4.3.2 Newton's Method

Here's our plan for finding a root r of the equation $f\left(t\right)=0:$ We start close to $t=r,$ say, at $t=t_0\mathrm{.}$ We find the slope of the graph at the point $(t_0,f\left(t_0\right))$ as $f^{\prime }\left(t_0\right)\mathrm{.}$ Our next guess will be the number $t_1$ at which a line of slope $f^{\prime }\left(t_0\right)$ crosses the horizontal axis. That number won't necessarily be the solution of $f\left(t\right)=0$ that we seek, but it will usually be much closer than $t_0$ was. Then we start over, replacing $t_0$ by $t_1$. The key to the success of this idea is the "much closer" that we achieve at each step, which usually makes it unnecessary to do more than a few steps.

In our example, $f\left(t\right)=e^{3t}-t^2\mathrm{,}$ so $f^{\prime }\left(t\right)=3e^{3t}-2t\mathrm{.}$ We start at the point $(-0.479,0.008)$, so our slope (see Figure 3) is

Now our calculation proceeds very much as before. Again we equate two expressions for the slope: the value, `1.670918828`, of the derivative and the slope of the line from the upper right point $(-0.479,0.008)$ to the point $(t_1,0)$ of intersection with the `t`-axis. Equating these two slope calculations, we find

When we clear the fraction and solve for $t_1,$ we get

This is a little better than before — the error is about `0.00012` — but it's still not correct to `10` places. On the other hand, if we start again from $t_1$ and calculate the next guess $t_2$, we get

Now we're getting somewhere. The error, when we compare this answer with that from a calculator's or computer's Root function, is about $4\times {10}^{-10},$ so we're already at `9`-place accuracy, with just two steps.

Let's outline the calculation in general. The slope of the graph of `f` at ${\displaystyle t=t_0}$ is $f^{\prime }\left(t_0\right)\mathrm{.}$ The point $(t_1,0)$ is also on the line through $(t_0,y_0)$ with the same slope. Thus, the slope also has the form "rise over run" or

where $y_0$ represents $f\left(t_0\right)$. Now we set our two expressions for slope equal:

And we solve for $t_1$:

This equation tells us how to get from $t_0$ to $t_1$: Subtract the ratio of the function value at $t_0$ to the slope at $t_0$. Look at Figure 4, and explain the calculation to yourself in these terms: The run is the rise divided by the slope. Be sure you understand where the minus sign comes from.

Now we get $t_2$ from $t_1$ in exactly the same way. Then $t_3$ from $t_2$, and so on. In fact, what we have here is an iterative formula for generating the numbers $t_1$, $t_2$, $t_3$, ... from the starting point $t_0$:

This formula is generally known as Newton's Method for solving equations.