1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams

3 A curve is defined for \(0 \leqslant x \leqslant \frac { \pi } { 4 }\) by the equation \(y = x \cos 2 x\), and is sketched below. \includegraphics[max width=\textwidth, alt={}, center]{6ce5aa0d-0a73-4bc4-aabc-314c0434e4f5-3_757_878_402_559}

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. The point \(A\), where \(x = \alpha\), on the curve is a stationary point.
   1. Show that \(1 - 2 \alpha \tan 2 \alpha = 0\).
   2. Show that \(0.4 < \alpha < 0.5\).
   3. Show that the equation \(1 - 2 x \tan 2 x = 0\) can be rearranged to become \(x = \frac { 1 } { 2 } \tan ^ { - 1 } \left( \frac { 1 } { 2 x } \right)\).
   4. Use the iteration \(x _ { n + 1 } = \frac { 1 } { 2 } \tan ^ { - 1 } \left( \frac { 1 } { 2 x _ { n } } \right)\) with \(x _ { 1 } = 0.4\) to find \(x _ { 3 }\), giving your answer to two significant figures.
3. Use integration by parts to find \(\int _ { 0 } ^ { 0.5 } x \cos 2 x \mathrm {~d} x\), giving your answer to three significant figures.

1

1. The curve with equation $$y = \frac { \cos x } { 2 x + 1 } , \quad x > - \frac { 1 } { 2 }$$ intersects the line \(y = \frac { 1 } { 2 }\) at the point where \(x = \alpha\).
   1. Show that \(\alpha\) lies between 0 and \(\frac { \pi } { 2 }\).
   2. Show that the equation \(\frac { \cos x } { 2 x + 1 } = \frac { 1 } { 2 }\) can be rearranged into the form $$x = \cos x - \frac { 1 } { 2 }$$
   3. Use the iteration \(x _ { n + 1 } = \cos x _ { n } - \frac { 1 } { 2 }\) with \(x _ { 1 } = 0\) to find \(x _ { 3 }\), giving your answer to three decimal places.
2. 1. Given that \(y = \frac { \cos x } { 2 x + 1 }\), use the quotient rule to find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
   2. Hence find the gradient of the normal to the curve \(y = \frac { \cos x } { 2 x + 1 }\) at the point on the curve where \(x = 0\).

1 The curve \(y = 3 ^ { x }\) intersects the curve \(y = 10 - x ^ { 3 }\) at the point where \(x = \alpha\).

1. Show that \(\alpha\) lies between 1 and 2 .
2. 1. Show that the equation \(3 ^ { x } = 10 - x ^ { 3 }\) can be rearranged into the form $$x = \sqrt [ 3 ] { 10 - 3 ^ { x } }$$
   2. Use the iteration \(x _ { n + 1 } = \sqrt [ 3 ] { 10 - 3 ^ { x _ { n } } }\) with \(x _ { 1 } = 1\) to find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.

2 For \(0 < x \leqslant 2\), the curves with equations \(y = 4 \ln x\) and \(y = \sqrt { x }\) intersect at a single point where \(x = \alpha\).

1. Show that \(\alpha\) lies between 0.5 and 1.5.
2. Show that the equation \(4 \ln x = \sqrt { x }\) can be rearranged into the form $$x = \mathrm { e } ^ { \left( \frac { \sqrt { x } } { 4 } \right) }$$
3. Use the iterative formula $$x _ { n + 1 } = \mathrm { e } ^ { \left( \frac { \sqrt { x _ { n } } } { 4 } \right) }$$ with \(x _ { 1 } = 0.5\) to find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.
4. Figure 1, on the page 3, shows a sketch of parts of the graphs of \(y = \mathrm { e } ^ { \left( \frac { \sqrt { x } } { 4 } \right) }\) and \(y = x\), and the position of \(x _ { 1 }\). On Figure 1, draw a cobweb or staircase diagram to show how convergence takes place, indicating the positions of \(x _ { 2 }\) and \(x _ { 3 }\) on the \(x\)-axis. \begin{figure}[h]
   \captionsetup{labelformat=empty} \caption{Figure 1} \includegraphics[alt={},max width=\textwidth]{d3c66c34-b09c-4223-8383-cf0a68419bf9-3_1285_1543_356_296}
   \end{figure}

