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OCR MEI Further Numerical Methods 2022 June Q4
8 marks Standard +0.8
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 \(\alpha\).
OCR MEI Further Numerical Methods 2022 June Q5
9 marks Standard +0.3
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.
FGHI
3\(n\)\(\mathrm { M } _ { \mathrm { n } }\)\(\mathrm { T } _ { \mathrm { n } }\)\(\mathrm { S } _ { 2 \mathrm { n } }\)
410.24366990.1479020
520.2306967
  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.
OCR MEI Further Numerical Methods 2022 June Q6
11 marks Standard +0.8
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.
\(r\)\(\mathrm { x } _ { \mathrm { r } }\)differenceratio
00
1-0.841471-0.84147
2-0.2877980.553673-0.65798
3-0.793885-0.50609-0.91405
4-0.3613790.432507-0.85461
5-0.763945-0.40257-0.93078
6-0.4044590.359486-0.89299
\begin{table}[h]
102-0.5963020.004626-0.95886
103-0.600738-0.00444-0.95911
104-0.5964840.004254-0.95887
105-0.600564-0.00408-0.95910
106-0.5966520.003912-0.95888
107-0.600404-0.00375-0.95909
108-0.5968060.003598-0.95889
\captionsetup{labelformat=empty} \caption{Fig. 6.1}
\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]
    \(r\)\(\mathrm { x } _ { \mathrm { r } }\)differenceratio
    00
    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.
OCR MEI Further Numerical Methods 2022 June Q7
14 marks Standard +0.8
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]
\(t\)0102030
\(M\)88.380.0578.778.85
\captionsetup{labelformat=empty} \caption{Fig. 7.1}
\end{table} A difference table for the data is shown in Fig. 7.2. \begin{table}[h]
\(t\)\(M\)\(\Delta M\)\(\Delta ^ { 2 } M\)\(\Delta ^ { 3 } M\)
088.3
1080.05
2078.7
3078.85
\captionsetup{labelformat=empty} \caption{Fig. 7.2}
\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.
OCR MEI Further Numerical Methods 2023 June Q1
7 marks Standard +0.8
1 You are given that \(\left( x _ { 1 } , y _ { 1 } \right) = ( 0.9,2.3 )\) and \(\left( x _ { 2 } , y _ { 2 } \right) = ( 1.1,2.7 )\).
The values of \(x _ { 1 }\) and \(x _ { 2 }\) have been rounded to \(\mathbf { 1 }\) decimal place.
  1. Determine the range of possible values of \(x _ { 2 } - x _ { 1 }\). The values of \(y _ { 1 }\) and \(y _ { 2 }\) have been chopped to \(\mathbf { 1 }\) decimal place.
  2. Determine the range of possible values of \(y _ { 2 } - y _ { 1 }\). You are given that \(m = \frac { y _ { 2 } - y _ { 1 } } { x _ { 2 } - x _ { 1 } }\).
  3. Determine the range of possible values of \(m\).
  4. Explain why your answer to part (c) is much larger than your answer to part (a) and your answer to part (b).
OCR MEI Further Numerical Methods 2023 June Q2
8 marks Standard +0.3
2 A car tyre has a slow puncture. Initially the tyre is inflated to a pressure of 34.5 psi . The pressure is checked after 3 days and then again after 5 days. The time \(t\) in days and the pressure, \(P\) psi, are shown in the table below. You are given that the pressure in a car tyre is measured in pounds per square inch (psi).
\(t\)035
\(P\)34.529.427.0
The owner of the car believes the relationship between \(P\) and \(t\) may be modelled by a polynomial.
  1. Explain why it is not possible to use Newton's forward difference interpolation method for these data.
  2. Use Lagrange's form of the interpolating polynomial to find an interpolating polynomial of degree 2 for these data. The car owner uses the polynomial found in part (b) to model the relationship between \(P\) and \(t\).
    Subsequently it is found that when \(t = 6 , P = 26.0\) and when \(t = 10 , P = 24.4\).
  3. Determine whether the owner's model is a good fit for these data.
  4. Explain why the model would not be suitable in the long term.
OCR MEI Further Numerical Methods 2023 June Q3
6 marks Standard +0.3
3 The diagram shows the graph of \(y = f ( x )\) for values of \(x\) from 1 to 3.5. \includegraphics[max width=\textwidth, alt={}, center]{4023e87c-34b1-4abd-9acc-ede5e4d68c7f-03_945_1248_312_244} The table shows some values of \(x\) and the associated values of \(y\).
