Student (Print)
Section
Last, First Middle
Student (Sign)
Student ID
Instructor
MATH 152,
Fall 2007
Common Exam 3
Test Form A
Solutions
Instructions:
You may not use notes, books, calculator or computer.
For Dept use Only:
1-11
/55
12
/15
13
/12
14
/12
15
/12
TOTAL
/106
Part I is multiple choice. There is no partial credit.
Mark the Scantron with a #2 pencil. For your own records,
also circle your choices in this exam. Scantrons will be collected
after 90 minutes and may not be returned.
Part II is work out. Show all your work. Partial credit will be given.
THE AGGIE CODE OF HONOR:
An Aggie does not lie, cheat or steal, or tolerate those who do.
1
Part I: Multiple Choice (5 points each)
There is no partial credit.
3n
lim
n→∞ 1 + 4 n
1. Compute
correct choice
a. 0
b. 3
4
1
1− 3
4
3
5
d.
1− 3
4
e. ∞
c.
lim
n→∞
n
3
1 + 4n
3n
4n
= lim
= 0 =0
n→∞ 1
1
+1
4n
∞
2. Compute
∑
n=1
2 n+1
5n
a. 4
b. 10
c. 4
correct choice
3
d. 10
3
e. The series diverges.
1+1
a = a1 = 2 1 = 4
5
5
4
5
=
= 4 = 4
2
5−2
3
1−
5
Geometric:
∞
∑
n=1
2 n+1
5n
r= 2
5
(converges)
∞
3. Compute S =
∑3 1/n − 3 1/n+1 
n=1
a. 3
correct choice
b. 2
c. 1
d. 0
e. The series diverges.
k
Sk =
∑3 1/n − 3 1/n+1  = 3 1 − 3 1/2  + 3 1/2 − 3 1/3  + ⋯ + 3 1/k − 3 1/k+1  = 3 1 − 3 1/k+1
n=1
S = lim S k = lim 3 1 − 3 1/k+1  = 3 1 − 3 0 = 3 − 1 = 2
k∞
k∞
2
∞
−1 n
n
∑
4. The series
n=1
is
a. absolutely convergent.
b. absolutely convergent but not convergent.
c. convergent but not absolutely convergent.
correct choice
d. divergent.
∞
∑
n=1
−1 n
n
is convergent because it is an alternating decreasing series and n∞
lim 1 = 0.
n
∞
The related absolute series is
p = 1 < 1. So
2
∑
n=1
∞
∑
n=1
−1 n
n
1
n
which is divergent because it is a p-series with
is convergent but not absolutely convergent.
5. Find the coefficient of x 6 in the Maclaurin series for fx = sinx 2 .
a. 0
1
24
c. − 1
6
d. − 1
2
e. 1
b.
correct choice
sin t = t − 1 t 3 + 1 t 5 + ⋯
6
5!
∞
6. Compute
∑
n=0
1
e2
b. −1
e2
c. 1
2
d. −e 2
a.
−2 n
n!
sin x 2 = x 2 − 1 x 6 + 1 x 10 + ⋯
6
5!
The coefficient of x 6 is − 1 .
6
HINT: Think about a known Maclaurin series.
correct choice
e. The series diverges.
∞
∑
n=0
xn = ex
n!
∞
So
∑
n=0
−2 n
= e −2 = 12
n!
e
3
7. Compute
cosx 3  − 1 + 1 x 6
2
lim
x→0
x 12
a. − 1
b.
c.
d.
e.
24
−1
4
1
4
1
6
1
24
correct choice
2
cosx 3  − 1 + 1 x 6
2
lim
= lim
12
x→0
x→0
x
∞
8. The series S =
∑
n=0
3n
1 + 4n
1−
4
x 3 
x 3 
+
⋯
2
4!
x
− 1 + 1 x6
2
12
4
= lim
x→0
x 3 
+⋯
24x 12
= 1
24
satisfies
a. S = 0
b. 0 < S < 4
correct choice
c. S = 4
d. S > 4
e. The series diverges.
∞
∑
n=0
3n
4n
is geometric.
∞
Since |r| < 1,
∑
n=0
∞
Also
∑
n=0
a = 1 and r = 3 .
4
3n =
1
=4
4n
1− 3
4
∞
So
∑
n=0
3n <
1 + 4n
∞
∑
n=0
3n = 4
4n
3 n > 0 because all terms are positive.
