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MathematicalPrependix
17
Read the components from these equations. E.g.
a
x
x
=
d
2
x=dt
2
. Forcylindrical l coordinates
the appropriate basis vectors conform to this system, with ^
r
pointing away y from the e origin and
^
perpendiculartothat.
~r
=
z^z
+
r^r
,and plane polarsimply omits the
z
. Nowtheunitvectors s are
functionsofposition,implyingthatasaparticlemovestheseunitvectorswillchange,andyouhaveto
usetheproductruletodierentiatetheterms. ^
z
isconstantsoitcausesnotrouble.
~v
_
z^z
+_
r^r
+
r
_
^r
The one newfeatureis the third term, andforthatyou need to notice that ^
r
is a a functionofthe
coordinate
,thoughnotof
z
or
r
. Toevaluatethisderivative,usethechainrule.
d^r
dt
=
d^r
d
d
dt
Therstofthesederivatives,
d
^
r=d
,isnowaproblemingeometry,andthere’saresultaboutdier-
entiatingvectorsthathaveaconstantmagnitude: thederivativeisperpendiculartotheoriginalvector.
Toshowthis,let
~u
beany vectorofconstantmagnitude. Thatis,
~u
.
~u
=
C
. Dierentiatethiswith
respecttoanything.
d
dt
~u
.
~u
=
dC
dt
=0=2
~u
.
d~u
dt
(0
:
38)
That’sallyouneed,becauseitsaysthatthederivativeiseitherzeroorisperpendicularto
~u
asclaimed.
d
^
r=d
isperpendicularto ^
r
. Itisinthe
^
direction. Now,what t isits magnitude? ? Asketch
answersthequestion. Thesketchwillalsoanswerthequestion:\Whyisitinthe+
^
directionandnot
along 
^
?"
^
r
(
)
^
r
(
+
)
^
r
Fig.0.11
Thethreevectors ^
r
(
),^
r
(
+
),and^
r
formanisoscelestriangle. Constructthebisector
ofthevertexangle,andyouimmediatelyseethatthelengthof^
r
is
^
r
=2
.
1
.
sin(
=
2)
As
!0,thesinebehavesas
=
2itself,sothequotient
^
r
=
!1.
d
^
r
d
=
^
;
andsimilarly
d
^
d
^
r
(0
:
39)
Nowbacktovelocityandacceleration.
~v
=
d~r=dt
_
z^z
+_
r^r
+
r
_
^r=z_^z+_r^r+r
_
^
(0
:
40)
Anotherderivative:
~a
=
d~v=dt
=
z
^
z
+
r
^
r
+_
r
_
^
r
+_
r
_
^
+
r
^
+
r
_
_
^
=
z^z
+
r^r
+_
r
d^r
d
_
+_
r
_
^
+
r
^
+
r
_
d
^
d
_
=
z
^
z
+^
r
r
r
_
2
+
^
r
+2_
r
_
(0
:
41)
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MathematicalPrependix
18
Hereyouneed
d
^
=d
,anotherderivativeofaunitvector,soit’sperpendicularto
^
. Howbigisit?
Youcandothesamesortofgeometryaswith^
r
,ornoticethat
^r
.
^
=0  !
d
d
^r
.
^
=0=
d^r
d
.
^
+^
r
.
d
^
d
=1+^
r
.
d
^
d
andthisestablishesthesignandmagnitudeof
d
^
=d
asinEq.(0.39).Increasingtheangle
alittle
bitrotatesanobjectalittlebitcounterclockwise.Thatrotates ^
r
toward
^
.Itrotates
^
toward ^
r
.
Thatthebasisvectorsvarywithpositionandarenotparalleltoeachotherasyoumovearound
thecoordinatesystemisafamiliarideainanothercontext: geography.
^
Up,
^
South,and
^
Eastarenot
parallelvectorsasyoumovearoundtheEarth.
Forsphericalcoordinatesthederivationscanbedonealongthesamelines,butwithalotmore
algebra. Itisnotworth h the troubletogothrough it,and youdon’tneed theresults asoften. . The
answeris(remembertodistinguishthesphericalcoordinate
r
fromthecylindricalone|they’respelled
thesame).
