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Autonomous System Summation

  1. Jan 10, 2012 #1
    This problem came up in a project I'm doing for work, and I don't have a very extensive math background so I don't know how to solve it. I would appreciate any help you guys could give me.

    X1 is a constant
    Y1 is a constant

    Xn = aXn-1 + bYn-1
    Yn = cXn-1 + dYn-1

    For all n, for some constants a,b,c,d (for my purposes, all between -1 and 0).

    What is the sum X1 + X2 + ... to infinity?

    I've tried summing it like an infinite series:
    (1/(1-a))(x+b((1/(1-d)y + cx))

    That gives me a (poor) approximation, but not the exact answer :-(.

    Thank you so much for your help in advance!
     
    Last edited: Jan 10, 2012
  2. jcsd
  3. Jan 10, 2012 #2
    You can rewrite the equation as follows...

    Xn = (a + b)Xn-1
    Xn = cXn-1

    ...where c = a + b. Note that every term is equal to the previous term multiplied by a constant, making this a geometric series, which only converges if |c| < 1. Any calculus textbook will discuss dealing with geometric series; the sum is given by...

    [tex] Sum = \frac{X_{1}}{1 - c}[/tex]
     
  4. Jan 10, 2012 #3
    I think i might have written it wrong. Xn is a function of both Xn-1 AND Yn-1. I've corrected the mistake in my post.

    I think it makes it a lot more complicated than a simple infinite series.
     
    Last edited: Jan 10, 2012
  5. Jan 10, 2012 #4
    Hey guys! I could still use some help.

    The logic I've been working on goes like this:

    1/(1-a) (x) = x + ax +a2x + ...

    1/(1-a) (x + by) = x + by + a2x + aby ...

    Which covers some, but not all of the terms created when you solve the problem by hand.

    I'm thinking it might be something like:

    [itex]\sum[/itex]X = 1/(1-a) ( x + bYn-1 )
    [itex]\sum[/itex]Yn-1 = 1/(1-d) (y + cXn-2)

    But I don't know how to express Xn-2 in terms of Xn.

    Please help! (or point me in the right direction)
     
  6. Jan 10, 2012 #5

    Stephen Tashi

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    Science Advisor

    To me, its simpler to look at it as a linear algebra problem. The recurrences amount to the matrix equation:

    [tex] \begin{pmatrix} X_n \\ Y_n \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} ^ n \begin{pmatrix} x_0 \\ y_0 \end{pmatrix} [/tex]

    For [itex] S_n = \sum_{i=1}^n X_i [/itex] and [itex] T_n = \sum_{i=1}^n Y_i [/itex]

    [tex] \begin{pmatrix} S_n \\ T_n \end{pmatrix} = \sum_{i=1}^n \begin{pmatrix} a & b \\ c & d \end{pmatrix}^i \begin{pmatrix} x_0 \\ y_0 \end{pmatrix} [/tex]

    To sum the matrix power series, we can try to find a matrix [itex] B [/itex] such that

    [tex] \begin{pmatrix} a & b \\ c & d \end{pmatrix} = B^{-1} \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix} B [/tex]

    Then

    [tex] \begin{pmatrix} S_n \\ T_n \end{pmatrix} = B^{-1} ( \sum_{i=1}^n \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix}^i ) B \begin{pmatrix} x_0 \\ y_0 \end{pmatrix} [/tex]

    [tex] = B^{-1} \begin{pmatrix} \sum_{i=1}^n \lambda_1^i & 0 \\ 0 & \sum_{i=1}^n \lambda_2^i \end{pmatrix} B \begin{pmatrix} x_0 \\ y_0 \end{pmatrix} [/tex]
     
  7. Jan 11, 2012 #6
    Thanks for the solution, Stephen!

    I actually cracked it yesterday by myself. My original logic turned out to be right:

    Since all terms in Xn-1 will be multiplied by a in Xn, we know that every term will be multiplied by a each step into infinity. Furthermore, we know that all new terms in X will come from b[itex]\Sigma[/itex]y (and the original X1. Therefore we know:

    [itex]\sum X=\frac{1}{1-a}\left ( x_{1}+b\sum Y \right )[/itex]

    By the same logic:

    [itex]\sum Y=\frac{1}{1-d}\left ( y_{1}+c\sum X \right )[/itex]

    Now, if we plug [itex]\Sigma[/itex]y into the formula for [itex]\Sigma[/itex]x, and solve for [itex]\Sigma[/itex]x, and then do some algebra, we get:

    [itex]\sum X=\frac{by_{1}+x_{1}-dx_{1}}{\left ( 1-d \right )\left ( 1-a \right )-bc}[/itex]

    Which turns out to be right, after I tested it.
     
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