Template
Forfatter:
Zaahir Ali
Sidst opdateret:
2 år siden
Licens:
Creative Commons CC BY 4.0
Resumé:
Notes Template
\begin
Opdag hvorfor 18 millioner mennesker verden rundt stoler på Overleaf med deres arbejde.
\begin
Opdag hvorfor 18 millioner mennesker verden rundt stoler på Overleaf med deres arbejde.
\documentclass{report}
\input{preamble}
\input{macros}
\input{letterfonts}
\title{\Huge{Some Class}\\Random Examples}
\author{\huge{Your Name}}
\date{}
\begin{document}
\maketitle
\newpage% or \cleardoublepage
% \pdfbookmark[<level>]{<title>}{<dest>}
\pdfbookmark[section]{\contentsname}{toc}
\tableofcontents
\pagebreak
\chapter{}
\section{Random Examples}
\dfn{Limit of Sequence in $\bs{\bbR}$}{Let $\{s_n\}$ be a sequence in $\bbR$. We say $$\lim_{n\to\infty}s_n=s$$ where $s\in\bbR$ if $\forall$ real numbers $\eps>0$ $\exists$ natural number $N$ such that for $n>N$ $$s-\eps<s_n<s+\eps\text{ i.e. }|s-s_n|<\eps$$}
\qs{}{Is the set ${x-}$axis${\setminus\{\text{Origin}\}}$ a closed set}
\sol We have to take its complement and check whether that set is a open set i.e. if it is a union of open balls
\nt{We will do topology in Normed Linear Space (Mainly $\bbR^n$ and occasionally $\bbC^n$)using the language of Metric Space}
\clm{Topology}{}{Topology is cool}
\ex{Open Set and Close Set}{
\begin{tabular}{rl}
Open Set: & $\bullet$ $\phi$ \\
& $\bullet$ $\bigcup\limits_{x\in X}B_r(x)$ (Any $r>0$ will do) \\[3mm]
& $\bullet$ $B_r(x)$ is open \\
Closed Set: & $\bullet$ $X,\ \phi$ \\
& $\bullet$ $\overline{B_r(x)}$ \\
& $x-$axis $\cup$ $y-$axis
\end{tabular}}
\thm{}{If $x\in$ open set $V$ then $\exists$ $\delta>0$ such that $B_{\delta}(x)\subset V$}
\begin{myproof}By openness of $V$, $x\in B_r(u)\subset V$
\begin{center}
\begin{tikzpicture}
\draw[red] (0,0) circle [x radius=3.5cm, y radius=2cm] ;
\draw (3,1.6) node[red]{$V$};
\draw [blue] (1,0) circle (1.45cm) ;
\filldraw[blue] (1,0) circle (1pt) node[anchor=north]{$u$};
\draw (2.9,0.4) node[blue]{$B_r(u)$};
\draw [green!40!black] (1.7,0) circle (0.5cm) node [yshift=0.7cm]{$B_{\delta}(x)$} ;
\filldraw[green!40!black] (1.7,0) circle (1pt) node[anchor=west]{$x$};
\end{tikzpicture}
\end{center}
Given $x\in B_r(u)\subset V$, we want $\delta>0$ such that $x\in B_{\delta} (x)\subset B_r(u)\subset V$. Let $d=d(u,x)$. Choose $\delta $ such that $d+\delta<r$ (e.g. $\delta<\frac{r-d}{2}$)
If $y\in B_{\delta}(x)$ we will be done by showing that $d(u,y)<r$ but $$d(u,y)\leq d(u,x)+d(x,y)<d+\delta<r$$
\end{myproof}
\cor{}{By the result of the proof, we can then show...}
\mlenma{}{Suppose $\vec{v_1}, \dots, \vec{v_n} \in \RR[n]$ is subspace of $\RR^n$.}
\mprop{}{$1 + 1 = 2$.}
\section{Random}
\dfn{Normed Linear Space and Norm $\boldsymbol{\|\cdot\|}$}{Let $V$ be a vector space over $\bbR$ (or $\bbC$). A norm on $V$ is function $\|\cdot\|\ V\to \bbR_{\geq 0}$ satisfying \begin{enumerate}[label=\bfseries\tiny\protect\circled{\small\arabic*}]
\item \label{n:1}$\|x\|=0 \iff x=0$ $\forall$ $x\in V$
\item \label{n:2} $\|\lambda x\|=|\lambda|\|x\|$ $\forall$ $\lambda\in\bbR$(or $\bbC$), $x\in V$
\item \label{n:3} $\|x+y\| \leq \|x\|+\|y\|$ $\forall$ $x,y\in V$ (Triangle Inequality/Subadditivity)
\end{enumerate}And $V$ is called a normed linear space.