~r
=
r
^
r
~v
=
_
~r
_
r^r
+
r
_
^
+
r
_
sin
^
~a
=
r
r
_
2
sin
2
r
_
2
^
r
+
r
+2_
r
_
r
_
2
sin
cos
^
+
r
sin
+2_
r
_
sin
+2
r
_
_
cos
^
(0
:
42)
Ingeographicalterms,
^
r
c
Up
^
S
c
outh
^
E
b
ast
Example
Circularmotionisafamiliarexamplefromintroductorycourses. If
z
=0and
r
=aconstant,the
equations(0.40)and(0.41)are
~v
=
r
_
^
and
~a
^
rr
_
2
+
^
r
^
r
v
2
r
+
^
dv
dt
(0
:
43)
Thelastformcomesfromusingthemagnitudeoftherstoftheseequationsfor
~v
,thatis
v
=
r
_
,and
itreproducesthefamiliarinwardradialaccelerationforcircularmotion(
r
_
2
=
v
2
=r
). Thetangential
componentis
r
=
d
(
r
_
)
=dt
=
dv=dt
Now waita minute! Ifyou u believe this manipulation,look
againmorecritically.Isthemotioncounterclockwiseorclockwiseanddoesitmatter?Is
_
positiveor
negative? Isthemagnitudeofavectorpositive? ? Whenyouexpressavectorintermsofcomponents,
isthecoecientoftheunitvectorthemagnitudeofthevectororacomponentofthevector?Answer:
thecorrectequationisnot
v
=
r
_
,but
v
=
r
_
,statingthatthephi-componentofthevelocityis
r
_
.
Gobackandmodifytheseequationsappropriately.
Example
x
=
x
0
,aconstant,
y
=
v
0
t
shouldhaveconstantvelocityandzeroacceleration,butthat’snotso
obviousifyouseeitinpolarcoordinates.
r
=
p
x
2
+
y
2
=
q
x
2
0
+
v
2
0
t
2
and
=tan
1
(
y=x
)=tan
1
v
0
t
x
0
~v
=^
r
_
r
+
^
r
_
=^
r
v
2
0
t
p
x
2
0
+
v
2
0
t
2
+
^
q
x
2
0
+
v
2
0
t
2
1
1+
v
0
t=x
0
2
.
v
0
x
0
=^
r
v
2
0
t
p
x
2
0
+
v
2
0
t
2
+
^
v
0
x
0
p
x
2
0
+
v
2
0
t
2
(0
:
44)
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MathematicalPrependix
19
Youcan see that this has the correctbehavior r at
t
= 0 and as
t
! 1. . Does s ithavethe correct
magnitude?Andwhatisitsderivative?
0.7ComplexAlgebra
Thissectionappearsinchapters3,4,6,andimplicitlymanyotherplaces.
Thereare some standard manipulationswith complexarithmetic thattake some practice. . Even n the
basic +, ,,andarenotexactly y whatyoulearnedinthird grade,so I’llstartwiththose. . The
standardcommutative,associative,anddistributivelawsapplytotherstthree,so
(7+2
i
)(6+3
i
)=(6+3
i
)(7+2
i
)=36+33
i
(1+2
i
)
(3+4
i
)(5+6
i
)
=
(1+2
i
)(3+4
i
)
(5+6
i
)=(1+2
i
)( 9+38
i
)= 85+20
i
(1+2
i
)
(3+4
i
)+(5+6
i
)
=(1+2
i
)(3+4
i
)+(1+2
i
)(5+6
i
)=
(1+2
i
)(8+10
i
)= 12+26
i
As fordivision,itisnomorecommutativeherethanitis forrealnumbers,butasimpletrick
allowsyoutosimplifysomeexpressions. Thecomplexconjugateofanumberisthenumberfoundby
changingthesignoftheimaginarypart.
z
=5+7
i
=)
z
=5 7
i
The
notation is acommononeforthisoperation,though 
z
is anothernotationthatmany prefer.
