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Worked solution

Relate an exponential recurrence to its two fixed points

The graph is y=ex−3xy=e^x-3x, whose roots are α<β\alpha<\beta.

(i) Find both roots to 3 decimal places.

A real sequence satisfies xn+1=13exnx_{n+1}=\frac13e^{x_n} for n≥1n\ge1.

(ii) Prove algebraically that, if it converges, its limit is α\alpha or β\beta.

(iii) Use a calculator to determine its behaviour for x1=0,1,2x_1=0,1,2.

(iv) By considering xn+1−xnx_{n+1}-x_n, prove that xn+1<xnx_{n+1}<x_n if α<xn<β\alpha<x_n<\beta, and xn+1>xnx_{n+1}>x_n if xn<αx_n<\alpha or xn>βx_n>\beta.

(v) Relate (iv) to the behaviours in (iii).

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Graph of e^x−3x: positive at x=0, crossing at two positive roots α<β, negative between the roots and positive outside them.
Graph of e^x−3x: positive at x=0, crossing at two positive roots α<β, negative between the roots and positive outside them.
Graph of e^x−3x: positive at x=0, crossing at two positive roots α<β, negative between the roots and positive outside them.

[2007/I/9]

CJC · Sequences and Series · Example 32

(i) Two roots

Answer

α≈0.619, β≈1.512\alpha\approx0.619,\ \beta\approx1.512 (3 d.p.).

Solve the graph equation on both sides of its minimum

ex−3x=0⟹α=0.619061…,β=1.512135…e^x-3x=0\quad\Longrightarrow\quad\alpha=0.619061\ldots,\quad\beta=1.512135\ldots

The calculator root values are ordered by alpha<beta. The positive y-intercept and one minimum give two positive crossings.

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(ii) Possible limits under the convergence assumption

Answer

If the sequence converges to a finite LL, then L=αL=\alpha or L=βL=\beta.

State the hypothesis before replacing terms by limits

xn→L,xn+1→L,L=13eLx_n\to L,\qquad x_{n+1}\to L,\qquad L=\frac13e^L

The source callout stresses that convergence is assumed here. The shifted sequence has the same limit, and the exponential is continuous.

Recognise the root equation

3L=eL⟹eL−3L=0⟹L∈{α,β}3L=e^L\quad\Longrightarrow\quad e^L-3L=0\quad\Longrightarrow\quad L\in\{\alpha,\beta\}

This identifies possible limits; it does not establish convergence for every initial value.

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(iii) Calculator behaviours

Answer

For x1=0x_1=0, it increases to α\alpha; for x1=1x_1=1, it decreases to α\alpha; for x1=2x_1=2, it increases and diverges to positive infinity.

Iterate using unrounded values

x1=0x1=1x1=2x20.3330.9062.463x30.4650.8253.913x40.5310.76116.69\begin{array}{c|rrr}&x_1=0&x_1=1&x_1=2\\\hline x_2&0.333&0.906&2.463\\x_3&0.465&0.825&3.913\\x_4&0.531&0.761&16.69\end{array}

The source table displays rounded early values. Repeated 0.619 entries are a display precision effect, not exact equality with the limit.

Compare the three starts with the two roots

The source reports increasing convergence from 0, decreasing convergence from 1, and increasing divergence from 2. The sign argument and interval reasoning in (iv)–(v) justify these observed patterns.

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(iv) Sign of the successive difference

Answer

xn+1<xnx_{n+1}<x_n between the roots, and xn+1>xnx_{n+1}>x_n outside them.

Express the difference as a positive rescaling of the given graph

xn+1−xn=13exn−xn=13(exn−3xn)\begin{aligned}&x_{n+1}-x_n\\ & =\frac13e^{x_n}-x_n\\ & =\frac13(e^{x_n}-3x_n)\end{aligned}

The source suggests entering y2=(e^x−3x)/3 on the calculator with y1=e^x−3x. Dividing vertical coordinates by 3 preserves roots and signs.

Use the negative region between the roots

α<xn<β⇒13(exn−3xn)<0⇒xn+1−xn<0⇒xn+1<xn\alpha<x_n<\beta\Rightarrow\frac13(e^{x_n}-3x_n)<0\Rightarrow x_{n+1}-x_n<0\Rightarrow x_{n+1}<x_n

The graph is below the axis between alpha and beta.

Use the positive regions outside the roots

xn<α or xn>β⇒13(exn−3xn)>0⇒xn+1>xnx_n<\alpha\ \text{or}\ x_n>\beta\Rightarrow\frac13(e^{x_n}-3x_n)>0\Rightarrow x_{n+1}>x_n

The graph is above the axis outside the two crossings. An algebraic sign check is also available: h″(x)=e^x>0 and h′(x)=e^x−3 has its only zero at ln3, so h decreases to one minimum then increases.

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The curves y₁=e^x−3x and y₂=(e^x−3x)/3 cross at the same α,β. Both are negative between those roots and positive outside; y₂ is vertically compressed by a factor of three.
The curves y₁=e^x−3x and y₂=(e^x−3x)/3 cross at the same α,β. Both are negative between those roots and positive outside; y₂ is vertically compressed by a factor of three.
The curves y₁=e^x−3x and y₂=(e^x−3x)/3 cross at the same α,β. Both are negative between those roots and positive outside; y₂ is vertically compressed by a factor of three.
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(v) Explain the three observed behaviours

Answer

0<α0<\alpha gives increase; α<1<β\alpha<1<\beta gives decrease; 2>β2>\beta gives increase. The first two converge to α\alpha and the third is unbounded.

Keep later iterates in the same relevant interval

0≤x<α⇒x<g(x)<α;α<x<β⇒α<g(x)<x;x>β⇒g(x)>x>β0\le x<\alpha\Rightarrow x<g(x)<\alpha;\qquad\alpha<x<\beta\Rightarrow\alpha<g(x)<x;\qquad x>\beta\Rightarrow g(x)>x>\beta

The source compares the three initial values with the roots. To extend each sign statement to every iterate, use that g(x)=e^x/3 is strictly increasing and fixes both roots.

Distinguish bounded increase from unbounded increase

Starts 0 and 1 are monotone and bounded on the appropriate side of alpha, so their finite limits must be alpha. Starting at 2 gives an increasing sequence always at least 2; any finite limit would exceed beta and contradict (ii), so it diverges to positive infinity.

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Solution notes

The source rounds roots and limit displays to 0.619 and 1.512; these are approximations, not exact roots. Its final sentence in (ii) must retain the initial “if convergent” hypothesis. Interval preservation and the boundedness argument in (v) are authored details that justify applying the source sign test repeatedly.