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

5 \includegraphics[max width=\textwidth, alt={}, center]{0af2714b-d3eb-4112-a869-eda5cf266cd8-08_410_977_274_543} The diagram shows the curve \(y = \frac { \sin 2 x } { x + 2 }\) for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\). The \(x\)-coordinate of the maximum point \(M\) is denoted by \(\alpha\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and show that \(\alpha\) satisfies the equation \(\tan 2 x = 2 x + 4\).
2. Show by calculation that \(\alpha\) lies between 0.6 and 0.7 .
3. Use the iterative formula \(x _ { n + 1 } = \frac { 1 } { 2 } \tan ^ { - 1 } \left( 2 x _ { n } + 4 \right)\) to find the value of \(\alpha\) correct to 3 decimal places. Give the result of each iteration to 5 decimal places.

5

1. By sketching a suitable pair of graphs, show that the equation $$\ln x = 2 - x ^ { 2 }$$ has exactly one root.
2. Verify by calculation that the root lies between 1.0 and 1.4 .
3. Use the iterative formula $$x _ { n + 1 } = \sqrt { } \left( 2 - \ln x _ { n } \right)$$ to determine the root correct to 2 decimal places, showing the result of each iteration.

2 The sequence of values given by the iterative formula $$x _ { n + 1 } = \frac { 1 } { 5 } \left( 4 x _ { n } + \frac { 306 } { x _ { n } ^ { 4 } } \right)$$ with initial value \(x _ { 1 } = 3\), converges to \(\alpha\).

1. Use this iterative formula to find \(\alpha\) correct to 3 decimal places, showing the result of each iteration.
2. State an equation satisfied by \(\alpha\), and hence show that the exact value of \(\alpha\) is \(\sqrt [ 5 ] { 306 }\).

3 The sequence of values given by the iterative formula $$x _ { n + 1 } = \frac { 3 x _ { n } } { 4 } + \frac { 2 } { x _ { n } ^ { 3 } }$$ with initial value \(x _ { 1 } = 2\), converges to \(\alpha\).

1. Use this iteration to calculate \(\alpha\) correct to 2 decimal places, showing the result of each iteration to 4 decimal places.
2. State an equation which is satisfied by \(\alpha\) and hence find the exact value of \(\alpha\).

6

1. By sketching a suitable pair of graphs, show that there is only one value of \(x\) that is a root of the equation \(x = 9 \mathrm { e } ^ { - 2 x }\).
2. Verify, by calculation, that this root lies between 1 and 2 .
3. Show that, if a sequence of values given by the iterative formula $$x _ { n + 1 } = \frac { 1 } { 2 } \left( \ln 9 - \ln x _ { n } \right)$$ converges, then it converges to the root of the equation given in part (i).
4. Use the iterative formula, with \(x _ { 1 } = 1\), to calculate the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

5

1. By sketching a suitable pair of graphs, show that the equation $$\sec x = 3 - x$$ where \(x\) is in radians, has only one root in the interval \(0 < x < \frac { 1 } { 2 } \pi\).
2. Verify by calculation that this root lies between 1.0 and 1.2.
3. Show that this root also satisfies the equation $$x = \cos ^ { - 1 } \left( \frac { 1 } { 3 - x } \right)$$
4. Use the iterative formula $$x _ { n + 1 } = \cos ^ { - 1 } \left( \frac { 1 } { 3 - x _ { n } } \right)$$ with initial value \(x _ { 1 } = 1.1\), to calculate the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

8 The constant \(a\), where \(a > 1\), is such that \(\int _ { 1 } ^ { a } \left( x + \frac { 1 } { x } \right) \mathrm { d } x = 6\).

1. Find an equation satisfied by \(a\), and show that it can be written in the form $$a = \sqrt { } ( 13 - 2 \ln a )$$
2. Verify, by calculation, that the equation \(a = \sqrt { } ( 13 - 2 \ln a )\) has a root between 3 and 3.5.
3. Use the iterative formula $$a _ { n + 1 } = \sqrt { } \left( 13 - 2 \ln a _ { n } \right)$$ with \(a _ { 1 } = 3.2\), to calculate the value of \(a\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

7 \includegraphics[max width=\textwidth, alt={}, center]{b9556031-871d-4dd3-9523-e3438a41339f-3_655_685_262_730} The diagram shows the curve \(y = x \mathrm { e } ^ { 2 x }\) and its minimum point \(M\).

