new font gfsdidot (+ olga)

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+%%  (c) copyright  2006, 2007
+%% Antonis Tsolomitis
+%% Department of Mathematics, University of the Aegean
+%%
+%%  This document can be redistributed and/or modified under the terms
+%%  of the LaTeX Project Public License Distributed from CTAN
+%%  archives in directory macros/latex/base/lppl.txt; either
+%%  version 1 of the License, or any later version.
+
+\documentclass{article}
+\usepackage[polutonikogreek,english]{babel}
+\usepackage[iso-8859-7]{inputenc}
+\usepackage{gfsdidot}
+
+%\renewcommand{\ttdefault}{hlst}
+
+%%%%% 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{gfsdidot} font family}
+\author{Antonis Tsolomitis\\
+Laboratory of Digital Typography\\ and Mathematical Software\\
+Department of Mathematics\\
+University of the Aegean}
+\date {\textsc{28} January \textsc{2006}}
+
+
+\begin{document}
+\maketitle
+
+\section{Introduction}
+The Didot family of the Greek Font Society was made available for free
+in autumn 2005. This font existed with a commercial license for many
+years before. Support for LaTeX and the babel package was prepared
+several years ago by the author and I.\ Vasilogiorgakis. With the
+free availability of the fonts I have modified the original package
+so that it reflects the changes occured in the latest releases by \textsc{gfs}.
+
+The package supports three encodings: OT1, T1 and LGR to the extend
+that the font themselves cover these. OT1 and LGR should be
+fairly complete. The greek part is to be used with the greek option of
+the Babel package.
+
+The fonts are loaded with
+
+\verb|\usepackage{gfsdidot}|.
+
+The fonts as released contain an italic version, but its greek part is
+just the roman in slanted form. To overcome this problem for the greek
+part, we use for italic another font by \textsc{gfs} called Olga.
+As far as the latin part is concerned the italic characters are taken 
+from Didot-Italic (Olga contains no latin characters).
+The package provides also a matching small caps shape for both latin
+and greek including old style numbers.
+
+Finally, the math symbols are taken from the pxfonts package except
+of course the characters that are already provided by Didot and Olga.
+The choice of pxfonts was made on the basis that the latin part of
+Didot is based on Palatino. Moreover, all Didot characters are scaled
+in the \verb|.fd| files by a factor of 1.04 in order to match the
+x-height of pxfonts.
+
+\section{Installation}
+
+Copy the contents of the subdirectory afm in
+texmf/fonts/afm/GFS/Didot/
+
+\medskip
+
+\noindent Copy the contents of the subdirectory doc in
+texmf/doc/latex/GFS/Didot/
+
+\medskip
+
+\noindent Copy the contents of the subdirectory enc in
+texmf/fonts/enc/dvips/GFS/Didot/
+
+\medskip
+
+\noindent Copy the contents of the subdirectory map in
+texmf/fonts/map/dvips/GFS/Didot/
+
+\medskip
+
+\noindent Copy the contents of the subdirectory tex in
+texmf/tex/latex/GFS/Didot/
+
+\medskip
+
+\noindent Copy the contents of the subdirectory tfm in
+texmf/fonts/tfm/GFS/Didot/
+
+\medskip
+
+\noindent Copy the contents of the subdirectory type1 in
+texmf/fonts/type1/GFS/Didot/
+
+\medskip
+
+\noindent Copy the contents of the subdirectory vf in
+texmf/fonts/vf/GFS/Didot/
+
+\medskip
+
+\noindent In your installations updmap.cfg file add the line
+
+\medskip
+
+\noindent Map gfsdidot.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 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{gfsdidot}
+\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 an upright italic shape and a slanted small caps shape
+using the standard mechanism provided by dvips. Upright italics
+are called with \verb|\uishape| and slanted small caps with \verb|\scslshape|.
