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12/7/14, 10:10 PM
REVIEW FOR THE EXAM 4 MATH 203 F 2014 (6655401)
Question
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35
Description
Basically practice with the previous Final. You should be fine.
Instructions
Important Review : Fall 2011: 6,7 and 12. Fall 2010: 6,7 and 12. Spring 2009: 6,7 and
11. Fall 2008: 4,5 and 9.
1.
Question Details
SEssCalc2 8.2.009. [2164888]
-
SEssCalc2 8.2.012. [2164799]
-
Determine whether the geometric series is convergent or divergent.
∞
(−8)n − 1
9n
n=1
convergent
divergent
If it is convergent, find its sum. (If the quantity diverges, enter DIVERGES.)
Solution or Explanation
Click to View Solution
2.
Question Details
Determine whether the geometric series is convergent or divergent.
∞
n=0
1
(
7 )n
convergent
divergent
If it is convergent, find its sum. (If the quantity diverges, enter DIVERGES.)
Solution or Explanation
Click to View Solution
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3.
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Question Details
SEssCalc2 8.2.021. [2164397]
-
SEssCalc2 8.2.027. [2377982]
-
Determine whether the series is convergent or divergent.
∞
arctan 3n
n=1
convergent
divergent
If it is convergent, find its sum. (If the quantity diverges, enter DIVERGES.)
Solution or Explanation
Click to View Solution
4.
Question Details
Determine whether the series is convergent or divergent by expressing sn as a telescoping sum (as in this example).
∞
n=1
15
n(n + 3)
convergent
divergent
If it is convergent, find its sum. (If the quantity diverges, enter DIVERGES.)
Solution or Explanation
Click to View Solution
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5.
12/7/14, 10:10 PM
Question Details
SEssCalc2 8.2.017. [2164194]
-
SEssCalc2 8.2.032.MI. [2329803]
-
SEssCalc2 8.3.006.MI. [2560428]
-
Determine whether the series is convergent or divergent.
∞
1 + 5n
8n
n=1
convergent
divergent
If it is convergent, find its sum. (If the quantity diverges, enter DIVERGES.)
Solution or Explanation
Click to View Solution
6.
Question Details
Express the number as a ratio of integers.
0.16 = 0.16161616
Solution or Explanation
0.16 = 0.16161616
a=
7.
is a geometric series with a =
16
1
and r =
. It converges to
100
100
a
16/100
16
=
=
.
1−r
1 − 1/100
99
Question Details
Use the Integral Test to determine whether the series is convergent or divergent.
∞
8
n=1
n9
Evaluate the following integral.
1
∞ 8
dx
x9
Since the integral
---Select---
is finite, the series is
---Select---
convergent .
Solution or Explanation
Click to View Solution
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8.
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Question Details
SEssCalc2 8.3.008. [2164272]
-
SEssCalc2 8.3.009. [2378305]
-
SEssCalc2 8.3.010. [2377835]
-
Use the Integral Test to determine whether the series is convergent or divergent.
∞
1
n+2
n=1
Evaluate the following integral.
∞
1
x+2
1
Since the integral
dx
---Select---
is not finite, the series is
---Select---
divergent .
Solution or Explanation
Click to View Solution
9.
Question Details
Use the Comparison Test to determine whether the series is convergent or divergent.
∞
n=1
n
5n3 + 3
converges
diverges
Solution or Explanation
Click to View Solution
10.
Question Details
Use the Comparison Test to determine whether the series is convergent or divergent.
∞
n=1
n4
4n5 − 2
converges
diverges
Solution or Explanation
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11.
12/7/14, 10:10 PM
Question Details
SEssCalc2 8.3.013. [2157500]
-
SEssCalc2 8.3.014. [2157446]
-
SEssCalc2 8.3.022. [2378192]
-
Determine whether the series is convergent or divergent.
1
1
1
1
+
+
+
+
4
9
16
25
1+
convergent
divergent
Solution or Explanation
Click to View Solution
12.
Question Details
Determine whether the series is convergent or divergent.