Whatistheproductofanumberanditscomplexconjugate?
z
=5+7
i; z
=5 7
i
=)
z
z
=(5 7
i
)(5+7
i
)=25+49+35
i
35
i
=74
z
z
is always realand positive: : (
a
+
ib
)(
a
ib
) =
a
2
+
b
2
is the squareofthe magnitude ofthe
complexnumber,thesquareof
p
a
2
+
b
2
.Howdoyouusethistomanipulatedivision?Rationalizethe
denominatorofaquotient.
1+2
i
3+4
i
=
(1+2
i
)(3 4
i
)
(3+4
i
)(3 4
i
)
=
11+2
i
25
(0
:
45)
Multiplyinganumberbyitscomplexconjugateresultsinareal,soyoucanmultiplythenumeratorand
denominatorofaquotientbythe complex conjugate ofthedenominatorin orderto bringtheresult
intoasimplerform. Ifyoueverwanttomakethenumeratorrealinstead,usethesameidea.
Example
Afewcasesofsuchmanipulation,simplifyingcomplexexpressions:
3 4
i
i
=
(3 4
i
)(2+
i
)
(2 
i
)(2+
i
)
=
10 5
i
5
=2 
i:
(3
i
+1)
2
1
i
+
3
i
2+
i
=( 8+6
i
)
(2+
i
)+3
i
(2 
i
)
(2 
i
)(2+
i
)
=( 8+6
i
)
5+7
i
5
=
2 26
i
5
:
i
3
+
i
10
+
i
i
2
+
i
137
+1
=
i
)+( 1)+
i
( 1)+(
i
)+(1)
=
1
i
=
i:
Whatisthegeometricinterpretationof
i
?Itisafactoritrotatesyouby90
.
z
=1+3
i
iz
i
2
z
i
3
z
iz
=
i
(1+3
i
)= 3+
i
i
2
z
=
i
( 3+
i
)= 1 3
i
i
3
z
=
i
( 1 3
i
)=3 
i
i
4
z
=
z
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MathematicalPrependix
20
Whatis
i
n
? Eachmultiplicationby
i
rotatesyouby90
inthecomplex plane,so
i
4
=1,and
i
217
=
i
4
.
54+1
=
i
.
Variousrootsof1orof 1orof
i
appearcommonly,andyouneedtheexponentialrepresentation,
Euler’sformula,tondthem.Thisis
x
+
iy
=
r
cos
+
ir
sin
=
re
i
x
y
(0
:
46)
Youcanderivethisequationfromtheseries(0.1). Put
i
intotheseriesfortheexponentialandcollect
therealandimaginarypieces,doneinsection3.2. Theresultis
e
i
=cos
+
i
sin
.
Specialcasesofthisequationsay
e
2
i
=1
;
e
i
= 1
;
e
i=
2
=
i;
e
2
ni
=1
Therearethreecuberootsofone,andallthatyouneedtondthemistheprecedingline.
1
1
=
3
=
e
2
ni
1
=
3
Take
n
tobeasuccessionofintegers
n
=0  ! ! 1
1
=
3
=1
n
=1  !
e
2
i
1
=
3
=
e
2
i=
3
=cos2
=
3+
i
sin2
=
3=( 1+
i
p
3)
=
2
n
=2  !
e
4
i
1
=
3
=
e
4
i=
3
=cos4
=
3+
i
sin4
=
3=( 1 
i
p
3)
=
2
Ifyoukeep goingto
n
=3
;
4
;
etc.orusenegativeintegers,you simply repeatthese threevalues. . A
pictureoftherootsshows themequallyspaced aroundtheunitcircle,exactly asdictatedbyEuler’s
equation,andthesamesortofpictureappearsforhigherrootstoo.
e
2
i=
3
e
2
i=
8
e
10
i=
8
The polarform of f complex x numbers uses the exponential l representation, , and here are some
examplesthatusethismanipulation.
p
i
=
e
i=
2
1
=
2
=
e
i=
4
=
1+
i
p
2
:
i
1+
i
3
=
p
2
e
i=
4
p
2
e
i=
4
!
3
=
e
i=
2
3
=
e
3
i=
2
=
i:
2
i
1+
i
p
3
25
=
2
e
i=
2
2
1
2
+
i
1
2
p
3
!
25
=
2
e
i=
2
2
ei=
3
!
