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\centerline{\bf Math 365 -- Monday 3/18/19 -- 8.2: Solving linear recurrence relations}
\begin{warmup}Recall that a solution for a sequence defined recursively is a closed formula that satisfies the recursion relation and the initial conditions. For example, the sequence
$$a_n = n a_{n-1},\ a_0 = 1 \qquad \text{ has solution } \qquad a_n=n!.$$
\begin{enumerate}
\item Consider the recurrence relation $a_n = 7a_{n-1}$.
\begin{enumerate}
\item Choose three different initial conditions (specific values for $a_0$), and find the corresponding solutions.
\item Write a general solution to this recurrence relation in terms of $a_0$ (without picking a specific value for $a_0$).
\end{enumerate}
\item For each of the following recursion relations, how many initial conditions are needed to determine a specific solution?
\begin{multicols}{2}
\begin{enumerate}
\item $a_n = 5 a_{n-1}$
\item $a_n = 3$
\item $a_n = 9 a_{n-2}$
\item $a_n = a_{n-1}^2$
\item $a_n = -2a_{n-1} - a_{n-2}$
\item $a_n = 7a_{n-1} - 6a_{n-2}$
\item $a_n = 8 a_{n-3}$
\item $a_n = 3a_{n-1} + 4 a_{n-2} - 12 a_{n-3}$
\item $a_n = a_{n-1}/n$
\item $a_n = a_{n-1} + a_{n-2} + n + 3$
\end{enumerate}
\end{multicols}
\item Factor the following polynomials
\begin{enumerate}
\item $x^2 - 2x + 1$
\item $x^2 + 5x + 6$
\item $x^2 - 9$
\item $x^2 + 2x +5$
\item $x^3 - 3x^2 + 3x -1$
\item $x^3-8$
\item $x^3 +5 x^2+8 x+ 4$
\item $x^3 - 3x^2 - 4 x + 12$
\item $x^4 - 2x^2 +1$
\end{enumerate}
\end{enumerate}
\end{warmup}
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\begin{try}\label{try:LRR1} For each of the following, decide if the recurrence relation is linear, homogeneous, and constant coefficient. If not, explain why it fails. If so, (i) give its degree, (ii) give its characteristic equation, and (iii) give the characteristic roots and their multiplicities.
\begin{enumerate}[(a)]
\item $a_n = 5 a_{n-1}$
\item $a_n = 3$
\item $a_n = 9 a_{n-2}$
\item $a_n = a_{n-1}^2$
\item $a_n = -2a_{n-1} - a_{n-2}$
\item $a_n = 7a_{n-1} - 6a_{n-2}$
\item $a_n = 8 a_{n-3}$
\item $a_n = 3a_{n-1} + 4 a_{n-2} - 12 a_{n-3}$
\item $a_n = a_{n-1}/n$
\item $a_n = a_{n-1} + a_{n-2} + n + 3$
\end{enumerate}
\end{try}
\vfill
\begin{try} For each of the recursion relations in Exercise \ref{try:LRR1} that were linear, homogeneous, and constant coefficient, decide which have characteristic equations with $k$ distinct roots. For those that do,
\begin{enumerate}[(i)]
\item write a general solution;
\item choose an example of initial conditions, and solve for specific $\alpha_i$'s; and
\item check your answer to the previous part by computing the first 5 terms of the sequence in two ways (recursively, and using your closed formula).
\end{enumerate}
\end{try}
\vfill
\begin{try} For each of the recursion relations in Exercise \ref{try:LRR1} that were linear, homogeneous, and constant coefficient, but had characteristic equations with repeated roots,
\begin{enumerate}[(i)]
\item write a general solution;
\item choose an example of initial conditions, and solve for specific $\alpha_{i,j}$'s; and
\item check your answer to the previous part by computing the first 5 terms of the sequence in two ways (recursively, and using your closed formula).
\end{enumerate}
\end{try}
\vfill
\begin{try}
Adapt the proof of Theorem 1 for $k=2$ to prove Theorem 2 for $k=2$. Namely, show that if
$$
a_n = c_1 a_{n-1} + c_2 a_{n-2}
$$
has a characteristic equation with a repeated root $r_0$, then $a_n = \alpha r_0^n + \beta n r_0^n$ is the general solution.
Outline/hints:
\begin{itemize}
\item Establish that the characteristic equation is $r^2 - c_1 r - c_2 =0$.
\item Justify that if $r_0$ is the only root of this equation, then actually $c_1 = 2r_0$ and $c_2 = -r_0^2$.
\item Use a similar computation as in class to show that $a_n = \alpha r_0^n + \beta n r_0^n$ is a solution to the recurrence for any constants $\alpha$ and $\beta$.
\item Use a similar computation as in class to show that if $\{a_n\}_{n \in \NN}$ is a solution, then it must be of the form $a_n = \alpha r_0^n + \beta n r_0^n$ (i.e.\ there are some $\alpha$ and $\beta$ that match your solution).
\end{itemize}
\end{try}
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