\(x\)1.522.5
\(y\)1.6821372.0943952.318559
  1. Use the forward difference method to calculate an approximation to \(\frac { \mathrm { dy } } { \mathrm { dx } }\) at \(x = 2\).
  2. Use the central difference method to calculate an approximation to \(\frac { \mathrm { dy } } { \mathrm { dx } }\) at \(x = 2\).
  3. On the copy of the diagram in the Printed Answer Booklet, show how the central difference method gives the approximation to \(\frac { \mathrm { dy } } { \mathrm { dx } }\) at \(x = 2\) which was found in part (b).
  4. Explain whether your answer to part (a) or your answer to part (b) is likely to give a better approximation to \(\frac { \mathrm { dy } } { \mathrm { dx } }\) at \(x = 2\).
OCR MEI Further Numerical Methods 2023 June Q4
4 marks Moderate -0.8
4 A spreadsheet is used to approximate \(\int _ { a } ^ { b } f ( x ) d x\) using the midpoint rule with 1 strip. The output is shown in the table below.
BCD
3\(x\)\(\mathrm { f } ( x )\)\(\mathrm { M } _ { 1 }\)
41.51.31037070.65518535
The formula in cell C4 is \(= \mathrm { B } 4 \wedge ( 1 / \mathrm { B } 4 )\).
The formula in cell D4 is \(= 0.5 ^ { * } \mathrm { C } 4\).
  1. Write the integral in standard mathematical notation. A graph of \(y = f ( x )\) is included in the diagram below. \includegraphics[max width=\textwidth, alt={}, center]{4023e87c-34b1-4abd-9acc-ede5e4d68c7f-04_789_1004_1199_235}
  2. Explain whether 0.65518535 is an over-estimate or an under-estimate of \(\int _ { a } ^ { b } f ( x ) d x\).
OCR MEI Further Numerical Methods 2023 June Q5
6 marks Standard +0.8
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.
OCR MEI Further Numerical Methods 2023 June Q6
6 marks Standard +0.8
6
    1. Calculate the relative error when \(\pi\) is chopped to \(\mathbf { 2 }\) decimal places in approximating $$\pi ^ { 2 } + 2 .$$
    2. Without doing any calculation, explain whether the relative error would be the same when \(\pi\) is chopped to 2 decimal places when approximating \(( \pi + 2 ) ^ { 2 }\). The table shows some spreadsheet output. The values of \(x\) in column A are exact.
      ABC
      1\(x\)\(10 ^ { x }\)\(\log _ { 10 } 10 ^ { x }\)
      2\(1 \mathrm { E } - 12\)1\(1.00001 \mathrm { E } - 12\)
      3\(1 \mathrm { E } - 11\)1\(9.99998 \mathrm { E } - 12\)
      The formula in cell B2 is \(= 10 ^ { \wedge } \mathrm { A } 2\).
      This has been copied down to cell B3.
      The formula in cell C2 is \(\quad =\) LOG(B2) .
      This formula has been copied down to cell C3.
    1. Write the value displayed in cell C 2 in standard mathematical notation.
    2. Explain why the values in cells C 2 and C 3 are neither zero nor the same as the values in cells A2 and A3 respectively.
OCR MEI Further Numerical Methods 2023 June Q7
6 marks Standard +0.3
7 The value of a function, \(\mathrm { y } = \mathrm { f } ( \mathrm { x } )\), and its gradient function, \(\frac { \mathrm { dy } } { \mathrm { dx } }\), when \(x = 2\), is given in Table 7.1. \begin{table}[h]
\captionsetup{labelformat=empty} \caption{Table 7.1}
\(x\)\(\mathrm { f } ( x )\)\(\frac { \mathrm { dy } } { \mathrm { dx } }\)
26- 2.8
\end{table}
  1. Determine the approximate value of the error when \(f ( 2 )\) is used to estimate \(f ( 2.03 )\). The Newton-Raphson method is used to find a sequence of approximations to a root, \(\alpha\), of the equation \(\mathrm { f } ( x ) = 0\). The spreadsheet output showing the iterates, together with some further analysis, is shown in Table 7.2. \begin{table}[h]
    \captionsetup{labelformat=empty} \caption{Table 7.2}
    ABCD
    1rXrdifferenceratio
    2012
    31-13.1165572-25.1165572
    421.7628327914.87939004-0.5924136
    532.180521570.417688780.02807163
    642.1824190240.0018974540.00454275
    752.182419066\(4.13985 \mathrm { E } - 08\)\(2.1818 \mathrm { E } - 05\)
