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September 12th, 2013, 08:58 PM   #1
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Recurrence relation

Given this recurrence relation:
[attachment=0:2jibvwh8]recurrence.jpg[/attachment:2jibvwh8]

with K>0, a parameter.

If initial condition is K/4, show that the sequence converges to K/2 (unique positive equilibrium point).

Can I show that using Banach fixed-point theorem?

Please help me.
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September 13th, 2013, 02:58 PM   #2
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Re: Recurrence relation

Quote:
Originally Posted by Joselynn
Given this recurrence relation:
[attachment=0:qsb8tfpt]recurrence.jpg[/attachment:qsb8tfpt]

with K>0, a parameter.

If initial condition is K/4, show that the sequence converges to K/2 (unique positive equilibrium point).

Can I show that using Banach fixed-point theorem?

Please help me.
At equilibrium x = 2x - 2x^2/K, or x = 2x^2/K. Solve for x. x = 0 or x = K/2.
Need to verify that of x starts at K/4 it won't converge to 0.
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September 14th, 2013, 12:52 AM   #3
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Re: Recurrence relation

As mathman has shown, computing the equilibrium point is not too hard. If you wish to confirm the uniqueness of this fixed point using Banach's contraction principle, you need to do a little further checking up. The function is differentiable and by the mean value theorem, it's a strict contraction whenever . Observe that . Thus
.
With little difficulty using your initial condition, you can verify that . Let .
The function is a strict contraction since the absolute value of the derivative of on the interior of A is less than 1. Hence, by Banach's contraction principle, the fixed point in A is unique since A is a complete metric space. Any sequence defined by will converge to the unique fixed point of in , also by Banach's contraction principle.
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