Finding the limit of a recurrence equation.

[iironiic]

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

What I figured out so far:

Spoiler

Let A = \lim_{n \rightarrow \infty} 2^n a_n.

When a = 0, A = 0.

When a = \frac{1}{2}, A = \frac{\pi^2}{9}.

When a = 1, A = \frac{\pi^2}{4}.

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

When a = 2, A = \pi^2.

I'm still trying to figure it out. Any insight on recurrence equations or limits would be greatly appreciated! Thanks!

 

[alexfloo]

Here's a tip: if the recurrence is a_{n+1}=f(a_n) 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.)