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516 • How to Solve Series Problems
by the p-test (with p = 1/2). So the modified series diverges to ∞; we con-
clude that the original series diverges to −∞.
2. If some terms are positive and some terms are negative, try the
nth term test first. That is, check that the terms tend to 0 as n → ∞;
otherwise you know right away that the series diverges. For example,
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hypotenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f(
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f(
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h)
x + h
F (
x + h) − F (x)
f(
x)
1
2
y = sin(
x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−
1
y = ln
|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(x)
a
b
a + ε
ε
Z
b
a+ε
f(x) dx
small
even smaller
y = g(x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
a
8
1
2
3
4
5
6
7
8
n
a
n
x
y
y = f(x)
∞
X
n=1
(−1)
n
n
2
diverges because the limit of the terms (−1)
n
n
2
isn’t zero. (In fact, the limit
doesn’t exist, because the sequence oscillates wildly between bigger and bigger
positive and negative numbers.) There’s no need to use any other tests here.
3. If some terms are positive and some terms are negative, and the
terms converge to 0 as n → ∞, next try the absolute convergence
test:
if
∞
X
n=1
|a
n
| converges, then so does
∞
X
n=1
a
n
.
In this case, we say that the sequence is absolutely convergent or that it
converges absolutely. For example, the series
∞
X
n=1
sin(n)
n
2
converges absolutely, since we already saw on page 513 above that
∞
X
n=1
|sin(n)|
n
2
converges. So, given a series with positive and negative terms that doesn’t
obviously fail the nth term test, you should always check to see if the series is
absolutely convergent. If it’s absolutely convergent, then the series converges.
On the other hand, if it doesn’t converge absolutely, don’t give up on it—go
on to the next step.
4. If the series doesn’t converge absolutely, try the alternating series
test. As we saw in Section 22.5.4 of the previous chapter,
if the absolute values of the terms of an alternating series
decrease to 0 monotonically as n → ∞, the series converges.
So there are actually three things to check if you want to use the test on a
series
P
∞
n=1
a
n
:
• the terms a
n
alternate between positive and negative (that is, the signs
of the terms are, in order, +, −, +, −, . . ., or perhaps −, +, −, +, . . .);
• the quantities |a
n
| tend to 0 as n gets large; that is,
lim
n→∞
|a
n
| = 0;
Section 23.7: How to Deal with Series with Negative Terms • 517
• the absolute values of the terms |a
n
| are decreasing in n (so the un-
derlying sequence is getting smaller and smaller, in terms of absolute
value).
If all three of these properties are true, then the series converges. Note: you
should always try the absolute convergence test first. If the series
converges absolutely, do not use the alternating series test! Also,
notice that the second property is just the nth term test in disguise; this
follows from the fact that lim
n →∞
|a
n
| = 0 if and only if lim
n →∞
a
n
= 0. So even if
you forget to try the nth term test first, you have to do it anyway as part of
the alternating series test.
Here’s a classic example:
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hypotenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f(
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f(
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h)
x + h
F (
x + h) − F (x)
f(
x)
1
2
y = sin(
x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−
1
y = ln
|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(x)
a
b
a + ε
ε
Z
b
a+ε
f(x) dx
small
even smaller
y = g(x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
a
8
1
2
3
4
5
6
7
8
n
a
n
x
y
y = f(x)
∞
X
n=1
(−1)
n
n
.
The absolute version is
P
∞
n=1
1/n, which diverges by the p-test; so our series
does not converge absolutely. Let’s dive straight in to the alternating series
test. We need to check the three properties. First, is the series alternating?
Yes. A series is automatically alternating if it has terms which look like
(−1)
n
or (−1)
n+1
multiplied by a positive number. In this case, the nth
term is (−1)
n
multiplied by the positive number 1/n. How about the second
property? We need to show that
lim
n→∞
(−1)
n
n
= 0.
This is obviously true, since |(−1)
n
/n| = 1/n. As for the third property,
we need to show that {|(−1)
n
/n|} is a decreasing sequence. This is pretty
straightforward, again since |(−1)
n
/n| = 1/n and we know that 1/n is de-
creasing in n. So the alternating series test applies and shows that the original
series
∞
X
n=1
(−1)
n
n
converges. Since we’ve already checked that it doesn’t converge absolutely,
we know that it converges conditionally.
On the other hand, consider the series
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hypotenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f(
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f(
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h)
x + h
F (
x + h) − F (x)
f(
x)
1
2
y = sin(
x)
π
−
π
−1
−2
y =
1
x
y = x
2
1
2
1
−1
y = ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(x)
a
b
a + ε
ε
Z
b
a+ε
f(x) dx
small
even smaller
y = g(x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
a
8
1
2
3
4
5
6
7
8
n
a
n
x
y
y = f(x)
∞
X
n=1
(−1)
n
n
2
.
The absolute version is
P
∞
n=1
1/n
2
, which converges by the p-test. So the
above series converges absolutely, and there’s no need to waste time with the
alternating series test.
