Question

help. Let a1 be a positive real number. Define a sequence an recursively by an+1 = a^2n -1. Show that an does not converges to some non-zero value?

Answer 1

I can't open the document, but if the question is written exactly as that, then it isn't necessarily true. If \[x=\frac{1+\sqrt5}{2}\]then it converges to itself.

Answer 2

If you start at something greater than that, then it will diverge, and if it starts smaller, then it will converge to 0.

Answer 3

\[a _{n}\]

Answer 4

\[a ^{2_{}}_{n}\]

Answer 5

-1

Answer 6

that what i meant to write for a sub n

Answer 7

Ah. So \[a_{n+1}=a_{n-1}^2?\]

Answer 8

yes

Answer 9

Well that isn't necessarily true either. If \(a_1=1\), then it converges to 1.

Answer 10

How do I prove that?

Answer 11

apart from a1 being 1. how do I prove that it does not converge to a non-zero value?
@KIngGeorge

Answer 12

@KingGeorge

Answer 13

Well, if \(0<a_1<1\), then what can you say about \(a_2=a_1^2\)? You know that \(a_2<a_1\). Similarly, if \(1<a_1\), then \(a_2>a_1\). Finally, since the square root function isn't bounded, you know that if \(a_1>1\), then the sequence you described must diverge.