Univariate Rational Interpolation I The art of reading between the lines in a numerical table (Thiele, 1909) Oliver Salazar Celis November 3, 2004
Overview [1] this seminar: develop the theory implementation problems connection with continued fractions
The Problem (1) [2] Denote R(l, m) the rational functions of degrees l in the numerator m in the denominator Given l, m, n N with n = l + m. Given n + 1 interpolation points (x 0, f 0 ), (x 1, f 1 ),..., (x n, f n ) with x k x j for k j {0,..., n}
The Problem (2) [3] Construct a rational function r l,m (x) R(l, m) with p, q irreducible a i, b i R and q 0 such that r l,m (x) = p(x) q(x) = a 0 + a 1 x +... a l x l b 0 + b 1 x +... b m x m (1) r l,m (x j ) = f j for every j {0,..., n} (2)
The Problem (3) [4] To solve (2) consider the modified interpolation problem the linearized interpolation problem for every j {0,..., n}: or f j q(x j ) p(x j ) = 0. (3) a 0 + a 1 x j +... + a l x l j (b 0 + b 1 x j +... + b m x m j )f j = 0. (4)
The Problem (4) [5] which is a homogeneous system of linear equations 1 x 0... x l 0 f 0 f 0 x 0... f 0 x l 0 1 x 1... x l 1 f 1 f 1 x 1... f 1 x l 1............ 1 x n... x l n f n f n x n... f n x l n a 0 a 1. a l b 0 b 1. b m = 0. (5)
The Problem (4) [6] Consisting of n + 1 = l + m + 1 equations l + m + 2 unknown a 0,..., a l, b 0,..., b m. hence (3) always has a non-trivial (i.e. 0 ) solution!
Remarks (1) [7] Theorem 1. If polynomials p 1, q 1 and p 2, q 2 both satisfy (3) then p 1 q 2 = p 2 q 1. A solution p/q of (3) may be reducible to p 0 /q 0. Theorem 1: all solutions have the same irreducible form p 0 /q 0.
Rational Interpolant [8] for p, q satisfying (3) we define the rational interpolant of order (l, m) for f as with r l,m (x) = p 0 q 0 (x) (6) p 0 /q 0 the irreducible form of p/q q 0 (x) normalized such that q 0 (x 0 ) = 1 for every l, m N such a rational interpolant for f exists.
The Table Property (1) [9] The rational interpolants of order (l, m) for f can be ordered in a table r 0,0 r 0,1 r 0,2... r 1,0 r 1,1 r 1,2... r 2,0 r 2,1... r 3,0. r 3,1....
The Table Property (2) [10] with r i,0 the polynomial interpolants for f r 0,i the inverse polynomial interpolants for 1/f An entry of the table is called normal if it occurs only once in that table.
Now the real problem.. (1) [11] to solve (2) we solved (3) f(x i ) = r l,m (x i ) = p 0 q 0 f i q(x i ) p(x i ) = 0 i = 0,..., n i = 0,..., n trivial remark: (2) implies (3) but what about (3) implies (2)?
Example [12] Let x 0 1 2 f 0 3 3 then r 1,1 = p q (x) = 3x p 0 (x) = 3 x q 0 1 and clearly p 0 (x 0 ) f 0 q 0 (0, 0) is called an unattainable point.
Now the real problem.. Poles (2) [13] Let p/q be a solution of (3) if q(x i ) = 0 for some i 0,..., n then from (3) also p(x i ) = 0 thus q and p have a common factor (x x i ) hence p q = (x x i)p (x x i )q = p q but for q(x i ) = 0, any f i solves (3)!
Methods [14] Various equivalent ways to calculate an explicit solution. The direct method : solve (3) by (5) 1 x 0... x l 0 f 0 f 0 x 0... f 0 x l 0 1 x 1... x l 1 f 1 f 1 x 1... f 1 x l 1............ 1 x n... x l n f n f n x n... f n x l n a 0 a 1. a l b 0 b 1. b m = 0.
The Direct Method [15] Disadvantages (5) is not singular for x i x j with i j {0,..., n} but when n increases in [x 0, x n ], the system becomes quasi singular. hard(er) to tell anything about solubility unattainable points
Thiele Interpolating Continued Fractions [16] Consider the staircase of normal rational interpolants T 0 = {r 0,0, r 1,0, r 1,1, r 2,1,...}. (7) calculate coefficients d i (i 0) such that the convergents of d 0 + x x 0 d 1 + x x 1 d 2 + x x 2 d 3 +... (8) are the subsequent elements of T 0.
