Question

Find Fourier series expansion for the function f(x)={x+4 for 0<x<π,-x-π for -π<x<0}

Answer 1

 f(x)={x+4 for 0<x<π,

-x-π for -π<x<0}

The Fourier series expansion of any function f(x) is given by

f(x) = $\dfrac{ a_0}{2}$ + $a_1 cosx$ + $a_2 cos2x$ + ........ + $b_1 sin x + b_2 sin 2x$ + ............ (1)

where,

a₀ = $\dfrac{1}{\pi}$ $\int_{-\pi}^\pi$ f(x) dx

a₀ = $\dfrac{1}{\pi}$ ($\int^0_{-\pi}$ ( - x - $\pi$ ) dx + $\int^\pi_{0}$ (x + 4))dx

a₀ = $\dfrac{1}{\pi}$[ $(\dfrac{-x^2}{2} - 4x)_{-\pi}^0$ +$(\dfrac{x^2}{2} + 4x)^\pi_0$ ]

a₀ = $\pi$ ...........................................(2)

a_n = $\dfrac{1}{\pi}$ $\int_{-\pi}^\pi$ f(x) cos nx dx

a_n = $\dfrac{1}{\pi}$ ($\int^0_{-\pi}$ ( - x - $\pi$ ) dx + $\int^\pi_{0}$ (x + 4)) cos nx dx

a_n = $\dfrac{1}{\pi}$ $(\dfrac{(-x-\pi) sin nx}{n} - \dfrac{cos nx}{n^2})^0_{-\pi}$ + $(\dfrac{(x + 4)sin nx}{n^2} + \dfrac{cos nx}{n^2})^\pi_0$

a_n = $\dfrac{(-1)^n - 1}{\pi*n^2}$ ...............................................(2)

From equation(2) we have

a₁ = $\dfrac{-2}{\pi}$ , a₂ = 0 , a₃ = $\dfrac{-2}{9\pi}$

b_n = $\dfrac{1}{\pi}$ $\int_{-\pi}^\pi$ f(x) sin nx dx

b_n = $\dfrac{1}{\pi}$ ($\int^0_{-\pi}$ ( - x - $\pi$ ) dx + $\int^\pi_{0}$ (x + 4)) sin nx dx

b_n = $\dfrac{1}{\pi}$ $(\dfrac{(-x-\pi) sin nx}{n} - \dfrac{cos nx}{n^2})^0_{-\pi}$ + $(\dfrac{(x + 4)sin nx}{n^2} + \dfrac{cos nx}{n^2})^\pi_0$

b_n = $\dfrac{1}{n\pi}$ $[\pi(1 + (-1)^n - 4(1 - (-1)^n)]$ ............................(3)

From equation (3) we have

b₁ = $\dfrac{-8}{\pi}$ , b₂ = 1 , b₃ = $\dfrac{-8}{3\pi}$

f(x) = $\dfrac{\pi}{2}$ + $\dfrac{-2cos x}{\pi} + \dfrac{-2cos3x}{9\pi}$ + ......... + $\dfrac{-8 sin x}{\pi} + sin 2x + \dfrac{-8 sin 3x}{3\pi}$ + ........