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**Mathstud28** Let $\displaystyle \mathcal{F}(x)=x~\forall{x}\in(0.2\pi)$. Also let $\displaystyle \mathcal{F}(x)$ posses the charcteristic that $\displaystyle \mathcal{F}(x)\in{P}_{2\pi}\implies\mathcal{F}(x+2 \pi)=\mathcal{F}(x)~\forall{x}\in\mathbb{R}$. Now note that $\displaystyle \mathcal{F}(x)\in\mathcal{C}\left(x\ne{2\pi\mathbb {N}}\right)$, and that $\displaystyle \mathcal{F}(x)\in{D^1}$ or in other words it posses a left and right hand limit at all points. So now that we have established these conditions we can see that $\displaystyle \mathcal{F}(x)\to{S_n(x)}~\forall{x}\ne{2\pi\mathb b{N}}$, where $\displaystyle S_n(x)$ is the Fourier Series given by

$\displaystyle S_n(x)=\frac{A_0}{2}+\sum_{n=1}^{\infty}\bigg[A_n\cos(nx)+B_n\sin(nx)\bigg]$

Where the coefficents are given by:

$\displaystyle \begin{aligned}A_0&=\frac{1}{\pi}\int_0^{2\pi}\mat hcal{F}(x)dx\\

&=\frac{1}{\pi}\int_0^{2\pi}xdx\\

&=2\pi

\end{aligned}$

$\displaystyle \begin{aligned}A_n&=\frac{1}{\pi}\int_0^{2\pi}\mat hcal{F}(x)\cos(nx)dx\\

&=\frac{1}{\pi}\int_0^{2\pi}x\cos(nx)dx\\

&=0\quad\forall{n}\in\mathbb{N}

\end{aligned}$

$\displaystyle \begin{aligned}A_n&=\frac{1}{\pi}\int_0^{2\pi}\mat hcal{F}(x)\sin(nx)dx\\

&=\frac{1}{\pi}\int_0^{2\pi}x\sin(nx)dx\\

&=\frac{-2}{n}\quad\forall{n}\in\mathbb{N}

\end{aligned}$

So we can see that

$\displaystyle S_n(x)=\pi-2\sum_{n=1}^{\infty}\frac{\sin(nx)}{n}$

And that

$\displaystyle \forall{x}\in(0,2\pi)~~x=\pi-2\sum_{n=1}^{\infty}\frac{\sin(nx)}{n}$