Question

The equation x^3 − x −1 = 0 has a positive root in the interval ] 1,2[ . Write a fixed point
iteration method and show that it converges. Starting with initial approximation x0 = 1.5
find the root of the equation correct to three decimal places.

Answer 1

Question #65623, Math / Other

The equation $x^3 - x - 1 = 0$ has a positive root in the interval $1,2$. Write a fixed point iteration method and show that it converges. Starting with initial approximation $x_0 = 1.5$ find the root of the equation correct to three decimal places.

Answer.

$$x_{n+1} = g(x_n)$$

Let

$$x_{n+1} = \sqrt[3]{x_n + 1}.$$ $$\frac{dg}{dx} = \frac{1}{3^{\sqrt[3]{(x+1)^2}}}, \quad \frac{dg}{dx} < 1 \quad \text{when } x \in [1,2].$$

So fixed point method converges.

Thus, the root is $x = 1.325$.

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