Linear equations 105
5A = [6, 1; 4, -8];
6b = [13;0];
7x solve = A\b;
8
19 saveas(h,‘s12c4p2 solution figure1.eps’,‘psc2’)
20
21 %make the zoom figure
22 small = 0.001;
23 axis([x solve(1)-small,x solve(1)+small,x solve(2)-small,…
34
35
36 function [x plot,y plot,err plot,k]=jacobi(A,x,b)
37 x old = x;
38 err = 100;
49 x plot(k) = x(1);
50 y plot(k) = x(2);
51 end
The output files are:
106 Linear equations
1.5 2 2.5 3 3.5 4 4.5 5
0
0.5
1
1.5
2
2.5
x
y
Solution t o s12c4p2
1.999 1.9992 1.9994 1.9996 1.9998 2 2.0002 2.0004 2.0006 2.0008 2.001
0.999
0.9992
0.9994
0.9996
0.9998
1
1.0002
1.0004
1.0006
1.0008
1.001
x
y
Solution to s12c 4p2
0.9996
0.9998
1
1.0002
1.0004
1.0006
1.0008
1.001
x
y
Solution to s12c 4p2
(2.82) The files for this problem are in the folder s10c2p1 matlab.
The Matlab script is:
Linear equations 107
1function s10c2p1
2close all
p and q = 0.
12 output(3,p) = size(find(A),1); %find the number of …
non-zero elements
13 b = rand(n,1); %generate the forcing function
14 tic;
25 figure
26 plot(output(1,:),output(3,:),‘o’,‘MarkerSize’,8)
27 xlabel(‘Value of p’,‘FontSize’,14)
28 ylabel(‘Number of non-zero elements in A’,‘FontSize’,14)
29 title(‘Solution to s10c2p1’,‘FontSize’,14)
of the matrix
40 j end = min(n,i+p);
41 for j = i:j end
42 A(i,j) = rand();
43 end
The output files are:
100101102103
10−3
10−2
10−1
Value f or p
T ime f or s ol ut ion ( s e c on ds )
Solution to s 10c2p1
100101102103
10−3
10−2
10−1
Value f or p
T ime f or s ol ut ion ( s e c on ds )
Solution to s 10c2p1
(2.83) The files for this problem are contained in the folder s12c4p3 matlab.
The Matlab script is:
1function s12c4p3
2clc
3close all
13 w = 0.1*z;
14 w plot(z) = w;
15 fprintf(‘\n*****************************************\n’)
16 fprintf(‘\n\n Starting calculation for w = %3.1f \n’,w)
17 k = 0;
28 end
29 r = (b(i) – s)/A(i,i);
30 x(i) = x(i) + w*r;
31 end
32 err = norm(A*x-b);
43 plot(w plot,n iter,‘-ok’)
44 xlabel(‘Relaxation parameter, $w$’,‘FontSize’,14),
45 ylabel(‘Solution to s12c4p3’,‘FontSize’,14)
46 saveas(h,‘s12c4p3 solution figure.eps’,‘psc2’)
The output file is:
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2
0
20
40
60
80
100
120
140
160
180
Re lax ation parame t er, w
Solution t o s12c4p3
110 Linear equations
(2.84) The files for this problem are contained in the folder s11c3p1 matlab.
The Matlab script is:
1function s11c3p1
2clc
3close all
14
15
16 for k = 1:npts
17 n=5*k;
18 n output(k) = n; %store the matrix size
29 end
30 b(i,1) = i*n;
31 end
32
33
44 end
Linear equations 111
45 end
46
47 if diag check >0
58 x = zeros(n,1);
59 iterations = 0;
60 err = 1000;
61
62 while err >tol
73 end
74
75
76 %check the solution
77 err = norm(A*x-b);
87 end
88 end
89 %save the number of iterations and the error
90 iter output(k) = iterations;
91 err output(k) = err;
101 title(‘Solution to s11c3p1’,‘FontSize’,14)
102 saveas(h,‘s11c3p1 solution figure.eps’,‘psc2’)
103
104 %output the data in the correct format
105 write output(:,1) = n output;
Linear equations 113
23 115,0,8.6687e-09,47
24 120,0,5.9007e-09,51
(2.85) The files for this problem are contained in the folder s10c3p1 matlab.
