1.09d Newton-Raphson method

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]

By drawing suitable graphs, show that \(x - 1 = \cos x\) has only one root. Starting with \(x_0 = 1\), use the Newton-Raphson method to find the value of this root correct to two decimal places. [6]

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 $$1 + 5x - x^4 = 0$$ has a positive root \(\alpha\).

1. Show that \(\alpha\) lies between 1 and 2. [2]
2. Use the iterative sequence based on the arrangement $$x = \sqrt[4]{1+5x}$$ with starting value 1.5 to find \(\alpha\) correct to two decimal places. [3]
3. Use the Newton-Raphson method to find \(\alpha\) correct to six decimal places. [6]

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]

You are given that \(f(x) = \ln(2x - 5) + 2x^2 - 30\), for \(x > 2.5\).

1. Show that \(f(x) = 0\) has a root \(\alpha\) in the interval \([3.5, 4]\). [2]

A student takes 4 as the first approximation to \(\alpha\). Given \(f(4) = 3.099\) and \(f'(4) = 16.67\) to 4 significant figures,

1. apply the Newton-Raphson procedure once to obtain a second approximation for \(\alpha\), giving your answer to 3 significant figures. [2]
2. Show that \(\alpha\) is the only root of \(f(x) = 0\). [2]

A curve \(C\) has equation \(y = f(x)\) where $$f(x) = x + 2\ln(e - x)$$

1. 1. Show that the equation of the normal to \(C\) at the point where \(C\) crosses the \(y\)-axis is given by $$y = \left(\frac{e}{2-e}\right)x + 2$$ [6 marks]
   2. Find the exact area enclosed by the normal and the coordinate axes. Fully justify your answer. [3 marks]

1. The equation \(f(x) = 0\) has one positive root, \(\alpha\).
   1. Show that \(\alpha\) lies between 2 and 3 Fully justify your answer. [3 marks]
   2. Show that the roots of \(f(x) = 0\) satisfy the equation $$x = e - e^{-\frac{x}{2}}$$ [2 marks]
   3. Use the recurrence relation $$x_{n+1} = e - e^{-\frac{x_n}{2}}$$ with \(x_1 = 2\) to find the values of \(x_2\) and \(x_3\) giving your answers to three decimal places. [2 marks]
   4. Figure 1 below shows a sketch of the graphs of \(y = e - e^{-\frac{x}{2}}\) 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 marks] \includegraphics{figure_1}

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]

A curve \(C\) has equation \(y = f(x)\) where $$f(x) = x + 2\ln(e - x)$$

1. 1. Show that the equation of the normal to \(C\) at the point where \(C\) crosses the \(y\)-axis is given by $$y = \left(\frac{e}{2-e}\right)x + 2$$ [6 marks]
   2. Find the exact area enclosed by the normal and the coordinate axes. Fully justify your answer. [3 marks]

1. The equation \(f(x) = 0\) has one positive root, \(\alpha\).
   1. Show that \(\alpha\) lies between 2 and 3 Fully justify your answer. [3 marks]
   2. Show that the roots of \(f(x) = 0\) satisfy the equation $$x = e - e^{\frac{x}{2}}$$ [2 marks]
   3. Use the recurrence relation $$x_{n+1} = e - e^{\frac{x_n}{2}}$$ with \(x_1 = 2\) to find the values of \(x_2\) and \(x_3\) giving your answers to three decimal places. [2 marks]
   4. Figure 1 below shows a sketch of the graphs of \(y = e - e^{\frac{x}{2}}\) 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 marks] \includegraphics{figure_9}

\includegraphics{figure_5} The functions f(x) and g(x) are defined for \(x \geqslant 0\) by \(\text{f}(x) = \frac{x}{x^2 + 3}\) and \(\text{g}(x) = \text{e}^{-2x}\). The diagram shows the curves \(y = \text{f}(x)\) and \(y = \text{g}(x)\). The equation \(\text{f}(x) = \text{g}(x)\) has exactly one real root \(\alpha\).

1. Show that \(\alpha\) satisfies the equation \(\text{h}(x) = 0\), where \(\text{h}(x) = x^2 + 3 - x\text{e}^{2x}\). [2]
2. Hence show that a Newton-Raphson iterative formula for finding \(\alpha\) can be written in the form $$x_{n+1} = \frac{x_n^2(1 - 2\text{e}^{2x_n}) - 3}{2x_n - (1 + 2x_n)\text{e}^{2x_n}}.$$ [5]
3. Use this iterative formula, with a suitable initial value, to find \(\alpha\) correct to 3 decimal places. Show the result of each iteration. [3]
4. In this question you must show detailed reasoning. Find the exact value of \(x\) for which \(\text{fg}(x) = \frac{2}{13}\). [6]

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

1. Show that the Newton-Raphson iterative formula for finding this root can be written as $$x_{n+1} = \frac{2x_n^3 - x_n^2 - 10}{3x_n^2 - 2x_n - 5}.$$ [3]
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. [1]
3. Use a change of sign method to show that \(\alpha = -2.533\) is correct to 4 significant figures. [3]
4. Explain why the Newton-Raphson method with initial value \(x_1 = -1\) would not converge to \(\alpha\). [2]

Let \(y = (x - 1)\left(\frac{2}{x^2} + t\right)\) define \(y\) as a function of \(x\) (\(x > 0\)), for each value of the real parameter \(t\).

1. When \(t = 0\),
   1. determine the set of values of \(x\) for which \(y\) is positive and an increasing function, [3]
   2. locate the stationary point of \(y\), and determine its nature. [2]
2. It is given that \(t = 2\) and \(y = -2\).
   1. Show that \(x\) satisfies \(f(x) = 0\), where \(f(x) = x^3 + x - 1\). [1]
   2. Prove that \(f\) has no stationary points. [2]
   3. Use the Newton-Raphson method, with \(x_0 = 1\), to find \(x\) correct to 4 significant figures. [4]

1. Use a change of sign to verify that the equation \(\cos x - x = 0\) has a root \(\alpha\) between \(x = 0.7\) and \(x = 0.8\). [2]
2. Sketch, on a single diagram, the curve \(y = \cos x\) and the line \(y = x\) for \(0 \leqslant x \leqslant \frac{1}{2}\pi\), giving the coordinates of all points of intersection with the coordinate axes. [2]

An iteration of the form \(x_{n+1} = \cos(x_n)\) is to be used to find \(\alpha\).

1. By considering the gradient of \(y = \cos x\), show that this iteration will converge. [3]
2. On a copy of your sketch from part (ii), illustrate how this iteration converges to \(\alpha\). [2]
3. Use a change of sign to verify that \(\alpha = 0.7391\) to 4 decimal places. [2]