Section 4.13 : Newton's Method
4. Use Newton’s Method to find the root of \(2{x^2} + 5 = {{\bf{e}}^x}\) accurate to six decimal places in the interval \(\left[ {3,4} \right]\).
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Start SolutionFirst, recall that Newton’s Method solves equation in the form \(f\left( x \right) = 0\) and so we’ll need move everything to one side. Doing this gives,
\[f\left( x \right) = 2{x^2} + 5 - {{\bf{e}}^x}\]Note that we could have just as easily gone the other direction. All that would have done was change the signs on the function and derivative evaluations in the work below. The final answers however would not be changed.
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, \({x_{\,0}} = 3.5\). 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 that 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 2The first iteration through the formula for \({x_{\,1}}\) is,
\[{x_{\,1}} = {x_{\,0}} - \frac{{f\left( {{x_{\,0}}} \right)}}{{f'\left( {{x_{\,0}}} \right)}} = 3.5 - \frac{{ - 3.615451959}}{{ - 19.11545196}} = 3.310862334\] Show Step 3The second iteration through the formula for \({x_{\,2}}\) is,
\[{x_{\,2}} = {x_{\,1}} - \frac{{f\left( {{x_{\,1}}} \right)}}{{f'\left( {{x_{\,1}}} \right)}} = 3.310862334 - \frac{{ - 0.4851319992}}{{ - 14.16530146}} = 3.276614422\]We’ll need to keep going because even the first decimal is not correct yet.
Show Step 4The second iteration through the formula for \({x_{\,3}}\) is,
\[{x_{\,3}} = {x_{\,2}} - \frac{{f\left( {{x_{\,2}}} \right)}}{{f'\left( {{x_{\,2}}} \right)}} = 3.276614422 - \frac{{ - 0.0135463486}}{{ - 13.37949281}} = 3.275601951\]At this point we are accurate to two decimal places so we need to continue.
Show Step 5The second iteration through the formula for \({x_{\,4}}\) is,
\[{x_{\,4}} = {x_{\,3}} - \frac{{f\left( {{x_{\,3}}} \right)}}{{f'\left( {{x_{\,3}}} \right)}} = 3.275601951 - \frac{{ - 0.00001152056596}}{{ - 13.356740003}} = 3.275601089\]At this point we are accurate to 6 decimal places which is what we were asked to do and so we can officially stop and we can estimate that the root in the interval is,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{x \approx 3.275601089}}\]Using computational aids we found that the actual root in this interval is \(3.27560108884732\). Note that this wasn’t actually asked for in the problem and is only given for comparison purposes and it does look like Newton’s Method did a pretty good job as this is identical to the final iteration that we did.