\documentclass{article}
\usepackage[polutonikogreek,english]{babel}
\usepackage[iso-8859-7]{inputenc}
\usepackage{txfontsb}
\newcommand{\uishape}{\relax}
\newcommand{\tabnums}{\relax}
\newcommand{\textfrac}[2]{\ensuremath\frac{#1}{#2}}

%%%%% Theorems and friends
\newtheorem{theorem}{Θεώρημα}[section]         
\newtheorem{lemma}[theorem]{Λήμμα}
\newtheorem{proposition}[theorem]{Πρόταση}
\newtheorem{corollary}[theorem]{Πόρισμα}
\newtheorem{definition}[theorem]{Ορισμός}
\newtheorem{remark}[theorem]{Παρατήρηση}
\newtheorem{axiom}[theorem]{Αξίωμα}
\newtheorem{exercise}[theorem]{Άσκηση}


%%%%% Environment ``proof''
\newenvironment{proof}[1]{{\textit{Απόδειξη:}}}{\ \hfill$\Box$}
\newenvironment{hint}[1]{{\textit{Υπόδειξη:}}}{\ \hfill$\Box$}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


\title{The \textsc{txfontsb} package}
\author{Antonis Tsolomitis\\
Laboratory of Digital Typography\\ and Mathematical Applications\\
Department of Mathematics\\
University of the Aegean}
\date {\textsc{20} December \textsc{2009}}


\begin{document}
\maketitle

\section{Introduction}


The txfontsb package is an extension of the txfonts package.
Mainly it adds two things:
\begin{itemize}
\item true small caps and old style numbers for the OT1 encoding
  (through the \verb|\textsc| and \verb|\scshape| commands), and an
  oblique small caps shape (through the \verb|\textscsl| and
  \verb|\scslshape| commands).
\item Greek support (LGR encoding) supporting monotonic and polytonic
  systems through the Babel package. This also includes true small
  caps for the Greek letters.
\end{itemize}

The greek shapes are based on the Free Font of \textsc{gnu}. However
since Babel composes the Greek accented characters using the ligature
mechanism, we had to modify the original GNU fonts, and this is the
reason that they have been renamed as FreeSerifb (instead of FreeSerif).

Moreover, kerning information has been added for Greek letters that was
missing from the original FreeSerif font.

The fonts are loaded with

\verb|\usepackage{txfontsb}|.

The package has two one option:
\begin{itemize}
\item the option \verb|Upsionalt|  uses an
alternative shape for the Greek capital and small capital Upsilon.
\end{itemize}


\newcommand{\textscy}[1]{{\fontfamily{txryc}\fontshape{sc}\selectfont#1}}
\font\Ualt gtimesysc6a at 10 pt
\greektext
\begin{center}
\begin{tabular}{c|cccc}
\ &\textlatin{Kerned (default)} & \textlatin{Unkerned}
&\textlatin{Kerned SC (default)} & \textlatin{Unkerned}\\\hline
\ &\ &\ &\ &\ \\[-2ex]
\textlatin{Default Upsilon} &ΑΫΛΟΣ &Α{Ϋ}ΛΟΣ &\textsc{Αϋλος} &\textsc{Α{ϋ}λος}\\
\textlatin{Upsilonalt} &{\Ualt ΑΫΛΟΣ} &{\Ualt Α{Ϋ}ΛΟΣ} &\textscy{Αϋλος} &\textscy{Α{ϋ}λος}\\
\end{tabular}
\end{center}
\latintext





\section{Installation}

Copy the contents of the subdirectory afm in
texmf/fonts/afm/GNU/FreeFont/FreeSerifb/

\medskip

\noindent Copy the contents of the subdirectory doc in
texmf/doc/latex/GNU/FreeFont/FreeSerifb/

\medskip

\noindent Copy the contents of the subdirectory enc in
texmf/fonts/enc/dvips/GNU/FreeFont/FreeSerifb/

\medskip

\noindent Copy the contents of the subdirectory map in
texmf/fonts/map/dvips/GNU/FreeFont/FreeSerifb/

\medskip

\noindent Copy the contents of the subdirectory tex in
texmf/tex/latex/GNU/FreeFont/FreeSerifb/

