Questions Further Numerical Methods

1	A	B	C	D	E	F
1	a	f(a)	$b$	f(b)	c	$\mathrm { f } ( c )$
2	2	-0.6109	3	6.08554	2.5	1.43249
3	2	-0.6109	2.5	1.43249	2.25	0.17524
4	2	-0.6109	2.25	0.17524	2.125	-0.2677
5	2.125	-0.2677	2.25	0.17524	2.1875	-0.0598
6

The formula in cell B2 is $\quad = \mathrm { EXP } ( \mathrm { A } 2 ) - \mathrm { A } 2 ^ { \wedge } 2 - \mathrm { A } 2 - 2$.

Write down the equation whose root is being found.
Write down a suitable formula for cell E2. The formula in cell A3 is $$= \mathrm { IF } ( \mathrm {~F} 2 < 0 , \mathrm { E } 2 , \mathrm {~A} 2 )$$ .
Write down a similar formula for cell C3.
Complete row 6 of the table on the copy in the Printed Answer Booklet.
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.
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

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}

$x$	1.5	1	2
$\mathrm { f } ( x )$	0.84089	1	1.18921

\end{table}

Explain why it is not possible to use the central difference method to calculate an estimate of $\frac { \mathrm { dy } } { \mathrm { dx } }$ when $x = 1$.

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}

$h$	0.8	0.4	0.2	0.1	0.05	0.025	0.0125	0.00625
approximation	0.130452	0.138647	0.143381	0.145942	0.147277	0.147959	0.148304	0.148477
difference		0.008195	0.004734	0.002561	0.001335	0.000682	0.000345	0.000173
ratio			0.577633	0.541099	0.521186	0.510762	0.505424	0.502723

\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.
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

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}

$n$	$\mathrm { M } _ { n }$	$\mathrm {~T} _ { n }$	$\mathrm {~S} _ { 2 n }$
1	0.612547	1
2	0.639735

\end{table}

Complete the copy of Table 7.1 in the Printed Answer Booklet. Give your answers correct to 5 decimal places.
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 $\mathrm { S } _ { 2 n }$ difference ratio
1
2
4 0.674353 -0.0209
8 0.665199 -0.00915 0.438059
16 0.661297 -0.0039 0.426286
32 0.659675 -0.00162 0.415762
64 0.659015 -0.00066 0.406785
\end{table}
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.
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

1 Fig. 1 shows some spreadsheet output. \begin{table}[h]

	A
1	1E-17
2	1E-17
3	1E-29

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

\end{table}

Write the value displayed in cell A3 in standard mathematical notation. The formula in cell A3 is
$= \mathrm { A } 2 - \mathrm { A } 1$
Explain why the value displayed in cell A3 is non zero.
Write down the value of the number stored in cell A2 to the highest precision possible.
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

2 Fig. 2 shows 3 values of $x$ and the associated values of a function, $\mathrm { f } ( x )$. \begin{table}[h]

$x$	1	2	5
$\mathrm { f } ( x )$	5	16.6	76.6

\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

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]

\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.

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.
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 .
Determine whether Winston's model or Piotre's model is a better fit for the data.

OCR MEI Further Numerical Methods 2020 November Q4

4 marks

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]
A B C
1 $n$ $T _ { n }$ $S _ { 2 n }$
2 1 2.130135
3 2 2.149378
4 4 2.134751 2.129862
5 8 2.131084
\captionsetup{labelformat=empty} \caption{Fig. 4}
\end{table}
Write down an efficient formula for cell C 4.
Find the value of $S _ { 4 }$, giving your answer correct to 6 decimal places.
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]
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

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]

