CAIE P3 2002 June — Question 4 5 marks

Exam BoardCAIE
ModuleP3 (Pure Mathematics 3)
Year2002
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFixed Point Iteration
TypeFind equation satisfied by limit
DifficultyModerate -0.3 This is a straightforward fixed point iteration question requiring routine application of the formula (part i) and simple algebraic manipulation to find the limit equation by setting x_{n+1} = x_n = α (part ii). The equation 3α³ = 2α² + 2 is easily derived and solved, making this slightly easier than average despite being a two-part question.
Spec1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams

4 The sequence of values given by the iterative formula $$x _ { n + 1 } = \frac { 2 } { 3 } \left( x _ { n } + \frac { 1 } { x _ { n } ^ { 2 } } \right)$$ with initial value \(x _ { 1 } = 1\), converges to \(\alpha\).
  1. Use this formula to find \(\alpha\) correct to 2 decimal places, showing the result of each iteration.
  2. State an equation satisfied by \(\alpha\), and hence find the exact value of \(\alpha\).

AnswerMarks Guidance
(i) Use the formula correctly at least onceM1
State \(\alpha = 1.26\) as final answerA1
Show sufficient iterations to justify \(\alpha = 1.26\) to 2d.p., or show there is a sign change in the interval \((1.255, 1.265)\)A1 Total: 3 marks
(ii) State any suitable equation in one unknown e.g. \(x = \frac{3}{2}\left(x+\frac{1}{x^2}\right)\)B1
State exact value of \(\alpha\) (or \(x\)) is \(\sqrt[3]{2}\) or \(2^{\frac{1}{3}}\)B1 Total: 2 marks
**(i)** Use the formula correctly at least once | M1 |
State $\alpha = 1.26$ as final answer | A1 |
Show sufficient iterations to justify $\alpha = 1.26$ to 2d.p., or show there is a sign change in the interval $(1.255, 1.265)$ | A1 | **Total: 3 marks**

**(ii)** State any suitable equation in one unknown e.g. $x = \frac{3}{2}\left(x+\frac{1}{x^2}\right)$ | B1 |
State exact value of $\alpha$ (or $x$) is $\sqrt[3]{2}$ or $2^{\frac{1}{3}}$ | B1 | **Total: 2 marks**

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4 The sequence of values given by the iterative formula

$$x _ { n + 1 } = \frac { 2 } { 3 } \left( x _ { n } + \frac { 1 } { x _ { n } ^ { 2 } } \right)$$

with initial value $x _ { 1 } = 1$, converges to $\alpha$.\\
(i) Use this formula to find $\alpha$ correct to 2 decimal places, showing the result of each iteration.\\
(ii) State an equation satisfied by $\alpha$, and hence find the exact value of $\alpha$.

\hfill \mbox{\textit{CAIE P3 2002 Q4 [5]}}