Half range cosine series: πx − x²

Half-range cosine series on (0, π): f(x) = a₀/2 + Σ aₙ cos nx, with

a₀ = (2/π) ∫₀^π f(x) dx,    aₙ = (2/π) ∫₀^π f(x) cos nx dx

For f(x) = πx − x²:

a₀ = (2/π) [ πx²/2 − x³/3 ]₀^π = (2/π)(π³/6) = π²/3

For aₙ, integrate by parts twice:

∫₀^π (πx − x²) cos nx dx = [ (πx − x²) sin nx / n ]₀^π − (1/n) ∫₀^π (π − 2x) sin nx dx
                         = 0 − (1/n) · π(1 + (−1)ⁿ)/n

(using ∫₀^π (π − 2x) sin nx dx = [−(π − 2x) cos nx/n]₀^π − (2/n)∫₀^π cos nx dx = π(1 + (−1)ⁿ)/n.)

aₙ = (2/π) · [ −π (1 + (−1)ⁿ)/n² ] = −2 (1 + (−1)ⁿ)/n²

So aₙ = 0 for odd n and aₙ = −4/n² for even n = 2m, i.e. −1/m².

Answer:

πx − x² = π²/6 − Σ_{m=1}^∞ cos 2mx / m²
        = π²/6 − cos 2x − (cos 4x)/4 − (cos 6x)/9 − ...,    0 < x < π

(Check at x = 0: 0 = π²/6 − Σ 1/m² = π²/6 − π²/6 ✓.)