Linearization and Newton’s Method. I. Linearization A.) Def. – If f is differentiable at x = a,...
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Transcript of Linearization and Newton’s Method. I. Linearization A.) Def. – If f is differentiable at x = a,...
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