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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.
    • The relative error \(\mathrm { R } _ { \mathrm { A } }\) when \(A\) is used to approximate \(a\).
    • The relative error \(\mathrm { R } _ { \mathrm { B } }\) when \(B\) is used to approximate \(b\).
    • Calculate the relative error \(\mathrm { R } _ { \mathrm { C } }\) when \(\mathrm { C } = \frac { \mathrm { A } } { \mathrm { B } }\) is used to approximate \(\mathrm { c } = \frac { \mathrm { a } } { \mathrm { b } }\).
    • Comment on the relationship between \(R _ { A } , R _ { B }\) and \(R _ { C }\).
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.
    • \(\lambda = 0.5\)
    • \(\lambda = - 0.4\)
OCR MEI Further Numerical Methods 2020 November Q6
10 marks Standard +0.3
6 Fig. 6.1 shows the graph of \(y = \mathrm { e } ^ { 3 x } - 11 x - 0.5\) for \(- 0.5 \leqslant x \leqslant 1\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{87bb8eb7-b725-48b0-b32b-0bfce624cd91-08_576_881_315_333} \captionsetup{labelformat=empty} \caption{Fig. 6.1}
\end{figure} The equation \(\mathrm { e } ^ { 3 x } - 11 x - 0.5 = 0\) has two roots, \(\alpha\) and \(\beta\), such that \(\alpha < \beta\). Dennis is going to use the method of interval bisection with starting values denoted by \(a\) and \(b\).
  1. Explain why the method of interval bisection starting with \(a = 0\) and \(b = 1\) may not be used to find either \(\alpha\) or \(\beta\). Dennis uses the method of interval bisection starting with \(a = 0\) and \(b = 0.5\) to find \(\alpha\). Some spreadsheet output is shown in Fig. 6.2. \begin{table}[h]
    ABCDEF
    1af(a)bf(b)\(x _ { \text {new } }\)\(\mathrm { f } \left( x _ { \text {new } } \right)\)
    200.50.5-1.518310.25-1.133
    300.50.25-1.1330.125-0.42
    400.50.125-0.420010.06250.01873
    50.06250.018730.125-0.420010.09375-0.2065
    \captionsetup{labelformat=empty} \caption{Fig. 6.2}
    \end{table} Dennis states that the formula in cell B2 is $$= \operatorname { EXP } \left( 3 ^ { * } \mathrm {~A} 1 \right) - 11 \mathrm {~A} 2 - 0.5$$ Dennis has made two errors.
  2. Write a correct version of Dennis's formula for cell B2. The formula in cell A3, which is correct, is
    = IF(F2 > 0, E2, A2)
  3. Write a suitable formula for cell C3.
  4. Use the information in Fig. 6.2 to
    • find the value of \(\alpha\) as accurately as possible,
    • state the maximum possible error in this estimate.
    Liren uses a different method to find a sequence of estimates of the value of \(\beta\) using a spreadsheet. The output, together with some further analysis, is shown in Fig. 6.3. \begin{table}[h]
    ABCD
    1\(x\)f(x)differenceratio
    20.4-1.5799
    30.6-1.0504
    40.996718.4245
    50.64398-0.6809-0.35273
    60.67036-0.40260.026378-0.0748
    70.708520.083860.0381641.44682
    80.70194-0.0075-0.00658-0.1724
    90.70248-0.00010.00054-0.082
    100.70249\(1.8 \mathrm { E } - 07\)\(8.88 \mathrm { E } - 06\)0.01646
    \captionsetup{labelformat=empty} \caption{Fig. 6.3}
    \end{table} The formula in cell A4 is $$= ( \mathrm { A } 2 * \mathrm {~B} 3 - \mathrm { A } 3 * \mathrm {~B} 2 ) / ( \mathrm { B } 3 - \mathrm { B } 2 )$$
  5. State the method being used.
  6. Explain what the values in column D tell you about the order of convergence of this sequence of estimates. Liren states that \(\beta = 0.70249\) correct to 5 decimal places.
