Question

The Fourier cosine series for an even function f(x) is given by

\(f(x) = {a_0} + \sum\limits_{n = 1}^\infty {{a_n}\cos (nx)} \)

The value of the coefficient a₂ for the function f(x) = cos²(x) in [0,π] is

1. -0.5
2. 0.0
3. 0.5
4. 1.0

Answer 1

Correct Answer - Option 3 : 0.5

Concept:

The Fourier series for the function f(x) in the interval α < x < α + 2π is given by

\(f\left( x \right) = \frac{{{a_o}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}\cos nx + \mathop \sum \limits_{n = 1}^\infty {b_n}\sin nx\)

where

\({a_o} = \frac{1}{\pi }\mathop \smallint \limits_\alpha ^{\alpha + 2\pi } f\left( x \right)dx;\;{a_n} = \frac{1}{\pi }\mathop \smallint \limits_\alpha ^{\alpha + 2\pi } f\left( x \right)\cos nxdx;\;{b_n} = \frac{1}{\pi }\mathop \smallint \limits_\alpha ^{\alpha + 2\pi } f\left( x \right)\sin nxdx\)

When f is an even periodic function of period 2L, then its Fourier series contains only cosine (include possibly, the constant term) terms.

\(f\left( x \right) = \frac{{{a_o}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}\frac{{\cos n\pi x}}{L}\)

When f is an odd periodic function of period 2L, then its Fourier series contains only sine terms.

\(f\left( x \right) = \mathop \sum \limits_{n = 1}^\infty {b_n}\sin \frac{{n\pi x}}{L}\)

Calculation:

 f(x) = cos2(x)

\(f(x) = \frac{{1 + \cos 2x}}{2} = \frac{1}{2} + \frac{{\cos 2x}}{2}\)

On comparing with: \(f\left( x \right) = \frac{{{a_o}}}{2} + \mathop \sum \limits_{n = 1}^\infty {a_n}\frac{{\cos n\pi x}}{L}\)

a₀ = 1

a₁ = 0

a₂ = 1/2