<
Chapter
3
Series
This
chapter
intro
duces
the
T
a
ylo
r
p
olynomial,
which
is
a
useful
to
ol
for
app
ro
ximating
functions
that
cannot
b
e
evaluated
with
a
rithmetic.
Lik
e
with
the
derivative
and
integral
b
efo
re
it,
we
would
lik
e
to
send
the
erro
r
in
these
app
ro
ximations
to
0
.
This
requires
us
to
tak
e
a
new
kind
of
limit
called
a
series.
W
e
will
develop
the
to
ols
to
wo
rk
with
series,
with
the
ultimate
goal
of
defining
and
utilizing
T
aylo
r
series.
Contents
3.1
T
a
ylo
r
Polynomials
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
170
3.2
Sequences
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
188
3.3
Series
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
198
3.4
P
o
w
er
Series
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
218
3.5
T
a
ylo
r
Series
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
.
227
Section
3.1
T
a
ylo
r
P
olynomials
Goals:
1
App
ro
ximate
a
function
with
a
T
aylo
r
polynomial.
2
Compute
error
bounds
fo
r
a
T
aylo
r
polynomial.
When
learning
algebra
and
trigonometry
,
w
e
lea
rn
to
use
exact
values
like
√
7
instead
of
decimal
app
ro
ximations,
like
2
.
646
.
This
prevents
us
from
introducing
erro
rs
into
our
calculations.
Ho
w
ever,
there
a
re
also
advantages
to
appro
ximation.
Decimal
app
ro
ximations
give
us
a
much
b
etter
sense
of
the
size
of
a
number
than
ln
873
or
e
9
5
.
(Which
of
these
is
larger?)
Unfo
rtunately
arithmetic
do
es
not
give
us
metho
ds
fo
r
appro
ximating
many
quantities.
Ideally
,
we
w
ould
like
a
metho
d
of
appro
ximation
whose
accuracy
is
limited
only
by
ho
w
much
time
we
wish
to
sp
end
computing.
An
example
of
this
is
long
division.
We
can
compute
as
many
decimal
places
of
32
13
as
we
want,
getting
closer
and
closer
to
the
exact
value.
Of
course,
long
division
can
only
appro
ximate
fractions.
The
metho
d
we
will
develop
in
this
section
is
called
a
T
a
ylo
r
p
olynomial.
It
gives
us
a
wa
y
to
app
ro
ximate
otherwise
incomputable
functions.
The
beginning
point
is
the
tangent
line.
The
tangent
line
was
the
motivation
for
developing
the
derivative,
but
its
greatest
b
enefit
is
not
geometric.
The
tangent
line
app
ro
ximates
the
values
of
a
function
near
the
p
oint
of
tangency
.
While
the
function
may
b
e
difficult
to
evaluate,
the
equation
of
the
tangent
line
is
linea
r.
We
can
evaluate
it
by
hand.
Question
3.1.1
Ho
w
Can
We
Improve
on
a
Linea
rization?
F
o
rmula
The
linearization
or
tangent
line
to
a
function
f
(
x
)
at
a
has
the
equation.
L
(
x
)
=
f
(
a
)
+
f
′
(
a
)(
x
−
a
)
By
design
f
and
L
have
1
Equal
values
at
a
.
2
Equal
first
derivatives
at
a
.
This
means
that
fo
r
values
of
x
near
a
,
L
(
x
)
and
f
(
x
)
will
have
simila
r
values.
L
(
x
)
,
which
is
easy
to
compute,
can
b
e
used
as
an
app
ro
ximation
of
f
(
x
)
.
As
x
travels
aw
a
y
from
a
and
y
=
f
(
x
)
curves
a
w
a
y
from
its
tangent
line,
this
method
will
lose
accuracy
.
We
could
make
a
b
etter
app
ro
ximation,
if
w
e
could
match
second,
third,
fourth
derivatives
of
f
(
x
)
.
A
line
cannot
do
that,
but
a
p
olynomial
can.
170
Question
3.1.2
What
Is
a
T
a
ylo
r
Polynomial?
A
p
olynomial
that
mimics
the
first
n
derivatives
of
a
function
is
called
a
T
aylo
r
p
olynomial.
Here
is
the
formal
definition.
Definition
The
n
th
T
a
ylo
r
p
olynomial
of
f
(
x
)
at
x
=
a
is
a
degree
n
p
olynomial
that
shares
the
value
and
first
n
derivatives
of
f
at
x
=
a
.
Its
fo
rmula
is
T
n
(
x
)
=
n
X
k
=0
f
(
k
)
(
a
)
k
!
(
x
−
a
)
k
.
Rema
rks
The
variable
is
x
.
f
(
k
)
(
a
)
is
not
a
function
but
a
number.
f
(0)
is
the
zeroth
derivative,
meaning
f
(0)
(
a
)
=
f
(
a
)
.
0!
is
defined
to
b
e
1
.
Example
3.1.3
Computing
a
T
aylo
r
Polynomial
a
Find
the
degree
3
T
aylo
r
p
olynomial
of
y
=
√
x
at
x
=
4
.
b
Use
it
to
estimate
√
5
.
171
Example
3.1.3
Computing
a
T
aylo
r
Polynomial
Solution
a
W
e
will
apply
the
equation
of
the
T
a
ylo
r
p
olynomial
where
a
=
4
and
n
=
3
.
Examining
the
fo
rmula
shows
we
need
to
know
the
value
of
first
three
derivatives
of
f
(
x
)
at
a
=
4
.
f
(
x
)
=
x
1
/
2
f
(4)
=
2
f
′
(
x
)
=
1
2
x
−
1
/
2
f
′
(4)
=
1
4
f
′′
(
x
)
=
−
1
4
x
−
3
/
2
f
′′
(4)
=
−
1
32
f
′′′
(
x
)
=
3
8
x
−
5
/
2
f
′′′
(4)
=
3
256
W
e
can
plug
these
into
the
summation
fo
rmula:
T
3
(
x
)
=
3
X
k
=0
f
(
k
)
(4)
k
!
(
x
−
4)
k
=
f
(4)
0!
(1)
+
f
′
(4)
1!
(
x
−
4)
+
f
′′
(4)
2!
(
x
−
4)
2
+
f
′′′
(4)
3!
(
x
−
4)
3
=
2
+
1
4
1
(
x
−
4)
−
1
32
2
(
x
−
4)
2
+
3
256
6
(
x
−
4)
3
=
2
+
1
4
(
x
−
4)
−
1
64
(
x
−
4)
2
+
1
512
(
x
−
4)
3
b
T
o
appro
ximate
√
5
,
notice
√
5
=
f
(5)
and
f
(5)
≈
T
3
(5)
.
T
3
(5)
=
2
+
1
4
(5
−
4)
−
1
64
(5
−
4)
2
+
1
512
(5
−
4)
3
=
2
+
1
4
(1)
−
1
64
(1)
+
1
512
(1)
=
1024
512
+
128
512
−
8
512
+
1
512
=
1145
512
172
Example
3.1.4
W
riting
a
Sum
in
Σ
Notation
As
our
T
aylo
r
polynomials
get
longer,
we
w
ould
like
to
condense
them
into
P
notation.
P
a
rt
of
the
challenge
is
cho
osing
an
exp
ression
that
will
p
roduce
all
the
terms
of
our
sum.
Write
each
of
the
follo
wing
sums
in
Σ
notation.
a
4
+
7
+
10
+
13
+
16
+
19
+
22
b
2
+
6
+
18
+
54
+
162
+
486
c
−
3
+
4
−
5
+
6
−
7
+
8
−
9
+
10
d
1
4
+
√
2
9
+
√
3
16
+
2
25
+
√
5
36
Solution
a
The
terms
increase
by
3
each
time.
Rep
eated
addition
is
multiplication,
in
this
case
3
k
plus
some
sta
rting
value.
Starting
with
index
k
=
0
is
convenient,
be
cause
3(0)
=
0
at
the
sta
rting
value.
4
+
7
+
10
+
13
+
16
+
19
+
22
=
6
X
k
=0
4
+
3
k
b
The
terms
a
re
multiplied
by
3
each
time.
Rep
eated
multiplication
is
exp
onentiation,
in
this
case
3
k
times
some
sta
rting
value.
Sta
rting
with
index
k
=
0
is
convenient,
b
ecause
3
0
=
1
at
the
sta
rting
value.
2
+
6
+
18
+
54
+
162
+
486
=
5
X
k
=0
(2)(3
k
)
c
The
absolute
values
of
this
sum
could
just
b
e
the
values
of
the
index
variable.
T
o
create
an
alternating
+
and
−
pattern,
we
can
multiply
by
(
−
1)
k
.
−
3
+
4
−
5
+
6
−
7
+
8
−
9
+
10
=
10
X
k
=3
(
−
1)
k
k
d
In
a
fraction,
w
e
can
mo
del
the
numerato
r
and
denominator
sepa
rately
.
1
4
+
√
2
9
+
√
3
16
+
2
25
+
√
5
36
=
5
X
k
=1
√
k
(
k
+
1)
2
173
Example
3.1.5
A
T
aylo
r
Polynomial
in
P
Notation
W
rite
the
10
th
degree
T
aylo
r
Polynomial
for
f
(
x
)
=
1
x
2
centered
at
x
=
3
.
Solution
Computing
10
derivatives
seems
excessive,
so
we
will
compute
4
and
try
to
find
a
pattern.
We’ll
write
f
(
x
)
=
x
−
2
and
apply
the
pow
er
rule.
f
(
x
)
=
x
−
2
f
′
(
x
)
=
−
2
x
−
3
f
′′
(
x
)
=
6
x
−
4
f
′′′
(
x
)
=
−
24
x
−
5
f
(4)
(
x
)
=
120
x
−
6
W
e
observe
The
sign
of
these
derivatives
is
alternating,
which
we
can
model
with
a
(
−
1)
k
.
The
co
efficients
look
lik
e
a
factorial
pattern,
but
offset.
F
o
r
example
when
k
=
2
we
obtain
3!
.
W
e
mo
del
this
with
(
k
+
1)!
.
The
exp
onent
of
x
decreases
by
the
same
amount
each
step.
W
e
mo
del
it
with
−
2
−
k
.
This
suggests
a
general
formula
for
the
k
th
derivative.
f
(
k
)
(
x
)
=
(
−
1)
k
(
k
+
1)!
x
−
2
−
k
W
e
plug
x
=
3
into
f
(
k
)
(
x
)
and
assemble
the
T
aylo
r
Polynomial:
T
10
(
x
)
=
10
X
k
=0
(
−
1)
k
(
k
+
1)!3
−
2
−
k
k
!
(
x
−
3)
k
Question
3.1.6
Ho
w
Accurate
Is
the
T
aylo
r
Polynomial?
An
appro
ximation
is
much
more
useful,
if
w
e
can
put
a
b
ound
on
its
erro
r.
W
e
will
present
an
error
b
ound
theorem
called
“T
aylo
r’s
Inequalit
y
.”
T
aylo
r
p
olynomials
a
re
effective
app
ro
ximations
because
they
try
to
match
the
values
and
rates
of
change
of
the
original
function.
In
o
rder
to
mak
e
a
ca
reful
a
rgument,
we
begin
with
the
basic
p
rincipal
that
w
e
can
compa
re
functions
using
the
values
of
their
derivatives.
174
Theo
rem
Let
f
and
g
be
differentiable
functions.
Consider
an
interval
[
a,
b
]
,
and
suppose
f
(
a
)
=
g
(
a
)
.
1
If
f
′
(
x
)
=
g
′
(
x
)
on
[
a,
b
]
,
then
f
(
x
)
=
g
(
x
)
on
[
a,
b
]
2
If
f
′
(
x
)
<
g
′
(
x
)
on
[
a,
b
]
,
then
f
(
x
)
<
g
(
x
)
on
(
a,
b
]
Reasoning
Intuitive
If
tw
o
functions
start
at
the
same
value
at
a
,
then
the
one
that
grows
faster
will
have
a
higher
value
at
b
.
F
o
rmal
The
F
undamental
Theo
rem
of
Calculus
sa
ys
f
(
x
)
−
f
(
a
)
=
Z
x
a
f
′
(
t
)
dt
g
(
x
)
−
g
(
a
)
=
Z
x
a
g
′
(
t
)
dt.
La
rger
functions
have
la
rger
integrals.
Figure:
Two
functions
with
a
common
value
at
a
:
f
(
x
)
with
a
smaller
derivative
and
g
(
x
)
with
a
la
rger
derivative.
Notation
Given
a
function
f
(
x
)
and
its
n
th
T
aylo
r
p
olynomial
T
n
(
x
)
centered
at
a
,
the
remainder
at
x
is
R
n
(
x
)
=
f
(
x
)
−
T
n
(
x
)
175
Question
3.1.6
Ho
w
Accurate
Is
the
T
aylo
r
Polynomial?
If
we
are
using
T
n
(
x
)
to
appro
ximate
f
(
x
)
,
R
n
(
x
)
=
−
erro
r
of
T
n
(
x
)
.
W
e
should
b
e
very
interested
in
kno
wing
the
value
of
R
n
(
x
)
.
We
will
use
our
derivative
comparison
theo
rem
to
make
tw
o
arguments
1
If
f
(
n
+1)
(
x
)
is
a
constant
M
,
then
w
e
can
compute
R
n
(
x
)
exactly
.
2
If
|
f
(
n
+1)
(
x
)
|
≤
M
then
the
error
in
1
is
the
wo
rst-case
scenario.
Theo
rem
If
f
(
n
+1)
(
x
)
is
a
constant
M
on
[
a,
b
]
,
then
f
(
x
)
=
T
n
+1
(
x
)
=
T
n
(
x
)
+
M
(
n
+
1)!
(
x
−
a
)
n
+1
.
Beginning
with
our
assumption
ab
out
the
(
n
+
1)
th
derivatives
and
the
equality
of
the
n
th
derivatives
at
a
,
we
can
use
our
derivative
comparison
theo
rem
to
equate
the
n
th
derivatives
on
[
a.b
]
.
W
e
can
use
that
equality
to
equate
the
(
n
−
1)
th
derivatives
on
[
a,
b
]
.
W
e
continue
this
reasoning
until
we
conclude
that
the
functions
a
re
equal.
d
dx
n
+1
f
(
x
)
=
d
dx
n
+1
T
n
+1
(
x
)
=
M
on
[
a,
b
]
d
dx
n
f
(
a
)
=
d
dx
n
T
n
+1
(
a
)
d
dx
n
f
(
x
)
=
d
dx
n
T
n
+1
(
x
)
on
[
a,
b
]
d
dx
n
−
1
f
(
a
)
=
d
dx
n
−
1
T
n
+1
(
a
)
d
dx
n
−
1
f
(
x
)
=
d
dx
n
−
1
T
n
+1
(
x
)
on
[
a,
b
]
d
dx
f
(
a
)
=
d
dx
T
n
+1
(
a
)
d
dx
f
(
x
)
=
d
dx
T
n
+1
(
x
)
on
[
a,
b
]
f
(
a
)
=
T
n
+1
(
a
)
f
(
x
)
=
T
n
+1
(
x
)
on
[
a,
b
]
Because
derivatives
and
values
of
a
T
aylo
r
p
olynomial
match
the
function
Rema
rk
This
theorem
tells
us
that
when
f
(
n
+1)
(
x
)
is
a
constant
M
,
R
n
(
x
)
=
f
(
x
)
−
T
n
(
x
)
=
M
(
n
+1)!
(
x
−
a
)
n
+1
But
what
if
f
(
n
+1)
(
x
)
is
not
a
constant?
In
this
case
w
e
will
settle
fo
r
a
b
ound
on
f
(
n
+1)
(
x
)
.