25
=
e
i=
6
25
=
e
i
(4+1
=
2)
=
i
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MathematicalPrependix
21
Anotherapplication of Euler’s formula a is s to ordinary trigonometry. . What t happens when you
multiplytwocomplexnumbersexpressedinpolarform?
z
1
z
2
=
r
1
e
i
1
r
2
e
i
2
=
r
1
r
2
e
i
(
1
+
2
)
(0
:
47)
Puttingitintowords,youmultiplythemagnitudesandaddtheanglesinpolarform.
Fromthisyoucanimmediatelydeducesomeofthecommontrigonometricidentities.UseEuler’s
formulaintheprecedingequationandwriteoutthetwosides.
r
1
(cos
1
+
i
sin
1
)
r
2
(cos
2
+
i
sin
2
)=
r
1
r
2
cos(
1
+
2
)+
i
sin(
1
+
2
)
Thefactors
r
1
and
r
2
cancel. Nowmultiplythetwobinomialsontheleftandmatchtherealandthe
imaginarypartstothecorrespondingtermsontheright. Theresultisthepairofequations
cos(
1
+
2
)=cos
1
cos
2
sin
1
sin
2
sin(
1
+
2
)=cos
1
sin
2
+sin
1
cos
2
(0
:
48)
and you have amuch simplerthanusualderivation of these common identities. . Youcandosimilar
manipulationsforothertrigonometricidentities,andinsomecasesyouwillencounterrelationsforwhich
there’sreallynootherwaytogettheresult. Thatiswhyyouwillndthatinphysicsapplicationswhere
you mightuse sines orcosines (oscillations,waves)noone uses anythingbut complexexponentials.
Getusedtoit.
Theimportantapplicationsofcomplexnumbersinthistextappearwhenyouwanttodierentiate
complexfunctions,especiallytheexponential.
d
dx
e
ix
=
ie
ix
=
d
dx
cos
x
+
i
sin
x
= sin
x
+
i
cos
x
andyoucaneasilyseethatthesecondandthefourthformsagree.Doanotherderivativeandyouget
d
2
dx
2
e
ix
=
i
2
e
ix
e
ix
sothisfunction
e
ix
satisestheharmonicoscillatorequation,thesubjectofchapterthree.
Therearesomepracticeexercises oncomplexalgebraattheendofthischapter,butformore
examplesseechapterthreeofMathematicalTools,mentionedinthebibliographyonpageiii.
0.8Separationofvariables
Thissectionappearsinchapters2,3,4,andinanotherversion,inchapter7.
The subject of dierential equations is large enough that you can make aprofession of it and still
not exhaust the e subject, but t in this text, when you solve dierential equations, there are justtwo
methodsthatshowupwithanyregularity.\Separationofvariables"isone.\Linearconstantcoecient
equations"is the other(nextsection). . Afterthatthere e are afewequations such asEq.(6.10) that
standontheirown,andyoucanwaituntilyougettheretondoutaboutthem.
Adierentialequationisanequationrelatingafunctionandoneormoreofitsderivatives,and
~
F
=
m~a
isthissemester’sdierentialequation. Thersttoolinyourkitisseparationofvariables,and
itiseasiesttounderstandifyoustartwithanexampleortwo. Let
c
beaconstant.
dx
dt
=
c
2
+
x
2
!
dx
c
2+
x
2
=
dt
!
Z
dx
c
2+
x
2
=
Z
dt
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MathematicalPrependix
22
Therstoftheseisthedierentialequationtobesolved. Itisarstorderequation,meaningthatit
isarelationbetweenthefunction
x
andonlytherstderivative
dx=dt
. Therearetwovariableshere,
theindependentvariable
t
,andthedependentvariable
x
.Youcan’tsimplyintegratethiswithrespect
to
t
becausetherightsideisafunctionof
x
,andthatisan(unknown)functionofthevariable
t
. To
separatevariablesputallthe
x
’sononesideoftheequationandallthe
t
’sontheother. Thesecond
equationdoesthis.Itisnowsetupforintegration.
Nowdotheintegral,atrigsubstitutionworks:
x
=
c
tan
.
dx
=
c
sec
2
d
!
