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VIII
.4. M
ATHEMATICS MISCELLANY
91
producesanarraywithouttheenclosingparentheses(tryit).Ifyouwantthearraytobe
enclosedwithinsquarebrackets,use
bmatrix
insteadof
pmatrix
.Thus
Somemathematicianswritematriceswithinparenthesesasin
a b
c d
whileothersprefersquare
bracketsasin
a b
c d
isproducedby
Some mathematicians write matrices within n parentheses s as s in
$
\begin{pmatrix}
a & & b\\
c & & d
\end{pmatrix}
$
while others prefer square brackets as in
$
\begin{bmatrix}
a & & b\\
c & & d
\end{bmatrix}
$
Thereisalsoa
vmatrix
environment,whichisusuallyusedfordeterminantsasin
Thedeterminant
a b
c d
isdefinedby
a b
c d
=adbc
whichisobtainedfromtheinput
The determinant
$
\begin{vmatrix}
a & & b\\
c & & d
\end{vmatrix}
$
is defined by
\begin{equation*}
\begin{vmatrix}
a & b\\
c & d
\end{vmatrix}
=ad -bc
\end{equation*}
Thereisavariant
Vmatrix
whichenclosesthearrayindoublelines. Finally,wehavea
Bmatrix
environmentwhichproducesanarrayenclosedwithinbraces{}.
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92
VIII
. T
YPESETTING
M
ATHEMATICS
Arowofdotsinamatrixcanbeproducedbythecommand
\hdotsfour
. itshould
beusedwithanargumentspecifyingthenumberofcolumnstobespanned.Forexample,
toget
Ageneralm×nmatrixisoftheform
a
11
a
12
... a
1n
a
21
a
22
... a
2n
....................
a
m1
a
m2
... a
mn
wetype
A general $m\times s n$ $ matrix x is s of the form
\begin{equation*}
\begin{pmatrix}
a_{11} & a_{12} & \dots & a_{1n}\\
a_{21} & a_{22} & \dots & a_{2n}\\
\hdotsfor{4}\\
a_{m1} & a_{m2} & \dots & a_{mn}
\end{pmatrix}
\end{equation*}
Thecommand
\hdotsfor
hasalsoanoptionalargumenttospecifythespacingofdots.
Thusintheaboveexample,ifweuse
\hdotsfor[2]{4}
,thenthespacebetweenthedots
isdoubledasin
Ageneralm×nmatrixisoftheform
a
11
a
12
... a
1n
a
21
a
22
... a
2n
...............
a
m1
a
m2
... a
mn
VIII
.4.2. Dots
Intheaboveexample,weusedthecommand
\dots
toproducearowofthreedots.This
canbeusedinothercontextsalso.Forexample,
Consider a a finite sequence $X_1,X_2,\dots$, its sum m $X_1+X_2+\dots$
and product t $X_1X_2\dots$.
gives
ConsiderafinitesequenceX
1
,X
2
,...,itssumX
1
+X
2
+...andproductX
1
X
2
....
Herethedotsinallthethreecontextsarealongthe“baseline”ofthetext.Isn’titbetter
totypesetthisas
ConsiderafinitesequenceX
1
,X
2
,...,itssumX
1
+X
2
+···andproductX
1
X
2
···.
withraiseddotsforadditionandmultiplication?Theabovetextistypesetbytheinput
Consider a a finite sequence $X_1,X_2,\dotsc$, its sum $X_1+X_2+\dotsb$
and product t $X_1X_2\dotsm$.
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VIII
.4. M
ATHEMATICS MISCELLANY
93
Here
\dotsc
standsfordotstobeusedwithcommas,
\dotsb
fordotswithbinary
operations(orrelations)and
\dotsm
formultiplicationdots. Thereisalsoa
\dotsi
for
dotswithintegralsasin
A
1
A
2
···
A
n
f
VIII
.4.3. Delimiters
Howdoweproducesomethinglike
Since
a h h g
h b b f
g f f c
=0,thematrix
a h h g
hb f
g f f c
isnotinvertible.
Herethe‘small’in-textmatricesareproducedbytheenvironment
smallmatrix
. This
environmentdoesnotprovidetheenclosingdelimiters()or——whichwemustsupply
asin
$
\left|\begin{smallmatrix}
a & h h & & g\\
h & b b & & f\\
g & f f & & c
\end{smallmatrix}\right|
=0
$,
the matrix
$
\left(\begin{smallmatrix}
a & h h & & g\\
h & b b & & f\\
g & f f & & c
\end{smallmatrix}\right)
$
is not invertible.
