MATH ASSIGNMENT 07 SOLUTIONS. 8.1 Following is census data showing the population of the US between 1900 and 2000:

Transcription

1 MATH ASSIGNMENT 07 SOLUTIONS 8.1 Following is census data showing the population of the US between 1900 and 2000: Years after 1900 Population in millions (a) Plot population versus years after 1900 using Matlab by entering the years after 1900 into a vector x and the populations into a vector y and giving the command plot(x,y, o ). Find the fifth degree polynomial that passes through each data point and determine its value in the year Plot this polynomial on the same graph as the population data (but with the x-axis now extending from 0 through 120). Do you think that this is a reasonable way to estimate the population of the US in future years? (To see a demonstration of other methods of extrapolation, type census in Matlab.) Turn in your plot and the value that you estimated for the population in the year 2020 along with your explanation as to why this is or is not a good way to estimate the population. Solution. There are lots of ways to determine this fifth degree polynomial. Here is one based on the Vandermonde matrix: x = 0:20:100; y = [76; 105.7; 131.7; 179.3; 226.5; 281.4]; plot(x, y, 'o'); coeffs = vander(x)\y; yrs = 0:120; hold on; plot(yrs, polyval(coeffs, yrs)); The above code generates the following plot: 1

2 Using this polynomial, together with polyval we see that the extrapolated population in the year 2020 is 459.6, more than double the population in This seems unreasonable, as using this model, the population would be doubling too fast. That is, instead of doubling roughly every 60 years as indicated by the measurements, the population would more than double in less than 40 years, and then more than double over the next 20 years (the extrapolated value for the year 2040 would be million). (b) Write down the Lagrange form of the second-degree polynomial that interpolates the population in the years 1900, 1920, and Solution. The Lagrange form of the interpolating polynomial is (x 20)(x 40) (x 0)(x 40) (x 0)(x 20) p(x) = (0 20)(0 40) (20 0)(20 40) (40 0)(40 20) = (x 20)(x 40) x(x 40) + x(x 20) Gathering like terms, we have p(x) = x x + 76 (c) Determine the coefficients of the Newton form of the interpolants of degrees 0, 1, and 2, that interpolate the first one, two, and three data points, respectively. Verify that the second-degree polynomial that you construct here is identical to that in part (b). Solution. The Newton forms of the interpolants are p 0 (x) = f[0] p 1 (x) = f[0] + f[0, 20](x 0) p 2 (x) = f[0] + f[0, 20](x 0) + f[0, 20, 40](x 0)(x 20) 2

3 where the coefficients are the divided differences f[0], f[0, 20], and f[0, 20, 40]. We can compute the divided differences using the following table: f[0] = 76 f[20] = f[0, 20] = f[20] f[0] = f[40] = f[20, 40] = f[40] f[20] = 26 f[0, 20, 40] = f[20,40] f[0,20] = So, the Newton form of the interpolating polynomial is p 2 (x) = x 3.7 x(x 20). 800 Combining like terms we have p 2 (x) = x x2 which is exactly the polynomial we found in part (b). 8.2 The secant method for finding a root of a function f(x) fits a first-degree polynomial (i.e. a straight line) through the points (x k 1, f(x k 1 )) and (x k, f(x k )) and takes the root of this polynomial as the next approximation x k+1. Another rootfinding algorithm, called Muller s method, fits a quadratic through the three points, (x k 2, f(x k 2 )), (x k 1, f(x k 1 )), and (x k, f(x k )), and takes the root of this quadratic that is closest to x k as the next approximation x k+1. Write down a formula for this quadratic. Suppose f(x) = x 3 2, x 0 = 0, x 1 = 1, and x 2 = 2. Find x 3. Solution. A formula for this quadratic can be given in Lagrange form: (t x k 1 )(t x k ) p(t) = f(x k 2 ) (x k 2 x k 1 )(x k 2 x k ) (t x k 2 )(t x k ) + f(x k 1 ) (x k 1 x k 2 )(x k 1 x k ) + f(x k ) (t x k 2)(t x k 1 ) (x k x k 2 )(x k x k 1 ). Supposing that f(x) = x 3 2, we have three points (0, 2), (1, 1), (2, 6) and so 2 p(t) = (t 1)(t 2) (0 1)(0 2) 1 + (t 0)(t 2) (1 0)(1 2) 6 + (t 0)(t 1) (2 0)(2 1) = 1(t 2 3t + 2) + (t 2 2t) + 3(t 2 t) = 3t 2 2t 2. The roots of p(t) and therefore the candidate values for x 3 are x 3 = 2 ± = 1 ± 7. 3

