f^(χ1)=1/2n, f^(χ2)=−1/2n, f^(χ3)=1/2n, f^(χ4)=−1/2n and for j=1,…,n/2−1: f^(ψj)=12n(0110) . Thus the Fourier inversion formula yields: f(x)=12n(χ1(x)−χ2(x)+χ3(x)−χ4(x))+1n∑j=1n/2−1tr((0110)Ψj(x))