Questions - OCR MEI Further Numerical Methods

	A	B	C
1	$x$	$\mathrm { f } ( x )$
2	1	0.367879441	0.367879441
3	2	0.018315639	0.38619508
4	3	0.00012341	0.38631849
5	4	$1.12535 \mathrm { E } - 07$	0.386318602
6	5	$1.38879 \mathrm { E } - 11$	0.386318602

\captionsetup{labelformat=empty} \caption{Fig. 1}

\end{table} The formula in cell B2 is ==EXP(-(A2\^{}2)) and equivalent formulae are in cells B3 to B6. The formula in cell C 2 is $= \mathrm { B } 2$.
The formula in cell C3 is $\quad = \mathrm { C } 2 + \mathrm { B } 3$. Equivalent formulae are in cells C4 to C6.

Use sigma notation to express the formula in cell C5 in standard mathematical notation.
Explain why the same value is displayed in cells C 5 and C 6. Now suppose that the value in cell C2 is chopped to 3 decimal places and used to approximate the value in cell C2.
Calculate the relative error when this approximation is used. Suppose that the values in cells B4, B5 and B6 are chopped to 3 decimal places and used as approximations to the original values in cells B4, B5 and B6 respectively.
Explain why the relative errors in these approximations are all the same.

OCR MEI Further Numerical Methods 2019 June Q2

2 Fig. 2.1 shows the graph of $y = x ^ { 2 } \mathrm { e } ^ { 2 x } - 5 x ^ { 2 } + 0.5$. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{4838f71e-a1d0-4695-89d2-c7ebb47edd77-4_757_545_315_248} \captionsetup{labelformat=empty} \caption{Fig. 2.1}

\end{figure} There are three roots of the equation $x ^ { 2 } \mathrm { e } ^ { 2 x } - 5 x ^ { 2 } + 0.5 = 0$. The roots are $\alpha , \beta$ and $\gamma$, where $\alpha < \beta < \gamma$.

Explain why it is not possible to use the method of false position with $x _ { 0 } = 0$ and $x _ { 1 } = 1$ to find $\beta$ and $\gamma$. The graph of the function indicates that the root $\gamma$ lies in the interval [0.6, 0.8]. Fig. 2.2 shows some spreadsheet output using the method of false position using these values as starting points. \begin{table}[h]

	A	B	C	D	E	F
1	a	f(a)	b	f(b)	approx
2	0.6	-0.10476	0.8	0.469941	0.636457	-0.07876
3	0.636457	-0.07876	0.8	0.469941	0.659931	-0.04748
4	0.659931	-0.04748	0.8	0.469941	0.672783	-0.0249
5	0.672783	-0.0249	0.8	0.469941	0.679184	-0.01211
6	0.679184	-0.01211	0.8	0.469941	0.682218	-0.00567
7	0.682218	-0.00567	0.8	0.469941	0.683623	-0.00261
8	0.683623	-0.00261	0.8	0.469941	0.684266	-0.00119
9	0.684266	-0.00119	0.8	0.469941	0.684559	-0.00054
10	0.684559	-0.00054	0.8	0.469941	0.684692	-0.00025
11	0.684692	-0.00025	0.8	0.469941	0.684753	-0.00011
12	0.684753	-0.00011	0.8	0.469941	0.68478	$- 5.1 \mathrm { E } - 05$

\captionsetup{labelformat=empty} \caption{Fig. 2.2}

\end{table}

Without doing any further calculation, write down the smallest possible interval which is certain to contain $\gamma$.
State what is being calculated in column F. The formula in cell A3 is $\quad = \operatorname { IF } ( \mathrm { F } 2 < 0 , \mathrm { E } 2 , \mathrm {~A} 2 )$.
Explain the purpose of this formula in the application of the method of false position. The method of false position uses the same formula for obtaining new approximations as the secant method.
Explain how the method of false position differs from the secant method.
Give one advantage and one disadvantage of using the method of false position instead of the secant method.

OCR MEI Further Numerical Methods 2019 June Q3

3 In the first week of an outbreak of influenza, 9 patients were diagnosed with the virus at a medical practice in Pencaster. Records were kept of $y$, the total number of patients diagnosed with influenza in week $n$. The data are shown in Fig. 3. \begin{table}[h]

$n$	1	2	3	4	5
$y$	9	32	63	96	125

\captionsetup{labelformat=empty} \caption{Fig. 3}

\end{table}

Complete the difference table in the Printed Answer Booklet.
Explain why a cubic model is appropriate for the data.
Use Newton's method to find the interpolating polynomial of degree 3 for these data. In both week 6 and week 7 there were 145 patients in total diagnosed with influenza at the medical practice.
Determine whether the model is a good fit for these data.
Determine the maximum number of weeks for which the model could possibly be valid.