$x$	1.45	1.55
$g ( x )$	1.49468	1.49949

\captionsetup{labelformat=empty} \caption{Fig. 5.1}

\end{table}

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 $\beta \approx 4.4$.
Obtain the iterative formula $x _ { n + 1 } = g \left( x _ { n } \right) = \frac { \sqrt [ 3 ] { x _ { n } ^ { X _ { n } } + 25 } } { 2 }$.
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 $\alpha$.
Starting with $x _ { 0 } = 1.5$, use the iterative formula to find $x _ { 1 } , x _ { 2 } , x _ { 3 } , x _ { 4 } , x _ { 5 }$, and $x _ { 6 }$.
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]
$n$ $x _ { n }$
0 4.5
1 4.81826433
2 6.27473453
3 23.2937196
4 $2.0654 \mathrm { E } + 10$
5 \#NUM!
\captionsetup{labelformat=empty} \caption{Fig. 5.2}
\end{table}
Explain why \#NUM! is displayed in the cell for $x _ { 5 }$.
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 $x _ { 2 }$.
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

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 $b$.

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]

	A	B	C	D	E	F
1	a	f(a)	b	f(b)	$x _ { \text {new } }$	$\mathrm { f } \left( x _ { \text {new } } \right)$
2	0	0.5	0.5	-1.51831	0.25	-1.133
3	0	0.5	0.25	-1.133	0.125	-0.42
4	0	0.5	0.125	-0.42001	0.0625	0.01873
5	0.0625	0.01873	0.125	-0.42001	0.09375	-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.

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)
Write a suitable formula for cell C3.

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]

	A	B	C	D
1	$x$	f(x)	difference	ratio
2	0.4	-1.5799
3	0.6	-1.0504
4	0.99671	8.4245
5	0.64398	-0.6809	-0.35273
6	0.67036	-0.4026	0.026378	-0.0748
7	0.70852	0.08386	0.038164	1.44682
8	0.70194	-0.0075	-0.00658	-0.1724
9	0.70248	-0.0001	0.00054	-0.082
10	0.70249	$1.8 \mathrm { E } - 07$	$8.88 \mathrm { E } - 06$	0.01646

\captionsetup{labelformat=empty} \caption{Fig. 6.3}

\end{table} The formula in cell A4 is $$= ( \mathrm { A } 2 * \mathrm {~B} 3 - \mathrm { A } 3 * \mathrm {~B} 2 ) / ( \mathrm { B } 3 - \mathrm { B } 2 )$$

State the method being used.
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.
Determine whether Liren is correct.

OCR MEI Further Numerical Methods 2020 November Q7

7 Fig. 7.1 shows two values of $x$ and the associated values of $\mathrm { f } ( x )$. \begin{table}[h]

$x$	3	3.5
$\mathrm { f } ( x )$	6.082763	4.596194

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

\end{table}

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]
$x$ 3 3.00001 3.0001 3.001 3.01 3.1
$\mathrm { f } ( x )$ 6.082763 6.08274 6.082541 6.08054 6.060454 5.848846
\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]
$h$ 0.1 0.01 0.001 0.0001 0.00001
estimate -2.339165 -2.230883 -2.220532 -2.219501 -2.219398
difference 0.1082815 0.010352 0.0010307 0.000103
ratio 0.095602 0.099567 0.0999568
\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.
Explain whether or not Tommy is correct.
Use extrapolation to determine the value of the gradient of $\mathrm { f } ( x )$ at $x = 3$ as accurately as possible, justifying the precision quoted.
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

1. Determine the relative error when
  - 1.414214 is used to approximate $\sqrt { 2 }$,
$1.414214 ^ { 2 }$ is used to approximate 2.
(ii) Write down the relationship between your answers to part (a)(i).
Fig. 1 shows some spreadsheet output.

	A	B	C
1	2	1.414214	2

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

OCR MEI Further Numerical Methods 2021 November Q2

2 The table shows some values of $x$ and the associated values of $\mathrm { f } ( x )$.

$x$	1	2	3	4	5
$\mathrm { f } ( x )$	- 0.65	- 0.35	1.77	5.71	11.47

Complete the difference table in the Printed Answer Booklet.
Explain why the data may be interpolated by a polynomial of degree 2.
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

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.

