bbolker
commented
6 years ago First, you should be careful with notation. |f'(N*)|<1 means |f'(N)| (as a general expression in terms of N) evaluated at N is <1.
In the homogeneous LUDD case, N*=0. Then it's easy to see that
f(N*+e) = f(N*)+ef'(N*) + ...
f(e) = 0 + eR + ...
~ eRMore generally (and including the non-homogeneous or affine LUDD case): start with f(N+e) = f(N) + ef'(N) + ... Then change variables so we are considering the distance from the equilibrium, i.e. N^ = N-N. Subtracting N* from both sides of the equation:
f(N*+e)-N* = f(N*) -N* + ef'(N*) + ...The left-hand side is the change in N^. The right-hand side simplifies/approximates to ef'(N) (the first two terms cancel and the later terms are dropped), leaving us with (change in N^ = `N^ f'(N)`) when N^ is small.
Does that help?
xiaokev
Hello Dr. Bolker, I was looking over the notes from the start of class and I'm having trouble seeing the general condition for UDD stability |f'(N)|<1. In class you explained it using Taylor's theorem so f(N+e) = f(N)+ ef'(N) +... but since f'(N) = R in LUDD you would get f(N+e) = f(N)+ eR +... = N+eR. Wouldn't this mean if you move away from the equilibrium by e the population would be N+eR? How would |R|<1 cause the taylor expansion to be equal to N* on the right?
I was thinking that since f(N+e) is the population at say time t then N(t+1) = f(N(t)) = f(N+e) then the next time step would give N(t+2) =f(N(t+1)) = R(f(N+e)) = RN + eR^2 = N* (when you move further from time t) I'm not sure if this logic works as having population on both axis trips me up sometimes.