1. Find the exact coordinates of \(M\).
2. Show that the curve intersects the line \(y = 20\) at the point whose \(x\)-coordinate is the root of the equation $$x = \frac { 1 } { 2 } \ln \left( \frac { 20 } { x } \right)$$
3. Use the iterative formula $$x _ { n + 1 } = \frac { 1 } { 2 } \ln \left( \frac { 20 } { x _ { n } } \right)$$ with initial value \(x _ { 1 } = 1.3\), to calculate the root correct to 2 decimal places, giving the result of each iteration to 4 decimal places.

7

1. By sketching a suitable pair of graphs, show that the equation $$\mathrm { e } ^ { 2 x } = 2 - x$$ has only one root.
2. Verify by calculation that this root lies between \(x = 0\) and \(x = 0.5\).
3. Show that, if a sequence of values given by the iterative formula $$x _ { n + 1 } = \frac { 1 } { 2 } \ln \left( 2 - x _ { n } \right)$$ converges, then it converges to the root of the equation in part (i).
4. Use this iterative formula, with initial value \(x _ { 1 } = 0.25\), to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
5. By differentiating \(\frac { \cos x } { \sin x }\), show that if \(y = \cot x\) then \(\frac { \mathrm { d } y } { \mathrm {~d} x } = - \operatorname { cosec } ^ { 2 } x\).
6. By expressing \(\cot ^ { 2 } x\) in terms of \(\operatorname { cosec } ^ { 2 } x\) and using the result of part (i), show that $$\int _ { \frac { 1 } { 4 } \pi } ^ { \frac { 1 } { 2 } \pi } \cot ^ { 2 } x \mathrm {~d} x = 1 - \frac { 1 } { 4 } \pi$$
7. Express \(\cos 2 x\) in terms of \(\sin ^ { 2 } x\) and hence show that \(\frac { 1 } { 1 - \cos 2 x }\) can be expressed as \(\frac { 1 } { 2 } \operatorname { cosec } ^ { 2 } x\). Hence, using the result of part (i), find $$\int \frac { 1 } { 1 - \cos 2 x } \mathrm {~d} x$$

7

1. By sketching a suitable pair of graphs, show that the equation $$\mathrm { e } ^ { 2 x } = 14 - x ^ { 2 }$$ has exactly two real roots.
2. Show by calculation that the positive root lies between 1.2 and 1.3.
3. Show that this root also satisfies the equation $$x = \frac { 1 } { 2 } \ln \left( 14 - x ^ { 2 } \right) .$$
4. Use an iteration process based on the equation in part (iii), with a suitable starting value, to find the root correct to 2 decimal places. Give the result of each step of the process to 4 decimal places.
5. Express \(4 \sin \theta - 6 \cos \theta\) in the form \(R \sin ( \theta - \alpha )\), where \(R > 0\) and \(0 ^ { \circ } < \alpha < 90 ^ { \circ }\). Give the exact value of \(R\) and the value of \(\alpha\) correct to 2 decimal places.
6. Solve the equation \(4 \sin \theta - 6 \cos \theta = 3\) for \(0 ^ { \circ } \leqslant \theta \leqslant 360 ^ { \circ }\).
7. Find the greatest and least possible values of \(( 4 \sin \theta - 6 \cos \theta ) ^ { 2 } + 8\) as \(\theta\) varies.

3 The sequence \(x _ { 1 } , x _ { 2 } , x _ { 3 } , \ldots\) defined by $$x _ { 1 } = 1 , \quad x _ { n + 1 } = \frac { 1 } { 2 } \sqrt [ 3 ] { } \left( x _ { n } ^ { 2 } + 6 \right)$$ converges to the value \(\alpha\).

1. Find the value of \(\alpha\) correct to 3 decimal places. Show your working, giving each calculated value of the sequence to 5 decimal places.
2. Find, in the form \(a x ^ { 3 } + b x ^ { 2 } + c = 0\), an equation of which \(\alpha\) is a root.

