Apply iteration to find root (pure fixed point)

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.

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.

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

1. Use the iterative formula $$x _ { n + 1 } = g \left( x _ { n } \right) \text { where } g \left( x _ { n } \right) = \sqrt { \cosh \left( x _ { n } - 2 \right) } \text {, }$$ starting with \(x _ { 0 } = 1\), to find \(\alpha\) correct to \(\mathbf { 3 }\) decimal places. The diagram shows the part of the graphs of \(\mathrm { y } = \mathrm { x }\) and \(\mathrm { y } = \mathrm { g } ( \mathrm { x } )\) for \(0 \leqslant x \leqslant 7\). \includegraphics[max width=\textwidth, alt={}, center]{83a06341-74e9-4f47-9104-e8e0259e7dfa-3_760_657_753_246}
2. Explain why the iterative formula used to find \(\alpha\) cannot successfully be used to find \(\beta\), even if \(x _ { 0 }\) is very close to \(\beta\).
3. Use the relaxed iteration $$\mathrm { x } _ { \mathrm { n } + 1 } = ( 1 - \lambda ) \mathrm { x } _ { \mathrm { n } } + \lambda \mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right) ,$$ with \(\lambda = - 0.21\) and \(x _ { 0 } = 6.4\), to find \(\beta\) correct to \(\mathbf { 3 }\) decimal places. In part (c) the method of relaxation was used to convert a divergent sequence of approximations into a convergent sequence.
4. State one other application of the method of relaxation.

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

1. Use the Newton-Raphson method with \(x _ { 0 } = 1\) to obtain \(\beta\) correct to \(\mathbf { 6 }\) decimal places. Fig. 6.1 shows part of the graph of \(y = 0.5 \ln x - x ^ { 2 } + x + 1\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{945883ad-c153-4c51-83d3-978e4c769ed5-06_1112_1156_529_354} \captionsetup{labelformat=empty} \caption{Fig. 6.1}
   \end{figure}
2. On the copy of Fig. 6.1 in the Printed Answer Booklet, illustrate the Newton-Raphson method working to obtain \(x _ { 1 }\) from \(x _ { 0 } = 1\). Beth is trying to find \(\alpha\) correct to 6 decimal places.
3. Suggest a reason why she might choose the Newton-Raphson method instead of fixed point iteration. Beth tries to find \(\alpha\) using the Newton-Raphson method with a starting value of \(x _ { 0 } = 0.5\). Her spreadsheet output is shown in Fig. 6.2. \begin{table}[h]

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

   \captionsetup{labelformat=empty} \caption{Fig. 6.2}
   \end{table}
4. Explain how the display \#NUM! has arisen in the cell for \(x _ { 2 }\). Beth decides to use the iterative formula $$x _ { n + 1 } = g \left( x _ { n } \right) = \sqrt { 0.5 \ln \left( x _ { n } \right) + x _ { n } + 1 }$$
5. Determine the outcome when Beth uses this formula with \(x _ { 0 } = 0.5\).
6. Use the relaxed iteration \(\mathrm { x } _ { \mathrm { n } + 1 } = ( 1 - \lambda ) \mathrm { x } _ { \mathrm { n } } + \lambda \mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right)\) with \(\lambda = - 0.041\) and \(x _ { 0 } = 0.5\) to obtain \(\alpha\) correct to \(\mathbf { 6 }\) decimal places.

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

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.
2. 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.

2. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of the vertical cross-section of the entrance to a tunnel. The width at the base of the tunnel entrance is 2 metres and its maximum height is 3 metres. The shape of the cross-section can be modelled by the curve with equation \(y = \mathrm { f } ( x )\) where $$f ( x ) = 3 \cos \left( \frac { \pi } { 2 } x ^ { 2 } \right) \quad x \in [ - 1,1 ]$$ A wooden door of uniform thickness 85 mm is to be made to seal the tunnel entrance.
Use Simpson's rule with 6 intervals to estimate the volume of wood required for this door, giving your answer in \(\mathrm { m } ^ { 3 }\) to 4 significant figures.

4 The diagram shows part of the graph of \(y = \cos x\), where \(x\) is measured in radians. \includegraphics[max width=\textwidth, alt={}, center]{6a6316e4-7b2d-4533-988a-4863d79ce668-05_609_846_294_607}

1. Use the copy of this diagram in the Printed Answer Booklet to find an approximate solution to the equation \(x = \cos x\).
2. Use an iterative method to find the solution to the equation \(x = \cos x\) correct to 3 significant figures. You should show your first, second and last two iterations, writing down all the figures on your calculator.

