Fixed Point Iteration

The equation \(x^3 - 5x + 3 = 0\) has a root between \(x = 0\) and \(x = 1\).

1. The equation can be rearranged into the form \(x = g(x)\) where \(g(x) = px^3 + q\). State the values of \(p\) and \(q\). [1]
2. By considering \(|g'(x)|\), show that the iterative form \(x_{n+1} = g(x_n)\) with a suitable starting value converges to the root between \(x = 0\) and \(x = 1\). [You are not required to find this root.] [2]

2 \includegraphics[max width=\textwidth, alt={}, center]{faf83d93-40b6-4557-bfd5-f94c67470dfa-2_449_639_388_753} The diagram shows the curve \(y = x ^ { 4 } + 2 x - 9\). The curve cuts the positive \(x\)-axis at the point \(P\).

1. Verify by calculation that the \(x\)-coordinate of \(P\) lies between 1.5 and 1.6.
2. Show that the \(x\)-coordinate of \(P\) satisfies the equation $$x = \sqrt [ 3 ] { \left( \frac { 9 } { x } - 2 \right) }$$
3. Use the iterative formula $$x _ { n + 1 } = \sqrt [ 3 ] { \left( \frac { 9 } { x _ { n } } - 2 \right) }$$ to determine the \(x\)-coordinate of \(P\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

It is given that there is exactly one value of \(x\), where \(0 < x < \pi\), that satisfies the equation $$3\tan 2x - 8\tan x = 4.$$

1. Show that \(t = \sqrt[3]{\frac{1}{2} + \frac{1}{3}t - \frac{1}{3}t^2}\), where \(t = \tan x\). [3]
2. Show by calculation that the value of \(t\) satisfying the equation in part (i) lies between 0.7 and 0.8. [2]
3. Use an iterative process based on the equation in part (i) to find the value of \(t\) correct to 4 significant figures. Use a starting value of 0.75 and show the result of each iteration. [3]
4. Solve the equation \(3\tan 4y - 8\tan 2y = 4\) for \(0 < y < \frac{1}{4}\pi\). [2]

The sequence \(u_1, u_2, u_3, \ldots\) is defined by $$u_{n+1} = k - \frac{24}{u_n} \quad u_1 = 2$$ where \(k\) is an integer. Given that \(u_1 + 2u_2 + u_3 = 0\)

1. show that $$3k^2 - 58k + 240 = 0$$ [3]
2. Find the value of \(k\), giving a reason for your answer. [2]
3. Find the value of \(u_3\) [1]

3 The following algorithm (J. M. Oudin, 1940) claims to compute the date of Easter Sunday in the Gregorian calendar system.
The algorithm uses the year, y, to give the month, m, and day, d, of Easter Sunday.
All variables are integers and all remainders from division are dropped. For example, 7 divided by 3 is 2 remainder 1 . The remainder is dropped, giving the answer 2. $$\begin{aligned} & c = y / 100 \\ & n = y - 19 \times ( y / 19 ) \\ & k = ( c - 17 ) / 25 \\ & i = c - ( c / 4 ) - ( c - k ) / 3 + ( 19 \times n ) + 15 \\ & i = i - 30 \times ( i / 30 ) \\ & i = i - ( i / 28 ) \times ( 1 - ( i / 28 ) \times ( 29 / ( i + 1 ) ) \times ( ( 21 - n ) / 11 ) ) \\ & j = y + ( y / 4 ) + i + 2 - c + ( c / 4 ) \\ & j = j - 7 \times ( j / 7 ) \\ & p = i - j \\ & m = 3 + ( p + 40 ) / 44 \\ & d = p + 28 - 31 \times ( m / 4 ) \end{aligned}$$ For example, for 2008: \(\mathrm { y } = 2008\) \(\mathrm { c } = 2008 / 100 = 20\) \(n = 2008 - 19 \times ( 2008 / 19 ) = 2008 - 19 \times ( 105 ) = 13\), etc.
Complete the calculation for 2008.

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 curve with equation $$y = 5 x - 2 \tan 2 x$$ has exactly one stationary point in the interval \(0 \leqslant x < \frac { 1 } { 4 } \pi\).
Find the coordinates of this stationary point, giving each coordinate correct to 3 significant figures.

