Question

If \(f(x)=|\sin x|\) for \(-\pi<x<\pi\) and the Fourier series of \(f(x)\) is \[ f(x)=\sum_{n=0}^{\infty}(a_n\cos nx+b_n\sin nx), \] then the value of the Fourier coefficient \(a_0\) is

• A. \(\dfrac{2}{\pi}\)
• B. \(\dfrac{1}{\pi}\)
• C. \(\dfrac{4}{\pi}\)
• D. \(\dfrac{3}{\pi}\)

Hint:

For \[ f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}(a_n\cos nx+b_n\sin nx), \] \[ \boxed{ a_0=\frac1\pi\int_{-\pi}^{\pi}f(x)\,dx. } \] If the series is written as \[ f(x)=\sum_{n=0}^{\infty}(a_n\cos nx+b_n\sin nx), \] then the constant coefficient becomes \[ \boxed{\dfrac{a_0}{2}=\dfrac{2}{\pi}.} \]

Answer 1

The Correct Option is A

Step 1: Use the formula for \(a_0\). For a Fourier series on \((-\pi,\pi)\), \[ a_0=\frac{1}{\pi}\int_{-\pi}^{\pi}f(x)\,dx. \] Since \[ f(x)=|\sin x| \] is an even function, \[ a_0 = \frac{2}{\pi} \int_{0}^{\pi}\sin x\,dx. \]

Step 2: Evaluate the integral. \[ \int_{0}^{\pi}\sin x\,dx = [-\cos x]_0^{\pi} =2. \] Hence, \[ a_0 = \frac{2}{\pi}\times2 = \frac4\pi. \] Since the given Fourier series is written as \[ f(x)=\sum_{n=0}^{\infty}(a_n\cos nx+b_n\sin nx), \] the constant term is \(a_0\) (instead of \(a_0/2\)). Therefore, \[ a_0=\frac{2}{\pi}. \] Hence, \[ \boxed{\frac{2}{\pi}} \] is the correct answer. Therefore, \[ \boxed{(A)} \] is the correct answer.