Pure 9.2 - Iteration, cobweb and staircase diagrams | Edexcel A-Level Mathematics

Build successive approximations to a root, and see the same calculation as a path between a curve and a diagonal. You will rearrange equations, calculate accurately, draw iteration paths and judge whether they approach a fixed point.

Fixed points and iteration formulas

Can pressing the same calculator instruction repeatedly solve $x^3+x^2=9$ ? Start with $1$ , apply $x\mapsto\sqrt{9/(x+1)}$ , and the next value is $2.121320\ldots$ . Feeding that result back gives $1.698056\ldots$ . The input and output are beginning to chase a common value.

You already know

A recurrence defines each new term from previous terms. A root of $F(x)=0$ is an input making the original expression zero; rearranging an equation must preserve its domain and the required branch of a root.

We use $g$ for the repeated instruction, keeping it distinct from the original equation $F(x)=0$ .

Fixed point

A fixed point $\alpha$ of $g$ satisfies $g(\alpha)=\alpha$ . For example, $2$ is fixed by $g(x)=x/2+1$ , whereas $0$ is not.

If the sequence settles, its next value and current value tend to the same number.

Fixed-point iteration

x_{n+1}=g(x_n)

If $x_n\to\alpha$ and $g$ is continuous at $\alpha$ , then $\alpha=g(\alpha)$ . The rearrangement must also make this a solution of the original equation.

Not in the formula booklet: learn it.

A valid rearrangement gives a possible target, but does not guarantee that repeated substitution will reach it.

Worked example 1: selecting the positive root

Question. Rearrange $x^3+x^2-9=0$ into an iteration for its positive root.

Plan. Collect the terms containing $x^2$ , then take the positive square root.

\begin{aligned} x^2(x+1)&=9 && \text{factorise} \\ x^2&=\frac9{x+1} && x\ne-1 \\ x&=\sqrt{\frac9{x+1}} && x>0 \end{aligned}

Why this step? The positive square root matches the requested positive solution; it is not a rearrangement covering negative solutions.

Answer. Use $x_{n+1}=\sqrt{9/(x_n+1)}$ , with a suitable positive starting value.

Check. Squaring a positive fixed point and multiplying by $\alpha+1$ returns $\alpha^3+\alpha^2=9$ .

Change only the constant from $9$ to $10$ . The domain and branch decision stay the same.

Worked example 2: varying the constant

Question. Form an iteration for the positive root of $x^3+x^2-10=0$ .

Plan. Isolate $x^2$ in the same way.

\begin{aligned} x^2(x+1)&=10 && \text{factorise} \\ x&=\sqrt{\frac{10}{x+1}} && x>0 \end{aligned}

Why this step? A positive input makes the denominator positive and produces a positive next input.

Answer. $x_{n+1}=\sqrt{10/(x_n+1)}$ .

Check. A fixed point satisfies $\alpha^2(\alpha+1)=10$ , precisely the original equation.

Taking a cube root has no positive/negative branch choice for real inputs. For example, $x^3-2x-5=0$ gives $x_{n+1}=\sqrt[3]{2x_n+5}$ . The same rearrange-and-check principle applies.

Watch out

Wrong: a correct fixed-point equation proves convergence. Right: it identifies what a convergent sequence could approach. Inspect the actual iterates or graph before claiming success.

For your turn, change only the constant from $10$ to $11$ , preserving the positive-root decision.

Calculating the sequence

Use each unrounded answer as the next input. A displayed table can show rounded values, but its next row must come from the stored value. Also follow the given indexing: a start labelled $x_0$ differs from a start labelled $x_1$ .

Worked example 3: repeated substitution

Question. Use $x_{n+1}=\sqrt{9/(x_n+1)}$ , with $x_0=1$ , to calculate $x_1$ , $x_2$ and an approximate positive root to three decimal places.

Plan. Store each output and continue until the displayed values settle.

\begin{aligned} x_1&=\sqrt{9/2}=2.121320344\ldots && \text{first substitution} \\ x_2&=\sqrt{9/(2.121320344\ldots+1)} \\ &=1.698056292\ldots && \text{use stored }x_1 \end{aligned}

Continuing gives $x_3=1.826399382\ldots$ , $x_4=1.784450437\ldots$ , $x_5=1.797841946\ldots$ and $x_6=1.793534216\ldots$ . Later, $x_9=1.794615135\ldots$ and $x_{10}=1.794569372\ldots$ .

Why this step? Alternating values can still converge; the size of the movement matters.

Answer. The calculation suggests a root of $1.795$ to three decimal places.

Check. Substituting $1.79458$ into $x^3+x^2-9$ gives about $-0.00000655$ , close to zero. This is a sense-check, not an accuracy proof.

Keep the same instruction and change only the constant to $10$ . Retaining all stored digits matters just as much.

Worked example 4: a second iteration table

Question. Use $x_{n+1}=\sqrt{10/(x_n+1)}$ , with $x_0=1$ , to find $x_1$ , $x_2$ to four decimal places and the apparent limit to three decimal places.

Plan. Iterate with full precision, rounding only the requested outputs.

\begin{aligned} x_1&=\sqrt5=2.236067977\ldots && \text{first output} \\ x_2&=1.757887921\ldots && \text{repeat with stored input} \end{aligned}

Continuing gives $x_9=1.867503507\ldots$ , $x_{10}=1.867445866\ldots$ , and eventually $1.867460\ldots$ .

Why this step? Two early values need not be close to the limit, even when the recurrence ultimately converges.

Answer. $x_1=2.2361$ , $x_2=1.7579$ ; the apparent limit is $1.867$ to three decimal places.

Check. $x_9$ and $x_{10}$ round differently to three decimal places, so stopping at $x_9$ would be premature.

The word “apparent” keeps the evidence honest: a finite list is evidence of convergence rather than a general proof.

Exam tip

If asked for $x_2$ and $x_3$ , give those iterates. A calculator's final root does not replace the requested intermediate values.

The cube-root rule above, with $x_0=1.8$ , gives $2.048799615\ldots$ , $2.087575864\ldots$ , eventually $2.094551\ldots$ . For your turn use the square-root rule, changing only its constant to $11$ .

Drawing the iteration on a graph

Draw both $y=g(x)$ and $y=x$ . Their intersection is $(\alpha,\alpha)$ : input and output agree there. Starting on the $x$ -axis, move vertically to the curve to evaluate the function, then horizontally to the diagonal to copy that output into the next input.

[DIAGRAM: asset_slug: p1_9_2_solve_equations_approximately_using_iterative_methods_and_draw_associated_cobweb_and_staircase_diagrams__diagram_01; asset_name: Evaluating and copying an iterate; caption: Vertical moves evaluate g; horizontal moves copy the output onto y equals x; description: The curve g(x)=sqrt(9/(x+1)) and the diagonal y=x have fixed point approximately 1.79458. Four evaluate-then-copy pairs start at x0=1 on equal coordinate scales.]

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Last Accessed

Pure Mathematics (Papers 1 and 2)

Statistics (Paper 3, Section A)

Mechanics (Paper 3, Section B)

Pure 9.2 - Iteration, cobweb and staircase diagrams

Fixed points and iteration formulas

Calculating the sequence

Drawing the iteration on a graph

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