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4.5: Linear Approximations, Differentials and Newton’s Method
For any function f (x), the tangent is a close approximation
of the function for some small distance from the tangent
point.
y
f  x  f  a
We call the equation of the
tangent the linearization of
the function.
0
xa
x

Start with the point/slope equation:
y  y1  m  x  x1 
x1  a
y1  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.
The linearization is the equation of the tangent line, and
you can use the old formulas if you like.

Important linearizations for x near zero:
f  x
1  x 
sin x
L  x
k
1  kx
x
cos x
1
tan x
x
1  x  1  x 
1
2
1
 1 x
2
This formula also leads to
non-linear approximations:
1
4 3


1
5
 1  5x   1  x
3
3
3
1  5x4  1  5x
4
4

Differentials:
When we first started to talk about derivatives, we said that
y
dy
becomes
when the change in x and change in
x
dx
y become very small.
dy can be considered a very small change in y.
dx can be considered a very small change in x.

Let y  f  x  be a differentiable function.
dx is an independent variable.
The differential dy is: dy  f   x  dx
The differential

Example: Consider a circle of radius 10. If the radius
increases by 0.1, approximately how much will the area
change?
A r
dA
dr
 2 r
dx
dx
2
dA  2 r dr
very small change in
r
very small change in A
dA  2   10   0.1
dA  2
(approximate change in area)

dA  2
(approximate change in area)
Compare to actual change:
New area:
 10.1  102.01
Old area:
 10   100.00
2
2
2.01
Error
.01
 .0049751  0.5%

Actual Answer 2.01
.01
Error
 .0001  0.01%

100
Original Area

Newton’s Method
1 2
Finding a root for: f  x   x  3
2
5
4
3
2
1
-4
-3
-2
-1
0
1
2
3
4
We will use
Newton’s Method to
find the root between
2 and 3.
-1
-2
-3

1 2
f  x  x  3
2
1.5
f  x  x
1.5
2
z 3
(not drawn to scale)
3
1 2
f  3   3  3  1.5
2
Guess:
mtangent  f   3  3
1.5
3
z
1.5
3
 2.5
(new guess)
3
1.5
z
3

1 2
f  x  x  3
2
1.5
f  x  x
2.5
1
2
f  2.5    2.5   3  .125
2
Guess:
2
z
3
mtangent  f   2.5  2.5
.125
2.5 
 2.45
2.5 (new guess)
.125
z
2.5

1 2
f  x  x  3
2
1.5
f  x  x
Guess: 2.45
f  2.45  .00125
z
2
3
mtangent  f   2.45  2.45
.00125
2.45 
 2.44948979592
2.45
.00125
z
2.45
(new guess)

Guess:
2.44948979592
f  2.44948979592  .00000013016
Amazingly close to zero!
This is Newton’s Method of finding roots. It is an example
of an algorithm (a specific set of computational steps.)
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 an iteration.

Guess:
2.44948979592
f  xn 
f  2.44948979592

.00000013016
 xn1  xn 
Newton’s Method:
f   xn 
Amazingly close to zero!
This is Newton’s Method of finding roots. It is an example
of an algorithm (a specific set of computational steps.)
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 an iteration.

This technique is sometimes called the Newton-Raphson
method.
Actually, the techniques that we will see in the next slides
were developed by an English mathematician named
Joseph Raphson, and published in 1690.
Isaac Newton developed a similar formula in 1671, but it
was not published until 1736 and is not as easy to use as
Raphson’s method.
Very little is known about Joseph Raphson, who lived
approximately 1648-1715. He was acquainted with
Newton, so he may have gotten the idea from Newton.
Anyway, Newton tends to get the credit.

3
Find where y  x  x crosses y  1 .
1  x3  x
n
0
1
xn
1
1.5
0  x 3  x  1 f  x   x3  x  1
f  xn 
1
.875
f  xn 
xn 1  xn 
f   xn 
f   xn 
2
1
1
 1.5
2
5.75
.875
1.5 
 1.3478261
5.75
2 1.3478261 .1006822 4.4499055
1.3252004 
3
f   x   3x 2  1
1.3252004
 1.3252004  1.0020584
1

There are some limitations to Newton’s method:
Looking for this root.
Bad guess.
Wrong root found
Failure to converge

We learn Newton’s method
of finding roots for historical
interest and to deepen our
appreciation of calculus.
Newton’s method is
sometimes tested on the AP
exam, and will therefore
sometimes appear on a test
in this class.
There are much easier
methods of finding roots
using our calculator.
Statue of Isaac Newton as a college student
in Trinity College Chapel at Cambridge
University, Cambridge, England

Example:
4
Approximate the negative root of: f  x   x  x  1
If you have the function
graphed, you can find the
roots by using:
F5
2
Math
Zero
Use the arrow keys to
select the lower and
upper bounds, and press
ENTER each time.
You would need to find
each root separately.

Example:
4
Approximate the negative root of: f  x   x  x  1
An even quicker way to find roots is to use the following
when in the home screen:
F2
4
Algebra
Zeros

zeros x 4  x  1, x
.724492

Press
ENTER
1.22074

Homework:
4.5a
4.5 p242 5,7,11
4.4 p228 45
3.6 p153 11,25,41,55
4.5b
4.5 p242 5,8,14,18,25,33
3.6 p153 17,33,47,63
4.5c
4.5 p242 19,27,36,50
3.6 p153 29; p156 1,2,3,4