Contributed by:

In this section, we will learn:

How to solve high degree equations using Newton’s method.

How to solve high degree equations using Newton’s method.

1.
4

APPLICATIONS OF DIFFERENTIATION

APPLICATIONS OF DIFFERENTIATION

2.
APPLICATIONS OF DIFFERENTIATION

4.8

Newton’s Method

In this section, we will learn:

How to solve high degree equations

using Newton’s method.

4.8

Newton’s Method

In this section, we will learn:

How to solve high degree equations

using Newton’s method.

3.
Suppose that a car dealer offers to sell

you a car for $18,000 or for payments of

$375 per month for five years.

You would like to know what monthly interest rate

the dealer is, in effect, charging you.

you a car for $18,000 or for payments of

$375 per month for five years.

You would like to know what monthly interest rate

the dealer is, in effect, charging you.

4.
INTRODUCTION Equation 1

To find the answer, you have to solve

the equation

48x(1 + x)60 - (1 + x)60 + 1 = 0

How would you solve such an equation?

To find the answer, you have to solve

the equation

48x(1 + x)60 - (1 + x)60 + 1 = 0

How would you solve such an equation?

5.
HIGH-DEGREE POLYNOMIALS

For a quadratic equation ax2 + bx + c = 0,

there is a well-known formula for the roots.

For third- and fourth-degree equations,

there are also formulas for the roots.

However, they are extremely complicated.

For a quadratic equation ax2 + bx + c = 0,

there is a well-known formula for the roots.

For third- and fourth-degree equations,

there are also formulas for the roots.

However, they are extremely complicated.

6.
HIGH-DEGREE POLYNOMIALS

If f is a polynomial of

degree 5 or higher, there is

no such formula.

If f is a polynomial of

degree 5 or higher, there is

no such formula.

7.
TRANSCENDENTAL EQUATIONS

Likewise, there is no formula that

will enable us to find the exact roots

of a transcendental equation such as

cos x = x.

Likewise, there is no formula that

will enable us to find the exact roots

of a transcendental equation such as

cos x = x.

8.
APPROXIMATE SOLUTION

We can find an approximate solution

by plotting the left side of the equation.

Using a graphing device, and after experimenting with

viewing rectangles, we produce the graph below.

We can find an approximate solution

by plotting the left side of the equation.

Using a graphing device, and after experimenting with

viewing rectangles, we produce the graph below.

9.
ZOOMING IN

We see that, in addition to the solution

x = 0, which doesn’t interest us, there is

a solution between 0.007 and 0.008

Zooming in shows

that the root is

approximately

0.0076

We see that, in addition to the solution

x = 0, which doesn’t interest us, there is

a solution between 0.007 and 0.008

Zooming in shows

that the root is

approximately

0.0076

10.
ZOOMING IN

If we need more accuracy, we could

zoom in repeatedly.

That becomes tiresome, though.

If we need more accuracy, we could

zoom in repeatedly.

That becomes tiresome, though.

11.
NUMERICAL ROOTFINDERS

A faster alternative is to use a

numerical rootfinder on a calculator

or computer algebra system.

If we do so, we find that the root, correct to

nine decimal places, is 0.007628603

A faster alternative is to use a

numerical rootfinder on a calculator

or computer algebra system.

If we do so, we find that the root, correct to

nine decimal places, is 0.007628603

12.
NUMERICAL ROOTFINDERS

How do those numerical rootfinders

They use a variety of methods.

Most, though, make some use of Newton’s method,

also called the Newton-Raphson method.

How do those numerical rootfinders

They use a variety of methods.

Most, though, make some use of Newton’s method,

also called the Newton-Raphson method.

13.
NEWTON’S METHOD

We will explain how the method works,

for two reasons:

To show what happens inside a calculator or

computer

As an application of the idea of linear approximation

We will explain how the method works,

for two reasons:

To show what happens inside a calculator or

computer

As an application of the idea of linear approximation

14.
NEWTON’S METHOD

The geometry behind Newton’s method

is shown here.

The root that we

are trying to find

is labeled r.

The geometry behind Newton’s method

is shown here.

The root that we

are trying to find

is labeled r.

15.
NEWTON’S METHOD

We start with a first approximation x1,

which is obtained by one of the following

Guessing

A rough sketch

of the graph of f

A computer-

generated graph

of f

We start with a first approximation x1,

which is obtained by one of the following

Guessing

A rough sketch

of the graph of f

A computer-

generated graph

of f

16.
NEWTON’S METHOD

Consider the tangent line L to the curve

y = f(x) at the point (x1,f(x1)) and look at

the x-intercept of L, labeled x2.

