Question

If any function is even, in Fourier series it contains

• A. Only b\(_n\)
• B. Only a\(_n\)
• C. Both a\(_0\) and a\(_n\)
• D. Only a\(_0\)

Hint:

Fourier Series for Even/Odd Functions. Even function \(f(-x)=f(x)\): \(b_n=0\), series contains only constant (\(a_0\)) and cosine (\(a_n\)) terms. Odd function \(f(-x)=-f(x)\): \(a_0=0, a_n=0\), series contains only sine (\(b_n\)) terms.

Answer 1

The Correct Option is C

The Fourier series for a function \(f(x)\) defined over an interval, say \([-\pi, \pi]\), is given by: $$ f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} (a_n \cos(nx) + b_n \sin(nx)) $$ where the coefficients are calculated as: $$ a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) dx $$ $$ a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) dx $$ $$ b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) dx $$ If the function \(f(x)\) is even, meaning \(f(-x) = f(x)\): - The product \(f(x)\sin(nx)\) is an odd function (Even \(\times\) Odd = Odd).
The integral of an odd function over a symmetric interval like \([-\pi, \pi]\) is zero.
Therefore, \(b_n = 0\) for all \(n \ge 1\).
- The product \(f(x)\cos(nx)\) is an even function (Even \(\times\) Even = Even).
The integral of an even function over a symmetric interval is non-zero (unless f(x) itself leads to cancellation) and equals \( 2 \times \) the integral over half the interval.
Thus, \(a_n\) (for \(n \ge 1\)) can be non-zero.
- Similarly, \(a_0\) involves the integral of \(f(x)\) (an even function), which can be non-zero.
Therefore, the Fourier series for an even function contains only the constant term (\(a_0/2\)) and the cosine terms (\(a_n \cos(nx)\)).
It contains both \(a_0\) and \(a_n\) coefficients (while \(b_n\) are zero).
Option (3) is the best description.