In section 11.9, we were able to find power series representations for a certain restricted class of functions. Here, we investigate more general problems

In section 11.9, we were able to find power series

representations for a certain restricted class of

functions.

Here, we investigate more general problems.

Which functions have power series representations?

How can we find such representations?

INFINITE SEQUENCES AND SERIES

11.10

Taylor and Maclaurin Series

INFINITE SEQUENCES AND SERIES

In this section, we will learn:

How to find the Taylor and Maclaurin Series of a

function and to multiply and divide a power series.

TAYLOR & MACLAURIN SERIES

We start by supposing that f is any function that

can be represented by a power series 2

0 1 2

3 43 4

( ) ( ) ( )

( ) ( ) ... | |

f x c c x a c x a

c x a c x a x a R

Equation 1


Let us try to determine what the coefficients cn

must be in terms of f. 2

0 1 2

3 43 4

0

( ) ( ) ( )

( ) ( ) ... | |

Now notice that,

( )

f x c c x a c x a

c x a c x a x a R

f a c


By Theorem 2 in Section 11.9, we can differentiate

the series in Equation 1 term by term:

21 2 3

34

1

( ) 2 ( ) 3 ( )

4 ( ) ... | |

Now notice that,

( )

f x c c x a c x a

c x a x a R

f a c

Equation 2


Now, we differentiate both sides of Equation 2 and

obtain:

2 3

24

2

( ) 2 2 3 ( )

3 4 ( ) ... | |

Now notice that,

( ) 2

f x c c x a

c x a x a R

f a c

Equation 3


Let us apply the procedure one more time.


Differentiation of the series in Equation 3 gives:

3 4

25

3

( ) 2 3 2 3 4 ( )

3 4 5 ( ) ... | |

Now notice that,

( ) 2 3

f x c c x a

c x a x a R

f a c

Equation 4


By now, you can see the pattern.

If we continue to differentiate and substitute x = a, we obtain:

( ) ( ) 2 3 4 !nn nf a nc n c


Solving the equation for the nth coefficient cn, we

get:

( ) ( )

!

n

n

f ac

n


The formula remains valid even for n = 0 if we

adopt the conventions that 0! = 1 and f (0) = f .

Thus, we have proved the following theorem.


If f has a power series representation (expansion)

at a, that is, if

then its coefficients are given by:

Theorem 5

0

( ) ( ) | |nn

n

f x c x a x a R

( ) ( )

!

n

n

f ac

n


Substituting this formula for cn back into the series,

we see that if f has a power series expansion at a,

then it must be of the following form.

Equation 6

TAYLOR SERIES Equation 6

( )

0

2

3

( )( ) ( )

!

( ) ( )( ) ( ) ( )

1! 2!( )

( )3!

nn

n

f af x x a

n

f a f af a x a x a

f ax a

This series, Equation 6, is called the Taylor series

of the function f at a (or about a or centered at a).

TAYLOR SERIES

For the special case a = 0, the Taylor series

becomes:( )

0

2

(0)( )

!

'(0) ''(0)(0)

1! 2!

nn

n

ff x x

n

f ff x x

Equation 7

MACLAURIN SERIES

This case arises frequently enough that it is given

the special name Maclaurin series.

Equation 7

( )

0

2

(0)( )

!

'(0) ''(0)(0)

1! 2!

nn

n

ff x x

n

f ff x x


The Taylor series is named after the English

mathematician Brook Taylor (1685–1731).

The Maclaurin series is named for the Scottish

mathematician Colin Maclaurin (1698–1746).

This is despite the fact that the Maclaurin series is really just a special case of the Taylor series.

MACLAURIN SERIES

Maclaurin series are named after Colin Maclaurin

because he popularized them in his calculus

textbook Treatise of Fluxions published in 1742.


We have shown that if, f can be represented as a

power series about a, then f is equal to the sum of

its Taylor series.

However, there exist functions that are not equal to the sum of their Taylor series.

An example is given in Exercise 70.

Note


Find the Maclaurin series of the function f (x) = ex

and its radius of convergence.

Example 1


If f (x) = ex, then f (n)(x) = e x. So, f (n)(0) = e0 = 1 for

all n.

Hence, the Taylor series for f at 0 (that is, the Maclaurin series) is:

Example 1

( ) 2 3

0 0

(0)1

! ! 1! 2! 3!

n nn

n n

f x x x xx

n n


To find the radius of convergence, we let an = xn/n!

