Interpolation. TANA09 Lecture 7. Error analysis for linear interpolation. Linear Interpolation. Suppose we have a table x x 1 x 2...

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1 TANA9 Lecture 7 Interpolation Suppose we have a table x x x... x n+ Interpolation Introduction. Polynomials. Error estimates. Runge s phenomena. Application - Equation solving. Spline functions and interpolation. End point conditions. Error estimate. f(x) f f... f n+ How to estimate f(x) för x x x n+? Definition A polynomial p(x) interpolates a function f(x) at x, x,..., x n+, if p(x i ) = f(x i ), i =,,..., n+. Questions Degree of the polynomial? How to compute the polynomial? Error estimate? 4 december 8 Sida / 3 4 december 8 Sida / 3 Linear Interpolation Error analysis for linear interpolation f(x) f f p (x) Lemma Let the function values f and f have errors f i ε. If linear interpolation is used we have the error estimate R XF ε. x x Theorem The linear polynomial p (x) that interpolates two points (x, f ) and (x, f ) is given by p (x) = f + x x x x (f f ). Theorem Let p (x) be the linear polynomial that interpolates f(x) at x and x. Then R T = f(x) p (x)= f (ξ) (x x )(x x ), x <ξ<x. or R T Ch, where h=x x. 4 december 8 Sida 3 / 3 4 december 8 Sida 4 / 3

2 Polynomial interpolation Theorem For n+ points (x i, f i ) there is a unique polynomial p n (x), of degree n, that interpolates the given points, i.e. p n (x i ) = f i, i=,,..., n+. Newton s interpolation formula Find a polynomial p n (x) that interpolates the table x x x... x n+ f(x) f f... f n+ The ansatz Theorem Let p n (x) be the polynomial, of degree n, that interpolates f(x) in the points x, x,..., x n+. Then f(x) p n (x) = f(n+) (ξ(x)) (x x ) (x x n+ ). (n+)! The function f(x) has to have continuous derivatives. How to organize the computations? 4 december 8 Sida 5 / 3 p n (x)=c +c (x x )+c (x x )(x x )+...+c n (x x ) (x x n ), leads to simple calculations. The interpolation conditions give p n (x ) = c = f, and p n (x ) = c + c (x x ) = f,= c = (f c )/(x x ). Every new condition p n (x i ) = f i gives one new c i. 4 december 8 Sida 6 / 3 Error estimate Remark In practice f (ξ) is unknown. Instead use an additional point (x n+, f n+ ) and compute a new interpolating polynomial p n+ (x). Approximate f (n+) (ξ(x)) p (n+) n+ (ξ(x)) = n!c n+. Lemma Let p n (x) be a polynomial interpolating f(x) at n+ points. We can estimate the error f(x) p n (x) = R T p n+ (x) p n (x). Example Let f(x) = e x/ cos(x/7)+(x.) /, and suppose we have a table x..5.7 f(x) Make use of the table to estimate f(x), for x =.35, by linear interpolation and estimate the error. Question How to do this using Matlab? What if we want to use quadratic interpolation? Remark If Newtons formula is used it is simple to carry out the calculations. 4 december 8 Sida 7 / 3 4 december 8 Sida 8 / 3

3 Find polynomials p (x) and p (x) by >> x=[.5.7];, y=[ ]; >> p = polyfit( x(:), y(:), ); >> p = polyfit( x(:3), y(:3), ); Quadratic interpolation require 3 points for p (x) and one more for estimating R T. Comnpute the polynomials on a dense grid and plot by >> xx=-.:.:.8; y=polyval(p,xx); plot(xx,y)..5.8 x Left p (x), p (x) and f(x). Right Error f(x) p (x) and R T p (x) p (x). Here R T.38. Left p (x), p 3 (x) and f(x). Right Error f(x) p (x) and R T p 3 (x) p (x). Here R T. 4. What if we increse the degree of the polynomial? 4 december 8 Sida 9 / 3 4 december 8 Sida / 3 Runge s phenomena Application - Equation solving.5 Problem Find the root x to the equation f(x) =. In an iterative method we have a sequence x, x, x,...,x n and want the next iterate x n+. How to compute?.5 Method Find a quadratic polynomial p(x) interpolating the table x x n x n x n f(x) f n f n f n Polynomials of degree n = 4, 6, and that interpolates f(x) = /(+x ). The maximnum error grows as the degree increses. In practice only low degree polynomials are used. Compute the two roots x i, i =, of p(x). Pick x n+ as the root with the smallest function value f(x i ). 4 december 8 Sida / 3 4 december 8 Sida / 3

4 In Matlab x=[ ];fvals=f(x); while abs(fvals(end))>^-5 p=polyfit(x(end-:end),fvals(end-:end),); r=roots(p);fr=f(r); if abs(fr())<abs(fr()) x=[x,r()];fvals=[fvals,fr()]; else x=[x,r()];fvals=[fvals,fr()]; end fprintf(, %6.4f %8.e\n, x(end),fvals(end)) end Remark A quadratic polynomial has two roots. Two function evaluations in each step. Very fast convergence but a step can fail. 4 december 8 Sida 3 / 3 Example Solve f(x) = cos(x/)+ e x/5 x/ 4x =. k x k f(x k ) Remark Looks like quadratic convergence! How to avoid two function evaluations? 4 december 8 Sida 4 / 3 Method Let y i = f(x i ) and x i = f (y i ). Find a polynomial quadratic polynomial p(y) interpolating Pick x n+ = p(). In Matlab y f n f n f n f (y) x n x n x n x=[ ];fvals=f(x); while abs(fvals(end))>^-5 p=polyfit(fvals(end-:end),x(end-:end),); x=[x,polyval(p,)];fvals=[fvals,f(x(end))]; fprintf(, %6.4f %8.e\n, x(end),fvals(end)) end Example Solve f(x) = cos(x/)+ e x/5 x/ 4x = using Inverse quadratic interpolation k x k f(x k ) Remark One function evaluation in each step. The interpolation step doesn t always work. This is the main method forfzero. 4 december 8 Sida 5 / 3 Remark Still very fast convergence! 4 december 8 Sida 6 / 3

