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 July 1st, 2012, 06:47 PM #1 Member   Joined: Jan 2012 Posts: 72 Thanks: 0 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.
July 1st, 2012, 07:25 PM   #2
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Re: Convergence of a recurrence relation

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.
Attached Images
 convergence.jpg (12.2 KB, 360 views)

 July 1st, 2012, 08:02 PM #3 Senior Member   Joined: Feb 2009 From: Adelaide, Australia Posts: 1,519 Thanks: 3 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.
 July 1st, 2012, 08:10 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs 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.

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