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526 • Taylor Polynomials, Taylor Series, and Power Series
In other words,
e
x
= 1 + x +
x
2
2
+
x
3
6
+
e
c
24
x
4
.
We have changed our approximation into an exact equation, but we don’t
know what c is! Still, we do get something very useful from this, because we
know that c lies between 0 and x. For example, if you put x = −1/10 once
again, you get
e
−1/10
∼
=
1 −
1
10
+
1/100
2
−
1/1000
6
+
e
c
24
(1/10000),
which reduces to
e
−1/10
=
5429
6000
+
e
c
240000
.
This time, we know c lies between 0 and x = −1/10, so we actually have
−1/10 < c < 0. Since e
c
is increasing in c, it’s clearly biggest when c is as big
as possible; this means that c would have to be 0, and so e
c
can’t be bigger
than e
0
= 1. So the error term is at most 1/240000. In other words, when
we write e
−1/10
∼
=
5249/6000, we know that the approximation is accurate to
better than 1/240000, which is about 0.0000041667. (Compare this with the
actual value of the difference in Section 24.1 above.)
We’ll look at some examples of using Taylor’s Theorem in Section 25.3 in
the next chapter. Now it’s time to check out power and Taylor series.
24.2 Power Series and Taylor Series
Here’s another fact:
e
x
= 1 + x +
x
2
2!
+
x
3
3!
+
x
4
4!
+
x
5
5!
+ ···
for all real numbers x. You might notice that it looks similar to the approxima-
tion at the beginning of Section 24.1 above, but there are two big differences.
First, we’re no longer dealing with an approximation, and second, there’s an
infinite series on the right-hand side! Whenever you have an infinite series,
you’ve got to be careful.
So, let’s see if we can understand what the above equation actually means.
Suppose we start with the right-hand side,
1 + x +
x
2
2!
+
x
3
3!
+
x
4
4!
+
x
5
5!
+ ··· .
This looks like a polynomial, but it isn’t, since there’s no highest-degree term.
It just keeps on going forever. In fact, it’s an example of a power series. If you
replace x by any particular value, you get a regular old series. For example,
if x = −1/10, you get the series
1 −
1
10
+
1/100
2!
−
1/1000
3!
+
1/10000
4!
−
1/100000
5!
+ ··· ,
which you could rewrite as
1 −
1
10
+
1
100 × 2!
−
1
1000 × 3!
+
1
10000 × 4!
−
1
100000 × 5!
+ ··· .
Section 24.2.1: Power series in general • 527
This series might converge, or it might diverge. So which is it? The answer is
that it converges, and what’s more, we even know that it converges to e
−1/10
.
That’s the power of knowing that our above equation is valid for any real x:
e
x
= 1 + x +
x
2
2!
+
x
3
3!
+
x
4
4!
+
x
5
5!
+ ··· .
It just means that if you plug any particular x into the right-hand side, you
get a series which converges to the number e
x
. We’ll prove that this is actually
true in Section 24.2.3 below; in the meantime, here are some more examples
of what happens when you plug in a few values of x, one at a time:
x = 2 : 1 + 2 +
2
2
2!
+
2
3
3!
+
2
4
4!
+
2
5
5!
+ ··· converges to e
2
,
x = −5 : 1 − 5 +
5
2
2!
−
5
3
3!
+
5
4
4!
−
5
5
5!
+ ··· converges to e
−5
,
x = 0 : 1 + 0 + 0 + 0 + 0 + ··· converges to 1.
I could give you a million more examples—actually, infinitely more. This
single power series gives us information about infinitely many regular series,
one for each value of x. By the way, it’s pretty obvious that the last series
above converges to 1. There’s something special about setting x = 0: it makes
all the terms vanish except for the constant term. We’ll address this point
soon; first, let’s look at general power series.
24.2.1 Power series in general
A power series about x = 0 is an expression of the form
a
0
+ a
1
x + a
2
x
2
+ a
3
x
3
+ a
4
x
4
+ ··· ,
where the numbers a
n
are fixed constants. Even though a power series isn’t a
polynomial, we’ll still refer to a
n
as the coefficient of x
n
in the power series.
The above series can also be written using sigma notation as
∞
X
n=0
a
n
x
n
.
In our example from the previous section, the series is
1 + x +
x
2
2!
+
x
3
3!
+
x
4
4!
+
x
5
5!
+ ···
which can be written in sigma notation as
∞
X
n=0
1
n!
x
n
.
