1.09c - Fixed point iteration and cobweb diagrams | OCR A-Level Mathematics A

Can you turn an awkward cubic into a calculation that improves itself? Under 1.09c, you will rearrange an equation, compute successive approximations and draw the staircase or cobweb that explains their movement. You will also check what the resulting decimal actually establishes.

An equation becomes a recurrence

Try repeatedly taking the cube root of twice your last answer plus $3$ , starting with $2$ . The outputs begin $1.912931$ , $1.896935$ , $1.893967$ : they seem to settle. Which equation are they solving?

You already know

A recurrence defines each new term from earlier terms. Rearrangements must preserve the equation and its domain. A real cube root is defined for negative as well as positive inputs.

The number being approached should reproduce itself under the rule.

Fixed point and iterate

A fixed point $a$ of $g$ satisfies $g(a)=a$ . An iteration applies the rule once; its output is an iterate. In $x_{n+1}=g(x_n)$ , the subscript counts steps, not powers.

If a sequence converges to $a$ and $g$ is continuous there, taking limits in the recurrence gives $a=g(a)$ . That implication does not promise convergence.

Fixed point iteration

Rearrange the equation as $x=g(x)$ , choose a start $x_0$ , then calculate

x_{n+1}=g(x_n),\qquad n=0,1,2,\ldots.

Check every fixed point against the initial equation and domain.

Not in the formula sheet: learn it.

Begin with a rearrangement that makes the leading power easy to undo.

Worked example 1: constructing the rule

Question. Rearrange $x^3-2x-3=0$ into a cube-root iteration and calculate $x_1$ from $x_0=2$ .

Plan. Isolate the cube before taking a real cube root.

\begin{aligned} x^3&=2x+3 && \text{move terms}\\ x&=\sqrt[3]{2x+3} && \text{real cube root}\\ x_{n+1}&=\sqrt[3]{2x_n+3} && \text{turn equality into a rule}\\ x_1&=\sqrt[3]{7}\approx1.912931 \end{aligned}

Why a cube root? Cubing reverses this step for every real input, so no root is lost.

Answer. Use $x_{n+1}=\sqrt[3]{2x_n+3}$ ; the first iterate is $1.912931$ to six decimal places.

Check. Cubing $x_1$ gives $7$ , the value of $2x_0+3$ , not $2x_1+3$ .

Change only the constant from $3$ to $5$ , keeping $x_0=2$ .

Worked example 2: changing the constant

Question. Construct the same type of iteration for $x^3-2x-5=0$ , starting at $x_0=2$ , and calculate $x_1$ .

Plan. Isolate $x^3$ and evaluate the first output.

\begin{aligned} x^3&=2x+5\\ x_{n+1}&=\sqrt[3]{2x_n+5}\\ x_1&=\sqrt[3]{9}\approx2.080084 \end{aligned}

Why keep the start unchanged? The changed output now comes entirely from the changed constant.

Answer. Use $x_{n+1}=\sqrt[3]{2x_n+5}$ , with $x_1=2.080084$ to six decimal places.

Check. Cubing the first output gives $9=2x_0+5$ .

The related equation $x^3-2x-4=0$ has fixed point $2$ ; from $1$ , its first cube-root iterate is $\sqrt[3]{6}\approx1.817121$ .

Equivalent algebra can behave very differently. Rearranging the first cubic as $x=(x^3-3)/2$ gives $2,2.5,6.3125,\ldots$ . Both rules have the same fixed points, but one start does not produce the same sequence.

Watch out

Wrong: “A valid rearrangement must converge.” Right: algebraic equivalence preserves solutions; it does not guarantee a useful sequence.

Make the constant one unit larger again, keeping the same start.

Compute a reproducible sequence

The rule uses the previous unrounded value, even when your written table displays fewer digits. This small distinction prevents avoidable drift.

Worked example 3: using the previous output

Question. Use $x_{n+1}=\sqrt[3]{2x_n+3}$ , $x_0=2$ , to find $x_1,x_2,x_3$ .

Plan. Retain full precision between successive evaluations.

\begin{aligned} x_1&=\sqrt[3]{7}=1.912931183\ldots\\ x_2&=\sqrt[3]{2(1.912931183\ldots)+3}=1.896935259\ldots\\ x_3&=\sqrt[3]{2(1.896935259\ldots)+3}=1.893967062\ldots \end{aligned}

Why retain the hidden digits? The rounded printed output is not the number stored for the next step.

Answer. To six decimal places: $1.912931,\ 1.896935,\ 1.893967$ .

Check. The next four displayed values are $1.893415,\ 1.893313,\ 1.893294,\ 1.893290$ ; the sequence is settling near $1.89329$ .

Change only the starting value to $1.5$ . The recurrence stays the same.

Worked example 4: a different start

Question. Calculate the first three iterates from $x_0=1.5$ .

Plan. Enter $1.5$ , then repeatedly evaluate the cube root of $2$ times ANS plus $3$ .

\begin{aligned} x_1&=\sqrt[3]{6}=1.817120593\ldots\\ x_2&=\sqrt[3]{2x_1+3}=1.879015742\ldots\\ x_3&=\sqrt[3]{2x_2+3}=1.890630836\ldots \end{aligned}

Why use ANS? It carries the full preceding result into every occurrence of the input.

Answer. To six decimal places: $1.817121,\ 1.879016,\ 1.890631$ .

Check. These approach the same neighbourhood from below, whereas the start $2$ approached from above.

A spreadsheet does the same job: put the start in cell A2, calculate the cube root of $2$ times A2 plus $3$ in A3, then fill down with a relative reference to the preceding row. Keep a step-number column. Use a real cube-root operation if the argument can be negative; software fractional powers do not always handle negative inputs as intended.

Exam tip

State the recurrence, start and enough successive iterates to show your method. Retain calculator precision, then give the requested final accuracy. Set radians if the iteration contains trigonometric functions.

For comparison, the exact recurrence $x_{n+1}=(x_n+8)/3$ from $1$ gives $3,11/3,35/9$ . Each output supplies the next input, just as in the cubic calculation.

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1.09c - Fixed point iteration and cobweb diagrams

An equation becomes a recurrence

Compute a reproducible sequence

Continue this lesson.

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