MAT 266 Test 3 SOLUTIONS, FORM A
1. [20 points] Write out the first four terms of the Taylor series for f (x) = sin(4x) at x =
Solution: The formula for the terms of the Taylor series of f (x) at x = a is
π
.
3
∞
X
f (n) (a)
(x − a)n .
n!
n=0
Hence, we need to find the first few derivatives at x = 3:
n
f (n) (x)
f (n)
π
3
√
0
sin(4x)
1
4 cos(4x)
2
−16 sin(4x)
3
−64 cos(4x)
−
3
2
1
= −2
2
√
√
3
−16 · −
=8 3
2
1
−64 · − = 32
2
4·−
Hence the first few terms of the Taylor series are
√
π 8 3 π 2 32 π 3
3
−2 x−
+
x−
x−
−
+
+ ···
2
3
2
3
6
3
√
or
−0.866 − 2(x − 1.0472) + 6.928(x − 1.0472)2 + 5.3333(x − 1.0472)3 + · · ·
Grading: +10 points for the derivatives, +5 points for the general formula, +5 points for writing out the formula.
2
MAT 266 Test 3 SOLUTIONS, FORM A
∞
X
3
.
2. Consider the series
i
2
i=1
a. [15 points] Write out the first three partial sums of this series.
Solution: Letting a =
3
, the first three partial sums are
2i
s1 = a1 =
3
2
s2 = a1 + a2 =
or
1.5
9
4
s3 = a1 + a2 + a3 =
or
21
8
2.25
or
2.625
Grading: +5 points for each. Grading for common mistakes: −10 points for only s3 .
b. [15 points] Does this series converge? If so, to what?
1
an+1
1
= , so this series is a geometric series. The common ratio r = is less
an
2
2
3/2
than 1 in absolute value, so it converges to the first term divided by 1 − r:
= 3.
1 − 1/2
Grading: +5 points for recognizing the series as being geometric, +5 points for the formula,
+5 points for substitution. Grading for common mistakes: −10 points for saying that the sequence
3
converges to 0; −3 points for not showing any work; −5 points for r = ; −8 points for using the
2
Ratio Test.
Solution: Note that
3
MAT 266 Test 3 SOLUTIONS, FORM A
∞
X
4n − 8
converges, diverges, or is
3. [20 points] Use the Ratio Test to determine whether the series
5n−1
n=0
inconclusive.
4n − 8
. Then
5n−1
n−1
an+1 1 4n − 4
1
1
= lim 4(n + 1) − 8 · 5
L = lim = lim ·
= ·1= .
(n+1)−1
n→∞
n→∞
n→∞
an
4n − 8
5 4n − 8
5
5
5
Solution: Let an =
Since L < 1, the series converges.
Grading: +5 points for setting up the limit of the ratio; +5 points for simplifying the limit;
+5 points for calculating the limit; +5 points for the statement of convergence.
4
MAT 266 Test 3 SOLUTIONS, FORM A
4. [15 points] Does the sequence an =
5n + 5n
converge? If so, to what?
2 · 5n − 7n
Solution: You need to determine whether lim an is a real number (in which case, an converges to
n→∞
that number), or something else (in which case, an diverges). One way is to calculate
1
5x + 5x L0 H
ln 5 · 5x + 5 L0 H
ln 5 · ln 5 · 5x
1
= lim
=== lim
=== lim
= .
x
x
x→∞ 2
x→∞ 2 · 5 − 7x
x→∞ 2 · ln 5 · 5 − 7
x→∞ 2 · ln 5 · ln 5 · 5x
2
lim
1
.
2
Grading for common mistakes: −10 points for doing the Ratio Test; −3 points for not showing
any work.
Hence, an converges to
5. [15 points] The Maclaurin series for
1
is
1 − 3x
1 + 3x + 9x2 + 27x3 + 81x4 + 243x5 + · · · ,
1
3
which has a Radius of Convergence of . Find the Maclaurin series for
and its radius of
3
(1 − 3x)2
convergence.
Solution: This involves manipulation of power series. Note that
3
d
1
=
, so you
(1 − 3x)2
dx 1 − 3x
only need to dfferentiate the power series:
1
= 1 + 3x + 9x2 + 27x3 + 81x4 + 243x5 + · · ·
1 − 3x
1
d d
=
1 + 3x + 9x2 + 27x3 + 81x4 + 243x5 + · · ·
dx 1 − 3x
dx
3
= 3 + 9 x2 + 81 x2 + 324 x3 + 1215 x4 + · · ·
(1 − 3x)2
Since differentiating does not change the radius of convergence, the radius of convergence of the
1
new power series is .
3
Grading: +5 points for recognizing how to manipulate the series, +5 points for integrating,
+5 points for giving the radius of convergence. Grading for common mistakes: −3 points for multiplying by −3; −8 points for squaring both sides.
