Hello MathExchange community !

I am working on some "simple" numerical methods to solve 4th degrees and below equations. To make it easier I am working on the $[0, 1]$ interval and I know for sure that I will find only one and always one solution.

I use n as the number of correct decimals I want to judge that my answer is fairly accurate, so I put $\epsilon$ = $\dfrac{1}{10^n}$

As I am not sure that the Newton's method will correctly converge, I decided to stop if I reached $n$ iterations without finding a correct answer.

I am having some hard time finding how to chose a good starting value for $x_0$ so I decided to chose $0.5$ but I would like to know a good way to do it.

Also I have an example telling me that for :

$x^4- 5x^3 + 6x^2 - 1 = 0$
Using $x_0 = 0.5$

I should end up with something similar with : iteration $1 x = 0.52272727272727270709$
iteration $2 x = 0.5227400035140858936$
$f(x) = -3.3e-11$

When I find a different ($x = 0.522572235129513$) answer by myself with over 162 iterations (slower than the dichotomy method)

I would like some help to understand the concept of this method. Thanks in advance.


The Newton-Raphson iteration is given by:

$$x_{n+1} = x_n - \dfrac{f(x_n)}{f'(x_n)} = x_n - \dfrac{x_n^4-5 x_n^3+6 x_n^2-1}{4 x_n^3-15 x_n^2+12 x_n}$$

  • $x_0 = 0.5$
  • $x_1 = 0.5 - (-0.0227273) = 0.522727$
  • $x_2 = 0.522727 - (-0.0000127308) = 0.52274$
  • $\begingroup$ I guess you made a mistake for x2 that should be equal to +0.52274 but anyway thanks, you pointed correctly my mistake on the "derivation" part, thank you sir. $\endgroup$ – soueuls Jan 14 '14 at 21:24

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