1. Chapter 14 Curve Fitting : Polynomial Interpolation Gab Byung Chae

3. You’ve got a problem Estimation of intermediate values between data points

4. 14.1 Introduction to Interpolation

5. (2) MATLAB Function : polyfit and polyval Only one straight line that connects two points. Only one parabola connects a set of three points.

6. 14.1.1 Determining polynomial coefficients Interpolating polynomial of solve Ill-conditioned system (Vandermonde matrices)

7. 4.1.2 MATLAB Functions >> format long >> T = [300 400 500]; >> density = [0.616 0.525 0.457]; >> p = polyfit(T, density,2) >> d = polyval(p,350)

8. 14.2 Newton Interpolating polynomial

9. Graphical depiction of linear interpolation. The shaded areas indicate the similar triangles used to derive the Newton linear -interpolation formula [Eq. (14.5)]. Figure 14.2

10. FDD = finite divided difference The second FDD

11. Two linear interpolations to estimate ln 2. Note how the smaller interval provides a better estimate. Figure 14.3

13. Example 14.3 Problem : f(x) = ln x –x 1 = 1 f(x 1 ) =0 –x 2 = 4 f(x 2 ) = 1.386294 –x 3 = 6 f(x 3 ) = 1.791759 Solution : –b 1 = 0 –b 2 = (1.386294 – 0) /(4-1) = 0.4620981 –b 3 = f 2 (x) = 0 + 0.4620981(x-1) – 0.0518731(x-1)(x-4) f 2 (x) = 0 + 0.4620981(x-1) – 0.0518731(x-1)(x-4)

14. The use of quadratic interpolation to estimate ln 2. The linear interpolation from x = 1 to 4 is also included for comparison. Figure 14.4

15. Newton Interpolation  Divided-difference table : Adding (3,14) and (4,22) <- a 1 <- a 2 <- a 3

16. Graphical depiction of the recursive nature of finite divided differences. This representation is referred to as a divided difference table. Figure 14.5

17. Example 14.4 Problem : f(x) = ln x –x 1 = 1 f(x 1 ) =0 –x 2 = 4 f(x 2 ) = 1.386294 –x 3 = 6 f(x 3 ) = 1.791759 –x 4 = 5 f(x 4 ) = 1.609438 f 3 (x) = b1 + b2(x-x1)+ b3(x-x1)(x-x2)+b4(x-x1)(x- x2)(x-x3) b3(x-x1)(x-x2)+b4(x-x1)(x- x2)(x-x3)

18. Solution : –b 1 = f(x 1 ) = 0 –b 2 = f[x 2, x 1 ] = (1.386294 – 0) /(4-1) = 0.4620981 –f[x 3,x 2 ] = (1.791759 – 1.386294) /(6-4) = 0.2027326 –f[x 4,x 3 ] = (1.609438 – 1.791759) /(5-6) = 0.1823216 –b 3 = f[x 3,x 2,x 1 ] = (0.2027326 – 0.4620981) /(6-1) = -0.05187311 = -0.05187311 –f[x 4,x 3,x 2 ] = (0.1823216 – 0.2027326) /(5-4) = -0.02041100 = -0.02041100 –b 4 =f[x 4,x 3,x 2,x 1 ] = (-0.02041100 + 0.05187311) /(5-1) = 0.007865529 = 0.007865529

19. f 3 (x) = 0 + 0.4620981(x-1) - 0.05187311(x-1)(x-4) - 0.05187311(x-1)(x-4) + 0.007865529(x-1)(x- 4)(x-6) + 0.007865529(x-1)(x- 4)(x-6)

20. function yint = Newtint(x,y,xx) % yint = Newtint(x,y,xx): % Newton interpolation. Uses an (n - 1)-order Newton % interpolating polynomial based on n data points (x, y) % to determine a value of the dependent variable (yint) % at a given value of the independent variable, xx. % input: % x = independent variable % y = dependent variable % xx = value of independent variable at which % interpolation is calculated % output: % yint = interpolated value of dependent variable

