Let's solve each part step by step. --- ## **(a) Sketch the periodic extension of \( f(x) \) over \( [-3\pi, 3\pi] \)** ### **Step 1: Understand \( f(x) \) on \( [-\pi, \pi] \)** Given: \( f(x) = x + 2 \) So on \( [-\pi, \pi] \), it is a straight line with slope 1 and y-intercept 2. - At \( x = -\pi \): \( f(-\pi) = -\pi + 2 \) - At \( x = \pi \): \( f(\pi) = \pi + 2 \) ### **Step 2: Periodic Extension** Since we are to sketch the periodic extension, repeat the function every \( 2\pi \): \[ f(x + 2\pi) = f(x) \] So, from \( -3\pi \) to \( 3\pi \), the function will repeat three times. - Interval 1: \( -3\pi \) to \( -\pi \): Same shape, shifted left by \( 2\pi \) - Interval 2: \( -\pi \) to \( \pi \): Original function - Interval 3: \( \pi \) to \( 3\pi \): Same shape, shifted right by \( 2\pi \) **Sketch:** A sawtooth wave, with each "tooth" rising from \( -\pi + 2 \) to \( \pi + 2 \), then jumping down and repeating. The jump occurs at every \( x = (2k + 1)\pi \). --- ## **(b) Fourier Series on \( (-\pi, \pi) \)** Fourier series: \[ f(x) = a_0 + \sum_{m=1}^\infty [a_m \cos(mx) + b_m \sin(mx)] \] ### **Step 1: Compute \( a_0 \)** \[ a_0 = \frac{1}{2\pi} \int_{-\pi}^\pi f(x) dx = \frac{1}{2\pi} \int_{-\pi}^\pi (x+2) dx \] \[ = \frac{1}{2\pi} \left[\int_{-\pi}^\pi x dx + \int_{-\pi}^\pi 2 dx \right] \] - \(\int_{-\pi}^\pi x dx = 0\) (since it's odd function over symmetric interval) - \(\int_{-\pi}^\pi 2 dx = 2 \times (2\pi) = 4\pi\) \[ a_0 = \frac{1}{2\pi}(0 + 4\pi) = 2 \] --- ### **Step 2: Compute \( a_m \)** \[ a_m = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \cos(mx) dx = \frac{1}{\pi} \int_{-\pi}^\pi (x+2) \cos(mx) dx \] \[ = \frac{1}{\pi} \left[ \int_{-\pi}^\pi x\cos(mx) dx + \int_{-\pi}^\pi 2\cos(mx) dx \right] \] - \(\int_{-\pi}^\pi x\cos(mx) dx = 0\) (odd function over symmetric interval) - \(\int_{-\pi}^\pi 2\cos(mx) dx = 2 \int_{-\pi}^\pi \cos(mx) dx = 0\) (cosine integral over full period, \(m \geq 1\)) So, \[ a_m = 0 \] --- ### **Step 3: Compute \( b_m \)** \[ b_m = \frac{1}{\pi} \int_{-\pi}^\pi f(x) \sin(mx) dx = \frac{1}{\pi} \int_{-\pi}^\pi (x+2) \sin(mx) dx \] \[ = \frac{1}{\pi} \left[ \int_{-\pi}^\pi x\sin(mx) dx + \int_{-\pi}^\pi 2\sin(mx) dx \right] \] - \(\int_{-\pi}^\pi 2\sin(mx) dx = 2 \int_{-\pi}^\pi \sin(mx) dx = 0\) (sine integral over full period) - \(\int_{-\pi}^\pi x\sin(mx) dx\): Use the antiderivative provided, \[ \int x\sin(mx) dx = \frac{\sin(mx) - xm\cos(mx)}{m^2} \] Evaluate from \( -\pi \) to \( \pi \): \[ \int_{-\pi}^\pi x\sin(mx) dx = \left[ \frac{\sin(mx) - xm\cos(mx)}{m^2} \right]_{-\pi}^{\pi} \] Plug in \( x = \pi \) and \( x = -\pi \): - \(\sin(m\pi) = 0\), \(\cos(m\pi) = (-1)^m\) So, \[ = \frac{0 - \pi m (-1)^m}{m^2} - \left( \frac{0 - (-\pi)m(-1)^m}{m^2} \right) \] \[ = \frac{ -\pi m (-1)^m }{ m^2 } - \frac{ \pi m (-1)^m }{ m^2 } \] \[ = \frac{ -2 \pi m (-1)^m }{ m^2 } = -2\pi \frac{ (-1)^m }{ m } \] Therefore, \[ b_m = \frac{1}{\pi} \left[ -2\pi \frac{ (-1)^m }{ m } \right] = -2 \frac{ (-1)^m }{ m } \] --- ## **Final Answers** ### **(a) Sketch:** Periodic sawtooth function, each period from \( x = -\pi \) to \( \pi \) is \( f(x) = x + 2 \), repeated over \( [-3\pi, 3\pi] \). ### **(b) Fourier Series:** \[ f(x) = 2 + \sum_{m=1}^\infty \left[ 0 \cdot \cos(mx) - 2\frac{(-1)^m}{m} \sin(mx) \right] \] Or, simplified: \[ \boxed{ f(x) = 2 - 2 \sum_{m=1}^\infty \frac{(-1)^m}{m} \sin(mx) } \] **Constants:** - \( a_0 = 2 \) - \( a_m = 0 \) for \( m \geq 1 \) - \( b_m = -2\frac{(-1)^m}{m} \) for \( m \geq 1 \) --- If you need the sketch, let me know!