    \end{table}
    1. Explain what the values in column D tell you about the order of convergence of this sequence of approximations.
    2. Without doing any further calculation, state the value of \(\alpha\) as accurately as you can, justifying the precision quoted.
OCR MEI Further Numerical Methods 2023 June Q8
8 marks Standard +0.3
8 The graph of \(\mathrm { y } = 0.2 \cosh \mathrm { x } - 0.4 \mathrm { x }\) for values of \(x\) from 0 to 3.32 is shown on the graph below. \includegraphics[max width=\textwidth, alt={}, center]{4023e87c-34b1-4abd-9acc-ede5e4d68c7f-08_988_1561_312_244} The equation \(0.2 \cosh x - 0.4 x = 0\) has two roots, \(\alpha\) and \(\beta\) where \(\alpha < \beta\), in the interval \(0 < x < 3\). The secant method with \(x _ { 0 } = 1\) and \(x _ { 1 } = 2\) is to be used to find \(\beta\).
  1. On the copy of the graph in the Printed Answer Booklet, show how the secant method works with these two values of \(x\) to obtain an improved approximation to \(\beta\). The spreadsheet output in the table below shows the result of applying the secant method with \(x _ { 0 } = 1\) and \(x _ { 1 } = 2\).
    IJKLM
    2\(r\)\(\mathrm { x } _ { \mathrm { r } }\)f(x)\(\mathrm { X } _ { \mathrm { r } + 1 }\)\(\mathrm { f } \left( \mathrm { x } _ { \mathrm { r } + 1 } \right)\)
    301-0.09142-0.0476
    412-0.04763.085290.95784
    523.085290.957842.05134-0.0298
    632.05134-0.02982.08259-0.0181
    742.08259-0.01812.130420.00155
    852.130420.001552.12664\(- 7 \mathrm { E } - 05\)
  2. Write down a suitable cell formula for cell J4.
  3. Write down a suitable cell formula for cell L4.
  4. Write down the most accurate approximation to \(\beta\) which is displayed in the table.
  5. Determine whether your answer to part (d) is correct to 5 decimal places. You should not calculate any more iterates.
  6. It is decided to use the secant method with starting values \(x _ { 0 } = 1\) and \(\mathrm { x } _ { 1 } = \mathrm { a }\), where \(a > 1\), to find \(\alpha\). State a suitable value for \(a\).
OCR MEI Further Numerical Methods 2023 June Q9
9 marks Challenging +1.2
9 The trapezium rule is used to calculate 3 approximations to \(\int _ { 0 } ^ { 1 } \sqrt [ 3 ] { \sinh ( x ) } \mathrm { d } x\) with 1,2 and 4 strips respectively. The results are shown in Table 9.1. \begin{table}[h]
\captionsetup{labelformat=empty} \caption{Table 9.1}
\(n\)\(\mathrm {~T} _ { n }\)
10.52764369
20.66617652
40.72534275
\end{table}
  1. Use these results to determine two approximations to \(\int _ { 0 } ^ { 1 } \sqrt [ 3 ] { \sinh ( x ) } \mathrm { d } x\) using Simpson's rule.
  2. Use your answers to part (a) to state the value of \(\int _ { 0 } ^ { 1 } \sqrt [ 3 ] { \sinh ( x ) } \mathrm { d } x\) as accurately as you can, justifying the precision quoted. Table 9.2 shows some further approximations found using the trapezium rule, together with some analysis of these approximations. \begin{table}[h]
    \captionsetup{labelformat=empty} \caption{Table 9.2}
    \(n\)\(\mathrm { T } _ { n }\)differenceratio
    10.5276437
    20.66617650.138533
    40.72534270.0591660.42709
    80.74988210.0245390.41475
    160.75988580.0100040.40766
    320.76392210.0040360.40348
    640.76554040.0016180.40095
    \end{table}
  3. Explain what can be deduced about the order of the method in this case.
  4. Use extrapolation to obtain the value of \(\int _ { 0 } ^ { 1 } \sqrt [ 3 ] { \sinh ( x ) } \mathrm { d } x\) as accurately as you can, justifying the precision quoted.