10 \includegraphics[max width=\textwidth, alt={}, center]{e3942549-bfc0-432a-bf49-7d01d44af01a-7_579_764_255_651} The diagram shows the graph of \(\mathrm { f } ( x ) = \ln ( 3 x + 1 ) - x\), which has a stationary point at \(x = \alpha\). A student wishes to find the non-zero root \(\beta\) of the equation \(\ln ( 3 x + 1 ) - x = 0\) using the Newton-Raphson method.

1. (a) Determine the value of \(\alpha\).
   (b) Explain why the Newton-Raphson method will fail if \(\alpha\) is used as the initial value.
2. Show that the Newton-Raphson iterative formula for finding \(\beta\) can be written as $$x _ { n + 1 } = \frac { 3 x _ { n } - \left( 3 x _ { n } + 1 \right) \ln \left( 3 x _ { n } + 1 \right) } { 2 - 3 x _ { n } } .$$
3. Apply the iterative formula in part (ii) with initial value \(x _ { 1 } = 1\) to find the value of \(\beta\) correct to 5 significant figures. You should show the result of each iteration.
4. Use a change of sign method to verify that the value of \(\beta\) found in part (iii) is correct to 5 significant figures.

5 A student is using the algorithm below to find an approximate value of \(\sqrt { 2 }\).
Line 10 Let \(A = 1 , B = 3 , C = 0\) Line \(20 \quad\) Let \(D = 1 , E = 2 , F = 0\) Line 30 Let \(G = B / E\) Line \(40 \quad\) Let \(H = G ^ { 2 }\) Line 50 If \(( H - 2 ) ^ { 2 } < 0.0001\) then go to Line 130
Line 60 Let \(C = 2 B + A\) Line 70 Let \(A = B\) Line 80 Let \(B = C\) Line 90 Let \(F = 2 E + D\) Line 100 Let \(D = E\) Line 110 Let \(E = F\) Line 120 Go to Line 30
Line 130 Print ' \(\sqrt { 2 }\) is approximately', \(B / E\) Line 140 Stop
Trace the algorithm.

6 A student is finding a numerical approximation for the area under a curve.
The algorithm that the student is using is as follows:
Line 10 Input \(A , B , N\) Line 20 Let \(T = 0\) Line 30 Let \(D = A\) Line \(40 \quad\) Let \(H = ( B - A ) / N\) Line \(50 \quad\) Let \(E = H / 2\) Line 60 Let \(T = T + A ^ { 3 } + B ^ { 3 }\) Line \(70 \quad\) Let \(D = D + H\) Line 80 If \(D = B\) then go to line 110
Line 90 Let \(T = T + 2 D ^ { 3 }\) Line 100 Go to line 70
Line \(110 \quad\) Print 'Area \(=\), \(T \times E\) Line 120 End
Trace the algorithm in the case where the input values are:

1. \(A = 1 , B = 5 , N = 2\);
2. \(A = 1 , B = 5 , N = 4\).

5 A student is using the following algorithm with different values of \(X\).

1. Trace the algorithm, giving your answers to three decimal places where appropriate:
   1. in the case where the input value of \(X\) is 2 ;
   2. in the case where the input value of \(X\) is - 6 .
2. Another student used the same algorithm but omitted LINE 70. Describe the outcome for this student.

13 The function f is defined by $$\mathrm { f } ( x ) = \arccos x \text { for } 0 \leq x \leq a$$ The curve with equation \(y = \mathrm { f } ( x )\) is shown below. \includegraphics[max width=\textwidth, alt={}, center]{6a03a035-ff32-4734-864b-a076aa9cbec0-18_842_837_550_603} 13

1. State the value of \(a\) 13
2. (i) On the diagram above, sketch the curve with equation $$y = \cos x \text { for } 0 \leq x \leq \frac { \pi } { 2 }$$ and
   sketch the line with equation $$y = x \text { for } 0 \leq x \leq \frac { \pi } { 2 }$$ 13 (b) (ii) Explain why the solution to the equation $$x - \cos x = 0$$ must also be a solution to the equation $$\cos x = \arccos x$$ Question 13 continues on the next page 13
3. Use the Newton-Raphson method with \(x _ { 0 } = 0\) to find an approximate solution, \(x _ { 3 }\), to the equation $$x - \cos x = 0$$ Give your answer to four decimal places. \includegraphics[max width=\textwidth, alt={}, center]{6a03a035-ff32-4734-864b-a076aa9cbec0-21_2491_1716_219_153}

\includegraphics{figure_1} Figure 1 shows a sketch of the curve with equation \(y = e^{-x} - 1\).

1. Copy Fig. 1 and on the same axes sketch the graph of \(y = \frac{1}{2}|x - 1|\). Show the coordinates of the points where the graph meets the axes. [2]

The \(x\)-coordinate of the point of intersection of the graphs is \(\alpha\).

