Lattice rules and sequences for non periodic smooth integrands

Transcription

1 Lattice rules and sequences for non periodic smooth integrands Department of Computer Science KU Leuven, Belgium Joint work with Josef Dick (UNSW) and Friedrich Pillichshammer (Linz) MCQMC 2012 February 13 17, 2012 Sydney, NSW Australia

2 Outline 1 Lattice rules and sequences 2 Basis functions 3 Method I: Symmetrization 4 Method II: Tent transform 5 Numerical examples

3 Multivariate integration Multivariate integration by lattice rules Approximate the s-dimensional integral I(f ) := f (x) dx [0,1] s by an N -point (rank-1) lattice rule Q(f ; g, N ) := 1 N N 1 n=0 with good generating vector g Z s N. f ({ gn }) N Aim: Non periodic functions and lattice rules. Result: Function space H(K cos ).

4 Preliminaries Imagery of good lattice rules and sequences fixed lattice rules (a) rank-1 rule (b) Fibonacci lattice (c) rank-2 copy rule lattice sequence in base 3 (d) 3 3 seq points (e) 64 seq points (f) 3 4 seq points

5 Preliminaries Error of integration for lattice rule For f with Fourier series representation f (x) = h Z s ˆf (h) exp(2πi h x), ˆf (h) = [0,1] s f (x) exp( 2πi h x) dx, we have Q(f ; g, N ) I(f ) = 1 ˆf (h) N h Z s = 0 h Z s h g 0 (mod N ) N 1 n=0 ˆf (h). exp(2πi (h g)n/n ) ˆf (0) The error is given as a sum, h 0, over the dual lattice: L := {h Z s : h g 0 (mod N )} Z s. See Sloan & Joe (1994), Sloan & Kachoyan (1987), Niederreiter (1992).

6 The classical Korobov space A reproducing kernel Hilbert space... Traditional setting for lattice rules: Korobov space. Absolutely convergent Fourier series representation f (x) = h Z s ˆf (h) exp(2πi h x). Then f H(K α ), for α > 1/2, if with r α (h) := f 2 K α := h Z s ˆf (h) 2 r α (h) < { s 1, if hj = 0, r α (h j ), r α (h j ) := h j 2α, otherwise. j=1

7 The classical Korobov space... and its worst case error The worst case error of QMC integration using a point set P = {x 0, x 1,..., x N 1 } is defined as e(h(k ); P) := sup f H(K ) f K 1 f (x) dx 1 [0,1] s N N 1 n=0 f (x n ). For Korobov space using a lattice rule P this can be written e(h(k α ); P) 2 = r α (h). 0 h Z s h g 0 (mod N ) Construction of rules with e(h(k α ); P) = O(N α ) using CBC. (Korobov, Kuo, Sloan, Joe, Dick,... )

8 The classical Korobov space So all is nice? Are your functions periodic? Thought so... Classical solution: randomly shifted spaces (create shift-invariant kernel); or periodization and symmetrization; or tent transform (bakers transform). (Kuo, Sloan, Joe, Hickernell (2002), Zaremba, Korobov,... )

9 Building reproducing kernels Kernels build from orthogonal bases If {ϕ k (x)} k is an ONB w.r.t. a specified inner product, then K (x, y) = k ϕ k (x) ϕ k (y), (*) conditions apply to make it a RKHS. E.g. could take {φ k (x)} k is an ONB w.r.t. L 2 ([0, 1]) and take K α (x, y) = k r α (k) φ k (x) φ k (y), then (f, g) Kα = k f (k) ĝ(k), with f (k) = f (x) φ r α (k) k (x) dx. [0,1]

10 Korobov space Korobov space What is the basis of the Korobov space of real functions? K (x, y) = h Z = 1 + = h Z k=1 exp(2πi h(x y)) h 2α exp(2πi hx) exp(2πi hy) h α h α 2 cos(2πkx) cos(2πky) h 2α + k=1 So we have, w.r.t. L 2 ([0, 1]): { } { } { } 1 2 cos(2πkx) 2 sin(2πkx) k=1 k=1 2 sin(2πkx) sin(2πky) h 2α

11 Korobov space Basis of Korobov space (periodic)

12 Korobov space A half-period cosine space Suppose we want to take the L 2 ([0, 1]) ONB: { } { } 1 2 cos(πkx) and define f (k) := [0,1] f (x) ( 2) k 0 where k 0 counts the non-zero entries, i.e., k 0 is 1 if k 0 and 0 otherwise. Why? Well, for starters: x = π 2 k=1 k odd k=1 cos(πkx) dx, cos(πkx) k 2.

