\documentclass{article}
\usepackage[polutonikogreek,english]{babel}
\usepackage[iso-8859-7]{inputenc}

\usepackage{epigrafica}



%%%%% 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}


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\newenvironment{proof}[1]{{\textit{Απόδειξη:}}}{\ \hfill$\Box$}
\newenvironment{hint}[1]{{\textit{Υπόδειξη:}}}{\ \hfill$\Box$}
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\title{The \textsc{epigrafica} font family}
\author{Antonis Tsolomitis\\
Laboratory of Digital Typography\\ and Mathematical Software\\
Department of Mathematics\\
University of the Aegean}
\date {\textsc{27} May \textsc{2006}}


\begin{document}
\maketitle



\section{Introduction}
The Epigrafica family is a derivative work of the MgOpenCosmetica
fonts which has been made available by Magenta Ltd
(\texttt{http://www.magenta.gr}) 
under the \textsc{gpl} license.

This is the initial release of Epigrafica and supports only
monotonic Greek, and the OT1 and T1 partially. Polytonic and full OT1
and T1 support is under development. However, basic latin is supported.


 The greek part is to be used with the greek option of
the Babel package.

The fonts are loaded with

\verb|\usepackage{epigrafica}|.

The package provides a true small caps font although not provided by
the source fonts from Magenta. However, the text figures are currently
under development. In addition to this there have been several
enhancements both to glyph coverage and to some buggy splines (for
example,
in O, Q and others) 


Finally, the math symbols are taken from the pxfonts package. 



\section{Installation}

Copy the contents of the subdirectory afm in
texmf/fonts/afm/source/public/Epigrafica/

\medskip

\noindent Copy the contents of the subdirectory doc in
texmf/doc/latex/Epigrafica/

\medskip

\noindent Copy the contents of the subdirectory enc in
texmf/fonts/enc/dvips/public/Epigrafica/

\medskip

\noindent Copy the contents of the subdirectory map in
texmf/fonts/map/dvips/Epigrafica/

\medskip

\noindent Copy the contents of the subdirectory tex in
texmf/tex/latex/Epigrafica/

\medskip

\noindent Copy the contents of the subdirectory tfm in
texmf/fonts/tfm/public/Epigrafica/

\medskip

\noindent Copy the contents of the subdirectory type1 in
texmf/fonts/type1/public/Epigrafica/

\medskip

\noindent Copy the contents of the subdirectory vf in
texmf/fonts/vf/public/Epigrafica/

\medskip

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

\medskip

\noindent Map epigrafica.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 pxfonts.

\section{Usage}

As said in the introduction the package covers both english and
greek. Greek covers only monotonic for the moment. 

For example, the preample

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

will be the correct setup for articles in Greek.

\bigskip

\subsection{Transformations by \texttt{dvips}}

Other than the shapes provided by the fonts themselves, this package
provides a slanted shape
using the standard mechanism provided by dvips. 



\subsection{Euro}

Euro is also available in LGR enconding. \verb|\textgreek{\euro}|
gives \textgreek{\euro}. 


\section{Samples}

The next two pages provide samples in english 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 \textlatin{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:}
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.\\
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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