MATH 152, SPRING 2012
COMMON EXAM III - VERSION A Solutions
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PART I: Multiple Choice (4 pts each)
1. What is the intersection of the sphere (x − 2)2 + (y − 1)2 + (z − 3)2 = 16 with the xy-plane?
(a) The circle (x − 2)2 + (y − 1)2 = 16.
(b) The circle (x − 2)2 + (y − 1)2 = 1.
(c) The point (2, 1, 0).
(d) The circle (x − 2)2 + (y − 1)2 = 7. Correct Answer
(e) The sphere does not intersect the xy-plane.
Solution: The sphere intersects the xy-plane when z = 0. Substituting z = 0 into the equation
(x − 2)2 + (y − 1)2 + (z − 3)2 = 16 yields (x − 2)2 + (y − 1)2 + (0 − 3)2 = 16, hence (x − 2)2 + (y − 1)2 = 7.
2. If we apply the Ratio Test to the series
∞ (−4)n n2
P
, which of the following statements is true?
5n+1
n=1
an+1 = − 4 < 1, therefore the series converges.
(a) lim n→∞
an 5
an+1 4
= < 1, therefore the series converges. Correct Answer
(b) lim n→∞
an 5
an+1 = 4 < 1, therefore the series converges.
(c) lim n→∞
an 25
an+1 = ∞ > 1, therefore the series diverges.
(d) lim n→∞
an an+1 = 0 < 1, therefore the series converges.
(e) lim n→∞
an n+1
an+1 (−4)n (−4)(n + 1)2 5n+1 (n + 1)2 5n+1 = lim (−4)
Solution: lim =
lim
n→∞
an n→∞ 5n+2
(−4)n (n2 ) n→∞ 5n+1 (5)
(−4)n (n2 ) (−4)(n + 1)2 4
= < 1, therefore the series converges absolutely, therefore converges.
= lim 5
n→∞
5n2
3. What does it mean to say
∞
P
an is convergent?
n=1
(a) lim an = 0.
n→∞
(b) lim an exists.
n→∞
(c) There are numbers A and A′ such that A ≤ an ≤ A′ for all n.
(d) There are numbers B and B ′ such that B ≤ a1 + a2 + ... + an ≤ B ′ for all n.
(e) lim (a1 + a2 + ... + an ) exists. Correct Answer
n→∞
Solution: The partial sum is sn = a1 + a2 + ... + an and
∞
P
n=1
1
an = lim sn .
n→∞
4. Which of the following is a unit vector in the direction of b − 3a, where a = h1, −2, 2i and b = h5, 0, −3i?
6 9
2
(a) − , − ,
11 11 11
2 6
9
Correct Answer
(b)
, ,−
11 11 11
(c) h2, 6, −9i
14
2
18
(d) − √
, −√
,√
524
524 524
2
18
14
,√
, −√
(e) √
524
524
524
Solution: b − 3a = h5, 0, −3i − 3 h1, −2, 2i = h2, 6, −9i. Divide
by the magnitude
to make this vector a unit vector.
√
9
2 6
h2, 6, −9i
.
=
, ,−
Since | h2, 6, −9i | = 4 + 36 + 81 = 11, u =
11
11 11 11
5. Represent
1
as a power series about 0 and give the interval of convergence.
1 + 4x2
(a)
∞
X
1
1
=
(−4)n x2n , where |x| < . Correct Answer
1 + 4x2
2
n=0
(b)
∞
X
1
1
=
(−4)n x2n , where |x| < .
1 + 4x2
4
n=0
(c)
∞
X
1
=
(−4)n x2n , where |x| < 4.
1 + 4x2
n=0
(d)
∞
X
1
1
=
4n x2n , where |x| < .
1 + 4x2
2
n=0
(e)
∞
X
1
1
=
4n x2n , where |x| < .
1 + 4x2
4
n=0
Solution:
∞
∞
X
X
1
1
2 n
=
(−4x
)
=
(−4)n x2n , where | − 4x2 | < 1 , hence |x| < .
1 + 4x2
2
n=0
n=0
6. Find the cosine of the angle, θ, between the vectors a = h2, 0, 1i and b = h1, −2, 1i.
4
(a) cos(θ) = √
30
5
(b) cos(θ) = √
30
3
(c) cos(θ) = √
12
5
(d) cos(θ) = √
12
3
(e) cos(θ) = √ Correct Answer
30
Solution: cos(θ) =
h2, 0, 1i · h1, −2, 1i
3
a·b
=
=√
|a||b|
| h2, 0, 1i || h1, −2, 1i |
30
2
7. Find the 15th derivative at x = 0 for f (x) =
(−1)n
xn .
n
n=0 7 (n + 5)
∞
P
(a) f (15) (0) = 0
15!
715 (20)
15!
