Newtons metode

Eit gjennomarbeidd døme

Hans Georg Schaathun

September 2016

$$x^3 - x - 1 = 0$$

  1. Finn \(x_0\)
  2. Finn \(x_{n+1}\) som funksjon av \(x_n\)
  3. Rekn ut \(x_1,x_2,\ldots\)

$$y = x^3 - x - 1 $$

$$y' = 3x^2 - 1 $$

$$x_{n+1} = x_n - \frac{y_n}{y_n'}$$

$$x_{n+1} = x_n - \frac{x_n^3-x_n-1}{3x_n^2-1}$$

$$x_0 = 1{,}5$$$$y_0 = 0{,}875$$
$$x_1 = 1{,}347\,826\,086\,96$$$$y_1 = 0{,}100\,682\,173\,091$$
$$x_2 = 1{,}325\,200\,398\,95$$$$y_2 = 0{,}002\,058\,361\,917$$
$$x_3 = 1{,}324\,718\,174\,00$$$$y_3 = 0{,}000\,000\,924\,378$$
$$x_4 = 1{,}324\,717\,957\,24$$$$y_4 = 0{,}000\,000\,000\,000$$
$$x_5 = 1{,}324\,717\,957\,24$$$$y_5 = 0{,}000\,000\,000\,000$$