1.09e Iterative method failure: convergence conditions

2 Let \(\mathrm { f } ( x ) = 2 x ^ { 3 } - 5 x ^ { 2 } + 4\).

1. Show that if a sequence of values given by the iterative formula $$x _ { n + 1 } = \sqrt { \frac { 4 } { 5 - 2 x _ { n } } }$$ converges, then it converges to a root of the equation \(\mathrm { f } ( x ) = 0\).
2. The equation has a root close to 1.2 . Use the iterative formula from part (a) and an initial value of 1.2 to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

9. $$\frac { \mathrm { d } y } { \mathrm {~d} x } = y \mathrm { e } ^ { x ^ { 2 } } .$$ It is given that \(y = 0.2\) at \(x = 0\).

1. Use the approximation \(\frac { y _ { 1 } - y _ { 0 } } { h } \approx \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) _ { 0 }\), with \(h = 0.1\), to obtain an estimate of the value of \(y\) at \(x = 0.1\).
2. Use your answer to part (a) and the approximation \(\frac { y _ { 2 } - y _ { 0 } } { 2 h } \approx \left( \frac { \mathrm {~d} y } { \mathrm {~d} x } \right) _ { 1 }\), with \(h = 0.1\), to obtain an estimate of the value of \(y\) at \(x = 0.2\). Gives your answer to 4 decimal places.
   (Total 5 marks)

8 The iteration \(x _ { n + 1 } = \frac { 1 } { \left( x _ { n } + 2 \right) ^ { 2 } }\), with \(x _ { 1 } = 0.3\), is to be used to find the real root, \(\alpha\), of the equation \(x ( x + 2 ) ^ { 2 } = 1\).

1. Find the value of \(\alpha\), correct to 4 decimal places. You should show the result of each step of the iteration process.
2. Given that \(\mathrm { f } ( x ) = \frac { 1 } { ( x + 2 ) ^ { 2 } }\), show that \(\mathrm { f } ^ { \prime } ( \alpha ) \neq 0\).
3. The difference, \(\delta _ { r }\), between successive approximations is given by \(\delta _ { r } = x _ { r + 1 } - x _ { r }\). Find \(\delta _ { 3 }\).
4. Given that \(\delta _ { r + 1 } \approx \mathrm { f } ^ { \prime } ( \alpha ) \delta _ { r }\), find an estimate for \(\delta _ { 10 }\).

6 It is given that \(\mathrm { f } ( x ) = 1 - \frac { 7 } { x ^ { 2 } }\).

1. Use the Newton-Raphson method, with a first approximation \(x _ { 1 } = 2.5\), to find the next approximations \(x _ { 2 }\) and \(x _ { 3 }\) to a root of \(\mathrm { f } ( x ) = 0\). Give the answers correct to 6 decimal places. [3]
2. The root of \(\mathrm { f } ( x ) = 0\) for which \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\) are approximations is denoted by \(\alpha\). Write down the exact value of \(\alpha\).
3. The error \(e _ { n }\) is defined by \(e _ { n } = \alpha - x _ { n }\). Find \(e _ { 1 } , e _ { 2 }\) and \(e _ { 3 }\), giving your answers correct to 5 decimal places. Verify that \(e _ { 3 } \approx \frac { e _ { 2 } ^ { 3 } } { e _ { 1 } ^ { 2 } }\).

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.

5 You are given that the equation \(x ^ { 3 } + 4 x ^ { 2 } + x - 1 = 0\) has a root, \(\alpha\), where \(- 1 < \alpha < 0\).

1. Show that the Newton-Raphson iterative formula for this equation can be written in the form $$x _ { n + 1 } = \frac { 2 x _ { n } ^ { 3 } + 4 x _ { n } ^ { 2 } + 1 } { 3 x _ { n } ^ { 2 } + 8 x _ { n } + 1 } .$$
2. Using the initial value \(x _ { 1 } = - 0.7\), find \(x _ { 2 }\) and \(x _ { 3 }\) and find \(\alpha\) correct to 5 decimal places.
3. The diagram shows a sketch of the curve \(y = x ^ { 3 } + 4 x ^ { 2 } + x - 1\) for \(- 1.5 \leqslant x \leqslant 1\). \includegraphics[max width=\textwidth, alt={}, center]{a80eb21f-c273-4b65-8617-16cdee783305-3_602_926_749_566} Using the copy of the diagram in your answer book, explain why the initial value \(x _ { 1 } = 0\) will fail to find \(\alpha\).

