Polynomial Interpolation

Transcription

1 Polyomial Iterpolatio 1 Iterpolatig Polyomials the iterpolatio problem 2 Lagrage Iterpolatio a basis of Lagrage polyomials 3 Neville Iterpolatio the value problem Neville s algorithm a Julia fuctio MCS 471 Lecture 7(a) Numerical Aalysis Ja Verschelde, 2 July 2018 Numerical Aalysis (MCS 471) Polyomial Iterpolatio L-7(a) 2 July / 21

2 Polyomial Iterpolatio 1 Iterpolatig Polyomials the iterpolatio problem 2 Lagrage Iterpolatio a basis of Lagrage polyomials 3 Neville Iterpolatio the value problem Neville s algorithm a Julia fuctio Numerical Aalysis (MCS 471) Polyomial Iterpolatio L-7(a) 2 July / 21

3 the iterpolatio problem Ofte we have data collected from some difficult fuctio f (x). With iterpolatio we ca represet the data by a polyomial. Iput: (x i, f i = f (x i )), i = 0, 1,...,, + 1 data poits, x i x j, for all i j, distict values for x. Output: p(x) a polyomial of degree at most so that for all i = 0, 1,..., : p(x i ) = f i. The polyomial p iterpolates the fuctio f (x) at the iterpolatio poits x i, i = 0, 1,...,. We say that p is the iterpolatig polyomial for the fuctio f (x) at x i. Two questios: 1 Is there a uique solutio to the iterpolatio problem? 2 How to efficietly compute the iterpolatig polyomial? Numerical Aalysis (MCS 471) Polyomial Iterpolatio L-7(a) 2 July / 21

4 the coefficiet problem The coefficiet problem asks to compute the coefficiets of p: p(x) = c x + c 1 x c 2 x 2 + c 1 x + c 0 so that p(x i ) = f i for i = 0, 1,...,. Observe that we have + 1 data poits (x i, f i ) give o iput; ad + 1 coefficiets c i, i = 0, 1,..., to compute. Theorem If all iterpolatio poits are mutually distict: x i x j, for all i j, the the polyomial iterpolatio problem has a uique solutio. We prove this by settig up the iterpolatio coditios. Numerical Aalysis (MCS 471) Polyomial Iterpolatio L-7(a) 2 July / 21

5 the iterpolatio coditios For p(x) = c x + + c 2 x 2 + c 1 x + c 0 the coditios p(x i ) = f i, for i = 0, 1,..., lead to a liear system of + 1 equatios: c x 0 + c 1x c 2 x c 1x 0 + c 0 = f 0 c x 1 + c 1x c 2 x c 1x 1 + c 0 = f 1 c x 2 + c 1x c 2 x c 1x 2 + c 0 = f 2. c x 1 + c 1x c 2x c 1x 1 + c 0 = f 1 c x + c 1 x c 2 x 2 + c 1 x + c 0 = f The ukows are the coefficiets c i of the polyomial p. Numerical Aalysis (MCS 471) Polyomial Iterpolatio L-7(a) 2 July / 21

6 the liear system i matrix otatio c x0 + c 1x c 2 x0 2 + c 1x 0 + c 0 = f 0 c x1 + c 1x c 2 x1 2 + c 1x 1 + c 0 = f 1 c x2 + c 1x c 2 x2 2 + c 1x 2 + c 0 = f 2. c x 1 + c 1x c 2x c 1x 1 + c 0 = f 1 c x + c 1 x c 2 x 2 + c 1 x + c 0 = f 1 x 0 x0 2 x 1 0 x0 1 x 1 x1 2 x 1 1 x1 1 x 2 x2 2 x 1 2 x x 1 x 2 1 x x x 2 x 1 x 1 x c 0 c 1 c 2. c 1 c = f 0 f 1 f 2. f 1 f Numerical Aalysis (MCS 471) Polyomial Iterpolatio L-7(a) 2 July / 21

7 the Vadermode matrix The liear system has a uique solutio the determiat of V (x 0, x 1, x 2,..., x 1, x ) = 1 x 0 x0 2 x 1 0 x0 1 x 1 x1 2 x 1 1 x1 1 x 2 x2 2 x 1 2 x x 1 x 2 1 x x x 2 x 1 x 1 x is differet from zero. Defiitio The matrix V (x 0, x 1, x 2,..., x 1, x ) is the Vadermode matrix for the poits x 0, x 1, x 2,..., x 1, x. Numerical Aalysis (MCS 471) Polyomial Iterpolatio L-7(a) 2 July / 21