Let’s see a couple more examples. First, let’s look at
PSfrag replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shadow
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror (
y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y = (
x − 1)
2
−
1
x
Same height
−
x
Same length,
opposite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y = 2
x
y = 10
x
y = 2
−
x
y = log
2
(
x)
4
3 units
mirror (
x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
= 0 radians
θ
hypotenuse
opposite
adjacent
0 (
≡ 2π)
π
2
π
3
π
2
I
II
III
IV
θ
(
x, y)
x
y
r
7
π
6
reference angle
reference angle =
π
6
sin +
sin −
cos +
cos −
tan +
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this angle is
5
π
6
clockwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = cos(
x)
−
π
2
π
2
y = tan(
x), −
π
2
< x <
π
2
0
−
π
2
π
2
y = tan(
x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y = sec(
x)
y = csc(
x)
y = cot(
x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y = tan
−
1
(x)
π
2
π
y =
sin(
x)
x
, x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 < x < 0
.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x + 2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangent at x = a
b
tangent at x = b
c
tangent at x = c
y = x
2
tangent
at x = −
1
u
v
uv
u + ∆
u
v + ∆
v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f (
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y = sin(
x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin(
x)
x
π
2
π
1
−
1
x = 0
a = 0
x > 0
a > 0
x < 0
a < 0
rest position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y = ln(
x)
y = cosh(
x)
y = sinh(
x)
y = tanh(
x)
y = sech(
x)
y = csch(
x)
y = coth(
x)
1
−
1
y = f(
x)
original function
inverse function
slope = 0 at (
x, y)
slope is infinite at (
y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y = sin(
x)
y = sin(
x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y = sin
−
1
(x)
y = cos(
x)
π
π
2
y = cos
−
1
(x)
−
π
2
1
x
α
β
y = tan(
x)
y = tan(
x)
1
y = tan
−
1
(x)
y = sec(
x)
y = sec
−
1
(x)
y = csc
−
1
(x)
y = cot
−
1
(x)
1
y = cosh
−
1
(x)
y = sinh
−
1
(x)
y = tanh
−
1
(x)
y = sech
−
1
(x)
y = csch
−
1
(x)
y = coth
−
1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y = 1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y = sin(
x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Local maximum
Local minimum
Horizontal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2 +
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2 +
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing fence
new fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallow
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f (
x)
11
y = L
(x)
y = f(
x)
F
P
a
a + ∆
x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
df
∆
x
a
b
y = f(
x)
true zero
starting approximation
better approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y = 1
a
b
y = sin(
x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
= 2
width of each interval =
2
n
−
2
1
3
0
I
II
III
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
II
IIa
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
av
y = f
av
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h)
x + h
F (x + h) − F (x)
f(x)
1
2
y = sin(x)
π
−π
−1
−2
y =
1
x
y = x
2
1
2
1
−1
y = ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(x)
a
b
a + ε
ε
Z
b
a+ε
f(x) dx
small
even smaller
y = g(x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
a
8
1
2
3
4
5
6
7
8
n
a
n
x
y
y = f(x)
∞
X
n=1
(−1)
n
sin
1
n
.
This is very similar to an example we looked at on page 513 above, except
that we now have an extra factor of (−1)
n
. The first thing to do with our
518 • How to Solve Series Problems
series is to check whether or not it is absolutely convergent. The absolute
version is
∞
X
n=1
(−1)
n
sin
1
n
=
∞
X
n=1
sin
1
n
.
We saw in that previous example that the quantity sin(1/n) is nonnegative
when n ≥ 1, so this justifies dropping the absolute value signs. We also saw
that this last series diverges, so our original series above doesn’t converge
absolutely. On the other hand, the terms of this series are clearly alternating,
and we’ve already seen that they go to 0 as n → ∞ since sin(1/n) does. Now
consider |a
n
|, which is just sin(1/n). Does this decrease in n? You could
differentiate sin(1/x) with respect to x to get −cos(1/x)/x
2
, and show that
this is negative whenever x ≥ 1; or you could just argue that 1/n decreases in
n, and sin(x) is increasing in x when x is near 0, so sin(1/n) decreases in n.
Either way, we have now verified all three properties, so the alternating series
test shows that our series converges. Since it doesn’t converge absolutely, it
must converge conditionally.
One final example. Consider the series
∞
X
n=1
(−1)
n
1 +
1
n
n
.
The series is certainly alternating, but what is the limit of the nth term? We
need it to be 0 for the series to have any hope at all of converging. That seems
problematic in this case—by the boxed limit at the end of Section 22.1.2 of
the previous chapter with k replaced by 1, we have
lim
n→∞
1 +
1
n
n
= e
1
= e,
so the alternating version of this sequence (with the (−1)
n
term in there)
oscillates between numbers close to e and −e. This means that
lim
n→∞
(−1)
n
1 +
1
n
n
does not exist (DNE).
Since this limit isn’t 0 (it doesn’t even exist!), the nth term test shows that
the original series
∞
X
n=1
(−1)
n
1 +
1
n
n
must diverge. Don’t fall into the trap of using the alternating series test here
to conclude that the series converges!
So as you have seen, this series business isn’t too easy. What’s more, we’re
going to have to use these techniques again when we look at power series and
Taylor series in the next chapter, so you really need to understand the stuff
in this chapter or there will be a large chunk of material coming up that will
elude you. Of course, doing as many problems as you can will really help.
C h a p t e r 24
Introduction to Taylor Polynomials, Taylor Series,
and Power Series
We now come to the important topics of power series and Taylor polynomials
and series. In this chapter, we’ll see a general overview of these topics. The
following two chapters will deal with problem-solving techniques in the context
of the material in this chapter. Here’s what we’ll look at first:
• approximations, Taylor polynomials, and a Taylor approximation theo-
rem;
• how good our approximations are, and the full Taylor Theorem;
• the definition of power series;
• the definition of Taylor series and Maclaurin series; and
• convergence issues involving Taylor series.
24.1 Approximations and Taylor Polynomials
Here’s a nice fact: for any real number x, we have
e
x
∼
=
1 + x +
x
2
2
+
x
3
6
.
