Newton-Raphson convergence failure

A question is this type if and only if it asks to explain why Newton-Raphson will not converge or why a particular starting value cannot be used, typically involving analysis of f'(x) = 0 or geometric interpretation.

7 questions · Standard +0.5

1.09d Newton-Raphson method
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Edexcel F1 2023 January Q4
8 marks Challenging +1.2
4. $$f ( x ) = 1 - \frac { 1 } { 8 x ^ { 4 } } + \frac { 2 } { 7 \sqrt { x ^ { 7 } } } \quad x > 0$$ The equation \(\mathrm { f } ( x ) = 0\) has a single root, \(\alpha\), that lies in the interval \([ 0.15,0.25 ]\)
    1. Determine \(\mathrm { f } ^ { \prime } ( x )\)
    2. Explain why 0.25 cannot be used as an initial approximation for \(\alpha\) in the Newton-Raphson process.
    3. Taking 0.15 as a first approximation to \(\alpha\) apply the Newton-Raphson process once to \(\mathrm { f } ( x )\) to obtain a second approximation to \(\alpha\) Give your answer to 3 decimal places.
  2. Use linear interpolation once on the interval \([ 0.15,0.25 ]\) to find another approximation to \(\alpha\) Give your answer to 3 decimal places.
OCR FP2 2008 January Q5
9 marks Standard +0.8
5 \includegraphics[max width=\textwidth, alt={}, center]{15dd10f9-73d4-4107-bb45-7866f5470572-3_606_890_815_630} The diagram shows the curve with equation \(y = x \mathrm { e } ^ { - x } + 1\). The curve crosses the \(x\)-axis at \(x = \alpha\).
  1. Use differentiation to show that the \(x\)-coordinate of the stationary point is 1 . \(\alpha\) is to be found using the Newton-Raphson method, with \(\mathrm { f } ( x ) = x \mathrm { e } ^ { - x } + 1\).
  2. Explain why this method will not converge to \(\alpha\) if an initial approximation \(x _ { 1 }\) is chosen such that \(x _ { 1 } > 1\).
  3. Use this method, with a first approximation \(x _ { 1 } = 0\), to find the next three approximations \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\). Find \(\alpha\), correct to 3 decimal places.
AQA Paper 2 Specimen Q4
6 marks Standard +0.3
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]
WJEC Unit 3 2023 June Q4
8 marks Standard +0.3
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]
WJEC Unit 3 2024 June Q8
7 marks Standard +0.3
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]
SPS SPS FM Pure 2021 June Q2
6 marks Moderate -0.3
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]
SPS SPS FM Pure 2025 June Q11
11 marks Challenging +1.2
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]