Apply iteration to find root

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 \(\delta _ { 10 }\).

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.

2 It is given that \(\alpha\) is the only real root of the equation \(x ^ { 5 } + 2 x - 28 = 0\) and that \(1.8 < \alpha < 2\).

1. The iteration \(x _ { n + 1 } = \sqrt [ 5 ] { 28 - 2 x _ { n } }\), with \(x _ { 1 } = 1.9\), is to be used to find \(\alpha\). Find the values of \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving the answers correct to 7 decimal places.
2. The error \(e _ { n }\) is defined by \(e _ { n } = \alpha - x _ { n }\). Given that \(\alpha = 1.8915749\), correct to 7 decimal places, evaluate \(\frac { e _ { 3 } } { e _ { 2 } }\) and \(\frac { e _ { 4 } } { e _ { 3 } }\). Comment on these values in relation to the gradient of the curve with equation \(y = \sqrt [ 5 ] { 28 - 2 x }\) at \(x = \alpha\).
3. Prove that the derivative of \(\sin ^ { - 1 } x\) is \(\frac { 1 } { \sqrt { 1 - x ^ { 2 } } }\).
4. Given that $$\sin ^ { - 1 } 2 x + \sin ^ { - 1 } y = \frac { 1 } { 2 } \pi$$ find the exact value of \(\frac { \mathrm { d } y } { \mathrm {~d} x }\) when \(x = \frac { 1 } { 4 }\).
5. By means of a suitable substitution, show that $$\int \frac { x ^ { 2 } } { \sqrt { x ^ { 2 } - 1 } } \mathrm {~d} x$$ can be transformed to \(\int \cosh ^ { 2 } \theta \mathrm {~d} \theta\).
6. Hence show that \(\int \frac { x ^ { 2 } } { \sqrt { x ^ { 2 } - 1 } } \mathrm {~d} x = \frac { 1 } { 2 } x \sqrt { x ^ { 2 } - 1 } + \frac { 1 } { 2 } \cosh ^ { - 1 } x + c\).

1 It is given that \(\mathrm { f } ( x ) = x ^ { 2 } - \sin x\).

1. The iteration \(x _ { n + 1 } = \sqrt { \sin x _ { n } }\), with \(x _ { 1 } = 0.875\), is to be used to find a real root, \(\alpha\), of the equation \(\mathrm { f } ( x ) = 0\). Find \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\), giving the answers correct to 6 decimal places.
2. The error \(e _ { n }\) is defined by \(e _ { n } = \alpha - x _ { n }\). Given that \(\alpha = 0.876726\), correct to 6 decimal places, find \(e _ { 3 }\) and \(e _ { 4 }\). Given that \(\mathrm { g } ( x ) = \sqrt { \sin x }\), use \(e _ { 3 }\) and \(e _ { 4 }\) to estimate \(\mathrm { g } ^ { \prime } ( \alpha )\).

9 The equation \(10 x - 8 \ln x = 28\) has a root \(\alpha\) in the interval [3,4]. The iteration \(x _ { n + 1 } = \mathrm { g } \left( x _ { n } \right)\), where \(\mathrm { g } ( x ) = 2.8 + 0.8 \ln x\) and \(x _ { 1 } = 3.8\), is to be used to find \(\alpha\).

1. Find the value of \(\alpha\) correct to 5 decimal places. You should show the result of each step of the iteration to 6 decimal places.
2. Illustrate this iteration by means of a sketch.
3. The difference, \(\delta _ { r }\), between successive approximations is given by \(\delta _ { r } = x _ { r + 1 } - x _ { r }\). Find \(\delta _ { 3 }\).
4. Given that \(\delta _ { n + 1 } \approx \mathrm {~g} ^ { \prime } ( \alpha ) \delta _ { n }\), for all positive integers \(n\), estimate the smallest value of \(n\) such that \(\delta _ { n } < 10 ^ { - 6 } \delta _ { 1 }\). \section*{OCR}

3 Complete this table to show the next 3 values of the iteration $$x _ { n + 1 } = k x _ { n } \left( 1 - x _ { n } \right)$$ in the case when \(k = 3.2\) and \(x _ { 0 } = 0.5\). Give your answers to calculator accuracy.