Also, the closer x is to 0, the better the approximation.
Let’s play around with this for a little bit. Start off with x = 0. Actually,
both sides are then equal to 1, so the approximation is perfect! What about
when x isn’t 0? Let’s try x = −1/10. The above equation says that
e
−1/10
∼
=
1 −
1
10
+
1/100
2
−
1/1000
6
,
which simplifies to
e
−1/10
∼
=
5429
6000
.
My calculator says that e
−1/10
is equal to 0.9048374180 (to ten decimal
places), while 5429/6000 is equal to 0.9048333333 (also to ten decimal places).
520 • Taylor Polynomials, Taylor Series, and Power Series
These numbers are pretty close to each other! In fact, the difference is only
about 0.0000040847.
How on earth did I come up with the polynomial 1 + x + x
2
/2 + x
3
/6?
It’s clearly not just any old polynomial; it’s specially related to e
x
. Rather
than concentrate on e
x
itself, let’s get a little more general and consider other
functions. Also, there’s nothing special about the degree 3 of our polynomial:
we could have used any degree. So let’s start with degree 1 and see what
happens.
24.1.1 Linearization revisited
Let’s say we have some function f which is very smooth, so that it can be
repeatedly differentiated as many times as you like without causing any prob-
lems. Here’s a question we asked back in Section 13.2 of Chapter 13: what is
the equation of the line which best approximates the curve y = f(x) near the
point (a, f(a))? The answer to this question is that the line we’re looking for
is the tangent line to the curve at the point (a, f (a)), and its equation is
y = f (a) + f
0
(a)(x − a).
This is precisely the linearization of f at x = a. The right-hand side is a
polynomial of degree 1. In the picture below, the tangent line to the curve
y = f (x) at x = a is drawn in, and looks like a pretty lousy approximation to
the whole curve:
PSfrag
replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shado
w
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror
(y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y =
(x − 1)
2
−
1
x
Same
height
−
x
Same
length,
opp
osite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y =
2
x
y =
10
x
y =
2
−x
y =
log
2
(x)
4
3
units
mirror
(x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
=
0 radians
θ
hypotenuse
opp
osite
adjacen
t
0
(≡ 2π)
π
2
π
3
π
2
I
I
I
I
II
IV
θ
(
x, y)
x
y
r
7
π
6
reference
angle
reference
angle =
π
6
sin
+
sin −
cos
+
cos −
tan
+
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this
angle is
5π
6
clo
ckwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
cos(x)
−
π
2
π
2
y =
tan(x), −
π
2
<
x <
π
2
0
−
π
2
π
2
y =
tan(x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y =
sec(x)
y =
csc(x)
y =
cot(x)
y = f (
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y =
tan
−1
(x)
π
2
π
y =
sin
(x)
x
,
x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 <
x < 0.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f (
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x +
2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f (
x)
a
tangen
t at x = a
b
tangen
t at x = b
c
tangen
t at x = c
y = x
2
tangen
t
at x = −
1
u
v
uv
u +
∆u
v +
∆v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f (
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f(
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y =
sin(x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin
(x)
x
π
2
π
1
−
1
x =
0
a =
0
x
> 0
a
> 0
x
< 0
a
< 0
rest
position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f (
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y =
ln(x)
y =
cosh(x)
y =
sinh(x)
y =
tanh(x)
y =
sech(x)
y =
csch(x)
y =
coth(x)
1
−
1
y = f (
x)
original
function
in
verse function
slop
e = 0 at (x, y)
slop
e is infinite at (y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
sin(x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y =
sin
−1
(x)
y =
cos(x)
π
π
2
y =
cos
−1
(x)
−
π
2
1
x
α
β
y =
tan(x)
y =
tan(x)
1
y =
tan
−1
(x)
y =
sec(x)
y =
sec
−1
(x)
y =
csc
−1
(x)
y =
cot
−1
(x)
1
y =
cosh
−1
(x)
y =
sinh
−1
(x)
y =
tanh
−1
(x)
y =
sech
−1
(x)
y =
csch
−1
(x)
y =
coth
−1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y =
1
2
(2
, 3)
y = f (
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Lo
cal maximum
Lo
cal minimum
Horizon
tal point of inflection
1
e
y = f
0
(
x)
y = f(
x) = x ln(x)
−
1
e
?
