Solutions to the
Practice Sets
Practice Set A
Problem 1
(a)
1111 - 345
ans =
766
(b)
format long; [exp(14), 382801*pi]
ans =
1.0e+006 *
1.20260428416478 1.20260480938683
The second number is bigger.
(c)
[2709/1024, 10583/4000, 2024/765, sqrt(7)]
ans =
2.64550781250000 2.64575000000000 2.64575163398693
2.64575131106459
The third, that is, 2024/765, is the best approximation.
235
236 Solutions to the Practice Sets
Problem 2
(a)
cosh(0.1)
ans =
1.00500416805580
(b)
log(2)
ans =
0.69314718055995
(c)
atan(1/2)
ans =
0.46364760900081
format short
Problem 3
[x, y, z] = solve(’3*x + 4*y + 5*z = 2’, ’2*x - 3*y + 7*z = -
1’, ’x - 6*y + z = 3’, ’x’, ’y’, ’z’)
x =
241/92
y =
-21/92
z =
-91/92
Now we’ll check the answer.
A = [3, 4, 5; 2, -3, 7; 1, -6, 1]; A*[x; y; z]
ans =
[ 2]
Practice Set A 237
[ -1]
[ 3]
It checks!
Problem 4
[x, y, z] = solve(’3*x - 9*y + 8*z = 2’, ’2*x - 3*y + 7*z = -
1’, ’x - 6*y + z = 3’, ’x’, ’y’, ’z’)
x =
39/5*y+22/5
y =
y
z =
-9/5*y-7/5
We get a one-parameter family of answers depending on the variable y. In
fact the three planes determined by the three linear equations are not
independent, because the first equation is the sum of the second and third.
The locus of points that satisfy the three equations is not a point, the
intersection of three independent planes, but rather a line, the intersection
of two distinct planes. Once again we check.
B = [3, -9, 8; 2, -3, 7; 1, -6, 1]; B*[x; y; z]
ans =
[ 2]
[ -1]
[ 3]
Problem 5
syms x y; factor(xˆ4 - yˆ4)
ans =
(x-y)*(x+y)*(x^2+y^2)
238 Solutions to the Practice Sets
Problem 6
(a)
simplify(1/(1 + 1/(1 + 1/x)))
ans =
(x+1)/(2*x+1)
(b)
simplify(cos(x)ˆ2 - sin(x)ˆ2)
ans =
2*cos(x)^2-1
Let’s try simple instead.
[better, how] = simple(cos(x)ˆ2 - sin(x)ˆ2)
better =
cos(2*x)
how =
combine
Problem 7
3ˆ301
ans =
4.1067e+143
sym(’3’)ˆ301
ans =
4106744371757651279739780821462649478993910868760123094144405
7023510699153249722978140061846706682416475145332179398212844
0538198297087323698003
But note the following:
sym(’3ˆ301’)
Practice Set A 239
ans =
3^301
This does not work because sym, by itself, does not cause an evaluation.
Problem 8
(a)
solve(’8*x + 3 = 0’, ’x’)
ans =
-3/8
(b)
vpa(ans, 15)
ans =
-.375000000000000
(c)
syms p q; solve(’xˆ3 + p*x + q = 0’, ’x’)
ans =
[
1/6*(-108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3)-2*p/(-
108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3)]
[ -1/12*(-108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3)+p/(-
108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3)+1/2*i*3^(1/2)*(1/6*(-
108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3)+2*p/(-
108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3))]
[ -1/12*(-108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3)+p/(-
108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3)-1/2*i*3^(1/2)*(1/6*(-
108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3)+2*p/(-
108*q+12*(12*p^3+81*q^2)^(1/2))^(1/3))]
(d)
ezplot(’exp(x)’); hold on; ezplot(’8*x - 4’); hold off
240 Solutions to the Practice Sets
8 x−4
40
20
–20
–40
–60
–6 –4 –2 0 2 4 6
x
fzero(inline(’exp(x) - 8*x + 4’), 1)
ans =
0.7700
fzero(inline(’exp(x) - 8*x + 4’), 3)
ans =
2.9929
Problem 9
(a)
ezplot(’xˆ3 - x’, [-4 4])
3
x −x
60
40
20
–20
–40
–60
–4 –3 –2 –1 0 1 2 3 4
x
Practice Set A 241
(b)
ezplot(’sin(1/xˆ2)’, [-2 2])
2
sin(1/x )
0.8
0.6
0.4
0.2
–0.2
–0.4
–0.6
–0.8
–1
–2 –1.5 –1 –0.5 0 0.5 1 1.5 2
x
X = -2:0.1:2;
plot(X, sin(1./X.ˆ2))
Warning: Divide by zero.
