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 Punch July 1st, 2012 06:47 PM

Convergence of a recurrence relation

Given $x_{n+1}=\frac{1}{3}e^{x_n}$. Use a calculator to determine the behaviour of the sequence for each of the cases $x_1=0, x_1=1, x_1=2$

I had no problems doing $x_1=0$ and $x_1=1$.
For $x_1=0$, the series increases and converges to 0.619 while for $x_1=1$, the series decreases and converges to 0.619.

However, when solving for $x_1=2$, all i got was error on the calculator screen.

 MarkFL July 1st, 2012 07:25 PM

Re: Convergence of a recurrence relation

1 Attachment(s)
I observed the same behavior you did, and for $x_1=2$ the series diverged.

For $x_1=0$ it increased and converged to 0.619061286736

For $x_1=1$ it decreased and converged to 0.619061286736

I tried finding the closed form via Newton's method to create a first order IVP, but wound up with an integral without an anti-derivative expressible in elementary terms. :(

I found by setting $x_{n+1}=x_n\$ that the value the series converged to in the first two cases is one of the solutions to:

$3x=e^x$

The other solution is about 1.5121345516578424739.

Here is a plot:

[attachment=0:5ucqk92t]convergence.jpg[/attachment:5ucqk92t]

The larger root seems to be the division between convergence and divergence for the given recurrence relation when used as the starting value.

 aswoods July 1st, 2012 08:02 PM

Re: Convergence of a recurrence relation

Solving for fixed points,
$x=\frac13e^x\;\therefore\; -xe^{-x}=-\frac13\;\therefore\;-x=\mathrm{W}(-\frac13)\;\therefore\;x=-\mathrm{W}(-\frac13)$
where W(z) is the Lambert W function, which can be assigned two values there, 0.619061 and 1.51213.
The first is stable and the second is unstable; any starting value greater than 1.51213 will result in divergence.

 MarkFL July 1st, 2012 08:10 PM

Re: Convergence of a recurrence relation

In fairness to [color=#0040FF]aswoods[/color], I was editing my post to include essentially the same information (albeit not as succinctly) in his post at the same time he was posting. :wink:

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