A20 - Numerical solutions by iteration

Iteration feeds an answer back into the same calculation. You will build suitable rules, retain enough accuracy and check whether repeated values really approach a solution.

Rearranging an equation for iteration (Higher)

Can a calculation find a root that does not factorise neatly? For $x^3-x-2=0$ , trying $x=1.5$ gives $-0.125$ , so 1.5 is close but not a root. Repeated substitution can improve that starting guess.

You already know

Rearranging keeps an equation's solutions when each step is valid. Substitution means evaluating an expression using the given input, with brackets where needed.

An iteration uses the output of one calculation as the input of the next. Its notation keeps track of this order.

Iteration and fixed point

In $x_{n+1}=g(x_n)$ , $x_n$ is the current value and $x_{n+1}$ is the next. The starting value is $x_0$ . A fixed point satisfies $x=g(x)$ , so the rule leaves it unchanged.

The subscript counts steps; it is not multiplication. An iteration rule is not on the formula sheet: derive it from the rearranged original equation, or use the rule supplied in the question. At a fixed point, the rule must recover the original equation.

Worked example 1: a cube-root rule

Question. Rearrange $x^3-x-2=0$ for iteration. Starting at $x_0=1.5$ , calculate $x_1$ and $x_2$ to 6 decimal places.

Plan. Isolate $x^3$ , take the cube root, then reuse each output.

\begin{aligned} x^3&=x+2 &&\text{add }x+2\\ x&=\sqrt[3]{x+2} &&\text{take the cube root}\\ x_{n+1}&=\sqrt[3]{x_n+2} &&\text{make the repeated rule}\\ x_1&=\sqrt[3]{1.5+2}\approx1.518294 &&\text{use the start}\\ x_2&=\sqrt[3]{x_1+2}\approx1.520935 &&\text{reuse the unrounded output} \end{aligned}

Why take a cube root? It makes the next $x$ -value an explicit calculation from the previous one.

Answer. $x_1=1.518294$ , $x_2=1.520935$ to 6 decimal places.

Check. A fixed point has $x=\sqrt[3]{x+2}$ ; cubing recovers $x^3-x-2=0$ .

Change only the original equation's constant from 2 to 3. The same construction gives a different sequence of estimates.

Worked example 2: changing the equation's constant

Question. Rearrange $x^3-x-3=0$ for iteration. Use $x_0=1.5$ to calculate the first two outputs to 6 decimal places.

Plan. Isolate the cube again and feed the first full output into the second calculation.

\begin{aligned} x^3&=x+3 &&\text{rearrange}\\ x_{n+1}&=\sqrt[3]{x_n+3} &&\text{take the cube root}\\ x_1&=\sqrt[3]{4.5}\approx1.650964 &&\text{substitute }1.5\\ x_2&=\sqrt[3]{x_1+3}\approx1.669223 &&\text{use unrounded }x_1 \end{aligned}

Why change the rule as well? Keeping $+2$ would still target the previous equation.

Answer. $x_1=1.650964$ , $x_2=1.669223$ .

Check. Cubing $x=\sqrt[3]{x+3}$ gives the required equation.

A rearrangement can restrict which roots it finds. From $x^2=x+2$ , the rule $x_{n+1}=\sqrt{x_n+2}$ always outputs a non-negative number. From 1 its first two outputs are $\sqrt3$ and $\sqrt{2+\sqrt3}\approx1.931852$ . It may approach the positive root 2, but cannot approach the other root $-1$ . Inputs must also satisfy $x_n+2\geq0$ .

Watch out

A correct fixed-point equation does not guarantee a useful iteration. Check the starting value, any square-root or denominator restrictions and the later behaviour.

Change only the cubic's constant again. A calculator may be used.

Intermediate accuracy and reported precision (Higher)

Store the full output between steps. Rounding the value you report is different from rounding the value you reuse.

On a calculator, use the previous answer or a stored memory value. Check the first two entries by hand to catch a wrong bracket or an unchanged input. Label $x_0$ as the start: calling it $x_1$ shifts every later label.

For the first cubic rule, the table displays 9 decimal places while the calculations retain more.

$n$	$x_n$
0	1.500000000
1	1.518294486
2	1.520935263
3	1.521315698
4	1.521370489
5	1.521378379
6	1.521379516

The last two outputs both round to 1.5214 at 4 decimal places. That is useful evidence, but not proof of accuracy. Slow change or premature rounding can also produce repeated displayed values.

Exact fractions make this distinction easy to see.

Worked example 3: exact values between steps

Question. Use $x_{n+1}=(x_n+4)/3$ , $x_0=1$ , to find $x_1,x_2,x_3$ . Give fractions, then report $x_3$ to 3 decimal places.

Plan. Retain fractions at every step and round only the requested final output.

\begin{aligned} x_1&=\frac{1+4}{3}=\frac53 &&\text{first substitution}\\ x_2&=\frac{5/3+4}{3}=\frac{17}9 &&\text{reuse the exact fraction}\\ x_3&=\frac{17/9+4}{3}=\frac{53}{27} &&\text{repeat the rule} \end{aligned}

Why keep the fractions? They prevent rounding error from becoming the next input.

Answer. $\frac53,\frac{17}9,\frac{53}{27}$ ; $x_3=1.963$ to 3 decimal places.

Check. $3x_3-4=\frac{53}9-\frac{36}9=\frac{17}9=x_2$ , reversing the last step.

Change only the constant in the numerator from 4 to 6.

Worked example 4: the same accuracy habit

Question. Use $x_{n+1}=(x_n+6)/3$ , $x_0=1$ , for three steps. Give exact fractions and $x_3$ to 3 decimal places.

Plan. Feed each exact fraction into the next calculation.

\begin{aligned} x_1&=\frac{1+6}{3}=\frac73 &&\text{start at 1}\\ x_2&=\frac{7/3+6}{3}=\frac{25}9 &&\text{reuse }x_1\\ x_3&=\frac{25/9+6}{3}=\frac{79}{27} &&\text{reuse }x_2 \end{aligned}

Why not enter 2.333? It replaces $\frac73$ by a different number.

Answer. $\frac73,\frac{25}9,\frac{79}{27}$ ; $x_3=2.926$ .

Check. $3x_3-6=\frac{79}9-\frac{54}9=\frac{25}9=x_2$ .

Rounding every cubic output to 1 decimal place creates a false fixed value: $1.518294\ldots$ becomes 1.5, then the same calculation returns 1.5 again. Yet $1.5^3-1.5-2=-0.125$ . Repeating a rounded number has not solved the equation.

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A20 - Numerical solutions by iteration

Rearranging an equation for iteration (Higher)

Intermediate accuracy and reported precision (Higher)

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