The Matlab script for this problem is:
1function s10c3p1
2
3close all
14 epsilon out(i) = epsilon;%store epsilon
15 condition out(i) = cond(A);%compute the condition number
16 x=A\b;%solve the system
17 error out(i) = abs(epsilon*x(3) – 1);%compute the error. …
the true value is 1/epsilon
28 xlabel(‘Condition Number’,‘FontSize’,14)
29 ylabel(‘Relative error in $x 3$‘,‘FontSize’,14)
30 title(‘Solution to s10c3p1’,‘FontSize’,14)
114 Linear equations
31 saveas(h,‘s10c3p1 solution figure2.eps’,‘psc2’)
10−15 10−10 10−5 100
102
104
106
108
1010
1012
1014
1016
1018
e p sil on
C ondit ion N umbe r
Solution t o s10c3p1
1021041061081010 1012 1014 1016 1018
10−16
10−14
10−12
10−10
10−8
10−6
10−4
10−2
100
C ondit ion N umbe r
Re lative e rror in x3
Solution t o s10c3p1
x(1) =2−23
(3.2) The general form of newtons method is
xk+1=xk−f(xk)
(3.3) The general form of newtons method is
xk+1=xk−f(xk)
x
(3.4) Newton’s method is
x(k+1) =x(k)−f(x(k))
f′(x(k))
For this problem, we have
(3.6) f′=2xso the Newton’s method scheme is
x(k+1) =1
2x(k)
For nsteps of Newton’s method
(3.7) For Newton’s method, the derivative is
f′=cos x−(cos x−xsin x)=xsin x
So we have
x(1)
n+1≈(x∗
n+π)+cos(x∗
n)
sin(x∗
n)
132 Nonlinear equations
(3.8) Find the solution of f(x)=e−x−x(with initial guess x=1)
(3.9) Determining the function requires recognizing that log in Matlab is the natural log:
xln x−x=0
x(ln x−1) =0
The real roots are x=0 and x=e, so the largest real root is x=e=2.178.
(3.12) The result for quadratic convergence using a Taylor series is
|xk+1−x∗|=
f′′(x(k))
2f′(x(k))
(x(k)−x∗)2
which we wrote as
f′′(x(k))
(3.13) J=“3x2−2y
y x #
(3.14)
x2x3−2x1x1x3+2x2x1x2
(3.15) (a) The steady state is the line x=0.
(b) The Jacobian is
J=“y x
Jhx(1) −x(0)i=−R
For a linear system of equations, we have an initial residual
Jhx(1) −x(0)i=−hJx(0) −bi
(3.17) The change in the unknowns on the first step is
δ=x(1) −x(0) =“1
1#
The residual for step 0 is
R=“4b+4
(3.18) For this initial guess, the residual is
R=
ǫ
ǫ
ǫ
The Jacobian is
x3+2x1x2x2
1x1+1
1 1 +ǫ1
1 1 5
1 1 +ǫ1
δ2
δ3
=−
ǫ
ǫ
You only need to do the first step of Gauss elimination on the augmented matrix
1 (1 +ǫ)22+ǫ−ǫ
1 1 5 −ǫ
is consistent. In other words, assume that
δ=
−ǫ
0
0
Then you need to show that
−ǫ
ǫ
(3.19) Starting from the quadratic Taylor series approximation
f(x∗)≈f(x)+f′(x)(x∗−x)+f′′(x)(x∗−x)2
if we let x∗be close to the root then the RHS goes to zero and we can change the the
variables to match our iterative scheme
(3.20) The files for the problem are contained in the folder s10c4p2 matlab.
The exact solution is obtained by factoring x2(x−1) =0, whereupon the roots are
x(3) =0.077
x(5) =0.0668
It takes k=91 iterations to get x<0.01 using this Matlab code:
x(1) =0.1−0.13−0.12
=0.0471
x(2) =0.0229
x(4) =0.0056
The scheme converges after 4 iterations.
The Matlab script is:
1function s10c4p2
2clc
12 disp(x) %print the first few values of x
13 end
14 end
15 disp(‘Total number of iterations’)
16 disp(i)
(3.21) The files for this problem are contained in the folder s10c4p1 matlab.
The Matlab script is:
1function s10c4p1
2close all
3set(0,‘defaulttextinterpreter’,‘latex’)
14 while abs(func(x)) >0.00001
15 x = x -func(x)/der(x);
16 end
17 root out(npts) = x; %record the root
18 npts = npts + 1; %update the counter for data output
28
29
30 function out = func(x)
31 out = sin(pi*x);
32
The three roots in our interval are x=0, 1 and 2. If the initial guess for Newton’s
(3.22) The files for this problem are in the folder s10c4p3 matlab.
The residual for the problem is
x3−x2
z−πsin[π(x−z)] −2ey−x
z2+πsin[π(x−b)]