University of Houston, Department of Mathematics Numerical Analysis, Fall 2005

Transcription

1 4 Interpolation 4.1 Polynomial interpolation Problem: LetP n (I), n ln, I := [a,b] lr, be the linear space of polynomials of degree n on I, P n (I) := { p n : I lr p n (x) = n i=0 a i x i, a i lr, 0 i n }. We have dim P n (I) = n+1, since any p n P n (I) is uniquely determined by the coefficients a i, 0 i n. y p(x) x 0 x 1 x 2 x 3 x 4 x 5 x 6 x Interpolation problem: Given n + 1 points (x i,f i ), 0 i n, find p n P n (I) such that (+) p n (x i ) = f i, 0 i n. Definition 4.1 Interpolating polynomial The points x i (resp. f i, 0 i n, ) are called nodal points (resp. nodal values) of the interpolation problem. The n + 1 equations (+) are called the interpolating conditions. A polynomial p n P n (I) satisfying (+) is called an interpolating polynomial.

2 4.1.1 Existence and Uniqueness In explicit form, the interpolating conditions are given as follows: n j=0 a j x j i = f i, 0 i n. They represent a linear system in the coefficients a j, 0 j n,: ( ) 1 x 0 x 2 0 x n 0 1 x 1 x 2 1 x n 1 1 x n x 2 n x n n } {{ } =: V Definition 4.2 Vandermonde s determinant The coeffizient matrix V in ( ) is called Vandermonde s matrix. Its determinant is referred to as Vandermonde s determinant. a 0 a 1 a n = f 0 f 1 f n.

3 Lemma 4.3 Vandermonde s determinant Vandermonde s determinant is given by: Proof: We consider V n (x) := det V = 0 j<k n (x k x j ). 1 x 0 x 2 0 x n 0 1 x 1 x 2 1 x n 1 1 x n 1 x 2 n 1 x n n 1 1 x x 2 x n Expansion of the determinant w.r.t. the (n + 1,n + 1)-st element x n results in: det V n (x) = det V n 1 (x n 1 ) }{{} leading coeff. x n + q n 1, q n 1 P n 1 (lr)..

4 We have: det V n (x) = det V n 1 (x n 1 ) }{{} leading coeff. On the other hand: x n + q n 1, q n 1 P n 1 (lr). det V n (x) P n (lr), det V n (x j ) = 0, 0 j n 1, = det V n (x) = det V n 1 (x n 1 ) n 1 (x x j ) = j=0 det V = det V n (x n ) = det V n 1 (x n 1 ) n 1 (x n x j ) = j=0 = det V n 2 (x n 2 ) n 2 (x n 1 x j ) n 1 (x n x j ) = =... = det V(x 1 ) }{{} =x 1 x 0 ) j=0 1 j<k n j=0 (x k x j ).

5 Theorem 4.4 Existence and uniqueness of the interpolating polynomial In case of mutually disjoint nodal points, i.e., x i x j, 0 i j n, there exists a uniquely determined interpolating polynomial p n P n (I). Proof: There holds det 0 j<k n (x k x j ) 0. Remark 4.5 Computation of the interpolating polynomial The computation of the polynomial by the solution of the linear system with the Vandermonde matrix as the coefficient matrix is not practicable, since the amount of computational work is too large. Moreover, the linear system is in general ill conditioned.

6 4.1.2 Lagrangian representation of the interpolating polynomial 1 y L i (x) x i 1 x i x i+ 1 Definition 4.6 Lagrangian fundamental polynomials The polynomials L i (x) = x x j j i x i x j are called Lagrangian fundamental polynomials. x In case of mutually disjoint nodal points x i, 0 i n, a basis of P n (lr) is given by the polynomials L i P n (lr), L i (x j ) = δ ij, 0 i,j n. Obviously, there holds: L i (x j ) = 0, i j = L i (x) = α n (x x j) j i L i (x i ) = α n (x i x j ) = 1 = α n = 1/( (x i x j )) j i j i

7 Theorem 4.7 Lagrangian representation of the interpolating polynomial In case of mutually disjoint nodal points x i, 0 i n, the interpolating polynomial admits the representation p n (x) = n i=0 f i L i (x). Remark 4.8 Practical computation of the interpolating polynomial The computation of the interpolating polynomial by the Lagrangian representation formula is not practicable, since the addition of a pair (x n+1,f n+1 ) requires that all fundamental polynomials have to be recomputed.

8 4.1.3 Newton s representation of the interpolating polynomial Idea: Recursive construction Let p 0 P 0 (lr) be the interpolating polynomial of degree 0 with respect to (x 0,f 0 ), i.e., p 0 (x) = f 0. Ansatz for p 1 P 1 (lr) with respect to (x 0,f 0 ), (x 1,f 1 ): Determination of q 1 : p 1 (x) = p 0 (x) + q 1 (x), q 1 P 1 (lr). p 1 (x 0 ) = f 0 + q 1 (x 0 ) = f 0 = q 0 (x 0 ) = 0, = q 1 (x) = b 1 (x x 0 ) = p 1 (x) = f 0 + b 1 (x x 0 ). We obtain b 1 from the condition p 1 (x 1 ) = f 1. Ansatz for p 2 P 2 (lr) with respect to (x 0,f 0 ), (x 1,f 1 ), (x 2,f 2 ): Determination of q 2 : p 2 (x) = p 1 (x) + q 2 (x), q 2 P 2 (lr). p 2 (x 0 ) = f 0 + q 2 (x 0 ) = f 0 = q 2 (x 0 ) = 0, p 2 (x 1 ) = f 1 + q 2 (x 1 ) = f 1 = q 2 (x 1 ) = 0 = q 2 (x) = b 2 (x x 0 ) (x x 1 ).

