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Math Help - Number of hares and rabbits

  1. #1
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    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?
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  2. #2
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    Re: Number of hares and rabbits

    Quote Originally Posted by whyyie View Post
    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.
    Last edited by HallsofIvy; June 9th 2012 at 06:58 AM.
    Thanks from Deveno and whyyie
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    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.
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    Re: Number of hares and rabbits

    Quote Originally Posted by whyyie View Post
    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}
    Thanks from whyyie
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