ForTheresa, Sophie,Alexandra,Jennifer,andErika
Prefae
Polynomialspervade mathematis, and muh that is beautiful in mathe-
matis is related to polynomials. Virtually every branh of mathematis,
from algebrai number theory and algebrai geometry to applied analy-
sis,Fourieranalysis,andomputersiene,hasitsorpusoftheoryarising
from the study of polynomials. Historially, questions relating to polyno-
mials,forexample,thesolutionofpolynomialequations,gaveriseto some
of themostimportantproblems of theday. Thesubjetis nowmuh too
largetoattemptanenylopedioverage.
Thebodyofmaterialwehoosetoexploreonernsprimarilypolyno-
mialsastheyariseinanalysis,andthetehniquesofthebookareprimarily
analyti.Whiletheonnetingthreadisthepolynomial,thisisananalysis
book.The polynomialsand rational funtions we are onerned with are
almostexlusivelyofasinglevariable.
We assumeat most aseniorundergraduatefamiliarity with real and
omplexanalysis (indeed in most plaes muh less is required). However,
the material is often tersely presented, with muh mathematis explored
intheexerises,someofwhiharequitehard,manyofwhiharesupplied
withopioushints,somewithompleteproofs.Welloverhalfthematerial
in thebook is presented in the exerises.The reader is enouraged to at
least browse through these. We havebeenmuh inuened byPolyaand
Szeg}o's lassi \Problemsand Theorems in Analysis"in ourapproah to
theexerises.(ThoughunlikePolyaandSzeg}owehosetoinorporatethe
hintswiththeexerises.)
Thebookismostlyself-ontained.Thetext,withouttheexerises,pro-
videsanintrodutionto thematerial,but muh oftherihnessisreserved
for theexerises.We haveattempted to highlight the parts ofthe theory
andthetehniqueswendmostattrative.So,forexample,Muntz'slovely
haraterizationofwhenthespanofasetofmonomialsisdenseisexplored
insomedetail.Thisresultepitomizesthebestofthesubjet:anattrative
andnontrivialresultwithseveralattrativeandnontrivialproofs.
Thereareexellentbooksonorthogonalpolynomials,Chebyshevpoly-
nomials,Chebyshevsystems,andthegeometryofpolynomials,tonamebut
afewofthetopisweover,anditisnotourintenttorewriteanyofthese.
Ofneessityandtaste,someofthismaterialispresented,andwehaveat-
tempted to provide someaess to these bodies of mathematis. Muh of
the material in the later hapters is reent and annot befound in book
formelsewhere.
Studentswhowishtostudy fromthisbookare enouragedtosample
widely from the exerises. This is denitely \hands on" material. There
is too muh material for a single semester graduate ourse, though suh
aourse maybebased onSetions 1.1 through5.1, plus aseletion from
later setionsand appendies. Mostof thematerial after Setion 5.1 may
beread independently.
Not all objets labeled with \E" are exerises. Some are examples.
Sometimes no question is asked beause none is intended. Oasionally
exerisesinlude astatementlike,\foraproofsee::: ";thisisusuallyan
indiationthatthereaderisnotexpetedto provideaproof.
Someoftheexerisesarelongbeausetheypresentabodyofmaterial.
Examplesofthisinlude E.11ofSetion2.1 onthetransnitediameterof
a set and E.11 of Setion 2.3 on the solvability of the moment problem.
Someof theexerisesare quitetehnial. Someof thetehnial exerises,
likeE.4ofSetion2.4,areinluded,indetail,beausetheypresentresults
thatarehardto aesselsewhere.
Aknowledgments
We would like to thank Dik Askey, Weiyu Chen, Carl de Boor, Karl
Dilher, JensHappe, Andras Kroo,Doron Lubinsky, Gua-HuaMin, Paul
Nevai, Allan Pinkus, Jozsef Szabados, Vilmos Totik, and Rihard Varga.
Theirthoughtfulandhelpfulommentsmadethisabetterbook.Wewould
alsoliketothankJudithBorwein,MariaFeElder,andChiaraVeronesifor
theirexpertassistanewiththepreparationofthemanusript.
Contents
PREFACE vii
CHAPTER1 IntrodutionandBasiProperties 1
1.1 Polynomialsand RationalFuntions 1
1.2 TheFundamentalTheoremof Algebra 11
1.3 ZerosoftheDerivative 18
CHAPTER2 SomeSpeialPolynomials 29
2.1 ChebyshevPolynomials 29
2.2 OrthogonalFuntions 41
2.3 OrthogonalPolynomials 57
2.4 Polynomialswith NonnegativeCoeÆients 79
CHAPTER3 ChebyshevandDesartesSystems 91
3.1 ChebyshevSystems 92
3.2 Desartes Systems 100
3.3 ChebyshevPolynomialsinChebyshevSpaes 114
3.4 Muntz-LegendrePolynomials 125
3.5 ChebyshevPolynomialsinRationalSpaes 139
CHAPTER4 DensenessQuestions 154
4.1 VariationsontheWeierstrass Theorem 154
4.2 Muntz's Theorem 171
4.3 UnboundedBernstein Inequalities 206
4.4 Muntz Rationals 218
CHAPTER5 BasiInequalities 227
5.1 ClassialPolynomialInequalities 227
5.2 Markov'sInequalityforHigherDerivatives 248
5.3 InequalitiesforNormsofFators 260
CHAPTER6 InequalitiesinMuntzSpaes 275
6.1 Inequalitiesin Muntz Spaes 275
6.2 NondenseMuntzSpaes 303
CHAPTER7 InequalitiesforRationalFuntionSpaes 320
7.1 InequalitiesforRationalFuntionSpaes 320
7.2 InequalitiesforLogarithmiDerivatives 344
APPENDIXA1 AlgorithmsandComputationalConerns 356
APPENDIXA2 OrthogonalityandIrrationality 372
APPENDIXA3 AnInterpolationTheorem 382
APPENDIXA4 InequalitiesforGeneralizedPolynomialsinL
p
392
APPENDIXA5 InequalitiesforPolynomialswithConstraints 417
BIBLIOGRAPHY 448
NOTATION 467
INDEX 473
1
Introdution and Basi Properties
Overview
ThemostbasiandimportanttheoremonerningpolynomialsistheFun-
damental Theorem of Algebra. This theorem, whih tells us that every
polynomial fators ompletely over the omplex numbers, is the starting
point for this book. Someof the intriate relationships betweenthe loa-
tionofthezerosofapolynomialanditsoeÆientsareexploredinSetion
2.Theequallyintriaterelationshipsbetweenthezerosofapolynomialand
the zerosof itsderivativeor integralare the subjet of Setion 1.3. This
hapterservesasageneralintrodutiontothebodyoftheoryknownasthe
geometryofpolynomials.HighlightsofthishapterinludetheFundamen-
tal Theorem of Algebra, the Enestrom-Kakeyatheorem, Luas' theorem,
andWalsh'stwo-irletheorem.
1.1PolynomialsandRationalFuntions
The fous for this book is the polynomial of asingle variable. This is an
extended notionof the polynomial,as wewill see later, but themostim-
portant examplesare thealgebrai and trigonometri polynomials,whih
wenowdene.Theomplex(n+1)-dimensionalvetorspae ofalgebrai
polynomialsof degree at most n with omplex oeÆients is denoted by
P
.
IfC denotestheset ofomplexnumbers,then
(1:1:1) P
n :=
(
p:p(z)= n
X
k =0 a
k z
k
; a
k 2C
)
:
WhenwerestritourattentiontopolynomialswithrealoeÆientswewill
usethenotation
(1:1:2) P
n :=
(
p:p(z)= n
X
k =0 a
k z
k
; a
k 2R
)
;
whereR is theset ofrealnumbers.Rationalfuntions oftype(m;n)with
omplexoeÆientsarethendened by
(1:1:3) R
m;n :=
p
q :p2P
m
;q2P
n
;
whiletheirrealousinsaredenoted by
(1:1:4) R
m;n :=
p
q :p2P
m
;q2P
n
:
Thedistintionbetweentherealand omplexasesispartiularlyimpor-
tantforrationalfuntions(seeE.4).
