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

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.

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] \end{table}

1. 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 \(\beta \approx 4.4\).
2. Obtain the iterative formula \(x _ { n + 1 } = g \left( x _ { n } \right) = \frac { \sqrt [ 3 ] { x _ { n } ^ { X _ { n } } + 25 } } { 2 }\).
3. 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 \(\alpha\).
4. Starting with \(x _ { 0 } = 1.5\), use the iterative formula to find \(x _ { 1 } , x _ { 2 } , x _ { 3 } , x _ { 4 } , x _ { 5 }\), and \(x _ { 6 }\).
5. 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}
6. Explain why \#NUM! is displayed in the cell for \(x _ { 5 }\).
7. 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 \(x _ { 2 }\).
8. 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.

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

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

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.

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

1. 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

2. What feature of column B shows that this application of the secant method has been successful?
3. 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).
4. (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.
5. Explain why the values in cells F6 and F7 are not 0 . [Question 7 is printed overleaf.]

1. The temperature, \(\theta ^ { \circ } \mathrm { C }\), of coffee in a cup, \(t\) minutes after the cup of coffee is put in a room, is modelled by the differential equation

$$\frac { \mathrm { d } \theta } { \mathrm {~d} t } = - k ( \theta - 20 )$$ where \(k\) is a constant.
The coffee has an initial temperature of \(80 ^ { \circ } \mathrm { C }\) Using \(k = 0.1\)

1. use two iterations of the approximation formula \(\left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) _ { 0 } = \frac { y _ { 1 } - y _ { 0 } } { h }\) to estimate the temperature of the coffee 3 minutes after it was put in the room. The coffee in a different cup, which also had an initial temperature of \(80 ^ { \circ } \mathrm { C }\) when it was put in the room, cools more slowly.
2. Use this information to suggest how the value of \(k\) would need to be changed in the model.

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1. Julie decides to start a business breeding rabbits to sell as pets.

Initially she buys 20 rabbits. After \(t\) years the number of rabbits, \(R\), is modelled by the differential equation $$\frac { \mathrm { d } R } { \mathrm {~d} t } = 2 R + 4 \sin t \quad t > 0$$ Julie needs to have at least 40 rabbits before she can start to sell them.
Use two iterations of the approximation formula $$\left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) _ { n } \approx \frac { y _ { n + 1 } - y _ { n } } { h }$$ to find out if, according to the model, Julie will be able to start selling rabbits after 4 months.

1. The value, V hundred pounds, of a particular stock thours after the opening of trading on a given day is modelled by the differential equation

$$\frac { d V } { d t } = \frac { V ^ { 2 } - t } { t ^ { 2 } + t V } \quad 0 < t < 8.5$$ A trader purchases \(\pounds 300\) of the stock one hour after the opening of trading.
Use two iterations of the approximation formula \(\left( \frac { \mathrm { dy } } { \mathrm { dx } } \right) _ { 0 } \approx \frac { \mathrm { y } _ { 1 } - \mathrm { y } _ { 0 } } { \mathrm {~h} }\) to estimate, to the nearest \(\pounds\), the value of the trader's stock half an hour after it was purchased.

1. A community is concerned about the rising level of pollutant in its local pond and applies a chemical treatment to stop the increase of pollutant.

The concentration, \(x\) parts per million (ppm), of the pollutant in the pond water \(t\) days after the chemical treatment was applied, is modelled by the differential equation $$\frac { \mathrm { d } x } { \mathrm {~d} t } = \frac { 3 + \cosh t } { 3 x ^ { 2 } \cosh t } - \frac { 1 } { 3 } x \tanh t$$ When the chemical treatment was applied the concentration of pollutant was 3 ppm .

1. Use the iteration formula $$\left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) _ { n } \approx \frac { \left( y _ { n + 1 } - y _ { n } \right) } { h }$$ once to estimate the concentration of the pollutant in the pond water 6 hours after the chemical treatment was applied.
2. Show that the transformation \(u = x ^ { 3 }\) transforms the differential equation (I) into the differential equation $$\frac { \mathrm { d } u } { \mathrm {~d} t } + u \tanh t = 1 + \frac { 3 } { \cosh t }$$
3. Determine the general solution of equation (II)
4. Hence find an equation for the concentration of pollutant in the pond water \(t\) days after the chemical treatment was applied.
5. Find the percentage error of the estimate found in part (a) compared to the value predicted by the model, stating if it is an overestimate or an underestimate.

