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Article

Some Results on Coefficient Estimate Problems for Four-Leaf-Type Bounded Turning Functions

1
School of Mathematical Sciences, Yangzhou Polytechnic College, Yangzhou 225009, China
2
Faculty of Humanities and Social Sciences, Guangzhou Civil Aviation College, Guangzhou 510403, China
*
Author to whom correspondence should be addressed.
Mathematics 2024, 12(12), 1875; https://doi.org/10.3390/math12121875
Submission received: 14 April 2024 / Revised: 7 June 2024 / Accepted: 7 June 2024 / Published: 16 June 2024

Abstract

:
Let B T 4 l denote a subclass of bounded turning functions connected with a four-leaf-type domain. The goal of the study is to probe into the bounds of coefficients | b 6 | , | b 7 | , | b 8 | , the bounds of the logarithmic coefficients, and the third-order determinants | H 3 , 1 | , | H 3 , 2 | , | H 3 , 3 | for the functions in this class.

1. Introduction and Definitions

Let H represent a family of mappings of the following kind, denoted by g
g ( ξ ) = ξ + n = 2 b n ξ n
in the unit disc U = { ξ C : | z | < 1 } . We refer to S as a subfamily of H , which consists of univalent functions in U .
For each function g H and positive integers i and n, the Hankel determinant H i , n g is defined by
H i , n g : = b n b n + 1 b n + i 1 b n + 1 b n + 2 b n + i b n + i 1 b n + i b n + 2 i 2 ( b 1 = 1 ) .
The Hankel determinant H i , n ( g ) was introduced by Pommerrenke [1,2]. The third Hankel determinants with n 3 are as follows:
H 3 , 1 ( g ) = b 3 b 5 + 2 b 2 b 3 b 4 b 3 3 b 4 2 b 2 2 b 5 ,
H 3 , 2 ( g ) = b 4 ( b 3 b 5 b 4 2 ) b 5 ( b 2 b 5 b 3 b 4 ) + b 6 ( b 2 b 4 b 3 2 ) ,
H 3 , 3 ( g ) = b 5 ( b 4 b 6 b 5 2 ) b 6 ( b 3 b 6 b 4 b 5 ) + b 7 ( b 3 b 5 b 4 2 ) .
For the first time, the bounds of the third-order Hankel determinant | H 3 , 1 ( g ) | for the families of S * of starlike functions, K of convex functions and R of functions with bounded turning were investigated by Babalola [3]. Recently, the sharp bounds of the Hankel determinant H 3 , 1 ( g ) of subclasses of analytic functions were obtained by many authors [4,5,6,7,8,9,10].
For g S , let F g ( ξ ) be defined as the
F g ( ξ ) = l o g g ( ξ ) ξ = 2 n = 1 δ n ξ n ,
where δ n = δ n ( g ) is referred to as the logarithmic coefficients of g. In the theory of univalent functions, these coefficients play an important role for different estimates. The problem of the best upper bounds for δ n is still open. In fact, even the proper order of magnitude is still not known. It is known, however, for the starlike functions that the best bound is | δ n | 1 n ( n 1 ) , and that this is not true in general [11].
