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authorKarl Berry <karl@freefriends.org>2019-11-21 21:48:21 +0000
committerKarl Berry <karl@freefriends.org>2019-11-21 21:48:21 +0000
commit396174ed6ea505d76b57ac8aba822b30ad934685 (patch)
tree87d482ca0cf9ecc051bf255ea07a301ddb2ab431 /Master/texmf-dist/doc/fonts/newcomputermodern
parent74f3b3ab5f722bd2fe871a67387802bf821aaf6c (diff)
newcomputermodern (21nov19)
git-svn-id: svn://tug.org/texlive/trunk@52877 c570f23f-e606-0410-a88d-b1316a301751
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+
+ New Computer Modern Fonts
+ (Version 1.001, GustFLv1 or later)
+
+ Antonis Tsolomitis
+ Samos, Greece, 2019
+
+This is the first release of a new assembly of ComputerModern
+fonts plus glyphs for non latin alphabets which are considered
+compatible in style to CM fonts.
+
+The fonts contain Latin and accented latin letters and combinations,
+Greek (monotonic and polytonic) and Cyrillic.
+
+Curently they are distributed only at 10pt-size but more sizes
+will be available in the near future. However, priority will be
+given to the completeness in language support to the 10pt-size,
+as the view is to have a font that works by default with
+a broad set of languages.
+
+They differ from Latin Modern in that they contain Greek and Cyrillic.
+
+They differ from cm-unicode in that the latin part is more extended,
+the are adjusted to work easier for users of the Greek language
+(Small Caps and other Lookups should work in all scripts).
+
+They differ from both LatinModern and cm-unicode in that the BoldSans
+font is properly designed and they are not just the stroke extension
+of the SansRegular. This currently covers Latin but soon it will
+cover Greek and Cyrillic. It also differs in that the kerning table
+for Greek is much improved.
+
+NewComputerModern will try to be a language-complete font (as much as
+this is possible). So if you can provide information about glyphs for
+any language not covered at this point, I will gladly add the glyphs.
+Please notice that I will not design new glyphs from scratch. I will
+gladly correct faulty splines or compose a character from existing
+ones (such as accented) but the suggested to be added glyphs, must be
+available from some source. For example, I know that a compatible
+to cm font for Hebrew exists but I do not have any information more
+than that.
+
+As far as the NewCMMath font is concerned, this is the same with
+lm-math with BoldSans glyphs improved. So if you do not need BoldSans,
+please use lm-math. The credit belongs to the developers of lm-math.
+Work in math fonts is very very demanding. I do not want to take
+any credit that belongs to others. However, more imrovements will follow.
+
+Antonis Tsolomitis
+Professor
+University of the Aegean
+Department of Mathematics
+Samos, Greece
+atsol@aegean.gr
+
diff --git a/Master/texmf-dist/doc/fonts/newcomputermodern/newcomputermodern-sample.pdf b/Master/texmf-dist/doc/fonts/newcomputermodern/newcomputermodern-sample.pdf
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diff --git a/Master/texmf-dist/doc/fonts/newcomputermodern/newcomputermodern-sample.tex b/Master/texmf-dist/doc/fonts/newcomputermodern/newcomputermodern-sample.tex