Z
c
sec
2
d
c
2
+
c
2
tan
2
=
Z
1
c
d
=
1
c
=
1
c
tan
1
x
c
=
t
+
D
andthesolutionis
x
(
t
)=
c
tan
c
(
t
+
D
). Withaninitialconditionssuchas
x
(0)=
x
0
youhave
x
(0)=
x
0
=
c
tan(
cD
)  !
D
=
1
c
tan
1
x
0
=c
!
x
(
t
)=
c
tan
ct
+tan
1
(
x
0
=c
)
Checkthelastexpression:
x
(0)=
c
tan
tan
1
(
x
0
=c
)
=
x
0
Neverassumethatyouhaven’tmadea
mistake. Astimeincreases,
x
(
t
)increases,so(
c
2
+
x
2
)increases,so
dx=dt
increases,sotheslopeof
thecurve
x
versus
t
getsbiggerandbigger|that’showthetangentof
t
behaves.
Thismethodlookslikesuchaspecialone;thecombinationoffactorsthatwillletyoudothis
seemssoimprobablethatitcan’tworkveryoften.True. But,ithappensinenoughimportantspecial
casesthatyouhavetoknowaboutitandlearntorecognizewhenitcanapply.
1:
dN=dt
N;
2:
d
2
x
dt
2
!
2
x;
3:
t
dx
dt
=
x
+
;
4:
t
dx
dt
+
tx
=
x
(0
:
49)
Equations1,3,and4areseparable,butnot2,thoughinchapters2and3youwillseesomemanipu-
lationsthatwilldigaseparableequationoutofeventhatone.
Wait,couldn’tyoumanipulatethe second ofthese tobe
d2x
x
!
2
dt
2
and integrate? ? No!
There’snosuchmathematicsasthis,sodon’ttry.
Forotherexamplesofthismethod,lookatEqs.(2.13),(2.17),(2.23),(3.55).
0.9Constant CoecientODEs
Thissortofdierentialequationshowsupofteninthiscourse,startinginchaptertwo,andcommonly
afterthat. Itlookslike
3
d
2
x
dt
2
4
dx
dt
+7
x
=0
or
d
3
x
dt
3
+
d
2
x
dt
2
+
dx
dt
+
x
=
A
cos
!t
Thedependentvariablecanhaveanynumberofderivatives,butitappearsjusttotherstpower,no
x
2
or
x
dx
dt
orsin(
kx
).Thatmakestheseequationlinear.Thatthecoecientofthe
x
’sareconstantsmake
theseconstantcoecientlinearequations.Thattherstonehasonlytermsin
x
oritsderivativesmakes
ithomogeneous andthatthesecondonehasanextratermwithno
x
atallmakesitinhomogeneous.
Theprecisedenitionofhomogeneousisthatifyoumultiplythevariable
x
byaconstant
,thenthe
wholeexpressionismultipliedbysomepowerof
,i.e.
n
. Here
n
=1.
Therstcase,thelinearconstantcoecienthomogeneousone,hasasimplesolution. Allyou
havetonoticeisthatthederivativeofanexponentialisanexponential,andtryasolution
x
(
t
)=
Ae
t
.
3
d
2
x
dt
2
4
dx
dt
+7
x
=0  !3
A
2
e
t
4
Ae
t
+7
Ae
t
=0
Ae
t
[3
2
4
+7]=0
MathematicalPrependix
23
Sinceneither
A
northeexponentialarezero,thatleaves3
2
4
+7=0,apolynomialequationwith
tworoots,givingtwosolutionstotheequation. Becauseyouaretryingtoundotwoderivativestoget
x
youwillsomehowgettwoarbitraryconstants. Thekeypropertyoflinearhomogeneousequationsis
thatthesumoftwosolutionsisasolution,sothefullsolutiontothisequationis
A
1
e
1
t
+
A
2
e
2
t
;
where
1
;
2
=
2
i
p
17

3
Howdoyouhandletheinhomogeneouscaseexampleabove? Anexponentialwon’tworkhere.
