\section{(Adams Methods)}
Give explicit values for the $\alpha$'s and $\beta$'s of the Adams-Bashforth and Adams-Moulton methods.

\subsection{Adams-Bashforth}
Adams-Bashforth methods are defined as follows
\begin{eqnarray}
y_{n+k} - y_{n+k-1} = h \sum_{j=0}^{k-1} \beta_{j} f_{n+j}
\end{eqnarray}
where the $\beta_{j}$ are uniquely determined by the fact that $p = k$.  We will also use the definition for the coefficients of the LTE.  These are
\begin{eqnarray}
C_{0} & = & \sum_{j=0}^{k}\alpha_{j} \\
C_{1} & = & \sum_{j=1}^{k}j\alpha_{j} - \sum_{j=0}^{k}\beta_{j} \\
C_{p} & = & \frac{1}{p!}\sum_{j=1}^{k}j^{p}\alpha_{j} - \frac{1}{(p-1)!}\sum_{j=1}^{k}j^{p-1}\beta_{j}
\end{eqnarray}
We will find the values for the coefficients of (11) for the values $k = 1,2,3,4$.

\subsubsection{$k=1$}
Here the Adams method becomes
\[y_{n+1} - y_{n} = h \beta_{0} f_{n}\]
Using the equations that define the LTE, we come up with the following equations
\begin{eqnarray*}
C_{0} & = & \alpha_{0} + \alpha_{1} = 0 \\
C_{1} & = & \alpha_{1} - (\beta_{0} + \beta_{1}) = 0
\end{eqnarray*}
From our method, we can tell that $\alpha_{0} = 1, \alpha_{1} = -1$ and since this method is explicit, we know that $\beta_{1} = 0$.  Which leaves us with the equation
\[\alpha_{1} - \beta_{0} = 0\]
Which tells us that $\beta_{0} = -1$.

\subsubsection{$k=2$}
The Adams method becomes
\[y_{n+2} - y_{n+1} = h \beta_{0} f_{n} + h \beta_{1} f_{n+1}\]
By again using the LTE equations, we get
\begin{eqnarray*}
C_{0} & = & \alpha_{0} + \alpha_{1} + \alpha_{2} = 0 \\
C_{1} & = & \alpha_{1} + 2\alpha_{2} - (\beta_{0} + \beta_{1} + \beta_{2}) = 0 \\
C_{2} & = & \frac{1}{2}(\alpha_{1} + 4\alpha_{2}) - (\beta_{1} + 2\beta_{2}) = 0
\end{eqnarray*}
This gives us the system of equations
\[ \left[ \begin{array}{rrrrrr}
	1 & 1 & 0 & 0 & 0 & 0 \\
	0 & 1 & 2 & -1 & -1 & -1 \\
	0 & 1/2 & 2 & 0 & -1 & 2 \end{array} \right]
  \left( \begin{array}{c} \alpha_{0} \\ \alpha_{1} \\ \alpha_{2} \\ \beta_{0} \\ \beta_{1} \\ \beta_{2} \end{array} \right) = \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right) \]
For this step, we can say that $\alpha_{0} = 0, \alpha_{1} = -1, \alpha_{2} = 1, \beta_{2} = 0$.  Which leaves us with the system
\begin{eqnarray*}
-1 - \beta_{0} - \beta_{1} & = & 0 \\
-1/2 - \beta_{1} & = & 0
\end{eqnarray*}
Which gives us that $\beta_{0} = 1/2, \beta_{1} = -1/2$.


