Iterative formula from rearrangement

OCR C3 Q8

10 marks Standard +0.3

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.

OCR FP2 2013 January Q8

13 marks Standard +0.3

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 $4.1 < \beta < 4.2$.

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.

OCR FP2 2012 June Q4

9 marks Standard +0.3

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

\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 $\alpha$.
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 $\alpha$.
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.

Pre-U Pre-U 9794/1 Specimen Q9

16 marks Challenging +1.8

9 The cubic polynomial $x ^ { 3 } + a x ^ { 2 } + b x + c$, where $a , b$ and $c$ are real, is denoted by $\mathrm { p } ( x )$.

Give a reason why the equation $\mathrm { p } ( x ) = 0$ has at least one real root.
Suppose that the curve with equation $y = \mathrm { p } ( x )$ has a local minimum point and a local maximum point with $y$-coordinates $y _ { \text {min } }$ and $y _ { \text {max } }$ respectively.
1. Prove that if $y _ { \text {min } } y _ { \text {max } } < 0$, then the equation $\mathrm { p } ( x ) = 0$ has three real roots.
2. Comment on the number of distinct real roots of the equation $\mathrm { p } ( x ) = 0$ in the case $y _ { \text {min } } y _ { \text {max } } = 0$.
3. Suppose instead that the equation $\mathrm { p } ( x ) = 0$ has only one real root for all values of $c$. Prove that $a ^ { 2 } \leqslant 3 b$.
4. The iterative scheme $$x _ { n + 1 } = \frac { 2 x _ { n } ^ { 3 } + 1 } { 3 x _ { n } ^ { 2 } + 1 } , \quad x _ { 0 } = 0$$ converges to a root of the cubic equation $\mathrm { p } ( x ) = 0$.
  (a) Find $\mathrm { p } ( x )$.
  (b) Find the limit of the iteration, correct to 4 decimal places.
5. Determine the rate of convergence of the iterative scheme.