ThesetoftrigonometripolynomialsT
n
isdened by
(1:1:5) T
n :=
(
t:t():=
n
X
k = n a
k e
ik
; a
k 2C
)
:
A real trigonometri polynomialof degree at mostn is an element of T
n
takingonlyrealvaluesontherealline.Wedenote byT
n
thesetofallreal
trigonometripolynomialsofdegreeat mostn. Other haraterizationsof
T
n
aregiveninE.9.Note thatifz:=e i
,then anarbitraryelementofT
n
isoftheform
(1:1:6) z
n 2n
X
k =0 b
k z
k
; b
k 2C
and so manypropertiesof trigonometri polynomialsredue to the study
ofalgebraipolynomialsoftwiethedegreeontheunitirlein C:
The most basi theorem of this book, and arguably the most basi
nonelementary theorem of mathematis, is the Fundamental Theorem of
ofP
n nP
n 1
)hasexatlynomplexzerosountedaordingtotheirmul-
tipliities.
Theorem1.1.1 (Fundamental TheoremofAlgebra). If
p(z):=
n
X
i=0 a
i z
i
; a
i
2C; a
n 6=0;
thenthereexist
1
;
2
;::: ;
n
2C suhthat
p(z)=a
n n
Y
i=1
(z
i ):
Here the multipliity of the zero at
i
is the number of times it is
repeated.So,forexample,
(z 1) 3
(z+i) 2
is a polynomial of degree5 with azero of multipliity 3 at 1and witha
zeroof multipliity2at i:Thepolynomial
p(z):=
n
X
i=0 a
i z
i
; a
i
2C; a
n 6=0
isalledmoniifitsleadingoeÆienta
n
equals1.Therearemanyproofs
of the Fundamental Theorem of Algebra based on elementary properties
of omplexfuntions (see Theorem 1.2.1and E.4 ofSetion 1.2). We will
explorethistheoremmoresubstantiallyinthenextsetionofthishapter.
Comments,Exerises,andExamples.
The importane of the solutionof polynomial equationsin the historyof
mathematisishardtooverestimate.TheGreeksofthelassialperiodun-
derstood quadratiequations(atleastwhenbothroots werepositive)but
ouldnotsolveubis.Theexpliitsolutionsoftheubiandquartiequa-
tionsinthesixteenthenturywereduetoNioloTartaglia(a1500{1557),
Ludovio Ferrari (1522{1565),andSipione del Ferro (a1465{1526)and
were popularized bythe publiationin 1545 ofthe \ArsMagna"of Giro-
lamoCardano(1501{1576).Theexatprioritiesarenotentirelylear,but
del Ferro probably has the strongest laim on the solution of the ubi.
These disoveries gave western mathematis an enormous boost in part
beause they represented one of the rst really major improvements on
Greekmathematis.Theimpossibilityofndingthezerosofapolynomial
ofdegreeatleast5,ingeneral,byaformulaontainingadditions,subtra-
tions,multipliations,divisions,andradialswouldawaitNielsHenrikAbel
(1802{1829)and his 1824 publiationof \On theAlgebrai Resolution of
Equations."Indeed,somuhalgebra,inludingGaloistheory,analysis,and
partiularly omplexanalysis,is born outofthese ideasthatit ishard to
imaginehowtheowofmathematismighthaveproeededwithoutthese
E.1 ExpliitSolutions.
a℄ QuadratiEquations. Verifythatthequadratipolynomialx 2
+bx+
haszerosat
b p
b 2
4
2
;
b+ p
b 2
4
2
:
b℄ Cubi Equations. Verify that the ubi polynomial x 3
+bx+ has
zerosat
+;
+
2
+i p
3
2
;
+
2
i p
3
2
;
where
= 3 s
2 +
r
2
4 +
b 3
27
and
= 3 s
2 r
2
4 +
b 3
27 :
℄ Show that an arbitrary ubi polynomial, x 3
+ax 2
+bx+; an be
transformedintoaubipolynomialasin partb℄ byatransformationx7!
ex+f.
d℄ Observethatifthepolynomialx 3
+bx+hasthreedistintrealzeros,
thenand areneessarilynonrealandhene4b 3
+27 2
isnegative.So,
inthissimplestofasesoneisforedtodealwithomplexnumbers(whih
was aserioustehnialproblemin thesixteenthentury).
e℄ QuartiEquations. The quarti polynomial x 4
+ax 3
+bx 2
+x+d
haszerosat
a
4 +
R
2
2
;
a
4 +
R
2
2
;
where
R= r
a 2
4
b+y;
y isanyrootoftheresolventubi
y 3
by 2
+(a+4d)y a 2
d+4bd 2
;
and
; = r
3a 2
4 R
2
2b
4ab 8 a 3
4R
; R6=0;
while
; = r
3a 2
4
2b2 p
y 2
4d; R=0:
Theseunwieldyequationsarequiteusefulinonjuntionwithanysymboli
E.2 Newton'sIdentities. Write
(x
1 )(x
2
)(x
n )=x
n
1 x
n 1
+
2 x
n 2
+( 1) n
n :
TheoeÆients
k
are, bydenition,the elementary symmetri funtions
inthevariables
1
;:::;
n :
a℄ For positiveintegersk,let
s
k :=
k
1 +
k
2
++ k
n :
Provethat
s
k
=( 1) k +1
k
k +( 1)
k k 1
X
j=1 ( 1)
j
k j s
j
; kn
and
s
k
=( 1) k +1
k 1
X
j=k n ( 1)
j
k j s
j
; k>n:
Here,and inwhatfollows,anemptysumisunderstoodtobe0.
Apolynomial ofn variables isafuntion that isapolynomialineah
of itsvariables. A symmetri polynomial of n variables is apolynomialof
nvariablesthatisinvariantunderanypermutationofthevariables.
b℄ Showbyindutionthatanysymmetripolynomialinnvariables(with
integeroeÆients)maybewrittenuniquelyasapolynomial(withinteger
oeÆients)in theelementarysymmetrifuntionsf
1
;f
2
;:::;f
n .
Hint:For asymmetripolynomialf innvariables,let
(f):=(
1
;
2
;:::;
n
);
1
2
n 0
if
f(x
1
;x
2
;:::;x
n )=
1
X
1
=0 2
X
2
=0
n
X
n
=0
1
;
2
;:::;
n x
1
1 x
2
2 x
n
n
and
1
;
2
;:::;
n 6=0.If
(f)=(
1
;
2
;:::;
n
) and (g)=(e
1
;e
2
;:::;e
n );
thenlet(f)<(g)if
j e
j
foreahjwithastritinequalityforatleast
oneindex.Thisgivesa(partial)wellorderingofsymmetripolynomialsin
nvariables, thatis, everysetof symmetripolynomialsin nvariableshas
℄ Showthat
1+ p
5
2
k
!0 (mod1):
(By onvergene to zero (mod1) we mean that the quantity approahes
integralvalues.)
Hint:Considertheintegers
s
k :=
k
1 +
k
2
;
where
1 :=
1
2 (1+
p
5)and
2 :=
1
2 (1
p
5). ut
d℄ Findanotheralgebraiintegerwiththepropertythat
k
!0 (mod1):
Suh numbers are alled Salem numbers (see Salem [63℄). It is an open
problemwhetheranynonalgebrainumbers>1satisfy k
!0 (mod 1).
E.3 NormsonP
n . P
n
isavetorspaeofdimensionn+1overR: Hene
P
n
equipped with any norm is isomorphi to the Eulidean vetorspae
R n+1
, and these norms are equivalent to eah other. Similarly, P
n is a
vetorspaeofdimensionn+1overC: HeneP
n
equippedwithanynorm
isisomorphito theEulideanvetorspaeC n+1
, sothese normsarealso
equivalenttoeahother.Let
p
n (x):=
n
X
k =0 a
k x
k
; a
k 2R:
SomeommonnormsonP
n andP
n are
kpk
A :=sup
x2A
jp(x)j supremumnorm
:=kpk
L
1 (A)
L
1 norm
kpk
Lp(A) :=
Z
A jp(t)j
p
dt
1=p
L
p
norm;p1
kpk
l
1
:=max
k fja
k
jg l
1 norm
kpk
l
p :=
n
X
k =0 ja
k j
p
1=p
l
p
norm;p1:
IntherstaseAmustontainn+1distintpoints.Intheseond aseA
a℄ Conludethat thereexistonstantsC
1 ,C
2 ,andC
3
dependingonlyon
nsothat
kp 0
k
[ 1;1℄
C
1 kpk
[ 1;1℄
;
n
X
i=0 ja
i jC
2 kpk
[ 1;1℄
;
kpk
[ 1;1℄
C
3 kpk
L2[ 1;1℄
foreveryp2P
n
,and,inpartiular, foreveryp2P
n .
These inequalities will be revisited in detail in later hapters, where
preiseestimatesaregivenintermsofn.
b℄ Showthat there exist extremal polynomialsfor eah of theabove in-
equalities.That is,forexample,
sup
06=p2P
n kp
0
k
[ 1;1℄
kpk
[ 1;1℄
isahieved.
