Real zeros of 2F1 hypergeometric polynomials

0.12 MB·English

by  D. Dominici

Real zeros of F hypergeometric polynomials
2 1
DDominici1
DepartmentofMathematics,StateUniversityofNewYorkatNewPaltz,1HawkDr.Suite9,NewPaltz,NY12561-2443,USA
SJJohnston
DepartmentofMathematicalSciences,UniversityofSouthAfrica,POBox392,UNISA0003,SouthAfrica
KJordaan2
3 DepartmentofMathematicsandAppliedMathematics,UniversityofPretoria,LynnwoodRoad,Pretoria,0002,SouthAfrica
1
0
2

n
a
J Abstract

1
Weuseamethodbasedonthedivisionalgorithmtodetermineallthevaluesoftherealparametersbandcforwhichthe
2
hypergeometricpolynomials2F1( n,b;c;z)haven real, simplezeros. Furthermore,we usethe quasi-orthogonality
−
] ofJacobipolynomialstodeterminetheintervalsonthereallinewherethezerosarelocated.
A
C Keywords: Orthogonalpolynomials,zeros,hypergeometricpolynomials.
2000MSC:33C05,33C45,42C05.
.
h
t
a
m
[ 1. Introduction

1
The F hypergeometricfunctionisdefinedby(cf. [1])
v 2 1
1
7 F (a,b;c;z)=1+ ∞ (a)k(b)k zk, z <1,
7 2 1 (c) k! | |
4 Xk=1 k
1. wherea, bandcarecomplexparameters, c<N0 = 0,1,2,... and
− { }
0
3 α(α+1)...(α+k 1) , k N,
1 (α)k =1 − , k =∈ 0, α,0
v: 
Xi isPochhammer’ssymbol.Thisseriesconvergeswhen z <1andalsowhenz=1providedthatRe(c a b)>0and
| | − −
whenz = 1providedthatRe(c a b+1) > 0. Whenoneofthenumeratorparametersisequaltoanonpositive
ar integer,say−a= n, n N ,thes−eries−terminatesandthefunctionisapolynomialofdegreeninz.
0
− ∈
The problem of describing the zeros of the polynomials F ( n,b;c;z) when b and c are complex arbitrary
2 1
−
parameters, has not been solved. Even when b and c are both real, the only cases that have been fully analyzed
imposeadditionalrestrictionsonbandc. Recentpublications(cf.[4],[6],[7],[8],[11]and[13])consideredthezero
locationofspecialclassesof F ( n,b;c;z)withrestrictionsonthe parameters bandc. Resultsonthe asymptotic
2 1
−
zerodistributionofcertainclassesof F ( n,b;c;z)havealsoappeared(cf. [5],[10],[14],[15]and[27]).
2 1
−
Emailaddresses:[email protected](DDominici),[email protected](SJJohnston),[email protected](K
Jordaan)
1ResearchbythisauthorwassupportedbyaHumboldtResearchFellowshipforExperiencedResearchersfromtheAlexandervonHumboldt
Foundation.
2ResearchbythisauthorwaspartiallysupportedbytheNationalResearchFoundationundergrantnumber2054423.
PreprintsubmittedtoJournalofComputationalandAppliedMathematics January22,2013
Differenttypes of F ( n,b;c;z)have well-established connectionswith classical orthogonalpolynomials, no-
2 1
−
tably the Jacobi polynomialsand the Gegenbauerorultrasphericalpolynomials(cf. [1]). Forthe rangesof the pa-
rameterswherethesepolynomialsareorthogonal,informationaboutthezerosof F ( n,b;c;z)followsimmediately
2 1
−
fromclassicalresults(cf. [1], [28]). Theasymptoticzero distributionof F ( n,b;c;z)whenbandc dependonn
2 1
−
canbededucedfromrecentresultsbyKuijlaars,Mart´ınez-Finkelshtein,Mart´ınez-Gonza´lezandOrive(cf.[20],[21],
[22], [23])onthe asymptoticzerodistributionofJacobipolynomialsP(α,β)(x) whentheparameters αandβdepend
n
onn. Conversely,ifthedistributionofthezerosof F ( n,b;c;z)isknown,thisleadstoinformationaboutthezero
2 1
−
distribution of other special functions (cf. [6]). This makes knowledge of the zero distribution of F ( n,b;c;z)
2 1
−
extremelyvaluable.