So this is a power series with coefficients given by a
n
= 1/n! for each non-
negative integer n. Notice that x is the only true variable here; n is just a
dummy variable which goes away when you actually expand the sum. An
528 • Taylor Polynomials, Taylor Series, and Power Series
even simpler power series than the above one, written in expanded form and
also using sigma notation, is
1 + x + x
2
+ x
3
+ x
4
+ ··· =
∞
X
n=0
x
n
.
In this series, the coefficients a
n
are all equal to 1. Hopefully you can recognize
this as a geometric series with first term 1 and ratio x.
We often want to write an equation like
f(x) = a
0
+ a
1
x + a
2
x
2
+ a
3
x
3
+ a
4
x
4
+ ···
for some given range of x. This means that when you plug in one of the allowed
values of x, the power series becomes a regular old series which converges to
the value f(x). For example, we have said (but not yet proved) that
e
x
= 1 + x +
x
2
2!
+
x
3
3!
+
x
4
4!
+
x
5
5!
+ ···
for all x. On the other hand, when we looked at how to find the sum of a
geometric progression in Section 22.2 of Chapter 22, we saw that
1 + r + r
2
+ r
3
+ r
4
+ ··· =
∞
X
n=0
r
n
=
1
1 − r
provided that −1 < r < 1.
Let’s replace r by x:
1 + x + x
2
+ x
3
+ x
4
+ ··· =
∞
X
n=0
x
n
=
1
1 − x
provided that −1 < x < 1.
That is, we are claiming that
1
1 − x
= 1 + x + x
2
+ x
3
+ x
4
+ ···
when −1 < x < 1. If you replace x by any such number, you get a regular
series on the right and the value it converges to on the left. On the other
hand, what if x > 1 or x ≤ −1? The left-hand side makes sense, but the
right-hand side doesn’t since the series diverges for these values of x. (Both
sides are undefined if x is actually equal to 1.)
Something nice happens to the power series
a
0
+ a
1
x + a
2
x
2
+ a
3
x
3
+ a
4
x
4
+ ···
when you set x = 0: all the terms vanish except for the a
0
at the beginning,
so the series automatically converges (to a
0
, of course!). This doesn’t tell us
anything about whether the series converges for any other value of x. For
example, the geometric series only converges when −1 < x < 1, while we’ll
show in Section 26.1.2 in Chapter 26 that the following power series only
converges when x = 0:
∞
X
n=0
n!x
n
Section 24.2.2: Taylor series and Maclaurin series • 529
Now 0 is a pretty funky number, admittedly, but it doesn’t need to be more
special than the rest of the real numbers. Let’s transfer this special property
over to some other number a. All we have to do is replace x by (x − a). So
here is the general expression for a power series about x = a:
a
0
+ a
1
(x − a) + a
2
(x − a)
2
+ a
3
(x − a)
3
+ a
4
(x − a)
4
+ ··· .
In sigma notation, this looks like
∞
X
n=0
a
n
(x − a)
n
.
This series converges for sure when x = a, since all the terms except a
0
vanish.
The number a is called the center of the power series. When would you want
to consider a power series with a center other than 0? One example might be
if you wanted to find a power series which converges to ln(x). This quantity
isn’t defined at x = 0, so it would be silly to try to find a power series about
x = 0 which converges to ln(x). On the other hand, we can find a power series
with center 1 which converges to ln(x), at least for some values of x. Indeed,
at the end of Section 26.2.1 of Chapter 26, we’ll see that the equation
∞
X
n=1
(−1)
n−1
n
(x − 1)
n
= ln(x)
is valid for −1 < (x − 1) < 1, that is, for 0 < x < 2. (It’s actually even true
for x = 2:
∞
X
n=1
(−1)
n−1
n
= 1 −
1
2
+
1
3
−
1
4
+
1
5
− ··· = ln(2).
This isn’t so easy to prove, however!)
24.2.2 Taylor series and Maclaurin series
In the previous section, we saw that a general power series about x = a is
given (using sigma notation and also in expanded form) by
∞
X
n=0
a
n
(x − a)
n
= a
0
+ a
1
(x −a) + a
2
(x −a)
2
+ a
3
(x − a)
3
+ a
4
(x −a)
4
+ ··· .
This converges for x = a, and might converge for other values of x. In
Section 26.1.2 of Chapter 26, we’ll look at some methods for finding which
values of x make the series converge. We could then plug in all these values of
x one at a time, find what the series converges to in each case, and call that
f(x). So, starting with a power series, we have defined a function.
