§8.7--Taylor and Maclaurin Series - Furmanmath.furman.edu/~mwoodard/math151/docs/sec8_7.pdfTaylor’s Theorem Overview In the last section we learned that some functions can be expressed

§8.7–Taylor and Maclaurin Series

Mark Woodard

Furman U

Spring 2008

Mark Woodard (Furman U) §8.7–Taylor and Maclaurin Series Spring 2008 1 / 23

Outline

1 Taylor’s Theorem

2 When does a function equal its Taylor series?

3 Important examples: ex , sin(x), cos(x)

4 Newton’s binomial theorem

5 Some uses of Taylor series

6 Multiplication and division of series


Taylor’s Theorem

Overview

In the last section we learned that some functions can be expressed aspower series. In this section we explore a general method for expressing afunction as a power series:

f (x) =∞∑

n=0

cn(x − a)n = c0 + c1(x − a)1 + c2(x − a)2 + · · ·

Some questions:

Which functions have power series representations?

If a function can be represented by a power series, what are thecoefficients {cn}?Why does it matter? Why is it helpful for a function to berepresented by a power series?


Taylor’s Theorem

Overview


f (x) =∞∑

n=0

cn(x − a)n = c0 + c1(x − a)1 + c2(x − a)2 + · · ·

Some questions:




Taylor’s Theorem

Overview


f (x) =∞∑

n=0

cn(x − a)n = c0 + c1(x − a)1 + c2(x − a)2 + · · ·

Some questions:


If a function can be represented by a power series, what are thecoefficients {cn}?

Why does it matter? Why is it helpful for a function to berepresented by a power series?


Taylor’s Theorem

Overview


f (x) =∞∑

n=0

cn(x − a)n = c0 + c1(x − a)1 + c2(x − a)2 + · · ·

Some questions:




Taylor’s Theorem

Theorem (The form of the series)

Suppose that

f (x) =∞∑

n=0

cn(x − a)n = c0 + c1(x − a)1 + c2(x − a)2 + · · ·

in some symmetric interval about the point a, then the coefficients mustbe of the form

cn =f (n)(a)

n!, n ≥ 0.

Thus, if f is represented by a power series, it must assume the form:

f (x) =∞∑

n=0

f (n)(a)

n!(x − a)n.


Taylor’s Theorem

Proof.

Suppose f (x) = c0 + c1(x − a)1 + c2(x − a)2 + c3(x − a)3 + · · · insome interval about the point a.

Evaluating both sides at x = a shows that f (a) = c0; thus, c0 = f (a).

Differentiate both sides: f ′(x) = c1 + 2c2(x − a)1 + 3c3(x − a)2 + · · · .Evaluate both sides at x = a and note that f ′(a) = c1.

Likewise f ′′(x) = 2c2 + 3 · 2(x − a)1 + · · · . Upon evaluating bothsides at x = a, we see that f ′′(a) = 2c2 or c2 = f ′′(a)/2!.

Differentiating and evaluating once more leads to c3 = f ′′′(a)/3!.

Continue in the same manner.


Taylor’s Theorem

Proof.








Taylor’s Theorem

Proof.








Taylor’s Theorem

Proof.








Taylor’s Theorem

Proof.








Taylor’s Theorem

Proof.








Taylor’s Theorem

Proof.








Taylor’s Theorem

Definition

Given a function f and a point a, the series

f (x) =∞∑

n=0

f (n)(a)

n!(x − a)n

is called a Taylor series for f centered at a.

When a = 0, the series is called a Maclaurin series.


Taylor’s Theorem

Definition


f (x) =∞∑

n=0

f (n)(a)

n!(x − a)n




Taylor’s Theorem

Definition


f (x) =∞∑

n=0

f (n)(a)

n!(x − a)n




Taylor’s Theorem

Problem

Find the Maclaurin series for ex . What is its radius of convergence?

Solution

If f (x) = ex , then f (n)(x) = ex for all n ≥ 0.

Thus the coefficients for the Maclaurin series are

cn =f (n)(0)

n!=

1

n!

The Maclaurin series is thus

∞∑n=0

1

n!xn = 1 + x +

x2

2!+

x3

3!+ · · ·

It is easy to see that the radius of convergence is R = +∞.


Taylor’s Theorem

Problem


Solution



cn =f (n)(0)

n!=

1

n!


∞∑n=0

1

n!xn = 1 + x +

x2

2!+

x3

3!+ · · ·



Taylor’s Theorem

Problem


Solution



cn =f (n)(0)

n!=

1

n!