y = f(
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2
+
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln
(x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2
+
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing
fence
new
fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallo
w
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f(
x)
11
y = L
(x)
y = f (
x)
F
P
a
a +
∆x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
d
f
∆
x
a
b
y = f (
x)
true
zero
starting
approximation
b
etter approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f (
x)
1
2
y = x
5
0
−
2
y =
1
a
b
y =
sin(x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
=
2
width
of each interval =
2
n
−
2
1
3
0
I
I
I
I
II
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
I
I
I
Ia
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f (
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
a
v
y = f
a
v
c
A
M
0
1
2
a
b
x
t
y = f(
t)
F (
x )
y = f(
t)
F (
x + h )
x + h
F (
x + h) − F (x)
f(
x)
1
2
y =
sin(x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−1
y = ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f (x)
a
b
a + ε
ε
Z
b
a+ε
f(x) dx
small
even smaller
y = g(x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
a
8
1
2
3
4
5
6
7
8
n
a
n
x
y
y = f (x)
(a, f(a))
a
Nevertheless, it is a pretty good approximation to the curve near (a, f(a)). In
fact, let’s zoom in near (a, f(a)):
PSfrag
replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shado
w
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror
(y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y =
(x − 1)
2
−
1
x
Same
height
−
x
Same
length,
opp
osite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y =
2
x
y =
10
x
y =
2
−x
y =
log
2
(x)
4
3
units
mirror
(x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
=
0 radians
θ
hypotenuse
opp
osite
adjacen
t
0
(≡ 2π)
π
2
π
3
π
2
I
I
I
I
II
IV
θ
(
x, y)
x
y
r
7
π
6
reference
angle
reference
angle =
π
6
sin
+
sin −
cos
+
cos −
tan
+
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this
angle is
5π
6
clo
ckwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
cos(x)
−
π
2
π
2
y =
tan(x), −
π
2
<
x <
π
2
0
−
π
2
π
2
y =
tan(x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y =
sec(x)
y =
csc(x)
y =
cot(x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y =
tan
−1
(x)
π
2
π
y =
sin(
x)
x
,
x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 <
x < 0.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f(
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x +
2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangen
t at x = a
b
tangen
t at x = b
c
tangen
t at x = c
y = x
2
tangen
t
at x = −
1
u
v
uv
u +
∆u
v +
∆v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f(
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y =
sin(x)
y = x
x
A
B
O
1
C
D
sin(
x)
tan(
x)
y =
sin
(x)
x
π
2
π
1
−
1
x =
0
a =
0
x
> 0
a
> 0
x
< 0
a
< 0
rest
position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y =
ln(x)
y =
cosh(x)
y =
sinh(x)
y =
tanh(x)
y =
sech(x)
y =
csch(x)
y =
coth(x)
1
−
1
y = f(
x)
original
function
in
verse function
slop
e = 0 at (x, y)
slop
e is infinite at (y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
sin(x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y =
sin
−1
(x)
y =
cos(x)
π
π
2
y =
cos
−1
(x)
−
π
2
1
x
α
β
y =
tan(x)
y =
tan(x)
1
y =
tan
−1
(x)
y =
sec(x)
y =
sec
−1
(x)
y =
csc
−1
(x)
y =
cot
−1
(x)
1
y =
cosh
−1
(x)
y =
sinh
−1
(x)
y =
tanh
−1
(x)
y =
sech
−1
(x)
y =
csch
−1
(x)
y =
coth
−1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y =
1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Lo
cal maximum
Lo
cal minimum
Horizon
tal point of inflection
1
e
y = f
0
(
x)
y = f(
x) = x ln(x)
−
1
e
?
y = f (
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2
+
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2
+
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing
fence
new
fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallo
w
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f(
x)
11
y = L
(x)
y = f(
x)
F
P
a
a +
∆x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
d
f
∆
x
a
b
y = f(
x)
true
zero
starting
approximation
b
etter approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y =
1
a
b
y =
sin(x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
=
2
width
of each interval =
2
n
−
2
1
3
0
I
I
I
I
II
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
I
I
I
Ia
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
a
v
y = f
a
v
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (x )
y = f (t)
F (x + h )
x + h
F (x + h) − F (x)
f(x)
1
2
y = sin(x)
π
−π
−1
−2
y =
1
x
y = x
2
1
2
1
−1
y = ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(x)
a
b
a + ε
ε
Z
b
a+ε
f(x) dx
small
even smaller
y = g(x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
a
8
1
2
3
4
5
6
7
8
n
a
n
x
y
y = f(x)
(a, f(a))
a
Section 24.1.2: Quadratic approximations • 521
Now you can see that there’s not that much difference between the tangent
line and the curve y = f(x). The more we zoom in near x = a, the smaller
this difference becomes.
24.1.2 Quadratic approximations
Why stick to lines, though? Let’s ask the same question we did at the begin-
ning of the previous section, but with parabolas instead. Here is our question:
what is the equation of the quadratic which best approximates the curve
y = f (x) near (a, f(a))? Using the same function as in the picture above,
here’s a guess as to what the quadratic should look like:
PSfrag
replacements
(
a, b)
[
a, b]
(
a, b]
[
a, b)
(
a, ∞)
[
a, ∞)
(
−∞, b)
(
−∞, b]
(
−∞, ∞)
{
x : a < x < b}
{
x : a ≤ x ≤ b}
{
x : a < x ≤ b}
{
x : a ≤ x < b}
{
x : x ≥ a}
{
x : x > a}
{
x : x ≤ b}
{
x : x < b}
R
a
b
shado
w
0
1
4
−
2
3
−
3
g(
x) = x
2
f(
x) = x
3
g(
x) = x
2
f(
x) = x
3
mirror
(y = x)
f
−
1
(x) =
3
√
x
y = h
(x)
y = h
−
1
(x)
y =
(x − 1)
2
−
1
x
Same
height
−
x
Same
length,
opp
osite signs
y = −
2x
−
2
1
y =
1
2
x − 1
2
−
1
y =
2
x
y =
10
x
y =
2
−x
y =
log
2
(x)
4
3
units
mirror
(x-axis)
y = |
x|
y = |
log
2
(x)|
θ radians
θ units
30
◦
=
π
6
45
◦
=
π
4
60
◦
=
π
3
120
◦
=
2
π
3
135
◦
=
3
π
4
150
◦
=
5
π
6
90
◦
=
π
2
180
◦
= π
210
◦
=
7
π
6
225
◦
=
5
π
4
240
◦
=
4
π
3
270
◦
=
3
π
2
300
◦
=
5
π
3
315
◦
=
7
π
4
330
◦
=
11
π
6
0
◦
=
0 radians
θ
hypotenuse
opp
osite
adjacen
t
0
(≡ 2π)
π
2
π
3
π
2
I
I
I
I
II
IV
θ
(
x, y)
x
y
r
7
π
6
reference
angle
reference
angle =
π
6
sin
+
sin −
cos
+
cos −
tan
+
tan −
A
S
T
C
7
π
4
9
π
13
5
π
6
(this
angle is
5π
6
clo
ckwise)
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
cos(x)
−
π
2
π
2
y =
tan(x), −
π
2
<
x <
π
2
0
−
π
2
π
2
y =
tan(x)
−
2π
−
3π
−
5
π
2
−
3
π
2
−
π
−
π
2
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
y =
sec(x)
y =
csc(x)
y =
cot(x)
y = f(
x)
−
1
1
2
y = g(
x)
3
y = h
(x)
4
5
−
2
f(
x) =
1
x
g(
x) =
1
x
2
etc.
