Notes 4.5 Linearization and Newton’s

Method

I. LinearizationA.) Def. – If f is differentiable at x = a, then the

approximating function

is the LINEARIZATION of f at x = a.

B.)

( ) ( ) ( )( )L x f a f a x a

The approximation ( ) ( ) is the STANDARD

LINEAR APPROXIMATION of at . The point

= is the CENTER of the approximation.

f x L x

f a

x a

C.) Note: This is just like

just different notation!!!

D.) Graphically:

0 0y y m x x

x0 x a

f x f aWe call the equation of the tangent the linearization of the function.

E.) Ex.- Find the linearization at x = 0 of

( ) 1f x x

1( )

2 1f x

x

1 1

(0)22 1 0

f

(0) 1 0 1f

1( ) 1 ( 0)

2L x x

1( ) 1

2L x x

Act. - Approx.

F.) How accurate is it?

Approx. Value Actual Value

11.1 .1 1 1.05

2 1.1 1.048804 0.00119

11.05 .05 1 1.025

2 1.05 1.0246951 0.0003049

1.0005 1.00025 1.0005 1.00024995 0.00000003

II. DifferentialsA.) Def.: Let y = f (x) be a differentiable function. The

DIFFERENTIAL dx is an independent variable. The DIFFERENTIAL dy = f’(x)dx, where dy is dependent upon the values of f’(x)dx.

B.) Ex. – Find the differential dy in each of the following:

22.) secy x4 21.) 3 2y x x

31.) 4 6dy

x xdx

34 6dy x x dx

2.) 2 sec sec tandy

x x xdx

22sec tandy

x xdx

22sec tandy x xdx

( )Note: If 0, then ( )

dy f x dxdx f x

dx dx

C.) Ex. – Find dy and evaluate dy for the given values of x and dx in the following:

0.45dy

3 3 , 2, 0.05y x x x dx

2 3 3dy

xdx

23 3dy x dx

23 2 3 0.05dy

III. Estimating ChangeA.) Graphically:

B.) Let y = f (x) be a differentiable function at x = a. The APPROXIMATE change in the value of f when x changes from x to x + a is

( )df f a dx

Estimated Change

Absolute Change

Actual Change

( ) ( )f f a dx f a ( )df f a dx

Relative Change( )

f

f a

( )

df

f a

Percent Change 100( )

f

f a

100

( )

df

f a

C.) Ex.- Consider a circle of radius 10. If the radius increases by 0.1, approximately how much will the area change?

2A r

2 dA r dr

2dA

rdr

very small change in A

very small change in r

2 10 0.1dA

2dA (approximate change in area)

2dA (approximate change in area)

Compare to actual change:

New area:

Old area:

210.1 102.01

210 100.00

2.01

.01

2.01

Error in Estimate

Original Area

Error in Estimate

Actual Answer.0049751 0.5%

0.01%.0001.01

100

2.01 2 .01

Notes 4.5 – Error Analysis

I. Examples The side of a square is measured with a

possible percentage error of ±5%. Use differentials to estimate the possible percentage error in the area of the square.

? when .05dA dx

A x

2 A x

2 dA xdx

2

2

dA xdx

A x

2

dA xdx

A A

2

dA dx

A x

2(.05) 10% dA

A

2 dA dx

A x

Therefore, the possible percentage error in the measurement of the area of the square will be between ±10%

The area of a circle is to be computed from a measured value of its diameter. Estimate the maximum permissible percentage error in the measurement if the percentage error in the area must be kept within 1%.

? when .01dD dA

D A

2

2

DA

2

4

DA

2dA D dD

2D dD

dA

A A

2

2

4

D dDdA

A D

2dA dD

A D

.01 2dD

D

.005dD

D

Therefore, the percentage error in the measurement of the diameter must be within ±.5%

IV. Newton’s MethodA.) Newton’s Method is an algorithm for finding roots.

Newton’s Method: 1

nn n

n

f xx x

f x

guessnx 1 next guessnx

It is sometimes called the Newton-Raphson method

This is a recursive algorithm because a set of steps are repeated with the previous answer put in the next repetition. Each repetition is called is called an iteration.

213

2f x x

We will use Newton’s Method to find the root between 2 and 3.

B.) Example: Finding a root for:

Guess: 3

213 3 3 1.5

2f

1.5

tangent 3 3m f

213

2f x x

f x x

z

1.5

1.53

z

1.5

3z

1.53 2.5

3

(not drawn to scale)

(new guess)

Guess: 2.5

212.5 2.5 3 .125

2f

.125

tangent 2.5 2.5m f

213

2f x x

f x x

z

.125

2.5z

.1252.5 2.45

2.5

(new guess)

Guess: 2.45

2.45 .00125f

.00125 tangent 2.45 2.45m f

213

2f x x

f x x

z.00125

2.45z

.001252.45 2.44948979592

2.45 (new guess)

Guess: 2.44948979592

2.44948979592 .00000013016f

Amazingly close to zero!

nx nf xn nf x 1

nn n

n

f xx x

f x

Find where crosses .3y x x 1y 31 x x 30 1x x 3 1f x x x 23 1f x x

0 1 1 21

1 1.52

1 1.5 .875 5.75.875

1.5 1.34782615.75

2 1.3478261 .1006822 4.4499055 1.3252004

31.3252004 1.3252004 1.0020584 1

There are some limitations to Newton’s method:

Wrong root found

Looking for this root.

Bad guess.Failure to converge