Question Number 35619 by abdo mathsup 649 cc last updated on 21/May/18
$${let}\:{f}\left({x}\right)\:=\:{x}\mid{x}\mid\:\:{odd}\:\mathrm{2}\pi\:{periodic} \\ $$$${developp}\:{f}\:{at}\:{fourier}\:{serie}\:. \\ $$
Commented by abdo mathsup 649 cc last updated on 25/May/18
![f(x)=Σ_(n=1) ^∞ a_n sin(nx) with a_n = (2/T)∫_([T]) f(x)sin(nx) dx=(2/(2π)) ∫_(−π) ^π x∣x∣sin(nx)dx =(2/π) ∫_0 ^π x^2 sin(nx)dx ⇒(π/2)a_n = ∫_0 ^π x^2 sin(nx)dx by parts ∫_0 ^π x^2 sin(nx)dx = [−(x^2 /n)cos(nx)]_0 ^π −∫_0 ^π ((−1)/n)cos(nx)2xdx =−((π^2 (−1)^n )/n) +(2/n) ∫_0 ^π x cos(nx)dx but ∫_0 ^π x cos(nx)dx =[(x/n)sin(nx)]_0 ^π −∫_0 ^π (1/n)sin(nx)dx =−(1/n)[−(1/n)cos(nx)]_0 ^π =(1/n^2 ){ (−1)^n −1} (π/2) a_n = −(((−1)^n π^2 )/n) +(2/n^3 ){ (−1)^(n ) −1} ⇒ a_n = ((−2π^2 (−1)^n )/(πn)) +(4/(πn^3 )){ (−1)^n −1} x∣x∣ =−2πΣ_(n=1) ^∞ (((−1)^n )/n) sin(nx) +(4/π)Σ_(n=1) ^∞ (((−1)^n −1)/n^3 )](https://www.tinkutara.com/question/Q35919.png)
$${f}\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:{a}_{{n}} \:{sin}\left({nx}\right)\:{with} \\ $$$${a}_{{n}} \:=\:\frac{\mathrm{2}}{{T}}\int_{\left[{T}\right]} \:{f}\left({x}\right){sin}\left({nx}\right)\:{dx}=\frac{\mathrm{2}}{\mathrm{2}\pi}\:\int_{−\pi} ^{\pi} \:{x}\mid{x}\mid{sin}\left({nx}\right){dx} \\ $$$$=\frac{\mathrm{2}}{\pi}\:\int_{\mathrm{0}} ^{\pi} \:\:{x}^{\mathrm{2}} \:{sin}\left({nx}\right){dx}\:\Rightarrow\frac{\pi}{\mathrm{2}}{a}_{{n}} \:=\:\int_{\mathrm{0}} ^{\pi} \:{x}^{\mathrm{2}} \:{sin}\left({nx}\right){dx} \\ $$$${by}\:{parts}\:\int_{\mathrm{0}} ^{\pi} \:{x}^{\mathrm{2}} {sin}\left({nx}\right){dx} \\ $$$$=\:\left[−\frac{{x}^{\mathrm{2}} }{{n}}{cos}\left({nx}\right)\right]_{\mathrm{0}} ^{\pi} \:−\int_{\mathrm{0}} ^{\pi} \:\:\frac{−\mathrm{1}}{{n}}{cos}\left({nx}\right)\mathrm{2}{xdx} \\ $$$$=−\frac{\pi^{\mathrm{2}} \left(−\mathrm{1}\right)^{{n}} }{{n}}\:\:+\frac{\mathrm{2}}{{n}}\:\int_{\mathrm{0}} ^{\pi} \:{x}\:{cos}\left({nx}\right){dx}\:{but} \\ $$$$\int_{\mathrm{0}} ^{\pi} \:\:\:{x}\:{cos}\left({nx}\right){dx}\:=\left[\frac{{x}}{{n}}{sin}\left({nx}\right)\right]_{\mathrm{0}} ^{\pi} \:−\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{1}}{{n}}{sin}\left({nx}\right){dx} \\ $$$$=−\frac{\mathrm{1}}{{n}}\left[−\frac{\mathrm{1}}{{n}}{cos}\left({nx}\right)\right]_{\mathrm{0}} ^{\pi} \:=\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\left\{\:\left(−\mathrm{1}\right)^{{n}} \:−\mathrm{1}\right\} \\ $$$$\frac{\pi}{\mathrm{2}}\:{a}_{{n}} \:=\:−\frac{\left(−\mathrm{1}\right)^{{n}} \:\pi^{\mathrm{2}} }{{n}}\:\:+\frac{\mathrm{2}}{{n}^{\mathrm{3}} }\left\{\:\left(−\mathrm{1}\right)^{{n}\:} \:−\mathrm{1}\right\}\:\Rightarrow \\ $$$${a}_{{n}} \:\:=\:\frac{−\mathrm{2}\pi^{\mathrm{2}} \left(−\mathrm{1}\right)^{{n}} }{\pi{n}}\:\:+\frac{\mathrm{4}}{\pi{n}^{\mathrm{3}} }\left\{\:\left(−\mathrm{1}\right)^{{n}} \:−\mathrm{1}\right\} \\ $$$${x}\mid{x}\mid\:=−\mathrm{2}\pi\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}\:{sin}\left({nx}\right)\:+\frac{\mathrm{4}}{\pi}\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} −\mathrm{1}}{{n}^{\mathrm{3}} } \\ $$
Commented by abdo mathsup 649 cc last updated on 25/May/18
$${x}\mid{x}\mid=−\mathrm{2}\pi\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}{sin}\left({nx}\right)\:+\frac{\mathrm{4}}{\pi}\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} −\mathrm{1}}{{n}^{\mathrm{3}} }{sin}\left({nx}\right). \\ $$