{nx}\right)\:\:\Rightarrow\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\:=\mathrm{2}\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} }\mathrm{cos}\left(\mathrm{nx}\right)\:+\mathrm{C} \\ $$$$\mathrm{x}=\mathrm{0}\:\Rightarrow\mathrm{0}\:=\mathrm{2}\:\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} }\:+\mathrm{c}\:\:=\mathrm{2}.\left(\mathrm{2}^{\mathrm{1}−\mathrm{2}} −\mathrm{1}\right)\xi\left(\mathrm{2}\right)+\mathrm{C} \\ $$$$=\mathrm{2}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:+\mathrm{C}\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:+\mathrm{C}\:\Rightarrow\mathrm{C}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:\Rightarrow \\ $$$$\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}\:=\mathrm{2}\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} }\mathrm{cos}\left(\mathrm{nx}\right)+\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:\Rightarrow\mathrm{2}\Sigma\left(....\right)=\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{2}}−\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:\Rightarrow \\ $$$$\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} }\mathrm{cos}\left(\mathrm{nx}\right)\:=\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{4}}−\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$$\mathrm{x}=\mathrm{1}\:\Rightarrow\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} }\mathrm{cos}\left(\mathrm{n}\right)\:=\frac{\mathrm{1}}{\mathrm{4}}−\frac{\pi^{\mathrm{2}} }{\mathrm{12}} \\ $$$$\mathrm{x}=\frac{\pi}{\mathrm{3}}\:\Rightarrow\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} }\mathrm{cos}\left(\mathrm{n}\frac{\pi}{\mathrm{3}}\right)=\frac{\pi^{\mathrm{2}} }{\mathrm{36}}−\frac{\pi^{\mathrm{2}} }{\mathrm{12}}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{36}}−\frac{\mathrm{3}\pi^{\mathrm{2}} }{\mathrm{36}}\:=−\frac{\pi^{\mathrm{2}} }{\mathrm{18}} \\ $$
Commented by mnjuly1970 last updated on 15/Feb/21
$${very}\:{nice}\:{sir}\:{max}.. \\ $$