$\bullet $ Same definition works with $V$ a vector space over $\bbC$ (again $\|\cdot\|\to\bbR_{\geq 0}$) where \ref{n:2} becomes $\|\lambda x\|=|\lambda|\|x\|$ $\forall$ $\lambda\in\bbC$, $x\in V$, where for $\lambda=a+ib$, $|\lambda|=\sqrt{a^2+b^2}$ }
\ex{$\bs{p-}$Norm}{\label{pnorm}$V={\bbR}^m$, $p\in\bbR_{\geq 0}$. Define for $x=(x_1,x_2,\cdots,x_m)\in\bbR^m$ $$\|x\|_p=\Big(|x_1|^p+|x_2|^p+\cdots+|x_m|^p\Big)^{\frac1p}$$(In school $p=2$)}
\textbf{Special Case $\bs{p=1}$}: $\|x\|_1=|x_1|+|x_2|+\cdots+|x_m|$ is clearly a norm by usual triangle inequality. \par
\textbf{Special Case $\bs{p\to\infty\ (\bbR^m$ with $\|\cdot\|_{\infty})}$}: $\|x\|_{\infty}=\max\{|x_1|,|x_2|,\cdots,|x_m|\}$\\
For $m=1$ these $p-$norms are nothing but $|x|$.
Now exercise
\qs{}{\label{exs1}Prove that triangle inequality is true if $p\geq 1$ for $p-$norms. (What goes wrong for $p<1$ ?)}
\sol{\textbf{For Property \ref{n:3} for norm-2} \subsubsection*{\textbf{When field is $\bbR:$}} We have to show\begin{align*}
& \sum_i(x_i+y_i)^2\leq \left(\sqrt{\sum_ix_i^2} +\sqrt{\sum_iy_i^2}\right)^2 \\
\implies & \sum_i (x_i^2+2x_iy_i+y_i^2)\leq \sum_ix_i^2+2\sqrt{\left[\sum_ix_i^2\right]\left[\sum_iy_i^2\right]}+\sum_iy_i^2 \\
\implies & \left[\sum_ix_iy_i\right]^2\leq \left[\sum_ix_i^2\right]\left[\sum_iy_i^2\right]
\end{align*}So in other words prove $\langle x,y\rangle^2 \leq \langle x,x\rangle\langle y,y\rangle$ where
$$\langle x,y\rangle =\sum\limits_i x_iy_i$$
\begin{note}
\begin{itemize}
\item $\|x\|^2=\langle x,x\rangle$
\item $\langle x,y\rangle=\langle y,x\rangle$
\item $\langle \cdot,\cdot\rangle$ is $\bbR-$linear in each slot i.e. \begin{align*}
\langle rx+x',y\rangle=r\langle x,y\rangle+\langle x',y\rangle \text{ and similarly for second slot}
\end{align*}Here in $\langle x,y\rangle$ $x$ is in first slot and $y$ is in second slot.
\end{itemize}
\end{note}Now the statement is just the Cauchy-Schwartz Inequality. For proof $$\langle x,y\rangle^2\leq \langle x,x\rangle\langle y,y\rangle $$ expand everything of $\langle x-\lambda y,x-\lambda y\rangle$ which is going to give a quadratic equation in variable $\lambda $ \begin{align*}
\langle x-\lambda y,x-\lambda y\rangle & =\langle x,x-\lambda y\rangle-\lambda\langle y,x-\lambda y\rangle \\
& =\langle x ,x\rangle -\lambda\langle x,y\rangle -\lambda\langle y,x\rangle +\lambda^2\langle y,y\rangle \\
& =\langle x,x\rangle -2\lambda\langle x,y\rangle+\lambda^2\langle y,y\rangle
\end{align*}Now unless $x=\lambda y$ we have $\langle x-\lambda y,x-\lambda y\rangle>0$ Hence the quadratic equation has no root therefore the discriminant is greater than zero.
\subsubsection*{\textbf{When field is $\bbC:$}}Modify the definition by $$\langle x,y\rangle=\sum_i\overline{x_i}y_i$$Then we still have $\langle x,x\rangle\geq 0$}
\section{Algorithms}
\begin{algorithm}[H]
\KwIn{This is some input}
\KwOut{This is some output}
\SetAlgoLined
\SetNoFillComment
\tcc{This is a comment}
\vspace{3mm}
some code here\;
$x \leftarrow 0$\;
$y \leftarrow 0$\;
\uIf{$ x > 5$} {
x is greater than 5 \tcp*{This is also a comment}
}
\Else {
x is less than or equal to 5\;
}
\ForEach{y in 0..5} {
$y \leftarrow y + 1$\;
}
\For{$y$ in $0..5$} {
$y \leftarrow y - 1$\;
}
\While{$x > 5$} {
$x \leftarrow x - 1$\;
}
\Return Return something here\;
\caption{what}
\end{algorithm}
\end{document}