	C	D	E	F	G	H	I	J
4	a	f(a)	b	$\mathrm { f } ( b )$	$x _ { \text {new } }$	$\mathrm { f } \left( x _ { \text {new } } \right)$	difference	ratio
5	1	-1.8248	2	17.2899	1.09547	-1.80507
6	1.09547	-1.80507	2	17.2899	1.18097	-1.75418	0.08551
7	1.18097	-1.75418	2	17.2899	1.25641	-1.66246	0.07544	0.88229
8	1.25641	-1.66246	2	17.2899	1.32164	-1.52781	0.06523	0.86458
9	1.32164	-1.52781	2	17.2899	1.37672	-1.35706	0.05508	0.84439
10	1.37672	-1.35706	2	17.2899	1.42208	-1.1642	0.04536	0.8236
11	1.42208	-1.1642	2	17.2899	1.45853	-0.96616	0.03646	0.80376
12	1.45853	-0.96616	2	17.2899	1.48719	-0.77825	0.02866	0.78598
13	1.48719	-0.77825	2	17.2899
14

The formula in cell D5 is $\quad = \mathrm { SINH } \left( \mathrm { C5 } ^ { \wedge } 2 \right) - \mathrm { C5 } ^ { \wedge } 3 - 2$.

Write down the equation which is being solved. The formula in cell C 6 is $\quad = \mathrm { IF } ( \mathrm { H } 5 < 0 , \mathrm { G } 5 , \mathrm { C } 5 )$.
Write down a similar formula for cell E6.
Calculate the values which would be displayed in cells G13 and G14 to find further approximations to the root.
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

4 The table shows some values of $x$ and the associated values of $\mathrm { f } ( x )$.

$x$	4	4.0001	4.001	4.01	4.1
$\mathrm { f } ( x )$	4	4.0002386	4.0023871	4.0239468	4.2472072

Calculate four estimates of the derivative of $\mathrm { f } ( x )$ at $x = 4$.
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

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.

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 $\pounds 90.23$.
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.
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 $\pounds 103.24$.
Explain how such a large error could arise. The next week Kareem buys $n$ items.
Write down a formula for the maximum possible error when Kareem uses his model to estimate the total cost of his shopping.
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

6 The equation $0.5 \ln x - x ^ { 2 } + x + 1 = 0$ has two roots $\alpha$ and $\beta$, such that $0 < \alpha < 1$ and $1 < \beta < 2$.

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}
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.
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]
$r$ $\mathrm { x } _ { \mathrm { r } }$
0 0.5
1 - 0.40343
2 \#NUM!
\captionsetup{labelformat=empty} \caption{Fig. 6.2}
\end{table}
Explain how the display \#NUM! has arisen in the cell for $x _ { 2 }$. Beth decides to use the iterative formula $$x _ { n + 1 } = g \left( x _ { n } \right) = \sqrt { 0.5 \ln \left( x _ { n } \right) + x _ { n } + 1 }$$
Determine the outcome when Beth uses this formula with $x _ { 0 } = 0.5$.
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

6 marks

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]

$n$	$T _ { n }$	difference	ratio
1	0.43634681
2	0.5580694	0.121723
4	0.60199843	0.043929	0.36089
8	0.61787073	0.015872	0.36132
16	0.62357601	0.005705	0.35945
32	0.62561716	0.002041	0.35777

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

\end{table}

Write down the value of $h$ used to find the approximation 0.62357601 .
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.

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]

	M	N	O	P	Q	R
1	$n$	$T _ { n }$	$M _ { n }$	$S _ { 2 n }$	difference	ratio
2	1	0.43634681	0.679792	0.5986436
3	2	0.5580694	0.64592745	0.61664144	0.018
4	4	0.60199843	0.63374304	0.6231615	0.00652	0.362269
5	8	0.61787073	0.62928129	0.62547777	0.00232	0.355253
6	16	0.62357601	0.62765831	0.62629755	0.00082	0.35392
7	32	0.62561716	0.62707259

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

\end{table}

Write down an efficient spreadsheet formula for calculating $S _ { 16 }$.
Determine the missing values in row 7.
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

Solve the following simultaneous equations. $$\begin{aligned} & x + \quad y = 1
& x + 0.99 y = 2 \end{aligned}$$
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.

