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- 2069 Chaitra · 5 marks
Half range sine series: x
Short answer
With L=2, bₙ = ∫₀² x sin(nπx/2) dx = 4(-1)^{n+1}/(nπ). The half-range sine series is x = (4/π) Σ_{n=1}∞ [(-1)^{n+1}/n] sin(nπx/2), 0<x<2.
Express f(x) = x as a half-range sine series in 0 < x < 2.
For a function on (0, L) the half range sine series is f(x) = Σ bₙ sin(nπx/L), with bₙ = (2/L)∫₀ᴸ f(x) sin(nπx/L) dx. Here L = 2.
Coefficient for f(x) = x:
bₙ = (2/2)∫₀² x sin(nπx/2) dx
= [ −x cos(nπx/2)·(2/nπ) ]₀² + (2/nπ)∫₀² cos(nπx/2) dx
= −(4/nπ) cos nπ + (2/nπ)·(2/nπ)[sin(nπx/2)]₀²
= −4(−1)ⁿ/(nπ) + 0
= 4(−1)ⁿ⁺¹/(nπ)
Series
x = (4/π) Σ (n=1 to ∞) [(−1)ⁿ⁺¹/n] sin(nπx/2), 0 < x < 2
= (4/π)[ sin(πx/2) − (1/2) sin(πx) + (1/3) sin(3πx/2) − … ]
Check: at x = 1 the series gives (4/π)(1 − 1/3 + 1/5 − …) = (4/π)(π/4) = 1 = x.
✓ Answer checked by Claude Opus 5.5 (high effort)
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