9.2 Simple iterative methods, cobweb and staircase diagrams — Practise

Work through these on your own. Keep the full value in your calculator, and write down four decimal places unless the question says otherwise.

The questions and the answers below come from a set that is checked by computer every time the book is built.

Use $x_{n+1} = \dfrac{x_n + 2}{3}$ with $x_0 = 0$ . Find $x_3$ .
Use $x_{n+1} = \dfrac{x_n + 2}{3}$ with $x_0 = 0$ . Find $x_4$ .
Use $x_{n+1} = \dfrac{1}{x_n + 1}$ with $x_0 = 1$ . Find $x_3$ .
Use $x_{n+1} = \dfrac{1}{x_n^2 + 1}$ with $x_0 = 1$ . Find $x_3$ .
Find the fixed point of the recurrence $x_{n+1} = \dfrac{x_n + 2}{3}$ .
Find the fixed points of the recurrence $x_{n+1} = \dfrac{1}{x_n + 1}$ . Give the answers in surd form.
For $g(x) = \dfrac{x + 2}{3}$ , find $g'(x)$ .
For $g(x) = \dfrac{1}{x + 1}$ , find $g'(x)$ .
Use $x_{n+1} = 3x_n - 2$ with $x_0 = 0$ . Find $x_3$ .
Use $x_{n+1} = \dfrac{x_n + 2}{3}$ with $x_0 = 0$ . Find $x_5$ , and state whether it is within $0.01$ of the fixed point $1$ .
The equation $x^3 + 2x - 5 = 0$ can be rearranged to $x = \dfrac{5}{x^2 + 2}$ . Starting from $x_0 = 1$ , find $x_1$ .
For $g(x) = 2 - \dfrac{x}{2}$ , find $g'(x)$ and state whether the iteration gives a cobweb or a staircase.

Answers and concise solutions

1. The value of x3

Answer: Apply the rule three times: $x_1 = \dfrac{2}{3}$ , $x_2 = \dfrac{8}{9}$ and $x_3 = \dfrac{26}{27} = 0.9630$ . The values climb towards the fixed point $x = 1$ .

2. The value of x4

Answer: Continuing from $x_3 = \dfrac{26}{27}$ gives $x_4 = \dfrac{80}{81} = 0.9877$ . Each step closes two thirds of the gap to $1$ .

3. The value of x3

Answer: The values are $x_1 = \dfrac{1}{2}$ , $x_2 = \dfrac{2}{3}$ and $x_3 = \dfrac{3}{5} = 0.6$ . They approach the fixed point $0.6180\ldots$ from both sides, so the diagram is a cobweb.

4. The value of x3

Answer: The values are $x_1 = \dfrac{1}{2}$ , $x_2 = \dfrac{4}{5}$ and $x_3 = \dfrac{25}{41} = 0.6098$ . They move to either side of the root $0.6823$ and settle slowly.

5. The fixed point

Answer: A fixed point satisfies $x = \dfrac{x + 2}{3}$ . Multiply by $3$ : $3x = x + 2$ , so $2x = 2$ and $x = 1$ . The iterates in the earlier questions approach this value.

6. The fixed points

Answer: A fixed point satisfies $x = \dfrac{1}{x + 1}$ . Multiply by $x + 1$ : $x^2 + x - 1 = 0$ . The quadratic formula gives $x = \dfrac{-1 \pm \sqrt{5}}{2}$ , so the values are $0.6180\ldots$ and $-1.6180\ldots$ The positive value is the one the recurrence from $x_0 = 1$ approaches.

7. The derivative

Answer: Differentiating gives $g'(x) = \dfrac{1}{3}$ . The magnitude is less than $1$ everywhere, so the iteration converges. A positive gradient means the path is a staircase.

8. The derivative

Answer: Write $g(x) = (x + 1)^{-1}$ and differentiate: $g'(x) = -(x + 1)^{-2} = -\dfrac{1}{(x + 1)^2}$ . At the fixed point $x = 0.6180$ this is about $-0.382$ , so $|g'| < 1$ and the iteration converges. A negative gradient gives a cobweb.

9. The value of x3

Answer: The values are $x_1 = -2$ , $x_2 = -8$ and $x_3 = -26$ . The fixed point is $x = 1$ but $g'(x) = 3 > 1$ , so the iteration diverges and each step moves further away.

10. The value of x5

Answer: $x_5 = \dfrac{242}{243} = 0.99588\ldots$ , which is within $0.01$ of $1$ . The difference from $1$ is $\dfrac{1}{243} = 0.0041\ldots$ , so the value is close enough for many purposes. Agreement to a stated accuracy is the practical stopping rule, not a proof of exactness.

11. The value of x1

Answer: Substitute $x_0 = 1$ : $x_1 = \dfrac{5}{1^2 + 2} = \dfrac{5}{3} = 1.6667$ . The fixed point of this rearrangement is the root of the original equation.

12. The derivative

Answer: $g'(x) = -\dfrac{1}{2}$ , which has magnitude $\dfrac{1}{2} < 1$ , so the iteration converges. The negative gradient makes the values alternate on either side of the fixed point, so the diagram is a cobweb. The fixed point is $x = \dfrac{4}{3}$ .