Compare iteration convergence

A question is this type if and only if it asks to test two different iterative formulae and determine which converges and which fails.

5 questions · Standard +0.6

1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams
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CAIE P3 2005 November Q4
7 marks Standard +0.3
4 The equation \(x ^ { 3 } - x - 3 = 0\) has one real root, \(\alpha\).
  1. Show that \(\alpha\) lies between 1 and 2 . Two iterative formulae derived from this equation are as follows: $$\begin{aligned} & x _ { n + 1 } = x _ { n } ^ { 3 } - 3 \\ & x _ { n + 1 } = \left( x _ { n } + 3 \right) ^ { \frac { 1 } { 3 } } \end{aligned}$$ Each formula is used with initial value \(x _ { 1 } = 1.5\).
  2. 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.
Edexcel Paper 1 2018 June Q4
4 marks Standard +0.3
  1. The curve with equation \(y = 2 \ln ( 8 - x )\) meets the line \(y = x\) at a single point, \(x = \alpha\).
    1. Show that \(3 < \alpha < 4\)
    \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{b5f50f17-9f1b-4b4c-baf3-e50de5f2ea9c-08_666_1061_445_502} \captionsetup{labelformat=empty} \caption{Figure 2}
    \end{figure} Figure 2 shows the graph of \(y = 2 \ln ( 8 - x )\) and the graph of \(y = x\).
    A student uses the iteration formula $$x _ { n + 1 } = 2 \ln \left( 8 - x _ { n } \right) , \quad n \in \mathbb { N }$$ in an attempt to find an approximation for \(\alpha\).
    Using the graph and starting with \(x _ { 1 } = 4\)
  2. determine whether or not this iteration formula can be used to find an approximation for \(\alpha\), justifying your answer.
OCR MEI Further Numerical Methods 2022 June Q4
8 marks Standard +0.8
4 Fig. 4.1 shows part of the graph of \(y = e ^ { x } - x ^ { 2 } - x - 1.1\). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{f8c78ab8-7daa-4d6f-9bb5-093a46c590b8-04_805_789_299_274} \captionsetup{labelformat=empty} \caption{Fig. 4.1}
\end{figure} The equation \(\mathrm { e } ^ { x } - x ^ { 2 } - x - 1.1 = 0\) has a root \(\alpha\) such that \(1 < \alpha < 2\).
Ali is considering using the Newton-Raphson method to find \(\alpha\). Ali could use a starting value of \(x _ { 0 } = 1\) or a starting value of \(x _ { 0 } = 2\).
  1. Without doing any calculations, explain whether Ali should use a starting value of \(x _ { 0 } = 1\) or a starting value of \(x _ { 0 } = 2\), or whether using either starting value would work equally well. Ali is also considering using the method of fixed point iteration to find \(\alpha\). Ali could use a starting value of \(x _ { 0 } = 1\) or a starting value of \(x _ { 0 } = 2\). Fig. 4.2 shows parts of the graphs of \(y = x\) and \(y = \ln \left( x ^ { 2 } + x + 1.1 \right)\). \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{f8c78ab8-7daa-4d6f-9bb5-093a46c590b8-04_819_1011_1818_255} \captionsetup{labelformat=empty} \caption{Fig. 4.2}
    \end{figure}
  2. Without doing any calculations, explain whether Ali should use a starting value of \(x _ { 0 } = 1\) or a starting value of \(x _ { 0 } = 2\) or whether either starting value would work equally well. Ali used one of the above methods to find a sequence of approximations to \(\alpha\). These are shown, together with some further analysis in the associated spreadsheet output in Fig. 4.3. \begin{table}[h]
    MNO
    \(r\)\(\mathrm { X } _ { \mathrm { r } }\)
    402
    511.879008- 0.121
    621.858143- 0.0210.172
    731.857565\(- 6 \mathrm { E } - 04\)0.028
    841.857564\(- 4 \mathrm { E } - 07\)\(8 \mathrm { E } - 04\)
    951.857564\(- 2 \mathrm { E } - 13\)\(6 \mathrm { E } - 07\)
    \captionsetup{labelformat=empty} \caption{Fig. 4.3}
    \end{table} The formula in cell N5 is =M5-M4
    and the formula in cell O6 is =N6/N5
    equivalent formulae are in cells N6 to N9 and O7 to O9 respectively.
  3. State what is being calculated in the following columns of the spreadsheet.
    1. Column N
    2. Column O
  4. Explain whether the values in column O suggest that Ali used the Newton-Raphson method or the iterative formula \(\mathrm { x } _ { \mathrm { n } + 1 } = \ln \left( \mathrm { x } _ { \mathrm { n } } ^ { 2 } + \mathrm { x } _ { \mathrm { n } } + 1.1 \right)\) to find this sequence of approximations to \(\alpha\).
Pre-U Pre-U 9794/2 2014 June Q11
12 marks Challenging +1.2
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 \(10 ^ { - 10 } \left| e _ { 1 } \right|\).
CAIE P3 2017 November Q3
6 marks Standard +0.3
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]