176
Theo
rem
[T
aylo
r’s
Inequality]
If
f
(
n
+1)
(
t
)
≤
M
for
all
x
b
etw
een
a
and
b
,
then
for
all
x
b
etw
een
a
and
b
,
|
R
n
(
x
)
|
≤
M
(
n
+
1)!
(
x
−
a
)
n
+1
T
o
prove
T
aylo
r’s
Inequality
,
we
compare
the
derivatives
of
f
(
x
)
with
the
wo
rst-case
scenario
w
(
x
)
=
T
n
(
x
)
+
M
(
n
+1)!
(
x
−
a
)
n
+1
.
The
derivatives
w
(
k
)
(
a
)
are
the
same
as
T
(
k
)
n
(
a
)
and
f
(
k
)
(
a
)
for
0
≤
k
≤
n
,
and
d
dx
n
+1
w
(
x
)
=
M
.
d
dx
n
+1
f
(
x
)
≤
d
dx
n
+1
w
(
a
)
=
M
on
[
a,
b
]
d
dx
n
f
(
a
)
=
d
dx
n
w
(
a
)
d
dx
n
f
(
x
)
≤
d
dx
n
w
(
x
)
on
[
a,
b
]
d
dx
n
−
1
f
(
a
)
=
d
dx
n
−
1
w
(
a
)
d
dx
n
−
1
f
(
x
)
≤
d
dx
n
−
1
w
(
x
)
on
[
a,
b
]
d
dx
f
(
a
)
=
d
dx
w
(
a
)
d
dx
f
(
x
)
≤
d
dx
w
(
x
)
on
[
a,
b
]
f
(
a
)
=
w
(
a
)
f
(
x
)
≤
w
(
x
)
on
[
a,
b
]
Because
M
is
a
bound
on
d
dx
n
+1
f
(
x
)
Because
derivatives
and
values
of
a
T
aylo
r
p
olynomial
match
the
function
T
o
finish
the
a
rgument
we
need
to
1
Pro
duce
a
low
er
b
ound
for
f
using
w
(
x
)
=
T
n
(
x
)
−
M
(
n
+1)!
(
x
−
a
)
n
+1
.
2
Solve
the
inequality
b
ounds
for
R
n
(
x
)
.
T
n
(
x
)
−
M
(
n
+
1)!
(
x
−
a
)
n
+1
≤
f
(
x
)
≤
T
n
(
x
)
+
M
(
n
+
1)!
(
x
−
a
)
n
+1
−
M
(
n
+
1)!
(
x
−
a
)
n
+1
≤
R
n
(
x
)
≤
M
(
n
+
1)!
(
x
−
a
)
n
+1
3
Rep
eat
for
intervals
of
the
form
[
b,
a
]
.
These
w
o
rk
the
same
wa
y
with
a
sign
reversed.
177
Example
3.1.7
A
T
aylo
r
Appro
ximation
Erro
r
Bound
Let
f
(
x
)
=
sin
x
.
a
Give
a
general
fo
rm
for
the
n
th
T
a
ylo
r
p
olynomial
for
f
at
x
=
0
.
b
Find
a
b
ound
on
f
(
n
)
(
x
)
for
each
n
.
c
What
happ
ens
to
the
error
b
ound
as
x
increases
but
n
sta
ys
the
same?
d
What
happ
ens
to
the
error
b
ound
as
n
increases
but
x
sta
ys
the
same?
e
What
do
es
this
tell
us
ab
out
the
relationship
b
etw
een
the
T
n
(
x
)
appro
ximations
and
f
(
x
)
?
Solution
a
F
o
r
the
T
aylo
r
p
olynomial
formula,
we
need
to
compute
the
derivatives
of
f
(
x
)
.
f
(
x
)
=
sin
x
f
(0)
=
0
f
′
(
x
)
=
cos
x
f
′
(0)
=
1
f
′′
(
x
)
=
−
sin
x
f
′′
(0)
=
0
f
′′′
(
x
)
=
−
cos
x
f
′′′
(0)
=
−
1
f
(4)
(
x
)
=
sin
x
f
(4)
(0)
=
0
f
(5)
(
x
)
=
cos
x
f
(5)
(0)
=
1
.
.
.
.
.
.
In
order
to
write
a
general
T
aylo
r
p
olynomial,
we
would
need
a
general
expression
for
f
(
k
)
(0)
.
The
pattern
is
obvious,
but
trying
to
exp
ress
it
as
a
formula
is
much
more
difficult.
The
solution
is
a
trick
wo
rth
rememb
ering:
Since
the
even
derivatives
a
re
zero,
those
terms
do
not
app
ear
in
our
T
aylo
r
polynomials.
Since
w
e
want
to
only
have
o
dd
terms
in
our
summation,
we
can
let
our
index
variable
b
e
k
,
but
our
exp
onents
in
each
term
b
e
2
k
+
1
.
Thus
as
k
go
es
from
0
to
n
,
the
summation
will
include
only
the
o
dd
terms
x
1
through
x
2
n
+1
.
We
can
produce
the
following
chart
to
w
o
rk
out
our
coefficients:
k
f
(2
k
+1)
(0)
0
1
1
−
1
2
1
3
−
1
.
.
.
.
.
.
178
This
is
an
easier
pattern
to
express:
f
(2
k
+1)
(0)
=
(
−
1)
k
No
w
we
are
ready
to
write
a
fo
rmula.
Since
w
e
intend
to
sum
from
k
=
0
to
k
=
n
,
w
e
are
actually
producing
the
(2
n
+
1)
th
T
aylo
r
p
olynomial.
T
2
n
+1
(
x
)
=
n
X
k
=0
f
(2
k
+1)
(0)
(2
k
+
1)!
x
2
k
+1
=
n
X
k
=0
(
−
1)
k
(2
k
+
1)!
x
2
k
+1
These
a
re
the
o
dd
degree
T
aylo
r
p
olynomials,
but
what
ab
out
the
even
numb
ered
ones?
Since
T
2
n
(
x
)
is
just
T
2
n
−
1
(
x
)
plus
the
2
n
th
term,
and
the
2
n
th
term
is
zero,
w
e
can
write
T
2
n
(
x
)
=
n
−
1
X
k
=0
(
−
1)
k
(2
k
+
1)!
x
2
k
+1
b
Given
the
chart
ab
ove,
we
can
see
that
the
derivatives
are
sines
and
cosines.
These
a
re
b
ounded
ab
ove
by
1
and
below
by
−
1
.
Since
T
aylo
r’s
inequality
requires
a
b
ound
of
the
form
|
f
(
n
+1)
(
x
)
|
≤
M
,
w
e
write
|
f
(
n
+1)
(
x
)
|
≤
1
And
luckily
,
thus
w
o
rks
for
all
x
and
all
n
.
c
T
a
ylo
r’s
Inequality
says
that
|
R
n
(
x
)
|
≤
1
(
n
+1)!
x
n
+1
.
As
x
go
es
to
∞
,
this
b
ound
go
es
to
∞
as
w
ell.
This
mak
es
sense,
since
T
n
(
x
)
is
polynomial,
while
the
function
it
is
app
ro
ximating
stays
b
et
w
een
−
1
and
1
.
d
When
n
increases
x
n
+1
increases
b
y
a
factor
of
x
.
On
the
other
hand,
(
n
+
1)!
increases
by
a
facto
r
of
n
+
2
.
As
n
increases
without
b
ound,
(
n
+
1)!
gro
ws
faster
than
x
n
+1
and
their
ratio
app
roaches
0
.
e
Any
T
n
(
x
)
will
eventually
become
inaccurate
outside
a
certain
distance
from
0
.
On
the
other
hand,
if
we
want
to
appro
ximate
sin(
x
)
for
a
pa
rticula
r
x
,
we
can
make
T
n
(
x
)
have
as
small
an
erro
r
as
we
want
by
cho
osing
sufficiently
large
n
.
179
Example
3.1.7
A
T
aylo
r
Appro
ximation
Erro
r
Bound
Click to Load Applet
Figure:
f
(
x
)
=
sin
x
appro
ximated
by
its
T
a
ylo
r
p
olynomials,
T
n
(
x
)
Main
Ideas
In
order
to
understand
ho
w
the
error
changes
as
n
increases,
we
need
to
have
an
expression
for
f
(
n
)
(
x
)
.
W
e
can
cho
ose
M
to
b
e
the
largest
value
of
|
f
(
n
+1)
|
on
the
interval
[
a,
x
]
.
This
ma
y
not
be
the
value
of
|
f
(
n
+1)
(
a
)
|
.
In
general,
T
aylo
r
p
olynomials
will
b
ecome
less
accurate
the
fa
rther
you
get
from
a
.
W
e
can
often
mitigate
this
inaccuracy
by
cho
osing
larger
values
of
n
.
The
(
n
+
1)!
in
T
a
ylo
r’s
Inequalit
y
might
suggest
that
as
n
increases,
the
erro
r
in
the
n
th
T
aylo
r
p
olynomial
must
shrink
to
w
ar
d
0
.
Ho
wever,
this
is
not
the
case.
Some
functions
are
not
well
estimated
b
y
their
T
aylo
r
p
olynomial.
180
Example
f
(
x
)
=
(
0
if
x
≤
0
e
−
1
x
if
x
>
0
f
(
k
)
(0)
=
0
for
all
k
.
So
the
T
aylo
r
p
olynomial
at
x
=
0
is
T
n
(
x
)
=
n
X
k
=0
0
x
k
.
No
matter
how
large
n
gets,
T
n
(
x
)
will
not
get
any
closer
to
f
(
x
)
for
any
x
>
0
.
Ho
w
can
this
happ
en,
given
T
aylo
r’s
Inequalit
y?
The
derivatives
of
f
get
bigger
and
bigger.
M
gro
ws
so
fast
that
the
error
R
n
(
x
)
gets
no
smaller
even
with
an
(
n
+
1)!
in
the
denominato
r
of
T
aylo
r’s
Inequalit
y
.
Click to Load Applet
Figure:
A
function
whose
derivative
b
ounds
grow
facto
rially
Despite
examples
lik
e
this,
it
turns
out
that
T
aylo
r
p
olynomials
often
do
a
go
o
d
job
of
app
ro
ximating
functions.
F
o
r
numerical
computations,
an
appro
ximation
is
go
o
d
enough.
F
o
r
more
theo
retical
situ-
ations,
w
e
would
lik
e
to
let
n
go
to
∞
so
that
the
erro
r
go
es
to
0
and
we
can
use
the
p
olynomial
as
an
exact
replacement
of
the
function.
Unfo
rtunately
,
with
infinitely
many
terms,
w
e
no
longer
have
a
p
olynomial
at
all.
Instead
we
have
an
object
that
w
e
will
call
a
T
aylo
r
series.
We
will
develop
the
tools
to
define
and
w
o
rk
with
T
aylo
r
series
over
the
course
of
this
chapter.
181
Section
3.1
Exercises
Summa
ry
Questions
Q1
Why
do
we
use
T
aylo
r
p
olynomials?
Q2
Why
is
there
a
denominator
of
k
!
in
the
formula
for
a
T
a
ylo
r
p
olynomial?
Q3
Explain
why
we’d
alwa
ys
rather
center
a
T
aylo
r
p
olynomial
for
y
=
ln
x
at
x
=
1
.
Q4
What
properties
make
a
T
aylo
r
p
olynomial
T
n
(
x
)
a
b
etter
app
ro
ximation
of
f
(
x
)
?
3.1.1
Q5
Supp
ose
we
use
the
linearization
of
f
(
x
)
=
3
√
x
at
x
=
8
to
appro
ximate
3
√
6
.
a
What
is
the
relationship
b
etw
een
f
(
x
)
=
3
√
x
and
3
√
6
?
b
Supp
ose
L
(
x
)
is
the
the
linearization
of
f
(
x
)
at
x
=
8
.
W
ould
you
exp
ect
L
(6)
to
overesti-
mate
or
underestimate
3
√
6
?
Explain
in
a
sentence
or
tw
o.
Q6
Supp
ose
you
were
lo
cked
in
a
ro
om
with
only
a
p
encil
and
paper
and
asked
to
compute
the
first
ten
decimal
places
of
the
following
numb
ers:
4
17
√
7
e
Which
could
you
compute?
F
o
r
the
ones
y
ou
can
compute,
how
would
you
do
it?
182
3.1.2
Q7
Is
a
tangent
line
a
T
aylo
r
p
olynomial?
Q8
Supp
ose
T
4
(
x
)
is
the
T
aylo
r
p
olynomial
fo
r
f
(
x
)
centered
at
x
=
10
.
List
what
information
T
4
(
x
)
and
f
(
x
)
have
in
common,
being
as
sp
ecific
as
p
ossible.
Q9
If
f
(
x
)
is
a
decreasing
function,
what
can
you
sa
y
about
the
co
efficients
of
any
T
aylo
r
p
olynomial
of
f
(
x
)
?
Q10
Supp
ose
f
(
x
)
has
a
T
aylo
r
p
olynomial
T
4
(
x
)
=
5
+
3(
x
−
2)
−
1
6
(
x
−
2)
2
+
2(
x
−
2)
4
a
What
is
f
(2)
?
b
Is
f
increasing
or
decreasing
at
x
=
2
?
c
Is
f
concave
up
o
r
concave
down
at
x
=
2
?
3.1.3
Q11
Let
f
(
x
)
=
e
x
.
a
Find
the
degree
8
T
aylo
r
p
olynomial
of
y
=
f
(
x
)
centered
at
x
=
0
.
b
Ho
w
could
you
use
this
to
estimate
the
value
of
e
?
c
Can
you
use
sigma
notation
to
write
a
general
form
fo
r
the
degree
n
T
a
ylo
r
p
olynomial
of
y
=
e
x
?
Q12
Let
f
(
x
)
=
ln
x
a
W
rite
the
5
th
T
a
ylo
r
p
olynomial
of
f
(
x
)
at
x
=
1
.
b
Use
your
p
olynomial
to
app
roximate
ln
2
.
183
Section
3.1
Exercises
Q13
W
rite
the
10
th
T
a
ylo
r
p
olynomial
for
f
(
x
)
=
cos
x
centered
at
x
=
π
.
Q14
W
rite
the
4
th
T
a
ylo
r
p
olynomial
for
f
(
x
)
=
1
x
2
centered
at
x
=
5
.
3.1.4
Q15
W
rite
each
of
the
following
sums
in
Σ
notation.
a
15
−
45
+
105
−
315
+
945
b
24
+
19
+
14
+
9
+
4
−
1
−
6
c
1
8
+
1
18
+
1
50
+
1
72
+
1
98
Q16
W
rite
each
of
the
following
sums
in
Σ
notation.
a
11
−
13
+
15
−
17
+
19
−
21
+
23
b
384
+
192
+
96
+
48
+
24
+
12
+
6
c
2
10
+
3
100
+
4
1000
+
5
10000
3.1.5
Q17
W
rite
an
expression
in
Σ
notation
for
the
53
rd
T
aylo
r
p
olynomial
of
f
(
x
)
=
ln
x
centered
at
x
=
1
Q18
W
rite
an
expression
in
Σ
notation
for
the
15
th
T
a
ylo
r
p
olynomial
of
f
(
x
)
=
e
x
centered
at
x
=
0
Q19
W
rite
an
expression
in
Σ
notation
for
the
100
th
T
aylo
r
p
olynomial
of
f
(
x
)
=
cos
x
centered
at
x
=
0
Q20
W
rite
an
expression
in
Σ
notation
fo
r
the
71
st
T
aylo
r
polynomial
of
f
(
x
)
=
1
x
2
centered
at
x
=
10
184
3.1.6
Q21
Why
don’t
we
have
any
theorems
for
a
lo
w
er
b
ound
for
erro
r?
Give
your
answer
in
a
few
sentences.