Theorthogonalityofthepolynomials F ( n,b;c;z)giveninthenexttheoremfollowsfromtheorthogonalityof
2 1
−
theJacobipolynomials(cf. [25, p. 257-261])andcanalso be proveddirectlyusingtheRodrigues’formulaforthe
polynomials F ( n,b;c;z)(cf. [1,p. 99])aswasdonein[9]and[21].
2 1
−
Theorem1(cf.[9]). Letn N ,b, c Rand c<N . Then F ( n,b;c;z)isthenthdegreeorthogonalpolynomial
0 0 2 1
∈ ∈ − −
forthen-dependentpositiveweightfunction zc 1(1 z)b c n ontheintervals
− − −
| − |
(i) ( ,0)forc>0andb<1 n;
−∞ −
(ii) (0,1)forc>0andb>c+n 1;
−
(iii) (1, )forc+n 1<b<1 n.
∞ − −
Asaconsequenceoforthogonality,weknowthatforeachn,thenzerosof F ( n,b;c;z)arereal,simpleandlie
2 1
−
intheintervaloforthogonalityforthecorrespondingrangesoftheparameters(see,forexample,[12],Theorem4)as
illustratedinFigure1.
b
G
1
b=c+n 1
−
c
b=1 n
−
G
3 G
2
Figure1:Valuesofbandcforwhich2F1( n,b;c;z)isorthogonalandhasnrealsimplezerosintheintervals(0,1),( ,0)and(1, )areindicated
− −∞ ∞
byregions 1, 2and 3respectively.
G G G
In his classical paper (cf. [19]), Felix Klein obtained results on the precise number of zeros of F (a,b;c;z)
2 1
thatlie in eachof the intervals( ,0),(0,1)and (1, )by generalizingearlierresultsofHilbert(cf. [17]). These
−∞ ∞
Hilbert-Kleinformulasarevalidforhypergeometricfunctionsandnotonlyforpolynomials. Szego¨ recapturedthese
results for the special case of Jacobi polynomials P(α,β)(x), which have a representation as F ( n,b;c;z), in the
n 2 1
−
intervals( , 1),( 1,1)and(1, )(cf. [28],p.145,Theorem6.72). Thenumberandlocationof therealzerosof
−∞ − − ∞
F ( n,b;c;z)forbandcrealcanbededucedasfollows.
2 1
−
2
Theorem2(cf. [11],Theorem3.2). Letn N,b, c Randc>0. Then,
∈ ∈
(i) Forb>c+n,allzerosof F ( n,b;c;z)arerealandlieintheinterval(0,1).
2 1
−
(ii) Forc<b<c+n,c+ j 1<b<c+ j, j=1,2,...,n, F ( n,b;c;z)has jrealzerosin(0,1).Theremaining
2 1
− −
(n j) zeros of F ( n,b;c;z) are all non-realif (n j) is even, while if (n j) is odd, F ( n,b;c;z) has
2 1 2 1
− − − − −
(n j 1)non-realzerosandoneadditionalrealzeroin(1, ).
− − ∞
(iii) For0 < b < c, allthezerosof F ( n,b;c;z)arenon-realifniseven,while ifnisodd, F ( n,b;c;z)has
2 1 2 1
− −
onerealzeroin(1, )andtheother(n 1)zerosarenon-real.
∞ −
(iv) For n < b < 0, j < b < j+1, j = 1,2,...,n, F ( n,b;c;z)has j realnegativezeros. The remaining
2 1
− − − −
(n j) zeros of F ( n,b;c;z) are all non-realif (n j) is even, while if (n j) is odd, F ( n,b;c;z) has
2 1 2 1
− − − − −
(n j 1)non-realzerosandoneadditionalrealzeroin(1, ).
− − ∞
(v) Forb< n,allzerosof F ( n,b;c;z)arerealandnegative.
