TITLE: How to find explicit formula for two recursions?
QUESTION [10 upvotes]: I have to find explicit solution for two intertwining recursions
$$\begin{align}
f(n)&=f(n-2)+2g(n-1) \\
g(n)&=g(n-2)+f(n-1)
\end{align}$$
for $f(0)=1, f(1)=0, g(0)=0 ,g(1)=1$. 
What techniques are commonly used for this types of problem? Thanks!

REPLY [2 votes]: $\newcommand{\angles}[1]{\left\langle{#1}\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace{#1}\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack{#1}\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\iff}{\Leftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\ol}[1]{\overline{#1}}
 \newcommand{\pars}[1]{\left({#1}\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert{#1}\right\vert}$

$\ds{%
\left.\begin{array}{rcl}
\ds{\mathrm{f}\pars{n}} & \ds{=} &
\ds{\mathrm{f}\pars{n - 2} + 2\mathrm{g}\pars{n-1}}
\\[1mm]
\ds{\mathrm{g}\pars{n}} & \ds{=} &
\ds{\mathrm{g}\pars{n - 2} +\mathrm{f}\pars{n - 1}}
\end{array}\right\rbrace
\quad\mbox{and}\quad
\left\lbrace\begin{array}{rclrcl}
\mathrm{f}\pars{0} & \ds{=} & \ds{1,} & \ds{\mathrm{f}\pars{1}} & \ds{=} & \ds{0}
\\[1mm]
\mathrm{g}\pars{0} & \ds{=} & \ds{0,} & \ds{\mathrm{g}\pars{1}} & \ds{=} & \ds{1}
\end{array}\right.}$


With $\ds{\mathbf{u}\pars{n} \equiv
          {\mathrm{f}\pars{n} \choose \mathrm{g}\pars{n}}}$:
\begin{align}
\mathbf{u}\pars{n} & =
\pars{\begin{array}{cc}\ds{0} & \ds{2}\\ \ds{1} & \ds{0} \end{array}}\mathbf{u}\pars{n - 1} +
\pars{\begin{array}{cc}\ds{1} & \ds{0}\\ \ds{0} & \ds{1}\end{array}}\mathbf{u}\pars{n - 2}\,,\qquad
\left\lbrace\begin{array}{rcl}
\ds{\mathbf{u}\pars{0}} & \ds{=} & \ds{1 \choose 0}
\\[1mm]
\ds{\mathbf{u}\pars{1}} & \ds{=} & \ds{0 \choose 1}
\end{array}\right.
\\[3mm] & = \bracks{%
{3 \over 2}\pars{\begin{array}{cc}\ds{0} & \ds{1}\\ \ds{1} & \ds{0} \end{array}}
+ \half\,\ic\pars{\begin{array}{cc}\ds{0} & \ds{-\ic}\\ \ds{\ic} & \ds{0} \end{array}}}\mathbf{u}\pars{n - 1} +
\pars{\begin{array}{cc}\ds{1} & \ds{0}\\ \ds{0} & \ds{1}\end{array}}\mathbf{u}\pars{n - 2}
\end{align}

\begin{align}
\mathbf{u}\pars{n} & =
\vec{a}\cdot\vec{\sigma}\,\,\mathbf{u}\pars{n - 1} +
\mathbf{u}\pars{n - 2}\,,\quad
\vec{a}\cdot\vec{\sigma} =
\pars{\begin{array}{cc}\ds{0} & \ds{2}\\ \ds{1} & \ds{0} \end{array}}\,,\quad
\left\lbrace\begin{array}{rcl}
\ds{\vec{a}\cdot\vec{\sigma}\,\mathbf{u}\pars{0}} & \ds{=} & \ds{\mathbf{u}\pars{1}}
\\[1mm]
\ds{\vec{a}\cdot\vec{\sigma}\,\mathbf{u}\pars{1}} & \ds{=} & \ds{2\mathbf{u}\pars{0}}
\end{array}\right.
\end{align}
where $\ds{\vec{a} \equiv {3 \over 2}\,\hat{x} + \half\,\ic\,\hat{y}}$ and $\ds{\braces{\sigma_{i}\ \mid\ i = 0,x,y,z}}$ are the Pauli Matrices.
Then,
\begin{align}
\sum_{n = 2}^{\infty}\mathbf{u}\pars{n}z^{n} & =
\vec{a}\cdot\vec{\sigma}\sum_{n = 2}^{\infty}\mathbf{u}\pars{n - 1}z^{n} +
\sum_{n = 2}^{\infty}\mathbf{u}\pars{n - 2}z^{n}
\\[3mm]
\sum_{n = 0}^{\infty}\mathbf{u}\pars{n}z^{n}
- \mathbf{u}\pars{0} - \mathbf{u}\pars{1}z
& =
\vec{a}\cdot\vec{\sigma}\,z
\bracks{\sum_{n = 0}^{\infty}\mathbf{u}\pars{n}z^{n} - \mathbf{u}\pars{0}} +
z^{2}\sum_{n = 0}^{\infty}\mathbf{u}\pars{n}z^{n}
\end{align}
which leads to
$$
\bracks{\pars{1 - z^{2}}\sigma_{0} - \vec{a}\cdot\vec{\sigma}\,z}
\sum_{n = 0}^{\infty}\mathbf{u}\pars{n}z^{n} =
\bracks{\sigma_{0} - \vec{a}\cdot\vec{\sigma}\,z}\mathbf{u}\pars{0} + z\mathbf{u}\pars{1} = \mathbf{u}\pars{0}
$$
Multiply both sides, by the left, with the matrix
$\ds{\bracks{\pars{1 - z^{2}}\sigma_{0} + \vec{a}\cdot\vec{\sigma}\,z}}$ $\ds{\pars{~\mbox{note that}\ \pars{\vec{a}\cdot\vec{\sigma}}^{2} = \vec{a}\cdot\vec{a} = 2~}}$:
\begin{align}
\bracks{\pars{1 - z^{2}}^{2} - 2z^{2}}
\sum_{n = 0}^{\infty}\mathbf{u}\pars{n}z^{n} & =
\bracks{\pars{1 - z^{2}}\sigma_{0} + \vec{a}\cdot\vec{\sigma}\,z}
\mathbf{u}\pars{0} =
\pars{1 - z^{2}}\mathbf{u}\pars{0} + z\,\mathbf{u}\pars{1}
\end{align}