OCR MEI Further Numerical Methods 2019 June Q4

4 Fig. 4 shows the graph of $y = x ^ { 5 } - 6 \sqrt { x } + 4$. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{4838f71e-a1d0-4695-89d2-c7ebb47edd77-6_867_700_317_246} \captionsetup{labelformat=empty} \caption{Fig. 4}

\end{figure} There are two roots of the equation $x ^ { 5 } - 6 \sqrt { x } + 4 = 0$. The roots are $\alpha$ and $\beta$, such that $\alpha < \beta$.

Show that $0 < \alpha < 1$ and $1 < \beta < 2$.
Obtain the Newton-Raphson iterative formula $$x _ { n + 1 } = x _ { n } - \frac { x _ { n } ^ { \frac { 11 } { 2 } } - 6 x _ { n } + 4 \sqrt { x _ { n } } } { 5 x _ { n } ^ { \frac { 9 } { 2 } } - 3 }$$
Use the iterative formula found in part (b) with a starting value of $x _ { 0 } = 1$ to obtain $\beta$ correct to 6 decimal places.
Use the iterative formula found in part (b) with a starting value of $x _ { 0 } = 0$ to find $x _ { 1 }$.
Give a geometrical explanation of why the Newton-Raphson iteration fails to find $\alpha$ in part (d).
Obtain the iterative formula $$x _ { n + 1 } = \left( \frac { x _ { n } ^ { 5 } + 4 } { 6 } \right) ^ { 2 }$$
Use the iterative formula found in part (f) with a starting value of $x _ { 0 } = 0$ to obtain $\alpha$ correct to 6 decimal places.

OCR MEI Further Numerical Methods 2019 June Q5

5 Fig. 5 shows spreadsheet output concerning the estimation of the derivative of a function $\mathrm { f } ( x )$ at $x = 2$ using the forward difference method. \begin{table}[h]

	A	B	C	D
1	h	estimate	difference	ratio
2	0.1	6.3050005
3	0.01	6.0300512	-0.274949
4	0.001	6.0030018	-0.027049	0.098379
5	0.0001	6.0003014	-0.0027	0.099835
6	0.00001	6.0000314	-0.00027	0.099983
7	0.000001	6.0000044	$- 2.7 \mathrm { E } - 05$	0.099994
8	1E-07	6.0000016	$- 2.71 \mathrm { E } - 06$	0.100352
9	1E-08	6.0000013	$- 3.02 \mathrm { E } - 07$	0.111457
10	1E-09	6.0000018	$4.885 \mathrm { E } - 07$	-1.61765
11	1E-10	6.0000049	$3.109 \mathrm { E } - 06$	6.363636
12	1E-11	6.0000005	$- 4.44 \mathrm { E } - 06$	-1.42857
13	1E-12	6.0005334	0.0005329	-120
14	1E-13	5.9952043	-0.005329	-10
15	1E-14	6.1284311	0.1332268	-25
16	1E-15	5.3290705	-0.799361	-6
17	1E-16	0	-5.329071	6.666667

\captionsetup{labelformat=empty} \caption{Fig. 5}

\end{table}

Write down suitable cell formulae for
- cell C3,
- cell D4.
- Explain what the entries in cells D4 to D8 tell you about the order of the convergence of the forward difference method.
- Write the entry in cell A10 in standard mathematical notation.
- Explain what the values displayed in cells D10 to D17 suggest about the values in cells B10 to B16.
- Write down the value of the derivative of $\mathrm { f } ( x )$ at $x = 2$ to an accuracy that seems justified, explaining your answer.
The formula in cell B2 is $\quad = ( \mathrm { LN } ( \mathrm { SQRT } ( \mathrm { SINH } ( ( 2 + \mathrm { A } 2 ) \wedge 3 ) ) ) - \mathrm { LN } ( \mathrm { SQRT } ( \mathrm { SINH } ( 2 \wedge 3 ) ) ) ) / \mathrm { A } 2$ and equivalent formulae are entered in cells B3 to B17.
Write $\mathrm { f } ( x )$ in standard mathematical notation. The value displayed in cell B17 is zero, even though the calculation results in a non-zero answer.
Explain how this has arisen.

