1.09d Newton-Raphson method

2. $$f ( x ) = \frac { 3 } { 2 } x ^ { 2 } + \frac { 4 } { 3 x } + 2 x - 5 , \quad x < 0$$ The equation \(\mathrm { f } ( x ) = 0\) has a single root \(\alpha\).

1. Show that \(\alpha\) lies in the interval \([ - 3 , - 2.5 ]\)
2. Taking - 3 as a first approximation to \(\alpha\), apply the Newton-Raphson procedure once to \(\mathrm { f } ( x )\) to obtain a second approximation to \(\alpha\). Give your answer to 3 decimal places.
3. Use linear interpolation once on the interval \([ - 3 , - 2.5 ]\) to find another approximation to \(\alpha\), giving your answer to 3 decimal places.

4. \(f ( x ) = x ^ { 3 } - 4 x ^ { 2 } + 5 x - 3\) The equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval ( 2,3 ).

1. Use linear interpolation on the end points of this interval to obtain an approximation for \(\alpha\).
2. Taking 2.5 as a first approximation to \(\alpha\), apply the Newton - Raphson procedure once to \(\mathrm { f } ( x )\) to obtain a second approximation to \(\alpha\). Give your answer to 2 decimal places.

1. $$f ( x ) = x ^ { 3 } - 3 x ^ { 2 } + 5 x - 4$$

1. Use differentiation to find \(\mathrm { f } ^ { \prime } ( x )\). The equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval \(1.4 < x < 1.5\)
2. Taking 1.4 as a first approximation to \(\alpha\), use the Newton-Raphson procedure once to obtain a second approximation to \(\alpha\). Give your answer to 3 decimal places.

9. The diagram shows the curve with equation \(y = 2 x - 3 \ln ( 2 x + 5 )\) and the normal to the curve at the point \(P ( - 2 , - 4 )\).

1. Find an equation for the normal to the curve at \(P\). The normal to the curve at \(P\) intersects the curve again at the point \(Q\) with \(x\)-coordinate \(q\).
2. Show that \(1 < q < 2\).
3. Show that \(q\) is a solution of the equation $$x = \frac { 12 } { 7 } \ln ( 2 x + 5 ) - 2 .$$
4. Use an iterative process based on the equation above with a starting value of 1.5 to find the value of \(q\) to 3 significant figures and justify the accuracy of your answer.

8. (i) Solve the equation $$\pi - 3 \cos ^ { - 1 } \theta = 0$$ (ii) 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 $$\cos ^ { - 1 } ( x - 1 ) = \sqrt { x + 2 }$$ (iii) show that \(0 < \alpha < 1\),
(iv) 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.

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

1. Show by calculation that the equation \(\mathrm { f } ( x ) = 0\) has a root in the interval \(0.8 < x < 0.9\).
2. Use the Newton-Raphson method, with a first approximation 0.8, to find the next approximation to this root. Give your answer correct to 3 decimal places.

5 \includegraphics[max width=\textwidth, alt={}, center]{15dd10f9-73d4-4107-bb45-7866f5470572-3_606_890_815_630} The diagram shows the curve with equation \(y = x \mathrm { e } ^ { - x } + 1\). The curve crosses the \(x\)-axis at \(x = \alpha\).

1. Use differentiation to show that the \(x\)-coordinate of the stationary point is 1 . \(\alpha\) is to be found using the Newton-Raphson method, with \(\mathrm { f } ( x ) = x \mathrm { e } ^ { - x } + 1\).
2. Explain why this method will not converge to \(\alpha\) if an initial approximation \(x _ { 1 }\) is chosen such that \(x _ { 1 } > 1\).
3. Use this method, with a first approximation \(x _ { 1 } = 0\), to find the next three approximations \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\). Find \(\alpha\), correct to 3 decimal places.

