research article faber polynomial coefficients of...

Research ArticleFaber Polynomial Coefficients of Classes ofMeromorphic Bistarlike Functions

Jay M Jahangiri1 and Samaneh G Hamidi2

1Department of Mathematical Sciences Kent State University Burton OH 44021-9500 USA2Department of Mathematics Brigham Young University Provo UT 84602 USA

Correspondence should be addressed to Jay M Jahangiri jjahangikentedu

Received 15 December 2014 Accepted 8 January 2015

Academic Editor A Zayed

Copyright copy 2015 J M Jahangiri and S G Hamidi This is an open access article distributed under the Creative CommonsAttribution License which permits unrestricted use distribution and reproduction in any medium provided the original work isproperly cited

Applying the Faber polynomial coefficient expansions to certain classes of meromorphic bistarlike functions we demonstrate theunpredictability of their early coefficients and also obtain general coefficient estimates for such functions subject to a given gapseries condition Our results improve some of the coefficient bounds published earlier

Let Σ be the family of functions 119892 of the form

119892 (119911) =1

119911+ 1198870+

infin

sum

119899=1

119887119899119911119899 (1)

that are univalent in the punctured unit disk D = 119911 0 lt

|119911| lt 1For the real constants 119860 and 119861 (0 le 119861 le 1 minus119861 le 119860 lt 119861)

let Σ(119860 119861) consist of functions 119892 isin Σ so that

1199111198921015840(119911)

119892 (119911)= minus

1 + 119860120593 (119911)

1 + 119861120593 (119911) (2)

where 120593(119911) = suminfin

119899=1119888119899119911119899 is a Schwarz function that is 120593(119911) is

analytic in the open unit disk |119911| lt 1 and |120593(119911)| le |119911| lt 1Note that |119888

119899| le 1 (Duren [1]) and the functions in Σ(119860 119861)

are meromorphic starlike in the punctured unit disk D (egsee Clunie [2] and Karunakaran [3]) It has been proved byLibera and Livingston [4] and Karunakaran [3] that |119887

119899| le

(119861 minus 119860)(119899 + 1) for 119892 isin Σ[119860 119861]

The coefficients of ℎ = 119892minus1 the inverse map of 119892 are

given by the Faber polynomial expansion (eg see Airaultand Bouali [5] or Airault and Ren [6 page 349])

ℎ (119908) = 119892minus1(119908) =

1

119908+

infin

sum

119899=0

119861119899119908119899

=1

119908minus 1198870minus sum

119899ge1

1

119899119870119899

119899+1(1198870 1198871 119887

119899) 119908119899

(3)

where 119908 isin D

119870119899

119899+1(1198870 1198871 119887

119899)

= 119899119887119899minus1

01198871+ 119899 (119899 minus 1) 119887

119899minus2

01198872

+1

2119899 (119899 minus 1) (119899 minus 2) 119887

119899minus3

0(1198873+ 1198872

1)

+119899 (119899 minus 1) (119899 minus 2) (119899 minus 3)

3119887119899minus4

0(1198874+ 311988711198872)

+ sum

119895ge5

119887119899minus119895

0119881119895

(4)

and 119881119895is a homogeneous polynomial of degree 119895 in the

variables 1198871 1198872 119887

119899

Hindawi Publishing CorporationInternational Journal of Mathematics and Mathematical SciencesVolume 2015 Article ID 161723 5 pageshttpdxdoiorg1011552015161723

2 International Journal of Mathematics and Mathematical Sciences

In 1923 Lowner [7] proved that the inverse of the Koebefunction 119896(119911) = 119911(1minus119911)

2 provides the best upper bounds forthe coefficients of the inverses of analytic univalent functionsAlthough the estimates for the coefficients of the inversesof analytic univalent functions have been obtained in asurprisingly straightforward way (eg see [8 page 104])the case turns out to be a challenge when the biunivalencycondition is imposed on these functions A function is saidto be biunivalent in a given domain if both the functionand its inverse are univalent there By the same token afunction is said to be bistarlike in a given domain if both thefunction and its inverse are starlike there Finding boundsfor the coefficients of classes of biunivalent functions datesback to 1967 (see Lewin [9]) The interest on the boundsfor the coefficients of subclasses of biunivalent functionspicked up by the publications [10ndash14] where the estimatesfor the first two coefficients of certain classes of biunivalentfunctions were provided Not much is known about thehigher coefficients of the subclasses biunivalent functions asAli et al [13] also declared that finding the bounds for |119886

119899|

119899 ge 4 is an open problem In this paper we use the Faberpolynomial expansions of the functions 119892 and ℎ = 119892

minus1 inΣ(119860 119861) to obtain bounds for their general coefficients |119886

119899| and

provide estimates for the early coefficients of these types offunctions

We will need the following well-known two lemmas thefirst of which can be found in [15] (also see Duren [1])

Lemma 1 Let 119901(119911) = 1 + suminfin

119899=1119901119899119911119899 be so that 119877119890(119901(119911)) gt 0

for |119911| lt 1 If 120572 ge minus12 then

100381610038161003816100381610038161199012+ 1205721199012

1

10038161003816100381610038161003816le 2 + 120572

100381610038161003816100381611990111003816100381610038161003816

2 (5)