OCR MEI Further Numerical Methods 2019 June Q6

6 The spreadsheet output in Fig. 6 shows approximations to $\int _ { 0 } ^ { 1 } x ^ { - \sqrt { x } } \mathrm {~d} x$ found using the midpoint rule, denoted by $M _ { n }$, and the trapezium rule, denoted by $T _ { n }$. \begin{table}[h]

	A	B	C
1	$n$	$M _ { n }$	$T _ { n }$
2	1	1.632527	1
3	2	1.641461	1.316263
4	4	1.623053	1.478862
5	8	1.610295	1.550957
6	16	1.604132	1.580626
7	32	1.601505	1.592379

\captionsetup{labelformat=empty} \caption{Fig. 6}

\end{table}

Write down an efficient spreadsheet formula for cell C3.
By first completing the table in the Printed Answer Booklet using the Simpson's rule, calculate the most accurate estimate of $\int _ { 0 } ^ { 1 } x ^ { - \sqrt { x } } \mathrm {~d} x$ that you can, justifying the precision quoted. \section*{END OF QUESTION PAPER}

OCR MEI Further Numerical Methods 2022 June Q2

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

$x$	2.75	3	3.25
$\mathrm { f } ( x )$	0.920799	1	1.072858

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.
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.
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.
Determine an estimate of the error when $\mathrm { f } ( 5 )$ is used to estimate $\mathrm { f } ( 5.024 )$.

OCR MEI Further Numerical Methods 2022 June Q3

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.

	C	D	E
4	$r$	$\mathrm { x } _ { \mathrm { r } }$	$\mathrm { f } \left( \mathrm { x } _ { \mathrm { r } } \right)$
5	0	- 1	- 0.4707963
6	1	- 0.8	- 0.0272952
7	2	- 0.787691	- 0.0193610
8	3	- 0.7576546	- 0.0020574
9	4	- 0.7540834	- 0.0001740
10	5
11	6

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.

State the method being used.
Use the values in the spreadsheet to calculate $x _ { 5 }$ and $x _ { 6 }$, giving your answers correct to 7 decimal places.
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 .
Explain why this is different to the value displayed in cell D9. The value displayed in cell E11 in Alex's spreadsheet is $- 1.4629 \mathrm { E } - 09$.
Write this value in standard mathematical notation.

OCR MEI Further Numerical Methods 2022 June Q4

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 $x _ { 0 } = 2$.

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}

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.

State what is being calculated in the following columns of the spreadsheet.
1. Column N
2. Column O
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

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.

	F	G	H	I
3	$n$	$\mathrm { M } _ { \mathrm { n } }$	$\mathrm { T } _ { \mathrm { n } }$	$\mathrm { S } _ { 2 \mathrm { n } }$
4	1	0.2436699	0.1479020
5	2	0.2306967

Write down a suitable spreadsheet formula for cell H 5 .
Complete the copy of the table in the Printed Answer Booklet, giving the values correct to 7 decimal places.
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

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.

$r$	$\mathrm { x } _ { \mathrm { r } }$	difference	ratio
0	0
1	-0.841471	-0.84147
2	-0.287798	0.553673	-0.65798
3	-0.793885	-0.50609	-0.91405
4	-0.361379	0.432507	-0.85461
5	-0.763945	-0.40257	-0.93078
6	-0.404459	0.359486	-0.89299

\begin{table}[h]

102	-0.596302	0.004626	-0.95886
103	-0.600738	-0.00444	-0.95911
104	-0.596484	0.004254	-0.95887
105	-0.600564	-0.00408	-0.95910
106	-0.596652	0.003912	-0.95888
107	-0.600404	-0.00375	-0.95909
108	-0.596806	0.003598	-0.95889

\captionsetup{labelformat=empty} \caption{Fig. 6.1}

\end{table}

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

OCR MEI Further Numerical Methods 2022 June Q7

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]

$t$	0	10	20	30
$M$	88.3	80.05	78.7	78.85

\captionsetup{labelformat=empty} \caption{Fig. 7.1}

\end{table} A difference table for the data is shown in Fig. 7.2. \begin{table}[h]

$t$	$M$	$\Delta M$	$\Delta ^ { 2 } M$	$\Delta ^ { 3 } M$
0	88.3

10	80.05

20	78.7

30	78.85

\captionsetup{labelformat=empty} \caption{Fig. 7.2}

\end{table}

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.
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 $\mathrm { t } = 50 , \mathrm { M } = 80.05$.
Determine whether the model is a good fit for these data.
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.
Refine the doctor's model to include a quartic term.
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

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.