∞∑n=0

1

n!xn = 1 + x +

x2

2!+

x3

3!+ · · ·



Taylor’s Theorem

Problem


Solution



cn =f (n)(0)

n!=

1

n!


∞∑n=0

1

n!xn = 1 + x +

x2

2!+

x3

3!+ · · ·



Taylor’s Theorem

Problem


Solution



cn =f (n)(0)

n!=

1

n!


∞∑n=0

1

n!xn = 1 + x +

x2

2!+

x3

3!+ · · ·



Taylor’s Theorem

Problem


Solution



cn =f (n)(0)

n!=

1

n!


∞∑n=0

1

n!xn = 1 + x +

x2

2!+

x3

3!+ · · ·



When does a function equal its Taylor series?

Definition

For each N ≥ 0, let

TN(x) =N∑

n=0

f (n)(a)

n!(x − a)n

This is called the Taylor polynomial of order N for f (x).

Let RN(x) = f (x)− TN(x). The function RN is called the remainder.

Theorem

If RN(x)→ 0 as N →∞, then

f (x) = limN→∞

TN(x) =∞∑

n=0

f (n)(a)

n!(x − a)n,

that is, f (x) is equal to its Taylor series.



Definition


TN(x) =N∑

n=0

f (n)(a)

n!(x − a)n



Theorem


f (x) = limN→∞

TN(x) =∞∑

n=0

f (n)(a)

n!(x − a)n,




Definition


TN(x) =N∑

n=0

f (n)(a)

n!(x − a)n



Theorem


f (x) = limN→∞

TN(x) =∞∑

n=0

f (n)(a)

n!(x − a)n,




Definition


TN(x) =N∑

n=0

f (n)(a)

n!(x − a)n



Theorem


f (x) = limN→∞

TN(x) =∞∑

n=0

f (n)(a)

n!(x − a)n,




Remark

Our ability to faithfully represent a function by its Taylor series hinges onour ability to show that RN tends to 0 as N →∞. Our next theorem givesan important way to represent RN .

Theorem (Taylor’s Theorem)

If f has N + 1 derivatives in an interval I that contains the number a, thenfor x ∈ I there is a number z strictly between x and a such that theremainder term in the Taylor series can be expressed as

RN(x) =f (N+1)(z)

(N + 1)!(x − a)N+1.



Remark

Our ability to faithfully represent a function by its Taylor series hinges onour ability to show that RN tends to 0 as N →∞. Our next theorem givesan important way to represent RN .

Theorem (Taylor’s Theorem)

If f has N + 1 derivatives in an interval I that contains the number a, thenfor x ∈ I there is a number z strictly between x and a such that theremainder term in the Taylor series can be expressed as

RN(x) =f (N+1)(z)

(N + 1)!(x − a)N+1.


Important examples: ex , sin(x), cos(x)

Theorem

For each x ∈ R, xn/n!→ 0 as n→∞.

Problem

Show that ex , sin(x) and cos(x) are represented by their Maclaurin seriesfor all x.



Theorem

For each x ∈ R, xn/n!→ 0 as n→∞.

Problem

Show that ex , sin(x) and cos(x) are represented by their Maclaurin seriesfor all x.



Solution (The exponential function)

Let f (x) = ex . Then f (n)(x) = ex for all n ≥ 0. In particularf (n)(0) = 1 for all n ≥ 0.

The Maclaurin series for ex is thus

∞∑n=0

1

n!xn

Now we need to show that the series represents ex . To this end, fixx ∈ R. Then

RN(x) =ez

(N + 1)!xN+1

It is clear that this tends to 0 as N →∞. Thus the series representsex for all x.



Solution (The sine function)

Let f (x) = sin(x). The function f and its derivatives will cycle asfollows: sin(x), cos(x), − sin(x), − cos(x), sin(x), etc.

The Maclaurin series for sin(x) is as follows:

x − x3

3!+

x5

5!− x7

7!+ · · · =

∞∑n=0

(−1)n

(2n + 1)!x2n+1.

Now we must show that the series represents the sine function. Fixx ∈ R and note that

RN(x) =f N+1(z)

(N + 1)!xN+1

The N + 1st derivative of f is one of the four functions listed above;thus, |RN(x)| ≤ |x |N+1/(N + 1)!. Since |RN(x)| → 0 as N →∞, wemay conclude that the series represents sine for all x ∈ R.






x − x3

3!+

x5

5!− x7

7!+ · · · =

∞∑n=0

(−1)n

(2n + 1)!x2n+1.


RN(x) =f N+1(z)

(N + 1)!xN+1







x − x3

3!+

x5

5!− x7

7!+ · · · =

∞∑n=0

(−1)n

(2n + 1)!x2n+1.


