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

4 You are given the equation \(( 2 x - 1 ) ^ { 2 } - \mathrm { e } ^ { x } = 0\).

1. Verify that 0 is a root of the equation. There are also two other roots, \(\alpha\) and \(\beta\), where \(0 < \alpha < \beta\).
2. The iterative formula \(x _ { r + 1 } = \ln \left( 2 x _ { r } - 1 \right) ^ { 2 }\) is to be used to find a root of the equation.
   1. Sketch the line \(y = x\) and the curve \(y = \ln ( 2 x - 1 ) ^ { 2 }\) on the same axes, showing the roots \(0 , \alpha\) and \(\beta\).
   2. By drawing a 'staircase' diagram on your sketch, starting with a value of \(x\) that is between \(\alpha\) and \(\beta\), show that this iteration does not converge to \(\alpha\).
   3. Using this iterative formula with \(x _ { 1 } = 3.75\), find the value of \(\beta\) correct to 3 decimal places.
   4. Using the Newton-Raphson method with \(x _ { 1 } = 1.6\), find the root \(\alpha\) of the equation \(( 2 x - 1 ) ^ { 2 } - \mathrm { e } ^ { x } = 0\) correct to 5 significant figures. Show the result of each iteration.

10.

1. Given that \(- \frac { \pi } { 2 } < \mathrm { g } ( x ) < \frac { \pi } { 2 }\), sketch the graph of \(y = \mathrm { g } ( x )\) where $$\mathrm { g } ( x ) = \arctan x , \quad x \in \mathbb { R }$$
2. Find the exact value of \(x\) for which $$3 g ( x + 1 ) - \pi = 0$$ The equation \(\arctan x - 4 + \frac { 1 } { 2 } x = 0\) has a positive root at \(x = \alpha\) radians.
3. Show that \(5 < \alpha < 6\) The iteration formula $$x _ { n + 1 } = 8 - 2 \arctan x _ { n }$$ can be used to find an approximation for \(\alpha\)
4. Taking \(x _ { 0 } = 5\), use this formula to find \(x _ { 1 }\) and \(x _ { 2 }\), giving each answer to 3 decimal places.

4 The parametric equations of a curve are $$x = \mathrm { e } ^ { 2 t - 3 } , \quad y = 4 \ln t$$ where \(t > 0\). When \(t = a\) the gradient of the curve is 2 .

1. Show that \(a\) satisfies the equation \(a = \frac { 1 } { 2 } ( 3 - \ln a )\).
2. Verify by calculation that this equation has a root between 1 and 2 .
3. Use the iterative formula \(a _ { n + 1 } = \frac { 1 } { 2 } \left( 3 - \ln a _ { n } \right)\) to calculate \(a\) correct to 2 decimal places, showing the result of each iteration to 4 decimal places.
   

6 \includegraphics[max width=\textwidth, alt={}, center]{c940af95-e291-402a-856c-9090d13163d5-3_627_689_264_726} The function f is defined for all real values of \(x\) by $$f ( x ) = \sqrt [ 3 ] { \frac { 1 } { 2 } x + 2 }$$ The graphs of \(y = \mathrm { f } ( x )\) and \(y = \mathrm { f } ^ { - 1 } ( x )\) meet at the point \(P\), and the graph of \(y = \mathrm { f } ^ { - 1 } ( x )\) meets the \(x\)-axis at \(Q\) (see diagram).

1. Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\) and determine the \(x\)-coordinate of the point \(Q\).
2. State how the graphs of \(y = \mathrm { f } ( x )\) and \(y = \mathrm { f } ^ { - 1 } ( x )\) are related geometrically, and hence show that the \(x\)-coordinate of the point \(P\) is the root of the equation $$x = \sqrt [ 3 ] { \frac { 1 } { 2 } x + 2 }$$
3. Use an iterative process, based on the equation \(x = \sqrt [ 3 ] { \frac { 1 } { 2 } x + 2 }\), to find the \(x\)-coordinate of \(P\), giving your answer correct to 2 decimal places.

6 The curve with equation \(y = \frac { 3 x + 4 } { x ^ { 3 } - 4 x ^ { 2 } + 2 }\) has a stationary point at \(P\). It is given that \(P\) is close to the point with coordinates \(( 2.4 , - 1.6 )\).

1. Find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and show that the \(x\)-coordinate of \(P\) satisfies the equation $$x = \sqrt [ 3 ] { \frac { 16 } { 3 } x + 1 }$$
2. By first using an iterative process based on the equation in part (i), find the coordinates of \(P\), giving each coordinate correct to 3 decimal places.

6 \includegraphics[max width=\textwidth, alt={}, center]{89e54367-bb83-483a-add5-0527b71a5cac-4_476_709_251_683} The diagram shows the curve with equation \(x = \ln \left( y ^ { 3 } + 2 y \right)\). At the point \(P\) on the curve, the gradient is 4 and it is given that \(P\) is close to the point with coordinates (7.5,12).

