Page Layout ⇒ frustrating formatting error

[gbux]

so, this is like latex day 2. Im getting a good handle and having a blast doing it. I have an assignment for my prob and stats class to make an outline of chapter 2 and thought this was a great time to learn latex. Im doing fairly well, and making use of every search bar I can lol. im at a part that im having a lot of trouble with. I have my title and everything, and it shows up all nicey nice, but then the author name also shows up on top of page 2 (which i removed to protect my good name lol) and then the title shows up again on top of page 3. also the margins are weird on every other page. heres my code:

Code: Select all

\documentclass[11pt]{amsart}
\usepackage{geometry}                % See geometry.pdf to learn the layout options. There are lots.
\geometry{letterpaper}                   % ... or a4paper or a5paper or ... 
%\geometry{landscape}                % Activate for for rotated page geometry
%\usepackage[parfill]{parskip}    % Activate to begin paragraphs with an empty line rather than an indent
\usepackage{graphicx}
\usepackage{amssymb}
\usepackage{epstopdf}
\usepackage{setspace} 
\DeclareGraphicsRule{.tif}{png}{.png}{`convert #1 `dirname #1`/`basename #1 .tif`.png}
\huge
\title{{\Huge Chapter 2 Study Guide}}
\large
\author{}
%\date{}                                           % Activate to display a given date or no date
\begin{document}

\begin{titlepage}

\end{titlepage}

\doublespace

\pagestyle{myheadings}

%\subsection{}\

\section*{\textbf{Section 2.3}}

 \pagestyle{myheadings}
 
 \textbf{Theorem 2.3.1-}

 P($A^{C}$)=1-P(A)
 
 \textbf{Theorem 2.3.2-}
 
 P($\varnothing$)=0
 
 \textbf{Theorem 2.3.3}

 If A$\subset$B, then P(A)$\leq$P(B).
 
\textbf{Theorem 2.3.4}
 
 For any Event A, P(A)$\leq$1
 
 \textbf{Theorem 2.3.5}
 
 Let $A_1$, $A_2$,.....,$A_N$ be events defined over S. If $A_i$$\cap$$A_j$=$\varnothing$ for i$ \not=$ then,
 
 P($\displaystyle\bigcup\limits_{i=1}^n A_i $)=$\displaystyle\sum\limits_{i=1}^n P(A_i) $
 
 \textbf{Theorem 2.3.6}
 
 P(A$\cup$B)=P(A) + P(B) - P(A$\cap$B)
   
  \section*{\textbf{Section 2.4}}
 
 \textbf{Equation 2.4.}
  \pagestyle{myheadings}
 \doublespace
 
  P(A$\mid$B)=$\frac {P(A\cap B)}{P(B)}$
 
 \textbf{2.4.3- Applying Conditional Probability to Higher-Order Intersections}
 
 P(A $\cap$ B$\cap$ C)=P(C$\mid$A$\cap$B$\cap$)P(A$\mid$B)P(A)
 
 leading to
 
 P($A_1\cap A_2\cap ...\cap A_n$)=P($A_n\mid A_1\cap A_2\cap ...\cap A_{n-1}$)P($A_n\mid 
 A_1\cap A_2\cap ...\cap A_{n-2}$)...      P($A_1\cap A_2$)P(A) 
 
\textbf{Theorem 2.4.1}
 
 B=($B\cap A_1)\cup (B\cap A_2)\cup ...\cup (B\cap A_n$)
 
 P(B)=P($B\cap A_1)+P(B\cap A_2)+...+P(B\cap A_n$)
 
 \textbf{Theorem 2.4.2}
 
 P($A_j\mid B$)=$\frac{P(B\mid A)P(A_j)}{P(B)}$


\section*{\textbf{Section 2.5}}
 \pagestyle{myheadings}
 
 \textbf{Deducing Independence}
 

 
 If P($A\cap B$)=P(A)P(B) then A and B are independent
 

\section*{\textbf{Section 2.6}}
 \pagestyle{myheadings}


\textbf{Multiplication Rule in Cominatorics}

 
 
 If operation A can be performed in m different ways and operation B can be performed in n different ways, the sequence (operation A, operation B) can be performed in m$\bullet$n different ways.

