Newton's method

Section 7.5 Newton's method

Our final application of differentiation is a very cool application of linear approximations. We study how repeated applications of linear approximations give rise to a very efficient numerical algorithm for finding roots of functions, known as “Newton's method”. We have already seen in Section 6.1 another method for finding roots of a function, namely the bisection method, which resulted from repeated applications of the Intermediate Value Theorem. It turns out that Newton's method is a much more efficient algorithm for finding roots of a function (however, Newton's method may sometime fail for some choices of initial conditions). In fact, one can use Newton's method to calculate square roots, such as

\sqrt{42},

$\sqrt{42}\text{,}$ to fairly good accuracy, in your head!

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Objectives

You should be able to:

Explain and illustrate Newton's method.
Apply Newton's method to find approximations of all roots of a given function correct to a given number of decimal places.
Explain and illustrate why Newton's method may fail for certain choices of initial conditions.

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Subsection 7.5.1 Instructional video

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Subsection 7.5.2 Key concepts

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Concept 7.5.1. Newton's method.

To find a root $r$ of a differentiable function $f(x)$ near an $x$ -value of $x_1\text{,}$ we apply Newton's Method. The idea is to replace the function by its linearization at $x_1\text{,}$ and then find the $x$ -intercept $x_2$ of the linearization, which is a first approximation of the root of $f(x)\text{.}$ We then repeat the process at $x_2\text{,}$ and so on, until we reach a desired numerical precision for the root of $f(x)\text{.}$ More precisely:

First we calculate $x_2\text{,}$ the $x$ -intercept of the tangent line of $f$ at $(x_1,f(x_1))$ (the linearization of $f$ at $x_1$ ). We get
$\begin{equation*} x_2=x_1-\frac{f(x_1)}{f'(x_1)}. \end{equation*}$
Then we find $x_3\text{,}$ the $x$ -intercept of the tangent line of $f$ at $(x_2,f(x_2))\text{,}$ and get
$\begin{equation*} x_3=x_2-\frac{f(x_2)}{f'(x_2)}. \end{equation*}$
We can repeat this process as many times as desired. At each step, the $x$ -intercept of the tangent line of $f$ at $(x_n, f(x_n))$ is given by
$\begin{equation*} x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}. \end{equation*}$
If we want an answer valid for $k$ decimal places, we stop the process when two successive values $x_n$ and $x_{n+1}$ have the exact same first $k$ decimals. We conclude that $x_{n+1}$ is then a root of $f(x)\text{,}$ up to a precision of $k>$ decimal places.

Note that Newton's method may sometimes converge very slowly. In fact, it can also fail, for a number of reasons, for instance if the starting point $x_1$ is not chosen appropriately.