Given:
The periodic function:
\(f(x) = \begin{cases}
x, & 0 < x < 1 \
0, & 1 < x < 2
\end{cases}\)
The Fourier series is given by:
\(f(x) = a_0 + \sum_{n=1}^\infty \left[a_n \cos(n\pi x) + b_n \sin(n\pi x)\right]\)
We need to find the Fourier coefficient \(b_n\).
Concept:
The Fourier coefficient \(b_n\) for a periodic function is given by:
\(b_n = \frac{2}{T}\int_0^T f(x) \sin\left(\frac{2\pi n x}{T}\right) dx\)
Here, \(T\) is the period of the function.
Calculation:
The period of the function \(f(x)\) is \(T = 2\).
\(b_n = \frac{2}{2} \int_0^2 f(x) \sin(n\pi x) dx\)
\(b_n = \int_0^2 f(x) \sin(n\pi x) dx\)
Split the integral according to the definition of \(f(x)\):
\(b_n = \int_0^1 x \sin(n\pi x) dx + \int_1^2 0 \cdot \sin(n\pi x) dx\)
\(b_n = \int_0^1 x \sin(n\pi x) dx\)
Use integration by parts:
Let \(u = x\) and \(dv = \sin(n\pi x) dx\).
Then, \(du = dx\) and \(v = -\frac{\cos(n\pi x)}{n\pi}\).
\(b_n = \left[-x \frac{\cos(n\pi x)}{n\pi}\right]_0^1 + \int_0^1 \frac{\cos(n\pi x)}{n\pi} dx\)
Evaluate the first term:
\(\left[-x \frac{\cos(n\pi x)}{n\pi}\right]_0^1 = -\frac{\cos(n\pi)}{n\pi} + 0 = -\frac{(-1)^n}{n\pi}\)
Now evaluate the second term:
\(\int_0^1 \frac{\cos(n\pi x)}{n\pi} dx = \frac{1}{n\pi} \int_0^1 \cos(n\pi x) dx\)
\(\int \cos(kx) dx = \frac{\sin(kx)}{k}\), so:
\(\int_0^1 \cos(n\pi x) dx = \left[\frac{\sin(n\pi x)}{n\pi}\right]_0^1\)
\(= \frac{\sin(n\pi)}{n\pi} - \frac{\sin(0)}{n\pi} = 0\)
Thus, the second term is zero.
So, \(b_n = -\frac{(-1)^n}{n\pi}\).
For \(n\) odd, \(b_n = -\frac{-1}{n\pi} = \frac{1}{n\pi}\).
For \(n\) even, \(b_n = -\frac{1}{n\pi}\).
Given \(n = 2\), \(b_2 = -\frac{2}{\pi^2}\).
∴ The Fourier coefficient \(b_n = -\frac{2}{\pi^2}\).