\medskip

\noindent Copy the contents of the subdirectory tfm in
texmf/fonts/tfm/GNU/FreeFont/FreeSerifb/

\medskip

\noindent Copy the contents of the subdirectory type1 in
texmf/fonts/type1/GNU/FreeFont/FreeSerifb/

\medskip

\noindent Copy the contents of the subdirectory vf in
texmf/fonts/vf/GNU/FreeFont/FreeSerifb/

\medskip

\noindent In your installations updmap.cfg file add the line

\medskip

\noindent Map gptimes.map

\medskip

Refresh your filename database and the map file database (for example, on Unix systems
run mktexlsr and then run the updmap script as root).

You are now ready to use the fonts provided that you have a relatively
modern installation that includes txfonts.

\section{Usage}

As said in the introduction the package covers both english (txfonts) and
greek. Greek covers polytonic too, through babel (read the
documentation
of the babel package and its greek option). 

For example, the preample

\begin{verbatim}
\documentclass{article}
\usepackage[english,greek]{babel}
\usepackage[iso-8859-7]{inputenc}
\usepackage{txfontsb}
\end{verbatim}

will be the correct setup for articles in Greek.

\bigskip

\section{Old style numbers}

Old style numbers are accesed with the \verb|\textsc| command:

\medskip

\noindent The command \verb|\textsc{0123456789}| gives \textsc{0123456789}.

\section{Samples}

The next two pages provide samples in english (just txfonts) and greek with math.


\newpage

Adding up these inequalities with respect to $i$, we get
\begin{equation} \sum c_i d_i \leq \frac1{p} +\frac1{q} =1\label{10}\end{equation}
since $\sum c_i^p =\sum d_i^q =1$.\hfill$\Box$

In the case $p=q=2$
the above inequality is also called the 
\textit{Cauchy-Schwartz inequality}.

Notice, also, that by formally defining $\left( \sum |b_k|^q\right)^{1/q}$ to be
$\sup |b_k|$ for $q=\infty$, we give sense to (9) for all 
$1\leq p\leq\infty$.


A similar inequality is true for functions instead of sequences with the sums 
being substituted by integrals.

\medskip

\textbf{Theorem} {\itshape Let $1<p<\infty$ and let $q$ be such that $1/p +1/q =1$. Then, 
for all functions $f,g$ on an interval $[a,b]$ 
such that the integrals $\int_a^b |f(t)|^p\,dt$, $\int_a^b |g(t)|^q\,dt$ and
$\int_a^b |f(t)g(t)|\,dt$ exist \textup{(}as Riemann integrals\textup{)},
we have 
\begin{equation}
\int_a^b |f(t)g(t)|\,dt\leq 
\biggl(\int_a^b |f(t)|^p\,dt\biggr)^{1/p}
\biggl(\int_a^b |g(t)|^q\,dt\biggr)^{1/q} .
\end{equation}
}

Notice that if the Riemann integral $\int_a^b f(t)g(t)\,dt$ also exists, then 
from the inequality $\left|\int_a^b f(t)g(t)\,dt\right|\leq 
\int_a^b |f(t)g(t)|\,dt$ follows that
\begin{equation}
\left|\int_a^b f(t)g(t)\,dt\right|\leq 
\biggl(\int_a^b |f(t)|^p\,dt\biggr)^{1/p}
\biggl(\int_a^b |g(t)|^q\,dt\biggr)^{1/q} .
\end{equation}

  

\textit{Proof:} Consider a partition of the interval $[a,b]$ in $n$ equal 
subintervals with endpoints
$a=x_0<x_1<\cdots<x_n=b$. Let $\Delta x=(b-a)/n$.
We have
\begin{eqnarray}
\sum_{i=1}^n |f(x_i)g(x_i)|\Delta x &\leq& 
\sum_{i=1}^n |f(x_i)g(x_i)|(\Delta x)^{\frac1{p}+\frac1{q}}\nonumber\\
&=&\sum_{i=1}^n \left(|f(x_i)|^p \Delta x\right)^{1/p} \left(|g(x_i)|^q 
\Delta x\right)^{1/q}.\label{functionalHolder1}\\ \nonumber
\end{eqnarray}