OCR MEI Further Numerical Methods 2024 June Q1
4 marks Standard +0.8
1 The table shows some values of \(x\), together with the associated values of a function, \(\mathrm { f } ( x )\).
\(x\)1.922.1
\(\mathrm { f } ( x )\)0.58420.63090.6753
  1. Use the information in the table to calculate the most accurate estimate of \(f ^ { \prime } ( 2 )\) possible.
  2. Calculate an estimate of the error when \(f ( 2 )\) is used as an estimate of \(f ( 2.05 )\).
OCR MEI Further Numerical Methods 2024 June Q2
6 marks Standard +0.3
2 You are given that \(a = \tanh ( 1 )\) and \(b = \tanh ( 2 )\). \(A\) is the approximation to \(a\) formed by rounding \(\tanh ( 1 )\) to 1 decimal place. \(B\) is the approximation to \(b\) formed by rounding \(\tanh ( 2 )\) to 1 decimal place.
  1. Calculate the following.
OCR MEI Further Numerical Methods 2024 June Q3
6 marks Standard +0.8
3 The equation \(x ^ { 2 } - \cosh ( x - 2 ) = 0\) has two roots, \(\alpha\) and \(\beta\), such that \(\alpha < \beta\).
  1. Use the iterative formula $$x _ { n + 1 } = g \left( x _ { n } \right) \text { where } g \left( x _ { n } \right) = \sqrt { \cosh \left( x _ { n } - 2 \right) } \text {, }$$ starting with \(x _ { 0 } = 1\), to find \(\alpha\) correct to \(\mathbf { 3 }\) decimal places. The diagram shows the part of the graphs of \(\mathrm { y } = \mathrm { x }\) and \(\mathrm { y } = \mathrm { g } ( \mathrm { x } )\) for \(0 \leqslant x \leqslant 7\). \includegraphics[max width=\textwidth, alt={}, center]{83a06341-74e9-4f47-9104-e8e0259e7dfa-3_760_657_753_246}
  2. Explain why the iterative formula used to find \(\alpha\) cannot successfully be used to find \(\beta\), even if \(x _ { 0 }\) is very close to \(\beta\).
  3. Use the relaxed iteration $$\mathrm { x } _ { \mathrm { n } + 1 } = ( 1 - \lambda ) \mathrm { x } _ { \mathrm { n } } + \lambda \mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right) ,$$ with \(\lambda = - 0.21\) and \(x _ { 0 } = 6.4\), to find \(\beta\) correct to \(\mathbf { 3 }\) decimal places. In part (c) the method of relaxation was used to convert a divergent sequence of approximations into a convergent sequence.
  4. State one other application of the method of relaxation.
OCR MEI Further Numerical Methods 2024 June Q4
10 marks Moderate -0.5
4 Between 1946 and 2012 the mean monthly maximum temperature of the water surface of a lake in northern England has been recorded by environmental scientists. Some of the data are shown in Table 4.1. \begin{table}[h]
\captionsetup{labelformat=empty} \caption{Table 4.1}
MonthMayJuneJulyAugustSeptember
\(t =\) Time in months01234
\(T =\) Mean temperature in \({ } ^ { \circ } \mathrm { C }\)8.813.215.415.413.3
\end{table} Table 4.2 shows a difference table for the data. \begin{table}[h]
\captionsetup{labelformat=empty} \caption{Table 4.2}
\(t\)\(T\)\(\Delta T\)\(\Delta T ^ { 2 }\)
08.8
113.2
215.4
315.4
413.3
\end{table}
  1. Complete the copy of the difference table in the Printed Answer Booklet.
  2. Explain why a quadratic model may be appropriate for these data.
  3. Use Newton's forward difference interpolation formula to construct an interpolating polynomial of degree 2 for these data. This polynomial is used to model the relationship between \(T\) and \(t\). Between 1946 and 2012 the mean monthly maximum temperature of the water surface of the lake was recorded as \(8.9 ^ { \circ } \mathrm { C }\) for October and \(7.5 ^ { \circ } \mathrm { C }\) for November.