Q22
Supp
ose
y
ou
a
re
using
T
aylo
r
p
olynomials
of
f
(
x
)
centered
at
x
=
0
to
appro
ximate
f
(
−
3)
.
Ho
w
ever,
fo
r
each
k
,
the
best
b
ound
you
can
put
on
f
(
k
)
(
x
)
on
[
−
3
,
0]
is
k
!
4
k
.
Will
you
b
e
able
to
guarantee
a
goo
d
app
ro
ximation
of
f
(
−
3)
this
wa
y?
Explain.
Q23
Supp
ose
the
fourth
derivative
of
f
(
x
)
is
f
(4)
(
x
)
=
e
x
3
.
Suppose
w
e
have
written
T
4
(
x
)
,
the
degree
4
T
aylo
r
p
olynomial
of
f
(
x
)
centered
at
x
=
1
.
What
can
you
sa
y
ab
out
the
difference
b
et
w
een
T
4
(5)
and
f
(5)
?
Be
sp
ecific
and
justify
your
answer
with
a
computation.
Y
ou
do
not
need
to
simplify
any
arithmetic
in
your
calculations.
Q24
Sk
etch
a
graph
of
y
=
e
x
and
several
tangent
lines.
On
which
part
of
the
graph
do
the
tangent
lines
app
ea
r
to
app
ro
ximate
the
function
b
etter?
Do
es
T
a
ylo
r’s
Inequalit
y
confirm
this
observa-
tion?
Explain.
3.1.7
Q25
Here
is
the
degree
3
T
aylo
r
p
olynomial
of
f
(
x
)
=
√
x
centered
at
x
=
4
:
T
3
(
x
)
=
2
+
1
4
(
x
−
4)
−
1
64
(
x
−
4)
2
+
1
512
(
x
−
4)
3
a
Which
derivative
will
let
you
b
ound
the
erro
r
of
this
app
ro
ximation?
b
Can
you
put
a
b
ound
on
this
derivative
that
holds
fo
r
all
x
?
c
Can
you
put
a
b
ound
on
this
derivative
that
holds
fo
r
x
in
the
interval
[4
,
5]
?
d
What
error
b
ound
does
this
suggest
for
using
T
3
(5)
to
appro
ximate
√
5
?
Q26
Let
f
(
x
)
=
3
√
x
.
a
W
rite
the
degree
2
T
aylo
r
p
olynomial
of
f
centered
at
x
=
8
.
b
If
you
wanted
to
use
the
T
a
ylo
r
p
olynomial
to
app
ro
ximate
3
√
10
,
how
would
y
ou
do
that?
185
Section
3.1
Exercises
c
What
b
ound
could
y
ou
place
on
the
erro
r
in
the
app
ro
ximation
in
b
?
Q27
Let
f
(
x
)
=
e
x
.
a
W
rite
the
degree
5
T
aylo
r
p
olynomial
of
f
centered
at
0
.
b
Ho
w
could
we
use
this
p
olynomial
to
appro
ximate
1
√
e
?
c
Pro
duce
an
error
b
ound
for
your
appro
ximation
in
b
.
Q28
Let
f
(
x
)
=
xe
x
.
a
Compute
the
T
aylo
r
p
olynomial
T
3
(
x
)
for
f
(
x
)
centered
at
x
=
0
.
b
Compute
the
theoretical
error
b
ound
for
T
3
(2)
.
c
Explain
the
difficulties
that
would
arise
from
this
error
b
ound,
if
your
goal
is
to
appro
ximate
f
(2)
by
hand.
Can
you
resolve
them?
Q29
Let
f
(
x
)
=
cos
3
x
a
W
rite
the
degree
4
T
aylo
r
p
olynomial
of
f
centered
at
x
=
0
.
b
Ho
w
would
you
use
that
T
aylo
r
p
olynomial
to
app
ro
ximate
the
value
of
cos
3
π
4
?
c
What
b
ound
can
y
ou
place
on
the
erro
r
of
such
an
app
ro
ximation?
Q30
Consider
the
graph
of
y
=
f
(
x
)
b
elow.
186
a
Supp
ose
y
ou
w
anted
to
produce
the
second
degree
T
a
ylo
r
polynomial
of
f
centered
at
a
=
−
1
.
Indicate
whether
the
constant
term
and
each
coefficient
w
ould
b
e
p
ositive
or
negative.
Provide
evidence
for
your
answer.
b
W
ould
T
2
(4)
underestimate
or
overestimate
f
(4)
?
Explain.
Synthesis
and
Extension
Q31
Let
f
(
x
)
=
x
3
−
3
x
+
5
.
a
W
rite
an
expression
for
T
3
(
x
)
,
the
T
aylo
r
p
olynomial
centered
at
x
=
2
.
b
What
can
you
say
ab
out
th
erro
r
R
3
(
x
)
for
any
x
?
c
What
relationship
do
es
this
suggest
b
etw
een
f
(
x
)
and
T
3
(
x
)
?
d
Can
you
verify
this
relationship
algebraically?
e
Conjecture
a
general
relationship
betw
een
p
olynomial
functions
and
certain
T
aylo
r
p
olyno-
mials.
Can
you
use
T
a
ylo
r’s
inequality
to
justify
y
our
conjecture?
187
Section
3.2
Sequences
Goals:
1
Use
notation
to
describe
the
terms
of
an
infinite
sequence.
2
Calculate
the
limit
of
an
infinite
sequence.
Sequences
are
the
first
step
in
our
development
of
T
a
ylo
r
series.
While
they
app
ear
to
have
little
in
common
with
p
olynomials
of
infinite
degree,
they
a
re
the
scaffolding
on
which
such
objects
a
re
built.
Question
3.2.1
What
Is
a
Sequence?
A
sequence
is
an
o
rdered
set
of
numb
ers.
If
this
set
is
infinite,
we
can
most
rigorously
define
it
by
giving
a
general
formula
fo
r
the
n
th
term
fo
r
some
index
variable
n
.
Here
a
re
three
different
notations
fo
r
the
same
sequence.
1
2
,
2
3
,
3
4
,
4
5
.
.
.
n
n
+
1
∞
n
=1
a
n
=
n
n
+
1
Example
The
first
three
terms
of
n
2
2
n
∞
n
=0
a
re
0
2
2
0
=
0
1
2
2
1
=
1
2
2
2
2
2
=
1
Question
3.2.2
What
Is
the
Limit
of
a
Sequence?
Definition
If
we
can
make
the
elements
of
a
sequence
a
n
a
rbitra
rily
close
to
some
numb
er
L
by
considering
only
n
ab
ove
a
certain
number,
then
we
write
lim
n
→∞
a
n
=
L
and
we
say
the
sequence
converges
to
L
.
If
a
n
do
es
not
converge
to
any
such
L
then
w
e
say
it
diverges
.
188
Rema
rks
The
first
few
or
even
the
first
thousand
terms
of
a
sequence
have
no
b
ea
ring
on
the
limit.
W
e
only
care
that
w
e
can
eventually
get
close
to
L
.
“Arbitra
rily
close”
means
any
level
of
closeness
than
anyone
could
ask
for.
Eventually
the
sequence
must
b
e
within
1
100
of
L
,
and
1
1000
and
1
1000000
.
Click to Load Applet
Figure:
A
sequence
converging
to
L
=
3
Example
3.2.3
Computing
a
Limit
Calculate
lim
n
→∞
n
n
+
1
1
2
,
2
3
,
3
4
,
4
5
.
.
.
189
Example
3.2.3
Computing
a
Limit
Solution
W
riting
the
first
few
terms
suggests
that
this
sequence
approaches
1
.
T
o
see
that,
we
can
measure
the
distance
to
1
:
1
−
a
n
=
1
−
n
n
+
1
=
1
n
+
1
W
e
can
make
this
smaller
than
any
p
ositive
numb
er.
Fo
r
instance
to
make
a
n
within
1
1000
of
1
,
we
can
consider
only
n
>
1000
.
We
conclude
lim
n
→∞
n
n
+
1
=
1
Click to Load Applet
Figure:
The
sequence
n
n
+1
converges
to
L
=
1
.
Question
3.2.4
Ho
w
Are
Limits
of
Sequences
and
Functions
Related?
The
definition
of
lim
n
→∞
a
n
should
lo
ok
familia
r.
The
definition
of
the
limit
of
a
function
is
similar.
In
fact,
the
limit
of
a
f
(
x
)
as
x
→
∞
has
a
nearly
identical
construction,
except
that
n
must
b
e
an
integer,
while
x
can
b
e
any
real
numb
er.
The
following
theorem
lets
us
use
that
connection
to
evaluate
limits.
190
Theo
rem
Supp
ose
fo
r
a
sequence
a
n
,
there
is
a
function
f
(
x
)
such
that
f
(
n
)
=
a
n
fo
r
all
n
(or
at
least
all
n
sufficiently
large).
If
lim
x
→∞
f
(
x
)
=
L
w
e
can
conclude
that
lim
n
→∞
a
n
=
L.
Example
3.2.5
Sequence
Limits
Using
F
unctions
Find
limits
of
the
following
sequences:
a
lim
n
→∞
2
n
n
+
3
b
lim
n
→∞
1
n
3
c
lim
n
→∞
e
−
n
d
lim
n
→∞
n
2
e
n
e
lim
n
→∞
(
−
1)
n
Solution
W
e
will
use
x
to
denote
a
real
numb
er
variable
and
n
to
denote
natural
numb
ers.
a
lim
x
→∞
2
x
x
+
3
=
2
,
so
lim
n
→∞
2
n
n
+
3
=
2
.
b
lim
x
→∞
1
x
3
=
0
,
so
lim
n
→∞
1
n
3
=
0
.
c
lim
x
→∞
e
−
x
=
0
so
lim
n
→∞
e
−
n
=
0
.
191
Example
3.2.5
Sequence
Limits
Using
F
unctions
d
lim
x
→∞
x
2
e
x
can
b
e
evaluated
with
L’hˆ
opital’s
rule.
lim
x
→∞
x
2
e
x
=
lim
x
→∞
2
x
e
x
∞
∞
fo
rm,
L’hˆ
opital’s
again
=
lim
x
→∞
2
e
x
=
0
so
lim
n
→∞
n
2
e
n
=
0
e
f
(
x
)
=
(
−
1)
x
is
not
well
defined
fo
r
real
numbers
so
w
e
can’t
use
its
limit.
Instead
examine
the
sequence
directly
.
The
sequence
has
the
form
−
1
,
1
,
−
1
,
1
,
−
1
,
1
,
−
1
,
1
,
.
.
.
This
do
es
not
approach
arbitra
rily
close
to
any
numb
er.
No
matter
how
many
ea
rly
terms
w
e
disrega
rd,
there
will
alwa
ys
b
e
terms
remaining
that
a
re
not
close
to
1
,
o
r
not
close
to
−
1
or
not
close
to
any
other
numb
er.
Thus
a
n
=
(
−
1)
n
diverges.
The
following
limit
laws
for
sequences
should
lo
ok
familiar.
They
mirror
the
la
ws
for
limits
of
functions.
Theo
rem
[Limit
Laws]
If
lim
n
→∞
a
n
=
K
and
lim
n
→∞
b
n
=
L
then
the
follo
wing
sequences
converge
with
the
following
limits:
lim
n
→∞
(
a
n
+
b
n
)
=
K
+
L
lim
n
→∞
(
a
n
−
b
n
)
=
K
−
L
lim
n
→∞
(
a
n
b
n
)
=
K
L
If
L
=
0
,
then
lim
n
→∞
a
n
b
n
=
K
L
F
o
r
any
constant
c
,
lim
n
→∞
ca
n
=
cK
192
Synthesis
3.2.6
Indeterminate
Fo
rms
with
F
acto
rials
W
e
will
encounter
sequences
of
the
form
a
n
=
b
n
c
n
.
If
b
n
o
r
c
n
b
oth
go
to
0
o
r
±∞
,
then
any
attempt
to
use
lim
n
→∞
a
n
=
lim
x
→∞
f
(
x
)
w
ould
require
l’Hˆ
opital’s
rule.
Dominance
W
e
say
f
(
x
)
dominates
g
(
x
)
if
lim
x
→∞
f
(
x
)
g
(
x
)
=
±∞
.
We
write
f
(
x
)
>>
g
(
x
)
Even
if
y
ou
include
a
constant
multiple
o
r
add
multiple
functions
together,
the
dominant
function
will
outgro
w
any
combination
of
dominated
ones.
We
have
already
established
an
order
of
dominance
using
l’Hˆ
opital’s
rule:
exp
onential
(larger
base
>>
smaller
base)
>>
p
olynomial
(larger
degree
>>
smaller)
>>
ro
ot
(smaller
p
ow
er
>>
la
rger)
>>
loga
rithm
(smaller
base
>>
larger)
But
n
!
is
not
a
differentiable
function.
W
e
cannot
analyze
it
using
l’Hˆ
opital’s
rule.
Where
do
es
it
fit
in
the
domincance
pecking
o
rder?
Theo
rem
As
n
→
∞
,
n
!
will
eventually
dominate
any
exponential
function
(and
thus
any
p
olynomial,
ro
ot
or
loga
rithm).
W
e
will
not
provide
a
formal
p
roof,
but
here
is
a
useful
thought
exp
eriment.
Supp
ose
we
compa
re
n
!
to
63
n
.
At
first
63
n
gro
ws
faster,
multiplying
by
63
every
time
w
e
increase
n
.
How
ever,
when
n
is
greater
than
63
,
n
!
is
multiplying
by
a
higher
numb
er.
When
n
reaches
one
billion,
63
n
increases
b
y
a
facto
r
of
63
every
step,
while
n
!
increases
by
a
factor
of
1
,
000
,
000
,
000
.
By
this
p
oint
n
!
is
much
larger
and
growing
much
faster.
193
Section
3.2
Exercises
Summa
ry
Questions
Q1
Why
do
we
use
n
instead
of
x
as
an
index
for
a
sequence?
Q2
Describ
e
three
different
w
a
ys
of
denoting
a
sequence.
Q3
When
is
the
limit
of
a
sequence
equal
to
the
limit
of
a
function?
Q4
If
a
n
=
b
n
+
1000
for
1
≤
n
≤
2000000
,
what
do
es
that
tell
us
about
the
limits
lim
n
→∞
a
n
and
lim
n
→∞
b
n
?
3.2.1
Q5
Find
a
general
expression
for
a
n
,
the
n
th
term
of
the
following
sequences.
Use
this
to
write
the
sequences
using
b
oth
other
types
of
notation.
a
{
2
,
5
,
10
,
17
,
26
,
37
,
50
,
.
.
.
}
b
3
2
,
−
3
4
,
3
8
,
−
3
16
,
3
32
,
.
.
.
c
1
2
,
1
6
,
1
12
,
1
20
,
1
30
,
.
.
.
Q6
What
is
the
fourth
term
in
the
sequence
{
n
3
−
5
n
}
∞
n
=3
?
194
3.2.2
Q7
Sho
w
using
the
definition
of
the
limit
of
a
sequence
that
lim
n
→∞
sin
n
n
2
=
0
.
Q8
Sho
w
using
the
definition
of
the
limit
of
a
sequence
that
lim
n
→∞
2
n
−
1
2
n
=
1
.
Q9
A
sequence
is
increasing
if
every
term
is
la
rger
than
the
previous
term.
Must
an
increasing
sequence
alwa
ys
diverge?
Explain.
Q10
A
sequence
is
alternating
if
its
terms
alternate
b
etw
een
p
ositive
and
negative
values.
Is
it
possible
that
the
limit
of
an
alternating
sequence
exists?
What
would
its
value
have
to
b
e?