  7. Determine whether Liren is correct.
OCR MEI Further Numerical Methods 2020 November Q7
11 marks Challenging +1.2
7 Fig. 7.1 shows two values of \(x\) and the associated values of \(\mathrm { f } ( x )\). \begin{table}[h]
\(x\)33.5
\(\mathrm { f } ( x )\)6.0827634.596194
\captionsetup{labelformat=empty} \caption{Fig. 7.1}
\end{table}
  1. Use the forward difference method to calculate an estimate of the gradient of \(\mathrm { f } ( x )\) at \(x = 3\), giving your answer correct to 4 decimal places. Fig. 7.2 shows some spreadsheet output with additional values of \(x\) and the associated values of \(\mathrm { f } ( x )\). \begin{table}[h]
    \(x\)33.000013.00013.0013.013.1
    \(\mathrm { f } ( x )\)6.0827636.082746.0825416.080546.0604545.848846
    \captionsetup{labelformat=empty} \caption{Fig. 7.2}
    \end{table} These values have been used to produce a sequence of estimates of the gradient of \(\mathrm { f } ( x )\) at \(x = 3\), together with some further analysis. This is shown in the spreadsheet output in Fig. 7.3. \begin{table}[h]
    \(h\)0.10.010.0010.00010.00001
    estimate-2.339165-2.230883-2.220532-2.219501-2.219398
    difference0.10828150.0103520.00103070.000103
    ratio0.0956020.0995670.0999568
    \captionsetup{labelformat=empty} \caption{Fig. 7.3}
    \end{table} Tommy states that the differences between successive estimates is decreasing so rapidly that the order of convergence of this sequence of estimates is much faster than first order.
  2. Explain whether or not Tommy is correct.
  3. Use extrapolation to determine the value of the gradient of \(\mathrm { f } ( x )\) at \(x = 3\) as accurately as possible, justifying the precision quoted.
  4. Calculate an estimate of the absolute error when \(\mathrm { f } ( 3 )\) is used as an approximation to \(\mathrm { f } ( 3.02 )\).
OCR MEI Further Numerical Methods 2021 November Q1
5 marks Standard +0.3
1
    1. Determine the relative error when
      • 1.414214 is used to approximate \(\sqrt { 2 }\),
  2. \(1.414214 ^ { 2 }\) is used to approximate 2.
    (ii) Write down the relationship between your answers to part (a)(i).
  3. Fig. 1 shows some spreadsheet output.
    4. \begin{table}[h]
    ABC
    121.4142142
    \captionsetup{labelformat=empty} \caption{Fig. 1}
    \end{table} The formula in cell B1 is = SQRT (A1)
    and the formula in cell C 1 is \(\quad = \mathrm { B } 1 \wedge 2\).
    Ben evaluates \(1.414214 ^ { 2 }\) on his calculator and obtains 2.000001238 . He states that this shows that the value displayed in cell C1 is wrong. Explain whether Ben is correct.
OCR MEI Further Numerical Methods 2021 November Q2
6 marks Challenging +1.2
2 The table shows some values of \(x\) and the associated values of \(\mathrm { f } ( x )\).
\(x\)12345
\(\mathrm { f } ( x )\)- 0.65- 0.351.775.7111.47
  1. Complete the difference table in the Printed Answer Booklet.
  2. Explain why the data may be interpolated by a polynomial of degree 2.
  3. Use Newton's forward difference interpolation formula to obtain a polynomial of degree 2 for the data.
OCR MEI Further Numerical Methods 2021 November Q3
7 marks Standard +0.3
3 The method of False Position is used to find a sequence of approximations to the root of an equation. The spreadsheet output showing these approximations, together with some further analysis, is shown below.