Consider the tangent line L to the curve

y = f(x) at the point (x1,f(x1)) and look at

the x-intercept of L, labeled x2.

17.
NEWTON’S METHOD

Here’s the idea behind the method.

The tangent line is close to the curve.

So, its x-intercept, x2 , is close to the x-intercept

of the curve

(namely, the root r

that we are seeking).

As the tangent is

a line, we can easily

find its x-intercept.

Here’s the idea behind the method.

The tangent line is close to the curve.

So, its x-intercept, x2 , is close to the x-intercept

of the curve

(namely, the root r

that we are seeking).

As the tangent is

a line, we can easily

find its x-intercept.

18.
NEWTON’S METHOD

To find a formula for x2 in terms of x1,

we use the fact that the slope of L is f’(x1).

So, its equation is:

y - f(x1) = f’(x1)(x - x1)

To find a formula for x2 in terms of x1,

we use the fact that the slope of L is f’(x1).

So, its equation is:

y - f(x1) = f’(x1)(x - x1)

19.
SECOND APPROXIMATION

As the x-intercept of L is x2, we set y = 0

and obtain: 0 - f(x1) = f’(x1)(x2 - x1)

If f’(x1) ≠ 0, we can solve this equation for x2:

f ( x1 )

x2 = x1 −

f '( x1 )

We use x2 as a second approximation to r.

As the x-intercept of L is x2, we set y = 0

and obtain: 0 - f(x1) = f’(x1)(x2 - x1)

If f’(x1) ≠ 0, we can solve this equation for x2:

f ( x1 )

x2 = x1 −

f '( x1 )

We use x2 as a second approximation to r.

20.
THIRD APPROXIMATION

Next, we repeat this procedure with x1

replaced by x2, using the tangent line at

This gives a third approximation:

f ( x2 )

x3 = x2 −

f '( x2 )

Next, we repeat this procedure with x1

replaced by x2, using the tangent line at

This gives a third approximation:

f ( x2 )

x3 = x2 −

f '( x2 )

21.
SUCCESSIVE APPROXIMATIONS

If we keep repeating this process,

we obtain a sequence of approximations

x1, x2, x3, x4, . . .

If we keep repeating this process,

we obtain a sequence of approximations

x1, x2, x3, x4, . . .

22.
SUBSEQUENT APPROXIMATION Equation/Formula 2

In general, if the nth approximation is xn and

f’(xn) ≠ 0, then the next approximation is

given by:

f ( xn )

xn +1 = xn −

f '( xn )

In general, if the nth approximation is xn and

f’(xn) ≠ 0, then the next approximation is

given by:

f ( xn )

xn +1 = xn −

f '( xn )

23.
If the numbers xn become closer and

closer to r as n becomes large, then

we say that the sequence converges to r

and we write:

lim xn =r

n→ ∞

closer to r as n becomes large, then

we say that the sequence converges to r

and we write:

lim xn =r

n→ ∞

24.
The sequence of successive approximations

converges to the desired root for functions of

the type illustrated in the previous figure.

However, in certain circumstances, it may

not converge.

converges to the desired root for functions of

the type illustrated in the previous figure.

However, in certain circumstances, it may

not converge.

25.
Consider the situation shown here.

You can see that x2 is a worse approximation

than x1.

This is likely to be

the case when

f’(x1) is close to 0.

You can see that x2 is a worse approximation

than x1.

This is likely to be

the case when

f’(x1) is close to 0.

26.
It might even happen that an approximation

falls outside the domain of f, such as x3.

Then, Newton’s method fails.

In that case,

a better initial

approximation x1

should be chosen.

falls outside the domain of f, such as x3.

Then, Newton’s method fails.

In that case,

a better initial

approximation x1

should be chosen.

27.
See Exercises 31–34 for specific

examples in which Newton’s method

works very slowly or does not work

at all.

examples in which Newton’s method

works very slowly or does not work

at all.

28.
NEWTON’S METHOD Example 1

Starting with x1 = 2, find the third

approximation x3 to the root of

the equation

x3 – 2x – 5 = 0

Starting with x1 = 2, find the third

approximation x3 to the root of

the equation

x3 – 2x – 5 = 0

29.
NEWTON’S METHOD Example 1

We apply Newton’s method with

f(x) = x3 – 2x – 5 and f’(x) = 3x2 – 2

Newton himself used this equation to illustrate

his method.

