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Question Number 217676 by Ikbal last updated on 18/Mar/25

F i n d a r a l r o o t o f t h e e q u a t i o n x^{3} - x - 1 = 0 b y f i x e d p o i n t i t e r a t i o n m e t h o d

Answered by SdC355 last updated on 18/Mar/25

z^{3} - z - 1 = 0

z_{n + 1} = z_{n} - \frac{f (z_{n})}{f^{(1)} (z_{n})}

z_{n + 1} = z_{n} - \frac{z_{n}^{3} - z_{n} - 1}{3 z_{n}^{2} - 1}

z_{1} = 1

z_{2} = \frac{3}{2} \approx 1.5

z_{3} = \frac{3}{2} - \frac{\frac{27}{8} - \frac{12}{8} - \frac{8}{8} = \frac{7}{8}}{3 (\frac{9}{4}) - 1 = \frac{27}{4} - \frac{4}{4} = \frac{23}{4}} = \frac{7}{2 \cdot 23} = \frac{7}{46}

z_{3} = \frac{31}{23} \approx 1.34782608696 \dots .

z_{4} = \frac{71749}{54142} \approx 1.32520039895 \dots . .

⋮

and \lim_{n \to \infty} z_{n + 1} is convergence

\lim_{n \to \infty} z_{n + 1} = \frac{1}{3}^{3} \sqrt{\frac{27}{2} - \frac{3 \sqrt{69}}{2}} +^{3} \sqrt{\frac{9 + \sqrt{69}}{18}}

and \lim_{n \to \infty} z_{n + 1} \approx 1.32471795724475 \dots . .

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