\subsubsection{$k=3$}
Here, the Adams method becomes
\[y_{n+3} - y_{n+2} = h \beta_{0} f_{n} + h \beta_{1} f_{n+1} + h \beta_{2} f_{n+2}\]
Using the LTE equations, we get
\begin{eqnarray*}
C_{0} & = & \alpha_{0} + \alpha_{1} + \alpha_{2} + \alpha_{3} = 0 \\
C_{1} & = & \alpha_{1} + 2\alpha_{2} + 3\alpha_{3} - (\beta_{0} + \beta_{1} + \beta_{2} + \beta_{3}) = 0 \\
C_{2} & = & \frac{1}{2}(\alpha_{1} + 4\alpha_{2} + 9\alpha_{3}) - (\beta_{1} + 2\beta_{2} + 3\beta_{3}) = 0 \\
C_{3} & = & \frac{1}{6}(\alpha_{1} + 8\alpha_{2} + 27\alpha_{3}) - \frac{1}{2}(\beta_{1} + 4\beta_{2} + 9\beta_{3}) = 0
\end{eqnarray*}
We know that $\alpha_{0} = \alpha_{1} = 0, \alpha_{2} = -1, \alpha_{3} = 1, \beta_{3} = 0$.  From the remaining conditions we get the following system of equations
\[\begin{array}{ccccccc}
	\beta_{0} & + & \beta_{1} & + & \beta_{2} & = & 1 \\
	          & + & \beta_{1} & + & 2\beta_{2} & = & 5/2 \\
	          & + & 1/2\beta_{1} & + & 2\beta_{2} & = & 19/6 \\
\end{array}\]
Which tells us that $\beta_{0} = 5/12, \beta_{1} = -4/3, \beta_{2} = 23/12$.

\subsubsection{$k=4$}
Here, the Adams method becomes
\[y_{n+4} - y_{n+3} = h \beta_{0} f_{n} + h \beta_{1} f_{n+1} + h \beta_{2} f_{n+2} + h \beta_{3} f_{n+3}\]
Using the LTE equations, we get
\begin{eqnarray*}
C_{0} & = & \alpha_{0} + \alpha_{1} + \alpha_{2} + \alpha_{3} + \alpha_{4} = 0 \\
C_{1} & = & \alpha_{1} + 2\alpha_{2} + 3\alpha_{3} + 4\alpha_{4} - (\beta_{0} + \beta_{1} + \beta_{2} + \beta_{3} + \beta_{4}) = 0 \\
C_{2} & = & \frac{1}{2}(\alpha_{1} + 4\alpha_{2} + 9\alpha_{3} + 16\alpha_{4}) - (\beta_{1} + 2\beta_{2} + 3\beta_{3} + 4\beta_{4}) = 0 \\
C_{3} & = & \frac{1}{6}(\alpha_{1} + 8\alpha_{2} + 27\alpha_{3} + 64\alpha_{4}) - \frac{1}{2}(\beta_{1} + 4\beta_{2} + 9\beta_{3} + 16\beta_{4}) = 0 \\
C_{4} & = & \frac{1}{24}(\alpha_{1} + 16\alpha_{2} + 81\alpha_{3} + 256\alpha_{4}) - \frac{1}{6}(\beta_{1} + 8\beta_{2} + 27\beta_{3} + 64\beta_{4}) = 0
\end{eqnarray*}
We know that $\alpha_{0} = \alpha_{1} = 0, \alpha_{2} = 0, \alpha_{3} = -1, \alpha_{4} = 1, \beta_{4} = 0$.  From the remaining conditions we get the following system of equations
\[\begin{array}{ccccccccc}
	\beta_{0} & + & \beta_{1} & + & \beta_{2} & + & \beta_{3} & = & 1 \\
	          & + & \beta_{1} & + & 2\beta_{2} & + & 3\beta_{3} & = & 7/2 \\
	          & + & 1/2\beta_{1} & + & 2\beta_{2} & + & 9/2\beta_{3} & = & 37/6 \\
	          & + & 1/6\beta_{1} & + & 8/6\beta_{2} & + & 27/6\beta_{3} & = & 175/24 \\
\end{array}\]
And we have $\beta_{0} = -3/8, \beta_{1} = 37/24, \beta_{2} = -59/24, \beta_{3} = 55/24$.

\subsection{Adams-Moulton}
Adams-Moulton methods are defined as follows
\begin{eqnarray}
y_{n+k} - y_{n+k-1} = h \sum_{j=0}^{k} \beta_{j} f_{n+j}
\end{eqnarray}
where the $\beta_{j}$ are uniquely determined by the fact that $p = k+1$.  We will again be using eqn (12),(13),(14) for the LTE in order to determine the coefficients.

