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

2 A curve satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \sin 2 x$$ where the angle \(2 x\) is measured in radians.
Starting at the point \(( 0.5,1 )\) on the curve, use a step-by-step method with a step length of 0.1 to estimate the value of \(y\) at \(x = 0.7\). Give your answer to three significant figures.
(6 marks)

2 A curve satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \log _ { 10 } x$$ Starting at the point \(( 2,3 )\) on the curve, use a step-by-step method with a step length of 0.2 to estimate the value of \(y\) at \(x = 2.4\). Give your answer to three decimal places.

4 Answer the whole of this question on the insert provided. The sketch shows the curve with equation \(y = \mathrm { F } ( x )\) and the line \(y = x\). The equation \(x = \mathrm { F } ( x )\) has roots \(x = \alpha\) and \(x = \beta\) as shown.

1. Use the copy of the sketch on the insert to show how an iteration of the form \(x _ { n + 1 } = \mathrm { F } \left( x _ { n } \right)\), with starting value \(x _ { 1 }\) such that \(0 < x _ { 1 } < \alpha\) as shown, converges to the root \(x = \alpha\).
2. State what happens in the iteration in the following two cases.
   1. \(x _ { 1 }\) is chosen such that \(\alpha < x _ { 1 } < \beta\).
   2. \(x _ { 1 }\) is chosen such that \(x _ { 1 } > \beta\). \section*{Jan 2006} 4
   3. \includegraphics[max width=\textwidth, alt={}, center]{0ec9c4ff-8622-4dda-a000-6ffe36f38023-03_873_1259_274_484}
   4. (a) \(\_\_\_\_\) (b) \(\_\_\_\_\) \section*{Jan 2006}

5

1. The function \(y ( x )\) satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$ where $$\mathrm { f } ( x , y ) = x \ln x + \frac { y } { x }$$ and $$y ( 1 ) = 1$$
   1. Use the Euler formula $$y _ { r + 1 } = y _ { r } + h \mathrm { f } \left( x _ { r } , y _ { r } \right)$$ with \(h = 0.1\), to obtain an approximation to \(y ( 1.1 )\).
   2. Use the formula $$y _ { r + 1 } = y _ { r - 1 } + 2 h \mathrm { f } \left( x _ { r } , y _ { r } \right)$$ with your answer to part (a)(i) to obtain an approximation to \(y ( 1.2 )\), giving your answer to three decimal places.
2. 1. Show that \(\frac { 1 } { x }\) is an integrating factor for the first-order differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } - \frac { 1 } { x } y = x \ln x$$
   2. Solve this differential equation, given that \(y = 1\) when \(x = 1\).
   3. Calculate the value of \(y\) when \(x = 1.2\), giving your answer to three decimal places.

1 The function \(y ( x )\) satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$ where $$\mathrm { f } ( x , y ) = \ln \left( 1 + x ^ { 2 } + y \right)$$ and $$y ( 1 ) = 0.6$$

1. Use the Euler formula $$y _ { r + 1 } = y _ { r } + h \mathrm { f } \left( x _ { r } , y _ { r } \right)$$ with \(h = 0.05\), to obtain an approximation to \(y ( 1.05 )\), giving your answer to four decimal places.
2. 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 \(h = 0.05\), to obtain an approximation to \(y ( 1.05 )\), giving your answer to four decimal places.

2 The function \(y ( x )\) satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$ where $$\mathrm { f } ( x , y ) = \sqrt { x ^ { 2 } + y ^ { 2 } + 3 }$$ and $$y ( 1 ) = 2$$

1. Use the Euler formula $$y _ { r + 1 } = y _ { r } + h \mathrm { f } \left( x _ { r } , y _ { r } \right)$$ with \(h = 0.1\), to obtain an approximation to \(y ( 1.1 )\), giving your answer to four decimal places.
2. 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 \(h = 0.1\), to obtain an approximation to \(y ( 1.1 )\), giving your answer to four decimal places.

