9.3 Newton–Raphson and other recurrence relations — Practise with help

Work each question closed book first. Write down the derivative before you substitute, and keep the full value in your calculator.

Warm-up: remember

Differentiate $f(x) = x^3 - 4x + 1$ .
For $f(x) = x^2 - 3x$ , work out $f(2)$ and $f'(2)$ .

Answers

$f'(x) = 3x^2 - 4$ .
$f(2) = 4 - 6 = -2$ and $f'(x) = 2x - 3$ , so $f'(2) = 1$ .

If the first part was hard, revise differentiating powers in Section 7.3. If the second was hard, revise function notation in Section 2.9.

Practise with help

1. For $f(x) = x^3 + x - 1$ , write down $f'(x)$ and use Newton–Raphson with $x_0 = 0.7$ to find $x_1$ to four decimal places.

Hint

The step needs the derivative at the starting value, so differentiate first.

Next step

$f'(x) = 3x^2 + 1$ , so $f(0.7) = 0.043$ and $f'(0.7) = 2.47$ .

Full working

f(0.7) = 0.7^3 + 0.7 - 1 = 0.043, \qquad f'(0.7) = 3(0.49) + 1 = 2.47.

x_1 = 0.7 - \frac{0.043}{2.47} = 0.6826.

2. Continue the iteration in question 1 to find $x_2$ . Give the root to three decimal places.

Hint

Use $x_1$ in the same formula. Carry the full value from the last step, not the rounded one.

Next step

$f(0.6825911) = 0.0006312$ and $f'(0.6825911) = 2.3977918$ .

Full working

x_2 = 0.6825911 - \frac{0.0006312}{2.3977918} = 0.6823279.

The value has settled, so the root is $0.682$ to three decimal places.

3. Use Newton–Raphson with $x_0 = 2$ to find $x_1$ and $x_2$ for $x^3 - 2x - 5 = 0$ .

Hint

Differentiate $f(x) = x^3 - 2x - 5$ term by term.

Next step

$f(2) = -1$ and $f'(2) = 10$ , so the first step adds $0.1$ .

Full working

f'(x) = 3x^2 - 2, \qquad x_1 = 2 - \frac{-1}{10} = 2.1.

Then $f(2.1) = 0.061$ and $f'(2.1) = 11.23$ , so

x_2 = 2.1 - \frac{0.061}{11.23} = 2.0946.

4. Show that $x_{n+1} = \dfrac{1}{2}\left(x_n + \dfrac{7}{x_n}\right)$ is Newton–Raphson for $x^2 - 7 = 0$ . Use it with $x_0 = 3$ to find $x_2$ to four decimal places.

Hint

Work out $x - \dfrac{f(x)}{f'(x)}$ for $f(x) = x^2 - 7$ , and compare it with the recurrence.

Next step

$f'(x) = 2x$ , so $x - \dfrac{x^2 - 7}{2x} = \dfrac{x^2 + 7}{2x}$ .

Full working

x - \frac{f(x)}{f'(x)} = x - \frac{x^2 - 7}{2x} = \frac{2x^2 - x^2 + 7}{2x} = \frac{1}{2}\left(x + \frac{7}{x}\right).

This is the given recurrence. From $x_0 = 3$ :

x_1 = \frac{1}{2}\left(3 + \frac{7}{3}\right) = 2.6667, \qquad x_2 = \frac{1}{2}\left(2.6667 + \frac{7}{2.6667}\right) = 2.6458.

5. Explain why Newton–Raphson fails for $f(x) = x^2 - 9$ when the starting value is $x_0 = 0$ .

Hint

Work out the derivative at the starting value.

Next step

$f'(x) = 2x$ , so $f'(0) = 0$ .

Full working

The formula divides by $f'(x_n)$ . At $x_0 = 0$ the derivative is $f'(0) = 0$ , so the first step would divide by zero. The tangent at $x_0 = 0$ is horizontal and never meets the axis, so there is no crossing point to use.

6. For $f(x) = x^3 - 2x + 2$ with $x_0 = 0$ , work out $x_1$ , $x_2$ and $x_3$ . Describe what the sequence does.

Hint

Find $f(0)$ and $f'(0)$ first, then apply the formula.

Next step

$f(0) = 2$ and $f'(0) = -2$ , so $x_1 = 0 - \dfrac{2}{-2} = 1$ .

Full working

x_1 = 0 - \frac{2}{-2} = 1.

Then $f(1) = 1$ and $f'(1) = 1$ , so

x_2 = 1 - \frac{1}{1} = 0, \qquad x_3 = 0 - \frac{2}{-2} = 1.

The values repeat $1, 0, 1, 0, \ldots$ The method does not settle, even though every step was correct.

7. For $x^3 + x - 1 = 0$ , start both methods from $x_0 = 0.7$ : Newton–Raphson and the simple iteration $x_{n+1} = \dfrac{1}{x_n^2 + 1}$ . Find $x_2$ for each. Which method is closer to the root $0.6823$ ?

Hint

The Newton–Raphson values are in question 2. For the simple iteration, substitute into $\dfrac{1}{x^2 + 1}$ .

Next step

For the simple iteration, $x_1 = \dfrac{1}{0.49 + 1} = 0.6711$ .

Full working

Newton–Raphson reached $x_2 = 0.6823279$ .

The simple iteration gives

x_1 = \frac{1}{0.7^2 + 1} = 0.6711, \qquad x_2 = \frac{1}{0.6711^2 + 1} = 0.6894.

Newton–Raphson is closer after two steps. It uses the gradient as well as the height, so it closes the gap faster. The simple iteration still works, because $|g'| < 1$ at the root, but it needs more steps.