\begin{align}
\imp\quad\sum_{n = 0}^{\infty}\mathbf{u}\pars{n}z^{n} & =
{\pars{1 - z^{2}}\mathbf{u}\pars{0} + z\,\mathbf{u}\pars{1} \over
z^{4} - 4z^{2} + 1}
\\[3mm]
\iff\quad & \left\lbrace\begin{array}{rcl}
\ds{\mathrm{f}\pars{n}} & \ds{=} &
\ds{\bracks{z^{n}}\pars{{1 - z^{2} \over z^{4} - 4z^{2} + 1}}}
\\[1mm]
\ds{\mathrm{g}\pars{n}} & \ds{=} &
\ds{\bracks{z^{n}}\pars{{z \over z^{4} - 4z^{2} + 1}}}
\end{array}\right.
\end{align}
The zeros of $\ds{w^{2} - 4w + 1 = 0}$ are given by
$\ds{r_{\pm} = 2 \pm \root{3}}$ with $\ds{r_{-} \approx 0.2679}$ and
$\ds{r_{+} = 1/r_{-} \approx 3.7321}$ such that
\begin{align}
{1 \over z^{4} - 4z^{2} + 1} & =
{1 \over \pars{z^{2} - r_{-}}\pars{z^{2} - r_{+}}} =
{1 \over r_{+} - r_{-}}\pars{{1 \over z^{2} - r_{+}} - {1 \over z^{2} - r_{-}}}
\\[3mm] & = 
{1 \over 2\root{3}}
\pars{{1/r_{-} \over 1 - z^{2}/r_{-}} - {1/r_{+} \over 1 - z^{2}/r_{+}}}
=
{1 \over 2\root{3}}
\pars{{r_{+} \over 1 - r_{+}z^{2}} - {r_{-} \over 1 - r_{-}z^{2}}}
\\[3mm] & =
{1 \over 2\root{3}}
\sum_{n = 0}^{\infty}c_{n}z^{2n}\,,
\qquad
c_{n} \equiv r_{+}^{n + 1} - r_{-}^{n + 1}\,,\quad\verts{z} < r_{-}^{1/2}
\end{align}
Also,
\begin{align}
{1 - z^{2} \over z^{4} - 4z^{2} + 1} & =
\sum_{n = 0}^{\infty}c_{n}z^{2n} - \sum_{n = 0}^{\infty}c_{n}z^{2n + 2} =
\sum_{n = 0}^{\infty}c_{n}z^{2n} - \sum_{n = 1}^{\infty}c_{n - 1}z^{2n} =
c_{0} + \sum_{n = 1}^{\infty}\pars{c_{n} - c_{n - 1}}z^{2n}
\\[3mm]
{z \over z^{4} - 4z^{2} + 1} & = \sum_{n = 0}^{\infty}c_{n}z^{2n + 1}
\end{align}
Finally,
\begin{align}
\color{#f00}{\mathrm{f}\pars{n}} & = \color{#f00}{%
\left\lbrace\begin{array}{lcl}
\ds{{1 \over 2\root{3}}\,c_{0} = 1} & \mbox{if} & \ds{n = 0}
\\[1mm]
\ds{{1 \over 2\root{3}}\,\pars{c_{n/2} - c_{n/2 - 1}}} & \mbox{if} &
\ds{n}\ \mbox{is}\ even
\\[1mm]
\ds{0} && \mbox{otherwise}
\end{array}\right.}
\\[3mm]
\color{#f00}{\mathrm{g}\pars{n}} & = \color{#f00}{%
\left\lbrace\begin{array}{lcl}
\ds{{1 \over 2\root{3}}\,c_{0} = 1} & \mbox{if} & \ds{n = 1}
\\[1mm]
\ds{{1 \over 2\root{3}}\,\bracks{c_{\pars{n - 1}/2} - c_{\pars{n - 1}/2 - 1}}}
& \mbox{if} & \ds{n}\ \mbox{is}\ odd
\\[1mm]
\ds{0} && \mbox{otherwise}
\end{array}\right.}
\end{align}
$\ds{\color{#f00}{\mbox{where}\ c_{n} = r_{+}^{n + 1} - r_{-}^{n + 1}\,,\quad r_{\pm} = 2 \pm \root{3}}}$.