9 General case: Let p n 1 P n 1 (lr) be the interpolating polynomial with respect to (x i,f i ), 0 i n 1. Ansatz for p n P n (lr) with respect to (x i,f i ), 0 i n 1: Determination of q n : p n (x) = p n 1 (x) + q n (x), q n P n (lr). p n (x i ) = f i + q n (x i ) = f i = q n (x i ) = 0, 0 i n 1 = q n (x) = b n (x x 0 ) (x x 1 )...(x x n 1 ). In principle, b n can be obtained from the condition: p n (x n ) = p n 1 (x n ) + q n (x n ) = f n. However, this approach is not practicable, since we must determine p n 1 and evaluate it in x n. On the other hand, we have: p n (x) = b n x n + n 1 c i x i. b n is the leading coefficient, i.e., the coefficient of x n in p n. Most wanted: Formula for the computation of the leading coefficients. i=0

10 Definition 4.9 Newton s representation of the interpolating polynomial The representation of the interpolating polynomial given by p n (x) = b 0 + b 1 (x x 0 ) b n (x x 0 )... (x x n 1 ) is called Newton s representation. Recursive computation of the leading coefficients: Let p ij P j i (lr), 0 i < j n, be the interpolating polynomials with respect to (x i,f i ), (x i+1,f i+1 ),..., (x j,f j ). Ansatz for p ij : Determination of r 1,s 1 : p ij (x) = r 1 (x) p i,j 1 (x) + s 1 (x) p i+1,j (x), r 1,s 1 P 1 (lr). p ij (x i ) = r 1 (x i ) p i,j 1 (x i ) + s 1 (x i ) p i+1,j (x i ) = f i Set r 1 (x i ) = 1, s 1 (x i ) = 0, p ij (x j ) = r 1 (x j ) p i,j 1 (x j ) + s 1 (x j ) p i+1,j (x j ) = f j Set r 1 (x j ) = 0, s 1 (x j ) = 1 = r 1 (x) = x j x, s 1 (x) = x x i. x j x i x j x i

11 Theorem 4.10 Recursion formula for the interpolating polynomials The polynomials p ij P j i (lr), 0 i < j n, interpolating in (x k,f k ), i k j, satisfy the recursion p ij (x) = x j x p i,j 1 (x) + x x i p i+1,j (x). x j x i x j x i Corollary 4.11 Recursion formula for the leading coefficients The leading coefficients f[x i,x i+1,...,x j ] of the interpolating polynomials p ij P j i (lr),0 i < j n, satisfy the recursion ( ) f[x i,x i+1,...,x j ] = f[x i+1,...,x j ] f[x i,x i+1,...,x j 1 ] x j x i where f[x i ] := f i, 0 i n. Example 4.12 (i = 0, j = 1) f[x 0,x 1 ] = f[x 1] f[x 0 ] x 1 x 0 = f 1 f 0 x 1 x 0. Definition 4.13 Divided differences The quantities defined recursively by means of ( ) are called divided differences. In particular, f[x i,...,x j ] is said to be the (j i)-th divided difference.

12 Remark 4.14 Recursive scheme for the divided differences x 0 f 0 = f[x 0 ] f[x 0,x 1 ] x 1 f 1 = f[x 1 ] f[x 0,x 1,x 2 ] f[x 1,x 2 ] x 2 f 2 = f[x 2 ] f[x 0,...,x n ] x n 2 f n 2 = f[x n 1 ] f[x n 2,x n 1 ] x n 1 f n 1 = f[x n 1 ] f[x n 2,x n 1,x n ] f[x n 1,x n ] x n f n = f[x n ]

13 Lemma 4.15 Newton s representation with divided differences Since the coefficients b i, 0 i n, of Newton s interpolating polynomial correspond to the divided differences f[x 0,...,x i ], we obtain the following representation ( ) p n (x) = f[x 0 ] + f[x 0,x 1 ] (x x 0 ) f[x 0,...,x n ] (x x 0 )... (x x n 1 ) Evaluation of interpolating polynomials (i) Horner s scheme We reformulate ( ) according to p n (x) = ( f[x 0,...,x n ] (x x n 1 ) + f[x 0,...,x n 1 ] (x x n 2 ) f[x 0,x 1 ] (x x 0 ) ) + f[x 0 ]. This leads to Horner s scheme: Compute f[x 0,...,x n ] (x x n 1 ) + f[x 0,...,x n 1 ] Multiply by (x x n 2 ) and add f[x 0,..., x n 2 ] Multiply by (x x n 3 ) and add f[x 0,..., x n 3 ] etc. Disadvantage: Explicit knowledge of Newton s interpolating polynomial is required. Too much computational work in case of evaluation at a single point.

14 (ii) Algorithm of Aitken-Neville Idea: Application of the recursion formula for the interpolating polynomials: x 0 f 0 = p 00 (x) p 01 (x) x 1 f 1 = p 11 (x) p 02 (x) p 01 (x) x 2 f 2 = p 22 (x) p 0n (x) x n 2 f n 2 = p n 2,n 2 (x) p n 2,n 1 (x) x n 1 f n 1 = p n 1,n 1 (x) p n 2,n (x) p n 1,n (x) x n f n = p nn (x)

15 4.1.5 Remainder term and error estimate Problem: Consider the approximation of a continuous function f : I := [a,b] lr by the polynomial p n P n (I) interpolating in x i [a,b], 0 i n,. What can be said about the accuracy of this approximation? Definition 4.16 Remainder term, interpolation error Assume f C([a,b]) and let p n P n ([a,b]) be the polynomial interpolating f in (x i,f(x i )), x i [a,b], 0 i n. The function R n := f p n is called the remainder of the interpolation. The value of the remainder in x [a,b], x x i, 0 i n, is referred to as the interpolation error.

16 Theorem 4.17 Representation of the remaining term Let f C n+1 ([a,b]) and assume that p n P n ([a,b]) is the polynomial interpolating f in (x i,f(x i )), x i [a,b],0 i n. For any x [a,b] there exists ξ I x := [min(x 0,...,x n,x),max(x 0,...,x n,x) ] such that ( ) R n (x) = f(x) p n (x) = f(n+1) (ξ) (n + 1)! ω n+1 (x), ω n+1 (x) := (x x 0 ) (x x 1 )... (x x n ). Proof: W.r.o.g. let x x i, 0 i n. We consider the auxiliary function F(t) := f(t) p n (t) f(x) p n(x) ω n+1 (t), t [a,b]. ω n+1 (x) Obviously, we have F C n+1 ([a,b]), and there holds = F has at least n + 2 zeroes in [a,b]. F(x i ) = 0, 0 i n, F(x) = 0.

