the Fourier coefficients are sampled in the order j1, j2, . . ., then f(x) = j∈Nd 0 f(j)φj(x) = ∞ i=1 f(ji)φji (x), f(x) = n i=1 f(ji)φji f − f 2 = ∞ i=n+1 f(ji) 2 γ−2 ji · γ2 ji sup j∈Nd 0 f(j) γj ∞ i=n+1 γ2 ji = sup j∈Nd 0 f(j) γj inferred from what’s observed ∞ j∈Nd 0 γ2 j − n i=1 γ2 ji →0 as n→∞ γj = γj1,1 · · · γjd,d, γj,k = 1, j = 0, γ1,k jp , j > 0, ∞ j∈Nd 0 γ2 j = d k=1 [1 + ζ(2p)γ1,k], γ1,k unknown Set j1 = 0, j1 = (1, 0 . . . , 0), . . . , jd+1 = (0, . . . , 0, 1), sample f(j1), . . . , f(jd+1 ), and set γ1,1 = f(1, 0, . . . , 0)/f(0) , γ1,2 = f(0, 1, 0, . . . , 0)/f(0) , . . . , γ1,d = f(0, . . . , 0, 1)/f(0) Increment i, choose ji with the next largest γj, and sample f(ji), until error bound is small. 11/14