Unusual convergence behaviour of certain rational interpolants - PowerPoint PPT Presentation

Unusual convergence behaviour of certain rational interpolants Joris Van Deun

Outline Chebyshev polynomials and rational functions Barycentric interpolation Conformal maps Strange convergence 2/30

Outline Chebyshev polynomials and rational functions Barycentric interpolation Conformal maps Strange convergence 3/30

Chebyshev polynomials Using the Joukowski map 1 0.8 x = 1 � z + 1 � 0.6 = J ( z ) , 2 z 0.4 0.2 the Chebyshev polynomial of 0 the first kind T n can be −0.2 −0.4 defined as −0.6 � � T n ( x ) = 1 z n + 1 −0.8 . −1 z n 2 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 4/30

Chebyshev rational functions ◮ Take n poles { α 1 , α 2 , . . . , α n } ◮ outside [ − 1 , 1], ◮ real or complex conjugate, ◮ possibly infinite. ◮ Put β k = J − 1 ( α k ) for k = 1 , . . . , n , such that | β k | < 1. ◮ Define the (finite) Blaschke product B n ( z ) = z − β 1 β 1 z · · · · · z − β n β n z . 1 − ¯ 1 − ¯ ◮ If all poles α k at infinity, then all β k = 0 and B n ( z ) = z n . 5/30

Chebyshev rational functions 1 Define 0.8 0.6 T n ( x ) = 1 � 1 � 0.4 B n ( z ) + , 2 B n ( z ) 0.2 0 then T n ( x ) is a [ n / n ] rational −0.2 function with poles −0.4 −0.6 { α 1 , . . . , α n } that attract the −0.8 zeros. −1 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 6/30

Outline Chebyshev polynomials and rational functions Barycentric interpolation Conformal maps Strange convergence 7/30

Polynomial Lagrange interpolation ◮ Let n + 1 distinct interpolation points x j , j = 0 , . . . , n , be given, together with function values f j = f ( x j ). ◮ The polynomial p n of degree n that interpolates f at the points x j is given by n � p n ( x ) = f j l j ( x ) j =0 where the Lagrange polynomials l j ( x ) are given by � n k =0 , k � = j ( x − x k ) l j ( x ) = k =0 , k � = j ( x j − x k ) . � n 8/30

Barycentric formula The above Lagrange formula for p n can be rewritten in barycentric form as n w j � f j x − x j j =0 p n ( x ) = , n w j � x − x j j =0 where the barycentric weights w j are given by 1 w j = k � = j ( x j − x k ) . � 9/30

Rational interpolation If this relation between weights w j and nodes x k is changed, then ◮ the interpolation property still holds, but ◮ the interpolant becomes a rational function of degree at most n in numerator and denominator. For a given set of poles and interpolation points, we can easily determine the corresponding weights. 10/30

Interpolation in Chebyshev zeros ◮ Polynomial interpolant in the zeros of T n +1 ( x ) = cos (2 j + 1) π x j = cos θ (p) , j 2 n + 2 w j = ( − 1) j sin θ (p) = ( − 1) j � 1 − x 2 j . j ◮ Rational interpolant in the zeros of T n +1 ( x ) with poles { α 1 , . . . , α n } x j = cos θ (r) j , � w j = ( − 1) j sin θ (r) α 2 n k − 1 H n +1 ( x j ) , with H n ( x ) = 1 j � . α k − x n k =1 11/30

Example Take 0 10 p ( x ) f ( x ) = sinh p ( x ) −5 10 with p ( x ) = π ω ( x 2 − c 2 ) −10 10 and ω = 0 . 01 and c = 0 . 6. polynomial Poles are at rational −15 10 0 20 40 60 80 � c 2 ± ki ω, ± k = 1 , 2 , . . . 12/30

Outline Chebyshev polynomials and rational functions Barycentric interpolation Conformal maps Strange convergence 13/30

Convergence rate ◮ Assume that f is analytic in the interior of and on an ellipse C ρ with foci at ± 1 and sum of major and minox axes equal to 2 ρ . Then | f ( x ) − p n ( x ) | = O ( ρ − n ) where p n is the polynomial interpolating f at the zeros of T n +1 . ◮ If f has singularities close to [ − 1 , 1], then ρ ≈ 1 and convergence is slow. 14/30

Transformed Chebyshev points ◮ Let g : D 1 → D 2 be a conformal map such that g ([ − 1 , 1]) = [ − 1 , 1]. ◮ Let f : D 2 → C be such that f ◦ g : D 1 → C is analytic inside and on an ellipse C ˜ ρ . ◮ Put x j = g ( y j ) where y j are the zeros of T n +1 . ◮ Put w j � n x − x j f j j =0 r n ( x ) = , � n w j j =0 x − x j where f j = f ( x j ) and w j = ( − 1) j � 1 − x 2 j . 15/30