1+
1
2
2
+
1
3
3
+
1
4
4
+
1
5
5
convergent
divergent
Solution or Explanation
Click to View Solution
13.
Question Details
Determine whether the series is convergent or divergent.
∞
n=1
2 + 4n
3n
convergent
divergent
Solution or Explanation
Click to View Solution
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14.
12/7/14, 10:10 PM
Question Details
SEssCalc2 8.4.009. [2157366]
-
Test the series for convergence or divergence.
∞
(−1)n + 1
2n5
n=1
converges
diverges
If the series is convergent, use the Alternating Series Estimation Theorem to determine how many terms we need to add in
order to find the sum with an error less than 0.00005. (If the quantity diverges, enter DIVERGES.)
6 terms
Solution or Explanation
Click to View Solution
15.
Question Details
SEssCalc2 8.4.021. [2378124]
-
Determine whether the series is absolutely convergent, conditionally convergent, or divergent.
∞
n=0
(−15)n
n!
absolutely convergent
conditionally convergent
divergent
Solution or Explanation
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16.
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Question Details
SEssCalc2 8.4.023. [2158247]
-
Determine whether the series is absolutely convergent, conditionally convergent, or divergent.
∞
n=0
(−1)n
6n + 1
absolutely convergent
conditionally convergent
divergent
Solution or Explanation
bn =
1
> 0 for n ≥ 0, {bn} is decreasing for n ≥ 0, and lim bn = 0, so
6n + 1
n→∞
Series Test. To determine absolute convergence, choose an =
an
1/n
6n + 1
= lim
= lim
= 6 > 0, so
n
n → ∞ bn
n → ∞ 1/(6n + 1)
n→∞
1
to get
n
∞
lim
∞
harmonic series. Thus, the series
n=0
17.
(−1)n
6n + 1
n=1
∞
n=0
(−1)n converges by the Alternating
6n + 1
1
diverges by the Limit Comparison Test with the
6n + 1
is conditionally convergent.
Question Details
SEssCalc2 8.4.034. [2158055]
-
Determine whether the series is absolutely convergent, conditionally convergent, or divergent.
∞
n=2
−2n
n+1
4n
absolutely convergent
conditionally convergent
divergent
Solution or Explanation
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18.
12/7/14, 10:10 PM
Question Details
SEssCalc2 8.4.026.MI. [2163954]
-
SEssCalc2 8.4.028. [2158312]
-
SEssCalc2 8.4.029. [2158227]
-
Determine whether the series is absolutely convergent, conditionally convergent, or divergent.
∞
n=1
n!
88n
absolutely convergent
conditionally convergent
divergent
Solution or Explanation
Click to View Solution
19.
Question Details
Determine whether the series is absolutely convergent, conditionally convergent, or divergent.
∞
n=1
sin 3n
6n
absolutely convergent
conditionally convergent
divergent
Solution or Explanation
Click to View Solution
20.
Question Details
Determine whether the series is absolutely convergent, conditionally convergent, or divergent.
∞
n=1
cos(nπ/9)
n!
absolutely convergent
conditionally convergent
divergent
Solution or Explanation
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21.
12/7/14, 10:10 PM
Question Details
SEssCalc2 8.5.003. [2163917]
-
Find the radius of convergence, R, of the series.
∞
8(−1)nnxn
n=1
R=
Find the interval, I, of convergence of the series. (Enter your answer using interval notation.)
I=
Solution or Explanation
If an = 8(−1)nnxn, then
an + 1
an
lim
n→∞
8(−1)n + 1(n + 1)xn + 1
= lim
8(−1)nnxn
n→∞
= lim
(−1)
n→∞
n+1
x = lim
n
n→∞
1+
1
|x| = |x|. By the Ratio Test, the
n
∞
(−1)nnxn converges when |x| < 1, so the radius of convergence R = 1. Now we'll check the endpoints, that is,
series
n=1
∞
x = ±1. Both series
∞
(−1)nn(±1)n =
n=1
(∓1)nn diverge by the Test for Divergence since
n=1
lim |(∓1)nn| = ∞. Thus
n→∞
the interval of convergence is I = (−1, 1).