11 The daily world production of oil can be modelled using $$V = 10 + 100 \left( \frac { t } { 30 } \right) ^ { 3 } - 50 \left( \frac { t } { 30 } \right) ^ { 4 }$$ where \(V\) is volume of oil in millions of barrels, and \(t\) is time in years since 1 January 1980. 11

1. 1. The model is used to predict the time, \(T\), when oil production will fall to zero.
      Show that \(T\) satisfies the equation $$T = \sqrt [ 3 ] { 60 T ^ { 2 } + \frac { 162000 } { T } }$$ 11
      1. (ii) Use the iterative formula \(T _ { n + 1 } = \sqrt [ 3 ] { 60 T _ { n } { } ^ { 2 } + \frac { 162000 } { T _ { n } } }\), with \(T _ { 0 } = 38\), to find the values of \(T _ { 1 } , T _ { 2 }\), and \(T _ { 3 }\), giving your answers to three decimal places.
         11
   2. (iii) Explain the relevance of using \(T _ { 0 } = 38\) 11
   3. From 1 January 1980 the daily use of oil by one technologically developing country can be modelled as $$V = 4.5 \times 1.063 ^ { t }$$ Use the models to show that the country's use of oil and the world production of oil will be equal during the year 2029.
      [0pt] [4 marks] \(12 \quad \mathrm { p } ( x ) = 30 x ^ { 3 } - 7 x ^ { 2 } - 7 x + 2\)

7 The equation \(x ^ { 2 } = x ^ { 3 } + x - 3\) has a single solution, \(x = \alpha\) 7

1. By considering a suitable change of sign, show that \(\alpha\) lies between 1.5 and 1.6
   [0pt] [2 marks]
   7
2. Show that the equation \(x ^ { 2 } = x ^ { 3 } + x - 3\) can be rearranged into the form $$x ^ { 2 } = x - 1 + \frac { 3 } { x }$$ 7
3. Use the iterative formula $$x _ { n + 1 } = \sqrt { x _ { n } - 1 + \frac { 3 } { x _ { n } } }$$ with \(x _ { 1 } = 1.5\), to find \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to four decimal places.
   7
4. Hence, deduce an interval of width 0.001 in which \(\alpha\) lies.

8. \begin{figure}[h] \end{figure} The heart rate of a horse is being monitored.
The heart rate \(H\), measured in beats per minute (bpm), is modelled by the equation $$H = 32 + 40 \mathrm { e } ^ { - 0.2 t } - 20 \mathrm { e } ^ { - 0.9 t }$$ where \(t\) minutes is the time after monitoring began.
Figure 4 is a sketch of \(H\) against \(t\). \section*{Use the equation of the model to answer parts (a) to (e).}

1. State the initial heart rate of the horse. In the long term, the heart rate of the horse approaches \(L \mathrm { bpm }\).
2. State the value of \(L\). The heart rate of the horse reaches its maximum value after \(T\) minutes.
3. Find the value of \(T\), giving your answer to 3 decimal places.
   (Solutions based entirely on calculator technology are not acceptable.) The heart rate of the horse is 37 bpm after \(M\) minutes.
4. Show that \(M\) is a solution of the equation $$t = 5 \ln \left( \frac { 8 } { 1 + 4 \mathrm { e } ^ { - 0.9 t } } \right)$$ Using the iteration formula $$t _ { n + 1 } = 5 \ln \left( \frac { 8 } { 1 + 4 \mathrm { e } ^ { - 0.9 t _ { n } } } \right) \quad \text { with } \quad t _ { 1 } = 10$$
5. 1. find, to 4 decimal places, the value of \(t _ { 2 }\)
   2. find, to 4 decimal places, the value of \(M\)

4

1. Show that the equation \(x ^ { 3 } - 6 x + 2 = 0\) has a root between \(x = 0\) and \(x = 1\).
2. Use the iterative formula \(x _ { n + 1 } = \frac { 2 + x _ { n } ^ { 3 } } { 6 }\) with \(x _ { 0 } = 0.5\) to find this root correct to 4 decimal places, showing the result of each iteration.

3

1. Given that \(X \sim \operatorname { Po } ( 5 )\), find \(\mathrm { P } ( X > 6 \mid X > 3 )\).
2. Given that \(Y \sim \operatorname { Po } ( \lambda )\) and \(\mathrm { P } ( Y \leqslant 1 ) = \frac { 1 } { 2 }\), show that \(\lambda\) satisfies the equation \(\lambda = \ln \{ 2 ( 1 + \lambda ) \}\).
3. Starting with a suitable approximation from the table of cumulative Poisson probabilities, use iteration to find \(\lambda\) correct to 3 decimal places.

