Solution.

Part 1

Hence, \(\displaystyle {f(x) = a_0 + \sum ^{\infty }_{n=1}a_n \cos n x}\)

\begin {align*} a_0 &= \frac {1}{\pi } \int ^{\pi }_0 f(x)\,dx = \frac {1}{\pi }\int ^{\pi }_0 x\,dx\\\\ & = \frac {\pi }{2}\\ \end {align*}

\begin {align*} a_n & = \frac {2}{\pi }\int ^{\pi }_0 f(x) \cos nx \,dx\\\\ & = \frac {2}{\pi }\int ^{\pi }_0 x \cos n x\,dx\\\\ & = \frac {2}{n^2\pi }\Big [\cos nx + nx\sin nx \Big ]^{\pi }_0\\\\ & = \frac {2}{n^2\pi }\Big [\cos n\pi + n\pi \sin n\pi - (1-0)\Big ]\\\\ & =\frac {2}{n^2\pi }\big [(-1)^n-1\big ] \end {align*}

\[a_n = \begin {cases} 0, & \text {if}\hspace {0.3cm} n \hspace {0.3cm}\text {is even}\\\\ \dfrac {-4}{n^2\pi }, & \text {if}\hspace {0.3cm} n \hspace {0.3cm} \text {is odd}\\ \end {cases} \]

\begin {align*} f(x) & = a_0 + \sum ^{\infty }_{n=1} a_n \cos nx\\ & = a_0 + a_1\cos x + a_2 \cos 2x + a_3\cos 3x+ \cdots \cdots \cdots \\\\ & = \frac {\pi }{2} - \frac {4}{\pi }\cos x - \frac {4}{9\pi }\cos 3x - \frac {4}{25\pi }\cos 5x - \cdots \cdots \\\\ & = \frac {\pi }{2}-\frac {4}{\pi }\Big (\cos x + \frac {\cos 3x}{9} + \frac {\cos 5x}{25}+ \cdots \cdots + \frac {\cos (2n-1)x}{(2n-1)^2}+\cdots \\ \end {align*}

Find the sum \(1 + \dfrac {1}{9}+ \dfrac {1}{25}+ \cdots \cdots \cdots \)

Let \(x=0 \implies f(0) = \dfrac {0+ 0}{2} = 0\)

\[f(0) = \frac {\pi }{2} - \frac {4}{\pi }\Big (1 + \dfrac {1}{9}+ \dfrac {1}{25}+ \cdots \cdots \cdots \Big )\hspace {0.3cm} \text {from the series}\]

\[\frac {-\pi }{2}= \frac {-4}{\pi }\Big (1 + \dfrac {1}{9}+ \dfrac {1}{25}+ \cdots \cdots \cdots \Big )\]

\[\therefore \hspace {0.5cm} 1 + \dfrac {1}{9}+ \dfrac {1}{25}+ \cdots \cdots + \cdots = \frac {\pi ^2}{8}\]

Part 2

\(P = 2L = 6 \implies L = 3\)

\(f(x)\) is odd and so \(a_n = 0\hspace {0.3cm}\forall n = 0,1,2,\cdots \cdots \)

Hence we get Fourier Sine series

\[f(x) = \sum ^{\infty }_{n =1} b_n \sin \frac {n\pi }{2}x\]

\begin {align*} b_n & = \frac {2}{L}\int ^L_0 f(x) \sin \frac {n\pi }{L}x\,dx\\\\ & = \frac {2}{3}\int ^3_0x \sin \frac {n\pi }{3}x\,dx\\\\ & = \frac {2}{3}\Big [\frac {-3}{n\pi }x\cos \frac {n\pi }{3}x\Big |^3_0 + \frac {3}{n\pi }\int ^3_0 \cos \frac {n\pi }{x}\,dx\Big ]\\\\ & = \frac {-6(-1)^n}{n\pi }\hspace {0.3cm},\hspace {0.3cm} n = 1,2,\cdots \cdots \\ \end {align*}

\[\implies \hspace {0.5cm} b_n = \begin {cases} \dfrac {-6}{n\pi } & \text {if}\hspace {0.2cm} n \hspace {0.2cm} \text {is even}\\\\ \dfrac {6}{n\pi } & \text {if}\hspace {0.2cm} n \hspace {0.2cm} \text {is odd}\\ \end {cases} \]

\begin {align*} \therefore \hspace {0.5cm} f(x) & = \sum ^{\infty }_{n=1}b_n \sin \frac {nx}{3}\\\\ & = \frac {6}{\pi }\sin \frac {\pi }{3} x - \frac {6}{2\pi }\sin \frac {2\pi }{3} x + \frac {6}{3\pi }\sin \pi x - \frac {6}{4\pi }\sin \frac {4\pi }{3} x + \cdots \cdots \ \end {align*}