22.
Question Details
SEssCalc2 8.5.005. [2164442]
-
Find the radius of convergence, R, of the series.
∞
n=1
xn
2n − 1
R=
Find the interval, I, of convergence of the series. (Enter your answer using interval notation.)
I=
Solution or Explanation
If an =
xn , then lim
2n − 1
n→∞
∞
Ratio Test, the series
n=1
∞
comparison with
n=1
an+1
an
xn + 1 · 2n − 1 = lim
n → ∞ 2n + 1
xn
n→∞
= lim
2n − 1
|x| = lim
2n + 1
n→∞
2 − 1/n
|x| = |x|. By the
2 + 1/n
xn
converges when |x| < 1, so R = 1. When x = 1, the series
2n − 1
1
1
1
1
since
>
and
2n
2n − 1
2n
2
∞
series. When x = −1, the series
n=1
(−1)n
2n − 1
∞
n=1
∞
n=1
1
diverges by
2n − 1
1
diverges since it is a constant multiple of the harmonic
n
converges by the Alternating Series Test. Thus, the interval of convergence is
[−1, 1).
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23.
12/7/14, 10:10 PM
Question Details
SEssCalc2 8.5.007.MI. [2164498]
-
SEssCalc2 8.5.016. [2164633]
-
SEssCalc2 8.5.018. [2164408]
-
Find the radius of convergence, R, of the series.
∞
n=1
xn + 4
5n!
R=
Find the interval, I, of convergence of the series. (Enter your answer using interval notation.)
I=
Solution or Explanation
Click to View Solution
24.
Question Details
Find the radius of convergence, R, of the series.
∞
n=0
n
(−1)n (x − 3)
6n + 1
R=
Find the interval, I, of convergence of the series. (Enter your answer using interval notation.)
I=
Solution or Explanation
Click to View Solution
25.
Question Details
Find the radius of convergence, R, of the series.
∞
n=1
n
4n
(x + 3)n
R=
Find the interval, I, of convergence of the series. (Enter your answer using interval notation.)
I=
Solution or Explanation
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26.
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Question Details
SEssCalc2 8.5.011. [2163967]
-
SEssCalc2 8.6.003.MI. [2164461]
-
Find the radius of convergence, R, of the series.
∞
(−1)n n=2
xn
4n ln n
R=
Find the interval, I, of convergence of the series. (Enter your answer using interval notation.)
I=
Solution or Explanation
Click to View Solution
27.
Question Details
Find a power series representation for the function.
f(x) =
1
5+x
∞
f(x) =
n=0
Determine the interval of convergence. (Enter your answer using interval notation.)
Solution or Explanation
Our goal is to write the function in the form
f(x) =
1
5+x
=
1
1
·
1− − x
5
5
=
1
5
=
1
, and then represent the function as a sum of a power series.
1−r
1
5
∞
−
n=0
∞
x n
5
(−1)n
n=0
x n
=
5
∞
n=0
(−1)n
xn
5n + 1
with −
x
5
< 1,
x
5
< 1,
So R = 5, and I = (−5, 5).
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28.
12/7/14, 10:10 PM
Question Details
SEssCalc2 8.6.007. [2164304]
-
SEssCalc2 8.6.013. [3184335]
-
Find a power series representation for the function.
f(x) =
x
36 + x2
∞
f(x) =
n=0
Determine the interval of convergence. (Enter your answer using interval notation.)
Solution or Explanation
Click to View Solution
29.
Question Details
(a) Use differentiation to find a power series representation for
f(x) =
1
(9 + x)2
.
∞
f(x) =
n=0
What is the radius of convergence, R?
R=
9
(b) Use part (a) to find a power series for
f(x) =
1
(9 + x)3
.
∞
f(x) =
n=0
What is the radius of convergence, R?
R=
9
(c) Use part (b) to find a power series for
f(x) =
x2
(9 + x)3
.
∞
f(x) =
n=2
What is the radius of convergence, R?