RN(x) =f N+1(z)

(N + 1)!xN+1







x − x3

3!+

x5

5!− x7

7!+ · · · =

∞∑n=0

(−1)n

(2n + 1)!x2n+1.


RN(x) =f N+1(z)

(N + 1)!xN+1







x − x3

3!+

x5

5!− x7

7!+ · · · =

∞∑n=0

(−1)n

(2n + 1)!x2n+1.


RN(x) =f N+1(z)

(N + 1)!xN+1




Solution (The cosine function)

The analysis for the cosine is almost identical to that of the sinefunction. The Maclaurin series is

1− x2

2!+

x4

4!− x6

6!+ · · · =

∞∑n=0

(−1)n

(2n)!x2n.

This series represents cosine for all x ∈ R. The proof is almostidentical to that given above for sine.





1− x2

2!+

x4

4!− x6

6!+ · · · =

∞∑n=0

(−1)n

(2n)!x2n.






1− x2

2!+

x4

4!− x6

6!+ · · · =

∞∑n=0

(−1)n

(2n)!x2n.




Problem

Find the sum of the series S =2

1!+

4

2!+

8

3!+

16

4!+ · · ·.

Solution

We can see that this is close to the series for e2:

e2 =∞∑

n=0

2n

n!= 1 +

2

1!+

22

2!+

23

3!+ · · · = 1 + S .

Thus S = e2 − 1.



Problem

Find the sum of the series S =2

1!+

4

2!+

8

3!+

16

4!+ · · ·.

Solution

We can see that this is close to the series for e2:

e2 =∞∑

n=0

2n

n!= 1 +

2

1!+

22

2!+

23

3!+ · · · = 1 + S .

Thus S = e2 − 1.


Newton’s binomial theorem

Definition (Falling factorial)

Given x ∈ R and an integer n ≥ 1, let

(x)n = (x)(x − 1)(x − 2) · · · (x − n + 1)︸︷︷︸n terms

Let (x)0 = 1 for completeness.

Definition (Binomial coefficients)

Given x ∈ R and an integer n ≥ 0, let(x

n

)=

(x)n

n!



Definition (Falling factorial)

Given x ∈ R and an integer n ≥ 1, let

(x)n = (x)(x − 1)(x − 2) · · · (x − n + 1)︸︷︷︸n terms

Let (x)0 = 1 for completeness.

Definition (Binomial coefficients)

Given x ∈ R and an integer n ≥ 0, let(x

n

)=

(x)n

n!



Problem

Evaluate(−1/2

3

).

Solution(−1/2

3

)=

(−1/2)3

3!=

(−1/2)(−3/2)(−5/2)

6=−15

48= − 5

16



Problem

Evaluate(−1/2

3

).

Solution(−1/2

3

)=

(−1/2)3

3!=

(−1/2)(−3/2)(−5/2)

6=−15

48= − 5

16



Problem

Let k ∈ R and let f (x) = (1 + x)k . Show that the Maclaurin series for f is

∞∑n=0

(k

n

)xn

Find the radius of convergence of the series.

Theorem (Newton’s binomial theorem)

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

(1 + x)k =∞∑

n=0

(k

n

)xn.



Problem

Let k ∈ R and let f (x) = (1 + x)k . Show that the Maclaurin series for f is

∞∑n=0

(k

n

)xn

Find the radius of convergence of the series.

Theorem (Newton’s binomial theorem)

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

(1 + x)k =∞∑

n=0

(k

n

)xn.



Problem

Find the Maclaurin series for√

1− x. For which x does the seriesrepresent the function?

Solution

We have

√1− x = (1 + (−x))1/2 =

∞∑n=0

(1/2

n

)(−x)n =

∞∑n=0

(−1)n

(1/2

n

)xn.

This is true for all | − x | < 1 or for |x | < 1.



Problem

Find the Maclaurin series for√

1− x. For which x does the seriesrepresent the function?

Solution

We have

√1− x = (1 + (−x))1/2 =

∞∑n=0

(1/2

n

)(−x)n =

∞∑n=0

(−1)n

(1/2

n

)xn.

This is true for all | − x | < 1 or for |x | < 1.



Problem

Find the Maclaurin series for 1/√

9 + x. For which x does the seriesrepresent the function?

Solution

We have

1/√

9 + x = (9 + x)−1/2 =1

3

(1 + (x/9)

)−1/2

=∞∑

n=0

1

3

(−1/2

n

)(x/9)n

=∞∑

n=0

1

3 · 9n

(−1/2

n

)xn.

