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

1 The function \(y ( x )\) satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$ where $$\mathrm { f } ( x , y ) = x + \ln ( 1 + y )$$ and $$y ( 2 ) = 1$$ Use the improved Euler formula $$y _ { r + 1 } = y _ { r } + \frac { 1 } { 2 } \left( k _ { 1 } + k _ { 2 } \right)$$ where \(k _ { 1 } = h \mathrm { f } \left( x _ { r } , y _ { r } \right)\) and \(k _ { 2 } = h \mathrm { f } \left( x _ { r } + h , y _ { r } + k _ { 1 } \right)\) and \(h = 0.2\), to obtain an approximation to \(y ( 2.2 )\), giving your answer to four decimal places.

1 The function \(y ( x )\) satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$ where $$\mathrm { f } ( x , y ) = \sqrt { ( 2 x ) } + \sqrt { y }$$ and $$y ( 2 ) = 9$$ Use the improved Euler formula $$y _ { r + 1 } = y _ { r } + \frac { 1 } { 2 } \left( k _ { 1 } + k _ { 2 } \right)$$ where \(k _ { 1 } = h \mathrm { f } \left( x _ { r } , y _ { r } \right)\) and \(k _ { 2 } = h \mathrm { f } \left( x _ { r } + h , y _ { r } + k _ { 1 } \right)\) and \(h = 0.25\), to obtain an approximation to \(y ( 2.25 )\), giving your answer to two decimal places.

10 \includegraphics[max width=\textwidth, alt={}, center]{05bec6d6-b526-4b6f-86f3-39aa38cbf5c6-6_405_661_251_703} The diagram shows a sector \(A O B\) of a circle with centre \(O\) and radius 6 cm .
The angle \(A O B\) is \(\theta\) radians.
The area of the segment bounded by the chord \(A B\) and the \(\operatorname { arc } A B\) is \(7.2 \mathrm {~cm} ^ { 2 }\).

1. Show that \(\theta = 0.4 + \sin \theta\).
2. Let \(\mathrm { F } ( \theta ) = 0.4 + \sin \theta\). By considering the value of \(\mathrm { F } ^ { \prime } ( \theta )\) where \(\theta = 1.2\), explain why using an iterative method based on the equation in part (a) will converge to the root, assuming that 1.2 is sufficiently close to the root.
3. Use the iterative formula \(\theta _ { n + 1 } = 0.4 + \sin \theta _ { n }\) with a starting value of 1.2 to find the value of \(\theta\) correct to 4 significant figures.
   You should show the result of each iteration.
4. Use a change of sign method to show that the value of \(\theta\) found in part (c) is correct to 4 significant figures.

7 Two students, Anna and Ben, are starting a revision programme. They will both revise for 30 minutes on Day 1. Anna will increase her revision time by 15 minutes for every subsequent day. Ben will increase his revision time by \(10 \%\) for every subsequent day.

1. Verify that on Day 10 Anna does 94 minutes more revision than Ben, correct to the nearest minute. Let Day \(X\) be the first day on which Ben does more revision than Anna.
2. Show that \(X\) satisfies the inequality \(X > \log _ { 1.1 } ( 0.5 X + 0.5 ) + 1\).
3. Use the iterative formula \(x _ { n + 1 } = \log _ { 1.1 } \left( 0.5 x _ { n } + 0.5 \right) + 1\) with \(x _ { 1 } = 10\) to find the value of \(X\). You should show the result of each iteration.
4. 1. Give a reason why Anna's revision programme may not be realistic.
   2. Give a different reason why Ben's revision programme may not be realistic.

10 \includegraphics[max width=\textwidth, alt={}, center]{38b515c2-4764-4b51-a1f5-9b48d46610f0-7_545_659_255_244} The diagram shows a sector \(O A B\) of a circle with centre \(O\) and radius \(O A\). The angle \(A O B\) is \(\theta\) radians. \(M\) is the mid-point of \(O A\). The ratio of areas \(O M B : M A B\) is 2:3.

1. Show that \(\theta = 1.25 \sin \theta\). The equation \(\theta = 1.25 \sin \theta\) has only one root for \(\theta > 0\).
2. This root can be found by using the iterative formula \(\theta _ { n + 1 } = 1.25 \sin \theta _ { n }\) with a starting value of \(\theta _ { 1 } = 0.5\).

