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Half range sine series: eˣ

Short answer

On (0,1), the half-range sine series coefficient bn = 2∫_0^1 e^x sin(nπx) dx = 2nπ[1-(-1)^n e]/(1+n^2π^2). Hence e^x = 2π Σ_{n=1}^∞ [n(1-(-1)^n e)/(1+n^2π^2)] sin(nπx), 0<x<1.

Obtain the half range sine series for f(x) = eˣ in 0 < x < 1.

Half-range sine series on (0, l) with l = 1:

f(x) = Σ bₙ sin nπx,    bₙ = 2 ∫₀¹ eˣ sin nπx dx

Use ∫ e^(ax) sin(bx) dx = e^(ax)(a sin bx − b cos bx)/(a² + b²) with a = 1, b = nπ:

∫₀¹ eˣ sin nπx dx = [ eˣ (sin nπx − nπ cos nπx) / (1 + n²π²) ]₀¹
                  = [ −nπ e (−1)ⁿ + nπ ] / (1 + n²π²)
                  = nπ [1 − (−1)ⁿ e] / (1 + n²π²)

bₙ = 2nπ [1 − (−1)ⁿ e] / (1 + n²π²)

Answer:

eˣ = 2π Σ_{n=1}^∞ [ n (1 − (−1)ⁿ e) / (1 + n²π²) ] sin nπx,   0 < x < 1

First coefficients: b₁ = 2π(1 + e)/(1 + π²) ≈ 2.15, b₂ = 4π(1 − e)/(1 + 4π²) ≈ −0.53.

✓ Answer checked by Claude Opus 5.5 (high effort)

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