Show that the equation f(x) = x3 - 2x - 1 has a root between x = 1 and x = 2.
(Total for Question 1 is 1 mark)
2
Use the fixed-point iteration x_{n+1} = (xn + 3/xn)/2 to approximate √3. Start with x0 = 2. Find x1 and x2, each to 3 decimal places.
(Total for Question 2 is 3 marks)
3
Use the iteration x_{n+1} = (xn + 2/xn)/2 to approximate √2. Start with x0 = 1.5. Find x2 to 3 decimal places.
(Total for Question 3 is 3 marks)
4
Show that the polynomial f(x) = x4 - 3x + 1 has a root between x = 0 and x = 1.
(Total for Question 4 is 1 mark)
5
Consider the equation x3 + 4x - 10 = 0 and the iteration x_{n+1} = (10 - xn3)/4. Starting with x0 = 1.5, calculate x1 and x2. From these two iterations, state whether the sequence appears to be converging.
(Total for Question 5 is 3 marks)
6
Use the iteration x_{n+1} = cos(xn) with x0 = 1. Find x2 and x3 to 3 decimal places. From the trend of values, state the value the sequence appears to be approaching (to 3 d.p.).
(Total for Question 6 is 4 marks)
7
Use the Newton-Raphson method to find a root of f(x) = x3 - x - 2. Start with x0 = 1.5. Perform two iterations and give your result x2 to 4 decimal places.
(Total for Question 7 is 4 marks)
8
The equation x2 + x - 6 = 0 has a positive root. Rearrange it to the fixed-point form x = 6/(x + 1). Starting with x0 = 1.5, perform two iterations using x_{n+1} = 6/(xn + 1). Give x2 to 3 decimal places and say whether the values appear to be approaching the positive root.
(Total for Question 8 is 3 marks)
9
Show that f(x) = x3 - 2x - 5 has a root between x = 2 and x = 3. Then use the iteration x_{n+1} = (5 + 2xn)1/3 with x0 = 2.5 to compute x1 and x2. Comment briefly on whether this rearrangement seems to produce convergence.
(Total for Question 9 is 4 marks)
10
Use Newton-Raphson to approximate the cube root of 7 by finding a root of f(x) = x3 - 7. Start with x0 = 2. Perform two iterations and give x2 to 4 decimal places.
(Total for Question 10 is 4 marks)
Mark scheme · 7.10D Iteration: Fluency and Exam Drill
Question 1
B1 f(1) = -2 and f(2) = 3 so signs differ, hence a root in (1, 2) cao
Answer: f(1) = 1 - 2 - 1 = -2, f(2) = 8 - 4 - 1 = 3. Since f(1) < 0 and f(2) > 0 there is a root between 1 and 2 (by intermediate value theorem).
Question 2
M1 substitute x0 = 2 into x_{n+1} = (xn + 3/xn)/2 and calculate x1
B1 conclude that the sequence does not appear to be converging (it oscillates) oe
Answer: x1 = (10 - 1.53)/4 = 6.625/4 = 1.65625; x2 = (10 - 1.656253)/4 approx 1.36415863. The values jump (1.5 -> 1.65625 -> 1.36416), so the sequence appears to oscillate and not settle (not converging).
Question 6
M1 calculate x1 = cos(1) and x2 = cos(x1)
M1 calculate x3 = cos(x2)
M1 give x2 and x3 to 3 dp correctly
A1 state the sequence appears to approach 0.739 (awrt) cao
Answer: x1 = cos(1) approx 0.5403023059; x2 = cos(0.5403023059) approx 0.8575532158 -> 0.858; x3 = cos(0.8575532158) approx 0.6542897905 -> 0.654. The sequence appears to approach approximately 0.739 (the fixed point of cos x).
Question 7
M1 state Newton formula x_{n+1} = xn - f(xn)/f'(xn) with f' = 3x2 - 1
A1 x2 = 1.765 (awrt); values appear to be moving toward the positive root x = 2 (comment) oe
Answer: x1 = 2.4; x2 approx 1.764706 -> 1.765. The iterations move around the true positive root 2 and appear to be approaching it (oscillating but converging).
A1 conclude whether iteration appears to converge (values move toward ~2.094...) cao
Answer: f(2) = -1, f(3) = 16 so a root lies between 2 and 3. x1 = 101/3 approx 2.154435; x2 approx 9.308871/3 approx 2.104. The values move toward about 2.094; the rearrangement appears to produce convergence.
Question 10
M1 state Newton formula x_{n+1} = xn - (xn3 - 7)/(3xn2)