Youwillnotget
A
cos
!t
outofitinordertomatchtheright-handside. Thesumoftwosolutionsis
nolongerasolution. But,thereisonesimplication: Ifyou(temporarily)throwawaytheinhomoge-
neousterm(
A
cos
!t
),youcansolvetheremaininghomogeneouspartoftheequationwithasimple
exponential. O.k.yougetacubicequation,butit’sonly y apolynomialequationsotherearewaysto
handleit. Thispartialsolutionwillhavethreearbitraryconstants. . Nowifsomehowyoucanndany
onesolutiontothewholeequationthetrickistoaddthetwopartialsolutions.
d
3
x
hom
dt
3
+
d
2
x
hom
dt
2
+
dx
hom
dt
+
x
hom
=0
;
withthreearbitraryconstants
d
3
x
inh
dt
3
+
d
2
x
inh
dt
2
+
dx
inh
dt
+
x
inh
=
A
cos
!t;
withnone
Then
x
(
t
)=
x
inh
(
t
)+
x
hom
(
t
). Howdoyouverifythis? Plugintotheoriginalequationandwatchit
work.
Fortheproblemsyouencounterin this book,ndingthespecial, inhomogeneoussolution will
notbedicult,andlateryouwillseesomegeneralmethodsforndingsuchsolutionsevenwhenitis
dicult.
0.10Matrices
Thissectionappearsinchapters4,8,10.
Justasyouhavecomponentsofvectorswithrespecttoabasisyou will havecomponents ofcertain
typesofvector-valued functions. . You u have(
v
x
;v
y
;v
z
)or(
v
r
;v
;v
) with threecomponents fora
vector. Animportantsortoffunction(alinear,vector-valuedfunctionofavectorvariable)appearsin
describingtheangularmomentumofarigidbody. Italsoappearsindescribingdielectricpropertiesof
acrystal. Andindescribingrotationsofvectors. And
:::
. Anyway,ittoohascomponents(ninethis
time)andtheseformmatrices.Thedevelopmentoftheseideas,showingthereasonfortheodd-looking
rulesthatmatricesobey,canwaituntilthey’reneededinsection 8.2. Forthemomentthiswillbea
summaryofsomeruleswithoutanydiscussionofthereasonsthattheyarethewaytheyare.
Forthemomentthenamatrixisasquarearrayofnumbers. Theycanberectangulartoo,but
nothere. Theycanbeadded,multiplied,divided,evenexponentiated.
a b
c d
+
e f
g h
=
a
+
e b
+
f
c
+
g d
+
h
(0
:
50)
andofcoursesubtractionjustchangesallthe+signsto . . Whatmatrixplaystheroleofzerosothat
addingitchangesnothing?Anarrayofallzeroes.
Isaidthatthereareninecomponentsandtheseobjectshaveonlyfour. Ifyouknoweverything
about22arrays,theextensionto33iseasy. Justaswithwitheithermechanicsorcalculus,the
step from one dimensionto two is thebigone. . Afterthatthesteptothree e dimensions oreven
N
dimensionsisrelatively small. . Besides,it’seasiertowritetheseandtheytakeonly y about 8/
27
ofthe
arithmetictomanipulatethem.
MathematicalPrependix
24
Multiplicationobeys
a b
c d

e f
g h
=
ae
+
bg af
+
bh
ce
+
dg cf
+
dh
(0
:
51)
Yourunacrossthe rows of therst matrix and down thecolumns ofthe second matrix in orderto
constructtheentriesintheproduct. Justasthereisazeromatrixforaddition,thereisaunitmatrix
formultiplication. Whatisit? Whatentriesinthe e rstfactorof(0.51)make theproductequalthe
arrayof
e;f;g;h
,therebyreproducingthesecondfactor? Forthetopleftentryoftheproduct,
ae
+
bg
=
e
forall
e
andforall
g
=)
a
=1
; b
=0
Thismakesthetoprightentryworktoo. Similarlyforthebottomentriesyouneedtohave
c
=0and
d
=1.Thatmakestheidentitymatrix
(
I
)=
1 0
0 1
The orderofmultiplication matters,and multiplication isnotcommutative. . You u can howevercheck
thespecialcaseshowingthatthisidentitymatrixfunctionsjustaswellastherighthandfactorasit
doesontheleft.
