1.09d Newton-Raphson method

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]

6 The following flow chart has been written to find a root of the cubic equation \(x ^ { 3 } + A x ^ { 2 } + B x + C = 0\), given a starting value \(X\) that is thought to be near the root. \includegraphics[max width=\textwidth, alt={}, center]{ccb12789-cd5f-40dc-9f10-f8bb45399580-8_1410_1648_324_212}

1. Work through the algorithm, recording the values of \(X , Y , Z\) and \(W\) each time they change, for the equation \(x ^ { 3 } - 4 x ^ { 2 } + 5 x + 1 = 0\), with a starting value of \(X = 0\).
2. Show what happens when the algorithm is used for the equation \(x ^ { 3 } - 4 x ^ { 2 } + 5 x + 1 = 0\), with a starting value of \(X = 1\).
3. Show what happens when the algorithm is used for the equation \(x ^ { 3 } - 4 x ^ { 2 } + 5 x + 1 = 0\), with a starting value of \(X = - 1\).
4. Identify a possible problem with using this algorithm.

7 The curve \(y = \left( x ^ { 2 } - 2 \right) \ln x\) has one stationary point which is close to \(x = 1\).

1. Show that the \(x\)-coordinate of this stationary point satisfies the equation \(2 x ^ { 2 } \ln x + x ^ { 2 } - 2 = 0\).
2. Show that the Newton-Raphson iterative formula for finding the root of the equation in part (a) can be written in the form \(x _ { n + 1 } = \frac { 2 x _ { n } ^ { 2 } \ln x _ { n } + 3 x _ { n } ^ { 2 } + 2 } { 4 x _ { n } \left( \ln x _ { n } + 1 \right) }\).
3. Apply the Newton-Raphson formula with initial value \(x _ { 1 } = 1\) to find \(x _ { 2 }\) and \(x _ { 3 }\).
4. Find the coordinates of this stationary point, giving each coordinate correct to \(\mathbf { 3 }\) decimal places.

5. \(\mathrm { f } ( x ) = \frac { 1 } { 3 } x ^ { 3 } - 4 x - 2\) a. Show that the equation \(\mathrm { f } ( x ) = 0\) can be written in the form \(x = \pm \sqrt { a + \frac { b } { x } }\), and state the values of the integers \(a\) and \(b\). \(\mathrm { f } ( x ) = 0\) has one positive root, \(\alpha\).
The iterative formula \(x _ { n + 1 } = \sqrt { a + \frac { b } { x _ { n } } } , \quad x _ { 0 } = 4\) is used to find an approximation value for \(\alpha\).
b. Calculate the values of \(x _ { 1 } , x _ { 2 } , x _ { 3 }\) and \(x _ { 4 }\) to 4 decimal places.
c. Explain why for this question, the Newton-Raphson method cannot be used with \(x _ { 1 } = 2\).

6. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of part of the curve with equation $$f ( x ) = 4 \cos 2 x - 2 x + 1 \quad x > 0$$ and where \(x\) is measured in radians.
The curve crosses the \(x\)-axis at the point \(A\), as shown in figure 1 .
Given that \(x\)-coordinate of \(A\) is \(\alpha\) a. show that \(\alpha\) lies between 0.7 and 0.8 Given that \(x\)-coordinates of \(B\) and \(C\) are \(\beta\) and \(\gamma\) respectively and they are two smallest values of \(x\) at which local maxima occur
b. find, using calculus, the value of \(\beta\) and the value of \(\gamma\), giving your answers to 3 significant figures.
c. taking \(x _ { 0 } = 0.7\) or 0.8 as a first approximation to \(\alpha\), apply the Newton-Raphson method once to \(\mathrm { f } ( x )\) to obtain a second approximation to \(\alpha\). Show, your method and give your answer to 2 significant figures.

