epigrafica update (3oct09)

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+\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}
+
+
+%%%%% Environment ``proof''
+\newenvironment{proof}[1]{{\textit{Απόδειξη:}}}{\ \hfill$\Box$}
+\newenvironment{hint}[1]{{\textit{Υπόδειξη:}}}{\ \hfill$\Box$}
+%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
+
+
+
+
+
+
+\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}
+
+%%% Local Variables: 
+%%% mode: latex
+%%% TeX-master: t
+%%% End: