1.09c Simple iterative methods: x_{n+1} = g(x_n), cobweb and staircase diagrams

7.

1. Sketch the curve with equation \(y = \ln x\).
2. Show that the tangent to the curve with equation \(y = \ln x\) at the point ( \(\mathrm { e } , 1\) ) passes through the origin.
3. Use your sketch to explain why the line \(y = m x\) cuts the curve \(y = \ln x\) between \(x = 1\) and \(x = \mathrm { e }\) if \(0 < m < \frac { 1 } { \mathrm { e } }\). Taking \(x _ { 0 } = 1.86\) and using the iteration \(x _ { n + 1 } = \mathrm { e } ^ { \frac { 1 } { 3 } x _ { n } }\),
4. calculate \(x _ { 1 } , x _ { 2 } , x _ { 3 } , x _ { 4 }\) and \(x _ { 5 }\), giving your answer to \(x _ { 5 }\) to 3 decimal places. The root of \(\ln x - \frac { 1 } { 3 } x = 0\) is \(\alpha\).
5. By considering the change of sign of \(\ln x - \frac { 1 } { 3 } x\) over a suitable interval, show that your answer for \(x _ { 5 }\) is an accurate estimate of \(\alpha\), correct to 3 decimal places.

   7. continued

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2. \(\quad \mathrm { f } ( x ) = x ^ { 3 } - 2 x - 5\).

1. Show that there is a root \(\alpha\) of \(\mathrm { f } ( x ) = 0\) for \(x\) in the interval \([ 2,3 ]\). The root \(\alpha\) is to be estimated using the iterative formula $$x _ { n + 1 } = \sqrt { \left( 2 + \frac { 5 } { x _ { n } } \right) } , \quad x _ { 0 } = 2$$
2. Calculate the values of \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to 4 significant figures.
3. Prove that, to 5 significant figures, \(\alpha\) is 2.0946.

1. In the interval \(2 < x < 3\), the equation

$$6 - x ^ { 2 } \cos \left( \frac { x } { 5 } \right) = 0 , \text { where } x \text { is measured in radians }$$ has exactly one root \(\alpha\).
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1. Starting with the interval [2,3], use interval bisection twice to find an interval of width 0.25 which contains \(\alpha\).
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2. Use linear interpolation once on the interval [2,3] to find an approximation to \(\alpha\). Give your answer to 2 decimal places.
   

6. $$f ( x ) = \tan \left( \frac { x } { 2 } \right) + 3 x - 6 , \quad - \pi < x < \pi$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval \([ 1,2 ]\).
2. Use linear interpolation once on the interval \([ 1,2 ]\) to find an approximation to \(\alpha\). Give your answer to 2 decimal places.

$$\mathrm { f } ( x ) = 2 \sin \left( x ^ { 2 } \right) + x - 2 , \quad 0 \leqslant x < 2 \pi$$

1. Show that \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) between \(x = 0.75\) and \(x = 0.85\) The equation \(\mathrm { f } ( x ) = 0\) can be written as \(x = [ \arcsin ( 1 - 0.5 x ) ] ^ { \frac { 1 } { 2 } }\).
2. Use the iterative formula $$x _ { n + 1 } = \left[ \arcsin \left( 1 - 0.5 x _ { n } \right) \right] ^ { \frac { 1 } { 2 } } , \quad x _ { 0 } = 0.8$$ to find the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\), giving your answers to 5 decimal places.
3. Show that \(\alpha = 0.80157\) is correct to 5 decimal places.

4

1. Sketch the graphs of \(y = 1 + \mathrm { e } ^ { 2 x }\) and \(y = | x - 4 |\) on the same diagram.
2. The two graphs meet at the point \(P\) .
   Show that the \(x\)-coordinate of \(P\) satisfies the equation \(x = \frac { 1 } { 2 } \ln ( 3 - x )\) . \includegraphics[max width=\textwidth, alt={}, center]{18aea465-b5b0-48f0-970a-e9ede1dc9370-06_2716_38_109_2012}
3. Use an iterative formula, based on the equation in part (b), to find the \(x\)-coordinate of \(P\) correct to 3 significant figures. Use an initial value of 0.45 and give the result of each iteration to 5 significant figures.

