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authorNorbert Preining <norbert@preining.info>2019-09-02 13:46:59 +0900
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+\sectiontitle{Mathematical Formulae using \LaTeX}
+\label{la-math}
+\subsectiontitle{Mathematics Mode}
+In order to obtain a mathematical formula using \TeX, one must
+enter {\it mathematics mode} before the formula and leave it
+afterwards. Mathematical formulae can occur either embedded in text
+or else displayed on a separate line. When a formula occurs within
+the text of a paragraph one should place a \verb?$? sign before and
+after the formula, in order to enter and leave mathematics mode.
+Thus to obtain a sentence like
+\begin{quotation}
+\small
+Let $f$ be the function defined by $f(x) = 3x + 7$, and
+let $a$ be a positive real number.
+\end{quotation}
+one should type
+\begin{quote}
+\begin{verbatim}
+Let $f$ be the function defined by $f(x) = 3x + 7$, and
+let $a$ be a positive real number.
+\end{verbatim}
+\end{quote}
+In particular, note that even mathematical expressions consisting
+of a single character, like $f$ and $a$ in the example above, are
+placed within \verb?$? signs. This is to ensure that they are set
+in italic type, as is customary in mathematical typesetting.
+
+\begin{quotation}
+\footnotesize
+\LaTeX\ also allows you to use \verb?\(? and \verb?\)? to mark
+the beginning and the end respectively of a mathematical formula
+embedded in text. Thus
+\begin{quote}
+Let \( f \) be the function defined by \( f(x) = 3x + 7 \).
+\end{quote}
+may be produced by typing
+\begin{quote}
+\begin{verbatim}
+Let \( f \) be the function defined by \( f(x) = 3x + 7 \).
+\end{verbatim}
+\end{quote}
+However this use of \verb?\(?$\ldots$\verb?\)? is only permitted
+in \LaTeX, whereas \verb?$?$\ldots$\verb?$? is more standard, and is
+used in other dialects of \TeX, such as Plain \TeX\ and \AmSTeX.
+\end{quotation}
+
+ In order to obtain an mathematical formula or equation which
+is displayed on a line by itself, one places \verb?\[? before and
+\verb?\]? after the formula. Thus to obtain
+\begin{quotation}
+\small
+The product of two first degree polynomials is a quadratic
+polynomial. For example, if $f(x) = 3x + 7$ and $g(x) = x + 4$
+then
+\[ f(x)g(x) = 3x^2 + 19x +28. \]
+The converse does not hold for polynomials over the field of
+real numbers. However if we consider polynomials over the
+complex field then every polynomial factorizes as a product
+of first degree polynomials, by the Fundamental Theorem of Algebra.
+\end{quotation}
+one would type
+\begin{quote}
+\begin{verbatim}
+The product of two first degree polynomials is a quadratic
+polynomial. For example, if $f(x) = 3x + 7$ and $g(x) = x + 4$
+then
+\[ f(x)g(x) = 3x^2 + 19x +28. \]
+The converse does not hold for polynomials over the field of
+real numbers. However if we consider polynomials over the
+complex field then every polynomial factorizes as a product
+of first degree polynomials, by the Fundamental Theorem of Algebra.
+\end{verbatim}
+\end{quote}
+
+\begin{quotation}
+\footnotesize
+It is also possible to use \verb?$$? in place of both \verb?\[ ?
+and \verb?\]? to mark the beginning and end of a displayed
+mathematical equation. Thus
+$$f(x)g(x) = 3x^2 + 19x +28.$$
+can be produced by typing
+\begin{quote}
+\begin{verbatim}
+$$f(x)g(x) = 3x^2 + 19x +28.$$
+\end{verbatim}
+\end{quote}
+Indeed this the method of producing displayed equations in
+dialects of \TeX\ other than \LaTeX, such as Plain \TeX\ and
+\AmSTeX.
+\end{quotation}
+
+ \LaTeX\ provides facilities for the automatic numbering of
+displayed equations. If you want an numbered equation then you
+use \verb?\begin{equation}? and \verb?\end{equation}? instead
+of using \verb?\[ ? and \verb?\]? . Thus
+\begin{quote}
+\begin{verbatim}
+The product of two first degree polynomials is a quadratic
+polynomial. For example, if $f(x) = 3x + 7$ and $g(x) = x + 4$
+then
+\begin{equation}
+f(x)g(x) = 3x^2 + 19x +28.
+\end{equation}
+\end{verbatim}
+\end{quote}
+produces
+\begin{quote}
+The product of two first degree polynomials is a quadratic
+polynomial. For example, if $f(x) = 3x + 7$ and $g(x) = x + 4$
+then
+\begin{equation}
+f(x)g(x) = 3x^2 + 19x +28.
+\end{equation}
+\end{quote}
+
+\subsectiontitle{Characters in Mathematics Mode}
+All the characters on the keyboard have their standard meaning
+in mathematics mode, with the exception of the characters
+\begin{verbatim}
+ # $ % & ~ _ ^ \ { } '
+\end{verbatim}
+Letters are set in italic type. In mathematics mode the character
+\verb?'? has a special meaning: typing \verb?$f' + g''$?
+produces $f' + g''$. When in mathematics mode the spaces you type
+between letters and other symbols do not affect the spacing of
+the final result, since \TeX\ determines the spacing of characters
+in formulae by its own internal rules. Thus \verb?$x ( y + z )$?
