Question Number 50702 by pooja24 last updated on 19/Dec/18
$${write}\:{the}\:{fourier}\:{series}\:{of}\: \\ $$$${f}\left({x}\right)={x}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{0}\leqslant{x}\leqslant\mathrm{2} \\ $$
Commented by maxmathsup by imad last updated on 19/Dec/18
![i think 0≤x≤2π if f is 2π periodic odd we have 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 sin(nx)dx =(2/π) ∫_0 ^π xsin(nxdx ⇒(π/2)a_n =∫_0 ^π xsin(nx)dx by parts ∫_0 ^π x sin(nx)dx =[−(x/n)cos(nx)]_0 ^π +∫_0 ^π (1/n) cos(nx)dx =−(π/n)(−1)^n +(1/n)[(1/n)sin(nx)]_0 ^π =((π(−1)^(n−1) )/n) ⇒(π/2)a_n =(π/n)(−1)^(n−1) ⇒ a_n =(2/n)(−1)^(n−1) ⇒ f(x) =Σ_(n=1) ^∞ (2/n)(−1)^(n−1) sin(nx).](https://www.tinkutara.com/question/Q50759.png)
$${i}\:{think}\:\:\mathrm{0}\leqslant{x}\leqslant\mathrm{2}\pi\:\:\:{if}\:{f}\:{is}\:\mathrm{2}\pi\:{periodic}\:{odd}\:{we}\:{have} \\ $$$${f}\left({x}\right)=\sum_{{n}=\mathrm{1}} ^{\infty} \:{a}_{{n}} {sin}\left({nx}\right)\:\:{with}\:{a}_{{n}} =\frac{\mathrm{2}}{{T}}\:\int_{\left[{T}\right]} \:{f}\left({x}\right)\:{sin}\left({nx}\right){dx} \\ $$$$=\frac{\mathrm{2}}{\mathrm{2}\pi}\:\int_{−\pi} ^{\pi} \:{x}\:{sin}\left({nx}\right){dx}\:=\frac{\mathrm{2}}{\pi}\:\int_{\mathrm{0}} ^{\pi} \:{xsin}\left({nxdx}\:\Rightarrow\frac{\pi}{\mathrm{2}}{a}_{{n}} =\int_{\mathrm{0}} ^{\pi} \:{xsin}\left({nx}\right){dx}\right. \\ $$$${by}\:{parts}\:\int_{\mathrm{0}} ^{\pi} \:{x}\:{sin}\left({nx}\right){dx}\:=\left[−\frac{{x}}{{n}}{cos}\left({nx}\right)\right]_{\mathrm{0}} ^{\pi} \:+\int_{\mathrm{0}} ^{\pi} \:\frac{\mathrm{1}}{{n}}\:{cos}\left({nx}\right){dx} \\ $$$$=−\frac{\pi}{{n}}\left(−\mathrm{1}\right)^{{n}} \:\:+\frac{\mathrm{1}}{{n}}\left[\frac{\mathrm{1}}{{n}}{sin}\left({nx}\right)\right]_{\mathrm{0}} ^{\pi} \:=\frac{\pi\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} }{{n}}\:\Rightarrow\frac{\pi}{\mathrm{2}}{a}_{{n}} =\frac{\pi}{{n}}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:\Rightarrow \\ $$$$ \\ $$$${a}_{{n}} =\frac{\mathrm{2}}{{n}}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} \:\Rightarrow\:{f}\left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{2}}{{n}}\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {sin}\left({nx}\right). \\ $$