Classical Orthogonal Polynomials
Dimitar K. Dimitrov y
Departamento de Ci^encias de Computac~ao e Estatstica, IBILCE, UniversidadeEstadual
Paulista, 15054-000 S~ao Jose do Rio Preto, SP, Brazil
Abstract. Let P (;)
n
; n =0;1;:::; ; > 1 be the Jacobi polynomials, orthogonal on ( 1;1)
with respect to the weight function (1 x)
(1+x)
. Denote by x
n;k
(;);k = 1;:::;n the zeros of
P (;)
n
enumerated in decreasingorder. Lower boundfor x
n;n
(;) and upper boundfor x
n;1
(;) are
obtained as a consequence of the electrostatic interpretation of the zeros of P (;)
n
. The results are
applied then toobtain bounds for the extreme zeros of Laguerre andHermite polynomials.
1 Introduction
One of the important reasons forthe interest in the zeros of Jacobi polynomials is the fact that
x
n;k
(;);k = 1;:::;n admit a very interesting electrostatic interpretation. They are the points of
equilibriumofnunitchargesin( 1;1) intheeldgenerated bycharges(+1)=2at 1and (+1)=2
at 1, wherethe charges repelleach other according to the lawof logarithmic potential. It isworth
mentioning that this electrostatic eld is not the eld of point charges but of long straight wires
perpendicularto the real axis [9]. Thisbeautiful interpretation is dueto Stieltjes[10 , 11 , 12 ]. Szego
[13 , Section6.7]provedthattheequilibriumisaminimumoftheenergyofthedescribedeld. Werefer
to thereview papers[14 , 15 ]forthehistoricalbackground andother resultsconcerningelectrostatics
of zeros oforthogonal polynomials.
Thus, anyresult aboutthebehaviourof x
n;k
(;) provides informationaboutthe above electro-
staticeld. Inthispaper,theoppositeapproachissuggested. WeemploytheaboveresultsofStielties
and Szegoto obtainboundsfortheextreme zeros ofJacobipolynomials.
Foreverypositiveintegernandforanyreal > 1and > 1,thefunctionB(n;;)isdened
by
B(n;;) = ( 2
2
) 2
+
(4n(n+++1)+(+)(++2))
4n(n+++1)+2(++2) ( ) 2
:
ResearchsupportedbytheBrazilianfoudationCNPqunderGrant300645/95-3andtheBulgarianScienceFoundation
underGrantMM-414.
y
OnleavefromTheUniversityofRousse,Bulgaria
x
n;n
(;)>
2
2
p
B(n;;)
4n(n+++1)+(+)(++2)
(1.1)
and
x
n;1
(;)<
2
2
+ p
B(n;;)
4n(n+++1)+(+)(++2)
(1.2)
hold.
Asconsequences weobtainboundsforthe extremezeroof Laguerre andHermite polynomials.
Corollary 1 Leth
n;1
bethe largest zero of the Hermite polynomial H
n
(x). Then
h
n;1
p
2n 1: (1.3)
Corollary 2 Letx
n;n
() andx
n;1
() bethe smallest andlargest zero Laguerrepolynomial L ()
n (x),
respectively. Then for any > 1 the inequalities
x
n;n
()2n++1 q
(2n++1) 2
+1 2
(1.4)
and
x
n;1
()2n++1+ q
(2n++1) 2
+1 2
(1.5)
hold.
Although the bounds obtained in Theorem 1 and Corollaries 1 and 2 are not the sharpest one,
theyareasymptoticalysharpinthesencethatasndiverges,theybehavelikethebestknownbounds.
SharperboundsfortheextremezerosofJacobipolynomialsaregivenbyElbert,LaforgiaandRodono
[3 ]andbyIsmailandLi[7]. ThebestlowerboundforthesmallestzeroofLaguerrepolynomialisthat
obtainedbyIsmailandLi[7 ]and thebestupperboundforthelargestzeroofLaguerre polynomialis
givenbySzego[13 ,Theorem6.32]. Inequality(1.3)isslightlysharperthantheinequalityh
n;1
<
p
2n,
which wasobtained recentlybyIfantis andSiafarikas[6 ].
2 Proof of Theorem 1
Consider theelectostatic eld,described above. Charges (+1)=2 and (+1)=2 are distributed
along long wires wich are perpendicular to the real axes and intersect it at 1 and 1, respectively.
