Numerical Analysis Homework 6 Solutions

Homework 6 Solutions Section 5.2, 1,2,3,7,10, Computer Problems 5.2.6, 7 1. 5.2.1: Apply the Composite Trapezoidal Rule with m = 1, 2, 4 panels to approximate the integral. Compute the error by comparing the correct value. (a) (b) Z 1 Z π/2 1 3 0 • m = 1, the area is 0.5. Error: 0.1666.. • m = 2, the area is 0.375. Error: 0.04166.. • m = 4, the area is 0.34375, Error: 0.0104166... x2 dx = cos(x) dx = 1 0 • m = 1, the area is: 0.785398.., Error: 0.2146018.. • m = 2, the area is: 0.948059.., Error: 0.0519405 • m = 4, the area is: 0.987115.., Error: 0.012884 (c) Z 1 ex dx = e − 1 ≈ 1.7272 0 • m = 1, the area is: 1.85914.., Error: 0.140859 • m = 2, the area is: 1.75391.., Error: 0.035649 • m = 4, the area is: 1.72722.., Error: 0.008940 2. (5.2.2) Same integrals as above, but with the Composite Midpoint Rule. (a) (b) Z 1 Z π/2 1 3 0 • m = 1, the area is 0.25. Error: 0.0833.. • m = 2, the area is 0.3125. Error: 0.020833.. • m = 4, the area is 0.328125, Error: 0.00520833... x2 dx = cos(x) dx = 1 0 • m = 1, the area is: 1.11072, Error: 0.11072 • m = 2, the area is: 1.02617, Error: 0.026172 • m = 4, the area is: 1.006454, Error: 0.006454 (c) Z 1 ex dx = e − 1 ≈ 1.7272 0 • m = 1, the area is: 1.64872.., Error: 0.06956 • m = 2, the area is: 1.70051.., Error: 0.01776 • m = 4, the area is: 1.71381.., Error: 0.0044665 3. (5.2.3) Same integrals as above, but with Composite Simpson’s Rule. (a) (b) Z 1 Z π/2 1 3 0 • m = 1, the area is 1/3. Error: 0 • m = 2, the area is 1/3. Error: 0 • m = 4, the area is 1/3, Error: 0 x2 dx = cos(x) dx = 1 0 • m = 1, the area is: 1.002279, Error: 0.0022798 • m = 2, the area is: 1.0001345, Error: 0.0001345 • m = 4, the area is: 1.00000829, Error: 0.00000829 (c) Z 1 ex dx = e − 1 ≈ 1.7272 0 • m = 1, the area is: 1.718861, Error: 0.000579 • m = 2, the area is: 1.783188, Error: 0.0000370 • m = 4, the area is: 1.718284, Error: 0.00000232 4. Problem 5.2.7: Find the precision of the given formulas for 2 3 4 Z 1 f (x) dx. We note that −1 this integral applied to 1, x, x , x , x is 2, 0, 2/3, 0, 2/5, respectively. • f (1) + f (−1): If f (x) = 1, we get an area of 2. If f (x) = x, we get an area of 0. If f (x) = x2 , we do not get 2/3. The degree of precision is therefore 1. • (2/3)(f (−1) + f (0) + f (1)). If f (x) = 1, we get an area of 2. If f (x) = x, we get 0. If f (x) = x2 , we get 4/3. The degree of precision is again 1. √ √ • f (−1/ 3) + f (1/ 3): If f (x) = 1, we get an area of 2. If f (x) = x, we get 0. If f (x) = x2 , we get 2/3. If f (x) = x3 , we get 0. If f (x) = x4 , we get 2/9. The degree of precision is 3. 5. Problem 5.2.10: Find c1 , c2 , c3 so that the rule: Z 0 1 f (x) dx ≈ c1 f (0) + c2 f (1/2) + c3 f (1) has the highest degree of precision possible. If f (x) = 1, the equation becomes: 1 = c1 + c2 + c3 If f (x) = x, the equation becomes: 1 1 = c2 + c3 2 2 If f (x) = x2 , the equation becomes: 1 1 = c2 + c3 3 4 Solve the system of equations to get: c1 = 1 6 c2 = 2 3 c3 = 1 6 Which is also Simpson’s Rule. 6. Computer Problem 5.2.6 Create a script file in Matlab: %Script file for Computer Problem 5.2.7 f=inline(’sqrt(1+(3*x.^2).^2)’); A=compsimp(f,0,1,32) f=inline(’sqrt(1+(sec(x)).^4)’); A=compsimp(f,0,pi/4,32) f=inline(’sqrt(1+(1./(1+x.^2)).^2)’); A=compsimp(f,0,1,32) whose output is: A = 1.54786565377914 A = 1.27797806958318 A = 1.27797805913564 And where the Composite Simpson’s Rule is: function A=compsimp(f,a,b,m) %Approximates the integral of f over [a,b] using 2m+1 points. h=(b-a)/(2*m); x=linspace(a,b,2*m+1); y=f(x); A1=4*sum(y(2:2:(2*m))); A2=2*sum(y(3:2:(2*m-1))); A=(h/3)*(y(1)+y(2*m+1)+A1+A2); 7. Computer Problem 5.2.7 For this problem, again type up a script file. This will use Composite Trapezoidal Rule, which looks like: function A=comptrap(f,a,b,m) %Approximates the integral of f over [a,b] using m+1 points. h=(b-a)/m; x=linspace(a,b,m+1); y=f(x); if m>1 A=(h/2)*(y(1)+y(m+1)+2*sum(y(2:m))); else A=(h/2)*(y(1)+y(2)); end To call this function with h= b−a , 1 b−a , 2 b−a b−a , . . . , 22 28 note that in our function, h = (b − a)/m. Therefore, we will call this function for m = 1, 2, 22 , 23 , . . . , 28 . What are we supposed to see? The Composite Trapezoidal Rule (as written on page 240) could be written as: Z b f (x) dx − Trap(h) = kh2 or I − A(h) = kh2 a where I is the actual area, A is the approximate area (using the rule), K is some constant, and h = (b − a)/2n . Taking the log of both sides, ln |I − A| = 2 ln(h) + ln(k) so in the log-log plot, there should be a general linear shape with a slope of 2. There is a problem using the log-log plot, however- You can’t directly use the “basic fitting” tools. Better to manually plot the log of the domain and range, as shown. Once you get these basic point-plots, then you can determine the line of best fit using the “basic fitting tools”. I did that, then went back into the code and had Matlab plot the lines with the data. For easy reference, the script file is also included here for just f (x) = x2 . f=inline(’x.^2’); M=2.^[0:16]; k=length(M); for j=1:k A(j)=comptrap(f,0,1,M(j)); end figure(1) x=log(1./M); plot(x,log(abs((1/3)-A)),’r*’,x,2*x-1.8,’b-’) title(’Area for x^2’) where the code for the Composite Trapezoidal Rule is as given previously.