E.4 OnR
n;m .
a℄ R
n;m
isnotavetorspaebeauseitisnotlosed underaddition.
b℄ PartialFrationDeomposition. Letr
n;m 2R
n;m
beoftheform
p(x)
Q
m 0
k =1
(x
k )
mk
; p2P
n
;
k
distint; p(
k )6=0:
Thenthereisauniquerepresentationoftheform
r
n;m
(x)=q(x)+ m
0
X
k =1 mk
X
j=1 a
k ;j
(x
k )
j
; q2P
n m
; a
k ;j 2C
(ifm>n;thenP
n m
ismeantto bef0g).
Hint:Considerthetypeanddimensionofexpressionsoftheaboveform. ut
℄ Showthatif
r
n;m 2R
n;m
;
then
Re(r
n;m
())2R
n+m;2m :
Thisisanimportantobservationbeauseinsomeproblemsarationalfun-
tioninR
anbehavemorelikeanelementofR
2n;2n
thanR
n;n .
E.5 Horner'sRule.
a℄ Wehave
n
X
i=0 a
i x
i
=(((a
n x+a
n 1 )x+a
n 2
)x++a
1 )x+a
0 :
Soeverypolynomialofdegree nan beevaluated byusing at mostnad-
ditions and n multipliations. (The onverse is learly not true; onsider
x 2
n
.)
b℄ Show that everyrational funtion of type (n 1;n) an be put in a
formso thatitanbeevaluated byusingndivisions andnadditions.
E.6 LagrangeInterpolation. Letz
i andy
i
bearbitraryomplexnumbers
exeptthatthez
i
mustbedistint(z
i 6=z
j
; for i6=j).Let
l
k (z):=
Q
n
i=0;i6=k (z z
i )
Q
n
i=0;i6=k (z
k z
i )
; k=0;1;:::;n:
a℄ Showthatthereexistsauniquep2P
n
thattakesn+1speiedvalues
atn+1speiedpoints,that is,
p(z
i )=y
i
; i=0;1;:::;n:
Thisp2P
n
isoftheform
p(z)= n
X
k =0 y
k l
k (z)
andisalledtheLagrangeinterpolation polynomial.
Ifallthez
i and y
i
arereal,thenthisuniqueinterpolationpolynomial
isinP
n :
b℄ Let
!(z):=
n
Y
i=0 (z z
i ):
Showthat l
k
isoftheform
l
k (z)=
!(z)
(z z
k )!
0
(z
k )
and
p(z)= n
X
y
k
!(z)
(z z
k )!
0
(z
k )
:
℄ AnErrorEstimate. Assumethatthepointsz
i
2[a;b℄; i=0;1;:::;n;
are distint and f 2C n+1
[a;b℄ (that is, f is an n+1times ontinuously
dierentiable real-valued funtion on [a;b℄). Let p 2 P
n
be the Lagrange
interpolationpolynomialsatisfying
p(z
i )=f(z
i
); i=0;1;:::;n:
Showthat foreveryx2[a;b℄thereisapoint2(a;b)sothat
f(x) p(x)= 1
(n+1)!
f (n+1)
()!(x):
Hene
kf pk
[a;b℄
1
(n+1)!
kf (n+1)
k
[a;b℄
k!k
[a;b℄
:
Hint:Chooseso that':=f p w vanishesat x;that is,
:=(f(x) p(x))=!(x):
ThenrepeatedappliationsofRolle'stheoremyieldthat
' (n+1)
=f (n+1)
(n+1)!
hasazero in(a;b): ut
E.7 HermiteInterpolation.
a℄ Letz
i
2C; i=1;2;:::k,bedistint.Letm
i
;i=1;2;:::;k,bepositive
integerswithn+1:=
P
k
i=1 m
i
,andlet
y
i;j
2C; i=1;2;:::;k; j=0;1;:::;m
i 1
bexed.Showthatthereisauniquep2P
n
,alledtheHermiteinterpola-
tionpolynomial,sothat
p (j)
(z
i )=y
i;j
; i=1;2;:::;k; j=0;1;:::;m
i 1:
Ifall thez
i and y
i;j
arereal, thenthis uniqueinterpolationpolynomialis
inP
n 1 .
Hint:Useindution onn. ut
b℄ Assume that the points z
i
2 [a;b℄ are distint and f 2 C n
[a;b℄. Let
p2P
n 1
betheHermiteinterpolationpolynomialsatisfying
p(z
i )=f
(j)
(z
i
); i=1;2;:::;k; j=0;1;:::;m
i 1:
f(x) p(x)= 1
n!
f (n)
()!(x)
with
!(x):=
k
Y
i=1 (x x
i )
mi 1
:
Hene
kf pk
[a;b℄
1
n!
kf (n)
k
[a;b℄
k!k
[a;b℄
:
Hint:FollowthehintgivenforE.6℄. ut
Polynomialinterpolationandrelatedtopis arestudiedthoroughly in
Davis [75℄;Lorentz, Jetter,and Riemenshneider [83℄;and Szabados and
Vertesi[92℄.
E.8 On the Zeros of a p 2 P
n
. Show that ifp 2 P
n
, then the nonreal
zerosof pformonjugatepairs(thatis,ifzisazeroofp;thensois z).
E.9 FatorizationofTrigonometriPolynomials.
a℄ Showthat t2T
n
(ort2T
n
)ifandonlyiftisoftheform
t(z)=a
0 +
n
X
k =1 (a
k
oskz+b
k
sinkz); a
k
;b
k
2R (orC):
b℄ Show that if t2 T
n nT
n 1
; then there arenumbers z
1
;z
2
;::: ;z
2n and
06=2C suh that
t(z)= 2n
Y
j=1 sin
z z
j
2 :
Showalsothat thenonreal zerosz
j
oft formonjugatepairs.
E.10 NewtonInterpolationandInteger-ValuedPolynomials. Let k
f(x)
bedened indutivelyby
0
f(x):=f(x); f(x)= 1
f(x):=f(x+1) f(x)
and
k +1
f(x):=( k
f(x)); k=1;2;::::
Let
x
k
:=
x(x 1)(x k+1)
k!
:
a℄ Showthat x
k
isapolynomialofdegreek thattakesintegervaluesat
b℄ Let f be an m times dierentiable funtion on[a;a+m℄. Show that
thereisa2(a;a+m)suhthat
m
f(a)=f (m)
():
℄ Showthatifp2P
n
;then
p(x)= n
X
k =0
k
p(0)
x
k
:
d℄ Suppose p2P
n
isinteger-valuedatallintegers.Showthat
p(x)= n
X
k =0 a
k
x
k
forsomeintegersa
0
;a
1
;:::;a
n
:Note that this haraterizessuh polyno-
mials.
e℄ Showthatifp2P
n
takesintegervaluesatn+1onseutiveintegers,
thenptakesintegervaluesateveryinteger.
f℄ Suppose2Randn
isanintegerforeveryn2N:Usepartb℄toshow
that isanonnegativeinteger.
1.2The FundamentalTheoremofAlgebra
ThefollowingtheoremisaquantitativeversionoftheFundamental Theo-
remofAlgebraduetoCauhy[1829℄.Weoeraproofthatdoesnotassume
the Fundamental Theorem of Algebra, but does requiresome elementary
omplexanalysis.
Theorem1.2.1. The polynomial
p(z):=a
n z
n
+a
n 1 z
n 1
++a
0 2P
n
; a
n 6=0
has exatly n zeros. These all lie in the open disk of radius r enteredat
the origin,where
r:=1+ max
0k n 1 ja
k j
ja
n j
:
Proof. Wemaysupposethata
0
6=0,orwemayrstdivide byz k
forsome
g(x):=ja
0 j+ja
1
jx++ja
n 1
jx ja
n jx
satises g(0)>0and lim
x!1
g(x)= 1.Soby theintermediatevaluethe-
orem,g has azero in (0;1) (whih is, on onsidering (g(x)=x n
) 0
, in fat
unique).Letsbethiszero.Thenforjzj>s;
(1:2:1) jp(z) a
n z
n
jja
0 j+ja
1
zj++ja
n 1 z
n 1
j<ja
n z
n
j:
This,byRouhe'stheorem(seeE.1),showsthatp(z)anda
n z
n
haveexatly
thesamenumberofzeros,namely,n,in anydiskofradiusgreaterthans.
Itremainstoobservethat ifxr;theng(x)<0sos<r.Indeed,
g(x)ja
n jx
n
1+
max
k =0;:::;n 1 ja
k j
ja
n j
n 1
X
k =0 x
k n
!
