Apply iteration to find root

7

1. Sketch the graph of \(y = \tan ^ { - 1 } x\).
2. 1. By drawing a suitable straight line on your sketch, show that the equation \(\tan ^ { - 1 } x = 2 x - 1\) has only one root.
   2. Given that the root of this equation is \(\alpha\), show that \(0.8 < \alpha < 0.9\).
3. Use the iteration \(x _ { n + 1 } = \frac { 1 } { 2 } \left( \tan ^ { - 1 } x _ { n } + 1 \right)\) with \(x _ { 1 } = 0.8\) to find the value of \(x _ { 3 }\), giving your answer to two significant figures.

1 The curve \(y = x ^ { 3 } - x - 7\) intersects the \(x\)-axis at the point where \(x = \alpha\).

1. Show that \(\alpha\) lies between 2.0 and 2.1.
2. Show that the equation \(x ^ { 3 } - x - 7 = 0\) can be rearranged in the form \(x = \sqrt [ 3 ] { x + 7 }\).
3. Use the iteration \(x _ { n + 1 } = \sqrt [ 3 ] { x _ { n } + 7 }\) with \(x _ { 1 } = 2\) to find the values of \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to three significant figures.

1

1. The curve with equation $$y = \frac { \cos x } { 2 x + 1 } , \quad x > - \frac { 1 } { 2 }$$ intersects the line \(y = \frac { 1 } { 2 }\) at the point where \(x = \alpha\).
   1. Show that \(\alpha\) lies between 0 and \(\frac { \pi } { 2 }\).
   2. Show that the equation \(\frac { \cos x } { 2 x + 1 } = \frac { 1 } { 2 }\) can be rearranged into the form $$x = \cos x - \frac { 1 } { 2 }$$
   3. Use the iteration \(x _ { n + 1 } = \cos x _ { n } - \frac { 1 } { 2 }\) with \(x _ { 1 } = 0\) to find \(x _ { 3 }\), giving your answer to three decimal places.
2. 1. Given that \(y = \frac { \cos x } { 2 x + 1 }\), use the quotient rule to find an expression for \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
   2. Hence find the gradient of the normal to the curve \(y = \frac { \cos x } { 2 x + 1 }\) at the point on the curve where \(x = 0\).

1 The curve \(y = 3 ^ { x }\) intersects the curve \(y = 10 - x ^ { 3 }\) at the point where \(x = \alpha\).

1. Show that \(\alpha\) lies between 1 and 2 .
2. 1. Show that the equation \(3 ^ { x } = 10 - x ^ { 3 }\) can be rearranged into the form $$x = \sqrt [ 3 ] { 10 - 3 ^ { x } }$$
   2. Use the iteration \(x _ { n + 1 } = \sqrt [ 3 ] { 10 - 3 ^ { x _ { n } } }\) with \(x _ { 1 } = 1\) to find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.

3 The curve \(y = \cos ^ { - 1 } ( 2 x - 1 )\) intersects the curve \(y = \mathrm { e } ^ { x }\) at a single point where \(x = \alpha\).

1. Show that \(\alpha\) lies between 0.4 and 0.5 .
2. Show that the equation \(\cos ^ { - 1 } ( 2 x - 1 ) = \mathrm { e } ^ { x }\) can be written as \(x = \frac { 1 } { 2 } + \frac { 1 } { 2 } \cos \left( \mathrm { e } ^ { x } \right)\).
3. Use the iteration \(x _ { n + 1 } = \frac { 1 } { 2 } + \frac { 1 } { 2 } \cos \left( \mathrm { e } ^ { x _ { n } } \right)\) with \(x _ { 1 } = 0.4\) to find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.

2 For \(0 < x \leqslant 2\), the curves with equations \(y = 4 \ln x\) and \(y = \sqrt { x }\) intersect at a single point where \(x = \alpha\).

