MATH 100, Section 110 (CSP)
Week 8: Marked Homework Solutions
2010 Nov 04
1. [3 each] (a) f (x) = x5/2 , a = 4, n = 2.
f (x) = x5/2 ,
√
f (4) = 45/2 = ( 4)5 = 32,
(b) f (x) =
√
f 0 (x) = 25 x3/2 ,
f 0 (4) = 25 43/2
f 00 (x) = 15
x1/2 .
4
√
= 52 ( 4)3 = 20, f 00 (x) =
T2 (x) = 32 + 20(x − 4) +
15
(x
4
15 1/2
4
4
=
15
.
2
− 4)2 .
x, a = 4, n = 3.
f (x) = x1/2 ,
f 0 (x) = 21 x−1/2 ,
f (4) = 2,
f 0 (4) = 14 ,
f 00 (x) = − 14 x−3/2 ,
f (3) (x) =
1
f 00 (x) = − 32
,
T3 (x) = 2 + 14 (x − 4) −
1
(x
64
f (3) (x) = 83 x−5/2 .
− 4)2 +
1
(x
512
3
.
256
− 4)3 .
(c) f (x) = cos x, a = π/3, n = 5.
f (x) = cos x, f 0 (x) = − sin x, f 00 (x) = − cos x, f (3) (x) = sin x, f (4) (x) = cos x, f (5) (x) = − sin x.
√
√
√
f π3 = cos π3 = 12 , f 0 π3 = − 23 , f 00 π3 = − 21 , f (3) π3 = 23 , f (4) π3 = 21 , f (5) π3 = − 23 .
√
2 √
3
4 √
5
1
T5 (x) = 12 − 23 x − π3 − 14 x − π3 + 123 x − π3 + 48
x − π3 − 2403 x − π3 .
2. [5] Let f (x) = x5/2 . Taylor’s Formula with Remainder says that
(3.9)5/2 = f (3.9) = T1 (3.9) + R1 (3.9),
where the remainder (error) is (see the solution of 1(a) for f 00 (x))
R1 (3.9) =
f 00 (c)
(3.9 − 4)2 =
2!
15
8
√
c(0.01),
for some number c between 3.9 and 4. Taking absolute values, we get the same right hand
side
√
|R1 (3.9)| = 15
c(0.01).
8
√
Since c is positive and increases as c goes from 3.9 to 4, the largest possible value would
be if c = 4, and we get an upper bound for the absolute value of the error
√
|R1 (3.9)| ≤ 15
4(0.01) = 15
(0.01) = 0.0375.
8
4
Because the remainder R1 (3.9) > 0, we have f (3.9) > T1 (3.9), so the linear approximation
is less than the exact value (3.9)5/2 .
√
3. [5] Let f (x) = x. Taylor’s Formula with Remainder says that
√
4.2 = f (4.2) = T2 (4.2) + R2 (4.2),
1
where the remainder (error) is
R2 (4.2) =
f (3) (c)
(4.2 − 4)3 =
3!
1 0.008
√ ,
16 c5
for some number c between 4.2 and 4. Since the expression is already positive, we get
|R2 (4.2)| =
1 0.008
√ .
16 c5
√
√
Since c5 is positive and increases as c goes from 4 to 4.2, its reciprocal 1/ c5 is positive
and decreases as c goes from 4 to 4.2, and the largest possible value for |R1 (4.2)| would occur
if c = 4, so we get an upper bound for the absolute value of the error
|R2 (4.2)| ≤
1 0.008
√
16 45
=
1
64000
= 0.000015625.
The error is positive (R2 (4.2) > 0) and,
T2 (4.2) <
√
4.2.
Using a calculator (and 1(b) above) we get
T2 (4.2) = 2 + 41 (4.2 − 4) −
1
(4.2
64
− 4)2 = 2.049375,
f (4.2) =
√
4.2 = 2.049390153,
therefore
R2 (4.2) = f (4.2) − T2 (4.2) = 0.000015153,
which is positive (as predicted) and its absolute value is less than the upper bound (as
predicted).
4. We are required to find a positive integer n such that the absolute value of the remainder
satisfies
|Rn (23π/60)| < 5 × 10−6
(69o is 23π/60 radians). The Lagrange Remainder Formula gives
|Rn (23π/60)| =
|f (n+1) (c)| 23π π n+1
−3
(n + 1)! 60
where c is some number between 23π/60 and π/3. The derivatives of f (x) = cos x are
f 0 (x) = − sin x, f 00 (x) = − cos x, f (3) (x) = sin x, f (4) (x) = cos x, etc., so the absolute value
|f (n+1) (c)| is either | sin c| or | cos c|. Both of these are always less than or equal to 1 in
absolute value, so it is true that
|f (n+1) (c)| ≤ 1
for all n. Using this, we get an upper bound for the absolute value of the error for the Taylor
polynomial Tn (x):
π n+1
1
|Rn (23π/60)| ≤
.
(n + 1)! 20
2
Trying different values of n, we get
|R1 (23π/60)| ≤
|R3 (23π/60)| ≤
1 π2
≤ 0.013,
2! 202
1 π4
≤ 0.000026,
4! 204
|R2 (23π/60)| ≤
|R4 (23π/60)| ≤
1 π3
≤ 0.00065,
3! 203
1 π5
≤ 0.0000008 = 8 × 10−7 ,
5! 205
and we can stop at
n=4
since 8 × 10−7 < 5 × 10−6 . Therefore the 4th-degree Taylor polynomial T4 (x), of cos x at
π/3, is guaranteed to be within 5 × 10−6 of the exact value of cos 69o , when it is evaluated
at x = 23π/60. (You were not required to calculate T4 (23π/60), but it is approximately
0.3583686496, while cos(23π/60) = 0.3583679494, so the remainder is (approximately) −7 ×
10−7 which is indeed less in absolute value than 5 × 10−6 . It might be true that a lower value
of n would be accurate enough, but we can’t guarantee it unless we do more work, finding
more accurate upper bounds for |f (n+1) (c)|.)