1. Find \(\frac { \mathrm { d } x } { \mathrm {~d} y }\) in terms of \(y\).
2. Show that the \(y\)-coordinate of \(P\) satisfies the equation $$y = \frac { 12 y ^ { 2 } + 8 } { y ^ { 2 } + 2 }$$
3. By first using an iterative process based on the equation in part (ii), find the coordinates of \(P\), giving each coordinate correct to 3 decimal places.

4

1. Show by means of suitable sketch graphs that the equation $$( x - 2 ) ^ { 4 } = x + 16$$ has exactly 2 real roots.
2. State the value of the smaller root.
3. Use the iterative formula $$x _ { n + 1 } = 2 + \sqrt [ 4 ] { x _ { n } + 16 }$$ with a suitable starting value, to find the larger root correct to 3 decimal places.

5

1. It is given that \(k\) is a positive constant. By sketching the graphs of $$y = 14 - x ^ { 2 } \text { and } y = k \ln x$$ on a single diagram, show that the equation $$14 - x ^ { 2 } = k \ln x$$ has exactly one real root.
2. The real root of the equation \(14 - x ^ { 2 } = 3 \ln x\) is denoted by \(\alpha\).
   1. Find by calculation the pair of consecutive integers between which \(\alpha\) lies.
   2. Use the iterative formula \(x _ { n + 1 } = \sqrt { 14 - 3 \ln x _ { n } }\), with a suitable starting value, to find \(\alpha\). Show the result of each iteration, and give \(\alpha\) correct to 2 decimal places.

7 \includegraphics[max width=\textwidth, alt={}, center]{71e01d8f-d0ed-4f17-b7cd-6f5a93bbe329-3_428_751_703_641} The diagram shows the curve \(y = \mathrm { f } ( x )\), where f is the function defined for all real values of \(x\) by $$\mathrm { f } ( x ) = 3 + 4 \mathrm { e } ^ { - x }$$

1. State the range of f .
2. Find an expression for \(\mathrm { f } ^ { - 1 } ( x )\), and state the domain and range of \(\mathrm { f } ^ { - 1 }\).
3. The straight line \(y = x\) meets the curve \(y = \mathrm { f } ( x )\) at the point \(P\). By using an iterative process based on the equation \(x = \mathrm { f } ( x )\), with a starting value of 3 , find the coordinates of the point \(P\). Show all your working and give each coordinate correct to 3 decimal places.
4. How is the point \(P\) related to the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { f } ^ { - 1 } ( x )\) ?

6 \includegraphics[max width=\textwidth, alt={}, center]{33a2b09d-0df9-48d6-9ee9-e0a1ec345f41-3_524_720_246_676} The diagram shows the curve \(y = x ^ { 4 } - 8 x\).

1. By sketching a second curve on the copy of the diagram, show that the equation $$x ^ { 4 } + x ^ { 2 } - 8 x - 9 = 0$$ has two real roots. State the equation of the second curve.
2. The larger root of the equation \(x ^ { 4 } + x ^ { 2 } - 8 x - 9 = 0\) is denoted by \(\alpha\).
   1. Show by calculation that \(2.1 < \alpha < 2.2\).
   2. Use an iterative process based on the equation $$x = \sqrt [ 4 ] { 9 + 8 x - x ^ { 2 } } ,$$ with a suitable starting value, to find \(\alpha\) correct to 3 decimal places. Give the result of each step of the iterative process.

6 \includegraphics[max width=\textwidth, alt={}, center]{00a4be37-c095-4d9c-a1cd-d03b8ab1d411-3_553_579_274_726} The diagram shows the curve \(y = 8 \sin ^ { - 1 } \left( x - \frac { 3 } { 2 } \right)\). The end-points \(A\) and \(B\) of the curve have coordinates ( \(a , - 4 \pi\) ) and ( \(b , 4 \pi\) ) respectively.

1. State the values of \(a\) and \(b\).
2. It is required to find the root of the equation \(8 \sin ^ { - 1 } \left( x - \frac { 3 } { 2 } \right) = x\).
   1. Show by calculation that the root lies between 1.7 and 1.8.
   2. In order to find the root, the iterative formula $$x _ { n + 1 } = p + \sin \left( q x _ { n } \right) ,$$ with a suitable starting value, is to be used. Determine the values of the constants \(p\) and \(q\) and hence find the root correct to 4 significant figures. Show the result of each step of the iteration process.