9. \(\quad f ( x ) = 3 - e ^ { 2 x } , \quad x \in \mathbb { R }\).

1. State the range of f .
2. Find the exact value of \(\mathrm { ff } ( 0 )\).
3. Define the inverse function \(\mathrm { f } ^ { - 1 } ( x )\) and state its domain. Given that \(\alpha\) is the solution of the equation \(\mathrm { f } ( x ) = \mathrm { f } ^ { - 1 } ( x )\),
4. explain why \(\alpha\) satisfies the equation $$x = \mathrm { f } ^ { - 1 } ( x )$$
5. use the iterative formula $$x _ { n + 1 } = \mathrm { f } ^ { - 1 } \left( x _ { n } \right)$$ with \(x _ { 0 } = 0.5\) to find \(\alpha\) correct to 3 significant figures.

8. The curve \(C\) has the equation \(y = \sqrt { x } + \mathrm { e } ^ { 1 - 4 x } , x \geq 0\).

1. Find an equation for the normal to the curve at the point \(\left( \frac { 1 } { 4 } , \frac { 3 } { 2 } \right)\). The curve \(C\) has a stationary point with \(x\)-coordinate \(\alpha\) where \(0.5 < \alpha < 1\).
2. Show that \(\alpha\) is a solution of the equation $$x = \frac { 1 } { 4 } [ 1 + \ln ( 8 \sqrt { x } ) ]$$
3. Use the iterative formula $$x _ { n + 1 } = \frac { 1 } { 4 } \left[ 1 + \ln \left( 8 \sqrt { x _ { n } } \right) \right]$$ with \(x _ { 0 } = 1\) to find \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving the value of \(x _ { 4 }\) to 3 decimal places.
4. Show that your value for \(x _ { 4 }\) is the value of \(\alpha\) correct to 3 decimal places.
5. Another attempt to find \(\alpha\) is made using the iterative formula $$x _ { n + 1 } = \frac { 1 } { 64 } \mathrm { e } ^ { 8 x _ { n } - 2 }$$ with \(x _ { 0 } = 1\). Describe the outcome of this attempt.

A curve has the equation \(y = x ^ { 2 } - \sqrt { 4 + \ln x }\).

1. Show that the tangent to the curve at the point where \(x = 1\) has the equation $$7 x - 4 y = 11$$ The curve has a stationary point with \(x\)-coordinate \(\alpha\).
2. Show that \(0.3 < \alpha < 0.4\)
3. Show that \(\alpha\) is a solution of the equation $$x = \frac { 1 } { 2 } ( 4 + \ln x ) ^ { - \frac { 1 } { 4 } }$$
4. Use the iterative formula $$x _ { n + 1 } = \frac { 1 } { 2 } \left( 4 + \ln x _ { n } \right) ^ { - \frac { 1 } { 4 } }$$ with \(x _ { 0 } = 0.35\), to find \(\alpha\) correct to 5 decimal places.
   You should show the result of each iteration.

8. The population in thousands, \(P\), of a town at time \(t\) years after \(1 ^ { \text {st } }\) January 1980 is modelled by the formula $$P = 30 + 50 \mathrm { e } ^ { 0.002 t }$$ Use this model to estimate

1. the population of the town on \(1 ^ { \text {st } }\) January 2010,
2. the year in which the population first exceeds 84000 . The population in thousands, \(Q\), of another town is modelled by the formula $$Q = 26 + 50 \mathrm { e } ^ { 0.003 t }$$
3. Show that the value of \(t\) when \(P = Q\) is a solution of the equation $$t = 1000 \ln \left( 1 + 0.08 \mathrm { e } ^ { - 0.002 t } \right)$$
4. Use the iterative formula $$t _ { n + 1 } = 1000 \ln \left( 1 + 0.08 \mathrm { e } ^ { - 0.002 t _ { n } } \right)$$ with \(t _ { 0 } = 50\) to find \(t _ { 1 } , t _ { 2 }\) and \(t _ { 3 }\) and hence, the year in which the populations of these two towns will be equal according to these models.