3

1. The equation \(\mathrm { e } ^ { - x } - 2 + \sqrt { x } = 0\) has a single root, \(\alpha\).
   Show that \(\alpha\) lies between 3 and 4 .
2. Use the recurrence relation \(x _ { n + 1 } = \left( 2 - e ^ { - x _ { n } } \right) ^ { 2 }\), with \(x _ { 1 } = 3.5\), to find \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.
3. The diagram below shows parts of the graphs of \(y = \left( 2 - \mathrm { e } ^ { - x } \right) ^ { 2 }\) and \(y = x\), and a position of \(x _ { 1 }\). On the diagram, draw a staircase or cobweb diagram to show how convergence takes place, indicating the positions of \(x _ { 2 }\) and \(x _ { 3 }\) on the \(x\)-axis. \includegraphics[max width=\textwidth, alt={}, center]{063bbfa5-df49-44a1-8143-5e076397f63f-03_1100_1402_881_367}

2 A curve has equation \(y = 2 \ln ( 2 \mathrm { e } - x )\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. Find an equation of the normal to the curve \(y = 2 \ln ( 2 \mathrm { e } - x )\) at the point on the curve where \(x = \mathrm { e }\).
   [0pt] [4 marks]
3. The curve \(y = 2 \ln ( 2 \mathrm { e } - x )\) intersects the line \(y = x\) at a single point, where \(x = \alpha\).
   1. Show that \(\alpha\) lies between 1 and 3 .
   2. Use the recurrence relation $$x _ { n + 1 } = 2 \ln \left( 2 \mathrm { e } - x _ { n } \right)$$ with \(x _ { 1 } = 1\) to find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.
   3. Figure 1, on the opposite page, shows a sketch of parts of the graphs of \(y = 2 \ln ( 2 \mathrm { e } - x )\) and \(y = x\), and the position of \(x _ { 1 }\). On Figure 1, draw a cobweb or staircase diagram to show how convergence takes place, indicating the positions of \(x _ { 2 }\) and \(x _ { 3 }\) on the \(x\)-axis.
      [0pt] [2 marks] \section*{(c)(iii)} \begin{figure}[h]
      \captionsetup{labelformat=empty} \caption{Figure 1} \includegraphics[alt={},max width=\textwidth]{57412ec0-ad97-4418-8ba8-93f1f7d8aac1-05_864_1284_1802_386}
      \end{figure}

2 The curve with equation \(y = x ^ { x }\), where \(x > 0\), intersects the line \(y = 5\) at a single point, where \(x = \alpha\).

1. Show that \(\alpha\) lies between 2 and 3 .
2. Show that the equation \(x ^ { x } = 5\) can be rearranged into the form $$x = \mathrm { e } ^ { \left( \frac { \ln 5 } { x } \right) }$$
3. Use the iterative formula $$x _ { n + 1 } = \mathrm { e } ^ { \left( \frac { \ln 5 } { x _ { n } } \right) }$$ with \(x _ { 1 } = 2\) to find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.
4. 1. Use Simpson's rule with 7 ordinates ( 6 strips) to find an approximation to $$\int _ { 0.5 } ^ { 1.7 } \left( 5 - x ^ { x } \right) \mathrm { d } x$$ giving your answer to three significant figures.
   2. Hence find an approximation to \(\int _ { 0.5 } ^ { 1.7 } x ^ { x } \mathrm {~d} x\).

\(f ( x ) = \frac { x ^ { 4 } + x ^ { 3 } - 5 x ^ { 2 } - 9 } { x ^ { 2 } + x - 6 }\).