Suppose that we instead start off with some smooth function f . We’re
going to define a special power series about x = a by using all the derivatives
of f:
∞
X
n=0
f
(n)
(a)
n!
(x − a)
n
.
530 • Taylor Polynomials, Taylor Series, and Power Series
When you expand the sigma notation, this becomes
f(a) + f
0
(a)(x −a) +
f
00
(a)
2!
(x −a)
2
+
f
(3)
(a)
3!
(x −a)
3
+
f
(4)
(a)
4!
(x −a)
4
+ ··· .
The coefficients of this power series are given by a
n
= f
(n)
(a)/n!. The series
is called the Taylor series of f about x = a. So, starting with a function, we
have defined a power series.
Take a closer look at the definition of the Taylor series above. It should
look familiar. In fact, the formula is very similar to the definition of the Taylor
polynomial P
N
(x) from Section 24.1.3 above. The only difference is that the
sum doesn’t stop at n = N: it keeps on going to ∞. In other words, the
Taylor polynomial P
N
(x) is the Nth partial sum of the Taylor series.
We’ll explore the connection between Taylor polynomials and Taylor series
in the next section. First, we have just one more definition: the Maclaurin
series of f is just another name for the Taylor series of f about x = 0. So it’s
given by
∞
X
n=0
f
(n)
(0)
n!
x
n
,
or in expanded form by
f(0) + f
0
(0)x +
f
00
(0)
2!
x
2
+
f
(3)
(0)
3!
x
3
+
f
(4)
(0)
4!
x
4
+ ··· .
Whenever you see the words “Maclaurin series,” mentally replace them by
“Taylor series with a = 0” and you’ll do just fine.
24.2.3 Convergence of Taylor series
OK, let’s review the situation. We started out with a function f and a number
a, and we constructed the Taylor series of f about x = a:
∞
X
n=0
f
(n)
(a)
n!
(x − a)
n
.
This is a power series with center a, but it’s not just any old power series: it
encapsulates the values of all the derivatives of f at x = a. It would be really
cool if we could write
f(x) =
∞
X
n=0
f
(n)
(a)
n!
(x − a)
n
,
since then we’d know that the Taylor series converges for any x and also that
it converges to the original function value f(x). The problem is, the above
equation isn’t always valid. The series could diverge for some values of x, or
even every value of x (except x = a: as we’ve seen, a power series always
converges at its center). Even worse, the series could converge to something
other than f(x)! Luckily, in our examples, we’ll avoid this weird possibility.
∗
∗
I will just mention a classic example of a whacked-out Taylor series: if f(x) = e
−1/x
2
when x 6= 0, and we also define f(0) = 0, then all the derivatives of f at 0 are equal to 0,
so the Taylor series of f with center 0 is just 0. This is not the same thing as f(x) at all,
except when x = 0.
Section 24.2.3: Convergence of Taylor series • 531
So, how do you know if and when a Taylor series actually converges to its
underlying function? Start by writing
f(x) = P
N
(x) + R
N
(x),
as we did in Section 24.1.4 above. Remember,
P
N
(x) =
N
X
n=0
f
(n)
(a)
n!
(x − a)
n
and R
N
(x) =
f
(N+1)
(c)
(N + 1)!
(x − a)
N+1
.
This expresses f(x) as its approximate value P
N
(x) plus the error, or remain-
der, R
N
(x). Now here’s the clever part: we let N get larger and larger. This
should hopefully make the approximation P
N
(x) get closer and closer to the
true value f(x). This is the same thing as saying that hopefully the error
R
N
(x) gets smaller and smaller.
Let’s try to write down some equations to describe all this. Suppose that
for some x, we know that
lim
N→∞
R
N
(x) = 0.
In the equation f(x) = P
N
(x) + R
N
(x), take limits as N → ∞:
lim
N→∞
f(x) = lim
N→∞
P
N
(x) + lim
N→∞
R
N
(x) = lim
N→∞
P
N
(x).
Since f(x) doesn’t depend on N , the left-hand side is just f(x), so we know
that
f(x) = lim
N→∞
P
N
(x) = lim
N→∞
N
X
n=0
f
(n)
n!
(x − a)
n
=
∞
X
n=0
f
(n)
n!
(x − a)
n
.
So f (x) equals its Taylor series! In other words, if you want to prove that
a function equals its Taylor series at some number x , try to show
that R
N
(x ) → 0 as N → ∞.