2 1
− −
Thevaluesoftheparametersbandcforwhich F ( n,b;c;z)hasexactlynrealsimplezerosin(0,1)givenin
2 1
−
Theorem2 (i)and(ii) correspondto thosein Theorem1(ii) whilethe parametervaluesinTheorem1(i) thatensure
thatallthezerosof F ( n,b;c;z)arereal,simpleandnegativearethesameasthoseinTheorem2(iv)and(v).The
2 1
−
valuesofbandcinTheorem1(iii)forwhichnzerosarein(1, )canalsobeobtainedfromTheorem2(iv)and(v)
∞
usingthetransformation(cf. [1,p. 79,(2.3.14)])
(c b)
F ( n,b;c;z)= − n F ( n,b;1 n+b c;1 z)
2 1 2 1
− (c) − − − −
n
duetoPfaff.
AnaturalquestiontoaskiswhethertheparameterrangesinTheorems1and2aretheonlyvaluesofb,c Rfor
∈
which F ( n,b;c;z)havenrealsimplezeros. Inthispaper,weuseamethodthatdoesnotrelyonorthogonalityto
2 1
−
determinealltherealvaluesoftheparametersbandcforwhich F ( n,b;c;z)havenrealsimplezeros. Weapply
2 1
−
analgorithmwhichcountsthezerosofpolynomialswithrealcoefficientsandtheirmultiplicities. Wealsodetermine
theintervalswheretherealzerosarelocatedforthesevaluesofbandc.
2. Thealgorithm
Recallthatgiventwopolynomials f(x)andg(x),withdeg(f) deg(g),thereexistuniquepolynomialsq(x)and
≥
r(x) suchthat f(x) = q(x)g(x)+r(x) with deg(r) < deg(g). We willdenotethe leadingcoefficientofa polynomial
f(x)=a xn+a xn 1+ +a bylc(f)=a .
n n 1 − 0 n
− ···
Weusethefollowingalgorithm(cf. [24]).
Let f(x)bearealpolynomialwithdeg(f)=n 2. Define
≥
f (x):= f(x) and f (x):= f (x)
0 1 ′
andproceedfork Nasfollows:
∈
Ifdeg(f )>0performthedivisionof f by f toobtain
k k 1 k
−
f (x)=q (x)f (x) r (x).
k 1 k 1 k k
− − −
Define
r (x) ifr (x).0
f (x)= k k
k+1 f (x) ifr (x) 0
 k′ k ≡
andgeneratethesequenceofnumbersc1,c2,...where
lc(f )
k+1 ifr (x).0
k
ck =lc(fk 1) .
0 − ifrk(x) 0
≡
 3
When f isconstant,thealgorithmterminates.
k
Notethatthealgorithmmustterminate,sincethedegreesofthepolynomials f (x)decreaseoneachstep.
k
Thenwehavethefollowingtheoremwhichwewillapplyto F ( n,b;c;z).
2 1
−
Theorem3(cf. [24],Theorem10.5.7,p.339). Let f beapolynomialofdegreenwithrealcoefficients. Then f has
onlyrealzeros ifandonlyifthe abovealgorithmproducedn 1non-negativenumbersc ,...,c . Moreover, the
1 n 1
zerosof f areallrealandsimpleifandonlyifthenumbersc ,−...,c areallpositive. −
1 n 1
−
3. Mainresults
We shallassumethroughoutourdiscussionthat b,c R with b,c , 0, 1,..., n+1. Theassumptiononb is
∈ − −
madetoensurethat F ( n,b;c;z)isapolynomialofdegreen.
2 1
−
Proposition4. Letb,c R. Then,
∈
1. Thezerosof F ( 2,b;c;z)arerealandsimpleifandonlyifeither(seeFigure2):
2 1
−
(i) c< 1andc<b<0.
−
(ii) 1<c<0andb>0orb<c.
−
(iii) c>0 andb<0orc<b.
2. Thezerosof F ( 3,b;c;z)arerealandsimpleifandonlyifeither(seeFigure3):
2 1
−
(i) c< 2 and1+c<b< 1.
− −
(ii) 2<c< 1 and 1<b<1+c.
− − −
(iii) c> 1,c,0andb< 1orb>c+1.
− −
b b
b=c+1
b=c
-1 c -2 -1 c
-1
Fhaigsuornely2:reVaallusiemspolfebzearnodscforwhich2F1(−2,b;c;z) Fhaigsuornely3:reVaallusiemspolfebzearnodscforwhich2F1(−3,b;c;z)
Theorem5. Foranyintegern 4,thepolynomial F ( n,b;c;z)hasonlyrealandsimplezerosifandonlyif(c,b)
2 1
≥ −
belongstooneofthefourn-dependentregions ,..., definedby
1 4
R R
= c+n 2<b<2 n ,
1
R { − − }
= c> 1, b<2 n ,
2
R { − − }
= c> 1, b>n 2, b>c+n 2 ,
3
R { − − − }
= 1<c<0, c+n 2<b<n 2 .
