SECOND HANKEL DETERMINANT PROBLEM FOR SOME ANALYTIC FUNCTION CLASSES WITH CONNECTED K-FIBONACCI NUMBERS

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1 Ata Universitatis Apulensis ISSN: No. 5/01 pp doi: /j.aua SECOND HANKEL DETERMINANT PROBLEM FOR SOME ANALYTIC FUNCTION CLASSES WITH CONNECTED K-FIBONACCI NUMBERS H. Özlem Güney, J. Sokó l, S. İlhan Abstrat. In this paper, we determine upper bound for the seond Hankel determinant in some lasses of analyti funtions in the open unit dis onneted with k-fibonai numbers k > 0. For this purpose we apply properties of k-fibonai numbers to onsider seond Hankel determinant problem for the lass SL k and KSL k. The results presented in this paper have been shown to generalize and improve some reent work of Sokó l et al. [1]. 010 Mathematis Subjet Classifiation: Primary 30C5, seondary 30C0. Keywords: univalent funtions, onvex funtions, starlike funtions, subordination, k-fibonai numbers. 1. Introdution Let D = z : z < 1} be the unit dis in the omplex plane. The lass of all analyti funtions fz = z + a n z n in the open unit dis D with normalization f0 = 0, n= f 0 = 1 is denoted by A and the lass S A is the lass whih onsists of univalent funtions in D. We say that f is subordinate to F in D, written as f F, if and only if fz = F ωz for some analyti funtion ω, ωz z, z D. Reently, N. Yilmaz Özgür and J. Sokó l [1] introdued the lass SLk of starlike funtions onneted with k Fibonai numbers as the set of funtions f A whih is desribed in the following definition. Definition 1. Let k be any positive real number. The funtion f A belongs to the lass SL k if it satisfies the ondition that where p k z = zf z fz p kz, z D, τk z 1 kτ k z τk, τ z k = k k +, z D. 161

2 Now we define the lass KSL k as follows: Definition. Let k be any positive real number. The funtion f A belongs to the lass KSL k if it satisfies the ondition that where the funtion p k is defined in. 1 + zf z f z p kz, z D, 3 For k = 1, the lasses SL k and KSL k beome the lasses SL and KSL of shelllike funtions defined in [15], see also [16]. It was proved in [1] that funtions in the lass SL k are univalent in D. Moreover, the lass SL k is a sublass of the lass of starlike funtions S, even more, starlike of order kk + 1/ /. The name attributed to the lass SL k is motivated by the shape of the urve C = p k e it : t [0, π \ π} }. The urve C has a shell-like shape and it is symmetri with respet to the real axis. Its graphi shape, for k = 1, is given below in Fig.1. Im ɛ = 5 10 ν = ɛ ɛ ν 3ɛ ɛ 1 5ɛ Re For k, note that we have Fig. 1. p 1 e it : y = 5 x 5x 1 10x. 5 p k e ±i arosk / = kk + 1/, and so the urve C intersets itself on the real axis at the point w 1 = kk + 1/. Thus C has a loop interseting the real axis also at the point w = k + /k. 16

3 For k >, the urve C has no loops and it is like a onhoid, see for details [1]. Moreover, the oeffiients of p k are onneted with k-fibonai numbers. For any positive real number k, the k-fibonai number sequene } n=0 is defined reursively by F k,0 = 0, F k,1 = 1 and +1 = k + 1 for n 1. When k = 1, we obtain the well-known Fibonai numbers F n. It is known that the n th k-fibonai number is given by = k τ k n τk n k +, 5 where τ k = k k + /. If p k z = 1 + n=1 p k,nz n, then we have see also [1]. p k,n = τk n, n = 1,, 3,..., 6 Lemma 1. [1] If fz = z + a n z n belongs to the lass SL k, then we have n= a n τ k n 1, 7 where τ k = k k + /. Equality holds in 7 for the funtion z g k z = 1 kτ k z τk z = τ n 1 k z n n=1 = z + k k + k z + k k k k + 1 z 3 +. Let Pβ, 0 β < 1, denote the lass of analyti funtions p in D with p0 = 1 and Repz} > β. Espeially, we use P0 = P as β = 0. In [1], they proved the following theorem: Theorem. Let } be the sequene of k-fibonai numbers defined in. If where τ k = k k +, z D, then we have 1 + τk p k z = z 1 kτ k z τk = 1 + p n z n, 9 z n=1 p n = τk n, n = 1,, 3,