4. $$\mathrm { f } ( x ) = 3 \mathrm { e } ^ { x } - \frac { 1 } { 2 } \ln x - 2 , \quad x > 0 .$$

1. Differentiate to find \(\mathrm { f } ^ { \prime } ( x )\). The curve with equation \(y = \mathrm { f } ( x )\) has a turning point at \(P\). The \(x\)-coordinate of \(P\) is \(\alpha\).
2. Show that \(\alpha = \frac { 1 } { 6 } \mathrm { e } ^ { - \alpha }\). The iterative formula $$x _ { n + 1 } = \frac { 1 } { 6 } \mathrm { e } ^ { - x _ { n } } , x _ { 0 } = 1$$ is used to find an approximate value for \(\alpha\).
3. Calculate the values of \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving your answers to 4 decimal places.
4. By considering the change of sign of \(\mathrm { f } ^ { \prime } ( x )\) in a suitable interval, prove that \(\alpha = 0.1443\) correct to 4 decimal places.

3 The curve with equation $$y = 5 x - 2 \tan 2 x$$ has exactly one stationary point in the interval \(0 \leqslant x < \frac { 1 } { 4 } \pi\).
Find the coordinates of this stationary point, giving each coordinate correct to 3 significant figures.

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.

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.

5

1. By sketching a suitable pair of graphs, show that the equation \(\operatorname { cosec } x = 1 + \mathrm { e } ^ { - \frac { 1 } { 2 } x }\) has exactly two roots in the interval \(0 < x < \pi\).
2. The sequence of values given by the iterative formula $$x _ { n + 1 } = \pi - \sin ^ { - 1 } \left( \frac { 1 } { \mathrm { e } ^ { - \frac { 1 } { 2 } x _ { n } } + 1 } \right)$$ with initial value \(x _ { 1 } = 2\), converges to one of these roots.
   Use the formula to determine this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

2 Solve the equation $$\ln ( 1 + x ) = 1 + \ln x$$ giving your answer correct to 2 significant figures.

2 Solve the equation $$\ln ( 1 + x ) = 1 + \ln x$$ giving your answer correct to 2 significant figures.

4 You are given the equation \(( 2 x - 1 ) ^ { 2 } - \mathrm { e } ^ { x } = 0\).

1. Verify that 0 is a root of the equation. There are also two other roots, \(\alpha\) and \(\beta\), where \(0 < \alpha < \beta\).
2. The iterative formula \(x _ { r + 1 } = \ln \left( 2 x _ { r } - 1 \right) ^ { 2 }\) is to be used to find a root of the equation.
   1. Sketch the line \(y = x\) and the curve \(y = \ln ( 2 x - 1 ) ^ { 2 }\) on the same axes, showing the roots \(0 , \alpha\) and \(\beta\).
   2. By drawing a 'staircase' diagram on your sketch, starting with a value of \(x\) that is between \(\alpha\) and \(\beta\), show that this iteration does not converge to \(\alpha\).
   3. Using this iterative formula with \(x _ { 1 } = 3.75\), find the value of \(\beta\) correct to 3 decimal places.
   4. Using the Newton-Raphson method with \(x _ { 1 } = 1.6\), find the root \(\alpha\) of the equation \(( 2 x - 1 ) ^ { 2 } - \mathrm { e } ^ { x } = 0\) correct to 5 significant figures. Show the result of each iteration.

1. By sketching a suitable pair of graphs, show that the equation \(\cot 2x = \sec x\) has exactly one root in the interval \(0 < x < \frac{1}{2}\pi\). [2]
2. Show that if a sequence of real values given by the iterative formula $$x_{n+1} = \frac{1}{2}\tan^{-1}(\cos x_n)$$ converges, then it converges to the root in part (a). [1]

6

1. By sketching a suitable pair of graphs on the same diagram, show that the equation $$\ln x = 2 \mathrm { e } ^ { - x }$$ has exactly one root.
2. Verify by calculation that the root lies between 1.5 and 1.6.
3. Show that if a sequence of values given by the iterative formula $$x _ { n + 1 } = \mathrm { e } ^ { 2 \mathrm { e } ^ { - x _ { n } } }$$ converges, then it converges to the root of the equation in part (a).
4. Use the iterative formula in part (c) to determine the root correct to 3 significant figures. Give the result of each iteration to 5 significant figures.

10 \includegraphics[max width=\textwidth, alt={}, center]{c1fbc9ef-2dc6-43c3-bc58-179f683c9acf-18_471_686_276_717} The curve \(y = x \sqrt { \sin x }\) has one stationary point in the interval \(0 < x < \pi\), where \(x = a\) (see diagram).

