Help with contractive sequence proof

[Ed Quanta]

Can someone help me prove that the following sequence is contractive?

Let a1= 1 and an+1=1/5(an)^2 - 1


I understand the definition of a contractive sequence, but I am not sure how to formally prove this,

 

[HallsofIvy]

Okay, what is the definition of a "contractive sequence"? I know the definition of a "contractive mapping" but I don't recall seeing it applied to a sequence!

 

[wais]

any help would be appreciated.

Sure, I can help you with proving that this sequence is contractive. First, let's recall the definition of a contractive sequence:

A sequence (an) is contractive if there exists a constant c such that for all n, |an+1 - an| ≤ c|an - an-1|.

Now, let's look at the given sequence:

a1 = 1
an+1 = 1/5(an)^2 - 1

We can rewrite this as:

an+1 = 1/5(an)(an) - 1

Next, we can use the fact that an > 0 for all n (since a1 = 1 and each term is squared), to simplify the expression:

an+1 = 1/5(an)(an) - 1
= 1/5(an)(an - 5)

Now, we can see that for all n, an+1 < an, since (an - 5) is always negative. This means that the sequence is decreasing. Additionally, we can see that as n approaches infinity, an approaches 0.

Now, to prove that this sequence is contractive, we need to find a constant c such that:

|an+1 - an| ≤ c|an - an-1|

Since we know that an+1 < an, we can rewrite this as:

an - an+1 ≤ c(an - an-1)

Now, we can plug in the expression for an+1 and simplify:

an - an+1 = an - 1/5(an)(an - 5)
= 5an/5 - 1/5(an)(an - 5)
= 1/5(an)(5 - (an - 5))
= 1/5(an)(10 - an)

Substituting this back into our original inequality, we get:

1/5(an)(10 - an) ≤ c(an - an-1)

Simplifying further, we get:

10 - an ≤ c(5 - an)

Since an approaches 0 as n approaches infinity, we can disregard it in this inequality, giving us:

10 ≤ 5c

Therefore, we can choose c = 2 as our constant, since it satisfies the condition above. This means that for all n, |an+1 - an| ≤ 2|an -