+and \verb?$x(y+z)$? both produce $x ( y + z )$. You can
+also type carriage returns where necessary in your input file
+(e.g., if you are typing in a complicated formula with many
+Greek characters and funny symbols) and this will have no effect on
+the final result if you are in mathematics mode.
+
+\begin{quotation}
+\footnotesize
+To obtain the characters
+\[ \# \quad \$ \quad \% \quad \& \quad \_ \quad \{ \quad \} \]
+in mathematics mode, one should type
+\begin{verbatim}
+ \# \$ \% \& \_ \{ \} .
+\end{verbatim}
+To obtain $\backslash$ in mathematics mode, one may type
+\verb?\backslash?.
+\end{quotation}
+
+\subsectiontitle{Subscripts and Superscripts}
+Subscripts and superscripts are obtained using the special
+characters \verb?_? and \verb?^? respectively. Thus the
+expression $t^3 + x_1^2 - x_2$ is obtained by typing
+\verb?$t^3 + x_1^2 - x_2$?. When the subscript or superscript
+consists of more than one character then the characters involved
+should be enclosed in curly brackets. Thus to obtain the
+expression $u_{i,j}^{12}$ one would type
+{\verb?$u_{i,j}^{12}$?}.
+
+ It is immaterial whether one specifies the subscript before the
+superscript or vica versa. Thus \verb?$u_1^2$? and \verb?$u^2_1$?
+both produce $u_1^2$. However \TeX\ does not like it if you type
+\verb?$s_n_j$? since this could be interpreted either as
+$s_{n j}$ or as $s_{n_j}$. The first of these alternatives is
+obtained by typing \verb?$s_{n j}$?, the second by typing
+\verb?$s_{n_j}$?. A similar remark applies to superscripts.
+Incidentally, the second alternative illustrates the fact that
+one can obtain subscripts (or superscripts) on subscripts
+(or superscripts). However one should not go beyond this to
+try to obtain triple subscripts.
+
+\begin{quotation}
+\footnotesize
+It is sometimes necessary to obtain expressions such as
+$R_i{}^j{}_{kl}$ in which the exact positioning of the subscripts
+and superscripts is important (e.g., in papers on general relativity
+and tensor analysis). The way this is done is to include the
+`empty group' \verb?{}? at the appropriate places to enable the
+superscripts and subscripts to be aligned correctly. Thus to
+obtain $R_i{}^j{}_{kl}$ one would type
+\verb?$R_i{}^j{}_{kl}$?.
+\end{quotation}
+
+\subsectiontitle{Greek Letters}
+Greek letters are produced in mathematics mode by preceding the
+name of the letter by a backslash \verb?\?. Thus the Greek letters
+alpha~($\alpha$), pi~($\pi$) and chi~($\chi$) are obtained by
+typing \verb?\alpha?,\verb?\pi? and \verb?\chi? respectively.
+Thus the sentence
+\begin{quotation}
+\small
+The area $A$ of a circle of radius $r$ is given by the
+formula $A = \pi r^2$.
+\end{quotation}
+is obtained by typing
+\begin{quote}
+\begin{verbatim}
+The area~$A$ of a circle of radius~$r$ is given by the
+formula $A = \pi r^2$.
+\end{verbatim}
+\end{quote}
+Upper case Greek letters are obtained by making the first character
+of the name upper case. Thus $\Gamma$,$\Phi$ and $\Lambda$ are
+obtained by typing \verb?\Gamma?,\verb?\Phi? and \verb?\Lambda?.
+\begin{quotation}
+\footnotesize
+There is no special command for omicron: just use \verb?o?.
+\end{quotation}
+
+ Some Greek letters occur in variant forms. The variant forms
+are obtained by preceding the name of the Greek letter by `var'.
+The following table lists the usual form of these letters and
+the variant forms:-
+{\def\displayandname#1{\rlap{$\displaystyle\csname #1\endcsname$}%
+ \qquad {\tt \char92 #1}}
+\[ \vcenter{\halign{\displayandname{#}\hfil&&\qquad
+ \displayandname{#}\hfil\cr
+epsilon&varepsilon\cr
+theta&vartheta\cr
+pi&varpi\cr
+rho&varrho\cr
+sigma&varsigma\cr
+phi&varphi\cr}}$$}
+
+\subsectiontitle{Mathematical Symbols}
+There are numerous mathematical symbols that can be used in
+mathematics mode. These are obtained by typing an appropriate
+control sequence. These are listed in Appendix~\ref{la-mthcs}.
+For example \verb?\neq?, \verb?\leq? and \verb?\geq? produce
+$\neq$, $\leq$ and $\geq$ respectively, \verb?\infty? produces
+$\infty$, \verb?\times? and \verb?\div? produce $\times$ and
+$\div$, both \verb?\to? and \verb?\rightarrow? produce $\to$,
+\verb?\in? produces $\in$, \verb?\cup?, \verb?\cap?,
+\verb?\setminus? and \verb?\subset? produce $\cup$,$\cap$,
+$\setminus$ and $\subset$ respectively. The list seems endless.
+
+\subsectiontitle{Changing Fonts in Mathematics Mode}
+\ifx\selectfont\undefined
+One can change fonts in mathematics mode in exactly the same
+way as when typesetting ordinary text. For instance \verb?\rm?