17 We already know: F has at least n + 2 zeroes in [a,b]. y x1 x 1 x 2 x 2 x 3 x 3 x4 x A repeated application of Rolle s theorem yields: F has at least n + 1 zeroes in [a,b],..., F (n+1) has at least 1 zero in ξ [a,b]. F (n+1) (ξ) = f (n+1) (ξ) p (n+1) (ξ) } n {{ } = 0 = f(x) p n (x) = f(n+1) (ξ) (n+1)! ω n+1 (x). f(x) p n(x) ω n+1 (x) (n + 1)! = 0 Corollary 4.18 Representation of the remaining term using divided differences Under the assumptions of Theorem 4.17 there holds: f[x 0,...,x n,x] = f(n+1) (ξ) (n + 1)!, ξ [a,b].

18 Proof: There holds: p n (x) = p 0,n (x) = Setting x n+1 := x and f n+1 := f(x), we obtain: n i=0 f[x 0,...,x i ] i 1 (x x j ). f(x) = p 0,n+1 (x) = p 0,n (x) + f[x 0,...,x n,x] ω n+1 (x). Corollary 4.19 Generalized mean value theorem If f C n ([a,b]), then there exists ξ [ min(x 0,...,x n ),max(x 0,...,x n ) ] such that f[x 0,...,x n ] = f(n) (ξ) n! Corollary 4.21 Estimate of the remainder If f C n+1 ([a,b]) and if p n P n ([a,b]) denotes the polynomial interpolating f in x i [a,b], 0 i n,, then the following estimate holds true f (n+1) (ξ) f(x) p n (x) max ξ [a,b] (n + 1)! j=0. max ω n+1(x). x [a,b]

19 Problem: Chebyshev nodes The estimate f (n+1) (ξ) f(x) p n (x) max ξ [a,b] (n + 1)! max ω n+1(x). x [a,b] motivates to determine the nodes x i [a,b], 0 i n, such that ( ) max ω n+1(x) = max (x x 0)...(x x n ) = x [a,b] x [a,b] min x i [a,b] max (x x 0)...(x x n ) x [a,b] ( ) represents a minimax-problem whose solution is already known from the cg method Theorem 4.21 Chebyshev nodes The solution of the minimax-problem ( ) is given by the zeroes t i = cos( 2i + 1 2n + 2 π), 0 i n of the Chebyshev polynomial T n+1 (x) = cos((n + 1) arccos(x)) transformed to the interval [a,b]. We thus obtain the Chebyshev nodes x i = b a 2 t i + b + a 2, 0 i n.

20 4.1.6 Convergence of interpolating polynomials Remark 4.22 About Weierstrass theorem Weierstrass theorem states: If f : [a,b] lr is continuous, for any ε > 0 there exists a polynomial pε such that sup f(x) pε(x) ε. x [a,b] Weierstrass theorem does not imply convergence of the sequence of interpolating polynomials. Problem: Assume that f : [a,b] lr is a continuous function, x i := a + i b a n p n P n ([a,b]) be the polynomials interpolating in (x i,f(x i )), 0 i n., 0 i n, and let What can be said about the convergence of the sequence (p n ) n ln of the interpolating polynomials?

21 4 1 1 y 1 f(x) = 1/(1+x 2 ) 4 x (i) Counterexample by Runge (1901) f(x) = x 2, x [ 5, +5]. The sequence of interpolating polynomials (p n ) n ln with respect to equidistant nodes x i, 0 i n, converges for x 3.63 and diverges for 3.63 < x < 5. y f(x) = x x (ii) Counterexample by Bernstein (1912) f(x) = x, x [ 1, +1]. The sequence of interpolating polynomials (p n ) n ln diverges for all 0 < x < 1.

22 Definition 4.23 Nodal scheme Let x n,i, 0 i n, n ln 0, be the mutually different nodes of p n P n ([a,b]) such that p n (x n,i ) = f(x n,i ), 0 i n. Then, the scheme x 0,0 x 1,0 x 1,1 S : x n,0 x n,1 x n,n is called the nodal scheme of the sequence (p n ) n ln 0 of interpolating polynomials. Theorem 4.24 Theorem of Marcinkiewicz For any function f C([a,b]) there exists a nodal scheme S with x n,i [a,b], 0 i n,n ln 0, such that the sequence (p n ) n ln 0 of interpolating polynomials uniformly converges to f. Remark 4.25 About the theorem of Marcienkiewicz The proof of the theorem of Marcinkiewicz uses nonconstructive results from approximation theory and hence, it does not provide a practical method for the determination of the nodal scheme.

23 Theorem 4.26 Faber s theorem For any nodal scheme S with x n,i [a,b], 0 i n, n ln 0, there exists a continuous function f : [a,b] lr such that the sequence (p n ) n ln 0 of interpolating polynomials does not uniformly converge to f.

24 4.2 Spline interpolation Foundations Example 4.27 Interpolation by piecewise linear functions x 0 x 1 x 2 y x 3 x 4 x 5 x 6 s (x) x Assume that := {x 0,...,x n } is a grid of n + 1 mutually disjoint nodes a := x 0 < x 1 <... < x n =: b. Given the values f i, 0 i n, consider the interpolating function s with s [xi,x i+1 ] P 1 ([x i,x i+1 ]), 0 i n 1, s (x i ) = f i, 0 i n. Obviously, s is continuous, but not continuously differentiable. The critical points are the inner nodes x i, 1 i n 1. Idea: Use piecewise polynomials of a higher degree with better global smoothness properties.