4 We will use the following lemma for proving our main result. Lemma 3. [13] Let p P with pz = z + z +, then n, for n If 1 =, then pz p 1 z 1 + xz/1 xz with x = 1. Conversely, if pz p 1 z for some x = 1, then 1 = x. Furthermore, we have If 1 <, and 1 = 1, then pz p z, where p z = 1 + xwz + zwz + x 1 + xwz zwz + x, and x = 1, w = 1 1 and 1 = 1. Lemma. [9] Let p P with oeffiients n as above, then In 1976, Noonan and Thomas [10] stated the s th Hankel determinant for s 1 and q 1 as a q a q+1... a q+s 1 H s q = a q+1 a q a q+s a q+s 1 where a 1 = 1. This determinant has also been onsidered by several authors. For example, Noor [11] determined the rate of growth of H s q as q for funtions f in S with bounded boundary. Ehrenborg in [3] studied the Hankel determinant of exponential polynomials. The Hankel transform of an integer sequene and some of its properties were disussed by Layman in []. Also, several authors onsidered the ase s =. Espeially, H 1 = a 3 a is known as Fekete-Szegö funtional and this funtional is generalized to a 3 µa where µ is some real number []. Estimating for an upper bound of H 1 is known as the Fekete-Szegö problem. Raina and Sokó l onsidered Fekete-Szegö problem for the lass SL in [1] and for the lass SL k in [17]. In 1969, Keogh and Merkes [7] solved this problem for the lasses S and C. 16

5 The seond Hankel determinant is H = a a a 3. Janteng [5] found the sharp upper bound for H for univalent funtions whose derivative has positive real part. In [6] Janteng et al. obtained the bounds for H for the lasses S and C. Also, Sokó l et al. onsidered seond Hankel determinant problem for the lasses SL and KSL in [1].. The Seond Hankel Determinant Problem Let we prove the oeffiient bound of the funtion in the lass KSL k as follows: Theorem 5. If fz = z + a n z n belongs to the lass KSL k, then we have n= a n τ k n 1, 15 n where τ k = k k + /. Equality holds in 7 for the funtion f k z = τ k log 1 + z 1 τ k z. 16 Proof. A funtion f is in the lass KSL k if and only if the funtion gz = zf z 17 is in the lass SL k. The relations 17 follows 3. Therefore, if zf z = z + na n z n z D 1 n= belongs to the lass SL k, then from Lemma 1, we an write na n τ k n 1, whih implies 15. The equation 16 is suh that zf k z = g kz where the funtion g k is given in, and so from 17, it follows that f k KSL k. Also, by we have f k z = z + n= Consequently, the result 7 is sharp. τ k n 1 z n z D. 19 n In [17], Sokol et. al proved the following oeffiient bounds: 165

6 Theorem 6. If pz = 1 + p 1 z + p z + and 1 + τk pz p k z = z 1 kτ k z τk, τ z k = k k +, z D, then we have and p 1 p k + k + k k 0 k } k + k The above estimations are sharp. Now, our first main result Theorem 7 below gives an upper bound for the oeffiient p 3. Theorem 7. If pz = 1 + p 1 z + p z + and 1 + τk pz p k z = z 1 kτ k z τk, τ z k = k k +, z D, then we have } 3 k p 3 k 3 + 3k + k. The above estimation is sharp. Proof. If p p k, then there exists an analyti funtion w suh that wz z in D and pz = p k wz. Therefore, the funtion is in the lass P. It follows that hz = 1 + wz 1 wz = 1 + 1z + z + z D wz = 1z + 1 z

7 and p k wz = 1 + p k,1 1 z + + p k,3 1 z = 1 + p k,1 1 z z z + 1 p k, z + } z } + p k, 1 z p k,1 + 1 } 1 p k, z } 1 p k, p k,3 z 3 + = pz. From 6, we find the oeffiients p k,n of the funtion p k given by p k,n = τ n k. This shows the relevant onnetion p k with the sequene of k-fibonai numbers p k z = 1 + p k,n z n n=1 = 1 + F k,0 + F k, τ k z + F k,1 + F k,3 τ k z + = 1 + kτ k z + k + τ k z + k 3 + 3kτ 3 k z If pz = 1 + p 1 z + p z +, then by and 5, we have and p 3 = kτ k p = kτ k p 1 = kτ k 1, k + 1τk 7 + k τk + k3 + 3k 3 1τk 3. } We know that τ k k τ k = 1, 9 167

8 where τ k = k k +. Now taking absolute value of and using 9, we an write kτ k p 3 = k kτ k = k = k k5 + k 3 k τ k + k3 + 3k 3 1 τ 3 k kτ k k3 + 3k 3 1 k + 1τ k + k + k5 + k 3 } + k 1 τ k k + From and 5, we find that kk3 + 3k 1} 3. n N, τ k = τ n k x k,n, x k,n = 1 1, lim = τ k. 30 n Therefore, we have 1 p 3 = k k5 + k 3 k k5 + k 3 } + k τ n 1 k kx k,n + k k + + k 5 + k 3 kx k,n kk3 + 3k k 5 + k 3 } + kx k,n k k5 + k 3 k k5 + k 3 } + k τ k n 1 kx k,n + k k + + k 5 + k 3 kx k,n kk3 + 3k k 5 + k 3 + kx k,n k k5 + k 3 k k5 + k 3 } + k τ k n 1 + k x k,n + k k + + k 5 + k 3 kx k,n kk3 + 3k k 5 + k 3 + kx k,n 1 By 30, for suffiiently large n we have k, kk 3 + 3k k 5 + k 3 + kx k,n = k 5 + k 3 + kx k,n kk 3 + 3k and k, k k + + k 5 + k 3 kx k,n = k k + + k 5 + k 3 kx k,n. 16

