3.2 - Interpolation and Lagrange Polynomials

Transcription

1 3. - Interpolation and Lagrange Polynomials. Polynomial Interpolation: Problem: Givenn pairs of data points x i, y i,wherey i fx i, i 0,,...,n for some function fx, we want to find a polynomial P x of lowest possible degree for which P x i y i, i 0,,...,n (P x agrees with fx at x 0,...,x n ). The polynomial P x is said to interpolate the data x i, y i, i 0,,...,n and is called an interpolating polynomial. Graphically, P x is an approximation to fx and satisfies the conditions: P x i fx i, i 0,,...,n. For example, y x -0 y fx, ---y P x Obviously, for this example P x is not a good approximation to fx though P x satisfies the conditions: P x i y i for i 0,,. How to form P x? There are several ways to form P x. Let us start with a way we now already. Let P x a 0 a x a x...a x. We want to find coefficients a 0, a,..., a such that the following n equations hold: P x i a 0 a x a x...a x y i for i 0,,...,n. Rewrite these n equations in a matrix-vector form: (*) x 0 x 0 x 0 x x x x n x n x n a 0 a a a y 0 y y n. The vector a 0 a a a is a solution of the system (*) of n linear equations in unnowns.

2 Consider the case where n. Recall in Linear Algebra, we learned that this system has a unique solution if the coefficient matrix form: a 0 a a a x 0 x 0 x 0 x x x x n x n x n x 0 x 0 x 0 x x x x n x n x n is nonsingular and the solution vector is of the and can be solved by Gaussian-Elimination and Bacward-Substution. Example Let x 0 0, x 0., x 0.9, fx cosx. Find P x a 0 a x a x. Setch both y fx and y P x for x in 0,. y 0 y y n Solve a 0 a a from the following system: a 0 a cos a cos0.8 a 0 a cos a cos Hence, P x x x. y y fx, ---y P x In MatLab, we can complete this example as follows. xv[0;0.;0.9]; A[ones(3,) xv xv.^]; x

3 yvcos(xv.^); coeffvinv(a)*yv coeffv Now we setch the graphs of y cosx and y P x x x for x in 0,. xvnew0:0.0:; ycos(xvnew.^); ycoeffv()coeffv()*xvnewcoeffv(3)*xvnew.^; plot(xvnew,y, r-,xvnew,y, b-. );.3 red - y=cos(x ), blue -. y=p (x) If we want to add a pair of data 0.3,cos0.09, then we can go over the following steps. xv[0;0.3;0.;0.9]; A[ones(4,) xv xv.^ xv.^3]; yvcos(xv.^); coeffvinv(a)*yv coeffv Hence, P 3 x x x x 3.Nowwe setch the graphs of y cosx and y P 3 x as follows. Remember that xvnew and y are still there. We need to recompute y. ycoeffv()coeffv()*xvnewcoeffv(3)*xvnew.^coeffv(4)*xvnew.^3; plot(xvnew,y, r-,xvnew,y, b-. ); title( red - ycos(x^), blue -. yp_3(x) ) 3

4 red - y=cos(x ), blue -. y=p 3 (x) Clearly, P 3 x fits y cosx much better than P x does. Note that the system (*) of linear equations are getting more difficult to solve when n is large. Can P x be formed with less computation?. Lagrange Interpolating Polynomials: a. Lagrange Polynomials: For 0,,...,n, define L n, x n i0, i x x i x x i x x 0...x x x x...x x n x x 0...x x x x...x x n L n, x are called Lagrange polynomials. For example, let x i i, i 0,,,3. Then x 0 0, x, x andx 3 3, and x 0x x 3 L 3, x 0 3 xx x 3 L 3, x xx x xx x 3 Observe that Lagrange Polynomials have the following properties: i. L n, x L n, x x x 0...x x x x...x x n x x 0...x x x x...x x n ii. L n, x i 0fori. L n, x i x i x 0...x i x i...x i x x i x...x i x n x x 0...x x i...x x x x...x x n For example, L 3, andl 3, b. Lagrange Interpolating Polynomials: The polynomial 4