4
R {− − − }
4
b
c= 1
−
b=n 2
−
1 0 c
−
b=c+n 1 c=0
−
b=c+n 2
−
b=2 n
−
b=1 n
−
Figure4:Valuesofbandcforwhich2F1( n,b;c;z),n=4,5,... hasnrealsimplezeros
−
Theparametervalues(c,b) describedinTheorem5forwhich F ( n,b;c;z),n = 4,5,...
1 2 3 4 2 1
∈ R ∪R ∪R ∪R −
has n real simple zeros are illustrated by the grey and diagonally shaded regionsin Figure 4 with the grey regions
indicatingthoseparametervaluesthatextendtheresultsinTheorems1and2.
Next, we turn ourattention to the location of the zeros of F ( n,b;c;z)for those parametervalueslocated in
2 1
−
thegreyshadedregionsinFigure4wherethepolynomialsarenolongerorthogonalandthelocationoftherealzeros
cannotbeobtainedusingTheorems1and2. Forthesevaluesof bandc,thepolynomials F ( n,b;c;z)arequasi-
2 1
−
orthogonalof order1 and, in some cases, order2 (cf. [3] and [2]). Theorem3 in [2] and Theorem6 in [18] yield
informationonthezerolocationofquasi-orthogonalpolynomialswithnon-varyingweightfunctions.However,these
resultscannotbeappliedto F ( n,b;c;z)sincetheirweightfunctiondependsonn. Weuseinformationaboutthe
2 1
−
zerosofJacobipolynomialstoobtainthefollowingthreeresults.
Theorem6. Letn Nandb,c R. Then, F ( n,b;c;z)hasallitszerosrealandsimpleand(n 2)ofthemliein
2 1
∈ ∈ − −
(i) (0,1)for 1<c<0andc+n 2<b<c+n 1. Oneoftheremainingzerosliesin(1, )andtheotherone
− − − ∞
in( ,0).
−∞
(ii) (1, )for1 n< b< 2 nandb n+1<c <b n+2. Oneoftheremainingzerosliesin( ,0)andthe
∞ − − − − −∞
otheronein(0,1).
(iii) ( ,0)for 1 < c < 0and1 n < b < 2 n. Oneoftheremainingzerosliesin(1, )andtheotheronein
−∞ − − − ∞
(0,1).
Theorem6appliestotheparametervaluesillustratedinFigure5.
Theorem7. Letn Nandb,c R. Then, F ( n,b;c;z)hasallitszerosrealandsimpleand(n 1)ofthemliein
2 1
∈ ∈ − −
(i) (0,1)for 1<c<0andb>c+n 1. Theremainingzeroisnegative.
− −
(ii) (1, )for1 n<b<2 nandc<b n 1. Theremainingzeroisnegative.
∞ − − − −
(iii) ( ,0)for 1<c<0andb<1 n. Theremainingzeroliesin(0,1).
−∞ − −
TheparametervaluesdescribedinTheorem7areillustratedinFigure6.
Theorem8. Letn Nandb,c R. Then, F ( n,b;c;z)hasallitszerosrealandsimpleand(n 1)ofthemliein
2 1
∈ ∈ − −
(i) (0,1)forc>0andc+n 2<b<c+n 1. Theremainingzeroisintheinterval(1, ).
− − ∞
5
b
c= 1
−
b=n 2
−
1 0 c
−
b=c+n 1 c=0
−
b=c+n 2
−
b=2 n
−
b=1 n
−
Figure5:ValuesofbandccorrespondingtothosedescribedinTheorem6
(ii) (1, )forb<1 nandc+n 2<b<c+n 1. Theremainingzeroliesin(0,1).
∞ − − −
(iii) ( ,0)forc>0and1 n<b<2 n. Theremainingzeroliesin(1, ).
−∞ − − ∞
Figure7illustratestherangeoftheparametersbandcreferredtoinTheorem8.
4. Proofs
ProofofProposition4.