3.2.3
Q11
Consider
the
sequence
a
n
=
2
n
.
a
What
function
could
w
e
write
such
that
f
(
n
)
=
a
n
.
b
Do
es
lim
x
→∞
f
(
x
)
converge?
c
Do
es
the
theorem
equating
limits
of
functions
and
sequences
apply
to
this
function?
d
Can
we
argue
that
lim
n
→∞
2
n
diverges
anywa
y?
Q12
Consider
the
sequence
a
n
=
n
sin(
π
n
)
a
What
is
lim
x
→∞
x
sin(
π
x
)
?
b
Compute
the
first
few
values
a
1
,
a
2
,
a
3
,
and
a
4
.
c
What
is
lim
n
→∞
n
sin(
π
n
)
?
d
Do
es
this
contradict
one
of
our
theorems?
Explain.
195
Section
3.2
Exercises
3.2.4
Q13
Compute
lim
n
→∞
log
n
3
n
.
Q14
Compute
lim
n
→∞
n
2
n
.
Q15
Compute
lim
n
→∞
n
3
+
3
4
n
3
−
9
.
Q16
Compute
lim
n
→∞
sin
n
log
n
.
Q17
Compute
lim
n
→∞
e
n
√
n
.
Q18
Compute
lim
n
→∞
tan
−
1
n
.
3.2.5
Q19
Compute
lim
n
→∞
n
!
5
n
.
Q20
Compute
lim
n
→∞
n
4
+
3
n
+
1
n
!
.
Q21
Do
es
n
n
gro
w
faster
or
slow
er
than
n
!
?
Ex
plain.
Q22
Y
uran
kno
ws
that
lim
n
→∞
n
!
5
n
=
∞
b
ecause
n
!
growns
faster
than
a
n
.
Ho
w
ever,
he
thinks
he
can
mak
e
the
denominator
gro
wn
faster
than
the
numerator
if
he
uses
a
p
ro
duct
like
n
!
5
n
6
n
o
r
n
!
5
n
6
n
7
n
.
Will
he
eventually
obtain
a
non-infinite
limit
b
y
this
metho
d?
Explain
ho
w
you
know.
196
Synthesis
&
Extension
Q23
Supp
ose
w
e
have
a
sequence
a
n
=
(
f
(
n
)
if
n
≤
342
g
(
n
)
if
n
>
342
.
Which
of
the
follo
wing
could
help
us
evaluate
lim
n
→∞
a
n
?
lim
x
→∞
f
(
x
)
lim
x
→∞
g
(
x
)
Q24
Let
T
n
(
x
)
b
e
the
n
th
T
aylo
r
p
olynomial
of
f
(
x
)
=
ln
x
centered
at
x
=
1
.
a
W
rite
an
expression
for
T
n
(
x
)
using
Σ
notation.
b
W
rite
an
expression
for
the
error
b
ound
of
T
n
(
x
)
for
some
x
b
et
w
een
0
and
1
.
c
F
o
r
what
values
of
x
will
the
erro
r
b
ound
shrink
to
0
as
n
goes
to
∞
?
197
Section
3.3
Series
Goals:
1
Identify
partial
sums
of
a
series.
2
Recognize
harmonic
and
alternating
ha
rmonic
series.
3
Apply
the
divergence
test.
4
Evaluate
geometric
series.
5
Apply
the
ratio
test.
The
first
step
in
understanding
a
T
a
ylo
r
polynomial
of
infinite
degree
is
understanding
how
to
add
up
infinitely
many
of
anything.
This
p
roposition
is
mechanically
absurd.
Addition
is
an
op
eration
fo
r
t
w
o
numb
ers
at
a
time.
Adding
three
o
r
four
numb
ers
requires
us
to
add
tw
o
or
three
times.
Adding
infinitely
many
requires
us
to
add
infinitely
many
times,
something
no
one
has
time
to
do.
Y
et
there
are
some
intuitive
exercises
we
could
p
erform.
Supp
ose
we
lay
a
length
of
1
2
m
next
to
1
4
m
next
to
1
8
m.
If
we
continued
indefinitely
,
we
could
imagine
these
lengths
extending
an
entire
meter.
Figure:
One
meter
expressed
as
a
sum
of
infinitely
many
smaller
lengths
What
reasoning
could
w
e
use
to
mak
e
this
exercise
rigorous?
How
could
we
add
up
lengths
o
r
numb
ers
where
the
pattern
is
not
so
intuitive?
The
formal
object
that
do
es
this
is
called
a
series.
A
series
is
the
first
step
on
our
wa
y
to
push
the
T
aylo
r
p
olynomial
to
infinite
degree.
It
is
also
the
most
general.
While
we
are
concerned
with
one
sp
ecific
(and
very
useful)
type
of
series,
there
are
other
applications
wo
rth
exploring
as
well.
Question
3.3.1
What
Is
a
Series?
Y
ou
have
b
een
encountering
series
since
y
ou
first
learned
ab
out
decimals.
Y
ou
lik
ely
have
not
seen
a
rigorous
description
of
what
they
mean.
0
.
33333333
.
.
.
3
.
1415926
...
W
e
can
write
0
.
3333
.
.
.
=
3
10
+
3
100
+
3
1000
+
3
10000
+
·
·
·
o
r
3
.
1415
.
.
.
=
3
+
1
10
+
4
100
+
1
1000
+
5
10000
+
·
·
·
Y
ou
may
have
an
intuitive
s
ense
of
what
these
quantities
are,
but
what
do
es
it
mean
to
add
up
infinitely
many
numb
ers?
198
Definition
A
series
is
a
sum
of
the
form
∞
X
k
=1
a
k
where
a
k
is
an
infinite
sequence.
If
it
is
mo
re
convenient,
we
can
give
k
a
different
initial
value.
If
the
context
is
clear,
we
can
write
X
a
k
as
a
shorthand.
Example
0
.
33333
.
.
.
=
∞
X
k
=1
3
10
k
The
harmonic
series
is
∞
X
k
=1
1
k
This
tells
us
what
a
series
is
but
not
how
to
evaluate
it.
How
do
we
know
that,
for
example
0
.
333
.
.
.
=
1
3
?
W
e
evaluate
a
series
by
asso
ciating
it
with
a
sequence
of
partial
sums.
Definition
The
n
th
pa
rtial
sum
of
the
series
∞
X
k
=1
a
k
is
s
n
=
a
1
+
a
2
+
a
3
+
·
·
·
+
a
n
A
series
∞
X
k
=1
a
k
converges
to
L
if
lim
n
→∞
s
n
=
L.
A
series
that
does
not
converge
to
any
L
diverges
.
V
o
cabula
ry
Note
Do
not
confuse
a
sequence
with
a
series.
One
is
a
list
of
numbers.
The
other
is
the
sum
of
a
list
of
numb
ers.
199
Example
3.3.2
Computing
Pa
rtial
Sums
Consider
∞
X
k
=1
3
10
k
.
a
Compute
the
first
few
partial
sums
s
1
,
s
2
,
s
3
of
this
series.
b
Compute
lim
n
→∞
s
n
Solution
a
s
1
=
3
10
s
2
=
3
10
+
3
100
=
33
100
s
3
=
3
10
+
3
100
+
3
1000
=
333
1000
s
4
=
3
10
+
3
100
+
3
1000
+
3
10000
=
3333
10000
b
In
order
to
use
our
usual
metho
ds
of
limits,
we
would
need
an
algeb
raic
expression
for
s
n
.
It
isn’t
immediately
clear
ho
w
to
p
roduce
one.
Given
our
knowledge
of
decimals,
w
e
exp
ect
the
answ
er
to
b
e
1
3
.
We
will
use
this
as
a
hint.
W
e
exp
ect
1
3
−
s
n
to
approach
0
.
1
3
−
s
1
=
1
30
1
3
−
s
2
=
1
300
1
3
−
s
3
=
1
3000
1
3
−
s
4
=
1
30000
extrap
olating
suggests
1
3
−
s
n
=
1
3(10)
n
Assuming
this
pattern
holds,
we
have
lim
n
→∞
1
3
−
s
n
=
lim
n
→∞
1
3(10)
n
=
0
and
we
conclude
that
∞
X
k
=1
3
10
k
=
1
3
.
200
Main
Idea
Often,
we
can
sho
w
that
∞
X
k
=1
a
k
=
L
by
computing
L
−
s
n
and
seeing
that
it
converges
to
0
.
Click to Load Applet
Figure:
The
partial
sums
s
n
converging
to
L
=
1
3
Example
3.3.3
The
Harmonic
Series
W
e
have
seen
examples
of
series
in
which
the
terms
approach
0
as
k
→
∞
.
These
have
allow
ed
us
to
add
infinitely
many
terms
and
obtain
a
finite
sum.
Do
es
this
alwa
ys
w
o
rk?
No.
A
series
can
have
its
terms
approach
0
,
and
yet
the
pa
rtial
sums
go
to
∞
.
The
most
famous
example
of
this
is
the
harmonic
series:
∞
X
k
=1
1
k
.
Rather
than
computing
the
partial
sums
directly
(which
would
b
e
a
lot
of
computation)
w
e
will
compare
the
partial
sums
to
an
expression
that
is
easier
to
calculate.
We
will
replace
each
term
b
y
a
fraction
with
a
p
ow
er
of
2
in
the
denominato
r.
Here’s
what
we’ll
do
with
s
8
.
s
8
=
1
1
+
1
2
+
1
3
+
1
4
+
1
5
+
1
6
+
1
7
+
1
8
>
1
1
+
1
2
+
1
4
+
1
4
|
{z
}
1
2
+
1
8
+
1
8
+
1
8
+
1
8
|
{z
}
1
2
=1
+
1
2
+
1
2
+
1
2
Since
w
e
replaced
each
term
with
something
smaller
and
obtained
a
sum
of
5
2
,
w
e
can
conclude
that
s
8
>
5
2
.
Continuing
this
pattern,
the
terms
1
9
to
1
16
sum
to
more
than
1
2
so
s
16
>
6
2
.
In
general
we
can
201
Example
3.3.3
The
Harmonic
Series
mak
e
s
n
bigger
than
any
integer
c
by
setting
n
=
2
m
where
1
+
1
2
m
>
c.
This
tells
us
that
the
harmonic
series
diverges.
Question
3.3.4
What
Is
a
Geometric
Series?
The
tw
o
series
so
fa
r
that
we
have
b
een
able
to
evaluate
b
elonged
to
a
la
rger
family
.
These
are
the
geometric
series.
Definition
A
geometric
series
is
a
series
of
the
fo
rm
∞
X
k
=1
ar
k
−
1
.
a
is
the
initial
term.
r
is
the
common
ratio
b
etw
een
terms.
Example
∞
X
k
=1
1
2
k
−
1
=
1
+
1
2
+
1
4
+
1
8
+
·
·
·
∞
X
k
=1
3
10
1
10
k
−
1
=
3
10
+
3
100
+
3
1000
+
·
·
·
=
1
3
Unlik
e
many
other
series,
geometric
series
a
re
simple
enough
that
we
can
write
a
fo
rmula
fo
r
their
sum.
We
can
get
a
convenient
exp
ression
for
s
n
b
y
performing
a
cute
algebra
trick.
W
e’ll
multiply
s
n
b
y
r
and
subtract
r
s
n
from
s
n
.
Most
of
the
terms
cancel
and
w
e
obtain
an
equation
that
we
can
solve
fo
r
s
n
.
s
n
=
a
+
ar
+
ar
2
+
·
·
·
+
ar
n
−
1
−
r
s
n
=
−
ar
−
ar
2
−
ar
3
−
·
·
·
−
ar
n
(1
−
r
)
s
n
=
a
−
ar
n
s
n
=
a
(1
−
r
n
)
1
−
r
The
last
step
requires
that
1
−
r
=
0
,
since
w
e
cannot
divide
b
y
0
.
As
long
as
r
=
1
,
we
can
evaluate
the
series
by
taking
a
limit.
∞
X
k
=1
ar
k
−
1
=
lim
n
→∞
a
(1
−
r
n
)
1
−
r
202
T
o
evaluate
this
limit,
we
need
to
understand
the
b
ehavior
of
r
n
as
n
→
∞
If
−
1
<
r
<
1
then
higher
p
ow
ers
of
r
get
smaller
and
smaller
and
r
n
→
0
.
If
r
>
1
then
higher
pow
ers
of
r
get
larger
and
larger
and
r
n
→
∞
.
If
r
<
−
1
then
higher
pow
ers
of
r
get
larger
but
alternate
signs.
lim
n
→∞
r
n
do
es
not
exist.
If
r
=
1
then
the
series
is
a
+
a
+
a
+
a
+
a
+
·
·
·
,
then
s
n
=
an
which
diverges
to
±∞
,
dep
ending
on
the
sign
of
a
.
If
r
=
−
1
then
the
series
is
a
−
a
+
a
−
a
+
a
−
·
·
·
,
then
s
n
alternates
b
et
w
een
a
and
0
.
This
sequence
do
es
not
converge.
W
e
can
apply
the
ab
ove
to
completely
solve
the
problem
of
evaluating
a
geometric
series.
Our
result
is
the
following
theorem:
Figure:
The
partial
sums
of
P
ar
k
−
1
fo
r
various
r
Theo
rem
Geometric
series
have
the
following
partial
sums
s
n
=
n
X
k
=1
ar
k
−
1
=
a
(1
−
r
n
)
1
−
r
if
r
=
1
an
if
r
=
1
These
converge
to
a
1
−
r
when
|
r
|
<
1
and
diverge
when
|
r
|
≥
1
.
203
Example
3.3.5
Evaluating
Geometric
Series
Identify
a
and
r
in
the
following
geometric
series.
Then
evaluate
the
series.
a
2
3
+
4
15
+
8
75
+
·
·
·
b
∞
X
n
=2
3
n
c
0
.
999999
.
.
.
Solution
a
a
is
the
initial
term,
which
is
2
3
.
The
common
ratio
is
the
ratio
betw
een
any
tw
o
terms.
4
/
15
2
/
3
=
2
5
.
Since
|
r
|
<
1
,
the
sum
of
the
series
is
∞
X
k
=1
2
3
2
5
k
−
1
=
2
3
1
−
2
5
=
2
3
3
5
=
10
9
b
The
initial
term
of
this
series
is
9
.
The
common
ratio
is
3
.
Since
|
3
|
≥
1
,
∞
X
n
=2
3
n
diverges.
c
0
.
999999
.
.
.
=
9
10
+
9
100
+
9
1000
+
·
·
·
.
This
has
an
initial
term
of
9
10
and
a
common
ratio
of
1
10
.
|
r
|
<
1
so
0
.
999999
.
.
.
=
9
10
1
−
1
10
=
9
10
9
10
=
1
204
Question
3.3.6
What
Do
es
the
Size
of
a
k
T
ell
Us
Ab
out
P
a
k
?
The
discussion
of
the
geometric
series
suggests
that
certain
properties
of
a
series
make
convergence
imp
ossible.
Sp
ecifically
,
in
the
cases
in
which
the
terms
were
not
shrinking
to
0
,
the
partial
sums
w
ere
gro
wing
without
b
ound
or
oscillating.
This
intuition
can
b
e
fo
rmalized
in
the
following
theorem,
which
applies
to
more
than
just
geometric
series.
Theo
rem
[The
Divergence
T
est]
Let
a
k
b
e
a
sequence.
If
lim
k
→∞
a
k
=
0
,
then
the
series
∞
X
k
=1
a
k
diverges.
Rema
rk
The
divergence
test
do
es
not
tell
us
anything,
if
lim
k
→∞
a
k
=
0
.
The
series
might
converge,
and
it
might
not.
In
this
case
w
e
say
the
test
is
inconclusive
.