Then,

So, by the Ratio Test, the series converges for all x and the radius of convergence is R = .

11 ! | |

0 1( 1)! 1

nn

nn

a x n x

a n x n


The conclusion we can draw from Theorem 5 and

Example 1 is:

If ex has a power series expansion at 0, then

0 !

nx

n

xe

n


So, how can we determine whether e x does have a

power series representation?

Let us investigate the more general question:

Under what circumstances is a function equal to the sum of its Taylor series?


In other words, if f has derivatives of all orders,

when is the following true?

( )

0

( )( ) ( )

!

nn

n

f af x x a

n


As with any convergent series, this means that f (x)

is the limit of the sequence of partial sums.


In the case of the Taylor series, the partial sums

are:( )

0

2

( )

( )( ) ( )

!

'( ) ''( )( ) ( ) ( )

1! 2!

( )( )

!

ini

ni

nn

f aT x x a

i

f a f af a x a x a

f ax a

n

nTH-DEGREE TAYLOR POLYNOMIAL OF f AT a

Notice that Tn is a polynomial of degree n called

the nth-degree Taylor polynomial of f at a.


For instance, for the exponential function f (x) = ex,

the result of Example 1 shows that the Taylor

polynomials at 0 (or Maclaurin polynomials) with

n = 1, 2, and 3 are: 2

1 2

2 3

3

( ) 1 ( ) 12!

( ) 12! 3!

xT x x T x x

x xT x x


The graphs of the exponential function and those

three Taylor polynomials are drawn here.


In general, f (x) is the sum of its Taylor series if:

( ) lim ( )nn

f x T x

REMAINDER OF TAYLOR SERIES

If we let Rn(x) = f (x) – Tn(x)

so that f (x) = Tn(x) + Rn(x)

then Rn(x) is called the remainder of the Taylor

series.


If we can somehow show that , then

it follows that:

Therefore, we have proved the following.

lim ( ) 0nnR x

lim ( ) lim[ ( ) ( )]

( ) lim ( )

( )

n nn n

nn

T x f x R x

f x R x

f x


If f (x) = Tn(x) + Rn(x), where Tn is the nth-degree

Taylor polynomial of f at a and

for |x – a| < R, then f is equal to the sum of its

Taylor series on the interval |x – a| < R.

Theorem 8

lim ( ) 0nn

R x


In trying to show that for a specific

function f, we usually use the following fact.

lim ( ) 0nn

R x

TAYLOR’S INEQUALITY

If | f (n+1)(x)| M for |x – a| d, then the remainder

Rn(x) of the Taylor series satisfies the inequality

Theorem 9

1| ( ) | | | for | |( 1)!

nn

MR x x a x a d

n


In Section 11.11, we will explore the use of

Taylor’s Inequality in approximating functions.

Our immediate use of it is in conjunction with

Theorem 8.

Note


This is true because we know from Example 1 that the series ∑ xn/n! converges for all x, and so its nth term approaches 0.

lim 0 for every real number!

n

n

xx

n

Equation 10

In applying Theorems 8 and 9, it is often helpful to

make use of the following fact:


Prove that ex is equal to the sum of its Maclaurin

series.

If f(x) = ex, then f (n+1)(x) = ex for all n.

If d is any positive number and |x| d, then | f (n+1)(x)| = ex ed.

Example 2


So, Taylor’s Inequality, with a = 0 and M = ed,

says that:

Notice that the same constant M = ed works for every value of n.

1| ( ) | | | for | |( 1)!

dn

n

eR x x x d

n

Example 2


However, from Equation 10, we have:

It follows from the Squeeze Theorem that

and so for all values of x.

11 | |

lim | | lim 0( 1)! ( 1)!

d nn d

n n

e xx e

n n

lim ( ) 0nnR x

Example 2

lim | ( ) | 0nnR x


By Theorem 8, ex is equal to the sum of its

Maclaurin series, that is,

0

for all!

nx

n

xe x

n

E. g. 2—Equation 11


In particular, if we put x = 1 in Equation 11, we

obtain the following expression for the number e

as a sum of an infinite series:

0

1 1 1 11

! 1! 2! 3!n

en

Equation 12


Find the Taylor series for f(x) = ex at a = 2.

We have f (n)(2) = e2. So, putting a = 2 in the definition of a Taylor series

(Equation 6), we get:

Example 3

( ) 22 2

0 0

(2)( 2) ( 2)

! !

n

n n

f ex x

n n


Again it can be verified, as in Example 1, that the

radius of convergence is R = .