5 n P n x y L n, x y 0 L n,0 x y L n, x...y n L n,n x 0 is called the nth Lagrange interpolating polynomial. Observe that for i 0,,...,n n P n x i y L n, x i y i L n,i x i y i. 0 P n x is an nth polynomial that agrees with fx at x 0, x,..., x n. Theorem Suppose that x 0,..., x n are distinct numbers in the interval a, b and f n is continuous in a, b. For each x in a, b, there exists a number cx in a, b such that fx P n x fn cx x x 0 x x...x x n. n! Proof For x x i and x in n a, b, define ht ft P n t fx P n x 0 t x x x. Clearly, hx i 0, i 0,,...,n. Observe that hx 0. Since x x i, ht 0att x 0, x,..., x n, x. Since ht has n distinct zeros in a, b, by Rolle s Theorem, we now there exists a constant c in a, b such that h n c 0. h n t f n n! t 0 fx P n x x x 0 x x x x n h n c f n n! c 0 fx P n x x x 0 x x...x x n 0 R n x fx P n x fn c n! x x 0x x x x n. Recall in Calculus II, we studied nth degree Taylor polynomial at x c : and its error P Taylor x fc f c! x c f c! x c fn c n! x c n R Taylor x fn n! x cn where is between x and c. Note that the differences between a th degree Taylor polynomial and a th degree interpolating polynomial are: a. P Taylor x fx at only x x 0 and P interpolating x fx, atx x 0, x,...,x n. b. P Taylor requires nowledge of f, f,... but P interpolating requires fx 0, fx,...,fx n.recall in Calculus II, we studied nth degree Taylor polynomial at x c : and its error P Taylor x fc f c! x c f c! x c fn c n! x c n R Taylor x fn n! x cn where is between x and c. Note that the differences between a th degree Taylor polynomial and a th degree interpolating 5

6 polynomial are: a. P Taylor x fx at only x x 0 and P interpolating x fx, atx x 0, x,...,x n. b. P Taylor requires nowledge of f, f,... but P interpolating requires fx 0, fx,...,fx n. Example Let x 0 0, x 0., x 0.9, fx cosx. () Find the Lagrange interpolating polynomial P x and R x. () Use P x to approximate f0.45 and estimate the approximation error. (3) Approximate 0 cosx dx by 0 P xdx. () P x f0 x 0.x f0. xx f0.9 xx P x cos0.3 x 0.x 0.9 xx 0.9 cos0.8 xx y x () f x xsinx, y cosx, ---y P x f x sinx x cosx, f x xcosx 4xcosx 4x 3 sinx 43xcosx x 3 sinx where c is in 0,. R x 43ccosc c 3 sinc xx 0.x 0.9 f0.45 P 0.45 cos cos True error cos 0.45 P Find the approximation error: f x max x in 0, 43xcosx x 3 sinx 43 0

7 or chec out the graph of f x to have a better estimate of an upper bound of f x : y f x.5, for x in 0, R x (3) y 43xcosx x 3 sinx 0 P xdx 0 cos0.3 x 0.x 0.9 xx 0.9 cos0.8 xx 0. dx x x 0. dx cos0.3 x xdx 0 cos0.8 x xdx 0 cos cos Neville s Iterated Interpolation: If we need to add more points x n,... fromp n x to construct an interpolation polynomial with larger degree, can we construct an interpolation polynomial iteratively? The answer is yes. The method is called the Neville Method. Let m, m,..., m be distinct integers where 0 m i n for each i. LetP m,m...,m x be the Lagrange polynomial that agrees with f at the points: x m, x m,..., x m. For example, P,,4 x fx x x x x 4 x x x x 4 fx x x x x 4 x x x x 4 fx 4 x x x x x 4 x x 4 x is the Lagrange polynomial that agrees with fx at 3 points: x, x, x 4 : P,,4 x fx, P,,4 x fx, P,,4 x 3 fx 3. Theorem Letfbedefinedatx 0, x,..., x, and x j and x be two distinct numbers. Then P 0,,..., x x x jp 0,,...,j,j,..., x x x i P 0,,...,i,i,..., x x i x j. Proof We now that both polynomials P 0,,...,j,j,..., x and P 0,,...,i,i,..., x are degree. thedegreeofp 0,,..., x is. Chec if P 0,,..., x l fx l for l 0,,..., : since P 0,,...,j,j,..., x l fx l if l j and P 0,,...,i,i,..., x l fx l if l i For l 0,,...,, but l i and l j, So, 7

8 For l i : For l j : P 0,,..., x l x l x j P 0,,...,j,j,..., x l x l x i P 0,,...,i,i,..., x l x i x j x l x j fx l x l x i fx l x i x j x i x j x i x j fx l fx l P 0,,..., x i x i x j P 0,,...,j,j,..., x i x i x i P 0,,...,i,i,..., x i x i x j P 0,,...,j,j,..., x i fx i P 0,,..., x i x j x j P 0,,...,j,j,..., x j x j x i P 0,,...,i,i,..., x j x i x j P 0,,...,i,i,..., x j fx j So, P 0,,..., x the th degree interpolating polynomial. Example Consider x 0, x,..., x n. Then we can form P, x fx x x x x fx x x x x and P,4x fx x x 4 x x 4 fx 4 x x 4 x x 4 using x, x and x 4. Then we can combine P, x and P,4 x to form P,,4 x: P,,4 x x x 4P, x x x P,4 x x x 4 which is the Lagrange polynomial that agrees with fx at 3 points: x, x, x 4. Now given 0,,, e,, e 4, and 3, e 9, find P 0,,,3 x using P 0,, x and P,,3 x. P 0,, x x x 0 0 e x 0x 0 e 4 x 0x 0 x x e xx e 4 xx P,,3 x e x x 3 3 x x 3 e 4 3 x x e e x x 3 e 4 x x 3 e 9 x x P 0,,,3 x x 0P,,3x x 3P 0,, x 3 0 xp,,3x x 3P 0,, x Algorithm Neville s Iterated Interpolation: Given x i, fx i for i 0,,..., and, let P i fx i Steps of Computing P 0,,,..., x for a given x : (neville.m) 8