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

1. Use the Euler formula $$y _ { r + 1 } = y _ { r } + h \mathrm { f } \left( x _ { r } , y _ { r } \right)$$ with \(h = 0.1\), to obtain an approximation to \(y ( 2.1 )\).
2. Use the formula $$y _ { r + 1 } = y _ { r - 1 } + 2 h \mathrm { f } \left( x _ { r } , y _ { r } \right)$$ with your answer to part (a), to obtain an approximation to \(y ( 2.2 )\).

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

1. Use the Euler formula $$y _ { r + 1 } = y _ { r } + h \mathrm { f } \left( x _ { r } , y _ { r } \right)$$ with \(h = 0.1\), to obtain an approximation to \(y ( 3.1 )\), giving your answer to four decimal places.
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 \(h = 0.1\), to obtain an approximation to \(y ( 3.1 )\), 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 $$\mathrm { f } ( x , y ) = x + \sqrt { y }$$ 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 $$\mathrm { f } ( x , y ) = \frac { y - x } { y ^ { 2 } + x }$$ and $$y ( 1 ) = 2$$

1. Use the Euler formula $$y _ { r + 1 } = y _ { r } + h \mathrm { f } \left( x _ { r } , y _ { r } \right)$$ with \(h = 0.1\), to obtain an approximation to \(y ( 1.1 )\).
2. Use the formula $$y _ { r + 1 } = y _ { r - 1 } + 2 h \mathrm { f } \left( x _ { r } , y _ { r } \right)$$ with your answer to part (a), to obtain an approximation to \(y ( 1.2 )\), giving your answer to three decimal places.

1 It is given that \(y ( x )\) satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$ where $$\mathrm { f } ( x , y ) = \sqrt { 2 x + y }$$ and $$y ( 3 ) = 5$$

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 \(y ( 3.2 )\), giving your answer to four decimal places.
2. Use the formula $$y _ { r + 1 } = y _ { r - 1 } + 2 h \mathrm { f } \left( x _ { r } , y _ { r } \right)$$ with your answer to part (a), to obtain an approximation to \(y ( 3.4 )\), giving your answer to three decimal places.

2 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 $$y ( 1 ) = 2$$

1. Use the Euler formula $$y _ { r + 1 } = y _ { r } + h \mathrm { f } \left( x _ { r } , y _ { r } \right)$$ with \(h = 0.1\), to obtain an approximation to \(y ( 1.1 )\).
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 \(h = 0.1\), to obtain an approximation to \(y ( 1.1 )\), 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 $$\mathrm { f } ( x , y ) = \ln ( x + y )$$ and $$y ( 2 ) = 3$$ 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 \(y ( 2.1 )\), giving your answer to four decimal places.
(6 marks)

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

1. Use the Euler formula $$y _ { r + 1 } = y _ { r } + h \mathrm { f } \left( x _ { r } , y _ { r } \right)$$ with \(h = 0.1\), to obtain an approximation to \(y ( 3.1 )\), giving your answer to four decimal places.
2. Use the formula $$y _ { r + 1 } = y _ { r - 1 } + 2 h \mathrm { f } \left( x _ { r } , y _ { r } \right)$$ with your answer to part (a), to obtain an approximation to \(y ( 3.2 )\), 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 $$\mathrm { f } ( x , y ) = x + 3 + \sin y$$ and $$y ( 1 ) = 1$$

1. Use the Euler formula $$y _ { r + 1 } = y _ { r } + h \mathrm { f } \left( x _ { r } , y _ { r } \right)$$ with \(h = 0.1\), to obtain an approximation to \(y ( 1.1 )\), giving your answer to four decimal places.
2. Use the formula $$y _ { r + 1 } = y _ { r - 1 } + 2 h \mathrm { f } \left( x _ { r } , y _ { r } \right)$$ with your answer to part (a), to obtain an approximation to \(y ( 1.2 )\), 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 $$\mathrm { f } ( x , y ) = x + \ln ( 1 + y )$$ and $$y ( 2 ) = 1$$ 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 \(y ( 2.2 )\), 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 $$\mathrm { f } ( x , y ) = \sqrt { ( 2 x ) } + \sqrt { y }$$ and $$y ( 2 ) = 9$$ 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 \(h = 0.25\), to obtain an approximation to \(y ( 2.25 )\), giving your answer to two decimal places.

1 It is given that \(y ( x )\) satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$ where $$\mathrm { f } ( x , y ) = ( x - y ) \sqrt { x + y }$$ and $$y ( 2 ) = 1$$ 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 \(y ( 2.2 )\), giving your answer to three decimal places.

