# Thread: Number of hares and rabbits

1. ## Number of hares and rabbits

Suppose hk and rk denote the number of hares and rabbis respectively in the year k.
Assume that the number of those hares and rabbits changes over the years according to the following rule:

rk+1 = 10rk - 3hk
hk+1 = 4rk - 2hk

If there are 2 hares and 100 rabbits initially, what is the number of hares and rabbits in 2 years? in 20 years?

2. ## Re: Number of hares and rabbits

Originally Posted by whyyie
Suppose hk and rk denote the number of hares and rabbis respectively in the year k.
Assume that the number of those hares and rabbits changes over the years according to the following rule:

rk+1 = 10rk - 3hk
hk+1 = 4rk - 2hk

If there are 2 hares and 100 rabbits initially, what is the number of hares and rabbits in 2 years? in 20 years?
Where are you having difficulty? If you understand what the equations say, this can be done by basic arithmetic.

If there are 2 hares and 100 rabbits initially, that is, if $r_0= 100$ and $h_0= 2$, then $r_1= 10(100)- 3(2)= 1000- 6= 994$ and $h_1= 4(100)- 2(2)= 400- 4= 396$. Now, $r_2= 10(994)- 3(396)= 9940- 1188= 8752$ and $h_2= 4(994)- 2(396)= 3976- 794= 3184$. There will be 8752 rabbits and 3184 hares after two years.

You could continue that until k= 20 or look for a general solution of the form $h_n= \alpha^n$, $r_n=\beta^n$. Putting those into the two equations gives $\beta^{n+1}= 10\beta^n- 3\alpha^n$ and $\alpha^n= 4\beta^n- 2\alpha^n$. Dividing both sides of the first equation by $\alpha^n$ gives $\beta\left(\frac{\beta}{\alpha}\right)^{n+1}= 10\left(\frac{\beta}{\alpha}\right)^n- 3$ and dividing both sides of the second equation by $\alpha$ gives $\alpha= 4\left(\frac{\beta}{\alpha}\right)^n- 2$. We can solve that second equation for $\left(\frac{\beta}{\alpha}\right)^n$: $\left(\frac{\beta}{\alpha}\right)^n= \frac{\alpha- 2}{4}$. Now put that into the previous equation to get an equation for $\alpha$ and $\beta$.

By the way, are you sure you have copied the problem correctly? It seems strange that a large number of hares would reduce both rabbit and hare populations while a large number of rabbits would increase both. Typically, for competing species, x and y, the equations are $x_{n+1}= Ax_n- By_n$, $y_{n+1}= -Cx_n+ Dy_n$, reversing the signs.

3. ## Re: Number of hares and rabbits

This is an algebra question and all the while we were learning about matrices and vectors only so I thought that the equation given could be converted into matrix form. and based on the question given, i think that's involving recurrence relations which I think is unrelated as this is not taught in algebra class and of there are only minority(or should I say none) who are taking this course have been exposed to that.

And substituting the value inside 1 by 1 is not an efficient method. So perhaps I still thinking of way to convert the equation into matrix form and solve it.

4. ## Re: Number of hares and rabbits

Originally Posted by whyyie
This is an algebra question and all the while we were learning about matrices and vectors only so I thought that the equation given could be converted into matrix form. and based on the question given, i think that's involving recurrence relations which I think is unrelated as this is not taught in algebra class and of there are only minority(or should I say none) who are taking this course have been exposed to that.
Here is a matrix solution, where $N$ is the number of years.
$\left[ {\begin{array}{cc}{100}&2\end{array}} \right] \cdot {\left[ {\begin{array}{rr}{10}&4\\{ - 3}&{ - 2}\end{array}} \right]^N}$