25 Definition 4.28 Spline of degree m 1 Assume that := {x 0,...,x n } is a grid of n + 1 mutually disjoint nodes satisfying a := x 0 < x 1 <... < x n =: b. A function s : [a,b] lr is called a spline of degree m 1, m ln, (resp. of order m) with respect to, if (S 1 ) s C m 2 ([a,b]) (m 2), (S 2 ) s i := s [xi,x i+1 ] P m 1 ([x i,x i+1 ]), 0 i n 1. The splines of degree m 1 form a linear space which will be denoted by S m,. In particular, we obtain m = 1 m = 2 m = 3 m = 4 step functions polygons (linear splines) quadratic splines cubic splines

26 Remark 4.29 Dimension A spline of the order m is the m 1-fold integral of a spline of order 1 (step function): dim S 1, = n = dim S m, = n + m 1. Continuity and differentiability conditions in the inner nodes: s (k) i 1(x i ) = s (k) i (x i ), 1 i n 1, 0 k m 2. By m 1-fold integration we obtain: s i (x) = s i 1 (x) + γ i (x x i ) m 1, and hence, s (x) = Definition 4.30 Truncated power function The function given by s i 1 (x), x < x i s i 1 (x) + γ i (x x i ) m 1., x x i (x x i ) m 1 + := is called truncated power function. 0, x < x i (x x i ) m 1, x x i

27 Remark 4.31 Differentiation of truncated power functions d dx (x x i) + m 1 = (m 1) (x x i ) + m 2 (m 2), d dx (x x 0, x < x i) + = i (m = 2). 1, x x i Theorem 4.32 Basis of S m, The linear space S m,, m ln, of splines of order m satisfies: (+) S m, = span {1,x,...,xm 1, (x x 1 ) + m 1,..., (x x n 1 ) + m 1 }, (++) dim S m, = n + m 1. Proof: We have to verify the linear independence of the functions in (+). Assume that: s (x) = m 1 It has to be shown that k=0 a k x k + n 1 j=1 c j (x x j ) m 1 + = 0, for all x [a,b]. a k = 0, 0 k m 1, c j = 0, 1 j n 1.

28 Consider the linear functionals 1 G i (f) = (m 1)! [ f(m 1) (x i+0 ) f (m 1) (x i 0 ) ], 1 i n 1. Inserting s for f, we obtain: G i (s ) = G i ( m 1 a k x k ) } k=0 {{ } =0 It follows thats (x) = m 1 k=0 + n 1 j=1 c j G i ((x x j ) m 1 + ) }{{} =δ ij = c i = 0, 1 i n 1. a k x k = 0, x [a,b] = a k = 0, 0 k m 1. Remark 4.33 Aspects of the basis representation The evaluation of a spline in the basis representation s (x) = m 1 k=0 a k x k + n 1 j=1 c j (x x j ) m 1 + can not be recommended, since it is ill conditioned with respect to perturbations in c j, the basis functions are non-local and the coefficients a k,c j do not have a geometrical meaning.

29 4.3.2 The cubic spline interpolant Definition 4.34 Cubic spline interpolant Assume that := { a = x 0 < x 1 <... < x n = b } is a not necessarily equidistant partition of the interval [a,b] and f i lr,0 i n. A spline s S 4, is called a cubic spline interpolant, if ( ) s (x i ) = f i, 0 i n. Remark 4.35 The well-posedness of the interpolation problem A simple reasoning reveals that a cubic spline interpolant is not uniquely determined alone by the interpolating conditions ( ): Number of unknowns: 4 (unknowns per subinterval) n (number of subintervals) = 4n Number of determining equations: 3 (continuity and differentiability) (n 1) (Number of interior nodes) + (n + 1) interpolating conditions = 4n 2.

30 Definition 4.36 Complete, natural, and periodic cubic spline interpolant A cubic spline interpolant is called complete, natural, periodic, if s (x 0) = f 0, s (x n) = f n, s (x 0) = 0, s (x n) = 0, s (x 0) = s (x n), s (x 0) = s (x n). Remark 4.37 Existence, uniqueness, and computation The proof of the existence and uniqueness of the complete (natural, periodic) cubic spline interpolant is constructive and thus provides a method of computation. There are two techniques: Computation via the slopes, Computation via the moments.

31 (i) Computation via the slopes In each subinterval [x i,x i+1 ], 0 i n 1, the polynomials s i = s [xi,x i+1 ] admit a representation as solutions of the Hermite interpolation problem: Compute s i P 3 ([x i,x i+1 ]) as given by ( ) s i (x) = a 0 + a 1 (x x i ) + a 2 (x x i ) 2 + a 3 (x x i ) 3, such that (+) s i (x j ) = f j, i j i + 1, (++) s i(x j ) = m j, i j i + 1. Here, the slopes m i in the nodes are unknown and have to be computed by means of the determining equations. The coefficients a k, 0 k 3, in ( ) can be uniquely determined by the conditions (+) and (++). Setting h i := x i+1 x i, we obtain: ( ) s i (x) = f i + m i (x x i ) + [ 3 f i+1 f i h 2 i m i+1 + 2m i h i ] (x x i ) [ 2 f i+1 f i h 3 i + m i+1 + m i h 2 i ] (x x i ) 3.

32 The function given by ( ) s i (x) = f i + m i (x x i ) + [ 3 f i+1 f i h 2 i + [ 2 f i+1 f i h 3 i + m i+1 + m i h 2 i m i+1 + 2m i h i ] (x x i ) 2 + ] (x x i ) 3 satisfies s C 1 ([a,b]). The condition s C 2 ([a,b]) leads to Differentiation in ( ) results in ( ) s (2) i 1(x i ) = s (2) i (x i ), 1 i n 1. s (2) i (x) = 2 [ 3 f i+1 f i h 2 i m i+1 + 2m i h i ] + 6 [ 2 f i+1 f i h 3 i + m i+1 + m i h 2 i ] (x x i ). Hence, ( ) provides the equations h i m i (h i 1 + h i ) m i + h i 1 m i+1 = 3 [ h i 1 f i+1 f i h i + h i f i f i 1 h i 1 ].

33 If we divide in h i m i (h i 1 + h i ) m i + h i 1 m i+1 = 3 [ h i 1 f i+1 f i h i by h i 1 + h i > 0, we obtain: h i h i 1 + h i }{{} =: ρ i m i m i + h i 1 h i 1 + h i }{{} =: λ i m i+1 = 3 f i+1 f i [ h i 1 h i 1 + h i These are n 1 equations in the n + 1 unknowns m i, 0 i n. (i) Complete cubic spline interpolant + h i f i f i 1 h i 1 ]. + h i f i f i 1 ]. h i h }{{ i 1 } =: d i In case of a complete cubic spline interpolant, we additionally obtain: s (x 0) = f 0 s (x n) = f n = m 0 = f 0 m n = f n.

