OCR MEI Further Numerical Methods 2022 June — Question 4 8 marks

Exam BoardOCR MEI
ModuleFurther Numerical Methods (Further Numerical Methods)
Year2022
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicFixed Point Iteration
TypeCompare iteration convergence
DifficultyStandard +0.8 This question requires qualitative analysis of Newton-Raphson and fixed point iteration convergence using graphical information, without calculations. Students must understand how gradient affects Newton-Raphson behavior and how the slope of g(x) relative to y=x determines fixed point iteration convergence. This goes beyond routine application to require conceptual understanding of numerical methods, making it moderately challenging for Further Maths students.
Spec1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams1.09d Newton-Raphson method
4 Fig. 4.1 shows part of the graph of \(y = e ^ { x } - x ^ { 2 } - x - 1.1\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{f8c78ab8-7daa-4d6f-9bb5-093a46c590b8-04_805_789_299_274} \captionsetup{labelformat=empty} \caption{Fig. 4.1}
\end{figure} The equation \(\mathrm { e } ^ { x } - x ^ { 2 } - x - 1.1 = 0\) has a root \(\alpha\) such that \(1 < \alpha < 2\).
Ali is considering using the Newton-Raphson method to find \(\alpha\). Ali could use a starting value of \(x _ { 0 } = 1\) or a starting value of \(x _ { 0 } = 2\).
  1. Without doing any calculations, explain whether Ali should use a starting value of \(x _ { 0 } = 1\) or a starting value of \(x _ { 0 } = 2\), or whether using either starting value would work equally well. Ali is also considering using the method of fixed point iteration to find \(\alpha\). Ali could use a starting value of \(x _ { 0 } = 1\) or a starting value of \(x _ { 0 } = 2\). Fig. 4.2 shows parts of the graphs of \(y = x\) and \(y = \ln \left( x ^ { 2 } + x + 1.1 \right)\). \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{f8c78ab8-7daa-4d6f-9bb5-093a46c590b8-04_819_1011_1818_255} \captionsetup{labelformat=empty} \caption{Fig. 4.2}
    \end{figure}
  2. Without doing any calculations, explain whether Ali should use a starting value of \(x _ { 0 } = 1\) or a starting value of \(x _ { 0 } = 2\) or whether either starting value would work equally well. Ali used one of the above methods to find a sequence of approximations to \(\alpha\). These are shown, together with some further analysis in the associated spreadsheet output in Fig. 4.3. \begin{table}[h]
    MNO
    \(r\)\(\mathrm { X } _ { \mathrm { r } }\)
    402
    511.879008- 0.121
    621.858143- 0.0210.172
    731.857565\(- 6 \mathrm { E } - 04\)0.028
    841.857564\(- 4 \mathrm { E } - 07\)\(8 \mathrm { E } - 04\)
    951.857564\(- 2 \mathrm { E } - 13\)\(6 \mathrm { E } - 07\)
    \captionsetup{labelformat=empty} \caption{Fig. 4.3}
    \end{table} The formula in cell N5 is =M5-M4
    and the formula in cell O6 is =N6/N5
    equivalent formulae are in cells N6 to N9 and O7 to O9 respectively.
  3. State what is being calculated in the following columns of the spreadsheet.
    1. Column N
    2. Column O
  4. Explain whether the values in column O suggest that Ali used the Newton-Raphson method or the iterative formula \(\mathrm { x } _ { \mathrm { n } + 1 } = \ln \left( \mathrm { x } _ { \mathrm { n } } ^ { 2 } + \mathrm { x } _ { \mathrm { n } } + 1.1 \right)\) to find this sequence of approximations to \(\alpha\).
Question 4:
AnswerMarks Guidance
4(a) with π‘₯ = 1 the value of π‘₯ is further away
0 1
from Ξ± oe
with π‘₯ = 2 the value of π‘₯ is closer to Ξ±
0 1
so better to use π‘₯ = 2
AnswerMarks
0B1
B1
AnswerMarks
[2]2.2a
2.2a(initially) diverges starting at 1 B1 for one correct
allow converges more slowly with statement and B1 for 2nd
π‘₯ = 1 correct statement and
0
better to start at 2
AnswerMarks Guidance
4(b) can use either (but better to start at x = 2
because it’s nearer)
because the magnitude of the gradient of
(𝑦 =)ln(π‘₯2 +π‘₯+1.1) is less than 1 at the
AnswerMarks
point of intersection with y = x oeB1
B1
AnswerMarks
[2]2.4
2.2aallow gradient of the curve for
gradient of (𝑦 =)ln(π‘₯2 +π‘₯ +1.1)
AnswerMarks Guidance
4(c) (i)
𝑛+1 𝑛B1
[1]2.2a
4(c) (ii)
[1]2.2a
4(d) ratio of differences decreasing
so convergence faster than first order
which suggests she used Newton-Raphson
AnswerMarks
methodB1
B1
AnswerMarks
[2]2.2a
2.2bor ratio of differences not constant
so convergence not first order which means she didn’t use
fixed point iteration – she probably used Newton-Raphson
B0B0 for reasoning based on values in columns M or N
Question 4:
4 | (a) | with π‘₯ = 1 the value of π‘₯ is further away
0 1
from Ξ± oe
with π‘₯ = 2 the value of π‘₯ is closer to Ξ±
0 1
so better to use π‘₯ = 2
0 | B1
B1
[2] | 2.2a
2.2a | (initially) diverges starting at 1 B1 for one correct
allow converges more slowly with statement and B1 for 2nd
π‘₯ = 1 correct statement and
0
better to start at 2
4 | (b) | can use either (but better to start at x = 2
because it’s nearer)
because the magnitude of the gradient of
(𝑦 =)ln(π‘₯2 +π‘₯+1.1) is less than 1 at the
point of intersection with y = x oe | B1
B1
[2] | 2.4
2.2a | allow gradient of the curve for
gradient of (𝑦 =)ln(π‘₯2 +π‘₯ +1.1)
4 | (c) | (i) | difference between π‘₯ and π‘₯ oe
𝑛+1 𝑛 | B1
[1] | 2.2a
4 | (c) | (ii) | ratio of differences | B1
[1] | 2.2a
4 | (d) | ratio of differences decreasing
so convergence faster than first order
which suggests she used Newton-Raphson
method | B1
B1
[2] | 2.2a
2.2b | or ratio of differences not constant
so convergence not first order which means she didn’t use
fixed point iteration – she probably used Newton-Raphson
B0B0 for reasoning based on values in columns M or N
4 Fig. 4.1 shows part of the graph of $y = e ^ { x } - x ^ { 2 } - x - 1.1$.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{f8c78ab8-7daa-4d6f-9bb5-093a46c590b8-04_805_789_299_274}
\captionsetup{labelformat=empty}
\caption{Fig. 4.1}
\end{center}
\end{figure}