Consequently we have the following lemma in which wewill provide a short proof for the sake of completeness

Lemma 2 Consider the Schwarz function 120593(119911) = suminfin

119899=1119888119899119911119899

where |120593(119911)| lt 1 for |119911| lt 1 If 120574 ge 0 then

100381610038161003816100381610038161198882+ 1205741198882

1

10038161003816100381610038161003816le 1 + (120574 minus 1)

100381610038161003816100381611988811003816100381610038161003816

2 (6)

Proof Write

119901 (119911) =[1 + 120593 (119911)]

[1 minus 120593 (119911)] (7)

where 119901(119911) = 1+suminfin

119899=1119901119899119911119899 is so that Re(119901(119911)) gt 0 for |119911| lt 1

Comparing the corresponding coefficients of powers of 119911 in119901(119911) = [1 + 120593(119911)][1 minus 120593(119911)] shows that 119901

1= 21198881and 119901

2=

2(1198882+ 1198882

1)

By substituting for 1199011= 21198881and 119901

2= 2(1198882+ 1198882

1) in (5) we

obtain100381610038161003816100381610038162 (1198882+ 1198882

1) + 120572 (2119888

1)210038161003816100381610038161003816le 2 + 120572

1003816100381610038161003816211988811003816100381610038161003816

2 (8)

or100381610038161003816100381610038161198882+ (1 + 2120572) 119888

2

1

10038161003816100381610038161003816le 1 + 2120572

100381610038161003816100381611988811003816100381610038161003816

2 (9)

Now (6) follows upon substitution of 120574 = 1 + 2120572 ge 0 in theabove inequality

In the following theorem we will observe the unpre-dictability of the early coefficients of the functions 119892 and itsinverse map ℎ = 119892

minus1 in Σ[119860 119861] providing an estimate for thegeneral coefficients of such functions

Theorem 3 For 0 le 119861 le 1 and minus119861 le 119860 lt 119861 let the function119892 and its inverse map ℎ = 119892

minus1 be in Σ[119860 119861] Then

(119894)10038161003816100381610038161198870

1003816100381610038161003816 le

119861 minus 119860

radic2119861 minus 119860

119894119891 2119861 minus 119860 ge 1

119861 minus 119860 119900119905ℎ119890119903119908119894119904119890

(119894119894)10038161003816100381610038161198871

1003816100381610038161003816

le

119861 minus 119860

2minus1 minus 119860 minus 2119861

2 (119861 minus 119860)|1198870|2 119894119891 0 le 119860 le 1 minus 2119861

119861 minus 119860

2 119900119905ℎ119890119903119908119894119904119890

(119894119894119894)

10038161003816100381610038161003816100381610038161003816

1198871plusmn

2119861 minus 119860

2 (119861 minus 119860)1198872

0

10038161003816100381610038161003816100381610038161003816

le119861 minus 119860

2

(119894V) 10038161003816100381610038161198871198991003816100381610038161003816 le

119861 minus 119860

119899 + 1 119894119891 119887

119896= 0 119891119900119903 0 le 119896 le 119899 minus 1

(10)

Proof Consider the function 119892 isin Σ given by (1) Therefore(see [5 6])

1199111198921015840(119911)

119892 (119911)= minus1 minus

infin

sum

119899=0

119865119899+1

(1198870 1198871 1198872 119887

119899) 119911119899+1

(11)

where 119865119899+1

(1198870 1198871 1198872 119887

119899) is a Faber polynomial of degree

119899 + 1 We note that 1198651= minus1198870 1198652= 1198872

0minus 21198871 1198653= minus1198873

0+

311988711198870minus 31198872 1198654= 1198874

0minus 41198872

01198871+ 411988701198872+ 21198872

1minus 41198873 and 119865

5=

minus1198875

0+ 51198873

01198871minus 51198872

01198872minus 511988701198872

1+ 511988711198872+ 511988701198873minus 51198874 In general

(Bouali [16 page 52])

119865119899+1

(1198870 1198871 119887

119899)

= sum

1198941+21198942+sdotsdotsdot+(119899+1)119894

119899+1=119899+1

119860 (1198941 1198942 119894

119899+1) 1198871198941

01198871198942

1sdot sdot sdot 119887119894119899+1

119899

(12)

where

119860 (1198941 1198942 119894

119899+1)

= (minus1)(119899+1)+2119894

1+sdotsdotsdot+(119899+2)119894

119899+1

(1198941+ 1198942+ sdot sdot sdot + 119894

119899+1minus 1) (119899 + 1)

(1198941) (1198942) sdot sdot sdot (119894

119899+1)

(13)

International Journal of Mathematics and Mathematical Sciences 3

Similarly for the inverse map ℎ = 119892minus1 we have

119908ℎ1015840(119908)

ℎ (119908)= minus1 minus

infin

sum

119899=1

119865119899+1

(1198610 1198611 1198612 119861

119899) 119908119899+1

(14)

where119865119899+1

(1198610 1198611 1198612 119861


119899 + 1 given by

119865119899+1

= minus119899 (119899 minus (119899 + 1))