<ja
n jx
n
1+
max
k =0;:::;n 1 ja
k j
ja
n j
1
x 1
0
for
x1+ max
k =0;:::;n 1 ja
k j
ja
n j
:
u t
TheexatrelationshipbetweentheoeÆientsofapolynomialandthe
loation ofits zerosis veryompliated. Of ourse,the moreinformation
wehaveabouttheoeÆients,thebettertheresultsweanhopefor.The
followingprettytheorememphasizesthis:
Theorem1.2.2(Enestrom-Kakeya). If
p(z):=a
n z
n
+a
n 1 z
n 1
++a
0
with
a
0 a
1
a
n
>0;
thenall thezerosof plie outsidethe open unitdisk.
Proof. Consider
(1 z)p(z)=a
0 +(a
1 a
0
)z++(a
n a
n 1 )z
n
a
n z
n+1
:
Then
j(1 z)p(z)ja
0 [(a
0 a
1
)jzj++(a
n 1 a
n )jzj
n
+a
n jzj
n+1
℄:
Sine a
k a
k +1
0, the right-hand expression abovedereases as jzj in-
reases.Thus,forjzj<1,
j(1 z)p(z)j>a
0 [(a
0 a
1
)++(a
n 1 a
n )+a
n
℄=0;
andtheresultfollows. ut
Corollary1.2.3. Suppose
p(z):=a
n z
n
+a
n 1 z
n 1
++a
0
witha
k
>0for eahk. Then all thezerosofplie inthe annulus
r
1
:= min
k =0;:::;n 1 a
k
a
k +1
jzj max
k =0;:::;n 1 a
k
a
k +1
=:r
2 :
Proof. ApplyTheorem1.2.2top(r
1
z)andz n
p(r
2
=z). ut
This isatheme withmany variations,someof whih are explored in
theexerises.
Theorem1.2.4. Supposep>1;q>1, andp 1
+q 1
=1.Then the poly-
nomial h2P
n
of the form
h(z)=a
n z
n
+a
n 1 z
n 1
++a
0
; a
n 6=0
has allitszerosinthe diskfz2C :jzjrg,where
r:=
8
<
: 1+
n 1
X
j=0 ja
j j
p
ja
n j
p
!
q=p 9
=
; 1=q
:
Proof. SeeE.6. ut
Comments,Exerises,andExamples.
TheFundamentalTheoremofAlgebraappearstohavebeengivenitsname
byGauss,although theresultwas familiarlongbefore;itresisted rigorous
proofbyd'Alembert(1740),Euler(1749),andLagrange(1772).Itwasmore
ommonlyformulatedasarealtheorem,namely:everyrealpolynomialfa-
tors ompletely into real linear or quadrati fators. (This is an essential
resultfortheintegration ofrationalfuntions.) Girardhasalaimto pri-
ority offormulation. Inhis \InventionNouvelleen L'Algebra" of1629 he
wrote\everyequationof degreen hasas manysolutionsas the exponent
ofthehighestterm." Gaussgavetherstsatisfatoryproofin 1799in his
dotoraldissertation,andhegavethreemoreproofsduringhislifetime.His
rstproof,whiletitled\Anewproofthateveryrationalintegralfuntionof
onevariableanberesolvedintorealfatorsoftherstorseonddegree,"
was in fat therst more-or-lesssatisfatoryproof. Gauss'rstproof isa
geometri argumentthat the realand imaginaryparts ofapolynomial,u
andv,havethepropertythattheurvesu=0andv=0interset,andby
modern standardshassometopologialproblems. His third proofof 1816
jzj=r p(z)
p(z) dz
must vanish if p has no roots, whih leads to a ontradition and is a
genuinelyanalytiproof(seeBoyer[68℄,Burton [85℄,andGauss[1866℄).
An almost purely algebrai proof using Galois theory, but based on
ideasof Legendre,maybefoundin Stewart[73℄.
The\geometryof polynomials"is extensivelystudied in Marden [66℄
and Walsh [50℄, where most of the results of the setion and muh more
maybeaessed.SeealsoBarbeau [89℄andPolyaandSzeg}o[76℄.
Theorem 1.2.2is dueto Kakeya[12℄.It is aspeial aseof Corollary
1.2.3,dueto Enestrom[1893℄.TheEnestrom-Kakeyatheoremandrelated
mattersarestudied thoroughlyin Anderson,Sa,andVarga[79℄and [81℄
andin VargaandWu Wen-da[85℄,andanumberofinterestingproperties
areexplored.For example,itisshownintherstoftheabovepapersthat
thezerosofallpsatisfyingtheassumptionof Corollary1.2.3are densein
theannulusfz2C :r
1
jzjr
2 g.
E.1 BasiTheoremsinComplexAnalysis. Weolletafew ofthebasi
theorems of omplexanalysis that we need.(Proofs may befound in any
omplexvariablestext suhas Ahlfors[53℄or Ash[71℄.)
a℄ Cauhy's Integral Formula. Let D
r
:= fz2 C : jzj <rg. Suppose f
isanalytionD
r
andontinuousonthelosureD
r ofD
r
:LetD
r denote
theboundaryofD
r :Then
0= Z
Dr
f(t)dt;
f(z)= 1
2i Z
D
r f(t)
t z
dt; z2D
r
;
and
f (n)
(z)= n!
2i Z
D
r
f(t)
(t z) n+1
dt; z2D
r :
Unless otherwise speied, integration on a simple losed urve is taken
antilokwise. (We may replae D
r and D
r
by any simple losed urve
and its interior, respetively, though for most of our appliations irles
suÆe.)
b℄ Rouhe's Theorem. Suppose f and g are analyti inside and on a
simple losedpath (for most purposes wemayuse airle). If
jf(z) g(z)j<jf(z)j
for every z 2 ; then f and g have the same number of zeros inside
Afuntion analytionC isalledentire.
℄ Liouville'sTheorem. Aboundedentire funtionisonstant.
d℄ Maximum Priniple. An analyti funtion on an open set U C
assumesitsmaximummodulusonthe boundary.Moreover, iff isanalyti
andtakes atleasttwo distintvaluesonanopenonnetedsetU C; then
jf(z)j<sup
z2U
jf(z)j; z2U:
e℄ UniityTheorem. Supposef andg areanalytion anopen onneted
set U. Supposef andg agree on S, whereS is aninnite ompat subset
of U,then f andg agreeeverywhere onU:
E.2 Division.
a℄ Suppose pisapolynomialofdegreenandp()=0.Thenthere exists
apolynomialqofdegreen 1suhthat
p(x)=(x )q(x):
Hint:Considertheusualdivisionalgorithmforpolynomials. ut
b℄ Apolynomialofdegreenhasat mostnroots.
ThisistheeasierpartoftheFundamentalTheoremofAlgebra.Theremain-
ingontentisthateverynononstantpolynomialhasatleastone omplex
root.
Thenextexerisedevelopsthebasiomplexanalysistoolsmostlyfor
polynomialsonirles.Thepointofthisexeriseistonotethattheproofs
inthis asearepartiularlystraightforward.
E.3 PolynomialComplexAnalysis.
a℄ DedueCauhy'sintegralformulaforpolynomialsonirles.
Hint:Integratez n
onD
r
. ut
b℄ Ifp(z)=a
n Q
n
i=1 (z
i
);thenthenumberofindiesiforwhihj
i j<r
is
1
2i Z
D
r p
0
(z)
p(z) dz;
provided no
i
liesonD
r :
Hint:Wehave
p 0
(z)
p(z)
= n
X
i=1 1
z
i
and
1
2i Z
D
r dz
z
i
=
1 if j
i j<r
0 if j
i j>r:
℄ Dedue Rouhe'stheorem from partb℄ for polynomialsf and g given
bytheirfatorizations,andforirles.
Hint:Leth:=1+(g f)=f.Sofh=g and
1
2i Z
Dr g
0
(z)
g(z) dz=
1
2i Z
Dr f
0
(z)
f(z) dz+
1
2i Z
Dr h
0
(z)
h(z) dz:
Showthat thelastintegraliszerobyexpandingh 1
andapplyingb℄. ut
d℄ DedueE.1℄andE.1d℄ fromE.1a℄.
e℄ Observethat theuniitytheoreman besharpenedfor polynomialsas
follows.Ifp;q2P
n
andp(z)=q(z)forn+1distintvaluesofz2C; then
pandq areidential, that is, p(z)=q(z)foreveryz2C: Equivalently,a
polynomialp2P
n
is either identially 0or has at mostn zeros.(This is
trivial from the Fundamental Theorem of Algebra, but as in E.2, it does
notrequireit.)