1. The velocity \(v \mathrm {~ms} ^ { - 1 }\), of a raindrop, \(t\) seconds after it falls from a cloud, is modelled by the differential equation

$$\frac { \mathrm { d } v } { \mathrm {~d} t } = - 0.1 v ^ { 2 } + 10 \quad t \geqslant 0$$ Initially the raindrop is at rest.

1. Use two iterations of the approximation formula \(\left( \frac { \mathrm { d } y } { \mathrm {~d} x } \right) _ { n } \approx \frac { y _ { n + 1 } - y _ { n } } { h }\) to estimate the velocity of the raindrop 1 second after it falls from the cloud. Given that the initial acceleration of the raindrop is found to be smaller than is suggested by the current model,
2. refine the model by changing the value of one constant.

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]{a16ab26f-21fb-4a73-8b94-c16bef611bcb-7_524_714_274_246} The diagram shows the graph of \(y = - \tan ^ { - 1 } \left( \frac { 1 } { 2 } x - \frac { 1 } { 3 } \pi \right)\), which crosses the \(x\)-axis at the point \(A\) and the \(y\)-axis at the point \(B\).

1. Determine the coordinates of the points \(A\) and \(B\).
2. Give full details of a sequence of three geometrical transformations which transform the graph of \(y = \tan ^ { - 1 } x\) to the graph of \(y = - \tan ^ { - 1 } \left( \frac { 1 } { 2 } x - \frac { 1 } { 3 } \pi \right)\). The equation \(x = - \tan ^ { - 1 } \left( \frac { 1 } { 2 } x - \frac { 1 } { 3 } \pi \right)\) has only one root.
3. Show by calculation that this root lies between \(x = 0\) and \(x = 1\).
4. Use the iterative formula \(x _ { n + 1 } = - \tan ^ { - 1 } \left( \frac { 1 } { 2 } x _ { n } - \frac { 1 } { 3 } \pi \right)\), with a suitable starting value, to find the root correct to 3 significant figures. Show the result of each iteration.
5. Using the diagram in the Printed Answer Booklet, show how the iterative process converges to the root.

4. $$\mathrm { f } ( x ) = 3 \mathrm { e } ^ { x } - \frac { 1 } { 2 } \ln x - 2 , \quad x > 0 .$$

1. Differentiate to find \(\mathrm { f } ^ { \prime } ( x )\). The curve with equation \(y = \mathrm { f } ( x )\) has a turning point at \(P\). The \(x\)-coordinate of \(P\) is \(\alpha\).
2. Show that \(\alpha = \frac { 1 } { 6 } \mathrm { e } ^ { - \alpha }\). The iterative formula $$x _ { n + 1 } = \frac { 1 } { 6 } \mathrm { e } ^ { - x _ { n } } , \quad x _ { 0 } = 1$$ is used to find an approximate value for \(\alpha\).
3. Calculate the values of \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to 4 decimal places.
4. By considering the change of sign of \(\mathrm { f } ^ { \prime } ( x )\) in a suitable interval, prove that \(\alpha = 0.1443\) correct to 4 decimal places.

6 [Figure 1, printed on the insert, is provided for use in this question.]
The curve \(y = x ^ { 3 } + 4 x - 3\) intersects the \(x\)-axis at the point \(A\) where \(x = \alpha\).

1. Show that \(\alpha\) lies between 0.5 and 1.0.
2. Show that the equation \(x ^ { 3 } + 4 x - 3 = 0\) can be rearranged into the form \(x = \frac { 3 - x ^ { 3 } } { 4 }\).
   (1 mark)
3. 1. Use the iteration \(x _ { n + 1 } = \frac { 3 - x _ { n } { } ^ { 3 } } { 4 }\) with \(x _ { 1 } = 0.5\) to find \(x _ { 3 }\), giving your answer to two decimal places.
   2. The sketch on Figure 1 shows parts of the graphs of \(y = \frac { 3 - x ^ { 3 } } { 4 }\) 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.
      (3 marks)

6 [Figure 1, printed on the insert, is provided for use in this question.]
The curve \(y = x ^ { 3 } + 4 x - 3\) intersects the \(x\)-axis at the point \(A\) where \(x = \alpha\).

1. Show that \(\alpha\) lies between 0.5 and 1.0.
2. Show that the equation \(x ^ { 3 } + 4 x - 3 = 0\) can be rearranged into the form \(x = \frac { 3 - x ^ { 3 } } { 4 }\).
   (1 mark)
3. 1. Use the iteration \(x _ { n + 1 } = \frac { 3 - x _ { n } { } ^ { 3 } } { 4 }\) with \(x _ { 1 } = 0.5\) to find \(x _ { 3 }\), giving your answer to two decimal places.
      (3 marks)
   2. The sketch on Figure 1 shows parts of the graphs of \(y = \frac { 3 - x ^ { 3 } } { 4 }\) 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.
      (3 marks)

3 [Figure 1, printed on the insert, is provided for use in this question.]
The curve with equation \(y = x ^ { 3 } + 5 x - 4\) intersects the \(x\)-axis at the point \(A\), where \(x = \alpha\).

