Question Number 53963 by maxmathsup by imad last updated on 27/Jan/19
$${let}\:{f}\left({x}\right)\:=\:{x}\mid{x}\mid\:\:\:,\:\mathrm{2}\pi\:{periodic}\:\:{odd}\: \\ $$$${developp}\:{f}\:{at}\:{fourier}\:{serie}\:. \\ $$
Commented by maxmathsup by imad last updated on 31/Jan/19
![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 u =x^2 and v =sin(nx) we get ∫_0 ^π x^2 sin(nx)dx =[−(1/n)x^2 cos(nx)]_0 ^π −∫_0 ^π 2x(−(1/n)cos(nx))dx =−(π^2 /n)(−1)^n +(2/n) ∫_0 ^π x cos(nx)dx also by parts ∫_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 =−(π^2 /n)(−1)^n +(2/n^3 ){ (−1)^n −1} ⇒ a_n =−(2/π) (π^2 /n)(−1)^n +(4/(πn^3 )){ (−1)^n −1} =((2π)/n)(−1)^(n−1) +(4/(πn^3 )){ (−1)^n −1} ⇒ x ∣x∣ =2π Σ_(n=1) ^∞ (((−1)^(n−1) )/n)sin(nx) +(4/π) Σ_(n=1) ^∞ (((−1)^n −1)/n^3 ) sin(nx) ⇒ x∣x∣ =2π Σ_(n=1) ^∞ (((−1)^(n−1) )/n) sin(nx) −(8/π) Σ_(n=1) ^∞ ((sin(2n+1)x)/((2n+1)^3 )) .](https://www.tinkutara.com/question/Q54234.png)
$${f}\left({x}\right)\:=\:\sum_{{n}=\mathrm{1}} ^{\infty} \:{a}_{{n}} {sin}\left({nx}\right)\:\:{with}\:{a}_{{n}} =\frac{\mathrm{2}}{{T}}\:\int_{\left[{T}\right]} {f}\left({x}\right){sin}\left({nx}\right){dx} \\ $$$$=\frac{\mathrm{2}}{\mathrm{2}\pi}\:\int_{−\pi} ^{\pi} \:{x}\mid{x}\mid\:{sin}\left({nx}\right){dx}\:=\frac{\mathrm{2}}{\pi}\:\int_{\mathrm{0}} ^{\pi} \:{x}^{\mathrm{2}} {sin}\left({nx}\right){dx}\:\Rightarrow\frac{\pi}{\mathrm{2}}{a}_{{n}} =\int_{\mathrm{0}} ^{\pi} \:{x}^{\mathrm{2}} {sin}\left({nx}\right){dx} \\ $$$${by}\:{parts}\:{u}\:={x}^{\mathrm{2}} \:\:{and}\:{v}\:={sin}\left({nx}\right)\:\:{we}\:{get} \\ $$$$\int_{\mathrm{0}} ^{\pi} \:{x}^{\mathrm{2}} {sin}\left({nx}\right){dx}\:=\left[−\frac{\mathrm{1}}{{n}}{x}^{\mathrm{2}} {cos}\left({nx}\right)\right]_{\mathrm{0}} ^{\pi} \:−\int_{\mathrm{0}} ^{\pi} \:\mathrm{2}{x}\left(−\frac{\mathrm{1}}{{n}}{cos}\left({nx}\right)\right){dx} \\ $$$$=−\frac{\pi^{\mathrm{2}} }{{n}}\left(−\mathrm{1}\right)^{{n}} \:\:+\frac{\mathrm{2}}{{n}}\:\int_{\mathrm{0}} ^{\pi} \:{x}\:{cos}\left({nx}\right){dx}\:\:{also}\:\:{by}\:{parts} \\ $$$$\int_{\mathrm{0}} ^{\pi} \:{x}\:{cos}\left({nx}\right){dx}\:=\left[\frac{{x}}{{n}}{sin}\left({nx}\right)\right]_{\mathrm{0}} ^{\pi} \:−\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{1}}{{n}}{sin}\left({nx}\right){dx} \\ $$$$=−\frac{\mathrm{1}}{{n}}\left[−\frac{\mathrm{1}}{{n}}{cos}\left({nx}\right)\right]_{\mathrm{0}} ^{\pi} \:=\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\left(\left(−\mathrm{1}\right)^{{n}} −\mathrm{1}\right)\:\Rightarrow \\ $$$$\frac{\pi}{\mathrm{2}}\:{a}_{{n}} =−\frac{\pi^{\mathrm{2}} }{{n}}\left(−\mathrm{1}\right)^{{n}} \:+\frac{\mathrm{2}}{{n}^{\mathrm{3}} }\left\{\:\left(−\mathrm{1}\right)^{{n}} \:−\mathrm{1}\right\}\:\Rightarrow \\ $$$${a}_{{n}} =−\frac{\mathrm{2}}{\pi}\:\frac{\pi^{\mathrm{2}} }{{n}}\left(−\mathrm{1}\right)^{{n}} \:+\frac{\mathrm{4}}{\pi{n}^{\mathrm{3}} }\left\{\:\left(−\mathrm{1}\right)^{{n}} −\mathrm{1}\right\}\:=\frac{\mathrm{2}\pi}{{n}}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:+\frac{\mathrm{4}}{\pi{n}^{\mathrm{3}} }\left\{\:\left(−\mathrm{1}\right)^{{n}} \:−\mathrm{1}\right\}\:\Rightarrow \\ $$$${x}\:\mid{x}\mid\:=\mathrm{2}\pi\:\sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}{sin}\left({nx}\right)\:+\frac{\mathrm{4}}{\pi}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} −\mathrm{1}}{{n}^{\mathrm{3}} }\:{sin}\left({nx}\right)\:\:\Rightarrow \\ $$$${x}\mid{x}\mid\:=\mathrm{2}\pi\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:{sin}\left({nx}\right)\:−\frac{\mathrm{8}}{\pi}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{sin}\left(\mathrm{2}{n}+\mathrm{1}\right){x}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{3}} }\:\:. \\ $$