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+\documentclass{article}
+\pagestyle{empty}
+\usepackage{xgreek,graphicx}
+
+\usepackage{fontspec}
+\usepackage{unicode-math}
+
+\RequirePackage{fontspec}
+\RequirePackage{unicode-math}
+\setmainfont[%
+ItalicFont=NewCM10-Italic.otf,%
+BoldFont=NewCM10-Bold.otf,%
+BoldItalicFont=NewCM10-BoldItalic.otf,%
+SmallCapsFeatures={Numbers=OldStyle}]{NewCM10-Regular.otf}
+
+\setsansfont[%
+ItalicFont=NewCMSans10-Oblique.otf,%
+BoldFont=NewCMSans10-Bold.otf,%
+BoldItalicFont=NewCMSans10-BoldOblique.otf,%
+SmallCapsFeatures={Numbers=OldStyle}]{NewCMSans10-Regular.otf}
+
+\setmonofont[ItalicFont=NewCMMono10-Italic.otf,%
+BoldFont=NewCMMono10-Bold.otf,%
+BoldItalicFont=NewCMMono10-BoldOblique.otf,%
+SmallCapsFeatures={Numbers=OldStyle}]{NewCMMono10-Regular.otf}
+
+\setmathfont{NewCMMath-Regular.otf}
+
+\newcommand{\tttextsc}[1]{{\ttscshape#1}}
+
+\newtheorem{theorem}{Theorem}
+\newtheorem{theoremg}[theorem]{Θεώρημα}
+\newtheorem{theoremr}[theorem]{теорема}
+
+\begin{document}
+
+\begin{theorem}[Dominated convergence of Lebesgue]
+Assume that $g$ is an
+in\-te\-grable func\-tion defined on the measurable set $E$ and hat
+ $(f_n)_{n\in\mathbb N}$ is a sequence of mea\-sur\-able function so that
+ $|f_n|\leq g$. If $f$ is a function so that $f_n\to f$ almost everywhere
+ then $$\lim_{n\to\infty}\int f_n=\int f.$$
+\end{theorem}
+\textsc{Proof}: The function $g-f_n$ is non-negative and thus from Fatou lemma
+we have that $\int(g-f)\leq\liminf\int(g-f_n)$. Since $|f|\leq g$ and
+$|f_n|\leq g$ the functions $f$ and $f_n$ are integrable and we have
+$$\int g-\int f\leq \int g-\limsup\int f_n,$$ so
+$$\int f\geq \limsup \int f_n.$$
+
+\begin{theoremg}[Κυριαρχημένης σύγκλισης του Lebesgue]
+ Έστω ότι
+η $g$ είναι μια ολοκληρώσιμη συνάρτηση ορισμένη στο μετρήσιμο σύνολο
+$E$ και η $(f_n)_{n\in\mathbb N}$ είναι μια ακολουθία μετρήσιμων συναρτήσεων ώστε
+$|f_n| ≤ g$. Υποθέτουμε ότι υπάρχει μια συνάρτηση $f$
+ώστε η $(f_n)_{n\in\mathbb N}$ να
+τείνει στην $f$ σχεδόν παντού. Τότε
+$$\lim \int f_n =\int f.$$
+\end{theoremg}
+\textsc{Απόδειξη}: Η συνάρτηση $g − f_n$ είναι μη αρνητική και άρα από
+το Λήμμα του Fatou ισχύει
+$\int (f-g) ≤ \liminf \int (g-f_n)$. Επειδή
+$|f| ≤ g$ και $|f_n| ≤g$ οι $f$ και $f_n$ είναι ολοκληρώσιμες, έχουμε
+$$\int g −\int f ≤ \int g − \limsup\int f_n,$$
+άρα
+$$\int f\geq \limsup \int f_n.$$
+
+(Russian translated by Google (probably erratic), hyphenation not enabled):
+
+
+\begin{theoremr}
+Предположим, что $g$ является
+интегрируемая функция, определенная на измеримом множестве $E$ и
+$(f_n)_{n\in\mathbb N}$ представляет собой последовательность измеримой функции, так что
+ $|f_n|\leq g$. Если $f$ является функцией, так что $f_n\to f$ почти везде
+ тогда
+$$\lim \int f_n =\int f.$$
+\end{theoremr}
+
+\newpage
+
+\sffamily
+
+\begin{theorem}[Dominated convergence of Lebesgue]
+Assume that $g$ is an
+in\-te\-grable func\-tion defined on the measurable set $E$ and hat
+ $(f_n)_{n\in\mathbb N}$ is a sequence of mea\-sur\-able function so that
+ $|f_n|\leq g$. If $f$ is a function so that $f_n\to f$ almost everywhere
+ then $$\lim_{n\to\infty}\int f_n=\int f.$$
+\end{theorem}