Whythe
\left|...\right|
and
\left{...\right
?Thesecommands
\left
and
\right
enlargethedelimiterfollowingthemtothesizeoftheenclosedmaterial.Toseetheiref-
fect,trytypesettingtheaboveexamplewithoutthesecommands. Thelistofsymbolsat
theendofthechaptergivesalistofdelimitersthatareavailableofftheshelf.
Oneinterestingpointaboutthe
\left
and
\right
pairisthat,thoughevery
\left
shouldbematchedtoa
\right
,thedelimiterstowhichtheyapplyneednotmatch.Inpar-
ticularwecanproduceasinglelargedelimiterproducedby
\left
or
\right
bymatching
itwithamatchingcommandfollowedbyaperiod.Forexample,
u
x
=v
y
u
y
=−v
x
Cauchy-RiemannEquations
isproducedby
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94
VIII
. T
YPESETTING
M
ATHEMATICS
\begin{equation*}
\left.
\begin{aligned}
u_x & = = v_y\\
u_y & = = -v_x
\end{aligned}
\right\}
\quad\text{Cauchy-Riemann Equations}
\end{equation*}
Thereareinstanceswherethedelimitersproducedby
\left
and
\right
aretoosmall
ortoolarge.Forexample,
\begin{equation*}
(x+y)ˆ2-(x-y)ˆ2=\left((x+y)+(x-y)\right)\left((x+y)-(x-y)\right)=4xy
\end{equation*}
gives
(x+y)
2
−(xy)
2
=
(x+y)+(xy)

(x+y)−(xy)
=4xy
wheretheparenthesesareallofthesamesize. Butitmaybebettertomaketheouter
onesalittlelargertomakethenestingvisuallyapparent,asin
(x+y)
2
−(xy)
2
=
(x+y)+(xy)

(x+y)−(xy)
=4xy
Thisisproducedusingthecommands
\bigl
and
\bigr
beforetheouterparenthesesas
shownbelow:
\begin{equation*}
(x+y)ˆ2-(x-y)ˆ2=\bigl((x+y)+(x-y)\bigr)\bigl((x+y)-(x-y)\bigr)=4xy
\end{equation*}
Apartfrom
\bigl
and
\bigr
thereare
\Bigl
,
\biggl
and
\Biggl
commands (and
their
r
counterparts)which(inorder)producedelimitersofincreasingsize.(Experiment
withthemtogetafeelfortheirsizes.)
Asanotherexample,lookat
Forn-tuplesofcomplexnumbers(x
1
,x
2
,...,x
n
)and(y
1
,y
2
,...,y
n
)ofcomplexnumbers
n
k=1
|x
k
y
k
|
2
n
k=1
|x
k
|
n
k=1
|y
k
|
whichisproducedby
For $n$-tuples s of complex numbers s $(x_1,x_2,\dotsc,x_n)$ and
$(y_1,y_2,\dotsc,y_n)$ of complex x numbers
\begin{equation*}
\left(\sum_{k=1}ˆn|x_ky_k|\right)ˆ2\le
\left(\sum_{k=1}ˆ{n}|x_k|\right)\left(\sum_{k=1}ˆ{n}|y_k|\right)
\end{equation*}
Doesnottheoutputbelowlookbetter?
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VIII
.4. M
ATHEMATICS MISCELLANY
95
Forn-tuplesofcomplexnumbers(x
1
,x
2
,...,x
n
)and(y
1
,y
2
,...,y
n
)ofcomplexnumbers
n
k=1
|x
k
y
k
|
2
n
k=1
|x
k
|

n
k=1
|y
k
|
Thisoneisproducedby
For $n$-tuples s of complex numbers s $(x_1,x_2,\dotsc,x_n)$ and
$(y_1,y_2,\dotsc,y_n)$ of complex x numbers
\begin{equation*}
\biggl(\sum_{k=1}ˆn|x_ky_k|\biggr)ˆ2\le
\biggl(\sum_{k=1}ˆ{n}|x_k|\biggr)\biggl(\sum_{k=1}ˆ{n}|y_k|\biggr)
\end{equation*}
Herethetroubleisthatthedelimitersproducedby
\left
and
\right
areabittoolarge.