Example
3.3.7
Applying
the
Divergence
T
est
What
do
es
the
divergence
test
tell
us
about
each
of
the
follo
wing
series?
a
∞
X
k
=2
3
k
b
∞
X
k
=2
1
k
c
∞
X
k
=2
k
2
−
1
3
k
2
+
7
d
∞
X
k
=2
k
2
e
k
205
Example
3.3.7
Applying
the
Divergence
T
est
Solution
a
The
sequence
is
a
k
=
3
k
.
lim
k
→∞
3
k
=
∞
.
This
limit
is
not
0
,
so
b
y
the
divergence
test,
the
series
diverges.
b
The
sequence
is
a
k
=
1
k
.
lim
k
→∞
1
k
=
0
.
The
divergence
test
is
inconclusive.
It
cannot
tell
us
whether
this
series
diverges
or
converges.
By
our
earlier
wo
rk,
we
happ
en
to
kno
w
this
series
diverges.
c
The
sequence
is
a
k
=
k
2
−
1
3
k
2
+7
.
lim
k
→∞
k
2
−
1
3
k
2
+
7
=
1
3
.
This
limit
is
not
0
,
so
by
the
divergence
test,
the
series
diverges.
d
The
sequence
is
a
k
=
k
2
e
k
.
We
need
L’Hˆ
optial’s
rule
to
evaluate
the
limit.
lim
k
→∞
k
2
e
k
=
lim
k
→∞
2
k
e
k
still
∞
∞
fo
rm
=
lim
k
→∞
2
e
k
=
0
The
divergence
test
is
inconclusive.
It
cannot
tell
us
whether
this
series
diverges
or
converges.
It
turns
out
that
this
series
converges,
but
w
e
do
not
have
a
metho
d
to
verify
that
yet.
Question
3.3.8
What
Is
the
Ratio
T
est?
So
fa
r
we
have
tw
o
tests
to
determine
the
convergence
of
a
series.
One
test
is
very
sp
ecific,
applying
only
to
geometric
series.
The
other
is
very
imprecise.
The
divergence
test
is
often
inconclusive.
It
do
es
not
help
us
to
evaluate
a
series
at
all,
only
recognizing
some
series
that
diverge.
Unfortunately
,
these
shortcoming
are
typical
of
series
tests.
A
rigorous
study
of
infinite
series
requires
learning
almost
a
dozen
tests.
On
a
randomly
chosen
series,
most
of
these
tests
will
b
e
inconclusive,
and
none
of
them
will
give
a
numerical
value,
even
if
the
series
happens
to
converge.
Because
we
are
interested
in
extending
T
a
ylo
r
polynomials
to
have
infinitely
many
terms,
some
of
these
tests
are
much
more
useful
than
others.
The
most
useful
is
the
ratio
test,
though
it
is
still
no
help
in
evaluating
a
series
and
is
still
sometimes
inconclusive.
In
the
case
of
a
geometric
series,
P
ar
k
−
1
,
the
common
ratio
b
etw
een
terms
determines
whether
this
series
gro
ws
out
of
control,
or
whether
the
terms
shrink
quickly
enough
that
the
partial
sums
converge.
Even
when
a
series
is
not
geometric,
we
can
attempt
to
apply
similar
reasoning
to
determine
whether
it
converges.
A
non-geometric
series
do
es
not
have
a
constant
ratio.
The
ratio
b
etw
een
successive
terms
will
change
as
w
e
progress
through
them.
W
e
will
instead
compute
the
limit
of
these
ratios.
206
Theo
rem
[The
Ratio
T
est]
If
lim
k
→∞
a
k
+1
a
k
=
L
<
1
,
then
X
a
k
converges
absolutely
.
If
lim
k
→∞
a
k
+1
a
k
=
L
>
1
or
is
infinite,
then
X
a
k
is
divergent.
If
lim
k
→∞
a
k
+1
a
k
=
1
,
then
the
ratio
test
is
inconclusive.
Rema
rk
Converges
absolutely
is
a
term
fo
r
series
with
b
oth
p
ositive
and
negative
terms.
It
means
the
series
w
ould
converge,
even
if
the
signs
of
all
the
terms
were
all
p
ositive.
The
alternative
is
conditional
convergence
,
meaning
the
series’s
convergence
ma
y
require
the
p
ositive
and
negative
terms
partially
canceling
each
other
out.
Example
The
series
1
−
1
2
+
1
3
−
1
4
+
1
5
−
·
·
·
converges
(w
e
w
on’t
p
rove
this).
If
we
made
all
the
terms
p
ositive,
it
w
ould
b
e
the
ha
rmonic
series,
which
diverges.
This
series
converges
conditionally
,
not
absolutely
.
Absolute
versus
conditional
convergence
can
b
e
interesting
to
play
with.
Y
ou
may
see
references
to
it
in
other
math
b
o
oks,
but
we
won’t
have
any
further
use
for
it.
Example
3.3.9
Applying
the
Ratio
T
est
a
Do
es
∞
X
k
=1
(
−
1)
k
−
1
k
!
converge
or
diverge?
b
Do
es
∞
X
k
=1
2
k
k
2
converge
or
diverge?
c
Do
es
∞
X
k
=1
k
converge
or
diverge?
207
Example
3.3.9
Applying
the
Ratio
T
est
Solution
a
First
we
will
compute
and
simplify
the
ratio.
Then
we
will
tak
e
its
limit
and
dra
w
a
conclusion.
a
k
+1
a
k
=
(
−
1)
k
(
k
+1)!
(
−
1)
k
−
1
k
!
=
(
−
1)
k
k
!
(
−
1)
k
−
1
(
k
+
1)!
=
(
−
1)
k
(1)(2)(3)
·
·
·
(
k
)
(
−
1)
k
−
1
(1)(2)(3)
·
·
·
(
k
)(
k
+
1)
(
expand
the
factorials
)
=
(
−
1)
k
(
−
1)
k
−
1
(
k
+
1)
(
cancel
the
matching
facto
rs
)
=
−
1
k
+
1
(
cancel
k
−
1
pow
ers
of
−
1)
=
1
k
+
1
(
absolute
value
of
a
negative
numb
er
is
its
negatve
)
No
w
we
take
the
limit
lim
k
→∞
1
k
+
1
=
0
0
<
1
so
b
y
the
ratio
test,
∞
X
k
=1
(
−
1)
k
−
1
k
!
converges.
b
W
e
will
apply
the
ratio
test.
First
w
e
compute
the
ratio,
and
then
we
take
a
limit.
a
k
+1
a
k
=
2
k
+1
(
k
+1)
2
2
k
k
2
=
2
k
+1
k
2
2
k
(
k
2
+
2
k
+
1)
=
2
k
2
k
2
+
2
k
+
1
(
cancel
the
2
s
)
=
2
k
2
k
2
+
2
k
+
1
lim
k
→∞
2
k
2
k
2
+
2
k
+
1
=
2
2
>
1
so
b
y
the
ratio
test,
this
series
diverges.
208
c
W
e
will
apply
the
ratio
test.
First
w
e
compute
the
ratio,
and
then
we
take
a
limit.
a
k
+1
a
k
=
k
+
1
k
=
k
+
1
k
lim
k
→∞
k
+
1
k
=
1
Here
the
ratio
test
is
inconclusive.
It
cannot
tell
whether
this
series
converges
o
r
diverges.
Ho
w
ever,
w
e
can
probably
figure
this
out
another
wa
y
.
The
terms
of
this
series
are
increasing,
which
means
the
partial
sums
will
gro
w
faster
and
faster.
This
was
the
reasoning
b
ehind
the
divergence
test.
lim
k
→∞
k
=
∞
Since
lim
k
→∞
k
=
0
,
the
divergence
test
concludes
that
the
series
diverges.
Main
Ideas
When
applying
the
ratio
test,
b
e
sure
to
replace
every
k
with
k
+
1
for
the
a
k
+1
term.
F
amilia
rize
yourself
with
the
algebra
rules
that
allow
y
ou
to
simplify
ratios
of
exp
onentials
and
facto
rials.
Example
3.3.10
A
Strategy
for
Series
T
ests
209
Example
3.3.10
A
Strategy
for
Series
T
ests
Strategy
Given
the
three
wa
ys
we
have
to
test
for
divergence
and
convergence
and
the
relative
ease
of
applying
each,
here
is
a
reasonable
approach
to
testing
a
series.
Check
lim
n
→∞
a
n
by
dominance
Compute
a
n
+1
a
n
Compute
lim
n
→∞
a
n
+1
a
n
P
a
n
converges
P
a
n
diverges
Inconclusive,
lo
ok
up
another
test
not
zero
zero
hard
to
tell
constant
|
r
|
≥
1
constant
|
r
|
<
1
not
constant
<
1
>
1
=
1
hard
to
tell
Let’s
apply
our
strategy
to
see
what
w
e
can
tell
about
∞
X
n
=1
1
n
2
.
Solution
First
we’ll
check
that
the
terms
go
to
zero.
If
they
don’t
w
e
quickly
classify
this
as
a
divergent
series.
lim
n
→∞
1
n
2
=
0
They
do,
so
w
e
need
another
check.
No
w
we’ll
compute
the
ratio
b
etw
een
terms.
a
n
+1
a
n
=
1
(
n
+1)
2
1
n
2
=
n
2
n
2
+
2
n
+
1
This
is
not
a
constant;
it
dep
ends
on
n
.
Thus
a
n
is
not
a
geometric
series.
We’ll
try
the
ratio
test.
lim
n
→∞
a
n
+1
a
n
=
lim
n
→∞
n
2
n
2
+
2
n
+
1
=
lim
n
→∞
n
2
n
2
+
2
n
+
1
=
1
This
means
that
the
ratio
test
is
inconclusive.
We
do
not
know
whether
this
series
converges
or
diverges.
W
e
have
exhausted
all
our
tests.
If
w
e
w
ant
the
answer,
we
need
to
look
up
another
test.
210
Section
3.3
Exercises
Summa
ry
Questions
Q1
What
is
the
difference
b
etw
een
a
sequence
and
a
series?
Q2
Ho
w
do
we
evaluate
a
series?
Q3
What
is
a
geometric
series.
How
do
w
e
evaluate
one?
Q4
What
do
es
it
mean
to
say
that
a
series
test
is
inconclusive?
Q5
Ho
w
do
each
of
the
following
factors
b
ehave
in
the
ratio
a
k
+1
a
k
?
a
k
p
(
p
a
constant)
b
c
k
(
c
a
constant)
c
k
!
Q6
Ho
w
would
the
ratio
test
apply
to
a
geometric
series
X
ar
k
−
1
?
3.3.1
Q7
Give
a
more
common
name
for
each
of
the
follo
wing
series.
a
2
+
7
10
+
1
100
+
8
1000
+
2
10000
+
8
100000
+
·
·
·
b
6
10
+
6
100
+
6
1000
+
6
10000
+
·
·
·
Q8
Use
a
calculator
to
get
a
decimal
app
ro
ximation
of
25
33
and
write
it
as
a
series
of
fractions
with
p
o
w
ers
of
10
as
denominators.
211
Section
3.3
Exercises
3.3.2
Q9
Consider
the
series
∞
X
k
=1
1
k
(
k
+
1)
a
Compute
the
first
four
elements
in
the
series.
b
Compute
the
partial
sums:
s
1
,
s
2
,
s
3
,
s
4
.
c
What
do
the
pa
rtial
sums
app
ear
to
be
converging
to?
d
Can
you
use
algeb
ra
to
generalize
your
answer
to
2
to
s
n
?
Q10
Compute
the
first
3
partial
sums
of
∞
X
k
=1
k
+
1
k
2
.
Don’t
simplify
the
a
rithmetic.
Q11
Compute
the
first
four
partial
sums
of
∞
X
k
=1
(
−
1)
k
.
What
do
y
ou
think
this
suggests
ab
out
the
sum
of
the
series?
Q12
Compute
the
first
five
partial
sums
of
∞
X
k
=0
1
(
−
2)
k
.
Use
them
to
make
a
p
rediction
ab
out
the
value
of
the
series.
3.3.3
Q13
Give
an
example
of
an
n
such
that
you
know
the
n
th
partial
sum
of
the
ha
rmonic
series
is
greater
than
20
.
Q14
Mo
dify
our
argument
for
the
harmonic
series
to
sho
w
that
∞
X
k
=0
1
√
k
diverges?
212
3.3.4
Q15
Is
1
2
+
1
4
+
1
6
+
1
8
+
·
·
·
a
geometric
series?
Ho
w
can
you
tell?
Q16
Is
1
+
4
+
9
+
16
+
25
+
·
·
·
a
geometric
series?
Ho
w
can
you
tell?
Q17
The
first
tw
o
terms
of
a
geometric
series
are
5
and
7
.
5
.
What
is
the
third
term?
Q18
The
fifth
term
of
a
geometric
series
is
17
.
The
eigth
term
is
51
.
What
is
the
sixth
term?
3.3.5
Q19
Evaluate
∞
X
k
=0
5(0
.
3)
k
Q20
Evaluate
∞
X
k
=0
1
4
4
3
k
.
Q21
Evaluate
∞
X
j
=3
15
5
j
.
Q22
Evaluate
∞
X
k
=1
0
.
8
k
.
Q23
Evaluate
∞
X
k
=4
3
k
2
k
(18)
.
Q24
Evaluate
∞
X
k
=1
37
100
k
.
What
decimal
do
es
this
represent?
Q25
F
o
r
what
values
of
z
do
es
∞
X
k
=0
3
k
z
k
converge?
Q26
F
o
r
what
values
of
p
do
es
∞
X
k
=3
12
p
2
k
16
k
converge?
213
Section
3.3
Exercises
3.3.6
Q27
If
a
k
>
1
100
fo
r
all
k
,
then
what
can
y
ou
say
ab
out
the
value
of
s
n
=
n
X
k
=1
a
k
?
Q28
If
lim
k
→∞
a
k
=
1
100
,
use
the
definition
of
a
limit
and
the
reasoning
in
the
p
revious
exercise
to
sho
w
that
∞
X
k
=1
a
n
diverges.
3.3.7
Q29
What
do
es
the
divergence
test
say
ab
out
∞
X
k
=1
1
k
3
?
Q30
What
do
es
the
divergence
test
say
ab
out
∞
X
k
=1
k
2
+
1
5
k
2
+
3
k
?
Q31
What
do
es
the
divergence
test
say
ab
out
∞
X
k
=2
ln
k
?
Q32
What
do
es
the
divergence
test
say
ab
out
∞
X
k
=2
1
ln
k
?
3.3.8
Q33
Will
the
divergence
test
detect
every
series
that
“fails”
the
ratio
test
(
L
>
1
)?
Explain.
Q34
If
lim
n
→∞
an
+
1
a
n
do
es
not
exist,
the
ratio
test
is
inconclusive.
Give
examples
of
tw
o
series
where
this
limit
do
es
not
exist,
one
series
that
diverges
and
one
that
converges.
214
3.3.9
Q35
Apply
the
ratio
test
to
∞
X
k
=1
k
!
4
k
.
What
can
you
conclude?
Q36
Apply
the
ratio
test
to
∞
X
k
=1
k
5
k
(
k
+
1)
2
.
What
can
you
conclude?
Q37
Apply
the
ratio
test
to
∞
X
k
=1
(
−
1)
k
−
1
k
2
.
What
can
you
conclude?
Q38
Apply
the
ratio
test
to
∞
X
k
=1
(
−
8)
k
k
2
5
k
.
What
can
you
conclude?
Q39
Apply
the
ratio
test
to
∞
X
k
=1
k
2
4
k
.
What
can
you
conclude?
Q40
Apply
the
ratio
test
to
∞
X
k
=3
k
!
5
k
3
+
4
k
−
2
.
What
can
you
conclude?
Q41
Apply
the
ratio
test
to
∞
X
k
=1
√
k
+
1
k
2
What
can
you
conclude?
Q42
Apply
the
ratio
test
to
∞
X
k
=1
√
k
e
−
k
What
can
you
conclude?