1 + 4n
4
9. Find the equation of the sphere centered at 1, 2, 3 which passes through the
point −3, 0, −1.
a. x − 3 2 + y 2 + z + 1 2 = 6
b. x + 3 2 + y 2 + z − 1 2 = 36
c. x + 3 2 + y 2 + z − 1 2 = 6
d. x − 1 2 + y − 2 2 + z − 3 2 = 36
correct choice
e. x − 1 + y − 2 + z − 3 = 6
2
2
2
The radius is the distance from 1, 2, 3 to −3, 0, −1:
r = d 1, 2, 3, −3, 0, −1
=
1 − −3 2 + 2 − 0 2 + 3 − −1 2 =
42 + 22 + 42 = 6
The circle is x − 1 2 + y − 2 2 + z − 3 2 = 36
10. A vector ⃗
u has length 3. A vector ⃗v has length 4. The angle between them is 45°.
Find ⃗
u ⋅ ⃗v.
a. 3
b. 12
c. 12 2
d. 6
e. 6 2
correct choice
⃗
⃗ | |v⃗| cos θ = 3 ⋅ 4 ⋅ cos 45° = 12 = 6 2
u ⋅ ⃗v = |u
2
11. A triangle has vertices A = 1, 2, 3, B = 2, 3, 3 and C = 1, 3, 2. Find the angle at A.
a. 0°
b. 30°
c. 45°
d. 60°
correct choice
e. 90°
AB = B − A = 2, 3, 3 − 1, 2, 3 = 1, 1, 0
AB ⋅ AC = 0 + 1 + 0 = 1
cos θ =
AB ⋅ AC
AB AC
=
AB =
1
= 1
2
2 2
AC = C − A = 1, 3, 2 − 1, 2, 3 = 0, 1, −1
1+1+0 =
2
AC =
0+1+1 =
2
θ = 60°
5
Part II: Work Out (points indicated)
Show all your work. Partial credit will be given.
12. (15 points) For each series, determine if it is convergent or divergent.
Be sure to identify the Convergence Test and check out its hypotheses.
∞
a.
∑ ne
−n 2
n=2
Consider fx = x e
Integral Test:
Note:
The terms are positive,
f is decreasing because f ′ x = e
−x 2
f interpolates:
−x 2
− xe
−x 2
fn = n e
2x = e
−x 2
−n 2
and
1 − 2x 2  < 0 for x ≥ 2.
We integration using the substitution u = −x 2
du = −2x dx
∞
∞
∫ 2 x e −x2 dx = − 12 ∫ e u du = − 12 e −x2 2 = 12 e −4 converges
∞
∑ ne
So
−n 2
converges.
n=2
∞
b.
∑
n=2
n5 − 3
n6 + 1
Limit Comparison Test:
∞
Compare to
∑
n=2
Note:
1
n
which is a divergent harmonic series (p-series with p = 1).
The Comparison Test will not work because
n5 − 3 < n5 = 1
n
n6 + 1
n6
5
lim a n = n∞
lim n 6 − 3 ⋅ n = 1 which is finite and non-zero.
bn
n +1 1
n∞
∞
So
∑
n=2
n5 − 3
n6 + 1
diverges also.
6
∞
c.
∑−1 n n −n 1
n=2
n =1
n−1
oscillates between numbers close to 1 and −1.
(N-th Term) Test for Divergence:
n
n−1
So n∞
lim −1 n n
n−1
So −1 n
lim
n∞
does not exist.
∞
So
∑−1 n n −n 1
diverges.
n=2
∞
13. (12 points) Find the radius and interval of convergence of the series
∑
n=1
|x − 4| n+1
n + 1 3 n+1
|x − 4| n+1
|x − 4|
n 3 n = |x − 4| lim
n
ρ = n∞
lim aan+1
=
lim
=
n∞ n + 1 3 n+1 |x − 4| n
n
3 n∞ n + 1
3
|x − 4|
The series is absolutely convergent when ρ =
< 1 or |x − 4| < 3.
3
So the radius of convergence is R = 3 and the series is absolutely convergent on 1, 7.
