GCSE Maths revision · Higher: number and algebra

Iteration

Higher only

The idea

Iteration finds a solution by feeding each answer back into a formula: $x_{n+1} = f(x_n)$ . With a suitable formula and starting value, the values settle towards a root of the equation.

Start with the given $x_0$ .
Put it in the formula to get $x_1$ , then put $x_1$ in to get $x_2$ .
Keep full calculator values; use the ANS key.

Watch out Feed in the latest value every time, and do not round between steps.

See it worked

Worked exampleCalculator allowed3 marks

Use the iteration $x_{n+1} = \sqrt{10 + x_n}$ , starting with x₀ = 3. Work out x₃ to 3 decimal places.

$\sqrt{13}$
$\sqrt{10+\sqrt{13}}$
Substitute each new value into the same rule, retaining calculator precision.
x₁ = $\sqrt{13}$ = 3.605551… and $x_2 = \sqrt{10 + 3.605551\ldots} = 3.688570\ldots$ .
$x_3 = \sqrt{10 + 3.688570\ldots} = 3.699806\ldots$ , which rounds to 3.700.

Answer $3.700$

Your turn

Question 1Calculator allowed3 marks

The positive solution of $x^{3}$ = 4x + 9 can be estimated using $x_{n+1} = \sqrt{4 + 9/x_n}$ . Start with x₀ = 3 and carry out four iterations. Give x₄ to 3 decimal places.

Question 2Calculator allowed5 marks

The iteration is $x_{n+1}=\sqrt{10+x_n}$ with $x_0=3$ .

(a) Calculate x4, keeping unrounded values until giving your answer to 3 decimal places.

(b) Write the quadratic equation satisfied by a positive fixed point.

Question 3Calculator allowed5 marks

An account starts with £1300. At each year end, 4% interest is added and then £70 is withdrawn.

(a) Find the balance immediately after the third withdrawal, to the nearest penny.

(b) Using x for the current balance and y for the next balance, write the update formula.

In past papers

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