1. Since
2b b(b+1)
F ( 2,b;c;z)=1 z+ z2,
2 1 − − c c(c+1)
weseethat F ( 2,b;c;z)=0ifandonlyif
2 1
−
b(c+1) √b(c+1)(b c)
z= ± − .
b(b+1)
Hence,thezerosof F ( 2,b;c;z)arerealandsimpleifandonlyifb(c+1)(b c)>0.
2 1
− −
2. Thediscriminantof
3b 3b(b+1) b(b+1)(b+2)
F ( 3,b;c;z)=1 z+ z2 z3
2 1 − − c c(c+1) − c(c+1)(c+2)
b2(b+1)(b c 1)(b c)2
isgivenby∆ =108 − − − (cf. [16])andtherefore F ( 3,b;c;z)hasrealsimpleroots
3 c4(c+1)3(c+2)2 2 1 −
ifandonlyif∆ >0.
3
Thefollowingtwolemmaswillbeusedintheproofofourmainresult.
6
b
c= 1
−
1 0 c
−
b=c+n 1 c=0
−
b=c+n 2
−
b=2 n
−
b=1 n
−
Figure6:ValuesofbandccorrespondingtothosedescribedinTheorem7
Lemma9. Let
k n 2k n b 3 k b 1+c
αk,l = (cid:16) −2 (cid:17)l(cid:16) −4− − (cid:17)l(cid:16) − −2 (cid:17)l
k b 2 2k b 1 n k n 1 c
−2− l −4− − l − −2 − l
(cid:16) (cid:17) (cid:16) (cid:17) (cid:16) (cid:17)
andletthesequenceθ berecursivelydefinedby
k
θ =α θ , (1)
k+1 k,1 k 1
−
fork 2,...,n 2 ,with
∈{ − } nb (b c)(n 1)
θ = , θ = − − .
1 − c 2 c(b+n 1)
−
Then,
(b+1)(n+c) n 1
θ = α , k=1,2,... − , (2)
2k c(n+b+1) 1,k $ 2 %
and
nb n 2
θ2k+1 = α2,k , k=0,1,... − . (3)
− c $ 2 %
ProofofLemma9. Weprovetheresultbyinductiononk. Whenk =1,theright-handsideof(2)is
(b+1)(n+c)(cid:16)1−2n(cid:17)(cid:16)−n+4b+1(cid:17)(cid:16)c−2b(cid:17) = (b−c)(n−1)
c(n+b+1) b+1 n+b 1 n+c c(b+n 1)
− 2 − 4− − 2 −
(cid:16) (cid:17)(cid:16) (cid:17)(cid:16) (cid:17)
whichisθ asrequired.
2
We now assume the result is true for k = t and prove the result true for k = t +1. If we let k = t+1 on the
7
b
c= 1
−
1 0 c
−
b=c+n 1 c=0
−
b=c+n 2
−
b=2 n
−
b=1 n
−
Figure7:ValuesofbandccorrespondingtothosedescribedinTheorem8
right-handsideof(2),weobtain
(b+1)(n+c)
RHS = α
c(n+b+1) 1,t+1
= (bc(+n+1)b(n++1c)) (cid:16)b1+−21n ++tt(cid:17)(cid:16)c−2nb+c++t(cid:17)t(cid:16)−n+nb4++b11++t(cid:17)t α1,t since (a)k+1 =(a+k)(a)k
− 2 − 2 − 4−
(1 n+2t(cid:16))(c b+(cid:17)2(cid:16)t)( n (cid:17)b(cid:16) 1+4t) (cid:17)
=θ − − − − − bytheinductivehypothesis
2t( b 1+2t)( n c+2t)( n b+1+4t)
− − − − − −
=α θ
2t+1,1 2t
=θ from(1)
2t+2
andtheresultfollowsbyinduction.
Thesecondrelation(3)maybeprovedbyinductioninasimilarway.
Lemma10. Letn 4. Then,forallk 2,...,n 1 ,
≥ ∈{ − }
(n k)(n+c k)(b+1 k)(b c+1 k)
− − − − − (4)
(n+b+2 2k)(n+b 2k)(n+b+1 2k)2
− − −
and
(n 1)(n+c 1)(b c)
− − − (5)
(n+b 2)(n+b 1)2
− −
arepositiveifandonlyif(c,b) .
