Question 6 - A-Level Maths

OCR MEI Further Numerical Methods 2022 June — Question 6 11 marks

Exam Board	OCR MEI
Module	Further Numerical Methods (Further Numerical Methods)
Year	2022
Session	June
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Fixed Point Iteration
Type	Apply iteration to find root (pure fixed point)
Difficulty	Standard +0.8 This is a Further Maths numerical methods question requiring understanding of fixed point iteration convergence, error analysis from oscillating sequences, and relaxed iteration. Part (a) demands careful analysis of the oscillating convergence pattern to determine precision, while part (b) involves applying the relaxation formula. The topic is specialized (Further Maths) and requires more sophisticated numerical analysis than standard A-level, but the calculations themselves are straightforward once the concepts are understood.
Spec	1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams

6 Charlie uses fixed point iteration to find a sequence of approximations to the root of the equation \(\sin ^ { - 1 } ( x ) - x ^ { 2 } + 1 = 0\). Charlie uses the iterative formula \(\mathrm { x } _ { \mathrm { n } + 1 } = \mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right)\), where \(\mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right) = \sin \left( \mathrm { x } _ { \mathrm { n } } ^ { 2 } - 1 \right)\).
Two sections of the associated spreadsheet output, showing \(x _ { 0 }\) to \(x _ { 6 }\) and \(x _ { 102 }\) to \(x _ { 108 }\), are shown in Fig. 6.1.

\(r\)	\(\mathrm { x } _ { \mathrm { r } }\)	difference	ratio
0	0
1	-0.841471	-0.84147
2	-0.287798	0.553673	-0.65798
3	-0.793885	-0.50609	-0.91405
4	-0.361379	0.432507	-0.85461
5	-0.763945	-0.40257	-0.93078
6	-0.404459	0.359486	-0.89299

\begin{table}[h]

102	-0.596302	0.004626	-0.95886
103	-0.600738	-0.00444	-0.95911
104	-0.596484	0.004254	-0.95887
105	-0.600564	-0.00408	-0.95910
106	-0.596652	0.003912	-0.95888
107	-0.600404	-0.00375	-0.95909
108	-0.596806	0.003598	-0.95889

\captionsetup{labelformat=empty} \caption{Fig. 6.1}

\end{table}

Use the information in Fig. 6.1 to find the value of the root as accurately as you can, justifying the precision quoted. The relaxed iteration \(\mathrm { x } _ { \mathrm { n } + 1 } = ( 1 - \lambda ) \mathrm { x } _ { \mathrm { n } } + \lambda \mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right)\), with \(\lambda = 0.51\) and \(x _ { 0 } = 0\), is to be used to find the root of the equation \(\sin ^ { - 1 } ( x ) - x ^ { 2 } + 1 = 0\).
Complete the copy of Fig. 6.2 in the Printed Answer Booklet, giving the values of \(\mathrm { x } _ { \mathrm { r } }\) correct to 7 decimal places and the values in the difference column and ratio column correct to 3 significant figures. \begin{table}[h]
\(r\) \(\mathrm { x } _ { \mathrm { r } }\) difference ratio
0 0
1
2
3
4 -0.000192
5 \(- 1.99 \times 10 ^ { - 7 }\) 0.00103
6 \(- 1.82 \times 10 ^ { - 10 }\) 0.000914
\captionsetup{labelformat=empty} \caption{Fig. 6.2}
\end{table}
Write down the value of the root correct to 7 decimal places.
Explain why extrapolation could not be used in this case to find an improved approximation using this sequence of iterates. In this case the method of relaxation has been used to speed up the convergence of an iterative scheme.
Name another application of the method of relaxation.

Show mark scheme Show mark scheme source

Question 6:

Answer	Marks	Guidance
6	(a)	𝑟

𝑥 +𝑑 × or 𝑥 +𝑑 ×𝑟

1−𝑟

𝑥 = −0.596806,𝑑 = 0.003598,

𝑟 = −0.95889 𝑜𝑟−0.9589 𝑜𝑟−

0.959 𝑜𝑟−0.96

awrt −0.598567 to awrt−0.598568

−0.5986 or −0.599 is possible due to

Answer	Marks
increased accuracy from extrapolation oe	M1

A1

Answer	Marks
[4]	3.1a

2.1

1.1

Answer	Marks
2.2b	at least two values from information in table; allow sign errors

allow−0.60 is certain by comparison with 𝑥 oe

108 the final mark is only available if all other marks earned

if M0 allow SC1 for −0.60 is certain by comparison of

𝑥 and 𝑥 and confirmation by correct change of sign test

108 107

Answer	Marks	Guidance
6	(b)	r
r	difference	ratio

A1

Answer	Marks
[4]	1.1

1.1

Answer	Marks
1.1	two iterates seen to 3 or more dp

iterates

differences

ratios

allow M1A1A1A0 for correct values to greater precision

Answer	Marks	Guidance
0	0
1	‒0.4291502	‒0.429
2	‒0.5817139	‒0.153
3	‒0.5983772	‒0.0167
4	‒0.5985695	‒0.000192
5	‒0.5985697	‒1.99×10-7
6	‒0.5985697	‒1.82×10‒10
6	(c)	‒0.5985697 cao
[1]	2.2a
6	(d)	ratio of differences is not constant
[1]	2.4
6	(e)	to adapt an iterative scheme which does not
converge into one which will converge oe	B1
[1]	1.2

