MATH 128A, SUMMER 2010, HOMEWORK 2 SOLUTION
BENJAMIN JOHNSON
Homework 2: Due Wednesday, June 30
1.3; 1ab, 6b, 13b, 16
2.1; 2b, 13, 17
2.2; 1d, 4, 8, 17
section 1.3
1.
a. Use threedigit chopping arithmetic to compute the sum
∑
10
i
=1
(1
/i
2
) first by
1
1
+
1
4
+
· · ·
+
1
100
and then by
1
100
+
1
81
+
· · ·
+
1
1
. Which method is more accurate and why?
solution:
The sequence of partial answers obtained by computing the first sum is
h
1
.
00
,
1
.
25
,
1
.
36
,
1
.
42
,
1
.
46
,
1
.
48
,
1
.
50
,
1
.
51
,
1
.
52
,
1
.
53
i
. The sequence obtained by doing
the same sum in reverse order is
h
0
.
01
,
0
.
0223
,
0
.
0379
,
0
.
0583
,
0
.
0860
, .
0860
, .
0126
,
0
.
188
,
0
.
299
,
0
.
549
,
1
.
54
i
.
The second
sequence is more accurate because it is able to use more significant digits for all steps
of the computation due to the 1 not being in the first decimal position.
b. Write an algorithm to sum the finite series
∑
N
i
=1
x
i
in reverse order.
solution:
INPUT
N, x
1
, . . . , x
N
OUTPUT
∑
N
i
=1
x
i
(evaluated in reverse order)
Step 1 Set SUM=0.
Step 2 For i=N to 1 do (consider variables in reverse order)
set
SUM
=
SUM
+
x
i
.
Step 3 OUTPUT SUM;
STOP.
6. Find the rate of convergence of the following sequences as
n
→ ∞
.
b. lim
n
→∞
sin
1
n
2
= 0
solution:
We need to find upper and lower bounds for the sequence

sin
1
n
2

0

=
sin
1
n
2
. Using Taylor’s theorem to expand sin
x
about
x
0
= 0, we get sin
x
= sin 0 +
cos 0
·
x

sin 0
2
x
2

cos
c
6
x
3
=
x

x
3
cos
c
6
where
c
is some number between
x
and 0. Since

1
≤
cos
c
≤
1 for every
c
we can deduce that
x

x
3
6
≤
sin
x
≤
x
+
x
3
6
for every
x
.
Furthermore, for
x
∈
[0
,
1], we have
x
3
≤
x
and so
5
6
x
≤
sin
x
≤
7
6
x
. Since
1
n
2
∈
[0
,
1]
for every
n
, this bound applies to our sequence, that is:
5
6
·
1
n
2
≤
sin
1
n
2
≤
7
6
·
1
n
2
, which
shows that the rate of convergence of sin
1
n
2
is
O
(
1
n
2
)
.
13.
b. Make a table listing 1
/n
, 1
/n
2
, 1
/n
3
, and 1
/n
4
for
n
= 5, 10, 100, and 1000, and discuss
the varying rates of convergence of these sequences as
n
becomes large.
solution:
The sequences with higher exponents converge faster.
Date
: June 30, 2010.
1
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BENJAMIN JOHNSON
Table 1.
Table of values for 1
/n
, 1
/n
2
, etc. . .
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