Theinverseofamatrixisthatmatrixsuchthattheproductwiththeoriginalistheidentity.Set
therightsideofEq.(0.51)totheidentitymatrixandsolvethefourequationsinthefourunknowns
a;b;c;d
. I’lljustwritetheanswer,butyoushould d carryoutthealgebrasothattheresultis yours
andnotmine.
e f
g h
1
=
1
eh
fg
h
f
g
e
(0
:
52)
Multiply this by the originalmatrix and verify that you gettheidentity. . Itworksineitherorder,so
checkitbothways.
Thereisno common notationforamatrix as there isforvectors. . Inthe e lattercaseyou see
boldfacetypeoranarroworsometimesasquigglyunderline,butformatricestherearenostandards.
Sometimesaboldfacesansseriffontischosenforthispurpose,anditservesaswellasanythingelse
sothat’swhatIwillusehere.
A=
a b
c d
;
B=
e f
g h
;
then
AB=C=
ae
+
bg af
+
bh
ce
+
dg cf
+
dh
TheinversematrixasinEq.(0.52)obeys
BB
1
=B
1
B=I
ThestatementthatmatrixmultiplicationisnotcommutativeisAB6=BA.Youdohavetheassociative
lawthough: A(BC)=(AB)C.Alsothedistributivelaw: A(B+C)=AB+AC.
Simultaneousequations
Matricesappearinmanyinterestingandelegantcontexts. Theyalsoappearinmundanesettings,but
thesearenolessimportant.Howdoyousolvetwolinearequationsintwounknowns?
ax
+
by
=
p;
cx
+
dy
=
q:
multiplytherstby
d
andthesecondby
b
,thensubtract
dax
+
dby
=
dp;
bcx
+
bdy
=
bq
!
dax
bcx
=
dp
bq
!
x
=
dp
bq
da
bc
multiplytherstby
c
andthesecondby
a
,thensubtract
(0
:
53)
cax
+
cby
=
cp;
acx
+
ady
=
aq
!
cby
ady
=
cp
aq
!
y
=
cp
aq
bc
ad
MathematicalPrependix
25
Thisismatrixinversionindisguise.
a b
c d

x
y
=
p
q
or
Mx=p
MultiplybothsidesofthismatrixequationbytheinverseofMfromEq.(0.52).
M
1
Mx=x=M
1
p=
1
ad
bc
d
b
c a

p
q
=
x
y
Thisisexactlythesameastheprecedingexplicitsolutionfor
x
and
y
.Infact,thatexplicitsolutionis
howtheinversematrixisderived,sothiscomparisonisreallycircular.
Doesthisalwayswork?No.Youcan’tdividebyzero,andinEq.(0.53)Iignoredthatimportant
point.
dax
bcx
=
dp
bq
! (
da
bc
)
x
=
dp
bq
also
(
da
bc
)
y
=
aq
cp
(0
:
54)
Whatif
da
bc
=0?thentherightsidesoftheequationsmustbezero,otherwisethereisnosolution.
Youcanhaveasolutionif
p
and
q
arebothzeroorifaparticularcombinationof
p
,
q
,andtheelements
ofthematrixconspiretomaketherightsidezero.
ad
bc
=determinantofthematrix
Thedeterminantdeterminesthenatureofthesolutions(deterministicallyofcourse).
1. Ifthedeterminantisnon-zerothenthesolutionexistsandisunique.
2. Ifthedeterminantiszeroand
p
or
q
isnon-zerothereisnosolutionunlessspecial
circumstancesoccur;thenthereareaninnitenumberofsolutions.
3. Ifthedeterminantiszeroandboth
p
and
q
arezerothereareaninnitenumberof
solutions.