1. Using algebraic division, show that $$f ( x ) = x ^ { 2 } + A + \frac { B } { x + C }$$ where \(A , B\) and \(C\) are integers to be found.
2. By sketching two suitable graphs on the same set of axes, show that the equation \(\mathrm { f } ( x ) = 0\) has exactly one real root.
3. Use the iterative formula $$x _ { n + 1 } = 2 + \frac { 1 } { x _ { n } ^ { 2 } + 1 } ,$$ with a suitable starting value to find the root of the equation \(\mathrm { f } ( x ) = 0\) correct to 3 significant figures and justify the accuracy of your answer.

7.

1. Solve the equation $$\pi - 3 \arccos \theta = 0$$
2. Sketch on the same diagram the curves \(y = \arccos ( x - 1 ) , 0 \leq x \leq 2\) and \(y = \sqrt { x + 2 } , x \geq - 2\). Given that \(\alpha\) is the root of the equation $$\arccos ( x - 1 ) = \sqrt { x + 2 }$$
3. show that \(0 < \alpha < 1\),
4. use the iterative formula $$x _ { n + 1 } = 1 + \cos \sqrt { x _ { n } + 2 }$$ with \(x _ { 0 } = 1\) to find \(\alpha\) correct to 3 decimal places. END

8. The curve \(C\) has the equation \(y = \sqrt { x } + \mathrm { e } ^ { 1 - 4 x } , x \geq 0\).

1. Find an equation for the normal to the curve at the point \(\left( \frac { 1 } { 4 } , \frac { 3 } { 2 } \right)\). The curve \(C\) has a stationary point with \(x\)-coordinate \(\alpha\) where \(0.5 < \alpha < 1\).
2. Show that \(\alpha\) is a solution of the equation $$x = \frac { 1 } { 4 } [ 1 + \ln ( 8 \sqrt { x } ) ]$$
3. Use the iteration formula $$x _ { n + 1 } = \frac { 1 } { 4 } \left[ 1 + \ln \left( 8 \sqrt { x _ { n } } \right) \right]$$ with \(x _ { 0 } = 1\) to find \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving the value of \(x _ { 4 }\) to 3 decimal places.
4. Show that your value for \(x _ { 4 }\) is the value of \(\alpha\) correct to 3 decimal places.
5. Another attempt to find \(\alpha\) is made using the iteration formula $$x _ { n + 1 } = \frac { 1 } { 64 } \mathrm { e } ^ { 8 x _ { n } - 2 }$$ with \(x _ { 0 } = 1\). Describe the outcome of this attempt.

8. \begin{figure}[h] \end{figure} Figure 1 shows the curve with equation \(y = 2 x - 3 \ln ( 2 x + 5 )\) and the normal to the curve at the point \(P ( - 2 , - 4 )\).

1. Find an equation for the normal to the curve at \(P\). The normal to the curve at \(P\) intersects the curve again at the point \(Q\) with \(x\)-coordinate \(q\).
2. Show that \(1 < q < 2\).
3. Show that \(q\) is a solution of the equation $$x = \frac { 12 } { 7 } \ln ( 2 x + 5 ) - 2 .$$
4. Use the iterative formula $$x _ { n + 1 } = \frac { 12 } { 7 } \ln \left( 2 x _ { n } + 5 \right) - 2 ,$$ with \(x _ { 0 } = 1.5\), to find the value of \(q\) to 3 significant figures and justify the accuracy of your answer.

3. $$f ( x ) = x ^ { 2 } + 5 x - 2 \sec x , \quad x \in \mathbb { R } , \quad - \frac { \pi } { 2 } < x < \frac { \pi } { 2 } .$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root in the interval [1,1.5]. A more accurate estimate of this root is to be found using iterations of the form $$x _ { n + 1 } = \arccos \mathrm { g } \left( x _ { n } \right) .$$
2. Find a suitable form for \(\mathrm { g } ( x )\) and use this formula with \(x _ { 0 } = 1.25\) to find \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\). Give the value of \(x _ { 4 }\) to 3 decimal places. The curve \(y = \mathrm { f } ( x )\) has a stationary point at \(P\).
3. Show that the \(x\)-coordinate of \(P\) is 1.0535 correct to 5 significant figures.