 
 \textbf{Theorem 2.6.1}
 
 
 The number of permutations of length k that can be formed from a set of n distinct elements, repetitions not allowed, is denoted by the symbol ${}_nP_k$, where
 
 
${}_nP_k=n(n-1)(n-2)...(n-k+1)=\frac {n!}{(n-k)!}$


\textbf{Theorem 2.6.2}


The number of ways to arrange n objects, $n_1$ being of one kind, $n_2$ of a second kind,..., and $n_r$, of an rth kind, is 



$\frac {n!}{n_1!n_2!...n_r!}$


Where 


$\displaystyle\sum\limits_{i=1}^n n_i=n $


 \textbf{Theorem 2.6.3}


$\binom{n}{k}=\frac{n!}{k!(n-k)!}$

\section*{\textbf{Section 2.7}}
 \pagestyle{myheadings}

Finding Combinatorial Probability

This is as simple as $\frac {m}{n}$, where m is the number of outcomes in the field that you are looking for, and n is the total number outcomes

 
  \end{document}

[Stefan Kottwitz]

Hi,

welcome to the board!

Very good for a 2nd day.

I would not use the amsart class, if you don't plan to publish it with AMS. It's a special AMS style.

Instead, I would use article (well, I prefer scrartcl) and load the amsmath package. This gives you all AMS math features without restrictions or their predefined style.

Code: Select all

\documentclass[11pt]{article}
\usepackage{amsmath}

Stefan

[gbux]

oh baby, you gone done fixed it!

hehe thanks for that, only thing, I have a title page then page one and so on, but page one wont show its page number any ideas?

also, while trying to do \ begin{center} and \end{center} on my tag labeled

\section*{ {\textbf{Practice Problems}}}

results in this untitled.tex:186L latex error:somethings wrong- perhaps a missed placed \item

i have center tags around other things.. just weird

[gbux]

sorry forgot the code, but upon further investigation, page one is on the bottom of the title page, then the next page has no number, then the page after that is 2

oh yea and heres this

Code: Select all

\documentclass[11pt]{article}
\usepackage{amsmath}
\usepackage{geometry}                % See geometry.pdf to learn the layout options. There are lots.
\geometry{letterpaper}                   % ... or a4paper or a5paper or ... 
%\geometry{landscape}                % Activate for for rotated page geometry
%\usepackage[parfill]{parskip}    % Activate to begin paragraphs with an empty line rather than an indent
\usepackage{graphicx}
\usepackage{amssymb}
\usepackage{epstopdf}
\usepackage{setspace} 
\DeclareGraphicsRule{.tif}{png}{.png}{`convert #1 `dirname #1`/`basename #1 .tif`.png}
\huge
\title{{\Huge Chapter 2 Study Guide}}
\large
\author{gbux}
%\date{}                                           % Activate to display a given date or no date
\begin{document}
\maketitle
\begin{titlepage}

\end{titlepage}

\doublespace

\pagestyle{myheadings}

%\subsection{}\
\begin{center} 
\section*{\textbf{Theorems, Lemmas, and Axioms}}
\end{center}
\section*{\textbf{Section 2.3}}

 \pagestyle{myheadings}
 
 \textbf{Theorem 2.3.1-}

 P($A^{C}$)=1-P(A)
 
 \textbf{Theorem 2.3.2-}
 
 P($\varnothing$)=0
 
 \textbf{Theorem 2.3.3}

 If A$\subset$B, then P(A)$\leq$P(B).
 
\textbf{Theorem 2.3.4}
 
 For any Event A, P(A)$\leq$1
 
 \textbf{Theorem 2.3.5}
 
 Let $A_1$, $A_2$,.....,$A_N$ be events defined over S. If $A_i$$\cap$$A_j$=$\varnothing$ for i$ \not=$ then,
 
 P($\displaystyle\bigcup\limits_{i=1}^n A_i $)=$\displaystyle\sum\limits_{i=1}^n P(A_i) $
 
 \textbf{Theorem 2.3.6}
 
 P(A$\cup$B)=P(A) + P(B) - P(A$\cap$B)
   
  \section*{\textbf{Section 2.4}}
 
 \textbf{Equation 2.4.}
  \pagestyle{myheadings}
 \doublespace
 
  P(A$\mid$B)=$\frac {P(A\cap B)}{P(B)}$
 
 \textbf{2.4.3- Applying Conditional Probability to Higher-Order Intersections}
 
 P(A $\cap$ B$\cap$ C)=P(C$\mid$A$\cap$B$\cap$)P(A$\mid$B)P(A)
 
 leading to
 
 P($A_1\cap A_2\cap ...\cap A_n$)=P($A_n\mid A_1\cap A_2\cap ...\cap A_{n-1}$)P($A_n\mid 
 A_1\cap A_2\cap ...\cap A_{n-2}$)...      P($A_1\cap A_2$)P(A) 
 