Inverse Differences (1) [17] Theorem 2. if we take d i = ϕ i [x 0,..., x i ] then C n (x i ) = f(x i ) for i = 0,..., n with ϕ 0 [x] = f(x) ϕ 1 [x 0, x 1 ] = ϕ k [x 0,..., x k ] =. x 1 x 0 ϕ 0 [x 1 ] ϕ 0 [x 0 ] x k x k 1 ϕ k 1 [x 0,..., x k 2, x k ] ϕ k 1 [x 0,..., x k 2, x k 1 ]
Inverse Differences (2) [18] Proof f(x) = ϕ 0 [x] = ϕ 0 [x 0 ] + x x 0 ϕ 1 [x 0, x] x x 0 = ϕ 0 [x 0 ] + ϕ 1 [x 0, x 1 ] + x x 1 ϕ 2 [x 0,x 1,x] = ϕ 0 [x 0 ] + x x 0 ϕ 1 [x 0, x 1 ] + x x 1 ϕ 2 [x 0, x 1, x 2 ] +... + x x n 1 ϕ n [x 0,..., x n 1, x]
Remarks (1) [19] just like divided differences we can organize the inverse differences in a table ϕ 0 [x 0 ] ϕ 0 [x 1 ] ϕ 1 [x 0, x 1 ] ϕ 0 [x 2 ] ϕ 1 [x 0, x 2 ] ϕ 2 [x 0, x 1, x 2 ] ϕ 0 [x 3 ]. ϕ 1 [x 0, x 3 ]. ϕ 2 [x 0, x 1, x 3 ]. ϕ 0 [x n ] ϕ 1 [x 0, x n ] ϕ 2 [x 0, x 1, x n ]... ϕ n [x 0,..x n ]
Exercise [20] 1.(a) Compute the inverse differences of the following data x 0 1 2 f 0 3 3. (b) Compute the Thiele interpolating Continued fraction. 2.(a) Compute the inverse differences of the following data what happens? x 0 1 2 3 4 f 1 0 2 2 5.
General staircases (1) [21] We only considered normal staircases T 0 = {r 0,0, r 1,0, r 1,1, r 2,1,...}. (9) with 0 l m 1 Consider now general staircases T k = {r k,0, r k+1,0, r k+1,1, r k+2,1,...} for k 0. (10) with l m 2
General staircases (2) [22] We assume that l m otherwise interpolate (x i, 1 f i ) for i = 0,..., n and interchange the roles of l and m. if some f i are zero, say f 0,..., f z = 0 if z < l and r l,m = (x x 0 ) (x x z )R 1 with R 1 R(l z 1, m) (11) R 1 (x i ) = f j = f j (x i x 0 ) (x i x z ) for i = z + 1..., n (12)
General staircases (3) [23] construct a polynomial p 1 of degree p1 l m 1 = k such that p 1 (x i ) = f i for i = 0,..., k reformulate the problem as r l,m (x) = r k+1+m,m (x) = p 1 (x) + (x x 0 ) (x x k )r m,m (x) (13)
General staircases (4) [24] we have the conditions r l,m (x i ) = f i for i = 0,..., l + m (14) implying for r m,m r m,m (x i ) = f i = f i p 1 (x i ) (x i x 0 ) (x i x k ) for i = k +1,..., l +m (15)
Exercise [25] 1. What happens to (13) and (15) if we take p 1 l m = k?
Conclusion [26] Next time we will Generalize Thiele interpolating continued fractions. See how to handle/detect unattainable points. Discuss a pivot strategy for numerical stabilization.
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[5] H. Werner, Algorithm 51: A reliable and numerically stable program for rational interpolation of Lagrange data, Computing 31, pp. 269-286, 1983. [28] [6] J. Meinguet, On the solubility of the Cauchy interpolation problem, A. Talbot (Ed.), Approximation Theory, Academic Press, London, 535-600, 1970. [7] E. Meijering, A Chronology of Interpolation: From Ancient Astronomy to Modern Signal and Image Processing, Proc. IEEE 90, pp. 319-342, 2002 [8] L. Wuytack, On Some Aspects of the Rational Interpolation Problem, SIAM Journal on Numerical Analysis, Vol. 11, No. 1., pp. 52-60, Mar. 1974
[9] H. Werner and R. Schaback, Praktische Mathematik II, Springer-Verlag, Berlin Heidelberg, 1979 [29]