10
The diagram shows part of the curve \(\mathrm { f } ( x ) = \frac { \mathrm { e } ^ { x } } { 4 x ^ { 2 } - 1 } + 2\). The equation \(\mathrm { f } ( x ) = 0\) has a positive root \(\alpha\) close to \(x = 0.3\).

1. Explain why using the sign change method with \(x = 0\) and \(x = 1\) will fail to locate \(\alpha\).
2. Show that the equation \(\mathrm { f } ( x ) = 0\) can be written as \(x = \frac { 1 } { 4 } \sqrt { \left( 4 - 2 \mathrm { e } ^ { x } \right) }\).
3. Use the iterative formula \(x _ { n + 1 } = \frac { 1 } { 4 } \sqrt { \left( 4 - 2 \mathrm { e } ^ { x _ { n } } \right) }\) with a starting value of \(x _ { 1 } = 0.3\) to find the value of \(\alpha\) correct to \(\mathbf { 4 }\) significant figures, showing the result of each iteration.
4. An alternative iterative formula is \(x _ { n + 1 } = \mathrm { F } \left( x _ { n } \right)\), where \(\mathrm { F } \left( x _ { n } \right) = \ln \left( 2 - 8 x _ { n } ^ { 2 } \right)\). By considering \(\mathrm { F } ^ { \prime } ( 0.3 )\) explain why this iterative formula will not find \(\alpha\).

4. The curve with equation \(y = 2 + \ln ( 4 - x )\) meets the line \(y = x\) at a single point, \(x = \beta\).
a. Show that \(2 < \beta < 3\) \begin{figure}[h] \end{figure} Figure 2 shows the graph of \(y = 2 + \ln ( 4 - x )\) and the graph of \(y = x\).
A student uses the iteration formula $$x _ { n + 1 } = 2 + \ln \left( 4 - x _ { n } \right) , \quad n \in N ,$$ in an attempt to find an approximation for \(\beta\).
Using the graph and starting with \(x _ { 1 } = 3\),
b. determine whether the or not this iteration formula can be used to find an approximation for \(\beta\), justifying your answer.

8. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of part of the curve with equation \(y = \mathrm { f } ( x )\), where \(x \in R\), \(x > 0\) $$\mathrm { f } ( x ) = ( 0.5 x - 8 ) \ln ( x + 1 ) \quad 0 \leq x \leq A$$ a. Find the value of \(A\).
b. Find \(\mathrm { f } ^ { \prime } ( x )\) The curve has a minimum turning point at \(B\).
c. Show that the \(x\)-coordinate of \(B\) is a solution of the equation $$x = \frac { 17 } { \ln ( x + 1 ) + 1 } - 1$$ d. Use the iteration formula $$x _ { n + 1 } = \frac { 17 } { \ln \left( x _ { n } + 1 \right) + 1 } - 1$$ with \(x _ { 0 } = 5\) to find the values of \(x _ { 1 }\) and the value of \(x _ { 6 }\) giving your answers to three decimal places.

6. \begin{figure}[h] \end{figure} Figure 4 shows a sketch of the curve \(C\) with equation \(y = \mathrm { f } ( x )\), where $$f ( x ) = \frac { 2 x ^ { 2 } - x } { \sqrt { x } } - 2 \ln \left( \frac { x } { 2 } \right) , \quad x > 0$$ The curve has a minimum turning point at \(Q\), as shown in Figure 4.
a. Show that \(\mathrm { f } ^ { \prime } ( x ) = \frac { 6 x ^ { 2 } - x - 4 \sqrt { x } } { 2 x \sqrt { x } }\) b. Show that the \(x\)-coordinate of \(Q\) is the solution of $$x = \sqrt { \frac { x } { 6 } + \frac { 2 \sqrt { x } } { 3 } }$$ To find an approximation for the \(x\)-coordinate of \(Q\), the iteration formula $$x _ { n + 1 } = \sqrt { \frac { x _ { n } } { 6 } + \frac { 2 \sqrt { x _ { n } } } { 3 } }$$ is used.
c. Taking \(x _ { 0 } = 0.8\), find the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\). Give your answers to 3 decimal places.

