Polynomial Interpolation with n + 1 nodes
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1 Polynomial Interpolation with n + 1 nodes Given n + 1 distinct points (x 0, f (x 0 )), (x 1, f (x 1 )),, (x n, f (x n )), Interpolating polynomial of degree n P(x) = f (x 0 )L 0 (x) + f (x 1 )L 1 (x) + + f (x n )L n (x), with P(x 0 ) = f (x 0 ), P(x 1 ) = f (x 1 ),, P(x n ) = f (x n ), and Lagrangian polynomials L i (x) = j i x x j x i x j. For all i, j, L i (x j ) = { 1, if i = j, 0, if i j,
2 Polynomial Interpolation with n + 1 nodes Analysis: Estimate errors with polynomial interpolation Computation: How to compute P(x) numerically.
3 Polynomial Interpolation: Analysis Theorem: Assume that the nodal points x 0,, x n are mutually distinct, then the interpolating polynomial P(x) of degree n exists and is unique.
4 Polynomial Interpolation: Analysis Lemma: Let f (x) C n+1 [a, b] be such that f (x 1 ) = 0, f (x 2 ) = 0,, f (x n ) = 0, where x 1, x 2,, x n [a, b] are mutually distinct. Then there exists a ξ [min(x 1, x 2,, x n ), max(x 1, x 2,, x n )] such that f (n 1) (ξ) = 0.
5 Polynomial Interpolation Error Theorem: Suppose x 0, x 1,, x n are distinct numbers in the interval [a, b] and f C n+1 [a, b]. Then, for each x [a, b], a number ξ(x) between, x 0, x 1,, x n (hence (a, b)) exists with f (x) = P(x) + f (n+1) (ξ(x)) (n + 1)! where P(x) is the interpolating polynomial. (x x 0 )(x x 1 ) (x x n ),
6 Polynomial Interpolation Error: Proof If x = x 0, x 1,, x n, then error = 0 and theorem is true. Now let x be not equal to any node. Define function g for t [a, b] g(t) def = (f (t) P(t)) (f (x) P(x)) (t x 0)(t x 1 ) (t x n ) (x x 0 )(x x 1 ) (x x n ) = (f (t) P(t)) (f (x) P(x)) n j=0 (t x j ) (x x j ) C n+1 [a, b]. Then g(t) vanishes at n + 2 distinct points: g(x) = 0, g(x k ) = 0,, for k = 0, 1,, n. There must be a ξ between x and nodal points such that g (n+1) (ξ) = 0.
7 Polynomial Interpolation Error: Proof Since g (n+1) (ξ) = (f (t) P(t)) (n+1) t=ξ (f (x) P(x))( = f (n+1) (ξ) (f (x) P(x)) = 0 Therefore f (x) = P(x) + f (n+1) (ξ(x)) (n + 1)! (n + 1)! n j=0 (x x j) n j=0 (t x j ) (x x j ) )(n+1) t=ξ (x x 0 )(x x 1 ) (x x n ),
8 Polynomial Interpolation: Corollary Corollary: Suppose x 0, x 1,, x n are distinct numbers in the interval [a, b] and f is polynomial of degree at most n, then P(x) = f (x).
9 Secant Method: Order of Convergence Assume that the secant method p k+1 = p k f (p k)(p k p k 1 ), k = 1, 2,, f (p k ) f (p k 1 ) converges to the root p (f (p) = 0.) It follows that p k+1 p = p k p (f (p k) f (p))(p k p k 1 ) f (p k ) f (p k 1 ) = (p k p)(f (p k ) f (p k 1 )) (f (p k ) f (p))(p k p k 1 ) f (p k ) f (p k 1 ) ( p k p k 1 = f (p) f (p k ) (p ) k p)(f (p k ) f (p k 1 )) f (p k ) f (p k 1 ) (p k p k 1 )
10 Secant Method: Order of Convergence But the last expression P(p) def = f (p k ) + (p k p)(f (p k ) f (p k 1 )) (p k p k 1 ) is linear interpolation on nodes p k and p k 1, so there exists ξ k between p, p k and p k 1 with and therefore p k+1 p = f (p) P(p) = f (2) (ξ k ) 2 (p p k )(p p k 1 ), p k p k 1 f (p k ) f (p k 1 ) f (2) (ξ k ) (p p k )(p p k 1 ). 2
11 Secant Method: Order of Convergence = α Let α = > 1 2 be the golden ratio. Then α(α 1) = 1, and p k+1 p p k p α ( ) pk p α 1 p k 1 p α = p k p k 1 f (p k ) f (p k 1 ) f (2) (ξ k ) 2 f (2) (p) 2f (p), provided f (p) 0, or p k+1 p lim k p k p α = f (2) (p) 2f (p) 1 α. (the limit does exist.)