5 Lagrange Interpolation Lemma The unique polynomial of degree n that interpolates f(x) in the points x j, i =,,..., n is Definition Let x j, j =,..., n be distinct points. The Lagrange basis polynomial is l i (x) = Π j i x x j x i x j. p(x) = n f(x i )l i (x). i= Example Find the unique polynomial interpolating the table Remark We have l i (x i ) = and l i (x j ) = for i j. x i f i december 8 Sida 7 / 3 4 december 8 Sida 8 / Lemma Let p(x) be an interpolating polynomial of degree n. If the approximate function values f i are used then p(x) n l i (x) f i, i= wherel i (x) are the Lagrange basis polynomials Results The Lagrange basis polynomials of degree 3 and the interpolating polynomial. The error iso(h 4 ). Question How to generalize to D? Errors in the used function values? Example Let p(x) interpolate the table x i f i Estimate the maximum error due to f i being correctly rounded. 4 december 8 Sida 9 / 3 4 december 8 Sida / 3

6 Example We have m = 6 points (x i, y j ) and want to find the interpolating polynomial u I (x, y). Introduce basis functions l i such that l i (x j, y j ) = if i = j and zero otherwise. Then the interpolant is u I (x) = m u(x i, x i )l i (x). i= The error due to f i.5. Estimated using the Lagrange interpolation formula. Remark The error is exactly f i at the interpolation points x i. 4 december 8 Sida / 3 4 december 8 Sida / Two basis functions l (x, y) and l 3 (x, y). We use m = 6 interpolation points and basis functions, x, y, xy, x, y. The interpolant u I (x, y) and the error u I u, where u = 3+x y+cos(x)y was interpolated using m = 6 points. Remark Really simple and flexible method. The error terms are obtained by comparing with a Taylor series expansion. 4 december 8 Sida 3 / 3 4 december 8 Sida 4 / 3

7 Spline interpolation Linear splines Problem A function f(x) is known at certain nodes x, x,...,x n. How can we find an approximation s(x) f(x) on [x, x n ]? Solution Use linear interpolation on each subinterval [x i, x i+ ]. f f(x) f f 3 f 5 s(x) f 4 Definition A function s(x) is an interpolating linear spline with nodes x,..., x n is. s(x) is continuous on [x, x n ].. s(x) is a straight line on each subinterval [x i, x i+ ]. 3. s(x) interpolates f(x) at the nodes, i.e. s(x i ) = f(x i ). Theorem For an interpolating linear spline it holds that x x x 3 x 4 x 5 The theory for linear interpolation holds. f(x) s(x) M 8 h, where f (x) M och h = max x i+ x i. 4 december 8 Sida 5 / 3 4 december 8 Sida 6 / 3 Cubic splines Example Find the cubic spline s(x) interpolating the table f f f 3 f 5 s s f 4 s 3 s 4 x f(x).5 with end point conditions f () = and f () =. Solution Find polynomials s (x) och s (x). x x x 3 x 4 x 5 Definition A function s(x) is an interpolating cubic spline with nodes x,..., x n if. s(x), s (x), och s (x) are continuous on[x, x n ].. s(x) is a cubic polynomial on each [x i, x i+ ]. 3. s(x) interpolates f(x) at the nodes, i.e. s(x i ) = f(x i ). 4 december 8 Sida 7 / 3 f f 3 f s s x x x 3 The coefficients of s and s are found by solving a linear system of equations. 4 december 8 Sida 8 / 3

8 The resulting spline is { s (x)=.+.(x )+.75(x ) s(x)=.5(x ) 3, x, s (x)=.5+.75(x ).(x ) +.75(x ) 3, x, 3 End point conditions Theorem A cubic spline s(x), interpolating f(x) at nodes x,..., x n, is uniqely determined if we provide two end point conditions..5.5 s (x) s (x) Prove by comparing the number of unknowns with the number of conditions..5 We have the options Two continuous derivatives and correct slopes at the end points. Natural conditions s (x )=s (x n )=. Correct conditions s (x )=f (x ) and s (x n )=f (x n ). Periodic conditions s (x )=s (x n ). 4 december 8 Sida 9 / 3 4 december 8 Sida 3 / 3 Error estimate Theorem If correct end point conditions are used then s(x) f(x) Mh4, where h = max x i+ x i and M = max f (4) (x). In Matlab csape computes an interpolating cubic spline. >> pp = csape( x, y, complete, [ d, d ] ); where d and d are numerical values for the derivatives f (x ) and f (x n ). Compute the values of the spline usingppval Exempel Approximate f(x) = /(+x ), on [ 5, 5], with a cubic spline and correct end point conditions. How does the error depend on the number of nodes? The error behave as Ch 4. N h /4 /8 /6 Felet december 8 Sida 3 / Plot the function f(x) = /(+x ) and the cubic splines s(x), for N = 5, 9, and 7 nodes. 4 december 8 Sida 3 / 3