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Math Help - reccurence relations

  1. #1
    Junior Member
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    reccurence relations

    In my notes I have the expression:
    a_n = 3a_{n-1} + 2a_{n-2}

    with
    a_0 = 1, a_1 = 2

    but I can't see how we get a_0 = 1

    When I try I get
     3a_{0-1} + 2a_{0-2}<br />
= 3*-1 + 2*-2<br />
= -3 + -4<br />
= -7<br />

    so why is
     + 2a_{n-2}<br />
= 4


    to get -3 + 4 = 1

    obviously I am not understanding something...
    can you point out where I am going wrong please?
    Thanks
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  2. #2
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    Go to this web link. You can change values for f(1) click the equals sign. You will see how the recursion changes.

    Now why? Well we need to know how the initial values because we have two definiting terms.
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  3. #3
    Junior Member
    Joined
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    That's a great link, I'm sorry but I still don't get it.
    if I'm looking for f(4)
    then
    f(2) = 3a(2-1) + 2a(2-2)
    f(2) = 3*a(1) + 2*a(0) <---- where a(1) = 2 and a(0) = 1
    f(2) = 3*2 + 2*1
    f(2) = 8

    f(3) = 3a(3-1) + 2a(3-2)
    f(3) = 3*a(2) + 2*a(1) <----a(2) = 8, a(1) =2
    f(3) = 3*8 + 2*2
    f(3) = 28

    f(4) = 3a(4-1) + 2a(4-2)
    f(4) = 3*a(3) + 2*a(2) <-----a(3)=28, a(2) = 8
    f(4) = 3*28 + 2*8
    f(4) = 100

    is that right?
    (I think I was reading too deep into things when i was trying to figure out how a(0) = 1, I guess it's just a given because a(0) =1 and a(1) = 2 are initial values, we don't need to work out how to get them, we just use them to find higher values??)

    Just wandering if I have worked through this correctly or not?
    Thanks for your time
    Last edited by dunsta; October 13th 2010 at 05:26 PM.
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