Fourier series: x − x²

On (−1, 1) the period is 2 (half-period l = 1), so f(x) = a₀/2 + Σ (aₙ cos nπx + bₙ sin nπx) with a₀ = ∫₋₁¹ f dx, aₙ = ∫₋₁¹ f cos nπx dx, bₙ = ∫₋₁¹ f sin nπx dx.

Here x is odd and x² is even.

a₀ = ∫₋₁¹ (x − x²) dx = −2∫₀¹ x² dx = −2/3

aₙ = −2∫₀¹ x² cos nπx dx
   = −2[ x² sin nπx/(nπ) + 2x cos nπx/(nπ)² − 2 sin nπx/(nπ)³ ]₀¹
   = −4(−1)ⁿ/(n²π²)

bₙ = 2∫₀¹ x sin nπx dx = 2[ −x cos nπx/(nπ) + sin nπx/(nπ)² ]₀¹
   = 2(−1)ⁿ⁺¹/(nπ)

Fourier series

x − x² = −1/3 + Σ (n=1 to ∞) [ 4(−1)ⁿ⁺¹ cos nπx / (n²π²) + 2(−1)ⁿ⁺¹ sin nπx / (nπ) ]

= −1/3 + (4/π²)(cos πx − cos 2πx/4 + cos 3πx/9 − …) + (2/π)(sin πx − sin 2πx/2 + sin 3πx/3 − …)

(At the end points x = ±1 the periodic extension jumps, so the series gives the mean value −1 there.)