8. A curve has the equation \(y = x ^ { 2 } - \sqrt { 4 + \ln x }\).

1. Show that the tangent to the curve at the point where \(x = 1\) has the equation $$7 x - 4 y = 11$$ The curve has a stationary point with \(x\)-coordinate \(\alpha\).
2. Show that \(0.3 < \alpha < 0.4\)
3. Show that \(\alpha\) is a solution of the equation $$x = \frac { 1 } { 2 } ( 4 + \ln x ) ^ { - \frac { 1 } { 4 } }$$
4. Use the iteration formula $$x _ { n + 1 } = \frac { 1 } { 2 } \left( 4 + \ln x _ { n } \right) ^ { - \frac { 1 } { 4 } }$$ with \(x _ { 0 } = 0.35\), to find \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to 5 decimal places. END

1. The population in thousands, \(P\), of a town at time \(t\) years after \(1 ^ { \text {st } }\) January 1980 is modelled by the formula

$$P = 30 + 50 \mathrm { e } ^ { 0.002 t }$$ Use this model to estimate

1. the population of the town on \(1 { } ^ { \text {st } }\) January 2010,
2. the year in which the population first exceeds 84000 . The population in thousands, \(Q\), of another town is modelled by the formula $$Q = 26 + 50 \mathrm { e } ^ { 0.003 t }$$
3. Show that the value of \(t\) when \(P = Q\) is a solution of the equation $$t = 1000 \ln \left( 1 + 0.08 \mathrm { e } ^ { - 0.002 t } \right) .$$
4. Use the iteration formula $$t _ { n + 1 } = 1000 \ln \left( 1 + 0.08 \mathrm { e } ^ { - 0.002 t _ { n } } \right)$$ with \(t _ { 0 } = 50\) to find \(t _ { 1 } , t _ { 2 }\) and \(t _ { 3 }\) and hence, the year in which the populations of these two towns will be equal according to these models.
   

8. A curve has the equation \(y = \frac { \mathrm { e } ^ { 2 } } { x } + \mathrm { e } ^ { x } , \quad x \neq 0\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
   [0pt]
2. Show that the curve has a stationary point in the interval [1.3,1.4]. The point \(A\) on the curve has \(x\)-coordinate 2 .
3. Show that the tangent to the curve at \(A\) passes through the origin. The tangent to the curve at \(A\) intersects the curve again at the point \(B\).
   The \(x\)-coordinate of \(B\) is to be estimated using the iterative formula $$x _ { n + 1 } = - \frac { 2 } { 3 } \sqrt { 3 + 3 x _ { n } \mathrm { e } ^ { x _ { n } - 2 } }$$ with \(x _ { 0 } = - 1\).
4. Find \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\) to 7 significant figures and hence state the \(x\)-coordinate of \(B\) to 5 significant figures.

7. The functions \(f\) and \(g\) are defined by $$\begin{aligned} & \mathrm { f } : x \rightarrow | 2 x - 5 | , \quad x \in \mathbb { R } , \\ & \mathrm {~g} : x \rightarrow \ln ( x + 3 ) , \quad x \in \mathbb { R } , \quad x > - 3 \end{aligned}$$

1. State the range of f .
2. Evaluate fg(-2).
3. Solve the equation $$\operatorname { fg } ( x ) = 3$$ giving your answers in exact form.
4. Show that the equation $$\mathrm { f } ( x ) = \mathrm { g } ( x )$$ has a root, \(\alpha\), in the interval [3,4].
5. Use the iteration formula $$x _ { n + 1 } = \frac { 1 } { 2 } \left[ 5 + \ln \left( x _ { n } + 3 \right) \right]$$ with \(x _ { 0 } = 3\), to find \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to 4 significant figures.
6. Show that your answer for \(x _ { 4 }\) is the value of \(\alpha\) correct to 4 significant figures. \end{document}

