Mathematics 1220
Exam III Review Solutions
Fall 2006
• Taylor and Maclaurin Series (§10.8)
!
∞
X
x2 x3
1. 1 + x −
−
+... =
(−1)n
2!
3!
n=0
x2n+1
x2n
+
represents f (x) = sin x +
(2n)! (2n + 1)!
∞
∞
X
X
x2n
x2n+1
and cos x =
(−1)n
for all
cos x for all x since sin x =
(−1)n
(2n + 1)!
(2n)!
n=0
n=0
x.
3
2
4
+ e2 (x−2)
+ e2 (x−2)
2. e2 + e2 (x − 2) + e2 (x−2)
2!
3!
4!
3. For |x| < 1, (1 +
f (x) = 2
4.
=
∞
X
n=0
!
∞
X
1/2 2n
x .
2n
n=0
∞
X
x)1/2
!
!
∞
X
1/2 n
1/2 n
x and (1 − x)1/2 =
(−1)n
x , so
n
n
n=0
∞
∞
X
X
1
1
=
(−1)n (x−1)n and 2 =
(−1)n+1 n(x−1)n−1 =
(−1)n (n+1)(x−1)n
x n=0
x
n=1
n=0
for all x in the interval (0, 2).
5. ln x =
Z
1
x
∞
X
1
(x − 1)n+1
dt =
(−1)n
.
t
n+1
n=0
• Approximation by Taylor polynomials (§11.1)
2
3
1. g(x) = 3 + 9(x − 2) + 8 (x−2)
+ 6 (x−2)
+ R3 (x) where R3 (x) =
2!
3!
(4)
but g (c) = 0 for all c, so R3 (x) = 0.
g (4) (c)
4! (x
2
− 2)4 ,
3
4
2. The fourth order Taylor approximation to f (x) is 21 − x−1
+ (x−1)
− (x−1)
+ (x−1)
,
22
23
24
25
5
so f (1.2) is approximately 0.45455. The error term is R4 (x) = − (x−1)
(1+c)6 for some
5
5
(0.2)
(0.2)
c in [1, 1.2]. |R4 (1.2)| = | (1+c)
6 | ≤ | 26 | = 0.000005.
3. |Rn (0.3)| ≤
(0.3)n+1
(n+1)!
approximation to cos x
4.
Z
1.2
ln x dx =
Z
1
−3
(n+1)! ≤ 10
4
6
2
is 1− x2! + x4! − x6! ,
≤
∞
1.2 X
(−1)
0.8 n=1
0.8
n+1 (x
when n ≥ 6. The sixth order Taylor
so cos 0.3 is approximately 0.9553364875.
∞
n+1 1.2
X
− 1)n
n+1 (x − 1)
dx =
(−1)
≈ (x −
n
(n
+
1)n
n=1
0.8
1.2
(x − 1)3 (x − 1)4 (x − 1)5 (0.2)3
(0.2)5
1) −
+
−
−
= −0.0026986̄.
=−
6
12
20 0.8
3
10
2
• Numerical Integration (§11.2)
1.
2.
Z
1.2
0.8
Z 1.2
0.8
3
(0.4)
ln x dx ≈ −0.002786, |E8 | = | 12(8)
2 c2 | ≤
5
(0.4)3
12(8)2 (0.8)2
(0.4) 3!
ln x dx ≈ −0.002699, |E8 | = | 180(8)
4 c4 | ≤
3. |En | ≤
1
120n2
≤
10−3
if
n2
=
(0.4)5 3!
180(8)4 (0.8)4
≥ 9. Take n = 3, then
Z
0
0.5
2
1
7680
≈ 0.0001302
≈ 2.0345 × 10−7 .
ex dx ≈ 0.54795
2
4
3
) cos x
≤ 51 for x in
4. Let f (x) = sinx x . Then, f (4) (x) = (24−12x +x ) sin xx+(−24x+4x
5
1000
1
−5
4
[0, 1]. Therefore, |En | ≤ 900n4 ≤ 10 if n ≥ 9 , so we may take n ≥ 4. The
Z 1
sin x
dx ≈ 0.946087
parabolic rule with n = 4 gives
x
0
• Numerical Equation Solving (§11.2-11.3)
1. The root is approximately 0.60, accurate to 2 decimal places. It takes 6 iterations
to achieve this accuracy.
2. n ≥ 40
√
3. 2 is the positive root of the equation x2 − 2, which lies in the interval [1, 2].
We will take x1 = 1.5. Then, x4 = 1.4142135 is accurate to five decimal places.
4. We will approximate the root in the interval [2, 3] and take x1 = 1.2. Then,
x6 = 1.0893 is accurate to 3 decimal places.
5. Newton’s method isn’t applicable because x1/3 is not differentiable at 0.
6. The fixed point is 0.60, accurate to 2 decimal places.
7. tan x does not take [4, 4.5] into itself, so we can not even guarantee a fixed point
exists on this interval. Even if a fixed point exists in this interval, the derivative
sec2 x < 1 if and only if 1 < cos2 x if and only if sin2 x < 0, which occurs for no
x. Thus, the fixed point algorithm may not succeed.
8. Let x1 = 4.5. Then, x3 = 4.4934 approximates the fixed point accurately to 3
decimal places.
• Approximating solutions to differential equations (§11.5)
1. Below is a table of the results from Euler’s method:
n
0
1
2
3
4
5
xn
1
1.2
1.4
1.6
1.8
2
yn
2
2.4
2.976
3.80928
5.0282496
6.838419456
2. Below is a table of the results from Improved Euler’s method:
n
0
1
2
3
4
5
xn
1
1.2
1.4
1.6
1.8
2
ŷn
2
2.4
3.08512
4.119650304
5.71323403284
8.22678000201
yn
2
2.488
3.2184768
4.32820760064
6.04910294265
8.78329747273
3. Euler’s method says yn = yn−1 + hyn−1 = yn−1 (1 + h) for n ≥ 1. Therefore,
yn = yn−1 (1 + h) = [yn−2 (1 + h)](1 + h) = yn−2 (1 + h)2 . By repeating this
reduction a total of n times, we find
yn = yn−1 (1 + h) = yn−2 (1 + h)2 = · · · = y0 (1 + h)n .
4. Below is a table of the results from Improved Euler’s method:
n
0
1
2
3
4
5
6
7
8
9
10
xn
0
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
ŷn
2
3.2
5.696
10.13888
18.0472064
32.124027392
57.1807687578
101.781768389
181.171547732
322.485354964
574.023932837
yn
2
3.56
6.3368
11.279504
20.07751712
35.7379804736
63.613605243
113.232217333
201.553346853
358.764957398
638.601624168