4 Muller s method chooses the root nearest to the previous iterate as the next iterate. In this case, then, x 3 = Write down a divided-difference table and the Newton form of the interpolating polynomial for the following set of data. x 1 3/2 0 2 f(x) Show how this polynomial can be evaluated at a given point x by using four steps of a Horner s rule-like method. Solution. The divided-difference table for the above set of data looks like this: = 8 3/ = = 4 0 3/ = = = /2 2 1 The Newton form of the interpolating polynomial is therefore. p(x) = 2 + 8(x 1) + 4(x 1)(x 3/2) + 2(x 1)(x 3/2)(x 0). The Horner s rule-like method takes the following form here: y = 2 y = y(x 0) + 4 y = y(x 3/2) + 8 y = y(x 1) Determine the piecewise polynomial function { P 1 (x) if 0 x < 1, P (x) = P 2 (x) if 1 x 2, that is defined by the conditions P 1 (x) is linear. P 2 (x) is quadratic. P (x) and P (x) are continuous at x = 1. P (0) = 1, P (1) = 1, and P (2) = 0. Plot this function. Solution. As P 1 (x) is to be linear, then P 1 (x) = a 0 + a 1 x. In order to enforce that P (0) = 1, a 0 = 1 must be the case. Also, so that P (1) = 1, a = 1 must be the case, and so a 1 = 2. As P 2 (x) is to be quadratic, then P 2 (x) = b 0 + b 1 x + b 2 x 2. P (2) = 0 forces b 0 + 2b 1 + 4b 2 = 0. Also, P (1) = 1 means that b 0 + b 1 + b 2 = 1. P (x) being continuous at x = 1 means that b 1 + 2b 2 = 2. 4

5 These conditions result in the following linear system: b 0 b 1 = Solving this linear system gives b 0 = 4, b 1 = 8 and b 2 = 3. Thus, { 1 2x if 0 x < 1, P (x) = 4 8x + 3x 2 if 1 x 2, The following Matlab code generates the plot: b 2 f (x < 1).*(1-2*x)+(x>=1).*(4-8*x+3*x.ˆ2); x = 0:0.01:2; plot(x, f(x), 'LineWidth', 2); 8.10 Let f be a given function satisfying f(0) = 1, f(1) = 2, and f(2) = 0. A quadratic spline interpolant r(x) is defined as a piecewise quadratic that interpolates f at the nodes (x 0 = 0, x 1 = 1, and x 2 = 2) and whose first derivative is continuous throughout the interval. Find the quadratic spline interpolant of f that also satisfies r (0) = 0. [Hint: Start from the left subinterval.] Solution. r(x) is made up of quadratics on subintervals. Here, r 1 (x) = a 0 + a 1 x + a 2 x 2 must satisfy Clearly, then, r 1 (x) = 1 + x 2. a 0 = r 1 (0) = 1 a 1 = r 1(0) = 0 a 0 + a 1 + a 2 = r 1 (1) = 2 5