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

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 iterative formula $$x _ { n + 1 } = 2 - \mathrm { e } ^ { k x _ { n } ^ { 2 } }$$ with \(x _ { 0 } = 1.9\) and your value of \(k\), to find \(\alpha\) correct to 3 decimal places.
   You should show the result of each iteration.
3. Solve the equation \(\mathrm { f } ^ { \prime } ( x ) = 0\).

1. The functions \(f\) and \(g\) are defined by

$$\begin{aligned} & \mathrm { f } : x \rightarrow | 2 x - 5 | , \quad x \in \mathbb { R } , \\ & \mathrm {~g} : x \rightarrow \ln ( x + 3 ) , \quad x \in \mathbb { R } , \quad x > - 3 \end{aligned}$$

1. State the range of f .
2. Evaluate fg(-2).
3. Solve the equation $$\operatorname { fg } ( x ) = 3$$ giving your answers in exact form.
4. Show that the equation $$\mathrm { f } ( x ) = \mathrm { g } ( x )$$ has a root, \(\alpha\), in the interval [3,4].
5. Use the iterative formula $$x _ { n + 1 } = \frac { 1 } { 2 } \left[ 5 + \ln \left( x _ { n } + 3 \right) \right]$$ with \(x _ { 0 } = 3\), to find \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to 4 significant figures.
6. Show that your answer for \(x _ { 4 }\) is the value of \(\alpha\) correct to 4 significant figures.

8.

1. Solve the equation $$\pi - 3 \cos ^ { - 1 } \theta = 0$$
2. Sketch on the same diagram the curves \(y = \cos ^ { - 1 } ( x - 1 ) , 0 \leq x \leq 2\) and \(y = \sqrt { x + 2 } , x \geq - 2\). Given that \(\alpha\) is the root of the equation $$\cos ^ { - 1 } ( x - 1 ) = \sqrt { x + 2 }$$
3. show that \(0 < \alpha < 1\),
4. 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.
   You should show the result of each iteration.

7 \includegraphics[max width=\textwidth, alt={}, center]{d858728a-3371-4755-880c-54f96c5e5156-3_465_748_1133_717} The diagram shows the curve with equation \(y = \cos ^ { - 1 } x\).

1. Sketch the curve with equation \(y = 3 \cos ^ { - 1 } ( x - 1 )\), showing the coordinates of the points where the curve meets the axes.
2. By drawing an appropriate straight line on your sketch in part (i), show that the equation \(3 \cos ^ { - 1 } ( x - 1 ) = x\) has exactly one root.
3. Show by calculation that the root of the equation \(3 \cos ^ { - 1 } ( x - 1 ) = x\) lies between 1.8 and 1.9 .
4. The sequence defined by $$x _ { 1 } = 2 , \quad x _ { n + 1 } = 1 + \cos \left( \frac { 1 } { 3 } x _ { n } \right)$$ converges to a number \(\alpha\). Find the value of \(\alpha\) correct to 2 decimal places and explain why \(\alpha\) is the root of the equation \(3 \cos ^ { - 1 } ( x - 1 ) = x\).

3

1. It is given that \(a\) and \(b\) are positive constants. By sketching graphs of $$y = x ^ { 5 } \quad \text { and } \quad y = a - b x$$ on the same diagram, show that the equation $$x ^ { 5 } + b x - a = 0$$ has exactly one real root.
2. Use the iterative formula \(x _ { n + 1 } = \sqrt [ 5 ] { 53 - 2 x _ { n } }\), with a suitable starting value, to find the real root of the equation \(x ^ { 5 } + 2 x - 53 = 0\). Show the result of each iteration, and give the root correct to 3 decimal places.