E.4 TheFundamental Theorem of Algebra. Every nononstant polyno-
mial hasatleastone omplexzero.
Provethisdiretly fromLiouville'stheorem.
E.5 Pellet'sTheorem. Supposea
p
6=0; ja
p+1
j++ja
n
j>0, and
g(x):=ja
0 j+ja
1
jx++ja
p 1 jx
p 1
ja
p jx
p
+ja
p+1 jx
p+1
++ja
n jx
n
has exatlytwo positive zeross
1
<s
2 .Then
f(z):=a
n z
n
+a
n 1 z
n 1
++a
0 2P
n
hasexatlypzerosinthediskfz2C :jzjs
1
gandnozerosintheannulus
fz2C :s
1
<jzj<s
2 g.
Proof. Lets
1
<t<s
2
: Theng(t)<0;thatis,
n
X
j=0
j6=p ja
j jt
j
<ja
p jt
p
:
NowapplyRouhe'stheoremto thefuntions
F(z):=a
p z
p
and G(z):=
n
X
j=0 a
j z
j
:
u t
E.6 ProofofTheorem1.2.4.
a℄ Holder'sinequality(seeE.7ofSetion2.2)assertsthat
n
X
k =1 ja
k b
k j
n
X
k =1 ja
k j
p
!
1=p
n
X
k =1 jb
k j
q
!
1=q
;
wherep 1
+q 1
=1andp1.Soif
p(z):=a
n z
n
+a
n 1 z
n 1
++a
0 2P
n
;
then
n 1
X
k =1 ja
k jjzj
k
n 1
X
k =0 ja
k j
p
!
1=p
n 1
X
k =0 jzj
k q
!
1=q
:
b℄ Thus,forjzj>1;
jp(z)jja
n jjzj
n n 1
X
k =0 ja
k jjzj
k
ja
n jjzj
n 8
<
: 1
n 1
X
k =0
a
k
a
n
p
!
1=p
n 1
X
k =0 jzj
k q
jzj nq
!
1=q 9
=
;
ja
n jjzj
n 8
<
: 1
n 1
X
k =0
a
k
a
n
p
!
1=p
1
(jzj q
1) 1=q
9
=
; :
℄ Whenisthelastexpressionpositive?
E.7 TheNumber ofPositiveZerosofaPolynomial. Suppose
p(z):=
n
X
j=0 a
j z
j
hasmpositiverealroots.Then
m 2
2nlog ja
0 j+ja
1
j++ja
n j
p
ja
0 a
n j
!
:
ThisresultisduetoShurthoughtheproofmoreorlessfollowsErd}osand
Turan[50℄.ItrequiresusingMuntz's theoremfromChapter4.
a℄ Suppose
p(z)=a
n n
Y
(z r
k e
i
k
)
and
q(z):=
n
Y
k =1 (z e
ik
):
Notethat forjzj=1;
jz re i
j 2
jrj
jz e i
j:
Usethistodeduethat
jq(z)j 2
jp(z)j
2
ja
0 a
n j
ja
0 j+ja
1
j++ja
n j
p
ja
0 a
n j
!
2
wheneverjzj=1:
b℄ Sinephasmpositiverealrootsqhasmrootsat1:Usethehangeof
variablesx:=z+z 1
appliedto z n
q(z 1
)q(z)to showthat
kqk 2
fjzj=1g
min
fb
k g
k(z 1) m
(z n m
+b
n m 1 z
n m 1
++b
1 z+b
0 )k
2
fjzj=1g
min
fkg kx
m
(x n m
+
n m 1 x
n m 1
++
1 x+
0 )k
[0;4℄
=4 n
min
fd
k g
kx m
(x n m
+d
n m 1 x
n m 1
++d
1 x+d
0 )k
[0;1℄
4 n
p
2n+1 2n
n+m ;
wherethelastinequalityfollowsbyE.2℄ofSetion4.2.
℄ Showthat
log
4 n
p
2n+1 2n
n+m
!
m 2
=n
andnishtheproofofthemainresult.
1.3Zeros oftheDerivative
Themostbasiandimportanttheoremlinkingthezerosofthederivativeof
apolynomialtothezerosofthepolynomialisvariouslyattributedtoGauss,
Luas,Grae,andothers,but isusuallyalledLuas'theorem[1874℄.
Theorem1.3.1(Luas'Theorem). Let p2P
n
. All the zerosof p 0
areon-
tainedinthe losedonvexhullofthe set ofzerosofp.
The proof of this theorem follows immediately from the following
lemmabyonsideringtheintersetionofthehalfplanesontainingtheon-
Lemma1.3.2. Let p2P
n
. If phas allitszerosin alosedhalfplane, then
p 0
n
alsohas all itszerosin thesame losedhalfplane.
Proof. Ononsiderationoftheeetofthetransformationz7!z+;by
whih any losed halfplane may be mapped to H
l
:=fz : Re(z) 0g; it
suÆestoprovethelemmaundertheassumptionthatphasallitszerosin
H
l
:IfphasallitszerosinH
l
;then
p 0
(z)
p(z)
= n
X
k =1 1
z
k
;
k 2H
l :
Butifz2H
r
:=fz2C :Re(z)>0g;then
1
z
k 2H
r
foreah
k 2H
l
;
anditfollowsthat
n
X
k =1 1
z
k 2H
r :
Inpartiular,
n
X
k =1 1
z
k 6=0;
whihnishestheproof. ut
There is asharpeningof Luas' theorem for real polynomials formu-
lated by Jensen. We need to introdue the notion of Jensen irlesfor a
polynomialp2 P
n
. For p2P
n
thenonreal roots of pome in onjugate
pairs.Foreahsuhpair,+i, i,formtheirleenteredatwith
radiusjj.Sothisirleenteredonthex-axisathas+iand ion
theoppositeendsof itsperpendiulardiameter.Theolletionofallsuh
irlesarealledtheJensenirlesforp.
Theorem 1.3.3 (Jensen's Theorem). Let p 2 P
n
. Eah nonreal zero of p 0
lies inoron someJensenirle for p.
Theproof, whih is similarto theproofof Luas'theorem, isleft for
thereaderas E.3.
WestatethefollowingprettygeneralizationofLuas'theoremdue to
Walsh [21℄.The proof is left as E.4. Proofs an also befound in Marden
Theorem1.3.4(Walsh'sTwo-CirleTheorem). Supposep2P
n
has allits
n zeros in the disk D
1
with enter
1
and radius r
1
. Suppose q 2P
m has
allitsm zerosin the diskD
2
withenter
2
andradiusr
2 . Then
a℄ All the zeros of (pq) 0
lie in D
1 [D
2 [D
3
, where D
3
is the disk with
enter
3
andradiusr
3
given by
3 :=
n
2 +m
1
n+m
; r
3 :=
nr
2 +mr
1
n+m :
b℄ Suppose(n6=m).Thenallthezerosof(p=q) 0
lieinD
1 [D
2 [D
3
;where
D
3
isthe disk withenter
3
andradiusr
3
given by
3 :=
n
2 m
1
n m
; r
3 :=
nr
2 +mr
1
jn mj :
Comments,Exerises,andExamples.
Luasprovedhis theoremin1874,although itisaneasyandobviouson-
sequene of anearlier result of Gauss.Jensen's theorem is formulated in
Jensen[13℄andprovedinWalsh[20℄.Muhmoreonerningthegeometry
ofzerosofthederivativeanbefoundin Marden[66℄.
E.1 A Remark onLuas' Theorem. Showthat p 0
2P
n
hasazeroon
theboundaryoftheonvexhullofthezerosofpifandonlyifisamultiple
zeroof p.
E.2 Laguerre's Theorem. Supposep2P
n
has all itszerosin a disk D.
Let 2C. Letw beany zeroof
q(z):=np(z)+( z)p 0
(z)
(q isalledthe polarderivative ofpwith respet to).
a℄ If 2=D;thenwliesinD.
Hint: Considerr(z) := p(z)(z ) n
; where phas allits zeros in D and
62D.Then
r 0
(z)
r(z)
= p
0
(z)
p(z) +
n
z
andifq(w)=0withw2=D, thenr 0
(w)=0.Nowobservethatr isofthe
form
r(z)=s
1
z
; s2P
n
;
wheres 0
((w ) 1
)=0.Notethat 2=Dimpliesthat
e
D:=f(z ) 1
:z2Dg
isa disk.Then shas allitszerosin e
D andso doess 0
byLuas' theorem.