New convergence rate Theorem (Baltensperger, Berrut, No¨ el) With the definitions given before, it holds that ρ − n ) | f ( x ) − r n ( x ) | = O (˜ for every x ∈ [ − 1 , 1]. Construction of appropriate conformal map g usually based on knowledge about singularities of f . 16/30

“Blaschke” map ◮ Let x j denote the zeros of T n +1 and y j those of T n +1 , then 1 n +1 ◦ J − 1 )( x j ) = g − 1 ( x j ) n +1 y j = ( J ◦ B where B n +1 ( x ) is the Blaschke product and J ( x ) the Joukowski map. ◮ Map g ( x ) is implicitly defined. ◮ Requires n + 1 poles, why not take m < n + 1 “poles” in definition of g ? 17/30

Example 0.25 0.5 0.2 0.4 0.15 0.3 0.1 0.2 0.05 0.1 0 0 −0.05 −0.1 −0.1 −0.2 −0.15 −0.3 −0.2 −0.4 −0.25 −0.5 −2 −1 0 1 2 −2 −1 0 1 2 18/30

Outline Chebyshev polynomials and rational functions Barycentric interpolation Conformal maps Strange convergence 19/30

Example ◮ Suppose we interpolate x 2 f ( x ) = x 2 + a 2 which has simples poles at ± ia . ◮ This suggests taking m = 2 and α 1 = ia , α 2 = − ia in the definition of g ( x ), which yields ax g ( x ) = √ a 2 + 1 − x 2 . ◮ It is easily computed that x 2 ( f ◦ g )( x ) = a 2 + 1 20/30

Example with a = 0 . 01 0 10 −5 10 −10 10 −15 10 0 5 10 15 20 25 30 35 40 21/30

Example — different map ◮ It seems that we have exact interpolation when n = 2 k for k = 2 , 3 , 4 , . . . ◮ Now take m = 3 and α 1 = ia , α 2 = − ia and α 3 = ∞ in the definition of g ( x ). ◮ Explicit expression for g ( x ) no longer available, but it can be computed that ˜ ρ = 1 . 15. ◮ It seems that we have exact interpolation when n = 3 k for k = 2 , 3 , 4 , . . . ◮ What happens when n = mk ? 22/30

Example — different map 0 10 −5 10 −10 10 −15 10 0 5 10 15 20 25 30 35 40 23/30

Explanation ◮ For n = mk the interpolation points are the zeros of T n ( x ) = 1 � 1 � = 1 � 1 � B k m ( z ) + B mk ( z ) + . 2 B k m ( z ) 2 B mk ( z ) ◮ The barycentric interpolant r n − 1 ( x ) turns out to be n − 1 1 T n ( x ) � r n − 1 ( x ) = n ( x j )( x − x j ) H m ( x j ) f ( x j ) H m ( x ) T ′ j =0 only when n = mk . 24/30

Estimated poles ◮ Suppose we again interpolate x 2 f ( x ) = x 2 + a 2 . ◮ Now take the map ˜ ax g ( x ) = √ a 2 + 1 − x 2 ˜ corresponding to B m ( z ) = B 2 ( z ) with α 1 = i ˜ a and α 2 = − i ˜ a . 25/30

Example, a = 0 . 001 and ˜ a = 0 . 00116 0 10 −5 10 −10 10 −15 10 0 5 10 15 20 25 30 35 40 26/30

Convergence speed ◮ Example: | f ( x ) − r n − 1 ( x ) | ≤ max { a 2 , ˜ a 2 } 1 · |T n ( ia ) | = O ( B n ( ib )) , | a 2 − ˜ a 2 | for n = 2 k . ◮ General convergence result for meromorphic f with poles a j :   ∞ �  , | f ( x ) − r n − 1 ( x ) | = O B n ( b j )  j =1 for n = mk (result obtained using Mittag-Leffler expansions). 27/30

Final example Take 3 � 1 + 3 � x e − x . √ f ( x ) = 2 erf 2 2 ǫ 2.5 Entire, but for ǫ → 0 almost 2 discontinuous. 1.5 Take poles of [2 / 2] Pad´ e approximant to erf ( x ): 1 a 1 , 2 = ± 1 . 732 i 0.5 −1 −0.5 0 0.5 1 √ and multiply by 2 ǫ . 28/30

√ α 1 , 2 = ± 1 . 732 i 2 ǫ ; ǫ = 0 . 0001 1 10 0 10 −1 10 −2 10 −3 10 −4 10 −5 10 −6 10 0 10 20 30 40 50 60 29/30

√ α 1 , 2 = ± 1 . 732 i 2 ǫ ; α 3 = ∞ 1 10 0 10 −1 10 −2 10 −3 10 −4 10 −5 10 −6 10 0 10 20 30 40 50 60 30/30