Creation and
Manipulation of
Taylor Series
Dan Kennedy
Baylor School
Chattanooga, TN
dkennedy@baylorschool.org
The recommended impetus for this
session is free-response question 6 from
the 2007 BC exam:
Let f be the function given by f ( x)  e x .
2
(a) Write the first four nonzero terms and the general term of the
Taylor series for f about x = 0.
1  x 2  f ( x)
(b) Use your answer to part (a) to find lim
.
4
x0
x
(c) Write the first four nonzero terms of the Taylor series for

x
0
t 2
e dt about x = 0. Use the first two terms of your answer
to estimate

1/ 2
0
t 2
e dt .
(d) Explain why the estimate found in part (c) differs from the
1/ 2
1
t 2
actual value of  e dt by less than
.
0
200
The BC students in 2007
did not exactly ace this
one.
The mean score was 2.28 out of 9.
A mere 0.3% of the students earned 9’s.
Approximately 40.6% of the students
earned no points at all.
Let f be the function given by f ( x)  e
(a)
 x2
.
Write the first four nonzero terms and the general term of the
Taylor series for f about x = 0.
Bad idea:
Find the first three derivatives of f and
use them to build the series.
Good idea:
Plug
x
e .
x
2
into the Maclaurin series for
2
3
n
x
x
e  1 x   
2! 3!
x
 
n!
x
Plug in
e
x
2
x
2
:
 1 x
x 


2
2 n
n!


x  x 



2 2
2!
2 3
3!
which can be
simplified to…

1 x
2

x  x 



2 2
2!
4
6
x
x
2
 1 x   
2! 3!
2 3
x 


2 n

3!
n!
n 2n
(1) x


n!
2
x
Other popular simplifications of
x
n!
2n
x
n!
2n

n!
2n
x

n!


n
:
Let f be the function given by f ( x)  e
(b)
 x2
.
1  x 2  f ( x)
Use your answer to part (a) to find lim
.
4
x 0
x
Bad idea:
Forget part (a). Let’s use L’Hopital’s
1 x  e
lim
4
x 0
x
2
rule to find
 x2
.
Dissing the AP Committee is a no-no!
Good idea:
Do what you are told.
1  x  f ( x) 1
2
 4 1  x   series in (a)  
4
x
x
4
6



1
x
x
2
2
 4 1  x  1  x     
x 
2 3!


1
2
   x  stuff 
2
2
Thus:
 1  x  f ( x) 
 1

2
lim 
 lim    x  stuff  

4
x 0
x 0
x
 2



1

2
2
All this was worth 1 point out of 9.
Let f be the function given by f ( x)  e
 x2
.
(c) Write the first four nonzero terms of the Taylor series for

x
0
e  t dt about x = 0. Use the first two terms of your answer
2
to estimate

1/ 2
0
t 2
e dt .
Baaaad idea:
Start by antidifferentiating
t 2
e .
Good idea:
Import your series from part (a).
By the FTC,
of
e
x
2

x
0
t 2
is the antiderivative
e dt
that equals 0 at x = 0.
From part (a):
e
 x2
So:

x
0
4
6
x
x
 1 x   
2! 3!
2
3
(1) x

n!
n
5
7
x
x
x
e dt  C  x    
3 10 42
3
5
7
x
x
x
 x   
3 10 42
t 2
2n

Using the first two terms of this
series, we estimate:

1/ 2
0
1  1  1  11
e dt      
2  3  8  24
t 2
Notice that the next term of the
series approximation would have
been
5
1
 1  1 
   
 10  2  320
This number plays a big role in (d).
(d) Explain why the estimate found in part (c) differs from the
1/ 2
1
t 2
actual value of  e dt by less than
.
0
200
Quel
imbecile!
Baaad idea:
Try arguing your case using
the Lagrange error bound.
Good idea:
Use the bound associated with the
Alternating Series Test.
Be sure to justify that it applies here!
The series in (c) for
is an

1/ 2
0
t 2
e dt
alternating series of terms that
decrease in absolute value with a
limit of zero. Thus, the truncation
error after two terms is less than
the absolute value of the third term:

