3 Attachment(s)

Find the equilibrium for a logistic recurrence relation.

How would I go about showing the reccurence relation fits the logisitcal model and hecne finding the equilibrium?

say

http://quicklatex.com/cache3/ql_2fa0...84ee8f7_l3.png

I can see the first part seems like it fits but how do i got about making a constant figure like the one above into Attachment 22388?

Thanks.

Re: Find the equilibrium for a logistic recurrence relation.

Quote:

Originally Posted by

**alyosha2**

The images show two different equations...

$\displaystyle p_{n+1}-p_{n}=.24\ p_{n}-2000$ (1)

$\displaystyle p_{n} +1-p_{n}=.24\ p_{n}-2000$ (2)

Which is the 'right' equation?... (1) or (2)?...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

Re: Find the equilibrium for a logistic recurrence relation.

Oh, sorry. It's (1) I just messed the subscript up on (2).

Re: Find the equilibrium for a logistic recurrence relation.

Very well!...we write the difference equation as...

$\displaystyle \Delta_{n}= p_{n+1}-p_{n}= .24\ p_{n}-2000= f(p_{n})$ (1)

... and neceesary it is 'started' by an 'initial condition' $\displaystyle p_{0}=p$. The procedure for solving a difference equation of the type (1) is described in...

http://www.mathhelpforum.com/math-he...-i-188482.html

In Your case the f(x) has only one zero in $\displaystyle x_{0}=8333.33...$ but, because is $\displaystyle f^{'}(x_{0})>0$ , $\displaystyle x_{0}$ is a *repulsive fixed point* and that means that $\displaystyle x_{0}$ is a *unstable equilibrium point*. In other words, for $\displaystyle p=x_{0}$ the solution of (1) is the constant sequence $\displaystyle p_{n}=x_{0}$, but for all other p the solution will diverge...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

Re: Find the equilibrium for a logistic recurrence relation.

Quote:

Originally Posted by

**chisigma** Very well!...we write the difference equation as...

$\displaystyle \Delta_{n}= p_{n-1}-p_{n}= .24\ p_{n}-2000= f(p_{n})$ (1)

... and neceesary it is 'started' by an 'initial condition' $\displaystyle p_{0}=p$. The procedure for solving a difference equation of the type (1) is described in...

http://www.mathhelpforum.com/math-he...-i-188482.html
In Your case the f(x) has only one zero in $\displaystyle x_{0}=8333.33...$ but, because is $\displaystyle f^{'}(x_{0})>0$ , $\displaystyle x_{0}$ is a

*repulsive fixed point* and that means that $\displaystyle x_{0}$ is a

*unstable equilibrium point*. In other words, for $\displaystyle p=x_{0}$ the solution of (1) is the constant sequence $\displaystyle p_{n}=x_{0}$, but for all other p the solution will diverge...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

I'm sorry, I'm not really sure I understand. What I'm trying to do is convert the given recurrence relation into the form for the logistic model

http://quicklatex.com/cache3/ql_eb38...dc35743_l3.png

I didin't think you would need the first term but in this case it is $\displaystyle P_0$ = 25000.

thanks.

Re: Find the equilibrium for a logistic recurrence relation.

Quote:

Originally Posted by

**alyosha2** I'm sorry, I'm not really sure I understand. What I'm trying to do is convert the given recurrence relation into the form for the logistic model

http://quicklatex.com/cache3/ql_eb38...dc35743_l3.png
I didin't think you would need the first term but in this case it is $\displaystyle P_0$ = 25000.

thanks.

If I understand correctly You have the difference equation...

$\displaystyle p_{n+1}-p_{n}= .24\ p_{n}-2000$ (1)

... and Your task is to 'convert' it in the form...

$\displaystyle p_{n+1}-p_{n}= r\ p_{n}\ (1-\frac{p_{n}}{E})$ (2)

Isn't it?...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

Re: Find the equilibrium for a logistic recurrence relation.

Quote:

Originally Posted by

**chisigma** If I understand correctly You have the difference equation...

$\displaystyle p_{n+1}-p_{n}= .24\ p_{n}-2000$ (1)

... and Your task is to 'convert' it in the form...

$\displaystyle p_{n+1}-p_{n}= r\ p_{n}\ (1-\frac{p_{n}}{E})$ (2)

Isn't it?...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

yes, that's correct.

Re: Find the equilibrium for a logistic recurrence relation.

Quote:

Originally Posted by

**alyosha2** yes, that's correct.

All right!... the only problem is that doesn't exist any pair r and E which 'convert' the equation (1) into equation (2) or vice versa...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

Re: Find the equilibrium for a logistic recurrence relation.

Quote:

Originally Posted by

**chisigma** All right!... the only problem is that doesn't exist any pair r and E which 'convert' the equation (1) into equation (2) or vice versa...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

So it's impossible to find the equilibrium?

Re: Find the equilibrium for a logistic recurrence relation.

Quote:

Originally Posted by

**alyosha2** So it's impossible to find the equilibrium?

... that seems to be (Thinking)...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$