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4.5: Linear Approximations, Differentialsand Newtons Method

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For any function f ( x), the tangent is a close approximation of the function for some small distance from the tangent point.

y

x0 x a=

( ) ( ) f x f a=We call the equation of thetangent the linearization of the function.

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The linearization is the equation of the tangent line, and

you can use the old formulas if you like.

Start with the point/slope equation:

( )1 1 y y m x x =

1 x a=

( )1 y f a=

( )m f a=

( ) ( ) ( ) y f a f a x a =

( ) ( ) ( ) y f a f a x a

= +

( ) ( ) ( ) ( ) L x f a f a x a= + linearization of f at a

( ) ( ) f x L x is the standard linear approximation of f at a.

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Important linearizations for x near zero:

( )1k

x+

1 kx+

sin x

cos x

tan x

x

1

x

( )12

11 1 1

2 x x x+ = + +

( )( )

1

3 4 4 3

4 4

1 5 1 5

1 51 5 1

3 3

x x

x x

+ = +

+ = +

( ) f x ( ) L x

This formula also leads tonon-linear approximations:

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Differentials:

When we first started to talk about derivatives, we said that

becomes when the change in x and change in y become very small.

y x

dydx

dy can be considered a very small change in y.

dxcan be considered a very small change in

x.

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Let be a differentiable function.

The differential is an independent variable.The differential is:

( ) y f x=

dxdy ( )dy f x dx=

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Example: Consider a circle of radius 10. If the radiusincreases by 0.1, approximately how much will the areachange?

2 A r =

2dA r dr =

2dA dr

r dx dx

=

very small change in Avery small change in r

( )2 10 0.1dA =

2dA = (approximate change in area)

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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 Original Area

Error

Actual Answer .0049751 0.5%

0.01%.0001.01100

=

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Newtons Method

( ) 21 32

f x x= Finding a root for:

We will use NewtonsMethod to find theroot between 2 and 3.

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Guess: 3

( ) 213 3 3 1.52

f = =

1.5

( )tangent 3 3m f = =

( ) 21 32

f x x=

( ) f x x =

z

1.5

1.53

z = 1.5

3 z =1.5

3 2.53

=

(not drawn to scale)

(new guess)

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Guess: 2.5

( ) ( )212.5 2.5 3 .1252

f = =

1.5

( )tangent 2.5 2.5m f = =

( ) 21 32

f x x=

( ) f x x =

z

.1252.5

z =.1252.5 2.45

2.5 =

(new guess)

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Guess: 2.45

( )2.45 .00125 f =

1.5

( )tangent 2.45 2.45m f = =

( ) 21 32

f x x=

( ) f x x =

z

.001252.45

z =

.001252.45 2.449489795922.45 = (new guess)

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Guess: 2.44948979592

( )2.44948979592 .00000013016 f =

Amazingly close to zero!

This is Newtons Method of finding roots. It is an exampleof an algorithm (a specific set of computational steps.)

It is sometimes called the Newton-Raphson method

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

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This is Newtons Method of finding roots. It is an exampleof an algorithm (a specific set of computational steps.)

It is sometimes called the Newton-Raphson method

Guess: 2.44948979592

( )2.44948979592 .00000013016 f =

Amazingly close to zero!

Newtons Method: ( )( )1n

n nn

f x x x f x

+=

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

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n x ( )n f xn ( )n f x ( )( )1n

n nn

f x x x f x

+ =

Find where crosses .3 y x x= 1 y =

31 x x= 30 1 x x= ( ) 3 1 f x x x= ( ) 23 1 f 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

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There are some limitations to Newtons method:

Wrong root found

Looking for this root.

Bad guess.

Failure to converge

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Newtons method is built in to the Calculus Toolsapplication on the TI-89.

Of course if you have a TI-89, you could just use theroot finder to answer the problem.

The only reason to use the calculator for Newtons Methodis to help your understanding or to check your work.

It would not be allowed in a college course,on the AP exam or on one of my tests.

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APPS

Select and press .Calculus Tools ENTER

If you see thisscreen, press

, change themode settings asnecessary, andpressagain.

ENTER

APPS

Now lets do one on the TI-89:

( ) 3 1 f x x x= Approximate the positive root of:

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Now lets do one on the TI-89:

APPS

Select and press .Calculus Tools ENTER

Press (Deriv)F2

Press (Newtons Method)3

Enter the equation.

(You will have to unlockthe alpha mode.)Set the initial guess to 1.

Press .ENTER

( ) 3 1 f x x x= Approximate the positive root of:

Set the iterations to 3.

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Press to see the

summary screen.

ESC

Press to see

each iteration.

ENTER

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Press and thento return your

calculator to normal.

ESCHOME