Correct Answer
(c) f (15) (0) = − 15
7 (20)
15
(d) f (15) (0) = − 15
7 (20)
15
(e) f (15) (0) = 15
7 (20)
(b) f (15) (0) =
(−1)n
1
1
1
1
1
1
xn = −
x + 2 x2 − 3 x3 + ... − 15
x15 + 16
x16 + .... Now, if
n (n + 5)
7
5
7(6)
7
(7)
7
(8)
7
(20)
7
(21)
n=0
the degree on x is less than 15, the 15th derivative of these terms will be zero. If the degree on x is greater than 15,
the 15th derivative of these terms will still contain a power of x, and hence will equal 0 when x = 0. If the degree
15!
on x is 15, then f (15) (x15 ) = 15! Thus f 15 (0) = − 15
7 (20)
Solution: f (x) =
∞
P
8. Find the Taylor Polynomial of degree 2, T2 (x), for f (x) =
1
(x − 9)2
216
1
T2 (x) = 3 + (x − 9) −
6
1
T2 (x) = 3 + (x − 9) −
6
1
(x − 9)2
T2 (x) = −
108
None of these.
√
x centered at a = 9.
(a) T2 (x) = −
(b)
(c)
(d)
(e)
1
(x − 9)2 Correct Answer
216
1
(x − 9)2
108
Solution: T2 (x) = f (9) + f ′ (9)(x − 9) +
f ′′ (9)
(x − 9)2 .
2
1
1
(x − 9)2 .
T2 (x) = 3 + (x − 9) −
6
216
9. The series
∞
P
2
ne−n
n=1
(a) converges by the Integral Test. Correct Answer
(b) diverges by the Ratio Test.
(c) diverges by the Test for Divergence.
(d) diverges by the p-series test.
(e) converges by the Comparison Test with
∞
P
2
e−n .
n=1
2
Solution: Since xe−x is a continuous, positive, decreasing function on [1, ∞), we may apply the integral test:
Z
∞
∞
2
2
1
1
xe−x dx = − e−x = . Since the improper integral converges, so does the series.
2
2e
1
1
3
10. In three-dimensional space, R3 , the equation x2 + y 2 = 9 describes
(a) a parabola.
(b) a sphere.
(c) a cylinder. Correct Answer
(d) a circle.
(e) a plane.
Solution: In two-dimension, x2 + y 2 = 9 describes a circle. Thus, in three-dimension, as z varies, x2 + y 2 = 9
describes a cylinder.
11. The series
∞ (−1)n n
P
2
n=1 n + 1
(a) is absolutely convergent.
(b) is convergent but not absolutely convergent. Correct Answer
(c) is divergent.
(d) is absolutely divergent but not divergent.
(e) converges absolutely by the ratio test.
∞ (−1)n n
P
n
converges by the Alternating Series Test since the sequence
decreases
Solution: The series
2
n2 + 1
n=1 n + 1
∞
P
n
for convergence. Using the Limit Comparison
to zero. To test for absolute convergence, test the series
2
n=1 n + 1
n
∞ 1
∞
∞ (−1)n n
2+1
P
P
P
1
n
an
n
= 1 > 0. Since
Test, with bn = , lim
= lim
diverges, so does
. Hence
2
2
1
n→∞
n n→∞ bn
n=1 n
n=1 n + 1
n=1 n + 1
n
x
is continuous, positive and decreasing
is converget but not absolutely convergent. Another solution: Since 2
x +1
∞
P
n
diverges:
on [1, ∞), we may use the integral test to show
2+1
n
n=1
Z ∞
∞
1
x
2
= ∞.
dx
=
ln(x
+
1)
1
x2 + 1
2
1
12. The series
1
√ is
n+
e
n
n=1
∞
P
∞ 1
P
1
1
√
and
is a divergent geometric series.
>
n
en + n
en
n=1 e
∞ 1
P
1
1
is a convergent geometric series. Correct Answer
(b) convergent because n √ < n and
n
e + n
e
n=1 e
∞
P
1
1
1
√ is a convergent p series.
(c) convergent because n √ < √ and
e + n
n
n
n=1
∞
P 1
1
1
√ is a divergent p series.
(d) divergent because n √ < √ and
e + n
n
n
n=1
∞
P 1
1
1
√ is a divergent p series.
(e) divergent because n √ > √ and
e + n
n
n
n=1
∞ n
X
∞ 1
P
1
1
1
1
=
is a convergent geometric series (r = ), and n √ < n , it follows
Solution: Since
n
e
e
e
e
e
+
n
n=1
n=1
(a) divergent because
that
∞
X
1
1
√
, thus by the comparison test, since the larger series converges, so does the smaller series.
<
n
n n=1 en
n=1 e +
∞
P
4
∞ (−2)n
P
.
n!
n=0
13. Find the sum of the series
(a) e2
(b) cos(−2)
(c) sin(−2)
(d) 0
(e) e−2 Correct Answer
Solution: Since ex =
∞ (−2)n
∞ xn
P
P
, it follows that e−2 =
.
n!
n=0
n=0 n!
PART II: WORK OUT (48 points total)
Directions: Present your solutions in the space provided. Show all your work neatly and concisely and box your
final answer. You will be graded not merely on the final answer, but also on the quality and correctness of the work
leading up to it.