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}

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 \(\alpha\).

8 Kareem wants to solve the equation \(\sin 4 x + \mathrm { e } ^ { - x } + 0.75 = 0\). He uses his calculator to create the following table of values for \(\mathrm { f } ( x ) = \sin 4 x + \mathrm { e } ^ { - x } + 0.75\). He argues that because \(\mathrm { f } ( 6 )\) is the first negative value in the table, there is a root of the equation between 5 and 6 .

1. Comment on the validity of his argument. The diagram shows the graph of \(y = \sin 4 x + e ^ { - x } + 0.75\). \includegraphics[max width=\textwidth, alt={}, center]{4fac72cb-85cb-48d9-8817-899ef3f80a0f-07_538_1260_920_244}
2. Explain why Kareem failed to find other roots between 0 and 6 . Kareem decides to use the Newton-Raphson method to find the root close to 3 .
3. 1. Determine the iterative formula he should use for this equation.
   2. Use the Newton-Raphson method with \(x _ { 0 } = 3\) to find a root of the equation \(\mathrm { f } ( x ) = 0\). Show three iterations and give your answer to a suitable degree of accuracy. Kareem uses the Newton-Raphson method with \(x _ { 0 } = 5\) and also with \(x _ { 0 } = 6\) to try to find the root which lies between 5 and 6 . He produces the following tables.

      \(x _ { 0 }\)	5
      \(x _ { 1 }\)	3.97288
      \(x _ { 2 }\)	4.12125

      \(x _ { 0 }\)	6
      \(x _ { 1 }\)	6.09036
      \(x _ { 2 }\)	6.07110

4. 1. For the iteration beginning with \(x _ { 0 } = 5\), represent the process on the graph in the Printed Answer Booklet.
   2. Explain why the method has failed to find the root which lies between 5 and 6 .
   3. Explain how Kareem can adapt his method to find the root between 5 and 6 .

9 The equation \(x ^ { 3 } - x ^ { 2 } - 5 x + 10 = 0\) has exactly one real root \(\alpha\).

1. Show that the Newton-Raphson iterative formula for finding this root can be written as $$x _ { n + 1 } = \frac { 2 x _ { n } ^ { 3 } - x _ { n } ^ { 2 } - 10 } { 3 x _ { n } ^ { 2 } - 2 x _ { n } - 5 }$$
2. Apply the iterative formula in part (a) with initial value \(x _ { 1 } = - 3\) to find \(x _ { 2 } , x _ { 3 } , x _ { 4 }\) correct to 4 significant figures.
3. Use a change of sign method to show that \(\alpha = - 2.533\) is correct to 4 significant figures.
4. Explain why the Newton-Raphson method with initial value \(x _ { 1 } = - 1\) would not converge to \(\alpha\).

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 \(10 ^ { - 10 } \left| e _ { 1 } \right|\).

The equation \(x^3 = 3x + 7\) has one real root, denoted by \(\alpha\).

1. Show by calculation that \(\alpha\) lies between 2 and 3. [2]

Two iterative formulae, \(A\) and \(B\), derived from this equation are as follows: $$x_{n+1} = (3x_n + 7)^{\frac{1}{3}}, \quad (A)$$ $$x_{n+1} = \frac{x_n^3 - 7}{3}. \quad (B)$$ Each formula is used with initial value \(x_1 = 2.5\).

1. Show that one of these formulae produces a sequence which fails to converge, and use the other formula to calculate \(\alpha\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [4]

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]

\includegraphics{figure_5} The diagram shows the curve with equation \(y = f(x)\), where $$f(x) = 2x^3 - 9x^2 + 12x - 4.36.$$ The curve has turning points at \(x = 1\) and \(x = 2\) and crosses the \(x\)-axis at \(x = \alpha\), \(x = \beta\) and \(x = \gamma\), where \(0 < \alpha < \beta < \gamma\).