6 A curve has parametric equations $$x = \frac { 1 } { ( 2 t + 1 ) ^ { 2 } } , \quad y = \sqrt { } ( t + 2 )$$ The point \(P\) on the curve has parameter \(p\) and it is given that the gradient of the curve at \(P\) is - 1 .

1. Show that \(p = ( p + 2 ) ^ { \frac { 1 } { 6 } } - \frac { 1 } { 2 }\).
2. Use an iterative process based on the equation in part (i) to find the value of \(p\) correct to 3 decimal places. Use a starting value of 0.7 and show the result of each iteration to 5 decimal places.

6 \includegraphics[max width=\textwidth, alt={}, center]{48ab71ff-c37b-4e0b-b031-d99b0cf517a8-3_421_976_251_580} The diagram shows the curve \(y = \frac { \sin 2 x } { x + 2 }\) for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\). The \(x\)-coordinate of the maximum point \(M\) is denoted by \(\alpha\).

1. Find \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) and show that \(\alpha\) satisfies the equation \(\tan 2 x = 2 x + 4\).
2. Show by calculation that \(\alpha\) lies between 0.6 and 0.7 .
3. Use the iterative formula \(x _ { n + 1 } = \frac { 1 } { 2 } \tan ^ { - 1 } \left( 2 x _ { n } + 4 \right)\) to find the value of \(\alpha\) correct to 3 decimal places. Give the result of each iteration to 5 decimal places.

6

1. By sketching a suitable pair of graphs, show that the equation $$\cot x = 4 x - 2$$ where \(x\) is in radians, has only one root for \(0 \leqslant x \leqslant \frac { 1 } { 2 } \pi\).
2. Verify by calculation that this root lies between \(x = 0.7\) and \(x = 0.9\).
3. Show that this root also satisfies the equation $$x = \frac { 1 + 2 \tan x } { 4 \tan x }$$
4. Use the iterative formula \(x _ { n + 1 } = \frac { 1 + 2 \tan x _ { n } } { 4 \tan x _ { n } }\) to determine this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

6

1. By sketching a suitable pair of graphs, show that the equation $$3 \mathrm { e } ^ { x } = 8 - 2 x$$ has only one root.
2. Verify by calculation that this root lies between \(x = 0.7\) and \(x = 0.8\).
3. Show that this root also satisfies the equation $$x = \ln \left( \frac { 8 - 2 x } { 3 } \right)$$
4. Use the iterative formula \(x _ { n + 1 } = \ln \left( \frac { 8 - 2 x _ { n } } { 3 } \right)\) to determine this root correct to 3 decimal places. Give the result of each iteration to 5 decimal places.

4

1. By sketching a suitable pair of graphs, show that the equation $$3 \ln x = 15 - x ^ { 3 }$$ has exactly one real root.
2. Show by calculation that the root lies between 2.0 and 2.5.
3. Use the iterative formula \(x _ { n + 1 } = \sqrt [ 3 ] { } \left( 15 - 3 \ln x _ { n } \right)\) to find the root correct to 3 decimal places. Give the result of each iteration to 5 decimal places.

7 It is given that \(\int _ { 0 } ^ { a } \left( \frac { 1 } { 2 } \mathrm { e } ^ { 3 x } + x ^ { 2 } \right) \mathrm { d } x = 10\), where \(a\) is a positive constant.

1. Show that \(a = \frac { 1 } { 3 } \ln \left( 61 - 2 a ^ { 3 } \right)\).
2. Show by calculation that the value of \(a\) lies between 1.0 and 1.5.
3. Use an iterative formula, based on the equation in part (i), to find the value of \(a\) correct to 3 decimal places. Give the result of each iteration to 5 decimal places.