1. Show that \(\alpha\) lies between 0.5 and 1 .
2. Show that the equation \(x ^ { 3 } + 5 x - 4 = 0\) can be rearranged into the form $$x = \frac { 1 } { 5 } \left( 4 - x ^ { 3 } \right)$$
3. Use the iteration \(x _ { n + 1 } = \frac { 1 } { 5 } \left( 4 - x _ { n } { } ^ { 3 } \right)\) with \(x _ { 1 } = 0.5\) to find \(x _ { 3 }\), giving your answer to three decimal places.
4. The sketch on Figure 1 shows parts of the graphs of \(y = \frac { 1 } { 5 } \left( 4 - x ^ { 3 } \right)\) 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.

2 [Figure 1, printed on the insert, is provided for use in this question.]

1. 1. Sketch the graph of \(y = \sin ^ { - 1 } x\), where \(y\) is in radians. State the coordinates of the end points of the graph.
   2. By drawing a suitable straight line on your sketch, show that the equation $$\sin ^ { - 1 } x = \frac { 1 } { 4 } x + 1$$ has only one solution.
2. The root of the equation \(\sin ^ { - 1 } x = \frac { 1 } { 4 } x + 1\) is \(\alpha\). Show that \(0.5 < \alpha < 1\).
3. The equation \(\sin ^ { - 1 } x = \frac { 1 } { 4 } x + 1\) can be rewritten as \(x = \sin \left( \frac { 1 } { 4 } x + 1 \right)\).
   1. Use the iteration \(x _ { n + 1 } = \sin \left( \frac { 1 } { 4 } x _ { n } + 1 \right)\) with \(x _ { 1 } = 0.5\) to find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.
   2. The sketch on Figure 1 shows parts of the graphs of \(y = \sin \left( \frac { 1 } { 4 } x + 1 \right)\) 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.

4 [Figure 1, printed on the insert, is provided for use in this question.]

1. Use Simpson's rule with 5 ordinates (4 strips) to find an approximation to \(\int _ { 1 } ^ { 2 } 3 ^ { x } \mathrm {~d} x\), giving your answer to three significant figures.
2. The curve \(y = 3 ^ { x }\) intersects the line \(y = x + 3\) at the point where \(x = \alpha\).
   1. Show that \(\alpha\) lies between 0.5 and 1.5.
   2. Show that the equation \(3 ^ { x } = x + 3\) can be rearranged into the form $$x = \frac { \ln ( x + 3 ) } { \ln 3 }$$
   3. Use the iteration \(x _ { n + 1 } = \frac { \ln \left( x _ { n } + 3 \right) } { \ln 3 }\) with \(x _ { 1 } = 0.5\) to find \(x _ { 3 }\) to two significant figures.
   4. The sketch on Figure 1 shows part of the graphs of \(y = \frac { \ln ( x + 3 ) } { \ln 3 }\) 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.

3

1. It is given that the curves with equations \(y = 6 \ln x\) and \(y = 8 x - x ^ { 2 } - 3\) intersect at a single point where \(x = \alpha\).
   1. Show that \(\alpha\) lies between 5 and 6 .
   2. Show that the equation \(x = 4 + \sqrt { 13 - 6 \ln x }\) can be rearranged into the form $$6 \ln x + x ^ { 2 } - 8 x + 3 = 0$$
   3. Use the iterative formula $$x _ { n + 1 } = 4 + \sqrt { 13 - 6 \ln x _ { n } }$$ with \(x _ { 1 } = 5\) to find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.
2. A curve has equation \(y = \mathrm { f } ( x )\) where \(\mathrm { f } ( x ) = 6 \ln x + x ^ { 2 } - 8 x + 3\).
   1. Find the exact values of the coordinates of the stationary points of the curve.
   2. Hence, or otherwise, find the exact values of the coordinates of the stationary points of the curve with equation $$y = 2 \mathrm { f } ( x - 4 )$$

2 A curve satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = 2 ^ { x }$$ Starting at the point \(( 1,4 )\) on the curve, use a step-by-step method with a step length of 0.01 to estimate the value of \(y\) at \(x = 1.02\). Give your answer to six significant figures.