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Two functions in two variables

  1. Mar 4, 2006 #1
    Sorry, I don't know what category to put this in.

    Here's my problem:

    I have some corresponding values for two variables a and b.

    I also know that
    [tex]a_n = 2a_{n-1} + b_{n-1}[/tex]
    [tex]b_n = b_{n-1} + a_n[/tex]
    for all a and b.

    I need to find functions [tex]f_1[/tex] and [tex]f_2[/tex] in terms of n so that [tex]f_1(n) = a_n[/tex] and [tex]f_2(n) = b_n[/tex]

    I don't really know where to start. I tried recombining the first two equations in myriad ways but I didn't get anywhere there. I have figured out that this would be pretty tough without the sample values to generate the others, and that the answers will be in general form because n = 1 could refer to any of the pairs of values, although all of them can be generated from a single pair. What do I do?
  2. jcsd
  3. Mar 4, 2006 #2


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    Why not start by telling us your sample values?

    One way to do this problem if you know linear algebra is to express the equations a_n and b_n each in terms of a_(n-1) and b_(n-1). This is then a simple Markov chain problem. Luckily the matrix you get is diagonalizable, so its powers will yield an easy algebraic expression.
    Last edited: Mar 4, 2006
  4. Mar 4, 2006 #3
    Sorry, should have included that. my sample values are:
    a = 1 when b =1 and a = 3 when b = 4(I know the second is unneccessary). The entire sequence can be generated from these values, though this point is not neccessarily n = 1. This is why the answer should be in general form.
  5. Mar 4, 2006 #4
    1) [tex]a_{n} = 2a_{n-1}+b_{n-1}[/tex]
    2) [tex]b_{n} = b_{n-1} + a_{n}[/tex]

    2) => [tex]a_{n} = b_{n}-b_{n-1}[/tex]

    Put that into 1 to get

    [tex]b_{n}-b_{n-1} = 2a_{n-1}+b_{n-1}[/tex]
    3) [tex]b_{n} = 2(a_{n-1}+b_{n-1})[/tex]

    2) => [tex]a_{n-1} = b_{n-1}-b_{n-2}[/tex]

    3) => [tex]b_{n} = 2(b_{n-1}-b_{n-2}+b_{n-1}) = 4b_{n-1}-2b_{n-2}[/tex]

    You need 2 initial values for b, one you're given, the other you generate from your initial condition on a. Now you can just use the equation for b and equation 2 to get a's equation, which will also need 2 initial values, one you're given, the other you get from b's initial value.
  6. Mar 4, 2006 #5


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    The general equation looks irritating to work out by hand, involving a lot of sqrt 2 and large expressions, unless my calculator is missing a major simplification somewhere.
    Last edited: Mar 4, 2006
  7. Mar 4, 2006 #6
    Thanks for your help!(I really should have seen that. It must be getting too late) Unfortunately the nature of the answer shows this series is not what I am looking for. I'll keep trying!
  8. Mar 4, 2006 #7


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    What do you mean?
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