As in Example 2, we can verify that

lim ( ) 0nn

R x



Thus, 2

0

( 2) for all!

x n

n

ee x x

n



We have two power series expansions for ex, the

Maclaurin series in Equation 11 and the Taylor

series in Equation 13.

The first is better if we are interested in values of x near 0.

The second is better if x is near 2.


Find the Maclaurin series for sin x and prove that it

represents sin x for all x. The first few derivatives

are

Example 4

(4) (4)

( ) sin (0) 0

( ) cos (0) 1

( ) sin (0) 0

( ) cos (0) 1

( ) sin (0) 0

f x x f

f x x f

f x x f

f x x f

f x x f


As the derivatives repeat in a cycle of four, we can

write the Maclaurin series as follows:

2 3

3 5 7

2 1

0

'(0) ''(0) '''(0)(0)

1! 2! 3!

3! 5! 7!

( 1)(2 1)!

nn

n

f f ff x x x

x x xx

x

n

Example 4


Since f (n+1)(x) is sin x or cos x, we know that

| f (n+1)(x)| 1 for all x.

Example 4


So, we can take M = 1 in Taylor’s Inequality:

11 | |

| ( ) | | |( 1)! ( 1)!

nn

n

M xR x x

n n



By Equation 10, the right side of that inequality

approaches 0 as n → .

So, |Rn(x)| → 0 by the Squeeze Theorem.

It follows that Rn(x) → 0 as n → .

So, sin x is equal to the sum of its Maclaurin series by Theorem 8.

Example 4


We state the result of Example 4 for future

reference. 3 5 7

2 1

0

sin3! 5! 7!

( 1) for all(2 1)!

nn

n

x x xx x

xx

n

Equation 15


The figure shows the graph of sin x together with

its Taylor (or Maclaurin) polynomials

1

3

3

3 5

5

( )

( )3!

3! 5!

T x x

xT x x

x xT x


Notice that, as n increases, Tn(x) becomes a better

approximation to sin x.


Find the Maclaurin series for cos x.

We could proceed directly as in Example 4.

However, it’s easier to differentiate the Maclaurin series for sin x given by Equation 15, as follows.

Example 5

TAYLOR & MACLAURIN SERIES Example 5

3 5 7

2 4 6

4 62

cos (sin )

3! 5! 7!

3 5 71

3! 5! 7!

12! 4! 6!

dx x

dx

d x x xx

dx

x x x

x x x


The Maclaurin series for sin x converges for all x.

So, Theorem 2 in Section 11.9 tells us that the differentiated series for cos x also converges for all x.

Example 5


Thus,

2 4 6

2

0

cos 12! 4! 6!

( 1) for all(2 )!

nn

n

x x xx

xx

n



The Maclaurin series for ex, sin x, and cos x that we

found in Examples 2, 4, and 5 were discovered by

Newton.

These equations are remarkable because they say we know everything about each of these functions if we know all its derivatives at the single number 0.


Find the Maclaurin series for the function

f (x) = x cos x.

Instead of computing derivatives and substituting in Equation 7, it is easier to multiply the series for cos x (Equation 16) by x:

Example 6

2 2 1

0 0

cos ( 1) ( 1)(2 )! (2 )!

n nn

n n

x xx x x

n n


Represent f (x) = sin x as the sum of its Taylor

series centered at /3.

Example 7


Arranging our work in columns, we have:

3( ) sin

3 2

1( ) cos

3 2

3( ) sin

3 2

1( ) cos

3 2

f x x f

f x x f

f x x f

f x x f

Example 7


That pattern repeats indefinitely.

Example 7


Thus, the Taylor series at /3 is:

2

3

2 3

3 33 1! 3 2! 3

33! 3

3 1 3 1

2 2 1! 3 2 2! 3 2 3! 3

f ff x x

fx

x x x

Example 7


The proof that this series represents sin x for all x

is very similar to that in Example 4.

Just replace x by x – /3 in Equation 14.

Example 7


We can write the series in sigma notation if we

separate the terms that contain :

2

0

2 1

0

( 1) 3sin

2(2 )! 3

( 1)

2(2 1)! 3

nn

n

nn

n

x xn

xn

Example 7

3


We have two different series representations for

sin x, the Maclaurin series in Example 4 and the

Taylor series in Example 7.

It is best to use the Maclaurin series for values of x

near 0 and the Taylor series for x near π/3.