0
1
π
1
2
π
1
3
π
1
4
π
1
5
π
1
6
π
1
7
π
g(
x) = sin
1
x
1
0
−
1
L
10
100
200
y =
π
2
y = −
π
2
y =
tan
−1
(x)
π
2
π
y =
sin(
x)
x
,
x > 3
0
1
−
1
a
L
f(
x) = x sin (1/x)
(0 <
x < 0.3)
h
(x) = x
g(
x) = −x
a
L
lim
x
→a
+
f(x) = L
lim
x
→a
+
f(x) = ∞
lim
x
→a
+
f(x) = −∞
lim
x
→a
+
f(x) DNE
lim
x
→a
−
f(x) = L
lim
x
→a
−
f(x) = ∞
lim
x
→a
−
f(x) = −∞
lim
x
→a
−
f(x) DNE
M
}
lim
x
→a
−
f(x) = M
lim
x
→a
f(x) = L
lim
x
→a
f(x) DNE
lim
x
→∞
f(x) = L
lim
x
→∞
f(x) = ∞
lim
x
→∞
f(x) = −∞
lim
x
→∞
f(x) DNE
lim
x
→−∞
f(x) = L
lim
x
→−∞
f(x) = ∞
lim
x
→−∞
f(x) = −∞
lim
x
→−∞
f(x) DNE
lim
x →a
+
f(
x) = ∞
lim
x →a
+
f(
x) = −∞
lim
x →a
−
f(
x) = ∞
lim
x →a
−
f(
x) = −∞
lim
x →a
f(
x) = ∞
lim
x →a
f(
x) = −∞
lim
x →a
f(
x) DNE
y = f(
x)
a
y =
|
x|
x
1
−
1
y =
|
x + 2|
x +
2
1
−
1
−
2
1
2
3
4
a
a
b
y = x sin
1
x
y = x
y = −
x
a
b
c
d
C
a
b
c
d
−
1
0
1
2
3
time
y
t
u
(
t, f(t))
(
u, f(u))
time
y
t
u
y
x
(
x, f(x))
y = |
x|
(
z, f(z))
z
y = f(
x)
a
tangen
t at x = a
b
tangen
t at x = b
c
tangen
t at x = c
y = x
2
tangen
t
at x = −
1
u
v
uv
u +
∆u
v +
∆v
(
u + ∆u)(v + ∆v)
∆
u
∆
v
u
∆v
v∆
u
∆
u∆v
y = f(
x)
1
2
−
2
y = |
x
2
− 4|
y = x
2
− 4
y = −
2x + 5
y = g(
x)
1
2
3
4
5
6
7
8
9
0
−
1
−
2
−
3
−
4
−
5
−
6
y = f(
x)
3
−
3
3
−
3
0
−
1
2
easy
hard
flat
y = f
0
(
x)
3
−
3
0
−
1
2
1
−
1
y =
sin(x)
y = x
x
A
B
O
1
C
D
sin
(x)
tan
(x)
y =
sin
(x)
x
π
2
π
1
−
1
x =
0
a =
0
x
> 0
a
> 0
x
< 0
a
< 0
rest
position
+
−
y = x
2
sin
1
x
N
A
B
H
a
b
c
O
H
A
B
C
D
h
r
R
θ
1000
2000
α
β
p
h
y = g(
x) = log
b
(x)
y = f(
x) = b
x
y = e
x
5
10
1
2
3
4
0
−
1
−
2
−
3
−
4
y =
ln(x)
y =
cosh(x)
y =
sinh(x)
y =
tanh(x)
y =
sech(x)
y =
csch(x)
y =
coth(x)
1
−
1
y = f(
x)
original
function
in
verse function
slop
e = 0 at (x, y)
slop
e is infinite at (y, x)
−
108
2
5
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
y =
sin(x)
y =
sin(x), −
π
2
≤ x ≤
π
2
−
2
−
1
0
2
π
2
−
π
2
y =
sin
−1
(x)
y =
cos(x)
π
π
2
y =
cos
−1
(x)
−
π
2
1
x
α
β
y =
tan(x)
y =
tan(x)
1
y =
tan
−1
(x)
y =
sec(x)
y =
sec
−1
(x)
y =
csc
−1
(x)
y =
cot
−1
(x)
1
y =
cosh
−1
(x)
y =
sinh
−1
(x)
y =
tanh
−1
(x)
y =
sech
−1
(x)
y =
csch
−1
(x)
y =
coth
−1
(x)
(0
, 3)
(2
, −1)
(5
, 2)
(7
, 0)
(
−1, 44)
(0
, 1)
(1
, −12)
(2
, 305)
y =
1
2
(2
, 3)
y = f(
x)
y = g(
x)
a
b
c
a
b
c
s
c
0
c
1
(
a, f(a))
(
b, f(b))
1
2
1
2
3
4
5
6
0
−
1
−
2
−
3
−
4
−
5
−
6
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
3
π
5
π
2
2
π
3
π
2
π
π
2
y =
sin(x)
1
0
−
1
−
3π
−
5
π
2
−
2π
−
3
π
2
−
π
−
π
2
3
π
5
π
2
2
π
2
π
3
π
2
π
π
2
c
OR
Lo
cal maximum
Lo
cal minimum
Horizon
tal point of inflection
1
e
y = f
0
(
x)
y = f (
x) = x ln(x)
−
1
e
?
