Iterative methods

Many equations can't be solved exactly. Iteration finds a root to any accuracy by repeating a calculation.

First, find where a root is. If an expression is negative at one value of x and positive at another, its graph must cross 0 in between.

1. Substitute each value into the expression.
2. If the answers have different signs, there is a root between them.

Example: show that x³ − 4x − 2 = 0 has a root between x = 2 and x = 3.

1. x = 2: 8 − 8 − 2 = −2
2. x = 3: 27 − 12 − 2 = 13
3. There is a change of sign, so there is a root between 2 and 3

To iterate, the equation needs to be in the form x = (an expression in x).

1. Get one x term on its own.
2. Undo whatever is done to it, as when rearranging a formula.

Example: show that x³ − 4x − 2 = 0 can be written as x = ∛(4x + 2).

1. Add 4x + 2 to both sides: x³ = 4x + 2
2. Cube root both sides: x = ∛(4x + 2)

There can be more than one rearrangement: x³ − 4x − 2 = 0 also gives x = (x³ − 2)/4. A question will tell you which one to use.

An iteration formula such as xₙ₊₁ = ∛(4xₙ + 2) gives the next value from the current one. Start from the value x₀ that you're given.

1. Type x₀ into the calculator and press =.
2. Type the formula using Ans in place of xₙ: ∛(4Ans + 2).
3. Press = repeatedly. Each press gives the next value.
4. Stop when the values agree to the accuracy you need.

Example: use xₙ₊₁ = ∛(4xₙ + 2) with x₀ = 2 to find the root to 2 decimal places.

1. x₁ = ∛10 = 2.1544…
2. x₂ = 2.1979…
3. x₃ = 2.2098…
4. x₄ = 2.2131…
5. x₅ = 2.2140…
6. x₄ and x₅ both round to 2.21

Check: at 2.205 the expression is negative and at 2.215 it is positive, so the root is 2.21 to 2 d.p.