4.9 Newton’s method  (Page 2/7)

Finding a root of a polynomial

Use Newton’s method to approximate a root of $f\left(x\right)=x^3-3x+1$ in the interval $\left[1,2\right].$ 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 $\left(1,2\right).$ Therefore $x_0=2$ seems like a reasonable first approximation. To find the next approximation, we use [link] . Since $f\left(x\right)=x^3-3x+1,$ the derivative is $f^{\prime }\left(x\right)=3x^2-3.$ Using [link] with $n=1$ (and a calculator that displays $10$ digits), we obtain

$x_1=x_0-\frac{f\left(x_0\right)}{f\prime \left(x_0\right)}=2-\frac{f\left(2\right)}{f\prime \left(2\right)}=2-\frac{3}{9}\approx 1.666666667.$

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

$x_2=x_1=\frac{f\left(x_1\right)}{f\prime \left(x_1\right)}\approx 1.548611111.$

Continuing in this way, we obtain the following results:

$\begin{array}{c}{x}_1\approx 1.666666667\hfill \\ x_2\approx 1.548611111\hfill \\ x_3\approx 1.532390162\hfill \\ x_4\approx 1.532088989\hfill \\ x_5\approx 1.532088886\hfill \\ x_6\approx 1.532088886.\hfill \end{array}$

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 $x_n.$

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

Finding a square root

Use Newton’s method to approximate $\sqrt{2}$ ( [link] ). Let $f\left(x\right)=x^2-2,$ let $x_0=2,$ and calculate $x_1,x_2,x_3,x_4,x_5.$ (We note that since $f\left(x\right)=x^2-2$ has a zero at $\sqrt{2},$ the initial value $x_0=2$ is a reasonable choice to approximate $\sqrt{2}.)$

For $f\left(x\right)=x^2-2,f^{\prime }\left(x\right)=2x.$ From [link] , we know that

$\begin{array}{cc}\hfill x_n& =x_{n-1}-\frac{f\left(x_{n-1}\right)}{f\prime \left(x_{n-1}\right)}\hfill \\ & =x_{n-1}-\frac{x^2{}_{n-1}-2}{2x_{n-1}}\hfill \\ & =\frac{1}{2}{x}_{n-1}+\frac{1}{x_{n-1}}\hfill \\ & =\frac{1}{2}\left(x_{n-1}+\frac{2}{x_{n-1}}\right).\hfill \end{array}$

Therefore,

$\begin{array}{}\\ \\ x_1=\frac{1}{2}\left(x_0+\frac{2}{x_0}\right)=\frac{1}{2}\left(2+\frac{2}{2}\right)=1.5\hfill \\ x_2=\frac{1}{2}\left(x_1+\frac{2}{x_1}\right)=\frac{1}{2}\left(1.5+\frac{2}{1.5}\right)\approx \mathrm{1.416666667.}\hfill \end{array}$

Continuing in this way, we find that

$\begin{array}{c}{x}_1=1.5\hfill \\ x_2\approx 1.416666667\hfill \\ x_3\approx 1.414215686\hfill \\ x_4\approx 1.414213562\hfill \\ x_5\approx \mathrm{1.414213562.}\hfill \end{array}$

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 $\sqrt{2}\approx 1.414213562.$

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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 $F\left(x\right)=x-\left[\frac{f\left(x\right)}{f^{\prime }\left(x\right)}\right],$ we can rewrite [link] as $x_n=F\left(x_{n-1}\right).$ 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 $f^{\prime }$ is zero at $x_n,$ but $f\left(x_n\right)\ne 0.$ 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 $x_0,x_1,x_2\text{,…}$ 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.