12 Polynomial Interpolation, Error Bounds: Estimate maximum error in second order polynomial interpolation of function f (x) = 1 x over interval [2, 4] using nodes x 0 = 2, x 1 = 2.75, and x 2 = 4.
13 Polynomial Interpolation, Error Bounds: f (x) = P(x) + f (3) (ξ(x)) (x 2)(x 2.75)(x 4), 3! where P(x) is the interpolating polynomial on [2, 4]. f (3) (x) = 6x 4, f (3) (ξ(x)) 6 2 4, def g(x) = (x 2)(x 2.75)(x 4) = x 3 35x x 22. d g(x) d x = 3x x = 1 (3x 7)(2x 7). 2 g( 7 2 ) = 9 16 = max x [2,4] g(x). ( g( 7 25 ) = 3 f (x) P(x) 1 3! = < 9 16 )
14 Polynomial Interpolation: Experiments: Easy function: f (x) = e x on [ 1, 1]. f (n) (ξ) e for all ξ ( 1, 1). 7 nodal points (n = 6); random x points.
15 Polynomial Interpolation: Experiments: Easy function: f (x) = e x on [ 1, 1]. f (n) (ξ) e for all ξ ( 1, 1). 21 nodal points (n = 20); random x points.
16 Polynomial Interpolation: Experiments: Hard function: f (x) = x f (n) (ξ) = on [ 1, 1]. n! (1.1 + ξ) n+1 10n+1 n! for all ξ ( 1, 1). 7 nodal points (n = 6); random x points.
17 Polynomial Interpolation: Experiments: Hard function: f (x) = x f (n) (ξ) = on [ 1, 1]. n! (1.1 + ξ) n+1 10n+1 n! for all ξ ( 1, 1). 21 nodal points (n = 20); random x points.
18 Polynomial Interpolation: Experiments: Hardest function: f (x) = x 2 on [ 1, 1]. f (n) (ξ) can be very large for some ξ ( 1, 1). 7 nodal points (n = 6); random x points.
19 Polynomial Interpolation: Experiments: Hardest function: f (x) = x 2 on [ 1, 1]. f (n) (ξ) can be very large for some ξ ( 1, 1). 21 nodal points (n = 20); random x points.
20 Polynomial Interpolation: Experiments: Hardest function: f (x) = x 2 on [ 1, 1]. f (n) (ξ) can be very large for some ξ ( 1, 1). 81 nodal points (n = 80); random x points.
21 Polynomial Interpolation: Analysis and Computation Analysis: Estimate errors with polynomial interpolation Computation: How to compute P(x) numerically
22 Polynomial Interpolation: Neville s Method Let Q(x) interpolate f (x) at x 0, x 1,, x k, Let Q(x) interpolate f (x) at x 1,, x k, x k+1. Then P(x) def = (x x k+1)q(x) (x x 0 ) Q(x) x 0 x k+1 interpolates f (x) at x 0, x 1,, x k, x k+1 Build higher degree interpolating polynomial from lower degreed ones.