5. \(\mathrm { f } ( x ) = \frac { 1 } { 3 } x ^ { 3 } - 4 x - 2\) a. Show that the equation \(\mathrm { f } ( x ) = 0\) can be written in the form \(x = \pm \sqrt { a + \frac { b } { x } }\), and state the values of the integers \(a\) and \(b\). \(\mathrm { f } ( x ) = 0\) has one positive root, \(\alpha\).
The iterative formula \(x _ { n + 1 } = \sqrt { a + \frac { b } { x _ { n } } } , \quad x _ { 0 } = 4\) is used to find an approximation value for \(\alpha\).
b. Calculate the values of \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\) to 4 decimal places.
c. Explain why for this question, the Newton-Raphson method cannot be used with \(x _ { 1 } = 2\).

The curve with equation \(y = 2 \ln ( 8 - x )\) meets the line \(y = x\) at a single point, \(x = \alpha\).

1. Show that \(3 < \alpha < 4\) \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{b5f50f17-9f1b-4b4c-baf3-e50de5f2ea9c-08_666_1061_445_502} \captionsetup{labelformat=empty} \caption{Figure 2}
   \end{figure} Figure 2 shows the graph of \(y = 2 \ln ( 8 - x )\) and the graph of \(y = x\).
   A student uses the iteration formula $$x _ { n + 1 } = 2 \ln \left( 8 - x _ { n } \right) , \quad n \in \mathbb { N }$$ in an attempt to find an approximation for \(\alpha\).
   Using the graph and starting with \(x _ { 1 } = 4\)
2. determine whether or not this iteration formula can be used to find an approximation for \(\alpha\), justifying your answer.

8. \begin{figure}[h] \end{figure} A car stops at two sets of traffic lights.
Figure 2 shows a graph of the speed of the car, \(v \mathrm {~ms} ^ { - 1 }\), as it travels between the two sets of traffic lights. The car takes \(T\) seconds to travel between the two sets of traffic lights.
The speed of the car is modelled by the equation $$v = ( 10 - 0.4 t ) \ln ( t + 1 ) \quad 0 \leqslant t \leqslant T$$ where \(t\) seconds is the time after the car leaves the first set of traffic lights.
According to the model,

1. find the value of \(T\)
2. show that the maximum speed of the car occurs when $$t = \frac { 26 } { 1 + \ln ( t + 1 ) } - 1$$ Using the iteration formula $$t _ { n + 1 } = \frac { 26 } { 1 + \ln \left( t _ { n } + 1 \right) } - 1$$ with \(t _ { 1 } = 7\)
3. 1. find the value of \(t _ { 3 }\) to 3 decimal places,
   2. find, by repeated iteration, the time taken for the car to reach maximum speed.

1. 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 \(u _ { 1 } + 2 u _ { 2 } + u _ { 3 } = 0\)

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

11. \begin{figure}[h] \end{figure} Figure 8 shows a sketch of the curve \(C\) with equation \(y = x ^ { x } , x > 0\)

1. Find, by firstly taking logarithms, the \(x\) coordinate of the turning point of \(C\).
   (Solutions based entirely on graphical or numerical methods are not acceptable.) The point \(P ( \alpha , 2 )\) lies on \(C\).
2. Show that \(1.5 < \alpha < 1.6\) A possible iteration formula that could be used in an attempt to find \(\alpha\) is $$x _ { n + 1 } = 2 x _ { n } ^ { 1 - x _ { n } }$$ Using this formula with \(x _ { 1 } = 1.5\)
3. find \(x _ { 4 }\) to 3 decimal places,
4. describe the long-term behaviour of \(x _ { n }\)

7. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of the curve \(C\) with equation $$y = \frac { 4 x ^ { 2 } + x } { 2 \sqrt { x } } - 4 \ln x \quad x > 0$$

1. Show that $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 12 x ^ { 2 } + x - 16 \sqrt { x } } { 4 x \sqrt { x } }$$ The point \(P\), shown in Figure 1, is the minimum turning point on \(C\).
2. Show that the \(x\) coordinate of \(P\) is a solution of $$x = \left( \frac { 4 } { 3 } - \frac { \sqrt { x } } { 12 } \right) ^ { \frac { 2 } { 3 } }$$
3. Use the iteration formula $$x _ { n + 1 } = \left( \frac { 4 } { 3 } - \frac { \sqrt { x _ { n } } } { 12 } \right) ^ { \frac { 2 } { 3 } } \quad \text { with } x _ { 1 } = 2$$ to find (i) the value of \(x _ { 2 }\) to 5 decimal places,
   (ii) the \(x\) coordinate of \(P\) to 5 decimal places.

6. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of the curve with equation \(y = \mathrm { f } ( x )\), where $$\mathrm { f } ( x ) = ( 8 - x ) \ln x , \quad x > 0$$ The curve cuts the \(x\)-axis at the points \(A\) and \(B\) and has a maximum turning point at \(Q\), as shown in Figure 2.

1. Find the \(x\) coordinate of \(A\) and the \(x\) coordinate of \(B\).
2. Show that the \(x\) coordinate of \(Q\) satisfies $$x = \frac { 8 } { 1 + \ln x }$$
3. Show that the \(x\) coordinate of \(Q\) lies between 3.5 and 3.6
4. Use the iterative formula $$x _ { n + 1 } = \frac { 8 } { 1 + \ln x _ { n } } \quad n \in \mathbb { N }$$ with \(x _ { 1 } = 3.5\) to
   1. find the value of \(x _ { 5 }\) to 4 decimal places,
   2. find the \(x\) coordinate of \(Q\) accurate to 2 decimal places.

11 Fig. 11 shows the graph of \(y = x ^ { 2 } - 4 x + x \ln x\). \begin{figure}[h] \end{figure}

1. Show that the \(x\)-coordinate of the stationary point on the curve may be found from the equation \(2 x - 3 + \ln x = 0\).
2. Use an iterative method to find the \(x\)-coordinate of the stationary point on the curve \(y = x ^ { 2 } - 4 x + x \ln x\), giving your answer correct to 4 decimal places.

5 Fig. 5 shows part of the curve \(y = \operatorname { cosec } x\) together with the \(x\) - and \(y\)-axes. \begin{figure}[h] \end{figure}

1. For the section of the curve which is shown in Fig. 5, write down
   1. the equations of the two vertical asymptotes,
   2. the coordinates of the minimum point.
2. Show that the equation \(x = \operatorname { cosec } x\) has a root which lies between \(x = 1\) and \(x = 2\).
3. Use the iteration \(\mathrm { x } _ { \mathrm { n } + 1 } = \operatorname { cosec } \left( \mathrm { x } _ { \mathrm { n } } \right)\), with \(x _ { 0 } = 1\), to find
   1. the values of \(x _ { 1 }\) and \(x _ { 2 }\), correct to 5 decimal places,
   2. this root of the equation, correct to 3 decimal places.
4. There is another root of \(x = \operatorname { cosec } x\) which lies between \(x = 2\) and \(x = 3\). Determine whether the iteration \(\mathrm { x } _ { \mathrm { n } + 1 } = \operatorname { cosec } \left( \mathrm { x } _ { \mathrm { n } } \right)\) with \(x _ { 0 } = 2.5\) converges to this root.
5. Sketch the staircase or cobweb diagram for the iteration, starting with \(x _ { 0 } = 2.5\), on the diagram in the Printed Answer Booklet.

8 The diagram shows the curve \(y = \cos ^ { - 1 } x\) for \(- 1 \leqslant x \leqslant 1\). \includegraphics[max width=\textwidth, alt={}, center]{6890a681-2b7f-4853-a5f0-f88b7b435367-4_492_698_1640_671}

1. Write down the exact coordinates of the points \(A\) and \(B\).
2. The equation \(\cos ^ { - 1 } x = 3 x + 1\) has only one root. Given that the root of this equation is \(\alpha\), show that \(0.1 \leqslant \alpha \leqslant 0.2\).
3. Use the iteration \(x _ { n + 1 } = \frac { 1 } { 3 } \left( \cos ^ { - 1 } x _ { n } - 1 \right)\) with \(x _ { 1 } = 0.1\) to find the values of \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to three decimal places.

