Newton's Method/Example/x^3 - 2 x - 5 = 0
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Example of Use of Newton's Method
- $x^3 - 2 x - 5 = 0$
can be found by using Newton's Method.
Its approximate value is:
- $2 \cdotp 09455 \, 1$
This sequence is A007493 in the On-Line Encyclopedia of Integer Sequences (N. J. A. Sloane (Ed.), 2008).
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:
| \(\ds x_{n + 1}\) | \(=\) | \(\ds x_n - \frac { {x_n}^3 - 2 x_n - 5} {3 x_n^2 - 2}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {2 {x_n}^3 + 5} {3 {x_n}^2 - 2}\) |
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:
| \(\ds x_2\) | \(=\) | \(\ds \frac {2 {x_1}^3 + 5} {3 {x_1}^2 - 2}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {21} {10}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \mathbf 2.1\) | ||||||||||||
| \(\ds x_3\) | \(=\) | \(\ds \frac {2 {x_2}^3 + 5} {3 {x_2}^2 - 2}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {11 \, 761} {5 \, 615}\) | ||||||||||||
| \(\ds \) | \(\approx\) | \(\ds \mathbf {2.0945} 6 \, 8 \dots\) | ||||||||||||
| \(\ds x_4\) | \(=\) | \(\ds \frac {2 {x_3}^3 + 5} {3 {x_3}^2 - 2}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \frac {4 \, 138 \, 744 \, 325 \, 037} {1 \, 975 \, 957 \, 316 \, 495}\) | ||||||||||||
| \(\ds \) | \(\approx\) | \(\ds \mathbf {2.09455 \, 1481} 6 \, 98 \dots\) |
at $x_4$ we already have the root accurate to $9$ decimal places.
This process can be continued to obtain increasingly accurate results.
$\blacksquare$