9 Therefore, from 11, 1 and 13 we an write for suffiiently large n 1 p k k5 + k 3 k k5 + k 3 } + k τ k n 1 + kx k,n + k k + + k 5 + k 3 kx k,n + kk 3 + 3k k 5 + k 3 + kx k,n 1 k k + + k 5 + k 3 } kx k,n = k k5 + k 3 k k5 + k 3 } + k τ k n 1 + kx k,n + k k + + k 5 + k 3 kx k,n + k 5 + k 3 + kx k,n kk 3 + 3k 1 k k + + k 5 + k 3 } kx k,n = k k5 + k 3 k + kx k,n + k k + + k 5 + 6k 3 kx k,n 1 k k + + k 5 + k 3 } kx k,n 1 3. Denote 1 = y, fy = k5 + k 3 } + k τ k n 1 kx k,n + k k + + k 5 + 6k 3 kx k,n y k k + + k 5 + k 3 } kx k,n y 3, y [0, ]. It is easy to hek that f y > 0 for y [0, ] and for suffiiently large n. Sine then, for suffiiently large n, we have Therefore, we have max fy} = y [0,] k5 + k 3 + 3kx k,n kk 3 + 3k at y =. lim max fy} = n y [0,] k5 + k 3 + 3k τ k kk 3 + 3k = k + 1k 3 + 3k τ k kk 3 + 3k = k + 1 τ k kk 3 + 3k = k 3 + 3k τ k

10 Hene, we get lim n [ k + k5 + k 3 k + k 5 + k 3 + k 1 k5 + k 3 } k 1 3 τk n + kx k,n + k k + + k 5 + k 3 kx k,n + kk 3 + 3k k 5 + k 3 + kx k,n 1 k k + + k 5 + k 3 }] kx k,n 1 3 = k 3 + 3k τ k 3 = k 3 + 3k k } 3 + k whih shows that If we take p 3 k 3 + 3k } 3 k + k. hz = 1 + z 1 z = 1 + z + z +..., k } 3 then putting 1 = = 3 = in gives p 3 = k 3 + 3k + k and it shows that is sharp. It ompletes the proof. Conjeture. If pz = 1 + p 1 z + p z +, and p p, then p n τ k n, n = 1,, 3,..., where F k,0 = 0, F k,1 = 1 and +1 = k + 1 for n 1 is the k-fibonai sequene. This bound would be sharp for the funtion 5. This onjeture has been just verified for n = 3 in last Theorem 7, while for n = 1, it was proved in [17]. Theorem. If fz = z + a z +... belongs to SL k, then a a a 3 k + 6k } k + k. 31 Proof. For given f SL k, define pz = 1 + p 1 z + p z +, by zf z fz = pz 170

11 where p p. Hene zf z fz = 1 + a z + a 3 a z + 3a 3a a 3 + a 3 z 3 + = 1 + p 1 z + p z + and Therefore, a = p 1, a 3 = p 1 + p, a = p p 1p + p 3. 6 a a a 3 = 1 1 p 1 + p 1 p 3 3p. 3 Using Theorem 6 and Theorem 7, we obtain a a a 3 = 1 1 p 1 + p 1 p 3 3p 1 p1 + p 1 p p 1 k 1 k + k k + k k k + k 3 + 3k + k 1 +3k + k } k + k + 1 = k + 6k } k + k. } 3 Conjeture. If fz = z + a z +... belongs to SL k, then The bound is sharp. a a a 3 Theorem 9. If fz = z + a z +... belongs to KSL k, then a a a 3 3k + 9k + 36 } k + k } k + k.

12 Proof. For given f KSL k, define pz = 1 + p 1 z + p z +, by where p p in U. Hene 1 + zf z f z = pz = 1 + p 1z + p z +, 1+ zf z f z = 1+a z+6a 3 a z +1a 1a a 3 +a 3 z 3 + = 1+p 1 z+p z + and a = p 1, a 3 = p 1 + p, a = p p 1p + p 3. 6 Therefore, using Theorem 6 and Theorem 7, we obtain a a a 3 3k + 9k + 36 } k + k. Espeially, if we take k = 1 in Theorem and Theorem 9, we obtain the results of Sokól et al. in [1]as follows: Corollary 10. If fz = z + a z +... belongs to SL, then a a a Corollary 11. If fz = z + a z +... belongs to KSL, then } } a a a 3 9 Aknowledgement This researh has been supported with grant number FEN by DUBAP Dile University Coordination Committee of Sientifi Researh Projets. The authors would like to thank DUBAP for their supporting and the referees for the helpful suggestions. Referenes [1] Dziok J., Raina R.K., Sokó l J., Certain results for a lass of onvex funtions related to a shell-like urve onneted with Fibonai numbers, Comp. Math. with Appliations, 61011,

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14 Hatun Özlem Güney Department of Mathematis, Faulty of Siene, University of Dile, Diyarbakır, Turkey Janusz Sokó l Faulty of Mathematis and Natural Sienes University of Rzeszów, Rzeszów, Poland jsokol@ur.edu.pl Sedat İlhan Department of Mathematis, Faulty of Siene, University of Dile, Diyarbakır, Turkey sedati@dile.edu.tr 17