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### Section 4-13 : Newton's Method

5. Use Newton’s Method to find all the roots of \({x^3} - {x^2} - 15x + 1 = 0\) accurate to six decimal places.

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First, recall that Newton’s Method solves equation in the form \(f\left( x \right) = 0\) and so it is (hopefully) fairly clear that we have,

\[f\left( x \right) = {x^3} - {x^2} - 15x + 1\]Next, we are not given a starting value, \({x_{\,0}}\) and unlike Problems 3 & 4 above we are not even given an interval to use as a way to determine a good possible value of \({x_{\,0}}\). We are also not even told how many roots we need to find.

Of course, if we recall our Algebra skills we can see that we have a cubic polynomial and so there should be at most three distinct roots of the equation (there may be some that repeat and so we may not have three distinct roots…). Knowing this all we really need to do to get potential starting values is to do a quick sketch of the function.

In determining a proper range of \(x\) values just keep in mind what we know about limits at infinity. Because the largest power of \(x\) is odd in this case we know that as \(x \to \infty \) the graph should also be approaching positive infinity and as \(x \to - \infty \) the graph should be approaching negative infinity. So, we can start with a large range of \(x\)’s that gives the behavior we expect at the right/left ends of the graph and then narrow it down until we see the actual roots showing up on the graph.

Doing this gives,

So, it looks like we are going to have three roots here (*i.e.* the graph crosses the \(x\)-axis three times and so three roots…).

For each root we’ll use the graph to pick a value of \({x_{\,0}}\) that is close to the root we are after (we’ll go from left to right for the problem) and then run through Newton’s Method,

\[{x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}\]until the answers agree to six decimal places.

Note as well that unlike Problems 3 & 4 we are not going to put in all the function evaluations for this problem. We’ll leave that to you to check and verify our final answers for each iteration.

Show Step 2For the left most root let’s start with \({x_{\,0}} = - 3.5\). Here are the results of iterating through Newton’s Method for this root.

\[\begin{align*}{x_{\,1}} & = {x_{\,0}} - \frac{{f\left( {{x_{\,0}}} \right)}}{{f'\left( {{x_{\,0}}} \right)}} = - 3.443478261 & \hspace{0.5in} & {\mbox{No decimal places agree}}\\ {x_{\,2}} & = {x_{\,1}} - \frac{{f\left( {{x_{\,1}}} \right)}}{{f'\left( {{x_{\,1}}} \right)}} = - 3.442146902 & \hspace{0.5in} & {\mbox{Accurate to two decimal places}}\\ {x_{\,3}} & = {x_{\,2}} - \frac{{f\left( {{x_{\,2}}} \right)}}{{f'\left( {{x_{\,2}}} \right)}} = - 3.44214617 & \hspace{0.5in} & {\mbox{Accurate to six decimal places}}\end{align*}\]So, it looks like the estimate of the left most root is : \(\require{bbox} \bbox[2pt,border:1px solid black]{{x \approx - 3.44214617}}\).

Show Step 3For the middle root let’s start with \({x_{\,0}} = 0\). Be careful with this root. From the graph we may be tempted to just say the root is zero. However, as we’ll see the root is not zero. It is close to zero, but is not exactly zero!

Here are the results of iterating through Newton’s Method for this root.

\[\begin{align*}{x_{\,1}} & = {x_{\,0}} - \frac{{f\left( {{x_{\,0}}} \right)}}{{f'\left( {{x_{\,0}}} \right)}} = 0.06666666667 & \hspace{0.5in} & {\mbox{No decimal places agree}}\\ {x_{\,2}} & = {x_{\,1}} - \frac{{f\left( {{x_{\,1}}} \right)}}{{f'\left( {{x_{\,1}}} \right)}} = 0.06639231824 & \hspace{0.5in} & {\mbox{Accurate to three decimal places}}\\ {x_{\,3}} & = {x_{\,2}} - \frac{{f\left( {{x_{\,2}}} \right)}}{{f'\left( {{x_{\,2}}} \right)}} = 0.06639231426 & \hspace{0.5in} & {\mbox{Accurate to eight decimal places}}\end{align*}\]So, it looks like the estimate of the middle root is : \(\require{bbox} \bbox[2pt,border:1px solid black]{{x \approx 0.06639231426}}\).

Show Step 4For the right most root let’s start with \({x_{\,0}} = 4.5\). Here are the results of iterating through Newton’s Method for this root.

\[\begin{align*}{x_{\,1}} & = {x_{\,0}} - \frac{{f\left( {{x_{\,0}}} \right)}}{{f'\left( {{x_{\,0}}} \right)}} = 4.380952381 & \hspace{0.5in} & {\mbox{No decimal places agree}}\\ {x_{\,2}} & = {x_{\,1}} - \frac{{f\left( {{x_{\,1}}} \right)}}{{f'\left( {{x_{\,1}}} \right)}} = 4.375763556 & \hspace{0.5in} & {\mbox{Accurate to one decimal place}}\\ {x_{\,3}} & = {x_{\,2}} - \frac{{f\left( {{x_{\,2}}} \right)}}{{f'\left( {{x_{\,2}}} \right)}} = 4.375753856 & \hspace{0.5in} & {\mbox{Accurate to four decimal places}}\\ {x_{\,4}} & = {x_{\,3}} - \frac{{f\left( {{x_{\,3}}} \right)}}{{f'\left( {{x_{\,3}}} \right)}} = 4.375753856 & \hspace{0.5in} & {\mbox{Accurate to nine decimal places}}\end{align*}\]So, it looks like the estimate of the right most root is : \(\require{bbox} \bbox[2pt,border:1px solid black]{{x \approx 4.375753856}}\).

Show Step 5Using computational aids we found that the actual roots of this equation to be,

\[x = - 3.44214616993\hspace{0.5in}x = 0.0663923142603\hspace{0.5in}x = 4.37575385567\]Note that these weren’t actually asked for in the problem and are only given for comparison purposes.

As a final warning about Newton’s Method, be careful to not assume that you’ll get six (or better in some cases) decimal places of accuracy with just a few iterations.

These problems were chosen with the understanding that it would only take a few iterations of the method. There are problems and/or choices of \({x_{\,0}}\) for which it will take significantly more iterations to get any kind of real accuracy, provided the method even works for that equation and/or choice of \({x_{\,0}}\). Recall that we saw an example in the notes in which the method failed spectacularly.