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Math Help - Find the equilibrium for a logistic recurrence relation.

  1. #1
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    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



    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 Find the equilibrium for a logistic recurrence relation.-ql_bf5e677d64f5a7bc68e51ac64ca44ce9_l3.png?


    Thanks.
    Attached Thumbnails Attached Thumbnails Find the equilibrium for a logistic recurrence relation.-ql_0856732c0254b1e345c460be29ad2ede_l3.png   Find the equilibrium for a logistic recurrence relation.-ql_2fa058d3be07dc65ede9f409284ee8f7_l3.png  
    Last edited by alyosha2; September 24th 2011 at 02:13 AM.
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  2. #2
    MHF Contributor chisigma's Avatar
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    Re: Find the equilibrium for a logistic recurrence relation.

    Quote Originally Posted by alyosha2 View Post
    How would I go about showing the reccurence relation fits the logisitcal model and hecne finding the equilibrium?

    say



    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 Click image for larger version. 

Name:	ql_bf5e677d64f5a7bc68e51ac64ca44ce9_l3.png 
Views:	34 
Size:	853 Bytes 
ID:	22388?


    Thanks.
    The images show two different equations...

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

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

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

    Kind regards

    \chi \sigma
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  3. #3
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    Re: Find the equilibrium for a logistic recurrence relation.

    Oh, sorry. It's (1) I just messed the subscript up on (2).
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  4. #4
    MHF Contributor chisigma's Avatar
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    Re: Find the equilibrium for a logistic recurrence relation.

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

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

    ... and neceesary it is 'started' by an 'initial condition' 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 x_{0}=8333.33... but, because is f^{'}(x_{0})>0 , x_{0} is a repulsive fixed point and that means that x_{0} is a unstable equilibrium point. In other words, for p=x_{0} the solution of (1) is the constant sequence p_{n}=x_{0}, but for all other p the solution will diverge...

    Kind regards

    \chi \sigma
    Last edited by chisigma; September 24th 2011 at 06:38 AM.
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  5. #5
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    Re: Find the equilibrium for a logistic recurrence relation.

    Quote Originally Posted by chisigma View Post
    Very well!...we write the difference equation as...

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

    ... and neceesary it is 'started' by an 'initial condition' 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 x_{0}=8333.33... but, because is f^{'}(x_{0})>0 , x_{0} is a repulsive fixed point and that means that x_{0} is a unstable equilibrium point. In other words, for p=x_{0} the solution of (1) is the constant sequence p_{n}=x_{0}, but for all other p the solution will diverge...

    Kind regards

    \chi \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



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

    thanks.
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  6. #6
    MHF Contributor chisigma's Avatar
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    Re: Find the equilibrium for a logistic recurrence relation.

    Quote Originally Posted by alyosha2 View Post
    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



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

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

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

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

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

    Isn't it?...

    Kind regards

    \chi \sigma
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  7. #7
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    Re: Find the equilibrium for a logistic recurrence relation.

    Quote Originally Posted by chisigma View Post
    If I understand correctly You have the difference equation...

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

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

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

    Isn't it?...

    Kind regards

    \chi \sigma


    yes, that's correct.
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  8. #8
    MHF Contributor chisigma's Avatar
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    Re: Find the equilibrium for a logistic recurrence relation.

    Quote Originally Posted by alyosha2 View Post
    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

    \chi \sigma
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  9. #9
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    Re: Find the equilibrium for a logistic recurrence relation.

    Quote Originally Posted by chisigma View Post
    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

    \chi \sigma
    So it's impossible to find the equilibrium?
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  10. #10
    MHF Contributor chisigma's Avatar
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    Re: Find the equilibrium for a logistic recurrence relation.

    Quote Originally Posted by alyosha2 View Post
    So it's impossible to find the equilibrium?
    ... that seems to be ...

    Kind regards

    \chi \sigma
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