Theremainingnunitchargesaredistributedalonglongwireswhicharealsoperpendiculartothereal
axes and pass throughthe pointsx
1
;:::;x
n
,where 1<x
1
<:::<x
n
<1:Then the energyof the
eldis given by
L(x
1
;:::;x
n )=
+1
2 n
X
i=1 log
1
j1 x
i j
+ +1
2 n
X
i=1 log
1
j1+x
i j
+ X
1i<jn log
1
jx
j x
i j
:
T(x
1
;:::;x
n ):=
n
Y
i=1 (1 x
i )
(+1)=2
(1+x
i )
(+1)=2 Y
1i<jn jx
j x
i
j: (2.1)
ismaximal. Stiltjesand Szego's resultsaresummarizedinthe followingtheorem:
Theorem A ([13 , Theorem6.7.1]) Let > 1; > 1; and let fx
i
g; 1x
i
1, be a systemof
values for which the expression (2.1) becomes a maximum. Thenthe fx
i
g are the zeros of the Jacobi
polynomial P (;)
n
(x).
The proof is straightforward and uses thefact that y(x)=P (;)
n
(x) isthe unique(up to a constant
factor)nonzero polynomialsolutionofthedierentialequation
(1 x 2
)y 00
+( (++2)x)y 0
+n(n+++1)y=0: (2.2)
Moreprecisely,oneneeds thefollowingtheorem:
Theorem B ([13 ,Theorem4.2.2]) Let > 1 and > 1. Thedierential equation
(1 x 2
)y 00
+( (++2)x)y 0
+y=0;
where is a parameter, has a polynomial solution not identically zero if and only if has the form
n(n++ +1); n = 0;1;:::. This solution is const P (;)
n
(x), and no solution which is linearly
independent of P (;)
n
(x) can be a polynomial.
It is well-known (see [13 , p. 117]) and easy to see that for any nonnegative integer l, y(x) =
P (;)
n
(x)is asolutionof thedierentialequation
(1 x 2
)y (l +2)
+( (++2l+2)x)y (l +1)
+( n(n+++1) l(l+++1))y (l )
=0:
(2.3)
Proof of Theorem 1. Considerthe functionT 2
(x
1
;:::;x
n
) on thesimplex:=f 1<x
1
<:::<
x
n
<1g. Observe that
T 2
(x
1
;:::;x
n
) = (x
1
;:::;x
k 1
;x
k+1
;:::;x
n )(1 x
k )
+1
(1+x
k )
+1 Y
i6=k (x
k x
i )
2
= (x
1
;:::;x
k 1
;x
k+1
;:::;x
n )(1 x
k )
+1
(1+x
k )
+1
! 2
k (x
k ):
Here the function(x
1
;:::;x
k 1
;x
k+1
;:::;x
n
) =(xnx
k
) doesnot depend on x
k and !
k (x) :=
!(x)=(x x
k
),where !(x):=
Q
n
i=1 (x x
i ).
Dierentiatingthe latter identityand usingtheequalities ! (l +1)
(x
k
)=(l+1)!
(l )
k (x
k
); l=0;1;2,
we obtainconsecutively
@
@x
k T
2
(x
1
;:::;x
n
) = (xnx
k )
@
@x
k
(1 x
k )
+1
(1+x
k )
+1
! 2
k (x
k )
= (xnx
k
) (1 x
k )
(1+x
k )
!
k (x
k )
2(1 x 2
k )!
0
k (x
k
)+( (++2)x
k )!
k (x
k )
= (xnx
k
)(1 x
k )
(1+x
k )
! 0
(x
k )
(1 x 2
k )!
00
(x
k
)+( (++2)x
k )!
0
(x
k :
(2.4)
1
(xnx
k )
@ 2
@x 2
k T
2
(x
1
;:::;x
n ) =
@
@x
k
(1 x
k )
(1+x
k )
!
k (x
k )
2(1 x 2
k )!
0
k (x
k
)+( (++2)x
k )!
k (x
k )
+(1 x
k )
(1+x
k )
!
k (x
k )
2(1 x 2
k )!
00
k (x
k
)+( (++6)x
k )!
0
k (x
k )
(++2)!
k (x
k )g
=
@
@x
k
(1 x
k )
(1+x
k )
!
k (x
k )
(1 x 2
k )!
00
(x
k
)+( (++2)x
k )!
0
(x
k )
+(1 x
k )
(1+x
k )
! 0
(x
k )
n
(1 x 2
k )
2!