However,w62Dimplies(w ) 1
62 e
D, sos 0
((w ) 1
)6=0,aontradi-
tion. ut
b℄ Ifp(w)6=0;thenanyirle throughw and either passesthroughall
thezerosofp orseparatesthem.
E.3 ProofofJensen's Theorem. ProveTheorem1.3.3.
Hint:Suppose p2P
n nP
n 1
and denote thezerosof pby z
1
;z
2
;:::;z
n :
Then
p 0
(z)
p(z)
= n
X
k =1 1
z z
k :
Ifz
k
=
k +i
k with
k
;
k
2R; andz=x+iywithx;y2R;then
Im
1
x+iy
k i
k +
1
x+iy
k +i
k
=
2y((x
k )
2
+y 2
2
k )
((x
k )
2
+(y
k )
2
)((x
k )
2
+(y+
k )
2
)
;
andsooutsidealltheJensenirlesandothex-axis,
sign
Im
p 0
(z)
p
n (z)
= sign(y)6=0:
u t
E.4 ProofofWalsh'sTheorem. ProveTheorem 1.3.4.
a℄ ProveTheorem1.3.4a℄.
Hint:Letz
0
beazeroofp 0
q+q 0
poutsideD
1 andD
2 .Let
1 :=z
0 np(z
0 )
p 0
(z
0 )
and
2 :=z
0
mq(z
0 )
q 0
(z
0 )
(p 0
(z
0
)6=0andq 0
(z
0
)6=0byLuas' theorem).Observethat
1 2D
1 and
2 2D
2
byE.2,and
z
0
= n
2 +m
1
n+m :
u t
b℄ ProveTheorem1.3.4b℄.
Hint:Proeedasinthehinttoparta℄,startingfromazeroz
0 ofp
0
q q 0
p
outsideD
1 andD
2
. ut
℄ If in Theorem 1.3.4 a℄D
1 , D
2
, and D
3
aredisjoint, then D
1
ontains
n 1zeros,D
2
ontainsm 1zeros,andD
3
ontains1zeroof(pq) 0
:
Hint:Byaontinuityargumentwemayreduethegeneralasetothease
wherep(z)=(z
1 )
n
andq(z)=(z
2 )
m
. ut
d℄ IfinTheorem1.3.4b℄n=mandD
1 andD
2
aredisjoint,thenD
1 [D
2
0
E.5 RealZerosandPoles.
a℄ Ifallthezerosofp2P
n
arereal,thenallthezerosofp 0
n
arealsoreal.
b℄ Supposeallthezerosofbothp2P
n
andq2P
m
arereal, andallthe
zerosofp
n
aresmallerthananyofthezerosof q
n
.Showthatallthezeros
of(p=q) 0
arereal.
Hint:Considerthegraphof
(p=q) 0
(p=q)
= p
0
p q
0
q :
u t
DeneW(p), theWronskianofp,by
W(p)(z)=p(z)p 00
(z) (p 0
(z)) 2
=
p(z) p 0
(z)
p 0
(z) p 00
(z)
=p 2
(z)
p 0
(z)
p(z)
0
:
℄ Provethatifp2P
n
hasonlydistintrealzeros,thenW(p)hasnoreal
zeros.
InCraven,Csordas,andSmith[87℄ itisonjeturedthat, forp2P
n ,
thenumberofrealzerosofW(p)=p 2
doesnotexeedthenumberofnonreal
zerosof p(aquestiontheyattribute toGauss).
d℄ Let p 2 P
n
. Show that any real zero of W(p) lies in or on a Jensen
irleofp.
Proof.SeeDilher[91℄. ut
e℄ ShowthatLuas'theoremdoesnotholdforrationalfuntions.
Hint:Considerr(x)=x=(
2
x 2
). ut
Thenextexeriseisaweakform ofDesartes'ruleofsigns.
E.6 PositiveZeros of Muntz Polynomials. Suppose Æ
0
< Æ
1
< <Æ
n
and
f(x):=a
0 x
Æ0
+a
1 x
Æ1
++a
n x
Æn
; a
k 2R:
Showthat eitherf =0orf hasat mostnzerosin(0;1).
E.7 ApolarPolynomialsandSzeg}o'sTheorem. Twopolynomials
f(x):= n
X
k =0 a
k
n
k
x k
; a
n 6=0
and
g(x):= n
X
k =0 b
k
n
k
x k
; b
n 6=0
arealledapolarif
n
X
k =0 ( 1)
k
a
k b
n k
n
k
=0:
a℄ ATheoremofGrae[02℄. Supposethatf andgareapolarpolynomials.
If f has all itszerosin a(losedor open) disk D; then g has at leastone
zeroin D.
Hint: Let
1
;
2
;:::;
n and
1
;
2
;:::;
n
denote the zeros of f and g,
respetively.SupposethatthezerosofgarealloutsideD. Let
f
1
(x):=nf(x)+(
1 x)f
0
(x)
andfork=2;3;:::;n, let
f
k
(x):=(n k+1)f
k 1
(x)+(
k x)f
0
k (x):
Then,byE.2,eah f
k
hasallitszerosinD:Nowompute
f
n 1 (
n )=
n!
b
n
n
0
a
0 b
n
n
1
a
1 b
n 1
++( 1) n
n
n
a
n b
0
=0;
wherethevanishingfollowsbyapolarity.Thisisaontradition. ut
b℄ If f and g are apolar, then the losed onvex hull of the zeros of f
intersetsthelosedonvexhullofthezerosofg.
℄ A TheoremofSzeg}o[22℄. Suppose
f(x):=
n
X
k =0 a
k
n
k
x k
; a
n 6=0;
g(x):=
n
X
k =0 b
k
n
k
x k
; b
n 6=0;
and
h(x):=
n
X
k =0 a
k b
k
n
k
x k
:
Supposef has all itszeros inaloseddisk D, and g has zeros
1
;::: ;
n :
Then allthe zerosofhare ofthe form with 2D:
Hint:SupposeÆisazeroof h.Then
n
X
k =0 a
k b
k
n
k
Æ k
=0:
Sothepolynomial
r(x):=
n
X
k =0 ( 1)
k
n
k
b
k Æ
k
x n k
isapolartof,andthushasazeroinD.Butthen= Æ=
i
forsomei
siner(x)=x n
g( Æ=x): ut
E.8 Zeros of the Integral. Suppose p 2 P
n nP
n 1
has all its zeros in
D
1
:=fz2C :jzj1g:
a℄ Show that the polynomial q dened by q(x) :=
R
x
0
p(t)dt has all its
zerosin D
2
:=fz2C :jzj2g:
Hint:ApplyE.7℄.Take
f(x):=p(x); g(x):=
n
X
k =0
n
k
x k
k+1 :
Then
h(x)= 1
x Z
x
0
p(t)dt:
Notethat g(x)=(n+1) 1
x 1
((1+x) n+1
1) hasallitszerosin D
2 : ut
b℄ Showthat
q(x):=
Z
x
0 Z
tm
1
0 Z
tm
2
0
Z
t1
0
p(t)dtdt
1 dt
m 2 dt
m 1
hasall itszerosin D
r
m;n
:=fz2 C :jzjr
m;n
g,where r
m;n
m+1is
thezeroof
n
X
k =0
m+n
m+k
x k
withthelargestmodulus. Notethat q isthemth integralof pnormalized
sothat theonstantsofintegrationareallzero.
E.9 Grae'sComplexVersion ofRolle'sTheorem. Supposeand are
zerosofp2P
n nP
n 1
. Then p 0
has atleastonezerointhe disk
D(;r):=fz2C :jz jrg;
where
:=
+
2
and r:=
j j
2 ot
n :
Hint:Assume,withoutlossofgenerality,that= 1and=1:Let
p 0
(x)= n 1
X
k =0 a
k x
k
; thatis, p(x)=+ n 1
X
k =0 a
k x
k +1
k+1 :
ApplyE.7a℄.Note that
0=
p(1) p( 1)
2
=a
0 +
a
2
3 +
a
4
5 +:
So
f(z):=(z 1) n
(z+1) n
andp 0
areapolar. ut
E.10 CorollariesofSzeg}o'sTheorem. Suppose
f(z):=
n
0
a
0 +
n
1
a
1
z++
n
n
a
n z
n
;
g(z):=
n
0
b
0 +
n
1
b
1
z++
n
n
b
n z
n
;
and
h(z):=
n
0
a
0 b
0 +
n
1
a
1 b
1
z++
n
n
a
n b
n z
n
witha
n b
n 6=0.
a℄ If f has all its zeros in a onvex set S ontaining 0 and g has all its
zerosin[ 1;0℄;thenhhas allitsrootsinS:
b℄ If f andg haveall theirzerosin[ 1;0℄;thenso does h.
