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Half range sine series: x − x²

Short answer

Half-range sine series on (0,1): f(x) = Σ b_n sin nπx with b_n = 4(1−(−1)^n)/(n³π³), so b_n = 8/(n³π³) for odd n, 0 for even. Thus f(x) = (8/π³) Σ_{m=0}∞ sin((2m+1)πx)/(2m+1)³.

Find the half-range Fourier sine series for f(x) = x − x² in 0 < x < 1.

Half-range sine series on (0, 1) (l = 1): f(x) = Σ bₙ sin nπx with bₙ = 2 ∫₀¹ (x − x²) sin nπx dx.

Integrate by parts twice (u = x − x²):

∫₀¹ (x − x²) sin nπx dx = [ −(x − x²) cos nπx/(nπ) ]₀¹ + (1/(nπ)) ∫₀¹ (1 − 2x) cos nπx dx
 = 0 + (1/(nπ)) { [ (1 − 2x) sin nπx/(nπ) ]₀¹ + (2/(nπ)) ∫₀¹ sin nπx dx }
 = (2/(nπ)²) · (1 − (−1)ⁿ)/(nπ)
 = 2 (1 − (−1)ⁿ)/(nπ)³

bₙ = 4 (1 − (−1)ⁿ)/(n³π³)  →  8/(n³π³) for odd n, 0 for even n

Answer:

x − x² = (8/π³) [ sin πx/1³ + sin 3πx/3³ + sin 5πx/5³ + ... ],   0 < x < 1

(Check at x = 1/2: 1/4 = (8/π³)(1 − 1/27 + 1/125 − ...) = 0.2580 × 0.9689 ≈ 0.25 ✓.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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