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  "body": "\n## Definition\n\nLet $k \\ge 1$, let $a_0, \\ldots, a_k$ be complex numbers with $a_0 \\ne 0$, let $(f_n)_{n \\ge 0}$ be a sequence with a [[unilateral Z-transform|EPG6JPVJE7]] $F$, and let $c_0, \\ldots, c_{k-1}$ be complex numbers.\nConsider the [[linear recurrence with constant coefficients|0NDVHBVBXB]]\n$$a_0 y_{n+k} + a_1 y_{n+k-1} + \\cdots + a_k y_n = f_n \\quad (n \\ge 0), \\qquad y_j = c_j \\ (0 \\le j \u003c k).$$\nThe *Z-transform method* solves it in three steps.\n\n1. Apply the unilateral Z-transform and the [[shift property|JWH2KYCGBQ]], $\\mathcal{Z}\\{(y_{n+m})_{n \\ge 0}\\} = z^m \\left( Y(z) - \\sum_{j\u003cm} y_j z^{-j} \\right)$, to get $P(z) Y(z) = F(z) + Q(z)$, where $P(z) = \\sum_{m=0}^{k} a_{k-m} z^m$ and $Q$ is a polynomial built from $c_0, \\ldots, c_{k-1}$.\n2. Solve: $Y(z) = (F(z) + Q(z)) / P(z)$.\n3. Recover $(y_n)$ from $Y$ by the [[inverse Z-transform|PH61T3B7W7]], in practice by a [[partial fraction decomposition|030VFEREKG]] and known transform pairs.\n\n## Remarks\n\nExample: the Fibonacci recurrence $y_{n+2} = y_{n+1} + y_n$ with $y_0 = 0$, $y_1 = 1$.\nStep 1 gives $z^2 (Y - z^{-1}) = z Y + Y$, so $Y(z) = z/(z^2 - z - 1)$.\nWith $\\varphi = (1 + \\sqrt 5)/2$ and $\\psi = (1 - \\sqrt 5)/2$, one has $Y(z) = \\frac{1}{\\sqrt 5}\\left( \\frac{z}{z - \\varphi} - \\frac{z}{z - \\psi} \\right)$, and the pair $a^n \\leftrightarrow z/(z - a)$ gives $y_n = (\\varphi^n - \\psi^n)/\\sqrt 5$.\nNon-example: the recurrence $y_{n+1} = y_n^2$ is not linear; the transform of $(y_n^2)$ is not a function of $Y(z)$ alone, so step 1 gives no algebraic equation for $Y$.\nThe polynomial $P$ is the characteristic polynomial of the recurrence, so the poles of $Y$ that do not come from $F$ lie among its roots.\n",
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