34 For the remaining n 1 unknowns m i, 1 i n 1, we obtain the linear tridiagonal system 2 λ ρ 2 2 λ ρ n 2 2 λ n ρ n 1 2 m 1 m 2 m n 2 m n 1 = d 1 ρ 1 f 0 d 2 d n 2 d n 1 λ n 1 f n. In view of ρ i + λ i = h i h i 1 + h i + h i 1 h i 1 + h i = 1, 1 i n 1 the coefficient matrix is strictly diagonally dominant and hence, the linear system is uniquely solvable.

35 (ii) Natural cubic spline interpolant In this case, we have the additional boundary conditions s (x 0) = 0, s (x n) = 0. Consequently, we obtain the additional two equations 2 m 0 + m 1 = 3 f 1 f 0 h 0 =: d 0, m n m n = 3 f n f n 1 h n 1 =: d n. The coefficient matrix of the resulting linear system of equations also is a strictly diagonally dominant tridiagonal matrix.

36 (ii) Periodic cubic spline interpolant Considering the additional boundary conditions s (x 0) = s (x n), s (x n) = s (x n), the first equation readily implies m 0 = m n. The second equation yields 2 m m m n m n = 3 [ f 1 f 0 h 0 h 0 h n 1 h n 1 h 2 + f n f n 1 0 h 2 n 1 In view of m n = m 0, it follows that ( 2 h h n 1 ) m h 0 m h n 1 m n 1 = d 0. The remaining n 1 equations read as follows ] =: d 0. 1 m i 1 + ( ) m i + 1 m i+1 = ( ) d i =: d i, 1 i n 2, h i 1 h i 1 h i h i h i 1 h i 1 m m n 2 + ( ) m n 1 = ( + 1 ) d n 1 =: d n 1. h n 1 h n 2 h n 2 h n 1 h n 2 h n 1

37 The coefficient matrix of the resulting linear algebraic system is given by ( 2 h h n 1 ) h h n 1 1 h 0 ( 2 h h 1 ) h h n 3 ( 2 h n h n 2 ) h n h n h n 2 ( 2 h n h n 1 ) Obviously, the coefficient matrix also is strictly diagonally dominant. Theorem 4.38 Existence and uniqueness of the cubic spline interpolant The complete (resp. natural, periodic) cubic spline interpolant s S 4, is uniquely determined. The slopes m 1,...,m n 1 (resp. m 0,...,m n ) are the solutions of a linear algebraic system with a strictly diagonally dominant coefficient matrix..

38 (ii) Computation via the moments In each subinterval [x i,x i+1 ], 0 i n 1, the polynomials s i = s [xi,x i+1 ] admit the representation as the solution of the local interpolation problem: Compute s i P 3 ([x i,x i+1 ]) as given by such that ( ) s i (x) = a 0 + a 1 (x x i ) + a 2 (x x i ) 2 + a 3 (x x i ) 3, (+) s i (x j ) = f j, i j i + 1, (++) s i (x j ) = M j, i j i + 1. Here, the moments M i in the nodes are unknown and have to be computed by means of the determining equations. The coefficients a k, 0 k 3, in ( ) can be uniquely determined by the conditions (+) and (++). Setting h i := x i+1 x i, we obtain: ( ) s i (x) = (f i 1 6 h2 i M i ) x i+1 x h i + (f i h2 i M i+1 ) x x i h i + M i (x i+1 x) 3 6h i + M i+1 (x x i ) 3 6h i.

39 The condition s C 2 ([a,b]) gives Differentiating in ( ) s i (x) = (f i 1 6 h2 i M i ) x i+1 x h i and inserting the result into ( ) yields ( ) s i 1(x i ) = s i(x i ), 1 i n 1. h i 1 M i (h i 1 + h i ) M i + h i M i+1 = 6 ( f i+1 f i h i Division by (h i 1 + h i ) gives h i 1 h i 1 + h i }{{} =: Σ i M i M i + h i h i 1 + h i }{{} =: Λ i + (f i h2 i M i+1 ) x x i h i + M i (x i+1 x) 3 6h i + M i+1 (x x i ) 3 6h i. M i+1 = f i f i 1 h i 1 ), 1 i n 1. 6 ( f i+1 f i f i f i 1 ) h i 1 + h i h i h }{{ i 1 }, 1 i n 1. =: D i

40 For the complete (rep. natural, periodic) cubic spline interpolant we obtain the additional boundary conditions: (i) Complete spline interpolant (ii) Natural spline interpolant (iii) Periodic spline interpolant 2 M 0 + M 1 = 1 6 h 0 ( f 1 f 0 h 0 f 0), M n M n = 1 6 h n 1 (f n f n f n 1 h n 1 ). M 0 = 0, M n = 0. M 0 = M n, 1 6 h 0 (2 M 0 + M 1 ) h n 1 (M n M n ) = f 1 f 0 h 0 f n f n 1 h n 1.

41 Theorem 4.39 Computation via the moments Computing the complete (resp. natural, periodic) cubic spline interpolant via the moments M 1,...,M n 1 (resp. M 0,...,M n ), the moments can be obtained as the solution of a linear algebraic system with a strictly diagonally dominant coefficient matrix. Remark 4.40 The computation of the cubic spline interpolant When solving the linear algebraic systems for the slopes m i resp. the moments M i, in general we only obtain some approximations m i resp. M i. The first case results in s C 1 ([a,b]) with an approximate coincidence of the second derivatives, whereas in the second case we only get s C([a,b]) with an approximate coincidence of the first derivatives. Therefore, the computation via the slopes is preferable.

