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Subsection 1.2.1 The basics of Newton's method

Let's begin with a look at the example that Newton himself used to illustrate his method. Let \(f(x)=x^3 - 2x - 5\text{.}\) The graph of \(f\) shown in Figure 1.2.1 seems to indicate that \(f\) has a root just to the right of \(x=2\text{.}\)

Figure 1.2.1 Newton's example function for his method

Newton figured that, if the root of \(f\) is a little bigger than \(2\text{,}\) then he could write it as \(2+\Delta x\text{.}\) He then plugged this into \(f\) to get

\begin{align*} f(2+\Delta x) &= (2+\Delta x)^3 - 2(2+\Delta x) - 5\\ &= 8 + 3\times2^2\Delta x + 3\times2\Delta x^2 + \Delta x^3 -4-2\Delta x - 5\\ &= -1 +10\Delta x + 6\Delta x^2 + \Delta x^3\\ &\approx -1+10\Delta x. \end{align*}

In that last step, since \(\Delta x\) is small, he figures that higher powers of \(\Delta x\) are negligibly small. Thus, he figures that \(-1+10\Delta x \approx 0\) so that \(\Delta x \approx 1/10\text{.}\)

The point is that, if \(2\) is a good guess at a root of \(f\text{,}\) then \(2+1/10 = 2.1\) should be an even better guess. A glance at the graph seems to verify this. Of course, we could then repeat the process using \(2.1\) as the initial guess. We should get an even better estimate. The process can be repeated as many times as desired. This is the basis of iteration.

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