Fourier series: 2x − x²

For a function on (0, 3) taken with period 3 (so 2l = 3 and l = 3/2): f(x) = a₀/2 + Σ [aₙ cos(2nπx/3) + bₙ sin(2nπx/3)], where a₀ = (2/3)∫₀³ f dx, aₙ = (2/3)∫₀³ f cos(2nπx/3) dx, bₙ = (2/3)∫₀³ f sin(2nπx/3) dx.

For f(x) = 2x − x²:

Let k = 2nπ/3. At x = 3, kx = 2nπ, so sin kx = 0 and cos kx = 1.

a₀ = (2/3)∫₀³ (2x − x²) dx = (2/3)[x² − x³/3]₀³ = (2/3)(9 − 9) = 0

∫₀³ (2x − x²) cos kx dx = [(2x − x²) sin kx/k]₀³ − (1/k)∫₀³ (2 − 2x) sin kx dx
                        = 0 − (1/k)(6/k) = −6/k²
aₙ = (2/3)(−6/k²) = −4/k² = −9/(n²π²)

∫₀³ (2x − x²) sin kx dx = [−(2x − x²) cos kx/k]₀³ + (1/k)∫₀³ (2 − 2x) cos kx dx
                        = 3/k + 0
bₙ = (2/3)(3/k) = 2/k = 3/(nπ)

(Here ∫₀³ (2 − 2x) sin kx dx = 6/k and ∫₀³ (2 − 2x) cos kx dx = 0, by one more integration by parts.)

Fourier series

2x − x² = Σ (n=1 to ∞) [ −(9/(n²π²)) cos(2nπx/3) + (3/(nπ)) sin(2nπx/3) ], 0 < x < 3

Since a₀ = 0 there is no constant term.