1. Show that \(x = \alpha\) is a root of the equation \(x + 2e^{-x} - 3 = 0\). [3]
2. Show that \(-1 < \alpha < 0\). [2]

The iterative formula \(x_{n+1} = -\ln[\frac{1}{2}(3 - x_n)]\) is used to solve the equation \(x + 2e^{-x} - 3 = 0\).

1. Starting with \(x_0 = -1\), find the values of \(x_1\) and \(x_2\). [2]
2. Show that, to 2 decimal places, \(\alpha = -0.58\). [2]

1. Sketch, on the same set of axes, the graphs of $$y = 2 - e^{-x} \text{ and } y = \sqrt{x}.$$ [3] [It is not necessary to find the coordinates of any points of intersection with the axes.] Given that f(x) = \(e^{-x} + \sqrt{x} - 2\), \(x \geq 0\),
2. explain how your graphs show that the equation f(x) = 0 has only one solution, [1]
3. show that the solution of f(x) = 0 lies between \(x = 3\) and \(x = 4\). [2]

The iterative formula \(x_{n+1} = (2 - e^{-x_n})^2\) is used to solve the equation f(x) = 0.

1. Taking \(x_0 = 4\), write down the values of \(x_1\), \(x_2\), \(x_3\) and \(x_4\), and hence find an approximation to the solution of f(x) = 0, giving your answer to 3 decimal places. [4]

\includegraphics{figure_1} Figure 1 shows a sketch of the curve with equation \(y = e^{-x} - 1\).

1. Copy Fig. 1 and on the same axes sketch the graph of \(y = \frac{1}{2}|x - 1|\). Show the coordinates of the points where the graph meets the axes. [2]

The \(x\)-coordinate of the point of intersection of the graph is \(\alpha\).

1. Show that \(x = \alpha\) is a root of the equation \(x + 2e^{-x} - 3 = 0\). [3]
2. Show that \(-1 < \alpha < 0\). [2]

The iterative formula \(x_{n+1} = -\ln[\frac{1}{2}(3 - x_n)]\) is used to solve the equation \(x + 2e^{-x} - 3 = 0\).

1. Starting with \(x_0 = -1\), find the values of \(x_1\) and \(x_2\). [2]
2. Show that, to 2 decimal places, \(\alpha = -0.58\). [2]

The sequence defined by $$x_1 = 3, \quad x_{n+1} = \sqrt{31 - \frac{5}{2}x_n}$$ converges to the number \(\alpha\).

1. Find the value of \(\alpha\) correct to 3 decimal places, showing the result of each iteration. [3]
2. Find an equation of the form \(ax^3 + bx + c = 0\), where \(a\), \(b\) and \(c\) are integers, which has \(\alpha\) as a root. [3]

It is given that \(\alpha\) is the only real root of the equation \(x^3 + 2x - 28 = 0\) and that \(1.8 < \alpha < 2\).

1. The iteration \(x_{n+1} = \sqrt[3]{28 - 2x_n}\), with \(x_1 = 1.9\), is to be used to find \(\alpha\). Find the values of \(x_2\), \(x_3\) and \(x_4\), giving the answers correct to 7 decimal places. [3]
2. The error \(e_n\) is defined by \(e_n = \alpha - x_n\). Given that \(\alpha = 1.891 574 9\), correct to 7 decimal places, evaluate \(\frac{e_3}{e_2}\) and \(\frac{e_4}{e_3}\). Comment on these values in relation to the gradient of the curve with equation \(y = \sqrt[3]{28 - 2x}\) at \(x = \alpha\). [3]

It is given that \(f(x) = x^2 - \sin x\).

1. The iteration \(x_{n+1} = \sqrt{\sin x_n}\), with \(x_1 = 0.875\), is to be used to find a real root, \(\alpha\), of the equation \(f(x) = 0\). Find \(x_2, x_3\) and \(x_4\), giving the answers correct to 6 decimal places. [2]
2. The error \(e_n\) is defined by \(e_n = \alpha - x_n\). Given that \(\alpha = 0.876726\), correct to 6 decimal places, find \(e_3\) and \(e_4\). Given that \(g(x) = \sqrt{\sin x}\), use \(e_3\) and \(e_4\) to estimate \(g'(\alpha)\). [3]

The sequence \(u_1, u_2, u_3, \ldots\) is defined by $$u_{n+1} = k - \frac{24}{u_n} \quad u_1 = 2$$ where \(k\) is an integer. Given that \(u_1 + 2u_2 + u_3 = 0\)

1. show that $$3k^2 - 58k + 240 = 0$$ [3]
2. Find the value of \(k\), giving a reason for your answer. [2]
3. Find the value of \(u_3\) [1]