+changes to the $\rm roman$ font, \verb?\bf? changes to the
+$\bf boldface$ font and \verb?\mit? changes to the
+$math$ $italic$ font. The $math$ $italic$ font is automatically
+used in mathematics mode unless you explicitly change the font.
+In addition there is a `calligraphic' font which is obtained using
+the control sequence \verb?\cal?. {\it This font can only be used
+for uppercase letters.} These calligraphic letters have the form
+\else
+The $math$ $italic$ font is automatically
+used in mathematics mode unless you explicitly change the font.
+The rules for changing the font in mathematics mode are rather different
+to those applying when typesetting ordinary text.
+Firstly, any change only applies to the single character
+or symbol that follows.
+Secondly, to change a character to the
+$\mathrm{r}\mathrm{o}\mathrm{m}\mathrm{a}\mathrm{n}$ font,
+the control sequence \verb?\mathrm? must be used
+(rather than \verb?\rm?).
+Thirdly to change a character to the
+$\boldsymbol{b}\boldsymbol{o}\boldsymbol{l}\boldsymbol{d}%
+\boldsymbol{f}\boldsymbol{a}\boldsymbol{c}\boldsymbol{e}$
+font the control sequence \verb?\boldsymbol? must be used,
+{\em and in addition the \verb?amsbsy? style must be included
+in the \verb?\documentstyle? heading}.
+
+There is also a `calligraphic' font available in mathematics mode.
+This is obtained using the control sequence \verb?\cal?.
+{\it This font can only be used for uppercase letters.}
+These calligraphic letters have the form
+\fi
+\[ \cal{A}\cal{B}\cal{C}\cal{D}\cal{E}\cal{F}\cal{G}\cal{H}\cal{I}
+\cal{J}\cal{K}\cal{L}\cal{M}\cal{N}\cal{O}\cal{P}\cal{Q}\cal{R}
+\cal{S}\cal{T}\cal{U}\cal{V}\cal{W}\cal{X}\cal{Y}\cal{Z}. \]
+
+ The following example shows how fonts are changed in an
+example involving mathematics. To obtain
+\ifx\selectfont\undefined
+\begin{quotation}
+\small
+Let $\bf u$,$\bf v$ and $\bf w$ be three vectors in
+${\bf R}^3$. The volume~$V$ of the parallelepiped with
+corners at the points $\bf 0$,$\bf u$,$\bf v$,
+$\bf w$,${\bf u}+{\bf v}$,
+${\bf u}+{\bf w}$,${\bf v}+{\bf w}$
+and ${\bf u}+{\bf v}+{\bf w}$
+is given by the formula
+\[ V = ({\bf u} \times {\bf v}) \cdot {\bf w}. \]
+\end{quotation}
+one would type
+\begin{quote}
+\begin{verbatim}
+Let $\bf u$,$\bf v$ and $\bf w$ be three vectors
+in ${\bf R}^3$. The volume~$V$ of the parallelepiped with corners
+at the points $\bf 0$,$\bf u$,$\bf v$,
+$\bf w$,${\bf u}+{\bf v}$,
+${\bf u}+{\bf w}$,${\bf v}+{\bf w}$
+and ${\bf u}+{\bf v}+{\bf w}$
+is given by the formula
+\[ V = ({\bf u} \times {\bf v}) . {\bf w}. \]
+\end{verbatim}
+\end{quote}
+\else
+\begin{quotation}
+\small
+Let $\boldsymbol u$,$\boldsymbol v$ and $\boldsymbol w$ be three vectors in
+${\boldsymbol R}^3$. The volume~$V$ of the parallelepiped with
+corners at the points $\boldsymbol{0}$,$\boldsymbol{u}$,$\boldsymbol{v}$,
+$\boldsymbol{w}$,$\boldsymbol{u}+\boldsymbol{v}$,
+$\boldsymbol{u}+\boldsymbol{w}$,$\boldsymbol{v}+\boldsymbol{w}$
+and $\boldsymbol{u}+\boldsymbol{v}+\boldsymbol{w}$
+is given by the formula
+\[ V = (\boldsymbol{u} \times \boldsymbol{v}) \cdot \boldsymbol{w}. \]
+\end{quotation}
+one would type
+\begin{quote}
+\begin{verbatim}
+Let $\boldsymbol u$,$\boldsymbol v$ and $\boldsymbol w$ be three vectors
+in ${\boldsymbol R}^3$. The volume~$V$ of the parallelepiped with corners
+at the points $\boldsymbol{0}$,$\boldsymbol{u}$,$\boldsymbol{v}$,
+$\boldsymbol{w}$,$\boldsymbol{u}+\boldsymbol{v}$,
+$\boldsymbol{u}+\boldsymbol{w}$,$\boldsymbol{v}+\boldsymbol{w}$
+and $\boldsymbol{u}+\boldsymbol{v}+\boldsymbol{w}$
+is given by the formula
+\[ V = (\boldsymbol{u} \times \boldsymbol{v}) . \boldsymbol{w}. \]
+\end{verbatim}
+\end{quote}
+\fi
+
+\subsectiontitle{Standard Functions and Embedded Text}
+The names of certain standard functions and abbreviations are
+obtained by typing a backlash \verb?\? before the name. The
+complete list in \TeX\ is as follows:-
+
+\[ \vcenter{\halign{$\backslash${\tt #}&&\quad $\backslash${\tt #}\cr
+arccos&cos&csc&exp&ker&limsup&min&sinh\cr
+arcsin&cosh&deg&gcd&lg&ln&Pr&sup\cr
+arctan&cot&det&hom&lim&log&sec&tan\cr
+arg&coth&dim&inf&liminf&max&sin&tanh\cr}} \]
+
+ Names of functions and other abbreviations not in this list can be
+obtained by converting to the roman font. Thus one obtains
+${\rm Aut}(V)$ by typing \verb?${\rm Aut}(V)$?.