1 2 3 4
∈R ∪R ∪R ∪R
ProofofLemma10. Sincen 1>0and(n+b 1)2 >0foralln N,b R,weseethat(5)ispositiveifandonly
− − ∈ ∈
8
if
(c,b) c<1 n, b>c, b<2 n = or
1 1
∈{ − − } A ⊃R
c>1 n, b<c, b<2 n = or
2 2
∈{ − − } A ⊃R
c>1 n, b>c, b>2 n = ( )or
3 3 4
∈{ − − } A ⊃ R ∪R
c<1 n, b<c, b>2 n = .
∈{ − − } ∅
(n k)
Clearly − >0forallk 2,...,n 1 ,n Nandb R,b,,3 n,5 n,...,n 5,n 3.
(n+b+1 2k)2 ∈{ − } ∈ ∈ − − − −
−
Furthermoreb>n 2ifandonlyifn+b+2 2k>0,n+b 2k>0andb+1 k>0forallk 2,...,n 1 .
− − − − ∈{ − }
Hence,whenb>n 2,(4)willbepositiveforallk 2,...,n 1 ifandonlyif
− ∈{ − }
(c,b) b>c+k 1, c+n k >0, k =2,...,n 1 = b>c+n 2, c> 1 = or
3
∈{ − − − } { − − } R
b<c+k 1, c+n k <0, k =2,...,n 1 = b<c+1, c<2 n = or
∈{ − − − } { − } ∅
b<c+k 1, c+n k <0, k =2,...,l b>c+k 1, c+n k >0, k =l+1,...,n 1 = or
∈{ − − }∩{ − − − } ∅
b>c+k 1, c+n k >0, k =2,...,l
∈{ − − }∩
b<c+k 1, c+n k<0, k=l+1,...,n 1 b>n 2 = .
{ − − − }∩{ − } ∅
Similarly,b<2 nifandonlyifb+1 k<0,n+b+2 2k<0andn+b 2k<0fork 2,...,n 1 .Hence,
− − − − ∈{ − }
whenb<2 n,(4)willbepositiveforallk 2,...,n 1 ifandonlyif
− ∈{ − }
(c,b) b>c+k 1, c+n k <0, k =2,...,n 1 = b>c+n 2, c<2 n = or
4 1
∈{ − − − } { − − } A ⊃R
b<c+k 1, c+n k >0, k =2,...,n 1 = b<c+1, c> 1 = or
5 2
∈{ − − − } { − } A ⊃R
b>c+k 1, c+n k <0, k =2,...,l b<c+k 1, c+n k >0, k =l+1,...,n 1 = or
∈{ − − }∩{ − − − } ∅
b<c+k 1, c+n k >0, k =2,...,l b>c+k 1, c+n k <0, k =l+1,...,n 1 = .
∈{ − − }∩{ − − − } ∅
Fortheremainingcasewhere2 n<b<n 2or,morespecifically, 2<b+n 2k<0withb k+1<0forall
− − − − −
k 2,...,n 1 ,theonlynon-emptypossibilityisthat(4)ispositivefork =n 1ifandonlyifc> 1,b>c+n 2
∈{ − } − − −
andn 4<b<n 2whereas(4)ispositivefork 2,3,...,n 2 ifandonlyifc> 1,b>c+n 2andb>n 3.
− − ∈{ − } − − −
Hence,when2 n<b<n 2,(4)ispositiveforallk 2,...,n 1 ifandonlyif(c,b) .
4
− − ∈{ − } ∈R
Since = and = ,theresultfollows.
1 4 1 2 5 2
A ∩A R A ∩A R
ProofofTheorem5. Weapplythealgorithmtothepolynomial
f(z)= F ( n,b;c;z).
2 1
−
Wehave(cf.[25],p.69,ex.1)
nb
f (z)= f (z)= F ( n+1,b+1;c+1;z).
1 ′ 2 1
− c −
UsingRaimundasVidu˜nas’Maplepackageforcontiguousrelationsof F hypergeometricseries(cf. [29],[30]),we
2 1
obtain
1 c+n 1 (b c)(n 1)
f (z)= z − f (z) − − F ( n+2,b;c+1;z).
0 n − b+n 1! 1 − c(b+n 1) 2 1 −
− −
Thisrelationcaneasilybeverifiedbycomparingcoefficients.Thus,
(b c)(n 1)
f (z)=r (z)= − − F ( n+2,b;c+1;z).
