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Half range cosine series: x

Short answer

a₀ = (2/π)∫₀^π x dx = π. For n≥1, aₙ = (2/π)∫₀^π x cos nx dx gives 0 for even n, -4/(πn²) for odd n. Hence x = π/2 - (4/π)(cos x + cos 3x/3² + cos 5x/5² + ...), 0≤x≤π.

Obtain the half range cosine series for f(x) = x in the interval 0 ≤ x ≤ π.

Half-range cosine series on (0, π): f(x) = a₀/2 + Σ aₙ cos nx, with a₀ = (2/π)∫₀^π f dx and aₙ = (2/π)∫₀^π f cos nx dx.

For f(x) = x:

a₀ = (2/π) ∫₀^π x dx = (2/π)(π²/2) = π
aₙ = (2/π) ∫₀^π x cos nx dx
   = (2/π) [ x sin nx/n + cos nx/n² ]₀^π
   = (2/(πn²)) [ (−1)ⁿ − 1 ]

So aₙ = 0 for even n, and aₙ = −4/(πn²) for odd n.

Answer:

x = π/2 − (4/π) [ cos x/1² + cos 3x/3² + cos 5x/5² + ... ],   0 ≤ x ≤ π

(Check at x = 0: 0 = π/2 − (4/π)(π²/8) ✓, since Σ 1/(2k+1)² = π²/8.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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