000
(x
k )
3
+( (++6)x
k )
! 00
(x
k )
2
(++2)!
0
(x
k )g:
(2.5)
The functionT 2
ispositiveinandvanisheson theboundaryof. Thereforethemaximalvalue
isattainedatan intrinsicpointof . Thismeansthatthemaximumislocal. Now(2.4) andTheorem
Bcompletetheproofof TheoremA.
AnotherimplicationofthefactthatthemaximumofT 2
islocalanditisunique,isthat@ 2
T 2
=@x 2
k
arenegativefork =1;:::;nattheonlyextremalpoint(x
n;n
(;);:::;x
n;1
(;)). Since 1
(xnx
k )>
0 and (1 x 2
k )!
00
(x
k
)+( (++2)x
k )!
0
(x
k
)=0forx
k
=x
n;k
(;); k =1;:::;n, then
F(x
k
) := ! 0
(x
k )
2
3 (1 x
2
k )!
000
(x
k )+
1
2
( (++6)x
k )!
00
(x
k
) (++2)!
0
(x
k )
< 0; for x
k
=x
n;k
(;); k =1;:::;n:
We use (2.3) forl=1and l=0 consecutivelyinorder to simplifyF(x
k ):
6F(x
k
) = !
0
(x
k
) f [(+ 2)x
k
+ ]!
00
(x
k )
[4n(n+++1)+2(++2)]!
0
(x
k )g
= (!
0
(x
k ))
2
1 x 2
k
f [(+ 2)x
k
+ ][(++2)x
k
+ ]
[4n(n+++1)+2(++2)] (1 x 2
k ) :
Thereforetheinequalities
[(+ 2)x
k
+ ][(++2)x
k
+ ]
[4n(n+++1)+2(++2)] (1 x 2
k ) <0
hold for x
k
= x
n;k
(;); k = 1;:::;n. This latter statement is equivalent to the statement of the
theorem. 2
3 Bounds for the extreme zeros of Gegenbauer, Laguerre and
Hermite polynomials
Let
n;1
()bethelargest zeroof theGegenbauer (ultraspherical)polynomialC
n
. Therearethree
recent resultsaboutupperbounds for
n;1
(). We refer to [8 ] fora more detailedreview. Elbertand
n;1 ()<
s
n 2
+2n
n 2
+2n+ 2
+
for 0;
and Ifantisand Siafarikas[5 ] proved theinequality
n;1 ()<
s
n 2
+2n n
n 2
+2n+ 2
n
for >1=2:
Forsterand Petras[4 ] obtainedtheupperbound
n;1 ()<
s
n 2
+2n 1=2
n 2
+2n+ 2
1=4
for > 1=2: (3.1)
Substituting= = 1=2 in(1.2),weobtain
n;1 ()<
s
n 2
+2n++1=2
n 2
+2n+ 2
1=4
for > 1=2: (3.2)
Although our estimate (3.2) is obviously weaker than (3.1), its advantage is that it is an immediate
consequenceofanestimateforx
n;1
(;), whileallthepreviousresultsholdspecicallyforthelargest
zerosofultrasphericalpolynomials. Weemploytheinequality(3.1) inorderto giveaveryshortproof
of Corollary1.
Proof of Corollary 1. It waspointedoutbyElbertandLaforgia[2 ] thatforanypositiveinteger n
p
n;1
() !h
n;1
as !1:
Then multiplying(3.1) by p
andletting to diverge, we obtainthedesired result. 2
Proof of Corollary 2. LetA(n;;) be denedby
A(n;;)=4n(n+++1)+(+)(++2)+ 2
2
:
Then inequalities(1.1) and (1.2) yield
1 x
n;n
(;)<
A(n;;)+ p
B(n;;)
4n(n+++1)+(+)(++2)
(3.3)
and
1 x
n;1
(;)>
A(n;;) p
B(n;;)
4n(n+++1)+(+)(++2)
; (3.4)
respectively. On theotherhand, by(6.71.11) in[13 ],we have
lim
!1
2 (1 x
n;n
(;))=x
n;1
() (3.5)
and
lim
!1
2 (1 x
n;1
(;))=x
n;n
(): (3.6)
A(n;;)=2
(2n++1)+2n(n++1)+ 2
+
andB(n;;)isaquadraticpolynomialin withleadingcoeÆcient4 (2n++1) 2
+1 2
,then
therelations (3.3) and (3.5) imply(1.5). Similarly,(3.4) and (3.6) yield(1.4). 2
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