He chose x1 = 2 after some experimentation

because f(1) = -6, f(2) = -1 and f(3) = 16

We apply Newton’s method with

f(x) = x3 – 2x – 5 and f’(x) = 3x2 – 2

Newton himself used this equation to illustrate

his method.

He chose x1 = 2 after some experimentation

because f(1) = -6, f(2) = -1 and f(3) = 16

30.
NEWTON’S METHOD Example 1

Equation 2 becomes:

3

x −2 xn −5

n

xn + 1 = xn − 2

3xn −2

Equation 2 becomes:

3

x −2 xn −5

n

xn + 1 = xn − 2

3xn −2

31.
NEWTON’S METHOD Example 1

With n = 1, we have:

3

x −2 x1 −5

x2 = x1 − 1

2

3 x1 −2

3

2 −2(2) −5

=2 − 2

3(2) −2

=2.1

With n = 1, we have:

3

x −2 x1 −5

x2 = x1 − 1

2

3 x1 −2

3

2 −2(2) −5

=2 − 2

3(2) −2

=2.1

32.
NEWTON’S METHOD Example 1

With n = 2, we obtain:

3

x −2 x2 −5

x3 = x2 − 2

2

3 x2 −2

3

2.1 −2(2.1) −5

=2.1 − 2

3(2.1) −2

≈2.0946

It turns out that this third approximation x3 ≈ 2.0946

is accurate to four decimal places.

With n = 2, we obtain:

3

x −2 x2 −5

x3 = x2 − 2

2

3 x2 −2

3

2.1 −2(2.1) −5

=2.1 − 2

3(2.1) −2

≈2.0946

It turns out that this third approximation x3 ≈ 2.0946

is accurate to four decimal places.

33.
NEWTON’S METHOD

The figure shows the geometry behind the

first step in Newton’s method in the example.

As f’(2) = 10, the tangent line to y = x3 - 2x - 5 at (2, -1)

has equation

y = 10x – 21

So, its x-intercept

is x2 = 2.1

The figure shows the geometry behind the

first step in Newton’s method in the example.

As f’(2) = 10, the tangent line to y = x3 - 2x - 5 at (2, -1)

has equation

y = 10x – 21

So, its x-intercept

is x2 = 2.1

34.
NEWTON’S METHOD

Suppose that we want to achieve a given

accuracy—say, to eight decimal places—

using Newton’s method.

How do we know when to stop?

Suppose that we want to achieve a given

accuracy—say, to eight decimal places—

using Newton’s method.

How do we know when to stop?

35.
NEWTON’S METHOD

The rule of thumb that is generally used is that

we can stop when successive approximations

xn and xn+1 agree to eight decimal places.

A precise statement concerning accuracy in the method

will be given in Exercise 37 in Section 11.11

The rule of thumb that is generally used is that

we can stop when successive approximations

xn and xn+1 agree to eight decimal places.

A precise statement concerning accuracy in the method

will be given in Exercise 37 in Section 11.11

36.
ITERATIVE PROCESS

Notice that the procedure in going from n to

n + 1 is the same for all values of n.

It is called an iterative process.

This means that the method is particularly convenient

for use with a programmable calculator or a computer.

Notice that the procedure in going from n to

n + 1 is the same for all values of n.

It is called an iterative process.

This means that the method is particularly convenient

for use with a programmable calculator or a computer.

37.
NEWTON’S METHOD Example 2

6

Use Newton’s method to find 2 correct

to eight decimal places.

First, we observe that finding 6 2 is equivalent to

finding the positive root of the equation x6 – 2 = 0

So, we take f(x) = x6 – 2

Then, f’(x) = 6x5

6

Use Newton’s method to find 2 correct

to eight decimal places.

First, we observe that finding 6 2 is equivalent to

finding the positive root of the equation x6 – 2 = 0

So, we take f(x) = x6 – 2

Then, f’(x) = 6x5

38.
NEWTON’S METHOD Example 2

So, Formula 2 (Newton’s method)

6

x −2

n

xn +1 = xn − 5

6 xn

So, Formula 2 (Newton’s method)

6

x −2

n

xn +1 = xn − 5

6 xn

39.
NEWTON’S METHOD Example 2

Choosing x1 = 1 as the initial approximation,

we obtain: x2 ≈1.16666667

x3 ≈1.12644368

x4 ≈1.12249707

x5 ≈1.12246205

x6 ≈1.12246205

As x5 and x6 agree to eight decimal places, we

6

conclude that 2 ≈1.12246205 to eight decimal places.

