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Half range sine series: x − x²

Short answer

For f(x)=x-x^2 on (0,1), the half-range sine series has coefficients b_n = 8/(n^3 pi^3) for odd n (0 for even n), giving x - x^2 = (8/pi^3) [sin(pi x)/1^3 + sin(3pi x)/3^3 + sin(5pi x)/5^3 + ...].

Obtain the half range sine series for the function f(x) = x − x² for 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, with u = x − x² (u′ = 1 − 2x, u″ = −2) and ∫ sin nπx dx = −cos nπx/(nπ):

∫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 }
 = (1/(nπ)) (2/(nπ)) [ (1 − cos nπ)/(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

✓ Answer checked by Claude Opus 5.5 (high effort)

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