7

1. By sketching the curves \(y = x ( 2 x + 5 )\) and \(y = \cos ^ { - 1 } x\) (where \(y\) is in radians) in a single diagram, show that the equation \(x ( 2 x + 5 ) = \cos ^ { - 1 } x\) has exactly one real root.
2. Use the iterative formula $$x _ { n + 1 } = \frac { \cos ^ { - 1 } x _ { n } } { 2 x _ { n } + 5 } \text { with } x _ { 1 } = 0.25$$ to find the root correct to 3 significant figures. Show the result of each iteration correct to at least 4 significant figures.
3. Two new curves are obtained by transforming each of the curves \(y = x ( 2 x + 5 )\) and \(y = \cos ^ { - 1 } x\) by the pair of transformations:
   reflection in the \(x\)-axis followed by reflection in the \(y\)-axis.
   State an equation of each of the new curves and determine the coordinates of their point of intersection, giving each coordinate correct to 3 significant figures.

8 It is required to solve the equation \(\ln ( x - 1 ) - x + 3 = 0\).
You are given that there are two roots, \(\alpha\) and \(\beta\), where \(1.1 < \alpha < 1.2\) and \(4.1 < \beta < 4.2\).

1. The root \(\beta\) can be found using the iterative formula $$x _ { n + 1 } = \ln \left( x _ { n } - 1 \right) + 3$$
   1. Using this iterative formula with \(x _ { 1 } = 4.15\), find \(\beta\) correct to 3 decimal places. Show all your working.
   2. Explain with the aid of a sketch why this iterative formula will not converge to \(\alpha\) whatever initial value is taken.
   3. (a) Show that the Newton-Raphson iterative formula for this equation can be written in the form $$x _ { n + 1 } = \frac { 3 - 2 x _ { n } - \left( x _ { n } - 1 \right) \ln \left( x _ { n } - 1 \right) } { 2 - x _ { n } }$$ (b) Use this formula with \(x _ { 1 } = 1.2\) to find \(\alpha\) correct to 3 decimal places.

7 \includegraphics[max width=\textwidth, alt={}, center]{074597e7-5bb1-4249-9cfa-784974a6fd2b-3_531_1065_1208_539} The line \(y = x\) and the curve \(y = 2 \ln ( 3 x - 2 )\) meet where \(x = \alpha\) and \(x = \beta\), as shown in the diagram.

1. Use the iteration \(x _ { n + 1 } = 2 \ln \left( 3 x _ { n } - 2 \right)\), with initial value \(x _ { 1 } = 5.25\), to find the value of \(\beta\) correct to 2 decimal places. Show all your working.
2. With the help of a 'staircase' diagram, explain why this iteration will not converge to \(\alpha\), whatever value of \(x _ { 1 }\) (other than \(\alpha\) ) is used.
3. Show that the equation \(x = 2 \ln ( 3 x - 2 )\) can be rewritten as \(x = \frac { 1 } { 3 } \left( \mathrm { e } ^ { \frac { 1 } { 2 } x } + 2 \right)\). Use the NewtonRaphson method, with \(\mathrm { f } ( x ) = \frac { 1 } { 3 } \left( \mathrm { e } ^ { \frac { 1 } { 2 } x } + 2 \right) - x\) and \(x _ { 1 } = 1.2\), to find \(\alpha\) correct to 2 decimal places. Show all your working.
4. Given that \(x _ { 1 } = \ln 36\), explain why the Newton-Raphson method would not converge to a root of \(\mathrm { f } ( x ) = 0\).

4 It is given that the equation \(x ^ { 4 } - 2 x - 1 = 0\) has only one positive root, \(\alpha\), and \(1.3 < \alpha < 1.5\).

1. \includegraphics[max width=\textwidth, alt={}, center]{72a1330a-c6dc-4f3a-9b0e-333b099f4509-2_433_424_1119_817} The diagram shows a sketch of \(y = x\) and \(y = \sqrt [ 4 ] { 2 x + 1 }\) for \(x \geqslant 0\). Use the iteration \(x _ { n + 1 } = \sqrt [ 4 ] { 2 x _ { n } + 1 }\) with \(x _ { 1 } = 1.35\) to find \(x _ { 2 }\) and \(x _ { 3 }\), correct to 4 decimal places. On the copy of the diagram show how the iteration converges to \(\alpha\).
2. For the same equation, the iteration \(x _ { n + 1 } = \frac { 1 } { 2 } \left( x _ { n } ^ { 4 } - 1 \right)\) with \(x _ { 1 } = 1.35\) gives \(x _ { 2 } = 1.1608\) and \(x _ { 3 } = 0.4077\), correct to 4 decimal places. Draw a sketch of \(y = x\) and \(y = \frac { 1 } { 2 } \left( x ^ { 4 } - 1 \right)\) for \(x \geqslant 0\), and show how this iteration does not converge to \(\alpha\).
3. Find the positive root of the equation \(x ^ { 4 } - 2 x - 1 = 0\) by using the Newton-Raphson method with \(x _ { 1 } = 1.35\), giving the root correct to 4 decimal places.

