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Math Help - f(n) general term of recurrence relation

  1. #1
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    f(n) general term of recurrence relation

    I have to find the general term f(n), n>= 1, for the recurrence relation
    f(n) = 3f(n-1), with f(1) = 4

    I assume there is a formula or a set of rules to follow but I can not decifer it from my notes.
    So I used a bit of trial and error + algebra

    f(1) = 4
    f(2) = 3*f(2-1) = 3*f1 = 12
    f(3) = 3*f(3-1) = 3*f2 = 36
    f(4) = 3*f(4-1) = 3*f3 = 108

    so
    \frac{4}{3} * \frac{3}{1} = \frac{12}{3} = \frac{f(2)}{3} = f(1)
    \frac{36}{3} = \frac{f(3)}{3} = f(2)
    \frac{180}{3} = \frac{f(4)}{3} = f(3)

    therefore
    f(n) = \frac{f(n+1)}{3} <-- the general term

    is this ok?
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  2. #2
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    Yes, except that usually recurrence equations are written in the form f(n + 1) = F(f(n)) for some expression F.

    Edit: The above is incorrect. It is just the original equation in a different form. Apparently, you need the solution, i.e., the formula f(n) = F(n) for some expression F that consists of arithmetic operations and n but not f.

    Note that

    f(2) = 3 * f(1)
    f(3) = 3 * f(2) = 3^2 * f(1)
    f(4) = 3 * f(3) = 3 * 3^2 * f(1) = 3^3 * f(1)

    You can guess what f(n) is.
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  3. #3
    MHF Contributor Also sprach Zarathustra's Avatar
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    f(n) = 4\cdot 3^{n-3}
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  4. #4
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    f(n) = 3n-1 * f(1)
    or should it be
    f(n) = 3n-1 * 4

    ???
    thanks for the help emakarov
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  5. #5
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    This depends on the precise definition, but in my opinion, the solution (the general term) is a formula f(n) = F(n) for some expression F that consists of arithmetic operations and n, but not f.

    I also suppose you understand that it's 3^{n-1}, not 3n-1.
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