CS 257
Numerical Methods  Homework 7
November 2, 2006
1.
[1pt]
5.2 #7
Solution:
We need to find the smallest integer
N
such that the following relationship holds

1
12
(
π

0)
h
2
sin (
ξ
)
≤
10

12
We can bound the left side as follows

1
12
πh
2
sin (
ξ
)
≤
1
12
πh
2
sin (
ξ
)
Using the fact that
sin (
x
)
≤
1 and
h
=
π/N
we can write
π
3
12
N
2
≤
10

12
which implies
π
3
10
12
12
≈
1607437
.
8339534581
≤
N
Therefore we can chose
N
= 1607438
No matter how many intervals we use to compute
π
0
sin(
x
) with the trapezoid rule, we will always compute
something less than the true value. This is because sin is a strictly concave function on the interval [0
, π
]
2.
[1pt]
5.2 #12
Solution:
Like the previous problem, we can bound
sin (
x
)
≤
1 for
x
∈
[2
,
5].

1
12
(5

2)
h
2
sin (
ξ
)
≤
3
·
0
.
01
2
12
= 0
.
000025
3.
[1pt]
6.1 #1
Solution:
The approximate value is
1
0
1
1 +
x
2
≈
1
6
(
f
(0) + 4
*
f
(0
.
5) +
f
(1)) =
1
6
1 + 4
4
5
+
1
2
≈
0
.
7833333
The true value is arctan(1)

arctan(0) =
π/
4. The error is then
≈
0
.
002064.
4.
[1pt]
6.1 #2 a,b
Solution:
•
First we compute the second derivative of
f
f
(
x
)
=
sin
π
2
x
2
f
(
x
)
=
πx
cos
π
2
x
2
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Fall '05
 ThomasKerkhoven
 Derivative, #, Mathematical analysis, 1pt

Click to edit the document details