CDEFGHIJ
4af(a)b\(\mathrm { f } ( b )\)\(x _ { \text {new } }\)\(\mathrm { f } \left( x _ { \text {new } } \right)\)differenceratio
51-1.8248217.28991.09547-1.80507
61.09547-1.80507217.28991.18097-1.754180.08551
71.18097-1.75418217.28991.25641-1.662460.075440.88229
81.25641-1.66246217.28991.32164-1.527810.065230.86458
91.32164-1.52781217.28991.37672-1.357060.055080.84439
101.37672-1.35706217.28991.42208-1.16420.045360.8236
111.42208-1.1642217.28991.45853-0.966160.036460.80376
121.45853-0.96616217.28991.48719-0.778250.028660.78598
131.48719-0.77825217.2899
14
The formula in cell D5 is \(\quad = \mathrm { SINH } \left( \mathrm { C5 } ^ { \wedge } 2 \right) - \mathrm { C5 } ^ { \wedge } 3 - 2\).
  1. Write down the equation which is being solved. The formula in cell C 6 is \(\quad = \mathrm { IF } ( \mathrm { H } 5 < 0 , \mathrm { G } 5 , \mathrm { C } 5 )\).
  2. Write down a similar formula for cell E6.
  3. Calculate the values which would be displayed in cells G13 and G14 to find further approximations to the root.
  4. Explain what the values in column J tell you about
    • the order of convergence of this sequence of estimates,
    • the speed of convergence of this sequence of estimates.
OCR MEI Further Numerical Methods 2021 November Q4
6 marks Standard +0.3
4 The table shows some values of \(x\) and the associated values of \(\mathrm { f } ( x )\).
\(x\)44.00014.0014.014.1
\(\mathrm { f } ( x )\)44.00023864.00238714.02394684.2472072
  1. Calculate four estimates of the derivative of \(\mathrm { f } ( x )\) at \(x = 4\).
  2. Without doing any further calculation, state the value of \(f ^ { \prime } ( 4 )\) as accurately as you can, justifying the precision quoted.
OCR MEI Further Numerical Methods 2021 November Q5
9 marks Standard +0.3
5 When Nina does the weekly grocery shopping she models the total cost by adding up the cost of each item in her head as she goes along. To simplify matters she rounds the cost of each item to the nearest pound. One week Nina buys 48 items.
  1. Calculate the maximum possible error in Nina's model in this case. Nina estimated the total cost of her shopping to be \(\pounds 92\). The actual cost is \(\pounds 90.23\).
  2. Explain whether this is consistent with Nina's model. The next week her husband, Kareem, does the weekly shopping. He models the total cost by chopping the cost of each item to the nearest pound as he goes along. On this occasion Kareem buys 52 items.
  3. Calculate the expected error in Kareem's model in this case. Using his model Kareem estimates the total cost as \(\pounds 76\). The total cost of the shopping is \(\pounds 103.24\).
  4. Explain how such a large error could arise. The next week Kareem buys \(n\) items.
  5. Write down a formula for the maximum possible error when Kareem uses his model to estimate the total cost of his shopping.
  6. Explain how Kareem's model could be adapted so that his formula gives the same expected error as Nina's model when they are both used to estimate the total cost of the shopping.
OCR MEI Further Numerical Methods 2021 November Q6
12 marks Challenging +1.2
6 The equation \(0.5 \ln x - x ^ { 2 } + x + 1 = 0\) has two roots \(\alpha\) and \(\beta\), such that \(0 < \alpha < 1\) and \(1 < \beta < 2\).
  1. Use the Newton-Raphson method with \(x _ { 0 } = 1\) to obtain \(\beta\) correct to \(\mathbf { 6 }\) decimal places. Fig. 6.1 shows part of the graph of \(y = 0.5 \ln x - x ^ { 2 } + x + 1\). \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{945883ad-c153-4c51-83d3-978e4c769ed5-06_1112_1156_529_354} \captionsetup{labelformat=empty} \caption{Fig. 6.1}
    \end{figure}
  2. On the copy of Fig. 6.1 in the Printed Answer Booklet, illustrate the Newton-Raphson method working to obtain \(x _ { 1 }\) from \(x _ { 0 } = 1\). Beth is trying to find \(\alpha\) correct to 6 decimal places.
