Dynamic system model

by Philip Wong
Tags: dynamic, model
 PF Gold P: 960 In b) you only need to calculate the population after one year, i.e. with n = 1 calculate x1 = A1 x0 = A x0. No need for eigenvalues yet. Also, your characteristic equation (before you start using matlab) is not entirely correct.
P: 95
 Quote by Filip Larsen In b) you only need to calculate the population after one year, i.e. with n = 1 calculate x1 = A1 x0 = A x0. No need for eigenvalues yet. Also, your characteristic equation (before you start using matlab) is not entirely correct.
what do you mean by "Also, your characteristic equation (before you start using matlab) is not entirely correct."?

PF Gold
P: 960
Dynamic system model

 Quote by Philip Wong what do you mean by "Also, your characteristic equation (before you start using matlab) is not entirely correct."?
Just before you turn to matlab, you write the equation "landa^2 - 0.1landa - 0.33 =0", but the 0.1 should have been 0.5. I cannot see if this is a typo so I thought I would just mention it in case you went back to solving for the eigenvalues manually (its good to check with matlab, but you should have no trouble calculating the values manually too by solving the characteristic equation).

And, just for your information, the Greek letter is called lambda, not landa.
 P: 95 yes whenever I'm doing excerise or pratice I always do the question first manually before I check it with matlab, the only time I don't do manually it anymore is when I got so used to working with those type of questions that I could mentally think the answer out, and then I check it with matlab. And yes it was I typo, ahaha! It is 0.5 instead of 0.1
P: 95
 Quote by Filip Larsen In b) you only need to calculate the population after one year, i.e. with n = 1 calculate x1 = A1 x0 = A x0. No need for eigenvalues yet. Also, your characteristic equation (before you start using matlab) is not entirely correct.
x1 = A1 x0 = A x0
So x1= C1(-0.3765)^1(-0.8481; 0.3238) + C2(0.8765)^1(-0.7821;-0.6232)

but since lambda 2 is the domain value it will over-run lambda 1 in the long term, therefore the state vector after 1 year will be 0.8765xk?

or did I think too much, the state vector is simply :
x1= C1(-0.3765)^1(-0.8481; 0.3238) + C2(0.8765)^1(-0.7821;-0.6232)
 PF Gold P: 960 If this is for b) then you are still getting ahead of yourself. You only need to perform a simple matrix multiplication of A and x0 to get x1. In this model this is, by definition, how the state for one generation is advanced to the state for the next generation. If you were to calculate, say, the 10th generation you would have to do the multiplication 10 times to get to x10. By the way, it can be instructive to plot how the state develops over time, especially in order to prepare yourself for part d) and e).
P: 95
 Quote by Filip Larsen If this is for b) then you are still getting ahead of yourself. You only need to perform a simple matrix multiplication of A and x0 to get x1. In this model this is, by definition, how the state for one generation is advanced to the state for the next generation. If you were to calculate, say, the 10th generation you would have to do the multiplication 10 times to get to x10. By the way, it can be instructive to plot how the state develops over time, especially in order to prepare yourself for part d) and e).
aha! so x1 = [0 1.1; 0.3 0.5] *[80 20]
so say if I want to work out 10, would I do this:
[0^10,1.1; 0.3,0.5^10]*[80,20]
PF Gold
P: 960
 Quote by Philip Wong aha! so x1 = [0 1.1; 0.3 0.5] *[80 20]
Yes.

 so say if I want to work out 10, would I do this: [0^10,1.1; 0.3,0.5^10]*[80,20]
No. The definition of your model really says xi+1 = A xi, that is, to get state for generation i+1 you matrix multiply A with the state vector for generation i, i.e. the previous generation. So, if you know x9 you can calculate x10 = A x9, but since x9 likewise can be written based on the state vector for the previous generation you also have x9 = A x8 which can be inserted into the equation for x10 to give x10 = A (A x8) = (A A) x8, the last step because matrix multiplication is commutative. If you continue replacing x in this fashion until you get to to x0 you arrive at x10 = (A A ... 10 times ... A) x0 = A10 x0. Just like with scalar numbers, you can think of the (integer) power of a matrix as a shorthand notation of simply multiplying out the matrix as many times as the power says.

Your questions seem to indicate you have missed something fundamentally somewhere in your textbook. If the above explanation is all news to you, I recommend that you take some time to read up on the parts in your textbook that are relevant for this problem or you will most likely have a very tough time understanding the eigenvalues and eigenvectors.
P: 95
 Quote by Filip Larsen Yes. No. The definition of your model really says xi+1 = A xi, that is, to get state for generation i+1 you matrix multiply A with the state vector for generation i, i.e. the previous generation. So, if you know x9 you can calculate x10 = A x9, but since x9 likewise can be written based on the state vector for the previous generation you also have x9 = A x8 which can be inserted into the equation for x10 to give x10 = A (A x8) = (A A) x8, the last step because matrix multiplication is commutative. If you continue replacing x in this fashion until you get to to x0 you arrive at x10 = (A A ... 10 times ... A) x0 = A10 x0. Just like with scalar numbers, you can think of the (integer) power of a matrix as a shorthand notation of simply multiplying out the matrix as many times as the power says. Your questions seem to indicate you have missed something fundamentally somewhere in your textbook. If the above explanation is all news to you, I recommend that you take some time to read up on the parts in your textbook that are relevant for this problem or you will most likely have a very tough time understanding the eigenvalues and eigenvectors.
ar for the x10 I think I skip a bit too much steps along the way. When I think about it again now, if I want to get the x10 I would use diagnolisation and symmetry.
so for x10 would I do the following?
A = VDV^-1
[0 1.1; 0.3 0.5]^10 =
[-0.9461,-0.7821; 0.3238,-0.6232]*[-0.3765^10, 0; 0,0.8765^10]*[-0.7394, 0.9279; -0.3842, -1.1225]

which gives:
[0.0805, 0.2349; 0.0641, 0.1872]

since X10 = A^10 x0, then:
[0.0805, 0.2349; 0.0641, 0.1872]*[80;20]

which gives:
11.1340
8.8692

is this correct?
 PF Gold P: 960 That is correct.
 P: 95 yay! thanks for your help

 Related Discussions Mechanical Engineering 3 Introductory Physics Homework 0 Calculus 8 Engineering Systems & Design 0