8. \begin{figure}[h] \end{figure} A car stops at two sets of traffic lights.
Figure 2 shows a graph of the speed of the car, \(v \mathrm {~ms} ^ { - 1 }\), as it travels between the two sets of traffic lights. The car takes \(T\) seconds to travel between the two sets of traffic lights.
The speed of the car is modelled by the equation $$v = ( 10 - 0.4 t ) \ln ( t + 1 ) \quad 0 \leqslant t \leqslant T$$ where \(t\) seconds is the time after the car leaves the first set of traffic lights.
According to the model,

1. find the value of \(T\)
2. show that the maximum speed of the car occurs when $$t = \frac { 26 } { 1 + \ln ( t + 1 ) } - 1$$ Using the iteration formula $$t _ { n + 1 } = \frac { 26 } { 1 + \ln \left( t _ { n } + 1 \right) } - 1$$ with \(t _ { 1 } = 7\)
3. 1. find the value of \(t _ { 3 }\) to 3 decimal places,
   2. find, by repeated iteration, the time taken for the car to reach maximum speed.

1. A curve has equation \(y = \mathrm { f } ( x )\), where

$$\mathrm { f } ( x ) = \frac { 7 x \mathrm { e } ^ { x } } { \sqrt { \mathrm { e } ^ { 3 x } - 2 } } \quad x > \ln \sqrt [ 3 ] { 2 }$$

1. Show that $$\mathrm { f } ^ { \prime } ( x ) = \frac { 7 \mathrm { e } ^ { x } \left( \mathrm { e } ^ { 3 x } ( 2 - x ) + A x + B \right) } { 2 \left( \mathrm { e } ^ { 3 x } - 2 \right) ^ { \frac { 3 } { 2 } } }$$ where \(A\) and \(B\) are constants to be found.
2. Hence show that the \(x\) coordinates of the turning points of the curve are solutions of the equation $$x = \frac { 2 \mathrm { e } ^ { 3 x } - 4 } { \mathrm { e } ^ { 3 x } + 4 }$$ The equation \(x = \frac { 2 \mathrm { e } ^ { 3 x } - 4 } { \mathrm { e } ^ { 3 x } + 4 }\) has two positive roots \(\alpha\) and \(\beta\) where \(\beta > \alpha\) A student uses the iteration formula $$x _ { n + 1 } = \frac { 2 \mathrm { e } ^ { 3 x _ { n } } - 4 } { \mathrm { e } ^ { 3 x _ { n } } + 4 }$$ in an attempt to find approximations for \(\alpha\) and \(\beta\) Diagram 1 shows a plot of part of the curve with equation \(y = \frac { 2 \mathrm { e } ^ { 3 x } - 4 } { \mathrm { e } ^ { 3 x } + 4 }\) and part of the line with equation \(y = x\) Using Diagram 1 on page 42
3. draw a staircase diagram to show that the iteration formula starting with \(x _ { 1 } = 1\) can be used to find an approximation for \(\beta\) Use the iteration formula with \(x _ { 1 } = 1\), to find, to 3 decimal places,
4. 1. the value of \(x _ { 2 }\)
   2. the value of \(\beta\) Using a suitable interval and a suitable function that should be stated
5. show that \(\alpha = 0.432\) to 3 decimal places. Only use the copy of Diagram 1 if you need to redraw your answer to part (c). \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{0839eb5f-2850-4d77-baf7-a6557d71076e-42_736_812_372_143} \captionsetup{labelformat=empty} \caption{Diagram 1}
   \end{figure} \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{0839eb5f-2850-4d77-baf7-a6557d71076e-42_738_815_370_1114} \captionsetup{labelformat=empty} \caption{copy of Diagram 1}
   \end{figure}

3. $$\mathrm { f } ( x ) = x + \tan \left( \frac { 1 } { 2 } x \right) \quad \pi < x < \frac { 3 \pi } { 2 }$$ Given that the equation \(\mathrm { f } ( x ) = 0\) has a single root \(\alpha\)

1. show that \(\alpha\) lies in the interval [3.6, 3.7]
2. Find \(\mathrm { f } ^ { \prime } ( x )\)
3. Using 3.7 as a first approximation for \(\alpha\), apply the Newton-Raphson method once to obtain a second approximation for \(\alpha\). Give your answer to 3 decimal places.

