MAE 107 Computational Methods in Engineering

MAE 107 Computational Methods in Engineering 

Spring 2007 

Homework #2 Solutions 

Problem 1: Problem 4.7 on page 110 of the textbook. 

The function (correcting a typographical error in the text) 

has the derivative 

f ( x) 

= 

f ′ ( x) 

= 

1 

2 ( 1− 

3x 

) 

6x 

( ) 2 2 

1− 

3x 

Evaluating this derivative at x = 0.577 using double precision arithmetic, rounded 

to the nearest integer 

f ′ ( 0. 

577) 

= 

2, 

352, 

911 

Since the denominator is close to zero, we expect problems with subtractive 

cancellation when using limited precision arithmetic. 

Using three-digit arithmetic with chopping (normalized base 10): 

6x 

= 0. 

346× 

10 

2 

x = 0. 

332× 

10 

2 

3x 

0 

= 0. 

996× 

10 

2 

1− 

3x 

= 0. 

400× 

10 

2 ( 1− 

3x 

) 

f 

2 

−2 

−4 

= 0. 

160× 

10 

1 

0. 

346× 

10 

′ ( 0. 

577) 

= 

−4 

0. 

160× 

10 

1 

0 

In this case, the relative error is 

ε 

= 

2, 

352, 

911 

− 

2, 

352, 

911 

216, 

000 

6 

= 0. 

216× 

10 

= 

= 

0. 

908 or 90. 

8% 

216, 

000

Using four-digit arithmetic with chopping (normalized base 10): 

6x 

= 0. 

3462× 

10 

2 

x = 0. 

3329× 

10 

2 

3x 

0 

= 0. 

9987 × 10 

2 

1− 

3x 

= 0. 

1300 × 10 

2 ( 1− 

3x 

) 

f 

2 

−2 

−5 

= 0. 

1690× 

10 

1 

0. 

3462× 

10 

′ ( 0. 

577) 

= 

−4 

0. 

1690× 

10 

1 

0 

7 

= 0. 

2048× 

10 

And in this case the relative error is 

ε = 

2, 

352, 

911 

− 

2, 

352, 

911 

2, 

048, 

000 

= 

= 

0. 

1296 or 12. 

96% 

2, 

048, 

000 

Problem 2: Problem 4.13 on page 110 of the textbook. 

Equation (4.11) on page 94 of the textbook is 

1 

f ( xi 

1 ) f ( xi 

) f ′ ( xi 

)( xi 

1 xi 

) f ′ 

+ = + 

+ − + ( xi 

)( xi+ 

1 − xi 

) 

2! 

and for the function 

we have 

2 

f ( x) 

= ax + bx + c 

f ′ ( x) 

= 2ax 

+ b 

. 

f ′ 

( x) 

= 2a 

Therefore, equation (4.11) becomes 

f ( 

x 

i+ 

1 

) = ax 

= ax 

= ax 

2 

i 

2 

i 

2 

i+ 

1 

2a 

2 

+ bxi 

+ c + ( 2axi 

+ b)( 

xi+ 

1 − xi 

) + ( xi+ 

1 − xi 

) 

2! 

+ bx + c + 2ax 

x 

2 

− 2ax 

+ bx − bx 

2 

+ ax − 2ax 

x 

i 

+ bx 

i+ 

1 

+ c 

i 

i+ 

1 

i 

i+ 

1 

i 

2 

i+ 

1 

i 

i+ 

1 

+ ax 

2 

i

MAE 107 Spring 2007 

HW 2 Problem 3 

Contents 

bisect.m 

bisect.m 

Define f(x), and plot it 

Use bisect.m to compute root of f(x) 

Use a smaller interval to see if number of iterations is reduced 

type bisect.m 

% Function to find the root of function 'f' near between [x1, x2] 

% using bisection method 

function xm = bisect(f, x1, x2) 

itmax = 100; % set maximum number of iterations 

accuracy = 1e-4; % set required accuracy 

prev_xm = NaN; % used for computing estimated error 

disp(sprintf('Desired accuracy = %e\n', accuracy)); 

for iter = 1:itmax 

% calculate midpoint of dcurrent domain 

xm = (x1 + x2) / 2; 

end 

disp( sprintf(... 

'Iter#=%2d, x-range=(%+6.4f,%+6.4f), xm=%+8.6f, f(xm)=%+8.6e, Estim error=%+6.4e', ... 

iter, x1, x2, xm, f(xm), abs(xm - prev_xm)) ) 

% check if accuracy is acheived 

if (abs(f(xm)) < accuracy); break; end 

% see if root lies in left or right sub-domain? 

if (f(x1) * f(xm) > 0) % lies in right half 

x1 = xm; 

else % lies in left half 

x2 = xm; 

end 

prev_xm = xm; 

disp(sprintf('\nNumber of iterations needed = %d', iter)) 

end 

Define f(x), and plot it

f = @(x) (exp(x) - 2 + sin(2*x)) ; 

x = [-3:.01:3]; 

plot(x, f(x), x, zeros(size(x))); 

Use bisect.m to compute root of f(x) 

x = bisect(f, -3, +3) 

Desired accuracy = 1.000000e-04 

Iter#= 1, x-range=(-3.0000,+3.0000), xm=+0.000000, f(xm)=-1.000000e+00, Estim error= NaN 

