Third-order difference equation: limit x(k+2)/x(k+1)

Solve by the z-transform method, with $u(k)$ a unit step.

Step 1: Z-transform (zero initial conditions)

With $x(0) = x(1) = x(2) = 0$, all initial-condition terms vanish. The input transform is $U(z) = \dfrac{z}{z-1}$:

$$(z^3 - 2.2z^2 + 1.57z - 0.36)X(z) = \frac{z}{z-1}$$

Step 2: Factorise the characteristic polynomial

Trial roots: $P(0.5) = 0.125 - 0.55 + 0.785 - 0.36 = 0$, so $z = 0.5$ is a root. Dividing out gives $z^2 - 1.7z + 0.72 = (z-0.8)(z-0.9)$.

$$z^3 - 2.2z^2 + 1.57z - 0.36 = (z-0.5)(z-0.8)(z-0.9)$$ $$X(z) = \frac{z}{(z-1)(z-0.5)(z-0.8)(z-0.9)}$$

Step 3: Partial fractions of $X(z)/z$

$$\frac{X(z)}{z} = \frac{A}{z-1} + \frac{B}{z-0.5} + \frac{C}{z-0.8} + \frac{D}{z-0.9}$$ $$\begin{aligned} A &= \frac{1}{(0.5)(0.2)(0.1)} = 100 \\ B &= \frac{1}{(-0.5)(-0.3)(-0.4)} = -16.667 \\ C &= \frac{1}{(-0.2)(0.3)(-0.1)} = 166.667 \\ D &= \frac{1}{(-0.1)(0.4)(0.1)} = -250 \end{aligned}$$

Step 4: Closed-form solution

$$x(k) = 100 - 16.667(0.5)^k + 166.667(0.8)^k - 250(0.9)^k, \quad k = 0, 1, 2, \dots$$

The exact fractions are $-\tfrac{50}{3}$ and $\tfrac{500}{3}$.

Check

The recursion used is $x(k+3) = 2.2x(k+2) - 1.57x(k+1) + 0.36x(k) + 1$. For example, $x(4) = 2.2(1) + 1 = 3.2$.

The final value theorem gives $\lim_{z\to1}(z-1)X(z) = \dfrac{1}{0.5\times0.2\times0.1} = 100$, which equals $A$.

Step 5: Limiting value of $x(k+2)/x(k+1)$

The poles $0.5$, $0.8$ and $0.9$ are inside the unit circle, so every transient term dies out and $x(k) \to 100$:

$$\lim_{k\to\infty}\frac{x(k+2)}{x(k+1)} = \frac{100 + 0}{100 + 0} = 1$$

The ratio approaches 1 slowly, because the slowest mode $(0.9)^k$ dominates the transient.

Answer: $x(k) = 100 - 16.667(0.5)^k + 166.667(0.8)^k - 250(0.9)^k$, and $\lim_{k\to\infty} x(k+2)/x(k+1) = 1$.