42 Convergence properties We consider a sequence of grids n := { a = x n,0 < x n,1 <... < x n,mn = b }, consisting of m n + 1 mutually disjoint nodes. For given m n + 1 values f n,j, 0 j m n, denote by s n S 4, n the cubic spline interpolant. Problem: If f C([a,b]) and f n,j = f(x n,j ), 0 j m n, does the sequence (s n ) n ln converge uniformly on [a,b] to the function f? Definition 4.41 Minimal, maximal mesh width The quantities n := max x n,j+1 x n,j, n := min x n,j+1 x n,j 0 j m n 0 j m n are called maximal resp. minimal mesh width. Definition 4.42 Quasiuniformity A sequence ( n ) n ln of grids is said to be quasiuniform, if there exists a constant β > 0, independent of n, such that n / n β, n ln.

43 Theorem 4.43 Uniform convergence of the cubic spline interpolants Assume that f C([a,b]) and that ( bdelta n ) n ln is a quasiuniform sequence of grids. Denote by (s n ) n ln the associated sequence of complete (resp. natural, periodoc) spline interpolants. Then, there holds: lim n f s n = 0 für n 0, n Under additional smoothness properties of the function f, we obtain the following convergence properties: Theorem 4.44 Uniform convergence of the derivatives Assume that f C 4 ([a,b]) and that ( n ) n ln is a quasiuniform sequence of grids. Denote by (s n ) n ln the associated sequence of spline interpolants. Then, there exist constants C i, 0 i 3, independent of n and f, such that f (i) s (i) n C i β f (4) ( n ) 4 i, 0 i 3, where f (3) s (3) n := max 0 j m n 1 max x [x n,j,x n,j+1 ] f(3) (x) s (3) n (x).

44 The preceding theorem suggests that even better smoothness properties than f C 4 ([a,b]) imply a higher convergence order. This is not the case as shown by the following result: Theorem 4.44 Maximal order of convergence of the cubic spline interpolant Assume that f C([a,b]) and that ( n ) n ln is a quaiuniform sequence of grids. Denote by (s n ) n ln the associated sequence of complete (natural, periodic) cubic spline interpolants. Moreover, assume that there exists µ > 0 such that f s n C ( n ) 4+µ. Then, there holds f (4) (x) = 0, x [a,b], i.e., f P 3 ([a,b]).

45 4.2.4 Geometric properties of cubic spline interpolants r Parameterized curve: y : [a,b] lr, y C 2 ([a,b]) Curvature of the curve in x [a,b]: y κ(x) = (x), κ(x) = 1/r (1 + y (x) 2 ) 3/2 r: radius of the supporting circle For small y (x), i.e., y (x) 1 there holds: y (x) (1 + y (x) 2 ) 3/2 y (x) A suitable measure for the curvature of the curve is the L 2 -norm of the second derivative, i.e.: y 0,2 := ( b a (y (x)) 2 dx) 1/2.

46 2 Theorem 4.46 Geometric property Assume that (x i,f i ), 0 i n, are n + 1 pairs satisfying := {a = x 0 < x 1 <... < x n = b}. Moreover, let s S 4, be a cubic spline interpolant and y C2 ([a,b]) an arbitrary interpolating function such that Then, there holds: Proof: Setting y = s + (y s ), we have: ( ) b a (y ) 2 dx = b a [ s (x) (y (x) s (x)) ] b a = 0. (s )2 dx + 2 s 0,2 y 0,2. b a s (y s b ) dx + (y s )2 dx. For the second term on the right-hand side in ( ), partial integration yields: b a s (y s n 1 ) dx = i=0 = n 1 i=0 x i+1 x i s (y s n 1 ) dx = { [s (y s )] x i+1 x i i=0 a x i+1 x i s }{{} =const. (y s ) dx } { [s (y s )] x i+1 x i s (y s ) x i+1 x i } = [ s (y s ) ] b a = 0.

47 Corollary 4.47 Geometric property Assume that s S 4, is the complete (resp. natural, periodic) cubic spline interpolating in (x i,f i ), 0 i n. Then, for any interpolating function y C 2 ([a,b]), satisfying the same boundary conditions, there holds: s 0,2 y 0,2. Remark 4.48 Spline Flexible wooden ruler used as a drawing device fixed in the nodes x i,0 i n. Stationary state: State for which the bending energy is minimal: E = b a ( y (x) (1 + y (x) 2 ) 3/2 ) 2 dx The cubic spline interpolant approximately describes the stationary state of the drawing device

48 4.2.5 B-splines Construction of a basis of S m,, := {a = x 0 < x 1 <... < x n = b}, m ln, such that ( ) Evaluation with respect to the basis representation is well conditioned, ( ) Basis functions have minimal support in [a,b]. Definition 4.49 B-splines Assume that τ 1 τ 2... τ n is an arbitrary sequence of nodes and f(t) := (t x) + m 1, m ln, x lr. If τ i < τ i+m, 1 i n m, the B-splines N im,1 i n m, of order m,1 m n, are defined as follows: Remark 4.50 B-splines in case m = 1 N i1 ( ) N im (x) := (τ i+m τ i ) f[τ i,..., τ i+m ]. τi τ i+1 For m = 1 it follows that: N i1 (x) = (τ i+1 τ i ) f[τ i, τ i+1 ] = (τ i+1 τ i ) (τ i+1 x) 0 (τ + i x) 0 + τ i+1 τ = i = 1, τ i x τ i+1 0, else

49 The B-splines satisfy a recursion which can be deduced from ( ) by using the Leibniz-rule for divided differences: Lemma 4.51 Leibniz-rule for divided differences Assume g,h C n ([a,b]) and x i [a,b], 0 i n. Then, there holds: (gh)[x 0,...,x n ] = Theorem 4.52 Recursion for B-splines For m > 1, the B-splines N im satisfy the recursion ( ) N im (x) = n i=0 x τ i τ i+m 1 τ i N i,m 1 (x) + g[x 0,...,x i ] h[x i,...,x n ]. τ i+m x τ i+m τ i+1 N i+1,m 1 (x). Here, it has to be observed that in case of multiple nodes the corresponding terms on the right-hand side in ( ) are omitted according to the convention 0/0 = 0.

