Fourier series: x − x²

A function of period 2 on (−1, 1) has the series f(x) = a₀/2 + Σ (aₙ cos nπx + bₙ sin nπx), where a₀ = ∫₋₁¹ f dx, aₙ = ∫₋₁¹ f cos nπx dx, bₙ = ∫₋₁¹ f sin nπx dx (half-period l = 1).

Coefficients for f(x) = x − x² (x is odd, x² is even):

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

aₙ = −∫₋₁¹ x² cos nπx dx = −2∫₀¹ x² cos nπx dx
   = −2[ x² sin nπx/(nπ) + 2x cos nπx/(nπ)² − 2 sin nπx/(nπ)³ ]₀¹
   = −2·2(−1)ⁿ/(n²π²) = −4(−1)ⁿ/(n²π²)

bₙ = ∫₋₁¹ x sin nπx dx = 2∫₀¹ x sin nπx dx
   = 2[ −x cos nπx/(nπ) + sin nπx/(nπ)² ]₀¹
   = −2(−1)ⁿ/(nπ) = 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/2² + cos 3πx/3² − …) + (2/π)(sin πx − sin 2πx/2 + sin 3πx/3 − …)

At x = 0 the function is continuous with f(0) = 0, which gives 1/1² − 1/2² + 1/3² − … = π²/12, a useful check.