E.11 AnotherCorollaryofSzeg}o'sTheorem. If n
X
k =0 a
k z
k
hasallitszeros
in D
1
:=fz 2C :jzj1g; then so does n
X
k =0 a
k z
k
n
k
. In partiular, n
X
k =0 z
k
n
k
has allitszerosinD
1 :
TheresultsofthenextexerisewererstprovedbyM.Riesz(see,for
E.12 ConseutiveZerosofp forp2P
n
withRealZeros. Forapolyno-
mial
p(x):=
n
Y
i=1
(x
i );
1
<
2
<<
n
; n2
withonlyrealzeros,let
(p):= min
1in 1 (
i+1
i ):
ByRolle'stheorem
p 0
(x)=n n 1
Y
i=1
(x
i );
1
<
1
<
2
<
2
<<
n 1
<
n :
a℄ Supposen3.Provethat(p)<(p 0
):
Outline. It isrequiredto showthat
j
j 1
>(p) foreahj 2.Let
2j nbexed.Sine
f 0
(
j )
f(
j )
= f
0
(
j 1 )
f(
j 1 )
=0
wehave
n
X
i=1
1
(
j 1
i )(
j
i )
=0:
Nowletu
j :=
j
j 1
; v
j :=
j
j
:Alsoforeahi;letd
i :=
j
j i
;
e
i :=
j+i
j
: Thentheaboveanberewrittenas
j 1
X
i=1
1
(d
i u
j )(d
i +v
j )
+ 1
( u
j v
j )
+ n j
X
i=1
1
(e
i +u
j )(e
i v
j )
=0:
Dene
F(u;v):=
j 1
X
i=1
uv
(d
i u)(d
i +v)
+ n j
X
i=1
uv
(e
i +u)(e
i v)
:
Note that F is inreasingin eah variable (0u<d
1
; 0v <e
1 )and
observethat
F(u
j
;v
j )=1:
To prove the result, it suÆes to show that if u and v are nonnegative
Nowshowthat
F(u;v)uv j 1
X
i=1
1
(d
i u)(d
i+1 u)
+ n j
X
i=1
1
(e
i+1 v)(e
i v)
!
uv
(p) j 1
X
i=1
1
d
i u
1
d
i+1 u
+ n j
X
i=1
1
e
i v
1
e
i+1 v
!
<
uv
(p)
1
d
1 u
+ 1
e
1 v
uv
(p)
1
v +
1
u
1
wheneveruandv arenonnegativenumberssatisfyingu+v=(p): ut
b℄ Suppose n 3 and 2R. Show that (p 0
p) has only realzeros
and(p 0
p)>(p):
℄ What happenswhen phasonlyrealzerosbut theyare notneessarily
distint?
E.13 Fejer'sTheoremontheZerosofMuntzPolynomials. Thefollowing
prettyresultsofFejermayalsobefoundin PolyaandSzeg}o[76℄:
Supposethat(
k )
1
k =0
isaninreasingsequeneofnonnegativeintegers
with
0
=0.
a℄ Let
p(z):=
n
X
k =0 a
k z
k
; a
k
2C ; a
0 a
1 6=0:
Thenphasatleastone zeroz
0
2C sothat
jz
0 j
2
3
n
(
2
1 )(
3
1 )(
n
1 )
1=1
a
0
a
1
1=1
:
Outline.Wesaythatz
1
2C isnotlessthanz
2
2C ifjz
2 jjz
1
j:Studying
q(z):=z
n
p(z 1
);weneedtoshowthat thelargestzeroof
q(z)=a
0 x
n
+ n
X
k =1 a
k x
n
k
isnotlessthan
(
2
1 )(
3
1 )(
n
1 )
2
3
n
1=
1
a
1
a
0
1=
1
:
Weprovethisstatementbyindutiononn:Thestatementisobviouslytrue
forn=1:Nowassumethatthestatementistrueforn 1:Itfollowsfrom
Luas'theoremthatifqisapolynomialwithomplexoeÆients,thenthe
0
BytheaboveorollaryofLuas'theorem,itissuÆienttoprovethat
thelargestzeroof
z n
1 n+1
q 0
(z)=
n a
0 z
n
1
+ n 1
X
k =1 (
n
k )a
k z
n
1
k
isnotlessthan
(
2
1 )(
3
1 )(
n
1 )
2
3
n
1=1
a
1
a
0
1=1
:
However,this istruebytheindutivehypothesis. ut
b℄ Suppose
f(z)= 1
X
k =0 a
k z
k
; a
k 2C
is anentire funtion so that P
1
k =1 1=
k
<1; that is, the entire funtion
f satises the Fejer gap ondition. Show that there is a z
0
2 C so that
f(z
0 )=0:
Hint:Useparta℄. ut
2
Some Speial Polynomials
Overview
Chebyshevpolynomialsareintroduedandtheirentralroleinproblemsin
theuniformnormon[ 1;1℄isexplored.Sequenesoforthogonalfuntions
are thenexamined in somegenerality,although ourprimary interestis in
orthogonalpolynomials(andrationalfuntions). Thethird setionof this
hapter is onerned withorthogonal polynomials;it introdues themost
lassialofthese.Thesepolynomialssatisfymanyextremalproperties,sim-
ilartothoseoftheChebyshevpolynomials,butwithrespetto(weighted)
L
2
norms. The nal setion of the hapter deals with polynomials with
positiveoeÆientsinvarious bases.
2.1ChebyshevPolynomials
The ubiquitousChebyshev polynomials lie at the heart of many analyti
problems, partiularly problems in C[a;b℄, the spae of real-valued on-
tinuous funtions equipped with the uniform (supremum) norm, kk
[a;b℄
.
Throughout thisbook,foranyreal-or omplex-valuedfuntion f dened
on[a;b℄,
kfk
[a;b℄
:= sup jf(x)j:
TheChebyshevpolynomialsaredenedby
T
n
(x):=os(narosx); x2[ 1;1℄;
= 1
2
(x+ p
x 2
1) n
+(x p
x 2
1) n
; x2C ; (2.1.1)
= n
2 bn=2
X
k =0 ( 1)
k
(n k 1)!
k!(n 2k)!
(2x) n 2k
; x2C :
These elementary equivalenes are left for the reader (see E.1). The nth
Chebyshevpolynomialhasthefollowingequiosillationpropertyon[ 1;1℄.
There existn+1points
i
2[ 1;1℄with 1=
n
<
n 1
<<
0
=1
sothat
(2:1:2) T
n (
j
)=( 1) n j
kT
n k
[ 1;1℄
=( 1) n j
; j=0;1;:::;n:
InotherwordsT
n 2P
n
takesthevalueskT
n k
[ 1;1℄
withalternatingsign
themaximum possiblenumberof timeson [ 1;1℄:(These extremepoints
arejustthepointsos(k=n),k=0;1;:::;n.)TheChebyshevpolynomial
T
n
satisesthefollowingextremalproperty:
Theorem2.1.1. Wehave
min
p2P
n 1 kx
n
p(x)k
[ 1;1℄
=k2 1 n
T
n k
[ 1;1℄
=2 1 n
;
wherethe minimum isuniquelyattainedby p(x)=x n
2 1 n
T
n (x):
Proof. Observethat, whiletheminimumistakenoverP
n 1
;weneedonly
onsiderp2P
n 1
;sinetakingtherealpartofap2P
n 1
anonlyimprove
theestimate.FromtheaboveformulasforT
n
wehave
2 1 n
T
n (x)=x
n
+s(x); s2P
n 1 :
Nowsupposethere existsq2P
n 1 with
(2:1:3) kx
n
q(x)k
[ 1;1℄
<2 1 n
:
Then
2 1 n
T
n
(x) (x n
q(x))=s(x)+q(x)2P
n 1
hanges sign betweenanytwoonseutiveextrema ofT
n
; heneit has at
leastnzerosin( 1;1);andthusitmustvanishidentially.Thisontradits
(2.1.3),andwearedoneuptoprovinguniqueness(thisisleftas E.2). ut
Comments,Exerises,andExamples.
TheChebyshevpolynomialsT
n
arenamedaftertheversatileRussianmath-
ematiian, P. L. Chebyshev (1821{1894).The T omes from the spelling
Thebyhef (or some suh variant; there are many in the literature). A
wealth of information on these polynomials may be found in Rivlin [90℄.
Throughout later setions of this book the Chebyshev polynomials will
keepreurring. The initial exerisesexplore elementary properties of the
Chebyshevpolynomials.
