a) Consider the periodic function f(x) defined by : f(x)= ) 0,&-pi lt xlt 0 x,&0leqslant xlt pi i.Sketch the function in the interval [-3pi ,3pi ] ii. Find the Fourier series expansion for f(x) Hence sum the series: 1+(1)/(3^2)+(1)/(5^2)+(1)/(7^2)+...

Let's correct and complete the Fourier series expansion for \( f(x) \) and sum the series.<br /><br />### i. Sketch the function in the interval \([-3\pi, 3\pi]\)<br /><br />The function \( f(x) \) is defined as:<br />\[ f(x) = \begin{cases} <br />0 & \text{for } -\pi < x < 0 \\<br />x & \text{for } 0 \leq x < \pi <br />\end{cases} \]<br /><br />This means \( f(x) \) is zero for \( -\pi < x < 0 \) and \( f(x) = x \) for \( 0 \leq x < \pi \). To sketch this:<br /><br />- For \( -3\pi \leq x < -\pi \), \( f(x) = 0 \).<br />- For \( -\pi < x < 0 \), \( f(x) = 0 \).<br />- For \( 0 \leq x < \pi \), \( f(x) = x \).<br />- For \( \pi \leq x < 3\pi \), \( f(x) = 0 \).<br /><br />### ii. Find the Fourier series expansion for \( f(x) \)<br /><br />The Fourier series expansion of a periodic function \( f(x) \) with period \( 2\pi \) is given by:<br />\[ f(x) = a_0 + \sum_{n=1}^{\infty} a_n \cos(nx) + b_n \sin(nx) \]<br /><br />Where the coefficients \( a_n \) and \( b_n \) are given by:<br />\[ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx \]<br />\[ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) \, dx \]<br /><br />For \( f(x) \) defined as:<br />\[ f(x) = \begin{cases} <br />0 & \text{for } -\pi < x < 0 \\<br />x & \text{for } 0 \leq x < \pi <br />\end{cases} \]<br /><br />#### Compute \( a_0 \):<br />\[ a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \, dx \]<br />Since \( f(x) = 0 \) for \( -\pi < x < 0 \):<br />\[ a_0 = \frac{1}{\pi} \int_{0}^{\pi} x \, dx = \frac{1}{\pi} \left[ \frac{x^2}{2} \right]_{0}^{\pi} = \frac{1}{\pi} \cdot \frac{\pi^2}{2} = \frac{\pi}{2} \]<br /><br />#### Compute \( a_n \):<br />\[ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) \, dx \]<br />Since \( f(x) = 0 \) for \( -\pi < x < 0 \):<br />\[ a_n = \frac{1}{\pi} \int_{0}^{\pi} x \cos(nx) \, dx \]<br />\[ a_n = \frac{1}{\pi} \left[ x \frac{\sin(nx)}{n} \right]_{0}^{\pi} - \frac{1}{\pi} \int_{0}^{\pi} \frac{\sin(nx)}{n} \, dx \]<br />\[ a_n = \frac{1}{\pi} \left[ \frac{\pi \sin(n\pi)}{n} - \frac{0 \sin(n\pi)}{n} \right] - \frac{1}{\pi n} \int_{0}^{\pi} \sin(nx) \, dx \]<br />\[ a_n = \frac{1}{\pi} \cdot 0 - \frac{1}{\pi n} \left[ -\frac{\cos(nx)}{n} \right]_{0}^{\pi} \]<br />\[ a_n = \frac{1}{\pi n} \left[ \frac{\cos(n\pi)}{n} - \frac{\cos(0)}{n} \right] \]<br />\[ a_n = \frac{1}{\pi n} \left[ \frac{(-1)^n - 1}{n} \right] \]<br />\[ a_n = \frac{(-1)^n - 1}{\pi n^2} \]<br /><br />#### Compute \( b_n \):<br />\[ b_n = \frac{1}{\pi}