23 Neville s Method: Proof Let Q(x) interpolate f (x) at x 0, x 1,, x k, Let Q(x) interpolate f (x) at x 1,, x k, x k+1. for 1 j k, P(x) def = (x x k+1)q(x) (x x 0 ) Q(x) x 0 x k+1 P(x j ) = (x j x k+1 )Q(x j ) (x j x 0 ) Q(x j ) x 0 x k+1 = (x j x k+1 )f (x j ) (x j x 0 )f (x j ) x 0 x k+1 = f (x j ).
24 Neville s Method: Proof Let Q(x) interpolate f (x) at x 0, x 1,, x k, Let Q(x) interpolate f (x) at x 1,, x k, x k+1. for j = 0, k + 1, P(x) def = (x x k+1)q(x) (x x 0 ) Q(x) x 0 x k+1 P(x 0 ) = (x 0 x k+1 )Q(x 0 ) (x 0 x 0 ) Q(x 0 ) x 0 x k+1 = f (x 0 ), P(x k+1 ) = (x k+1 x k+1 )Q(x k+1 ) (x k+1 x 0 ) Q(x k+1 ) x 0 x k+1 = f (x k+1 ). Thus P(x) interpolates f (x) at x 0, x 1,, x k, x k+1.
25 Neville s Method, general case Let Q(x) interpolate f (x) at all of x 0, x 1,, x k, x k+1 but x j, Let Q(x) interpolate f (x) at all of x 0, x 1,, x k, x k+1 but x i, i j. Then P(x) def = (x x j)q(x) (x x i ) Q(x) x i x j interpolates f (x) at all of x 0, x 1,, x k, x k+1
26 Neville s Method, recursive Step 1: compute Q i,i+1 (x), linear polynomial interpolating f (x) at x i, x i+1, i = 0, 1,, n 1. Step 2: compute Q i,i+1,,j+1 (x), polynomial of degree j i + 1, interpolating f (x) at x i, i + 1,, x j+1, from Q i,i+1,,j (x) and Q i+1,i+2,,j+1 (x), for i = 0, 1,, n 1, and j = i + 1, n.
27 Neville s Method, pseudo-code
28 Divided Differences Given n + 1 distinct points (x 0, f (x 0 )), (x 1, f (x 1 )),, (x n, f (x n )), Interpolating polynomial of degree n It follows P(x) = a 0 + a 1 (x x 0 ) + a 2 (x x 0 )(x x 1 ) + + a n (x x 0 )(x x 1 ) (x x n 1 ), with P(x 0 ) = f (x 0 ), P(x 1 ) = f (x 1 ),, P(x n ) = f (x n ). a 0 = P(x 0 ) = f (x 0 ), P(x) f (x 0 ) x x 0 = a 1 + a 2 (x x 1 ) + + a n (x x 1 ) (x x n 1 ).
29 Divided Differences Define f [x i ] = f (x i ), f [x i, x i+1 ] = f (x i+1) f (x i ) x i+1 x i, for all i Let x = x 1 in P(x) f (x 0 ) x x 0 = a 1 + a 2 (x x 1 ) + + a n (x x 1 ) (x x n 1 ). It follows that a 1 = P(x 1) f (x 0 ) = f [x 0, x 1 ] x 1 x 0 P(x) P(x 0 ) x x 0 = f [x 0, x 1 ] + a 2 (x x 1 ) + +a n (x x 1 ) (x x n 1 ) P[x 0, x] P[x 0, x 1 ] x x 1 = a a n (x x 2 ) (x x n 1 ).