  4. Determine whether the model is a good fit for the temperatures recorded in October and November. A scientist recorded the mean monthly maximum temperature of the water surface of the lake in 2022. Some of the data are shown in Table 4.3. \begin{table}[h]
    \captionsetup{labelformat=empty} \caption{Table 4.3}
    MonthMayJuneJulyAugustSeptember
    \(t =\) Time in months01234
    \(T =\) Mean temperature in \({ } ^ { \circ } \mathrm { C }\)10.314.716.916.914.8
    \end{table}
  5. Adapt the polynomial found in part (c) so that it can be used to model the relationship between \(T\) and \(t\) for the data in Table 4.3.
OCR MEI Further Numerical Methods 2024 June Q5
10 marks Moderate -0.8
5 The root of the equation \(\mathrm { f } ( x ) = 0\) is being found using the method of interval bisection. Some of the associated spreadsheet output is shown in the table below.
1ABCDEF
1af(a)\(b\)f(b)c\(\mathrm { f } ( c )\)
22-0.610936.085542.51.43249
32-0.61092.51.432492.250.17524
42-0.61092.250.175242.125-0.2677
52.125-0.26772.250.175242.1875-0.0598
6
The formula in cell B2 is \(\quad = \mathrm { EXP } ( \mathrm { A } 2 ) - \mathrm { A } 2 ^ { \wedge } 2 - \mathrm { A } 2 - 2\).
  1. Write down the equation whose root is being found.
  2. Write down a suitable formula for cell E2. The formula in cell A3 is $$= \mathrm { IF } ( \mathrm {~F} 2 < 0 , \mathrm { E } 2 , \mathrm {~A} 2 )$$ .
  3. Write down a similar formula for cell C3.
  4. Complete row 6 of the table on the copy in the Printed Answer Booklet.
  5. Without doing any calculations, write down the value of the root correct to the number of decimal places which seems justified. You must explain the precision quoted.
  6. Determine how many more applications of the bisection method are needed such that the interval which contains the root is less than 0.0005 .
OCR MEI Further Numerical Methods 2024 June Q6
10 marks Standard +0.3
6 Table 6.1 shows some values of \(x\) and the associated values of a function, \(y = f ( x )\). \begin{table}[h]
\captionsetup{labelformat=empty} \caption{Table 6.1}
\(x\)1.512
\(\mathrm { f } ( x )\)0.8408911.18921
\end{table}
  1. Explain why it is not possible to use the central difference method to calculate an estimate of \(\frac { \mathrm { dy } } { \mathrm { dx } }\) when \(x = 1\).
  2. Use the forward difference method to calculate an estimate of \(\frac { \mathrm { dy } } { \mathrm { dx } }\) when \(x = 1\). A student uses the forward difference method to calculate a series of approximations to \(\frac { \mathrm { dy } } { \mathrm { dx } }\) when \(x = 2\) with different values of the step length, \(h\). These approximations are shown in Table 6.2, together with some further analysis. \begin{table}[h]
    \captionsetup{labelformat=empty} \caption{Table 6.2}
    \(h\)0.80.40.20.10.050.0250.01250.00625
    approximation0.1304520.1386470.1433810.1459420.1472770.1479590.1483040.148477
    difference0.0081950.0047340.0025610.0013350.0006820.0003450.000173
    ratio0.5776330.5410990.5211860.5107620.5054240.502723
    \end{table}
    1. Explain what the ratios of differences tell you about the order of the method in this case.
    2. Comment on whether this is unusual.
  4. Determine the value of \(\frac { \mathrm { dy } } { \mathrm { dx } }\) when \(x = 2\) as accurately as possible. You must justify the precision quoted.
