>>882
 |cos(x)| は周期πを持つらしい・・・

 f(x) = (1/2)cos(x) + (1/2)|cos(x)|
    = (1/2)cos(x) + (1/π)Σ[k=-∞,∞] (-1)^(k-1)/[(2k)^2 -1]・cos(2kx)
    = (1/2)cos(x) + (1/π){1 + 2Σ[k=1,∞] (-1)^(k-1)/[(2k)^2 -1]・cos(2kx)},