Differentiating (5), we have
ξ g ( ξ ) g ( ξ ) ξ g ( ξ ) = 2 δ 1 + 4 δ 2 ξ + 6 δ 3 ξ 2 + 8 δ 4 ξ 3 + 10 δ 5 ξ 4 + · · · .
From (1), we obtain
ξ g ( ξ ) g ( ξ ) ξ g ( ξ ) = b 2 + ( 2 b 3 b 2 2 ) ξ + ( 3 b 4 3 b 2 b 3 + b 2 3 ) ξ 2 + ( 4 b 5 4 b 2 b 4 2 b 3 2 + 4 b 2 2 b 3 b 2 4 ) ξ 3 + ( 5 b 6 5 b 2 b 5 5 b 3 b 4 + 5 b 4 b 2 2 + 5 b 2 b 3 2 5 b 2 3 b 3 + b 2 5 ) ξ 4 + · · · .
By comparing the coefficients, we achieve
δ 1 = 1 2 b 2 , δ 2 = 1 2 ( b 3 1 2 b 2 2 ) , δ 3 = 1 2 ( b 4 b 2 b 3 + 1 3 b 2 3 ) , δ 4 = 1 2 ( b 5 b 4 b 2 + b 3 b 2 2 1 2 b 3 2 1 4 b 2 4 ) , δ 5 = 1 2 ( b 6 b 2 b 5 b 3 b 4 + b 4 b 2 2 + b 2 b 3 2 b 2 3 b 3 + 1 5 b 2 5 ) .
In 2022, Sunthrayuth et al. [12] introduced a subclass of bounded turning functions associated with a four-leaf function defined by
B T 4 l = g S : g ( ξ ) 1 + 5 6 ξ + 1 6 ξ 5 ,
and obtained Kruskal inequality, the bounds of the coefficient inequalities and the two-order Hankel determinant of bounded turning class B T 4 l .
Utilizing the estimates of the coefficients of the Schwartz function, we study the third-order Hankel determinants | H 3 , 1 ( g ) | , | H 3 , 2 ( g ) | and | H 3 , 3 ( g ) | for the class B T 4 l ; also, we obtain the bounds for the logarithmic coefficients for g B T 4 l .
Let ω 0 be the family of Schwarz functions. Thus, the functions w ω 0 can be expressed as a power series:
w ( ξ ) = n = 1 c n ξ n .
Lemma 1
(see [13]). Suppose a function w is a member of ω 0 . Then,
c 2 1 c 1 2 , | c 3 | 1 | c 1 | 2 | c 2 | 2 1 + | c 1 | , | c 4 | 1 | c 1 | 2 | c 2 | 2 ,
| c 5 | 1 | c 1 | 2 | c 2 | 2 | c 3 | 2 1 + | c 1 | , | c 6 | 1 | c 1 | 2 | c 2 | 2 | c 3 | 2 ,
| c 7 | 1 | c 1 | 2 | c 2 | 2 | c 3 | 2 | c 4 | 2 1 + | c 1 | .
Lemma 2
(see [14,15]). Suppose a function w is a member of ω 0 . Then,
| c 1 c 3 c 2 2 | 1 | c 1 | 2 , | c 2 c 4 c 3 2 | ( 1 c 1 2 ) 2 , | c 3 c 5 c 4 2 | 1 c 1 2 .
Lemma 3
(see [12]). If g B T 4 l , then
| b 2 | 5 12 , | b 3 | 5 18 , | b 4 | 5 24 , | b 5 | 1 6 .
Lemma 4
(see [11]). If w ω 0 , then c n 1 ( n 1 ) .
Lemma 5
(see [16]). Let w ω 0 ; then, for γ C , we have
c 2 + γ c 1 2 m a x 1 , γ
Lemma 6
(see [17]). If w ω 0 . Then, we obtain
| c 3 + γ c 1 c 2 + ς c 1 3 | 1 ,
where ( γ , ς ) D 1 D 2 , with
D 1 = { ( γ , ς ) R 2 : | γ | 1 2 , 1 ς 1 } ,
D 2 = { ( γ , ς ) R 2 : 1 2 | γ | 2 , 4 27 ( | γ | + 1 ) 3 ( | γ | + 1 ) ς 1 } .
Lemma 7
(see [18]). Let w ω 0 and γ C ; we obtain
c 4 + 2 γ c 1 c 3 + γ c 2 2 + 3 γ 2 c 1 2 c 2 + γ 3 c 1 4 m a x { 1 , | γ | 3 } .
Lemma 8
(see [19]). If w ω 0 , then for all γ C , we have
| c 5 + 2 γ c 1 c 4 + 2 γ c 2 c 3 + 3 γ 2 c 1 c 2 2 + 3 γ 2 c 1 2 c 3 + 4 γ 3 c 1 3 c 2 + γ 4 c 1 5 | 1 .