3

1. Show that the equation \(x ^ { 2 } - \ln x - 2 = 0\) has a solution between \(x = 1\) and \(x = 2\).
2. Find an approximation to that solution using the iteration \(x _ { n + 1 } = \sqrt { 2 + \ln x _ { n } }\), giving your answer correct to 2 decimal places.

11 The cubic equation \(x ^ { 3 } - 2 x ^ { 2 } + 4 x - 7 = 0\) has a single root \(\alpha\), close to 1.9 , which can be found using an iteration of the form \(x _ { n + 1 } = \mathrm { F } \left( x _ { n } \right)\). Three possible functions that can be used for such an iteration are $$\mathrm { F } _ { 1 } ( x ) = \frac { 7 } { 4 } + \frac { 1 } { 2 } x ^ { 2 } - \frac { 1 } { 4 } x ^ { 3 } , \quad \mathrm {~F} _ { 2 } ( x ) = \sqrt [ 3 ] { 2 x ^ { 2 } - 4 x + 7 } , \quad \mathrm {~F} _ { 3 } ( x ) = \frac { 7 - 4 x } { x ^ { 2 } - 2 x }$$

1. Differentiate each of these functions with respect to \(x\).
2. Without performing any iterations, and using \(x = 1.9\), show that an iterative process based on only two of the given functions will converge. Determine which one will do so more rapidly. The sequence of errors, \(e _ { n }\), is such that \(e _ { n + 1 } \approx \mathrm {~F} ^ { \prime } ( \alpha ) e _ { n }\).
3. Using the iteration from part (ii) with the most rapid convergence, estimate the number of iterations required to reduce the magnitude of the error from \(\left| e _ { 1 } \right|\) in the first term to less than \(10 ^ { - 10 } \left| e _ { 1 } \right|\).

5

1. Show that the equation \(\sin x - x + 1 = 0\) has a root between 1.5 and 2 .
2. Use the iteration \(x _ { n + 1 } = 1 + \sin x _ { n }\), with a suitable starting value, to find that root correct to 2 decimal places.
3. Sketch the graphs of \(y = \sin x\) and \(y = x - 1\), on the same set of axes, for \(0 \leqslant x \leqslant \pi\).
4. Explain why the equation \(\sin x - x + 1 = 0\) has no root other than the one found in part (ii). [1]

6 \includegraphics[max width=\textwidth, alt={}, center]{f4b66aaa-16b9-4b15-b3f5-b9657fe98274-3_545_557_269_794} The diagram shows a sector \(A O B\) of a circle, centre \(O\) and radius \(r\). Angle \(A O B\) is \(\theta\) radians. The point \(C\) lies on \(O B\), and \(A C\) is perpendicular to \(O B\). The area of the triangle \(A O C\) is equal to the area of the segment bounded by the chord \(A B\) and the \(\operatorname { arc } A B\).

1. Show that \(\theta = \sin \theta ( 1 + \cos \theta )\). The equation \(\theta = \sin \theta ( 1 + \cos \theta )\) has only one positive root.
2. Use an iterative process based on this equation to find the value of the root correct to 3 significant figures. Use a starting value of 1 and show the result of each iteration. Use a change of sign to verify that the value you have found is correct to 3 significant figures.

9 The cubic polynomial \(x ^ { 3 } + a x ^ { 2 } + b x + c\), where \(a , b\) and \(c\) are real, is denoted by \(\mathrm { p } ( x )\).

1. Give a reason why the equation \(\mathrm { p } ( x ) = 0\) has at least one real root.
2. Suppose that the curve with equation \(y = \mathrm { p } ( x )\) has a local minimum point and a local maximum point with \(y\)-coordinates \(y _ { \text {min } }\) and \(y _ { \text {max } }\) respectively.
   1. Prove that if \(y _ { \text {min } } y _ { \text {max } } < 0\), then the equation \(\mathrm { p } ( x ) = 0\) has three real roots.
   2. Comment on the number of distinct real roots of the equation \(\mathrm { p } ( x ) = 0\) in the case \(y _ { \text {min } } y _ { \text {max } } = 0\).
   3. Suppose instead that the equation \(\mathrm { p } ( x ) = 0\) has only one real root for all values of \(c\). Prove that \(a ^ { 2 } \leqslant 3 b\).
   4. The iterative scheme $$x _ { n + 1 } = \frac { 2 x _ { n } ^ { 3 } + 1 } { 3 x _ { n } ^ { 2 } + 1 } , \quad x _ { 0 } = 0$$ converges to a root of the cubic equation \(\mathrm { p } ( x ) = 0\).
      (a) Find \(\mathrm { p } ( x )\).
      (b) Find the limit of the iteration, correct to 4 decimal places.
   5. Determine the rate of convergence of the iterative scheme.