1 It is given that \(y ( x )\) satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$ where $$\mathrm { f } ( x , y ) = \frac { \ln ( x + y ) } { \ln y }$$ and $$y ( 6 ) = 3$$ 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 \(h = 0.4\), to obtain an approximation to \(y ( 6.4 )\), giving your answer to three decimal places.
[0pt] [5 marks]

1 It is given that \(y ( x )\) satisfies the differential equation $$\frac { \mathrm { d } y } { \mathrm {~d} x } = \mathrm { f } ( x , y )$$ where $$\mathrm { f } ( x , y ) = \frac { x + y ^ { 2 } } { x }$$ and $$y ( 2 ) = 5$$

1. Use the Euler formula $$y _ { r + 1 } = y _ { r } + h \mathrm { f } \left( x _ { r } , y _ { r } \right)$$ with \(h = 0.05\), to obtain an approximation to \(y ( 2.05 )\).
2. Use the formula $$y _ { r + 1 } = y _ { r - 1 } + 2 h \mathrm { f } \left( x _ { r } , y _ { r } \right)$$ with your answer to part (a), to obtain an approximation to \(y ( 2.1 )\), giving your answer to three significant figures.
   [0pt] [3 marks]

12 JANUARY 2005
Afternoon
1 hour 30 minutes

• This insert should be used to answer Questions 4 and 7.
• Write your name, centre number and candidate number in the spaces provided at the top of this page.
• Write your answers to Questions 4 and 7 in the spaces provided in this insert, and attach it to your answer booklet.

4

1. 	\(A\)	\(B\)	C	D	\(E\)	\(F\)	G	\(H\)
   A	-	4	2	3	-	-	-	-
   \(B\)	4	-	1	-	3	-	-	-
   C	2	1	-	2	-	6	5	-
   \(D\)	3	-	2	-	-	-	4	-
   E	-	3	-	-	-	8	-	7
   \(F\)	-	-	6	-	8	-	-	8
   \(G\)	-	-	5	4	-	-	-	9
   \(H\)	-	-	-	-	7	8	9	-

2. B \(E\) \(C\) F
   • \(H\) \(A\) •
   • \({ } ^ { \text {F } }\)
   H D
   G
3. \(\_\_\_\_\)
4. \(\_\_\_\_\)
5. \(\_\_\_\_\) 7 (a) (i) \includegraphics[max width=\textwidth, alt={}, center]{197624b2-ca67-4bad-9c2c-dc68c10be0fd-11_191_1179_269_482} Do not cross out your working values (temporary labels) \includegraphics[max width=\textwidth, alt={}, center]{197624b2-ca67-4bad-9c2c-dc68c10be0fd-11_871_1557_612_335} Shortest route from \(A\) to \(E =\) \(\_\_\_\_\) Length = \(\_\_\_\_\) Shortest route from \(A\) to \(J =\) \(\_\_\_\_\) Length = \(\_\_\_\_\)
6. Length of route \(=\) \(\_\_\_\_\) Vertices visited in order \(\_\_\_\_\)
7. Explanation \(\_\_\_\_\) (b) \(\_\_\_\_\) Length = \(\_\_\_\_\)

7 In this question, the function INT( \(X\) ) is the largest integer less than or equal to \(X\). For example, $$\begin{aligned} & \operatorname { INT } ( 3.6 ) = 3 , \\ & \operatorname { INT } ( 3 ) = 3 , \\ & \operatorname { INT } ( - 3.6 ) = - 4 . \end{aligned}$$ Consider the following algorithm.

1. Apply the algorithm with the inputs \(B = 2\) and \(N = 5\). Record the values of \(F , G , H , C\) and \(N\) each time Step 9 is reached.
2. Explain what happens when the algorithm is applied with the inputs \(B = 2\) and \(N = - 5\).
3. Apply the algorithm with the inputs \(B = 10\) and \(N = 37\). Record the values of \(F , G , H , C\) and \(N\) each time Step 9 is reached. What are the output values when \(B = 10\) and \(N\) is any positive integer?

1 The flow chart shows an algorithm for which the input is a three-digit positive integer. \includegraphics[max width=\textwidth, alt={}, center]{43fe5fd5-4b98-4c3a-90ca-a1bd5cf065fe-2_1294_1493_356_328}

1. Trace through the algorithm using the input \(A = 614\) to show that the output is 297 . Write down the values of \(A , B , C\) and \(D\) in each pass through the algorithm.
2. What is the output when \(A = 616\) ?
3. Explain why the counter \(C\) is needed.