Background Known Results Results Under Construction References
Possible Algorithm
If 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.
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