8 The curve with equation \(y = \frac { \sinh x } { x ^ { 2 } }\), for \(x > 0\), has one turning point.

1. Show that the \(x\)-coordinate of the turning point satisfies the equation \(x - 2 \tanh x = 0\).
2. Use the Newton-Raphson method, with a first approximation \(x _ { 1 } = 2\), to find the next two approximations, \(x _ { 2 }\) and \(x _ { 3 }\), to the positive root of \(x - 2 \tanh x = 0\).
3. By considering the approximate errors in \(x _ { 1 }\) and \(x _ { 2 }\), estimate the error in \(x _ { 3 }\). (You are not expected to evaluate \(x _ { 4 }\).)

6 It is given that \(\mathrm { f } ( x ) = 1 - \frac { 7 } { x ^ { 2 } }\).

1. Use the Newton-Raphson method, with a first approximation \(x _ { 1 } = 2.5\), to find the next approximations \(x _ { 2 }\) and \(x _ { 3 }\) to a root of \(\mathrm { f } ( x ) = 0\). Give the answers correct to 6 decimal places. [3]
2. The root of \(\mathrm { f } ( x ) = 0\) for which \(x _ { 1 } , x _ { 2 }\) and \(x _ { 3 }\) are approximations is denoted by \(\alpha\). Write down the exact value of \(\alpha\).
3. The error \(e _ { n }\) is defined by \(e _ { n } = \alpha - x _ { n }\). Find \(e _ { 1 } , e _ { 2 }\) and \(e _ { 3 }\), giving your answers correct to 5 decimal places. Verify that \(e _ { 3 } \approx \frac { e _ { 2 } ^ { 3 } } { e _ { 1 } ^ { 2 } }\).

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.

7 The curve with equation $$y = \frac { x } { \cosh x }$$ has one stationary point for \(x > 0\).

1. Show that the \(x\)-coordinate of this stationary point satisfies the equation \(x \tanh x - 1 = 0\). The positive root of the equation \(x \tanh x - 1 = 0\) is denoted by \(\alpha\).
2. Draw a sketch showing (for positive values of \(x\) ) the graph of \(y = \tanh x\) and its asymptote, and the graph of \(y = \frac { 1 } { x }\). Explain how you can deduce from your sketch that \(\alpha > 1\).
3. Use the Newton-Raphson method, taking first approximation \(x _ { 1 } = 1\), to find further approximations \(x _ { 2 }\) and \(x _ { 3 }\) for \(\alpha\).
4. By considering the approximate errors in \(x _ { 1 }\) and \(x _ { 2 }\), estimate the error in \(x _ { 3 }\).

1.(i) $$f ( x ) = x ^ { 3 } + 4 x - 6$$

1. Show that the equation \(\mathrm { f } ( x ) = 0\) has a root \(\alpha\) in the interval[1,1.5]
2. Taking 1.5 as a first approximation,apply the Newton Raphson process twice to \(\mathrm { f } ( x )\) to obtain an approximate value of \(\alpha\) .Give your answer to 3 decimal places. Show your working clearly.
   (ii) $$g ( x ) = 4 x ^ { 2 } + x - \tan x$$ where \(x\) is measured in radians. The equation \(\mathrm { g } ( x ) = 0\) has a single root \(\beta\) in the interval[1.4,1.5]
   Use linear interpolation on the values at the end points of this interval to obtain an approximation to \(\beta\) .Give your answer to 3 decimal places.

5 The equation $$x ^ { 3 } - 5 x + 3 = 0$$ may be solved by the Newton-Raphson method. Successive approximations to a root are denoted by \(x _ { 1 } , x _ { 2 } , \ldots , x _ { n } , \ldots\).

1. Show that the Newton-Raphson formula can be written in the form \(x _ { n + 1 } = \mathrm { F } \left( x _ { n } \right)\), where $$\mathrm { F } ( x ) = \frac { 2 x ^ { 3 } - 3 } { 3 x ^ { 2 } - 5 }$$
2. Find \(\mathrm { F } ^ { \prime } ( x )\) and hence verify that \(\mathrm { F } ^ { \prime } ( \alpha ) = 0\), where \(\alpha\) is any one of the roots of equation (A).
3. Use the Newton-Raphson method to find the root of equation (A) which is close to 2 . Write down sufficient approximations to find the root correct to 4 decimal places.