30 Divided Differences recursive, f [x i, x i+1,, x j+1 ] = f [x i+1,, x j+1 ] f [x i, x i+1,, x j ] x j+1 x i, then in we have P(x) = a 0 + a 1 (x x 0 ) + a 2 (x x 0 )(x x 1 ) + + a n (x x 0 )(x x 1 ) (x x n 1 ), a 0 = f [x 0 ] a 1 = f [x 0, x 1 ] a 2 = f [x 0, x 1, x 2 ].. a n 1 = f [x 0, x 1,, x n ]
31 Newton Divided Difference Example
32 Newton Divided Difference Example
33 Coefficients are numbers on top of all f [ ] columns
34 Newton Divided Difference Theorem: Suppose that f C n [a, b] and x 0, x 1,, x n are distinct nodes in [a, b]. Then a number ξ (a, b) exists such that f [x 0, x 1,, x n ] = f (n) (ξ). n! Proof: Let g(x) = f (x) P(x), where P(x) = a 0 + a 1 (x x 0 ) + a 2 (x x 0 )(x x 1 ) + + a n (x x 0 )(x x 1 ) (x x n 1 ). Since g(x j ) = f (x j ) P(x j ) = 0, j = 0, 1,, n, it follows that a number ξ (a, b) exists such that g (n) (ξ) = 0. But g (n) (ξ) = f (n) (ξ) P (n) (ξ) = f (n) (ξ) a n n! = f (n) (ξ) f [x 0, x 1,, x n ] n! = 0.
35 Polynomial Interpolation: Experiments: Easy function: f (x) = e x on [ 1, 1]. f (n) (ξ) e for all ξ ( 1, 1). 7 nodal points (n = 6); random x points.
36 Divided difference coefficients for e x on [ 1, 1] 10 0 Divided Difference Coefficients, semilog scale
37 Polynomial Interpolation: Experiments: Hardest function: f (x) = x 2 on [ 1, 1]. f (n) (ξ) can be very large for some ξ ( 1, 1). 21 nodal points (n = 20); random x points.
38 Divided difference coefficients for 1/(0.2 + x 2 ) on [ 1, 1] 10 4 Divided Difference Coefficients, semilog scale
39 Divided difference, book version
40 Divided difference, O(n) memory
41 Interpolation via NDD
42 Forward Differences Let x 1 = x 0 + h, x 2 = x 0 + 2h f [x 0, x 1 ] = f [x 0, x 0 + h] = f (x 0 + h) f (x 0 ) h f [x 0, x 1, x 2 ] = f [x 0, x 0 + h, x 0 + 2h] = f [x 1, x 2 ] f [x 0, x 1 ] 2h = f (x 2) + f (x 0 ) 2f (x 1 ) 2h 2 (first order FD), (second order FD)
43 Forward Differences Let x 1 = x 0 + h, x 2 = x 0 + 2h f [x 0, x 1 ] = f [x 0, x 0 + h] = f (x 0 + h) f (x 0 ) h f [x 0, x 1, x 2 ] = f [x 0, x 0 + h, x 0 + 2h] = f [x 1, x 2 ] f [x 0, x 1 ] 2h = f (x 2) + f (x 0 ) 2f (x 1 ) 2h 2 (first order FD), (second order FD)
44 Backward Differences Let x n 1 = x n h, x n 2 = x n 2h f [x n 1, x n ] = f [x n h, x n ] = f (x n) f (x n h) (1st order BD), h f [x n 2, x n 1, x n ] = f [x n 2h, x n h, x n ] = f [x n h, x n ] f [x n 2h, x n h] 2h = f (x n) + f (x n 2 ) 2f (x n 1 ) 2h 2 (2nd order BD)
45 Backward Differences Let x n 1 = x n h, x n 2 = x n 2h f [x n 1, x n ] = f [x n h, x n ] = f (x n) f (x n h) (1st order BD), h f [x n 2, x n 1, x n ] = f [x n 2h, x n h, x n ] = f [x n h, x n ] f [x n 2h, x n h] 2h = f (x n) + f (x n 2 ) 2f (x n 1 ) 2h 2 (2nd order BD)
46 Double nodes (I) Given 2 distinct points (x 0, f (x 0 )), (x 1, f (x 1 )) Interpolating polynomial of degree 1 P(x) = a 0 + a 1 (x x 0 ) with P(x 0 ) = f (x 0 ), P(x 1 ) = f (x 1 ), Where a 0 = f (x 0 ), a 1 = f (x 1) f (x 0 ) x 1 x 0.