Notice that the third Taylor polynomial T3 in the

figure is a good approximation to sin x near /3 but

not as good near 0.


Compare it with the third Maclaurin polynomial T3

in the earlier figure, where the opposite is true.


The power series that we obtained by indirect

methods in Examples 5 and 6 and in Section 11.9

are indeed the Taylor or Maclaurin series of the

given functions.


That is because Theorem 5 asserts that, no matter

how a power series representation

f (x) = cn(x – a)n

is obtained, it is always true that cn = f (n)(a)/n!

In other words, the coefficients are uniquely determined.


Find the Maclaurin series for f(x) = (1 + x)k, where

k is any real number.

Example 8


Arranging our work in columns, we have:

1

2

3

( ) ( )

( ) (1 ) (0) 1

'( ) (1 ) '(0)

''( ) ( 1)(1 ) ''(0) ( 1)

'''( ) ( 1)( 2)(1 ) '''(0) ( 1)( 2)

( 1) ( 1)(1 ) (0) ( 1) ( 1)

k

k

k

k

n k n n

f x x f

f x k x f k

f x k k x f k k

f x k k k x f k k k

f k k k n x f k k k n

Example 8

BINOMIAL SERIES

Thus, the Maclaurin series of f (x) = (1 + x)k is:

This series is called the binomial series.

( )

0 0

(0) ( 1) ( 1)

! !

nn n

n n

f k k k nx x

n n

Example 8


If its nth term is an, then

1

1( 1) ( 1)( ) !

( 1)! ( 1) ( 1)

1| |

| | | | | | as11 1

n

n

n

n

a

a

k k k n k n x n

n k k k n x

kk n n

x x x nn

n

Example 8


Therefore, by the Ratio Test, the binomial series

converges if |x| < 1 and diverges if |x| > 1.

Example 8

BINOMIAL COEFFICIENTS.

The traditional notation for the coefficients in the

binomial series is:

These numbers are called the binomial coefficients.

( 1)( 2) ( 1)

!

k k k k k n

n n


The following theorem states that (1 + x)k is equal

to the sum of its Maclaurin series.

It is possible to prove this by showing that the remainder term Rn(x) approaches 0.

That, however, turns out to be quite difficult.

The proof outlined in Exercise 71 is much easier.

THE BINOMIAL SERIES

If k is any real number and |x| < 1, then

Theorem 17

0

2

3

(1 )

( 1)1

2!( 1)( 2)

3!

k n

n

kx x

n

k kkx x

k k kx


Though the binomial series always converges

when |x| < 1, the question of whether or not it

converges at the endpoints, 1, depends on the

value of k.

It turns out that the series converges at 1 if -1 < k 0 and at both endpoints if k 0.


Notice that, if k is a positive integer and n > k, then

the expression for contains a factor (k – k).

So, for n > k.

This means that the series terminates and reduces to the ordinary Binomial Theorem when k is a positive integer.

0kn

kn


Find the Maclaurin series for the function

and its radius of convergence.

Example 9

1( )

4f x

x


We write f (x) in a form where we can use the

binomial series:

1/ 2

1 1

44 1

4

1 11

2 42 1

4

x x

x

x

Example 9


Using the binomial series with k = –½ and with x

replaced by –x/4, we have:

1/ 2

12

0

1 11

2 44

1

2 4

n

n

x

x

x

n

Example 9


2312 2

33 512 2 2

3 51 12 2 2 2

1 11

2 2 4 2! 4

3! 4

1

! 4

n

x x

x

n x

n

Example 9


We know from Theorem 17 that this series converges when |–x/4| < 1, that is, |x| < 4.

So, the radius of convergence is R = 4.

2 32 3

1 1 1 3 1 3 51

2 8 2!8 3!8

1 3 5 (2 1)

!8n

n

x x x

nx

n

Example 9


For future reference, we collect some important

Maclaurin series that we have derived in this

section and Section 11.9, in the following table.

IMPORTANT MACLAURIN SERIES

2 3

0

2 3

0

11 1

1

1! 1! 2! 3!

n

n

nx

n

x x x x Rx

x x x xe R

n

Table 1


2 1 3 5 7

0

2 2 4 6

0

2 1 3 5 71

0

sin ( 1)(2 1)! 3! 5! 7!

cos ( 1) 1(2 )! 2! 4! 6!

tan ( 1) 12 1 3 5 7

nn

n

nn

n

nn

n

x x x xx x R

n

x x x xx R

n

x x x xx x R

n

Table 1


2

0

3

( 1)(1 ) 1

2!