50 Proof: Setting g(t) := (t x) and h(t) := (t x) m 2 + and using the Leibniz-rule, we obtain N im (x) = (τ i+m τ i ) (gh)[τ i,..., τ i+m ] = (τ i+m τ i ) m k=0 g[τ i,..., τ i+k ] h[τ i+k,..., τ i+m ]. In view of the linearity of g, all divided differences of higher than first order disappear: N im (x) = (τ i+m τ i ) { g[τ i ] h[τ i,..., τ i+m ] + g[τ i, τ i+1 ] h[τ }{{} i+1,..., τ i+m ] } = =1 = (τ i+m τ i ) { (τ i x) h[τ i+1,..., τ i+m ] h[τ i,..., τ i+m 1 ] + h[τ i+1,..., τ i+m ] } = τ i+m τ i = = x τ i (τ i+m 1 τ i ) h[τ i,..., τ i+m 1 ] + τ i+m 1 τ i x τ i N i,m 1 (x) + τ i+m 1 τ i τ i+m x τ i+m τ i+1 N i+1,m 1 (x). τ i+m x τ i+m τ i+1 (τ i+m τ i+1 ) h[τ i+1,..., τ i+m ] =

51 Lemma 4.53 Properties of B-splines The B-splines N im, 1 i n m, as defined by ( ) have the following properties: Corollary 4.54 Multiple nodes Assume that τ i is a k-fold node, i.e., (B1) supp N im [τ i, τ i+m ], (B2) N im (x) 0, x lr, (B3) N im [τ j,τ j+1 ] P m 1 ([τ j, τ j+1 ]). τ i 1 < τ i = τ i+1 =... = τ i+k 1 < τ i+k. Then, N im is at least (m k 1) times continuously differentiable in x = τ i and N im(x) = (m 1) [ N i,m 1 (x) τ i+m 1 τ i N i+1,m 1(x) τ i+m τ i+1 ].

52 Construction of a basis of S m, For the grid := {a = x 0 <... < x n = b} we define extended nodes T = {τ i } 2m+n 1 i=1 with m-fold nodes in a = x 0 and b = x n according to : a = x 0 < x 1 <... < x n = b T : τ 1 =... = τ m < τ m+1 <... < τ m+n =... = τ 2m+n 1 In view of (B3) we have N im [xj,x j+1 ] P m 1 ([x j,x j+1 ]), 0 j n 1, and Corollary 4.54 implies that N im is (m 2) times continuously differentiable in x = x j, 1 j n 1,. Hence, we get: N im S m,, 1 i n + m 1. The linear independence of the N im remains to be shown.

53 Lemma 4.55 Marsden identity Let T = {τ i } n+2m 1 i=1 be the extended sequence of nodes and for s lr assume that ϕ im (s) := m 1 (τ i+j s). Then, for all x [a,b] the so-called Marsden identity holds true: ( ) j=1 (x s) m 1 = n+m 1 i=1 ϕ im (s) N im (x). Proof: Induction on m. Corollary 4.56 Partition of unity The B-splines N im, 1 i n + m 1, form a partition of unity on [a,b] according to (P1) P m 1 ([a,b]) span {N 1m,...,N n+m 1,m }, (P2) n+m 1 i=1 N im (x) = 1, x [a,b].

54 Proof: Taking into account the Marsden identity ( ) (x s) m 1 = n+m 1 i=1 ϕ im (s) N im (x) for the l-th derivative of the function f(s) := (x s) m 1 it follows that f (l) (0) = (m 1)(m 2)...(m l) ( 1) l x m l 1 = n+m 1 In particular, for m l 1 = k we obtain: x k = ( 1) m k 1 (m 1)...(k + 1) For k = 0, the assertion (P2) follows in view of ϕ (m 1) im n+m 1 i=1 i=1 ϕ (m k 1) im (0) N im (x). ϕ (l) im (0) N im (x). (s) = ( m 1 (τ i+j s)) (m 1) = [ ( 1) m 1 s m ] (m 1) = ( 1) m 1 (m 1)!. j=1

55 Theorem 4.57 Linear independence of the B-splines The B-splines N im, 1 i n + m 1, are linear independent, i.e., n+m 1 i=1 c i N im (x) = 0, x [a,b] = c i = 0, 1 i n + m 1. Proof: Consider [x j,x j+1 ] = [τ m+j, τ m+j+1 ], 0 j n 1. Since N im [τ m+j,τ m+j+1 ] = 0, i / {j + 1,...,j + m}, it follows that (+) j+m i=j+1 c i N im (x) = 0, x [x j,x j+1 ]. In view of (P1), p 0 P m 1 ([a,b]), p 0 0, can be written as a linear combination of the B-splines N im,1 i n + m 1,. Since p 0 [xj,x j+1 ] is uniquely determined by m = dim P m 1 ([a,b]) values, taking into account (+), the N im, j + 1 i j + m, must be linearly independent, i.e., we must have c i = 0, j + 1 i j + m.

56 Korollar 4.58 de Boor points Every spline s S m, admits the unique representation s = n+m 1 i=1 d i N im. The coefficients d i, 1 i n + m 1, are called the de Boor points of s. Theorem 4.59 Recursive evaluation in the B-spline representation Let s S m, and x [a,b]. Then, s (x) can be recursively computed according to s (x) = n+m 1 i=j+1 d j i(x) N i,m j (x), where d 0 i (x) := d i and for j > 0 there holds: d j i(x) := x τ i τ i+m j τ d j 1 i i (x) + τ i+m j x τ i+m j τ d j 1 i i 1(x), τ i+m j τ i 0, else. In particular, we have s(x) = d m 1 i (x), x [τ i, τ i+1 ]. Proof: Induction argument.

