1MA1 · A20AlgebraHigherGrades 7–9

Iteration

Q: Does iteration always work?

No. Some rearrangements converge to a root and others diverge. Exam questions give formulas that converge.

Q: Why does a change of sign prove a root exists?

A continuous graph cannot get from below the axis to above it without crossing the axis.

Q: What is x_0?

The starting value. Some questions call it x_1; follow the notation in the question.

Some equations, like $x^{3} + 2x - 5 = 0$ , cannot be solved with a formula you know. $\textbf{Iteration}$ finds an approximate solution by repeatedly feeding an answer back into a formula, getting closer each time. It is how calculators and computers solve most equations in practice.

40 min Sequences & Functions

The big picture

Two ideas work together. A change of sign shows a root exists in an interval: if a continuous expression is negative at one value and positive at another, it must pass through zero in between. An iterative formula then homes in on that root.

Edexcel iteration questions follow a standard pattern worth learning: show that the equation has a root between two values, show that it rearranges to a given form, use the iterative formula to find $x_{1}$ , $x_{2}$ , $x_{3}$ , and finally justify an answer to a given accuracy.

What you'll learn

Show that an equation has a root in an interval using a change of sign
Rearrange an equation into an iterative form $x = \mathrm{g}(x)$
Use an iterative formula $x_{n+1} = \mathrm{g}(x_{n})$ with a calculator
Justify a solution to a given degree of accuracy

Change of sign

Rewrite the equation as "expression $= 0$ " and substitute the two endpoints of the interval.

If one result is negative and the other positive, the expression must equal zero somewhere between them, so a root lies in the interval.

Example 1Show that

x^{3} + 2x - 5 = 0

has a solution between

x = 1

and

x = 2

1At

x = 1

1 + 2 - 5 = -2

2At

x = 2

8 + 4 - 5 = 7

3There is a change of sign.

AnswerSince the value changes from negative to positive, there is a solution between 1 and 2.

Your turnSame method, new numbers — try it before moving on.

For $f(x) = x^3 + 2x - 5$ , work out $f(1)$ .

Your answer

Show worked solution

Solution

1 $1 + 2 - 5 = -2$ .
2(And $f(2) = 7$ — a change of sign.)

Answer: $-2$

Tip — The concluding sentence is part of the answer: "there is a change of sign, so there is a root between 1 and 2".

Rearranging into iterative form

An iterative formula has the form $x = \mathrm{g}(x)$ . Rearrange the equation so that $x$ is on its own on one side, even though $x$ also appears on the other.

For $x^{3} + 2x - 5 = 0$ : $x^{3} = 5 - 2x$ , so $x = \sqrt[3]{5 - 2x}$ .

Questions usually give the target form, so work towards it one step at a time.

Example 1Show that

x^{2} - 5x + 2 = 0

can be written as

x = \sqrt{5x - 2}

1Add

5x - 2

to both sides:

x^{2} = 5x - 2

2Take the square root.

Answer

x = \sqrt{5x - 2}

, as required.

Using the iterative formula

Write the rearrangement as $x_{n+1} = \mathrm{g}(x_{n})$ and start from the given $x_{0}$ (or $x_{1}$ ).

On a calculator, type the starting value and press equals, then type the formula using ANS in place of $x_{n}$ . Each press of equals gives the next value.

Write down each iterate to the accuracy the question requests, but keep full calculator values when continuing.

Example 1Using

x_{n+1} = \sqrt[3]{5 - 2x_{n}}

with

x_{0} = 1.5

, find

x_{1}

x_{2}

and

x_{3}

to 5 decimal places.

x_{1} = \sqrt[3]{5 - 3} = 1.25992

x_{2} = \sqrt[3]{5 - 2(1.25992\dots)} = 1.35361

x_{3} = 1.31862

Answer

x_{1} = 1.25992

x_{2} = 1.35361

x_{3} = 1.31862

Your turnSame method, new numbers — try it before moving on.

Using $x_{n+1} = \sqrt[3]{5 - 2x_n}$ with $x_0 = 1.3$ , find $x_1$ to 5 decimal places.

Your answer

To 5 decimal places.

Show worked solution

Solution

1 $x_1 = \sqrt[3]{5 - 2.6} = \sqrt[3]{2.4}$ .
2 $= 1.338866\ldots$

Answer: $x_1 = 1.33887$

Why this works

The iterates jump above and below the root — $1.26, 1.35, 1.32, 1.33, \dots$ — closing in on about $1.328$ . Not every rearrangement behaves this well: $x_{n+1} = \dfrac{5 - x_{n}^{3}}{2}$ from the same equation runs away from the root entirely.