+\begin{quotation}
+\footnotesize
+Note that if one were to type simply \verb?$Aut(V)$? one
+would obtain $Aut(V)$, because \TeX\ has treated
+\verb?Aut? as the product of three quantities $A$,$u$ and $t$ and
+typeset the formula accordingly.
+\end{quotation}
+
+The recommended way to obtain ordinary text in displayed mathematical
+formulae is to use \verb?\mbox?. Thus one obtains
+\[ M^\bot = \{ f \in V' : f(m) = 0 \mbox{ for all } m \in M \}. \]
+by typing
+\begin{quote}
+\begin{verbatim}
+\[ M^\bot = \{ f \in V' : f(m) = 0 \mbox{ for all } m \in M \}. \]
+\end{verbatim}
+\end{quote}
+Note the blank spaces before and after the words `for all' in the above
+example. Had we typed
+\begin{quote}
+\begin{verbatim}
+\[ M^\bot = \{ f \in V' : f(m) = 0 \mbox{for all} m \in M \}. \]
+\end{verbatim}
+\end{quote}
+we would have obtained
+\[ M^\bot = \{ f \in V' : f(m) = 0 \mbox{for all} m \in M \}. \]
+
+\begin{quotation}
+\footnotesize
+One can use \verb?\hbox? as an alternative to \verb?\mbox? in
+mathematical formulae. Indeed \verb?\mbox? is specific to
+\LaTeX, whereas \verb?\hbox? is used in Plain \TeX\ and in
+other dialects of \TeX. Also \verb?\hbox? and \verb?\mbox?
+only differ in their behaviour when used to begin a paragraph
+of ordinary text, and in particular behave in an identical
+manner when used in a mathematical formula.
+\end{quotation}
+
+\subsectiontitle{Fractions,Roots and Ellipsis}
+Fractions of the form
+\[ \frac{\mbox{\it numerator}}{\mbox{\it denominator}} \]
+are obtained in \LaTeX\ using the construction
+\begin{quote}
+\verb?\frac{?{\it numerator\verb?}{?denominator}\verb?}?.
+\end{quote}
+For example, to obtain
+\begin{quotation}
+\small
+The function $f$ is given by
+\[ f(x) = 2x + \frac{x - 7}{x^2 + 4} \]
+for all real numbers $x$.
+\end{quotation}
+one would type
+\begin{quote}
+\begin{verbatim}
+The function $f$ is given by
+\[ f(x) = 2x + \frac{x - 7}{x^2 + 4} \]
+for all real numbers $x$.
+\end{verbatim}
+\end{quote}
+
+ To obtain square roots one uses the control sequence
+\verb?\sqrt?. For example, $\sqrt{x^2 + y^2}$ is produced
+by typing \verb?$\sqrt{x^2 + y^2}$?. In \LaTeX, an $n$th
+root is produced using
+\begin{quote}
+\verb?\sqrt[n]{?{\it expression}\verb?}?.
+\end{quote}
+Thus $\sqrt[3]{x + 3y}$ is produced in \LaTeX\ by typing
+\verb?$\sqrt[3]{x + 3y}$?
+
+Ellipsis (three dots) is produced in mathematics mode using
+the control sequences \verb?\cdots? and \verb?\ldots?. A
+low ellipsis, such as $(x_1,x_2,\ldots ,x_n)$, is produced by
+typing
+\begin{quote}
+\begin{verbatim}
+$(x_1,x_2,\ldots ,x_n)$.
+\end{verbatim}
+\end{quote}
+A centred ellipsis, such as $x_1 + x_2 + \cdots + x_n$ is produced
+by typing
+\begin{quote}
+\begin{verbatim}
+$x_1 + x_2 + \cdots + x_n$.
+\end{verbatim}
+\end{quote}
+
+\subsectiontitle{Accents in Mathematics Mode}
+The control sequences \verb?\underline?, \verb?\overline?,
+ \verb?\hat?, \verb?\check?, \verb?\tilde?, \verb?\acute?,
+\verb?\grave?, \verb?\dot?, \verb?\ddot?, \verb?\breve?,
+\verb?\bar? and \verb?\vec? produce underlining, overlining,
+and various accents, {\it but only in mathematics mode}.
+For example, $\tilde c$ is produced by \verb?$\tilde{c}$?.