57 Remark 4.60 de Boor algorithm The value s (x) = di m 1 (x) is obtained by a successive convex combination of the de Boor points according to the following scheme d 1 d0 i m+1(x) d d 3 d 2 6 d4 d 5 d 1 i m+2(x) d 0 i m+2(x) d m 1 i d 0 i 1(x) d 1 i (x) d 0 i (x) (x)

58 Theorem 4.61 Condition of the evaluation in the B-spline representation There exists a constant D m > 0, depending only on the order m, such that ( ) D m max d i n+m 1 1 i n+m 1 i=1 d i N im max 1 i n+m 1 d i. Remark 4.62 About the condition of the evaluation in the B-spline representation According to ( ), perturbations in s (x) = i d i N im (x) of a spline s S m, as well as in the coefficients d i can be mutually estimated. Therefore, the evaluation in the B-spline representation is well conditioned. For this reason, the basis N im, 1 i n + m 1, is referred to as a well conditioned basis.

59 4.3 Trigonometric interpolation Discrete Fourier transformation Reminder from calculus: Fourier transformation and Fourier series The Fourier transform of a 2π-periodic function f : lr C is a function ˆf := Ff : Z C with ( ) ˆf(k) = 1 2π f(t) exp( ikt) dt, k Z. 2π 0 The inverse mapping f = F 1ˆf is given by the Fourier series ( ) f(t) = + ˆf(k) exp(ikt). k= Choosing an equidistant partition of [0,2π) with respect to the nodes t j := 2πj/n,0 j n 1 n ln, and using the trapezoidal rule ( ) yields: 2π 0 ˆf(k). = 1 n g(t) dt. = 2π n n 1 j=0 n 1 j=0 g(t j ), f(t j ) exp( 2πijk/n).

60 Definition 4.63 Discrete Fourier transformation Assume that f : lr C is 2π-periodic, t j := 2πj/n, 0 j n 1, n ln, and f j := f(t j ), 0 j n 1. Then, the function F n : C n C n, (f j ) (c j ) c k = 1 n n 1 j=0 f j exp( 2πijk/n), 0 k n 1 is called the discrete Fourier transformation. The inverse mapping Fn 1 ( ) f j = n 1 c k exp(2πijk/n), 0 j n 1. k=0 Remark 4.64 Truncated Fourier series If n = 2m + 1, then setting c k = c m+k, m k < 0, we obtain: f j = +m k= m c k exp(2πijk/n). Hence, ( ) can be interpreted as a truncated Fourier series. is given by

61 Definition 4.65 Trigonometric polynomial Let c k C, 0 k n 1, n ln. Then T n 1 (t) = n 1 is called a trigonometric polynomial. k=0 c k exp(ikt), t [0,2π) Definition 4.66 Trigonometric interpolation problem Assume that f : lr C is 2π-periodic, t j := 2πj/n, 0 j n 1, n ln, and f j = f(t j ), 0 j n 1. The computation of a trigonometric polynomial T n 1 such that ( ) T n 1 (t j ) = f j, 0 j n 1, is called a trigonometric interpolation problem. Theorem 4.57 Solution of the trigonometric interpolation problem The trigonometric interpolation problem ( ) is uniquely solvable. The coefficients c k, 0 k n 1, are given by c k = 1 n n 1 j=0 f j exp( 2πijk/n)

62 Proof: Denoting by ω j = exp(2πij/n), 0 j n 1 the n-th unit roots, ( ) is equivalent to the complex polynomial interpolation problem: Compute T n 1 (z) = n 1 k=0 c k z k, z = exp(it) such that T n 1 (ω j ) = f j, 0 j n 1. Since ω j ω l for 0 j l n 1, the interpolation problem is uniquely solvable. The representation of the coefficients according to c k = 1 n follows from the orthogonality property 1 n n 1 j=0 n 1 j=0 f j exp( 2πijk/n) ω k j ω l j = δ kl.

63 4.3.2 Fast Fourier Transform (FFT) The Fast Fourier Transform is the efficient implementation of the discrete Fourier transform ( ) resp. of the inverse mapping in case n = 2 p, p ln. ( ) c k = 1 n n 1 j=0 f j = n 1 k=0 f j exp( 2πijk/n), 0 k n 1 c k exp(2πijk/n), 0 j n 1 Explanation of the FFT for ( ): Computational work in case of a naive approach: O(n 2 ) complex additions, multiplications, evaluation of trigonometric functions Idea for a reduction of the computational work: Let n = 2m, m = 2 p 1. Partition ( ) ( ) w.r.t. k into even (2k) and odd (2k + 1) terms, ( ) w.r.t. j into the lower half (0,...,m 1) and into the upper half (m,...,2m 1).

64 For 0 j m 1, the partition leads to: (+) f j = m 1 c 2k exp(2πijk/m) + exp(πij/m) m 1 k=0 (++) f j+m = m 1 k=0 k=0 c 2k exp(2πijk/m) exp(πij/m) m 1 k=0 c 2k+1 exp(2πijk/m), c 2k+1 exp(2πijk/m). Since both partial sums in (+) also appear in (++): Reduction of the computational work by 50 %, but additionally m additions resp. subtractions, multiplications and evaluations of trigonometric functions ( working costs ). Recursive application of the splitting yields the following partition consisting of p steps f j 0 j 2 p 1 1.step: f j f j+2 p 1 0 j 2 p step: f j f j+2 p 2 f j+2 p 1 f j+2 p 1 +2 p 2 0 j 2p 2 1 p-th step: f 0 f 1 f 2 n 2 f 2 n 1

65 Computational work (only sum of the working costs): p n/2 additions, subtractions and multiplications n 2 + n n 2 p = n p i=1 2 i 2n trigonometric functions Since n = 2 p and hence, p = log 2 n = O(n log 2 n) essential operations Practical implementation of the FFT: Input: c k, 0 k n 1 in array c[0,...,n 1] Output: f j, 0 j n 1 in the same array, i.e., the c-array is overwritten Realization: Sorting and combination steps 1. Sorting step: Sort all components c k, 0 k n 1, according to the principle even before odd : p p p 1

66 p-th combination step: where f j = ĉ j + exp(πij/m) ĉ j+m, f j+m = ĉ j exp(πij/m) ĉ j+m, 0 j m 1, ĉ j = m 1 k=0 c k exp(2πijk/m), ĉ j+m = m 1 f j and f j+m are stored in ĉ j and ĉ j+m. k=0 c k+m exp(2πijk/m), 0 j m 1. ^c j c^ j+m x exp( πij/m) + butterfly diagram j j+m