POP

G7.10 Iteration

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01 — Explanation

What problem does iteration actually solve?

Some equations, like $x^3+2x-7=0$ , can't be solved exactly by factorising or a formula. Iteration is a repeatable calculator process that homes in on an approximate solution (root) to a high accuracy. A full question has four distinct parts, each testing a different skill — you need all four to score well.

02 — Explanation

How do I show a root lies in a given interval?

Define $f(x)$ by setting the equation to zero, then substitute both ends of the interval. If the sign changes, the graph must have crossed the $x$ -axis somewhere between them.

Example: $f(x)=x^3+2x-7$ . $f(1)=1+2-7=-4$ (negative), $f(2)=8+4-7=+3$ (positive).

Exam Tip: The calculation alone isn't enough — you must write a concluding sentence: "change of sign, so a root lies between $x=1$ and $x=2$ ." Missing this line is the most common way to lose the final mark here.

03 — Explanation

How do I show an equation rearranges into a given iterative form?

Isolate the term the question wants you to end up with, then apply the matching inverse operation — showing every line.

$x^3+2x-7=0 \Rightarrow x^3=7-2x \Rightarrow x=\sqrt[3]{7-2x}$ .

This becomes the iterative formula $x_{n+1}=\sqrt[3]{7-2x_n}$ .

04 — Explanation

How do I actually run the iteration on a calculator?

Type the starting value $x_0$ and press $=$ to store it as "Ans". Type the formula using the Ans button in place of $x_n$ , then press $=$ repeatedly — each press feeds the previous output straight back in as the next input.

x_0=2 \to x_1=\sqrt[3]{7-2(2)}=\sqrt[3]3\approx1.442250 \to x_2\approx1.602523 \to x_3\approx1.559794

Exam Tip: Never round an intermediate value and retype it — this is the single most damaging error in the topic. Using the Ans button carries the full, unrounded value forward automatically; manual re-entry compounds rounding error through every later step.

05 — Explanation

How do I prove my rounded answer is accurate to a given number of decimal places?

Repeat the change-of-sign test from earlier, but now on the tiny interval either side of your rounded answer — using the original $f(x)$ , never the iterative $g(x)$ .

Worked exam-style question: The iteration above converges to $\alpha\approx1.564$ . Show this is correct to 3 d.p.

Bounds: $1.5635$ and $1.5645$ .
$f(1.5635)\approx-0.005$ (negative); $f(1.5645)\approx+0.003$ (positive).
Change of sign between the bounds, so $\alpha=1.564$ to 3 d.p.

06 — Explanation

How do I avoid mixing up f(x) and g(x)?

They have strictly separate jobs:

Function	Used for
$f(x)$ (original, $=0$ )	the first change-of-sign test; the final accuracy proof
$g(x)$ (rearranged, $x_{n+1}=g(x_n)$ )	generating $x_1, x_2, x_3,\dots$ on the calculator

Substituting bounds into $g(x)$ instead of $f(x)$ for the accuracy proof is a common and costly mix-up.

07 — Common mistakes

Common mistakes to avoid

Premature rounding between iteration steps instead of using the Ans button.
Using the iterative formula $g(x)$ for the change-of-sign proof (it must be $f(x)$ ).
Omitting the written conclusion after a correct pair of sign calculations.
Writing too few decimal places during the iteration, which can mask whether the sequence has actually converged.