1. Show that \(\alpha\) lies between 0.5 and 1.5.
2. Show that the equation \(4 \ln x = \sqrt { x }\) can be rearranged into the form $$x = \mathrm { e } ^ { \left( \frac { \sqrt { x } } { 4 } \right) }$$
3. Use the iterative formula $$x _ { n + 1 } = \mathrm { e } ^ { \left( \frac { \sqrt { x _ { n } } } { 4 } \right) }$$ with \(x _ { 1 } = 0.5\) to find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.
4. Figure 1, on the page 3, shows a sketch of parts of the graphs of \(y = \mathrm { e } ^ { \left( \frac { \sqrt { x } } { 4 } \right) }\) 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. \begin{figure}[h]
   \captionsetup{labelformat=empty} \caption{Figure 1} \includegraphics[alt={},max width=\textwidth]{d3c66c34-b09c-4223-8383-cf0a68419bf9-3_1285_1543_356_296}
   \end{figure}

3

1. The equation \(\mathrm { e } ^ { - x } - 2 + \sqrt { x } = 0\) has a single root, \(\alpha\).
   Show that \(\alpha\) lies between 3 and 4 .
2. Use the recurrence relation \(x _ { n + 1 } = \left( 2 - e ^ { - x _ { n } } \right) ^ { 2 }\), with \(x _ { 1 } = 3.5\), to find \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.
3. The diagram below shows parts of the graphs of \(y = \left( 2 - \mathrm { e } ^ { - x } \right) ^ { 2 }\) and \(y = x\), and a position of \(x _ { 1 }\). On the diagram, draw a staircase or cobweb diagram to show how convergence takes place, indicating the positions of \(x _ { 2 }\) and \(x _ { 3 }\) on the \(x\)-axis. \includegraphics[max width=\textwidth, alt={}, center]{063bbfa5-df49-44a1-8143-5e076397f63f-03_1100_1402_881_367}

2 A curve has equation \(y = 2 \ln ( 2 \mathrm { e } - x )\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
2. Find an equation of the normal to the curve \(y = 2 \ln ( 2 \mathrm { e } - x )\) at the point on the curve where \(x = \mathrm { e }\).
   [0pt] [4 marks]
3. The curve \(y = 2 \ln ( 2 \mathrm { e } - x )\) intersects the line \(y = x\) at a single point, where \(x = \alpha\).
   1. Show that \(\alpha\) lies between 1 and 3 .
   2. Use the recurrence relation $$x _ { n + 1 } = 2 \ln \left( 2 \mathrm { e } - x _ { n } \right)$$ with \(x _ { 1 } = 1\) to find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.
   3. Figure 1, on the opposite page, shows a sketch of parts of the graphs of \(y = 2 \ln ( 2 \mathrm { e } - x )\) 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.
      [0pt] [2 marks] \section*{(c)(iii)} \begin{figure}[h]
      \captionsetup{labelformat=empty} \caption{Figure 1} \includegraphics[alt={},max width=\textwidth]{57412ec0-ad97-4418-8ba8-93f1f7d8aac1-05_864_1284_1802_386}
      \end{figure}

2 The curve with equation \(y = x ^ { x }\), where \(x > 0\), intersects the line \(y = 5\) at a single point, where \(x = \alpha\).

1. Show that \(\alpha\) lies between 2 and 3 .
2. Show that the equation \(x ^ { x } = 5\) can be rearranged into the form $$x = \mathrm { e } ^ { \left( \frac { \ln 5 } { x } \right) }$$
3. Use the iterative formula $$x _ { n + 1 } = \mathrm { e } ^ { \left( \frac { \ln 5 } { x _ { n } } \right) }$$ with \(x _ { 1 } = 2\) to find the values of \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers to three decimal places.
4. 1. Use Simpson's rule with 7 ordinates ( 6 strips) to find an approximation to $$\int _ { 0.5 } ^ { 1.7 } \left( 5 - x ^ { x } \right) \mathrm { d } x$$ giving your answer to three significant figures.
   2. Hence find an approximation to \(\int _ { 0.5 } ^ { 1.7 } x ^ { x } \mathrm {~d} x\).

1. The root of the equation \(\mathrm { f } ( x ) = 0\), where

$$f ( x ) = x + \ln 2 x - 4$$ is to be estimated using the iterative formula \(x _ { n + 1 } = 4 - \ln 2 x _ { n }\), with \(x _ { 0 } = 2.4\).

1. Showing your values of \(x _ { 1 } , x _ { 2 } , x _ { 3 } , \ldots\), obtain the value, to 3 decimal places, of the root.
2. By considering the change of sign of \(\mathrm { f } ( x )\) in a suitable interval, justify the accuracy of your answer to part (a).

4. $$\mathrm { f } ( x ) = x ^ { 3 } + x ^ { 2 } - 4 x - 1$$ The equation \(\mathrm { f } ( x ) = 0\) has only one positive root, \(\alpha\).