1. The Newton-Raphson method is to be used to find the roots of the equation \(f(x) = 0\), with \(x_1 = k\).
   1. To which root, if any, would successive approximations converge in each of the cases \(k < 0\) and \(k = 1\)? [2]
   2. What happens if \(1 < k < 2\)? [2]
2. Sketch the curve with equation \(y^2 = f(x)\). State the coordinates of the points where the curve crosses the \(x\)-axis and the coordinates of any turning points. [4]

A particle of mass 4 kg moves horizontally in a straight line. At time \(t\) seconds the velocity of the particle is \(v\) m s\(^{-1}\) The following horizontal forces act on the particle: • a constant driving force of magnitude 1.8 newtons • another driving force of magnitude \(30\sqrt{t}\) newtons • a resistive force of magnitude \(0.08v^2\) newtons When \(t = 70\), \(v = 54\) Use Euler's method with a step length of 0.5 to estimate the velocity of the particle after 71 seconds. Give your answer to four significant figures. [6 marks]

A function \(f\) with domain \((-\infty,\infty)\) is defined by \(f(x) = 6x^3 + 35x^2 - 7x - 6\).

1. Determine the number of roots of the equation \(f(x) = 0\) in the interval \([-1, 1]\). [2]
2. Use the Newton-Raphson method to find a root of the equation \(f(x) = 0\). Starting with \(x_0 = 1\),
   1. write down the value of \(x_1\),
   2. determine the value of the root correct to one decimal place. [4]
3. It is suggested that another iterative sequence $$x_{n+1} = \sqrt{\frac{7x_n + 6 - 6x_n^3}{35}},$$ starting with \(x_0 = -3\), could be used to find a root of the equation \(f(x) = 0\). Explain why this method fails. [2]

The function \(f\) is defined by $$f(x) = x^3 + 4x^2 - 3x - 1.$$

1. Show that the equation \(f(x) = 0\) has a root in the interval \([0, 1]\). [1]
2. Using the Newton-Raphson method with \(x_0 = 0 \cdot 8\),
   1. write down in full the decimal value of \(x_1\) as given in your calculator,
   2. determine the value of this root correct to six decimal places. [4]
3. Explain why the Newton-Raphson method does not work if \(x_0 = \frac{1}{3}\). [2]

The equation \(x^3 - 3x + 1 = 0\) has three real roots.

1. Show that one of the roots lies between \(-2\) and \(-1\) [2 marks]
2. Taking \(x_1 = -2\) as the first approximation to one of the roots, use the Newton-Raphson method to find \(x_2\), the second approximation. [3 marks]
3. Explain why the Newton-Raphson method fails in the case when the first approximation is \(x_1 = -1\) [1 mark]

Fig. 15 shows the graph of \(f(x) = 2x + \frac{1}{x} + \ln x - 4\). \includegraphics{figure_11}

1. Show that the equation $$2x + \frac{1}{x} + \ln x - 4 = 0$$ has a root, \(\alpha\), such that \(0.1 < \alpha < 0.9\). [2]
2. Obtain the following Newton-Raphson iteration for the equation in part (i). $$x_{r+1} = x_r - \frac{2x_r^3 + x_r + x_r^2(\ln x_r - 4)}{2x_r^2 - 1 + x_r}$$ [3]
3. Explain why this iteration fails to find \(\alpha\) using each of the following starting values.
   1. \(x_0 = 0.4\) [2]
   2. \(x_0 = 0.5\) [2]
   3. \(x_0 = 0.6\) [2]

The equation \(x^3 - 5x + 3 = 0\) has a root between \(x = 0\) and \(x = 1\).

1. The equation can be rearranged into the form \(x = g(x)\) where \(g(x) = px^3 + q\). State the values of \(p\) and \(q\). [1]
2. By considering \(|g'(x)|\), show that the iterative form \(x_{n+1} = g(x_n)\) with a suitable starting value converges to the root between \(x = 0\) and \(x = 1\). [You are not required to find this root.] [2]