GCSE Mathematics (9-1)

14.5 Iteration and numerical methods — Practise

Work through these independently. Choose a method that fits each question. Questions marked H are Higher only.

  1. H Use xn+1=12(xn+6/xn)x_{n+1}=\frac12(x_n+6/x_n) with x0=2x_0=2. Find x1x_1.
  2. H Use xn+1=12(xn+6/xn)x_{n+1}=\frac12(x_n+6/x_n) with x0=2x_0=2. Find x2x_2.
  3. H Use xn+1=12(xn+6/xn)x_{n+1}=\frac12(x_n+6/x_n) with x0=2x_0=2. Find x3x_3 to 3 decimal places.
  4. H Use xn+1=2+1/xnx_{n+1}=2+1/x_n with x0=2x_0=2. Find x2x_2.
  5. H Use xn+1=3+2/xnx_{n+1}=3+2/x_n with x0=2x_0=2. Find x2x_2.
  6. H Use xn+1=12(xn+6/xn)x_{n+1}=\frac12(x_n+6/x_n) with x0=3x_0=3. Find x2x_2.
  7. H Use xn+1=xn/2+1x_{n+1}=x_n/2+1 with x0=8x_0=8. Find x3x_3.
  8. H Use xn+1=1+2/xnx_{n+1}=1+2/x_n with x0=2x_0=2. Find x3x_3.
  9. H Which method fits each part, and why? (a) Solve y=2x+1y=2x+1 and y=3x−4y=3x-4. (b) Use xn+1=2+1/xnx_{n+1}=2+1/x_n with x0=2x_0=2 to find x2x_2.
Answers and concise solutions

1. First iteration

Answer: x1=2.5x_1=2.5.

2. Second iteration

Answer: x2=2.45x_2=2.45.

3. Third iteration, 3 decimal places

Answer: x3≈2.449x_3\approx2.449.

4. Two iterations

Answer: x1=2.5x_1=2.5 and x2=2.4x_2=2.4.

5. Iteration with a different rule

Answer: x1=4x_1=4 and x2=3.5x_2=3.5.

6. Iteration from a new start value

Answer: x1=2.5x_1=2.5 and x2=2.45x_2=2.45.

7. Three steps of iteration

Answer: x1=5x_1=5, x2=3.5x_2=3.5, and x3=2.75x_3=2.75.

8. Three iterations with fractions

Answer: x1=2x_1=2, x2=2x_2=2, and x3=2x_3=2.

9. Method choice: solve or iterate

Answer: Use substitution for (a): the equations are linear, so set 2x+1=3x−42x+1=3x-4, giving x=5x=5 and y=11y=11. Use iteration for (b): the rule defines each term from the last, so x1=5/2x_1=5/2 and x2=12/5=2.4x_2=12/5=2.4.

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