9.2 Simple iterative methods, cobweb and staircase diagrams — Practise with help

Work each question closed book first. Carry the full value in your calculator and write down four decimal places.

Warm-up: remember

Differentiate $g(x) = \dfrac{2x + 1}{4}$ .
Make $x$ the subject of $4x + 1 = 9$ .

Answers

$g'(x) = \dfrac{2}{4} = \dfrac{1}{2}$ .
Subtract $1$ : $4x = 8$ . Divide by $4$ : $x = 2$ .

If the first part was hard, revise differentiating a linear expression in Section 7.3. If the second was hard, revise rearranging a formula in Section 2.5.

Practise with help

1. Rearrange $x^2 + x - 3 = 0$ into the form $x = \sqrt{3 - x}$ .

Hint

Move the $x^2$ term to the other side first. That leaves $x$ alone on the left.

Next step

From $x^2 + x - 3 = 0$ you get $x^2 = 3 - x$ . Take the positive square root of both sides.

Full working

x^2 + x - 3 = 0 \quad\Longrightarrow\quad x^2 = 3 - x \quad\Longrightarrow\quad x = \sqrt{3 - x}.

The positive root is the one that matches the solution near $1.3$ . The recurrence is $x_{n+1} = \sqrt{3 - x_n}$ .

2. Use $x_{n+1} = \sqrt{3 - x_n}$ with $x_0 = 1$ to find $x_1$ , $x_2$ and $x_3$ to four decimal places.

Hint

Substitute the previous value under the square root each time. Keep the full answer in the calculator.

Next step

$x_1 = \sqrt{3 - 1} = \sqrt{2}$ . Then substitute $1.4142$ to get $x_2$ .

Full working

x_1 = \sqrt{3 - 1} = \sqrt{2} = 1.4142,

x_2 = \sqrt{3 - 1.4142} = \sqrt{1.5858} = 1.2593, \qquad x_3 = \sqrt{3 - 1.2593} = \sqrt{1.7407} = 1.3194.

The values close in on the root $x = 1.3028$ , which is $\dfrac{\sqrt{13} - 1}{2}$ .

3. Rearrange $x^3 - 2x - 5 = 0$ into the form $x = (2x + 5)^{1/3}$ .

Hint

Move the term $-2x$ to the right of the equals sign, then take the cube root of both sides.

Next step

From $x^3 = 2x + 5$ you get $x = \sqrt[3]{2x + 5}$ , which is the same as $(2x + 5)^{1/3}$ .

Full working

x^3 - 2x - 5 = 0 \quad\Longrightarrow\quad x^3 = 2x + 5 \quad\Longrightarrow\quad x = \sqrt[3]{2x + 5}.

The recurrence is $x_{n+1} = (2x_n + 5)^{1/3}$ .

4. Use $x_{n+1} = (2x_n + 5)^{1/3}$ with $x_0 = 2$ to find $x_1$ , $x_2$ and $x_3$ to four decimal places.

Hint

Work out $2x_0 + 5$ first, then take the cube root. That value is the next input.

Next step

$x_1 = \sqrt[3]{2(2) + 5} = \sqrt[3]{9}$ . Use the cube-root key on your calculator.

Full working

x_1 = \sqrt[3]{9} = 2.0801, \qquad x_2 = \sqrt[3]{2(2.0801) + 5} = \sqrt[3]{9.1602} = 2.0924,

x_3 = \sqrt[3]{2(2.0924) + 5} = \sqrt[3]{9.1848} = 2.0942.

The root is $2.0946$ to four decimal places. The values are closing on it.

5. For $g(x) = \dfrac{x + 3}{4}$ , work out $g'(x)$ . Explain why the recurrence $x_{n+1} = g(x_n)$ converges, and say whether the diagram is a staircase or a cobweb.

Hint

Differentiate the expression. The fixed point is where $x = \dfrac{x + 3}{4}$ .

Next step

$g'(x) = \dfrac{1}{4}$ , which is positive and less than $1$ . A positive gradient gives a staircase.

Full working

Differentiating gives

g'(x) = \frac{1}{4}.

The fixed point solves $4x = x + 3$ , so $x = 1$ . At that point $|g'(1)| = \dfrac{1}{4} < 1$ , so the iteration converges to $1$ . The gradient is positive, so the path steps towards the fixed point from one side: a staircase.

6. The recurrence $x_{n+1} = 4x_n - 3$ has the fixed point $x = 1$ . Take $x_0 = 0$ and work out $x_1$ , $x_2$ and $x_3$ . Find $g'(x)$ and explain why the iteration diverges.

Hint

Substitute each value into $4x - 3$ . Compare the size of the numbers with the starting value $0$ .

Next step

$x_1 = 4(0) - 3 = -3$ , then $x_2 = 4(-3) - 3 = -15$ . The values are growing in size, not settling.

Full working

x_1 = -3, \qquad x_2 = 4(-3) - 3 = -15, \qquad x_3 = 4(-15) - 3 = -63.

Here $g(x) = 4x - 3$ , so $g'(x) = 4$ . At the fixed point $|g'(1)| = 4 > 1$ , which is the condition for divergence. Each step multiplies the distance from the fixed point by about $4$ , so the values run away.