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Finding the limit of a recurrence equation.

by iironiic
Tags: equation, limit, recurrence
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iironiic
#1
Feb25-12, 12:15 PM
P: 9
I have been working on a problem proposed in a math journal, and there is only one thing I need to figure out. Here it is:

Let [itex](a_n)[/itex] be a sequence defined by [itex]a_1 = a[/itex] and [itex]a_{n+1} = 2^n-\sqrt{2^n(2^n-a_n)}[/itex] for all [itex]0 \leq a \leq 2[/itex] and [itex]n \geq 1[/itex]. Find [itex]\lim_{n \rightarrow \infty} 2^n a_n[/itex] in terms of [itex]a[/itex].

What I figured out so far:

Spoiler

Let [itex]A = \lim_{n \rightarrow \infty} 2^n a_n[/itex].

When [itex]a = 0[/itex], [itex]A = 0[/itex].

When [itex]a = \frac{1}{2}[/itex], [itex]A = \frac{\pi^2}{9}[/itex].

When [itex]a = 1[/itex], [itex]A = \frac{\pi^2}{4}[/itex].

When [itex]a = \frac{3}{2}[/itex], [itex]A = \frac{4\pi^2}{9}[/itex].

When [itex]a = 2[/itex], [itex]A = \pi^2[/itex].


I'm still trying to figure it out. Any insight on recurrence equations or limits would be greatly appreciated! Thanks!
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alexfloo
#2
Feb25-12, 01:27 PM
P: 192
Here's a tip: if the recurrence is [itex]a_{n+1}=f(a_n)[/itex] where f is continuous, then the limit is a solution to a=f(a). Intuitively, this is because since points near the limit change little, we reason that the limit ought to remain fixed under the recurrence.

Your recurrence is not in that form (since f depends on n), but some trickery may be used to put it there, for instance by considering f(a,n) as a function of a, for fixed, large n. (It'll be tough to justify rigorously, but it may be a good starting point.)


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