9.3 Newton–Raphson and other recurrence relations — Practise

Work through these on your own. Write down the derivative before you substitute. Give 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.

For $f(x) = x^3 + x - 1$ , write down $f'(x)$ .
Use Newton–Raphson on $f(x) = x^3 + x - 1$ with $x_0 = 0.7$ . Find $x_1$ to four decimal places.
Continue the iteration to find $x_2$ for $f(x) = x^3 + x - 1$ from $x_0 = 0.7$ .
State the root of $x^3 + x - 1 = 0$ to four decimal places.
For $f(x) = e^x - 3x$ , write down $f'(x)$ .
Use Newton–Raphson on $e^x = 3x$ with $x_0 = 1.5$ . Find $x_1$ to four decimal places.
Continue the iteration on $e^x = 3x$ to find the root near $1.5$ to four decimal places.
Use $x_{n+1} = \dfrac{1}{2}\left(x_n + \dfrac{5}{x_n}\right)$ with $x_0 = 2$ . Find $x_3$ to four decimal places.
Continue the recurrence in the previous question to find $x_4$ to five decimal places.
The recurrence $x_{n+1} = \dfrac{1}{2}\left(x_n + \dfrac{7}{x_n}\right)$ starts from $x_0 = 3$ . Find $x_2$ as an exact fraction.
For $f(x) = x^2 - 4$ , work out $f'(0)$ . Explain why Newton–Raphson fails from $x_0 = 0$ .
For $f(x) = x^3 - 2x + 2$ with $x_0 = 0$ , find $x_2$ .
Use Newton–Raphson on $x^3 - 2x - 5 = 0$ with $x_0 = 2$ . Find $x_2$ .

Answers and concise solutions

1. The derivative

Answer: Differentiate term by term: $f'(x) = 3x^2 + 1$ . The Newton–Raphson step needs this at each estimate.

2. The value of x1

Answer: Here $f(0.7) = 0.043$ and $f'(0.7) = 2.47$ , so $x_1 = 0.7 - \dfrac{0.043}{2.47} = 0.6825911$ , which is $0.6826$ to four decimal places.

3. The value of x2

Answer: With $x_1 = 0.6825911$ , $f(x_1) = 0.0006312$ and $f'(x_1) = 2.3977918$ , so $x_2 = 0.6823279$ . The value has settled.

4. The root

Answer: The Newton–Raphson values settle on $0.6823278$ , so the root is $0.6823$ to four decimal places. Section 9.1 found the same root, but needed four halving steps to reach only two decimal places.

5. The derivative

Answer: The derivative of $e^x$ is $e^x$ , and the derivative of $3x$ is $3$ , so $f'(x) = e^x - 3$ .

6. The value of x1

Answer: Rearrange to $e^x - 3x = 0$ , so $f'(x) = e^x - 3$ . Then $f(1.5) = -0.0183109$ and $f'(1.5) = 1.4816891$ , so $x_1 = 1.5 + 0.0123581 = 1.5124$ to four decimal places.

7. The root

Answer: The values are $x_1 = 1.5123581$ , $x_2 = 1.5121346$ and $x_3 = 1.5121346$ , so the root is $1.5121$ to four decimal places. The equation also has a root near $0.6191$ , which a different starting value would find.

8. The value of x3

Answer: $x_1 = 2.25$ , $x_2 = 2.236111$ and $x_3 = 2.236068$ , which is $2.2361$ to four decimal places. The recurrence is Newton–Raphson applied to $x^2 - 5 = 0$ , so it finds $\sqrt{5}$ .

9. The value of x4

Answer: $x_4 = 2.2360680$ to five decimal places. This matches $\sqrt{5} = 2.2360680\ldots$ The recurrence is Newton–Raphson for $x^2 - 5 = 0$ , so it homes in on $\sqrt{5}$ .

10. The value of x2

Answer: $x_1 = \dfrac{1}{2}\left(3 + \dfrac{7}{3}\right) = \dfrac{8}{3}$ , so $x_2 = \dfrac{1}{2}\left(\dfrac{8}{3} + \dfrac{21}{8}\right) = \dfrac{127}{48} = 2.6458$ . This is Newton–Raphson for $x^2 - 7 = 0$ , so it finds $\sqrt{7}$ .

11. The value of the derivative at 0

Answer: $f'(x) = 2x$ and $f'(0) = 0$ . The Newton–Raphson step divides by $f'(x_0)$ , so the first step would divide by zero. The tangent at $x_0 = 0$ is horizontal and never meets the axis.

12. The value of x2

Answer: $x_1 = 1$ and $x_2 = 0$ , because $f(1) = 1$ and $f'(1) = 1$ . The values repeat $1, 0, 1, 0, \ldots$ The formula was applied correctly, but the method cycles and never settles.

13. The value of x2

Answer: $f'(x) = 3x^2 - 2$ , so $x_1 = 2 - \dfrac{-1}{10} = 2.1$ , and $x_2 = 2.1 - \dfrac{0.061}{11.23} = 2.0946$ to four decimal places.