Question Number 67378 by mathmax by abdo last updated on 26/Aug/19
$${let}\:{f}\left({x}\right)\:={x}^{\mathrm{3}} \:\:\:\:\:\:,\mathrm{2}\pi\:{periodic}\:{odd} \\ $$$${developp}\:{f}\:{at}\:{fourier}\:{serie}\: \\ $$
Commented by mathmax by abdo last updated on 30/Aug/19
![f(x) =Σ_(n=1) ^∞ a_n sin(nx) with a_n =(2/T)∫_([T]) x^3 sin)nx)dx =(2/(2π)) ∫_(−π) ^π x^3 sin(nx)dx =(2/π) ∫_0 ^π x^3 sin(nx)dx ⇒ (π/2)a_n = ∫_0 ^π x^3 sin(nx)dx =_(nx=t) ∫_0 ^(nπ) (t^3 /n^3 )sin(t)(dt/n) =(1/n^4 ) ∫_0 ^(nπ) t^3 sint dt by parts ∫_0 ^(nπ) t^3 sint dt=[−t^3 cost]_0 ^(nπ) +∫_0 ^(nπ) 3t^2 cost dt =−n^3 π^3 (−1)^n +3 ∫_0 ^(nπ) t^2 cost dt by parts again ∫_0 ^π t^2 cost dt =[t^2 sint]_0 ^(nπ) −∫_0 ^(nπ) 2t sint =2 ∫_0 ^(nπ) t (−sint)dt =[t cost]_0 ^(nπ) −∫_0 ^(nπ) cost dt =nπ(−1)^n −0 ⇒(π/2)a_n =(1/n^4 ){−n^3 π^3 (−1)^n +3nπ(−1)^n } ⇒ a_n =(2/(πn^4 )){−π^3 n^3 (−1)^n +3πn(−1)^n } =−((2π^2 )/n)(−1)^n +(6/n^3 )(−1)^n ⇒ x^3 =−2π^2 Σ_(n=1) ^∞ (((−1)^n )/n)sin(nx) +6 Σ_(n=1) ^∞ (((−1)^n )/n^3 )sin(nx)](https://www.tinkutara.com/question/Q67732.png)
$$\left.{f}\left.\left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:{a}_{{n}} {sin}\left({nx}\right)\:{with}\:{a}_{{n}} =\frac{\mathrm{2}}{{T}}\int_{\left[{T}\right]} \:\:{x}^{\mathrm{3}} {sin}\right){nx}\right){dx} \\ $$$$=\frac{\mathrm{2}}{\mathrm{2}\pi}\:\int_{−\pi} ^{\pi} \:{x}^{\mathrm{3}} \:{sin}\left({nx}\right){dx}\:=\frac{\mathrm{2}}{\pi}\:\int_{\mathrm{0}} ^{\pi} \:{x}^{\mathrm{3}} \:{sin}\left({nx}\right){dx}\:\Rightarrow \\ $$$$\frac{\pi}{\mathrm{2}}{a}_{{n}} =\:\int_{\mathrm{0}} ^{\pi} \:{x}^{\mathrm{3}} \:{sin}\left({nx}\right){dx}\:=_{{nx}={t}} \:\:\:\int_{\mathrm{0}} ^{{n}\pi} \frac{{t}^{\mathrm{3}} }{{n}^{\mathrm{3}} }{sin}\left({t}\right)\frac{{dt}}{{n}} \\ $$$$=\frac{\mathrm{1}}{{n}^{\mathrm{4}} }\:\int_{\mathrm{0}} ^{{n}\pi} \:{t}^{\mathrm{3}} \:{sint}\:{dt}\:\:\:{by}\:{parts}\:\:\int_{\mathrm{0}} ^{{n}\pi} \:{t}^{\mathrm{3}} \:{sint}\:{dt}=\left[−{t}^{\mathrm{3}} {cost}\right]_{\mathrm{0}} ^{{n}\pi} \\ $$$$+\int_{\mathrm{0}} ^{{n}\pi} \mathrm{3}{t}^{\mathrm{2}} {cost}\:{dt}\:=−{n}^{\mathrm{3}} \pi^{\mathrm{3}} \left(−\mathrm{1}\right)^{{n}} \:+\mathrm{3}\:\int_{\mathrm{0}} ^{{n}\pi} \:{t}^{\mathrm{2}} \:{cost}\:{dt} \\ $$$${by}\:{parts}\:{again}\:\:\:\int_{\mathrm{0}} ^{\pi} \:{t}^{\mathrm{2}} {cost}\:{dt}\:=\left[{t}^{\mathrm{2}} {sint}\right]_{\mathrm{0}} ^{{n}\pi} \:−\int_{\mathrm{0}} ^{{n}\pi} \mathrm{2}{t}\:{sint} \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{{n}\pi} \:{t}\:\left(−{sint}\right){dt}\:=\left[{t}\:{cost}\right]_{\mathrm{0}} ^{{n}\pi} −\int_{\mathrm{0}} ^{{n}\pi} \:{cost}\:{dt} \\ $$$$={n}\pi\left(−\mathrm{1}\right)^{{n}} −\mathrm{0}\:\Rightarrow\frac{\pi}{\mathrm{2}}{a}_{{n}} =\frac{\mathrm{1}}{{n}^{\mathrm{4}} }\left\{−{n}^{\mathrm{3}} \pi^{\mathrm{3}} \left(−\mathrm{1}\right)^{{n}} \:+\mathrm{3}{n}\pi\left(−\mathrm{1}\right)^{{n}} \right\}\:\Rightarrow \\ $$$${a}_{{n}} =\frac{\mathrm{2}}{\pi{n}^{\mathrm{4}} }\left\{−\pi^{\mathrm{3}} {n}^{\mathrm{3}} \left(−\mathrm{1}\right)^{{n}} +\mathrm{3}\pi{n}\left(−\mathrm{1}\right)^{{n}} \right\}\:=−\frac{\mathrm{2}\pi^{\mathrm{2}} }{{n}}\left(−\mathrm{1}\right)^{{n}} \:+\frac{\mathrm{6}}{{n}^{\mathrm{3}} }\left(−\mathrm{1}\right)^{{n}} \:\Rightarrow \\ $$$${x}^{\mathrm{3}} \:=−\mathrm{2}\pi^{\mathrm{2}} \sum_{{n}=\mathrm{1}} ^{\infty} \frac{\left(−\mathrm{1}\right)^{{n}} }{{n}}{sin}\left({nx}\right)\:+\mathrm{6}\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{3}} }{sin}\left({nx}\right) \\ $$$$ \\ $$