1. 數值方法 2008 Applied Mathematics, NDHU1  Lagrange polynomial  Polynomial interpolation Lecture 4II

2. 數值方法 2008 Applied Mathematics, NDHU2 poly Find the polynomial defined by given zeros poly returns coefficients of a polynomial with roots –5,0 and 5 poly([-5 0 5])

3. 數值方法 2008 Applied Mathematics, NDHU3 polyval Polynomial evaluation y stores a result of substituting elements in v to a polynomial defined by roots: 5, 0 and -5 v=linspace(-5,5); p= poly([-5 0 5]); y=polyval(p,v);

4. p 3 (x)=(x-5)(x+5)(x-0) 數值方法 2008 Applied Mathematics, NDHU4 x=[4 1 3 -4];y=polyval(p,x);hold on;plot(x,y,'ro') Given data, find an interpolating polynomial

5. Polynomial Interpolation Paired data Each pair (x i,y i ) specifies a point on a target polynomial y i = f(x i ) for all i Given paired data, find an interpolating polynomial 數值方法 2008 Applied Mathematics, NDHU5

6. Strategy Use a linear combination of Lagrange polynomials determined by all x i to express the interpolating polynomial 數值方法 2008 Applied Mathematics, NDHU6

7. 7 Lagrange polynomial n knots, n Lagrange polynomials

8. 數值方法 2008 Applied Mathematics, NDHU8 Mathematical expression

9. 數值方法 2008 Applied Mathematics, NDHU9 Proof

10. 數值方法 2008 Applied Mathematics, NDHU10 Product form

11. 數值方法 2008 Applied Mathematics, NDHU11 Lagrange polynomial Given n knots, Let L i denote the ith Lagrange polynomial L i is a polynomial of degree n-1 L i has n-1 roots: Normalization: L i satisfies L i (x i )=1

12. 數值方法 2008 Applied Mathematics, NDHU12 Implementation: n-1 roots x is a vector that consists of n distinct knots pi is a polynomial whose roots are all knots except for x i xzeros=[x(1:i-1) x(i+1:n)]; pi=poly(xzeros);

13. 數值方法 2008 Applied Mathematics, NDHU13 Implementation: Normalization c=polyval(pi,x(i)); pi=pi/c; Normalization condition

14. The ith Lagrange polynomial xzeros=[x(1:i-1) x(i+1:n)]; p_i=poly(xzeros); c=polyval(p_i,x(i)); p_i=p_i/c; STRATEGY I : use matlab functions to determine L i 數值方法 2008 Applied Mathematics, NDHU14

15. x_knot consists of n knots x_in collects input values Evaluate the ith Lagrange polynomial L i defined by knots in x_knot y= L i (x_in) y=lagrange_poly(x_in,x_knot,i) 數值方法 2008 Applied Mathematics, NDHU15 Evaluation of the ith Lagrange polynomial

16. 數值方法 2008 Applied Mathematics, NDHU16 STRATEGY II : Product form evaluation T=length(x_in); n=length(x_knot); y=ones(1,T); for j=1:n j~=i a=(x_in-x_knot(j))/(x_knot(i)-x_knot(j)); y=y.*a T EXIT

17. 數值方法 2008 Applied Mathematics, NDHU17 function y=lagrange_poly(x_in,x_knot,i) T=length(x_in); y=ones(1,T); n=length(x_knot); for j=1:n if j~=i a=(x_in-x_knot(j))/(x_knot(i)-x_knot(j)); y=y.*a; end return Direct Implementation of the product form

18. x_in=linspace(-2,3); y=lagrange_poly(x_in,[-1 0 1 2],1); plot(x_in,y); 數值方法 2008 Applied Mathematics, NDHU18 n=4,i=1 (strategy II)

19. 數值方法 2008 Applied Mathematics, NDHU19 n=4,i=1 (strategy I) x=[-1 0 1 2];i=1;n=length(x); xzeros=[x(1:i-1) x(i+1:n)]; p_i=poly(xzeros); p_i=pi/polyval(p_i,x(i)); x_in=linspace(-2,3); plot(x_in,polyval(p_i,x_in),'r');

20. x_in=linspace(-2,3); y=lagrange_poly(x_in,[-1 0 1 2],2); plot(v,y); 數值方法 2008 Applied Mathematics, NDHU20 n=4,i=2 (strategy II)

21. 數值方法 2008 Applied Mathematics, NDHU21 n=4,i=2 (strategy I) x=[-1 0 1 2];i=2;n=length(x); xzeros=[x(1:i-1) x(i+1:n)]; p_i=poly(xzeros); p_i=p_i/polyval(p_i,x(i)); x_in=linspace(-2,3); plot(v,polyval(p_i,x_in),'r');

22. Polynomial Interpolation Paired data Each pair (x i,y i ) specifies a point on a target polynomial y i = f(x i ) for all i Given paired data, find an interpolating polynomial 數值方法 2008 Applied Mathematics, NDHU22

23. 數值方法 2008 Applied Mathematics, NDHU23 Input x y A sample from an unknown function

24. 數值方法 2008 Applied Mathematics, NDHU24 Polynomial interpolation Given Find a polynomial that satisfies for all i

25. Interpolating polynomial is with degree n- 1 數值方法 2008 Applied Mathematics, NDHU25 Interpolating polynomial

26. 數值方法 2008 Applied Mathematics, NDHU26 Verification

27. 數值方法 2008 Applied Mathematics, NDHU27 function y_out=int_poly(x_in,x,y) for i=1:n y_out=zeros(size(x_in)); n=length(x); y_out=y_out+lagrange_poly(x_in,x,i).*y(i); EXIT

28. 數值方法 2008 Applied Mathematics, NDHU28 function y_out=int_poly(x_in,x,y) y_out=zeros(size(x_in)); n=length(x); for i=1:n y_out=y_out+lagrange_poly(x_in,x,i).*y(i); end return