I2 - Fixed-point iteration | AQA A-Level Mathematics

Rearrange equations into iterative rules, generate accurate calculator sequences, and construct staircase and cobweb diagrams. You will see why equivalent equations can produce very different sequences, and how to distinguish numerical evidence from a guaranteed root accuracy.

Equations as fixed points

Can two correct rearrangements of the same equation send you in opposite directions? For the positive root of $x^2-x-2=0$ , try repeatedly applying either $\sqrt{x+2}$ or $x^2-2$ . Both leave $2$ unchanged, but a small error near $2$ behaves very differently under the two rules.

You already know

A recurrence uses earlier terms to calculate later ones. Rearranging an equation must preserve its solutions on the domain you need; taking a positive square root can exclude a negative solution.

Write the equation as $x=g(x)$ . Feed a trial value into $g$ , then use the output as the next input. The supplied starting value is already the first term; do not apply the rule before recording it.

Simple iteration

x_{n+1}=g(x_n)\qquad n=1,2,3,\ldots

Choose or use the stated starting value $x_1$ , and check that each new value remains in the domain of $g$ .
Not in the formula booklet: learn it.

The value we hope to approach is one that the rule leaves unchanged.

Fixed point

A fixed point $\alpha$ satisfies $g(\alpha)=\alpha$ . It is the horizontal coordinate where $y=g(x)$ intersects $y=x$ . For example, $2$ is fixed by $g(x)=\sqrt{x+2}$ , since $g(2)=2$ ; $-1$ is not, since $g(-1)=1$ .

If the sequence tends to $\alpha$ and $g$ is continuous there, passing to the limit gives $\alpha=g(\alpha)$ . This explains why a convergent sequence can solve an equation, but does not establish that a chosen sequence converges. The fixed-point limit reasoning is not in the formula booklet: learn it.

Worked example 1: two rules for one positive root

Question. Rearrange $x^2-x-2=0$ into two iterative forms and identify any root excluded by taking a square root.

Plan. Isolate $x^2$ for one form and $x$ for the other, then check their fixed points.

\begin{aligned} x^2&=x+2 && \text{rearrange} \\ x&=\sqrt{x+2} && \text{for the positive root} \\ x&=x^2-2 && \text{isolate a different }x \\ x^2-x-2&=(x-2)(x+1) && \text{find the original roots} \end{aligned}

Why this step? The principal square root is nonnegative, so that rearrangement is equivalent to the original equation only when seeking its positive root.

Answer. The rules are $x_{n+1}=\sqrt{x_n+2}$ and $x_{n+1}=x_n^2-2$ . Both have fixed point $2$ ; only the second also has fixed point $-1$ .

Check. $\sqrt{2+2}=2$ and $2^2-2=2$ , whereas $\sqrt{-1+2}=1\ne-1$ .

Change only the constant term, from $-2$ to $-6$ . The algebra follows the same route, but the fixed points and excluded negative root change.

Worked example 2: a changed quadratic

Question. Obtain two iterative forms for $x^2-x-6=0$ and state which original root is excluded by the square-root form.

Plan. Make the two rearrangements and factorise the original quadratic to check the domain restriction.

\begin{aligned} x^2&=x+6 && \text{rearrange} \\ x&=\sqrt{x+6} && \text{positive-root form} \\ x&=x^2-6 && \text{alternative form} \\ (x-3)(x+2)&=0 && \text{original roots} \end{aligned}

Why this step? Checking the original roots prevents an algebraically convenient rule from silently losing a solution you were asked to find.

Answer. Use $x_{n+1}=\sqrt{x_n+6}$ for the positive root $3$ , or $x_{n+1}=x_n^2-6$ . The first excludes $-2$ .

Check. $\sqrt{3+6}=3$ , but $\sqrt{-2+6}=2\ne-2$ ; both $3$ and $-2$ satisfy the original quadratic.

These are equations for fixed points, not promises about convergence. The starting value and the chosen rearrangement still matter.

Watch out

Wrong: “Squaring $x=\sqrt{x+6}$ gives the quadratic, so the two equations have exactly the same roots.” Squaring loses the nonnegative restriction. Correct: they are equivalent for positive solutions; the square-root rule excludes the negative root.

