What is 9.2: Iterative methods?

A fixed-point iteration rewrites an equation as x=g(x) and generates approximations using x_n+1=g(x_n) from a chosen starting value.

How many marks are 9.2 questions worth?

9.2 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Which past exam papers has 9.2 appeared in?

We have not yet indexed a verified past-paper appearance for 9.2 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

9.2

Solve equations approximately using simple iterative methods; be able to draw associated cobweb and staircase diagrams.

Draft — not yet indexed

Iterative methods

Worked answers and methods for 9.2 on Edexcel A-level Maths 9MA0.

Explanation

A fixed-point iteration rewrites an equation as $x=g(x)$ and generates approximations using $x_{n+1}=g(x_n)$ from a chosen starting value.
Calculate successive values without premature rounding; on a graph of $y=g(x)$ and $y=x$ , move vertically to $y=g(x)$ and horizontally to $y=x$ to display each iteration.
Near a fixed point $\alpha$ , iterations usually converge when $|g'(\alpha)|<1$ : a positive gradient gives a staircase pattern and a negative gradient gives a cobweb pattern.
Different rearrangements of the same equation can converge at different rates or diverge; a common error is to assume that obtaining $x=g(x)$ automatically produces a useful recurrence.

A staircase diagram alternates between

y=g(x)

and

y=x

to display successive fixed-point iterates.

Worked example

Use $x_{n+1}=\sqrt{7-x_n}$ with $x_0=2$ to calculate $x_1$ , $x_2$ and $x_3$ to $4$ decimal places. State the equation satisfied by any limiting value.

1.Repeated substitution gives $x_1=\sqrt5=2.236067\ldots$ , $x_2=\sqrt{7-2.236067\ldots}=2.182643\ldots$ and $x_3=\sqrt{7-2.182643\ldots}=2.194847\ldots$ .
2.If the sequence tends to $L$ , then $L=\sqrt{7-L}$ , so $L^2+L-7=0$ .

Answer: $x_1=2.2361$ , $x_2=2.1826$ and $x_3=2.1948$ .; A limiting value satisfies $x^2+x-7=0$ .

Common mistakes

Don't draw horizontal moves to $y=g(x)$ and vertical moves to $y=x$ , reversing the cobweb construction.
Don't calculate iterates and fail to state the fixed-point equation or assess convergence from the iteration graph.

Exam tip

Show each substituted iterate to the requested accuracy, then set the limiting value equal to the iteration function.

Worked practice

Tier 1 · Easy

The recurrence

x_{n+1}=\frac{1}{2}\left(x_n+\frac{6}{x_n}\right)

is used with

x_0=2

. Find

x_1

and

x_2

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$x_1=2.5$ and $x_2=2.45$	2
Notes
Substitute $x_0=2$ to obtain $x_1=\frac12(2+6/2)=2.5$ . Then $x_2=\frac12(2.5+6/2.5)=\frac12(4.9)=2.45$ .

(2 marks)

Tier 2 · Standard

On a pair of axes, draw

y=x

and

y=2+\dfrac14x

. Starting at

x_0=0

on the

x

-axis, draw the first three steps of the fixed-point construction. State whether the construction is a staircase or a cobweb, and find its limiting value.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	The construction moves vertically to $y=2+x/4$ and horizontally to $y=x$ , giving $x_1=2$ , $x_2=2.5$ and $x_3=2.625$ . It is a convergent staircase. The limiting value is $\dfrac83$ .	4
Notes
The graphical steps follow $x_{n+1}=2+x_n/4$ : from $x_0=0$ , they give $x_1=2$ , $x_2=5/2$ and $x_3=21/8$ . The gradient $g'(x)=1/4$ is positive, so the path is a staircase; since $0<1/4<1$ , it converges. At the fixed point $L=2+L/4$ , so $L=8/3$ .

