Question 6 - A-Level Maths

OCR MEI Further Numerical Methods Specimen — Question 6 12 marks

Exam Board	OCR MEI
Module	Further Numerical Methods (Further Numerical Methods)
Session	Specimen
Marks	12
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Sign Change & Interval Methods
Type	Secant Method or False Position
Difficulty	Standard +0.8 This is a Further Maths numerical methods question requiring understanding of when the secant method fails, interpretation of spreadsheet output showing convergence behavior, and formula construction. Part (i) requires showing method failure (non-trivial analysis), while parts (ii)-(iii) are more routine. The multi-part nature and Further Maths context place it moderately above average difficulty.
Spec	1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams

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.]

Show mark scheme Show mark scheme source

Question 6:

Answer	Marks	Guidance
6	(i)	(cid:16)1(cid:117)(cid:16)1(cid:16)0(cid:117)(cid:16)1.38347

(cid:68)(cid:32)

(cid:16)1(cid:16)(cid:16)1.38347

(cid:32)2.601636

cos2.601636 = − 0.8577… and ln(− 0.8577…) is

Answer	Marks
undefined so the method breaks down	M1

A1

Answer	Marks
[3]	1.1

1.1

2.4

Answer	Marks	Guidance
6	(ii)	The values are approaching zero
[1]	2.3	N
6	(iii)	=(A2D2-C2B2)/(D2-B2) oe
[2]	3.1a
1.1	E
B1 if * omitted	must show application of secant

method

Answer	Marks	Guidance
6	(iv)	(A)

estimates of the root

= F2/F1 finds the ratio of differences of these estimates

= F2/F1^2 finds the ratio of differences of each

Answer	Marks
estimate to the square of the previous estimate	B1

C

B1

Answer	Marks
[3]	1.2

I

2.5

Answer	Marks	Guidance
2.4	M
6	(iv)	(B)

P

The values in column G suggest that convergence [of

secant method] is faster than 1st order

S

The values in column H suggest that the convergence

Answer	Marks
[of secant method] is slower than 2nd order	B1

B1

Answer	Marks
[2]	2.2b
2.2b	If B0B0, SC1 for “neither 1st

order nor 2nd order convergence”

Answer	Marks	Guidance
6	(v)	The values in E6, E7 and E8 are stored to greater
accuracy than is displayed oe	B1
[1]	2.4

Question 6:
6 | (i) | (cid:16)1(cid:117)(cid:16)1(cid:16)0(cid:117)(cid:16)1.38347
(cid:68)(cid:32)
(cid:16)1(cid:16)(cid:16)1.38347
(cid:32)2.601636
cos2.601636 = − 0.8577… and ln(− 0.8577…) is
undefined so the method breaks down | M1
A1
A1
[3] | 1.1
1.1
2.4
6 | (ii) | The values are approaching zero | B1
[1] | 2.3 | N | “getting smaller” is insufficient
6 | (iii) | =(A2*D2-C2*B2)/(D2-B2) oe | B2
[2] | 3.1a
1.1 | E
B1 if * omitted | must show application of secant
method
6 | (iv) | (A) | = E2 – E1 finds the differences between successive
estimates of the root
= F2/F1 finds the ratio of differences of these estimates
= F2/F1^2 finds the ratio of differences of each
estimate to the square of the previous estimate | B1
C
B1
B1
[3] | 1.2
I
2.5
2.4 | M
6 | (iv) | (B) | E
P
The values in column G suggest that convergence [of
secant method] is faster than 1st order
S
The values in column H suggest that the convergence
[of secant method] is slower than 2nd order | B1
B1
[2] | 2.2b
2.2b | If B0B0, SC1 for “neither 1st
order nor 2nd order convergence”
6 | (v) | The values in E6, E7 and E8 are stored to greater
accuracy than is displayed oe | B1
[1] | 2.4

Show LaTeX source

6 The secant method is to be used to solve the equation $x - \ln ( \cos x ) - 1 = 0$.
\begin{enumerate}[label=(\roman*)]
\item 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.

\begin{center}
\begin{tabular}{|l|l|l|l|l|l|}
\hline
 & A & B & C & D & E \\
\hline
1 & $x _ { n }$ & $\mathrm { f } \left( x _ { n } \right)$ & $x _ { n + 1 }$ & $\mathrm { f } \left( x _ { n + 1 } \right)$ & $x _ { n + 2 }$ \\
\hline
2 & 0 & -1 & 1 & 0.6156265 & 0.6189549 \\
\hline
3 & 1 & 0.6156265 & 0.6189549 & -0.175846 & 0.7036139 \\
\hline
4 & 0.6189549 & -0.1758461 & 0.7036139 & -0.025245 & 0.7178053 \\
\hline
5 & 0.7036139 & -0.0252451 & 0.7178053 & 0.0011619 & 0.7171808 \\
\hline
6 & 0.7178053 & 0.0011619 & 0.7171808 & - 7.4 E -06 & 0.7171848 \\
\hline
7 & 0.7171808 & -7.402E-06 & 0.7171848 & -2.16E-09 & 0.7171848 \\
\hline
8 & 0.7171848 & -2.16E-09 & 0.7171848 & 3.997 E -15 & 0.7171848 \\
\hline
\end{tabular}
\end{center}
\item What feature of column B shows that this application of the secant method has been successful?
\item 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.

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

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).
\item (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.
\item Explain why the values in cells F6 and F7 are not 0 .

[Question 7 is printed overleaf.]
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Numerical Methods  Q6 [12]}}

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