+For example, the code
+\begin{verbatim}
+{\itshape italics {\uishape upright italics} {\itshape italics again}
+\textgreek{{\itshape >'αβγδζξφψ`ω| {\uishape >'αβγδζξφψ`ω|} {\itshape >'αβγδζξφψ`ω|}}}}
+
+\textsc{small caps \textgreek{πεζοκεφαλαία} 0123456789} {\scslshape
+  \textgreek{πεζοκεφαλαία 0123456789}}
+\end{verbatim}
+will give 
+
+{\itshape italics {\uishape upright italics} {\itshape italics again}
+\textgreek{{\itshape >'αβγδζξφψ`ω| {\uishape >'αβγδζξφψ`ω|} {\itshape
+      >'αβγδζξφψ`ω|}}}}
+
+\textsc{small caps \textgreek{πεζοκεφαλαία} 0123456789} {\scslshape
+  \textgreek{πεζοκεφαλαία 0123456789}}
+
+\noindent The commands \verb|\textui{}| and \verb|\textscsl{}| are also provided.
+
+
+
+
+\subsection{Tabular numbers}
+
+Tabular numbers (of fixed width) are accessed with the command
+\verb|\tabnums{}|. Compare
+
+\begin{tabular}{ll}
+\verb+|0|1|2|3|4|5|6|7|8|9|+ & |0|1|2|3|4|5|6|7|8|9|\\
+\verb+\tabnums{|0|1|2|3|4|5|6|7|8|9|}+ & \tabnums{|0|1|2|3|4|5|6|7|8|9|}
+\end{tabular}
+
+
+\subsection{Text fractions}
+
+Text fractions are composed using the lower and upper numerals
+provided by the fonts, and are
+accessed with the command \verb|\textfrac{}{}|.
+For example, \verb|\textfrac{-22}{7}| gives \textfrac{-22}{7}.
+
+Precomposed fractions are provided too by \verb|\onehalf|,
+\verb|\onethird|, etc.
+
+
+\subsection{Additional characters}
+
+\begin{center}
+\begin{tabular}{|c|c|}\hline
+\verb|\textbullet| &\textbullet \\ \hline
+\verb|\textparagraph| &\textparagraph \\ \hline
+\verb|\textparagraphalt| & \textparagraphalt\\ \hline
+\verb|\careof| & \careof\\ \hline
+\verb|\numero| & \numero\\ \hline
+\verb|\estimated| & \estimated\\ \hline
+\verb|\whitebullet| & \whitebullet\\ \hline
+\verb|\textlozenge| & \textlozenge\\ \hline
+\verb|\eurocurrency| & \eurocurrency\\ \hline
+\verb|\interrobang| & \interrobang\\ \hline
+\verb|\textdagger| & \textdagger\\ \hline
+\verb|\textdaggerdbl| & \textdaggerdbl\\ \hline
+\verb|\yencurrency| & \yencurrency\\ \hline
+\end{tabular}
+\end{center}
+
+Euro is also available in LGR enconding. \verb|\textgreek{\euro}|
+gives \textgreek{\euro}. 
+
+\subsection{Alternate characters}
+
+In the greek encoding the initial theta is chosen
+automatically. Compare: \textgreek{θάλασσα} but \textgreek{Αθηνά}.
+Other alternate characters are not chosen automatically. 
+
+Olga provides a double lambda: \lambdadbl. This can be accessed with the
+command \verb|\lambdadbl| in textmode. 
+
+For example, in LGR encoding
+
+\medskip
+
+\verb|\textit{a\lambdadbl'a kat'allhlos metasjhmatism;oc}|
+
+\smallskip
+
+gives 
+
+\smallskip
+
+\textgreek{\textit{a\lambdadbl'a  kat'allhlos metasqhmatism'oc}}.
+
+
+\section{Problems}
+
+
+ The
+accents of the capital letters should hang in the left margin when such a letter starts a
+line.  \TeX\ and \LaTeX\ do not provide the tools for such a
+feature. However, this seems to be possible with 
+\textlatin{pdf\TeX}
+As this is work in progress, please be patient\ldots
+
+
+
+
+\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 $Riemann$ 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: