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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?

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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.

 September 14th, 2013, 12:52 AM #3 Senior Member   Joined: Dec 2012 Posts: 372 Thanks: 2 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 $f(x)= 2x - \dfrac{2x^2}{K}$ is differentiable and by the mean value theorem, it's a strict contraction whenever $|f'(x)| < 1$. Observe that $f'(x) = 2 - \dfrac{4x}{K}$. Thus $|f'(x)| < 1 \ \Leftrightarrow \ \dfrac{3K}{4} > x > \dfrac{K}{4}$. With little difficulty using your initial condition, you can verify that $(x_k)_k \subset [\frac{K}{4} , \frac{3K}{4}]$. Let $[\frac{K}{4} , \frac{3K}{4}]= A$. The function $f: A \rightarrow A$ is a strict contraction since the absolute value of the derivative of $f$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 $(x_k)_k \subseteq A$ defined by $x_{k+1}= f(x_k)$ will converge to the unique fixed point of $f$ in $A$, also by Banach's contraction principle.

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