Newton's Method & Linear Approximation

Home > Calculus 1 > Linear Approximation

Linear Approximation: Newton's Method | Linear Approximation | Accuracy | Practice Problems | Quiz

We've already examined how to use derivatives to find the tangent line of a function at a particular point. Here, we'll demonstrate some of the uses of tangent lines. Newton's Method is used to estimate the roots of a function, while linear approximation is used to estimate the value of a function at a particular point.

NEWTON'S METHOD: (top)

Recall that the roots of a function are the points where it equals zero; these are also the points at which the function's graph touch the x-axis. It can be difficult to find the roots of a given function without the aid of a calculator or a computer. Newton's method is a way of estimating these roots using tangent lines.

Since the ideas here will be much clearer if motivated by an example, let's find the roots of the equation (thereby solving the equation ) as we explain Newton's method. We'll always follow the same steps:

Use the Intermediate Value Theorem to locate an approximate for the root.
We'll start off by making an x|y chart to find where there's a change in sign in the function. The image at right shows such a chart for . Since there is a sign change between f(1) and f(2), we know by the Intermediate Value Theorem that there is a root between x=1 and x=2.
Find a point of tangency on the function to serve as the initial guess for the root.

We're just going to choose a point that could be the root. Since in this example we know there's a root somewhere between x=1 and x=2, let's guess x=1.5 as the root.
Let . Now we'll find the y-value of that root:
Thus, the point of tangency is (1.5, –.75).
Write the equation of the tangent line at that point.
To find the equation of the tangent line, simply find the derivative at the point of tangency, then use the equation y=mx+b to find the tangent line.

Thus, .
Find where the tangent line crosses the x-axis (since we don't know where the curve crosses the x-axis). This is an approximate of the root.

This is achieved by simply setting the equation of the tangent line equal to 0: , so
Thus, is our new aproximate for the root.

Repeat steps 2-4 until the root is found to the desired level of precision.

We can see how close this approximate is by finding the value of the function there. The closer the function is to 0, the more accurate our approximate is. (If the funciton equals exactly 0, then the root is exact).

In this case, f(1.75)=.0625; not bad, but not perfect, either. Suppose we want a greater level of precision than we have achieved thus far. We can simply repeat the procedure as many times as desired to get greater accuracy. Let's do it once more:

Find the point of tangency:
Here, it's (1.75, .0625)

Write the equation of tangent line:

Thus, .

Find where the the tangent line crosses the x-axis:

Thus, is our new approximate for the root.

Thus, we've found that has a root at approximately x=1.732. We could, of course, do the steps of Newton's method yet again to refine this answer even further, but since f(1.732)=-0.00176, which is very close to 0, there's no real reason to do so; furthermore, keeping track of all the decimals will start to make the computations more complicated.

Take a look at the graph of , at left. Notice that there are clearly two places where the graph crosses the x-axis. This also becomes apparent if we solve for the roots algebraically using a calculator:

Using Newton's method, however, we can only find one root at a time. This is because our first step is to use the Intermediate Value Theorem to find a single range of x-values where a root could exist. Consequently, to find all roots of a function using Newton's method, one must perform these steps at every sign change in the function.

LINEAR APPROXIMATION: (top)

Newton's method allows us to find approximate values of x where f(x)=0. We can also use tangent lines to estimate the value of a function at other x-values. This is known as linear approximation. Really, we just want to estimate "ugly" (non-linear) functions at "ugly" numbers.

As with Newton's method, discussion of how to perform linear approximation will be much clearer if accompanied by a motivating example. Thus, let's see how tangent lines can help us approximate . Again, we'll have a general procedure to follow:

Identify the "ugly function" and the "ugly number"
Here, the ugly function is , and the ugly number is 4.2
Find a nice number close to the ugly number, and write the equation of the tangent line at that nice number.

A good nice number is 4. It's nice because it's very easy to evaluate the function at the point: . Thus, the point of tangency is (4, 2).

To find the slope of the tangent line, we'll first take the derivative:
Since , and
Now, we'll use y=mx+b to find the tangent line: .

Thus, we have as the equation of the tanget line at the nice number, which we can use to estimate the ugly function.
Substitute the ugly number into the nice, linear function to approximate the ugly function at the ugly number.
In our example, the tangent line is , so for x=4.2, we find

Thus, y=2.05 is the exact y-value for x=4.2 on the tangent line, and it is an approximation of x=4.2 on the curve. (Using a calculator, we find that ≈2.0493).

A COMMENT ON ACCURACY: (top)

In the examples used to motivate discussion of Newton's method and linear approximation, we used a calculator to evaluate the quantities we had approximated. The point was to show the high level of accuracy achieved by these approximations, and, indeed, the approximations achieved by these methods can often be very good.

They can also be somewhat less good, however, and we'll take a brief moment to explain when such methods will do a good job of approximating curves and when they will do a poor job. The bottom line is this:

The more closely a tangent line matches its parent function, the better a job it will do of approximating the roots and y-values of its function.
Once stated, this principle seems rather obvious. Let's consider a few examples to drive the point home, though. The images below show the function y=5sin(x) with the tangent lines drawn in a two different points.

In the image on the left, the point of tangency is at (π/2, 5). Here, the tangent line does not approximate the curve particularly well. If one moves π/4 to the right of the POT, to 3π/4, a fairly large error is introduced. The actual y-value is 3.53, but the tangent line gives an estimated y-value of 5, a difference of 2.47.

In the image on the right, however, the point of tangency is at (π,0), and in this case, the tangent line does do a good job of approximating the curve. If we move π/4 to the left of the POT, to π/2, the difference between the y-values on the tangent line and the actual curve is small: the actual y-value is 3.53 and the tangent line gives an estimated y-value of 3.93, a difference of only .4.

It's also important to notice that how well a tangent line does of approximating a curve depends on how close the point of interest is to the point of tangency. Even for the graph above on the left, the tangent line does a very good job of approximating the curve for points very close to the point of tangency. For both graphs, however, the tangent line diverges wildy from the actual graph as one moves away from the point of tangency. Thus, when using linear approximation to estimate "ugly funtions at ugly numbers," it's important that the "nice number" for the point of tangency be chose close to the "ugly number" of interest.

PRACTICE PROBLEMS: (top)

QUIZ: (top)

This page last updated 8 August, 2008 3:25 PM