An alternating fifth-root iteration

Supplied graph of y=x⁵+2x−8 with axes, origin and root α.y = x⁵ + 2x − 8. Native mathematical paths and source labels.xy0αy = x⁵ + 2x − 8
Question diagram 1
Supplied graph of y=x⁵+2x−8 with axes, origin and root α.y = x⁵ + 2x − 8. Native mathematical paths and source labels.xy0αy = x⁵ + 2x − 8

Supplied graph of y=x⁵+2x−8 with axes, origin and root α.

The diagram shows y=x5+2x−8y=x^5+2x-8. Its root is denoted by α\alpha.

(i) Find α\alpha correct to 3 decimal places. mark

The real numbers xnx_n satisfy xn+1=(8−2xn)1/5x_{n+1}=(8-2x_n)^{1/5}, n≥1n\ge1.

(ii) Using (i), show that if the sequence converges, it converges to α\alpha. marks

(iii) Considering the graphs of y=xy=x and y=(8−2x)1/5y=(8-2x)^{1/5} on the same diagram, or otherwise, prove that if α<xn<4\alpha<x_n<4, then (a) 0<xn+1<α0<x_{n+1}<\alpha, (b) α<xn+2<xn\alpha<x_{n+2}<x_n. marks

(iv) Given x1=3x_1=3, use (iii) to obtain an inequality relating 0,α,x1,x2,x30,\alpha,x_1,x_2,x_3. With a graphic calculator, describe the sequence behaviour. marks

[2014/TJC/Promo/7]

Total marks: 8

Solution

(i)

Locate the root

The diagram shows y=x5+2x−8y=x^5+2x-8 with root α\alpha. Find α\alpha correct to 3 decimal places.

Explore this idea

Hint

  1. Use a graphing calculator, then check there is only one real root.

Answer · (i)

α=1.392\alpha=1.392 (3 d.p.).

Explore the idea

Where the iteration function meets the diagonal

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 = 3
Iterations drawn0
Try an example
u(n = 1)
3
u(n = 1)
3
Distance to α ≈ 1.392
1.6085
Key idea

Raising both sides of x=(8−2x)1/5x=(8-2x)^{1/5} to the fifth power gives x5+2x−8=0x^5+2x-8=0: the same root.

Ideas to try
The curve y=(8−2x)1/5y=(8-2x)^{1/5} meets y=xy=x exactly where x5+2x−8=0x^5+2x-8=0.
(ii)

Limit argument

Using (i), show that if the sequence converges, it converges to α\alpha.

Hint

  1. Take limits in xn+1=(8−2xn)1/5x_{n+1}=(8-2x_n)^{1/5}.

Answer · (ii)

Any limit solves L5+2L−8=0L^5+2L-8=0, hence L=αL=\alpha.

(iii)(a)

Flip across the root

Prove that if α<xn<4\alpha<x_n<4 then 0<xn+1<α0<x_{n+1}<\alpha.

Explore this idea

Hint

  1. Use that F(x)=(8−2x)1/5F(x)=(8-2x)^{1/5} is decreasing.

Answer · (iii)(a)

0<xn+1<α0<x_{n+1}<\alpha.

Explore the idea

A step from above lands below

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 = 3
Iterations drawn1
Try an example
u(n = 2)
1.1487
u(n = 1)
3
Distance to α ≈ 1.392
0.2428
Key idea

Because the curve is decreasing, a step across the fixed point always switches sides.

Ideas to try
Pick any start between α\alpha and 44. After one iteration the term is in (0,α)(0,\alpha).
(iii)(b)

Two steps move inward

Prove that if α<xn<4\alpha<x_n<4 then α<xn+2<xn\alpha<x_{n+2}<x_n.

Explore this idea

Hint

  1. Write t=xnt=x_n and y=xn+1=F(t)y=x_{n+1}=F(t).

  2. Show F(y)>αF(y)>\alpha, then F(y)<tF(y)<t.

Answer · (iii)(b)

α<xn+2<xn\alpha<x_{n+2}<x_n.

Explore the idea

Two steps: back above α, but closer

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 = 3
Iterations drawn2
Try an example
u(n = 3)
1.41649
u(n = 1)
3
Distance to α ≈ 1.392
0.02499
Key idea

Every other term sits above α\alpha and decreases; the others sit below and increase. They squeeze in on α\alpha.

Ideas to try
  1. Two iterations from 33 give 1.1491.149 then 1.4161.416: above α\alpha again, and below 33.
  2. The web is a tightening spiral because the curve is much flatter than the diagonal at α\alpha.
(iv)

Chain the inequalities

Given x1=3x_1=3, use (iii) to obtain an inequality relating 0,α,x1,x2,x30,\alpha,x_1,x_2,x_3. With a graphic calculator, describe the behaviour of the sequence.

Explore this idea

Hint

  1. Apply (a) then (b) with n=1n=1.

Answer · (iv)

0<x2≈1.149<α<x3≈1.416<x1=30<x_2\approx1.149<\alpha<x_3\approx1.416<x_1=3; oscillates toward α\alpha, odd terms decreasing and even terms increasing.

Explore the idea

From x₁ = 3 to 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 = 3
Iterations drawn7
Try an example
u(n = 8)
1.3915
u(n = 1)
3
Distance to α ≈ 1.392
0
Key idea

A decreasing iteration function gives an oscillating sequence; flatness at the fixed point makes the oscillation die out quickly.

Ideas to try
Open the terms plot: the points alternate, above then below the guide line at α\alpha, tightening each time.
Takeaway

Odd terms down, even terms up

For a decreasing map, the odd and even subsequences move in opposite directions toward the same fixed point.