Skip to main content

Half range sine series: x

Short answer

bₙ = (2/π)∫₀^π x sin nx dx = 2(−1)ⁿ⁺¹/n. The half-range sine series is x = 2 Σ_{n=1}^∞ [(−1)ⁿ⁺¹/n] sin nx.

Express f(x) = x as a half range sine series in 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.

bₙ = (2/π)∫₀^π x sin nx dx
   = (2/π)[ −x cos nx/n + sin nx/n² ]₀^π
   = (2/π)(−π cos nπ / n)
   = 2(−1)ⁿ⁺¹/n

Answer:

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

= 2( sin x − sin 2x/2 + sin 3x/3 − sin 4x/4 + … )

Check: at x = π/2 this gives 2(1 − 1/3 + 1/5 − …) = 2(π/4) = π/2.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Mathematics III chapter 4 questions →