7. Figure 2 Figure 2 shows a sketch of a triangle \(A B C\).
Given \(\overrightarrow { A B } = 2 \mathbf { i } + 3 \mathbf { j } + \mathbf { k }\) and \(\overrightarrow { B C } = \mathbf { i } - 9 \mathbf { j } + 3 \mathbf { k }\),
show that \(\angle B A C = 105.9 ^ { \circ }\) to one decimal place.

11. \begin{figure}[h] \end{figure} Figure 8 shows a sketch of the curve \(C\) with equation \(y = x ^ { x } , x > 0\)

1. Find, by firstly taking logarithms, the \(x\) coordinate of the turning point of \(C\).
   (Solutions based entirely on graphical or numerical methods are not acceptable.) The point \(P ( \alpha , 2 )\) lies on \(C\).
2. Show that \(1.5 < \alpha < 1.6\) A possible iteration formula that could be used in an attempt to find \(\alpha\) is $$x _ { n + 1 } = 2 x _ { n } ^ { 1 - x _ { n } }$$ Using this formula with \(x _ { 1 } = 1.5\)
3. find \(x _ { 4 }\) to 3 decimal places,
4. describe the long-term behaviour of \(x _ { n }\)

6. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of part of the curve with equation \(y = \mathrm { f } ( x )\) where $$f ( x ) = 8 \sin \left( \frac { 1 } { 2 } x \right) - 3 x + 9 \quad x > 0$$ and \(x\) is measured in radians.
The point \(P\), shown in Figure 2, is a local maximum point on the curve.
Using calculus and the sketch in Figure 2,

1. find the \(x\) coordinate of \(P\), giving your answer to 3 significant figures. The curve crosses the \(x\)-axis at \(x = \alpha\), as shown in Figure 2 .
   Given that, to 3 decimal places, \(f ( 4 ) = 4.274\) and \(f ( 5 ) = - 1.212\)
2. explain why \(\alpha\) must lie in the interval \([ 4,5 ]\)
3. Taking \(x _ { 0 } = 5\) as a first approximation to \(\alpha\), apply the Newton-Raphson method once to \(\mathrm { f } ( x )\) to obtain a second approximation to \(\alpha\). Show your method and give your answer to 3 significant figures.

6. \begin{figure}[h] \end{figure} Figure 1 shows a sketch of the curves with equations \(y = \mathrm { f } ( x )\) and \(y = \mathrm { g } ( x )\) where $$\begin{array} { l l } \mathrm { f } ( x ) = \mathrm { e } ^ { 4 x ^ { 2 } - 1 } & x > 0 \\ \mathrm {~g} ( x ) = 8 \ln x & x > 0 \end{array}$$

1. Find
   1. \(\mathrm { f } ^ { \prime } ( x )\)
   2. \(\mathrm { g } ^ { \prime } ( x )\) Given that \(\mathrm { f } ^ { \prime } ( x ) = \mathrm { g } ^ { \prime } ( x )\) at \(x = \alpha\)
2. show that \(\alpha\) satisfies the equation $$4 x ^ { 2 } + 2 \ln x - 1 = 0$$ The iterative formula $$x _ { n + 1 } = \sqrt { \frac { 1 - 2 \ln x _ { n } } { 4 } }$$ is used with \(x _ { 1 } = 0.6\) to find an approximate value for \(\alpha\)
3. Calculate, giving each answer to 4 decimal places,
   1. the value of \(x _ { 2 }\)
   2. the value of \(\alpha\)

6. \begin{figure}[h] \end{figure} Figure 2 shows a sketch of the curve with equation \(y = \mathrm { f } ( x )\), where $$\mathrm { f } ( x ) = ( 8 - x ) \ln x , \quad x > 0$$ The curve cuts the \(x\)-axis at the points \(A\) and \(B\) and has a maximum turning point at \(Q\), as shown in Figure 2.

