Newton's Method/Example/x^3 - 2 x - 5 = 0

Example of Use of Newton's Method
The real root of the cubic:
 * $x^3 - 2 x - 5 = 0$

can be found by using Newton's Method.

Its approximate value is:
 * $2 \cdotp 09455 \, 1$

Proof
We are to solve $\map f x = x^3 - 2 x - 5 = 0$ numerically.

By Newton's Method, we are to iterate the solution using:
 * $x_{n + 1} = x_n - \dfrac {\map f {x_n}} {\map {f'} {x_n}}$

We have $\map {f'} x = 3 x^2 - 2$.

Hence:

We can start with any number that is close to the solution.

We have:
 * $\map f 2 = -1 < 0$
 * $\map f 3 = 16 > 0$

By Intermediate Value Theorem a root exists between $2$ and $3$.

For convenience we set $x_1 = 2$.

Then:

at $x_4$ we already have the root accurate to $9$ decimal places.

This process can be continued to obtain increasingly accurate results.

Also see

 * Definition:Wallis's Number