y = f (
x) = x
3
y = g(
x) = x
4
x
f(
x)
−
3
−
2
−
1
0
1
2
1
2
3
4
+
−
?
1
5
6
3
f
0
(
x)
2 −
1
2
√
6
2
+
1
2
√
6
f
00
(
x)
7
8
g
00
(
x)
f
00
(
x)
0
y =
(
x − 3)(x − 1)
2
x
3
(
x + 2)
y = x ln(
x)
1
e
−
1
e
5
−
108
2
α
β
2 −
1
2
√
6
2
+
1
2
√
6
y = x
2
(
x − 5)
3
−
e
−
1/2
√
3
e
−
1/2
√
3
−
e
−3/2
e
−
3/2
−
1
√
3
1
√
3
−
1
1
y = xe
−
3x
2
/2
y =
x
3
− 6
x
2
+ 13x − 8
x
28
2
600
500
400
300
200
100
0
−
100
−
200
−
300
−
400
−
500
−
600
0
10
−
10
5
−
5
20
−
20
15
−
15
0
4
5
6
x
P
0
(
x)
+
−
−
existing
fence
new
fence
enclosure
A
h
b
H
99
100
101
h
dA/dh
r
h
1
2
7
shallo
w
deep
LAND
SEA
N
y
z
s
t
3
11
9
L
(11)
√
11
y = L
(x)
y = f(
x)
11
y = L
(x)
y = f(
x)
F
P
a
a +
∆x
f(
a + ∆x)
L
(a + ∆x)
f(
a)
error
d
f
∆
x
a
b
y = f(
x)
true
zero
starting
approximation
b
etter approximation
v
t
3
5
50
40
60
4
20
30
25
t
1
t
2
t
3
t
4
t
n
−2
t
n
−1
t
0
= a
t
n
= b
v
1
v
2
v
3
v
4
v
n
−1
v
n
−
30
6
30
|
v|
a
b
p
q
c
v(
c)
v(
c
1
)
v(
c
2
)
v(
c
3
)
v(
c
4
)
v(
c
5
)
v(
c
6
)
t
1
t
2
t
3
t
4
t
5
c
1
c
2
c
3
c
4
c
5
c
6
t
0
=
a
t
6
=
b
t
16
=
b
t
10
=
b
a
b
x
y
y = f(
x)
1
2
y = x
5
0
−
2
y =
1
a
b
y =
sin(x)
π
−
π
0
−
1
−
2
0
2
4
y = x
2
0
1
2
3
4
2
n
4
n
6
n
2(
n−2)
n
2(
n−1)
n
2
n
n
=
2
width
of each interval =
2
n
−
2
1
3
0
I
I
I
I
II
IV
4
y
dx
y = −
x
2
− 2x + 3
3
−
5
y = |−
x
2
− 2x + 3|
I
I
I
I
Ia
5
3
0
1
2
a
b
y = f (
x)
y = g(
x)
y = x
2
a
b
5
3
0
1
2
y =
√
x
2
√
2
2
2
dy
x
2
a
b
y = f(
x)
y = g(
x)
M
m
1
2
−
1
−
2
0
y = e
−
x
2
1
2
e
−
1/4
f
a
v
y = f
a
v
c
A
M
0
1
2
a
b
x
t
y = f (
t)
F (
x )
y = f (
t)
F (
x + h )
x + h
F (
x + h) − F (x)
f(
x)
1
2
y =
sin(x)
π
−
π
−
1
−
2
y =
1
x
y = x
2
1
2
1
−
1
y =
ln|x|
θ
a
x
a
x
p
a
2
− x
2
3
x
p
9 − x
2
p
x
2
+ a
2
x
a
p
x
2
+ 15
x
√
15
x
p
x
2
− a
2
a
x
p
x
2
− 4
2
x
−
p
x
2
− a
2
a
x
−
p
x
2
− 4
2
y = f(x)
a
b
a + ε
ε
Z
b
a+ε
f(x) dx
small
even smaller
y = g(x)
infinite area
finite area
1
y =
1
x
y =
1
x
p
, p < 1 (typical)
y =
1
x
p
, p > 1 (typical)
a
1
a
2
a
3
a
4
a
5
a
6
a
7
a
8
1
2
3
4
5
6
7
8
n
a
n
x
y
y = f(x)
(a, f(a))
a
It turns out that the formula for the quadratic which best approximates the
curve y = f(x) for x near a (that is, near the point (a, f (a)) on the curve) is
given by
y = f(a) + f
0
(a)(x − a) +
f
00
(a)
2
(x − a)
2
.
This is actually a quadratic in x, because if you expand (x −a)
2
, the highest
power of x floating around is x
2
. Still, it’s better to leave it in the above form
and say that it’s a “quadratic in (x −a).” Let’s call the quadratic P
2
; that is,
we set
P
2
(x) = f(a) + f
0
(a)(x − a) +
f
00
(a)
2
(x − a)
2
.