Choosing x1 = 1 as the initial approximation,

we obtain: x2 ≈1.16666667

x3 ≈1.12644368

x4 ≈1.12249707

x5 ≈1.12246205

x6 ≈1.12246205

As x5 and x6 agree to eight decimal places, we

6

conclude that 2 ≈1.12246205 to eight decimal places.

40.
NEWTON’S METHOD Example 3

Find, correct to six decimal places,

the root of the equation cos x = x.

We rewrite the equation in standard form: cos x – x = 0

Therefore, we let f(x) = cos x – x

Then, f’(x) = –sinx – 1

Find, correct to six decimal places,

the root of the equation cos x = x.

We rewrite the equation in standard form: cos x – x = 0

Therefore, we let f(x) = cos x – x

Then, f’(x) = –sinx – 1

41.
NEWTON’S METHOD Example 3

So, Formula 2 becomes:

cos xn −xn

xn +1 = xn −

−sin xn −1

cos xn −xn

= xn +

sin xn + 1

So, Formula 2 becomes:

cos xn −xn

xn +1 = xn −

−sin xn −1

cos xn −xn

= xn +

sin xn + 1

42.
NEWTON’S METHOD Example 3

To guess a suitable value for x1, we sketch

the graphs of y = cos x and y = x.

It appears they intersect at a point whose x-coordinate

is somewhat less than 1.

To guess a suitable value for x1, we sketch

the graphs of y = cos x and y = x.

It appears they intersect at a point whose x-coordinate

is somewhat less than 1.

43.
NEWTON’S METHOD Example 3

So, let’s take x1 = 1 as a convenient first

Then, remembering to put our calculator

in radian mode, we get: x2 ≈0.75036387

x3 ≈0.73911289

x4 ≈0.73908513

x5 ≈0.73908513

As x4 and x5 agree to six decimal places (eight, in fact),

we conclude that the root of the equation, correct to

six decimal places, is 0.739085

So, let’s take x1 = 1 as a convenient first

Then, remembering to put our calculator

in radian mode, we get: x2 ≈0.75036387

x3 ≈0.73911289

x4 ≈0.73908513

x5 ≈0.73908513

As x4 and x5 agree to six decimal places (eight, in fact),

we conclude that the root of the equation, correct to

six decimal places, is 0.739085

44.
NEWTON’S METHOD

Instead of using this rough sketch to get

a starting approximation for the method in

the example, we could have used the more

accurate graph that a calculator or computer

Instead of using this rough sketch to get

a starting approximation for the method in

the example, we could have used the more

accurate graph that a calculator or computer

45.
NEWTON’S METHOD

This figure suggests that we use

x1 = 0.75 as the initial approximation.

This figure suggests that we use

x1 = 0.75 as the initial approximation.

46.
NEWTON’S METHOD

Then, Newton’s method gives:

x2 ≈0.73911114

x3 ≈0.73908513

x4 ≈0.73908513

So we obtain the same answer as before—but

with one fewer step.

Then, Newton’s method gives:

x2 ≈0.73911114

x3 ≈0.73908513

x4 ≈0.73908513

So we obtain the same answer as before—but

with one fewer step.

47.
NEWTON’S METHOD VS. GRAPHING DEVICES

You might wonder why we bother at all

with Newton’s method if a graphing

device is available.

Isn’t it easier to zoom in repeatedly and

find the roots as we did in Section 1.4?

You might wonder why we bother at all

with Newton’s method if a graphing

device is available.

Isn’t it easier to zoom in repeatedly and

find the roots as we did in Section 1.4?

48.
NEWTON’S METHOD VS. GRAPHING DEVICES

If only one or two decimal places of accuracy

are required, then indeed the method is

inappropriate and a graphing device suffices.

However, if six or eight decimal places

are required, then repeated zooming

becomes tiresome.

If only one or two decimal places of accuracy

are required, then indeed the method is

inappropriate and a graphing device suffices.

However, if six or eight decimal places

are required, then repeated zooming

becomes tiresome.

49.
NEWTON’S METHOD VS. GRAPHING DEVICES

It is usually faster and more efficient

to use a computer and the method in

You start with the graphing device and finish

with the method.

It is usually faster and more efficient

to use a computer and the method in

You start with the graphing device and finish

with the method.