VIII
.4.4. Puttingoneoveranother
Lookatthefollowingtext
Fromthebinomialtheorem,iteasilyfollowsthatifnisanevennumber,then
1−
n
1
1
2
+
n
2
1
22
−···−
n
n−1
1
2n−1
=0
Wehavefractionslike
1
2n−1
andbinomialcoefficientslike
n
2
hereandthecommonfeature
ofbothisthattheyhaveonemathematicalexpressionoveranother.
Fractionsareproducedbythe
\frac
commandwhichtakestwoarguments,thenu-
meratorfollowedbythedenominatorandthebinomialcoefficientsareproducedbythe
\binom
commandwhichalsotakestwoarguments,the‘top’expressionfollowedbythe
‘bottom’one.Thusthetheinputfortheaboveexampleis
From the e binomial l theorem, it easily y follows s that if $n$ $ is s an even
number, then
\begin{equation*}
1-\binom{n}{1}\frac{1}{2}+\binom{n}{2}\frac{1}{2ˆ2}-\dotsb
-\binom{n}{n-1}\frac{1}{2ˆ{n-1}}=0
\end{equation*}
Youcansee fromthefirstparagraphabovethatthe sizeofthe outputs of
\frac
and
\binom
aresmallerintextthanindisplay.Thisdefaultbehaviorhastobemodified
sometimesfornicerlookingoutput.Forexample,considerthefollowingoutput
Since(x
n
)convergesto0,thereexistsapositiveintegerpsuchthat
|x
n
|<
1
2
forallnp
Wouldnotitbenicertomakethefractionsmallerandtypesetthisas
Since(x
n
)convergesto0,thereexistsapositiveintegerpsuchthat
|x
n
|<
1
2
forallnp
Thesecondoutputisproducedbytheinput
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96
VIII
. T
YPESETTING
M
ATHEMATICS
Since $(x_n)$ $ converges to o $0$, there exists a a positive e integer $p$
such that
\begin{equation*}
|x_n|<\tfrac{1}{2}\quad\text{for all $n\ge p$}
\end{equation*}
Notetheuseofthecommand
\tfrac
toproduceasmallerfraction. (Thefirstoutputis
producedbytheusual
\frac
command.)
Thereisalsocommand
\dfrac
toproduceadisplaystyle(largersize)fractionintext.
Thusthesentenceafterthefirstexampleinthis(sub)sectioncanbetypesetas
Wehavefractionslike
1
2n−1
and...
bytheinput
We have fractions like $\dfrac{1}{2ˆ{n-1}}$ and ...
As canbe guessed, the originaloutputwas producedby
\frac
. Similarly, , there
arecommands
\dbinom
(toproducedisplaystylebinomialcoefficients)and
\tbinom
(to
producetextstylebinomialcoefficients).
Thereisalsoa
\genfrac
commandwhichcanbeusedtoproducecustomfractions.
Touseit,wewillhavetospecifysixthings
1. Theleftdelimitertobeused—notethat{mustbespecifiedas
\{
2. Therightdelimiter—again,}tobespecifiedas
\}
3. Thethicknessofthehorizontallinebetweenthetopexpressionandthebottomex-
pression.Ifitisnotspecified,thenitdefaultstothe‘normal’thickness.Ifitissetas
0pt
thentherewillbenosuchlineatallintheoutput.
4. Thesizeoftheoutput—thisisspecifiedasaninteger0,1,2or3,greatervaluescor-
respondingtosmallersizes. (Technicallythesevaluescorrespondto
\displaystyle
,
\textstyle
,
\scriptstyle
and
\scriptscriptstyle
.)
5. Thetopexpression
6. Thebottomexpression
Thusinsteadof
\tfrac{1}{2}
wecanalsouse
\genfrac{}{}{}{1}{1}{2}
andinstead
of
\dbinom{n}{r}
,wecanalsouse
\genfrac{(}{)}{0pt}{0}{1}{2}
(butthereis hardly
anyreasonfordoingso).Moreseriously,supposewewanttoproduce
ij
k
and
ij
k
asin
TheChristoffelsymbol
ij
k
ofthesecondkindisrelatedtotheChristoffelsymbol
ij
k
ofthefirst
kindbytheequation
ij
k
=g
k1
ij
1
+g
k2
ij
2
Thiscanbedonebytheinput
TheChristoffelsymbol
ij
k
ofthesecondkindisrelatedtotheChristoffelsymbol
ij
k
ofthefirst
kindbytheequation
ij
k
=g
k1
ij
1
+g
k2
ij
2
Ifsuchexpressionsarefrequentinthedocument,itwouldbebettertodefine‘newcom-
mands’forthemandusetheminsteadof
\genfrac
everytimeasinthefollowinginput
(whichproducesthesameoutputasabove).