Iter#= 2, x-range=(+0.0000,+3.0000), xm=+1.500000, f(xm)=+2.622809e+00, Estim error=+1.5000e+00 

Iter#= 3, x-range=(+0.0000,+1.5000), xm=+0.750000, f(xm)=+1.114495e+00, Estim error=+7.5000e-01 

Iter#= 4, x-range=(+0.0000,+0.7500), xm=+0.375000, f(xm)=+1.366302e-01, Estim error=+3.7500e-01 

Iter#= 5, x-range=(+0.0000,+0.3750), xm=+0.187500, f(xm)=-4.274972e-01, Estim error=+1.8750e-01 





Iter#=10, x-range=(+0.3281,+0.3398), xm=+0.333984, f(xm)=+1.591389e-02, Estim error=+5.8594e-03 


Iter#=12, x-range=(+0.3281,+0.3311), xm=+0.329590, f(xm)=+2.866336e-03, Estim error=+1.4648e-03


Iter#=14, x-range=(+0.3281,+0.3289), xm=+0.328491, f(xm)=-3.987657e-04, Estim error=+3.6621e-04 




Number of iterations needed = 17 

x = 

0.3286 

Use a smaller interval to see if number of iterations is reduced 

x = bisect(f, -1, +1) 

Desired accuracy = 1.000000e-04 

Iter#= 1, x-range=(-1.0000,+1.0000), xm=+0.000000, f(xm)=-1.000000e+00, Estim error= NaN 












Number of iterations needed = 12 

x = 

0.3286 

Published with MATLAB® 7.3


Applying equation (6.6) to f(x)=x n a=0, the n th root of a real number can be calculated as the limit of 

x = k1 1 

n n−1 x a 

k n−1 

x 

k−1 

. 

The accuracy at each step can be calculated using |x k x k1 |. Starting from x(1)=2, k is increased up to the 

point that the accuracy is smaller that 10 6 . 

NewtonRaphson.m 

%========================================================================== 

% Problem 4: This program calculates the nth root of a real positive number 

% with accuracy of 10^6 using NewtonRaphson algorithm. 

%========================================================================== 

clc;clear;close all; 

n=5; % The value of 'n' to calculate the nth root 

a=243; % The number which its root is to be determined 

delta=10; % Residual at each step. It is initialized by a large number... 

% ... to start the while loop 

x(1)=2; % The initial guess of the root 

k=1; % Counter 

while (delta>10e6) %checks if the target accuracy is met 

end 

error(k)=abs(x(k)3); % Absolute error at each time step 

k=k+1; 

x(k)=(1/n)*((n1)*x(k1)+a/(x(k1)^(n1))); 

delta=abs(x(k)x(k1)); % Residuals at each step 

error(k)=x(k)3; % Absolute error of the last point (x(k)) 

disp('x(1)...x(k):');disp(x); % Listing the x(k)s 

plot(log(error(1:(k1))),log(error(2:k))); % Plotting ln(E(k+1)) versus ln(E(k)) 

axis equal; xlabel('ln(E^K)');ylabel('ln(E^{K+1})'); grid on; % Setting the plot 

attributes

Output: 

x(1)...x(k): 

2.0000 4.6375 3.8151 3.2815 3.0443 3.0013 3.0000 3.0000 

Published with MATLAB® 7.1 

As we know, taking the natural logarithm of a quadratic function y=c 1 x 2 +b gives ln(y)=2ln(x)+c 2 where c 1 

and c 2 are constants. Thus, plotting the quadratic function in the logarithmic scale leads to a straight line 

which makes an angle of tan 1 (2) with the horizontal axis. Based on this discussion, it is clearly seen in the 

plot above that the Error converges quadratically as expected by equation (6.7).


Polyroots.m 

%========================================================================== 

%Problem 5: This program plots the function f(x)=x^5x^43x^3+3x^24x+4 on 

%the interval (3,3) and finds its roots using the Matlab roots command. 

%========================================================================== 

clc;clear;close all; % Initializing the screen, variables and plotting windows 

x=[3:.1:3]; % Defining the 'x' vector from 3 to 3 with increments of 0.1 

for i=1:length(x) % Calculating the corresponding 'f' value for each point of 'x' 

end 

f(i)=x(i)^5x(i)^43*x(i)^3+3*x(i)^24*x(i)+4; 

plot(x,f,'linewidth',2);grid on; % Plotting the function 

ylabel('f(x)');xlabel('x'); 

root=roots([1 1 3 3 4 4]);hold on; % Finding the roots using the roots command 

display('Roots of the polynominal are :');root %Displaying the roots 

zer=zeros(1,length(root)); 

plot(real(root),zer,' ok','MarkerFaceColor','r','MarkerSize',8); %Plotting the real 

part of the roots as red circles 

%Note: the vector zer is defined as [0 0... 0] so then the last line... 

%...plots the points (root1,0), (root2,0)...(rootn,0) 

Output: 

Roots of the polynominal are : 

root = 

2.0000 

2.0000 

0.0000 + 1.0000i 

0.0000 1.0000i 

1.0000

Published with MATLAB® 7.1 

As the figure shows, the real roots (red circles) obtained from the roots command coincide with the points 

that the curve crosses the x axis. However, the red circle located at x=0 does not lay on the curve since it is 

only representing the real parts of the two imaginary roots 0+i and 0i.

MAE 107 Computational Methods in Engineering

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