Calculus
Homework # 4
Homework # 4
8.3 # 32: (4 pts, p. 584)
(a) Find the partial sum s10 of the series
∞
∑
1
. Estimate the error in using s10 as an
n4
n=1
approximation for the sum of the series.
(b) Use (4) with n = 10 to give an improved estimate of the sum.
(c) Find a value of n so that sn is within 0.00001 of the sum.
a: The partial sum of the series can be computed as:
n
1
2
3
4
5
6
7
8
9
10
1
n4
1.0
0.0625
0.0123456790
0.00390625
0.0016
0.0007716049
0.0004164931
0.0002441406
0.0001524158
0.0001
Partial Sum
1.0
1.0625
1.0748456790
1.0787519290
1.0803519290
1.0811235340
1.0815400271
1.0817841677
1.0819365835
1.0820365835
Using (3) the Remainder Estimate for the Integral Test gives:
∫ ∞
∫ ∞
f (x) dx ≤ Rn ≤
f (x) dx
n+1
∫ ∞
∫n ∞
1
1
dx ≤ R10 ≤
dx
4
4
11 x
10 x
[
]
[
]
1 ∞
1 ∞
− 3
≤ R10 ≤ − 3
3x 11
3x 10
1
1
≤ R10 ≤
3993
3000
2.50438 × 10−4 ≤ R10 ≤ 3.33333 × 10−4
Approximately
b: To use (4) for an improved estimate:
∫ ∞
∫ ∞
sn +
f (x) dx ≤ s ≤ sn +
f (x) dx
n+1
∫ ∞
∫n ∞
1
1
s10 +
dx ≤ s ≤ s10 +
dx
4
4
11 x
10 x
[
]
[
]
1 ∞
1 ∞
1.0820365835 + − 3
≤ s ≤ 1.0820365835 + − 3
3x 11
3x 10
1
Approximately
Calculus
Homework # 4
1
1
≤ s ≤ 1.0820365835 +
3993
3000
1.0822870218 ≤ s ≤ 1.0823699168
1.0820365835 +
c: To use the remainder estimate (3), use the right side of (3) and find an n so that the
upper bound is less than or equal to 0.00001. That is:
∫ ∞
1
Rn ≤
dx
4
x
n
[
]
1 ∞
0.00001 ≥ − 3
Integrating
3x n
1
Simplifying
0.00001 ≥ 3
3n
3n3 ≥ 100000
√
3 100000
≈ 32.1829794869
n≥
3
So n ≥ 33 for the remainder to be less than 0.00001.
As a comparison, to use the improved estimate (4), use the right and left sides of (4), and
find an n so that the difference between these two is less than or equal to 0.00001. That is:
(
) (
)
∫ ∞
∫ ∞
1
1
sn +
dx − sn +
dx ≤ 0.00001
4
4
n x
n+1 x
∫ ∞
∫ ∞
1
1
dx −
dx ≤ 0.00001 Simplifying
4
4
n+1 x
n x
∫ n+1
1
dx ≤ 0.00001 Subtracting integrals
x4
n
[
]
1 n+1
dx ≤ 0.00001 Integrating
− 3
3x n
1
1
−
≤ 0.00001 Evaluating
3
3n
3(n + 1)3
(n + 1)3 − n3
≤ 0.00001 Simplifying
3n3 (n + 1)3
3n2 + 3n + 1
≤ 0.00001
3n3 (n + 1)3
3n6 + 9n5 + 9n4 + 3n3 − 300000n2 − 300000n − 100000 ≥ 0
As a polynomial
This last polynomial has a value less than zero for n = 0 and a value greater than zero for n
large enough, such as n = 50. Using Descartes rule of signs (there is only one sign change in
the coefficients of the polynomial), there is only one positive root for this polynomial; only one
positive value for n which will make the polynomial equal to zero. Since the polynomial has a
negative value at n = 0 and a positive value for n = 50 this root must be between 0 and 50.
Using a bisection method (taking the average of two candidate roots), the value of this root
can be approximated as in this table:
2
Calculus
Homework # 4
Low Candidate
0
0
12.5
12.5
15.625
17.1875
High Candidate
50
25
25
18.75
18.75
18.75
New Candidate
25
12.5
18.75
15.625
17.1875
17.96875
New Polynomial Value
628775000.00
-36308740.23
41150606.54
-25444152.75
-2240704.99
16439620.86
From this, the value of the polynomial will be negative for n ≤ 17 and positive for n ≥ 18.
Therefore, the accuracy of the improved estimate will be within 0.00001 if n ≥ 18.
8.5 # 9: (3 pts, p. 597) Find the radius of convergence and the interval of convergence of the
series.
∞
∑
(−1)n
n=1
n2 x n
2n
Using the ratio test:
(−1)n+1 (n+1)2 xn+1 an+1 n+1
2
= lim lim 2 n
n→∞ an n→∞ (−1)n n2nx
(n + 1)2 x 1 = x
= lim 2
n→∞
2n2 Therefore the series will converge if |x| < 2 so 2 is the radius of convergence. Note that
∞
∑
when x = ±2 each term in this series will be
(−1)n n2 which by the alternating series test
0
fails to converge, so the series will not converge if x = ±2. This means that the interval of
convergence is (−2, 2).
8.6 # 3: (3 pts, p. 583) Find a power series representation for the function and determine the
interval of convergence.
f (x) =
1
1+x
There are several ways to do this including polynomial division, the binomial theorem for
the expression (1 + x)−1 (but not with Pascal’s triangle), and working backwards from a series
which when multiplied by 1 + x gives a product of 1. The way highlighted here comes from the
text:
1
1
=
1+x
1 − (−x)
∞
∑
(−x)n
=
See page 598 or 599 of the text
n=0
= 1 − x + x2 − x3 + x4 − x5 + · · ·
3
Calculus
Homework # 4
To find the interval of convergence, use the ratio test:
(−1)n+1 1 an+1 n+1
x
= lim lim n→∞ an n→∞ (−1)n 1n
x
1
= lim n→∞ x
So the radius of convergence is 1, that is |x| < 1. Then if x = −1 the partial sums of this series
will be either 0 or 1, but the sum of the series does not converge to a single value. If x = 1,
the partial sums of this series will be n + 1 where n was the upper limit for the partial sum so
that the sum of the full series is infinite. Therefore, the interval of convergence is (−1, 1).
4