21. % compute the finite divided differences in the form of a % difference table n = length(x); if length(y)~=n, error('x and y must be same length'); end b = zeros(n,n); % assign dependent variables to the first column of b. b(:,1) = y(:); % the (:) ensures that y is a column vector. for j = 2:n for i = 1:n-j+1 for i = 1:n-j+1 b(i,j) = (b(i+1,j-1)-b(i,j-1))/(x(i+j-1)-x(i)); b(i,j) = (b(i+1,j-1)-b(i,j-1))/(x(i+j-1)-x(i)); end endend % use the finite divided differences to interpolate xt = 1; yint = b(1,1); for j = 1:n-1 xt = xt*(xx-x(j)); xt = xt*(xx-x(j)); yint = yint+b(1,j+1)*xt; yint = yint+b(1,j+1)*xt;end

22. 14.3 Lagrange Interpolating polynomial weight coefficients The first order Lagrange interpolating polynomial Lagrange interpolating polynomial The n-th order Lagrange interpolating polynomial Lagrange interpolating polynomial

23. The second order Lagrange interpolating polynomial Lagrange interpolating polynomial

24. Example 14.5 Problem : Use a Lagrange interpolating polynomial of the first and second order to evaluate the density of unused motor oil at T = 15 o C –x 1 = 0 f(x 1 ) = 3.85 –x 2 = 20 f(x 2 ) = 0.800 –x 3 = 40 f(x 3 ) = 0.212 Solution : First-order at x=15: –f 1 (x)= Second-order : – f 2 (x)=

25. function yint = Lagrange(x,y,xx) % yint = Lagrange(x,y,xx): % Lagrange interpolation. Uses an (n - 1)-order Lagrange % interpolating polynomial based on n data points (x, y) % to determine a value of the dependent variable (yint) % at a given value of the independent variable, xx. % input: % x = independent variable % y = dependent variable % xx = value of independent variable at which the % interpolation is calculated % output: % yint = interpolated value of dependent variable 14.3.1 MATLAB M-file:Lagrange

26. n = length(x); if length(y)~=n, error('x and y must be same length'); end s = 0; for i = 1:n product = y(i); for j = 1:n if i ~= j product = product*(xx-x(j))/(x(i)-x(j)); end s = s+product; end yint = s;

27. 14.4 Inverse Interpolation (1,1) (3,3) (5,5) -> (?,2) Polynomial interpolation –Determine (n-1)-th order polynomial p(x) –solve p(t) = 2 Ex : (2, 0.5), (3,0.3333), and (4,0.25) –Find x so that f(x)=0.3 –f 2 (x) = 0.041667x 2 – 0.375x+1.08333 –Solve 0.3 = 0.041667x 2 – 0.375x+1.08333 for x –x = 5.704158 or 3.295842

28. Extrapolation is the process of estimating a value of f(x) that lies outside the range of the known base points. 14.5.1 Extrapolation

29. Illustration of the possible divergence of an extrapolated prediction. The extrapolation is based on fitting a parabola through the first three known points. Figure 14.10

30. Example 14.6 Problem : Fit a seventh-order polynomial to the first 8 points (1920 to 1990). Use it to compute the population in 2000 by extrapolationand compare your prediction with the actual result. Solution : >> t = [1920 :10:1990]; >> t = [1920 :10:1990]; >> pop = [106.46 123.08 132.12 152.27 180.67 205.05 227.23 249.46]; >> pop = [106.46 123.08 132.12 152.27 180.67 205.05 227.23 249.46]; >> p = polyfit(t, pop, 7) >> p = polyfit(t, pop, 7) Warning message …… Warning message ……

31. >> ts = (t-1955)/35; >> P = polyfit(ts, pop, 7); >> polyval(p, (2000-1955)/35) >>tt=linspace(1920,2000); >>pp=polyval(p, (tt-1955)/35); Plot(t,pop, ‘o’, tt, pp)

32. Use of a seventh-order polynomial to make a prediction of U.S. population in 2000 based on data from 1920 through 1990. Figure 14.11

33. 14.5.2 Oscillations Dangers of higher-order polynomial interpolation Ex : Ringe’s function

34. Comparison of Runge’s function (dashed line) with a fourth-order polynomial fit to 5 points sampled from the function. Figure 14.12

35. Comparison of Runge’s function (dashed line) with a tenth-order polynomial fit to 11 points sampled from the function. Figure 14.13