Equilibrium solutions in mathematical modelling

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SUMMARY

The discussion focuses on finding equilibrium solutions in mathematical modeling, specifically for the function Pk+1 = a(Pk-1). Participants emphasize that equilibrium points can be determined by setting Pk equal to Pk+1. The function, when expanded, results in a cubic polynomial, indicating the presence of multiple solutions influenced by various constants. Deriving the function and setting it to zero is a critical step in identifying these equilibrium points.

PREREQUISITES
  • Understanding of equilibrium points in mathematical modeling
  • Familiarity with polynomial functions and their properties
  • Knowledge of calculus, specifically differentiation
  • Ability to solve cubic equations
NEXT STEPS
  • Study the concept of equilibrium points in dynamic systems
  • Learn about polynomial expansion and its implications
  • Explore techniques for solving cubic equations
  • Review differentiation methods for finding critical points
USEFUL FOR

Mathematicians, students in mathematical modeling, and anyone interested in solving equilibrium problems in dynamic systems will benefit from this discussion.

klinklindeman
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Hi,
I don't want to list the exact problem I have to solve, as I would rather do it myself, all I would like is some guidance. The problem is a function like: Pk+1=a (Pk-1) with a few more constants and some brackets thrown about on the right hand side. I need to find all the equilibrium points. I am not exactly sure how to get started, and I have had a look in my textbook and couldn't find any examples similar to this. Do I just derive this function, and then let it equal zero to solve for the points?

Also, when I expand the function I will have a polynomial with Pk^3, so there will be multiple solutions I assume.

Thanks for the help!
 
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welcome to pf!

hi klinklindeman! welcome to pf! :smile:
klinklindeman said:
I need to find all the equilibrium points.

it's a little difficult to tell without seeing the question, but equilibrium usually means that you can put Pk = Pk+1
Also, when I expand the function I will have a polynomial with Pk^3, so there will be multiple solutions I assume.

not so much multiple solutions, as a solution with multiple constants :wink:
 

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