Near best rational approximation and spectral methods Joris Van - PowerPoint PPT Presentation

Near best rational approximation and spectral methods Joris Van Deun University of Antwerp Dept. Math. & Computer Science 20 May 2008 1 / 44

Part I Near best interpolation 2 / 44

Introduction A very old and very classical problem. . . Given a real, continuous function f ( x ) on [ − 1 , 1 ] , find a good polynomial approximation Possible solutions ◮ Best (minimax) polynomial approximation according to the norm � f � := � f � ∞ = − 1 ≤ x ≤ 1 | f ( x ) | max ◮ Polynomial least squares approximation ◮ Interpolating polynomial 3 / 44

Linear minimax approximation Problem Given linearly independent functions { ϕ k } find � � n � � � min � f ( x ) − a k ϕ k ( x ) � � � � a k k = 0 � Solution a k such that f − � a k ϕ k equi-oscillates, i.e. n + 2 extremal points of equal magnitude and alternating sign Example: minimax polynomial approximation Take ϕ k ( x ) = x k for k = 0 , 1 , . . . , n 4 / 44

Interpolating polynomial Take n + 1 points x 0 , x 1 , . . . , x n and construct polynomial p n ( x ) such that f ( x i ) = p n ( x i ) , i = 1 , 2 , . . . n Choice of interpolation points? It is well-known that f ( x ) − p n ( x ) = f ( n + 1 ) ( ξ ) ( n + 1 )! ( x − x 0 ) · · · ( x − x n ) where ξ depends on x and x 0 , x 1 , . . . , x n and f Try to choose x 0 , . . . , x n such that f − p n equi-oscillates . . . 5 / 44

Equi-oscillating polynomial on [ − 1 , 1 ] Find points x 0 , . . . , x n such that ( x − x 0 ) · · · ( x − x n ) equi-oscillates on [ − 1 , 1 ] ◮ Chebyshev polynomial 1 0.8 T n + 1 ( x ) = cos (( n + 1 ) arccos x ) 0.6 0.4 ◮ Zeros are given by 0.2 0 x k = cos π ( 2 k + 1 ) −0.2 2 ( n + 1 ) −0.4 −0.6 for k = 0 , . . . , n −0.8 −1 ◮ Interpolation in x k is near −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 best 6 / 44

Alternative interpretation From f ( x ) − p n ( x ) = f ( n + 1 ) ( ξ ) ( n + 1 )! ( x − x 0 ) · · · ( x − x n ) it follows that � f − p n � ≤ max − 1 ≤ t ≤ 1 f ( n + 1 ) ( t ) � ( x − x 0 ) · · · ( x − x n ) � ( n + 1 )! Minimising � ( x − x 0 ) · · · ( x − x n ) � over x 0 , . . . , x n leads to the Chebyshev zeros The unique monic polynomial of degree n + 1 which deviates least from zero in the infinity norm, is a scaled Chebyshev polynomial 7 / 44

How good is near best? Let f be a continuous function on [ − 1 , 1 ] , p n its polynomial interpolant in the Chebyshev zeros, and p ∗ n its best approximation on [ − 1 , 1 ] according to the infinity norm. Then � � 2 + 2 � f − p ∗ � f − p n � ≤ π log n n � ◮ If n < 10 5 we loose at most 1 digit ◮ If n < 10 66 we loose at most 2 digits If f is analytic in an ellipse with foci ± 1 and semimajor/minor axis lengths a ≥ 1 and b ≥ 0 , then � f − p n � = O (( a + b ) − n ) , n → ∞ 8 / 44

Rational generalisation What if f has singularities close to [ − 1 , 1 ] ? Example Take 1 f ( x ) = ε 2 + x 2 , 0 < ε ≪ 1 with poles at ± i ε Then � f − p n � = O (( 1 + ε ) − n ) Polynomial interpolation converges too slowly! 9 / 44

Near best fixed pole rational interpolation Let poles α 1 , . . . , α m be given (real or complex conjugate) and put π m ( x ) = ( x − α 1 ) · · · ( x − α m ) Then π m ( x ) = [ π m ( ξ ) f ( ξ )] ( n + 1 ) f ( x ) − p n ( x ) ( x − x 0 ) · · · ( x − x n ) ( n + 1 )! π m ( x ) when f ( x i ) = p n ( x i ) π m ( x i ) , i = 0 , 1 , . . . , n 10 / 44

Linear minimax approximation Problem Given linearly independent functions { ϕ k } find � � n � � � min � f ( x ) − a k ϕ k ( x ) � � � � a k k = 0 � Solution a k such that f − � a k ϕ k equi-oscillates, i.e. n + 2 extremal points of equal magnitude and alternating sign Example: minimax rational approximation Take ϕ k ( x ) = x k /π m ( x ) for k = 0 , 1 , . . . , n 11 / 44

Near best fixed pole rational interpolation Problem statement Given π m , find x 0 , . . . , x n with n + 1 ≥ m such that � q n + 1 /π m � is minimal, where q n + 1 ( x ) = ( x − x 0 ) · · · ( x − x n ) (equivalently: such that q n + 1 /π m equi-oscillates) History ◮ Special case studied by Markoff, 1884 ◮ General case solved by Bernstein, 1937 ◮ Discussed in Appendix A of Achieser’s “Theory of Approximation”, 1956 ◮ Only theoretical solution, no properties, computational aspects, . . . 12 / 44

Joukowski transformation J − 1 J � � x = J ( z ) = 1 z + 1 2 z � x 2 − 1 z = x − 13 / 44

