Note: Before using this module, teachers should read the 'Learning Activity' and the 'Goals for the Module'.
Assumption: Near the zero value, the tangent line to the function will cross the x-axis "near" the true axis crossing.
The procedure follows these steps:
1) Obtain a starting value
by estimating a convenient input value x* (the first approximation)
near the actual axis crossing. A graphing calculator is an excellent tool
for arriving at an initial input value.
2) Find the slope function
(derivative) and evaluate it at x* . This gives the slope of the
tangent line.
3) Evaluate the function
at x*. This gives a point (x*,f(x*)) on the tangent line.
4) The slope-intercept method
is used to find the y-intercept and also the slope- intercept equation
of the line.
5) When this equation is
set equal to zero and solved, we have the location at which the tangent
line crosses the x-axis. This is the second approximation x**
of the true x-axis intercept for the polynomial function.
6) Now use the calculated
x** as the input and repeat the process until the desired accuracy is obtained.
Learning Activity: This module teaches the Newton-Raphson Method for finding the x-axis crossing point of a polynomial function. The method is process-based, not formula-based.
To the student
Please read the statement of the
'Learning Activity' and the 'Goals for the Module'
Goals for the Module
1) Students should be able to evaluate a polynomial function.
2) Students should be able to take the derivative of a polynomial function.
3) Students should be able to form the equation of the tangent line and
find its x-intercept.
4) Students should realize that the equation of the tangent line really
does
cross the x-axis near the true x-intercept of the function provided:
a) The initial selection of x does not yield a zero slope in
the
derivative equation. A horizontal line will never cross the x-axis!
b) The guess is 'close enough' to the place where the function
crosses the x-axis.
NOTE: A graphing calculator can provide a basis for the initial x value.
Example 1: Where does the graph of the
curve f(x) = x^2 - 24 cross the x-axis?
We wish to calculate the x value at which the tangent line crosses the
x-axis.
To make the equation of any line, we need a slope m
and a y-intercept b .
y = mx + b
Choose x = 5 as the first estimate of the axis crossing.
The process proceeds as follows:
To find an x value such that f(x) = x^2 - 24 = 0 :
Let x = 5 be the first estimate of the location of the true
x-intercept.
f '(x) = 2x gives the slope of the tangent line to f(x) for any value
of x .
f '(5) =2(5) =10 The slope is 10 when
x equals 5.
m = 10
Now we need a point on the line.
Since f(5) = 25 -24 = 1 , the point (5,1) is on the line.
(The point is also on the curve!)
Using y = mx + b ,
1 = 10(5) + b
b = - 49
Now write the equation of the line in slope - intercept form.
y = 10x - 49
Set y = 0 to find the x-intercept of the equation.
x = 4.9 is the refinement of the first guess based on
the
Newton-Raphson method.
x = 4.9 becomes the input for the next step of the procedure!
In order to get an even closer approximation of the true axis crossing,
the process is now repeated using 4.9 as the input value.
NOTE: We know that the exact answer to the equation is
root 24 (whose
approximation is 4.89898.......). Not bad for one iteration of the process!!
An Application
of the Newton - Raphson Method
Find the x (domain) value at which the graph of
f(x) = x^3 crosses
the graph of f(x) = cos(x) .
As luck(?) would have it, the graphs cross at exactly the same place
(i.e. at the same x value) as the place where the function
{ x^3 - cos(x) }
crosses the x-axis! This looks like an obvious application of the
Newton-Raphson Method!!
First set [x^3 - cos(x)] = 0
and make a guess say x = 1.
Then find the derivative and evaluate it at x = 1
3x^2 + sin(x)
3(1) + sin(1) = 3+.8415 = 3.8415
Now we evaluate the function { x^3 - cos(x) } for x = 1.
The value is (1^3 - cos(1)) =(1 - .5403) = .4597 .
Using the point (1, .4597) and the slope 3.8415 we can write
the
equation of the tangent line as follows:
y = mx + b
.4597 =3.8415(1) + b
b = - 3.3818
y = 3.8415x - 3.3818
To find where this tangent line strikes the x-axis, set y
equal to zero.
0 = 3.8415x - 3.3818
3.8415x = 3.3818
x = .8803
Since the true value is known to be near .866 , we have found a pretty
good first approximation using the Newton - Raphson Method.
( Addendum:)
To the teacher
To the student
1) Does this module improve your understandind of the
Newton-Raphson Method?
2) Does the module make its point clearly?
3) Do the example and application make the topic clearer?
4) Can you now use the Newton-Raphson Method in an appropriate
setting?
Acknowledgment
I wish to express my gratitude to the Mellon Foundation and the
Associated Colleges of the South for making the production of this module
possible.