OCR MEI Further Numerical Methods 2024 June Q7
14 marks Standard +0.3
7 A student is using a spreadsheet to find approximations to \(\int _ { 0 } ^ { 1 } f ( x ) d x\) using the midpoint rule, the trapezium rule and Simpson's rule. Some of the associated spreadsheet output with \(n = 1\) and \(n = 2\), is shown in Table 7.1. \begin{table}[h]
\captionsetup{labelformat=empty} \caption{Table 7.1}
\(n\)\(\mathrm { M } _ { n }\)\(\mathrm {~T} _ { n }\)\(\mathrm {~S} _ { 2 n }\)
10.6125471
20.639735
\end{table}
  1. Complete the copy of Table 7.1 in the Printed Answer Booklet. Give your answers correct to 5 decimal places.
  2. State the value of \(\int _ { 0 } ^ { 1 } \mathrm { f } ( x ) \mathrm { d } x\) as accurately as possible. You must justify the precision quoted. The student calculates some more approximations using Simpson's rule. These approximations are shown in the associated spreadsheet output, together with some further analysis, in Table 7.2. The values of \(S _ { 2 }\) and \(S _ { 4 }\) have been blacked out, together with the associated difference and ratio. \begin{table}[h]
    \captionsetup{labelformat=empty} \caption{Table 7.2}
    n\(\mathrm { S } _ { 2 n }\)differenceratio
    1
    2
    40.674353-0.0209
    80.665199-0.009150.438059
    160.661297-0.00390.426286
    320.659675-0.001620.415762
    640.659015-0.000660.406785
    \end{table}
  3. The student checks some of her values with a calculator. She does not obtain 0.406785 when she calculates \(- 0.00066 \div ( - 0.00162 )\). Explain whether the value in the spreadsheet, or her value, is a more precise approximation to the ratio of differences in this case.
    1. State the order of convergence of the values in the ratio column. You must justify your answer.
    2. Explain what the values in the ratio column tell you about the order of the method in this case.
    3. Comment on whether this is unusual.
  5. Determine the value of \(\int _ { 0 } ^ { 1 } f ( x ) d x\) as accurately as you can. You must justify the precision quoted.
OCR MEI Further Numerical Methods 2020 November Q1
4 marks Standard +0.3
1 Fig. 1 shows some spreadsheet output. \begin{table}[h]
A
11E-17
21E-17
31E-29
\captionsetup{labelformat=empty} \caption{Fig. 1}
\end{table}
  1. Write the value displayed in cell A3 in standard mathematical notation. The formula in cell A3 is \(= \mathrm { A } 2 - \mathrm { A } 1\)
  2. Explain why the value displayed in cell A3 is non zero.
  3. Write down the value of the number stored in cell A2 to the highest precision possible.
  4. Explain why your answer to part (c) may be different to the actual value stored in cell A2.
OCR MEI Further Numerical Methods 2020 November Q2
5 marks Moderate -0.5
2 Fig. 2 shows 3 values of \(x\) and the associated values of a function, \(\mathrm { f } ( x )\). \begin{table}[h]
\(x\)125
\(\mathrm { f } ( x )\)516.676.6
\captionsetup{labelformat=empty} \caption{Fig. 2}
\end{table} Find a polynomial \(p ( x )\) of degree 2 to approximate \(\mathrm { f } ( x )\), giving your answer in the form \(p ( x ) = a x ^ { 2 } + b x + c\), where \(a\), \(b\) and \(c\) are constants to be determined.
OCR MEI Further Numerical Methods 2020 November Q3
7 marks Moderate -0.3
3 At Heathwick airport each passenger's luggage is weighed before being loaded into the hold of the aeroplane. Each weight is displayed digitally in kg to 1 decimal place. Some examples are given in Fig. 3. \begin{table}[h]
Weight (kg)
17.2
19.9
22.3
20.1
21.5
\captionsetup{labelformat=empty} \caption{Fig. 3}
\end{table} On each flight, the total weight of luggage is calculated to ensure compliance with health and safety regulations. Winston models this situation by assuming that the displayed weights are rounded to 1 decimal place, and that the total weight of luggage is calculated using the displayed values. On a flight to Athens, there are 154 items of passengers' luggage.
  1. Determine the maximum possible error, according to Winston's model, when the total weight of luggage is calculated for the flight to Athens. Piotre models this situation by assuming that the displayed weights are chopped to 1 decimal place, and that the total weight of luggage is calculated using the displayed values.
  2. Determine the maximum possible error, according to Piotre's model, when the total weight of luggage is calculated for the flight to Athens. A health and safety inspector notes that the total of the displayed weights is 3080.2 kg . However, when the luggage is all weighed together in the loading bay, the total weight is found to be 3089.44 kg .