(4 marks)

Tier 3 · Hard

For

x_{n+1}=3-0.4x_n

with

x_0=1

, find

x_1

x_4

. Determine the limiting value and state whether the graphical construction is a cobweb or a staircase, giving a reason.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$x_1=2.6$ , $x_2=1.96$ , $x_3=2.216$ and $x_4=2.1136$ . The limit is $\frac{15}{7}$ , and the construction is a convergent cobweb because $g'(x)=-0.4$ .	5
Notes
Apply the recurrence successively: $x_1=3-0.4=2.6$ , $x_2=3-0.4(2.6)=1.96$ , $x_3=3-0.4(1.96)=2.216$ and $x_4=3-0.4(2.216)=2.1136$ . A limit $L$ obeys $L=3-0.4L$ , hence $L=3/1.4=15/7$ . Since the constant gradient is negative, the iterates alternate across the fixed point, producing a cobweb; since $\|-0.4\|<1$ , it converges.

(5 marks)

Tier 1 · Easy

The recurrence

x_{n+1}=\sqrt{11-x_n}

is used with

x_0=3

. Find

x_1

and

x_2

, giving each answer to

4

decimal places.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$x_1=2.8284$ and $x_2=2.8586$	2
Notes
$x_1=\sqrt{11-3}=\sqrt8=2.828427\ldots$ . Using this unrounded value, $x_2=\sqrt{11-2.828427\ldots}=2.858596\ldots$ . Therefore $x_1=2.8284$ and $x_2=2.8586$ to $4$ decimal places.

(2 marks)

Tier 2 · Standard

Starting with

x_0=0

, use

x_{n+1}=1+\dfrac{6}{x_n+4}

to find

x_1

x_2

and

x_3

. Find the positive limiting value and state whether the associated construction is a staircase or a cobweb.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$x_1=\dfrac52$ , $x_2=\dfrac{25}{13}$ and $x_3=\dfrac{155}{77}$ . The positive limit is $2$ , and the construction is a convergent cobweb.	4
Notes
Successive substitution gives $x_1=1+6/4=2.5$ , $x_2=1+6/6.5=25/13=1.923076\ldots$ and $x_3=1+6/(25/13+4)=155/77=2.012987\ldots$ . A limit $L$ satisfies $(L-1)(L+4)=6$ , so $L^2+3L-10=0$ and the positive solution is $L=2$ . Here $g'(x)=-6/(x+4)^2$ , so $g'(2)=-1/6$ : the negative gradient gives a cobweb and its magnitude below $1$ gives convergence.

(4 marks)

Tier 3 · Hard

The positive root

\alpha

x^2+x-4=0

lies in

(1.5,1.6)

. Two proposed iterations are

x_{n+1}=4-x_n^2

and

x_{n+1}=\sqrt{4-x_n}

. Without carrying out a long sequence of iterations, determine which recurrence is locally convergent to

\alpha

. State whether its graphical construction is a staircase or a cobweb.

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$x_{n+1}=\sqrt{4-x_n}$ is locally convergent to $\alpha$ and gives a cobweb; $x_{n+1}=4-x_n^2$ is locally divergent.	5
Notes
Both fixed-point equations rearrange to $x^2+x-4=0$ . For $g_1(x)=4-x^2$ , $\|g_1'(\alpha)\|=2\alpha>3>1$ , so this recurrence is locally divergent. For $g_2(x)=\sqrt{4-x}$ , $g_2'(x)=-1/(2\sqrt{4-x})$ . At the fixed point $\sqrt{4-\alpha}=\alpha$ , so $\|g_2'(\alpha)\|=1/(2\alpha)<1/3<1$ and the recurrence is locally convergent. Its derivative is negative, so successive graphical steps alternate across the fixed point and form a cobweb.

(5 marks)

Tier 2 · Standard

The recurrence

x_{n+1}=\ln(x_n+6)

is used with

x_0=1

and is known to converge to a fixed point

\alpha

. Write down an equation of the form

f(x)=0

whose root is

\alpha

. Calculate

x_1

x_2

and

x_3

, giving each value to

4

decimal places. The values satisfy

x_0<x_1<x_2<x_3<\alpha

. Without using derivatives, explain why the associated fixed-point path is a staircase rather than a cobweb.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$e^x-x-6=0$ $x_1=1.9459$ , $x_2=2.0727$ and $x_3=2.0885$ . The path is a staircase because the iterates remain on the same side of $\alpha$ and increase towards it instead of alternating across it.	4
Notes
At a fixed point, $\alpha=\ln(\alpha+6)$ , hence $e^\alpha=\alpha+6$ and $e^x-x-6=0$ . Using unrounded values gives $x_1=\ln7=1.945910149\ldots$ , $x_2=\ln(6+x_1)=2.072657349\ldots$ and $x_3=\ln(6+x_2)=2.088482715\ldots$ , so the requested values are $1.9459$ , $2.0727$ and $2.0885$ . Since the iterates increase towards $\alpha$ while staying below it, the graphical steps approach the intersection from one side and form a staircase, not a cobweb.