5. “Maclaurin” means “Taylor with a = 0”. Here we use
x = 0.4
since we want ln(1 + x) = ln 1.4. We need to find a positive integer n such that the absolute
value of the remainder satisfies
|Rn (0.4)| < 0.001.
The Lagrange Remainder Formula gives
Rn (0.4) =
f (n+1) (c)
(0.4)n+1
(n + 1)!
where c is some number between 0.4 and 0. The derivatives of f (x) = ln(1 + x) are
f 0 (x) = (1 + x)−1 , f 00 (x) = (−1)(1 + x)−2 , f (3) (x) = (−1)(−2)(1 + x)−3 ,
f (4) (x) = (−1)(−2)(−3)(1 + x)−4 ,
etc., so
|f (n+1) (c)| = (1)(2)(3) · · · (n)(1 + c)−n−1 =
and
|Rn (0.4)| =
n!
(1 + c)n+1
|f (n+1) (c)|
1
|(0.4)n+1 | =
(0.4)n+1 .
|(n + 1)!|
(n + 1)(1 + c)n+1
For any positive integer n the expression (1 + c)n+1 is positive and increases as c increases
from 0 to 0.4, its reciprocal 1/(1 + c)n+1 is positive and decreases as c increases from 0 to 0.4,
so the maximum possible value of |Rn (0.4)| would occur if c = 0. We get the upper bound
|Rn (0.4)| ≤
(0.4)n+1
1
n+1
.
(0.4)
=
(n + 1)(1 + 0)n+1
n+1
3
Trying different values of n, we get
|R1 (0.4)| ≤
(0.4)2
=
2
80
,
1000
|R2 (0.4)| ≤
(0.4)3
= 0.0213333,
3
(0.4)4
(0.4)5
(0.4)6
= 0.0064, |R4 (0.4)| ≤
= 0.002048, |R5 (0.4)| ≤
= 0.000682667,
4
5
6
and we can stop at
n=5
|R3 (0.4)| ≤
since the upper bound is less than 0.001. Therefore the 5th-degree Maclaurin polynomial
T5 (x), of ln(1 + x), is guaranteed to be within 0.001 of ln(1 + 0.4), when it is evaluated
at x = 0.4. (T5 (0.4) = 0.3369813333, ln(1.4) = 0.3364722366, so the actual remainder
is ln(1.4) − T5 (0.4) = −0.0005090967, whose absolute value is indeed less than 0.001, as
predicted.)
6. [5 each] (a) f (x) = 8x5 − 5x4 − 20x3 , −1 ≤ x ≤ 2.
f 0 (x) = 40x4 − 20x3 − 60x2 = 20x2 (x + 1)(2x − 3).
Critical numbers: f 0 (x) = 0 iff x = −1, 0 or 3/2. The critical numbers in the open interval
(−1, 2) are
0, 23 .
Evaluating f at each of the critical numbers gives
,
f (0) = 0, f 23 = − 513
16
and at the end points gives
f (−1) = 7,
f (2) = 16.
Therefore the absolute maximum value is
16
(attained at the end point x = 2) and the absolute minimum value is
− 513
16
(attained at the critical number x = 3/2).
(b) g(x) = x1/3 − x2/3 , −1 ≤ x ≤ 1.
√
1 −2/3 2 −1/3 1 − 2 3 x
√
g (x) = x
− x
=
.
3
3
3
3 x2
√
Critical numbers: g 0 (x)
= 0 iff the numerator 1 − 2 3 x = 0, i.e. x = 1/8; g 0 (x) does not exist
√
3
iff the denominator 3 x2 = 0, i.e. x = 0. The critical numbers in the open interval (−1, 1)
are
0, 81 .
0
4
Evaluating g at the critical numbers gives
g(0) = 0,
1
8
g
= 41 ,
and at the end points gives
g(−1) = −2,
g(1) = 0.
The absolute maximum value is
1
4
(attained at the critical number x = 1/8), and the absolute minimum value is
−2
(attained at the end point x = −1).
(c) h(t) = t2/3 (t − 2)2 = t8/3 − 4t5/3 + 4t2/3 , −1 ≤ t ≤ 1.
20
8
4(2t − 1)(t − 2)
8
√
.
h0 (t) = t5/3 − t2/3 + t−1/3 =
3
3
3
33t
√
h0 (t) = 0 iff 4(2t − 1)(t − 2) = 0, i.e., t = 1/2 or 2. h0 (t) does not exist iff 3 3 t = 0, i.e.,
t = 0. The critical numbers in the open interval (−1, 1) are
1
.
2
0,
(The critical point t = 2 is outside the domain.) Evaluating h at the critical numbers gives
h(0) = 0,
h
1
2
9
= √
,
434
and evaluating at the end points gives
h(−1) = 9,
h(1) = 1.
so the absolute maximum value is
9
(attained at the end point t = −1), and the absolute minimum value is
0
(attained
at the critical number t = 0). (How can you tell without a calculator that 9 >
√
3
9/(4 4)?)
5