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\centerline{\bf Math 365 -- Wednesday 3/20/19 -- 8.2 and 8.4}
\begin{try} 
Consider the recurrence relation
$$a_n = 8a_{n-2} - 16 a_{n-4} + F(n).$$
\begin{enumerate}[(a)]
\item Write the associated homogeneous recursion relation and solve for its general solution $\{h_n\}$. 
\medskip

\item For each of the following $F(n)$, write the general form for the particular solution (\underline{don't solve} for the unknowns).
\begin{enumerate}[(i)]
\item $F(n) = n^3$
\item $F(n) = n 2^n$
\item $F(n) = (n^2 - 2)(-2)^n$
\item $F(n) = 2$
\item $F(n) = (-2)^n$
\item $F(n) = n^2 4^n$
\item $F(n) = n^4 2^n$
\end{enumerate}

\medskip

\item Find the general solution to 
$$a_n = 8a_{n-2} - 16 a_{n-4} + (-2)^n.$$

\item Find the general solution to 
$$a_n = 8a_{n-2} - 16 a_{n-4} + n^3.$$


\medskip

\item Pick an example of appropriate initial conditions for the sequence in part (c), and solve for the corresponding specific solution.  Check your answer by computing the first 6 terms of the sequence both recursively and using your closed formula.
\end{enumerate}

\end{try}


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\begin{try}\label{try:fundamental-series}~
\begin{enumerate}[(a)]
\item Explicitly compute the series for 
$$\frac{1}{(1-x)^3} \quad \text{ and } \quad \frac{1}{(1-x)^4}$$
by taking derivatives and rescaling appropriately. Conjecture what the general formula for the series for $1/(1-x)^n$. \medskip
\item Substitute $y = 3x$  into the series for $\frac{1}{1-y}$ to get the series for $\displaystyle\frac{1}{1-3x}$. What is the series for $1/(1-4x)$? What is the series for $1/(1+x)$?  \medskip
\item Substitute $y = x^3$ into the series for $\displaystyle\frac{1}{1-y}$ to get the series for $\displaystyle\frac{1}{1-x^3}$. What is the series for $1/(1-x^4)$? What is the series for $1/(1-2x^3)$?  \medskip
\item Use the fact that $\displaystyle\frac{d}{dx} (\ln(1-x)) = -\frac{1}{1-x}$ to compute the series for $\ln(1-x)$. (Integrate, and pick your ``$+C$'' to make it so that evaluating your series at $x=0$ matches correctly with evaluating $\ln(1-x)$.)
\end{enumerate}
\end{try}

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