2 The sequence defined by $$x _ { 1 } = 3 , \quad x _ { n + 1 } = \sqrt [ 3 ] { 31 - \frac { 5 } { 2 } x _ { n } }$$ converges to the number \(\alpha\).

1. Find the value of \(\alpha\) correct to 3 decimal places, showing the result of each iteration.
2. Find an equation of the form \(a x ^ { 3 } + b x + c = 0\), where \(a\), \(b\) and \(c\) are integers, which has \(\alpha\) as a root.

8 \includegraphics[max width=\textwidth, alt={}, center]{e0e2a26b-d4d6-46ea-ac12-a882f3465e5e-3_588_915_954_614} The diagram shows part of each of the curves \(y = e ^ { \frac { 1 } { 5 } x }\) and \(y = \sqrt [ 3 ] { } ( 3 x + 8 )\). The curves meet, as shown in the diagram, at the point \(P\). The region \(R\), shaded in the diagram, is bounded by the two curves and by the \(y\)-axis.

1. Show by calculation that the \(x\)-coordinate of \(P\) lies between 5.2 and 5.3.
2. Show that the \(x\)-coordinate of \(P\) satisfies the equation \(x = \frac { 5 } { 3 } \ln ( 3 x + 8 )\).
3. Use an iterative formula, based on the equation in part (ii), to find the \(x\)-coordinate of \(P\) correct to 2 decimal places.
4. Use integration, and your answer to part (iii), to find an approximate value of the area of the region \(R\). \includegraphics[max width=\textwidth, alt={}, center]{e0e2a26b-d4d6-46ea-ac12-a882f3465e5e-4_625_647_264_749} The function f is defined by \(\mathrm { f } ( x ) = \sqrt { } ( m x + 7 ) - 4\), where \(x \geqslant - \frac { 7 } { m }\) and \(m\) is a positive constant. The diagram shows the curve \(y = \mathrm { f } ( x )\).

1. A sequence of transformations maps the curve \(y = \sqrt { } x\) to the curve \(y = \mathrm { f } ( x )\). Give details of these transformations.
2. Explain how you can tell that f is a one-one function and find an expression for \(\mathrm { f } ^ { - 1 } ( x )\).
3. It is given that the curves \(y = \mathrm { f } ( x )\) and \(y = \mathrm { f } ^ { - 1 } ( x )\) do not meet. Explain how it can be deduced that neither curve meets the line \(y = x\), and hence determine the set of possible values of \(m\). [5]

3 The equation \(2 x ^ { 3 } + 4 x - 35 = 0\) has one real root.

1. Show by calculation that this real root lies between 2 and 3 .
2. Use the iterative formula $$x _ { n + 1 } = \sqrt [ 3 ] { 17.5 - 2 x _ { n } }$$ with a suitable starting value, to find the real root of the equation \(2 x ^ { 3 } + 4 x - 35 = 0\) correct to 2 decimal places. You should show the result of each iteration.

6

1. Given that \(\int _ { 0 } ^ { \mathrm { a } } \left( 6 \mathrm { e } ^ { 2 \mathrm { x } } + \mathrm { x } \right) \mathrm { dx } = 42\), show that \(\mathrm { a } = \frac { 1 } { 2 } \ln \left( 15 - \frac { 1 } { 6 } \mathrm { a } ^ { 2 } \right)\).
2. Use an iterative formula, based on the equation in part (i), to find the value of a correct to 3 decimal places. Use a starting value of 1 and show the result of each iteration.

4 The gradient of the curve \(y = \left( 2 x ^ { 2 } + 9 \right) ^ { \frac { 5 } { 2 } }\) at the point \(P\) is 100 .

1. Show that the \(x\)-coordinate of \(P\) satisfies the equation \(x = 10 \left( 2 x ^ { 2 } + 9 \right) ^ { - \frac { 3 } { 2 } }\).
2. Show by calculation that the \(x\)-coordinate of \(P\) lies between 0.3 and 0.4 .
3. Use an iterative formula, based on the equation in part (i), to find the \(x\)-coordinate of \(P\) correct to 4 decimal places. You should show the result of each iteration.

