OCR MEI Further Numerical Methods 2023 June — Question 5 6 marks

Exam BoardOCR MEI
ModuleFurther Numerical Methods (Further Numerical Methods)
Year2023
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFixed Point Iteration
TypeApply iteration to find root (pure fixed point)
DifficultyStandard +0.8 This is a Further Maths numerical methods question requiring understanding of fixed point iteration convergence criteria, application of relaxed iteration (a non-standard technique), and analysis of convergence order from numerical data. Part (a) requires graphical interpretation of convergence failure, part (b) is computational but with an unfamiliar relaxation parameter, and part (c) requires theoretical understanding of convergence rates. The combination of theory and application with a specialized technique places this above average difficulty.
Spec1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams
5 The equation \(3 - 2 \ln x - x = 0\) has a root near \(x = 1.8\).
A student proposes to use the iterative formula \(\mathrm { x } _ { \mathrm { n } + 1 } = \mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right) = 3 - 2 \ln \mathrm { x } _ { \mathrm { n } }\) to find this root.
The diagram shows the graphs of \(\mathrm { y } = \mathrm { x }\) and \(\mathrm { y } = \mathrm { g } ( \mathrm { x } )\) for values of \(x\) from - 2 to 6 . \includegraphics[max width=\textwidth, alt={}, center]{4023e87c-34b1-4abd-9acc-ede5e4d68c7f-05_913_917_502_233}
  1. With reference to the graph, explain why it might not be possible to use the student's iterative formula to find the root near \(x = 1.8\).
  2. Use the relaxed iteration \(\mathrm { x } _ { \mathrm { n } + 1 } = \lambda \mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right) + ( 1 - \lambda ) \mathrm { x } _ { \mathrm { n } }\), with \(\lambda = 0.475\) and \(x _ { 0 } = 2\), to determine the root correct to \(\mathbf { 6 }\) decimal places. A student uses the same relaxed iteration with the same starting value. Some analysis of the iterates is carried out using a spreadsheet, which is shown in the table below.
    \(r\)differenceratio
    0
    1- 0.1834898
    2- 0.00491370.02678
    3\(- 6.44 \mathrm { E } - 06\)0.00131
    4\(- 3.862 \mathrm { E } - 09\)0.0006
    5\(- 2.313 \mathrm { E } - 12\)0.0006
  3. Explain what the analysis tells you about the order of convergence of this sequence of approximations.
Question 5:
AnswerMarks Guidance
5(a) the gradient of g(x) at the root is less than
β€’1 oe so the iteration will not convergeB1 2.4
of [y =] x and [y =] g(x);
allow eg won’t converge because
[1]
AnswerMarks Guidance
5(b) 2
1.81651017847
1.81159652649
1.81159008685
1.81159008298
AnswerMarks
1.811590B1
B1
AnswerMarks
B11.1
2.1
AnswerMarks Guidance
2.2a gβ€²(1.8)
oe seen
1.816510… 𝐚𝐚𝐚𝐚𝐚𝐚 1.811596…
oe seen
1.8115900… 𝐚𝐚𝐚𝐚𝐚𝐚 1.8115900…
dependent on award of first two marks
if B1B0B0 allow B1 SCB2 for 1.811590 seen twice and 6 dp
accuracy confirmed by change of sign
[3]
AnswerMarks Guidance
5(c) ratio of differences appears to converge to a
constant [0.0006]
AnswerMarks
(which suggests) first order convergenceB1
B12.2a
2.2bignore comments on speed of convergence
B0 if spoiled by incorrect reasoning
[2]
Question 5:
5 | (a) | the gradient of g(x) at the root is less than
β€’1 oe so the iteration will not converge | B1 | 2.4 | must refer to gradient at root or gradient of g(x) at intersection
of [y =] x and [y =] g(x);
allow eg won’t converge because
[1]
5 | (b) | 2
1.81651017847
1.81159652649
1.81159008685
1.81159008298
1.811590 | B1
B1
B1 | 1.1
2.1
2.2a | |gβ€²(1.8)| > 1
oe seen
1.816510… 𝐚𝐚𝐚𝐚𝐚𝐚 1.811596…
oe seen
1.8115900… 𝐚𝐚𝐚𝐚𝐚𝐚 1.8115900…
dependent on award of first two marks
if B1B0B0 allow B1 SCB2 for 1.811590 seen twice and 6 dp
accuracy confirmed by change of sign
[3]
5 | (c) | ratio of differences appears to converge to a
constant [0.0006]
(which suggests) first order convergence | B1
B1 | 2.2a
2.2b | ignore comments on speed of convergence
B0 if spoiled by incorrect reasoning
[2]
5 The equation $3 - 2 \ln x - x = 0$ has a root near $x = 1.8$.\\
A student proposes to use the iterative formula $\mathrm { x } _ { \mathrm { n } + 1 } = \mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right) = 3 - 2 \ln \mathrm { x } _ { \mathrm { n } }$ to find this root.\\
The diagram shows the graphs of $\mathrm { y } = \mathrm { x }$ and $\mathrm { y } = \mathrm { g } ( \mathrm { x } )$ for values of $x$ from - 2 to 6 .\\
\includegraphics[max width=\textwidth, alt={}, center]{4023e87c-34b1-4abd-9acc-ede5e4d68c7f-05_913_917_502_233}
\begin{enumerate}[label=(\alph*)]
\item With reference to the graph, explain why it might not be possible to use the student's iterative formula to find the root near $x = 1.8$.
\item Use the relaxed iteration $\mathrm { x } _ { \mathrm { n } + 1 } = \lambda \mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right) + ( 1 - \lambda ) \mathrm { x } _ { \mathrm { n } }$, with $\lambda = 0.475$ and $x _ { 0 } = 2$, to determine the root correct to $\mathbf { 6 }$ decimal places.

A student uses the same relaxed iteration with the same starting value. Some analysis of the iterates is carried out using a spreadsheet, which is shown in the table below.

\begin{center}
\begin{tabular}{ | c | c | c | }
\hline
$r$ & difference & ratio \\
\hline
0 &  &  \\
\hline
1 & - 0.1834898 &  \\
\hline
2 & - 0.0049137 & 0.02678 \\
\hline
3 & $- 6.44 \mathrm { E } - 06$ & 0.00131 \\
\hline
4 & $- 3.862 \mathrm { E } - 09$ & 0.0006 \\
\hline
5 & $- 2.313 \mathrm { E } - 12$ & 0.0006 \\
\hline
\end{tabular}
\end{center}
\item Explain what the analysis tells you about the order of convergence of this sequence of approximations.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Numerical Methods 2023 Q5 [6]}}
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