f(x) = |x|

$\therefore f(-x)= |-x|= |x| = f(x)$

$\therefore f(x) \ $is even function

$\therefore b_n = 0$

Here, $l=\pi$

Now, $a_0 = \dfrac{2}{l} \displaystyle \int^1_0 f(x) dx$

$= \dfrac {2}{\pi} \displaystyle \int^\pi _0 |x| dx$

$\displaystyle =\dfrac {2}{\pi} \int^\pi_0 xdx$

$= \dfrac {2}{\pi}\left [ \dfrac {x^2}{2} \right ]^\pi_0$

$= \dfrac {1}{\pi} \Big [ \pi ^2 -0 \Big]$

$= \pi $

$a_n = \dfrac {2}{l} \displaystyle \int^1_0 f(x) \cos \dfrac {n \pi x}{l}dx $

$=\dfrac {2}{\pi} \displaystyle \int^\pi_0 |x| \cos \dfrac {n \pi x}{\pi}dx $

$= \dfrac {2}{\pi} \displaystyle \int^\pi_0 x \cos n x \ dx$

$= \dfrac {2}{\pi} \left [ x \cdot \dfrac {\sin nx}{n} - 1 \cdot \dfrac {-\cos nx}{n^2} \right ] $

$= \dfrac {2}{\pi} \left [ \left ( \pi \cdot \dfrac {\sin n \pi} { n} + \dfrac {\cos n \pi}{n^2} \right ) - \left ( 0+ \dfrac {\cos 0}{n^2} \right ) \right ]$

$= \dfrac {2}{\pi} \left [ 0 + \dfrac {(-1)^n}{n^2} - 0 - \dfrac {1}{n^2} \right ]$

$= \dfrac {2} {\pi n^2 } \Big [ (-1)^n - 1 \Big ] $

In Fourier Series,

$f(x)= \dfrac {a_0}{2} + \displaystyle \sum^\infty _ {n=1} a_n\cos \dfrac {n\pi x}{l} + \sum^\infty _{n=1} b_n, \sin \dfrac {n \pi x}{l}$

$\therefore |x| = \dfrac {\pi}{2} + \displaystyle \sum^{\infty}_{n=1} \dfrac {2}{\pi n^2} \Big [ (-1)^n -1 \Big] \cos \dfrac {n\pi x}{\pi} + 0 $

$\therefore |x| = \dfrac {\pi}{2}+ \dfrac {2} {\pi} \left ( \dfrac {-2 \cos x}{1^2} + 0 - \dfrac {2 \cos 3x}{3^2}+0 - \dfrac {2 \cos 5x}{5^2}+ \cdots \right ) $

$\therefore |x| = \dfrac {\pi}{2} - \dfrac {4}{\pi} \left ( \dfrac {\cos x} {l^2} + \dfrac {\cos 3x}{3^2} + \dfrac {\cos 5x}{5^2}+\cdots \right )$