Exercise · ZW-FIXED-POINT-ITERATION-010 · ★★★ Advanced

To solve \( x^3 - x - 1 = 0 \) with \( g(x) = \sqrt[3]{x + 1} \), does the iteration from \( x_0 = 1 \) converge? Justify using \( |g'(p)| \) at the real root \( p \approx 1.3247 \).

Choose one option.

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C. Yes, because \( |g'(p)| \approx 0.2 < 1 \)

Explanation

We have \( g'(x) = \frac{1}{3}(x+1)^{-2/3} \), and at \( p \approx 1.3247 \) it is about \( 0.2 \), so the iteration converges locally. The option claiming \( |g'(p)| \approx 1.5 \) overestimates the derivative by inverting the exponent.

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Review the concept: Fixed-point iteration

Iterative method that approximates a solution of the equation \( x = g(x) \) by generating the sequence \( x_{n+1} = g(x_n) \) from an initial value \( x_0 \).

\[x_{n+1} = g(x_n), \quad |g'(p)| < 1 \Rightarrow \text{convergencia local}\]

To solve \( x^3 - x - 1 = 0 \) with \( g(x) = \sqrt[3]{x + 1} \), does the iteration from \( x_0 = 1 \) converge? Justify using \( |g'(p)| \) at the real root \( p \approx 1.3247 \).

Explanation

Review the concept: Fixed-point iteration

More practice problems: fixed-point iteration