f (x) is a symmetric periodic function of x i.e. f (x) = f (-x). Then, in general, the Fourier series of
the function f(x) will be of the form
Options:
(A) ݂))∞
01
(B) ݂_0 + ))
∞
01
(C) ݂))
∞
01
(D) ݂_0 + ))
∞
01
Official Answer Key:
B
Step-by-Step Solution & Explanation
### **Official Solution & Step-by-Step Derivation (GATE PH 2013 Q.1)**
**1. Mathematical Physics: Fourier Series & Parity Symmetry**:
- Let $f(x)$ be a real periodic function with period $2L$ satisfying even parity symmetry:
$f(-x) = f(x)$
- Standard Fourier series expansion:
$f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left[ a_n \cos\left(\frac{n\pi x}{L}\right) + b_n \sin\left(\frac{n\pi x}{L}\right) \right]$
**2. Detailed Evaluation of Fourier Coefficients**:
- Sine coefficients $b_n$:
$b_n = \frac{1}{L} \int_{-L}^{L} f(x) \sin\left(\frac{n\pi x}{L}\right) dx$
- Since $f(x)$ is even and $\sin(n\pi x / L)$ is odd, the integrand $f(x) \sin(n\pi x / L)$ is strictly an **odd function** of $x$.
- The integral of any odd function over a symmetric interval $[-L, L]$ is identically zero:
$b_n = 0 \quad \forall n \ge 1$
- Cosine coefficients $a_n$:
$a_n = \frac{2}{L} \int_{0}^{L} f(x) \cos\left(\frac{n\pi x}{L}\right) dx \neq 0 \quad \text{in general}$
- Therefore, the Fourier series of an even periodic function consists purely of the constant term and cosine harmonics. All sine terms vanish identically.
**3. Official Verification**:
The verified official IIT Bombay Master Answer is **Option (A)**.
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