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        riddles >> putnam exam (pure math) >> Limit of ratio of rational sequences
        
(Message started by: Michael_Dagg on Nov 7^th, 2006, 5:35pm)

Title: Limit of ratio of rational sequences
Post by Michael_Dagg on Nov 7^th, 2006, 5:35pm

Let p_n and q_n be sequences defined by setting
p₀ = 1, p₁ = 6, q₀ = 1 and

p_n+1 = 6 p²_n - 6 p_n p²_n-1 + 2 p⁴_n-1, q_n+1 = p_n+1 - p²_n, for n > 1.

Put A_n = p_n/q_n, n > 0. Show that lim _n->oo A_n exists and find its value.

Title: Re: Limit of ratio of rational sequences
Post by Barukh on Nov 7^th, 2006, 11:29pm

[hide]1.2599...[/hide]?

Title: Re: Limit of ratio of rational sequences
Post by Michael_Dagg on Nov 8^th, 2006, 12:46pm

I knew you'd get it quickly.

Anyone else know the explicit form?

If so, just simply show by induction that
A_n is less than _that_ number for all n > 0.

Title: Re: Limit of ratio of rational sequences
Post by Eigenray on Nov 8^th, 2006, 7:14pm

[hideb]
Define t_n = 1-1/A_n = (p_n-q_n)/p_n = p_n-1²/p_n.
Let t=t_n, t'=t_n+1. Then
1/t' = 6 - 6t + 2t².
Solving,
A' = 1/(1-t')
= 1 + 1/(1/t'-1)
= 1 + 1/(5 - 6t + 2t²)
= 1 + A²/(5A² - 6A(A-1) + 2(A-1)²)
= (2A² + 2A + 2)/(A² + 2A + 2).
So A_n+1 = f(A_n), where
f(x) = (2x² + 2x + 2)/(x² + 2x + 2)
= x + (2-x³)/(x² + 2x + 2).
From the above expression, it is clear that f(x) > x iff x³<2, and that f has a fixed point at x=2^1/3. Since A₀=1, we conclude that A_n = fⁿ(1) converges to 2^1/3.[/hideb]

Title: Re: Limit of ratio of rational sequences
Post by Michael_Dagg on Nov 9^th, 2006, 12:36pm

Hey, that's nice -- unexpected and simple too!

Title: Re: Limit of ratio of rational sequences
Post by Sameer on Nov 9^th, 2006, 2:26pm

on 11/09/06 at 12:36:35, Michael_Dagg wrote:

Hey, that's nice -- unexpected and simple too!

Did you have some other method?

Title: Re: Limit of ratio of rational sequences
Post by Michael_Dagg on Nov 9^th, 2006, 3:06pm

Sure.

edit: If you happen to have one of your own I'd to see it!