R=
9
Solution or Explanation
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30.
12/7/14, 10:10 PM
Question Details
SEssCalc2 8.6.027. [2164534]
-
Evaluate the indefinite integral as a power series.
x 7 ln(1 + x) dx
∞
f(x) = C +
n=1
What is the radius of convergence R?
R=
1
Solution or Explanation
∞
From this example, ln(1 + x) =
n=1
∞
x7 ln(1 + x) dx = C +
n=1
n
(−1)n − 1 x
n
for |x| < 1, so x7 ln(1 + x) =
n=1
n+7
(−1)n − 1 x
n
and
n+8
(−1)n − 1 x
. R = 1 for the series for ln(1 + x), so R = 1 for the series representing
n(n + 8)
x7 ln(1 + x) as well. By this theorem, the series for
31.
∞
x7 ln(1 + x) dx also has R = 1.
Question Details
SEssCalc2 8.6.025.MI. [2164718]
-
Evaluate the indefinite integral as a power series.
t
dt
1 − t11
∞
C+
n=0
What is the radius of convergence R?
R=
1
Solution or Explanation
t
1
= t·
1 − t11
1 − t11
∞
t11
= t
n
n=0
∞
t11n + 1
=
n=0
∞
t
dt = C +
1 − t11
The series for
n=0
t11n + 2
11n + 2
1
converges when |t11| < 1
1 − t11
this theorem, the series for
|t| < 1, so R = 1 for that series and also the series for
t
. By
1 − t11
t
dt also has R = 1.
1 − t11
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Question Details
SEssCalc2 8.6.015.MI. [2560802]
-
SEssCalc2 8.7.014.MI. [2378005]
-
Find a power series representation for the function.
f(x) = ln(7 − x)
∞
f(x) = ln(7) −
n=1
Determine the radius of convergence, R.
R=
7
Solution or Explanation
f(x) = ln(7 − x)
dx
7−x
= −
= −
= −
dx
1− x
1
7
7
∞
1
7
x n
dx
7
n=0
1
7
= C−
∞
xn + 1
n=0
∞
= C−
n=1
7n(n + 1)
xn
n7n
Putting x = 0, we get C = ln 7. The series converges for
33.
Question Details
x
7
< 1,
|x| < 7, so R = 7.
Find the Taylor series for f(x) centered at the given value of a. [Assume that f has a power series expansion. Do not show that
Rn(x) → 0.]
f(x) =
10
,
x
a = −2
∞
f(x) =
n=0
Solution or Explanation
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34.
12/7/14, 10:10 PM
Question Details
SEssCalc2 8.7.028. [2163971]
-
Use a Maclaurin series in the table below to obtain the Maclaurin series for the given function.
f(x) = 8 cos
πx
5
∞
f(x) =
n=0
Solution or Explanation
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35.
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Question Details
SEssCalc2 8.7.029.MI. [2164366]
-
Use a Maclaurin series in the table below to obtain the Maclaurin series for the given function.
f(x) = 8ex + e8x
∞
n=0
Solution or Explanation
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36.
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Question Details
SEssCalc2 8.7.030. [2164081]
-
SEssCalc2 8.7.031.MI. [2164218]
-
SEssCalc2 8.8.001. [2164121]
-
Use a Maclaurin series in the table below to obtain the Maclaurin series for the given function.
f(x) = e6x + 8e−6x
∞
n=0
Solution or Explanation
Click to View Solution
37.
Question Details
Use a Maclaurin series in this table to obtain the Maclaurin series for the given function.
f(x) = 9x cos
1 2
x
4
∞
n=0
Solution or Explanation
Click to View Solution
38.
Question Details
(a) Find the Taylor polynomials up to degree 6 for f(x) = cos x centered at a = 0.
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T0 =
T1 =
T2 =
T3 =
T4 =
T5 =
T6 =
Graph f and these polynomials on a common screen.
