Question Number 45044 by maxmathsup by imad last updated on 07/Oct/18
$${let}\:{f}\left({x}\right)\:={x}^{\mathrm{2}} \:,\:{function}\:\mathrm{2}\pi\:{peridic}\:{even} \\ $$$$\left.\mathrm{1}\right)\:{developp}\:{f}\:{at}\:{fourier}\:{serie} \\ $$$$\left.\mathrm{2}\right){find}\:{the}\:{value}\:{of}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{4}} } \\ $$
Commented by maxmathsup by imad last updated on 08/Oct/18
![f is even ⇒f(x)=(a_0 /2) +Σ_(n=1) ^∞ a_n cos(nx) with a_n =(2/T)∫_([T]) f(x)cos(nx) =(2/(2π)) ∫_(−π) ^π x^2 cos(nx)dx =(2/π) ∫_0 ^π x^2 cos(nx) ⇒(π/2)a_n = ∫_0 ^π x^2 cos(nx) by psrts we get ∫_0 ^π x^2 cos(nx)dx =[(x^2 /n)sin(nx)]_0 ^π −∫_0 ^π ((2x)/n)sin(nx)dx =−(2/n) ∫_0 ^π x sin(nx)dx =−(2/n){ [−(x/n)cos(nx)]_0 ^π −∫_0 ^π −(1/n)cos(nx)dx} =(2/n^2 ) ( π(−1)^n ) −(2/n^2 )∫_0 ^π cos(nx)dx =((2π(−1)^n )/n^2 ) ⇒(π/2)a_n =((2π(−1)^n )/n^2 ) ⇒ a_n = ((4(−1)^n )/n^2 ) also a_0 =(2/π) ∫_0 ^π x^2 dx =(2/π) (π^3 /3) =((2π^2 )/3) ⇒(a_0 /2) =(π^2 /3) ⇒ ★x^2 =(π^2 /3) +4 Σ_(n=1) ^∞ (((−1)^n )/n^2 ) cos(nx)★ 2) the parseval formulae give (1/T)∫_([T]) f^2 (x)dx =((a_o /2))^2 +(1/2)Σ_(n≥1) (a_n ^2 +b_n ^2 ) ⇒ (1/(2π)) ∫_(−π) ^π x^4 dx =(π^4 /9) +8 Σ_(n=1) ^∞ (1/n^4 ) ⇒(1/π) ∫_0 ^π x^4 dx =(π^4 /9) +8 Σ_(n=1) ^∞ (1/n^4 ) ⇒ (π^4 /5) −(π^4 /9) =8Σ_(n=1) ^∞ (1/n^4 ) ⇒ Σ_(n=1) ^∞ (1/n^4 ) =(1/8)(((4π^4 )/(45))) = (π^4 /(90)) ★ Σ_(n=1) ^∞ (1/n^4 ) =(π^4 /(90)) ★](https://www.tinkutara.com/question/Q45073.png)
$${f}\:{is}\:{even}\:\Rightarrow{f}\left({x}\right)=\frac{{a}_{\mathrm{0}} }{\mathrm{2}}\:+\sum_{{n}=\mathrm{1}} ^{\infty} \:\:{a}_{{n}} {cos}\left({nx}\right)\:{with}\: \\ $$$${a}_{{n}} =\frac{\mathrm{2}}{{T}}\int_{\left[{T}\right]} {f}\left({x}\right){cos}\left({nx}\right)\:=\frac{\mathrm{2}}{\mathrm{2}\pi}\:\int_{−\pi} ^{\pi} \:{x}^{\mathrm{2}} {cos}\left({nx}\right){dx} \\ $$$$=\frac{\mathrm{2}}{\pi}\:\int_{\mathrm{0}} ^{\pi} \:{x}^{\mathrm{2}} {cos}\left({nx}\right)\:\Rightarrow\frac{\pi}{\mathrm{2}}{a}_{{n}} =\:\int_{\mathrm{0}} ^{\pi} \:{x}^{\mathrm{2}} {cos}\left({nx}\right)\:\:{by}\:{psrts}\:{we}\:{get} \\ $$$$\int_{\mathrm{0}} ^{\pi} \:{x}^{\mathrm{2}} {cos}\left({nx}\right){dx}\:=\left[\frac{{x}^{\mathrm{2}} }{{n}}{sin}\left({nx}\right)\right]_{\mathrm{0}} ^{\pi} \:−\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{2}{x}}{{n}}{sin}\left({nx}\right){dx} \\ $$$$=−\frac{\mathrm{2}}{{n}}\:\int_{\mathrm{0}} ^{\pi} \:{x}\:{sin}\left({nx}\right){dx}\:=−\frac{\mathrm{2}}{{n}}\left\{\:\left[−\frac{{x}}{{n}}{cos}\left({nx}\right)\right]_{\mathrm{0}} ^{\pi} \:−\int_{\mathrm{0}} ^{\pi} −\frac{\mathrm{1}}{{n}}{cos}\left({nx}\right){dx}\right\} \\ $$$$=\frac{\mathrm{2}}{{n}^{\mathrm{2}} }\:\left(\:\pi\left(−\mathrm{1}\right)^{{n}} \right)\:−\frac{\mathrm{2}}{{n}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\pi} \:{cos}\left({nx}\right){dx}\:=\frac{\mathrm{2}\pi\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\:\Rightarrow\frac{\pi}{\mathrm{2}}{a}_{{n}} =\frac{\mathrm{2}\pi\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\:\Rightarrow \\ $$$${a}_{{n}} =\:\frac{\mathrm{4}\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\:\:\:{also}\:{a}_{\mathrm{0}} =\frac{\mathrm{2}}{\pi}\:\int_{\mathrm{0}} ^{\pi} {x}^{\mathrm{2}} {dx}\:=\frac{\mathrm{2}}{\pi}\:\frac{\pi^{\mathrm{3}} }{\mathrm{3}}\:=\frac{\mathrm{2}\pi^{\mathrm{2}} }{\mathrm{3}}\:\Rightarrow\frac{{a}_{\mathrm{0}} }{\mathrm{2}}\:=\frac{\pi^{\mathrm{2}} }{\mathrm{3}}\:\Rightarrow \\ $$$$\bigstar{x}^{\mathrm{2}} \:=\frac{\pi^{\mathrm{2}} }{\mathrm{3}}\:+\mathrm{4}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} }\:{cos}\left({nx}\right)\bigstar \\ $$$$\left.\mathrm{2}\right)\:{the}\:{parseval}\:{formulae}\:{give}\:\frac{\mathrm{1}}{{T}}\int_{\left[{T}\right]} {f}^{\mathrm{2}} \left({x}\right){dx}\:=\left(\frac{{a}_{{o}} }{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}}\sum_{{n}\geqslant\mathrm{1}} \left({a}_{{n}} ^{\mathrm{2}} \:+{b}_{{n}} ^{\mathrm{2}} \right)\:\Rightarrow \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\pi}\:\int_{−\pi} ^{\pi} \:{x}^{\mathrm{4}} {dx}\:=\frac{\pi^{\mathrm{4}} }{\mathrm{9}}\:+\mathrm{8}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{4}} }\:\Rightarrow\frac{\mathrm{1}}{\pi}\:\int_{\mathrm{0}} ^{\pi} \:{x}^{\mathrm{4}} {dx}\:=\frac{\pi^{\mathrm{4}} }{\mathrm{9}}\:+\mathrm{8}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{4}} }\:\Rightarrow \\ $$$$\frac{\pi^{\mathrm{4}} }{\mathrm{5}}\:−\frac{\pi^{\mathrm{4}} }{\mathrm{9}}\:=\mathrm{8}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{1}}{{n}^{\mathrm{4}} }\:\Rightarrow\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}^{\mathrm{4}} }\:=\frac{\mathrm{1}}{\mathrm{8}}\left(\frac{\mathrm{4}\pi^{\mathrm{4}} }{\mathrm{45}}\right)\:=\:\frac{\pi^{\mathrm{4}} }{\mathrm{90}} \\ $$$$\bigstar\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}^{\mathrm{4}} }\:=\frac{\pi^{\mathrm{4}} }{\mathrm{90}}\:\bigstar \\ $$$$ \\ $$