2. The Bounds of the Third Hankel Determinant for g B T 4 l

Theorem 1.
If g B T 4 l , then
| b 6 | 5 36 , | b 7 | 5 42 , | b 8 | 5 48 .
The bounds are sharp.
Proof. 
For a function g B T 4 l , there exists a Schwarz function w ξ , such that
g ξ = 1 + 5 6 w ξ + 1 6 w 5 ξ ,
Comparing the coefficients, we yield
b 2 = 5 12 c 1 , b 3 = 5 18 c 2 , b 4 = 5 24 c 3 , b 5 = 1 6 c 4 ,
b 6 = 5 c 5 + c 1 5 36
b 7 = 5 c 6 + 5 c 1 4 c 2 42 ,
b 8 = 5 c 7 + 5 c 1 4 c 3 + 10 c 1 3 c 2 2 48 .
From (9) and Lemma 1, we have
b 6 1 36 5 | c 5 | + | c 1 | 5 1 36 [ 5 ( 1 | c 1 | 2 | c 2 | 2 | c 3 | 2 1 + | c 1 | ) + | c 1 | 5 ] = 1 36 ( 5 5 | c 1 | 2 + | c 1 | 5 5 | c 2 | 2 5 | c 3 | 2 1 + | c 1 | ) 1 36 ( 5 5 | c 1 | 2 + | c 1 | 5 ) 5 36 .
From (10) and Lemma 1, we achieve
b 7 1 42 5 | c 6 | + 5 | c 1 | 4 | c 2 | 1 42 5 ( 1 | c 1 | 2 | c 2 | 2 | c 3 | 2 ) + 5 | c 1 | 4 | c 2 | = 1 42 ( 5 5 | c 1 | 2 + 5 | c 1 | 4 | c 2 | 5 | c 2 | 2 5 | c 3 | 2 ) 1 42 5 5 | c 1 | 2 + 5 | c 1 | 4 ( 1 | c 1 | 2 ) = 1 42 5 5 | c 1 | 2 + 5 | c 1 | 4 5 | c 1 | 6 5 42 .
From (11) and Lemma 1, we obtain
b 8 1 48 5 | c 7 | + 5 | c 1 | 4 | c 3 | + 10 | c 1 | 3 | c 2 | 2 1 48 [ 5 ( 1 | c 1 | 2 | c 2 | 2 | c 3 | 2 | c 4 | 2 1 + | c 1 | ) + 5 | c 1 | 4 | c 3 | + 10 | c 1 | 3 | c 2 | 2 ] = 1 48 [ 5 5 | c 1 | 2 + ( 5 + 10 | c 1 | 3 ) | c 2 | 2 + 5 | c 1 | 4 | c 3 | 5 | c 3 | 2 5 | c 4 | 2 1 + | c 1 | ] 1 48 [ 5 5 | c 1 | 2 + ( 5 + 10 | c 1 | 3 ) | c 2 | 2 + 5 | c 1 | 4 ( 1 | c 1 | 2 | c 2 | 2 1 + | c 1 | ) ] = 1 48 [ 5 5 | c 1 | 2 + 5 | c 1 | 4 5 | c 1 | 6 + 5 5 | c 1 | + 10 | c 1 | 3 + 5 | c 1 | 4 1 + | c 1 | | c 2 | 2 ] .
Setting c = | c 1 | and d = | c 2 | , we obtain
b 8 1 48 ϵ 1 ( c , d )
where
ϵ 1 ( c , d ) = 5 5 c 2 + 5 c 4 5 c 6 + 5 5 c + 10 c 3 + 5 c 4 1 + c d 2 .
The critical points of ϵ 1 satisfy
ϵ 1 c = 10 c + 20 c 3 30 c 5 + 15 c 4 + 40 c 3 + 30 c 2 ( 1 + c ) 2 d 2 , ϵ 1 d = 10 10 c + 20 c 3 + 10 c 4 1 + c d = 0 .
Applying numerical computations, we have
c 0 = 0 , d 0 = 0 , c 1 = 0.8668 , d 1 = 0.7940 , c 2 = 0.8668 , d 2 = 0.7940 .
Thus, in ( 0 , 1 ) × ( 0 , 1 c 2 ) , there is no critical points which satisfies 0 d 1 c 2 .
For c = 0 ,
ϵ 1 ( 0 , d ) = 5 5 d 2 5 .
For d = 0 ,
