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Posted

Say you have a function [math]f(x)[/math]. Then the equilibrium points are those points x such that [math]f'(x) = 0[/math].

 

For example, [math]f(x) = x^2 + 3x + 2 \Rightarrow f'(x) = 2x + 3[/math]

 

So just the one equilibrium point at x = -3/2.

Posted

I am not sure about this. just following on from daves answer

 

If your working in fuctions of more variables [math]f(x_{1},x_{2},...,x_{n})[/math] then the eq point is the point [math](x_{1},x_{2},...,x_{n})[/math] where

 

[math]f_{x_1}=0,f_{x_2}=0,...,f_{x_n}=0[/math]

so u just get n simultaneous eq. for n variables.

Posted

A better way is to treat the variables x1 to xn as a column vector and get the derivative as a n cross 1 matrice then set it equal to the null matrice of same dimensions. Though in essence the same, it provides for a cleaner solution more appealing to the eye.

Posted

In practice, you're never going to be asked to do this. At most, you'll be asked for a 3D equilibrium point. For larger dimensions, it's better to use something like Mathematica to solve the equations for you.

Posted
In practice, you're never going to be asked to do this. At most, you'll be asked for a 3D equilibrium point. For larger dimensions, it's better to use something like Mathematica to solve the equations for you.

 

Speaking of Mathematica, I remember you praising it alot over SFN.

 

What's a better program overall: Mathematica or Matlab?

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