Recent content by 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?

Homework Statement Suppose that a mathematically inclined child plays with a basket containing an infinite subset of integers (with some repetitions). If an integer k is present in the basket then there are initially |k| copies of it. The child pulls out the integers from the basket at...

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...

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...

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

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...