Identify the changing parts. We see an alternating
±
, so we will get (

1)
k
or (

1)
k
+1
. If
k
= 0, the
term is positive, so we take (

1)
k
. In the denominator, we have a power of 2. For the 0th derivative,
the power is 2.
In the first derivative, the power is 3, so in each case, the power is one more than
the derivative we’re taking.
So we get 2
k
+2
in the denominator.
On top, we see factorials.
In the
second derivative, we have 3!. In the third derivative, it is 4!, so in general, it will be (
k
+ 1)!. So,
f
(
k
)
(1) =
(

1)
k
(
k
+1)!
2
k
+2
. Using the general formula for the Taylor formula, we get,
T
n
(
x
) =
n
X
k
=0
f
(
k
)
(1)
k
!
(
x

1)
k
=
n
X
k
=0
(

1)
k
(
k
+ 1)!
2
k
+2
·
1
k
!
(
x

1)
k
=
n
X
k
=0
(

1)
k
(
k
+ 1)
2
k
+2
(
x

1)
k
.
Example 19.4.
If
f
(
x
) = sin(2
x
)
, find
T
3
(
x
)
centered at
a
=
π/
4
, and then find
T
n
(
x
)
for any
n
.
Solution.
(a) To find
T
3
(
x
), first take three derivatives of
f
(
x
):
f
0
(
x
) = 2 cos(2
x
)
,
f
00
(
x
) =

2
2
sin(2
x
)
,
f
000
(
x
) =

2
3
cos(2
x
)
.
So at
a
=
π/
4, we have
f
(
π/
4) = sin(
π/
2) = 1
,
f
0
(
π/
4) = 0
,
f
00
(
π/
4) =

4
,
f
000
(
π/
4) = 0
,
since cos(
π/
2) = 0. So
T
3
(
x
) =
f
(
π/
4)+
f
0
(
π/
4)(
x

π/
4)+
f
00
(
π/
4)
2!
(
x

π/
4)
2
+
f
000
(
π/
4)
3!
(
x

π/
4)
3
= 1

4
2!
(
x

π/
4)
2
= 1

2(
x

π/
4)
2
.
(b) To find
T
n
(
x
), we need a formula for
f
(
n
)
(
π/
4). We have three derivatives above, and already we notice
that every other derivative will be 0. We can calculate a couple more derivatives:
f
(4)
(
x
) = 2
4
sin(2
x
)
,
f
(5)
(
x
) = 2
5
cos(2
x
)
,
f
(6)
(
x
) =

2
6
sin(2
x
)
,
and so on. So at
a
=
π/
4, the sequence of derivatives we have is:
1
,
0
,

2
2
,
0
,
2
4
,
0
,

2
6
,
0
,
. . . .
Now, you have to deal with the fact that we only have nonzero terms in the even slots (namely,
k
= 0
,
2
,
4
,
6
, . . .
). Even numbers have the form 2
k
, for
k
an integer, and odd numbers have the form
2
k
+ 1, where
k
is some integer. The only nonzero derivatives come from
f
(2
k
)
(
π/
4), so let’s focus on
the even terms:
1
,

2
2
,
2
4
,

2
6
,
. . . .
We see the alternating
±
, so we’re going to get a (

1)
k
or (

1)
k
+1
.
If
k
= 0, we have a positive
term, so that tells us to take the (

1)
k
. We then see we have 2 to an appropriate power. In the 0th
derivative, the power is 0. In the second derivative, the power is 2, in the fourth derivative the power
is 4, so in the 2
k
th derivative, the power will just be 2
k
. So
f
(2
k
)
(
π/
4) = (

1)
k
·
2
2
k
.
Again, all the odd derivatives are 0 at
π/
4. So instead of giving the polynomial
T
n
(
x
), we can give the
polynomials
T
2
n
(
x
) and
T
2
n
+1
(
x
). The formula for
T
2
n
(
x
) is
T
2
n
(
x
) =
f
(
π/
4) +
f
0
(
π/
4)(
x

π/
4) +
f
00
(
π/
4)
2!
(
x

π/
4)
2
+
. . .
+
f
(2
n
)
(
π/
4)
(2
n
)!
(
x

π/
4)
2
n
.
51
Math 31B Notes
Sudesh Kalyanswamy
All the odd terms are 0, and we have a formula for
f
(2
n
)
(
π/
4), namely
f
(2
n
)
(
π/
4) = (

1)
n
·
2
2
n
. So
T
2
n
(
x
) = 1

2(
x

π/
4)
2
+
. . .
+
(

1)
n
·
2
2
n
(2
n
)!
(
x

π/
4)
2
n
.
This only gives the even order Taylor polynomials.
We should also give the odd order ones.
But
the next Taylor expansion, namely
T
2
n
+1
(
x
), will be the same as
T
2
n
(
x
) since
T
2
n
+1
(
x
) =
T
2
n
(
x
) +
f
(2
n
+1)
(
π/
4)
(2
n
+1)!
(
x

π/
4)
2
n
+1
and
f
(2
n
+1)
(
π/
4) = 0, so
T
2
n
(
x
) =
T
2
n
+1
(
x
)
.
So this describes every Taylor polynomial.