1

§3.6 Newton’s Method.

The student will learn about

Newton’s method of approximating roots and tangent line approximations.

2

Introduction to Newton’s MethodSometimes we are presented with a problem which cannot be solved by simple algebraic means.

However, we will see that calculus through Newton’s Method gives us a way of finding approximate solutions.

For instance, if we needed to find the roots of the polynomial ,

we would find that the tried and true techniques just wouldn't work.

3x x 1 0

3

An Easier Example to StartLet’s start by computing the √5. This is of course easy with your calculator but stay with me for this.

First we rewrite the problem as an equation f (x) = x 2 – 5 = 0

Newton’s method is an iterative method. That means that you must first pick an initial value for the solution and then the method will yield a better value. The method may be repeated as often as necessary to get the accuracy needed.

What would be a good initial value for √5? OK we will use 2.

4

An Easier Example to Start

Before we continue let’s look at the method.

Consider the drawing. If x Is the root and x n is an approximation then x n + 1 is a better approximation.

Using the tangent line slope

n n 1 nn

n n 1 n n 1

y y ydy yf '(x )

dx x x x x x

And solving for x n + 1 yields nn 1 n

n

f (x )x x

f '(x )

5

Back to our ExampleWe were trying to find √5 using f (x) = x 2 – 5 = 0 and x n = x 0 = 2 and f ′ (x) = 2x.

Probably not too impressed! Let’s find x 2.

With just two iterations we have accuracy to 0.000043134.

nn 1 n

n

f (x )x x

f '(x )

1

1x 2 2.25

4

n

n 1 nn

f (x )x x

f '(x ) 01 0

0

f (x )or x x and

f '(x )

12 1

1

f (x )x x and

f '(x ) 2

0.0625x 2.25 2.23611111...

4.5

FACT: Newton described this method in a book he wrote in 1669!

6

Example 2Approximate the solution to cos x = x in the interval [0, 2].

First we rewrite the problem as f (x) = cos x – x = 0

We will let x 0 = 1 (midpoint of the interval) and we know f ′(x) = - sin x - 1

If we were to repeat the process we would get x 2 = 0.7391128909 x 3 = 0.7390851334 - accuracy to 9 places!

A bit tedious BUT if you know a little programming your calculator or computer can do this easily.

nn 1 n

n

f (x )x x

f '(x )

01 0

0

f (x ) cos(1) 1 0.5403023059 1x x 1 1 0.7503668679

f '(x ) sin(1) 1 0.8414709848 1

7

Example 3Approximate using x 0 = 1

OK it’s a silly example (Do you know the solution?) but stay with me while I make a point.

The computation is easy with x 0 = 1, x 1 = - 2, x 2 = 4, x 3 = - 8, x 4 = 16, etc.

So the method fails. But, it fails spectacularly!

nn 1 n

n

f (x )x x

f '(x )

1

3n

n 1 n n n n2

3n

xx x x 3x 2x

1x

3

3f (x) x 0

8

Failure

Newton's method makes no guarantee on convergence.

Indeed, convergence depends on the starting point and on the shape of the function.

9

Your Calculator

Calculators basically only know how to add and multiply.

Use x 0 = 2 and approximate √5 with two iterations.

nn 1 n

n

f (x )x x

f '(x )

2n

n 1 n nn n

x a 1 ax x x

2x 2 x

2f (x) x a 0 and then f '(x) 2x

So, how does it find ? Let’s use Newton.a

Notice that the operations involved in the iteration are addition and multiplication and you computer can do that!

10

Your Calculator

Use x 0 = 2 and approximate √5 with two iterations.

nn 1 n

n

f (x )x x

f '(x )

2n

n 1 n nn n

x a 1 ax x x

2x 2 x

1

1 5 9x 2 2.25

2 2 4

2

1 5x 2.25 2.23611.. . 2.236067977 .

2 2.25

11

Tangent Line ApproximationsFrom our definition of derivative we know that

Δy is the exact change in y

h 0

f(x h) f (x) f(x h) f (x)f '(x) lim

h h

If we multiply both sides of the above by h, we get

dy h f '(x) f(x h) f (x) y

When h is small.

dy = h · f ′ (x) is an approximation of Δy and is called the differential.

12

Tangent Line ApproximationsSummary

dy h f '(x) f(x h) f (x) y

Approximate change exact change

Another useful form:

f(x h) f (x) h f '(x)

13

Tangent Line Approximationsf(x h) f (x) h f '(x)

Let’s use this form for a practical problem.

Approximate √5 using the differential above.

5 4 1

With x = 4, h = 1, 1 1

2 21

f (x) x and f '(x) x2

f(x h) f (x) h f '(x)

f(4 1) f (4) 1 f '(4) 1 1

4 1 4 1 2 2.2542 4

Does this look familiar?

14

Summary.

• We learned how to use Newton’s Method to solve equations.

• We developed the approximation formula using the differential dy,

f (x + h) – f (x) ≈ h · f ′ (x) = dy

15

ASSIGNMENT

§3.6; Page 66; 1 - 9, odd.