The check changes only the constant term once more. State the recurrence as well as its domain consequence.

Calculator iteration and accuracy

A calculator's previous-answer function keeps the digits that your displayed table leaves out, so use the stored output for every new input.

For $x_{n+1}=\cos x_n$ , use radians. Enter $x_1$ , then apply cosine to the stored value repeatedly. Numerical iteration is permitted on an examination calculator; symbolic algebra and symbolic equation-solving facilities must be unavailable in examination mode. The recurrence and the requested iterates show the specified method; a root-solver's final output alone does not. Early rounding can change a later displayed digit.

Worked example 3: generating an oscillating sequence

Question. Use $x_{n+1}=\cos x_n$ , with $x_1=1$ and angles in radians, to find $x_2$ to $x_6$ , to six decimal places.

Plan. Apply cosine to the previous unrounded output, rounding only the table entries.

\begin{aligned} x_2&=\cos(1)\approx0.5403023059 && \text{first application} \\ x_3&=\cos(x_2)\approx0.8575532158 && \text{use stored }x_2 \\ x_4&=\cos(x_3)\approx0.6542897905 && \text{use stored }x_3 \\ x_5&=\cos(x_4)\approx0.7934803587 && \text{repeat} \\ x_6&=\cos(x_5)\approx0.7013687736 && \text{repeat} \end{aligned}

Why this step? Feeding $0.540302$ back into the rule would replace the true stored value by a rounded approximation at the very next calculation.

Answer. The requested values are $0.540302$ , $0.857553$ , $0.654290$ , $0.793480$ and $0.701369$ .

Check. They alternate about $0.739085\ldots$ , the value approached by continued iteration, rather than decreasing monotonically.

Change only the starting value, from $1$ to $0.5$ . The first few outputs change, so recalculate them; do not copy the previous sequence.

Worked example 4: another start, the same rule

Question. Starting with $x_1=0.5$ , use $x_{n+1}=\cos x_n$ in radians to find $x_2$ , $x_3$ and $x_4$ to six decimal places.

Plan. Store $0.5$ , apply the same recurrence three times, and retain all intermediate digits.

\begin{aligned} x_2&=\cos(0.5)\approx0.8775825619 && \text{new first output} \\ x_3&=\cos(x_2)\approx0.6390124942 && \text{stored input} \\ x_4&=\cos(x_3)\approx0.8026851007 && \text{stored input} \end{aligned}

Why this step? The subscript identifies the term: three applications starting from $x_1$ end at $x_4$ , not $x_3$ .

Answer. $x_2=0.877583$ , $x_3=0.639012$ , $x_4=0.802685$ , each to six decimal places.

Check. The values again alternate around the same candidate fixed point. Substitution gives $\cos(0.739085\ldots)\approx0.739085\ldots$ .

Repeated iteration suggests $0.739$ to three decimal places. To certify it, use $f(x)=\cos x-x$ : $f(0.7385)\approx0.0009791594>0$ and $f(0.7395)\approx-0.0006943896<0$ . Continuity gives a root inside these rounding boundaries. Also $f'(x)=-\sin x-1<0$ on $(0,1)$ , so this root is unique there. The sign-change and rounding test is not in the formula booklet: learn it.

The cosine example has another neat feature: any real first input gives $x_2\in[-1,1]$ , then $x_3\in[\cos1,1]$ . On $[-1,1]$ , $|g'(x)|=|\sin x|\le\sin1<1$ , so the rule contracts errors within an interval it preserves. This is much stronger evidence than watching a few decimal digits settle. The derivative of cosine and this contraction reasoning are not in the formula booklet: learn their meaning.

Watch out

Wrong: “ $x_7$ and $x_8$ are almost equal, so the root is accurate to that many places.” Close consecutive iterates alone do not bound the error from the root. Correct: use a justified error bound or a continuous sign-changing bracket inside the required rounding boundaries.

For the next calculation, change only the starting value again and keep radians selected.

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I2 - Fixed-point iteration

Equations as fixed points

Calculator iteration and accuracy

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