(b) Evaluate f and these polynomials at x = π/4, π/2, and π. (Round your answers to four decimal places.)
x
f
T0 = T1
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T2 = T3
T4 = T5
T6
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π
0.7071
1
0.6916
0.7074
0.7071
0.0000
1
-0.2337
0.0200
-0.0009
-1
1
-3.9348
0.1239
-1.2114
4
π
2
π
(c) Comment on how the Taylor polynomials converge to f(x).
As n
increases , Tn(x) is a good approximation to f(x) on a
---Select---
---Select---
larger and larger interval.
Solution or Explanation
(a)
n
f (n)(x)
f (n)(0)
0
cos x
1
1
1
−sin x
0
1
2
−cos x
−1
1−
1 2
x
2
3
sin x
0
1−
1 2
x
2
4
cos x
1
1−
1 2
1 4
x +
x
2
24
5
−sin x
0
1−
1 2
1 4
x +
x
2
24
6
−cos x
−1
1−
1 2
1 4
1 6
x +
x −
x
2
24
720
Tn(x)
(b)
x
π
4
π
2
π
f
T0 = T1
T2 = T3
T4 = T5
0.7071
T6
1
0.6916
0.7074
0.7071
0.0000
1
−0.2337
0.0200
−0.0009
−1
1
−3.9348
0.1239
−1.2114
(c) As n increases, Tn(x) is a good approximation to f(x) on a larger and larger interval.
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Question Details
SEssCalc2 8.8.003.MI. [2164795]
-
Find the Taylor polynomial T3(x) for the function f centered at the number a.
f(x) =
1
,
x
a=2
T3(x) =
Graph f and T3 on the same screen.
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Solution or Explanation
Click to View Solution
40.
Question Details
SEssCalc2 8.8.005. [2164259]
-
Find the Taylor polynomial T3(x) for the function f centered at the number a.
f(x) = cos x,
a = π/2
T3(x) =
Graph f and T3 on the same screen.
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Solution or Explanation
n f (n)(x) f (n)(π/2)
0
cos x
0
1
−sin x
−1
2
−cos x
0
3
sin x
1
3
T3(x) =
n=0
f (n)(π/2) x − π n
2
n!
= − x−
π + 1 x− π
2
6
3
2
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41.
12/7/14, 10:10 PM
Question Details
SEssCalc2 8.8.009.MI. [2164646]
-
Consider the following function.
f(x) =
x,
a = 4,
n = 2,
4 ≤ x ≤ 4.3
(a) Approximate f by a Taylor polynomial with degree n at the number a.
T2(x) =
(b) Use Taylor's Inequality to estimate the accuracy of the approximation f(x) ≈ Tn(x) when x lies in the given interval.
(Round your answer to six decimal places.)
|R2(x)| ≤
0.000053
(c) Check your result in part (b) by graphing |Rn(x)|.
Solution or Explanation
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42.
Question Details
SEssCalc2 8.8.007. [2164745]
-
Find the Taylor polynomial T3(x) for the function f centered at the number a.
f(x) = xe−4x,
a=0
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T3(x) =
Graph f and T3 on the same screen.
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Solution or Explanation
n
f (n)(x)
f (n)(0)
0
xe−4x
0
1
(1 − 4x)e−4x
1
2
(16x − 8)e−4x
−8
3 (48 − 64x)e−4x
3
T3(x) =
n=0
=
48
f (n)(0) xn
n!
0
· 1 + 1 x1 + −8 x2 − 48 x3 = x − 4x2 − 8x3
1
1
2
6
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Question Details
SEssCalc2 8.8.012. [2164132]
-
Consider the following function.
f(x) = sin x,
a = π/6,
n = 4,
0 ≤ x ≤ π/3
(a) Approximate f by a Taylor polynomial with degree n at the number a.
T4(x) =
(b) Use Taylor's Inequality to estimate the accuracy of the approximation f(x) ≈ Tn(x) when x lies in the given
interval. (Round your answer to six decimal places.)
|R4(x)| ≤
0.000328
(c) Check your result in part (b) by graphing |Rn(x)|.
Solution or Explanation
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Name (AID): REVIEW FOR THE EXAM 4 MATH 203 F 2014 (6655401)
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