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Half range sine series: x

Short answer

The half range sine series of f(x)=x on (0,2) has coefficients b_n = (4/(nπ))(-1)^(n+1). Therefore, x = 4/π Σ_{n=1}^∞ [(-1)^(n+1)/n] sin(nπx/2) for 0<x<2.

Find the half range Fourier sine series of the function f(x) = x in the interval (0, 2).

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 = 2.

bₙ = ∫₀² x sin(nπx/2) dx
   = [ −(2x/nπ) cos(nπx/2) ]₀² + (2/nπ)∫₀² cos(nπx/2) dx
   = −(4/nπ) cos nπ + 0
   = 4(−1)ⁿ⁺¹/(nπ)

Answer:

x = (4/π) Σ (n=1 to ∞) [(−1)ⁿ⁺¹/n] sin(nπx/2), 0 < x < 2

= (4/π)[ sin(πx/2) − ½ sin(πx) + ⅓ sin(3πx/2) − … ]

✓ Answer checked by Claude Opus 5.5 (high effort)

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