+\textsc{Proof}: The function $g-f_n$ is non-negative and thus from Fatou lemma
+we have that $\int(g-f)\leq\liminf\int(g-f_n)$. Since $|f|\leq g$ and
+$|f_n|\leq g$ the functions $f$ and $f_n$ are integrable and we have
+$$\int g-\int f\leq \int g-\limsup\int f_n,$$ so
+$$\int f\geq \limsup \int f_n.$$
+
+\begin{theoremg}[Κυριαρχημένης σύγκλισης του Lebesgue]
+ Έστω ότι
+η $g$ είναι μια ολοκληρώσιμη συνάρτηση ορισμένη στο μετρήσιμο σύνολο
+$E$ και η $(f_n)_{n\in\mathbb N}$ είναι μια ακολουθία μετρήσιμων συναρτήσεων ώστε
+$|f_n| ≤ g$. Υποθέτουμε ότι υπάρχει μια συνάρτηση $f$
+ώστε η $(f_n)_{n\in\mathbb N}$ να
+τείνει στην $f$ σχεδόν παντού. Τότε
+$$\lim \int f_n =\int f.$$
+\end{theoremg}
+\textsc{Απόδειξη}: Η συνάρτηση $g − f_n$ είναι μη αρνητική και άρα από
+το Λήμμα του Fatou ισχύει
+$\int (f-g) ≤ \liminf \int (g-f_n)$. Επειδή
+$|f| ≤ g$ και $|f_n| ≤g$ οι $f$ και $f_n$ είναι ολοκληρώσιμες, έχουμε
+$$\int g −\int f ≤ \int g − \limsup\int f_n,$$
+άρα
+$$\int f\geq \limsup \int f_n.$$
+
+(Russian translated by Google (probably erratic), hyphenation not enabled):
+
+
+\begin{theoremr}
+Предположим, что $g$ является
+интегрируемая функция, определенная на измеримом множестве $E$ и
+$(f_n)_{n\in\mathbb N}$ представляет собой последовательность измеримой функции, так что
+ $|f_n|\leq g$. Если $f$ является функцией, так что $f_n\to f$ почти везде
+ тогда
+$$\lim \int f_n =\int f.$$
+\end{theoremr}
+
+\newpage
+
+\ttfamily
+
+\begin{theorem}[Dominated convergence of Lebesgue]
+Assume that $g$ is an
+in\-te\-grable func\-tion defined on the measurable set $E$ and hat
+ $(f_n)_{n\in\mathbb N}$ is a sequence of mea\-sur\-able function so that
+ $|f_n|\leq g$. If $f$ is a function so that $f_n\to f$ almost everywhere
+ then $$\lim_{n\to\infty}\int f_n=\int f.$$
+\end{theorem}
+\textsc{Proof}: The function $g-f_n$ is non-negative and thus from Fatou lemma
+we have that $\int(g-f)\leq\liminf\int(g-f_n)$. Since $|f|\leq g$ and
+$|f_n|\leq g$ the functions $f$ and $f_n$ are integrable and we have
+$$\int g-\int f\leq \int g-\limsup\int f_n,$$ so
+$$\int f\geq \limsup \int f_n.$$
+
+\begin{theoremg}[Κυριαρχημένης σύγκλισης του Lebesgue]
+ Έστω ότι
+η $g$ είναι μια ολοκληρώσιμη συνάρτηση ορισμένη στο μετρήσιμο σύνολο
+$E$ και η $(f_n)_{n\in\mathbb N}$ είναι μια ακολουθία μετρήσιμων συναρτήσεων ώστε
+$|f_n| ≤ g$. Υποθέτουμε ότι υπάρχει μια συνάρτηση $f$
+ώστε η $(f_n)_{n\in\mathbb N}$ να
+τείνει στην $f$ σχεδόν παντού. Τότε
+$$\lim \int f_n =\int f.$$
+\end{theoremg}
+\textsc{Απόδειξη}: Η συνάρτηση $g − f_n$ είναι μη αρνητική και άρα από
+το Λήμμα του Fatou ισχύει
+$\int (f-g) ≤ \liminf \int (g-f_n)$. Επειδή
+$|f| ≤ g$ και $|f_n| ≤g$ οι $f$ και $f_n$ είναι ολοκληρώσιμες, έχουμε
+$$\int g −\int f ≤ \int g − \limsup\int f_n,$$
+άρα
+$$\int f\geq \limsup \int f_n.$$
+
+(Russian translated by Google (probably erratic), hyphenation not enabled):
+
+
+\begin{theoremr}
+Предположим, что $g$ является
+интегрируемая функция, определенная на измеримом множестве $E$ и
+$(f_n)_{n\in\mathbb N}$ представляет собой последовательность измеримой функции, так что
+ $|f_n|\leq g$. Если $f$ является функцией, так что $f_n\to f$ почти везде
+ тогда
+$$\lim \int f_n =\int f.$$
+\end{theoremr}
+
+
+\end{document}