47 Double nodes (I) Given 2 distinct points (x 0, f (x 0 )), (x 1, f (x 1 )) Interpolating polynomial of degree 1 Where P(x) = a 0 + a 1 (x x 0 ) with P(x 0 ) = f (x 0 ), P(x 1 ) = f (x 1 ), a 0 = f (x 0 ), a 1 = f (x 1) f (x 0 ) x 1 x 0. Now let x 1 x 0, we obtain (Taylor expansion) P(x) = a 0 + a 1 (x x 0 ), where a 0 = f (x 0 ), a 1 = f (x 0 ) def = f [x 0, x 0 ]. P(x 0 ) = f (x 0 ), P (x 0 ) = f (x 0 ).
48 Double nodes (II) Given 3 distinct points (x 0, f (x 0 )), (x 1, f (x 1 )), (x 2, f (x 2 )), Interpolating polynomial of degree 2 P(x) = a 0 + a 1 (x x 0 ) + a 2 (x x 0 )(x x 1 ), with P(x 0 ) = f (x 0 ), P(x 1 ) = f (x 1 ), P(x 2 ) = f (x 2 ), where we have a 0 = f [x 0 ], a 1 = f [x 0, x 1 ], a 2 = f [x 0, x 1, x 2 ] = f [x 1, x 2 ] f [x 0, x 1 ] x 2 x 0.
49 Double nodes (II) Given 3 distinct points (x 0, f (x 0 )), (x 1, f (x 1 )), (x 2, f (x 2 )), Interpolating polynomial of degree 2 P(x) = a 0 + a 1 (x x 0 ) + a 2 (x x 0 )(x x 1 ), with P(x 0 ) = f (x 0 ), P(x 1 ) = f (x 1 ), P(x 2 ) = f (x 2 ), where we have a 0 = f [x 0 ], a 1 = f [x 0, x 1 ], a 2 = f [x 0, x 1, x 2 ] = f [x 1, x 2 ] f [x 0, x 1 ] x 2 x 0. Now let x 2 = x 1, we obtain (mixed interpolation) f [x 1, x 2 ] = f (x 1 ) = f [x 1, x 1 ], a 2 = f [x 1, x 1 ] f [x 0, x 1 ] x 1 x 0 def = f [x 0, x 1, x 1 ] P(x) = a 0 + a 1 (x x 0 ) + a 2 (x x 0 )(x x 1 ) P(x 0 ) = f (x 0 ), P(x 1 ) = f (x 1 ) P (x 1 ) = f (x 1 ).
50 Double nodes (III) Given 4 distinct points (x 0, f (x 0 )), (x 1, f (x 1 )), (x 2, f (x 2 )), (x 3, f (x 3 )), Interpolating polynomial of degree 3 P(x) = a 0 +a 1 (x x 0 )+a 2 (x x 0 )(x x 1 )+a 3 (x x 0 )(x x 1 )(x x 2 ). where a 0 = f [x 0 ], a 1 = f [x 0, x 1 ], It follows a 2 = f [x 0, x 1, x 2 ] = f [x 1, x 2 ] f [x 0, x 1 ] x 2 x 0 a 3 = f [x 0, x 1, x 2, x 3 ] = f [x 1, x 2, x 3 ] f [x 0, x 1, x 2 ] x 3 x 0.
51 Double nodes (IV) Let x 1 = x 0, and x 3 = x 2. It follows that a 1 = f [x 0, x 1 ] = f [x 0, x 0 ] a 2 = f [x 1, x 2 ] f [x 0, x 1 ] x 2 x 0 = f [x 0, x 2 ] f [x 0, x 0 ] x 2 x 0 a 3 = f [x 0, x 1, x 2, x 3 ] = f [x 0, x 2, x 2 ] f [x 0, x 0, x 2 ] x 2 x 0. Hermite Interpolation: P(x) = a 0 + a 1 (x x 0 ) + a 2 (x x 0 ) 2 + a 3 (x x 0 ) 2 (x x 2 ) P(x 0 ) = f (x 0 ), P (x 0 ) = f (x 0 ), P(x 2 ) = f (x 2 ), P (x 2 ) = f (x 2 ).
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