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Section 4-13 : Newton's Method

3. Use Newton’s Method to find the root of \({x^4} - 5{x^3} + 9x + 3 = 0\) accurate to six decimal places in the interval \(\left[ {4,6} \right]\).

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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,

\[f\left( x \right) = {x^4} - 5{x^3} + 9x + 3\]

Next, we are not given a starting value, \({x_{\,0}}\), but we were given an interval in which the root exists so we may as well use the midpoint of this interval as our starting point or, \({x_{\,0}} = 5\). Note that this is not the only value we could use and if you use a different one (which is perfectly acceptable) then your values will be different from those here.

At this point all we need to do is 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.

Show Step 2

The first iteration through the formula for \({x_{\,1}}\) is,

\[{x_{\,1}} = {x_{\,0}} - \frac{{f\left( {{x_{\,0}}} \right)}}{{f'\left( {{x_{\,0}}} \right)}} = 5 - \frac{{48}}{{134}} = 4.641791045\] Show Step 3

The second iteration through the formula for \({x_{\,2}}\) is,

\[{x_{\,2}} = {x_{\,1}} - \frac{{f\left( {{x_{\,1}}} \right)}}{{f'\left( {{x_{\,1}}} \right)}} = 4.641791045 - \frac{{8.950542057}}{{85.85891882}} = 4.537543959\]

We’ll need to keep going because even the first decimal is not correct yet.

Show Step 4

The third iteration through the formula for \({x_{\,3}}\) is,

\[{x_{\,3}} = {x_{\,2}} - \frac{{f\left( {{x_{\,2}}} \right)}}{{f'\left( {{x_{\,2}}} \right)}} = 4.537543959 - \frac{{0.6329967413}}{{73.85993168}} = 4.528973727\]

At this point we are accurate to the first decimal place so we need to continue.

Show Step 5

The fourth iteration through the formula for \({x_{\,4}}\) is,

\[{x_{\,4}} = {x_{\,3}} - \frac{{f\left( {{x_{\,3}}} \right)}}{{f'\left( {{x_{\,3}}} \right)}} = 4.528973727 - \frac{{0.004066133005}}{{72.91199944}} = 4.52891796\]

At this point we are accurate to 4 decimal places so we need to continue.

Show Step 6

The fifth iteration through the formula for \({x_{\,5}}\) is,

\[{x_{\,5}} = {x_{\,4}} - \frac{{f\left( {{x_{\,4}}} \right)}}{{f'\left( {{x_{\,4}}} \right)}} = 4.52891796 - \frac{{1.714694911*{{10}^{ - 7}}}}{{72.90585006}} = 4.52891796\]

At this point we are accurate to 8 decimal places which is actually better than we asked and so we can officially stop and we can estimate that the root in the interval is,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{x \approx 4.52891796}}\]

Using computational aids we found that the actual root in this interval is 4.52891795729. Note that this wasn’t actually asked for in the problem and is only given for comparison purposes.