MATH 1321-004 Homework #2
due 2013-JAN-25, 11:35 a.m.
1
2
Example
Assignments
For all types of problems, you need to give a clear step- All numbers below refer to the textbook:
by-step derivation that identify the names of the relevant Calculus Concepts and Contexts, J. Stewart, 4th Ed.
theorems. In other words, you need to justify each step
• §8.5: 3, 8, 13, 18, 19, 23, 24, 27
of moving towards the solution by theorems, lemmas, or
previous knowledge.
• §8.6: 1, 2, 6, 8, 10, 12, 26, 28, 31, 37, 39, 40
Example 1. Prove the Maclaurin series of f (x) = sin(x)
• §8.7: 1, 6, 8, 10, 13, 14, 19, 22, 24, 27, 29, 30, 32,
converges ∀x ∈ R.
34, 40, 41, 42, 45, 46, 48, 53, 59, 60, 62, 64, 66
(n)
Proof. ∀x ∈ R, ∀n ∈ N, f (x) ≤ 1. Identify M = 1 and
|x|n+1
• Extra credit §8.7: 69
a = 0 in Taylor’s inequality, we have |Rn (x)| ≤ (n+1)! .
P∞ xn
By the ratio test, the series n=0 n! converges. Hence
Boldfaced problems weigh 2 points each and all other
|x|n+1
= 0 by Theorem 6 on page 570 of the problems 1 point each. A random subset of the assigned
limn→∞ (n+1)!
textbook. By the comparison test, limn→∞ |Rn (x)| = 0. problems with a total of 30 points will be graded. AdBy the absolute convergence theorem on page 588 of the ditional extra credit of 10% will be given to you if you
text, limn→∞ Rn (x) = 0. Hence the Maclaurin series of typeset your solutions in LATEX(ask me for the template
f (x) = sin(x) converges by Theorem 6 on the summary tex file if you plan to do this). You might also get partial
handout of §8.5 - §8.7.
extra credit for typeset solutions of some problems. You
R t
will also get extra credit if you find typos or errors in the
Example 2. Express the indefinite integral 1−t8 dt as
summary handouts.
R 1/2 t
a power series and then approximate 0 1−t
8 dt to six
Caution: please make sure that your handwriting is
decimal places.
recognizable, otherwise you only get partial credit for the
recognizable part.
1
Solution. By the power series of 1−x
,
Note: Taylor series is the most important math tool
for
engineers, therefore I have assigned correspondingly
∞
∞
X
X
1
t
8 n
8n+1
a fair number of problems. The next assignment will be
=
t
=
t
(t
)
=
t
.
1 − t8
1 − t8
much lighter.
n=0
n=0
Then, by the theorem of term-by-term integration,
Z
∞
X
t
t8n+2
dt
=
C
+
.
1 − t8
8n + 2
n=0
Setting t = 0 in the above equation yields C = 0.
1/2
Z 1/2
∞
X
t
t8n+2 dt = (
)
1 − t8
8n + 2 0
n=0
0
1/2
t2
t10
t18
=( +
+
+ ...)
2
10
18
0
1
1
≈ +
.
(1)
8 10240
By the theorem on remainder estimate for integral test,
(1) satisfies the accuracy requirement since
1
R1 =
≈ 2 × 10−7 < 10−6 .
18 × 218
1