OCR MEI Further Numerical Methods Specimen Q2

2 The following spreadsheet printout shows the bisection method being applied to the equation $\mathrm { f } ( x ) = 0$, where $\mathrm { f } ( x ) = \mathrm { e } ^ { x } - x ^ { 2 } - 2$. Some values of $\mathrm { f } ( x )$ are shown in columns B and D.

	A	B	C	D	E	F	G
1	a	$\mathrm { f } ( a )$	b	f(b)	$( a + b ) / 2$	$\mathrm { f } ( ( a + b ) / 2 )$	mpe
2	1	-0.28172	2	1.389056	1.5	0.231689	0.5
3	1	-0.28172	1.5	0.231689	1.25	-0.072157	0.25
4	1.25	-0.07216	1.5	0.231689	1.375	0.064452	0.125
5	1.25	-0.07216	1.375	0.064452	1.3125	-0.007206	0.0625
6	1.3125	-0.00721	1.375	0.064452	1.34375	0.027728	0.03125

The formula in cell A 3 is $= \mathrm { IF } ( \mathrm { F } 2 > 0$, A2, E2). State the purpose of this formula.
The formula in cell C 3 is $= \mathrm { IF } ( \mathrm { F } 2 > 0 , \ldots , \ldots )$. What are the missing cell references?
In which row is the magnitude of the maximum possible error (mpe) less than $5 \times 10 ^ { - 7 }$ for the first time?

OCR MEI Further Numerical Methods Specimen Q3

3 The equation $\sinh x + x ^ { 2 } - 1 = 0$ has a root, $\alpha$, such that $0 < \alpha < 1$.

Verify that the iteration $x _ { r + 1 } = \frac { 1 - \sinh x _ { r } } { x _ { r } }$ with $x _ { 0 } = 1$ fails to converge to this root.
Use the relaxed iteration $x _ { r + 1 } = ( 1 - \lambda ) x _ { r } + \lambda \left( \frac { 1 - \sinh x _ { r } } { x _ { r } } \right)$ with $\lambda = \frac { 1 } { 4 }$ and $x _ { 0 } = 1$ to find $\alpha$ correct to 6 decimal places.

OCR MEI Further Numerical Methods Specimen Q4

4 The table below gives values of a function $y = \mathrm { f } ( x )$.

$x$	0.2	0.3	0.35	0.4	0.45	0.5	0.6
$\mathrm { f } ( x )$	0.789922	0.754628	0.749199	0.749997	0.756257	0.767523	0.804299

Calculate three estimates of $\frac { \mathrm { d } y } { \mathrm {~d} x }$ at $x = 0.4$ using the central difference method.
State the value of $\frac { \mathrm { d } y } { \mathrm {~d} x }$ at $x = 0.4$ to an appropriate degree of accuracy. Justify your answer.

OCR MEI Further Numerical Methods Specimen Q5

5 A vehicle is moving in a straight line. Its velocity at different times is recorded and shown below. The velocities are recorded to 5 significant figures and the times may be assumed to be exact.

Time $( t$ seconds $)$	5	10	12	15
Velocity $( v$ metres per second $)$	5.1250	11.000	14.000	18.375

Given that the vehicle is modelled as a particle with constant mass, what assumption about the net force acting on the vehicle leads to a quadratic model?
Find Newton's interpolating polynomial of degree 2 to model this situation. Write your answer in the form $v = a t ^ { 2 } + b t + c$.
Comment on whether this model appears to be appropriate.
Use this model to find an approximation to the distance travelled over the interval $5 \leq t \leq 15$. Further investigation suggests that a cubic model may be more appropriate.
What technique would you use to fit a cubic model to the data in the table?

OCR MEI Further Numerical Methods Specimen Q6

6 The secant method is to be used to solve the equation $x - \ln ( \cos x ) - 1 = 0$.

Show that starting with $x _ { 0 } = - 1$ and $x _ { 1 } = 0$ leads to the method failing to find the root between $x = 0$ and $x = 1$. The spreadsheet printout shows the application of the secant method starting with $x _ { 0 } = 0$ and $x _ { 1 } = 1$. Successive approximations to the root are in column E.

	A	B	C	D	E
1	$x _ { n }$	$\mathrm { f } \left( x _ { n } \right)$	$x _ { n + 1 }$	$\mathrm { f } \left( x _ { n + 1 } \right)$	$x _ { n + 2 }$
2	0	-1	1	0.6156265	0.6189549
3	1	0.6156265	0.6189549	-0.175846	0.7036139
4	0.6189549	-0.1758461	0.7036139	-0.025245	0.7178053
5	0.7036139	-0.0252451	0.7178053	0.0011619	0.7171808
6	0.7178053	0.0011619	0.7171808	- 7.4 E -06	0.7171848
7	0.7171808	-7.402E-06	0.7171848	-2.16E-09	0.7171848
8	0.7171848	-2.16E-09	0.7171848	3.997 E -15	0.7171848

What feature of column B shows that this application of the secant method has been successful?

Write down a suitable spreadsheet formula to obtain the value in cell E2. Some analysis of convergence is carried out, and the following spreadsheet output is obtained.

	A	B	C	D	E	F	G	H
1	$x _ { n }$	$\mathrm { f } \left( x _ { n } \right)$	$x _ { n + 1 }$	$\mathrm { f } \left( x _ { n + 1 } \right)$	$x _ { n + 2 }$
2	0	-1	1	0.6156265	0.6189549	0.084659	0.167629	1.980053
3	1	0.6156265	0.6189549	-0.175846	0.7036139	0.0141913	-0.044	-3.10054
4	0.6189549	-0.1758461	0.7036139	-0.025245	0.7178053	-0.0006244	-0.00633	10.13727
5	0.7036139	-0.0252451	0.7178053	0.0011619	0.7171808	3.953 E -06	0.000292	73.83899
6	0.7178053	0.0011619	0.7171808	- 7.4 E -06	0.7171848	1.154 E -09	-1.8E-06
7	0.7171808	-7.402E-06	0.7171848	- 2.16 E -09	0.7171848	- 2.109 E -15
8	0.7171848	- 2.16 E -09	0.7171848	3.997 E -15	0.7171848

The formula in cell F2 is =E3-E2. The formula in cell G2 is =F3/F2. The formula in cell H2 is =F3/(F2\^{}2).

(A) Explain the purpose of each of these three formulae.
(B) Explain the significance of the values in columns G and H in terms of the rate of convergence of the secant method.
Explain why the values in cells F6 and F7 are not 0 . [Question 7 is printed overleaf.]

OCR MEI Further Numerical Methods Specimen Q7

7 Fig. 7 shows the graph of $y = \mathrm { f } ( x )$ for values of $x$ from 0 to 1 . \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{662c2d48-228a-4b94-a4b4-cdd31634ef21-6_693_673_390_696} \captionsetup{labelformat=empty} \caption{Fig. 7}

\end{figure} The following spreadsheet printout shows estimates of $\int _ { 0 } ^ { 1 } f ( x ) d x$ found using the midpoint and trapezium rules for different values of $h$, the strip width.

A	A	B	C
1	$h$	Midpoint	Trapezium
2	1	1.99851742	1.751283839
3	0.5	1.9638591	1.874900631
4	0.25	1.95135259	1.919379864
5	0.125	1.94682102	1.935366229

Without doing any further calculation, write down the smallest possible interval which contains the value of the integral. Justify your answer.
(A) - Calculate the ratio of differences, $r$, for the sequence of estimates calculated using the trapezium rule.
- Hence suggest a value for $r$ correct to 2 significant figures.
- Comment on your suggested value for $r$.
  (B) - Use extrapolation to find an improved approximation to the value of the integral.
- State the value of the integral to two decimal places.
- Explain why this precision is secure.
Using a similar approach with the sequence of estimates calculated using the midpoint rule, the approximation to the integral from extrapolation was found to be 1.94427 correct to 5 decimal places.
Andrea uses the extrapolated midpoint rule value and the value found in part (ii) ( $B$ ) to write down an interval which contains the value of the integral. Comment on the validity of Andrea's method.