Let’s do exactly this for f(x) = e
x
with a = 0. By adapting some stuff we
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
looked at in Section 24.1.4 above, you should be able to see that
P
N
(x) =
N
X
n=0
x
n
n!
= 1 + x +
x
2
2!
+
x
3
3!
+ ··· +
x
N
N!
,
and that
R
N
(x) =
e
c
(N + 1)!
x
N+1
for some c between x and 0. Now, we need to find the limit of R
N
(x) as
N → ∞ and show that it’s zero:
lim
N→∞
R
N
(x) = lim
N→∞
e
c
x
N+1
(N + 1)!
.
In Section 24.3 below, I’ll prove that
lim
N→∞
x
N+1
(N + 1)!
= 0
532 • Taylor Polynomials, Taylor Series, and Power Series
for any x. We have to be a little careful about the e
c
factor, since c depends
on N. The question is, how big can e
c
be? Remember that c is between 0
and x. If x is negative, the biggest e
c
could be is if c = 0, which means that
e
c
≤ 1. If x is positive, the biggest e
c
could be is if c = x, which means
that e
c
≤ e
x
. In either case, since x is fixed (that is, treated as constant),
we can write that 0 ≤ e
c
≤ C, where C is another constant. This is true
no matter what N is, even though c is wobbling around all over the place as
N is changing. Anyway, hopefully you believe this, in which case you might
believe that
0 ≤ e
c
|x|
N+1
(N + 1)!
≤ C
|x|
N+1
(N + 1)!
.
Now the left-hand and right-hand sides go to 0 as N tends to ∞, so we can
apply the sandwich principle to see that the middle quantity does too. So,
we’ve proved that
lim
N→∞
R
N
(x) = 0
for any real x. This means that we have finally proved that
e
x
= 1 + x +
x
2
2!
+
x
3
3!
+
x
4
4!
+
x
5
5!
+ ···
for all real x.
Let’s try to see everything in one self-contained example by finding the
Maclaurin series of f(x) = cos(x) and showing that it converges to f(x) for
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)
(a, f(a))
a
all x. First, we need to differentiate f over and over again, then plug in 0
for each derivative and see what happens. Well, when you differentiate cos(x)
with respect to x, over and over again, you get −sin(x), then −cos(x), then
sin(x), then cos(x), then −sin(x), then −cos(x), and so on. Clearly this cycle
will keep on going. When you plug in x = 0, the sin(x) terms go away, and
the ±cos(x) terms become ±1. So the sequence of numbers f
(n)
(0) looks like
this:
1, 0, −1, 0, 1, 0, −1, 0, 1, 0, −1, 0, . . ..
If you plug these numbers into the Maclaurin series formula
f(0)+f
0
(0)x+
f
00
(0)
2!
x
2
+
f
(3)
(0)
3!
x
3
+
f
(4)
(0)
4!
x
4
+
f
(5)
(0)
5!
x
5
+
f
(6)
(0)
6!
x
6
+··· ,
all the odd-degree terms go away and you get
1 −
1
2!
x
2
+
1
4!
x
4
−
1
6!
x
6
+ ··· ,
which you can rewrite more compactly as
1 −
x
2
2!
+
x
4
4!
−
x
6
6!
+ ··· .
This is the Maclaurin series for cos(x), or if you prefer, the Taylor series for
cos(x) about x = 0. To get the corresponding Taylor polynomials, all you
have to do is chop off the series at the right place. For example,
P
4
(x) = 1 −
1
2!
x
2
+
1
4!
x
4
.
Section 24.2.3: Convergence of Taylor series • 533
By the way, the formula for P
5
(x) is the same as for P
4
(x), since there’s no
fifth-degree term in the above Maclaurin series. This is a good example of
why we need the word “order”: P
5
is of order 5, but degree only 4.
Now, all that’s left is to prove that cos(x) actually equals its Maclaurin
series for all real x:
cos(x) = 1 −
x
2
2!
+
x
4
4!
−
x
6
6!
+ ··· .
To do this, we need to show that
lim
N→∞
R
N
(x) = 0.
We know that
R
N
(x) =
f
(N+1)
(c)
(N + 1)!
x
N+1
,
where c is between 0 and x. Let’s take absolute values:
|R
N
(x)| =
|f
(N+1)
(c)|
(N + 1)!
|x|
N+1
.
