Rearrange to iterative form - Fixed Point Iteration

11 marks Standard +0.3

The curve with equation $y = \ln 3x$ crosses the $x$-axis at the point $P (p, 0)$.

The normal to the curve at the point $Q$, with $x$-coordinate $q$, passes through the origin.

Show that $x = q$ is a solution of the equation $x^2 + \ln 3x = 0$. [4]
Show that the equation in part (b) can be rearranged in the form $x = \frac{1}{3}e^{-x^2}$. [2]
Use the iteration formula $x_{n+1} = \frac{1}{3}e^{-x_n^2}$, with $x_0 = \frac{1}{4}$, to find $x_1, x_2, x_3$ and $x_4$. Hence write down, to 3 decimal places, an approximation for $q$. [3]

10 marks Standard +0.2

$\text{f}(x) = x^3 + x^2 - 4x - 1$. The equation f(x) = 0 has only one positive root, $\alpha$.

Show that f(x) = 0 can be rearranged as $$x = \sqrt{\frac{4x+1}{x+1}}, \quad x \neq -1.$$ [2]

The iterative formula $x_{n+1} = \sqrt{\frac{4x_n+1}{x_n+1}}$ is used to find an approximation to $\alpha$.

Taking $x_1 = 1$, find, to 2 decimal places, the values of $x_2$, $x_3$ and $x_4$. [3]
By choosing values of $x$ in a suitable interval, prove that $\alpha = 1.70$, correct to 2 decimal places. [3]
Write down a value of $x_1$ for which the iteration formula $x_{n+1} = \sqrt{\frac{4x_n+1}{x_n+1}}$ does not produce a valid value for $x_2$. Justify your answer. [2]

11 marks Standard +0.3

The curve with equation $y = \ln 3x$ crosses the $x$-axis at the point $P(p, 0)$.

The normal to the curve at the point $Q$, with $x$-coordinate $q$, passes through the origin.

Show that $x = q$ is a solution of the equation $x^2 + \ln 3x = 0$. [4]
Show that the equation in part (b) can be rearranged in the form $x = \frac{1}{3}e^{-x^2}$. [2]
Use the iteration formula $x_{n + 1} = \frac{1}{3}e^{-x_n^2}$, with $x_0 = \frac{1}{3}$, to find $x_1, x_2, x_3$ and $x_4$. Hence write down, to 3 decimal places, an approximation for $q$. [3]

13 marks Standard +0.3

$$f(x) = 2x^2 + 3 \ln (2 - x), \quad x \in \mathbb{R}, \quad x < 2.$$

Show that the equation $f(x) = 0$ can be written in the form $$x = 2 - e^{kx^2},$$ where $k$ is a constant to be found. [3]

The root, $\alpha$, of the equation $f(x) = 0$ is $1.9$ correct to $1$ decimal place.

Use the iteration formula $$x_{n+1} = 2 - e^{kx_n^2},$$ with $x_0 = 1.9$ and your value of $k$, to find $\alpha$ to $3$ decimal places and justify the accuracy of your answer. [5]
Solve the equation $f'(x) = 0$. [5]

13 marks Challenging +1.2

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

Show that $t = \sqrt[3]{\frac{1}{2} + \frac{1}{3}t - \frac{1}{3}t^2}$, where $t = \tan x$. [3]
Show by calculation that the value of $t$ satisfying the equation in part (i) lies between 0.7 and 0.8. [2]
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]
Solve the equation $3\tan 4y - 8\tan 2y = 4$ for $0 < y < \frac{1}{4}\pi$. [2]

18 marks Standard +0.3

A curve $C$ has equation $y = f(x)$ where $$f(x) = x + 2\ln(e - x)$$

1. Show that the equation of the normal to $C$ at the point where $C$ crosses the $y$-axis is given by $$y = \left(\frac{e}{2-e}\right)x + 2$$ [6 marks]
2. Find the exact area enclosed by the normal and the coordinate axes. Fully justify your answer. [3 marks]

The equation $f(x) = 0$ has one positive root, $\alpha$.
1. Show that $\alpha$ lies between 2 and 3 Fully justify your answer. [3 marks]
2. Show that the roots of $f(x) = 0$ satisfy the equation $$x = e - e^{-\frac{x}{2}}$$ [2 marks]
3. Use the recurrence relation $$x_{n+1} = e - e^{-\frac{x_n}{2}}$$ with $x_1 = 2$ to find the values of $x_2$ and $x_3$ giving your answers to three decimal places. [2 marks]
4. Figure 1 below shows a sketch of the graphs of $y = e - e^{-\frac{x}{2}}$ and $y = x$, and the position of $x_1$ On Figure 1, draw a cobweb or staircase diagram to show how convergence takes place, indicating the positions of $x_2$ and $x_3$ on the $x$-axis. [2 marks] \includegraphics{figure_1}

3 marks Standard +0.3

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

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]
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]