3 The sequence defined by the iterative formula $$x _ { n + 1 } = \sqrt [ 3 ] { } \left( 17 - 5 x _ { n } \right)$$ with \(x _ { 1 } = 2\), converges to \(\alpha\).

1. Use the iterative formula to find \(\alpha\) correct to 2 decimal places. You should show the result of each iteration.
2. Find a cubic equation of the form $$x ^ { 3 } + c x + d = 0$$ which has \(\alpha\) as a root.
3. Does this cubic equation have any other real roots? Justify your answer.

A curve has the equation \(y = \frac { \mathrm { e } ^ { 2 } } { x } + \mathrm { e } ^ { x } , x \neq 0\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\).
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2. Show that the curve has a stationary point in the interval [1.3,1.4]. The point \(A\) on the curve has \(x\)-coordinate 2 .
3. Show that the tangent to the curve at \(A\) passes through the origin. The tangent to the curve at \(A\) intersects the curve again at the point \(B\).
   The \(x\)-coordinate of \(B\) is to be estimated using the iterative formula $$x _ { n + 1 } = - \frac { 2 } { 3 } \sqrt { 3 + 3 x _ { n } \mathrm { e } ^ { x _ { n } - 2 } }$$ with \(x _ { 0 } = - 1\).
4. Find \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\) to 7 significant figures and hence state the \(x\)-coordinate of \(B\) to 5 significant figures.

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 \(\delta _ { 10 }\).

4

1. Sketch, on the same diagram, the curves with equations \(y = \operatorname { sech } x\) and \(y = x ^ { 2 }\).
2. By using the definition of \(\operatorname { sech } x\) in terms of \(\mathrm { e } ^ { x }\) and \(\mathrm { e } ^ { - x }\), show that the \(x\)-coordinates of the points at which these curves meet are solutions of the equation $$x ^ { 2 } = \frac { 2 \mathrm { e } ^ { x } } { \mathrm { e } ^ { 2 x } + 1 } .$$
3. The iteration $$x _ { n + 1 } = \sqrt { \frac { 2 \mathrm { e } ^ { x _ { n } } } { \mathrm { e } ^ { 2 x _ { n } } + 1 } }$$ can be used to find the positive root of the equation in part (ii). With initial value \(x _ { 1 } = 1\), the approximations \(x _ { 2 } = 0.8050 , x _ { 3 } = 0.8633 , x _ { 4 } = 0.8463\) and \(x _ { 5 } = 0.8513\) are obtained, correct to 4 decimal places. State with a reason whether, in this case, the iteration produces a 'staircase' or a 'cobweb' diagram.

3 It is given that \(\mathrm { F } ( x ) = 2 + \ln x\). The iteration \(x _ { n + 1 } = \mathrm { F } \left( x _ { n } \right)\) is to be used to find a root, \(\alpha\), of the equation \(x = 2 + \ln x\).

1. Taking \(x _ { 1 } = 3.1\), find \(x _ { 2 }\) and \(x _ { 3 }\), giving your answers correct to 5 decimal places.
2. The error \(e _ { n }\) is defined by \(e _ { n } = \alpha - x _ { n }\). Given that \(\alpha = 3.14619\), correct to 5 decimal places, use the values of \(e _ { 2 }\) and \(e _ { 3 }\) to make an estimate of \(\mathrm { F } ^ { \prime } ( \alpha )\) correct to 3 decimal places. State the true value of \(\mathrm { F } ^ { \prime } ( \alpha )\) correct to 4 decimal places.
3. Illustrate the iteration by drawing a sketch of \(y = x\) and \(y = \mathrm { F } ( x )\), showing how the values of \(x _ { n }\) approach \(\alpha\). State whether the convergence is of the 'staircase' or 'cobweb' type.