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Half range sine series: eˣ
Short answer
e^x = (2/π) Σ_{n=1}^∞ { n[1−(−1)ⁿ e^π]/(n²+1) } sin nx. Coefficient bₙ = (2/π) n[1−(−1)ⁿ e^π]/(n²+1) found by integration.
Find the half range sine series for the function f(x) = eˣ for 0 < x < π.
The half range sine series on (0, π) is f(x) = Σ bₙ sin nx, with bₙ = (2/π)∫₀^π f(x) sin nx dx.
Coefficient for f(x) = eˣ. Use ∫ eᵃˣ sin nx dx = eᵃˣ(a sin nx − n cos nx)/(a² + n²) with a = 1:
bₙ = (2/π)∫₀^π eˣ sin nx dx
= (2/π)[ eˣ(sin nx − n cos nx)/(1 + n²) ]₀^π
= (2/π)·[ −n(−1)ⁿ e^π + n ]/(1 + n²)
= (2/π)·n[1 − (−1)ⁿ e^π]/(1 + n²)
Series
eˣ = (2/π) Σ (n=1 to ∞) { n[1 − (−1)ⁿ e^π] / (n² + 1) } sin nx, 0 < x < π
= (2/π)[ (1 + e^π)/2 · sin x + 2(1 − e^π)/5 · sin 2x + 3(1 + e^π)/10 · sin 3x + … ]
(The first terms follow by putting n = 1, 2, 3.) A numerical check at x = 1 with many terms gives 2.718, matching e¹.
✓ Answer checked by Claude Opus 5.5 (high effort)
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