3.3.10
Q43
Use
one
of
the
tests
from
this
section
to
deterine
whether
P
∞
k
=1
k
+1
k
converges.
Q44
Use
one
of
the
tests
from
this
section
to
deterine
whether
P
∞
k
=1
3(4
k
)
7
k
converges.
Q45
Use
one
of
the
tests
from
this
section
to
deterine
whether
P
∞
k
=1
ke
k
4
k
+1
converges.
Q46
Use
one
of
the
tests
from
this
section
to
deterine
whether
P
∞
k
=1
7
k
9
k
k
3
2
k
+1
converges.
215
Section
3.3
Exercises
Synthesis
&
Extension
Q47
In
a
paragraph
o
r
t
w
o,
explain:
How
is
evaluating
an
improper
integral
simila
r
to
evaluating
an
infinite
series.
How
are
they
different?
Q48
Supp
ose
w
e
have
a
sequence
a
n
such
that
lim
n
→∞
a
n
=
0
and
∞
X
n
=1
a
n
=
437
.
Supp
ose
we
then
increase
the
values
of
the
first
five
terms
of
a
n
b
y
10
,
000
each.
a
Explain
how
this
will
affect
the
value
of
lim
n
→∞
a
n
.
b
Explain
how
this
will
affect
the
value
of
∞
X
n
=1
a
n
.
Q49
Supp
ose
w
e
w
anted
to
appro
ximate
R
∞
0
1
e
x
dx
b
y
rectangles
of
length
∆
x
=
1
,
with
heights
measured
at
the
left
endp
oints.
a
What
are
the
a
reas
of
the
first
5
rectangles,
starting
from
x
=
0
?
b
Ho
w
many
rectangles
will
you
need
in
total?
c
Exp
ress
the
sum
of
the
areas
of
these
rectangles
as
a
series.
d
Do
es
this
series
converge?
T
o
what
value?
e
Do
es
your
series
over-
or
underestimate
the
true
value
of
the
integral?
Q50
Supp
ose
w
e
w
anted
to
appro
ximate
R
∞
1
1
x
2
dx
b
y
rectangles
of
length
∆
x
=
1
,
with
heights
measured
at
the
right
endp
oints.
a
What
are
the
a
reas
of
the
first
5
rectangles,
starting
from
x
=
1
?
b
Exp
ress
the
sum
of
the
areas
of
all
the
the
rectangles
you’ll
need
as
a
series.
c
Do
es
your
series
over-
or
underestimate
the
true
value
of
the
integral?
d
What
is
the
true
value
of
the
integral?
What
does
this
suggest
ab
out
whether
your
series
converges
or
diverges?
Q51
Supp
ose
that
a
discrete
random
variable
X
has
distribution
function
f
X
(
x
)
=
(
1
2
x
if
x
is
a
p
ositive
integer
0
otherwise
216
a
V
erify
that
f
X
(
x
)
is
a
valid
probabilit
y
distribution
function.
b
Compute
P
(
X
>
4)
.
c
Compute
E
[
X
]
(this
is
difficult).
Q52
Supp
ose
that
a
discrete
random
variable
X
has
distribution
function
f
X
(
x
)
=
(
1
x
−
1
x
+1
if
x
is
a
p
ositive
integer
0
otherwise
a
V
erify
that
f
X
(
x
)
is
a
valid
probabilit
y
distribution
function.
b
Compute
P
(3
≤
X
≤
5)
.
c
Explain
why
you
can’t
compute
E
[
X
]
.
217
Section
3.4
P
o
w
er
Series
Goals:
1
Use
series
tests
to
determine
for
what
values
of
x
a
p
ow
er
series
converges.
2
Identify
the
radius
of
convergence
of
a
pow
er
series.
3
Recognize
functions
that
can
b
e
rewritten
as
a
p
ow
er
series.
The
infinite
degree
p
olynomials
we
seek
to
define
are
series.
The
tools
we’ve
develop
ed
so
far
p
rovide
the
foundation
for
understanding
the
objects
we
want
to
construct,
but
there
is
mo
re
to
do.
A
p
olynomial
also
contains
a
variable.
In
this
section
we
deal
with
the
ramifications
of
including
a
variable
in
an
infinite
series.
Question
3.4.1
What
Is
a
P
o
w
er
Series?
So
fa
r
w
e
have
studied
infinite
series
of
numb
ers.
If
instead
of
just
numb
ers,
our
terms
include
va
riables,
then
we’ve
created
a
function.
Plugging
in
different
values
for
the
va
riable
gives
us
a
different
series
of
numb
ers.
Example
The
expression
1
+
x
+
x
2
+
x
3
+
·
·
·
b
ecomes
1
+
2
+
4
+
8
+
·
·
·
when
we
evaluate
it
at
x
=
2
.
It
b
ecomes
1
−
1
3
+
1
9
−
1
27
+
·
·
·
when
we
evaluate
it
at
x
=
−
1
3
.
Definition
An
infinite
series
of
the
form
∞
X
k
=0
c
k
(
x
−
a
)
k
is
called
a
pow
er
series
centered
at
a
.
It
is
a
function
of
x
whose
domain
is
all
values
of
x
that
mak
e
the
series
converge.
F
o
r
the
purp
oses
of
this
definition,
we
define
x
0
=
1
even
when
x
=
0
.
218
Example
3.4.2
A
Geometric
Series
as
a
Po
w
er
Series
Use
the
geometric
series
fo
rmula
to
write
f
(
x
)
=
1
1
−
x
as
a
pow
er
series
and
find
its
domain.
Solution
1
1
−
x
is
the
sum
of
a
geometric
series.
In
this
case,
the
initial
term
a
=
1
and
the
common
ratio
r
is
x
.
If
we
write
out
the
first
few
terms
w
e
obtain
1
+
x
+
x
2
+
x
3
+
·
·
·
.
W
e
see
this
is
a
p
ow
er
series
centered
at
0
.
The
coefficients
c
k
a
re
all
equal
to
1
.
We
could
write
it
as
P
∞
k
=0
x
k
.
The
domain
of
a
p
o
w
er
series
is
the
values
of
x
that
make
it
converge.
W
e
kno
w
that
this
geometric
series
converges
if
and
only
if
the
common
ratio
x
has
absolute
value
less
than
1
.
Those
values
of
x
,
the
op
en
interval
(
−
1
,
1)
,
a
re
the
domain
of
f
.
Example
3.4.3
The
Domain
of
a
Po
w
er
Series
What
is
the
domain
of
∞
X
k
=1
k
2
4
k
(
x
−
5)
k
?
Solution
The
domain
is
the
set
of
x
values
that
make
the
series
converge.
The
ratio
test
will
b
e
helpful
here.
The
ratio
b
etw
een
terms
is
a
k
+1
a
k
=
(
k
+1)
2
4
k
+1
(
x
−
5)
k
+1
k
2
4
k
(
x
−
5)
k
=
(
k
+
1)
2
4
k
(
x
−
5)
k
+1
k
2
4
k
+1
(
x
−
5)
k
=
(
k
2
+
2
k
+
1)(
x
−
5)
4
k
2
Notice
this
entire
computation
is
invalid
if
x
=
5
,
b
ecause
we
cannot
divide
by
0
.
We
can
examine
this
case
directly
.
If
x
=
5
then
every
term
of
the
series
is
0
,
and
the
series
converges.
Fo
r
the
rest
of
the
real
numb
ers,
w
e
compute
the
limit
as
k
→
∞
,
but
x
will
remain
in
the
result.
lim
k
→∞
(
k
2
+
2
k
+
1)(
x
−
5)
4
k
2
=
(
x
−
5)
4
lim
k
→∞
k
2
+
2
k
+
1
k
2
=
(
x
−
5)
4
219
Example
3.4.3
The
Domain
of
a
Po
w
er
Series
The
ratio
test
can
tell
us
whether
the
series
converges
for
some
values
of
x
.
If
(
x
−
5)
4
<
1
the
series
converges.
We
can
solve
fo
r
x
(
x
−
5)
4
<
1
x
−
5
<
4
(
since
4
>
0)
−
4
<
x
−
5
<
4
1
<
x
<
9
(
add
5
to
all
three
exp
ressions
)
On
the
other
hand,
if
(
x
−
5)
4
>
1
the
series
diverges.
Solving
for
x
follows
a
simila
r
procedure.
(
x
−
5)
4
>
1
x
−
5
>
4
(
since
4
>
0)
x
−
5
<
−
4
or
x
−
5
>
4
x
<
1
or
x
>
9
What
about
when
x
=
1
or
x
=
9
?
(
x
−
5)
4
=
1
so
the
ratio
test
is
indeterminate.
W
e
would
need
another
test
to
resolve
these
p
oints.
In
this
case,
w
e
are
lucky
.
If
x
=
9
the
series
b
ecomes
P
∞
k
=1
k
2
4
(4)
.
The
divergence
test
is
useful
here:
lim
k
→∞
k
2
=
∞
.
Since
the
terms
do
not
approach
0
,
the
series
diverges.
A
simila
r
argument
wo
rks
for
k
=
1
.
Main
Idea
The
ratio
test
is
usually
successful
in
finding
where
a
p
ow
er
series
converges.
Generally
it
is
inconclusive
at
only
tw
o
p
oints.
We
will
not
alw
a
ys
have
a
test
that
can
tell
us
whether
the
series
converges
at
these
p
oints.
Y
ou
may
notice
a
pattern
in
the
types
of
domains
we
have
computed
for
p
o
w
er
series.
That
pattern
is
formalized
in
the
theo
rem
below,
which
tells
us
that
the
domain
of
a
p
ow
er
series
must
tak
e
a
very
pa
rticula
r
form.
220
Theo
rem
Given
a
p
ow
er
series
∞
X
k
=0
c
k
(
x
−
a
)
k
centered
at
a
,
one
of
the
following
is
true.
1
The
series
converges
only
when
x
=
a
.
2
The
series
converges
when
x
is
any
real
numb
er.
3
There
is
a
radius
of
convergence
R
such
that
a
The
series
converges
when
|
x
−
a
|
<
R
,
and
b
The
series
diverges
when
|
x
−
a
|
>
R
.
In
case
3
,
the
inequalit
y
|
x
−
a
|
<
R
solves
to
a
−
R
<
x
<
a
+
R
,
which
means
the
domain
is
an
interval
centered
at
a
and
extending
a
distance
R
to
either
side.
The
theo
rem
does
not
state
whether
this
is
a
closed,
op
en
o
r
half
open
interval.
This
reasoning
extends
intuitively
,
if
not
fo
rmally
,
to
the
other
cases.
1
can
the
thought
of
as
a
(closed)
interval
extending
distance
0
on
either
side.
2
w
ould
then
b
e
an
interval
extending
infinitely
on
either
side.
Click to Load Applet
Figure:
The
domain
|
x
−
a
|
<
R
of
a
p
ow
er
series.
Rema
rk
The
main
consequence
of
this
theo
rem
is
that
when
solving
for
the
domain
of
a
p
ow
er
series,
we
can
simplify
our
use
of
the
ratio
test.
The
interval
of
convergence
will
alwa
ys
b
e
the
solution
to
lim
k
→∞
ak
+
1
a
k
<
1
.
The
endp
oints
may
or
may
not
lie
in
the
domain.
The
p
oints
b
eyond
the
endp
oints
will
never
b
e
pa
rt
of
the
domain.
Question
3.4.4
Can
We
Integrate
o
r
Differentiate
a
Po
w
er
Series?
When
f
(
x
)
is
a
p
olynomial,
we
can
find
the
derivative
and
anti-derivative
of
f
(
x
)
by
computing
the
(anti-)derivative
of
each
term.
The
following
theorem
says
that
we
can
do
this
for
a
p
ow
er
series
too.
221
Question
3.4.4
Can
We
Integrate
o
r
Differentiate
a
Po
w
er
Series?
Theo
rem
If
f
(
x
)
is
the
p
o
w
er
series
∞
X
k
=0
c
k
(
x
−
a
)
k
and
f
(
x
)
has
radius
of
convergence
R
>
0
then
f
(
x
)
is
differentiable
and
continuous
on
the
interval
(
a
−
R
,
a
+
R
)
,
and
1
f
′
(
x
)
=
∞
X
k
=1
k
c
k
(
x
−
a
)
k
−
1
2
Z
f
(
x
)
dx
=
C
+
∞
X
k
=0
c
k
(
x
−
a
)
k
+1
k
+
1
Both
of
these
functions
also
have
radius
of
convergence
R
.
Rema
rk
Notice
that
we
remove
the
k
=
0
term
from
the
derivative.
The
derivative
of
that
term
is
0
,
but
0
c
0
(
x
−
a
)
−
1
is
undefined
at
x
=
a
.
Example
W
e
have
seen
that
1
1
−
x
=
∞
X
k
=0
x
k
on
the
interval
(
−
1
,
1)
.
F
rom
that
we
can
compute:
d
dx
∞
X
k
=0
x
k
=
∞
X
k
=1
k
x
k
−
1
Z
∞
X
k
=0
x
k
dx
=
∞
X
k
=0
x
k
+1
k
+
1
+
c
Both
have
domain
(
−
1
,
1)
.
222
Section
3.4
Exercises
Summa
ry
Questions
Q1
What
is
the
difference
b
etw
een
a
p
olynomial
and
a
p
ow
er
series?
Q2
What
test
is
useful
fo
r
establishing
the
domain
of
a
pow
er
series?
What
form
can
this
domain
have?
Q3
Ho
w
can
we
integrate
or
differentiate
a
p
ow
er
series?
Q4
Ho
w
do
es
differentiation
affect
the
radius
of
convergence
of
a
pow
er
series?
3.4.1
Q5
Use
Σ
notation
to
express
the
following
series
a
10
+
15
x
+
20
x
2
+
25
x
3
+
30
x
4
+
·
·
·
b
1
2
−
1
4
x
2
+
1
8
x
4
−
1
16
x
6
+
1
32
x
8
−
·
·
·
Q6
Use
Σ
notation
to
express
the
following
series
a
1
−
x
2
2
+
x
4
24
−
x
6
720
+
x
8
40640
−
·
·
·
b
x
3
+
4
x
4
+
9
x
5
+
16
x
6
+
25
x
7
+
·
223
Section
3.4
Exercises
3.4.2
Q7
Consider
f
(
x
)
=
1
1
−
4
x
2
.
a
If
f
is
the
sum
of
a
geometric
series,
what
is
r
?
b
W
rite
f
(
x
)
as
a
geometric
series
centered
at
0
.
c
What
is
the
domain
of
your
answer
in
b
?
Q8
W
rite
5
1
−
3(
x
−
2)
as
a
p
ow
er
series
centered
at
x
=
2
.
Q9
Can
the
p
o
w
er
series
p
(
x
)
=
∞
X
k
=1
k
3
4
k
(
x
+
7)
k
b
e
evaluated
using
the
sum
of
a
geometric
series
fo
rmula?
Explain.
Q10
Evaluate
f
(
x
)
=
∞
X
k
=3
1
5
k
(
x
−
2)
k
at
x
=
6
using
the
formula
fo
r
the
sum
of
a
geometric
series.
3.4.3
Q11
What
is
the
domain
of
∞
X
k
=1
2
k
(
x
−
3)
k
?
Q12
Compute
the
domain
of
∞
X
k
=0
(
x
+
2)
k
k
3
.
Q13
Compute
the
domain
of
∞
X
k
=0
1
4
k
(
x
−
6)
k
.
Q14
Compute
the
domain
of
∞
X
k
=0
x
k
k
!
.
Q15
Compute
the
radius
of
convergence
of
∞
X
k
=0
k
(
x
+
3)
k
.
What
interval
do
es
this
guarantee
the
series
converges
on?