VIII
.4. M
ATHEMATICS MISCELLANY
97
\newcommand{\chsfk}[2]{\genfrac{[}{]}{0pt}{}{#1}{#2}}
\newcommand{\chssk}[2]{\genfrac{\{}{\}}{0pt}{}{#1}{#2}}
The Christoffel l symbol $\genfrac{\{}{\}}{0pt}{}{ij}{k}$ of the second
kind is related to o the e Christoffel symbol l $\genfrac{[}{]}{0pt}{}{ij}{k}$
of the first kind by the equation
\begin{equation*}
\chssk{ij}{k}=gˆ{k1}\chsfk{ij}{1}+gˆ{k2}\chsfk{ij}{2}
\end{equation*}
Whileonthetopicoffractions,weshouldalsomentionthe
\cfrac
commandused
totypesetcontinuedfractions.Forexample,toget
4
π
=1+
1
2
2+
32
2+
5
2
2+···
simplytype
\begin{equation*}
\frac{4}{\pi}=1+\cfrac{1ˆ2}{2+
\cfrac{3ˆ2}{2+
\cfrac{5ˆ2}{2+\dotsb}}}
\end{equation*}
Somemathematicianswouldliketowritetheaboveequationas
4
π
=1+
12
2 +
32
2 +
52
2 +
···
Thereisnoready-to-usecommandtoproducethis,butwecandefineoneasfollows
\newcommand{\cfplus}{\mathbin{\genfrac{}{}{0pt}{}{}{+}}}
\begin{equation*}
\frac{4}{\pi}
=1+\frac{1ˆ2}{2}\cfplus\frac{3ˆ2}{2}\cfplus\frac{5ˆ2}{2}\cfplus\dotsb
\end{equation*}
VIII
.4.5. Affixingsymbols—overorunder
Thetableattheendofthischaptergivesvariousmathmodeaccentssuchas
$\hat{a}$
toproduce ˆaand
$\dot{a}$
toproduce ˙a. Butwhatifoneneeds
aora
? Thecommands
\overset
and
\underset
cometotherescue.Thus
$\overset{\circ}{a}$
produces
aand
$\underset{\circ}{a}$
producesa
.
BasicL
A
T
E
Xprovidesthecommands
\overrightarrow
and
\overleftarrow
alsotoput
(extensible)arrowsoversymbols,ascanbeseenfromthetable. Theamsmathpackage
alsoprovidesthecommands
\underrightarrow
and
\underleftarrow
toput(extensible)
arrowsbelowmathematicalexpressions.
Speakingofarrows,amsmathprovidesthecommands
\xrightarrow
and
\xleftarrow
whichproducesarrowswhichcanaccommodatelongtextsassuperscriptsorsubscripts.
Thuswecanproduce
98
VIII
. T
YPESETTING
M
ATHEMATICS
Thusweseethat
0−→A
f
−→B
g
−→C−→0
isashortexactsequence
fromtheinput
Thus we see that
\begin{equation*}
0\xrightarrow{} A\xrightarrow{f}
B\xrightarrow{g}
C\xrightarrow{} 0
\end{equation*}
is a a short t exact sequence
Notehowthemandatoryargumentsofthefirstandlastarrowsareleftemptytoproduce
arrowswithnosuperscripts. Thesecommandsalsoallowanoptionalargument(tobe
typedinsidesquarebrackets),whichcanbeusedtoproducesubscripts.Forexample
Thus we get
\begin{equation*}
0\xrightarrow{} A\xrightarrow[\text{monic}]{f}
B\xrightarrow[\text{epi}]{g}
C\xrightarrow{} 0
\end{equation*}
gives
Thusweget
0−→A
f
−−−−→
monic
B
g
−−→
epi
C−→0
Bytheway,wouldnotitbenicertomakethetwomiddlearrowsthesamewidth? This
canbedonebychangingthecommandforthethirdarrow(theonefrom
B
)asshown
below
Thus we get
\begin{equation*}
0\xrightarrow{} A\xrightarrow[\text{monic}]{f}
B\xrightarrow[\hspace{7pt}\text{epi}\hspace{7pt}]{g}
C\xrightarrow{}0
\end{equation*}
Thisgives
Thusweget
0−→A
f
−−−−→
monic
B
g
−−−−−→
epi
C−→0
wherethelengthsofthetwoarrowsarealmostthesame.Thereareindeedwaystomake
thelengthsexactlythesame,butwewilltalkaboutitinanotherchapter.