14. a.) (5 pts) Using the known Maclaurin series for sin x, find the Maclaurin series for sin(x3 ).
∞
∞ X
∞ (−1)n x2n+1
P
P
(−1)n x6n+3
, thus sin(x3 ) =
.
Solution: sin x =
n=0 n=0 (2n + 1)!
n=0 (2n + 1)!
sin(x3 ) − x3
.
x→0
x9
b.) (5 pts) Using the Maclaurin series for sin(x3 ), compute lim
(−1)n x6n+3
x9
x15
x21
3
− x3
x
−
+
−
+ ... − x3
sin(x ) − x
(2n + 1)!
3!
5!
7!
=
lim
=
lim
Solution: lim
x→0
x→0
x→0
x9
x9
x9
x6
x12
1
−
+ ...
x9 − +
1
1
3!
5!
7!
= lim
=− =−
x→0
x9
3!
6
3
3
2
15. (5 pts) Using the known Maclaurin series for ex , find a Maclaurin series for e−x .
∞ xn
∞ (−1)n x2n
P
P
2
Solution: ex =
, thus e−x =
.
n!
n=0 n!
n=0
2
b.) (5 pts) Expand the Maclaurin series for e−x found above to degree 4. Use this 4th degree Maclaurin polynomial
Z 1
2
e−x dx. Do not simplify.
to estimate
0
∞ (−1)n x2n
P
x4
≈ 1 − x2 + .
n!
2
n=0
1
Z 1
Z 1
x3
1
x5 1
23
x4
2
−x2
dx = x −
+
1−x +
Thus
e
dx ≈
= 1 − 3 + 10 = 30 .
2
3
10
0
0
0
2
Solution: e−x =
16. (10 pts) Find a power series about 0 for f (x) = ln(6x + 5). What is the associated radius of convergence?
n+1
n X
∞
∞ 6x 6x
6
d
6
6
6X
=
−
=
(−1)n
xn , where < 1.
Solution:
(ln(6x + 5)) =
= 6x
dx
6x + 5
5 n=0
5
5
5
n=0
5 1+
5
n+1
n+1 n+1
Z X
∞
∞
X
6
6
x
5
(−1)n
Thus, ln(6x + 5) =
xn dx = C +
(−1)n
, where |x| < . Choosing x = 0 to
5
5
n
+
1
6
n=0
n=0
n+1 n+1
∞
P
x
5
6
solve for C yields C = ln 5. Thus ln(6x + 5) = ln 5 +
,R= .
(−1)n
5
n
+
1
6
n=0
5
∞ (2x − 1)n
P
. Be sure to test the endpoints for convergence.
n4n
n=1
an+1 (2x − 1)(n) 2x − 1 (2x − 1)n+1
n4n .
Solution: Using the Ratio Test, lim = lim
= lim
=
n→∞
an n→∞ (n + 1)4n+1 (2x − 1)n n→∞ (n + 1)4 4 ∞
an+1 P
< 1, then the series
The Ratio Test states if lim an converges. Hence
n→∞
an
n=1
2x − 1 < 1, so |2x − 1| < 4, thus x − 1 < 2. Solve this inequality for x: − 3 < x < 5 .
4 2
2
2
Test the endpoints of the interval for convergence :
∞
X
∞ (−4)n
(−1)n
3 P
=
which converges by the Alternating Series Test since the sequence
Test x = − :
2 n=1 n4n
n
n=1
1
decreases to zero.
n
∞
X
∞ (4)n
1
5 P
=
which diverges by the p-series test. Thus the interval of convergence is
Test x = :
2 n=1 n4n
n
n=1
17. (10 pts) Find the interval of convergence of the series
−
3
5
≤x< .
2
2
18. Consider the series
∞ (−1)n
P
.
2
n=1 2n + 1
a.) (4 pts) Prove the series is absolutely convergent.
∞
∞
∞
X
∞ 1
∞ (−1)n P
P
1
1
1X 1
= P
=
. Since
is a convergent p-series, it follows
Solution:
<
2n2 + 1 2
2
2
2
2n
2 n=1 n
n=1 2n + 1
n=1 n
n=1
n=1
∞
∞ (−1)n
∞
P
P
P
1
1
converges, thus
also converges by the comparison test. Hence
is absolutely
that
2
2
2
n=1 2n + 1
n=1 2n + 1
n=1 2n
convergent.
∞ (−1)n
P
b.) (2 pts) Find s2 , the sum of the first 2 terms, to approximate
.
2
n=1 2n + 1
Solution: S2 =
2 (−1)i
P
1 1
2
=− + =−
2
3 9
9
i=1 2i + 1
c.) (2 pts) Find an upper bound on |R2 |, the absolute value of the remainder, in using s2 to approximate
Solution: |R2 | ≤ |a3 |, where an =
(−1)n
1
. Thus |R2 | ≤
.
2n2 + 1
19
6
∞ (−1)n
P
.
2
n=1 2n + 1