\subsubsection{$k = 1$}
When $k=1$, the Adams-Moulton method becomes
\[y_{n+1} - y_{n} = h \beta_{0} y_{n} + h\beta_{1} y_{1}\]
Using the equations defining the LTE, we get
\begin{eqnarray*}
C_{0} & = & \alpha_{0} + \alpha_{1} \\
C_{1} & = & \alpha_{1} - (\beta_{0} + \beta_{1}) \\
C_{2} & = & \frac{1}{2}\alpha_{1} - \beta_{1}
\end{eqnarray*}
We know that for $k = 1$, $\alpha_{0} = -1, \alpha_{1} = 1$, which gives us the system of equations
\[\begin{array}{rrrrr}
	\beta_{0} & + & \beta_{1} & = & 1 \\
	& & \beta_{1} & = & 1/2
\end{array}\]
So, $\beta_{0} = \beta_{1} = 1/2$.

\subsubsection{$k = 2$}
When $k=2$, the Adams-Moulton method becomes
\[y_{n+2} - y_{n+1} = h \beta_{0} y_{n} + h\beta_{1} y_{1} + h\beta_{2} y_{2}\]
Using the equations defining the LTE, we get
\begin{eqnarray*}
C_{0} & = & \alpha_{0} + \alpha_{1} + \alpha_{2} = 0\\
C_{1} & = & \alpha_{1} + 2\alpha_{2} - (\beta_{0} + \beta_{1} + \beta_{2}) = 0\\
C_{2} & = & \frac{1}{2}(\alpha_{1} + 4\alpha_{2}) - (\beta_{1} + 2\beta_{2}) = 0\\
C_{3} & = & \frac{1}{6}(\alpha_{1} + 8\alpha_{2}) - \frac{1}{2}(\beta_{1} + 4\beta_{2}) = 0
\end{eqnarray*}
We know that for $k = 2$, $\alpha_{0} = 0, \alpha_{1} = -1, \alpha_{2} = 1$.  Which gives us the system
\[\begin{array}{rrrrrrr}
	\beta_{0} & + & \beta_{1} & + & \beta_{2} & = & 1 \\
	& & \beta_{1} & + & 2\beta_{2} & = & 3/2 \\
	& & \frac{1}{2}\beta_{1} & + & 2\beta_{2} & = & 7/6
\end{array}\]
So, $\beta_{0} = -1/12, \beta_{1} = 2/3, \beta_{2} = 5/12$.

\subsubsection{$k = 3$}
When $k=3$, the Adams-Moulton method becomes
\[y_{n+3} - y_{n+2} = h \beta_{0} y_{n} + h\beta_{1} y_{1} + h\beta_{2} y_{2} + h\beta_{3} y_{3}\]
Using the equations defining the LTE, we get
\begin{eqnarray*}
C_{0} & = & \alpha_{0} + \alpha_{1} + \alpha_{2} + \alpha_{3} = 0\\
C_{1} & = & \alpha_{1} + 2\alpha_{2} + 3\alpha_{3} - (\beta_{0} + \beta_{1} + \beta_{2} + \beta_{3}) = 0\\
C_{2} & = & \frac{1}{2}(\alpha_{1} + 4\alpha_{2} + 9\alpha_{3}) - (\beta_{1} + 2\beta_{2} + 3\beta_{3}) = 0\\
C_{3} & = & \frac{1}{6}(\alpha_{1} + 8\alpha_{2} + 27\alpha_{3}) - \frac{1}{2}(\beta_{1} + 4\beta_{2} + 9\beta_{3}) = 0 \\
C_{4} & = & \frac{1}{24}(\alpha_{1} + 16\alpha_{2} + 81\alpha_{3}) - \frac{1}{6}(\beta_{1} + 8\beta_{2} + 27\beta_{3}) = 0
\end{eqnarray*}
For $k=3$, $\alpha_{0} = \alpha_{1} = 0, \alpha_{2} = -1, \alpha_{3} = 1$, and we have the system
\[\begin{array}{rrrrrrrrr}
	\beta_{0} & + & \beta_{1} & + & \beta_{2} & + & \beta_{3} & = & 1 \\
	& & \beta_{1} & + & 2\beta_{2} & + & 3\beta_{3} & = & 5/2 \\
	& & 1/2\beta_{1} & + & 2\beta_{2} & + & 9/2\beta_{3} & = & 19/6 \\
	& & 1/6\beta_{1} & + & 8/6\beta_{2} & + & 27/6\beta_{3} & = & 65/24
\end{array}\]
So, $\beta_{0} = -47/12, \beta_{1} = 155/16, \beta_{2} = -57/8, \beta_{3} = 113/48$.

