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