119899 (119899 minus 2119899)119861119899

0

minus119899 (119899 minus (119899 + 1))

(119899 minus 2) (119899 minus (2119899 minus 1))119861119899minus2

01198611

minus119899 (119899 minus (119899 + 1))


01198612

minus119899 (119899 minus (119899 + 1))


0

sdot (1198613+119899 minus (2119899 minus 3)

21198612

1) minus sum

119895ge5

119861119899minus119895

0119870119895

(15)

where 119870119895is a homogeneous polynomial of degree 119895 in the

variables 1198611 1198612 119861

119899minus1and

1198610= minus1198870 119861

119899= minus

1

119899119870119899

119899+1(1198870 1198871 119887

119899) (16)

Since both 119892 and its inverse map ℎ = 119892minus1 are in Σ[119860 119861] the

Faber polynomial expansion yields (also see Duren [1 pages118-119])

1199111198921015840(119911)

119892 (119911)= minus

1 + 119860120593 (119911)

1 + 119861120593 (119911)

= minus1 +

infin

sum

119899=1

(119860 minus 119861)119870minus1

119899(1198881 1198882 119888

119899 119861) 119911119899

(17)

119908ℎ1015840(119908)

ℎ (119908)= minus

1 + 119860120595 (119908)

1 + 119861120595 (119908)

= minus1 +

infin

sum

119899=1

(119860 minus 119861)119870minus1

119899(1198891 1198892 119889

119899 119861) 119908

119899

(18)


119899=1119888119899119911119899 and 120595(119908) = sum

infin

119899=1119889119899119908119899 are two

Schwarz functions that is |120593(119911)| lt 1 for |119911| lt 1 and |120595(119908)| lt1 for |119908| lt 1

In general (see Airault [17] or Airault and Bouali [5]) thecoefficients 119870119901

119899(1198961 1198962 119896

119899 119861) are given by

119870119901

119899(1198961 1198962 119896

119899 119861)

=119901

(119901 minus 119899)119899119896119899

1119861119899minus1

+119901

(119901 minus 119899 + 1) (119899 minus 2)119896119899minus2

11198962119861119899minus2

+119901


11198963119861119899minus3

+119901


1

sdot [1198964119861119899minus4

+119901 minus 119899 + 3

21198962

2119861]

+119901


1

sdot [1198965119861119899minus5

+ (119901 minus 119899 + 4) 11989621198963119861]

+ sum

119895ge6

119896119899minus119895

1119883119895

(19)


variables 1198962 1198963 119896

119899

Comparing the corresponding coefficients of (11) and (17)implies

minus119865119899+1

(1198870 1198871 1198872 119887

119899) = (119860 minus 119861)119870

minus1

119899+1(1198881 1198882 119888

119899+1 119861)

(20)

Similarly comparing the corresponding coefficients of(14) and (18) gives

minus 119865119899+1

(1198610 1198611 1198612 119861

119899)

= (119860 minus 119861)119870minus1

119899+1(1198891 1198892 119889

119899+1 119861)

(21)

Substituting 119899 = 0 119899 = 1 and 119899 = 2 in (16) (20) and (21)respectively yields

1198870= (119860 minus 119861) 119888

1

minus1198870= (119860 minus 119861) 119889

1

(22)

21198871minus 1198872

0= (119861 minus 119860) (119888

2

1119861 minus 1198882)

21198871+ 1198872

0= (119860 minus 119861) (119889

2

1119861 minus 1198892)

(23)

Taking the absolute values of either equation in (22) weobtain |119887

0| le 119861 minus 119860 Obviously from (22) we note that

1198881= minus1198891 Solving the equations in (23) for 1198872

0and then adding

them gives

21198872

0= (119861 minus 119860) (1198882 + 119889

2minus 1198611198882

1minus 1198611198892

1) (24)


Now in light of (22) we conclude that

21198872

0= (119861 minus 119860) (119888

2+ 1198892minus

2119861

(119861 minus 119860)21198872

0) (25)

Once again solving for 11988720and taking the square root of both

sides we obtain

100381610038161003816100381611988701003816100381610038161003816 le

radic(119861 minus 119860)

2(1198882+ 1198892)

2 (2119861 minus 119860)le

119861 minus 119860


(26)

Now the first part of Theorem 3 follows since for 2119861 minus 119860 gt 1

it is easy to see that

119861 minus 119860


lt 119861 minus 119860 (27)

Adding the equations in (23) and using the fact that 1198881= minus1198891

we obtain

41198871= (119861 minus 119860) [(119889

2minus 1198882) minus (119889

2

1minus 1198882

1) 119861] = (119861 minus 119860) (119889

2minus 1198882)

(28)

Dividing by 4 and taking the absolute values of both sidesyield

100381610038161003816100381611988711003816100381610038161003816 le

119861 minus 119860

2 (29)

On the other hand from the second equations in (22) and(23) we obtain

21198871= (119861 minus 119860) (119889

2+ 1198601198892

1minus 2119861119889

2

1) (30)

