Skip to main content

Half range sine series: x²

Short answer

With L=3, bₙ = (2/3)∫₀³ x² sin(nπx/3)dx = 18(−1)ⁿ⁺¹/(nπ) − 36(1−(−1)ⁿ)/(n³π³). The half-range sine series is x² = Σ_{n=1}^∞ bₙ sin(nπx/3).

Obtain the half range sine series for the function f(x) = x² in the interval (0, 3).

The half range sine series on (0, L) is f(x) = Σ bₙ sin(nπx/L), with bₙ = (2/L)∫₀ᴸ f(x) sin(nπx/L) dx. Here L = 3.

Coefficient. Integrate by parts with u = x², dv = sin(nπx/3)dx; repeated integration gives

bₙ = (2/3)∫₀³ x² sin(nπx/3) dx
   = (2/3)[ −x²(3/nπ) cos(nπx/3) + 2x(3/nπ)² sin(nπx/3) + 2(3/nπ)³ cos(nπx/3) ]₀³
   = (2/3)[ −(27/nπ)(−1)ⁿ + (54/n³π³)((−1)ⁿ − 1) ]
   = 18(−1)ⁿ⁺¹/(nπ) − 36(1 − (−1)ⁿ)/(n³π³)
  • n even: bₙ = −18/(nπ)
  • n odd: bₙ = 18/(nπ) − 72/(n³π³)

Series

x² = Σ (n=1 to ∞) [ 18(−1)ⁿ⁺¹/(nπ) − 36(1 − (−1)ⁿ)/(n³π³) ] sin(nπx/3), 0 < x < 3

First terms: b₁ = 18/π − 72/π³ ≈ 3.41, b₂ = −9/π ≈ −2.86.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Mathematics III chapter 4 questions →