8 It is required to solve the equation \(\ln ( x - 1 ) - x + 3 = 0\).
You are given that there are two roots, \(\alpha\) and \(\beta\), where \(1.1 < \alpha < 1.2\) and \(4.1 < \beta < 4.2\).

1. The root \(\beta\) can be found using the iterative formula $$x _ { n + 1 } = \ln \left( x _ { n } - 1 \right) + 3$$
   1. Using this iterative formula with \(x _ { 1 } = 4.15\), find \(\beta\) correct to 3 decimal places. Show all your working.
   2. Explain with the aid of a sketch why this iterative formula will not converge to \(\alpha\) whatever initial value is taken.
   3. (a) Show that the Newton-Raphson iterative formula for this equation can be written in the form $$x _ { n + 1 } = \frac { 3 - 2 x _ { n } - \left( x _ { n } - 1 \right) \ln \left( x _ { n } - 1 \right) } { 2 - x _ { n } }$$ (b) Use this formula with \(x _ { 1 } = 1.2\) to find \(\alpha\) correct to 3 decimal places.

7

1. Sketch the graph of \(y = \operatorname { coth } x\), and give the equations of any asymptotes.
2. It is given that \(\mathrm { f } ( x ) = x \tanh x - 2\). Use the Newton-Raphson method, with a first approximation \(x _ { 1 } = 2\), to find the next three approximations \(x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\) to a root of \(\mathrm { f } ( x ) = 0\). Give the answers correct to 4 decimal places.
3. If \(\mathrm { f } ( x ) = 0\), show that \(\operatorname { coth } x = \frac { 1 } { 2 } x\). Hence write down the roots of \(\mathrm { f } ( x ) = 0\), correct to 4 decimal places.

7 \includegraphics[max width=\textwidth, alt={}, center]{074597e7-5bb1-4249-9cfa-784974a6fd2b-3_531_1065_1208_539} The line \(y = x\) and the curve \(y = 2 \ln ( 3 x - 2 )\) meet where \(x = \alpha\) and \(x = \beta\), as shown in the diagram.

1. Use the iteration \(x _ { n + 1 } = 2 \ln \left( 3 x _ { n } - 2 \right)\), with initial value \(x _ { 1 } = 5.25\), to find the value of \(\beta\) correct to 2 decimal places. Show all your working.
2. With the help of a 'staircase' diagram, explain why this iteration will not converge to \(\alpha\), whatever value of \(x _ { 1 }\) (other than \(\alpha\) ) is used.
3. Show that the equation \(x = 2 \ln ( 3 x - 2 )\) can be rewritten as \(x = \frac { 1 } { 3 } \left( \mathrm { e } ^ { \frac { 1 } { 2 } x } + 2 \right)\). Use the NewtonRaphson method, with \(\mathrm { f } ( x ) = \frac { 1 } { 3 } \left( \mathrm { e } ^ { \frac { 1 } { 2 } x } + 2 \right) - x\) and \(x _ { 1 } = 1.2\), to find \(\alpha\) correct to 2 decimal places. Show all your working.
4. Given that \(x _ { 1 } = \ln 36\), explain why the Newton-Raphson method would not converge to a root of \(\mathrm { f } ( x ) = 0\).

4 It is given that the equation \(x ^ { 4 } - 2 x - 1 = 0\) has only one positive root, \(\alpha\), and \(1.3 < \alpha < 1.5\).

