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- 2073 Shrawan · 5 marks
Half range sine series: x²
Short answer
Take L=3. Integration by parts twice gives bₙ = 18(-1)^{n+1}/(nπ) + 36[(-1)ⁿ - 1]/(n³π³). The half-range sine series is x² = Σ_{n=1}∞ bₙ sin(nπx/3), 0<x<3.
Obtain the half range Fourier sine series for the function f(x) = x² in the interval (0, 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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