Justifying accuracy

To show a root is $1.33$ correct to 2 decimal places, check the bounds of that rounding: substitute $1.325$ and $1.335$ into the original expression.

A change of sign between the bounds proves the root lies in $1.325 \le x < 1.335$ , which rounds to $1.33$ .

Example 1Show that a root of

x^{3} + 2x - 5 = 0

1.33

to 2 decimal places.

1At

x = 1.325

-0.0238

2At

x = 1.335

0.0493

3Change of sign.

AnswerThe root lies between 1.325 and 1.335, so it is

1.33

to 2 d.p.

Think like an examiner

Show both substituted values and state the change of sign in words.
Write each iterate with its subscript: $x_{1}$ , $x_{2}$ , $x_{3}$ .
Use the full calculator value of each iterate for the next step.

Iteration

x_{n+1} = \mathrm{g}(x_{n})

Iterative formula.

\mathrm{f}(a) < 0 < \mathrm{f}(b)

Change of sign: root between a and b.

\text{test } x \pm 0.005

Justify 2 d.p. accuracy.

Watch out for these

Substituting into the rearranged formula to show a change of sign.

Use the original equation written as expression

= 0

Rounding each iterate before using it.

Continue with the unrounded value; round only what you write down.

Concluding a root without a statement.

Say "change of sign, so a root lies in the interval".

Checking

1.33

and

1.34

to justify 2 d.p.

Check the rounding bounds

1.325

and

1.335

Stretch yourself

Use $x_{n+1} = \sqrt{5x_{n} - 2}$ with $x_{0} = 4$ to find $x_{3}$ to 4 decimal places. Explain which root of $x^{2} - 5x + 2 = 0$ the iteration is approaching, and find that root exactly.

Hint — Iterate three times. Then solve the quadratic with the formula and compare.

Reveal solution

Solution

x_{1} = \sqrt{18} = 4.2426

x_{2} = \sqrt{5(4.2426\dots) - 2} = 4.3833

x_{3} = 4.4628

. The values are increasing towards the larger root. By the quadratic formula,

x = \dfrac{5 \pm \sqrt{17}}{2}

, giving

4.5616

and

0.4384

. So the iteration approaches

x = \dfrac{5 + \sqrt{17}}{2} \approx 4.5616

; further iterates

4.5071, 4.5316, \dots

confirm it.

\blacksquare

Questions students ask

Does iteration always work?

No. Some rearrangements converge to a root and others diverge. Exam questions give formulas that converge.

Why does a change of sign prove a root exists?

A continuous graph cannot get from below the axis to above it without crossing the axis.

What is

x_{0}

The starting value. Some questions call it

x_{1}

; follow the notation in the question.

Key takeaways

A change of sign in a continuous expression shows a root in an interval.
Rearrange $\mathrm{f}(x) = 0$ into $x = \mathrm{g}(x)$ .
Iterate with $x_{n+1} = \mathrm{g}(x_{n})$ using the ANS key.
Keep full accuracy between iterations.
Justify accuracy by checking a change of sign at the rounding bounds.

Exam-style questions

Practise it the way it is examined

3 original questions · 8 marks, written to match the style and mark allocation of the real papers. Work on paper, then open the mark scheme and tick the marks you earned — method marks count even if the final answer slips.

Question 1Grade 72 marks

Show that $x^3 - 3x - 5 = 0$ has a solution between $x = 2$ and $x = 3$ .[2]
Show mark scheme
- M1 $f(2) = -3$ and $f(3) = 13$
- A1Change of sign, so a solution lies between 2 and 3

Question 2Grade 83 marks

(a)Show that $x^3 - 3x - 5 = 0$ can be written as $x = \sqrt[3]{3x + 5}$ .[1]
Show mark scheme (a)
- B1 $x^3 = 3x + 5$ then cube root both sides
(b)Using $x_{n+1} = \sqrt[3]{3x_n + 5}$ with $x_0 = 2$ , find $x_1$ , $x_2$ and $x_3$ to 4 decimal places.[2]
Show mark scheme (b)
- M1 $x_1 = 2.2240$
- A1 $x_2 = 2.2684$ , $x_3 = 2.2770$

Question 3Grade 93 marks

Show that $2.28$ is a solution of $x^3 - 3x - 5 = 0$ correct to 2 decimal places.[3]
Show mark scheme
- M1Evaluates $f(2.275)$ and $f(2.285)$
- A1 $f(2.275) = -0.050\ldots$ and $f(2.285) = 0.075\ldots$
- C1Change of sign, so the root rounds to 2.28

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