(4 marks)

Tier 3 · Hard

The recurrence

x_{n+1}=\dfrac{3}{x_n+1}

is used with

x_0=1

. Use exact arithmetic to find

x_1

x_2

x_3

and

x_4

. Find the positive fixed point. Use the pattern in these four iterates to identify the type of fixed-point diagram and state whether convergence is indicated.

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$x_1=\dfrac32$ , $x_2=\dfrac65$ , $x_3=\dfrac{15}{11}$ and $x_4=\dfrac{33}{26}$ . The positive fixed point is $\dfrac{-1+\sqrt{13}}2$ ; the construction is a cobweb and the iterates appear to converge.	5
Notes
Exact substitution gives $x_1=3/2$ , $x_2=3/(5/2)=6/5$ , $x_3=3/(11/5)=15/11$ and $x_4=3/(26/11)=33/26$ . A fixed point $L$ satisfies $L=3/(L+1)$ , so $L^2+L-3=0$ and the positive solution is $(-1+\sqrt{13})/2=1.3027\ldots$ . The iterates lie alternately above and below this value, with decreasing deviations over the four steps, so the graphical path is a converging cobweb.

(5 marks)

Tier 3 · Hard

A fixed-point iteration has the linear form

x_{n+1}=a+bx_n

. On its fixed-point diagram, the first vertical step from

x_0=1

reaches

(1,1.8)

, and the next vertical step reaches

(1.8,2.04)

. Find

a

and

b

. Calculate

x_3

, find the limiting value, and classify the diagram as a convergent staircase or a convergent cobweb.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$a=1.5$ and $b=0.3$ , so $x_{n+1}=1.5+0.3x_n$ . $x_3=2.112$ , the limit is $\dfrac{15}{7}$ , and the diagram is a convergent staircase.	5
Notes
The two vertical steps show that $g(1)=1.8$ and $g(1.8)=2.04$ . Hence $a+b=1.8$ and $a+1.8b=2.04$ . Subtracting gives $0.8b=0.24$ , so $b=0.3$ and $a=1.5$ . The second ordinate is $x_2=2.04$ , so $x_3=1.5+0.3(2.04)=2.112$ . A limit $L$ satisfies $L=1.5+0.3L$ , giving $L=15/7$ . Since $0<b<1$ , the steps approach the fixed point from one side as a convergent staircase.

(5 marks)

Q10

Tier 3 · Hard

10.

The iteration

x_{n+1}=2-e^{-x_n}

starts from

x_0=1

. Use unrounded values in your working. Calculate

x_1

x_6

, giving each value to

6

decimal places, and find the least value of

n

for which

|x_{n+1}-x_n|<0.0005

. Hence estimate the fixed point to

4

decimal places and state the type of fixed-point diagram.

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$x_1=1.632121$ , $x_2=1.804485$ , $x_3=1.835441$ , $x_4=1.840457$ , $x_5=1.841255$ and $x_6=1.841382$ . The least value is $n=5$ , since $\|x_6-x_5\|=0.000126668\ldots<0.0005$ but $\|x_5-x_4\|=0.000798258\ldots>0.0005$ . The fixed point is approximately $1.8414$ , and the diagram is a convergent staircase.	6
Notes
Repeated substitution, retaining the unrounded value each time, gives $x_1=1.632120558\ldots$ , $x_2=1.804485465\ldots$ , $x_3=1.835440893\ldots$ , $x_4=1.840456855\ldots$ , $x_5=1.841255113\ldots$ and $x_6=1.841381782\ldots$ . The first consecutive difference below $0.0005$ is therefore $\|x_6-x_5\|$ , so $n=5$ . Continuing the iteration gives the fixed point $1.841405660\ldots$ , hence $1.8414$ to $4$ decimal places. Here $g'(x)=e^{-x}$ is positive and is less than $1$ for these positive iterates, so the path is a convergent staircase.

(6 marks)

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