8 there is a uique solutio Let V = V (x 0, x 1, x 2,..., x 1, x ), the determiat of V is ( ) det(v ) = xi x j. i=0 j=i+1 Example for = 3: det(v ) = (x 0 x 1 )(x 0 x 2 )(x 0 x 3 )(x 1 x 2 )(x 1 x 3 )(x 2 x 3 ). Observe deg(det(v )) = ( + 1)/2 ad det(v ) 0 if x i x j for i j. Theorem The solutio to the iterpolatio problem is uique if x i x j for i j. Numerical Aalysis (MCS 471) Polyomial Iterpolatio L-7(a) 2 July / 21

9 Polyomial Iterpolatio 1 Iterpolatig Polyomials the iterpolatio problem 2 Lagrage Iterpolatio a basis of Lagrage polyomials 3 Neville Iterpolatio the value problem Neville s algorithm a Julia fuctio Numerical Aalysis (MCS 471) Polyomial Iterpolatio L-7(a) 2 July / 21

10 Lagrage iterpolatio Give are + 1 iterpolatio poits (x i, f i ), i = 0, 1,...,, where for all i j: x i x j. The Lagrage iterpolatig polyomial has the form where p(x) = l 0 (x)f 0 + l 1 (x)f (x 1 ) + + l (x)f, l i (x j ) = { 1 if i = j 0 if i j. I this form, we have that p(x i ) = f i, i = 0, 1,...,. Defiitio For + 1 mutually distict poits x i, i = 0, 1,...,, ( x xj the i-th Lagrage polyomial is l i (x) = x i x j j = 0 j i ). Numerical Aalysis (MCS 471) Polyomial Iterpolatio L-7(a) 2 July / 21

11 the Lagrage iterpolatig polyomial The i-th Lagrage polyomial l i (x) is l i (x) = j = 0 j i ( x xj x i x j Example: = 3, i = 1: l 1 (x 0 ) = 0, l 1 (x 1 ) = 1, l 1 (x 2 ) = 0, l 1 (x 3 ) = 0. l 1 (x) = (x x 0)(x x 2 )(x x 3 ) (x 1 x 0 )(x 1 x 2 )(x 1 x 3 ). The solutio of the iterpolatio problem is uique as well, so the form for l 1 (x) is uique. The Lagrage iterpolatig polyomial is p(x) = i=0 j = 0 j i ( x xj x i x j ) f i. This is coveiet if oly the f i s chage while the x i s stays the same. ). Numerical Aalysis (MCS 471) Polyomial Iterpolatio L-7(a) 2 July / 21

12 a exercise Exercise 1: Show that l i (x) = 1. i=0 Cosider the iterpolatio data (x i, f i = 1), i = 0, 1,..., ad remember that the iterpolatio problem has a uique solutio. Numerical Aalysis (MCS 471) Polyomial Iterpolatio L-7(a) 2 July / 21

13 Polyomial Iterpolatio 1 Iterpolatig Polyomials the iterpolatio problem 2 Lagrage Iterpolatio a basis of Lagrage polyomials 3 Neville Iterpolatio the value problem Neville s algorithm a Julia fuctio Numerical Aalysis (MCS 471) Polyomial Iterpolatio L-7(a) 2 July / 21

14 the value problem Iput: (x i, f i = f (x i )), i = 0, 1,...,, + 1 data poits, x i x j, for all i j, distict values for x, x is the value for some x. Output: p(x ) a the value of the iterpolatig polyomial at x. Theorem Let p i = f i, for i = 1, 2,..., ad ( x ) ( x x ) x 0 p 0,1,..., = p 0,..., 1 + p 1,..., x 0 x x x 0 the p 0,1,..., = p(x ). Numerical Aalysis (MCS 471) Polyomial Iterpolatio L-7(a) 2 July / 21