Ratio Test:
|a n | =
|x − 4| n
n 3n
−1 n
x − 4 n .
n 3n
|a n+1 | =
We check the endpoints:
∞
x = 1:
∑
n=1
∞
x = 7:
∑
n=1
∞
−1 n
1 − 4 n =
n 3n
∑
−1 n
7 − 4 n =
n 3n
∑
n=1
∞
1
n
which is a divergent harmonic series.
−1 n
n
which is convergent by the Alternating Series Test.
n=1
So the interval of convergence is 1 < x ≤ 7 or 1, 7.
7
14. (12 points)
a. Find the 2 nd degree Taylor polynomial, T 2 x, for fx =
1
x
at x = 4.
3
f ′ x = − 1 x −3/2 = −1 3
f ′′ x = 3 x −5/2 =
2
4
4 x5
2 x
3
f4 = 1 = 1
f ′ 4 = −1 3 = −1
f ′′ 4 =
= 3
5
2
16
128
4
4 4
2 4
T 2 x = f4 + f ′ 4x − 4 + 1 f ′′ 4x − 4 2 = 1 − 1 x − 4 + 3 x − 4 2
2
2
16
256
fx =
1 = x −1/2
x
b. Use this polynomial to approximate
1 . Do not simplify.
6
1 = f6 ≈ T 2 6 = 1 − 1 6 − 4 + 3 6 − 4 2
2
16
256
6
= 1 − 1 + 3 = 0. 42188
2
8
64
Exact value:
OK to stop here
1 = 0. 40825
6
c. Estimate the error in your approximation in part (b). Do not simplify. Justify your answer.
METHOD 1: This is the beginning of an alternating series.
So the error in the approximation is less than the absolute value of the next term.
f ′′′ x = −15 x −7/2 = −157
f ′′′ 4 = −157 = −15
8
1024
8 x
8 4
The cubic term in T 3 is 1 f ′′′ 4x − 4 3 = 1 −15 x − 4 3 = −5 x − 4 3 .
6
6 1024
2048
1
−5 6 − 4 3 = −5
The next term in
is obtained by setting x = 6 or
2048
256
6
|E| < 5 ≈ 1. 95 × 10 −2
256
METHOD 2: The Taylor Remainder Theorem says:
If you use T n x, the Taylor polynomial of degree n, to approximate the function, fx,
then the remainder, R n x = fx − T n x, is bounded by |R n x| ≤
Mx − a n+1
n + 1!
where a is the center and M ≥ f n+1 c for all c between x and a.
In this case, n = 2, a = 4, x = 6 and f ′′′ x = −15 x −7/2 = −157 .
8
8 x
15
On the interval 4, 6, the largest value of
occurs at x = 4.
8 x7
So we take M =
15 = 15 .
7
1024
8 4
Then |R 2 x| ≤
3
15 6 − 4 = 5 .
1024 3!
256
8
15. (12 points)
2
a. Find a Maclaurin series for e −x .
∞
et =
∑
n=0
tn = 1 + t + t2 + t3 + t4 + ⋯
n!
2
6
24
∞
e
−x 2
=
∑
n=0
−x 2 
n!
b. Write the integal
∫0
0.1
e
−x 2
dx =
∞
n
∑
=
n=0
∫0
0.1
∞
∫0 ∑
0.1
n=0
−3
= 0. 1 − 10
3
4
6
8
−1 n x 2n
= 1 − x 2 + x − x + x − ⋯ (alternating series)
n!
2
6
24
e −x dx as an infinite series using summation notation.
2
−1 n x 2n
n!
−5
∞
dx =
∑
n=0
−7
+ 10 − 10 + ⋯
10
42
−1 n x 2n+1
n! 2n + 1
0.1
∞
=
∑
n=0
0
−1 n 0. 1 2n+1
n! 2n + 1
(alternating series)
c. Use the series from part (b) to approximate the integal
∫0
0.1
e −x dx to within ± 10 −5 .
2
Explain how you know your estimate is correct to within ± 10 −5 .
Keeping 2 terms:
∫0
0.1
2
e −x dx ≈ 0. 1 − . 001 ≈ 0. 1 −. 0003333 =. 0996667
3
The integral is expressed as an alternating series.
When an alternating series is approximated by using the first k terms
the error is less than the absolute value of the k + 1-th term.
−5
In this case the error is less than 10 = 10 −6 < 10 −5 .
10
9