Beep, Beep!

A stretch of desert is populated by two species of animals, roadrunners and coyotes, who are engaged in an endless game of rivalry and mischief. The populations r ( t ) r(t) and c ( t ) c(t) of roadrunners and coyotes t t years from now can be modelled by

r ( t + 1 ) = 3 r ( t ) 2 c ( t ) c ( t + 1 ) = 5 r ( t ) 3 c ( t ) 100 \begin{aligned} r(t+1)&=&3r(t)-2c(t) \\ c(t+1)&=&5r(t)-3c(t)-100 \end{aligned}

If there are 120 roadrunners and 110 coyotes initially (at time t = 0 t=0 ), how many roadrunners will there be 1001 years later (at time t = 1001 t=1001 )?


The answer is 140.

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1 solution

Ashwin K
Mar 15, 2016

This is an interesting equation :

It is given, r(0) = 120, c(0) = 110.
then, r(1) = 3(120) - 2(110) = 140, c(1) = 5(120) - 3(110) - 100 = 170.
then, r(2) = 3(140) - 2(170) = 80,  c(2) = 5(140) - 3(170) - 100 = 90.
then, r(3) = 3(80)  - 2(90)  = 60,  c(3) = 5(80)  - 3(90)  - 100 = 30.
then, r(4) = 3(60)  - 2(30)  = 120, c(4) = 5(60)  - 3(30)  - 100 = 110.

From above equations it is known that r(0) = r(4) & c(0) = c(4) . So, it repeats after a period of 4 and so r ( 1001 ) = r ( 1 ) = 140 \boxed{r(1001) = r(1) = 140} .

Exactly, the system is periodic with a period of 4. (+1) This is happening because the eigenvalues of the matrix A = [ 3 2 5 3 ] A= \begin{bmatrix} 3 & -2\\5 & -3\end{bmatrix} are ± i \pm i , so that A 4 = I 2 A^4=I_2 since i 4 = 1 i^4=1 .

Otto Bretscher - 5 years, 3 months ago

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Sir...how do we find eigen values in general.

ASHWIN K - 5 years, 3 months ago

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They are the roots of the characteristic polynomial, det ( λ I n A ) \det(\lambda I_n-A) . In our case, this polynomial is det [ λ 3 2 5 λ + 3 ] = λ 2 + 1 \det\begin{bmatrix} \lambda-3 & 2 \\ -5 & \lambda+3 \end{bmatrix}=\lambda^2+1 , with roots ± i \pm i .

Otto Bretscher - 5 years, 3 months ago

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@Otto Bretscher Wow...That's really interesting. I always wonder how this works. Thanks a lot sir. Is it possible to get periods other than 4?

ASHWIN K - 5 years, 3 months ago

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@Ashwin K Yes, it's really not that complicated, and this is a very, very central concept throughout math and physics.

Equivalently, the eigenvalues of A A are the complex numbers λ \lambda such that A v = λ v A\mathbf{v}=\lambda\mathbf{v} (*) for some non-zero vector v \mathbf{v} . Those vectors v \mathbf{v} are the eigenvectors with eigenvalue λ \lambda . You can write equation (*) as ( λ I n A ) v = 0 (\lambda I_n-A)\mathbf{v}=\mathbf{0} , showing that the matrix λ I n A \lambda I _n-A is singular and its determinant is zero... thus the condition I gave in my last post.

Otto Bretscher - 5 years, 3 months ago

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