Math 1152 Assignment 8 Solutions
DUE DATE: Mar.24,2017.
Page 370
2)
0
Z
(x + 1)3 dx
−1
with n=5.
P=[0, -0.2, -0.4, -0.6, -0.8, -1]
Midpoints are -0.1, -0.3, -0.5, -0.7, -0.9.
Width=0.2
f (x) = (x + 1)3
M5 = (0.2)[(0.9)3 + (0.7)3 + (0.5)3 + (0.3)3 + (0.1)3 ] = 0.245 =
49
200
8)
Z
4
2
2
√ dx
x
with n=5.
P=[2, 2.4, 2.8, 3.2, 3.6, 4]
Midpoints are 2.2, 2.6, 3, 3.4, 3.8.
2
2
2
2
2 M5 = (0.4) √
+√
+√ +√
+√
= 2.3416313
2.2
2.6
3
3.4
3.8
Z
2
4
2
√ dx = 2
x
Z
2
4
√
4
4
x−1/2 dx = 2(2x1/2 )2 = 4x1/2 2 = 8 − 4 2 = 2.3431
12) Trapezoidal Rule
Z
π/2
sinxdx
0
with n=4.
T4 =
π/2 0
1
1
[ + 0.3827 + √ + 0.9239 + ] = 0.98713
4 2
2
2
1
k
xk
f(xk )
0
0
0
1
π/8
2
π/4
0.3827
√
1/ 2
3 3π/8 0.9239
π/2
4
1
14) Trapezoidal Rule
Z
1
(1 − e−x )dx
−1
with n=4.
k
xk
f(xk )
0
-1
-1.71828
1 -0.5
-0.64872
2
0
0
3
0.5
0.39346
4
1
0.632120
0.632120 2 −1.71828
+ (−0.64872) + 0 + 0.39346 +
= −0.399169
4
2
2
T4 =
Z
1
−1
1
(1 − e−x )dx = (x + e−x )−1 = 1 + e−1 + 1 − e1 = −0.3504118
20)
Z
8
2
1
dt
lnt
Midpoint Error Rule
K(b − a)3
= 10−3
24n2
|f 00 (x)| ≤ K
Note that,
f0 =
f 00 =
−1
t(ln2 t)
(lnt)2 + 2(lnt)
lnt + 2
lnt
2
= 2 3 = 2 3 + 2 3
2
2
t(lnt) )
t ln t
t ln t t ln t
2
Hence, K=2.
K(b − a)3
= 10−3
24n2
2(8 − 2)3
= 10−3
24n2
n = 134.16
Thus, n=135.
22)
Z
2
sinxdx
0
Trapezoidal Error Rule
K(b − a)3
= 10−4
2
12n
f (x) = sinx
Note that,
f 0 = cosx
f 00 = −sinx
Hence, in this case we can use k=2, or |f 00 (x)| ≤ k = 1. Thus, we have,
K(2 − 0)3
= 10−4
3
12n
If we use k=1,
(2 − 0)3
= 10−4
2
12n
4
= 10−4
12n2
Hence, n=58.
If we use k=2,
2(2 − 0)3
= 10−4
2
12n
8
= 10−4
12n2
Hence, n=82.
Page 78
3
6)
an = √
a0 = 1,
√
a1 = 1/ 2 = 0.7071,
1
n+1
√
a2 = 1/ 3 = 0.5773,
√
a3 = 1/ 4 = 1/2,
√
√
a4 = 1/ 5 = 0.4472, and a5 = 1/ 6 = 0.4082
14)
f (n) = 3e−0.1n
a1 = 3e−0.1 ,
a0 = 1,
a2 = 3e−0.2 ,
a3 = 3e−0.3 ,
a4 = 3e−0.4 , and a5 = 3e−0.5
22)
1 4 9 16 25
, , , ,
5 10 17 26 37
The given sequence is
an =
a5 =
24)
(n + 1)2
n2 + 4n + 5
36
49
64
81
, a6 = , a7 = , a8 =
50
65
82
101
π
π
π
π
π
sin , −sin , sin , −sin , sin
2
4
6
8
10
The given sequence is
an = (−1)n sin
a5 = −sin
π 2n + 2
π
π
π
π
, a6 = sin , a7 = −sin , a8 = sin
12
14
16
18
28)
1, 3, 5, 7, 9, ...
