OCR MEI Further Numerical Methods (Further Numerical Methods) 2022 June

2 The table shows some values of \(x\) and the associated values of \(y = f ( x )\).

1. Calculate an estimate of \(\frac { \mathrm { dy } } { \mathrm { dx } }\) at \(x = 3\) using the forward difference method, giving your answer correct to \(\mathbf { 5 }\) decimal places.
2. Calculate an estimate of \(\frac { \mathrm { dy } } { \mathrm { dx } }\) at \(x = 3\) using the central difference method, giving your answer correct to \(\mathbf { 5 }\) decimal places.
3. Explain why your answer to part (b) is likely to be closer than your answer to part (a) to the true value of \(\frac { \mathrm { dy } } { \mathrm { dx } }\) at \(x = 3\). When \(x = 5\) it is given that \(y = 1.4645\) and \(\frac { \mathrm { dy } } { \mathrm { dx } } = 0.1820\), correct to 4 decimal places.
4. Determine an estimate of the error when \(\mathrm { f } ( 5 )\) is used to estimate \(\mathrm { f } ( 5.024 )\).

3 The equation \(\mathrm { f } ( x ) = \sin ^ { - 1 } ( x ) - x + 0.1 = 0\) has a root \(\alpha\) such that \(- 1 < \alpha < 0\).
Alex uses an iterative method to find a sequence of approximations to \(\alpha\). Some of the associated spreadsheet output is shown in the table. The formula in cell D7 is $$= ( \mathrm { D } 5 * \mathrm { E } 6 - \mathrm { D } 6 * \mathrm { E } 5 ) / ( \mathrm { E } 6 - \mathrm { E } 5 )$$ and equivalent formulae are in cells D8 and D9.

1. State the method being used.
2. Use the values in the spreadsheet to calculate \(x _ { 5 }\) and \(x _ { 6 }\), giving your answers correct to 7 decimal places.
3. State the value of \(\alpha\) as accurately as you can, justifying the precision quoted. Alex uses a calculator to check the value in cell D9, his result is - 0.7540832686 .
4. Explain why this is different to the value displayed in cell D9. The value displayed in cell E11 in Alex's spreadsheet is \(- 1.4629 \mathrm { E } - 09\).
5. Write this value in standard mathematical notation.

4 Fig. 4.1 shows part of the graph of \(y = e ^ { x } - x ^ { 2 } - x - 1.1\). \begin{figure}[h] \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]

   		M	N	O
   	\(r\)	\(\mathrm { X } _ { \mathrm { r } }\)
   4	0	2
   5	1	1.879008	- 0.121
   6	2	1.858143	- 0.021	0.172
   7	3	1.857565	\(- 6 \mathrm { E } - 04\)	0.028
   8	4	1.857564	\(- 4 \mathrm { E } - 07\)	\(8 \mathrm { E } - 04\)
   9	5	1.857564	\(- 2 \mathrm { E } - 13\)	\(6 \mathrm { E } - 07\)

   \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 \(\alpha\).

5 Kai uses the midpoint rule, trapezium rule and Simpson's rule to find approximations to \(\int _ { \mathrm { a } } ^ { \mathrm { b } } \mathrm { f } ( \mathrm { x } ) \mathrm { dx }\), where \(a\) and \(b\) are constants. The associated spreadsheet output is shown in the table. Some of the values are missing.

1. Write down a suitable spreadsheet formula for cell H 5 .
2. Complete the copy of the table in the Printed Answer Booklet, giving the values correct to 7 decimal places.
3. Use your answers to part (b) to determine the value of \(\int _ { a } ^ { b } f ( x ) d x\) as accurately as you can, justifying the precision quoted.

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{table}[h] \end{table}

1. 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 \(\sin ^ { - 1 } ( x ) - x ^ { 2 } + 1 = 0\).
2. 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}
3. Write down the value of the root correct to 7 decimal places.
4. 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.
5. Name another application of the method of relaxation.

7 Sam decided to go on a high-protein diet. Sam's mass in \(\mathrm { kg } , M\), after \(t\) days of following the diet is recorded in Fig. 7.1. \begin{table}[h] \end{table} A difference table for the data is shown in Fig. 7.2. \begin{table}[h] \end{table}

1. Complete the copy of the difference table in the Printed Answer Booklet. Sam's doctor uses these data to construct a cubic interpolating polynomial to model Sam's mass at time \(t\) days after starting the diet.
2. Find the model in the form \(\mathrm { M } = \mathrm { at } ^ { 3 } + \mathrm { bt } ^ { 2 } + \mathrm { ct } + \mathrm { d }\), where \(a , b , c\) and \(d\) are constants to be determined. Subsequently it is found that when \(\mathrm { t } = 40 , \mathrm { M } = 78.7\) and when \(\mathrm { t } = 50 , \mathrm { M } = 80.05\).
3. Determine whether the model is a good fit for these data.
4. By completing the extended copy of Fig. 7.2 in the Printed Answer Booklet, explain why a quartic model may be more appropriate for the data.
5. Refine the doctor's model to include a quartic term.
6. Explain whether the new model for Sam's mass is likely to be appropriate over a longer period of time.