MATH1120 Calculus II
Solutions to Selected Exercises in Notes 13
November 25, 2013
∞
X
x3 x5
x2 x4
Exercise 1.1 Note that sin x = x− + −· · · =
an xn and cos x = 1− + −· · · =
3! 5!
2! 4!
n=0
∞
X
1
bn xn for any x. Here a0 = a2 = a4 = 0, a1 = 1, a3 = − 61 , a5 = 120
, b1 = b3 = b5 = 0,
n=0
b0 = 1, b2 = − 21 , b4 =
1
24 .
So by the series multiplication theorem,
sin x cos x =
∞
X
cn xn
for all x
n=0
where
c0 = a0 b0 = 0
c1 = a0 b1 + a1 b0 = 0 · 0 + 1 · 1 = 1
1
+1·0+0·1=0
c2 = a0 b2 + a1 b1 + a2 b0 = 0 · −
2
1
1
2
c3 = a0 b3 + a1 b2 + a2 b1 + a3 b0 = 0 · 0 + 1 · −
+0·0+ −
·1=−
2
6
3
1
1
1
+1·0+0· −
+ −
·0+0·1=0
c4 = a0 b4 + a1 b3 + a2 b2 + a3 b1 + a4 b0 = 0 ·
24
2
6
1
1
1
1
2
c5 = a0 b5 + a1 b4 + a2 b3 + a3 b2 + a4 b1 + a5 b0 = 0 · 0 + 1 ·
+0·0+ −
· −
+0·0+
·1=
24
6
2
120
15
It follows that
2x3 2x5
sin x cos x = x −
+
− ···
3
15
and multiplying by 2 gives
2 sin x cos x = 2x −
4x3 4x5
+
− ···
3
15
By substitution, we have
(2x)3 (2x)5
+
− ···
3!
5!
4x3 4x5
= 2x −
+
− ···
3
15
Indeed the first three non-zero terms of the power series of 2 sin x cos x and sin 2x are the
same.
sin 2x = 2x −
1
2
Exercise 1.2
(1)
1
= 1 + x + x2 + x3 + · · · for − 1 < x < 1
1−x
By the term-by-term differentiation theorem,
1
= 1 + 2x + 3x2 + · · · for − 1 < x < 1
(1 − x)2
Multiplying by x, we have
x
= x + 2x2 + 3x3 + · · · for − 1 < x < 1
(1 − x)2
1
Since
∈ (−1, 1), which is the interval within which the RHS converges to the
2
1
LHS, we can put x = and get
2
∞
1
X
1
2
3
n
2
=
+
+
+
·
·
·
=
2 = 2
2n
2 22 23
1− 1
n=1
2
(2) Let f (x) = −1 + 2x − 3x2 + 4x3 − 5x4 + · · · =
∞
X
(−1)n−1 (n + 1)xn . Applying the
n=0
Ratio Test, we have
|(−1)n (n + 2)xn+1 |
<1
n→∞ |(−1)n−1 (n + 1)xn |
n+2
lim
|x| < 1
n→∞ n + 1
|x| < 1
lim
Indeed the given series converges (absolutely) for |x| < 1. By the term-by-term
integration theorem, we have
Z x
Z x
Z x
Z x
Z x
Z x
F (x) =
f (t)dt =
−dt +
2tdt −
3t2 dt +
4t3 dt −
5t4 dt + · · ·
0
0
0
0
= −x + x2 − x3 + x4 − x5 + · · ·
−x
=
for |x| < 1
1+x
d
−x
1
0
∴ f (x) = F (x) =
=−
dx 1 + x
(1 + x)2
0
0
3
Exercise 2.1
∞
X
1
= 1 + x + x2 + x3 + · · · =
xn for x ∈ (−1, 1). By the series multiplication
(1)
1−x
n=0
theorem,
2−x
= (2 − x)(1 + x + x2 + x3 + · · · )
1−x
= (2 + 2x + 2x2 + 2x3 + · · · ) − (x + x2 + x3 + x4 + · · · )
= 2 + x + x2 + x3 + · · ·
∞
X
=2+
xn for x ∈ (−1, 1)
n=1
(2)
∞
X xn
x2 x3
e =1+x+
+
+ ··· =
for all x
2!
3!
n!
x
n=0
x3 x4
xex = x + x2 +
+
+ ···
2!
3!
∞
X
xn+1
=
for all x
n!
n=0
=
(3) By putting x =
∞
X
xk
(k − 1)!
k=1
t2 to
2
(re-indexing by letting n + 1 = k)
the Maclaurin series of ex , we have
(t2 )2 (t2 )3
+
+ · · · for all t
2!
3!
t4 t6
= 1 + t2 + + + · · ·
2! 3!