Taking the absolute values of both sides and applyingLemma 2 it follows that

100381610038161003816100381611988711003816100381610038161003816 le

1

2(119861 minus 119860) (

100381610038161003816100381610038161198892+ 1198601198892

1

10038161003816100381610038161003816+ 2119861

100381610038161003816100381611988911003816100381610038161003816

2)

le1

2(119861 minus 119860) (1 + (119860 minus 1)

100381610038161003816100381611988911003816100381610038161003816

2+ 2119861

100381610038161003816100381611988911003816100381610038161003816

2)

=119861 minus 119860


2 (119861 minus 119860)

100381610038161003816100381611988701003816100381610038161003816

2

(31)

This concludes the second part of Theorem 3 since for 0 lt

119860 lt 1 minus 2119861 we have119861 minus 119860


2 (119861 minus 119860)

100381610038161003816100381611988701003816100381610038161003816

2lt119861 minus 119860

2 (32)

Substituting (22) in (23) we obtain

21198871minus 1198872

0= (119861 minus 119860) (

119861

(119861 minus 119860)21198872

0minus 1198882)

21198871+ 1198872

0= (119860 minus 119861) (

119861

(119860 minus 119861)21198872

0minus 1198892)

(33)

Following a simple algebraic manipulation we obtain thecoefficient body

10038161003816100381610038161003816100381610038161003816

1198871plusmn

2119861 minus 119860

2 (119861 minus 119860)1198872

0

10038161003816100381610038161003816100381610038161003816

le119861 minus 119860

2 (34)

Finally for 119887119896= 0 0 le 119896 le 119899 minus 1 (20) yields

(119899 + 1) 119887119899= (119860 minus 119861) 119888

119899+1 (35)

Solving for 119887119899and taking the absolute values of both sides we

obtain

10038161003816100381610038161198871198991003816100381610038161003816 le

119861 minus 119860

119899 + 1 (36)

Remark 4 The estimate |1198870| le (119861 minus 119860)radic2119861 minus 119860 given by

Theorem 3(i) is better than that given in ([14Theorem 2(i)])

Remark 5 In ([3Theorem 1]) the bound |119887119899| le (119861minus119860)(119899+1)

was declared to be sharp for the coefficients of the function119892 isin Σ[119860 119861]The coefficient estimates |119887

0| le (119861minus119860)radic2119861 minus 119860

and |1198871| le (119861 minus 119860)2 minus ((1 minus 119860 minus 2119861)2(119861 minus 119860))|119887

0|2 given by

Theorem 3 show that the coefficient bound |119887119899| le (119861minus119860)(119899+

1) is not sharp for the meromorphic bistarlike functions thatis if both 119892 and its inverse map 119892

minus1 are in Σ[119860 119861] Findingsharp coefficient bound formeromorphic bistarlike functionsremains an open problem

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

References

[1] P L Duren Univalent Functions vol 259 of Grundlehren derMathematischenWissenschaften Springer New York NY USA1983

[2] J Clunie ldquoOn meromorphic schlicht functionsrdquo Journal of theLondon Mathematical Society vol 34 pp 215ndash216 1959

[3] V Karunakaran ldquoOn a class of meromorphic starlike functionsin the unit discrdquoMathematical Chronicle vol 4 no 2-3 pp 112ndash121 1976

[4] R J Libera and A E Livingston ldquoBounded functions with pos-itive real partrdquo Czechoslovak Mathematical Journal vol 22 pp195ndash209 1972

[5] H Airault and A Bouali ldquoDifferential calculus on the Faberpolynomialsrdquo Bulletin des Sciences Mathematiques vol 130 no3 pp 179ndash222 2006

[6] H Airault and J Ren ldquoAn algebra of differential operators andgenerating functions on the set of univalent functionsrdquo Bulletindes Sciences Mathematiques vol 126 no 5 pp 343ndash367 2002

[7] K Lowner ldquoUntersuchungen uber schlichte konforme Abbil-dungen des Einheitskreises IrdquoMathematische Annalen vol 89no 1-2 pp 103ndash121 1923

[8] J G Krzyz R J Libera and E Złotkiewicz ldquoCoefficients ofinverses of regular starlike functionsrdquo Annales UniversitatisMariae Curie-Skłodowska A vol 33 pp 103ndash110 1979

[9] M Lewin ldquoOn a coefficient problem for bi-univalent functionsrdquoProceedings of the American Mathematical Society vol 18 no 1pp 63ndash63 1967

[10] D A Brannan and T S Taha ldquoOn some classes of bi-univalent functionsrdquo Studia Universitatis Babes-Bolyai SeriesMathematica vol 31 no 2 pp 70ndash77 1986


[11] HM Srivastava A KMishra and P Gochhayat ldquoCertain sub-classes of analytic and bi-univalent functionsrdquo Applied Math-ematics Letters vol 23 no 10 pp 1188ndash1192 2010

[12] B A Frasin and M K Aouf ldquoNew subclasses of bi-univalentfunctionsrdquoAppliedMathematics Letters vol 24 no 9 pp 1569ndash1573 2011

