Question

Let \( g(x) = x^2 \), \( -\pi \le x \le \pi \). The coefficient of \( \cos(3x) \) in the Fourier series expansion of \(g(x)\) is:

• A. \( -4/9 \)
• B. \( -1/4 \)
• C. \( 1/16 \)
• D. \( 9/16 \)

Hint:

For Fourier series on \([-\pi, \pi]\): - If \(f(x)\) is even, all \(b_n=0\) and \(a_n = \frac{2}{\pi}\int_0^\pi f(x)\cos(nx)dx\). - If \(f(x)\) is odd, all \(a_n=0\) and \(b_n = \frac{2}{\pi}\int_0^\pi f(x)\sin(nx)dx\). This can save significant calculation time.

Answer 1

The Correct Option is A

Step 1: Find the formula for the Fourier coefficient \(a_n\). For a function \(f(x)\) on \([-\pi, \pi]\), the Fourier series is \( \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos(nx) + b_n \sin(nx)) \). The coefficient of \( \cos(nx) \) is \(a_n\), calculated by:\[ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} g(x) \cos(nx) dx \]We need \(a_3\).

Step 2: Set up the integral for \(a_n\) with \(g(x)=x^2\). Since \(g(x)=x^2\) and \(\cos(nx)\) are even functions, their product is also even. Simplify the integral:\[ a_n = \frac{2}{\pi} \int_{0}^{\pi} x^2 \cos(nx) dx \]

Step 3: Evaluate the integral using integration by parts. Use \( \int u dv = uv - \int v du \). Let \(u=x^2\) and \(dv = \cos(nx)dx\).\[ a_n = \frac{2}{\pi} \left[ x^2 \frac{\sin(nx)}{n} \right]_{0}^{\pi} - \frac{2}{\pi} \int_{0}^{\pi} 2x \frac{\sin(nx)}{n} dx \]The first term is zero. Use parts again for the second integral with \(u=2x\) and \(dv=\frac{\sin(nx)}{n}dx\).\[ a_n = -\frac{4}{n\pi} \left( \left[ x \left(-\frac{\cos(nx)}{n}\right) \right]_{0}^{\pi} - \int_{0}^{\pi} \left(-\frac{\cos(nx)}{n}\right) dx \right) \]\[ a_n = -\frac{4}{n\pi} \left( -\frac{\pi\cos(n\pi)}{n} - 0 + \left[ \frac{\sin(nx)}{n^2} \right]_{0}^{\pi} \right) \]The sine term evaluates to zero.\[ a_n = -\frac{4}{n\pi} \left( -\frac{\pi\cos(n\pi)}{n} \right) = \frac{4\cos(n\pi)}{n^2} \]

Step 4: Substitute \(n=3\) to find \(a_3\). Since \( \cos(n\pi) = (-1)^n \), \( a_n = \frac{4(-1)^n}{n^2} \).For \(n=3\):\[ a_3 = \frac{4(-1)^3}{3^2} = \frac{-4}{9} \]