Newton's Method for Finding Roots

The basic idea of Newton's Method is remarkably simple and graphic (Fig. 1):

at a point $(\mathrm{x}, \mathrm{f}(\mathrm{x}))$ on the graph of $\mathrm{f}$, the tangent line to the graph of $\mathrm{f}$ "points toward" a root of $\mathrm{f}$, a place where the graph touches the $\mathrm{x}$-axis.

If we want to find a root of $f$, all we need to do is pick a starting value $x_{0}$, go up or down to the point $\left(\mathrm{x}_{0}, \mathrm{f}\left(\mathrm{x}_{0}\right)\right)$ on the graph of $\mathrm{f}$, build a tangent line there, and follow the tangent line to where it crosses the $\mathrm{x}$-axis, say at $\mathrm{x}_{1}$.

If $x_{1}$ is a root of $f$, then we are done. If $x_{1}$ is not a root of $f$, then $x_{1}$ is usually closer to the root than $\mathrm{x}_{0}$ was, and we can repeat the process, using $\mathrm{x}_{1}$ as our new starting point. Newton's method is an iterative procedure, that is, the output from one application of the method becomes the starting point for the next application.

Let's start with a differentiable function $f(x)=x^{2}-5$, (Fig. 2) whose roots we already know, $\mathrm{x}=\pm \sqrt{5} \approx \pm 2.236067977$, and illustrate how Newton's method works. First we pick some value for $x_{0}$, say $x_{0}=4$ for this example, and move to the point $\left(x_{0}, f\left(x_{0}\right)\right)=$ $(4,11)$ on the graph of $\mathrm{f}$.

At $(4,11)$, the graph of $\mathrm{f}$ "points to" a location on the $\mathrm{x}$-axis which is closer to the root of $\mathrm{f}$ (Fig. 3). We can calculate this location on the $\mathrm{x}$-axis by finding the equation of the line tangent to the graph of $\mathrm{f}$ at the point $(4,11)$ and then finding where this tangent line intersects the $\mathrm{x}$-axis:

At the point $(4,11)$, the line tangent to $\mathrm{f}$ has slope $\mathrm{m}=\mathrm{f}^{\prime}(4)=2(4)=8$, so the equation of the tangent line is $\mathrm{y}$ $-11=8(x-4)$. Setting $y=0$, we can find where the tangent line crosses the $x$-axis:

$0-11=8(x-4)$, so $x=4-\frac{11}{8}=\frac{21}{8}=2.625$

Call this new value $x_{1}: x_{1}=2.625$.

The point $\mathrm{x}_{1}=2.625$ is closer to the actual root, but it certainly does not equal the actual root. If Newton's method stopped after one step with the estimate of $2.625$, it would not be very useful. Instead, we can use this new value for $\mathrm{x}, \mathrm{x}_{1}=2.625$, to repeat the procedure (Fig. 4$)$:

(i) move to the point $\left(\mathrm{x}_{1}, \mathrm{f}\left(\mathrm{x}_{1}\right)\right)=(2.625,1.890625)$ on the graph,

(ii) find the equation of the tangent line at the point $\left(\mathrm{x}_{1}, \mathrm{f}\left(\mathrm{x}_{1}\right)\right)$:

$y-1.890625=5.25(x-2.625)$

(iii) find the new value where the tangent line intersects the $\mathrm{x}$-axis and call it $\mathrm{x}_{2} \cdot\left(\mathrm{x}_{2}=2.262880952\right)$

When we continue repeating this process, (Fig. 5) using each new estimate for the root of $f(x)=x^{2}-5$ as the beginning point for calculating the next estimate, we get:

$\begin{array}{lll} \text { Beginning estimate: } & \mathrm{x}_{0}=4 & & \text { (} 0 \text { correct digits) } \\ \text { after } 1 \text { iteration: } & \mathrm{x}_{1}=\underline{2.625} & & \text { (} 1 \text { correct digit }) \\ \text { after } 2 \text { iterations: } & \mathrm{x}_{2}=\underline{2.262880952} & & \text { (} 2 \text { correct digits) } \\ \text { after 3 iterations: } & \mathrm{x}_{3}=\underline{2.236251252} & & \text { (} 4 \text { correct digits) } \\ \text { after 4 iterations: } & \mathrm{x}_{4}=\underline{2.236067985} & & \text { (} 8 \text { correct digits) } \end{array}$

It only took 4 iterations to get an approximation of $\sqrt{5}$ which is within $0.000000008$ of the exact value. One more iteration gives an approximation $\mathrm{x}_{5}$ which has 16 correct digits. If we start with $\mathrm{x}_{0}=-2$ (or any negative number), then the values of $\mathrm{x}_{\mathrm{n}}$ approach $-\sqrt{5} \approx-2.23606$

Fig. 6 shows the process for Newton's Method, starting with $\mathrm{x}_{0}$ and graphically finding the locations on the $\mathrm{x}$-axis of $\mathrm{x}_{1}, \mathrm{x}_{2}$, and $\mathrm{x}_{3}$.

Practice 1: Find where the tangent line to $\mathrm{f}(\mathrm{x})=\mathrm{x}^{3}+3 \mathrm{x}-1$ at $(1, \, 3)$ intersects the $\mathrm{x}$-axis.

Practice 2: A starting point and a graph of $\mathrm{f}$ are given in Fig. 7. Label the approximate locations of the next two points on the $\mathrm{x}$-axis which are found by Newton's method.