Me3 - Fixed point iteration and cobweb diagrams | OCR A-Level Mathematics B (MEI)

Can two equivalent rearrangements send a calculator in opposite directions? Under Me3, you will build a fixed point iteration, use the ANS key or a spreadsheet accurately, and connect the resulting sequence to a staircase or cobweb diagram. A final check will distinguish a numerical suggestion from a justified root.

An equation becomes a recurrence

For $x^3-x-4=0$ , try $x_{n+1}=\sqrt[3]{x_n+4}$ from $x_0=2$ . The first outputs are $1.817121$ and $1.798468$ . Yet the equivalent rearrangement $x=x^3-4$ sends the same start to $4$ , then $60$ . Correct algebra does not automatically make a good algorithm.

You already know

Rearranging an equation must preserve its solutions and domain. A real cube root reverses cubing for every real number.

The useful rule appears to be seeking an input that reproduces itself.

Fixed point

A fixed point $a$ satisfies $g(a)=a$ . Applying $g$ once is an iteration; the output is an iterate. The index in $x_n$ counts iterations, not powers.

If the iterates converge to $a$ and $g$ is continuous there, taking limits gives $a=g(a)$ . That explains what a limit solves; it does not guarantee that there is a limit.

Fixed point iteration

For a rearrangement $x=g(x)$ , use a stated initial value $x_0$ and the rule

x_{n+1}=g(x_n).

Check the fixed point against the equation and its domain.

Not in the formula sheet: learn it.

Show the algebra before introducing the subscripts.

Worked example 1: showing the rearrangement

Question. Show that $x^3-x-4=0$ can be written as $x=\sqrt[3]{x+4}$ , then find $x_1$ from $x_0=2$ .

Plan. Isolate the cube and take a real cube root.

\begin{aligned} x^3-x-4&=0\\ x^3&=x+4 && \text{add }x+4\\ x&=\sqrt[3]{x+4} && \text{take the real cube root}\\ x_1&=\sqrt[3]{2+4}=1.817120593\ldots \end{aligned}

Why show the intermediate line? It establishes the supplied rearrangement, rather than merely repeating it.

Answer. Use $x_{n+1}=\sqrt[3]{x_n+4}$ , with $x_1=1.817121$ to six decimal places.

Check. Cubing the first output gives $6$ , the value of $x_0+4$ .

Change only the constant to $5$ .

Worked example 2: a changed constant

Question. Construct a cube-root iteration for $x^3-x-5=0$ , starting at $2$ .

Plan. Repeat the reversible algebra with the new constant.

\begin{aligned} x^3&=x+5\\ x_{n+1}&=\sqrt[3]{x_n+5}\\ x_1&=\sqrt[3]{7}=1.912931183\ldots \end{aligned}

Why not retain the old rule? Its fixed points solve a different cubic.

Answer. The new rule gives $x_1=1.912931$ to six decimal places.

Check. Cubing $x_1$ gives $7=x_0+5$ .

For comparison, $x^3-2x-3=0$ gives $\sqrt[3]{2x+3}$ ; from $2$ , its first output is also $\sqrt[3]{7}$ . One matching iterate does not mean two equations have the same root.

Watch out

Wrong: “The rearrangements are equivalent, so their sequences agree.” Right: equivalent equations have the same solutions; different recurrence rules can produce different behaviour.

Change the constant once more.

Compute a reproducible sequence

The ANS key is useful because it retains digits that your written answer need not display. Every occurrence of the input must use the same preceding value.

Worked example 3: a reproducible calculation

Question. For $x_{n+1}=\sqrt[3]{x_n+4}$ , $x_0=2$ , find the first three iterates.

Plan. Enter $2$ , then repeatedly evaluate the real cube root of ANS plus $4$ .

\begin{aligned} x_1&=1.817120593\ldots\\ x_2&=\sqrt[3]{x_1+4}=1.798467893\ldots\\ x_3&=\sqrt[3]{x_2+4}=1.796543562\ldots \end{aligned}

Why use the stored value? Retyping $1.817121$ changes the next calculation slightly.

Answer. To six decimal places: $1.817121,\ 1.798468,\ 1.796544$ .

Check. The next values are $1.796345,\ 1.796324,\ 1.796322,\ 1.796322$ , suggesting a root near $1.796322$ .

Change only the start to $1.5$ .

Worked example 4: approaching from the other side

Question. Find the first three iterates for the same rule from $x_0=1.5$ .

Plan. Keep the recurrence fixed and carry full precision between outputs.

\begin{aligned} x_1&=\sqrt[3]{5.5}=1.765174168\ldots\\ x_2&=\sqrt[3]{x_1+4}=1.793098487\ldots\\ x_3&=\sqrt[3]{x_2+4}=1.795988855\ldots \end{aligned}

Why state the start? The rule alone does not determine a sequence.

Answer. To six decimal places: $1.765174,\ 1.793098,\ 1.795989$ .

Check. The values approach the same neighbourhood from below, instead of above.

For a spreadsheet, place the start in A2; in A3 calculate the cube root of A2 plus $4$ ; fill down so each row refers to the preceding one. Include iteration numbers and retain stored precision. For negative arguments use a real cube-root facility: a fractional-power calculation may not behave as expected in every program.

Exam tip

Show the recurrence, start and relevant iterates before the final rounded root. Use the requested accuracy. A table does not replace the intermediate algebra in a “Show that” rearrangement.

An exact comparison is $x_{n+1}=(x_n+10)/3$ from $1$ : the first values are $11/3,41/9,131/27$ . The preceding output becomes the next input in exactly the same way.

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Me3 - Fixed point iteration and cobweb diagrams

An equation becomes a recurrence

Compute a reproducible sequence

Continue this lesson.

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