1/ 2
0
11
1
1
e dt 


.
24 320 200
t 2
So let’s talk about series manipulation.
In the old days (pre-1989), the approach
to series in a calculus class was quite
different from what it is today (as with
many other topics).
1.
2.
3.
Convergence tests for series of
constants
Constructing Taylor series
Intervals of convergence
For example, consider BC-4 from 1979…
Let f be the function defined by f ( x ) 
1
.
1  2x
(a) Write the first four terms and the general term of the Taylor
series expansion of f(x) about x – 0.
(b) What is the interval of convergence for the series found in part (a)?
Justify your answer.
1
(c) Find the value of f at x =  . How many terms of the series are
4
 1
adequate for approximating f    with an error not exceeding
 4
one per cent? Justify your answer.
Notice that the series is geometric.
Here is the actual grading standard for
part (a), which was worth 5 points out of
15. Notice that it was assumed that
students would build the series using nth
derivatives.
Since it was 1979, they all did.
The grading standard for part (b)
assumed the standard Ratio Test with
endpoint analysis. There is no visible
acknowledgment that r  1 justifies
the answer in one step!
Today’s calculus students would (I hope)
1
a
recognize
as a variation of
,
1  2x
1 r
but the students in 1979 apparently did
not.
How did our students get better????
We need to reflect on these beneficial
changes occasionally … if only to
remind ourselves that the good old days
were actually not all that great.
What happened was a new emphasis on
series as functions … the real reason
for having them in the course.
In fact, many teachers are talking
about series as functions long before
they talk about convergence tests for
series of constants.
What follow are some of my favorite
student explorations …
 Consider the function f defined by the infinite
x 2 x3
xn
series f ( x)  1  x 

 
 .
2! 3!
n!
(a) Find f(0).
(b) Find f ( x) . What is interesting about it?
(c) What can you conclude about the function f ?
When they find that this function
is its own derivative, most
x
students will guess that it is e .
You hope someone
will realize
x
why it must be e .
a
 a  ar  ar 2  (geometric series)
1 r
For what r is this equation valid?
 Recall that
 Find a series for
1
. For what x is it valid?
2
1 x
 Use the previous series to get a series for tan 1 x .
For what x is it valid?
The third task is particularly rich.
Students might forget the constant of
integration. When reminded, they’ll
cheerfully add it. Then remind them
1
that they can use tan 0 to find it!
1
2
4
6
n 2n

1

x

x

x


(

1)
x 
2
1 x
3
5
7
2 n 1
x
x
x
1
n x
tan x  C  x      ( 1)

3 5 7
2n  1
1
tan 0  0  C  0
3
5
7
x
x
x
tan 1 x  x    
3 5 7
2 n 1
x
 ( 1) n

2n  1
This began with a series that was valid
for -1 < x < 1. An interesting postscript is
that we also get convergence at -1 and 1.
1 1 1
1
n
tan 1  1      ( 1)

3 5 7
2n  1
n 1
1 1 1
( 1)
1
tan ( 1)  1     

3 5 7
2n  1
1
11

53
2
3
13
15
1
Limit = tan
1
1
 Construct a fifth-degree polynomial P such that:
P (0)  2
P(0)  3
P(0)  5
P(0)  7
P (4) (0)  11
P (5) (0)  13
Students can generally do this. They will
discover that the polynomial is:
5 2 7 3 11 4 13 5
2  3x  x  x  x  x
2!
3!
4!
5!
Now they’re ready for the nicest
little exploration in the course!
 Construct a fifth-degree polynomial P such that P and its
first four derivatives match sin x and its first four
derivatives at x = 0. That is,
P (0)  sin(0)
P(0)  sin(0)
P(0)  sin(0)
P(0)  sin(0)
P (4) (0)  sin (4) (0)
P (5) (0)  sin (5) (0)
One of the most powerful visualizations
in mathematics is the spectacle of the
convergence of Taylor series. Here are
the Taylor polynomials for sin x about
x = 0:
yx
x3
y  x
3!
3
5
x x
y  x 
3! 5!
3
5
7
x x x
y  x  
3! 5! 7!
etc.
Here are their graphs, superimposed on
the graph of y = sin x:
Coming next year
from the College
Board:
The 2008-2009 AP Calculus
Focus Materials
The topic: Infinite Series!
dkennedy@baylorschool.org