Recursive sequence - show it does not converge to zero

[mathscott123]

Hi guys,

I'm new here at this forum, but I don't understand this problem.

Let a1 be a positive real number. Define a sequence an recursively by a(n+1) = (an)^2 - 1. Show that an does not converge to zero.

(Is there a1 such that the sequence an converges to some non-zero value?)

I'm not sure if this requires something with epsilon and delta.

 

[micromass]

Can you prove that the limit of the sequence must be a fixed point of $f(x)=x^2-1$? That is, it satisfies $x=x^2-1$.

 

[mathscott123]

Well I have this:

an = (an)^2 - 1
0 = (an)^2 - an - 1
an = 1 +/- sqrt (5)/2

I'm not sure what this tells me though

 

[micromass]

See my post 2. Prove that $x=\lim_n a_n$satisfies $x=x^2-1$.

 

[mathscott123]

a1 = real number > 0

a2 = (a1)^2 - 1 = (a1+1)(a1 - 1)

a3 = (a2)^2 - 1 = (a2 + 1)(a2-1)

a(n+1) = (an)^2 - 1 = (an+1)(an-1). This can only converge to 0 if and only if an equals either +1 or -1. However, that seems contradictory, doesn't it? How can something converge to 0 if it must converge to 1 or -1?

 

[mathscott123]

a1 = real number > 0

a2 = (a1)^2 - 1 = (a1+1)(a1 - 1)

a3 = (a2)^2 - 1 = (a2 + 1)(a2-1)

a(n+1) = (an)^2 - 1 = (an+1)(an-1). This can only converge to 0 if and only if an equals either +1 or -1. However, that seems contradictory, doesn't it? How can something converge to 0 if it must converge to 1 or -1?

 

[mathscott123]

Here is what I am thinking:

An+1=an^2-1
L=L^2-1
0=L^2-L-1
L=1+\- sqrt 5/2 if An converges

Therefore if it does converge it does not
converge to zero

Is that enough?