Example 5.3.1.

(a).

Find the Fourier series representation of \[f(x) = \begin {cases} 0, & -\pi \leq x < 0\\ \dfrac {x}{\pi }, & 0 \leq x \leq \pi \end {cases}\] with period \(2\pi .\)

Solution. \[a_n = \frac {1}{\pi }\int _{-\pi }^{\pi }f(x)\, \cos nx\, dx = \frac {1}{\pi }\int _{-\pi }^{\pi }\frac {x}{\pi }\, \cos nx\, dx = \frac {1}{\pi ^2}\int _{-\pi }^{\pi }x\, \cos nx\, dx.\] Let \(\, u = x, \, \hspace {0.3cm} dv = \cos nx\, dx, \hspace {0.3cm} du = dx, \, \hspace {0.3cm} v = \frac {1}{n}\, \sin x\) \begin {align*} a_n & = \frac {1}{\pi ^2}\left [\frac {x\sin x}{n}\big |^{\pi }_0 - \frac {1}{n}\int _0^{\pi }\sin nx\, dx\right ]\\ & = \frac {1}{\pi ^2\, n^2}[(-1)^n - 1]\, , \hspace {0.3cm} \cos nx = (-1)^n\\\\ & = \begin {cases} 0, & n = \, \text {even}\\ \dfrac {-2}{\pi ^2\, n^2}, & n = \, \text {odd} \end {cases} \end {align*}

\begin {align*} b_n = \frac {1}{\pi }\int _{\pi }^{\pi }f(x)\, \sin nx\, dx & = \frac {1}{\pi ^2}\int _0^{\pi }x\, \sin nx\, dx\\ & = \frac {(-1)^{n + 1}}{n\pi }\\ & = \begin {cases} \dfrac {-1}{\pi }, & n \, \, \text {is even}\\ \dfrac {1}{\pi }, & n\, \, \text {is odd}\\ \end {cases} \end {align*}

For \(n = 0\), \[a_0 = \frac {1}{\pi }\int _0^{\pi }f(x)\, dx = \frac {1}{\pi ^2}\int _0^{\pi }x\, dx = \frac {1}{2}.\] Thus \[f(x) \cong \frac {1}{4} - \frac {2}{\pi ^2}\sum _{n=1}^{\infty }\frac {1}{(2n - 1)^2}\, \cos [(2n-1)x] - \frac {1}{\pi }\sum ^{\infty }_{n=1}\frac {(-1)^n}{n}\,\sin nx.\] Since \(a_n\) is defined on odd hence we make \(n^2\) odd as \((2n-1)\). □

(b).

Find the Fourier series representation of \(f(x) = x^2\) on \([-\pi ,\pi ]\) with period \(2\pi \).

Solution.

\[a_0 = \frac {1}{\pi }\int _{-\pi }^{\pi } x^2 \, dx = \frac {2\pi ^2}{3}.\]

\[a_n = \frac {1}{\pi }\int _{-\pi }^{\pi } x^2\, \cos nx\, dx,\] let \(\, \, u= x^2, \hspace {0.3cm} du = 2x\, dx, \hspace {0.3cm} dv = \cos nx dx, \hspace {0.3cm} v = \dfrac {\sin nx}{n}\) \begin {align*} a_n & = \frac {x^2\sin nx}{\pi n}\big |^{\pi }_{-\pi } - \frac {2}{\pi n}\int _{-\pi }^{\pi } x\sin nx\, dx\\ & = \frac {-2}{n\pi }\int _{-\pi }^{\pi }x\, \sin nx\, dx\\ & = \frac {2x\, \cos nx}{\pi n^2}\big |_{-\pi }^{\pi } - \frac {2}{\pi n^2}\int _{-\pi }^{\pi }\cos nx\, dx\\ & = \frac {4\, \cos n\pi }{n^2}\\ & = \frac {4(-1)^n}{n^2}\, , \hspace {0.5cm} n = 1, \, 2, \, \cdots \cdots \end {align*}

\[b_n = \frac {1}{\pi }\int _{-\pi }^{\pi }x^2\, \sin nx\,dx = 0, \, \,\hspace {0.3cm} \text {since}\, \, x^2\, \sin x\,\, \text {is odd}.\]

Thus, \[x^2 \cong \frac {\pi ^2}{3} - 4\left (\cos x - \frac {\cos 2x}{2^2} + \frac {\cos 3x}{3^2} + \cdots \cdots \right ).\] □