Y MX B: Linearization and Newton's Method

Linearization and Newtons Method
Linearization
The line y mx b = + which is tangent to the curve ( ) y f x = is very close to the curve at points
close to the point of tangency.
Now writing the slope at a as ( )
( )
0
0
0
x x
x f y
x
y
x f
=
A
A
= '
Where (x, y) is a point very close to tangent point ( ) ( )
0 0
, x f x . Now solving for y we obtain
( ) ( )( )
0 0 0
y f x f x x x ' = + which gives us an equation which gives us the approximate value
of the function ( ) f x at a point close to x
0
.
This tangent line approximation is sometimes written as ( ) ( ) ( ) f a h f a f a h ' + ~ + Where
the distance x is away from a (remember h should be small).
Definition: Linearization
If f is differentiable at x=a, then the approximating function
( ) ( ) ( ) ( ) L x f a f a x a ' = +
Is the linearization of f at a
In English:
( ) ( ) ( ) ( ) required known known difference between required and known f f f ' = +
Example
Use linear approximation technique to estimate ( )
2
3
122 .
Solution:
Let our function ( )
2
3
f x x =
Therefore ( )
1
3
2
3
f x x
' =
So we require ( ) ( )
2
3
122 122 f =
We need a point close to 122, which we can easily find its
2
3
x value. So choose x=125
because ( )
2
3
125 125 25 f = = and ( )
1
3
2 2
125 125
3 15
f

' = =
The point a=125 is only h=3 units away from the required point x=122 (a small distance)
Thus
122 125 3 a h = =
( ) ( )( )
( ) ( ) ( )( )
( )
2
3
125 3 125 125 3
2
25 3
15
3
24
5
y f a f a x a
f f
' = +
' ~ +
= +
=
This value 24.6 is very close to the actual value of 24.59838.
Example
Approximate
7
sin
36
t | |
|
\ .
Solution:
7
36 6
t t
~ , thus
7
36 6 36
t t t
= + where , and
6 36
a h
t t
= =
6 36 6 6 36
0.5 cos
6 36
3
0.5
2 36
0.576
f f f
t t t t t
t t
t
| | | | | || |
' + ~ +
| | | |
\ . \ . \ .\ .
| || |
= +
| |
\ .\ .
| |
= +
|
\ .
~
Always pick a
value you can
evaluate without
a calculator.
Therefore 576 . 0
36
7
sin ~ |
.
|
\
| t
.
Example
Find the linearization of ( ) 4 f x x = + at 0 x = .
Solution
First, ( ) ( )
1
2
1
4
2
f x x

' = +
Now, ( ) 0 2 f = and ( )
1
0
4
f ' =
Thus,
( ) ( ) ( )( )
( )
1
2 0
4
1
2
4
L x f a f a x a
x
x
' = +
= +
= +
Notice that as we move away from 2, we lose accuracy.
Example
Find the linearization of ( ) ( ) sin f x x = at
4
x
t
= .
Solution
First, ( ) ( ) cos f x x ' =
Now,
1
sin
4 4 2
f
t t
| | | |
= =
| |
\ . \ .
and
1
cos
4 4 2
f
t t
| | | |
' = =
| |
\ . \ .
Thus,
( ) ( ) ( )( )
1 1
4 2 2
1 1
2 2 4 2
4 1
4 2 2
L x f a f a x a
x
x
x
t
t
t
' = +
| |
= +
|
\ .
= +
= +
Newtons Method
Newtons method is a mathematical technique that allows you to approximate a zero of a
function . Under ideal situations, the zeros of the linearizations rapidly converges to an
accurate approximation.
The first guess,
1
x , produces a point on the curve ( ) ( ) 1 1
, x f x . The tangent line from this
point to the x-axis produces the next approximation,
2
x .
Now finding the slope of this tangent line
( )
( ) ( )
( )
( )( ) ( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( )
( )
( )
( )
1 2
1
1 2
1
1 2
1 1 2 1
1 1 1 2 1
1 2 1 1 1
1 2 1 1 1
1 1 1
2
1
1
1
1
0
f x f x
f x
x x
f x
x x
f x x x f x
f x x f x x f x
f x x f x f x x
f x x f x x f x
f x x f x
x
f x
f x
x
f x
' =
' =
' ' =
' ' =
' ' =
'
=
'
=
'
The general term is
( )
( )
1
n
n n
n
f x
x x
f x
+
=
'
The procedure is as follows:
- Guess a first approximation to a solution of ( ) 0 f x =
- Repeatedly use the general term to obtain closer approximations
Note: if Newtons method seems to divert from an answer, then try another initial guess and
repeat.
Example
Determine
2
Solution
Let 2 x = then
2
2 x = which can be written as
2
2 0 x = a function that equals zero.
( )
( )
2
2
2
f x x
f x x
=
' =
{ }
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
( )
1
1
2 1
1
2
3 2
2
3
4 3
3
1 an initial guess
1
1
1
1
1
2
1.5
1.5
1.5
1.5
0.25
1.5
3
1.4166666666
1.4166666666
1.4166666666
1.4166666666
0.006944444
1.4166666666
2.833333333
1.414215
x
f x
x x
f x
f
f
f x
x x
f x
f
f
f x
x x
f x
f
f
=
=
'
=
'
=
=
=
'
=
'
=
~
=
'
=
'
=
~ 68
Therefore after only 4 iterations we have 2 1.41421568 ~
Example
Consider the function ( )
4
25 f x x x = + .
a) Show that ( ) 2 0 f < and ( ) 3 0 f > and thus the equation ( ) 0 f x = must have a root b that
satisfies 2<b<3 .
b) Use Newtons Method the approximate b.
c) Evaluate ( ) f b to confirm the accuracy of your answer.
Solution
a)
( )
7
25 2 2 2
4
=
+ = f
( )
59
25 3 3 3
4
=
+ = f
b)
( )
( ) 1 4
25
3
4
+ = '
+ =
x x f
x x x f
{ }
( )
( )
1
2
3
4
5
2.5
2.5
2.5
2.5
16.5625
2.5
63.5
2.239173
2.187366
2.185511
2.185509
x guess
f
x
f
x
x
x
=
=
'
=
=
=
=
=
c) ( ) 00002 . 0 185509 . 2 = f

Y MX B: Linearization and Newton's Method

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Linearization and Newtons Method

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