Alternation around a fixed point

u_1,u_2,u_3,\ldots

[2023/DHS/Prelim/II/2]

Total marks: 9

Solution

(a)(i)(A)

Describe the behaviour for c = 5

$u_{n+1}=3-0.5u_n$ with $u_1=c$ . Describe how the sequence behaves when $c=5$ .

Explore this idea

Hint

Find the fixed point $l=3-0.5l$ .
Compute a few terms and see which side of the fixed point they fall on.

Answer · (a)(i)(A)

5,0.5,2.75,1.625,\ldots

Explore the idea

Spiralling into the fixed point

Start at the chosen value on the horizontal axis, go up to the curve, across to the diagonal, and repeat. Each corner of the web is the next term.

Starting valueu₁ = 5

Iterations drawn5

Try an example

u(n = 6)

1.90625

u(n = 1)

Distance to l = 2

0.09375

Key idea

A slope of $-\frac12$ at the fixed point means each step reverses the error and halves it, so the terms alternate and converge.

Ideas to try

Add iterations one at a time. The web spirals in because the line has slope $-0.5$ , between $-1$ and $0$ .
Move the starting value. Wherever you start, the web ends at the same point.

(a)(i)(B)

Describe the behaviour for c = 2

Describe how the sequence behaves when $c=2$ .

Explore this idea

Hint

Check what happens when the starting value is the fixed point.

Answer · (a)(i)(B)

u_n=2

Explore the idea

Starting exactly at the fixed point

Start at the chosen value on the horizontal axis, go up to the curve, across to the diagonal, and repeat. Each corner of the web is the next term.

Starting valueu₁ = 2

Iterations drawn6

Try an example

u(n = 7)

u(n = 1)

Distance to l = 2

Key idea

A fixed point is an equilibrium: starting there changes nothing.

Ideas to try

With

c=2

the web is a single dot. Nudge the start by one notch and watch it spiral back.

(a)(ii)

Solve for c

Find the value of $c$ for which $2u_3=-5u_2$ .

Explore this idea

Hint

Write $u_2$ and $u_3$ in terms of $c$ .
Substitute in $2u_3=-5u_2$ and solve the linear equation.

Answer · (a)(ii)

c=9

Explore the idea

Where the relation holds

Start at the chosen value on the horizontal axis, go up to the curve, across to the diagonal, and repeat. Each corner of the web is the next term.

Starting valueu₁ = 9

Iterations drawn2

Try an example

u(n = 3)

3.75

u(n = 1)

Distance to l = 2

1.75

Key idea

The equation $2u_3=-5u_2$ needs $u_2$ and $u_3$ of opposite sign, so the starting value must be far from the fixed point.

Ideas to try

Set two iterations with $c=9$ . The sequence reads $9,\ -1.5,\ 3.75$ ; the third term is $-\frac{5}{2}$ times the second, as required.
Try $c=0$ for comparison: the two terms are $3$ and $1.5$ , with the wrong signs for this relation.

(b)(i)

Relate c and p

$v_1=p$ , $v_2=2$ and $v_{n+2}=2v_n+v_{n+1}-1$ . If $v_3+1=2u_2-u_1$ , find a relationship between $c$ and $p$ .

Hint

Compute $v_3$ in terms of $p$ .
Compute $2u_2-u_1$ in terms of $c$ and equate.

Answer · (b)(i)

c+p=2

(b)(ii)

Find p from v₅

Find the value of $p$ for which $v_5=77$ .

Explore this idea

Hint

Compute $v_3,v_4,v_5$ in terms of $p$ and solve.

Answer · (b)(ii)

p=12

Explore the idea

The sequence v for p = 12

With $p=12$ the first five terms are $12,\ 2,\ 25,\ 28,\ 77$ .

Indexn = 5

Try an example

term vₙ at n = 5

Key idea

Because each term depends on two earlier terms, $v_5$ is a linear expression in $p$ , and one equation fixes $p$ .

Ideas to try

Move to

n=3

v_3=2(12)+1=25

. Then compare

v_4=2(12)+4=28

and

v_5=6(12)+5=77

Takeaway

Solve the fixed point first

Find the value that maps to itself, then measure everything as a distance from it. A multiplier between $-1$ and $0$ gives an alternating approach.

Alternation around a fixed point

Solution

Describe the behaviour for c = 5

Find the fixed point

List the first terms

Shift to the error

Read a calculator table carefully

Spiralling into the fixed point

Describe the behaviour for c = 2

Substitute

Use the formula

Starting exactly at the fixed point

Solve for c

Express the terms

Form and solve the equation

Check

Where the relation holds

Relate c and p

Third term of the second sequence

Right-hand side in terms of c

Equate

Find p from v₅

Climb to the fifth term

Solve

The sequence v for p = 12

Solve the fixed point first