The equation $\mathrm { e } ^ { x } - x ^ { 2 } - x - 1.1 = 0$ has a root $\alpha$ such that $1 < \alpha < 2$.\\
Ali is considering using the Newton-Raphson method to find $\alpha$. Ali could use a starting value of $x _ { 0 } = 1$ or a starting value of $x _ { 0 } = 2$.
\begin{enumerate}[label=(\alph*)]
\item Without doing any calculations, explain whether Ali should use a starting value of $x _ { 0 } = 1$ or a starting value of $x _ { 0 } = 2$, or whether using either starting value would work equally well.

Ali is also considering using the method of fixed point iteration to find $\alpha$. Ali could use a starting value of $x _ { 0 } = 1$ or a starting value of $x _ { 0 } = 2$.

Fig. 4.2 shows parts of the graphs of $y = x$ and $y = \ln \left( x ^ { 2 } + x + 1.1 \right)$.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{f8c78ab8-7daa-4d6f-9bb5-093a46c590b8-04_819_1011_1818_255}
\captionsetup{labelformat=empty}
\caption{Fig. 4.2}
\end{center}
\end{figure}
\item Without doing any calculations, explain whether Ali should use a starting value of $x _ { 0 } = 1$ or a starting value of $x _ { 0 } = 2$ or whether either starting value would work equally well.

Ali used one of the above methods to find a sequence of approximations to $\alpha$. These are shown, together with some further analysis in the associated spreadsheet output in Fig. 4.3.

\begin{table}[h]
\begin{center}
\begin{tabular}{ | c | c | c | c | c | c | }
\hline
 &  & M & N & O &  \\
\hline
 & $r$ & $\mathrm { X } _ { \mathrm { r } }$ &  &  &  \\
\hline
4 & 0 & 2 &  &  &  \\
\hline
5 & 1 & 1.879008 & - 0.121 &  &  \\
\hline
6 & 2 & 1.858143 & - 0.021 & 0.172 &  \\
\hline
7 & 3 & 1.857565 & $- 6 \mathrm { E } - 04$ & 0.028 &  \\
\hline
8 & 4 & 1.857564 & $- 4 \mathrm { E } - 07$ & $8 \mathrm { E } - 04$ &  \\
\hline
9 & 5 & 1.857564 & $- 2 \mathrm { E } - 13$ & $6 \mathrm { E } - 07$ &  \\
\hline
\end{tabular}
\captionsetup{labelformat=empty}
\caption{Fig. 4.3}
\end{center}
\end{table}

The formula in cell N5 is =M5-M4\\
and the formula in cell O6 is =N6/N5\\
equivalent formulae are in cells N6 to N9 and O7 to O9 respectively.
\item State what is being calculated in the following columns of the spreadsheet.
\begin{enumerate}[label=(\roman*)]
\item Column N
\item Column O
\end{enumerate}\item Explain whether the values in column O suggest that Ali used the Newton-Raphson method or the iterative formula $\mathrm { x } _ { \mathrm { n } + 1 } = \ln \left( \mathrm { x } _ { \mathrm { n } } ^ { 2 } + \mathrm { x } _ { \mathrm { n } } + 1.1 \right)$ to find this sequence of approximations to $\alpha$.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Numerical Methods 2022 Q4 [8]}}
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