  3. Suggest a reason why she might choose the Newton-Raphson method instead of fixed point iteration. Beth tries to find \(\alpha\) using the Newton-Raphson method with a starting value of \(x _ { 0 } = 0.5\). Her spreadsheet output is shown in Fig. 6.2. \begin{table}[h]
    \(r\)\(\mathrm { x } _ { \mathrm { r } }\)
    00.5
    1- 0.40343
    2\#NUM!
    \captionsetup{labelformat=empty} \caption{Fig. 6.2}
    \end{table}
  4. Explain how the display \#NUM! has arisen in the cell for \(x _ { 2 }\). Beth decides to use the iterative formula $$x _ { n + 1 } = g \left( x _ { n } \right) = \sqrt { 0.5 \ln \left( x _ { n } \right) + x _ { n } + 1 }$$
  5. Determine the outcome when Beth uses this formula with \(x _ { 0 } = 0.5\).
  6. 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 \(\lambda = - 0.041\) and \(x _ { 0 } = 0.5\) to obtain \(\alpha\) correct to \(\mathbf { 6 }\) decimal places.
OCR MEI Further Numerical Methods 2021 November Q7
15 marks Standard +0.8
7 Sarah uses the trapezium rule to find a sequence of approximations to \(\int _ { 0 } ^ { 1 } \sqrt { \tanh ( x ) } \mathrm { d } x\).
Her spreadsheet output is shown in Fig. 7.1. \begin{table}[h]
\(n\)\(T _ { n }\)differenceratio
10.43634681
20.55806940.121723
40.601998430.0439290.36089
80.617870730.0158720.36132
160.623576010.0057050.35945
320.625617160.0020410.35777
\captionsetup{labelformat=empty} \caption{Fig. 7.1}
\end{table}
  1. Write down the value of \(h\) used to find the approximation 0.62357601 .
  2. Without doing any further calculation, state the value of \(\int _ { 0 } ^ { 1 } \sqrt { \tanh ( x ) } \mathrm { d } x\) as accurately as you
    can, justifying the precision quoted.
  3. Explain what the values in the ratio column tell you about the order of convergence of this sequence of approximations. Sarah carries out further work using the midpoint rule and Simpson's rule. Her results are shown in Fig. 7.2. \begin{table}[h]
    MNOPQR
    1\(n\)\(T _ { n }\)\(M _ { n }\)\(S _ { 2 n }\)differenceratio
    210.436346810.6797920.5986436
    320.55806940.645927450.616641440.018
    440.601998430.633743040.62316150.006520.362269
    580.617870730.629281290.625477770.002320.355253
    6160.623576010.627658310.626297550.000820.35392
    7320.625617160.62707259
    \captionsetup{labelformat=empty} \caption{Fig. 7.2}
    \end{table}
  4. Write down an efficient spreadsheet formula for calculating \(S _ { 16 }\).
  5. Determine the missing values in row 7.
  6. Use extrapolation to determine the value of \(\int _ { 0 } ^ { 1 } \sqrt { \tanh ( x ) } d x\) as accurately as you can, justifying
    the precision quoted.
    [0pt] [6]
OCR MEI Further Numerical Methods Specimen Q1
5 marks Standard +0.8
1
  1. Solve the following simultaneous equations. $$\begin{aligned} & x + \quad y = 1 \\ & x + 0.99 y = 2 \end{aligned}$$
  2. The coefficient 0.99 is correct to two decimal places. All other coefficients in the equations are exact. With the aid of suitable calculations, explain why your answer to part (i) is unreliable.