2 1 c(b+n 1) 2 1 −
−
Inthenextstep(k=2),weget
f (z)=q (z)f (z) r (z),
1 1 2 2
−
9
with
n(b+n 1)2(b+n 2) (n 2)(c+n 2) (n 1)(c+n 1)
q (z)= − − z+ − − − −
1 (n 1)(c+n 1)(b c)" b+n 3 − b+n 1 #
− − − − −
and
(b+n 1)(b 1 c)n(n 2)
r (z)= − − − − F ( n+3,b 1;c+1;z).
2 c(c+n 1)(b+n 3) 2 1 − −
− −
Setting
f (z)=θ F ( n+k,b+2 k;c+1;z), k 1,...,n 1 ,
k k 2 1
− − ∈{ − }
weseethatingeneralweneedacontiguousrelationoftheform
F ( n+k 1,b+3 k;c+1;z)
2 1
− − −
θ θ
=q (z) k F ( n+k,b+2 k;c+1;z) k+1 F ( n+k+1,b+1 k;c+1;z),
k−1 θk 1 2 1 − − − θk 1 2 1 − −
− −
fork 2,...,n 2 ,with
∈{ − } nb (b c)(n 1)
θ = , θ = − − .
1 − c 2 c(b+n 1)
−
Using Vidu˜nas’package,we obtain(1) fork = 2,3,... andfromLemma9we concludethat θ iswell defined
k
andnon-zeroforallk 1,...,n 1 whenc,0and(c,b) .Thus,
1 2 3 4
∈{ − } ∈R ∪R ∪R ∪R
lc(f )
c = k+1 , k N,
k lc(f ) ∈
k 1
−
whichimpliesthat
(b c)(n 1)( n+2) (b) (c) n! (n 1)(c+n 1)(b c)
c1 = c(−b+n−−1) (c−+1)n−n2−(2n−n2−)2!(−n)nn(b)n = (b−+n−2)(b−+n−1−)2
and,fork 2,...,n 1 ,
∈{ − }
θ ( n+k+1) (b+1 k) (n k+1)!(c+1)
ck = θk+1 − (c+1) n−k−1(n k −1)!n−k−1( n+k− 1) (b+3n−k+k1)
(kn−1 k)(n k+cn)−(kb−1 k−+1)−( c+1+b− k) − n−k+1 − n−k+1
= − − − − − .
(n 2k+b+2)(n 2k+b)(b 2k+1+n)2
− − −
FromLemma10,weknowthatc >0forallk 1,...,n 1 when(c,b) . Theresultnow
k 1 2 3 4
∈{ − } ∈R ∪R ∪R ∪R
followsfromTheorem3.
ProofofTheorem6. From[2],Corollary4(i),weknowthatfor 1<α<0and 1<β<0,theJacobipolynomials
P(nα−1,β−1)(x)haverealsimplezerosand(n 2)ofthemareinthein−terval( 1,1).T−hesmallestzeroissmallerthan 1
andthelargestzeroislargerthan1. Equiv−alently,thesame istrueforthe−zerosofJacobipolynomialsP(α,β)(x)wh−en
n
2<α< 1and 2<β< 1.
− − − −
(i) OneoftheconnectionsbetweenJacobipolynomialsandthepolynomials F ( n,b;c;z)isgivenby(cf. [25],
2 1
−
p. 254,eq. 3)
( 1)n(1+β) x+1
P(α,β)(x)= − n F n,1+α+β+n;1+β; (6)
n n! 2 1 − 2 !
whereα=b n candβ=c 1. Theconditions 2<β< 1and 2<α< 1areequivalentto 1<c<0
− − − − − − − −
and c+n 2 < b < c+n 1. Furthermore, the intervals ( 1,1), (1, ) and ( , 1) are transformed to
− − − ∞ −∞ −
(0,1), (1, ) and ( , 1) respectively under the linear mapping x = 2z 1. Thus, when c ( 1,0) and
∞ −∞ − − ∈ −
b (c+n 2,c+n 1), F ( n,b;c;z)hasn 2real,simplezerosintheinterval(0,1),onezeroin(1, )
2 1
an∈donezer−oin( ,−0)foreach−n N. − ∞
−∞ ∈
10