Question Number 82286 by mathmax by abdo last updated on 19/Feb/20
$$\left.\mathrm{1}\right)\:{find}\:{a}\:{and}\:{b}\:{wich}\:{verify}\:\:\int_{\mathrm{0}} ^{\pi} \left({at}^{\mathrm{2}} \:+{bt}\right){cos}\left({nx}\right)\:=\frac{\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$
Commented by john santu last updated on 22/Feb/20
$$\left(\mathrm{2}\right)\mathrm{sin}\:{x}\:=\:{x}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}!}\:+\:\frac{{x}^{\mathrm{5}} }{\mathrm{5}!}\:−... \\ $$$$\left(\mathrm{3}\right)\:\frac{\mathrm{sin}\:{x}}{{x}}\:=\:\underset{{n}\:=\:\mathrm{1}} {\overset{\infty} {\prod}}\:\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{n}^{\mathrm{2}} \pi^{\mathrm{2}} }\right)=\:\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\pi^{\mathrm{2}} }\right)\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{4}\pi^{\mathrm{2}} }\right)... \\ $$$$\left(\mathrm{2}\right)\:=\:\left(\mathrm{3}\right)\: \\ $$$$\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{3}!}\:+\:\frac{{x}^{\mathrm{4}} }{\mathrm{5}!}\:−\:\frac{{x}^{\mathrm{6}} }{\mathrm{7}!}+...\:=\:\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\pi^{\mathrm{2}} }\right)\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{\mathrm{4}\pi^{\mathrm{2}} }\right)... \\ $$$${coefficient}\:{term}\:{of}\:{x}^{\mathrm{2}} \:{must}\:{be}\:{same} \\ $$$$−\frac{\mathrm{1}}{\mathrm{3}!}\:=\:−\left(\frac{\mathrm{1}}{\pi^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{4}\pi^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{9}\pi^{\mathrm{2}} }+...\right) \\ $$$$−\frac{\mathrm{1}}{\mathrm{6}}\:=\:−\frac{\mathrm{1}}{\pi^{\mathrm{2}} }\:\underset{{n}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\:\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$$${therefore}\:\underset{{n}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\:\frac{\pi^{\mathrm{2}} }{\mathrm{6}}\:.\: \\ $$
Commented by mathmax by abdo last updated on 20/Feb/20
$${sir}\:{john}\:{can}\:{you}\:{prove}\:{the}\:{second}\:{line}\:{its}\:{not}\:{clear}... \\ $$
Commented by mind is power last updated on 20/Feb/20
$${nice}\:{But}\:{i}\:{think}\:\mathrm{1\&2}\:{are}\:{related}\:{quation} \\ $$
Answered by mind is power last updated on 20/Feb/20
![∫_0 ^π t^2 cos(nt)dt =−2∫_0 ^π ((tsin(nt))/n)dt =(2/n)[((tcos(nt))/n)]_0 ^π =((2π(−1)^n )/n^2 ) ∫tcos(nt)dt=−∫_0 ^π ((sin(nt))/n)dt=(([cos(nt)]_0 ^π )/n^2 )=(((−1)^n −1)/n^2 ) ⇒∫_0 ^π (at^2 +bt)cos(nt)dt=(((−1)^n )/n^2 )(2πa+b)−(b/n^2 )=(1/n^2 ) ⇒ { ((−b=1)),((2πa+b=0)) :}⇒b=−1,a=(1/(2π)) 2) let f(t)=(t^2 /(2π))−t 2π periodic odd f(x)=f(−x)paire f∈C_0 ]−π,π[ serie of fourier of f is S_n (f)=a_0 +Σ_(n≥1) a_n (f)cos(nx) a_0 =(1/(2π))∫_(−π) ^π f(x)dx=(1/π)∫_0 ^π ((x^2 /(2π))−x)dx =(π/6)−(π/2)=−(π/3) a_k =(1/π)∫_(−π) ^π f(x)cos(kx)dx=(2/π)∫_0 ^π f(x)cos(kx)dx by one a_k =(2/π)(1/k^2 ) S_n (f)(x)=−(π/3)+Σ_(n≥1) ((2cos(kx))/k^2 ) by direchlet f(0)=S_n (f)(0)⇒ 0=−(π/3)+(2/π)Σ_(k≥1) ((cos(k.0))/k^2 )⇒ (π/3)=(2/π)Σ_(k≥1) (1/k^2 )⇒Σ_(k≥1) (1/k^2 )=(π^2 /6)](Q82362.png)