15 Neville iterpolatio Let p i = f i, for i = 0, 1,..., ad ( ) ( ) x x x x0 p 0,1,..., (x) = p 0,..., 1 (x) + p 1,..., (x) x 0 x x x 0 the p 0,1,..., (x i ) = f i, i = 0, 1,...,. ( ) x0 x p 0,1,..., (x 0 ) = p 0,..., = f 0, aaloguous for x. x 0 x For x k, for k > 1 ad k < : ( ) ( ) xk x xk x 0 p 0,1,..., (x k ) = p 0,..., 1 (x k ) + p 1,..., (x k ) x 0 x }{{} x x 0 }{{} f k f k = x k x (x k x 0 ) f k x 0 x = f k Numerical Aalysis (MCS 471) Polyomial Iterpolatio L-7(a) 2 July / 21

16 Polyomial Iterpolatio 1 Iterpolatig Polyomials the iterpolatio problem 2 Lagrage Iterpolatio a basis of Lagrage polyomials 3 Neville Iterpolatio the value problem Neville s algorithm a Julia fuctio Numerical Aalysis (MCS 471) Polyomial Iterpolatio L-7(a) 2 July / 21

17 derivatio of Neville s algorithm For iterpolatio data (x 0, f 0 ), (x 1, f 1 ),..., (x, f ) ad a value x : p i = f i, for i = 0, 1,...,, p i,...,j = p(x ) is the value of the iterpolatig polyomial at x through (x i, f i ), (x i+1, f i+1 ),..., (x j, f j ). The values p i,...,j ca be orgaized i a triagular table. For example, for = 3: x 0 f 0 x 1 f 1 p 0,1 x 2 f 2 p 1,2 p 0,1,2 x 3 f 3 p 2,3 p 1,2,3 p 0,1,2,3 Observe that we may replace f 0 by p 0,1, f 1 by p 1,2, ad f 2 by p 2,3. Numerical Aalysis (MCS 471) Polyomial Iterpolatio L-7(a) 2 July / 21

18 Neville s algorithm For iterpolatio data (x 0, f 0 ), (x 1, f 1 ),..., (x, f ) ad a value x : p i = f i, for i = 0, 1,...,, p i,...,j = p(x ) is the value of the iterpolatig polyomial at x through (x i, f i ), (x i+1, f i+1 ),..., (x j, f j ). for i = 1, 2,..., do for j = 1, 2,..., i do p i j,...,i = (x x i )p i j,...,i 1 (x x i j )p i j+1,...,i x i j x i For efficiet memory usage, relabel p i j,...,i as p[i j], p i j,...,i 1 as p[i j] ad p i j+1,...,i as p[i j + 1]. The cost of Neville s algorithm is O( 2 ). Numerical Aalysis (MCS 471) Polyomial Iterpolatio L-7(a) 2 July / 21

19 Polyomial Iterpolatio 1 Iterpolatig Polyomials the iterpolatio problem 2 Lagrage Iterpolatio a basis of Lagrage polyomials 3 Neville Iterpolatio the value problem Neville s algorithm a Julia fuctio Numerical Aalysis (MCS 471) Polyomial Iterpolatio L-7(a) 2 July / 21

20 a Julia fuctio fuctio eville(x::array{float64,1},f::array{float64,1},xx::float64) # # implemets the iterpolatio algorithm of Neville # # ON ENTRY : # x abscisses, give as a colum vector; # f ordiates, give as a colum vector; # xx poit where to evaluate the iterpolatig # polyomial through (x[i],f[i]). # # ON RETURN : # p last row of Neville s table where p[1] is # the value of the iterpolator at xx. # # EXAMPLE : # x = [32.0, 22.2, 41.6, 10.1, 50.5] # f = [ , , , , ] # xx = 27.5 # p = eville(x,f,xx) # Numerical Aalysis (MCS 471) Polyomial Iterpolatio L-7(a) 2 July / 21

21 Neville s algorithm = legth(x) p = f dx = [0.0 for i=1:] for i=1: dx[i] = xx - x[i] ed for i=2: for j=2:i p[i-j+1] = (dx[i]*p[i-j+1] - dx[i-j+1]*p[i-j+2])/(x[i-j+1]-x[i]) ed ed retur p Numerical Aalysis (MCS 471) Polyomial Iterpolatio L-7(a) 2 July / 21