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.
Determine the range of possible values of $y _ { 2 } - y _ { 1 }$. You are given that $m = \frac { y _ { 2 } - y _ { 1 } } { x _ { 2 } - x _ { 1 } }$.
Determine the range of possible values of $m$.
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

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).

$t$	0	3	5
$P$	34.5	29.4	27.0

The owner of the car believes the relationship between $P$ and $t$ may be modelled by a polynomial.

Explain why it is not possible to use Newton's forward difference interpolation method for these data.
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 $t = 10 , P = 24.4$.
Determine whether the owner's model is a good fit for these data.
Explain why the model would not be suitable in the long term.

OCR MEI Further Numerical Methods 2023 June Q3

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 $y$.

$x$	1.5	2	2.5
$y$	1.682137	2.094395	2.318559

Use the forward difference method to calculate an approximation to $\frac { \mathrm { dy } } { \mathrm { dx } }$ at $x = 2$.
Use the central difference method to calculate an approximation to $\frac { \mathrm { dy } } { \mathrm { dx } }$ at $x = 2$.
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).
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 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.

	B	C	D
3	$x$	$\mathrm { f } ( x )$	$\mathrm { M } _ { 1 }$
4	1.5	1.3103707	0.65518535

The formula in cell C4 is $= \mathrm { B } 4 \wedge ( 1 / \mathrm { B } 4 )$.
The formula in cell D4 is $= 0.5 ^ { * } \mathrm { C } 4$.

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}
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

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}

With reference to the graph, explain why it might not be possible to use the student's iterative formula to find the root near $x = 1.8$.
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.
$r$ difference ratio
0
1 - 0.1834898
2 - 0.0049137 0.02678
3 $- 6.44 \mathrm { E } - 06$ 0.00131
4 $- 3.862 \mathrm { E } - 09$ 0.0006
5 $- 2.313 \mathrm { E } - 12$ 0.0006
Explain what the analysis tells you about the order of convergence of this sequence of approximations.

OCR MEI Further Numerical Methods 2023 June Q6

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.
  A B C
  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

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}

$x$	$\mathrm { f } ( x )$	$\frac { \mathrm { dy } } { \mathrm { dx } }$
2	6	- 2.8

\end{table}

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}

	A	B	C	D
1	r	Xr	difference	ratio
2	0	12
3	1	-13.1165572	-25.1165572
4	2	1.76283279	14.87939004	-0.5924136
5	3	2.18052157	0.41768878	0.02807163
6	4	2.182419024	0.001897454	0.00454275
7	5	2.182419066	$4.13985 \mathrm { E } - 08$	$2.1818 \mathrm { E } - 05$

\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 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 $\beta$.

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 $x _ { 1 } = 2$.

	I	J	K	L	M
2	$r$	$\mathrm { x } _ { \mathrm { r } }$	f(x)	$\mathrm { X } _ { \mathrm { r } + 1 }$	$\mathrm { f } \left( \mathrm { x } _ { \mathrm { r } + 1 } \right)$
3	0	1	-0.0914	2	-0.0476
4	1	2	-0.0476	3.08529	0.95784
5	2	3.08529	0.95784	2.05134	-0.0298
6	3	2.05134	-0.0298	2.08259	-0.0181
7	4	2.08259	-0.0181	2.13042	0.00155
8	5	2.13042	0.00155	2.12664	$- 7 \mathrm { E } - 05$

Write down a suitable cell formula for cell J4.
Write down a suitable cell formula for cell L4.
Write down the most accurate approximation to $\beta$ which is displayed in the table.
Determine whether your answer to part (d) is correct to 5 decimal places. You should not calculate any more iterates.
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 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}

$n$	$\mathrm {~T} _ { n }$
1	0.52764369
2	0.66617652
4	0.72534275

\end{table}

Use these results to determine two approximations to $\int _ { 0 } ^ { 1 } \sqrt [ 3 ] { \sinh ( x ) } \mathrm { d } x$ using Simpson's rule.