This series converges for |x/3| < 1 or |x | < 3.



Problem

Find the Maclaurin series for 1/√

9 + x. For which x does the seriesrepresent the function?

Solution

We have

1/√

9 + x = (9 + x)−1/2 =1

3

(1 + (x/9)

)−1/2

=∞∑

n=0

1

3

(−1/2

n

)(x/9)n

=∞∑

n=0

1

3 · 9n

(−1/2

n

)xn.

This series converges for |x/3| < 1 or |x | < 3.


Some uses of Taylor series

Problem

Express I =∫ 10 e−x2

dx as an infinite series. Approximate the integral towithin an error of .001.

Solution

We integrate term-by-term with the series for eu with u = −x2. Thus

I =∞∑

n=0

(−1)n

∫ 1

0

x2n

n!dx =

∞∑n=0

(−1)n 1

(2n + 1)n!= 1−1

3+

1

10− 1

42+· · ·

Since the series alternates, we can estimate the error using thealternating series test. The term 1/((2n + 1)n!) is first less than .001for n = 5. Thus

I ≈ 1− 1

3+

1

10− 1

42+

1

216= .747486772 . . .



Problem



Solution


I =∞∑

n=0

(−1)n

∫ 1

0

x2n

n!dx =

∞∑n=0

(−1)n 1

(2n + 1)n!= 1−1

3+

1

10− 1

42+· · ·


I ≈ 1− 1

3+

1

10− 1

42+

1

216= .747486772 . . .



Problem



Solution


I =∞∑

n=0

(−1)n

∫ 1

0

x2n

n!dx =

∞∑n=0

(−1)n 1

(2n + 1)n!= 1−1

3+

1

10− 1

42+· · ·


I ≈ 1− 1

3+

1

10− 1

42+

1

216= .747486772 . . .



Problem



Solution


I =∞∑

n=0

(−1)n

∫ 1

0

x2n

n!dx =

∞∑n=0

(−1)n 1

(2n + 1)n!= 1−1

3+

1

10− 1

42+· · ·


I ≈ 1− 1

3+

1

10− 1

42+

1

216= .747486772 . . .



Problem

Evaluate the limit

limx→0

sin(x)− x

x3

using the Maclaurin series for sin(x).

Solution

We can replace sin(x) by its Maclaurin series:

limx→0

sin(x)− x

x3= lim

x→0

(x − x3/6 + x5/120− · · ·

)− x

x3

= limx→0

(−1

6+

x2

120− · · ·

)= −1

6



Problem

Evaluate the limit

limx→0

sin(x)− x

x3

using the Maclaurin series for sin(x).

Solution

We can replace sin(x) by its Maclaurin series:

limx→0

sin(x)− x

x3= lim

x→0

(x − x3/6 + x5/120− · · ·

)− x

x3

= limx→0

(−1

6+

x2

120− · · ·

)= −1

6


Multiplication and division of series

Problem

Find the first three non-zero terms of the Maclaurin series for (1− x)−1ex .

Solution

We will multiply the power series and collect the terms up to x2:

1

1− xex = (1 + x + x2 + x3 + · · · )(1 + x + x2/2 + x3/6 + · · · )

= (1 + x + x2/2) + (x + x2) + (x2/2) + · · ·

= 1 + 2x +5

2x2 + · · ·



Problem

Find the first three non-zero terms of the Maclaurin series for (1− x)−1ex .

Solution

We will multiply the power series and collect the terms up to x2:

1

1− xex = (1 + x + x2 + x3 + · · · )(1 + x + x2/2 + x3/6 + · · · )

= (1 + x + x2/2) + (x + x2) + (x2/2) + · · ·

= 1 + 2x +5

2x2 + · · ·



Problem

Find the first three non-zero terms of the Maclaurin series for tan(x).

Solution

Using long division, we find that

tan(x) =sin(x)

cos(x)

=x − x3/3! + x5/5!− · · ·1− x2/2! + x4/4!− · · ·

= x +1

3x3 +

2

15x5 + · · ·



Problem

Find the first three non-zero terms of the Maclaurin series for tan(x).

Solution

Using long division, we find that

tan(x) =sin(x)

cos(x)

=x − x3/3! + x5/5!− · · ·1− x2/2! + x4/4!− · · ·

= x +1

3x3 +

2

15x5 + · · ·


Documents

§8.7--Taylor and Maclaurin Series - Furmanmath.furman.edu/~mwoodard/math151/docs/sec8_7.pdfTaylor’s Theorem Overview In the last section we learned that some functions can be expressed