13 Half-period cosine space Basis of half-period cosine space (non periodic)

14 Half-period cosine space Half-period cosine RKHS Define Set and 1 if h = 0, r α,β (h) := h/2 2α if h is even, h/2 2β if h is odd, K cos (x, y) = 1 + f 2 K cos = k=0 r α,β (k) 2 cos(πkx) cos(πky), k=1 f (k) 2 r α,β (k) = 2 h f ( h ) 2 0 r α,β (h). h Z

15 Half-period cosine space Further motivation for the cosine space... The unanchored Sobolev space of smoothness a = 1 has kernel K a (x, y) = 1 + B 2 ( x y ) + (x 1 2 )(y 1 2 ) 1 2 cos(πkx) cos(πky) = 1 + π 2 k 2. k=1 Compare with, taking α = β in r α,β, for cos kernel K cos (x, y) = 1 + k=1 2α 2 cos(πkx) cos(πky) 2 k 2α. Now take α = 1 and imagine product weights on K cos of the form γ j (2π) 2...

16 Half-period cosine space In pictures... = K a=1 K cos,γ = K a=1 K cos,γ

17 Definition Method I: The symmetrization operation Define sym u (x) = (y 1,..., y s ) with y j = We define the symmetrized lattice rule { 1 x j if j u, x j if j u. Q sym (f ; g, N ) := 1 N 1 2 s N n=0 u 1:s f ( ({ ng })) sym u. N Note that for k Z and x R: { 2 cos(2πk x) if k = 2k is even, cos(πkx) + cos(πk(1 x)) = 0 if k is odd. (Use of symmetrization: Korobov (1963), Genz & Malik (1983))

18 Analysis Error in cos space with symmetrization Proposition Let f be given as f (x) = f (h) ( 2) k 0 k {0,1,...} s Then Q sym (f ; g, N ) f (x) dx = [0,1] s s cos(πk j x j ). j=1 0 h Z s h g 0 (mod N ) ( 2) h 0 f ( 2h ).

19 Analysis Worst case error in cos space with symmetrization Worst case error is the same as in the Korobov space with smoothness α. The parameter β does not matter. Corollary e 2 (H(K cos ); P sym (g, N )) = e 2 (H(K α ); P(g, N )) = r α,β (2h). h Z s \{0} h g 0 (mod N ) We can construct lattice rules by (fast) CBC such that, δ > 0, e(h(k cos ); P sym (g, N )) = e(h(k α ); P(g, N )) C α,s,δ N α+δ C α,s,δ 2 sα M α+δ.

20 Method II: The tent transform Define φ(x) := 1 2x 1, and the tent transformed lattice point set { ({ ng }) } P φ (g, N ) := φ : 0 n < N. N Used by Hickernell (2002) for randomly shifted unanchored Sobolev spaces. NB: Here we don t shift! Note that for h Z and x R: cos(πhφ(x)) = cos(2πhx).

21 Error in cos space with tent transform Proposition Let f be given as f (x) = f (k)( 2) k 0 k {0,1,...} s s cos(πk j x j ), j=1 then the error for numerical integration using the tent transformed lattice point set P φ (g, N ) is given by Q φ (f ; g, N ) f (x) dx = [0,1] s 0 h Z s h g 0 (mod N ) ( 2) h 0 f ( h ).

22 Worst case error in the cos space with tent transform Corollary e 2 (H(K cos ); P φ (g, N )) = 0 h Z s h g 0 (mod N ) r α,β (h) and for α := min(α, β) and β := max(α, β) we have e(h(k α ); P(g, N )) e(h(k cos ); P φ (g, N )) 2 sβ e(h(k α ); P(g, N )). We can construct lattice rules by (fast) CBC such that, δ > 0, e(h(k cos ); P φ (g, N )) C α,β,s,δ N α +δ.

23 Unweighted Consider the function s f (x) = in 5 dimensions: 10 4 j=1 ( 1 + ( )) x2 j 2xj 5 + xj relative error lat latper latsym latbak number of function evaluations

24 Unweighted Consider the function s f (x) = in 10 dimensions: 10 4 j=1 ( 1 + ( )) x2 j 2xj 5 + xj relative error lat latper latsym latbak number of function evaluations

25 Weighted With weights γ j = 0.9 j : s ( ( )) f (x) = j x2 j 2xj 5 + xj 6 in 5 dimensions: 10 4 j= relative error lat latper latsym latbak Lattice rules and sequences for non periodic smooth number integrands of function evaluations

26 Weighted With weights γ j = 0.9 j : s ( ( )) f (x) = j x2 j 2xj 5 + xj 6 j=1 in 10 dimensions: relative error lat latper latsym latbak Lattice rules and sequences for non periodic smooth number integrands of function evaluations

27 Thank you for your attention.