  3. Determine whether Winston's model or Piotre's model is a better fit for the data.
OCR MEI Further Numerical Methods 2020 November Q4
10 marks Challenging +1.2
4
  1. Use the trapezium rule with 1 strip to calculate an estimate of \(\int _ { 1 } ^ { 2 } \sqrt { 1 + x ^ { 3 } } \mathrm {~d} x\), giving your
    answer correct to six decimal places.
    [0pt] [2]
    Fig. 4 shows some spreadsheet output containing further approximations to this integral using the trapezium rule, denoted by \(T _ { n }\), and Simpson's rule, denoted by \(S _ { 2 n }\). \begin{table}[h]
    ABC
    1\(n\)\(T _ { n }\)\(S _ { 2 n }\)
    212.130135
    322.149378
    442.1347512.129862
    582.131084
    \captionsetup{labelformat=empty} \caption{Fig. 4}
    \end{table}
  2. Write down an efficient formula for cell C 4.
  3. Find the value of \(S _ { 4 }\), giving your answer correct to 6 decimal places.
  4. Without doing any further calculation, state the value of \(\int _ { 1 } ^ { 2 } \sqrt { 1 + x ^ { 3 } } \mathrm {~d} x\) as accurately as
    possible, justifying the precision quoted.
    [0pt] [2]
  5. Use the fact that Simpson's rule is a fourth order method to obtain an improved approximation to the value of \(\int _ { 1 } ^ { 2 } \sqrt { 1 + x ^ { 3 } } \mathrm {~d} x\), stating the value of this integral to a precision which seems justified.
OCR MEI Further Numerical Methods 2020 November Q5
13 marks Standard +0.3
5 You are given that \(g ( x ) = \frac { \sqrt [ 3 ] { x ^ { x } + 25 } } { 2 }\). Fig. 5.1 shows two values of \(x\) and the associated values of \(\mathrm { g } ( x )\). \begin{table}[h]
\(x\)1.451.55
\(g ( x )\)1.494681.49949
\captionsetup{labelformat=empty} \caption{Fig. 5.1}
\end{table}
  1. Use the central difference method to calculate an estimate of \(\mathrm { g } ^ { \prime } ( 1.5 )\), giving your answer correct to 3 decimal places. The equation \(x ^ { x } - 8 x ^ { 3 } + 25 = 0\) has two roots, \(\alpha\) and \(\beta\), such that \(\alpha \approx 1.5\) and \(\beta \approx 4.4\).
  2. Obtain the iterative formula \(x _ { n + 1 } = g \left( x _ { n } \right) = \frac { \sqrt [ 3 ] { x _ { n } ^ { X _ { n } } + 25 } } { 2 }\).
  3. Use your answer to part (a) to explain why it is possible that the iterative formula \(x _ { n + 1 } = g \left( x _ { n } \right) = \frac { \sqrt [ 3 ] { x _ { n } ^ { X _ { n } } + 25 } } { 2 }\) may be used to find \(\alpha\).
  4. Starting with \(x _ { 0 } = 1.5\), use the iterative formula to find \(x _ { 1 } , x _ { 2 } , x _ { 3 } , x _ { 4 } , x _ { 5 }\), and \(x _ { 6 }\).
  5. Use your answer to part (d) to state the value of \(\alpha\) correct to 8 decimal places. Starting with \(x _ { 0 } = 4.5\) the same iterative formula is used in an attempt to find \(\beta\). The results are shown in Fig. 5.2. \begin{table}[h]
    \(n\)\(x _ { n }\)
    04.5
    14.81826433
    26.27473453
    323.2937196
    4\(2.0654 \mathrm { E } + 10\)
    5\#NUM!
    \captionsetup{labelformat=empty} \caption{Fig. 5.2}
    \end{table}
  6. Explain why \#NUM! is displayed in the cell for \(x _ { 5 }\).
  7. On the diagram in the Printed Answer Booklet, starting with \(x _ { 0 } = 4.5\), illustrate how the iterative formula works to find \(x _ { 1 }\) and \(x _ { 2 }\).
  8. Determine what happens when the relaxed iteration \(x _ { n + 1 } = ( 1 - \lambda ) x _ { n } + \lambda g \left( x _ { n } \right)\) is used to try to find \(\beta\) with \(x _ { 0 } = 4.5\), in each of the following cases.