+The effect of these accents on the letter $a$ is shown in
+the table below:
+\begin{quote}
+\begin{tabular}{ll}
+\verb?$\underline{a}$? & $\underline{a}$\\
+\verb?$\overline{a}$? & $\overline{a}$\\
+\verb?$\hat{a}$? & $\hat{a}$\\
+\verb?$\check{a}$? & $\check{a}$\\
+\verb?$\tilde{a}$? & $\tilde{a}$\\
+\verb?$\acute{a}$? & $\acute{a}$\\
+\verb?$\grave{a}$? & $\grave{a}$\\
+\verb?$\dot{a}$? & $\dot{a}$\\
+\verb?$\ddot{a}$? & $\ddot{a}$\\
+\verb?$\breve{a}$? & $\breve{a}$\\
+\verb?$\bar{a}$? & $\bar{a}$\\
+\verb?$\vec{a}$? & $\vec{a}$
+\end{tabular}
+\end{quote}
+You should bear in mind that when a character is underlined in
+a mathematical manuscript then it is normally typeset in
+bold face without any underlining. Underlining is used very
+rarely in print.
+
+\begin{quotation}
+\footnotesize
+The control sequences such as \verb?\'? and \verb?\"?, used
+to produce accents in ordinary text, may not be used in
+mathematics mode.
+\end{quotation}
+
+\subsectiontitle{Brackets and Norms}
+The frequently used left delimiters include $($, $[$ and $\{$,
+which are obtained by typing \verb?(?, \verb?[? and \verb?\{?
+respectively. The corresponding right delimiters are of
+course $)$, $]$ and $\}$, obtained by typing \verb?)?,
+\verb?]? and \verb?\}?. In addition $|$ and $\|$ are used as
+both left and right delimiters, and are obtained by typing
+\verb?|? and \verb?\|? respectively. For example, we obtain
+\begin{quotation}
+\small
+Let $X$ be a Banach space and let $f \colon B \to {\bf R}$
+be a bounded linear functional on $X$. The {\it norm} of
+$f$, denoted by $\|f\|$, is defined by
+\[ \|f\| = \inf \{ K \in [0,+\infty) :
+ |f(x)| \leq K \|x\| \mbox{ for all } x \in X \}. \]
+\end{quotation}
+by typing
+\begin{quote}
+\begin{verbatim}
+Let $X$ be a Banach space and let $f \colon B \to {\bf R}$
+be a bounded linear functional on $X$. The {\it norm} of
+$f$, denoted by $\|f\|$, is defined by
+\[ \|f\| = \inf \{ K \in [0,+\infty) :
+ |f(x)| \leq K \|x\| \mbox{ for all } x \in X \}. \]
+\end{verbatim}
+\end{quote}
+
+ Larger delimiters are sometimes required which have the
+appropriate height to match the size of the subformula which
+they enclose. Consider, for instance, the problem of typesetting
+the following formula:
+\[ f(x,y,z) = 3y^2 z \left( 3 + \frac{7x+5}{1 + y^2} \right). \]
+The way to type the large parentheses is to type \verb?\left(?
+for the left parenthesis and \verb?\right)? for the right
+parenthesis, and let \TeX\ do the rest of the work for you.
+Thus the above formula was obtained by typing
+\begin{quote}
+\begin{verbatim}
+\[ f(x,y,z) = 3y^2 z \left( 3 + \frac{7x+5}{1 + y^2} \right). \]
+\end{verbatim}
+\end{quote}
+If you type a delimiter which is preceded by \verb?\left? then
+\TeX\ will search for a corresponding delimiter preceded by
+\verb?\right? and calculate the size of the delimiters required
+to enclose the intervening subformula. One is allowed to balance
+a \verb?\left(? with a \verb?\right]? (say) if one desires: there
+is no reason why the enclosing delimiters have to have the same
+shape. One may also nest pairs of delimiters within one another:
+by typing
+\begin{quote}
+\begin{verbatim}
+\[ \left| 4 x^3 + \left( x + \frac{42}{1+x^4} \right) \right|. \]
+\end{verbatim}
+\end{quote}
+we obtain
+\[ \left| 4 x^3 + \left( x + \frac{42}{1+x^4} \right) \right|. \]
+
+\begin{quotation}
+\footnotesize
+By typing \verb?\left.? and \verb?\right.? one obtains
+{\it null delimiters} which are completely invisible. Consider,
+for example, the problem of typesetting
+\[ \left. \frac{du}{dx} \right|_{x=0}. \]
+We wish to make the vertical bar big enough to match the
+derivative preceding it. To do this, we suppose that the
+derivative is enclosed by delimiters, where the left delimiter
+is invisible and the right delimiter is the vertical line.
+The invisible delimiter is produced using \verb?\left.? and thus
+the whole formula is produced by typing
+\begin{verbatim}
+\[ \left. \frac{du}{dx} \right|_{x=0}. \]
+\end{verbatim}
+\end{quotation}
+
+\subsectiontitle{Multiline Formulae in \LaTeX}
+Consider the problem of typesetting the formula
+\begin{eqnarray*}
+\cos 2\theta & = & \cos^2 \theta - \sin^2 \theta \\
+ & = & 2 \cos^2 \theta - 1.
+\end{eqnarray*}
+It is necessary to ensure that the $=$ signs are aligned with one
+another. In \LaTeX, such a formula is typeset using the
+\verb?eqnarray*? environment. The above example was obtained by
+typing the lines
+\begin{quote}
+\begin{verbatim}
+\begin{eqnarray*}
+\cos 2\theta & = & \cos^2 \theta - \sin^2 \theta \\
+ & = & 2 \cos^2 \theta - 1.
+\end{eqnarray*}
+\end{verbatim}
+\end{quote}
+Note the use of the special character \verb?&? as an {it alignment
+tab}. When the formula is typeset, the part of the second line of
+the formula beginning with an occurrence of \verb?&? will be
+placed immediately beneath that part of the first line of the
+formula which begins with the corresponding occurrence of \verb?&?.