$$\int_{\mathrm{0}} ^{\pi} {t}^{\mathrm{2}} {cos}\left({nt}\right){dt} \\ $$$$=−\mathrm{2}\int_{\mathrm{0}} ^{\pi} \frac{{tsin}\left({nt}\right)}{{n}}{dt} \\ $$$$=\frac{\mathrm{2}}{{n}}\left[\frac{{tcos}\left({nt}\right)}{{n}}\right]_{\mathrm{0}} ^{\pi} =\frac{\mathrm{2}\pi\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} } \\ $$$$\int{tcos}\left({nt}\right){dt}=−\int_{\mathrm{0}} ^{\pi} \frac{{sin}\left({nt}\right)}{{n}}{dt}=\frac{\left[{cos}\left({nt}\right)\right]_{\mathrm{0}} ^{\pi} }{{n}^{\mathrm{2}} }=\frac{\left(−\mathrm{1}\right)^{{n}} −\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$$$\Rightarrow\int_{\mathrm{0}} ^{\pi} \left({at}^{\mathrm{2}} +{bt}\right){cos}\left({nt}\right){dt}=\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\left(\mathrm{2}\pi{a}+{b}\right)−\frac{{b}}{{n}^{\mathrm{2}} }=\frac{\mathrm{1}}{{n}^{\mathrm{2}} } \\ $$$$\Rightarrow\begin{cases}{−{b}=\mathrm{1}}\\{\mathrm{2}\pi{a}+{b}=\mathrm{0}}\end{cases}\Rightarrow{b}=−\mathrm{1},{a}=\frac{\mathrm{1}}{\mathrm{2}\pi} \\ $$$$\left.\mathrm{2}\right) \\ $$$${let}\:{f}\left({t}\right)=\frac{{t}^{\mathrm{2}} }{\mathrm{2}\pi}−{t}\:\:\mathrm{2}\pi\:{periodic}\:\:{odd}\:{f}\left({x}\right)={f}\left(−{x}\right){paire} \\ $$$$\left.{f}\in{C}_{\mathrm{0}} \right]−\pi,\pi\left[\right. \\ $$$${serie}\:{of}\:{fourier}\:{of}\:{f}\:{is} \\ $$$${S}_{{n}} \left({f}\right)={a}_{\mathrm{0}} +\underset{{n}\geqslant\mathrm{1}} {\sum}{a}_{{n}} \left({f}\right){cos}\left({nx}\right) \\ $$$${a}_{\mathrm{0}} =\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{−\pi} ^{\pi} {f}\left({x}\right){dx}=\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\pi} \left(\frac{{x}^{\mathrm{2}} }{\mathrm{2}\pi}−{x}\right){dx} \\ $$$$=\frac{\pi}{\mathrm{6}}−\frac{\pi}{\mathrm{2}}=−\frac{\pi}{\mathrm{3}} \\ $$$${a}_{{k}} =\frac{\mathrm{1}}{\pi}\int_{−\pi} ^{\pi} {f}\left({x}\right){cos}\left({kx}\right){dx}=\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\pi} {f}\left({x}\right){cos}\left({kx}\right){dx}\:{by}\:{one} \\ $$$${a}_{{k}} =\frac{\mathrm{2}}{\pi}\frac{\mathrm{1}}{{k}^{\mathrm{2}} } \\ $$$${S}_{{n}} \left({f}\right)\left({x}\right)=−\frac{\pi}{\mathrm{3}}+\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{2}{cos}\left({kx}\right)}{{k}^{\mathrm{2}} } \\ $$$${by}\:{direchlet}\:{f}\left(\mathrm{0}\right)={S}_{{n}} \left({f}\right)\left(\mathrm{0}\right)\Rightarrow \\ $$$$\mathrm{0}=−\frac{\pi}{\mathrm{3}}+\frac{\mathrm{2}}{\pi}\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{{cos}\left({k}.\mathrm{0}\right)}{{k}^{\mathrm{2}} }\Rightarrow \\ $$$$\frac{\pi}{\mathrm{3}}=\frac{\mathrm{2}}{\pi}\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{k}^{\mathrm{2}} }\Rightarrow\underset{{k}\geqslant\mathrm{1}} {\sum}\frac{\mathrm{1}}{{k}^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{6}} \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 20/Feb/20
$${thanks}\:{sir}. \\ $$
Commented by mind is power last updated on 21/Feb/20
$${withe}\:{pleasur} \\ $$