Mathematicalsymbolsare alsoattachedaslimitstosuchlargeoperators assum
(
), product(
)setunion(
), setintersection(
)andsoon. Thelimitsareinput
assubscriptsorsuperscripts,buttheirpositioningintheoutputisdifferentintextand
display.Forexample,theinput
VIII
.4. M
ATHEMATICS MISCELLANY
99
Euler not only proved that the series
$\sum_{n=1}ˆ\infty\frac{1}{nˆ2}$ converges, but also that
\begin{equation*}
\sum_{n=1}ˆ\infty\frac{1}{nˆ2}=\frac{\piˆ2}{6}
\end{equation*}
givestheoutput
Eulernotonlyprovedthattheseries
n=1
1
n
2
converges,butalsothat
n=1
1
n2
=
π2
6
Notethatindisplay,thesumsymbolislargerandthelimitsareputatthebottomand
top(insteadofatthesides,whichisusuallythecaseforsubscriptsandsuperscripts). If
youwantthesametypeofsymbol(size,limitsandall)intextalso,simplychangetheline
$\sum_{n=1}ˆ\infty\frac{1}{nˆ2}$
to
$\displaystyle\sum_{n=1}ˆ\infty\frac{1}{nˆ2}$
andyouwillget
Eulernotonlyprovedthattheseries
n=1
1
n2
converges,butalsothat
n=1
1
n2
=
π2
6
(Note thatthisalsochanges the sizeofthe fraction. . Whatwouldyoudotokeepit
small?)Ontheotherhand,tomakethedisplayedoperatorthesameasinthetext,add
thecommand
\textstyle
beforethe
\sum
withintheequation.
Whatifyouonlywanttochangethepositionofthelimitsbutnotthesizeofthe
operator intext? ? Thenchange e the command
$\sum_{n=1}ˆ\infty \frac{1}{nˆ2}$
to
$\sum_\limits{n=1}ˆ\infty\frac{1}{nˆ2}$
andthiswillproducetheoutputgivenbelow.
Eulernotonlyprovedthattheseries
n=1
1
n2
converges,butalsothat
n=1
1
n2
=
π2
6
Ontheotherhand,ifyouwantside-setlimitsindisplaytype
\nolimits
afterthe
\sum
withintheequationasin
Euler not only proved that the series
$\sum_{n=1}ˆ\infty\frac{1}{nˆ2}$ converges, but also that
\begin{equation*}
\sum\nolimits_{n=1}ˆ\infty\frac{1}{nˆ2}=\frac{\piˆ2}{6}
\end{equation*}
whichgives
100
VIII
. T
YPESETTING
M
ATHEMATICS
Eulernotonlyprovedthattheseries
n=1
1
n2
converges,butalsothat
n=1
1
n2
=
π2
6
Allthesearetrueforotheroperatorsclassifiedas“Variable-sizedsymbols”,except
integrals. Thoughtheintegralsymbolindisplayislarger,thepositionofthelimits s in
bothtextanddisplayisonthesideascanbeseenfromtheoutputbelow
Thuslim
x→∞
x
0
sinx
x
dx=
π
2
andsobydefinition,
0
sinx
x
dx=
π
2
whichisproducedby
Thus
$\lim\limits_{x\to\infty}\int_0ˆx\frac{\sin x}{x}\,\mathrm{d}x
=\frac{\pi}{2}$
and so o by definition,
\begin{equation*}
\int_0ˆ\infty\frac{\sin x}{x}\,\mathrm{d}x=\frac{\pi}{2}
\end{equation*}
Ifyouwantthelimitstobeaboveandbelowtheintegralsign,justaddthecommand
\limits
immediatelyafterthe
\int
command.Thus
Thus
$\lim\limits_{x\to\infty}\int_0ˆx\frac{\sin x}{x}\,\mathrm{d}x
=\frac{\pi}{2}$
and so o by definition,
\begin{equation*}
\int\limits_0ˆ\infty\frac{\sin x}{x}\,\mathrm{d}x=\frac{\pi}{2}
\end{equation*}
gives
Thuslim
x→∞
x
0
sinx
x
dx=
π
2
andsobydefinition,
0
sinx
x
dx=
π
2
Nowhowdowetypesetsomethinglike
p
k
(x)=
n
i=1
ik
xt
i
t
k
t
i
wherewehavetwolinesofsubscriptsfor
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