Fourier series: |sin x|

f(x) = |sin x| on (−π, π). It is even (|sin(−x)| = |sin x|), so bₙ = 0, and on (0, π), |sin x| = sin x.

Coefficients

a₀ = (2/π) ∫₀^π sin x dx = 4/π
a₁ = (2/π) ∫₀^π sin x cos x dx = (1/π)∫₀^π sin 2x dx = 0
aₙ = (1/π) ∫₀^π [ sin(n+1)x − sin(n−1)x ] dx    (n ≥ 2)
   = (1/π) [ (1 − (−1)ⁿ⁺¹)/(n+1) − (1 − (−1)ⁿ⁻¹)/(n−1) ]

For odd n, aₙ = 0. For even n,

aₙ = (1/π) [ 2/(n+1) − 2/(n−1) ] = −4/(π(n² − 1))

Series

Putting n = 2m:

|sin x| = 2/π − (4/π) Σ_{m=1}^∞ cos 2mx / (4m² − 1)
        = 2/π − (4/π) [ cos 2x/3 + cos 4x/15 + cos 6x/35 + ... ]

The period of |sin x| is π, which is why only even multiples of x appear.