1. Find the \(x\) coordinate of \(A\) and the \(x\) coordinate of \(B\).
2. Show that the \(x\) coordinate of \(Q\) satisfies $$x = \frac { 8 } { 1 + \ln x }$$
3. Show that the \(x\) coordinate of \(Q\) lies between 3.5 and 3.6
4. Use the iterative formula $$x _ { n + 1 } = \frac { 8 } { 1 + \ln x _ { n } } \quad n \in \mathbb { N }$$ with \(x _ { 1 } = 3.5\) to
   1. find the value of \(x _ { 5 }\) to 4 decimal places,
   2. find the \(x\) coordinate of \(Q\) accurate to 2 decimal places.

14 Fig. 14 shows a circle with centre O and radius \(r \mathrm {~cm}\). The chord AB is such that angle \(\mathrm { AOB } = x\) radians. The area of the shaded segment formed by AB is \(5 \%\) of the area of the circle. \begin{figure}[h] \end{figure}

1. Show that \(x - \sin x - \frac { 1 } { 10 } \pi = 0\). The Newton-Raphson method is to be used to find \(x\).
2. Write down the iterative formula to be used for the equation in part (a).
3. Use three iterations of the Newton-Raphson method with \(x _ { 0 } = 1.2\) to find the value of \(x\) to a suitable degree of accuracy.

8 Kareem wants to solve the equation \(\sin 4 x + \mathrm { e } ^ { - x } + 0.75 = 0\). He uses his calculator to create the following table of values for \(\mathrm { f } ( x ) = \sin 4 x + \mathrm { e } ^ { - x } + 0.75\). He argues that because \(\mathrm { f } ( 6 )\) is the first negative value in the table, there is a root of the equation between 5 and 6 .

1. Comment on the validity of his argument. The diagram shows the graph of \(y = \sin 4 x + e ^ { - x } + 0.75\). \includegraphics[max width=\textwidth, alt={}, center]{4fac72cb-85cb-48d9-8817-899ef3f80a0f-07_538_1260_920_244}
2. Explain why Kareem failed to find other roots between 0 and 6 . Kareem decides to use the Newton-Raphson method to find the root close to 3 .
3. 1. Determine the iterative formula he should use for this equation.
   2. Use the Newton-Raphson method with \(x _ { 0 } = 3\) to find a root of the equation \(\mathrm { f } ( x ) = 0\). Show three iterations and give your answer to a suitable degree of accuracy. Kareem uses the Newton-Raphson method with \(x _ { 0 } = 5\) and also with \(x _ { 0 } = 6\) to try to find the root which lies between 5 and 6 . He produces the following tables.

      \(x _ { 0 }\)	5
      \(x _ { 1 }\)	3.97288
      \(x _ { 2 }\)	4.12125

      \(x _ { 0 }\)	6
      \(x _ { 1 }\)	6.09036
      \(x _ { 2 }\)	6.07110

4. 1. For the iteration beginning with \(x _ { 0 } = 5\), represent the process on the graph in the Printed Answer Booklet.
   2. Explain why the method has failed to find the root which lies between 5 and 6 .
   3. Explain how Kareem can adapt his method to find the root between 5 and 6 .

11 Fig. 11.1 shows the curve with equation \(\mathrm { y } = \mathrm { g } ( \mathrm { x } )\) where \(\mathrm { g } ( x ) = x \sin x + \cos x\) and the curve of the gradient function \(\mathrm { y } = \mathrm { g } ^ { \prime } ( \mathrm { x } )\) for \(- 2 \pi \leqslant x \leqslant 2 \pi\). \begin{figure}[h] \end{figure}

1. Show that the \(x\)-coordinates of the points on the curve \(y = g ( x )\) where the gradient is 1 satisfy the equation \(\frac { 1 } { x } - \cos x = 0\). Fig. 11.2 shows part of the curve with equation \(y = \frac { 1 } { x } - \cos x\). \begin{figure}[h]
   \captionsetup{labelformat=empty} \caption{Fig. 11.2} \includegraphics[alt={},max width=\textwidth]{60e1e785-c34b-48ef-a63f-13a25fee186e-09_678_1363_424_239}
   \end{figure}
2. Use the Newton-Raphson method with a suitable starting value to find the smallest positive \(x\)-coordinate of a point on the curve \(y = x \sin x + \cos x\) where the gradient is 1 . You should write down at least the following.

