\sectiontitle{Mathematical Formulae using Plain \TeX} \label{pl-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. In order to obtain an mathematical formula or equation which is displayed on a line by itself, one places \verb?$$? before and after the formula. Thus to obtain \begin{quotation} \small The product of two first degree polynomials is a quadratic polynomial. For example, if f(x) = 3x + 7 and g(x) = x + 4 then$$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.$$The converse does not hold for polynomials over the field of real numbers. However if we consider polynomials over the complex field then every polynomial factorizes as a product of first degree polynomials, by the Fundamental Theorem of Algebra. \end{verbatim} \end{quote} Numbered equations are produced using the control sequence \verb?\eqno?. For example, if we type \begin{quote} \begin{verbatim}$$f(x)g(x) = 3x^2 + 19x +28.\eqno(15)$$\end{verbatim} \end{quote} we obtain$$f(x)g(x) = 3x^2 + 19x +28.\eqno(15)$$We obtain displayed equations with numbers on the left hand side by using \verb?\leqno? in place of \verb?\eqno?. Thus if we type \begin{quote} \begin{verbatim}$$f(x)g(x) = 3x^2 + 19x +28.\leqno(15)$$\end{verbatim} \end{quote} we obtain$$f(x)g(x) = 3x^2 + 19x +28.\leqno(15)$$\subsectiontitle{Characters in Mathematics Mode} All the characters on the keyboard have their standard meaning in mathematics mode, with the exception of the characters \begin{verbatim} #  % & ~ _ ^ \ { } ' \end{verbatim} Letters are set in italic type. In mathematics mode the character \verb?'? has a special meaning: typing \verb?f' + g''? produces f' + g''. When in mathematics mode the spaces you type between letters and other symbols do not affect the spacing of the final result, since \TeX\ determines the spacing of characters in formulae by its own internal rules. Thus \verb?x ( y + z )? and \verb?x(y+z)? both produce x ( y + z ). You can also type carriage returns where necessary in your input file (e.g., if you are typing in a complicated formula with many Greek characters and funny symbols) and this will have no effect on the final result if you are in mathematics mode. \begin{quotation} \footnotesize To obtain the characters$$\# \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}}$$} \subsectiontitle{Mathematical Symbols} There are numerous mathematical symbols that can be used in mathematics mode. These are obtained by typing an appropriate control sequence. These are listed in Appendix~\ref{pl-mthcs}. For example \verb?\neq?, \verb?\leq? and \verb?\geq? produce \neq, \leq and \geq respectively, \verb?\infty? produces \infty, \verb?\times? and \verb?\div? produce \times and \div, both \verb?\to? and \verb?\rightarrow? produce \to, \verb?\in? produces \in, \verb?\cup?, \verb?\cap?, \verb?\setminus? and \verb?\subset? produce \cup,\cap, \setminus and \subset respectively. The list seems endless. \subsectiontitle{Changing Fonts in Mathematics Mode} One can change fonts in mathematics mode in exactly the same way as when typesetting ordinary text. For instance \verb?\rm? changes to the \rm roman font, \verb?\bf? changes to the \bf boldface font and \verb?\mit? changes to the math italic font. The math italic font is automatically used in mathematics mode unless you explicitly change the font. In addition there is a calligraphic' font which is obtained using the control sequence \verb?\cal?. {\it This font can only be used for uppercase letters.} These calligraphic letters have the form$$\cal ABCDEFGHIJKLMNOPQRSTUVWXYZ.$$The following example shows how fonts are changed in an example involving mathematics. To obtain \begin{quotation} \small Let \bf u,\bf v and \bf w be three vectors in {\bf R}^3. The volume~V of the parallelepiped with corners at the points \bf 0,\bf u,\bf v, \bf w,\bf u+v,\bf u+w,\bf v+w and \bf u+v+w is given by the formula$$V = {\bf (u \times v) . w}.$$\end{quotation} one would type \begin{quote} \begin{verbatim} Let \bf u,\bf v and \bf w be three vectors in {\bf R}^3. The volume~V of the parallelepiped with corners at the points \bf 0,\bf u,\bf v, \bf w,\bf u+v,\bf u+w,\bf v+w and \bf u+v+w is given by the formula$$V = {\bf (u \times v) . w}.$$\end{verbatim} \end{quote} \subsectiontitle{Standard Functions and Embedded Text} The names of certain standard functions and abbreviations are obtained by typing a backlash \verb?\? before the name. The complete list in \TeX\ is as follows:-$$\vcenter{\halign{\backslash${\tt #}&&\quad$\backslash{\tt #}\cr arccos&cos&csc&exp&ker&limsup&min&sinh\cr arcsin&cosh°&gcd&lg&ln&Pr&sup\cr arctan&cot&det&hom&lim&log&sec&tan\cr arg&coth&dim&inf&liminf&max&sin&tanh\cr}}$$Names of functions and other abbreviations not in this list can be obtained by converting to the roman font. Thus one obtains {\rm Aut}(V) by typing \verb?{\rm Aut}(V)?. \begin{quotation} \footnotesize Note that if one were to type simply \verb?Aut(V)? one would obtain Aut(V), because \TeX\ has treated \verb?Aut? as the product of three quantities A,u and t and typeset the formula accordingly. \end{quotation} The recommended way to obtain ordinary text in displayed mathematical formulae is to use \verb?\hbox?. Thus one obtains$$M^\bot = \{ f \in V' : f(m) = 0 \hbox{ for all } m \in M \}.$$by typing \begin{quote} \begin{verbatim}$$M^\bot = \{ f \in V' : f(m) = 0 \hbox{ 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 \hbox{for all} m \in M \}.$$\end{verbatim} \end{quote} we would have obtained$$M^\bot = \{ f \in V' : f(m) = 0 \hbox{for all} m \in M \}.$$\subsectiontitle{Fractions,Roots and Ellipsis} Fractions of the form$${\hbox{\it numerator} \over \hbox{\it denominator}}$$are obtained in Plain \TeX\ using the construction \begin{quote} \verb?{?{\it numerator\verb? \over ?denominator}\verb?}?. \end{quote} For example, to obtain \begin{quotation} \small The function f is given by$$f(x) = 2x + {x - 7 \over 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 + {x - 7 \over 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}?. To produce roots of higher order in Plain \TeX\ one uses the construction \begin{quote} \verb?\root ?n\verb? \of ?{\it expression} \end{quote} to produce \root n \of{\hbox{\it expression}}. Thus typing \verb?\root 3 \of {x + 3y}? produces \root 3 \of {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\| \hbox{ 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\| \hbox{ 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 + {7x+5 \over 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 + {7x+5 \over 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 + {42 \over 1+x^4} \right) \right|.$$\end{verbatim} \end{quote} we obtain$$\left| 4 x^3 + \left( x + {42 \over 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. {du \over 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. {du \over dx} \right|_{x=0}.