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MAT 266 Test 3 SOLUTIONS, FORM B
1. [20 points] Write out the first four terms of the Taylor series for f (x) = cos(3x) at x =
Solution: The formula for the terms of the Taylor series of f (x) at x = a is
π
.
4
∞
X
f (n) (a)
(x − a)n .
n!
n=0
Hence, we need to find the first few derivatives at x = 3:
n
f (n) (x)
0
cos(3x)
1
−3 sin(3x)
2
−9 cos(3x)
3
27 sin(3x)
f (n)
π
3
1
−√
2
−3
√
2
9
√
2
27
√
2
Hence the first few terms of the Taylor series are
1
3 9 27 π
π 2
π 3
−√ − √ x −
+ √ x−
+ √ x−
+ ···
4
4
4
2
2
2 2
6 2
or
−0.7071 − 2.1213(x − 0.7854) + 3.1820(x − 7854)2 + 3.1820(x − 0.7854)3 + · · ·
Grading: +10 points for the derivatives, +5 points for the general formula, +5 points for writing out the formula.
2
MAT 266 Test 3 SOLUTIONS, FORM B
∞
X
2
.
2. Consider the series
i
3
i=1
a. [15 points] Write out the first three partial sums of this series.
Solution: Letting a =
2
, the first three partial sums are
3i
s1 = a1 =
2
3
s2 = a1 + a2 =
or
0.6667
8
9
s3 = a1 + a2 + a3 =
or
26
27
0.8889
or
0.9630
Grading: +5 points for each. Grading for common mistakes: −10 points for only s3 .
b. [15 points] Does this series converge? If so, to what?
1
an+1
1
= , so this series is a geometric series. The common ratio r = is less
an
3
3
2/3
than 1 in absolute value, so it converges to the first term divided by 1 − r:
= 1.
1 − 1/3
Grading: +5 points for recognizing the series as being geometric, +5 points for the formula,
+5 points for substitution. Grading for common mistakes: −10 points for saying that the sequence
2
converges to 0; −3 points for not showing any work; −5 points for r = ; −8 points for using the
3
Ratio Test.
Solution: Note that
3
MAT 266 Test 3 SOLUTIONS, FORM B
∞
X
3n+2
3. [20 points] Use the Ratio Test to determine whether the series
converges, diverges, or is
2n + 5
n=0
inconclusive.
3n+2
. Then
2n + 5
(n+1)+2
an+1 2n + 5
2n + 5
= lim 3
= lim 3 ·
L = lim ·
= 3 · 1 = 3.
n→∞
n→∞
an n→∞ 2(n + 1) + 5 3n+2
2n + 7
Solution: Let an =
Since L > 1, the series diverges.
Grading: +5 points for setting up the limit of the ratio; +5 points for simplifying the limit;
+5 points for calculating the limit; +5 points for the statement of divergence.
4
MAT 266 Test 3 SOLUTIONS, FORM B
4. [15 points] Does the sequence an =
3n + n
converge? If so, to what?
4 · 3n − n
Solution: You need to determine whether lim an is a real number (in which case, an converges to
n→∞
that number), or something else (in which case, an diverges). One way is to calculate
1
3x + x L0 H
ln 3 · 3x + 1 L0 H ln 3 · ln 3 · 3x
1
= lim
=== lim
===
= .
x
x→∞ 4
x→∞ 4 · 3 − x
x→∞ 4 · ln 3 · 3x − 1
4 · ln 3 · ln 3 · 3x
4
lim
1
.
4
Grading for common mistakes: −10 points for doing the Ratio Test; −3 points for not showing
any work.
Hence, an converges to
5. [15 points] The Maclaurin series for
4
is
1 + 4x
4 − 16x + 64x2 − 256x3 + 1024x4 + · · · ,
1
which has a Radius of Convergence of . Find the Maclaurin series for ln(1 + 4x) and its radius of
4
convergence.
Z
4
Solution: This involves manipulation of power series. Note that
dx = ln(1 + 4x) + C, so
1 + 4x
you only need to integrate the power series (and determine the constant of integration):
4
= 4 − 16x + 64x2 − 256x3 + 1024x4 + · · ·
1 + 4x
Z
Z
4
dx =
4 − 16x + 64x2 − 256x3 + 1024x4 + · · · dx
1 + 4x
64 3
1024 5
ln(1 + 4x) = C + 4 x − 8 x2 +
x − 64 x4 +
x + ···
3
5
To determine the value of C, substitute x = 0 to obtain ln(1 + 4 · 0) = C, or C = 0. Thus, the
Maclaurin (power) series for ln(1 + 4x) is
4 x − 8 x2 +
64 3
1024 5
x − 64 x4 +
x + ···.
3
5
Since integration does not change the radius of convergence, the radius of convergence of the
1
new power series is .
4
Grading: +5 points for recognizing how to manipulate the series, +5 points for integrating,
+5 points for giving the radius of convergence. Grading for common mistakes: −3 points for dividing by 4; −2 points for leaving a +C in the final answer.
5