Case#1isroutine. Yousolvesimultaneousequationsandyouexpecttondasolution. . The
secondcase isexceptional,anditcanbeusedtodeterminepropertiesoftheright-handside. . Itwill
showupindisguiseinsections7.10and10.7. Thethirdcaseisthemostcommonforthepurposesof
thisbook.Itmeansthatthetwoequationsyouaresolvingare
ax
+
by
=0
;
cx
+
dy
=0
but
ad
bc
=0
(0
:
55)
Ifforexample
d
and
b
are6=0,multiplytherstoftheseby
d
:
dax
+
dby
=0.Now
ad
=
bc
,sothis
equationisthesameas
bcx
+
bdy
=0or
cx
+
dy
=0. Thatmeansyoureallyhaveoneequationfor
thetwounknowns,nottwo. Thatinturnmeansthatyouhaveaninnitenumberofsolutions
x
and
y
fortheanswer. Onceyouhavefoundone,simplymultiply
x
and
y
byany constantandyouhave
another.Youcanunderstandthismostsimplybyagraphicalinterpretation.
ax
+
by
=0 isastraightlinethroughtheorigin.
andthisgraphrepresentsaninnitenumberofpossiblesolutions.
And howdoyou write \boldfacesans serif" on paper? ? Perhapsby y using\Blackboard Bold"
style:
ABCDEFGHIJKLMNOPQRSTUVWXYZ
.Thisisawaytofakeboldfacetypeinwriting.
IndexNotation
A
ij
isthesetofelementsofthematrixA. Theindices
i
and
j
runfromonetowhateverthesizeof
thematrixis(twointheseexamples).Therstindexspeciestherowandthesecondthecolumn.
A
row
;
column
=
A
ij
 !
A
11
A
12
A
21
A
22
rstrow
secondrow
MathematicalPrependix
26
Thesearethreenotationsforthesamethingbecauseyoudon’thavetothinkofthesubscripts
i
and
j
asparticularvalues. Itislikethecommonnotationforafunction,
f
(
x
). Youarenotsupposedto
thinkofthisassomeparticular\
x
"butasaplaceholderforanyvaluethatargumentcantakeon.* In
thisnotationmatrixadditionandmultiplicationare
A+B= !
A
ij
+
B
ij
=
C
ij
and
AB= !
N
X
j
=1
A
ij
B
jk
=
C
ik
Acolumnmatrixhasasingleindex
x
i
 !
x
1
x
2
and
Ax= !
X
k
A
jk
x
k
=
y
j
!
A
11
A
12
A
21
A
22

x
1
x
2
=
y
1
y
2
Thissortofproduct-and-sumoccurssooftenthattheconventionalnotationisagaintoomitthe
summationsymboljustasinEq.(0.28).Wheneveraproductappearswitharepeatedindexsummation
isimplied. Thetwosumsjustabovethenappearas
A
ij
B
jk
=
C
ik
and
A
jk
x
k
=
y
j
(0
:
56)
Inthiskindofmanipulationyouwillndthatarepeatedindexalwaysappearsasapair. Ifyounda
combinationsuchas
A
ij
B
jk
C
jn
thengobackandndyourmistake. Itshouldn’thappen.
0.11IterativeSolutions
Thismaterialisusedinchapters4and5.
Sometimesacomplicatedequationisreallyasimpleequationindisguise. Youjusthavetolookatit
therightway. Youcantakethequadraticequation
:
01
x
2
x
+1=0
;
andsolve:
x
=
1
p
0
:
96
=:
02
butreally,thisisalmostalinearequation:  
x
+1=0  !
x
=1,andthat’smucheasier. Whatif
youneedmoreaccuracy? Justrearrangetheequationtobe
x
=1+
:
01
x
2
;
thenanimprovedsolutionis
x
=1+
:
01(1)
2
=1
:
01
Ifthat’sstillnotgoodenoughthenyoucaniteratetheprocessuntilyou’retiredofit.
x
=1+
:
01(1
:
01)
2
=1
:
010201
andagain,
x
=1+
:
01(1
:
010201)
2
=1
:
01020506060401
ormaybeyouwanttodoitagaintoget1
:
0102051426447orjust1
:
0102051,andyoumaythendecide
that1
:
01wasprobablygoodenough.
* Mathematicians s will argue that this s is bad notation, , and that t you should think of
f
at the
functionand
f
(
x
)astheparticularvalueofthefunctionatthepoint
x
. Theyhaveapoint. Thatis
technicallythecorrectthingtodo,andmakingthatdistinctioncanhelpkeepyououtoftrouble,butit
iscumbersome,andthisgoodadviceisoftenignored. Thereisacaseinchaptereighthowever,where
Iwillraisethisissueagain,andthereIwillsidewiththemathematicians.
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