\end{verbatim} \end{quotation} \newskip\plaincentering \plaincentering=0pt plus 1000pt minus 1000pt \def\eqalign#1{\null\,\vcenter{\openup\jot\mathsurround=0pt \ialign{\strut\hfil\displaystyle{##}&\displaystyle{{}##}\hfil \crcr#1\crcr}}\,} \newif\ifdtp \def\mthdisplay{\global\dtptrue\openup\jot\mathsurround=0pt \everycr{\noalign{\ifdtp \global\dtpfalse \vskip-\lineskiplimit \vskip\normallineskiplimit \else \penalty\interdisplaylinepenalty \fi}}} \def\restindisp{\tabskip=0pt\everycr{}} % restore inside \mthdisplay \def\displaylines#1{\mthdisplay \halign{\hbox to\displaywidth{\restindisp\hfil\displaystyle##\hfil}\crcr #1\crcr}} \def\eqalignno#1{\mthdisplay \tabskip=\plaincentering \halign to\displaywidth{\hfil\restindisp\displaystyle{##}\tabskip=0pt &\restindisp\displaystyle{{}##}\hfil\tabskip=\plaincentering &\llap{\restindisp##}\tabskip=0pt\crcr #1\crcr}} \def\leqalignno#1{\mthdisplay \tabskip=\plaincentering \halign to\displaywidth{\hfil\restindisp\displaystyle{##}\tabskip=0pt &\restindisp\displaystyle{{}##}\hfil\tabskip=\plaincentering &\kern-\displaywidth\rlap{\restindisp##}\tabskip=\displaywidth\crcr #1\crcr}} \subsectiontitle{Multiline Formulae in Plain \TeX} Consider the problem of typesetting the formula\eqalign{\cos 2\theta &= \cos^2 \theta - \sin^2 \theta \cr &= 2 \cos^2 \theta - 1.\cr}It is necessary to ensure that the = signs are aligned with one another. The above example was obtained by typing typing the lines \begin{quote} \begin{verbatim}\eqalign{\cos 2\theta &= \cos^2 \theta - \sin^2 \theta \cr &= 2 \cos^2 \theta - 1.\cr}\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 the control sequence \verb?\cr? is placed at the end of each line 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 {1 \over 2} |\zeta - z| then|\zeta - z - h| \geq {1 \over 2} |\zeta - z|$$and hence$$\eqalign{ \left| {1 \over \zeta - z - h} - {1 \over \zeta - z} \right| & = \left| {(\zeta - z) - (\zeta - z - h) \over (\zeta - z - h)(\zeta - z)} \right| \cr & = \left| {h \over (\zeta - z - h)(\zeta - z)} \right| \cr & \leq {2 |h| \over |\zeta - z|^2}.\cr}$$\end{quotation} was obtained by typing \begin{quote} \begin{verbatim} If h \leq {1 \over 2} |\zeta - z| then$$|\zeta - z - h| \geq {1 \over 2} |\zeta - z|$$and hence$$\eqalign{ \left| {1 \over \zeta - z - h} - {1 \over \zeta - z} \right| & = \left| {(\zeta - z) - (\zeta - z - h) \over (\zeta - z - h)(\zeta - z)} \right| \cr & = \left| {h \over (\zeta - z - h)(\zeta - z)} \right| \cr & \leq {2 |h| \over |\zeta - z|^2}.\cr}\end{verbatim} \end{quote} Numbered multiline formulae are produced using the control sequence \verb?\eqalignno?. This works exactly like \verb?\eqalign?, but on each line for which you want an equation number you insert `\verb?&?{\it equation number}' immediately before the \verb?\cr?. Thus typing \begin{quote} \begin{verbatim}\eqalignno{\sin 2\theta &= 2\sin \theta \cos \theta,&(6)\cr \cos 2\theta &= \cos^2 \theta - \sin^2 \theta \cr &= 2 \cos^2 \theta - 1.&(7)\cr}$$\end{verbatim} \end{quote} produces$$\eqalignno{\sin 2\theta &= 2\sin \theta \cos \theta,&(6)\cr \cos 2\theta &= \cos^2 \theta - \sin^2 \theta \cr &= 2 \cos^2 \theta - 1.&(7)\cr}$$It is occasionally necessary to produce formulae such as$$|x| = \cases{ x &ifx \geq 0$;\cr -x &if$x < 0$.\cr}$$We use the control sequence \verb?\cases?. The above formula is obtained by typing \begin{quote} \begin{verbatim}$$|x| = \cases{ x &if$x \geq 0$;\cr -x &if$x < 0.\cr}\end{verbatim} \end{quote} Note the use of the alignment tab \verb?&?. Also note that the expression to the left of the alignment tab \verb?&? is a mathematical expression, processed in mathematics mode, whereas the expression to the right of the alignment tab \verb?&? is treated as ordinary text. Thus one must place \verb?? before and after any mathematical expression occurring to the right of the alignment tab \verb?&?. Note also the use of \verb?\cr? at the end of each line on the right hand side of the equation. \subsectiontitle{Matrices and other arrays in Plain \TeX} Matrices and other arrays are produced in Plain \TeX\ using the control sequences \verb?\matrix? and \verb?