Newton's Recursive Method for Solving Equations

We know that if we want to solve the equation

Then we want to find the point(s) where the graph of crosses the -axis. The -coordinates of these points will be all the solutions.

Method of Differential Approximation

Calculus provides a very convenient method for approximating functions, called differential approximation. We will explain the general procedure, and then show how it applies to Newton's simple problem: How to solve an equation.

Suppose given a function

and suppose we know that for the particular number that the value is another definite number: . We might then like to approximate the value of this function at a nearby point

That is, we might like to get a good estimate of when is small, or, put another way, when is close to .

Now, if we define the new function of

then this measures the difference between the actual value of at and the -coordinate of the point on the tangent line to the graph of at whose -coordinate is . To be sure that you understand that last sentence, which is a mouthful, we draw a picture. Let

The point on the tangent line

approximates the actual point

We may say that in general, for arbitrary , the point on the tangent line

approximates the actual point

Now, in the picture, is the length of the vertical segment between the horizontal lines.

From the definition of the derivative, we know that, assuming that is differentiable at the point

Question 1: Why is this so?

This means that the function of that appears to the right of the minus sign in the definition

is a very good approximation to the function

The new function is essentially , except that we measure the independent variable, h, from and the graph of is the tangent line when we measure its independent variable also from . This function is called the "differential approximation" to . We have said nothing new. We are simply reinterpreting the derivative.

Newton's Recursive Method

Newton said something new. As we shall see, Newton's Method leads to a sequence of numbers that may converge to a solution. The kind of sequence it defines is called recursive for reasons that we shall explain later. As we saw, there is another (simpler) sort of sequence called iterative and both are important to help give a clear concept of limit.

Now suppose that the number is not a solution (i.e. ), but that it is "close" to a solution. Then the point lies on the graph of f, and is "near" the -axis. If we draw the tangent line to the graph of at the point , we know from what we said above about differential approximation that this will be the graph of the function

and therefore

and
is "close to" when x is close to .

So it is plausible that the point where this tangent line crosses the -axis is close to the point we are after, a point where the graph crosses the -axis. We may find this point easily (assuming that ). It is the point

This new number is entirely determined by our first "guess" at a solution, together with the value of and of at the number . It may be a better guess than our first one, if the first one was a good guess, and so we nominate it our "second guess" :

This recursive procedure allows us to start at a reasonable guess at a solution to the equation

and then to pass to a "better" guess .

We may continue this process as long as we like, but of course, if our first guess was bad, there is no guarantee that the next step of the iteration will give a better number. And we must always be on the lookout for cases where . Further, if the solution point itself has a vertical tangent, then the successive iterations of this method may wander "chaotically" about the line, never converging to a solution. With that said, let's illustrate the method with an example.

Let us estimate . Thus we want to approximate a solution for the equation

A good first guess, our , might be , since we know that , so .

Now, according to the Method, the next guess will be

And so our answer is

Now do it again

so our answer is

This is approximately equal to 14.142135642135642

Exploration: Newton's Method

Let's do the experiment. Type in the yellow fields the function for which we seek a zero: and the initial guess for x that we call . Here, we will start with a bad guess = 11.

Next, click and you will see:

You will also see in the upper left MathEdit, some information: The step n = 0, x = 11, and f(x) = -79.

That is the 0th step. Now press . The tangent line is drawn and as you can see when we zoom in, by clicking In, then dragging a small rectangle around the region of interest,

This is the first step. The tangent line crosses the x-axis at nearly the same point the curve does. The system reports:

The new guess is approximated as 14.59090909090909 . You can see that the value of f on is 12.894628099173531 , already somewhat smaller than -79.

Now press again for the next step.

The system now reports the approximate value of to be .

14.149037099971679 . Two more steps give an approximation: 14.142135623731051

If we go to the calculator and type:

calc 14.142135623731051^2; (Enter)

We see: 200.00000000000284 to 15 place accuracy.

If we type:

calc sqrt(200); (Enter)

We see: 14.142135623730951 There is no need to go farther.

We might summarize the example above by saying that if we wanted to estimate and a first approximation was , then a next approximation would be

Question 1:
This is just the average of with and so is plausible. Try to devise a general formula for approximating by starting with and calculating (using two steps of Newton's Method instead of just one). Apply your formula to estimate the square root of 10, starting with =3.

Solution: First of all, we may simplify our recursion formula:

(1.a.6)

Now apply it again:

(1.a.7)

That is the approximation formula for is

starting with approximation .

Let's take it out for a spin. Suppose . Then the formula gives:

go to the calculator and type

calc sqrt(10); (Enter)

You see 3.162277660168379. So our answer is accurate to 1 part in 100,000. (5 decimal places) .

End of Solution

This recursion is Newton's Method. Try it on the previous page on Recursive Sequences. Type the information:

and you see that the recursion converges (to within 15 places) on the 5th step!

1 : 3
2 : 3.166666666666666
3 : 3.162280701754386
4 : 3.162277660169841
5 : 3.162277660168379
6 : 3.162277660168379
7 : 3.162277660168379

when we type

calc 3.162277660168379^2; (Enter)

we see: 9.999999999999998

That is not bad at all.

Question 2: Develop a formula for estimating cube roots of numbers in a similar fashion, using two steps of Newton's method. Apply your formula to estimate the cube root of 10, starting with =2.