9 x i P i x P i,i x P i,i,i x P i,i,i,i3 x P i,i,i,i3,i4 x x 0 P 0 x P P 0, x x P P, x P 0,, x x 3 P 3 P,3 x P,,3 x P 0,,,3 x x 4 P 4 P 3,4 x P,3,4 x P,,3,4 x P 0,,,3,4 x The algorithm stops and fx P 0,,,..., x whenever P 0,,,..., x P 0,,,..., x. An alternative way: x i P i x P i,i x P i,i,i x P i,i,i,i3 x P i,i,i,i3,i4 x x 0 P 0 x P P 0, x x P P 0, x P 0,, x x 3 P 3 P 0,3 x P 0,,3 x P 0,,,3 x x 4 P 4 P 0,4 x P 0,3,4 x P 0,,3,4 x P 0,,,3,4 x Example Suppose that x j j for j 0,,,3 and it is nown that P 0, x x, P, x 3x, and P,, Find P,3.5, P 0,, x and P 0,,,3,.5. P 0,, x x x P 0, x x x 0 P, x x 0 x x x x3x x i P i x P i,i x P i,i,i x P i,i,i,i3 x x 0 P 0 P P 0,.5.5 P P, P 0,, P 3 P,3.5 P,,3.5 4 P 0,,, P,,3 x x x P,3 x x x 3 P, x x 3 x P,3.5 x P,3x x 33x P 0,,.5.5 P 0,.5.5 0P,

10 P 0,,, P 0,,.5.5 0P,, Example Neville s method is used to approximate f0.4 as follows. Complete the table. x i P i x P i,i x P i,i,i x P i,i,i,i3 x P 0, P P, 0.4 P 0,, P, P,, P 0,,, P, P P P, P P P 0,, P 0, P, Chec: P 0,,, , P Example Let fx sinlnx, x 0.0, x.4, x.. () Estimate the approximation error R x of an interpolating polynomial P x over.0,.. () () Approximate f. and f.5, and estimate their errors. R x f c 3! x x.4x.,wherec is in,. f x coslnx x, f x sinlnx x x coslnx x f x coslnx x sinlnx x x xsinlnx coslnx coslnx sinlnx sinlnx coslnx x 3 x 4 sinlnx coslnx x 3sinlnx coslnx x 3 0

11 y y x f x R x f c 3! f x f y x x.4x. 3sinln cosln x x.4x () xv[;.4;.]; yvsin(log(xv)); [yout,yall]neville(xv,yv,.,) yout yall [yout,yall]neville(xv,yv,.5,) yout yall true error: sinlog true error: sinlog R. f c 3! 0.33 R.5 f c 3! f c f c

12 Exercises:. Givenx i,y i 0,,,,, where y i fx i for some fx, construct the polynomial P x that agrees with fx at x 0, x and x in the following two ways. () P x a 0 a x a x a standard form. () P x is a Lagrange interpolating polynomial.. Let x 0, x 0, x, and fx e x. () Find algebraically (eep e in P x) the Lagrange interpolating polynomial P x and its approximation error R x. () Use P x to approximate f and f. 3 (3) Compute the true approximation errors f P and f P 3. 3 (4) Estimate the approximation errors using R and R. 3 (5) Approximate e x dx by P xdx and find the true approximation error e x dx P xdx Let x 0 0, x 0., x 0.9, and fx x. () Find the Lagrange interpolating polynomial P x and its approximation error R x. () Use P x to approximate f0.45 and R 0.45 to estimate the approximation error. 4. Use Neville s method to approximate f0.5, giving the following table. Determine P f0.7. i x i P i P i,i P i,i,i Suppose we now that f0, f , f , f Use given data to approximate f0.43 by a Lagrange interpolating polynomial using Neville s method (MatLab program neville.m).. A census of the population of the US is taen every 0 years. The population, in thousands, from 940 to 000 is given below. year t population Use Neville s method (MatLab program neville.m) to estimate the populations in 975 and 005 and predict the population in Let fx x. Approximate f0.5 by P0.5 where Px is a Lagrangian interpolating polynomial. Suppose that Neville s Method is used to generate P0.5 iteratively as follows.

13 P P 0, 0.5 P 0,, 0.5 P 0,,,3 0.5 P 0,,,3,4 0.5 P 0,,,3,4,5 0.5 P 0,,,3,4,5, Let the stopping criterion be set as P P a. Determine the degree () of the polynomial used to approximate with b. If P 0,,,3,4,5, 0.5 is chosen from above sequence to approximate, what is? 3