Autonomous System Summation

1. Jan 10, 2012

arentmuskets

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 :-(.

Last edited: Jan 10, 2012
2. Jan 10, 2012

Number Nine

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...

$$Sum = \frac{X_{1}}{1 - c}$$

3. Jan 10, 2012

arentmuskets

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
4. Jan 10, 2012

arentmuskets

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:

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

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

5. Jan 10, 2012

Stephen Tashi

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

$$\begin{pmatrix} X_n \\ Y_n \end{pmatrix} = \begin{pmatrix} a & b \\ c & d \end{pmatrix} ^ n \begin{pmatrix} x_0 \\ y_0 \end{pmatrix}$$

For $S_n = \sum_{i=1}^n X_i$ and $T_n = \sum_{i=1}^n Y_i$

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

To sum the matrix power series, we can try to find a matrix $B$ such that

$$\begin{pmatrix} a & b \\ c & d \end{pmatrix} = B^{-1} \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix} B$$

Then

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

$$= B^{-1} \begin{pmatrix} \sum_{i=1}^n \lambda_1^i & 0 \\ 0 & \sum_{i=1}^n \lambda_2^i \end{pmatrix} B \begin{pmatrix} x_0 \\ y_0 \end{pmatrix}$$

6. Jan 11, 2012

arentmuskets

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$\Sigma$y (and the original X1. Therefore we know:

$\sum X=\frac{1}{1-a}\left ( x_{1}+b\sum Y \right )$

By the same logic:

$\sum Y=\frac{1}{1-d}\left ( y_{1}+c\sum X \right )$

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

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

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