Automatic algorithms

Transcription

1 Automatic algorithms Function approximation using its series decomposition in a basis Lluís Antoni Jiménez Rugama May 30, 2013

2 Index 1 Introduction Environment Linking H in Example and H out 2 Our algorithms Algorithm 1 Algorithm 2 3 Examples Example Algorithm 1 Example Algorithm 2

3 Index 1 Introduction Environment Linking H in Example and H out 2 Our algorithms Algorithm 1 Algorithm 2 3 Examples Example Algorithm 1 Example Algorithm 2

4 The problem Approximate a solution operator S applied over a set of functions: embedding, integration, derivation, etc.

5 The problem Approximate a solution operator S applied over a set of functions: embedding, integration, derivation, etc. Consider input and output spaces as separable Banach spaces so that we can decompose functions in the basis chosen.

6 The problem Approximate a solution operator S applied over a set of functions: embedding, integration, derivation, etc. Consider input and output spaces as separable Banach spaces so that we can decompose functions in the basis chosen. Assume we have all the coecients of the decomposition.

7 The problem Approximate a solution operator S applied over a set of functions: embedding, integration, derivation, etc. Consider input and output spaces as separable Banach spaces so that we can decompose functions in the basis chosen. Assume we have all the coecients of the decomposition. Design ecient algorithms.

8 The working spaces Input Output H in H out {u i } i I {v i } i I f = i I ˆf i u i f = i I f i v i f Hin = (ˆfi f )i I p Hout = ( fi in )i I p out

9 Relationship between spaces Applying S to the basis S(f ) = S(u i ) = λ i v i λ i ˆfi v i = f i v i, f H in i I i I To ensure that the solution operator is bounded, it is assumed that (λ i ) i I q <, q = p in p out max(p in p out, 0) (1)

10 Example for S : f f [ 1 f Hin = { f (x) 2 + f (x) } 2 u i (x) = 0 0 dx [ 1 ] 1/2 f Hout = f (x) 2 dx, ] 1/2, e2π 1ix, v i (x) = 2π e 1ix, 1 + 4π 2 i 2 λ i = 2π 1i 1 + 4π 2 i 2, i Z, p in = p out = 2, q =.

11 Index 1 Introduction Environment Linking H in Example and H out 2 Our algorithms Algorithm 1 Algorithm 2 3 Examples Example Algorithm 1 Example Algorithm 2

12 Denition of A n (f ) On the one hand we dene, A n (f ) = n λ ij ˆfij v ij f H in, n N, (2) j=1 ) S(f ) A n (f ) Hout = (λ ij ˆfij j n+1 (3) p out On the other hand, two bounds on the error S(f ) A n (f ) Hout will be the base for dening Algorithms 1 and 2: CONE CONDITION

13 Algorithm 1: Set up The aim is to approximate a weaker norm as a surrogate for a stronger norm: f = i I ˆf i u i H in f G = ( ˆfi ω i ) i I, f F = p in ( ˆfi ν i ω i ). i I p in

14 Algorithm 1: Error For non-adaptive algorithms we have the following error denition h(n) = sup 0 f F S(f ) A n (f ) Hout f F = ( ν ij ω ij λ ij )j n+1 q

15 Algorithm 1: Bounding the G-norm The approximation for the G-norm is: n G n (f ) = ˆf ij u ij = G i=1 ( ˆfij ω ij ) n j=1 p in (4)

16 Algorithm 1: Bounding the G-norm Assuming f lies in the specied cone of functions, dened by it follows that C τ = {f F : τ min f G f F τ f G }, (5) G n (f ) f G C n G n (f ) (6) [ 1 τ p in ( ] ν C n = ij )j n+1 p in 1/p in, 1 p in <, (7) 1, p in =.

17 Algorithm 1: Error bound Upper bound Therefore, S(f ) A n (f ) Hout C n G n (f )τ h(n) (8)

18 Algorithm 2: Set up The aim is to approximate the function depending on the decay of the series. Let 0 = n 0 < n 1 < n 2 <... be an ordered, unbounded sequence of integers. Dene the sums: ) nk σ k (f ) = (λ ij ˆfij, k = 1, 2,... (9) p out j=n k 1+1

19 Algorithm 2: Set up We also dene the cone of functions in H in coecients decay at a given rate: as those whose Fourier C = {f H in : σ k+r (f ) γ(r)σ k (f ), k N}. (10) where the sequence γ(r) has to verify lim n γ(n) = 0, γ(r) > 0 and that γ pout = (γ(r)) r=1 p out < (11)

20 Algorithm 2: Error bound Upper bound Thus, ) S(f ) A nk (f ) Hout = (λ ij ˆfij = j n k +1 n k+r r=1 j=n k+r 1+1 p out p out λ ij ˆfij 1/p out (12) = (σ k+r (f )) r=1 p out (γ(r)σ k (f )) r=1 p out = σ k (f ) γ pout

21 Index 1 Introduction Environment Linking H in Example and H out 2 Our algorithms Algorithm 1 Algorithm 2 3 Examples Example Algorithm 1 Example Algorithm 2

22 Algorithm 1: Numerical example Lets consider H in = H out = L 2 [0, 1], and S : f f the embedding operator with: [ 1 ] 1/2 f Hin = f Hout = f (x) 2 dx, u i (x) = v i (x) = e 2π 1ix, λ i = 1, i Z, p in = p out = 2, q =. In this case, our index set I = Z = {0, 1, 1, 2, 2,...} and we dene, ω ij = 1, ν 0 = 0, ν ij = 1 i j, i I \ {0}, i j = ( 1) j j, j N. 2 0

23 Algorithm 1: Results Figure: f (x) = x 2 x Error tolerance Real error

24 Algorithm 2: Numerical example We use the same settings than before adding: n 0 = 0, n k = 2k + 1, k N. Considering that ˆf ij = b i j for f (x) = conclude (b 2 1) b b cos(2πx), we σ 1 (f ) = 1 + 2b 2 σ k (f ) = 2b k, k N \ {1}.

25 Algorithm 2: Results Figure: f (x) = (b 2 1) b b cos(2πx) Error tolerance Real error