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Recurrence relations

  1. Apr 1, 2015 #1
    1. The problem statement, all variables and given/known data:
    I am just learning recurrence relations, and they are proving challenging.


    2. Relevant equations -- Let's have xn = 3xn-1 + 6xn-2.
    I wanted to look at it with two scenarios. The first is x0 = 1 and x1 = 3. The second is x1=3 and x2 = 4

    3. The attempt at a solution
    Is there a difference between having a staring point of x0and x1? That is the sticking point.

    As far as the solution is concerned, if I wanted to go to the 5th term, I can do the x1=3 and x2 = 4 scenario -- x1 = 1, x2 = 4, x3 = x3=30, x4 = 114, and x5= 522.

    The scenario with x0 is challenging to me. How can you have a recurrence relation where the n-1 term leaves you a negative number? How do you know where to start? Is it understood that, in the equation that is above, I would start with 3(x0(or 3 * 1) + 6(x1(6 * 3), with a sum of 21?

    Also, how would you turn this into a formula? How do you determine the degree? For this last question, I just need a point in the right direction, I'm not asking for someone to do the questions for me.
     
  2. jcsd
  3. Apr 1, 2015 #2

    Ray Vickson

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    The usual way of writing these things is as ##x_n = 3 x_{n-1} + 6x_{n-2} \; \text{for} \; \; n \geq 2##, with ##x_0=1, x_1 = 3##. In other words, the recursion does not apply when ##n = 0, 1##.

    If you wanted a formula you could try ##x_n = r^n## for some ##r##. Then, the recursion would give ##r^n = 3 r^{n-1}+ 6 r^{n-2}##, or ##r^2 = 3 r + 6##. This is a quadratic, and so has two roots, usually---in this case it does have two real roots. So, the general solution would be of the form
    [tex] x_n = A r_1^n + B r_2^n, \; n \geq 2 [/tex]
    where ##r_1, r_2## are the two roots. We can find values of ##A, B## either by (i) requiring that this expression hold as well for ##n = 0,1##; or (ii) computing ##x_2, x_3## from the recursion as well as from the above formula, giviing two equations for ##A,B##.
     
  4. Apr 1, 2015 #3
    Thank you, this is something to speak about w/ the prof, it is still not sinking in.
     
  5. Apr 1, 2015 #4

    rcgldr

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    In some cases, you can solve for negative n using the same recurrence relation. Given [itex]x_{n}[/itex] and [itex]x_{n-1}[/itex], you can solve for [itex]x_{n-2}[/itex]:
    [tex]
    x_n = 3x_{n-1} + 6x_{n-2} \\
    x_{n-2} = (x_n - 3x_{n-1}) / 6 \\
    x_1 = 3 \\
    x_0 = 0 \\
    x_{-1} = 1/2 \\
    x_{-2} = -1/4 \\
    x_{-3} = 5/24 \\
    x_{-4} = -7/48 \\
    x_{-5} = 31/288 \\
    ...
    [/tex]
     
    Last edited: Apr 2, 2015
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