∞ 2n
X
t
=
n!
et = 1 + t2 +
n=0
By the term-by-term integration theorem,
Z x
Z x
Z x
Z x 4 Z x 6
t
t
t2
2
e dt =
+
+ · · · for all x
dt +
t dt +
0
0
0
0 2!
0 3!
t3
t5
x7
=x+ +
+
+ ···
3
5 · 2! 7 · 3!
∞
X
x2n+1
=
(2n + 1)n!
n=0
4
Exercise 2.2
(1)
x
e =
cos x =
∞
X
xn
n=0
∞
X
n!
= an xn for all x
∞
X
x2n
(−1)
=
bn xn for all x
(2n)!
n
n=0
n=0
1
1
1
Here a0 = a1 = 1, a2 = , a3 = , b0 = 1, b1 = 0, b2 = − , b3 = 0. By the series
2
6
2
multiplication theorem,
! ∞
!
∞
X
X
x
n
n
e cos x =
an x
bn x
for all x
n=0
n=0
where
c0 = a0 b0 = 1 · 1 = 1
c1 = a0 b1 + a1 b0 = 1 · 0 + 1 · 1
1
1
+1·0+ ·1=0
c2 = a0 b2 + a1 b1 + a2 b0 = 1 · −
2
2
1
1
1
1
c3 = a0 b3 + a1 b2 + a2 b1 + a3 b0 = 1 · 0 + 1 · −
+ ·0+ ·1=
2
2
6
3
x3
+ · · · for all x.
3
(2) By applying a substitution to the Taylor series of tan−1 x, we have
So ex cos x = 1 + x +
(x2 )3 (x2 )5
+
− · · · for − 1 ≤ x2 ≤ 1 i.e. − 1 ≤ x ≤ 1
3
5
x6 x10
+
− ···
= x2 −
3
5
tan−1 x2 = x2 −
Here a0 = a1 = 0, a2 = 1, a3 = 0.
1
1
1
=
3+x
3 1 + x3
1
x x2 x3
x
=
1− +
−
+ ···
for − 1 < < 1 i.e. − 3 < x < 3
3
3
9
27
3
2
3
1 x x
x
= − +
−
+ ···
3 9 27 81
5
1
1
1
1
Here b0 = , b1 = − , b2 = , b3 = − . By the series multiplication theorem,
3
9
27
81
−1
2
6
tan x
1
x
1 x x2 x3
−1 2
2
= tan x ·
= x −
+ ···
− +
−
+ ···
3+x
3+x
3
3 9 27 81
x2 x3 x4
−
+
− ···
=
3
9
27
and the interval within which the RHS converges to the LHS is the intersection of
the two intervals [−1, 1] and (−3, 3), i.e. [−1, 1].
(3)
ln(1 + t) =
ln(1 + t)
=
t
∞
X
(−1)n−1
n=1
∞
X
(−1)n−1
n=1
tn
for − 1 < t ≤ 1
n
tn−1
for − 1 < t ≤ 1
n
By the term-by-term integration theorem,
Z x
∞ Z x
X
tn−1
ln(1 + t)
(−1)n−1
dt =
dt for − 1 < t ≤ 1
t
n
0
0
=
n=1
∞
X
(−1)n−1
n=1
xn
n2
x2 x3
+
− ···
4
9
Exercise 2.3 The Taylor polynomial used to approximate (x) = ex is of degree 4 and
centered at 0. So n = 4 and a = 0. By the Remainder Estimation Theorem,
M 21 4+1
1
≤
R4
2 (4 + 1)!
=x−
where M satisfies
1
M ≥ |f (4+1) (t)| = |et | for 0 < t <
2
1
1
1 e2
So M can be taken to be e 2 . Consequently R4
≤
.
2
3840
Exercise 2.4 By part (3) of Exercise 2.2,
Z x
∞
X
ln(1 + t)
xn
x2 x3
dt =
(−1)n−1 2 = x −
+
− · · · for − 1 < x ≤ 1
t
n
4
9
0
n=1
xn
Indeed for x ∈ [0, 1] the power series is alternating, with un = 2 , which satisfies all the
n
three conditions in the alternating series test. By the Alternating Series Error Estimation
6
Theorem,
|Error| = |sn − F (x)| ≤ un+1 =
If
xn+1
1
≤
2
(n + 1)
(n + 1)2
1
< 10−3 , then the error is bounded above by 10−3 .
(n + 1)2
1
< 10−3
(n + 1)2
(n + 1)2 > 1000
n + 1 > 31
n > 30
So a polynomial that approximates F (x) for x ∈ [0, 1] with an error of magnitude bounded
x2
x31
above by 10−3 can be taken to be x −
+ ··· +
.
4
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