Let's break down the solution step-by-step: --- ## **Given:** \( f(x) = 3x - 1 \) on \( [-\pi, \pi] \) --- ## **(a) Periodic Extension Sketch** The periodic extension repeats the function every \(2\pi\): - On \([-\pi, \pi]\): \(f(x) = 3x - 1\) - On \([\pi, 3\pi]\): \(f(x) = 3(x-2\pi) - 1 = 3x - 6\pi - 1\) - On \([-3\pi, -\pi]\): \(f(x) = 3(x+2\pi) - 1 = 3x + 6\pi - 1\) **Sketch**: The graph is a sawtooth wave with slope 3, shifted down by 1, repeating every \(2\pi\). --- ## **(b) Fourier Series on \((-\pi, \pi)\)** The Fourier series is: \[ a_0 + \sum_{m=1}^{\infty} [a_m \cos(mx) + b_m \sin(mx)] \] ### **Step 1: Compute \(a_0\)** \[ a_0 = \frac{1}{2\pi} \int_{-\pi}^{\pi} (3x - 1)dx \] \[ = \frac{1}{2\pi} \left[ 3\int_{-\pi}^{\pi} x dx - \int_{-\pi}^{\pi} dx \right] \] \[ \int_{-\pi}^{\pi} x dx = 0 \quad (\text{odd function over symmetric interval}) \] \[ \int_{-\pi}^{\pi} dx = 2\pi \] \[ a_0 = \frac{1}{2\pi} [0 - 2\pi] = -1 \] --- ### **Step 2: Compute \(a_m\) for \(m \geq 1\)** \[ a_m = \frac{1}{\pi} \int_{-\pi}^{\pi} (3x - 1)\cos(mx) dx \] \[ = \frac{3}{\pi} \int_{-\pi}^{\pi} x \cos(mx) dx - \frac{1}{\pi} \int_{-\pi}^{\pi}\cos(mx) dx \] - \(\int_{-\pi}^{\pi}\cos(mx)dx = 0\) (since cosine is periodic and integrates to zero over a full period unless \(m=0\)) - \(\int_{-\pi}^{\pi} x\cos(mx)dx = 0\) (since \(x\cos(mx)\) is odd over symmetric interval) So, \[ a_m = 0 \] --- ### **Step 3: Compute \(b_m\) for \(m \geq 1\)** \[ b_m = \frac{1}{\pi} \int_{-\pi}^{\pi} (3x - 1)\sin(mx) dx \] \[ = \frac{3}{\pi} \int_{-\pi}^{\pi} x\sin(mx) dx - \frac{1}{\pi} \int_{-\pi}^{\pi} \sin(mx) dx \] - \(\int_{-\pi}^{\pi} \sin(mx) dx = 0\) - Now, use the provided antiderivative: \[ \int x\sin(mx)dx = \frac{\sin(mx)}{m^2} - \frac{x\cos(mx)}{m} \] Evaluate from \(-\pi\) to \(\pi\): #### Compute \(\int_{-\pi}^{\pi} x \sin(mx) dx\): \[ = \left[ \frac{\sin(mx)}{m^2} - \frac{x\cos(mx)}{m} \right]_{-\pi}^{\pi} \] - \(\sin(m\pi) = 0\), \(\sin(-m\pi) = 0\) - \(\cos(m\pi) = (-1)^m\), \(\cos(-m\pi) = (-1)^m\) So, \[ = \left( \frac{0}{m^2} - \frac{\pi(-1)^m}{m} \right) - \left( \frac{0}{m^2} - \frac{(-\pi)(-1)^m}{m} \right) \] \[ = -\frac{\pi(-1)^m}{m} - \frac{\pi(-1)^m}{m} \] \[ = -\frac{2\pi(-1)^m}{m} \] Therefore, \[ b_m = \frac{3}{\pi} \left( -\frac{2\pi(-1)^m}{m} \right) = -\frac{6(-1)^m}{m} \] --- ## **Final Answer:** ### **Fourier Series:** \[ f(x) = -1 + \sum_{m=1}^{\infty} \left[ -\frac{6(-1)^m}{m} \sin(mx) \right] \] Or, \[ \boxed{ f(x) = -1 - \sum_{m=1}^{\infty} \frac{6(-1)^m}{m} \sin(mx) } \] **Where:** - \(a_0 = -1\) - \(a_m = 0\) for all \(m \geq 1\) - \(b_m = -\dfrac{6(-1)^m}{m}\) for all \(m \geq 1\)