Q16
Compute
the
radius
of
convergence
of
∞
X
k
=0
k
!
x
k
.
What
interval
do
es
this
guarantee
the
series
converges
on?
224
Q17
Compute
the
radius
of
convergence
of
∞
X
k
=1
4
k
3
k
(
x
−
5)
k
.
What
interval
do
es
this
gua
rantee
the
series
converges
on?
Q18
Supp
ose
you
are
told
that
a
given
p
o
w
er
series
p
(
x
)
centered
at
x
=
a
converges
at
x
=
−
4
and
diverges
at
x
=
−
7
.
a
If
a
=
1
,
what
can
you
say
ab
out
the
domain
of
p
(
x
)
?
b
What
are
all
of
the
the
p
ossible
values
of
a
?
Explain
y
our
reasoning
(briefly).
3.4.4
Q19
Compute
the
antiderivative
of
∞
X
k
=0
2
k
(
x
−
3)
k
.
Q20
Compute
the
derivative
of
∞
X
k
=0
(
x
+
2)
k
k
3
.
What
is
its
domain?
Q21
Compute
the
derivative
of
∞
X
k
=0
1
4
k
(
x
−
6)
k
.
What
is
its
domain?
Q22
Compute
the
antiderivative
of
∞
X
k
=0
x
k
k
!
.
Q23
What
is
the
domain
of
the
fifth
deriative
of
∞
X
k
=0
k
(
x
+
3)
k
?
Q24
Compute
the
radius
of
convergece
of
the
antiderivative
of
∞
X
k
=4
4
k
3
k
(
x
−
5)
k
.
225
Section
3.4
Exercises
Synthesis
&
Extension
Q25
Consider
the
p
ow
er
series
p
(
x
)
=
∞
X
k
=0
k
2
+
k
5
k
(
x
+
3)
k
.
a
Compute
the
domain
of
P
.
Y
ou
do
not
need
to
check
any
endp
oints
of
y
our
answer.
b
W
rite
an
expression
for
Z
p
(
x
)
dx.
Q26
Consider
the
series
S
=
∞
X
k
=1
k
2
k
.
a
Ho
w
is
S
related
to
the
p
o
w
er
series
p
(
x
)
=
∞
X
k
=1
k
x
k
−
1
2
k
.
b
Compute
the
an
avtiderivative
P
(
x
)
of
p
(
x
)
.
c
W
rite
P
(
x
)
as
ratio
F
(
x
)
,
using
the
sum
of
a
geometric
series
fo
rmula.
d
Compute
F
′
(1)
.
What
is
the
significance
of
this
value?
Q27
W
rite
a
p
ow
er
series
for
f
(
x
)
=
tan
−
1
x
by
Diffrentiating
f
(
x
)
W
riting
f
′
(
x
)
as
a
geometric
series
T
aking
an
antiderivative
of
the
geometric
series
226
Section
3.5
T
a
ylo
r
Series
Goals:
1
Use
a
combination
of
p
ow
er
series
and
algeb
ra
to
wo
rk
with
functions.
2
Integrate
and
differentiate
pow
er
series.
Our
goal
has
b
een
to
understand
how
to
extend
a
T
aylo
r
p
olynomial
to
have
infinite
degree.
We
a
re
no
w
ready
to
define
the
object
rigorously
.
In
general
w
e
will
not
kno
w
how
to
evaluate
T
a
ylo
r
series.
If
all
we
w
ant
to
do
is
appro
ximate
values,
they
offer
no
advantages
over
T
a
ylo
r
p
olynomials.
The
applications
of
T
aylo
r
series
are
more
abstract.
After
defining
these
objects,
w
e
collect
some
tricks
and
applications
for
wo
rking
with
them.
Question
3.5.1
What
Is
a
T
a
ylo
r
Series?
Definition
The
T
a
ylo
r
series
of
f
(
x
)
at
x
=
a
is
T
(
x
)
=
∞
X
k
=0
f
(
k
)
(
a
)
k
!
(
x
−
a
)
k
.
The
T
aylo
r
se
ries’s
notation
simply
sw
aps
an
n
for
an
∞
in
the
expression
of
a
T
aylo
r
p
olynomial.
If
w
e
w
anted
to
describ
e
the
mathematical
relationship
precisely
,
we
would
say
its
partial
sums
s
n
a
re
the
T
a
ylo
r
p
olynomials
T
n
(
x
)
of
f
at
x
=
a
.
Rema
rk
Several
mathematicians
contributed
to
the
discovery
of
T
aylo
r
series.
T
a
ylo
r
series
centered
at
x
=
0
w
ere
p
opularized
by
Colin
Maclaurin,
and
so
are
often
called
Maclaurin
series
.
This
definition
is
built
up
on
a
stack
of
more
general
definitions,
and
the
metho
ds
w
e
have
for
wo
rking
with
those
apply
here.
A
T
aylo
r
series
is
a
type
of
pow
er
series.
A
p
ow
er
series
is
a
type
of
series
A
series
is
equivalent
to
a
sequence
of
partial
sums.
This
list
should
make
us
feel
b
etter
ab
out
our
hard
w
o
rk
over
the
last
few
sections.
It
also
gives
us
info
rmation
that
helps
us
understand
T
aylo
r
series
b
etter.
Fo
r
example,
since
T
a
ylo
r
series
are
pow
er
series,
their
domains
a
re
also
intervals
of
radius
R
centered
at
a
.
227
Question
3.5.1
What
Is
a
T
a
ylo
r
Series?
Limitations
of
T
aylo
r
Series
T
a
ylo
r
p
olynomials
were
designed
to
appro
ximate
f
(
x
)
.
We
might
hop
e
that
T
(
x
)
would
b
e
the
p
erfect
app
ro
ximation,
that
T
(
x
)
and
f
(
x
)
a
re
equal.
Unfortunately
,
there
a
re
obstacles
to
this.
The
T
aylo
r
series
might
not
converge
for
all
x
.
The
T
aylo
r
p
olynomials
might
not
appro
ximate
f
(
x
)
very
w
ell
at
all.
Recall
our
example
f
(
x
)
=
(
0
if
x
≤
0
e
−
1
x
if
x
>
0
F
o
r
this
function
T
(
x
)
=
0
.
Example
3.5.2
W
riting
a
T
aylo
r
series
Let
f
(
x
)
=
e
x
a
Find
the
T
aylo
r
series
for
f
(
x
)
centered
at
x
=
0
.
b
On
what
interval
does
it
converge?
Solution
a
W
e
have
seen
previously
that
f
(
k
)
(
x
)
=
e
x
fo
r
all
k
and
thus
f
(
k
)
(0)
=
1
.
We
plug
this
into
the
T
a
ylo
r
series
formula.
T
(
x
)
=
∞
X
k
=0
1
k
!
x
k
b
A
T
aylo
r
series
is
a
p
ow
er
series.
We
will
use
the
ratio
test
to
identify
the
interval
of
convergence.
228
The
ratio
of
successive
terms
is
a
k
+1
a
k
=
1
(
k
+1)!
x
k
+1
1
k
!
x
k
=
k
!
x
k
+1
(
k
+
1)!
x
k
=
x
k
+
1
lim
k
→∞
x
k
+
1
=
0
This
limit
is
zero
no
matter
what
value
of
x
w
e
cho
ose.
Since
0
<
1
,
the
ratio
test
concludes
that
this
series
converges
fo
r
any
value
of
x
.
In
other
wo
rds,
the
domain
is
all
real
numb
ers.
Synthesis
3.5.3
Is
a
T
aylo
r
Series
Equal
to
the
Function
it
App
ro
ximates?
Let
f
(
x
)
=
ln
x
a
Find
a
pattern
in
the
derivatives
and
write
a
general
exp
ression
fo
r
the
k
th
derivative:
f
(
k
)
(
x
)
.
b
Use
your
answer
to
a
to
write
expressions
for
the
T
a
ylo
r
p
olynomials
T
n
(
x
)
and
the
T
aylo
r
series
T
(
x
)
of
ln
x
centered
at
1
.
Simplify
the
co
efficients
if
p
ossible.
c
What
do
es
the
ratio
test
tell
you
ab
out
where
T
(
x
)
converges?
d
If
we
wanted
to
apply
T
aylo
r’s
inequalit
y
to
T
n
(
x
)
,
we
would
need
to
kno
w
where
the
derivative
is
la
rgest
(in
absolute
value).
Where
is
the
(
n
+
1)
th
derivative
la
rgest
on
the
interval
[
x,
1]
?
(Here
0
<
x
<
1
).
e
Where
is
the
(
n
+
1)
th
derivative
largest
on
the
interval
[1
,
x
]
?
(Here
x
>
1
).
f
What
do
es
T
aylo
r’s
inequality
say
ab
out
where
R
n
(
x
)
→
0
as
n
→
∞
?
g
What
do
es
our
answ
er
to
the
previous
question
tell
us
ab
out
T
(
x
)
?
229
Synthesis
3.5.3
Is
a
T
aylo
r
Series
Equal
to
the
Function
it
App
ro
ximates?
Solution
a
Let’s
compute
some
derivatives
and
see
if
w
e
can
find
an
exp
ression
for
f
(
k
)
(
x
)
f
(
x
)
=
ln(
x
)
f
(1)
=
0
f
′
(
x
)
=
x
−
1
f
′
(1)
=
1
f
′′
(
x
)
=
−
x
−
2
f
′′
(1)
=
−
1
f
′′′
(
x
)
=
2
x
−
3
f
′′′
(1)
=
2
f
(4)
(
x
)
=
−
6
x
−
4
f
(4)
(1)
=
−
6
f
(5)
(
x
)
=
24
x
−
5
f
(5)
(1)
=
24
These
answers
lo
ok
lik
e
factorials,
but
they’re
shifted
by
1
.
They’re
also
alternating
signs,
which
w
e
can
mo
del
with
(
−
1)
k
,
except
that
the
even
p
ow
ers
are
negative.
The
p
ow
er
of
x
is
−
k
.
One
w
a
y
to
mo
del
this
is
f
(
k
)
(
x
)
=
(
−
1)
k
+1
(
k
−
1)!
x
−
k
.
b
Plugging
in
x
=
1
gives
f
(
k
)
(1)
=
(
−
1)
k
+1
(
k
−
1)!
except
at
k
=
0
.
Fo
r
that
case
we
compute
ln
1
=
0
.
This
means
w
e
can
leave
it
out
of
the
summation.
The
form
fo
r
the
remaining
terms
allo
ws
for
some
nice
simplification.
T
(
x
)
=
∞
X
k
=1
(
−
1)
k
+1
(
k
−
1)!
k
!
(
x
−
1)
k
=
∞
X
k
=1
(
−
1)
k
+1
k
(
x
−
1)
k
c
W
e’ll
apply
the
ratio
test
a
k
+1
a
k
=
(
−
1)
k
+2
(
k
+1)
(
x
−
1)
k
+1
(
−
1)
k
+1
k
(
x
−
1)
k
=
(
−
1)
k
+2
k
(
x
−
1)
k
+1
(
−
1)
k
+1
(
k
+
1)(
x
−
1)
k
=
−
k
(
x
−
1)
k
+
1
=
k
|
x
−
1
|
k
+
1
230
No
w
we’ll
solve
fo
r
when
the
limit
of
this
ratio
is
less
than
1
.
lim
k
→∞
k
|
x
−
1
|
k
+
1
<
1
|
x
−
1
|
lim
k
→∞
k
k
+
1
<
1
|
x
−
1
|
<
1
−
1
<
x
−
1
<
1
0
<
x
<
2
The
T
aylo
r
series
converges
on
the
interval
(0
,
2)
.
d
T
o
apply
T
a
ylo
r’s
inequalit
y
to
bound
|
R
n
(
x
)
|
.
We
need
a
bound
on
|
f
(
n
+1)
(
x
)
|
on
the
interval
from
1
to
x
.
Looking
back
at
our
ea
rlier
computation,
we
obtain
f
(
n
+1)
(
x
)
=
(
−
1)
n
+2
n
!
x
−
n
−
1
.
In
this
case
that
x
>
1
,
the
derivative
f
(
n
+1)
decreases
in
magnitude
from
x
to
1
so
it
is
la
rgest
at
x
.
We
can
use
M
=
n
!
x
−
n
−
1
.
e
In
this
case,
f
(
n
+1)
decreases
in
magnitude
from
1
to
x
so
it
is
largest
at
1
.
We
can
use
M
=
n
!
.
f
The
easier
case
is
x
≥
1
.
In
this
case
T
a
ylo
r’s
inequality
states
|
R
n
(
x
)
|
≤
n
!
(
n
+
1)!
(
x
−
1)
n
+1
≤
1
n
+
1
(
x
−
1)
n
+1
As
n
app
roaches
infinit
y
,
this
bound
goes
to
infinit
y
if
x
−
1
>
1
and
to
0
if
x
−
1
≤
1
.
In
the
case
that
0
<
x
<
1
,
we
need
some
clever
algebra
to
write
this
as
a
multiple
of
an
exp
onential.
|
R
n
(
x
)
|
≤
n
!
x
−
n
−
1
(
n
+
1)!
(
x
−
1)
n
+1
≤
1
n
+
1
x
−
1
x
n
+1
This
go
es
to
0
if
x
−
1
x
≤
1
and
infinity
otherwise.
Solving
this
(and
assuming
x
>
0
)
gives
x
≥
1
2
.
Putting
these
together,
we
can
state
that
the
error
bound
from
T
aylo
r’s
inequalit
y
app
roaches
0
as
we
takes
higher
degree
T
aylo
r
p
olynomials,
as
long
as
1
2
≤
x
≤
2
.
g
The
answ
er
to
the
previous
question
tells
us
that
T
(
x
)
converges
to
ln
x
on
1
2
,
2
,
since
the
error
b
ound
and
hence
the
erro
r
go
es
to
0
.
On
the
other
hand,
outside
this
interval,
the
erro
r
might
still
go
to
0
on
0
,
1
2
,
even
though
the
error
b
ound
do
es
not.
The
series
diverges
outside
(0
,
2)
so
it
cannot
converge
to
ln
x
there.
231
Synthesis
3.5.3
Is
a
T
aylo
r
Series
Equal
to
the
Function
it
App
ro
ximates?
Rema
rk
It
turns
out
that
T
(
x
)
=
ln
x
on
(0
,
2]
,
which
is
a
larger
interval
than
we
were
able
to
establish
using
T
a
ylo
r’s
inequalit
y
.
This
should
not
b
other
us.
T
aylo
r’s
inequality
p
ro
duces
a
b
ound
on
the
erro
r.
The
fact
that
the
b
ound
on
the
error
is
going
to
infinit
y
,
do
esn’t
me
an
the
actual
error
do
es.
In
this
case,
fo
r
x
b
etw
een
0
and
1
2
,
the
actual
erro
r
approaches
0
.
Click to Load Applet
Figure:
The
T
aylo
r
p
olynomials
app
roach
ln
x
only
on
(0
,
2]
.
Example
3.5.4
Mixing
T
aylo
r
Series
and
Algeb
ra
Let
f
(
x
)
=
x
2
sin
x
.
Compute
a
T
aylo
r
series
for
f
(
x
)
centered
at
x
=
0
.
Solution
W
e
could
try
to
wo
rk
out
a
pattern
in
the
derivatives
of
f
,
but
even
evaluating
at
x
=
0
the
computations
b
ecome
intractable.
f
′
(
x
)
=
2
x
sin
x
+
x
2
cos
x
f
′′
(
x
)
=
2
sin
x
+
4
x
cos
x
−
x
2
sin
x
f
′′′
(
x
)
=
6
cos
x
−
6
x
sin
x
−
x
2
cos
x
f
(4)
(
x
)
=
−
12
sin
x
−
8
x
cos
x
+
x
2
sin
x
232
Instead
w
e
can
write
the
T
aylo
r
series
fo
r
sin
x
.
