Fourier series: (π − x)²/4

f(x) = (π − x)²/4 on [0, 2π]; the period is 2π.

Coefficients

Put u = x − π, so u runs over (−π, π) and f = u²/4 (an even function of u). Also, since sin nπ = 0, cos nx = cos(nu + nπ) = (−1)ⁿ cos nu and sin nx = sin(nu + nπ) = (−1)ⁿ sin nu.

a₀ = (1/π) ∫₀^{2π} (π − x)²/4 dx = (1/(4π)) ∫_{−π}^{π} u² du = (1/(4π))(2π³/3) = π²/6
aₙ = (1/(4π)) ∫_{−π}^{π} u² cos(nu + nπ) du
   = ((−1)ⁿ/(4π)) · 4π(−1)ⁿ/n² = 1/n²
bₙ = (1/(4π)) ∫_{−π}^{π} u² sin(nu + nπ) du = ((−1)ⁿ/(4π)) ∫ u² sin nu du = 0   (odd integrand)

(using ∫_{−π}^{π} u² cos nu du = 4π(−1)ⁿ/n².)

Answer:

(π − x)²/4 = π²/12 + Σ_{n=1}^∞ (cos nx)/n²
           = π²/12 + cos x + (cos 2x)/4 + (cos 3x)/9 + ...,    0 ≤ x ≤ 2π

(Check at x = 0: π²/4 = π²/12 + Σ 1/n² gives Σ 1/n² = π²/6 ✓.)