All the derivatives of f are equal to either ±cos(x) or ±sin(x), so the quantity
|f
(N+1)
(c)| is either |cos(c)| or |sin(c)|. In either case, this quantity is less than
or equal to 1, so we have
0 ≤ |R
N
(x)| ≤
1
(N + 1)!
|x|
N+1
.
Once again, we’ll show in the next section that
lim
N→∞
x
N+1
(N + 1)!
= 0.
Now you can use the sandwich principle to show that
lim
N→∞
|R
N
(x)| = 0,
which means that also
lim
N→∞
R
N
(x) = 0.
We have proved that
cos(x) = 1 −
x
2
2!
+
x
4
4!
−
x
6
6!
+ ···
for all real x. Let’s celebrate by expressing the above series in sigma notation.
(What, isn’t that how you celebrate solving a tough problem?) Anyway, how
do you get only even powers of x? The answer is to use 2n instead of n (see
the end of Section 15.1 in Chapter 15 for a discussion of this sort of thing).
Since the factorial on the bottom matches the degree, we might guess that
the Maclaurin series can be written as
∞
X
n=0
x
2n
(2n)!
.
534 • Taylor Polynomials, Taylor Series, and Power Series
The problem is, this series doesn’t alternate. So insert a factor of (−1)
n
:
∞
X
n=0
(−1)
n
x
2n
(2n)!
.
If you expand this, you’ll find that it works. Here’s a summary of what we
found:
cos(x) =
∞
X
n=0
(−1)
n
x
2n
(2n)!
= 1 −
x
2
2!
+
x
4
4!
−
x
6
6!
+ ···
for all real x.
24.3 A Useful Limit
This section isn’t about power series at all—it just contains a proof of a limit
we needed twice in the previous section:
lim
N→∞
x
N+1
(N + 1)!
= 0
for any real number x. By letting n = N + 1 (think of it like a substitution
in an integral), this is exactly the same as showing that
lim
n→∞
x
n
n!
= 0
for any real number x. There are several ways to prove this last statement,
but here’s a sneaky way. Let me explain the logic I’m going to use, then
actually do it. I’m going to prove that the series
∞
X
n=0
x
n
n!
converges, regardless of what x is. (Yes, we “know” that it actually converges
to e
x
, but not until after we finish showing that the above limit is zero after
all!) Anyway, it doesn’t matter what the series converges to; simply knowing
that it converges is enough. Why? Because then the nth term x
n
/n! must
go to 0 as n goes to ∞, or else the nth term test would fail. That is, if the
terms didn’t go to 0 as n goes to ∞, then the series would diverge. So let’s
use the ratio test to show that the series converges for all x. Let’s fix x for
once and for all and, with a
n
= x
n
/n!, simply look at the limiting ratio:
L = lim
n→∞
a
n+1
a
n
= lim
n→∞
x
n+1
/(n + 1)!
x
n
/n!
= lim
n→∞
x
n+1
x
n
n!
(n + 1)!
.
Now we know that n!/(n + 1)! boils down to 1/(n + 1), so this last limit is
lim
n→∞
|x|
1
n + 1
= 0,
since |x| is fixed and 1/(n + 1) goes to 0. The limit L is 0, which is less than
1, so the series converges and we have, as a by-product, shown that the useful
limit is correctly stated above. By the way, the technique of fixing x and then
applying the ratio test to see whether the series converges for that particular
x will be used many times in Section 26.1.2 of Chapter 26.
C h a p t e r 25
How to Solve Estimation Problems
In the previous chapter, we showed how Taylor polynomials can be used to
estimate (or approximate, if you prefer) certain quantities. We also saw that
the remainder term could be used to get an idea of how good the approxima-
tion actually is. In this chapter, we’ll develop these techniques and look an
number of examples. So, here’s the plan for the chapter:
• a review of the most important facts about Taylor polynomials and
series;
• how to find Taylor polynomials and series;
• estimation problems; and
• a different method for analyzing the error.
25.1 Summary of Taylor Polynomials and Series
Here are the most important facts about Taylor polynomials and series, all of
which were developed in the previous chapter:
1. Of all the polynomials of degree N or less, the one which best approxi-
mates the smooth function f for x near a is called the Nth-order Taylor
polynomial about x = a, and 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
.
Using sigma notation, this can be written as
P
N
(x) =
N
X
n=0
f
(n)
(a)
n!
(x − a)
n
.
2. The polynomial P
N
has the same derivatives as f at x = a, up to and