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Section 4.13 : Newton's Method

2. Use Newton’s Method to determine \({x_{\,2}}\) for \(f\left( x \right) = x\cos \left( x \right) - {x^2}\) if \({x_{\,0}} = 1\)

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There really isn’t that much to do with this problem. We know that the basic formula for Newton’s Method is,

\[{x_{n + 1}} = {x_n} - \frac{{f\left( {{x_n}} \right)}}{{f'\left( {{x_n}} \right)}}\]

so all we need to do is run through this twice.

Here is the derivative of the function since we’ll need that.

\[f'\left( x \right) = \cos \left( x \right) - x\sin \left( x \right) - 2x\]

We just now need to run through the formula above twice.

Show Step 2

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

\[{x_{\,1}} = {x_{\,0}} - \frac{{f\left( {{x_{\,0}}} \right)}}{{f'\left( {{x_{\,0}}} \right)}} = 1 - \frac{{f\left( 1 \right)}}{{f'\left( 1 \right)}} = 1 - \frac{{ - 0.4596976941}}{{ - 2.301168679}} = 0.8002329432\]

Don’t forget that for us angles are always in radians so make sure your calculator is set to compute in radians.

Show Step 3

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

\[\begin{align*}{x_{\,2}} = {x_{\,1}} - \frac{{f\left( {{x_{\,1}}} \right)}}{{f'\left( {{x_{\,1}}} \right)}} & = 0.8002329432 - \frac{{f\left( {0.8002329432} \right)}}{{f'\left( {0.8002329432} \right)}}\\ & = 0.8002329432 - \frac{{ - 0.08297883948}}{{ - 1.478108132}} = 0.7440943985\end{align*}\] So, the answer for this problem is \(\require{bbox} \bbox[2pt,border:1px solid black]{{{x_{\,2}} = 0.7440943985}}\).

Although it was not asked for in the problem statement the actual root is 0.739085133215161. Note as well that this did require some computational aid to get and it not something that you can, in general, get by hand.