4.8: Newton’s Method

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Throughout this module, if something does not exist, write DNE in the answer box.

Recap Video

Take a look at the following video which recaps the ideas from the section.

Example Video

Below is a video showing a worked example.

Problems

Given a differentiable function $f(x)$ and an initial guess $x_0$ , the formula to get from one approximation of a root of $f(x)$ to another is given by

$x_{n+1} = x_n - \frac {f(x_n)}{f'(x_n)},$ assuming $f'(x_n) \neq 0$ .

(For this problem, you can use a calculator) Consider $f(x) = \sin (x)-x+1$ , whose graph is shown below.

If an initial guess for the root is $x_0 = 2$ , find $x_1$ .

The formula for Newton’s method is

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

$f'(x) = \answer {\cos (x)-1}.$ Therefore, the formula becomes

$x_{n+1} = x_n - \frac {\sin (x_n)-x_n+1}{\cos (x_n)-1}.$ Plugging in $n=0$ and using our initial value of $x_0=2$ gives (round your answer to five decimal places).

$x_1 = x_0 - \frac {\sin (x_0)-x_0+1}{\cos (x_0)-1} = \answer {1.93595}.$

(For this problem, you can use a calculator) Use $f(x) = x^2-3$ to approximate $\sqrt {3}$ to $3$ decimal places.

Note that $\sqrt {3}$ is a root of $f(x)$ . Let’s make an initial guess of $x_0=2$ , since this is close to $\sqrt {3}$ . The formula for Newton’s method is

$x_{n+1} = x_n - \frac {f(x_n)}{f'(x_n)}.$ Note that $f'(x) = \answer {2x}$ , so the formula becomes

$x_{n+1} = x_n - \frac {x_n^2-3}{2x_n}.$ We get the following table giving values of $x_n$ (round your answers to five decimal places):

$\begin {array}{c|c|c} n & x_n\\ \hline 0 & 2 \\ 1 &\answer {1.75} \\ 2 & \answer {1.73214} \\ 3 & \answer {1.73205} \\ 4 & \answer {1.73205} \\ \end {array}$ Using this, we can say that $\sqrt {3}$ to three decimal places is $\sqrt {3} \approx \answer {1.732}$ .

In the previous problem, we saw that for $f(x) = x^2-3$ and $x_0=2$ , we had $x_1 = 1.75$ . Which of the following describes how $x_1 = 1.75$ is obtained?

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