\includegraphics{figure_6} The diagram shows the curve with equation \(y = \frac{\ln(2x + 1)}{x + 3}\). The curve has a maximum point M.

1. Find an expression for \(\frac{dy}{dx}\). [2]
2. Show that the x-coordinate of M satisfies the equation \(x = \frac{x + 3}{\ln(2x + 1)} - 0.5\). [2]
3. Show by calculation that the x-coordinate of M lies between 2.5 and 3.0. [2]
4. Use an iterative formula based on the equation in part (b) to find the x-coordinate of M correct to 4 significant figures. Give the result of each iteration to 6 significant figures. [3]

It is given that \(\int_1^a \left(\frac{4}{1 + 2x} + \frac{3}{x}\right) dx = \ln 10\), where \(a\) is a constant greater than 1.

1. Show that \(a = \sqrt{90(1 + 2a)^{-2}}\). [5]
2. Use an iterative formula, based on the equation in (a), to find the value of \(a\) correct to 3 significant figures. Use an initial value of 1.7 and give the result of each iteration to 5 significant figures. [3]

\includegraphics{figure_5} The diagram shows part of the curve with equation \(y = \frac{x^3}{x + 2}\). At the point \(P\), the gradient of the curve is 6.

1. Show that the \(x\)-coordinate of \(P\) satisfies the equation \(x = \sqrt[3]{12x + 12}\). [4]
2. Show by calculation that the \(x\)-coordinate of \(P\) lies between 3.8 and 4.0. [2]
3. Use an iterative formula, based on the equation in part (a), to find the \(x\)-coordinate of \(P\) correct to 3 significant figures. Show the result of each iteration to 5 significant figures. [3]

It is given that \(\int_a^{a^2} \frac{10}{2x+1} dx = 7\), where \(a\) is a constant greater than \(1\).

1. Show that \(a = \sqrt[9]{0.5e^{1.4}(2a+1) - 0.5}\). [5]
2. Use an iterative formula, based on the equation in part (a), to find the value of \(a\) correct to \(3\) significant figures. Use an initial value of \(2\) and give the result of each iteration to \(5\) significant figures. [3]

It is given that the positive constant \(a\) is such that $$\int_{-a}^a (4e^{2x} + 5) dx = 100.$$

1. Show that \(a = \frac{1}{4}\ln(50 + e^{-2a} - 5a)\). [4]
2. Use the iterative formula \(a_{n+1} = \frac{1}{4}\ln(50 + e^{-2a_n} - 5a_n)\) to find \(a\) correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3]

The sequence of values given by the iterative formula $$x_{n+1} = \frac{4}{x_n^2} + \frac{2x_n}{3},$$ with initial value \(x_1 = 2\), converges to \(\alpha\).

1. Use this iterative formula to find \(\alpha\) correct to 3 decimal places. Give the result of each iteration to 5 decimal places. [3]
2. State an equation that is satisfied by \(\alpha\), and hence find the exact value of \(\alpha\). [2]

\includegraphics{figure_4} The diagram shows the curve with equation $$y = x^4 + 2x^3 + 2x^2 - 12x - 32.$$ The curve crosses the \(x\)-axis at points with coordinates \((\alpha, 0)\) and \((\beta, 0)\).

1. Use the factor theorem to show that \((x + 2)\) is a factor of $$x^4 + 2x^3 + 2x^2 - 12x - 32.$$ [2]
2. Show that \(\beta\) satisfies an equation of the form \(x = \sqrt[3]{p + qx}\), and state the values of \(p\) and \(q\). [3]
3. Use an iterative formula based on the equation in part (ii) to find the value of \(\beta\) correct to 4 significant figures. Give the result of each iteration to 6 significant figures. [3]