The given sequence is
an = 2n + 1
34)
1 −1 1 −1 1
,
, ,
, , ...
2 8 18 32 50
4
The given sequence is
an =
(−1)n
2(n + 1)2
40)
an =
a0 = 0,
2n
n+2
2
a1 = ,
3
a4 =
a2 = 1,
6
a3 = ,
5
8
4
=
6
3
2n
n
1
= 2 lim
= 2 lim
n→∞ n + 2
n→∞ n + 2
n→∞ 1 +
lim
2
n
Note that,
lim 1 = 1
n→∞
lim (1 +
n→∞
2
)=1+0=1
n
Thus,
1
2n
= 2 lim
n→∞ 1 +
n→∞ n + 2
lim
44)
1
a0 = ,
3
an =
(−1)n
n3 + 3
−1
,
4
a2 =
a1 =
a4 =
Note that, limn→∞
1
n3
2
n
1
=2
=2
1
1
,
11
a3 =
−1
,
30
1
67
= 0.
Thus,
(−1)n 1
lim 3
=0
= lim 3
n→∞ n + 3
n→∞ n + 3
48)
an = n2
a0 = 0,
a1 = 1,
a2 = 4,
5
a3 = 9, a4 = 16
lim an = lim n2 = ∞
n→∞
n→∞
Thus, an diverges as n goes to ∞, hence the limit does not exist.
50)
an =
a0 = 1,
1 n
1
1
a3 = , a4 =
8
16
1n = lim
2
n→∞
2
1
a2 = ,
4
1
a1 = ,
2
lim
1 n
n→∞
2n
Note that,
lim 1n = 1
n→∞
lim 2n = ∞
n→∞
Thus,
lim
n→∞
72)
1 n
2
= lim
n→∞
1n 2n
=0
2
1 lim
− 2
n→∞ n
n +1
Note that, limn→∞
1
n
= 0.
1/n2 1 2
1
limn→∞ (1/n2 )
2
− 2
= lim − lim
=
2
lim
−
= ...
lim
n→∞ n
n→∞ n
n→∞ 1 + 1/n2
n→∞ n
n +1
limn→∞ (1 + 1/n2 )
(limn→∞ n1 )(limn→∞ n1 )
1
(0)(0)
−
= 2(0) −
=0
1
1
n→∞ n
1 + (0)(0)
1 + (limn→∞ n )(limn→∞ n )
= 2 lim
76)
lim
n→∞
3n2 − 5 n2
Using the fact,
1 1 3n2 − 5
=
3
−
5
n2
n n
1
Note that, limn→∞ 3 = 3 and limn→∞ n = 0.
1 1 1 1 3n2 − 5 lim
= lim 3 − 5
= lim 3 − 5 lim
(
= ...
n→∞
n→∞
n→∞
n→∞
n2
n n
n n
6
1
1
= lim 3 − 5 lim × lim
= 3 − 5(0)(0) = 3
n→∞
n→∞ n
n→∞ n
78)
lim
n→∞
Note that,
n+2
n2 −4
lim
n→∞
=
n+2
(n−2)(n+2)
=
1
n−2
as well, limn→∞
1
n
= 0.
n+2 1
1/n
limn→∞ (1/n)
=
lim
=
lim
=
= ...
n→∞ n − 2
n→∞ 1 − 2/n
n2 − 4
limn→∞ (1 − 2/n)
=
limn→∞ (1/n)
0
=
=0
1 − 2 limn→∞ (1/n)
1 − (2)(0)
82)
n + 3−n n
lim
n→∞
−n
n+3
n
Note that,
n+2 n2 − 4
= 1 + ( 13 )n ( n1 )
As well, limn→∞ 1 = 1, limn→∞ ( 31 )n = 0, and limn→∞ ( n1 ) = 0.
h
h 1
i
n + 3−n 1 n 1i
1
1
n 1
lim
= lim 1+( ) ×
= lim 1+ lim ( ) ×
= lim 1+ lim ( )n × lim = ...
n→∞
n→∞
n→∞
n→∞
n→∞
n→∞
n→∞
n
3
n
3
n
3
n
= 1 + (0)(0) = 1
88)
an+1 = 4 − 2an
with a0 =
4
3
a1 = 43 , a2 = 34 , a3 = 34 , a4 = 43 , a5 =
4
3
7