Description of the method

The idea of the method is as follows: one starts with an initial guess which is reasonably close to the true root, then the function is approximated by its tangent line (which can be computed using the tools of calculus), and one computes the x-intercept of this tangent line (which is easily done with elementary algebra). This x-intercept will typically be a better approximation to the function's root than the original guess, and the method can be iterated.

Suppose $f : [a, b] \to \mathbb{R}$ is a differentiable function defined on the interval $[a, b]$ with values in the real numbers $\mathbb{R}$. The formula for converging on the root can be easily derived. Suppose we have some current approximation $x_n$. Then we can derive the formula for a better approximation, $x_{n+1}$ by referring to the diagram on the right. We know from the definition of the derivative at a given point that it is the slope of a tangent at that point.

That is

$$f'(x_{n}) = \frac{ \mathrm{rise} }{ \mathrm{run} } = \frac{ \ma... ..._{n} ) - 0 }{ x_{n} - x_{n+1} } = \frac{0 - f(x_{n})}{(x_{n+1} - x_{n})}\,\!.$$

Here, $f'$ denotes the derivative of the function $f$. Then by simple algebra we can derive

$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\,\!.$$

We start the process off with some arbitrary initial value $x_0$. (The closer to the zero, the better. But, in the absence of any intuition about where the zero might lie, a ''guess and check'' method might narrow the possibilities to a reasonably small interval by appealing to the intermediate value theorem.) The method will usually converge, provided this initial guess is close enough to the unknown zero, and that $f'(x_0) \neq 0$. Furthermore, for a zero of multiplicity $1$, the convergence is at least quadratic (see rate of convergence) in a neighbourhood of the zero, which intuitively means that the number of correct digits roughly at least doubles in every step. More details can be found in the analysis section below.

Example 6.17.1   Consider the problem of finding the positive number $x$ with $\cos(x) = x^3$. We can rephrase that as finding the zero of $f(x) = \cos(x) - x^3$. We have $f'(x) = -\sin(x) - 3x^2$. Since $\cos(x) \leq 1$ for all $x$ and $x^3 > 1$ for $x > 1$, we know that our zero lies between 0 and $1$. We try a starting value of $x_0 = 0.5$.

$$\begin{array}{lllll} x_1 & = & x_0 - \frac{f(x_0)}{f'(x_0)}... ...{f(x_5)}{f'(x_5)} & = & \underline{0.865474033102} \end{array}$$

The correct digits are underlined in the above example. In particular, $x_6$ is correct to the number of decimal places given. We see that the number of correct digits after the decimal point increases from $2$ (for $x_3$) to $5$ and $10$, illustrating the quadratic convergence.