Use the values from the spreadsheet output to calculate an approximation to the integral using Simpson's rule with $h = 0.125$. Give your answer to 5 decimal places. Approximations to the integral using Simpson's rule are given in the spreadsheet output below. The number of applications of Simpson's rule is given in column N.

N	O	P	Q
$n$	Simpson	differences	ratio
1	1.91610623	0.01810005	0.3584931
2	1.93420628	0.00648874	0.3556525
4	1.94069502	0.00230774	0.3544828
8	1.94300275	0.00081805	0.3539885
16	1.94382081	0.00028958	0.3537638
32	1.94411039	0.00010244	0.3536568
64	1.94421283	$3.623 \mathrm { E } - 05$
128	1.94424906

Use the spreadsheet output to find the value of the integral as accurately as possible. Justify the precision quoted. \section*{END OF QUESTION PAPER} OCR is committed to seeking permission to reproduce all third-party content that it uses in the assessment materials. OCR has attempted to identify and contact all copyright holders whose work is used in this paper. To avoid the issue of disclosure of answer-related information to candidates, all copyright acknowledgements are reproduced in the OCR Copyright Acknowledgements booklet. This is produced for each series of examinations and is freely available to download from our public website (\href{http://www.ocr.org.uk}{www.ocr.org.uk}) after the live examination series. If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correct its mistake at the earliest possible opportunity. For queries or further information please contact the Copyright Team, First Floor, 9 Hills Road, Cambridge CB2 1GE.
OCR is part of the Cambridge Assessment Group; Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge. }\section*{}

\(n\)	\(\mathrm { M } _ { n }\)	\(\mathrm {~T} _ { n }\)	\(\mathrm {~S} _ { 2 n }\)
1	0.612547	1
2	0.639735

1	A	B	C	D	E	F
1	a	f(a)	\(b\)	f(b)	c	\(\mathrm { f } ( c )\)
2	2	-0.6109	3	6.08554	2.5	1.43249
3	2	-0.6109	2.5	1.43249	2.25	0.17524
4	2	-0.6109	2.25	0.17524	2.125	-0.2677
5	2.125	-0.2677	2.25	0.17524	2.1875	-0.0598
6

\(h\)	0.8	0.4	0.2	0.1	0.05	0.025	0.0125	0.00625
approximation	0.130452	0.138647	0.143381	0.145942	0.147277	0.147959	0.148304	0.148477
difference		0.008195	0.004734	0.002561	0.001335	0.000682	0.000345	0.000173
ratio			0.577633	0.541099	0.521186	0.510762	0.505424	0.502723

n	\(\mathrm { S } _ { 2 n }\)	difference	ratio
1
2
4	0.674353	-0.0209
8	0.665199	-0.00915	0.438059
16	0.661297	-0.0039	0.426286
32	0.659675	-0.00162	0.415762
64	0.659015	-0.00066	0.406785

	A	B	C
1	\(n\)	\(T _ { n }\)	\(S _ { 2 n }\)
2	1		2.130135
3	2	2.149378
4	4	2.134751	2.129862
5	8	2.131084

\(n\)	\(x _ { n }\)
0	4.5
1	4.81826433
2	6.27473453
3	23.2937196
4	\(2.0654 \mathrm { E } + 10\)
5	\#NUM!

	A	B	C	D	E	F
1	a	f(a)	b	f(b)	\(x _ { \text {new } }\)	\(\mathrm { f } \left( x _ { \text {new } } \right)\)
2	0	0.5	0.5	-1.51831	0.25	-1.133
3	0	0.5	0.25	-1.133	0.125	-0.42
4	0	0.5	0.125	-0.42001	0.0625	0.01873
5	0.0625	0.01873	0.125	-0.42001	0.09375	-0.2065

	A	B	C	D
1	\(x\)	f(x)	difference	ratio
2	0.4	-1.5799
3	0.6	-1.0504
4	0.99671	8.4245
5	0.64398	-0.6809	-0.35273
6	0.67036	-0.4026	0.026378	-0.0748
7	0.70852	0.08386	0.038164	1.44682
8	0.70194	-0.0075	-0.00658	-0.1724
9	0.70248	-0.0001	0.00054	-0.082
10	0.70249	\(1.8 \mathrm { E } - 07\)	\(8.88 \mathrm { E } - 06\)	0.01646