Erd}os[39℄provedthatfort2T
n
withktk
R
1;thelengthofthegraph
ofton[0;2℄isthelongestifandonlyiftisoftheformt()=os(n+)
with some 2 R (see E.6). He onjetured that for any p 2 P
n with
kpk
[ 1;1℄
1;the maximum arlength isattained by thenth Chebyshev
polynomialT
n
.ThisisprovedinBojanov[82b℄.Kristiansen[79℄alsolaims
aproof. InE.9 thereduibilityof T
n
is onsidered,and in E.11the basi
propertiesofthetransnitediameterareestablished.
E.1 BasiProperties.
a℄ Establish the equivalene of the three representations of T
n
given in
equation(2:1:1):
Hint: osn = 1
2
[(os+isin) n
+((os isin) n
℄: To get the third
representation,useE.3b℄. ut
b℄ ThezerosofT
n
arepreiselythepoints
x
k
=os (2k 1)
2n
; k=1;2;:::;n:
℄ TheextremaofT
n
(x)in[ 1;1℄arepreiselythepoints
k
=os k
n
; k=0;1;:::;n:
d℄ Observethat thezerosofT
n andT
n+1
interlae,as dotheextrema.
E.2 UniquenessoftheMinimuminTheorem2.1.1. Provetheuniqueness
oftheminimuminTheorem2.1.1.
Hint:Assume thatq2P
n 1 and
kx n
q(x)k
[ 1;1℄
2 1 n
:
Then
h(x):=2 1 n
T
n
(x) Re(x n
q(x))
denes a polynomial from P
n 1
on R having at least n zeros (ounted
aordingtotheirmultipliities).Thus
2 1 n
T (x)=Re(x n
q(x)); x2R;
whih,togetherwiththepreviousinequality,impliesthatq(x)isrealwhen-
ever T
n
(x) = 1: NowE.6 of Setion 1.1 (Lagrangeinterpolation) yields
thatq hasrealoeÆients.Hene
2 1 n
T
n (x)=x
n
q(x); x2R:
u t
E.3 FurtherPropertiesofT
n .
a℄ Composition. ShowthatT
nm (x)=T
n (T
m (x)):
b℄ Three-TermReursion. Showthat
T
n
(x)=2xT
n 1 (x) T
n 2
(x); n=2;3;::::
℄ Verifythat
T
0 (x)=1
T
1 (x)=x
T
2
(x)=2x 2
1
T
3
(x)=4x 3
3x
T
4
(x)=8x 4
8x 2
+1
T
5
(x)=16x 5
20x 3
+5x:
Notethat T
n
isevenfornevenandoddfornodd.
d℄ Another Formula for T
n
: Show that T
n
(x) = osh(nosh 1
(x)) for
everyx2Rn[ 1;1℄:
e℄ Dierential Equation. Showthat
(1 x 2
)T 00
n
(x) xT 0
n (x)+n
2
T
n
(x)=0:
f℄ AnIdentity. Showthat
T
n (x)=
T 0
n+1 (x)
2n+2 T
0
n 1 (x)
2n 2 :
g℄ Orthogonality. Showthat
2
Z
1
1 T
n (x)T
m (x)dx
p
1 x 2
=Æ
n;m :=
0; n6=m
1; n=m>0:
h℄ GeneratingFuntion. Showthat
1 yx
1 2yx+y 2
= 1
X
n=0 T
n (x)y
n
; x2[ 1;1℄; jyj<1:
i℄ AnotherRepresentationofT
n
: Showthat
T
n (x)=
bn=2
X
k =0
n
2k
x n 2k
(x 2
1) k
:
j℄ AnotherIdentity. Showthat
T
n 1
2 (x+x
1
)
= 1
2 (x
n
+x n
):
E.4 Approximation tox k
on[0;1℄.
a℄ Let T
n
(x) = T
n
(2x 1) be the nth Chebyshev polynomialshifted to
theinterval[0;1℄:Suppose
T
n (x)=
n
X
k =0 b
k x
k
:
Showthat foreahk=0;1;:::;n;
min
j2R kx
k n
X
j=0
j6=k
j x
j
k
[0;1℄
=kb 1
k T
n k
[0;1℄
:
Hint:Proeedas intheproofofTheorem2.1.1anduseE.6ofSetion 1.3.
u t
b℄ Whydoesthisnothold forT
n
on[ 1;1℄?
E.5 A CompositionCharaterization. Suppose (p
n )
1
n=1
is asequene of
polynomialsofdegreenandforallpositiveintegersnandm
p
n Æp
m
=p
nm :
Thenthereexistsalineartransformationw(x)=x+ so that
wÆp
n Æw
1
=x n
; n=1;2;:::
or
wÆp
n Æw
1
=T
n
; n=1;2;::::
This resultis due to Blokand Thielman [51℄.Theproof outlined in this
a℄ Let
q(x):=a
0 +a
1 x+a
2 x
2
; a
2
6=0; and v(x):=
x
a
2 a
1
2a
2 :
Then
v 1
(q(v(x)))=x 2
+ with :=a
0 a
2 +(a
1
=2) (a 2
1
=4):
b℄ Let q(x) = a
0 +a
1 x +a
2 x
2
; a
2
6= 0: Then there is at most one
polynomialp
n
ofdegreeexatlynsothat
p
n
(q(x))=q(p
n (x)):
Hint:Bya℄wemayassumeq(x)=x 2
+:Nowsupposer,s2P
n nP
n 1
;
r(x 2
+)=r 2
(x)+
and
s(x 2
+)=s 2
(x)+:
Thenu:=r s2P
n 1
satises
u(x 2
+)=u(x)(r(x)+s(x))
from whih wededue, by omparing degrees onboth sides, that n =0:
(Notethattheaboveonditionsimplyrandsmoni.) ut
℄ Finish theproofoftheinitial statementofthisexerise.
Thisis aspeialaseofamoregeneraltheorem ofRitt[23℄that lassies
allrationalfuntions randsthatommuteinthesensethat rÆs=sÆr:
d℄ AnotherCompositionCharaterization. Supposep2P
n
hastheprop-
ertythatthelosureoftheset
I
p
:=fz2C :p [k ℄
(z)=0 forsome k=1;2;:::g
istheinterval[ 1;1℄;where p [k ℄
isthekthiterate ofp;that is,
p [1℄
:=p and p [k ℄
:=pÆp [k 1℄
for k=2;3;::::
Thenp(x)=T
n (x):
e℄ Let
r
n
(x)=tan(ntan 1
(x)):
Showthat r
n
isarationalfuntionin R
n;n
;andobservethat
r
n Ær
m
=r
nm :
E.6 Trigonometri Polynomials of Longest Ar Length. Theorem 5.1.3
(Bernstein-Szeg}oinequality)assertsthat
t 0
() 2
+n 2
t 2
()n 2
ktk 2
R
foreveryt2T
n
and2R: UsethistoprovethefollowingresultofErd}os.
For t 2 T
n
with ktk
R
1; the length of the graph of t on [0;2℄ is the
longestifandonlyifitisoftheformt()=os(n+)forsome2R:
Hint:Supposet2T
n
withktk
R
=1:Lets():=osn:If
1<t(
1 )=s(
2 )<1
holds,thenbytheBernstein-Szeg}oinequality(seealsoE.5of Setion5.1)
jt 0
(
1
)jn(1 t 2
(
1 ))
1=2
=n(1 s 2
(
2 ))
1=2
=js 0
(
2 )j;
and ifequalityholdsfor one pairof
1 ,
2
; thenit holdsforall pairs,and
t()os(n+)forsome2R:Supposet
n
()6os(n+):Let and
bemonotone ars of thegraphsof y = t() and y = s(); respetively,
withendpointsofeahhavingthesameordinatesy
1 andy
2
:Letjjandjj
bethelengthof and;respetively,andletj
x
jandj
x
jbethelengthof
theprojetionof and;respetively,onthex-axis.Showthat
jj<jj+(j
x j j
x j)
byapproximating and byapolygonalline orrespondingtoasubdivi-
sionoftheintervalwithendpoints,y
1 andy
2
onthey-axis. ut
E.7 MoniPolynomials withMinimalNormonan Interval.
a℄ Theuniquemonipolynomialp2P
n
minimizingkpk
[a;b℄
isgivenby
p(x)=2
b a
4
n
T
n
2x a b
b a
:
b℄ Let0<a<b.Find allmonipolynomialsp2P
n
minimizing
kpk
[ b; a℄[[a;b℄
:
(For twointervalsof dierentlengthsthis isamuh harderproblem. The
problem was originally due to Zolotarev andis solvedin termsof ellipti