1. Show that \(\tan a = - \frac { 1 } { 2 } a\).
2. Verify by calculation that \(a\) lies between 2 and 2.5.
3. Show that if a sequence of values in the interval \(0 < x < \pi\) given by the iterative formula \(x _ { n + 1 } = \pi - \tan ^ { - 1 } \left( \frac { 1 } { 2 } x _ { n } \right)\) converges, then it converges to \(a\), the root of the equation in part (a). [2]
4. Use the iterative formula given in part (c) to determine \(a\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.
   If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

1. Given that \(\int_0^a (6e^{2x} + x) \, dx = 42\), show that \(a = \frac{1}{2} \ln(15 - \frac{1}{6}a^2)\). [5]
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]

3 It is given that \(\int _ { 0 } ^ { a } \left( 3 \mathrm { e } ^ { 2 x } - 1 \right) \mathrm { d } x = 12\), where \(a\) is a positive constant.

1. Show that \(a = \frac { 1 } { 2 } \ln \left( 9 + \frac { 2 } { 3 } a \right)\).
2. Use an iterative formula, based on the equation in (a), to find the value of \(a\) correct to 4 significant figures. Use an initial value of 1 and give the result of each iteration to 6 significant figures. [3]

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.

1. Show that \(x\) satisfies the equation \(x = \frac { 1 } { 3 } ( \pi + \sin x )\).
2. Verify by calculation that \(x\) lies between 1 and 1.5.
3. Use an iterative formula based on the equation in part (i) to determine \(x\) correct to 3 decimal places. Give the result of each iteration to 5 decimal places.

1. Show that \(x\) satisfies the equation \(x = \frac { 1 } { 3 } ( \pi + \sin x )\).
2. Verify by calculation that \(x\) lies between 1 and 1.5.
3. Use an iterative formula based on the equation in part (i) to determine \(x\) correct to 3 decimal places. Give the result of each iteration to 5 decimal places.

5 \includegraphics[max width=\textwidth, alt={}, center]{d1377d66-73c8-4d97-9cae-d784b41fb0a8-2_519_800_1359_669} The diagram shows a circle with centre \(O\) and radius \(r\). The tangents to the circle at the points \(A\) and \(B\) meet at \(T\), and the angle \(A O B\) is \(2 x\) radians. The shaded region is bounded by the tangents \(A T\) and \(B T\), and by the minor \(\operatorname { arc } A B\). The perimeter of the shaded region is equal to the circumference of the circle.

1. Show that \(x\) satisfies the equation $$\tan x = \pi - x .$$
2. This equation has one root in the interval \(0 < x < \frac { 1 } { 2 } \pi\). Verify by calculation that this root lies between 1 and 1.3.
3. Use the iterative formula $$x _ { n + 1 } = \tan ^ { - 1 } \left( \pi - x _ { n } \right)$$ to determine the root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

3

1. Show that the equation \(x ^ { 2 } - \ln x - 2 = 0\) has a solution between \(x = 1\) and \(x = 2\).
2. Find an approximation to that solution using the iteration \(x _ { n + 1 } = \sqrt { 2 + \ln x _ { n } }\), giving your answer correct to 2 decimal places.

2 The equation \(x ^ { 3 } - 8 x - 13 = 0\) has one real root.

1. Find the two consecutive integers between which this root lies.
2. Use the iterative formula $$x _ { n + 1 } = \left( 8 x _ { n } + 13 \right) ^ { \frac { 1 } { 3 } }$$ to determine this root correct to 2 decimal places. Give the result of each iteration to 4 decimal places.

6. \begin{figure}[h] \end{figure} Figure 3 shows a sketch of curve \(C _ { 1 }\) with equation \(y = 5 \mathrm { e } ^ { x - 1 } + 3\) and curve \(C _ { 2 }\) with equation \(y = 10 - x ^ { 2 }\) The point \(P\) lies on \(C _ { 1 }\) and has \(y\) coordinate 18

1. Find the \(x\) coordinate of \(P\), writing your answer in the form \(\ln k\), where \(k\) is a constant to be found. The curve \(C _ { 1 }\) meets the curve \(C _ { 2 }\) at \(x = \alpha\) and at \(x = \beta\), as shown in Figure 3.
2. Using a suitable interval and a suitable function that should be stated, show that to 3 decimal places \(\alpha = 1.134\) The iterative equation $$x _ { n + 1 } = - \sqrt { 7 - 5 \mathrm { e } ^ { x _ { n } - 1 } }$$ is used to find an approximation to \(\beta\). Using this iterative formula with \(x _ { 1 } = - 3\)
3. find the value of \(x _ { 2 }\) and the value of \(\beta\), giving each answer to 6 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 \(\alpha\).