3 The equation $$x + ( 1 + 3 x ) ^ { \frac { 1 } { 4 } } = 0$$ has a single root, \(\alpha\).

1. Show that \(\alpha\) lies between - 0.33 and - 0.32 .
2. Show that the equation \(x + ( 1 + 3 x ) ^ { \frac { 1 } { 4 } } = 0\) can be rearranged into the form $$x = \frac { 1 } { 3 } \left( x ^ { 4 } - 1 \right)$$
3. Use the iteration \(x _ { n + 1 } = \frac { \left( x _ { n } ^ { 4 } - 1 \right) } { 3 }\) with \(x _ { 1 } = - 0.3\) to find \(x _ { 4 }\), giving your answer to three significant figures.

2 A curve is defined by the equation \(y = \left( x ^ { 2 } - 4 \right) \ln ( x + 2 )\) for \(x \geqslant 3\).
The curve intersects the line \(y = 15\) at a single point, where \(x = \alpha\).

1. Show that \(\alpha\) lies between 3.5 and 3.6.
2. Show that the equation \(\left( x ^ { 2 } - 4 \right) \ln ( x + 2 ) = 15\) can be arranged into the form $$x = \pm \sqrt { 4 + \frac { 15 } { \ln ( x + 2 ) } }$$ (2 marks)
3. Use the iteration $$x _ { n + 1 } = \sqrt { 4 + \frac { 15 } { \ln \left( x _ { n } + 2 \right) } }$$ with \(x _ { 1 } = 3.5\) to find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.
   (2 marks)

1

1. Use Simpson's rule with 7 ordinates (6 strips) to find an estimate for \(\int _ { 0 } ^ { 3 } 4 ^ { x } \mathrm {~d} x\).
2. A curve is defined by the equation \(y = 4 ^ { x }\). The curve intersects the line \(y = 8 - 2 x\) at a single point where \(x = \alpha\).
   1. Show that \(\alpha\) lies between 1.2 and 1.3.
   2. The equation \(4 ^ { x } = 8 - 2 x\) can be rearranged into the form \(x = \frac { \ln ( 8 - 2 x ) } { \ln 4 }\). Use the iterative formula \(x _ { n + 1 } = \frac { \ln \left( 8 - 2 x _ { n } \right) } { \ln 4 }\) with \(x _ { 1 } = 1.2\) to find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.
      (2 marks)

1

1. Show that the equation \(x ^ { 3 } - 6 x + 1 = 0\) has a root \(\alpha\), where \(2 < \alpha < 3\).
2. Show that the equation \(x ^ { 3 } - 6 x + 1 = 0\) can be rearranged into the form $$x ^ { 2 } = 6 - \frac { 1 } { x }$$ (1 mark)
3. Use the recurrence relation \(x _ { n + 1 } = \sqrt { 6 - \frac { 1 } { x _ { n } } }\), with \(x _ { 1 } = 2.5\), to find the value of \(x _ { 3 }\), giving your answer to four significant figures.
   (2 marks)

   \includegraphics[max width=\textwidth, alt={}]{b8614dd6-2197-40c3-a673-5bef3e3653a5-2_142_116_2560_157}

   \(\_\_\_\_\) \(\_\_\_\_\) \(\_\_\_\_\)

7

1. Sketch the graph of \(y = \tan ^ { - 1 } x\).
2. 1. By drawing a suitable straight line on your sketch, show that the equation \(\tan ^ { - 1 } x = 2 x - 1\) has only one root.
   2. Given that the root of this equation is \(\alpha\), show that \(0.8 < \alpha < 0.9\).
3. Use the iteration \(x _ { n + 1 } = \frac { 1 } { 2 } \left( \tan ^ { - 1 } x _ { n } + 1 \right)\) with \(x _ { 1 } = 0.8\) to find the value of \(x _ { 3 }\), giving your answer to two significant figures.

1 The curve \(y = x ^ { 3 } - x - 7\) intersects the \(x\)-axis at the point where \(x = \alpha\).

1. Show that \(\alpha\) lies between 2.0 and 2.1.
2. Show that the equation \(x ^ { 3 } - x - 7 = 0\) can be rearranged in the form \(x = \sqrt [ 3 ] { x + 7 }\).
3. Use the iteration \(x _ { n + 1 } = \sqrt [ 3 ] { x _ { n } + 7 }\) with \(x _ { 1 } = 2\) to find the values of \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to three significant figures.