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Finding a root of a polynomial

Use Newton’s method to approximate a root of f ( x ) = x 3 3 x + 1 in the interval [ 1 , 2 ] . Let x 0 = 2 and find x 1 , x 2 , x 3 , x 4 , and x 5 .

From [link] , we see that f has one root over the interval ( 1 , 2 ) . Therefore x 0 = 2 seems like a reasonable first approximation. To find the next approximation, we use [link] . Since f ( x ) = x 3 3 x + 1 , the derivative is f ( x ) = 3 x 2 3 . Using [link] with n = 1 (and a calculator that displays 10 digits), we obtain

x 1 = x 0 f ( x 0 ) f ( x 0 ) = 2 f ( 2 ) f ( 2 ) = 2 3 9 1.666666667 .

To find the next approximation, x 2 , we use [link] with n = 2 and the value of x 1 stored on the calculator. We find that

x 2 = x 1 = f ( x 1 ) f ( x 1 ) 1.548611111 .

Continuing in this way, we obtain the following results:

x 1 1.666666667 x 2 1.548611111 x 3 1.532390162 x 4 1.532088989 x 5 1.532088886 x 6 1.532088886 .

We note that we obtained the same value for x 5 and x 6 . Therefore, any subsequent application of Newton’s method will most likely give the same value for x n .

The function f(x) = x3 – 3x + 1 is drawn. It has roots between −2 and −1, 0 and 1, and 1 and 2.
The function f ( x ) = x 3 3 x + 1 has one root over the interval [ 1 , 2 ] .
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Letting x 0 = 0 , let’s use Newton’s method to approximate the root of f ( x ) = x 3 3 x + 1 over the interval [ 0 , 1 ] by calculating x 1 and x 2 .

x 1 0.33333333 , x 2 0.347222222

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Newton’s method can also be used to approximate square roots. Here we show how to approximate 2 . This method can be modified to approximate the square root of any positive number.

Finding a square root

Use Newton’s method to approximate 2 ( [link] ). Let f ( x ) = x 2 2 , let x 0 = 2 , and calculate x 1 , x 2 , x 3 , x 4 , x 5 . (We note that since f ( x ) = x 2 2 has a zero at 2 , the initial value x 0 = 2 is a reasonable choice to approximate 2 . )

For f ( x ) = x 2 2 , f ( x ) = 2 x . From [link] , we know that

x n = x n 1 f ( x n 1 ) f ( x n 1 ) = x n 1 x 2 n 1 2 2 x n 1 = 1 2 x n 1 + 1 x n 1 = 1 2 ( x n 1 + 2 x n 1 ) .

Therefore,

x 1 = 1 2 ( x 0 + 2 x 0 ) = 1 2 ( 2 + 2 2 ) = 1.5 x 2 = 1 2 ( x 1 + 2 x 1 ) = 1 2 ( 1.5 + 2 1.5 ) 1.416666667.

Continuing in this way, we find that

x 1 = 1.5 x 2 1.416666667 x 3 1.414215686 x 4 1.414213562 x 5 1.414213562.

Since we obtained the same value for x 4 and x 5 , it is unlikely that the value x n will change on any subsequent application of Newton’s method. We conclude that 2 1.414213562 .

The function y = x2 – 2 is drawn. A dashed line comes up from x0 = 2, and a tangent line is drawn down from there. It touches x1 = 1.5, which is near x* = the square root of 2.
We can use Newton’s method to find 2 .
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Use Newton’s method to approximate 3 by letting f ( x ) = x 2 3 and x 0 = 3 . Find x 1 and x 2 .

x 1 = 2 , x 2 = 1.75

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When using Newton’s method, each approximation after the initial guess is defined in terms of the previous approximation by using the same formula. In particular, by defining the function F ( x ) = x [ f ( x ) f ( x ) ] , we can rewrite [link] as x n = F ( x n 1 ) . This type of process, where each x n is defined in terms of x n 1 by repeating the same function, is an example of an iterative process    . Shortly, we examine other iterative processes. First, let’s look at the reasons why Newton’s method could fail to find a root.

Failures of newton’s method

Typically, Newton’s method is used to find roots fairly quickly. However, things can go wrong. Some reasons why Newton’s method might fail include the following:

  1. At one of the approximations x n , the derivative f is zero at x n , but f ( x n ) 0 . As a result, the tangent line of f at x n does not intersect the x -axis. Therefore, we cannot continue the iterative process.
  2. The approximations x 0 , x 1 , x 2 ,… may approach a different root. If the function f has more than one root, it is possible that our approximations do not approach the one for which we are looking, but approach a different root (see [link] ). This event most often occurs when we do not choose the approximation x 0 close enough to the desired root.
  3. The approximations may fail to approach a root entirely. In [link] , we provide an example of a function and an initial guess x 0 such that the successive approximations never approach a root because the successive approximations continue to alternate back and forth between two values.

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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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