5 The sequence of values given by the iterative formula \(x _ { n + 1 } = \frac { 6 + 8 x _ { n } } { 8 + x _ { n } ^ { 2 } }\) with initial value \(x _ { 1 } = 2\) converges to \(\alpha\).

1. Use the iterative formula to find the value of \(\alpha\) correct to 4 significant figures. Give the result of each iteration to 6 significant figures.
2. State an equation satisfied by \(\alpha\) and hence determine the exact value of \(\alpha\).

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 \(\sqrt [ 5 ] { 306 }\).

1. Show that the \(x\)-coordinate of \(Q\) satisfies the equation \(x = \frac { 9 } { 8 } - \frac { 1 } { 2 } \mathrm { e } ^ { - 2 x }\).
2. Use an iterative formula based on the equation in part (i) to find the \(x\)-coordinate of \(Q\) correct to 3 significant figures. Give the result of each iteration to 5 significant figures.

6 A curve has equation \(y = x ^ { 3 } \mathrm { e } ^ { 0.2 x }\) where \(x \geqslant 0\). At the point \(P\) on the curve, the gradient of the curve is 15 .

1. Show that the \(x\)-coordinate of \(P\) satisfies the equation \(x = \sqrt { \frac { 75 \mathrm { e } ^ { - 0.2 x } } { 15 + x } }\).
2. Use the equation in part (a) to show by calculation that the \(x\)-coordinate of \(P\) lies between 1.7 and 1.8.
3. Use an iterative formula, based on the equation in part (a), to find the \(x\)-coordinate of \(P\) correct to 4 significant figures. Give the result of each iteration to 6 significant figures.

9 \includegraphics[max width=\textwidth, alt={}, center]{ccadf73b-16f5-463a-8f69-1394839d5325-3_481_483_1434_831} The diagram shows the curves \(y = x \cos x\) and \(y = \frac { k } { x }\), where \(k\) is a constant, for \(0 < x \leqslant \frac { 1 } { 2 } \pi\). The curves touch at the point where \(x = a\).

1. Show that \(a\) satisfies the equation \(\tan a = \frac { 2 } { a }\).
2. Use the iterative formula \(a _ { n + 1 } = \tan ^ { - 1 } \left( \frac { 2 } { a _ { n } } \right)\) to determine \(a\) correct to 3 decimal places. Give the result of each iteration to 5 decimal places.
3. Hence find the value of \(k\) correct to 2 decimal places.

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

1. $$f ( x ) = \sec x + 3 x - 2 , \quad - \frac { \pi } { 2 } < x < \frac { \pi } { 2 }$$

1. Show that there is a root of \(\mathrm { f } ( x ) = 0\) in the interval \([ 0.2,0.4 ]\)
2. Show that the equation \(\mathrm { f } ( x ) = 0\) can be written in the form $$x = \frac { 2 } { 3 } - \frac { 1 } { 3 \cos x }$$ The solution of \(\mathrm { f } ( x ) = 0\) is \(\alpha\), where \(\alpha = 0.3\) to 1 decimal place.
3. Starting with \(x _ { 0 } = 0.3\), use the iterative formula $$x _ { n + 1 } = \frac { 2 } { 3 } - \frac { 1 } { 3 \cos x _ { n } }$$ to calculate the values of \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\), giving your answers to 4 decimal places.
4. State the value of \(\alpha\) correct to 3 decimal places.

6 \includegraphics[max width=\textwidth, alt={}, center]{dd7b2aee-4318-48e8-97c0-541e47f2e83a-2_551_567_1416_788} In the diagram, \(A\) is a point on the circumference of a circle with centre \(O\) and radius \(r\). A circular arc with centre \(A\) meets the circumference at \(B\) and \(C\). The angle \(O A B\) is \(\theta\) radians. The shaded region is bounded by the circumference of the circle and the arc with centre \(A\) joining \(B\) and \(C\). The area of the shaded region is equal to half the area of the circle.

1. Show that \(\cos 2 \theta = \frac { 2 \sin 2 \theta - \pi } { 4 \theta }\).
2. Use the iterative formula $$\theta _ { n + 1 } = \frac { 1 } { 2 } \cos ^ { - 1 } \left( \frac { 2 \sin 2 \theta _ { n } - \pi } { 4 \theta _ { n } } \right)$$ with initial value \(\theta _ { 1 } = 1\), to determine \(\theta\) correct to 2 decimal places, showing the result of each iteration to 4 decimal places. \(7 \quad\) Let \(\mathrm { f } ( x ) = \frac { 2 x ^ { 2 } - 7 x - 1 } { ( x - 2 ) \left( x ^ { 2 } + 3 \right) }\).
3. Express \(\mathrm { f } ( x )\) in partial fractions.
4. Hence obtain the expansion of \(\mathrm { f } ( x )\) in ascending powers of \(x\), up to and including the term in \(x ^ { 2 }\).