+Also \verb?\\? is used to separate the lines of the formula.
+
+Although we have placed corresponding occurrences of \verb?&?
+beneath one another in the above example, it is not necessary to
+do this in the input file. It was done in the above example merely
+to improve the appearance (and readability) of the input file.
+
+ The more complicated example
+\begin{quotation}
+\small
+If $h \leq \frac{1}{2} |\zeta - z|$ then
+\[ |\zeta - z - h| \geq \frac{1}{2} |\zeta - z| \]
+and hence
+\begin{eqnarray*}
+\left| \frac{1}{\zeta - z - h} - \frac{1}{\zeta - z} \right|
+& = & \left|
+\frac{(\zeta - z) - (\zeta - z - h)}{(\zeta - z - h)(\zeta - z)}
+\right| \\ & = &
+\left| \frac{h}{(\zeta - z - h)(\zeta - z)} \right| \\
+ & \leq & \frac{2 |h|}{|\zeta - z|^2}.
+\end{eqnarray*}
+\end{quotation}
+was obtained by typing
+\begin{quote}
+\begin{verbatim}
+If $h \leq \frac{1}{2} |\zeta - z|$ then
+\[ |\zeta - z - h| \geq \frac{1}{2} |\zeta - z| \]
+and hence
+\begin{eqnarray*}
+\left| \frac{1}{\zeta - z - h} - \frac{1}{\zeta - z} \right|
+& = & \left|
+\frac{(\zeta - z) - (\zeta - z - h)}{(\zeta - z - h)(\zeta - z)}
+\right| \\ & = &
+\left| \frac{h}{(\zeta - z - h)(\zeta - z)} \right| \\
+ & \leq & \frac{2 |h|}{|\zeta - z|^2}.
+\end{eqnarray*}
+\end{verbatim}
+\end{quote}
+
+ The asterisk in \verb?eqnarray*? is put there to suppress the
+automatic equation numbering produced by \LaTeX. If you wish for
+an automatically numbered multiline formula, you should use
+\verb?\begin{eqnarray}? and \verb?\end{eqnarray}?.
+
+\subsectiontitle{Matrices and other arrays in \LaTeX}
+Matrices and other arrays are produced in \LaTeX\ using the
+{\bf array} environment. For example, suppose that we wish to
+typeset the following passage:
+\begin{quotation}
+\small
+The {\em characteristic polynomial} $\chi(\lambda)$ of the
+$3 \times 3$~matrix
+\[ \left( \begin{array}{ccc}
+a & b & c \\
+d & e & f \\
+g & h & i \end{array} \right) \]
+is given by the formula
+\[ \chi(\lambda) = \left| \begin{array}{ccc}
+\lambda - a & -b & -c \\
+-d & \lambda - e & -f \\
+-g & -h & \lambda - i \end{array} \right|. \]
+\end{quotation}
+This passage is produced by the following input:
+\begin{quote}
+\begin{verbatim}
+The {\em characteristic polynomial} $\chi(\lambda)$ of the
+$3 \times 3$~matrix
+\[ \left( \begin{array}{ccc}
+a & b & c \\
+d & e & f \\
+g & h & i \end{array} \right) \]
+is given by the formula
+\[ \chi(\lambda) = \left| \begin{array}{ccc}
+\lambda - a & -b & -c \\
+-d & \lambda - e & -f \\
+-g & -h & \lambda - i \end{array} \right|. \]
+\end{verbatim}
+\end{quote}
+First of all, note the use of \verb?\left? and \verb?\right?
+to produce the large delimiters around the arrays. As we have
+already seen, if we use
+$$\hbox{\verb?\left(?} \qquad \ldots \qquad
+ \hbox{\verb?\right)?}$$
+then the size of the parentheses is chosen to match the subformula
+that they enclose. Next note the use of the alignment tab
+character \verb?&? to separate the entries of the matrix and
+the use of \verb?\\? to separate the rows of the matrix, exactly
+as in the construction of multiline formulae described above.
+We begin the array with \verb?\begin{array}? and end it with
+\verb?\end{array}?. The only thing left to explain, therefore,
+is the mysterious \verb?{ccc}? which occurs immediately after
+\verb?\begin{array}?.
+ Now each of the \verb?c?'s in \verb?{ccc}? represents a
+column of the matrix and indicates that the entries of the
+column should be {\em centred}. If the \verb?c? were replaced by
+\verb?l? then the corresponding column would be typeset with
+all the entries flush {\em left}, and \verb?r? would produce a
+column with all entries flush {\em right}. Thus
+\begin{quote}
+\begin{verbatim}
+\[ \begin{array}{lcr}
+\mbox{First number} & x & 8 \\
+\mbox{Second number} & y & 15 \\
+\mbox{Sum} & x + y & 23 \\
+\mbox{Difference} & x - y & -7 \\
+\mbox{Product} & xy & 120 \end{array} \]
+\end{verbatim}
+\end{quote}
+produces
+\begin{quotation}
+\small
+\[ \begin{array}{lcr}
+\mbox{First number} & x & 8 \\
+\mbox{Second number} & y & 15 \\
+\mbox{Sum} & x + y & 23 \\
+\mbox{Difference} & x - y & -7 \\
+\mbox{Product} & xy & 120 \end{array} \]
+\end{quotation}
+
+ We can use the array environment to produce formulae such as
+\[ |x| = \left\{ \begin{array}{ll}
+ x & \mbox{if $x \geq 0$};\\
+ -x & \mbox{if $x < 0$}.\end{array} \right. \]
+Note that both columns of this array are set flush left. Thus we
+use \verb?{ll}? immediately after \verb?\begin{array}?. The large
+curly bracket is produced using \verb?\left\{?. However this
+requires a corresponding \verb?\right? delimiter to match it.