0.8
0.6
0.4
0.2
–0.2
–0.4
–0.6
–0.8
–1
–2 –1.5 –1 –0.5 0 0.5 1 1.5 2
This picture is incomplete. Let’s see what happens if we refine the mesh.
X = -2:0.01:2; plot(X, sin(1./X.ˆ2))
Warning: Divide by zero.
242 Solutions to the Practice Sets
0.8
0.6
0.4
0.2
–0.2
–0.4
–0.6
–0.8
–1
–2 –1.5 –1 –0.5 0 0.5 1 1.5 2
Because of the wild oscillation near x = 0, neither plot nor ezplot
gives a totally accurate graph of the function.
(c)
ezplot(’tan(x/2)’, [-pi pi]); axis([-pi, pi, -10, 10])
tan(x/2)
10
–2
–4
–6
–8
–10
–3 –2 –1 0 1 2 3
x
(d)
X = -2:0.05:2;
plot(X, exp(-X.ˆ2), X, X.ˆ4 - X.ˆ2)
Practice Set A 243
12
10
–2
–2 –1.5 –1 –0.5 0 0.5 1 1.5 2
Problem 10
Let’s plot 2x and x 4 and look for points of intersection. We plot them first
with ezplot just to get a feel for the graph.
ezplot(’xˆ4’); hold on; ezplot(’2ˆx’); hold off
2x
50
45
40
35
30
25
20
15
10
−6 −4 −2 0 2 4 6
x
Note the large vertical range. We learn from the plot that there are no points
of intersection between 2 and 6 or −6 and −2; but there are apparently two
points of intersection between −2 and 2. Let’s change to plot now and focus
on the interval between −2 and 2. We’ll plot the monomial dashed.
X = -2:0.1:2; plot(X, 2.ˆX); hold on; plot(X, X.ˆ4, ’--’);
hold off
244 Solutions to the Practice Sets
16
14
12
10
0
−2 −1.5 −1 −0.5 0 0.5 1 1.5 2
We see that there are points of intersection near −0.9 and 1.2. Are there any
other points of intersection? To the left of 0, 2x is always less than 1, whereas
x 4 goes to infinity as x goes to −∞. However, both x 4 and 2x go to infinity as
x goes to ∞, so the graphs may cross again to the right of 6. Let’s check.
X = 6:0.1:20; plot(X, 2.ˆX); hold on; plot(X, X.ˆ4, ’--’);
hold off
5
x 10
12
10
0
6 8 10 12 14 16 18 20
We see that they do cross again, near x = 16. If you know a little calculus,
you can show that the graphs never cross again (by taking logarithms, for
example), so we have found all the points of intersection. Now let’s use
fzero to find these points of intersection numerically. This command looks
for a solution near a given starting point. To find the three different points of
Practice Set A 245
intersection we will have to use three different starting points. The graphical
analysis above suggests appropriate starting points.
r1 = fzero(inline(’2ˆx - xˆ4’), -0.9)
r2 = fzero(inline(’2ˆx - xˆ4’), 1.2)
r3 = fzero(inline(’2ˆx - xˆ4’), 16)
r1 =
-0.8613
r2 =
1.2396
r3 =
16
Let’s check that these “solutions” satisfy the equation.
subs(’2ˆx - xˆ4’, ’x’, r1)
subs(’2ˆx - xˆ4’, ’x’, r2)
subs(’2ˆx - xˆ4’, ’x’, r3)
ans =
2.2204e-016
ans =
-8.8818e-016
ans =
0
So r1 and r2 very nearly satisfy the equation, and r3 satisfies it exactly. It is
easily seen that 16 is a solution. It is also interesting to try solve on this
equation.
symroots = solve(’2ˆx - xˆ4 = 0’)
symroots =
[ -4*lambertw(-1/4*log(2))/log(2)]
[ 16]
[ -4*lambertw(-1/4*i*log(2))/log(2)]
246 Solutions to the Practice Sets
[ -4*lambertw(1/4*log(2))/log(2)]
[ -4*lambertw(1/4*i*log(2))/log(2)]
In fact we get the three real solutions already found and two complex solutions.
double(symroots)
ans =
1.2396
16.0000
-0.1609 + 0.9591i
-0.8613
-0.1609 - 0.9591i
Only the real solutions correspond to points where the graphs intersect.
Practice Set B
Problem 1
(a)
[X, Y] = meshgrid(-1:0.1:1, -1:0.1:1); contour(X, Y, 3*Y +
Y.ˆ3 - X.ˆ3, ’k’)
0.8
0.6
0.4
0.2
−0.2
−0.4
−0.6
−0.8
−1
−1 −0.8 −0.6 −0.4 −0.2 0 0.2 0.4 0.6 0.8 1
[X, Y] = meshgrid(-10:0.1:10, -10:0.1:10); contour(X, Y, 3*Y
+ Y.ˆ3 - X.ˆ3, ’k’)
