A-level Maths revision · Pure, year 2

Fixed-point iteration

The idea

Rearrange f(x)=0f(x) = 0 into x=g(x)x = g(x), then iterate xn+1=g(xn)x_{n+1} = g(x_n). If the values settle, they approach a root. On a graph of y=xy = x and y=g(x)y = g(x), the steps form a staircase or a cobweb.

  1. Start with the given x1x_1.
  2. Substitute into gg to get x2x_2, then x3x_3, keeping full calculator accuracy.
  3. An iteration converges only if ∣g′(x)∣<1|g'(x)| < 1 near the root.

Watch out Not every rearrangement converges: some move away from the root.

See it worked

Worked example

Use xn+1=2xn+53x_{n+1} = \sqrt[3]{2x_n + 5} with x1=2x_1 = 2 to find x2x_2, x3x_3 and x4x_4 to 4 decimal places.

  1. x2=93=2.0801x_2 = \sqrt[3]{9} = 2.0801.
  2. x3=9.16023=2.0924x_3 = \sqrt[3]{9.1602} = 2.0924.
  3. x4=2.0942x_4 = 2.0942.

Answer 2.08012.0801, 2.09242.0924, 2.09422.0942

Your turn

Question 1Calculator allowed3 marks

Use xn+1=xn+3x_{n+1} = \sqrt{x_n + 3} with x1=2x_1 = 2 to find x3x_3 to 4 decimal places.

Question 2Non-calculator3 marks

The iteration xn+1=xn+3x_{n+1} = \sqrt{x_n + 3} converges to a root of x2−x−3=0x^2 - x - 3 = 0. Find the root exactly.

Question 3Non-calculator3 marks

Use xn+1=1+1xnx_{n+1} = 1 + \frac{1}{x_n} with x1=1x_1 = 1 to find x4x_4 exactly.

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