My Math Forum Recurrence relation

 Real Analysis Real Analysis Math Forum

September 12th, 2013, 08:58 PM   #1
Newbie

Joined: Sep 2013

Posts: 7
Thanks: 0

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?

Attached Images
 recurrence.jpg (9.9 KB, 219 views)

September 13th, 2013, 02:58 PM   #2
Global Moderator

Joined: May 2007

Posts: 6,728
Thanks: 689

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.

 Tags recurrence, relation

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Dragonkiller Linear Algebra 2 May 15th, 2012 10:49 AM roguebyte Applied Math 1 April 14th, 2012 11:39 PM fe phi fo Applied Math 3 December 4th, 2011 09:22 AM tuzzi-i Real Analysis 1 October 6th, 2007 10:25 AM roguebyte Advanced Statistics 1 December 31st, 1969 04:00 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top