1. Show that \(\mathrm { f } ( x ) = 0\) can be rearranged as $$x = \sqrt { \left( \frac { 4 x + 1 } { x + 1 } \right) } , x \neq - 1$$ The iterative formula \(x _ { n + 1 } = \sqrt { \left( \frac { 4 x _ { n } + 1 } { x _ { n } + 1 } \right) }\) is used to find an approximation to \(\alpha\).
2. Taking \(x _ { 1 } = 1\), find, to 2 decimal places, the values of \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\).
3. By choosing values of \(x\) in a suitable interval, prove that \(\alpha = 1.70\), correct to 2 decimal places.
4. Write down a value of \(x _ { 1 }\) for which the iteration formula \(x _ { n + 1 } = \sqrt { \left( \frac { 4 x _ { n } + 1 } { x _ { n } + 1 } \right) }\) does not produce a valid value for \(x _ { 2 }\). Justify your answer.

8. The curve with equation \(y = \ln 3 x\) crosses the \(x\)-axis at the point \(P ( p , 0 )\).

1. Sketch the graph of \(y = \ln 3 x\), showing the exact value of \(p\). The normal to the curve at the point \(Q\), with \(x\)-coordinate \(q\), passes through the origin.
2. Show that \(x = q\) is a solution of the equation \(x ^ { 2 } + \ln 3 x = 0\).
3. Show that the equation in part (b) can be rearranged in the form \(x = \frac { 1 } { 3 } \mathrm { e } ^ { - x ^ { 2 } }\).
4. Use the iteration formula \(x _ { n + 1 } = \frac { 1 } { 3 } \mathrm { e } ^ { - x _ { n } ^ { 2 } }\), with \(x _ { 0 } = \frac { 1 } { 3 }\), to find \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\). Hence write down, to 3 decimal places, an approximation for \(q\).

7. (a) Solve the equation $$\pi - 3 \arccos \theta = 0$$ (b) Sketch on the same diagram the curves \(y = \arccos ( x - 1 ) , 0 \leq x \leq 2\) and \(y = \sqrt { x + 2 } , x \geq - 2\). Given that \(\alpha\) is the root of the equation $$\arccos ( x - 1 ) = \sqrt { x + 2 }$$ (c) show that \(0 < \alpha < 1\),
(d) use the iterative formula $$x _ { n + 1 } = 1 + \cos \sqrt { x _ { n } + 2 }$$ with \(x _ { 0 } = 1\) to find \(\alpha\) correct to 3 decimal places. END

6. $$f ( x ) = 2 x ^ { 2 } + 3 \ln ( 2 - x ) , \quad x \in \mathbb { R } , \quad x < 2 .$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) can be written in the form $$x = 2 - \mathrm { e } ^ { k x ^ { 2 } } ,$$ where \(k\) is a constant to be found. The root, \(\alpha\), of the equation \(\mathrm { f } ( x ) = 0\) is 1.9 correct to 1 decimal place.
2. Use the iteration formula $$x _ { n + 1 } = 2 - \mathrm { e } ^ { k x _ { n } ^ { 2 } } ,$$ with \(x _ { 0 } = 1.9\) and your value of \(k\), to find \(\alpha\) to 3 decimal places and justify the accuracy of your answer.
3. Solve the equation \(\mathrm { f } ^ { \prime } ( x ) = 0\).

3 The equation \(\mathrm { f } ( x ) = \sin ^ { - 1 } ( x ) - x + 0.1 = 0\) has a root \(\alpha\) such that \(- 1 < \alpha < 0\).
Alex uses an iterative method to find a sequence of approximations to \(\alpha\). Some of the associated spreadsheet output is shown in the table. The formula in cell D7 is $$= ( \mathrm { D } 5 * \mathrm { E } 6 - \mathrm { D } 6 * \mathrm { E } 5 ) / ( \mathrm { E } 6 - \mathrm { E } 5 )$$ and equivalent formulae are in cells D8 and D9.

1. State the method being used.
2. Use the values in the spreadsheet to calculate \(x _ { 5 }\) and \(x _ { 6 }\), giving your answers correct to 7 decimal places.
3. State the value of \(\alpha\) as accurately as you can, justifying the precision quoted. Alex uses a calculator to check the value in cell D9, his result is - 0.7540832686 .
4. Explain why this is different to the value displayed in cell D9. The value displayed in cell E11 in Alex's spreadsheet is \(- 1.4629 \mathrm { E } - 09\).
5. Write this value in standard mathematical notation.