5 Fig. 5 shows part of the curve \(y = \operatorname { cosec } x\) together with the \(x\) - and \(y\)-axes. \begin{figure}[h] \end{figure}

1. For the section of the curve which is shown in Fig. 5, write down
   1. the equations of the two vertical asymptotes,
   2. the coordinates of the minimum point.
2. Show that the equation \(x = \operatorname { cosec } x\) has a root which lies between \(x = 1\) and \(x = 2\).
3. Use the iteration \(\mathrm { x } _ { \mathrm { n } + 1 } = \operatorname { cosec } \left( \mathrm { x } _ { \mathrm { n } } \right)\), with \(x _ { 0 } = 1\), to find
   1. the values of \(x _ { 1 }\) and \(x _ { 2 }\), correct to 5 decimal places,
   2. this root of the equation, correct to 3 decimal places.
4. There is another root of \(x = \operatorname { cosec } x\) which lies between \(x = 2\) and \(x = 3\). Determine whether the iteration \(\mathrm { x } _ { \mathrm { n } + 1 } = \operatorname { cosec } \left( \mathrm { x } _ { \mathrm { n } } \right)\) with \(x _ { 0 } = 2.5\) converges to this root.
5. Sketch the staircase or cobweb diagram for the iteration, starting with \(x _ { 0 } = 2.5\), on the diagram in the Printed Answer Booklet.

10 The function \(\mathrm { f } ( x )\) is defined by \(\mathrm { f } ( x ) = x ^ { 4 } + x ^ { 3 } - 2 x ^ { 2 } - 4 x - 2\).

1. Show that \(x = - 1\) is a root of \(\mathrm { f } ( x ) = 0\).
2. Show that another root of \(\mathrm { f } ( x ) = 0\) lies between \(x = 1\) and \(x = 2\).
3. Show that \(\mathrm { f } ( x ) = ( x + 1 ) \mathrm { g } ( x )\), where \(\mathrm { g } ( x ) = x ^ { 3 } + a x + b\) and \(a\) and \(b\) are integers to be determined.
4. Without further calculation, explain why \(\mathrm { g } ( x ) = 0\) has a root between \(x = 1\) and \(x = 2\).
5. Use the Newton-Raphson formula to show that an iteration formula for finding roots of \(\mathrm { g } ( x ) = 0\) may be written $$x _ { n + 1 } = \frac { 2 x _ { n } ^ { 3 } + 2 } { 3 x _ { n } ^ { 2 } - 2 }$$ Determine the root of \(\mathrm { g } ( x ) = 0\) which lies between \(x = 1\) and \(x = 2\) correct to 4 significant figures.

6. (a) Use the derivative of \(\cos x\) to prove that $$\frac { \mathrm { d } } { \mathrm {~d} x } ( \sec x ) = \sec x \tan x$$ The curve \(C\) has the equation \(y = \mathrm { e } ^ { 2 x } \sec x , - \frac { \pi } { 2 } < x < \frac { \pi } { 2 }\).
(b) Find an equation for the tangent to \(C\) at the point where it crosses the \(y\)-axis.
(c) Find, to 2 decimal places, the \(x\)-coordinate of the stationary point of \(C\).

