MATH 128A, SUMMER 2009: HOMEWORK 2 SOLUTIONS
2.1
10. This can be solved by hand or with a computer:
bisection_table(@(x)(x+2)*(x+1)^2*x*(x1)^3*(x2), a, b, .1, 10)
In each case it only takes one or two iterations for the bisection method to ﬁnd an interval containing
only one of the roots.
(a) 0
(b) 0
(c) 2
(d)

2
15. To guarantee an accuracy of 10

4
after
n
iterations, we can make sure that (2

1)
/
2
n
≤
10

4
. This
happens when
n
≥
14.
bisection(@(x)x^3x1, 1 , 2, 1e4, 14)
outputs 1
.
3248.
2.2
3. These tables were created with MATLAB code such as the following, for (a):
fixedpoint_table(@(x)(20*x+21/x^2)/21, 1, 1e6, 10); %similar for (b)(d)
(a) 3rd fastest
(b) Fastest
n 
p_n  Deltap(n1)
n 
p_n  Deltap(n1)


1 
+1.95238095 
+0.95238095
1 
+7.66666667 
+6.66666667
2 
+2.12175427 
+0.16937332
2 
+5.23020374 
2.43646293
3 
+2.24284969 
+0.12109542
3 
+3.74269692 
1.48750682
4 
+2.33483967 
+0.09198998
4 
+2.99485357 
0.74784335
5 
+2.40709338 
+0.07225371
5 
+2.77702223 
0.21783134
6 
+2.46505929 
+0.05796591
6 
+2.75904187 
0.01798036
7 
+2.51224346 
+0.04718418
7 
+2.75892418 
0.00011769
8 
+2.55105710 
+0.03881363
8 
+2.75892418 
0.00000001
9 
+2.58323777 
+0.03218067
10 
+2.61008144 
+0.02684368
(c) Not convergent to 21^(1/3)
(d) 2nd fastest
n 
p_n  Deltap(n1)
n 
p_n  Deltap(n1)


1 
+0.00000000 
1.00000000
1 
+4.58257569 
+3.58257569
2 
+0.00000000 
+0.00000000
2 
+2.14069514 
2.44188055
3 
+3.13207559 
+0.99138045
4 
+2.58936653 
0.54270907
5 
+2.84782227 
+0.25845575
6 
+2.71552125 
0.13230102
7 
+2.78088509 
+0.06536384
8 
+2.74800884 
0.03287626
9 
+2.76439809 
+0.01638926
10 
+2.75619128 
0.00820681
7. Since the range of sin is [

1
,
1], the range of
g
(
x
) is [
π

0
.
5
,π
+ 0
.
5]. In particular,
g
(
x
)
∈
[0
,
2
π
]
for any
x
∈
[0
,
2
π
].
g
0
(
x
) = 0
.
25 cos(
x/
2), so the range of
g
0
(
x
) is [

.
25
,.
25]. In particular,

g
0
(
x
)
 ≤
k
= 1
/
4
<
1 for
any
x
∈
[0
,
2
π
].
Date
: Due Thursday 7/02.
1