6 Charlie uses fixed point iteration to find a sequence of approximations to the root of the equation \(\sin ^ { - 1 } ( x ) - x ^ { 2 } + 1 = 0\). Charlie uses the iterative formula \(\mathrm { x } _ { \mathrm { n } + 1 } = \mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right)\), where \(\mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right) = \sin \left( \mathrm { x } _ { \mathrm { n } } ^ { 2 } - 1 \right)\).
Two sections of the associated spreadsheet output, showing \(x _ { 0 }\) to \(x _ { 6 }\) and \(x _ { 102 }\) to \(x _ { 108 }\), are shown in Fig. 6.1. \begin{table}[h] \end{table}

1. Use the information in Fig. 6.1 to find the value of the root as accurately as you can, justifying the precision quoted. The relaxed iteration \(\mathrm { x } _ { \mathrm { n } + 1 } = ( 1 - \lambda ) \mathrm { x } _ { \mathrm { n } } + \lambda \mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right)\), with \(\lambda = 0.51\) and \(x _ { 0 } = 0\), is to be used to find the root of the equation \(\sin ^ { - 1 } ( x ) - x ^ { 2 } + 1 = 0\).
2. Complete the copy of Fig. 6.2 in the Printed Answer Booklet, giving the values of \(\mathrm { x } _ { \mathrm { r } }\) correct to 7 decimal places and the values in the difference column and ratio column correct to 3 significant figures. \begin{table}[h]

   \(r\)	\(\mathrm { x } _ { \mathrm { r } }\)	difference	ratio
   0	0
   1
   2
   3
   4		-0.000192
   5		\(- 1.99 \times 10 ^ { - 7 }\)	0.00103
   6		\(- 1.82 \times 10 ^ { - 10 }\)	0.000914

   \captionsetup{labelformat=empty} \caption{Fig. 6.2}
   \end{table}
3. Write down the value of the root correct to 7 decimal places.
4. Explain why extrapolation could not be used in this case to find an improved approximation using this sequence of iterates. In this case the method of relaxation has been used to speed up the convergence of an iterative scheme.
5. Name another application of the method of relaxation.

5 The equation \(3 - 2 \ln x - x = 0\) has a root near \(x = 1.8\).
A student proposes to use the iterative formula \(\mathrm { x } _ { \mathrm { n } + 1 } = \mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right) = 3 - 2 \ln \mathrm { x } _ { \mathrm { n } }\) to find this root.
The diagram shows the graphs of \(\mathrm { y } = \mathrm { x }\) and \(\mathrm { y } = \mathrm { g } ( \mathrm { x } )\) for values of \(x\) from - 2 to 6 . \includegraphics[max width=\textwidth, alt={}, center]{4023e87c-34b1-4abd-9acc-ede5e4d68c7f-05_913_917_502_233}

1. With reference to the graph, explain why it might not be possible to use the student's iterative formula to find the root near \(x = 1.8\).
2. Use the relaxed iteration \(\mathrm { x } _ { \mathrm { n } + 1 } = \lambda \mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right) + ( 1 - \lambda ) \mathrm { x } _ { \mathrm { n } }\), with \(\lambda = 0.475\) and \(x _ { 0 } = 2\), to determine the root correct to \(\mathbf { 6 }\) decimal places. A student uses the same relaxed iteration with the same starting value. Some analysis of the iterates is carried out using a spreadsheet, which is shown in the table below.

   \(r\)	difference	ratio
   0
   1	- 0.1834898
   2	- 0.0049137	0.02678
   3	\(- 6.44 \mathrm { E } - 06\)	0.00131
   4	\(- 3.862 \mathrm { E } - 09\)	0.0006
   5	\(- 2.313 \mathrm { E } - 12\)	0.0006

3. Explain what the analysis tells you about the order of convergence of this sequence of approximations.

3 The equation \(x ^ { 2 } - \cosh ( x - 2 ) = 0\) has two roots, \(\alpha\) and \(\beta\), such that \(\alpha < \beta\).