[13] R M Ali S K Lee V Ravichandran and S SupramaniamldquoCoefficient estimates for bi-univalent Ma-Minda starlike andconvex functionsrdquo Applied Mathematics Letters vol 25 no 3pp 344ndash351 2012

[14] S G Hamidi S A Halim and J M Jahangiri ldquoFaber poly-nomial coefficient estimates for meromorphic bi-starlike func-tionsrdquo International Journal of Mathematics and MathematicalSciences vol 2013 Article ID 498159 4 pages 2013

[15] M Jahangiri ldquoOn the coefficients of powers of a class of Bazile-vic functionsrdquo Indian Journal of Pure and Applied Mathematicsvol 17 no 9 pp 1140ndash1144 1986

[16] A Bouali ldquoFaber polynomials Cayley-Hamilton equationand Newton symmetric functionsrdquo Bulletin des SciencesMathematiques vol 130 no 1 pp 49ndash70 2006

[17] H Airault ldquoRemarks on Faber polynomialsrdquo InternationalMathematical Forum vol 3 no 9ndash12 pp 449ndash456 2008

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


In 1923 Lowner [7] proved that the inverse of the Koebefunction 119896(119911) = 119911(1minus119911)

2 provides the best upper bounds forthe coefficients of the inverses of analytic univalent functionsAlthough the estimates for the coefficients of the inversesof analytic univalent functions have been obtained in asurprisingly straightforward way (eg see [8 page 104])the case turns out to be a challenge when the biunivalencycondition is imposed on these functions A function is saidto be biunivalent in a given domain if both the functionand its inverse are univalent there By the same token afunction is said to be bistarlike in a given domain if both thefunction and its inverse are starlike there Finding boundsfor the coefficients of classes of biunivalent functions datesback to 1967 (see Lewin [9]) The interest on the boundsfor the coefficients of subclasses of biunivalent functionspicked up by the publications [10ndash14] where the estimatesfor the first two coefficients of certain classes of biunivalentfunctions were provided Not much is known about thehigher coefficients of the subclasses biunivalent functions asAli et al [13] also declared that finding the bounds for |119886

119899|

119899 ge 4 is an open problem In this paper we use the Faberpolynomial expansions of the functions 119892 and ℎ = 119892

minus1 inΣ(119860 119861) to obtain bounds for their general coefficients |119886

119899| and

provide estimates for the early coefficients of these types offunctions

We will need the following well-known two lemmas thefirst of which can be found in [15] (also see Duren [1])

Lemma 1 Let 119901(119911) = 1 + suminfin

119899=1119901119899119911119899 be so that 119877119890(119901(119911)) gt 0

for |119911| lt 1 If 120572 ge minus12 then

100381610038161003816100381610038161199012+ 1205721199012

1

10038161003816100381610038161003816le 2 + 120572

100381610038161003816100381611990111003816100381610038161003816

2 (5)

Consequently we have the following lemma in which wewill provide a short proof for the sake of completeness

Lemma 2 Consider the Schwarz function 120593(119911) = suminfin

119899=1119888119899119911119899

where |120593(119911)| lt 1 for |119911| lt 1 If 120574 ge 0 then

100381610038161003816100381610038161198882+ 1205741198882

1

10038161003816100381610038161003816le 1 + (120574 minus 1)

100381610038161003816100381611988811003816100381610038161003816

2 (6)

Proof Write

119901 (119911) =[1 + 120593 (119911)]

[1 minus 120593 (119911)] (7)

where 119901(119911) = 1+suminfin

119899=1119901119899119911119899 is so that Re(119901(119911)) gt 0 for |119911| lt 1

Comparing the corresponding coefficients of powers of 119911 in119901(119911) = [1 + 120593(119911)][1 minus 120593(119911)] shows that 119901

1= 21198881and 119901

2=

2(1198882+ 1198882

1)

By substituting for 1199011= 21198881and 119901

2= 2(1198882+ 1198882

1) in (5) we

obtain100381610038161003816100381610038162 (1198882+ 1198882

1) + 120572 (2119888

1)210038161003816100381610038161003816le 2 + 120572

1003816100381610038161003816211988811003816100381610038161003816

2 (8)

or100381610038161003816100381610038161198882+ (1 + 2120572) 119888

2

1

10038161003816100381610038161003816le 1 + 2120572

100381610038161003816100381611988811003816100381610038161003816

2 (9)

Now (6) follows upon substitution of 120574 = 1 + 2120572 ge 0 in theabove inequality

In the following theorem we will observe the unpre-dictability of the early coefficients of the functions 119892 and itsinverse map ℎ = 119892

minus1 in Σ[119860 119861] providing an estimate for thegeneral coefficients of such functions

Theorem 3 For 0 le 119861 le 1 and minus119861 le 119860 lt 119861 let the function119892 and its inverse map ℎ = 119892

minus1 be in Σ[119860 119861] Then

(119894)10038161003816100381610038161198870

1003816100381610038161003816 le

119861 minus 119860


119894119891 2119861 minus 119860 ge 1

119861 minus 119860 119900119905ℎ119890119903119908119894119904119890

(119894119894)10038161003816100381610038161198871

1003816100381610038161003816

le

119861 minus 119860


2 (119861 minus 119860)|1198870|2 119894119891 0 le 119860 le 1 minus 2119861

119861 minus 119860

2 119900119905ℎ119890119903119908119894119904119890

(119894119894119894)

