The next application that we’ll take a look at in this
chapter is an important application that is used in many areas. If you’ve been following along in the chapter
to this point it’s quite possible that you’ve gotten the impression that many
of the applications that we’ve looked at are just made up by us to make you
work. This is unfortunate because all of
the applications that we’ve looked at to this point are real applications that
really are used in real situations. The
problem is often that in order to work more meaningful examples of the
applications we would need more knowledge than we generally have about the
science and/or physics behind the problem.
Without that knowledge we’re stuck doing some fairly simplistic examples
that often don’t seem very realistic at all and that makes it hard to
understand that the application we’re looking at is a real application.
That is going to change in this section. This is an application that we can all
understand and we can all understand needs to be done on occasion even if we
don’t understand the physics/science behind an actual application.
In this section we are going to look at a method for
approximating solutions to equations. We
all know that equations need to be solved on occasion and in fact we’ve solved
quite a few equations ourselves to this point.
In all the examples we’ve looked at to this point we were able to
actually find the solutions, but it’s not always possible to do that exactly
and/or do the work by hand. That is where
this application comes into play. So,
let’s see what this application is all about.
Let’s suppose that we want to approximate the solution to and let’s also suppose that we have somehow
found an initial approximation to this solution say, x0. This initial
approximation is probably not all that good and so we’d like to find a better
approximation. This is easy enough to
do. First we will get the tangent line
to at x0.
Now, take a look at the graph below.
The blue line (if you’re reading this in color anyway…) is
the tangent line at x0. We can see that this line will cross the x-axis much closer to the actual
solution to the equation than x0
does. Let’s call this point where the
tangent at x0 crosses the x-axis x1 and we’ll use this point as our new approximation to
So, how do we find this point? Well we know it’s coordinates, ,
and we know that it’s on the tangent line so plug this point into the tangent
line and solve for x1 as
So, we can find the new approximation provided the
derivative isn’t zero at the original approximation.
Now we repeat the whole process to find an even better
approximation. We form up the tangent
line to at x1
and use its root, which we’ll call x2,
as a new approximation to the actual solution.
If we do this we will arrive at the following formula.
This point is also shown on the graph above and we can see
from this graph that if we continue following this process will get a sequence
of numbers that are getting very close the actual solution. This process is called Newton’s Method.
Here is the general Newton’s
This should lead to the question of when do we stop? How many times do we go through this
process? One of the more common stopping
points in the process is to continue until two successive approximations agree
to a given number of decimal places.
Before working any examples we should address two
issues. First, we really do need to be
solving in order for Newton’s Method to be applied. This isn’t really all that much of an issue
but we do need to make sure that the equation is in this form prior to using
Secondly, we do need to somehow get our hands on an initial
approximation to the solution (i.e.
we need somehow).
One of the more common ways of getting our hands on is to sketch the graph of the function and use
that to get an estimate of the solution which we then use as . Another common method is if we know that
there is a solution to a function in an interval then we can use the midpoint
of the interval as .
Let’s work an example of Newton’s Method.
Example 1 Use
Newton’s Method to determine an approximation to the solution to that lies in the interval [0,2]. Find the approximation to six decimal
First note that we weren’t given an initial guess. We were however, given an interval in which
to look. We will use this to get our
initial guess. As noted above the
general rule of thumb in these cases is to take the initial approximation to
be the midpoint of the interval. So,
we’ll use as our initial guess.
Next, recall that we must
have the function in the form . Therefore, we first rewrite the equation
We can now write down the general formula for Newton’s
Method. Doing this will often simplify
up the work a little so it’s generally not a bad idea to do this.
Let’s now get the first approximation.
At this point we should point out that the phrase “six
decimal places” does not mean just get x1
to six decimal places and then stop.
Instead it means that we continue until two successive approximations
agree to six decimal places.
Given that stopping condition we clearly need to go at
least one step farther.
Alright, we’re making progress. We’ve got the approximation to 1 decimal
place. Let’s do another one, leaving
the details of the computation to you.
We’ve got it to three decimal places. We’ll need another one.
And now we’ve got two approximations that agree to 9
decimal places and so we can stop. We
will assume that the solution is approximately .
In this last example we saw that we didn’t have to do too
many computations in order for Newton’s
Method to give us an approximation in the desired range of accuracy. This will not always be the case. Sometimes it will take many iterations
through the process to get to the desired accuracy and on occasion it can fail
The following example is a little silly but it makes the
point about the method failing.
Example 2 Use
to find the approximation to the solution to
Yes, it’s a silly example.
Clearly the solution is ,
but it does make a very important point.
Let’s get the general formula for Newton’s method.
In fact we don’t really need to do any computations
here. These computations get farther
and farther away from the solution, ,with
each iteration. Here are a couple of
computations to make the point.
So, in this case the method fails and fails spectacularly.
So, we need to be a little careful with Newton’s method. It will usually quickly find an approximation
to an equation. However, there are times
when it will take a lot of work or when it won’t work at all.