Our
ea
rlier
w
o
rk
gave
us
an
expression
for
the
T
a
ylo
r
p
olynomials
and
show
ed
that
their
error
go
es
to
0
as
the
degree
goes
to
infinity
.
sin
x
=
∞
X
k
=0
(
−
1)
k
)
(2
k
+
1)!
x
2
k
+1
W
e
can
obtain
an
expression
fo
r
x
2
sin
x
by
multiplying
b
oth
sides
by
x
2
.
Since
w
e’re
only
multiplying
b
y
a
p
ow
er
of
x
,
the
resulting
series
will
still
b
e
a
p
ow
er
series
centered
at
0
.
x
2
sin
x
=
x
2
∞
X
k
=0
(
−
1)
k
)
(2
k
+
1)!
x
2
k
+1
=
∞
X
k
=0
(
−
1)
k
)
(2
k
+
1)!
x
2
k
+3
Main
Idea
When
constructing
a
T
aylo
r
series
fo
r
f
(
x
)
=
x
k
g
(
x
)
centered
at
0
,
construct
the
T
aylo
r
series
of
g
(
x
)
,
and
then
distribute
the
x
k
.
Example
3.5.5
Integrating
a
T
aylo
r
Series
Let
f
(
x
)
=
e
x
2
.
a
W
rite
a
T
aylo
r
p
olynomial
T
4
(
x
)
for
f
(
x
)
at
x
=
0
.
b
Find
a
b
etter
w
a
y
to
produce
the
T
a
ylo
r
series
for
f
(
x
)
.
c
Compute
a
T
aylo
r
series
for
Z
e
x
2
dx
.
233
Example
3.5.5
Integrating
a
T
aylo
r
Series
Solution
a
W
e
will
compute
the
first
four
derivatives
of
f
(
x
)
.
W
e
will
need
the
chain
rule
and
later
the
p
ro
duct
rule.
f
(
x
)
=
e
x
2
f
(0)
=
1
f
′
(
x
)
=
2
xe
x
2
f
′
(0)
=
0
f
′′
(
x
)
=
2
e
x
2
+
4
x
2
e
x
2
f
′′
(0)
=
2
f
′′′
(
x
)
=
12
xe
x
2
+
8
x
3
e
x
2
f
′′′
(0)
=
0
f
(4)
(
x
)
=
12
e
x
2
+
48
x
2
e
x
2
+
16
x
4
e
x
2
f
(4)
(0)
=
12
W
e
can
plug
these
values
into
our
T
4
(
x
)
formula.
T
4
(
x
)
=
1
0!
x
0
+
0
1!
x
1
+
2
2!
x
2
+
0
3!
x
3
+
12
4!
x
4
=
1
+
x
2
+
1
2
x
4
W
e
can
see
that
our
derivative
calculations
would
quickly
get
out
of
hand
as
we
take
higher
order
derivatives.
Even
if
there
is
a
discernible
pattern,
it
might
take
more
computation
to
determine
it.
b
A
b
etter
approach
is
to
start
with
a
simpler
T
a
ylo
r
series
that
w
e
know.
e
x
=
∞
X
k
=0
1
k
!
x
k
e
x
2
is
a
comp
osition
of
e
x
and
x
2
,
so
we
will
plug
in
x
2
fo
r
x
in
our
e
x
T
a
ylo
r
series.
e
x
2
=
∞
X
k
=0
1
k
!
(
x
2
)
k
=
∞
X
k
=0
1
k
!
x
2
k
c
T
a
ylo
r
series
are
also
pow
er
saeries.
By
our
theorem
on
p
o
w
er
series,
w
e
can
integrate
term
by
term.
Z
e
x
2
dx
=
∞
X
k
=0
=
1
k
!(2
k
+
1)
x
2
k
+1
+
c
Note
that
R
e
x
2
dx
is
not
a
function
we
can
exp
ress
algebraically
or
compute.
A
T
aylo
r
series
gives
us
some
w
a
y
to
rep
resent
this
function,
but
we
shouldn’t
b
e
to
o
satisfied.
If
we
actually
wanted
to
evaluate
it,
the
b
est
we
could
do
is
appro
ximate
it
with
a
partial
sum.
234
Click to Load Applet
Figure:
The
graph
of
e
x
2
,
R
e
x
2
dx
,
and
the
pa
rtial
sums
of
its
T
a
ylo
r
series.
Main
Ideas
Comp
ositions
of
functions
can
b
e
comp
osed
through
T
aylo
r
series.
T
a
ylo
r
series
allow
us
to
integrate
functions
that
are
otherwise
imp
ossible
to
integrate.
Application
3.5.6
Euler’s
Fo
rmula
Recall
i
is
an
imaginary
numb
er
that
satisfies
i
2
=
−
1
.
a
Find
an
expression
for
f
(
x
)
=
e
ix
.
b
W
rite
your
answer
in
terms
of
the
T
aylo
r
series
for
sin
x
and
cos
x
.
c
W
rite
tw
o
different
exp
ressions
for
e
i
2
x
.
How
is
this
equation
useful?
235
Application
3.5.6
Euler’s
Fo
rmula
Solution
a
W
e
can
express
e
ix
b
y
replacing
x
b
y
ix
in
the
T
a
ylo
r
series
for
e
x
.
T
(
x
)
=
∞
X
k
=0
1
k
!
(
ix
)
k
b
T
o
make
much
sense
of
this,
we
should
try
to
simplify
i
k
.
i
0
=
1
i
4
=
1
i
1
=
i
i
5
=
i
i
2
=
−
1
.
.
.
i
3
=
−
i
W
e
can
write
out
the
terms
of
T
(
x
)
as
follo
ws:
T
(
x
)
=
1
+
ix
−
1
2
x
2
−
1
3!
ix
3
+
1
4!
x
4
+
1
5!
ix
5
−
1
6!
x
6
−
1
7!
ix
7
The
terms
with
a
factor
of
i
are
the
T
a
ylo
r
series
for
sin
x
multiplied
by
i
.
The
terms
without
a
facto
r
of
i
a
re
the
T
aylo
r
series
fo
r
cos
x
.
We
can
write
e
ix
=
cos
x
+
i
sin
x
c
One
wa
y
to
write
this
would
b
e
to
substitute
2
x
for
x
:
e
i
2
x
=
cos
2
x
+
i
sin
2
x
Another
wa
y
would
b
e
to
square
our
original
formula.
e
i
2
x
=
e
ix
2
=
(cos
x
+
i
sin
x
)
2
=
cos
2
x
+
2
i
cos
x
sin
x
−
sin
2
x
(
i
2
=
−
1)
Setting
these
equal
to
each
other,
we
note
that
for
tw
o
complex
numb
ers
to
b
e
equal,
their
real
pa
rts
must
b
e
equal
and
their
imaginary
parts
must
be
equal.
cos
2
x
+
i
sin
2
x
=
cos
2
x
+
2
i
cos
x
sin
x
−
sin
2
x
cos
2
x
=
cos
2
x
−
sin
2
x
and
sin
2
x
=
2
cos
x
sin
x
These
are
the
double
angle
formulas
for
sine
and
cosine.
236
W
e
can
tak
e
higher
p
ow
ers
of
e
ix
to
p
roduce
triple
or
quadruple
angle
fo
rmulas.
This
converts
a
difficult
geometry
problem
into
something
a
high
scho
ol
algebra
student
could
compute.
Rema
rk
Y
ou
w
ould
expect
a
relationship
like
this
to
b
e
very
famous,
and
it
is.
e
ix
=
cos
x
+
i
sin
x
is
called
Euler’s
F
o
rmula
.
In
addition
to
trigonometric
formulas,
it
gives
us
insight
into
the
complex
numb
ers.
This
connection
betw
een
an
exponential
and
a
p
erio
dic
function
is
so
p
ow
erful
that
it
is
used
in
such
concrete
applications
as
electrical
engineering
and
signal
p
rocessing.
Section
3.5
Exercises
Summa
ry
Questions
Q1
Ho
w
can
we
b
e
sure
that
a
T
aylo
r
series
converges
to
the
function
it
is
appro
ximating?
Q2
What
is
the
domain
of
a
T
aylo
r
series?
Q3
Ho
w
can
we
p
roduce
the
T
aylo
r
series
for
x
n
f
(
x
)
or
f
(
x
n
)
?
Where
do
es
the
center
need
to
b
e
fo
r
the
result
to
b
e
a
T
a
ylo
r
series?
Q4
What
is
a
Maclaurin
series?
3.5.1
Q5
If
we
w
anted
to
compute
a
decimal
appro
ximation
of
ln(1
.
25)
by
hand,
would
the
T
aylo
r
p
olyno-
mial
or
the
T
a
ylor
series
b
e
mo
re
useful?
Q6
If
T
(
x
)
is
a
T
aylo
r
series
centered
at
x
=
a
,
what
are
the
p
ossible
fo
rms
that
the
domain
of
T
(
x
)
could
take?
237
Section
3.5
Exercises
3.5.2
Q7
Ho
w
would
the
T
a
ylor
series
of
f
(
x
)
=
e
x
change
if
we
centered
it
at
x
=
1
instead
of
x
=
0
?
Q8
Let
T
(
x
)
b
e
the
T
a
ylo
r
series
of
f
(
x
)
=
e
x
centered
at
0
.
Verify
that
T
′
(
x
)
=
T
(
x
)
.
Q9
W
rite
a
T
aylo
r
series
of
f
(
x
)
=
1
x
centered
at
4
.
Q10
W
rite
a
T
aylo
r
series
of
f
(
x
)
=
1
x
2
centered
at
−
5
.
Q11
W
rite
a
T
aylo
r
series
of
f
(
x
)
=
cos
x
centered
at
0
.
Q12
W
rite
a
T
aylo
r
series
of
f
(
x
)
=
sin
x
centered
at
0
.
3.5.3
Q13
Sho
w
that
the
T
aylo
r
series
of
f
(
x
)
=
e
x
centered
at
x
=
0
is
equal
to
f
(
x
)
for
all
real
numb
ers
x
.
Q14
Sho
w
that
the
T
aylo
r
series
of
f
(
x
)
=
sin
x
centered
at
x
=
0
is
equal
to
f
(
x
)
fo
r
all
real
numbers
x
.
Q15
Sho
w
that
the
T
aylo
r
series
of
f
(
x
)
=
1
x
centered
at
4
is
equal
to
f
(
x
)
for
all
x
in
the
interval
(2
,
6)
.
Q16
Supp
ose
for
a
function
f
we
a
re
able
to
place
a
b
ound
of
k
!
3
k
on
the
k
th
derivative
of
f
over
any
interval.
Fo
r
what
values
of
x
can
we
conclude
that
T
(
x
)
,
the
T
aylo
r
series
centered
at
2
,
is
equal
to
f
(
x
)
?
Q17
W
e
didn’t
have
a
series
test
to
determine
whether
∞
X
k
=1
(
−
1)
k
+1
k
converges.
How
do
es
our
analysis
of
the
T
aylo
r
series
of
ln
x
allows
us
to
conclude
that
this
series
converges?
Hint:
what
is
T
(2)
?
Q18
F
o
r
a
general
function
f
and
its
T
aylo
r
p
olynomials
and
series,
how
are
the
following
sets
of
p
oints
related?
Do
es
every
numb
er
b
elonging
to
one
of
these
sets
belong
to
one
of
the
others?
The
set
of
numbers
x
where
T
(
x
)
converges.
The
set
of
numbers
x
where
|
R
n
(
x
)
|
→
0
as
n
→
∞
.
The
set
of
numbers
where
f
(
x
)
=
T
(
x
)
.
238
3.5.4
Q19
W
rite
a
T
aylo
r
series
for
f
(
x
)
=
x
5
cos
x
centered
at
x
=
0
.
Q20
W
rite
a
T
aylo
r
series
for
f
(
x
)
=
x
3
e
x
centered
at
x
=
0
.
Q21
Can
w
e
use
our
T
aylo
r
series
for
f
(
x
)
=
ln
x
centered
at
1
to
write
a
T
aylo
r
series
fo
r
g
(
x
)
=
x
2
ln
x
?
Explain.
Q22
W
rite
a
T
aylo
r
series
for
f
(
x
)
=
(
x
+5)
3
x
2
centered
at
−
5
.
3.5.5
Q23
Let
g
(
x
)
be
an
antiderivative
of
e
x
3
.
Write
the
T
aylo
r
series
for
g
(
x
)
centered
at
x
=
0
.
Q24
Let
g
(
x
)
be
an
antiderivative
of
cos(
x
2
)
.
Write
the
T
aylo
r
series
for
g
(
x
)
centered
at
x
=
0
.
Q25
Let
f
(
x
)
=
cos
x
.
Let
T
(
x
)
b
e
the
T
aylo
r
series
of
f
centered
at
x
=
0
.
Compute
T
′′
(
x
)
.
Why
do
es
your
answer
mak
e
sense?
Q26
W
rite
the
T
aylo
r
series
for
f
(
x
)
=
1
x
centered
at
1
.
V
erify
that
one
of
its
antiderivatives
is
a
T
a
ylo
r
series
for
ln
x
.
3.5.6
Q27
Rewrite
our
formula
for
cos(2
x
)
to
be
entirely
in
terms
of
cos
x
.
Q28
Use
Euler’s
formula
to
compute
a
fo
rmula
for
cos
3
x
in
terms
of
cos
x
and
sin
x
.
Q29
Acco
rding
to
Euler’s
fo
rmula,
what
is
the
value
of
e
2
π
i
?
Q30
Use
the
T
a
ylor
series
of
ln
x
centered
at
x
=
1
to
compute
ln(1
+
i
)
.
Do
you
think
this
series
converges?
239
Section
3.5
Exercises
Synthesis
&
Extension
Q31
Let
h
(
x
)
=
1
x
2
.
a
Compute
the
T
aylo
r
polynomial
T
3
centered
at
x
=
4
.
b
If
you
wanted
to
use
your
T
a
ylo
r
p
olynomial
from
a
to
appro
ximate
1
2
.
5
2
,
what
b
ound
would
T
a
ylo
r’s
inequality
put
on
the
error?
Don’t
simplify
the
arithmetic.
c
What
do
es
the
ratio
test
tell
y
ou
ab
out
the
domain
of
the
T
aylo
r
series
of
h
(
x
)
centered
at
x
=
4
?
Q32
Let
X
b
e
a
normal
random
variable
with
mean
0
and
standard
deviation
1
.
Write
a
series
whose
value
is
P
(0
≤
X
≤
1)
.
Q33
Supp
ose
we
produce
the
T
aylo
r
series
T
(
x
)
for
some
f
(
x
)
centered
at
x
=
10
.
a
If
the
T
aylo
r
series
converges
at
x
=
5
,
must
it
also
converge
at
x
=
7
?
Explain.
b
If
the
errors
of
the
T
a
ylo
r
p
olynomials
T
n
(2)
converge
0
as
n
go
es
to
∞
fo
r
some
x
,
must
T
(2)
converge?
If
T
(2)
converges,
must
the
errors
converge
to
0
?
c
If
you
wanted
to
app
roximate
f
(7)
as
accurately
as
p
ossible,
which
would
b
e
more
useful,
a
T
a
ylo
r
p
olynomial
or
a
T
aylo
r
series?
Q34
Supp
ose
w
e
have
a
function
f
(
x
)
and
tw
o
different
numb
ers
a
and
b
.
Supp
ose
further
that
the
T
a
ylo
r
series
for
f
(
x
)
centered
at
a
is
equal
to
the
T
aylo
r
series
for
f
(
x
)
centered
at
b
.
What
can
you
say
ab
out
the
domain
of
this
T
aylo
r
series?
240
>
Back to Contents