Question 6:
6 | (a) | 𝑟
𝑥 +𝑑 × or 𝑥 +𝑑 ×𝑟
1−𝑟
𝑥 = −0.596806,𝑑 = 0.003598,
𝑟 = −0.95889 𝑜𝑟−0.9589 𝑜𝑟−
0.959 𝑜𝑟−0.96
awrt −0.598567 to awrt−0.598568
−0.5986 or −0.599 is possible due to
increased accuracy from extrapolation oe | M1
A1
A1
A1
[4] | 3.1a
2.1
1.1
2.2b | at least two values from information in table; allow sign errors
allow−0.60 is certain by comparison with 𝑥 oe
108
the final mark is only available if all other marks earned
if M0 allow SC1 for −0.60 is certain by comparison of
𝑥 and 𝑥 and confirmation by correct change of sign test
108 107
6 | (b) | r | x
r | difference | ratio | M1
A1
A1
A1
[4] | 1.1
1.1
1.1
1.1 | two iterates seen to 3 or more dp
iterates
differences
ratios
allow M1A1A1A0 for correct values to greater precision
0 | 0
1 | ‒0.4291502 | ‒0.429
2 | ‒0.5817139 | ‒0.153 | 0.356
3 | ‒0.5983772 | ‒0.0167 | 0.109
4 | ‒0.5985695 | ‒0.000192 | 0.0115
5 | ‒0.5985697 | ‒1.99×10-7 | 0.00103
6 | ‒0.5985697 | ‒1.82×10‒10 | 0.000914
6 | (c) | ‒0.5985697 cao | B1
[1] | 2.2a
6 | (d) | ratio of differences is not constant | B1
[1] | 2.4
6 | (e) | to adapt an iterative scheme which does not
converge into one which will converge oe | B1
[1] | 1.2

Show LaTeX source

6 Charlie uses fixed point iteration to find a sequence of approximations to the root of the equation $\sin ^ { - 1 } ( x ) - x ^ { 2 } + 1 = 0$.

Charlie uses the iterative formula $\mathrm { x } _ { \mathrm { n } + 1 } = \mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right)$, where $\mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right) = \sin \left( \mathrm { x } _ { \mathrm { n } } ^ { 2 } - 1 \right)$.\\
Two sections of the associated spreadsheet output, showing $x _ { 0 }$ to $x _ { 6 }$ and $x _ { 102 }$ to $x _ { 108 }$, are shown in Fig. 6.1.

\begin{center}
\begin{tabular}{|l|l|l|l|}
\hline
$r$ & $\mathrm { x } _ { \mathrm { r } }$ & difference & ratio \\
\hline
0 & 0 &  &  \\
\hline
1 & -0.841471 & -0.84147 &  \\
\hline
2 & -0.287798 & 0.553673 & -0.65798 \\
\hline
3 & -0.793885 & -0.50609 & -0.91405 \\
\hline
4 & -0.361379 & 0.432507 & -0.85461 \\
\hline
5 & -0.763945 & -0.40257 & -0.93078 \\
\hline
6 & -0.404459 & 0.359486 & -0.89299 \\
\hline
\end{tabular}
\end{center}

\begin{table}[h]
\begin{center}
\begin{tabular}{|l|l|l|l|}
\hline
102 & -0.596302 & 0.004626 & -0.95886 \\
\hline
103 & -0.600738 & -0.00444 & -0.95911 \\
\hline
104 & -0.596484 & 0.004254 & -0.95887 \\
\hline
105 & -0.600564 & -0.00408 & -0.95910 \\
\hline
106 & -0.596652 & 0.003912 & -0.95888 \\
\hline
107 & -0.600404 & -0.00375 & -0.95909 \\
\hline
108 & -0.596806 & 0.003598 & -0.95889 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Fig. 6.1}
\end{center}
\end{table}
\begin{enumerate}[label=(\alph*)]
\item Use the information in Fig. 6.1 to find the value of the root as accurately as you can, justifying the precision quoted.

The relaxed iteration $\mathrm { x } _ { \mathrm { n } + 1 } = ( 1 - \lambda ) \mathrm { x } _ { \mathrm { n } } + \lambda \mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right)$, with $\lambda = 0.51$ and $x _ { 0 } = 0$, is to be used to find the root of the equation $\sin ^ { - 1 } ( x ) - x ^ { 2 } + 1 = 0$.
\item Complete the copy of Fig. 6.2 in the Printed Answer Booklet, giving the values of $\mathrm { x } _ { \mathrm { r } }$ correct to 7 decimal places and the values in the difference column and ratio column correct to 3 significant figures.

\begin{table}[h]
\begin{center}
\begin{tabular}{|l|l|l|l|}
\hline
$r$ & $\mathrm { x } _ { \mathrm { r } }$ & difference & ratio \\
\hline
0 & 0 &  &  \\
\hline
1 &  &  &  \\
\hline
2 &  &  &  \\
\hline
3 &  &  &  \\
\hline
4 &  & -0.000192 &  \\
\hline
5 &  & $- 1.99 \times 10 ^ { - 7 }$ & 0.00103 \\
\hline
6 &  & $- 1.82 \times 10 ^ { - 10 }$ & 0.000914 \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Fig. 6.2}
\end{center}
\end{table}
\item Write down the value of the root correct to 7 decimal places.
\item Explain why extrapolation could not be used in this case to find an improved approximation using this sequence of iterates.

In this case the method of relaxation has been used to speed up the convergence of an iterative scheme.
\item Name another application of the method of relaxation.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Numerical Methods 2022 Q6 [11]}}

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