\pmatrix?. For example, suppose that we wish to typeset the following passage: \begin{quotation} \small The {\it characteristic polynomial} \chi(\lambda) of the 3 \times 3~matrix\left( \matrix{ a & b & c \cr d & e & f \cr g & h & i \cr} \right)$$is given by the formula$$\chi(\lambda) = \left| \matrix{ \lambda - a & -b & -c \cr -d & \lambda - e & -f \cr -g & -h & \lambda - i \cr} \right|.$$\end{quotation} This passage is produced by the following input: \begin{quote} \begin{verbatim} The {\it characteristic polynomial} \chi(\lambda) of the 3 \times 3~matrix$$\left( \matrix{ a & b & c \cr d & e & f \cr g & h & i \cr} \right)$$is given by the formula$$\chi(\lambda) = \left| \matrix{ \lambda - a & -b & -c \cr -d & \lambda - e & -f \cr -g & -h & \lambda - i \cr} \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?\cr? at the end of each row of the matrix, exactly as in the construction of multiline formulae described above. Since matrices delimited by parentheses are common, Plain \TeX\ provides the control sequence \verb?\pmatrix? to construct them. Thus\pmatrix{\lambda - a & -b & -c \cr -d & \lambda - e & -f \cr -g & -h & \lambda - i \cr}.$$may be obtained by typing \begin{quote} \begin{verbatim}$$\pmatrix{\lambda - a & -b & -c \cr -d & \lambda - e & -f \cr -g & -h & \lambda - i \cr}.\end{verbatim} \end{quote} Note that \verb?\pmatrix? behaves exactly like \verb?\matrix?, except that there is no need to use \verb?\left(? and \verb?\right)? to produce the parentheses around the matrix, since these are automatically produced by \verb?\pmatrix?. More complicated arrays can be produced in Plain \TeX\ with comparative ease using \verb?\halign? (see Chapter 22 of the \TeX book). \subsectiontitle{Derivatives, Limits, Sums and Integrals} The expressions{du \over dt} \hbox{ and } {d^2 u \over dx^2}$$are obtained by typing \verb?{du \over dt}? and \verb?{d^2 u \over dx^2}? respectively. The mathematical symbol \partial is produced using \verb?\partial?. Thus to obtain partial derivatives such as$${\partial u \over \partial t} \hbox{ and } {\partial^2 u \over \partial x^2}$$one types \verb?{\partial u \over \partial t}? and \verb?{\partial^2 u \over \partial x^2}? respectively. To obtain mathematical expressions such as$$\lim_{x \to +\infty} \hbox{, } \inf_{x > s} \hbox{ 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} {3x^2 +7 \over x^2 +1} = 3.$$we type \begin{quote} \begin{verbatim}$$\lim_{x \to 0} {3x^2 +7x^3 \over 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 = {1 \over 2} n (n+1).$$is obtained by typing \begin{quote} \begin{verbatim}$$\sum_{k=1}^n k^2 = {1 \over 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 {2x\,dx \over 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 {2x\,dx \over 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{\partial \psi \over \partial t} = {-\hbar^2 \over 2m} \left( {\partial^2 \over \partial x^2} + {\partial^2 \over \partial y^2} + {\partial^2 \over \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.$$A simple calculation using the Schr\"{o}dinger wave equation shows that$${d \over 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$$represents the probability that the particle is to be found within the region~V at time~t. \end{quotation} One would typeset this in Plain \TeX\ by typing \begin{quote} \begin{verbatim} In non-relativistic wave mechanics, the wave function \psi({\bf r},t) of a particle satisfies the {\it Schr\"{o}dinger Wave Equation}$$i\hbar{\partial \psi \over \partial t} = {-\hbar^2 \over 2m} \left( {\partial^2 \over \partial x^2} + {\partial^2 \over \partial y^2} + {\partial^2 \over \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.$$A simple calculation using the Schr\"{o}dinger wave equation shows that$${d \over 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}