localghost wrote:If the problem is solved, then please mark the topic (not the last post) accordingly as written in Section 3 of the
Board Rules (to be read before posting). Otherwise tell us what is missing.
Best regards and welcome to the board
Thorsten
OK, here, finally, is my the code I ended up with. Since I handed in the assignment, it's not going to improve anymore:
Code: Select all
Adding the two equations gives:
\begin{equation*}\label{eq:add}
\begin{aligned}
c_1 &= \frac{y_1 - 2y_2 + y_3}{2 \Delta t^2} \\
c_2 &= \frac{y_3-y_1}{2 \Delta t} \\
c_3 &= y_2
\end{aligned}
\end{equation*}
Code: Select all
\begin{equation*}\label{eq:vecC}
\begin{alignedat}{3}
c_1 &= \frac{\left| V_1 \right|}{\left| V \right|} &&= \frac{-\Delta t (y_1-2y_2+y_3)}{-2 \Delta t^3} &&= \frac{(y_1-2y_2+y_3)}{2 \Delta t^2} \\
c_2 &= \frac{\left| V_2 \right|}{\left| V \right|} &&= \frac{\Delta t^2(y_1-y_3)}{-2 \Delta t^3} &&= \frac{y_3-y_1}{2 \Delta t} \\
c_3 &= \frac{\left| V_3 \right|}{\left| V \right|} &&= \frac{-2\Delta t^3y_2}{-2 \Delta t^3} &&= y_2
\end{alignedat}
\end{equation*}
Code: Select all
The vertex occurs at $t=-b/2a$:
\begin{equation*}\label{eq:third}
\begin{alignedat}{2}
t &= -\frac{c_2}{2c_1} &&= \frac{-\frac{y_3-y_1}{2 \Delta t}}{2\frac{(y_1-2y_2+y_3)}{2 \Delta t^2}}+t_2 \\
&&&= \frac{y_1-y_3}{2 \Delta t} \cdot \frac{\Delta t^2}{y_1-2y_2+y_3}+t_2 \\
&&&= \frac{\Delta t(y_1-y_3)}{2(y_1-2y_2+y_3)}+t_2
\end{alignedat}
\end{equation*}
But I still have general spacing issues with Latex. For example, I have to manually add spacing to the following nested enumerated lists, and I can't find any way to tighten up white space where I don't want it (e.g. before the first list starts). I need to get the spacing right without jumping through all kinds of hoops. Can I?
Code: Select all
\subsection*{23.1:} For each of the following relations, please answer these questions:
\begin{enumerate}
\item Is it a function? If not, explain why.
\item If yes, what are it's domain and range?
\item Is the function one-to-one? If not, explain why.
\item If yes, what is the inverse function? \\[\baselineskip]
\begin{enumerate}
\item $f=\set{(1,2),(3,4)}$ \\
\A \begin{enumerate}
\item Yes
\item dom $f=\set{1,3}$, im $f=\set{2,4}$
\item Yes
\item $f^{-1}=\set{(2,1),(4,3)}$ \\[\baselineskip]
\end{enumerate}
\item $f=\setof{(x,y)}{x,y \in \mathbb{Z}, y=2x}$ \\
\A \begin{enumerate}
\item Yes
\item dom $f=\mathbb{Z}$, im $f=\mathbb{E}=\set{\text{even integers}}$
\item Yes
\item $f^{-1}=\setof{(2x,x)}{x \in \mathbb{E}}$ \\[\baselineskip]
\end{enumerate}
\item $f=\setof{(x,y)}{x,y \in \mathbb{Z}, x+y=0}$ \\
\A \begin{enumerate}
\item Yes
\item dom $f=\mathbb{Z}$, im $f=\mathbb{Z}$
\item Yes
\item $f^{-1} \colon x=-y$ (basically, the same as $f$). \\[\baselineskip]
\end{enumerate}
\item $f=\setof{(x,y)}{x,y \in \mathbb{Z}, xy=0}$ \\
\A \begin{enumerate}
\item No. $(0,1) \in f$ and $(0,2) \in f$. \\[\baselineskip]
\end{enumerate}
\item $f=\setof{(x,y)}{x,y \in \mathbb{Z}, y=x^2}$ \\
\A \begin{enumerate}
\item Yes
\item dom $f=\mathbb{Z}$, im $f=\mathbb{Z^+}=\set{\text{non-negative integers}}$
\item No. $(1,1) \in f$ and $(-1,1) \in f$. \\[\baselineskip]
\end{enumerate}
\item $f=\emptyset$ \\
\A \begin{enumerate}
\item Frankly, this doesn't look like a relation (a set of ordered pairs), so I doubt it's a function. \\[\baselineskip]
\end{enumerate}
\item $f=\setof{(x,y)}{x,y \in \mathbb{Q}, x^2+y^2=1}$ \\
\A \begin{enumerate}
\item No. $(0,1) \in f$ and $(0,-1) \in f$. \\[\baselineskip]
\end{enumerate}
\item $f=\setof{(x,y)}{x,y \in \mathbb{Z}, x|y}$ \\
\A \begin{enumerate}
\item No. $(2,2) \in f$ and $(2,4) \in f$. \\[\baselineskip]
\end{enumerate}
\item $f=\setof{(x,y)}{x,y \in \mathbb{N}, x|y \text{ and }y|x}$ \\
\A \begin{enumerate}
\item Yes. $f=\setof{(x,x)}{x \in \mathbb{Z^+}}$
\item dom $f=\mathbb{Z^+}$, im $f=\mathbb{Z^+}$
\item Yes
\item $f^{-1}=f$. \\[\baselineskip]
\end{enumerate}
\item $f=\setof{(x,y)}{x,y \in \mathbb{N}, \dbinom xy=1}$ \\
\A \begin{enumerate}
\item No. $\dbinom 20 = \dbinom 22 = 1$.
\end{enumerate}
\end{enumerate}
\end{enumerate}
