Rearrange an equation into x = g(x), then feed each answer back in until the values settle on a root.
What you will learn
- Use an iteration formula x₍ₙ₊₁₎ = g(xₙ)
- Iterate to a given accuracy
- Show which equation an iteration solves
- Check a root by substitution
Feed the answer back in
x³ + 2x − 7 = 0 can be rearranged to x = ∛(7 − 2x).
That suggests the iteration
x₍ₙ₊₁₎ = ∛(7 − 2xₙ)
Start with x₀ = 1: x₁ = ∛5 = 1.7100, x₂ = ∛3.5800 = 1.5299, x₃ = 1.6143, ... The values close in on the root, 1.5946... On a calculator: type 1, press =, then type ∛(7 − 2Ans) and press = again and again. Round each value to 4 decimal places when you write it down, but let Ans carry the full value into the next step.
To give a root to 3 decimal places, keep going until two values in a row round to the same 3 d.p. number.
Iterations can be simple too. x₍ₙ₊₁₎ = 0.5xₙ + 3 with x₀ = 2: x₁ = 0.5 × 2 + 3 = 4, and x₂ = 0.5 × 4 + 3 = 5.
x₍ₙ₊₁₎ = ½(xₙ + N/xₙ) closes in on √N. With N = 10 and x₀ = 3: x₁ = ½(3 + 10/3) = 3.1667, and x₂ = 3.1623, which is √10 to 4 d.p.
The rest of this lesson and its 7 practice questions are locked.
In this lesson
- Feed the answer back in
- Why it works, and checking
- Showing a rearrangement
- Try it: Watch an iteration settle
7 practice questions, marked as you go; about 25 minutes in all.
In this unit: Proof and iteration
Algebraic proof and counterexamples, then finding roots that do not come out exactly: changes of sign and iteration. Higher tier.