10038161003816100381610038161003816100381610038161003816

1198871plusmn

2119861 minus 119860

2 (119861 minus 119860)1198872

0

10038161003816100381610038161003816100381610038161003816

le119861 minus 119860

2

(119894V) 10038161003816100381610038161198871198991003816100381610038161003816 le

119861 minus 119860

119899 + 1 119894119891 119887

119896= 0 119891119900119903 0 le 119896 le 119899 minus 1

(10)

Proof Consider the function 119892 isin Σ given by (1) Therefore(see [5 6])

1199111198921015840(119911)

119892 (119911)= minus1 minus

infin

sum

119899=0

119865119899+1

(1198870 1198871 1198872 119887

119899) 119911119899+1

(11)

where 119865119899+1

(1198870 1198871 1198872 119887


119899 + 1 We note that 1198651= minus1198870 1198652= 1198872

0minus 21198871 1198653= minus1198873

0+

311988711198870minus 31198872 1198654= 1198874

0minus 41198872

01198871+ 411988701198872+ 21198872

1minus 41198873 and 119865

5=

minus1198875

0+ 51198873

01198871minus 51198872

01198872minus 511988701198872

1+ 511988711198872+ 511988701198873minus 51198874 In general

(Bouali [16 page 52])

119865119899+1

(1198870 1198871 119887

119899)

= sum

1198941+21198942+sdotsdotsdot+(119899+1)119894

119899+1=119899+1

119860 (1198941 1198942 119894

119899+1) 1198871198941

01198871198942

1sdot sdot sdot 119887119894119899+1

119899

(12)

where

119860 (1198941 1198942 119894

119899+1)

= (minus1)(119899+1)+2119894

1+sdotsdotsdot+(119899+2)119894

119899+1

(1198941+ 1198942+ sdot sdot sdot + 119894

119899+1minus 1) (119899 + 1)

(1198941) (1198942) sdot sdot sdot (119894

119899+1)

(13)



119908ℎ1015840(119908)


infin

sum

119899=1

119865119899+1

(1198610 1198611 1198612 119861

119899) 119908119899+1

(14)

where119865119899+1

(1198610 1198611 1198612 119861


119899 + 1 given by

119865119899+1

= minus119899 (119899 minus (119899 + 1))

119899 (119899 minus 2119899)119861119899

0

minus119899 (119899 minus (119899 + 1))


01198611

minus119899 (119899 minus (119899 + 1))


01198612

minus119899 (119899 minus (119899 + 1))


0

sdot (1198613+119899 minus (2119899 minus 3)

21198612

1) minus sum

119895ge5

119861119899minus119895

0119870119895

(15)


variables 1198611 1198612 119861

119899minus1and

1198610= minus1198870 119861

119899= minus

1

119899119870119899

119899+1(1198870 1198871 119887

119899) (16)



1199111198921015840(119911)

119892 (119911)= minus

1 + 119860120593 (119911)

1 + 119861120593 (119911)

= minus1 +

infin

sum

119899=1

(119860 minus 119861)119870minus1

119899(1198881 1198882 119888

119899 119861) 119911119899

(17)

119908ℎ1015840(119908)

ℎ (119908)= minus

1 + 119860120595 (119908)

1 + 119861120595 (119908)

= minus1 +

infin

sum

119899=1

(119860 minus 119861)119870minus1

119899(1198891 1198892 119889

119899 119861) 119908

119899

(18)


119899=1119888119899119911119899 and 120595(119908) = sum

infin

119899=1119889119899119908119899 are two



119899(1198961 1198962 119896

119899 119861) are given by

119870119901

119899(1198961 1198962 119896

119899 119861)

=119901

(119901 minus 119899)119899119896119899

1119861119899minus1

+119901


11198962119861119899minus2

+119901


11198963119861119899minus3

+119901


1

sdot [1198964119861119899minus4

+119901 minus 119899 + 3

21198962

2119861]

+119901


1

sdot [1198965119861119899minus5

+ (119901 minus 119899 + 4) 11989621198963119861]

+ sum

119895ge6

119896119899minus119895

1119883119895

(19)


variables 1198962 1198963 119896

119899


minus119865119899+1

(1198870 1198871 1198872 119887

119899) = (119860 minus 119861)119870

minus1

119899+1(1198881 1198882 119888

119899+1 119861)

(20)


minus 119865119899+1

(1198610 1198611 1198612 119861

119899)

= (119860 minus 119861)119870minus1

119899+1(1198891 1198892 119889

119899+1 119861)

(21)


1198870= (119860 minus 119861) 119888

1

minus1198870= (119860 minus 119861) 119889

1

(22)

21198871minus 1198872

0= (119861 minus 119860) (119888

2

1119861 minus 1198882)

21198871+ 1198872

0= (119860 minus 119861) (119889

2

1119861 minus 1198892)

(23)