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}

$n$	$\mathrm { T } _ { n }$	difference	ratio
1	0.5276437
2	0.6661765	0.138533
4	0.7253427	0.059166	0.42709
8	0.7498821	0.024539	0.41475
16	0.7598858	0.010004	0.40766
32	0.7639221	0.004036	0.40348
64	0.7655404	0.001618	0.40095

\end{table}

Explain what can be deduced about the order of the method in this case.
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

1 The table shows some values of $x$, together with the associated values of a function, $\mathrm { f } ( x )$.

$x$	1.9	2	2.1
$\mathrm { f } ( x )$	0.5842	0.6309	0.6753

Use the information in the table to calculate the most accurate estimate of $f ^ { \prime } ( 2 )$ possible.
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

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.

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

3 The equation $x ^ { 2 } - \cosh ( x - 2 ) = 0$ has two roots, $\alpha$ and $\beta$, such that $\alpha < \beta$.

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}
Explain why the iterative formula used to find $\alpha$ cannot successfully be used to find $\beta$, even if $x _ { 0 }$ is very close to $\beta$.
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.
State one other application of the method of relaxation.

OCR MEI Further Numerical Methods 2024 June Q4

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}

Month	May	June	July	August	September
$t =$ Time in months	0	1	2	3	4
$T =$ Mean temperature in ${ } ^ { \circ } \mathrm { C }$	8.8	13.2	15.4	15.4	13.3

\end{table} Table 4.2 shows a difference table for the data. \begin{table}[h]

\captionsetup{labelformat=empty} \caption{Table 4.2}

$t$	$T$	$\Delta T$	$\Delta T ^ { 2 }$
0	8.8

1	13.2

2	15.4

3	15.4

4	13.3

\end{table}

Complete the copy of the difference table in the Printed Answer Booklet.
Explain why a quadratic model may be appropriate for these data.
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.
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}
Month May June July August September
$t =$ Time in months 0 1 2 3 4
$T =$ Mean temperature in ${ } ^ { \circ } \mathrm { C }$ 10.3 14.7 16.9 16.9 14.8
\end{table}
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.

	F	G	H	I
3	\(n\)	\(\mathrm { M } _ { \mathrm { n } }\)	\(\mathrm { T } _ { \mathrm { n } }\)	\(\mathrm { S } _ { 2 \mathrm { n } }\)
4	1	0.2436699	0.1479020
5	2	0.2306967

	A	B	C
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	\(9.99998 \mathrm { E } - 12\)

\(x\)	\(\mathrm { f } ( x )\)	\(\frac { \mathrm { dy } } { \mathrm { dx } }\)
2	6	- 2.8

	A	B	C
1	\(n\)	\(M _ { n }\)	\(T _ { n }\)
2	1	1.632527	1
3	2	1.641461	1.316263
4	4	1.623053	1.478862
5	8	1.610295	1.550957
6	16	1.604132	1.580626
7	32	1.601505	1.592379

	C	D	E
4	\(r\)	\(\mathrm { x } _ { \mathrm { r } }\)	\(\mathrm { f } \left( \mathrm { x } _ { \mathrm { r } } \right)\)
5	0	- 1	- 0.4707963
6	1	- 0.8	- 0.0272952
7	2	- 0.787691	- 0.0193610
8	3	- 0.7576546	- 0.0020574
9	4	- 0.7540834	- 0.0001740
10	5
11	6

\(t\)	\(M\)	\(\Delta M\)	\(\Delta ^ { 2 } M\)	\(\Delta ^ { 3 } M\)
0	88.3

10	80.05

20	78.7

30	78.85

	B	C	D
3	\(x\)	\(\mathrm { f } ( x )\)	\(\mathrm { M } _ { 1 }\)
4	1.5	1.3103707	0.65518535

\(n\)	\(\mathrm {~T} _ { n }\)
1	0.52764369
2	0.66617652
4	0.72534275

\(n\)	\(\mathrm { T } _ { n }\)	difference	ratio
1	0.5276437
2	0.6661765	0.138533
4	0.7253427	0.059166	0.42709
8	0.7498821	0.024539	0.41475
16	0.7598858	0.010004	0.40766
32	0.7639221	0.004036	0.40348
64	0.7655404	0.001618	0.40095

\(t\)	\(T\)	\(\Delta T\)	\(\Delta T ^ { 2 }\)
0	8.8

1	13.2

2	15.4

3	15.4

4	13.3

\(r\)	\(\mathrm { x } _ { \mathrm { r } }\)	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

\(r\)	difference	ratio
0
1	- 0.1834898
2	- 0.0049137	0.02678
3	\(- 6.44 \mathrm { E } - 06\)	0.00131
4	\(- 3.862 \mathrm { E } - 09\)	0.0006
5	\(- 2.313 \mathrm { E } - 12\)	0.0006

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