\(x\)	3	3.00001	3.0001	3.001	3.01	3.1
\(\mathrm { f } ( x )\)	6.082763	6.08274	6.082541	6.08054	6.060454	5.848846

\(h\)	0.1	0.01	0.001	0.0001	0.00001
estimate	-2.339165	-2.230883	-2.220532	-2.219501	-2.219398
difference	0.1082815	0.010352	0.0010307	0.000103
ratio	0.095602	0.099567	0.0999568

	C	D	E	F	G	H	I	J
4	a	f(a)	b	\(\mathrm { f } ( b )\)	\(x _ { \text {new } }\)	\(\mathrm { f } \left( x _ { \text {new } } \right)\)	difference	ratio
5	1	-1.8248	2	17.2899	1.09547	-1.80507
6	1.09547	-1.80507	2	17.2899	1.18097	-1.75418	0.08551
7	1.18097	-1.75418	2	17.2899	1.25641	-1.66246	0.07544	0.88229
8	1.25641	-1.66246	2	17.2899	1.32164	-1.52781	0.06523	0.86458
9	1.32164	-1.52781	2	17.2899	1.37672	-1.35706	0.05508	0.84439
10	1.37672	-1.35706	2	17.2899	1.42208	-1.1642	0.04536	0.8236
11	1.42208	-1.1642	2	17.2899	1.45853	-0.96616	0.03646	0.80376
12	1.45853	-0.96616	2	17.2899	1.48719	-0.77825	0.02866	0.78598
13	1.48719	-0.77825	2	17.2899
14

\(r\)	\(\mathrm { x } _ { \mathrm { r } }\)
0	0.5
1	- 0.40343
2	\#NUM!

	M	N	O	P	Q	R
1	\(n\)	\(T _ { n }\)	\(M _ { n }\)	\(S _ { 2 n }\)	difference	ratio
2	1	0.43634681	0.679792	0.5986436
3	2	0.5580694	0.64592745	0.61664144	0.018
4	4	0.60199843	0.63374304	0.6231615	0.00652	0.362269
5	8	0.61787073	0.62928129	0.62547777	0.00232	0.355253
6	16	0.62357601	0.62765831	0.62629755	0.00082	0.35392
7	32	0.62561716	0.62707259

	A	B	C	D	E	F	G
1	a	\(\mathrm { f } ( a )\)	b	f(b)	\(( a + b ) / 2\)	\(\mathrm { f } ( ( a + b ) / 2 )\)	mpe
2	1	-0.28172	2	1.389056	1.5	0.231689	0.5
3	1	-0.28172	1.5	0.231689	1.25	-0.072157	0.25
4	1.25	-0.07216	1.5	0.231689	1.375	0.064452	0.125
5	1.25	-0.07216	1.375	0.064452	1.3125	-0.007206	0.0625
6	1.3125	-0.00721	1.375	0.064452	1.34375	0.027728	0.03125

Time \(( t\) seconds \()\)	5	10	12	15
Velocity \(( v\) metres per second \()\)	5.1250	11.000	14.000	18.375

	A	B	C	D	E
1	\(x _ { n }\)	\(\mathrm { f } \left( x _ { n } \right)\)	\(x _ { n + 1 }\)	\(\mathrm { f } \left( x _ { n + 1 } \right)\)	\(x _ { n + 2 }\)
2	0	-1	1	0.6156265	0.6189549
3	1	0.6156265	0.6189549	-0.175846	0.7036139
4	0.6189549	-0.1758461	0.7036139	-0.025245	0.7178053
5	0.7036139	-0.0252451	0.7178053	0.0011619	0.7171808
6	0.7178053	0.0011619	0.7171808	- 7.4 E -06	0.7171848
7	0.7171808	-7.402E-06	0.7171848	-2.16E-09	0.7171848
8	0.7171848	-2.16E-09	0.7171848	3.997 E -15	0.7171848

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