0and then adding

them gives

21198872

0= (119861 minus 119860) (1198882 + 119889

2minus 1198611198882

1minus 1198611198892

1) (24)



21198872

0= (119861 minus 119860) (119888

2+ 1198892minus

2119861

(119861 minus 119860)21198872

0) (25)


sides we obtain

100381610038161003816100381611988701003816100381610038161003816 le

radic(119861 minus 119860)

2(1198882+ 1198892)

2 (2119861 minus 119860)le

119861 minus 119860


(26)



119861 minus 119860


lt 119861 minus 119860 (27)


we obtain

41198871= (119861 minus 119860) [(119889

2minus 1198882) minus (119889

2

1minus 1198882

1) 119861] = (119861 minus 119860) (119889

2minus 1198882)

(28)


100381610038161003816100381611988711003816100381610038161003816 le

119861 minus 119860

2 (29)


21198871= (119861 minus 119860) (119889

2+ 1198601198892

1minus 2119861119889

2

1) (30)


100381610038161003816100381611988711003816100381610038161003816 le

1

2(119861 minus 119860) (

100381610038161003816100381610038161198892+ 1198601198892

1

10038161003816100381610038161003816+ 2119861

100381610038161003816100381611988911003816100381610038161003816

2)

le1

2(119861 minus 119860) (1 + (119860 minus 1)

100381610038161003816100381611988911003816100381610038161003816

2+ 2119861

100381610038161003816100381611988911003816100381610038161003816

2)

=119861 minus 119860


2 (119861 minus 119860)

100381610038161003816100381611988701003816100381610038161003816

2

(31)




2 (119861 minus 119860)

100381610038161003816100381611988701003816100381610038161003816

2lt119861 minus 119860

2 (32)


21198871minus 1198872

0= (119861 minus 119860) (

119861

(119861 minus 119860)21198872

0minus 1198882)

21198871+ 1198872

0= (119860 minus 119861) (

119861

(119860 minus 119861)21198872

0minus 1198892)

(33)


10038161003816100381610038161003816100381610038161003816

1198871plusmn

2119861 minus 119860

2 (119861 minus 119860)1198872

0

10038161003816100381610038161003816100381610038161003816

le119861 minus 119860

2 (34)


(119899 + 1) 119887119899= (119860 minus 119861) 119888

119899+1 (35)


obtain

10038161003816100381610038161198871198991003816100381610038161003816 le

119861 minus 119860

119899 + 1 (36)







0|2 given by






References


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119908ℎ1015840(119908)


infin

sum

119899=1

119865119899+1

(1198610 1198611 1198612 119861

119899) 119908119899+1

(14)

where119865119899+1

(1198610 1198611 1198612 119861


119899 + 1 given by

119865119899+1

= minus119899 (119899 minus (119899 + 1))

119899 (119899 minus 2119899)119861119899

0

minus119899 (119899 minus (119899 + 1))


01198611

minus119899 (119899 minus (119899 + 1))


01198612

minus119899 (119899 minus (119899 + 1))


0

sdot (1198613+119899 minus (2119899 minus 3)

21198612

1) minus sum

119895ge5

119861119899minus119895

0119870119895

(15)


variables 1198611 1198612 119861

119899minus1and

1198610= minus1198870 119861

119899= minus

1

119899119870119899

119899+1(1198870 1198871 119887

119899) (16)



1199111198921015840(119911)

119892 (119911)= minus

1 + 119860120593 (119911)

1 + 119861120593 (119911)

= minus1 +

infin

sum

119899=1

(119860 minus 119861)119870minus1

119899(1198881 1198882 119888

119899 119861) 119911119899

(17)

119908ℎ1015840(119908)

ℎ (119908)= minus

1 + 119860120595 (119908)

1 + 119861120595 (119908)

= minus1 +

infin

sum

119899=1

(119860 minus 119861)119870minus1

119899(1198891 1198892 119889

119899 119861) 119908

119899

(18)


119899=1119888119899119911119899 and 120595(119908) = sum

infin

119899=1119889119899119908119899 are two



119899(1198961 1198962 119896

119899 119861) are given by

119870119901

119899(1198961 1198962 119896

119899 119861)

=119901

(119901 minus 119899)119899119896119899

1119861119899minus1

+119901


11198962119861119899minus2

+119901


11198963119861119899minus3

+119901


1

sdot [1198964119861119899minus4

+119901 minus 119899 + 3

21198962

2119861]

+119901


1

sdot [1198965119861119899minus5

+ (119901 minus 119899 + 4) 11989621198963119861]

+ sum

119895ge6

119896119899minus119895

1119883119895

(19)


variables 1198962 1198963 119896

119899


minus119865119899+1

(1198870 1198871 1198872 119887

119899) = (119860 minus 119861)119870

minus1

119899+1(1198881 1198882 119888

119899+1 119861)

(20)


minus 119865119899+1

(1198610 1198611 1198612 119861

119899)

= (119860 minus 119861)119870minus1

119899+1(1198891 1198892 119889

119899+1 119861)

(21)


1198870= (119860 minus 119861) 119888

1

minus1198870= (119860 minus 119861) 119889

1

(22)

