Creation and Manipulation of Taylor Series

Dan KennedyBaylor 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 2

( ) xf x e . (a) Write the first four nonzero terms and the general term of the Taylor series for f about x = 0.

(b) Use your answer to part (a) to find 2

40

1 ( )limx

x f x

x

.

(c) Write the first four nonzero terms of the Taylor series for

2

0

x te dt about x = 0. Use the first two terms of your answer

to estimate 21/ 2

0

te dt .

(d) Explain why the estimate found in part (c) differs from the

actual value of 21/ 2

0

te dt by less than 1

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 2

( ) xf x e . (a) 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 into the Maclaurin series for 2x

.xe

2 3

12! 3! !

nx x x xe x

n

2

3

2

22 2

212! 3!

!

n

xx

xe

n

x

x

Plug in 2x :

which can be simplified to…

2 32 2 2

4 6 2

2

2

12! 3!

( 1)1

2! 3! !

!

n

n n

x x xx

n

x x xx

n

Other popular simplifications of 2

:!

nx

n

2

!

nx

n

2

!

nx

n

2

!

nx

n

Let f be the function given by 2

( ) xf x e .

(b) Use your answer to part (a) to find 2

40

1 ( )limx

x f x

x

.

Bad idea:

Forget part (a). Let’s use L’Hopital’s

rule to find

22

40

1lim .

x

x

x e

x

Dissing the AP Committee is a no-no!

Good idea:

Do what you are told.

22

4 4

4 62 2

4

2

1 ( ) 11 series in (a)

11 1

2 3!

1stuff

2

x f xx

x x

x xx x

x

x

2

240 0

1 ( ) 1lim lim stuff

2

1

2

x x

x f xx

x

Thus:

All this was worth 1 point out of 9.

Let f be the function given by 2

( ) xf x e . (c) Write the first four nonzero terms of the Taylor series for

2

0

x te dt about x = 0. Use the first two terms of your answer

to estimate 21/ 2

0

te dt .

Baaaad idea:

Start by antidifferentiating 2

.te

Good idea:

Import your series from part (a).

By the FTC, is the antiderivative

of that equals 0 at x = 0.

2

0

x te dt2xe

24 6 2

2 ( 1)1

2! 3! !

n nx x x x

e xn

From part (a):

2

5

3 5

3 7

7

0

3

3 10

4

2

10 2

4

x t x x xe dt C

x x xx

x

So:

Using the first two terms of this series, we estimate:

21/ 2

0

1 1 1

2 3 8

11

24te dt

Notice that the next term of the series approximation would have been

51

320

1 1

10 2

This number plays a big role in (d).

(d) Explain why the estimate found in part (c) differs from the

actual value of 21/ 2

0

te dt by less than 1

200.

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!

Quel

imbecile!

The series in (c) for

is an

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:

21/ 2

0

te dt

21/ 2

0

11 1 1.

24 320 200te dt

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. Convergence tests for series of constants 2. Constructing Taylor series3. Intervals of convergence

For example, consider BC-4 from 1979…

Let f be the function defined by 1

( )1 2

f xx

.

(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.

(c) Find the value of f at x = 1

4 . How many terms of the series are

adequate for approximating 1

4f

with an error not exceeding

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 justifies the answer in one step!1r

Today’s calculus students would (I hope)

recognize as a variation of ,

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.

1

1 2x 1

a

r

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

series 2 3

( ) 1 .2! 3! !

nx x xf x x

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 students will guess that it is .

You hope someone will realize why it must be .

xe

xe

Recall that 2

1

aa ar ar

r

(geometric series)

For what r is this equation valid?

Find a series for 2

1

1 x . For what x is it valid?

Use the previous series to get a series for 1tan 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 that they can use tan 0 to find it!

1

2 4 6 22

3 5 7 2

3 5 7 2 11

11

1

tan ( 1)3

11 ( 1)

1

tan ( 1)3 5

5 7 2

7 2 1

tan 0 0 0

1

n n

nn

nn

x x x xx

x x x xx C x

n

C

x x x xx x

n

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

11

1 1 1 1tan 1 1 ( 1)

3 5 7 2 1

1 1 1 ( 1)tan ( 1) 1

3 5 7 2 1

n

n

n

n

1

1

3

2

3

1

5

13

15

Limit = tan 11

Construct a fifth-degree polynomial P such that:

(4)

(5)

(0) 2

(0) 3

(0) 5

(0) 7

(0) 11

(0) 13

P

P

P

P

P

P

Students can generally do this. They will discover that the polynomial is:

2 3 4 55 7 11 132 3

2! 3! 4! 5!x x x x x

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,

(4) (4)

(5) (5)

(0) sin(0)

(0) sin (0)

(0) sin (0)

(0) sin (0)

(0) sin (0)

(0) sin (0)

P

P

P

P

P

P

Now they’re ready for the nicest little exploration in the course!

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:

3

3 5

3 5 7

3!

3! 5!

3! 5! 7!.

x

x x

x x x

etc

y x

y x

y x

y x

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