# Finding the limit of a recurrence equation.

by iironiic
Tags: equation, limit, recurrence
 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 $(a_n)$ 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!
 P: 192 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.)

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