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4.9 Newton’s method  (Page 3/7)

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A function is drawn with two roots, labeled root sought and root found. A point x0 is chosen such that when the tangent of x0 is taken, even though it is nearer to the root sought, the tangent points to the root found.
If the initial guess x 0 is too far from the root sought, it may lead to approximations that approach a different root.

When newton’s method fails

Consider the function f ( x ) = x 3 − 2 x + 2 . Let x 0 = 0 . Show that the sequence x 1 , x 2 ,… fails to approach a root of f .

For f ( x ) = x 3 − 2 x + 2 , the derivative is f ′ ( x ) = 3 x 2 − 2 . Therefore,

x 1 = x 0 − f ( x 0 ) f ′ ( x 0 ) = 0 − f ( 0 ) f ′ ( 0 ) = − 2 −2 = 1 .

In the next step,

x 2 = x 1 − f ( x 1 ) f ′ ( x 1 ) = 1 − f ( 1 ) f ′ ( 1 ) = 1 − 1 1 = 0 .

Consequently, the numbers x 0 , x 1 , x 2 ,… continue to bounce back and forth between 0 and 1 and never get closer to the root of f which is over the interval [ −2 , −1 ] (see [link] ). Fortunately, if we choose an initial approximation x 0 closer to the actual root, we can avoid this situation.

The function f(x) = x3 – 2x + 2 is drawn, which has a root between −2 and −1. The tangent from x = 0 goes to x = 1, and the tangent from x = 1 goes to x = 0.
The approximations continue to alternate between 0 and 1 and never approach the root of f .
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For f ( x ) = x 3 − 2 x + 2 , let x 0 = −1.5 and find x 1 and x 2 .

x 1 ≈ − 1.842105263 , x 2 ≈ − 1.772826920

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From [link] , we see that Newton’s method does not always work. However, when it does work, the sequence of approximations approaches the root very quickly. Discussions of how quickly the sequence of approximations approach a root found using Newton’s method are included in texts on numerical analysis.

Other iterative processes

As mentioned earlier, Newton’s method is a type of iterative process. We now look at an example of a different type of iterative process.

Consider a function F and an initial number x 0 . Define the subsequent numbers x n by the formula x n = F ( x n − 1 ) . This process is an iterative process that creates a list of numbers x 0 , x 1 , x 2 ,… , x n ,… . This list of numbers may approach a finite number x * as n gets larger, or it may not. In [link] , we see an example of a function F and an initial guess x 0 such that the resulting list of numbers approaches a finite value.

Finding a limit for an iterative process

Let F ( x ) = 1 2 x + 4 and let x 0 = 0 . For all n ≥ 1 , let x n = F ( x n − 1 ) . Find the values x 1 , x 2 , x 3 , x 4 , x 5 . Make a conjecture about what happens to this list of numbers x 1 , x 2 , x 3 … , x n ,… as n → ∞ . If the list of numbers x 1 , x 2 , x 3 ,… approaches a finite number x * , then x * satisfies x * = F ( x * ) , and x * is called a fixed point of F .

If x 0 = 0 , then

x 1 = 1 2 ( 0 ) + 4 = 4 x 2 = 1 2 ( 4 ) + 4 = 6 x 3 = 1 2 ( 6 ) + 4 = 7 x 4 = 1 2 ( 7 ) + 4 = 7.5 x 5 = 1 2 ( 7.5 ) + 4 = 7.75 x 6 = 1 2 ( 7.75 ) + 4 = 7.875 x 7 = 1 2 ( 7.875 ) + 4 = 7.9375 x 8 = 1 2 ( 7.9375 ) + 4 = 7.96875 x 9 = 1 2 ( 7.96875 ) + 4 = 7.984375.

From this list, we conjecture that the values x n approach 8 .

[link] provides a graphical argument that the values approach 8 as n → ∞ . Starting at the point ( x 0 , x 0 ) , we draw a vertical line to the point ( x 0 , F ( x 0 ) ) . The next number in our list is x 1 = F ( x 0 ) . We use x 1 to calculate x 2 . Therefore, we draw a horizontal line connecting ( x 0 , x 1 ) to the point ( x 1 , x 1 ) on the line y = x , and then draw a vertical line connecting ( x 1 , x 1 ) to the point ( x 1 , F ( x 1 ) ) . The output F ( x 1 ) becomes x 2 . Continuing in this way, we could create an infinite number of line segments. These line segments are trapped between the lines F ( x ) = x 2 + 4 and y = x . The line segments get closer to the intersection point of these two lines, which occurs when x = F ( x ) . Solving the equation x = x 2 + 4 , we conclude they intersect at x = 8 . Therefore, our graphical evidence agrees with our numerical evidence that the list of numbers x 0 , x 1 , x 2 ,… approaches x * = 8 as n → ∞ .

The function F(x) = (1/2)x + 4 is graphed along with y = x. From x0, which appears to be at the origin, a line is drawn to the function F(x) at x1 = F(x0). Then a line is drawn to the right from here to the line y = x, at which point a line is drawn up to x2 = F(x1). Then a line is drawn to the right from here to the line y = x, at which point a line is drawn up to x3 = F(x2). Continuing this process would converge on the two lines’ intersection point at x* = 8.
This iterative process approaches the value x * = 8 .
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Source:  OpenStax, Calculus volume 1. OpenStax CNX. Feb 05, 2016 Download for free at http://cnx.org/content/col11964/1.2
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