1 The function \(y ( x )\) satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$ where $$\mathrm { f } ( x , y ) = x + \sqrt { y }$$ and $$y ( 3 ) = 4$$ Use the improved Euler formula $$y _ { r + 1 } = y _ { r } + \frac { 1 } { 2 } \left( k _ { 1 } + k _ { 2 } \right)$$ where \(k _ { 1 } = h \mathrm { f } \left( x _ { r } , y _ { r } \right)\) and \(k _ { 2 } = h \mathrm { f } \left( x _ { r } + h , y _ { r } + k _ { 1 } \right)\) and \(h = 0.1\), to obtain an approximation to \(y ( 3.1 )\), giving your answer to three decimal places.

1 The function \(y ( x )\) satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$ where $$\mathrm { f } ( x , y ) = \frac { x ^ { 2 } + y ^ { 2 } } { x + y }$$ and $$y ( 1 ) = 3$$

1. Use the Euler formula $$y _ { r + 1 } = y _ { r } + h \mathrm { f } \left( x _ { r } , y _ { r } \right)$$ with \(h = 0.2\), to obtain an approximation to \(y ( 1.2 )\).
2. Use the improved Euler formula $$y _ { r + 1 } = y _ { r } + \frac { 1 } { 2 } \left( k _ { 1 } + k _ { 2 } \right)$$ where \(k _ { 1 } = h \mathrm { f } \left( x _ { r } , y _ { r } \right)\) and \(k _ { 2 } = h \mathrm { f } \left( x _ { r } + h , y _ { r } + k _ { 1 } \right)\) and \(h = 0.2\), to obtain an approximation to \(y ( 1.2 )\), giving your answer to four decimal places.

1 The function \(y ( x )\) satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$ where $$\mathrm { f } ( x , y ) = \frac { x ^ { 2 } + y ^ { 2 } } { x + y }$$ and $$y ( 1 ) = 3$$

1. Use the Euler formula $$y _ { r + 1 } = y _ { r } + h \mathrm { f } \left( x _ { r } , y _ { r } \right)$$ with \(h = 0.2\), to obtain an approximation to \(y ( 1.2 )\).
2. Use the improved Euler formula $$y _ { r + 1 } = y _ { r } + \frac { 1 } { 2 } \left( k _ { 1 } + k _ { 2 } \right)$$ where \(k _ { 1 } = h \mathrm { f } \left( x _ { r } , y _ { r } \right)\) and \(k _ { 2 } = h \mathrm { f } \left( x _ { r } + h , y _ { r } + k _ { 1 } \right)\) and \(h = 0.2\), to obtain an approximation to \(y ( 1.2 )\), giving your answer to four decimal places.

2 A curve satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \log _ { 10 } x$$ Starting at the point \(( 2,3 )\) on the curve, use a step-by-step method with a step length of 0.2 to estimate the value of \(y\) at \(x = 2.4\). Give your answer to three decimal places.

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.

4. The curve with equation \(y = 2 + \ln ( 4 - x )\) meets the line \(y = x\) at a single point, \(x = \beta\).
a. Show that \(2 < \beta < 3\) \begin{figure}[h] \end{figure} Figure 2 shows the graph of \(y = 2 + \ln ( 4 - x )\) and the graph of \(y = x\).
A student uses the iteration formula $$x _ { n + 1 } = 2 + \ln \left( 4 - x _ { n } \right) , \quad n \in N ,$$ in an attempt to find an approximation for \(\beta\).
Using the graph and starting with \(x _ { 1 } = 3\),
b. determine whether the or not this iteration formula can be used to find an approximation for \(\beta\), justifying your answer.

6 The curve with equation \(y = \frac { 6 } { x ^ { 2 } }\) intersects the line \(y = x + 1\) at the point \(P\).

1. Verify by calculation that the \(x\)-coordinate of \(P\) lies between 1.4 and 1.6.
2. Show that the \(x\)-coordinate of \(P\) satisfies the equation $$x = \sqrt { } \left( \frac { 6 } { x + 1 } \right)$$
3. Use the iterative formula $$x _ { n + 1 } = \sqrt { } \left( \frac { 6 } { x _ { n } + 1 } \right)$$ with initial value \(x _ { 1 } = 1.5\), to determine the \(x\)-coordinate of \(P\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places.