9 The equation \(10 x - 8 \ln x = 28\) has a root \(\alpha\) in the interval [3,4]. The iteration \(x _ { n + 1 } = \mathrm { g } \left( x _ { n } \right)\), where \(\mathrm { g } ( x ) = 2.8 + 0.8 \ln x\) and \(x _ { 1 } = 3.8\), is to be used to find \(\alpha\).

1. Find the value of \(\alpha\) correct to 5 decimal places. You should show the result of each step of the iteration to 6 decimal places.
2. Illustrate this iteration by means of a sketch.
3. The difference, \(\delta _ { r }\), between successive approximations is given by \(\delta _ { r } = x _ { r + 1 } - x _ { r }\). Find \(\delta _ { 3 }\).
4. Given that \(\delta _ { n + 1 } \approx \mathrm {~g} ^ { \prime } ( \alpha ) \delta _ { n }\), for all positive integers \(n\), estimate the smallest value of \(n\) such that \(\delta _ { n } < 10 ^ { - 6 } \delta _ { 1 }\). \section*{OCR}

6 It is given that the equation \(3 x ^ { 3 } + 5 x ^ { 2 } - x - 1 = 0\) has three roots, one of which is positive.

1. Show that the Newton-Raphson iterative formula for finding this root can be written $$x _ { n + 1 } = \frac { 6 x _ { n } ^ { 3 } + 5 x _ { n } ^ { 2 } + 1 } { 9 x _ { n } ^ { 2 } + 10 x _ { n } - 1 } .$$
2. A sequence of iterates \(x _ { 1 } , x _ { 2 } , x _ { 3 } , \ldots\) which will find the positive root is such that the magnitude of the error in \(x _ { 2 }\) is greater than the magnitude of the error in \(x _ { 1 }\). On the graph given in the Printed Answer Book, mark a possible position for \(x _ { 1 }\).
3. Apply the iterative formula in part (i) when the initial value is \(x _ { 1 } = - 1\). Describe the behaviour of the iterative sequence, illustrating your answer on the graph given in the Printed Answer Book.
4. A sequence of approximations to the positive root is given by \(x _ { 1 } , x _ { 2 } , x _ { 3 } , \ldots\). Successive differences \(x _ { r } - x _ { r - 1 } = d _ { r }\), where \(r \geqslant 2\), are such that \(d _ { r } \approx k \left( d _ { r - 1 } \right) ^ { 2 }\) where \(k\) is a constant. Show that \(d _ { 4 } \approx \frac { d _ { 3 } ^ { 3 } } { d _ { 2 } ^ { 2 } }\) and demonstrate this numerically when \(x _ { 1 } = 1\).
5. Find the value of the positive root correct to 5 decimal places.

3 Complete this table to show the next 3 values of the iteration $$x _ { n + 1 } = k x _ { n } \left( 1 - x _ { n } \right)$$ in the case when \(k = 3.2\) and \(x _ { 0 } = 0.5\). Give your answers to calculator accuracy.

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

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 \(y ( 2.1 )\).
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), to obtain an approximation to \(y ( 2.2 )\).

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

1. Use the Euler formula $$y _ { r + 1 } = y _ { r } + h \mathrm { f } \left( x _ { r } , y _ { r } \right)$$ with \(h = 0.2\), to obtain an approximation to \(y ( 1.2 )\).
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 \(h = 0.2\), to obtain an approximation to \(y ( 1.2 )\), giving your answer to four decimal places.

1 It is given that \(y ( x )\) satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$ where $$\mathrm { f } ( x , y ) = \sqrt { 2 x + y }$$ and $$y ( 3 ) = 5$$

1. Use the Euler formula $$y _ { r + 1 } = y _ { r } + h \mathrm { f } \left( x _ { r } , y _ { r } \right)$$ with \(h = 0.2\), to obtain an approximation to \(y ( 3.2 )\), giving your answer to four decimal places.
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), to obtain an approximation to \(y ( 3.4 )\), giving your answer to three decimal places.

2 The function \(y ( x )\) satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$ where $$\mathrm { f } ( x , y ) = \frac { x ^ { 2 } + y ^ { 2 } } { x y }$$ 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 \(y ( 1.1 )\).
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.

1 The function \(y ( x )\) satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$ where $$\mathrm { f } ( x , y ) = \ln ( x + y )$$ and $$y ( 2 ) = 3$$ 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 \(y ( 2.1 )\), giving your answer to four decimal places.
(6 marks)

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

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 \(y ( 3.1 )\), giving your answer to four decimal places.
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), to obtain an approximation to \(y ( 3.2 )\), 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 $$\mathrm { f } ( x , y ) = x + 3 + \sin y$$ 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 \(y ( 1.1 )\), giving your answer to four decimal places.
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), to obtain an approximation to \(y ( 1.2 )\), giving your answer to three decimal places.