+We therefore use the {\em null delimiter} \verb?\right.?
+discussed earlier. This delimiter is invisible. We can
+therefore obtain the above formula by typing
+\begin{quote}
+\begin{verbatim}
+\[ |x| = \left{ \begin{array}{ll}
+ x & \mbox{if $x \geq 0$};\\
+ -x & \mbox{if $x < 0$}.\end{array} \right. \]
+\end{verbatim}
+\end{quote}
+
+\subsectiontitle{Derivatives, Limits, Sums and Integrals}
+The expressions
+\[ \frac{du}{dt} \mbox{ and } \frac{d^2 u}{dx^2} \]
+are obtained in \LaTeX\ by typing \verb?\frac{du}{dt}?
+and \verb?\frac{d^2 u dx^2}? respectively. The mathematical
+symbol $\partial$ is produced using \verb?\partial?. Thus
+partial derivatives such as
+$\displaystyle\frac{\partial u}{\partial t}$ and
+$\displaystyle\frac{\partial^2 u}{\partial x^2}$ are obtained
+in \LaTeX\ by typing
+\begin{quote}
+\verb?\frac{\partial u}{\partial t}? and
+\verb?\frac{\partial^2 u}{\partial x^2}?.
+\end{quote}
+
+To obtain mathematical expressions such as
+\[ \lim_{x \to +\infty} \mbox{, } \inf_{x > s} \mbox{ and } \sup_K \]
+in displayed equations we type \verb?\lim_{x \to +\infty}?,
+\verb?\inf_{x > s}? and \verb?\sup_K? respectively. Thus to obtain
+\[ \lim_{x \to 0} \frac{3x^2 +7}{x^2 +1} = 3. \]
+(in \LaTeX) we type
+\begin{quote}
+\begin{verbatim}
+\[ \lim_{x \to 0} \frac{3x^2 +7x^3}{x^2 +5x^4} = 3. \]
+\end{verbatim}
+\end{quote}
+
+To obtain a summation sign such as
+\[ \sum_{i=1}^{2n} \]
+we type \verb?\sum_{i=1}^{2n}?. Thus
+\[ \sum_{k=1}^n k^2 = \frac{1}{2} n (n+1). \]
+is obtained by typing
+\begin{quote}
+\begin{verbatim}
+\[ \sum_{k=1}^n k^2 = \frac{1}{2} n (n+1). \]
+\end{verbatim}
+\end{quote}
+
+ We now discuss how to obtain {\it integrals} in mathematical
+documents. A typical integral is the following:
+\[ \int_a^b f(x)\,dx. \]
+This is typeset using
+\begin{quote}
+\begin{verbatim}
+\[ \int_a^b f(x)\,dx. \]
+\end{verbatim}
+\end{quote}
+The integral sign $\int$ is typeset using the control sequence
+\verb?\int?, and the {\it limits of integration} (in this case
+$a$ and $b$) are treated as a subscript and a superscript on the
+integral sign. It remains to describe the purpose of the \verb?\,?
+occurring immediately before the \verb?dx?. This is the means of telling
+\TeX\ to put extra space before the $d$. This is necessary to
+produce the correct appearance.
+
+ Most integrals occurring in mathematical documents begin with
+an integral sign and contain one or more instances of \verb?d?
+followed by another (Latin or Greek) letter, as in $dx$, $dt$,
+and $d\theta$. To obtain the correct appearance one should put
+extra space before the $d$, using \verb?\,?. Thus
+\[ \int_0^{+\infty} x^n e^{-x} \,dx = n!. \]
+\[ \int \cos \theta \,d\theta = \sin \theta. \]
+\[ \int_{x^2 + y^2 \leq R^2} f(x,y)\,dx\,dy
+ = \int_{\theta=0}^{2\pi} \int_{r=0}^R
+ f(r\cos\theta,r\sin\theta) r\,dr\,d\theta. \]
+and
+\[ \int_0^R \frac{2x\,dx}{1+x^2} = \log(1+R^2). \]
+are obtained by typing
+\begin{quote}
+\begin{verbatim}
+\[ \int_0^{+\infty} x^n e^{-x} \,dx = n!. \]
+\end{verbatim}
+\end{quote}
+\begin{quote}
+\begin{verbatim}
+\[ \int \cos \theta \,d\theta = \sin \theta. \]
+\end{verbatim}
+\end{quote}
+\begin{quote}
+\begin{verbatim}
+\[ \int_{x^2 + y^2 \leq R^2} f(x,y)\,dx\,dy
+ = \int_{\theta=0}^{2\pi} \int_{r=0}^R
+ f(r\cos\theta,r\sin\theta) r\,dr\,d\theta. \]
+\end{verbatim}
+\end{quote}
+and
+\begin{quote}
+\begin{verbatim}
+\[ \int_0^R \frac{2x\,dx}{1+x^2} = \log(1+R^2). \]
+\end{verbatim}
+\end{quote}
+respectively.