Now, let’s gather some nice facts about P
2
:
1. Plug x = a into the above equation for P
2
(x) and you easily see that
P
2
(a) = f(a). So the values of P
2
and f match when x = a. In fact,
since the zeroth derivative of a function is just the function itself, you
might say that the values of the zeroth derivatives of P
2
and f match
when x = a.
2. Now differentiate P
2
to see that P
0
2
(x) = f
0
(a) + f
00
(a)(x −a). Again, if
you plug in x = a, you see that P
0
2
(a) = f
0
(a). The values of the first
derivatives P
0
2
and f
0
also match when x = a.
3. Differentiate once more to get P
00
2
(x) = f
00
(a). When x = a, this becomes
P
00
2
(a) = f
00
(a), so even the values of the second derivatives match when
x = a.
522 • Taylor Polynomials, Taylor Series, and Power Series
4. On the other hand, since f
00
(a) is constant, P
000
2
(x) = 0 for all x. The
same is true for all higher derivatives. (After all, P
2
is a quadratic,
and the third and higher derivatives of any quadratic must be zero
everywhere!)
So P
2
shares the zeroth, first, and second derivatives with f at x = a; but the
third and higher derivatives of P
2
are always 0. You might say that P
2
is the
distillation of all the information about f at x = a up to and including the
second derivative.
Here’s another nice fact about P
2
: if you ignore the last term on the right-
hand side of the above equation for P
2
(x), you just get f(a) + f
0
(a)(x − a).
This is exactly the linearization from the previous section. So you can think of
the last term
1
2
f
00
(a)(x −a)
2
as a so-called second-order correction term. This
means that we should actually be able to do a better job of approximation
than just by using the tangent line. The second-order correction term helps
us get even closer to the curve, at least for x near a. (An exception to this
occurs when f
00
(a) = 0, in which case P
2
is actually the linearization and we
haven’t gotten any closer.)
24.1.3 Higher-degree approximations
Let’s continue the same pattern, except that we’ll use some arbitrary degree
N instead of just 1 or 2. So, here’s our question: which polynomial of degree
N or less gives the best approximation to f(x) for values of x near a? The
answer is provided by the following theorem.
A Taylor approximation theorem: if f is smooth at x = a,
then of all the polynomials of degree N or less, the one which
best approximates f(x) for x near a is given by
P
N
(x) = f(a) + f
0
(a)(x − a) +
f
00
(a)
2!
(x − a)
2
+
f
(3)
(a)
3!
(x − a)
3
+ ··· +
f
(N)
(a)
N!
(x − a)
N
.
In sigma notation, the formula looks like this:
P
N
(x) =
N
X
n=0
f
(n)
(a)
n!
(x − a)
n
.
In this formula, remember that 0! = 1, that f
(0)
(a) means the same thing as
f(a) (zero derivatives), and that f
(1)
(a) means the same thing as f
0
(a) (one
derivative).
We call the polynomial P
N
the Nth-order Taylor polynomial of f(x) at
x = a. Note that the degree of P
N
might be less than N; for example, if
f
(N)
(a) = 0, then the last term in the above sum vanishes and the degree of
P
N
could be at most N −1. This is why we call it an Nth-order Taylor poly-
nomial, not an Nth-degree Taylor polynomial. (By the way, the polynomial
P
N
(x) is sometimes written as P
N
(x; a) to emphasize that you get a different
polynomial for each choice of N and a. I’ll just write P
N
(x), since we’re only
dealing with one choice of a at a time anyway.)
Section 24.1.4: Taylor’s Theorem • 523
Once again, the important property of P
N
is that
P
(n)
N
(a) = f
(n)
(a)
for all n = 0, 1, . . . , N. That is, the values of all the derivatives of P
N
and
f match when x = a, up to and including the Nth derivative; but all higher
derivatives of P
N
must be zero everywhere. The function P
N
is the distillation
of all the information about f which comes from its derivatives up to order
N at x = a.
Of course, when N = 1, we just get P
1
(x) = f(a) + f
0
(a)(x − a), which
is the linearization of f at x = a, and when N = 2, we recover the formula
for P
2
(x) from the previous section. Let’s see how this works for f (x) = e
x
,
setting a = 0. By the above formula with N = 3 and a = 0, we have
P
3
(x) = f(0) + f
0
(0)x +
f
00
(0)
2!
x
2
+
f
(3)
(0)
3!
x
3
.
Luckily, all the derivatives of e
x
with respect to x are just e
x
, so we can see
that f(0), f
0
(0), f
00
(0), and f
(3)
(0) are all e
0
, which is equal to 1. Since 2! = 2
and 3! = 6, the above formula becomes
P
3
(x) = 1 + x +
1
2
x
2
+
1
6
x
3
.
This is exactly the cubic polynomial from the beginning of Section 24.1 above!
Of all degree three or lower polynomials, this one gives the best approximation
for e
x
when x is near 0. Why 0? Because that’s the value we chose for a. If
we chose a different value of a, we’d get a different polynomial which would
approximate e
x
really well for x near a. By getting rid of the cubic term
x
3
/6, we can see that P
2
(x) = 1 + x + x
2
/2; then by throwing away the
quadratic term x
2
/2, we get the linearization P
1
(x) = 1 + x. Another way to
look at this is that P
2
(x) improves upon P
1
(x) by adding on the second-order
correction term x
2
/2, while P
3
(x) improves upon P
2
(x) by adding on a third-
order correction term x
3
/6. Each time you increase N by 1, you are making
the approximation better by adding on another correction term.