7 \includegraphics[max width=\textwidth, alt={}, center]{f334f6e4-2a60-4647-8b37-e48937c85747-4_721_691_269_726} The diagram shows a particle \(P\) of mass 0.5 kg attached to the highest point \(A\) of a fixed smooth sphere by a light elastic string. The sphere has centre \(O\) and radius 1.2 m . The string has natural length 0.6 m and modulus of elasticity \(6.86 \mathrm {~N} . P\) is released from rest at a point on the surface of the sphere where the acute angle \(A O P\) is at least 0.5 radians.

1. (a) For the case angle \(A O P = \alpha , P\) remains at rest. Show that \(\sin \alpha = 2.8 \alpha - 1.4\).
   (b) Use the iterative formula $$\alpha _ { n + 1 } = \frac { \sin \alpha _ { n } } { 2.8 } + 0.5 ,$$ with \(\alpha _ { 1 } = 0.8\), to find \(\alpha\) correct to 2 significant figures.
2. Given instead that angle \(A O P = 0.5\) radians when \(P\) is released, find the speed of \(P\) when angle \(A O P = 0.8\) radians, given that \(P\) is at all times in contact with the surface of the sphere. State whether the speed of \(P\) is increasing or decreasing when angle \(A O P = 0.8\) radians.

1 A curve passes through the point \(( 0,1 )\) and satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \sqrt { 1 + x ^ { 2 } }$$ Starting at the point \(( 0,1 )\), use a step-by-step method with a step length of 0.2 to estimate the value of \(y\) at \(x = 0.4\). Give your answer to five decimal places.

1 A curve passes through the point ( 1,3 ) and satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = 1 + x ^ { 3 }$$ Starting at the point ( 1,3 ), use a step-by-step method with a step length of 0.1 to estimate the \(y\)-coordinate of the point on the curve for which \(x = 1.3\). Give your answer to three decimal places.
(No credit will be given for methods involving integration.)

4 Fig. 4 shows the graph of \(y = x ^ { 5 } - 6 \sqrt { x } + 4\). \begin{figure}[h] \end{figure} There are two roots of the equation \(x ^ { 5 } - 6 \sqrt { x } + 4 = 0\). The roots are \(\alpha\) and \(\beta\), such that \(\alpha < \beta\).

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

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. 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.

1. State the method being used.
2. Use the values in the spreadsheet to calculate \(x _ { 5 }\) and \(x _ { 6 }\), giving your answers correct to 7 decimal places.
3. 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 .
4. Explain why this is different to the value displayed in cell D9. The value displayed in cell E11 in Alex's spreadsheet is \(- 1.4629 \mathrm { E } - 09\).
5. Write this value in standard mathematical notation.

4 Fig. 4.1 shows part of the graph of \(y = e ^ { x } - x ^ { 2 } - x - 1.1\). \begin{figure}[h] \end{figure} The equation \(\mathrm { e } ^ { x } - x ^ { 2 } - x - 1.1 = 0\) has a root \(\alpha\) such that \(1 < \alpha < 2\).
Ali is considering using the Newton-Raphson method to find \(\alpha\). Ali could use a starting value of \(x _ { 0 } = 1\) or a starting value of \(x _ { 0 } = 2\).

