Iterative methods

Algebra

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An iteration is given by $x_{n+1} = \sqrt{3x_{n} + 1}$, with $x_{0} = 1$. Find $x_{1}$.

2 marks
Using the same iteration, find $x_{2}$ to $3$ decimal places.

2 marks
Show that $x^{2} + 4x - 1 = 0$ can be rearranged to $x = \dfrac{1}{x + 4}$.

3 marks
Use $x_{n+1} = \dfrac{1}{x_{n} + 4}$ with $x_{0} = 0$. Find $x_{1}$, $x_{2}$ and $x_{3}$ to $4$ decimal places where needed.

3 marks
Continue the iteration until you can give the positive solution of $x^{2} + 4x - 1 = 0$ to $3$ decimal places.

4 marks
Show that $x = 5 - \dfrac{2}{x}$ is a rearrangement of $x^{2} - 5x + 2 = 0$.

4 marks
Use $x_{n+1} = 5 - \dfrac{2}{x_{n}}$ with $x_{0} = 5$. Find $x_{1}$, $x_{2}$ and $x_{3}$ to $4$ decimal places.

5 marks
Use one further iteration, then write the solution of $x^{2} - 5x + 2 = 0$ given by this iteration to $3$ decimal places. Check it by substitution.

6 marks

Mark scheme

Answer. $2$
- $3 \times 1 + 1 = 4$
- $\sqrt{4} = 2$
Answer. $2.646$
- $x_{2} = \sqrt{7}$
- $\sqrt{7} = 2.64575\ldots = 2.646$ to $3$ d.p.
Answer. $x(x + 4) = 1$
- $x^{2} + 4x = 1$
- $x(x + 4) = 1$
- $x = \dfrac{1}{x + 4}$
Answer. $0.25$, $0.2353$ and $0.2361$
- $x_{1} = \dfrac{1}{4} = 0.25$
- $x_{2} = \dfrac{4}{17} = 0.2353$ to $4$ d.p.
- $x_{3} = \dfrac{17}{72} = 0.2361$ to $4$ d.p.
Answer. $0.236$
- The values settle at $-2 + \sqrt{5}$.
- $-2 + \sqrt{5} = 0.236067\ldots$
- $0.236$ to $3$ d.p.
Answer. $x^{2} = 5x - 2$
- Multiply by $x$: $x^{2} = 5x - 2$.
- $x^{2} - 5x + 2 = 0$
Answer. $4.6000$, $4.5652$ and $4.5619$
- $x_{1} = 5 - \dfrac{2}{5} = 4.6000$
- $x_{2} = 5 - \dfrac{2}{4.6} = 4.5652$
- $x_{3} = 4.5619$ to $4$ d.p.
Answer. $4.562$
- $x_{4} = 4.5616$ to $4$ d.p.
- Both $x_{3}$ and $x_{4}$ round to $4.562$ to $3$ d.p.
- $4.562^{2} - 5 \times 4.562 + 2 \approx 0.002$, so the rounded value is close to a root.