1. Use the iterative formula $$x _ { n + 1 } = g \left( x _ { n } \right) \text { where } g \left( x _ { n } \right) = \sqrt { \cosh \left( x _ { n } - 2 \right) } \text {, }$$ starting with \(x _ { 0 } = 1\), to find \(\alpha\) correct to \(\mathbf { 3 }\) decimal places. The diagram shows the part of the graphs of \(\mathrm { y } = \mathrm { x }\) and \(\mathrm { y } = \mathrm { g } ( \mathrm { x } )\) for \(0 \leqslant x \leqslant 7\). \includegraphics[max width=\textwidth, alt={}, center]{83a06341-74e9-4f47-9104-e8e0259e7dfa-3_760_657_753_246}
2. Explain why the iterative formula used to find \(\alpha\) cannot successfully be used to find \(\beta\), even if \(x _ { 0 }\) is very close to \(\beta\).
3. Use the relaxed iteration $$\mathrm { x } _ { \mathrm { n } + 1 } = ( 1 - \lambda ) \mathrm { x } _ { \mathrm { n } } + \lambda \mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right) ,$$ with \(\lambda = - 0.21\) and \(x _ { 0 } = 6.4\), to find \(\beta\) correct to \(\mathbf { 3 }\) decimal places. In part (c) the method of relaxation was used to convert a divergent sequence of approximations into a convergent sequence.
4. State one other application of the method of relaxation.

6 The equation \(0.5 \ln x - x ^ { 2 } + x + 1 = 0\) has two roots \(\alpha\) and \(\beta\), such that \(0 < \alpha < 1\) and \(1 < \beta < 2\).

1. Use the Newton-Raphson method with \(x _ { 0 } = 1\) to obtain \(\beta\) correct to \(\mathbf { 6 }\) decimal places. Fig. 6.1 shows part of the graph of \(y = 0.5 \ln x - x ^ { 2 } + x + 1\). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{945883ad-c153-4c51-83d3-978e4c769ed5-06_1112_1156_529_354} \captionsetup{labelformat=empty} \caption{Fig. 6.1}
   \end{figure}
2. On the copy of Fig. 6.1 in the Printed Answer Booklet, illustrate the Newton-Raphson method working to obtain \(x _ { 1 }\) from \(x _ { 0 } = 1\). Beth is trying to find \(\alpha\) correct to 6 decimal places.
3. Suggest a reason why she might choose the Newton-Raphson method instead of fixed point iteration. Beth tries to find \(\alpha\) using the Newton-Raphson method with a starting value of \(x _ { 0 } = 0.5\). Her spreadsheet output is shown in Fig. 6.2. \begin{table}[h]

   \(r\)	\(\mathrm { x } _ { \mathrm { r } }\)
   0	0.5
   1	- 0.40343
   2	\#NUM!

   \captionsetup{labelformat=empty} \caption{Fig. 6.2}
   \end{table}
4. Explain how the display \#NUM! has arisen in the cell for \(x _ { 2 }\). Beth decides to use the iterative formula $$x _ { n + 1 } = g \left( x _ { n } \right) = \sqrt { 0.5 \ln \left( x _ { n } \right) + x _ { n } + 1 }$$
5. Determine the outcome when Beth uses this formula with \(x _ { 0 } = 0.5\).
6. Use the relaxed iteration \(\mathrm { x } _ { \mathrm { n } + 1 } = ( 1 - \lambda ) \mathrm { x } _ { \mathrm { n } } + \lambda \mathrm { g } \left( \mathrm { x } _ { \mathrm { n } } \right)\) with \(\lambda = - 0.041\) and \(x _ { 0 } = 0.5\) to obtain \(\alpha\) correct to \(\mathbf { 6 }\) decimal places.

3 The equation \(\sinh x + x ^ { 2 } - 1 = 0\) has a root, \(\alpha\), such that \(0 < \alpha < 1\).

1. Verify that the iteration \(x _ { r + 1 } = \frac { 1 - \sinh x _ { r } } { x _ { r } }\) with \(x _ { 0 } = 1\) fails to converge to this root.
2. Use the relaxed iteration \(x _ { r + 1 } = ( 1 - \lambda ) x _ { r } + \lambda \left( \frac { 1 - \sinh x _ { r } } { x _ { r } } \right)\) with \(\lambda = \frac { 1 } { 4 }\) and \(x _ { 0 } = 1\) to find \(\alpha\) correct to 6 decimal places.

10. The equation $$1 + 5 x - x ^ { 4 } = 0$$ has a positive root \(\alpha\).

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