# Homework Solutions, Chapter 1

Shared by:
Categories
-
Stats
views:
32
posted:
2/23/2010
language:
English
pages:
2
Document Sample

```							Homework Solutions, Chapter 1

6.17 ∀n ≥ 0, e−n2 ≤ e−n . So by the comparison test, if                                       e−n converges,
−n2
then so does                e       . Since
∞
dx e−x = 0 ≤ ∞

both series converge.

9.10 limn→∞ n/(n − 1) = 1 = 0. Therefore                                            (−1)n n/(n − 1) does not
converge.

10.8 Ratio test:
an+1       n
=−         → −1
an     (n + 1)
√
So     (−1)n xn / n converges for −1 < x < 1. The endpoints give series:
√
• x = 1: (−1)n / n converges by the alternating series test.
√
• x = −1: 1/ n = ζ( 1 ) = ∞.
2

So the interval of convergence is −1 < x ≤ 1.

13.12 Expand series within the integral:
x                          x               1 4 1 8               1      1 9
dt cos t2 =                dt 1 −         t + t − . . . = x − x5 +     x − ...
0                          0                   2!    4!              10    216
Or, more generally,
∞                ∞
x                        x            (−1)n t4n        (−1)n x4n+1
dt cos t2 =              dt                 =
0                        0            n=0   (2n)!     n=0 (4n + 1)(2n)!

14.2 For parts (a) and (b), use a program such as Mathematica, or look
up values for the ζ function on the web. For (c), The error is

EN =               an x n ≤          |an xn |
N +1               N +1

Phys 382 HW# 1 Solutions                                                                                  1
Since ∀n > N, |an+1 | ≤ |an |, we can replace each an with aN +1 :
∞
1
EN ≤ |aN +1 xN +1 |       |xn | = |aN +1 xN +1 |
0                            1 − |x|

15.28 Do a binomial expansion with p = − 1 :
2

1 v2 3 v4              1      3 v4
E = mc2 1 +        + 4 + . . . = mc2 + mv 2 + m 2 + . . .
2 c2 8 c               2      8 c

∞
15.31 The area is A = π       21/(n ln n)2 . Since ∀n > 2, 1/(n ln n)2 ≤ 1/n2 ,
and n−2 converges, A < ∞. The wire length is L = 2π ∞ 1/(n ln n). Try
2
an integral test, for ∞ dx 1/(x ln x). Change variables to u = ln x, resulting
in ∞ du/u = ln ∞ = ∞. So L = ∞.

16.23 Calling the series s, observe that
π − πs = sin π

Since sin π = 0, 1 − s = 0 ⇒ s = 1.

Phys 382 HW# 1 Solutions                                                     2

```
Related docs