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Question Number 50702 by pooja24 last updated on 19/Dec/18

$$writethefourierseriesof$$$$f\left(x\right)=x0⩽x⩽2$$

Commented by maxmathsup by imad last updated on 19/Dec/18

$$ithink0⩽x⩽2\pi iffis2\pi periodicoddwehave$$$$f\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}{a}_{n}sin\left(nx\right)with{a}_n=\frac{2}{T}{\int }_{\left[T\right]}f\left(x\right)sin\left(nx\right)dx$$$$=\frac{2}{2\pi }{\int }_{-\pi }^{\pi }xsin\left(nx\right)dx=\frac{2}{\pi }{\int }_0^{\pi }xsin(nxdx\Rightarrow \frac{\pi }{2}{a}_n={\int }_0^{\pi }xsin\left(nx\right)dx$$$$byparts{\int }_0^{\pi }xsin\left(nx\right)dx={[-\frac{x}{n}cos\left(nx\right)]}_0^{\pi }+{\int }_0^{\pi }\frac{1}{n}cos\left(nx\right)dx$$$$=-\frac{\pi }{n}{(-1)}^n+\frac{1}{n}{\left[\frac{1}{n}sin\left(nx\right)\right]}_0^{\pi }=\frac{\pi {(-1)}^{n-1}}{n}\Rightarrow \frac{\pi }{2}{a}_n=\frac{\pi }{n}{(-1)}^{n-1}\Rightarrow$$$$$$$$a_n=\frac{2}{n}{(-1)}^{n-1}\Rightarrow f\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{2}{n}{(-1)}^{n-1}sin\left(nx\right).$$

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