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Half range sine series: x²

Short answer

For l=3, b_n = (2/3)∫_0^3 x² sin(nπx/3) dx = 18(-1)^(n+1)/(nπ) + 36[(-1)^n - 1]/(n³π³). Thus x² = ∑_{n=1}^∞ [18(-1)^(n+1)/(nπ) + 36((-1)^n - 1)/(n³π³)] sin(nπx/3).

Express f(x) = x² as a half-range sine series in 0 < x < 3.

Half-range sine series on (0, 3): l = 3, f(x) = Σ bₙ sin(nπx/3), with

bₙ = (2/3) ∫₀³ x² sin(nπx/3) dx

Let k = nπ/3. Integrating by parts twice:

∫₀³ x² sin kx dx = [ −x² cos kx/k + 2x sin kx/k² + 2 cos kx/k³ ]₀³
                 = −9(−1)ⁿ/k + 2[(−1)ⁿ − 1]/k³
                 = −27(−1)ⁿ/(nπ) + 54[(−1)ⁿ − 1]/(nπ)³

(at x = 3, kx = nπ so sin kx = 0, cos kx = (−1)ⁿ.)

bₙ = (2/3) × above = 18(−1)ⁿ⁺¹/(nπ) + 36[(−1)ⁿ − 1]/(n³π³)

For odd n: bₙ = 18/(nπ) − 72/(n³π³). For even n: bₙ = −18/(nπ).

Answer:

x² = Σ_{n=1}^∞ { 18(−1)ⁿ⁺¹/(nπ) + 36[(−1)ⁿ − 1]/(n³π³) } sin(nπx/3),   0 < x < 3
   = 3.407 sin(πx/3) − 2.865 sin(2πx/3) + 1.824 sin(πx) − 1.432 sin(4πx/3) + ...

✓ Answer checked by Claude Opus 5.5 (high effort)

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