3 A particle of mass 4 kg moves along the \(x\)-axis. At time \(t\) seconds the particle is \(x \mathrm {~m}\) from the origin O and has velocity \(v \mathrm {~ms} ^ { - 1 }\). A driving force of magnitude \(20 t \mathrm { t } ^ { - t } \mathrm {~N}\) is applied in the positive \(x\) direction. Initially \(v = 2\) and the particle is at O .

1. Find, in either order, the impulse of the force over the first 3 seconds and the velocity of the particle after 3 seconds. From time \(t = 3\) a resistive force of magnitude \(\frac { 1 } { 2 } t \mathrm {~N}\) and the driving force are applied until the particle comes to rest.
2. Show that, after the resistive force is applied, the only time at which the resultant force on the particle is zero is when \(t = \ln 40\). Hence find the maximum velocity of the particle during the motion.
3. Given that the time \(T\) seconds at which the particle comes to rest is given by the equation \(T = \sqrt { 121 - 80 \mathrm { e } ^ { - T } ( 1 + T ) }\), without solving the equation deduce that \(T \approx 11\).
4. Use a numerical method to find \(T\) correct to 4 decimal places.

8

1. The equation $$x ^ { 3 } + 2 x ^ { 2 } + x - 100000 = 0$$ has one real root. Taking \(x _ { 1 } = 50\) as a first approximation to this root, use the Newton-Raphson method to find a second approximation, \(x _ { 2 }\), to the root.
2. 1. Given that \(S _ { n } = \sum _ { r = 1 } ^ { n } r ( 3 r + 1 )\), use the formulae for \(\sum _ { r = 1 } ^ { n } r ^ { 2 }\) and \(\sum _ { r = 1 } ^ { n } r\) to show that $$S _ { n } = n ( n + 1 ) ^ { 2 }$$
   2. The lowest integer \(n\) for which \(S _ { n } > 100000\) is denoted by \(N\). Show that $$N ^ { 3 } + 2 N ^ { 2 } + N - 100000 > 0$$
3. Find the value of \(N\), justifying your answer.

5 The diagram below (not to scale) shows a part of a curve \(y = \mathrm { f } ( x )\) which passes through the points \(A ( 2 , - 10 )\) and \(B ( 5,22 )\).

1. 1. On the diagram, draw a line which illustrates the method of linear interpolation for solving the equation \(\mathrm { f } ( x ) = 0\). The point of intersection of this line with the \(x\)-axis should be labelled \(P\).
   2. Calculate the \(x\)-coordinate of \(P\). Give your answer to one decimal place.
2. 1. On the same diagram, draw a line which illustrates the Newton-Raphson method for solving the equation \(\mathrm { f } ( x ) = 0\), with initial value \(x _ { 1 } = 2\). The point of intersection of this line with the \(x\)-axis should be labelled \(Q\).
   2. Given that the gradient of the curve at \(A\) is 8 , calculate the \(x\)-coordinate of \(Q\). Give your answer as an exact decimal. \includegraphics[max width=\textwidth, alt={}, center]{f9345653-d426-4350-bf1d-901506211078-3_876_1063_1779_523}

7 The equation $$24 x ^ { 3 } + 36 x ^ { 2 } + 18 x - 5 = 0$$ has one real root, \(\alpha\).

1. Show that \(\alpha\) lies in the interval \(0.1 < x < 0.2\).
2. Starting from the interval \(0.1 < x < 0.2\), use interval bisection twice to obtain an interval of width 0.025 within which \(\alpha\) must lie.
3. Taking \(x _ { 1 } = 0.2\) as a first approximation to \(\alpha\), use the Newton-Raphson method to find a second approximation, \(x _ { 2 }\), to \(\alpha\). Give your answer to four decimal places.
   (4 marks)

1 The equation $$x ^ { 3 } - x ^ { 2 } + 4 x - 900 = 0$$ has exactly one real root, \(\alpha\). Taking \(x _ { 1 } = 10\) as a first approximation to \(\alpha\), use the Newton-Raphson method to find a second approximation, \(x _ { 2 }\), to \(\alpha\). Give your answer to four significant figures.
(3 marks)