5

1. Given that \(\int _ { 0 } ^ { a } \left( 3 \mathrm { e } ^ { \frac { 1 } { 2 } x } + 1 \right) \mathrm { d } x = 10\), show that the positive constant \(a\) satisfies the equation $$a = 2 \ln \left( \frac { 16 - a } { 6 } \right)$$
2. Use the iterative formula \(a _ { n + 1 } = 2 \ln \left( \frac { 16 - a _ { n } } { 6 } \right)\) with \(a _ { 1 } = 2\) to find the value of \(a\) correct to 3 decimal places. Give the result of each iteration to 5 decimal places.

6 The equation of a curve is \(y = \frac { 3 x ^ { 2 } } { x ^ { 2 } + 4 }\). At the point on the curve with positive \(x\)-coordinate \(p\), the gradient of the curve is \(\frac { 1 } { 2 }\).

1. Show that \(p = \sqrt { } \left( \frac { 48 p - 16 } { p ^ { 2 } + 8 } \right)\).
2. Show by calculation that \(2 < p < 3\).
3. Use an iterative formula based on the equation in part (i) to find the value of \(p\) correct to 4 significant figures. Give the result of each iteration to 6 significant figures.

5 The equation of a curve is \(y = 6 x \mathrm { e } ^ { \frac { 1 } { 3 } x }\). At the point on the curve with \(x\)-coordinate \(p\), the gradient of the curve is 40 .

1. Show that \(p = 3 \ln \left( \frac { 20 } { p + 3 } \right)\).
2. Show by calculation that \(3.3 < p < 3.5\).
3. Use an iterative formula based on the equation in part (i) to find the value of \(p\) correct to 3 decimal places. Give the result of each iteration to 5 decimal places.

4 The sequence of values given by the iterative formula $$x _ { n + 1 } = \frac { 2 x _ { n } ^ { 2 } + x _ { n } + 9 } { \left( x _ { n } + 1 \right) ^ { 2 } }$$ with \(x _ { 1 } = 2\), converges to \(\alpha\).

1. Find the value of \(\alpha\) correct to 2 decimal places, giving the result of each iteration to 4 decimal places.
2. Determine the exact value of \(\alpha\).

3

1. By sketching a suitable pair of graphs, show that the equation $$x ^ { 3 } = 11 - 2 x$$ has exactly one real root.
2. Use the iterative formula $$x _ { n + 1 } = \sqrt [ 3 ] { } \left( 11 - 2 x _ { n } \right)$$ to find the root correct to 4 significant figures. Give the result of each iteration to 6 significant figures.

4 \includegraphics[max width=\textwidth, alt={}, center]{873a104f-e2e2-49bb-b943-583769728fbb-06_355_839_260_653} The diagram shows the curve with equation \(y = \frac { 5 \ln x } { 2 x + 1 }\). The curve crosses the \(x\)-axis at the point \(P\) and has a maximum point \(M\).

1. Find the gradient of the curve at the point \(P\).
2. Show that the \(x\)-coordinate of the point \(M\) satisfies the equation \(x = \frac { x + 0.5 } { \ln x }\).
3. Use an iterative formula based on the equation in part (ii) to find the \(x\)-coordinate of \(M\) correct to 4 significant figures. Show the result of each iteration to 6 significant figures.

6 It is given that \(\int _ { 0 } ^ { a } \left( 1 + \mathrm { e } ^ { \frac { 1 } { 2 } x } \right) ^ { 2 } \mathrm {~d} x = 10\), where \(a\) is a positive constant.

1. Show that \(a = 2 \ln \left( \frac { 15 - a } { 4 + \mathrm { e } ^ { \frac { 1 } { 2 } a } } \right)\).
2. Use the equation in part (i) to show by calculation that \(1.5 < a < 1.6\).
3. Use an iterative formula based on the equation in part (i) to find the value of \(a\) correct to 3 significant figures. Give the result of each iteration to 5 significant figures.

4 The sequence of values given by the iterative formula $$x _ { n + 1 } = \frac { 2 } { 3 } \left( x _ { n } + \frac { 1 } { x _ { n } ^ { 2 } } \right)$$ with initial value \(x _ { 1 } = 1\), converges to \(\alpha\).

1. Use this formula to find \(\alpha\) correct to 2 decimal places, showing the result of each iteration.
2. State an equation satisfied by \(\alpha\), and hence find the exact value of \(\alpha\).