\subsubsection{$k = 4$}
When $k=4$, the Adams-Moulton method becomes
\[y_{n+4} - y_{n+3} = h \beta_{0} y_{n} + h\beta_{1} y_{1} + h\beta_{2} y_{2} + h\beta_{3} y_{3} + h\beta_{4} y_{4}\]
Using the equations defining the LTE, we get \small
\begin{eqnarray*}
C_{0} & = & \alpha_{0} + \alpha_{1} + \alpha_{2} + \alpha_{3} + \alpha_{4} = 0\\
C_{1} & = & \alpha_{1} + 2\alpha_{2} + 3\alpha_{3} + 4\alpha_{4} - (\beta_{0} + \beta_{1} + \beta_{2} + \beta_{3} + \alpha_{4}) = 0\\
C_{2} & = & \frac{1}{2}(\alpha_{1} + 4\alpha_{2} + 9\alpha_{3} + 16\alpha_{4}) - (\beta_{1} + 2\beta_{2} + 3\beta_{3} + 4\beta_{4}) = 0\\
C_{3} & = & \frac{1}{6}(\alpha_{1} + 8\alpha_{2} + 27\alpha_{3} + 64\alpha_{4}) - \frac{1}{2}(\beta_{1} + 4\beta_{2} + 9\beta_{3} + 16\beta_{4}) = 0 \\
C_{4} & = & \frac{1}{24}(\alpha_{1} + 16\alpha_{2} + 81\alpha_{3} + 256\alpha_{4}) - \frac{1}{6}(\beta_{1} + 8\beta_{2} + 27\beta_{3} + 64\beta_{4}) = 0 \\
C_{5} & = & \frac{1}{120}(\alpha_{1} + 32\alpha_{2} + 243\alpha_{3} + 1024\alpha_{4}) - \frac{1}{24}(\beta_{1} + 16\beta_{2} + 81\beta_{3} + 256\beta_{4}) = 0
\end{eqnarray*} \normalsize
For $k=4$, $\alpha_{0} = \alpha_{1} = \alpha_{2} = 0, \alpha_{3} = -1, \alpha_{4} = 1$, and we have the system
\[\begin{array}{rrrrrrrrrrr}
	\beta_{0} & + & \beta_{1} & + & \beta_{2} & + & \beta_{3} & + & \beta_{4} & = & 1 \\
	& & \beta_{1} & + & 2\beta_{2} & + & 3\beta_{3} & + & 4\beta_{4} & = & 7/2 \\
	& & 1/2\beta_{1} & + & 2\beta_{2} & + & 9/2\beta_{3} & + & 8\beta_{4} & = & 37/6 \\
	& & 1/6\beta_{1} & + & 8/6\beta_{2} & + & 27/6\beta_{3} & + & 64/6\beta_{4} & = & 65/24 \\
	& & 1/24\beta_{1} & + & 16/24\beta_{2} & + & 81/24\beta_{3} & + & 256/24\beta_{4} & = & 781/24
\end{array}\]
So, $\beta_{0} = -199/720, \beta_{1} = 413/360, \beta_{2} = -28/15, \beta_{3} = 683/360, \beta_{4} = 71/720$.