1. \includegraphics[max width=\textwidth, alt={}, center]{72a1330a-c6dc-4f3a-9b0e-333b099f4509-2_433_424_1119_817} The diagram shows a sketch of \(y = x\) and \(y = \sqrt [ 4 ] { 2 x + 1 }\) for \(x \geqslant 0\). Use the iteration \(x _ { n + 1 } = \sqrt [ 4 ] { 2 x _ { n } + 1 }\) with \(x _ { 1 } = 1.35\) to find \(x _ { 2 }\) and \(x _ { 3 }\), correct to 4 decimal places. On the copy of the diagram show how the iteration converges to \(\alpha\).
2. For the same equation, the iteration \(x _ { n + 1 } = \frac { 1 } { 2 } \left( x _ { n } ^ { 4 } - 1 \right)\) with \(x _ { 1 } = 1.35\) gives \(x _ { 2 } = 1.1608\) and \(x _ { 3 } = 0.4077\), correct to 4 decimal places. Draw a sketch of \(y = x\) and \(y = \frac { 1 } { 2 } \left( x ^ { 4 } - 1 \right)\) for \(x \geqslant 0\), and show how this iteration does not converge to \(\alpha\).
3. Find the positive root of the equation \(x ^ { 4 } - 2 x - 1 = 0\) by using the Newton-Raphson method with \(x _ { 1 } = 1.35\), giving the root correct to 4 decimal places.

5 You are given that the equation \(x ^ { 3 } + 4 x ^ { 2 } + x - 1 = 0\) has a root, \(\alpha\), where \(- 1 < \alpha < 0\).

1. Show that the Newton-Raphson iterative formula for this equation can be written in the form $$x _ { n + 1 } = \frac { 2 x _ { n } ^ { 3 } + 4 x _ { n } ^ { 2 } + 1 } { 3 x _ { n } ^ { 2 } + 8 x _ { n } + 1 } .$$
2. Using the initial value \(x _ { 1 } = - 0.7\), find \(x _ { 2 }\) and \(x _ { 3 }\) and find \(\alpha\) correct to 5 decimal places.
3. The diagram shows a sketch of the curve \(y = x ^ { 3 } + 4 x ^ { 2 } + x - 1\) for \(- 1.5 \leqslant x \leqslant 1\). \includegraphics[max width=\textwidth, alt={}, center]{a80eb21f-c273-4b65-8617-16cdee783305-3_602_926_749_566} Using the copy of the diagram in your answer book, explain why the initial value \(x _ { 1 } = 0\) will fail to find \(\alpha\).

6 It is given that the equation \(3 x ^ { 3 } + 5 x ^ { 2 } - x - 1 = 0\) has three roots, one of which is positive.

1. Show that the Newton-Raphson iterative formula for finding this root can be written $$x _ { n + 1 } = \frac { 6 x _ { n } ^ { 3 } + 5 x _ { n } ^ { 2 } + 1 } { 9 x _ { n } ^ { 2 } + 10 x _ { n } - 1 } .$$
2. A sequence of iterates \(x _ { 1 } , x _ { 2 } , x _ { 3 } , \ldots\) which will find the positive root is such that the magnitude of the error in \(x _ { 2 }\) is greater than the magnitude of the error in \(x _ { 1 }\). On the graph given in the Printed Answer Book, mark a possible position for \(x _ { 1 }\).
3. Apply the iterative formula in part (i) when the initial value is \(x _ { 1 } = - 1\). Describe the behaviour of the iterative sequence, illustrating your answer on the graph given in the Printed Answer Book.
4. A sequence of approximations to the positive root is given by \(x _ { 1 } , x _ { 2 } , x _ { 3 } , \ldots\). Successive differences \(x _ { r } - x _ { r - 1 } = d _ { r }\), where \(r \geqslant 2\), are such that \(d _ { r } \approx k \left( d _ { r - 1 } \right) ^ { 2 }\) where \(k\) is a constant. Show that \(d _ { 4 } \approx \frac { d _ { 3 } ^ { 3 } } { d _ { 2 } ^ { 2 } }\) and demonstrate this numerically when \(x _ { 1 } = 1\).
5. Find the value of the positive root correct to 5 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 $$\mathrm { f } ( x , y ) = ( x - y ) \sqrt { x + y }$$ 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]