+
+ In some multiple integrals (i.e., integrals containing more than
+one integral sign) one finds that \TeX\ puts too much space
+between the integral signs. The way to improve the appearance of
+of the integral is to use the control sequence \verb?\!? to
+remove a thin strip of unwanted space. Thus, for example, the
+multiple integral
+\[ \int_0^1 \! \int_0^1 x^2 y^2\,dx\,dy. \]
+is obtained by typing
+\begin{quote}
+\begin{verbatim}
+\[ \int_0^1 \! \int_0^1 x^2 y^2\,dx\,dy. \]
+\end{verbatim}
+\end{quote}
+Had we typed
+\begin{quote}
+\begin{verbatim}
+\[ \int_0^1 \int_0^1 x^2 y^2\,dx\,dy. \]
+\end{verbatim}
+\end{quote}
+we would have obtained
+\[ \int_0^1 \int_0^1 x^2 y^2\,dx\,dy. \]
+
+ A particularly noteworthy example comes when we are
+typesetting a multiple integral such as
+\[ \int \!\!\! \int_D f(x,y)\,dx\,dy. \]
+Here we use \verb?\!? three times to obtain suitable spacing
+between the integral signs. We typeset this integral using
+\begin{quote}
+\begin{verbatim}
+\[ \int \!\!\! \int_D f(x,y)\,dx\,dy. \]
+\end{verbatim}
+\end{quote}
+Had we typed
+\begin{quote}
+\begin{verbatim}
+\[ \int \int_D f(x,y)\,dx\,dy. \]
+\end{verbatim}
+\end{quote}
+we would have obtained
+\[ \int \int_D f(x,y)\,dx\,dy. \]
+
+ The following (reasonably complicated) passage exhibits a
+number of the features which we have been discussing:
+\begin{quotation}
+\small
+ In non-relativistic wave mechanics, the wave function
+$\psi({\bf r},t)$ of a particle satisfies the
+{\it Schr\"{o}dinger Wave Equation}
+\[ i\hbar\frac{\partial \psi}{\partial t}
+ = \frac{-\hbar^2}{2m} \left(
+ \frac{\partial^2}{\partial x^2}
+ + \frac{\partial^2}{\partial y^2}
+ + \frac{\partial^2}{\partial z^2}
+ \right) \psi + V \psi. \]
+It is customary to normalize the wave equation by
+demanding that
+\[ \int \!\!\! \int \!\!\! \int_{{\bf R}^3}
+ \left| \psi({\bf r},0) \right|^2\,dx\,dy\,dz = 1. \]
+A simple calculation using the Schr\"{o}dinger wave
+equation shows that
+\[ \frac{d}{dt} \int \!\!\! \int \!\!\! \int_{{\bf R}^3}
+ \left| \psi({\bf r},t) \right|^2\,dx\,dy\,dz = 0, \]
+and hence
+\[ \int \!\!\! \int \!\!\! \int_{{\bf R}^3}
+ \left| \psi({\bf r},t) \right|^2\,dx\,dy\,dz = 1 \]
+for all times~$t$. If we normalize the wave function in this
+way then, for any (measurable) subset~$V$ of ${\bf R}^3$ and
+time~$t$,
+\[ \int \!\!\! \int \!\!\! \int_V
+ \left| \psi({\bf r},t) \right|^2\,dx\,dy\,dz \]
+represents the probability that the particle is to be found
+within the region~$V$ at time~$t$.
+\end{quotation}
+One would typeset this in \LaTeX\ by typing
+\begin{quote}
+\begin{verbatim}
+ In non-relativistic wave mechanics, the wave function
+$\psi({\bf r},t)$ of a particle satisfies the
+{\it Schr\"{o}dinger Wave Equation}
+\[ i\hbar\frac{\partial \psi}{\partial t}
+ = \frac{-\hbar^2}{2m} \left(
+ \frac{\partial^2}{\partial x^2}
+ + \frac{\partial^2}{\partial y^2}
+ + \frac{\partial^2}{\partial z^2}
+ \right) \psi + V \psi. \]
+It is customary to normalize the wave equation by
+demanding that
+\[ \int \!\!\! \int \!\!\! \int_{{\bf R}^3}
+ \left| \psi({\bf r},0) \right|^2\,dx\,dy\,dz = 1. \]
+A simple calculation using the Schr\"{o}dinger wave
+equation shows that
+\[ \frac{d}{dt} \int \!\!\! \int \!\!\! \int_{{\bf R}^3}
+ \left| \psi({\bf r},t) \right|^2\,dx\,dy\,dz = 0, \]
+and hence
+\[ \int \!\!\! \int \!\!\! \int_{{\bf R}^3}
+ \left| \psi({\bf r},t) \right|^2\,dx\,dy\,dz = 1 \]
+for all times~$t$. If we normalize the wave function in this
+way then, for any (measurable) subset~$V$ of ${\bf R}^3$ and
+time~$t$,
+\[ \int \!\!\! \int \!\!\! \int_V
+ \left| \psi({\bf r},t) \right|^2\,dx\,dy\,dz \]
+represents the probability that the particle is to be found
+within the region~$V$ at time~$t$.
+\end{verbatim}
+\end{quote}
+