Near best fixed pole rational interpolation Solution ◮ Let { α 1 , . . . , α m } denote zeros of π m ◮ Put β k = J − 1 ( α k ) for k = 1 , . . . , m ◮ Define B m by B m ( z ) = z − β 1 1 − β 1 z · · · z − β m 1 − β m z Then � � T n ( x ) = 1 1 z n − m B m ( z ) + z n − m B m ( z ) 2 is a rational function in x of the form q n ( x ) /π m ( x ) . The interpolation points x 0 , . . . , x n are the zeros of T n + 1 ( x ) . 14 / 44

Equi-oscillating rational function on [ − 1 , 1 ] Example π m ( x ) = � m / 2 k = 1 ( x 2 + k 2 ω 2 ) where ω = 0 . 1 1 0.4 0.8 0.3 0.6 0.2 0.4 0.2 0.1 0 0 −0.2 −0.1 −0.4 −0.2 −0.6 −0.8 −0.3 −1 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 −0.4 −1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1 T n ( x ) Poles & zeros Note Poles attract zeros (see later: electrostatic interpretation) 15 / 44

Why bother? Can we not just do rational interpolation in the (polynomial) Chebyshev points (zeros of Chebyshev polynomial T n )? ◮ If α 1 , . . . , α m correspond to poles of f close to the interval, then � π m f − p n � will be small (enlarging the ellipse of analyticity) ◮ However, dividing by π m can destroy this advantage and � f − p n /π m � may not be small ◮ If poles gather near the interior of the interval, Chebyshev zeros are useless ◮ Application: differential equations with interior layers 16 / 44

Example Let π x /ω f ( x ) = sinh ( π x /ω ) This function has simple poles at ± i k ω for k = 1 , 2 , . . . ◮ Interpolate by p n − 1 in zeros of T n ◮ Interpolate by p n − 1 /π n − 2 ◮ in zeros of T n ◮ in zeros of T n Plot interpolation error � f − p n − 1 � and � f − p n − 1 /π n − 2 � for the case ω = 0 . 01 17 / 44

Graph of f ( x ) 1 0 . 75 0 . 5 0 . 25 0 − 1 − 0 . 5 0 0 . 5 1 18 / 44

Interpolation error as function of n 0 10 −2 10 −4 10 −6 10 −8 10 −10 10 −12 10 −14 10 0 10 20 30 40 50 60 70 80 90 100 19 / 44

Interpolation error as function of n 0 10 −2 10 −4 10 −6 10 −8 10 −10 10 −12 10 −14 10 0 10 20 30 40 50 60 70 80 90 100 19 / 44

Interpolation error as function of n 0 10 −2 10 −4 10 −6 10 −8 10 −10 10 −12 10 −14 10 0 10 20 30 40 50 60 70 80 90 100 19 / 44

Part II Properties 20 / 44

Properties of T n and T n Definition � � T n ( x ) = 1 z n + 1 z n 2 � � T n ( x ) = 1 1 z n − m B m ( z ) + z n − m B m ( z ) 2 Orthogonality property � 1 dx T j ( x ) T k ( x ) √ 1 − x 2 = 0 , j � = k − 1 � 1 dx √ T j ( x ) T k ( x ) 1 − x 2 = 0 , j � = k , j , k ≥ m − 1 21 / 44

Properties of T n and T n Three term recurrence It is well known that T n + 1 ( x ) = 2 xT n ( x ) − T n − 1 ( x ) for n = 1 , 2 , . . . Writing T n ( x ) = q n ( x ) π n ( x ) where π n ( x ) = ( x − α 1 ) · · · ( x − α n ) , n ≤ m = π m ( x ) , n > m we can extend the definition for T n to n < m using the theory of orthogonal rational functions 22 / 44

They satisfy the recurrence relation � 1 − x /α n − 1 � x T n ( x ) = A n + B n T n − 1 ( x ) 1 − x /α n 1 − x /α n 1 − x / ¯ α n − 2 + C n T n − 2 ( x ) 1 − x /α n for n = 1 , 2 , . . . with T 0 = 1 and T − 1 = 0 The recurrence coefficients A n , B n and C n are known explicitly 23 / 44

Explicit formulas for the recurrence coefficients A n = 2 ( 1 − β n β n − 1 )( 1 − | β n − 1 | 2 ) ( 1 + β 2 n − 1 )( 1 + β 2 n ) B n = − ( 1 − | β n − 1 | 2 )( β n + ¯ β n − 2 ) + ( β n − 1 + ¯ β n − 1 )( 1 − β n ¯ β n − 2 ) n )( 1 − β n − 1 ¯ ( 1 + β 2 β n − 2 ) C n = − ( 1 − β n ¯ β n − 1 )( 1 + ¯ β 2 n − 2 ) ( 1 − β n − 1 ¯ β n − 2 )( 1 + β 2 n ) 24 / 44

Interpolation points as eigenvalues From the three term recurrence it follows immediately that the zeros of T n ( x ) are the eigenvalues of   0 1 1 1 0   2 2   ... ...       ... 1    2  1 0 2 Explicitly: x k = cos π ( 2 k + 1 ) , k = 0 , 1 , . . . , n − 1 2 n 25 / 44

Interpolation points as eigenvalues The zeros of T n ( x ) are also the generalised eigenvalues of the matrix pencil ( J n , J n D n − S n + I n ) , where   − B 1 1  0  A 1 A 1 1 − C 2 − B 2 1   α 1   A 2 A 2 A 2   J n = , D n =   ... ... ... ...           1 − C n − B n α n − 1 A n A n   0    ℑ ( α 1 ) C 3    S n = 2i | α 1 | 2 A 3   ...       ℑ ( α n − 2 ) C n | α n − 2 | 2 A n 26 / 44