What would be the best way to remove the numbering in the following document ("Note 1.1", "Definition 1.2"), but keep the same formatting? I've read that it has something to do with using the \newtheorem* command; however, I keep getting error messages if I add "*" to the preamble.
Thank you.
\documentclass{amsart}
\usepackage{amssymb}
\usepackage{amsthm}
\usepackage[mathscr]{euscript}
\usepackage{pgfplots}
\usepackage{tikz}
\usepackage{marginnote}
% Theorem Styles
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{corollary}[theorem]{Corollary}
% Definition Styles
\theoremstyle{definition}
\newtheorem{definition}[theorem]{Definition}%[section]
\newtheorem{example}[theorem]{Example}%[section]
\theoremstyle{remark}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{note}[theorem]{Note}
\newtheorem{question}[theorem]{Question}%[section]
\input {xy}
\xyoption{all}
\newcommand{\R}{\mathbb{R}}
%\usepackage{parskip}
\begin{document}
%+Title
\title{October 9, 2015}
\maketitle
%-Title
\section{Differentiable Functions}
The curve $ \{ (x,y) \ : \ y=f(x) \}$ where $f:\R \to \R$ can be approximated with a tangent line.
The surface $ \{ (x,y,z) \ : \ z=f(x,y) \}$ where $f:\R^2 \to \R$ can be approximated with a tangent plane.
In single-variable calculus, we defined a function $f:\R \to \R$ to be differentiable at $x_0$ if the following limit existed:
\begin{equation*}
f'(x_0)=\lim_{h \to 0} \frac{f(x_0+h)-f(x_0)}{h}
\end{equation*}
Notice that this expression represents the slope of the secant line:
\begin{center}
\pgfplotsset{
compat=1.12,
standard/.style={
axis lines=middle,
enlarge x limits=0.15,
enlarge y limits=0.15,
every axis x label/.style={at={(current axis.right of origin)},anchor=north west},
every axis y label/.style={at={(current axis.above origin)},anchor=north east},
}
}
\begin{tikzpicture}
\begin{axis}[
standard,
xlabel = $x$,
ylabel = {$y$},
xtick={\empty},
ytick={\empty},
extra x ticks={1,4},
extra x tick labels={$x_0$,$x_0+h$},
]
\addplot [
domain=0:5,
samples=100,
color=black,
]
{x^2};
\addplot [
domain=0:5,
samples=100,
color=black,
]
{5*x-4};
\end{axis}
\end{tikzpicture}
\end{center}
\begin{note}
$f(x)=\|x\|$ is not differentiable at $x=0$ because $\lim_{h \to 0}\frac{f(0+h)-f(0)}{h}$ does not exist:
\begin{gather*}
\lim_{h \to 0^+} \frac{f(0+h)-f(0)}{h}=\lim_{h \to 0^+} \frac{h}{h}=1 \\
\lim_{h \to 0^-} \frac{f(0+h)-f(0)}{h}=\lim_{h \to 0^-} \frac{h}{h}=-1
\end{gather*}
\end{note}
Suppose that $f'(x_0)$ exists. We then know that:
\begin{gather*}
\lim_{h \to 0} \frac{f(x_0+h)-f(x_0)}{h}-f'(x_0)=0 \\
\lim_{h \to 0} \frac{f(x_0+h)-f(x_0)-hf'(x_0)}{h}=0
\end{gather*}
We can use $f(x_0)+hf'(x_0)$ as a \textit{linear approximation} of $f(x_0+h)$:
\begin{center}
\begin{tikzpicture}
\begin{axis}[
% standard,
clip=true,
clip mode=individual,
restrict y to domain=0:2.3,
axis lines=middle,
xlabel = $x$,
ylabel = {$y$},
xtick={\empty},
ytick={\empty},
extra x ticks={1,3},
extra x tick labels={$x_0$,$x_0+h$},
extra y ticks={1, 1.73, 2},
extra y tick labels={$f(x_0)$, $f(x_0+h)$, $f(x_0+h)+h'(x_0+h)$},
ymax=3,
xmax=5
]
\addplot [
domain=0:4.5,
samples=100,
color=black,
]
{(1/2)*x+(1/2)} node [above,pos=1] {$f(x_0)+hf'(x_0)$};
\addplot [
domain=0.1:4,
samples=100,
color=black,
]
{sqrt(x)};
\addplot [dashed,
domain=0.1:4,
samples=2,
]
{1};
\addplot [dashed,
domain=0:4,
samples=2,
]
{sqrt(3)};
\addplot [dashed,
domain=0:4,
samples=2,
]
{2};
\addplot[mark=*] coordinates {(1,1)};
\addplot[mark=*] coordinates {(3,1.73)};
\addplot[mark=*] coordinates {(3,2)};
\end{axis}
\end{tikzpicture}
\end{center}
\marginnote{\textbf{Error term $\epsilon(h)$}}
From our graph, we can formulate an expression for the error $\epsilon(h)$ between our linear approximation and the function itself:
\begin{equation*}
\epsilon(h)=\underbrace{f(x_0+h)}_{\text{Value of function}}-\underbrace{\left(f(x_0)+hf'(x_0)\right)}_{\text{Linear Approximation}}
\end{equation*}
We propose that $f$ is differentiable at $x_0$ if $\frac{\epsilon(h)}{h}\to0.$
\medskip
$ \therefore f(x_0)+hf'(x_0)=f(x_0+h)$.
\medskip
We will use this as a guide to form our definition for a differentiable function for multiple variables.
\marginnote{\textbf{Definition of Differentiable at $x_0$}}[1.5cm]
\begin{definition} Let $f:U \to \R^n $ be a function and $ \vec{x_0} \in U$. We say $f$ is \textbf{differentiable} at $ \vec{x_0}$ if $\exists$ a linear map $L_{\vec{x_0}}:\R^m \to \R^n$ such that:
\begin{equation*}
\lim_{\vec{h} \to \vec{0}} \frac{\Vert{f(\vec{x_0}+\vec{h})-f(\vec{x_0})-L_{\vec{x_0}}(\vec{h})}\Vert}{\Vert{\vec{h}}\Vert}=0
\end{equation*}
\end{definition}
Our job to show $f$ is differentiable at $\vec{x_0}$ will be to prove that this linear map does indeed exist.
\end{document}
Best Answer
You are close except on small thing. The (optional) argument given inside
[...]in definitions like\newtheorem{proposition}[theorem]{Proposition}say that the proposition shares the same counter as theorem. When you don't use number at all this is meaningless and things get messed up. So remove all arguments inside[...]in the definitions as below:Now all of them will not get a number.