The Taylor approximation theorem actually depends on Taylor’s Theorem,
which we’ll look at in the next section. There’s something ambiguous about
the statement of the approximation theorem, as well: what on earth does it
mean to be the “best approximation,” anyway? We’ll explore this more in the
next section, but the real answer is contained in Section A.7 of Appendix A,
along with the proof of the theorem.
24.1.4 Taylor’s Theorem
In Section 24.1 above, we saw that
e
x
∼
=
1 + x +
x
2
2
+
x
3
6
.
In particular, we noticed that when x = −1/10, the above approximation
becomes
e
−1/10
∼
=
1 −
1
10
+
1/100
2
−
1/1000
6
=
5429
6000
.
524 • Taylor Polynomials, Taylor Series, and Power Series
How good is this approximation? One way to measure this is to consider the
difference between the true quantity e
−1/10
and the approximation 5429/6000.
We’ll call this quantity the error in the approximation, since it shows how
wrong we are to use our approximation instead of the true value. Here’s what
the error is in our case:
error = true value − approximate value = e
−1/10
−
5429
6000
.
If the error is small, then the approximation is good. In Section 24.1, we saw
that the difference was 0.0000040847 to 10 decimal places—but we needed
to use a calculator, which defeats the entire purpose of doing our own ap-
proximation. Remember that the number the calculator gives you is also an
approximation! Besides, how do you think the calculator works? It probably
finds its approximation of e
−1/10
using a Taylor polynomial.
What we’d really like is another formula for the error. That’s where Tay-
lor’s theorem comes in. Rather than just specializing to the case of e
x
, let’s
get more general once again. We’re dealing with a smooth function f and its
Nth-order Taylor polynomial about x = a; as we saw in the previous section,
this polynomial is given by
P
N
(x) =
N
X
n=0
f
(n)
(a)
n!
(x − a)
n
.
We want to use the value of P
N
(x) to approximate the true value f(x), so we
consider the error term, which is the difference between the true value and
the approximate value:
R
N
(x) = f(x) − P
N
(x).
Actually, R
N
(x) is called the Nth-order error term; it’s also referred to as the
Nth-order remainder term, since it’s all that remains when you take P
N
(x)
away from f(x). As promised above, Taylor’s Theorem gives an alternative
formula for R
N
(x):
Taylor’s Theorem: the Nth-order remainder term R
N
(x) about x = a is
R
N
(x) =
f
(N+1)
(c)
(N + 1)!
(x − a)
N+1
where c is some number which lies between x and a.
Note that the number c depends on what x and N are, and cannot be deter-
mined in general! Since f(x) = P
N
(x) + R
N
(x), we can write the whole kit
and caboodle as
f(x) =
N
X
n=0
f
(n)
(a)
n!
(x − a)
n
+
f
(N+1)
(c)
(N + 1)!
(x − a)
N+1
.
This seems pretty nasty. And what on earth is with this number c, anyway?
Actually, we’ve seen something like this before. Take a look back at our
Section 24.1.4: Taylor’s Theorem • 525
discussion of Mean Value Theorem (MVT) in Section 11.3 of Chapter 11.
The MVT says that if f is smooth enough on an interval [a, b], then there is
a number c in [a, b] (which cannot be determined in general) such that
f
0
(c) =
f(b) − f(a)
b − a
.
If you replace b by x and solve for f(x), you get
f(x) = f (a) + f
0
(c)(x − a),
where c is between a and x. Now let’s look back to the last equation in
Taylor’s Theorem and put N = 0. What is P
0
(x)? It’s just f(a). How about
R
0
(x)? According to Taylor’s Theorem,
R
0
(x) =
f
(1)
(c)
1!
(x − a)
1
= f
0
(c)(x − a),
where c is between x and a. Hey, so Taylor’s Theorem (with N = 0) says that
f(x) = P
0
(x) + R
0
(x) = f(a) + f
0
(c)(x − a),
which is exactly what the MVT says! So, Taylor’s Theorem is basically the
Mean Value Theorem on steroids. By the way, the reason we say that c is
between x and a instead of writing a ≤ c ≤ x is that x might actually be less
than a, so then we would have x ≤ c ≤ a.
Now let’s put N = 1 instead of N = 0. The main formula in the box above
becomes
f(x) = f (a) + f
0
(a)(x − a) +
f
00
(c)
2!
(x − a)
2
= L(x) + R
1
(x);
here L(x) = f(a) + f
0
(a)(x − a) is the linearization of f about x = a, and
R
1
(x) =
1
2
f
00
(c)(x − a)
2
is the first-order error term. This agrees with the
formula for the error term r(x) which we gave in Section 13.2.4 of Chapter 13.
We still have to go back to our approximation for e
x
. When we wrote
e
x
∼
=
1 + x +
x
2
2
+
x
3
6
,
we now understand that this is just saying that e
x
∼
=
P
3
(x), where P
3
is the
third-order Taylor polynomial of f(x) = e
x
about x = 0. Taylor’s Theorem
above says that R
3
is then given by
R
3
(x) =
f
(4)
(c)
4!
x
4
,
where c is between 0 and x. (I just plugged N = 3 and a = 0 into the formula
for R
N
(x) in the box above.) Since any derivative of e
x
(with respect to x) is
e
x
, we know that f
(4)
(c) = e
c
; also 4! = 24, so we actually have
R
3
(x) =
e
c
24
x
4
.