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Half range sine series: πx − x²

Short answer

Half-range sine series: f(x) = Σ b_n sin nx with b_n = 4(1−(−1)^n)/(π n^3), non-zero for odd n. First three terms: (8/π)(sin x + sin 3x/27 + sin 5x/125).

Expand f(x) = πx − x² in a half range sine series in the interval (0, π) up to the first three terms.

Half-range sine series on (0, π): f(x) = Σ bₙ sin nx, bₙ = (2/π) ∫₀^π (πx − x²) sin nx dx.

Integrate by parts twice:

∫₀^π (πx − x²) sin nx dx = [ −(πx − x²) cos nx / n ]₀^π + (1/n) ∫₀^π (π − 2x) cos nx dx
                         = 0 + (1/n) [ (π − 2x) sin nx/n ]₀^π + (2/n²) ∫₀^π sin nx dx
                         = (2/n²) (1 − (−1)ⁿ)/n
                         = 2 (1 − (−1)ⁿ)/n³

bₙ = (2/π) · 2 (1 − (−1)ⁿ)/n³ = 4 (1 − (−1)ⁿ)/(π n³)

So bₙ = 0 for even n, and bₙ = 8/(π n³) for odd n.

πx − x² = (8/π) [ sin x/1³ + sin 3x/3³ + sin 5x/5³ + ... ]

First three terms:

πx − x² ≈ (8/π) [ sin x + (1/27) sin 3x + (1/125) sin 5x ]
        ≈ 2.546 sin x + 0.0943 sin 3x + 0.0204 sin 5x

(Check at x = π/2: exact value π²/4 = 2.467; three terms give 2.546 − 0.0943 + 0.0204 = 2.472 ✓.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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