Question

1. Consider the function f(x)=x^3-2 using Newton's method. Take x0=1.5 for the starting value.
using Newton's method. Take x0=1.5 for the starting value.
For each method, present the results in the form of table:
Column1:n (step)
Column2:xn (approximation)
Column3: ( ) n f x
Column4:| xn - xn-1 | (error)

Answer 1

Conditions

1. Consider the function $f(x) = x^3 - 2$ using Newton's method. Take $x_0 = 1.5$ for the starting value.

using Newton's method. Take $x_0 = 1.5$ for the starting value.

For each method, present the results in the form of table:

- Column 1: n (step)

- Column 2: xn (approximation)

- Column 3: ( ) n f x

- Column 4: | xn - xn-1 | (error)

Solution

In numerical analysis, Newton's method (also known as the Newton-Raphson method), named after Isaac Newton and Joseph Raphson, is a method for finding successively better approximations to the roots (or zeroes) of a real-valued function.

$$x: f(x) = 0.$$

The algorithm is first in the class of Householder's methods, succeeded by Halley's method. The method can also be extended to complex functions and to systems of equations.

The Newton-Raphson method in one variable is implemented as follows:

Given a function $f$ defined over the reals $x$, and its derivative $f'$, we begin with a first guess $x_0$ for a root of the function $f$. Provided the function satisfies all the assumptions made in the derivation of the formula, a better approximation $x_1$ is

$$x_1 = x_0 - \frac{f(x_0)}{f'(x_0)}.$$

Geometrically, $(x_1, 0)$ is the intersection with the $x$-axis of a line tangent to $f$ at $(x_0, f(x_0))$.

The process is repeated as

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

until a sufficiently accurate value is reached.

For our example:

Answer:

As we see, after $6^{\text{th}}$ iteration the root has been found, and it's approximately equal to:

$x = 1.2599210498949$