21198871minus 1198872

0= (119861 minus 119860) (119888

2

1119861 minus 1198882)

21198871+ 1198872

0= (119860 minus 119861) (119889

2

1119861 minus 1198892)

(23)




0and then adding

them gives

21198872

0= (119861 minus 119860) (1198882 + 119889

2minus 1198611198882

1minus 1198611198892

1) (24)



21198872

0= (119861 minus 119860) (119888

2+ 1198892minus

2119861

(119861 minus 119860)21198872

0) (25)


sides we obtain

100381610038161003816100381611988701003816100381610038161003816 le

radic(119861 minus 119860)

2(1198882+ 1198892)

2 (2119861 minus 119860)le

119861 minus 119860


(26)



119861 minus 119860


lt 119861 minus 119860 (27)


we obtain

41198871= (119861 minus 119860) [(119889

2minus 1198882) minus (119889

2

1minus 1198882

1) 119861] = (119861 minus 119860) (119889

2minus 1198882)

(28)


100381610038161003816100381611988711003816100381610038161003816 le

119861 minus 119860

2 (29)


21198871= (119861 minus 119860) (119889

2+ 1198601198892

1minus 2119861119889

2

1) (30)


100381610038161003816100381611988711003816100381610038161003816 le

1

2(119861 minus 119860) (

100381610038161003816100381610038161198892+ 1198601198892

1

10038161003816100381610038161003816+ 2119861

100381610038161003816100381611988911003816100381610038161003816

2)

le1

2(119861 minus 119860) (1 + (119860 minus 1)

100381610038161003816100381611988911003816100381610038161003816

2+ 2119861

100381610038161003816100381611988911003816100381610038161003816

2)

=119861 minus 119860


2 (119861 minus 119860)

100381610038161003816100381611988701003816100381610038161003816

2

(31)




2 (119861 minus 119860)

100381610038161003816100381611988701003816100381610038161003816

2lt119861 minus 119860

2 (32)


21198871minus 1198872

0= (119861 minus 119860) (

119861

(119861 minus 119860)21198872

0minus 1198882)

21198871+ 1198872

0= (119860 minus 119861) (

119861

(119860 minus 119861)21198872

0minus 1198892)

(33)


10038161003816100381610038161003816100381610038161003816

1198871plusmn

2119861 minus 119860

2 (119861 minus 119860)1198872

0

10038161003816100381610038161003816100381610038161003816

le119861 minus 119860

2 (34)


(119899 + 1) 119887119899= (119860 minus 119861) 119888

119899+1 (35)


obtain

10038161003816100381610038161198871198991003816100381610038161003816 le

119861 minus 119860

119899 + 1 (36)







0|2 given by






References


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











21198872

0= (119861 minus 119860) (119888

2+ 1198892minus

2119861

(119861 minus 119860)21198872

0) (25)


sides we obtain

100381610038161003816100381611988701003816100381610038161003816 le

radic(119861 minus 119860)

2(1198882+ 1198892)

2 (2119861 minus 119860)le

119861 minus 119860


(26)



119861 minus 119860


lt 119861 minus 119860 (27)


we obtain

41198871= (119861 minus 119860) [(119889

2minus 1198882) minus (119889

2

1minus 1198882

1) 119861] = (119861 minus 119860) (119889

2minus 1198882)

(28)


100381610038161003816100381611988711003816100381610038161003816 le

119861 minus 119860

2 (29)


21198871= (119861 minus 119860) (119889

2+ 1198601198892

1minus 2119861119889

2

1) (30)


100381610038161003816100381611988711003816100381610038161003816 le

1

2(119861 minus 119860) (

100381610038161003816100381610038161198892+ 1198601198892

1

10038161003816100381610038161003816+ 2119861

100381610038161003816100381611988911003816100381610038161003816

2)

le1

2(119861 minus 119860) (1 + (119860 minus 1)

100381610038161003816100381611988911003816100381610038161003816

2+ 2119861

100381610038161003816100381611988911003816100381610038161003816

2)

=119861 minus 119860


2 (119861 minus 119860)

100381610038161003816100381611988701003816100381610038161003816

2

(31)




2 (119861 minus 119860)

100381610038161003816100381611988701003816100381610038161003816

2lt119861 minus 119860

2 (32)


21198871minus 1198872

0= (119861 minus 119860) (

119861

(119861 minus 119860)21198872

0minus 1198882)

21198871+ 1198872

0= (119860 minus 119861) (

119861

(119860 minus 119861)21198872

0minus 1198892)

(33)


10038161003816100381610038161003816100381610038161003816

1198871plusmn

2119861 minus 119860

2 (119861 minus 119860)1198872

0

10038161003816100381610038161003816100381610038161003816

le119861 minus 119860

2 (34)


(119899 + 1) 119887119899= (119860 minus 119861) 119888

119899+1 (35)


obtain

10038161003816100381610038161198871198991003816100381610038161003816 le

119861 minus 119860

119899 + 1 (36)







0|2 given by






References


























Volume 2014




Journal of











Journal of


Function Spaces






Algebra
























Volume 2014




Journal of











Journal of


Function Spaces






Algebra









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