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Math Help - Recurrence relation.. does this look correct?

  1. #1
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    Recurrence relation.. does this look correct?

    I have to find the recurrence relation by the sequence

    a(x) = x - (-1)^x

    initial conditions are x(0) = -1 x(1) = 2

    A(x) - a(x-1) = x-(-1)^x (x-1) + (-1)^x-1
    A(x) = x-(-1)^x (x-1) + (-1)^x-1 + a(x-1)
    A(x) = a(x-1) 2*(-1)^x +1

    i believe I've done this correctly.. but i just wanted to make sure.. any help would be appreciated! danke
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  2. #2
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    Recurrence relation

    Hello JoeCrow
    Quote Originally Posted by JoeCrow View Post
    I have to find the recurrence relation by the sequence

    a(x) = x - (-1)^x

    initial conditions are x(0) = -1 x(1) = 2

    A(x) - a(x-1) = x-(-1)^x (x-1) + (-1)^x-1
    A(x) = x-(-1)^x (x-1) + (-1)^x-1 + a(x-1)
    A(x) = a(x-1) 2*(-1)^x +1

    i believe I've done this correctly.. but i just wanted to make sure.. any help would be appreciated! danke
    If I re-write your working to make it easier to read, I get:

    a_x - a_{x-1} = x-(-1)^x - (x-1) + (-1)^{x-1}

     a_{x} = x-(-1)^x - (x-1) + (-1)^{x-1} + a_{x-1}

    a_x = a_{x-1} - 2\cdot(-1)^x +1

    This is OK, but I'm not sure it's exactly what is wanted, because there's still a (-1)^x term in the equation.

    How about this? The sequence is

    -1,
    2, 1, 4, 3, 6, 5, 8, ...

    which is really just two sequences interwoven. In each of the separate sequences, each term is 2 more than its predecessor. So you can simply write the recurrence relation:

    a_{n+2} = a_n + 2, a_0 = -1, a_1 = 2

    Grandad
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