\textbf{Theorem 2.4.1}
 
 B=($B\cap A_1)\cup (B\cap A_2)\cup ...\cup (B\cap A_n$)
 
 P(B)=P($B\cap A_1)+P(B\cap A_2)+...+P(B\cap A_n$)
 
 \textbf{Theorem 2.4.2}
 
 P($A_j\mid B$)=$\frac{P(B\mid A)P(A_j)}{P(B)}$


\section*{\textbf{Section 2.5}}
 \pagestyle{myheadings}
 
 \textbf{Deducing Independence}
 

 
 If P($A\cap B$)=P(A)P(B) then A and B are independent
 

\section*{\textbf{Section 2.6}}
 \pagestyle{myheadings}


\textbf{Multiplication Rule in Cominatorics}

 
 
 If operation A can be performed in m different ways and operation B can be performed in n different ways, the sequence (operation A, operation B) can be performed in m$\bullet$n different ways.

 
 \textbf{Theorem 2.6.1}
 
 
 The number of permutations of length k that can be formed from a set of n distinct elements, repetitions not allowed, is denoted by the symbol ${}_nP_k$, where
 
 
${}_nP_k=n(n-1)(n-2)...(n-k+1)=\frac {n!}{(n-k)!}$


\textbf{Theorem 2.6.2}


The number of ways to arrange n objects, $n_1$ being of one kind, $n_2$ of a second kind,..., and $n_r$, of an rth kind, is 



$\frac {n!}{n_1!n_2!...n_r!}$


Where 


$\displaystyle\sum\limits_{i=1}^n n_i=n $


 \textbf{Theorem 2.6.3}


$\binom{n}{k}=\frac{n!}{k!(n-k)!}$

\section*{\textbf{Section 2.7}}
 \pagestyle{myheadings}

Finding Combinatorial Probability

This is as simple as $\frac {m}{n}$, where m is the number of outcomes in the field that you are looking for, and n is the total number outcomes

\\begin{center}
\section*{{\textbf{Definitions}}}
\end{center}

\section*{{\textbf{Section 2.2}}}

Experiment-is a procedure that 
	1) can be repeated, theoretically a million times
	2)has a well-defined set of possible outcomes.  
Sample Space-All the possible outcomes of an experiment

\section*{{\textbf{Section 2.3}}}

 Probability-the chance that something will happen
 
 Probability function- a formula used to find out the chance a given event will happen
 
\section*{ {\textbf{Section 2.4}}}
 
 Conditional probability-Probability that is revised to take into account the occurrence of other events
 
 Partitioning: $A_1, A_2, ..., A_n$ partition s if every outcome in the sample space belongs to one and only one of the $A_i$'s
 
 \section*{ {\textbf{Section 2.5}}}
 
 independence- one the occurrence (or non occurrence) of one event has no impact on another event
 
 \section*{ {\textbf{Section 2.6}}}

 Combinatorics-The branch of mathematics that studies counting, combination and permutations
 
 Permutation-Is a rearrangement of elements of a set 
 
 \section*{ {\textbf{Section 2.7}}}
 
  \section*{ {\textbf{Practice Problems}}}
 
  \section*{ {\textbf{Section 2.2}}}
 
  \section*{ {\textbf{Section 2.6}}}
  
  A chemical engineer 
  \end{document}

[Stefan Kottwitz]

\maketitle is sufficient. Remove \begin{titlepage}...\end{titlepage} because this ruins the desired page style, as it removed the page number since title pages commonly don't have page numbers. Then also the second page would be numbered.

Furthermore,

Code: Select all

\begin{center}
\section*{\textbf{Theorems, Lemmas, and Axioms}}
\end{center}

is bad because your sections are usually left aligned, not centered. I suggest make it consistently, remove the center environment. This would also not be a suitable way for getting centered headings.

You also don't need \textbf within a \section argument, as section headings are already bold. Generally, a good LaTeX document doesn't need \textbf and similar direct font commands within the text. Such commands are used for defining style macros in the preamble.

Stefan