6 On [1, 2], r(x) = r 2 (x) = b 0 + b 1 x + b 2 x 2. In order for r(x) to be a differentiable interpolant, r 2 must satisfy the following b 0 + b 1 + b 2 = r 2 (1) = 2 b 0 + 2b 1 + 4b 2 = r 2 (2) = 0 b 1 + 2b 2 = r 2(1) = r 1(1) = 2 Solving this linear system gives r 2 (x) = x 4x 2. The quadratic spline interpolant r(x) then, is { 1 + x 2 if 0 x < 1, r(x) = x 4x 2 if 1 x Show that the following function is a natural cubic spline through the points (0, 1), (1, 1), (2, 0), and (3, 10): 1 + x x 3 if 0 x < 1, s(x) = 1 2(x 1) 3(x 1) 2 + 4(x 1) 3 if 1 x < 2, 4(x 2) + 9(x 2) 2 3(x 2) 3 if 2 x 3. Solution. Let s consider s 1 (x) = 1 + x x 3 s 2 (x) = 1 2(x 1) 3(x 1) 2 + 4(x 1) 3 s 3 (x) = 4(x 2) + 9(x 2) 2 3(x 2) 3. In order for s to be a natural cubic spline for these points, all of the following must hold: s 1 (0) = 1, s 1 (1) = 1. Note that s 1 (0) = 1 and s 1 (1) = = 1. s 2 (1) = 1, s 2 (2) = 0. Here, s 2 (1) = 1 and s 2 (2) = = 0. s 3 (2) = 0, s 3 (3) = 10. Now s 3 (2) = 0 and s 3 (3) = = 10. s 1(1) = s 2(1). Here, s 1(x) = 1 3x 2 so s 1(1) = 1 3 = 2. Also, s 2(x) = 2 6(x 1) + 12(x 1) 2 and so s 2(1) = 2. s 2(2) = s 3(2). With s 2(x) as above, s 2(2) = = 4. Also, s 3(x) = 4+18(x 2) 9(x 2) 2 and so s 3(2) = 4. s 1(1) = s 2(1). s 1(x) = 6x so s 1(x) = 6. Also, s 2(x) = (x 1) and so s 2(1) = 6 as desired. s 2(2) = s 3(2). With s 2(x) as above, s 2(2) = = 18. Also, s 3(x) = 18 18(x 2) so s 3(2) = 18. 6

7 s 1(0) = 0. s 1(0) = 0 is clear. s 3(3) = 0. s 3(3) = = 0. Therefore s(x) is a natural cubic spline. (2) A secret formula for eternal youth, f(x), was discovered by Dr. Quick, who has been working in our biotech company. However, Dr. Quick has disappeared and is rumored to be negotiating with a rival organization. From the notes that Dr. Quick left behind in his hasty exit it is clear that f(0) = 0, f(1) = 2, and that f[x 0, x 1, x 2 ] = 1 for any three points x 0, x 1, x 2. Find f(x). Solution. Consider a divided difference table for the points 0, 1, x: x f(x) f(x) 2 x 1 f(x) 2 x 1 2 x 0 The last entry in the lower right hand corner is f[0, 1, x] and by hypothesis, this is 1, so f(x) 2 x 1 After some algebra, f(x) = x 2 + x. 2 = x. (3) Suppose we want to approximate the function e x on the interval [0, 1] by using polynomial interpolation with x 0 = 0, x 1 = 1/2, and x 2 = 1. Let p 2 (x) denote the interpolating polynomial. (a) Find an upper bound for the error magnitude max 0 x 1 ex p 2 (x). Solution. Recall that the error can be bounded by max 0 x 1 ex p 2 (x) 1 3! max f (3) (t) max (s 0)(s 1/2)(s 1) = e 0 t 1 0 s 1 6 α where α is the maximal value assumed by r(s) = s(s 0.5)(s 1) on the interval [0, 1]. It is not hard to establish that r(s) assumes its maximal absolute value at both 1 ± and that α = An upper bound for the error then is max 0 x 1 ex p 2 (x) e (b) Find the interpolating polynomial using your favorite technique. Solution. Using divided differences, we obtain f[0] = 1, f[0, 1/2] = 2(e 1 2 1), and f[0, 1/2, 1] = 2(e 1 2 1) 2 and a Newton form of the polynomial, then is p 2 (x) = 1 + 2e 1/2 x + (e 2e 1/2 )x(x 1/2) (c) Plot the function e x and the interpolant you found, both on the same figure, using the commands plot. 7

8 Solution. The following Matlab code produces the plot: p 1+2*(exp(0.5)-1).*x+2*(exp(0.5)-1).ˆ2.*x.*(x-0.5); f exp(x); x = 0:0.01:1; figure; plot(x, f(x)); hold on; plot(x, p(x), 'r.'); And the resulting plot is as follows: (d) Plot the error magnitude e x p 2 (x) on the interval using logarithmic scale (the command semilogy) and verify by inspection that it is below the bound you found in part (a). Solution. The following Matlab code produces the plot: p 1+2*(exp(0.5)-1).*x+2*(exp(0.5)-1).ˆ2.*x.*(x-0.5); f exp(x); x = 0:0.01:1; semilogy(x, abs(p(x)-f(x))); % Plot a dashed line at the upper bound bnd = exp(1)./(36*sqrt(12)); 8

9 hold on semilogy(x, bnd*ones(size(x)), 'k--'); And the resulting plot is as follows: 9