An exponential denominator creates a stable limit

Supplied graph of y=x/2ˣ−3x/4 with axes, origin and positive root α.y = x/2ˣ − 3x/4. Native mathematical paths and source labels.xy0α
Question diagram 1
Supplied graph of y=x/2ˣ−3x/4 with axes, origin and positive root α.y = x/2ˣ − 3x/4. Native mathematical paths and source labels.xy0α

Supplied graph of y=x/2ˣ−3x/4 with axes, origin and positive root α.

The diagram shows y=x2x−34xy=\frac{x}{2^x}-\frac34x. The two roots of x2x−34x=0\frac{x}{2^x}-\frac34x=0 are 0,α0,\alpha.

(i) Find α\alpha exactly. marks

A sequence of positive real numbers satisfies xn+1=xn2xn+14xnx_{n+1}=\frac{x_n}{2^{x_n}}+\frac14x_n, n≥1n\ge1.

(ii) Considering xn+1−xnx_{n+1}-x_n, show xn+1>xnx_{n+1}>x_n if 0<xn<α0<x_n<\alpha. marks

(iii) For 0<x1<α0<x_1<\alpha, use (i),(ii) to find the exact limit. marks

[2011/YJC/Prelim/I/1]

Total marks: 7

Solution

(i)

Exact root

The diagram shows y=x2x−34xy=\frac x{2^x}-\frac34x with roots 00 and α\alpha. Find α\alpha exactly.

Explore this idea

Hint

  1. Factor out xx.

  2. Solve the exponential equation.

Answer · (i)

α=ln⁡(4/3)/ln⁡2\alpha=\ln(4/3)/\ln2.

Explore the idea

Two fixed points: 0 and α

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 = 0.1
Iterations drawn6
Try an example
u(n = 7)
0.23178
u(n = 1)
0.1
Distance to α ≈ 0.415
0.18325
Key idea

The roots of x2x−34x\frac x{2^x}-\frac34x are the fixed points of the recurrence.

Ideas to try
The curve touches the diagonal at the origin and again at α\alpha. These are the roots of the increment.
(ii)

Increment sign

Considering xn+1−xnx_{n+1}-x_n, show that xn+1>xnx_{n+1}>x_n if 0<xn<α0<x_n<\alpha.

Explore this idea

Hint

  1. Write the step as the graph in (i) evaluated at xnx_n.

Answer · (ii)

xn+1−xn=xn(2−xn−3/4)>0x_{n+1}-x_n=x_n(2^{-x_n}-3/4)>0 for 0<xn<α0<x_n<\alpha.

Explore the idea

A positive increment on (0, α)

Open the increment plot: the graph is above the axis for 0<x<α0<x<\alpha and below it afterwards.

Starting valueu₁ = 0.1
Iterations drawn1
Try an example
u(n = 2)
0.1183
u(n = 1)
0.1
Distance to 0
0.1183
Key idea

The increment changes sign at the fixed points, so terms are pushed toward α\alpha from both sides and away from 00.

Ideas to try
  1. With one iteration from 0.20.2 the term rises. From 0.80.8 it falls.
  2. Move the start across α≈0.415\alpha\approx0.415 in the increment plot and watch the dot cross the axis.
(iii)

The exact limit

For 0<x1<α0<x_1<\alpha, use (i) and (ii) to find the exact limit of the sequence.

Explore this idea

Hint

  1. Show the sequence increases and stays below α\alpha.

  2. Use the fixed-point roots from (i).

Answer · (iii)

Limit α=ln⁡(4/3)/ln⁡2\alpha=\ln(4/3)/\ln2.

0<x1<α0<x_1<\alpha; the sequence stays below α\alpha.

Explore the idea

Climbing to α

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 = 0.1
Iterations drawn10
Try an example
u(n = 11)
0.31737
u(n = 1)
0.1
Distance to α ≈ 0.415
0.09766
Key idea

The origin repels (slope 54\frac54) and α\alpha attracts (slope about 0.780.78), so every start in (0,α)(0,\alpha) ends at α\alpha.

Ideas to try
  1. Start tiny: the web leaves the origin slowly, then speeds up and settles at α\alpha.
  2. Use the distance readout. It shrinks toward zero as iterations are added.
Takeaway

Bounded and monotone

Show the terms move one way and cannot pass a barrier, then use the fixed-point equation to name the limit.