( 1)( 2)1

3!

k n

n

k k kx x kx x

n

k k kx R

Table 1

USES OF TAYLOR SERIES

One reason Taylor series are important is that they

enable us to integrate functions that we could not

previously handle.


In fact, in the introduction to this chapter, we

mentioned that Newton often integrated functions

by first expressing them as power series and then

integrating the series term by term.


The function f (x) = ex2 can not be integrated by

techniques discussed so far.

Its antiderivative is not an elementary function (see Section 7.5).

In the following example, we use Newton’s idea to integrate this function.


a. Evaluate as an infinite series.

b. Evaluate correct to within an error of

0.001

Example 10

1 2

0

xe dx

2xe dx


First, we find the Maclaurin series for f(x) = e-x2

It is possible to use the direct method.

However, let’s find it simply by replacing x with –x2 in the series for ex given in Table 1.

Example 10 a


Thus, for all values of x,

Example 10 a

22

0

2

0

2 4 6

( )

!

( 1)!

1 ...1! 2! 3!

nx

n

nn

n

xe

n

x

n

x x x


Now, we integrate term by term:

This series converges for all x because the original series

for e-x2 converges for all x.

22 4 6 2

3 5 7

2 1

1 ( 1)1! 2! 3! !

3 1! 5 2! 7 3!

( 1)(2 1) !

nx n

nn

x x x xe dx dx

n

x x xC x

x

n n

Example 10 a


The FTC gives:

Example 10 b

2

13 5 7 91

00

1 1 1 13 10 42 216

1 1 1 13 10 42 216

3 1! 5 2! 7 3! 9 4!

1

1

0.7475

x x x x xe dx x


The Alternating Series Estimation Theorem shows

that the error involved in this approximation is less

than

1 10.001

11 5! 1320

Example 10 b


Another use of Taylor series is illustrated in the

next example.

The limit could be found with l’Hospital’s Rule.

Instead, we use a series.


Evaluate

Using the Maclaurin series for ex, we have the following result.

Example 11

20

1lim

x

x

e x

x


This is because power series are continuous functions.

2 3

2 20 0

2 3 4

20

2 3

0

1 11! 2! 3!1

lim lim

2! 3! 4!lim

1 1lim

2 3! 4! 5! 2

x

x x

x

x

x x xx

e x

x x

x x x

x

x x x

Example 11

MULTIPLICATION AND DIVISION OF POWER SERIES

If power series are added or subtracted, they

behave like polynomials.

Theorem 8 in Section 11.2 shows this.

In fact, as the following example shows, they can also be multiplied and divided like polynomials.

MULTIPLICATION AND DIVISION OF POWER SERIES

In the example, we find only the first few terms.

The calculations for the later terms become tedious.

The initial terms are the most important ones.

MULTIPLICATION AND DIVISION

Find the first three nonzero terms in the

Maclaurin series for:

a. ex sin x

b. tan x

Example 12


Using the Maclaurin series for ex and sin x in Table

1, we have:

Example 12 a

2 3 3

sin 11! 2! 3! 3!

x x x x xe x x


We multiply these expressions, collecting like

terms just as for polynomials:

Example 12 a

2 31 12 6

316

2 3 41 12 6

3 41 16 6

2 313

1 x x x

x x

x x x x

x x

x x x


Thus,

Example 12 a

2 313sinxe x x x x


Using the Maclaurin series in Table 1, we have:

Example 12 b

3 5

2 4

sin 3! 5!tancos

12! 4!

x xxx

xx xx


We use a procedure like long division:

Example 12 b

3 51 23 15

2 4 3 51 1 1 12 24 6 120

3 51 12 24

3 51 13 30

3 51 13 6

516

1

x x x

x x x x x

x x x

x x

x x

x


Thus,

Example 12 b

3 51 23 15tan x x x x


Although we have not attempted to justify the

formal manipulations used in Example 12, they are

legitimate.

There is a theorem that states the following:

Suppose both f(x) = cnxn and g(x) = bnxn converge for |x| < R and the series are multiplied as if they were polynomials.

Then, the resulting series also converges for |x| < R and represents f(x)g(x).


For division, we require b0 0.

The resulting series converges for sufficiently small x.


Documents

In section 11.9, we were able to find power series representations for a certain restricted class of functions. Here, we investigate more general problems