\newpage\greektext


% $\bullet$ Μήκος τόξου καμπύλης 

% \begin{proposition}\label{chap2:sec1:prop 23}
% Έστω $\gamma$ καμπύλη με παραμετρική εξίσωση $x=g(t)$, $y=f(t)$,
% $t\in [a,\,b]$ αν $g'$, $f'$ συνεχείς στο $[a,\,b]$ τότε η
% $\gamma$ έχει μήκος $S=L(\gamma)=\int_a^b \sqrt{g'(t)^2+f'(t)^2}
% dt$.
% \end{proposition}

\textbullet\ Εμβαδόν επιφάνειας από περιστροφή\\

\begin{proposition}\label{chap2:sec1:prop23-2}
Έστω $\gamma$ καμπύλη με παραμετρική εξίσωση $x=g(t)$, $y=f(t)$,
$t\in [a,\,b]$ αν $g'$, $f'$ συνεχείς στο $[a,\,b]$ τότε το
εμβαδόν από περιστροφή της $\gamma$ γύρω από τον $xx'$ δίνεται \\
$Β=2\pi\int_a^b |f(t)| \sqrt{g'(t)^2+f^{\prime}(t^2)} dt$. \\ Αν η
$\gamma$ δίνεται από την $y=f(x)$, $x\in [a,\,b]$ τότε
$Β=2\pi\int_a^b |f(t)| \sqrt{1+f'(x)^2} dx$
\end{proposition}

\textbullet\ Όγκος στερεών από περιστροφή\\ Έστω $f :
[a,\,b]\rightarrow \mathbb{R}$ συνεχής και $R=\{f, Ox,x=a,x=b\}$
είναι ο όγκος από περιστροφή του γραφήματος της $f$ γύρω από τον
$Ox$ μεταξύ των ευθειών $x=a$, και $x=b$, τότε $V=\pi\int_a^b f
(x)^2 dx$

\textbullet\ Αν $f,g : [a,\,b]\rightarrow \mathbb{R}$ και $0\leq
g(x)\leq f(x)$ τότε ο όγκος στερεού που παράγεται από περιστροφή
των γραφημάτων των $f$ και $g$, $R=\{f,g, Ox,x=a,x=b\}$ είναι \\
$V=\pi\int_a^b\{ f (x)^2-g(x)^2\} dx$.

\textbullet\ Αν $x=g(t)$, $y=f(t)$, $t=[t_1,\,t_2]$ τότε
$V=\pi\int_{t_1}^{t_2}\{ f (t)^2 g'(t)\} dt$ για $g(t_1)=a$,
$g(t_2)=b$.


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Ασκήσεις}\label{chap2:sec2}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\begin{exercise}\label{chap2:ex1}
Να εκφραστεί το παρακάτω όριο ως ολοκλήρωμα $Riemann$ κατάλ\-ληλης
συνάρτησης\\
$$\lim_{n\rightarrow\infty} \frac{1}{n}\sum_{k=1}^{n}\sqrt[n]{e^k} $$
\end{exercise}
%%%%%%%%%
\textit{Υπόδειξη:}
Πρέπει να σκεφτούμε μια συνάρτηση της οποίας γνωρίζουμε ότι υπάρχει το ολοκλήρωμα.
 Τότε παίρνουμε μια διαμέριση $P_n$ και δείχνουμε π.χ.\ ότι το $U(f,P_n)$ είναι η ζητούμενη σειρά.

\bigskip

%%%%%%%%%%%%%%
\textit{Λύση:}
Έχουμε ότι
\begin{eqnarray}\frac{1}{n}\sum_{k=1}^{n}\sqrt[n]{e^k} =
\frac{1}{n}\sqrt[n]{e}+\frac{1}{n}\sqrt[n]{e^2}+\cdots +
\frac{1}{n}\sqrt[n]{e^n}\nonumber\\
=\frac{1}{n}e^{\frac{1}{n}}+\frac{1}{n}e^{\frac{2}{n}}+\cdots+\frac{1}{n}e^{\frac{n}{n}}\nonumber
\end{eqnarray}

\end{document}
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