Question

A periodic function \( f(x) \), with period 2, is defined as \[ f(x) = \begin{cases} -1 - x & \text{for } -1 \leq x&t;0 \\ 1 - x & \text{for } 0 \leq x \leq 1 \end{cases} \] The Fourier series of this function contains _________

• A. Both cos (\( n \pi x \)) and sin (\( n \pi x \)) where \( n = 1, 2, 3, \dots \)
• B. Only sin (\( n \pi x \)) where \( n = 1, 2, 3, \dots \)
• C. Only cos (\( n \pi x \)) where \( n = 1, 2, 3, \dots \)
• D. Only cos (\( 2n \pi x \)) where \( n = 1, 2, 3, \dots \)

Hint:

For odd periodic functions, the Fourier series contains only sine terms, while for even functions, it contains only cosine terms.

Answer 1

The Correct Option is B

Step 1: Fourier series for periodic functions.
The Fourier series for a periodic function consists of sine and cosine terms. Since the given function is odd (i.e., \( f(-x) = -f(x) \)), its Fourier series will only contain sine terms. Step 2: Analyzing the options.
- (A) Incorrect, since the function is odd, the Fourier series will not contain cosine terms.
- (B) Correct, since the function is odd, the Fourier series contains only sine terms.
- (C) Incorrect, cosine terms are excluded for odd functions.
- (D) Incorrect, this suggests a different form for cosine terms, but we need only sine terms for an odd function. Step 3: Conclusion.
Thus, the correct answer is (B) Only sin (\( n \pi x \)) where \( n = 1, 2, 3, \dots \).