1. Without doing any calculations, explain whether Ali should use a starting value of \(x _ { 0 } = 1\) or a starting value of \(x _ { 0 } = 2\), or whether using either starting value would work equally well. Ali is also considering using the method of fixed point iteration to find \(\alpha\). Ali could use a starting value of \(x _ { 0 } = 1\) or a starting value of \(x _ { 0 } = 2\). Fig. 4.2 shows parts of the graphs of \(y = x\) and \(y = \ln \left( x ^ { 2 } + x + 1.1 \right)\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{f8c78ab8-7daa-4d6f-9bb5-093a46c590b8-04_819_1011_1818_255} \captionsetup{labelformat=empty} \caption{Fig. 4.2}
   \end{figure}
2. Without doing any calculations, explain whether Ali should use a starting value of \(x _ { 0 } = 1\) or a starting value of \(x _ { 0 } = 2\) or whether either starting value would work equally well. Ali used one of the above methods to find a sequence of approximations to \(\alpha\). These are shown, together with some further analysis in the associated spreadsheet output in Fig. 4.3. \begin{table}[h]

   		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.
3. State what is being calculated in the following columns of the spreadsheet.
   1. Column N
   2. Column O
4. Explain whether the values in column O suggest that Ali used the Newton-Raphson method or the iterative formula \(\mathrm { x } _ { \mathrm { n } + 1 } = \ln \left( \mathrm { x } _ { \mathrm { n } } ^ { 2 } + \mathrm { x } _ { \mathrm { n } } + 1.1 \right)\) to find this sequence of approximations to \(\alpha\).

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. \begin{table}[h] \end{table}

1. Use the information in Fig. 6.1 to find the value of the root as accurately as you can, justifying the precision quoted. The relaxed iteration \(\mathrm { x } _ { \mathrm { n } + 1 } = ( 1 - \lambda ) \mathrm { x } _ { \mathrm { n } } + \lambda \mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right)\), with \(\lambda = 0.51\) and \(x _ { 0 } = 0\), is to be used to find the root of the equation \(\sin ^ { - 1 } ( x ) - x ^ { 2 } + 1 = 0\).
2. Complete the copy of Fig. 6.2 in the Printed Answer Booklet, giving the values of \(\mathrm { x } _ { \mathrm { r } }\) correct to 7 decimal places and the values in the difference column and ratio column correct to 3 significant figures. \begin{table}[h]

   \(r\)	\(\mathrm { x } _ { \mathrm { r } }\)	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}
3. Write down the value of the root correct to 7 decimal places.
4. Explain why extrapolation could not be used in this case to find an improved approximation using this sequence of iterates. In this case the method of relaxation has been used to speed up the convergence of an iterative scheme.
5. Name another application of the method of relaxation.

5 The equation \(3 - 2 \ln x - x = 0\) has a root near \(x = 1.8\).
A student proposes to use the iterative formula \(\mathrm { x } _ { \mathrm { n } + 1 } = \mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right) = 3 - 2 \ln \mathrm { x } _ { \mathrm { n } }\) to find this root.
The diagram shows the graphs of \(\mathrm { y } = \mathrm { x }\) and \(\mathrm { y } = \mathrm { g } ( \mathrm { x } )\) for values of \(x\) from - 2 to 6 . \includegraphics[max width=\textwidth, alt={}, center]{4023e87c-34b1-4abd-9acc-ede5e4d68c7f-05_913_917_502_233}

1. With reference to the graph, explain why it might not be possible to use the student's iterative formula to find the root near \(x = 1.8\).
2. Use the relaxed iteration \(\mathrm { x } _ { \mathrm { n } + 1 } = \lambda \mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right) + ( 1 - \lambda ) \mathrm { x } _ { \mathrm { n } }\), with \(\lambda = 0.475\) and \(x _ { 0 } = 2\), to determine the root correct to \(\mathbf { 6 }\) decimal places. A student uses the same relaxed iteration with the same starting value. Some analysis of the iterates is carried out using a spreadsheet, which is shown in the table below.

   \(r\)	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

3. Explain what the analysis tells you about the order of convergence of this sequence of approximations.

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 \(x _ { 1 } = 2\).

   	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	\(- 7 \mathrm { E } - 05\)

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 \(a\).