ϵ 1 ( c , 0 ) = 5 5 c 2 + 5 c 4 5 c 6 5 .
For d = 1 c 2 ,
ϵ 1 ( c , 1 c 2 ) = 5 c 2 + 10 c 3 5 c 4 15 c 5 + 5 c 7 = η 1 ( c ) η 1 ( 0.7202 ) = 2.5799 .
Thus, we have
| b 8 | 5 48 .
These bounds can be determined by the following extremal functions:
g 1 ( ξ ) = 0 ξ ( 1 + 5 6 t 5 + 1 6 t 25 ) d t = ξ + 5 36 ξ 6 + 1 156 ξ 26 + · · · ,
g 2 ( ξ ) = 0 ξ ( 1 + 5 6 t 6 + 1 6 t 30 ) d t = ξ + 5 42 ξ 7 + 1 186 ξ 31 + · · · ,
g 3 ( ξ ) = 0 ξ ( 1 + 5 6 t 7 + 1 6 t 35 ) d t = ξ + 5 48 ξ 8 + 1 216 ξ 36 + · · · .
Theorem 2.
If g B T 4 l , then
H 2 , 3 ( g ) 0.044516 .
Proof. 
Let g B T 4 l .
From (8), we receive
H 2 , 3 g = | b 3 b 5 b 4 2 | = 5 1728 16 c 2 c 4 15 c 3 2 = 5 1728 | 15 ( c 2 c 4 c 3 2 ) + c 2 c 4 | .
Utilizing inequality, Lemma 1 and Lemma 2 in (12), we obtain
H 2 , 3 g 5 1728 [ 15 | c 2 c 4 c 3 2 | + | c 2 | | c 4 | ] 5 1728 [ 15 ( 1 | c 1 | 2 ) 2 + | c 2 | ( 1 | c 1 | 2 | c 2 | 2 ) ] = 5 1728 [ 15 30 | c 1 | 2 + 15 | c 1 | 4 + ( 1 | c 1 | 2 ) | c 2 | | c 2 | 3 ) ] 5 1728 [ 15 30 | c 1 | 2 + 15 | c 1 | 4 + ( 1 | c 1 | 2 ) | c 2 | | c 2 | 3 ] .
Let | c 1 | = c and | c 2 | = d , we obtain
H 2 , 3 g 5 1728 ϵ 2 ( c , d ) .
Consider
ϵ 2 c = 60 c + 60 c 3 2 c d = 0 , ϵ 2 d = 1 c 2 3 d 2 = 0 .
Applying numerical computations, we obtain
c 1 = 0 , d 1 = 0.5774 , c 2 = 0 , d 2 = 0.5774 , c 3 = 1 , d 3 = 0 , c 4 = 1 , d 4 = 0 , c 5 = 0.9998 , d 5 = 0.0111 , c 6 = 0.9998 , d 6 = 0.0111 .
Thus, there is no critical point in ( 0 , 1 ) × ( 0 , 1 c 2 ) .
(1)
For d = 0 ,
ϵ 2 ( c , 0 ) = 15 30 c 2 + 15 c 4 15 .
(2)
For c = 0 ,
ϵ 2 ( 0 , d ) = 15 + d d 3 ϵ 2 ( 0 , 3 3 ) = 2 3 + 135 9 = 15.384888 .
(3)
For d = 1 c 2 ,
ϵ 2 ( c , 1 c 2 ) = 15 29 c 2 + 13 c 4 + c 6 15 .
Therefore, we yield
H 2 , 3 ( g ) 10 3 + 675 15552 = 0.044516 .
The proof of Theorem 2 is completed.  □
Theorem 3.
If g B T 4 l , then
H 3 , 1 ( g ) 3025 46656 = 0.064836 .
Proof. 
Assume that g B T 4 l . From (8) and (2), we achieve
H 3 , 1 g = 1 46656 2250 c 1 c 2 c 3 1000 c 2 3 2025 c 3 2 + 2160 c 2 c 4 1350 c 1 2 c 4 = 1 46656 1000 c 2 ( c 1 c 3 c 2 2 ) + 1250 c 1 c 2 c 3 + 2025 ( c 2 c 4 c 3 2 ) + 135 c 2 c 4 1350 c 1 2 c 4
By applying the triangle inequality, Lemma 1 and Lemma 2 in (13), we receive
46656 H 3 , 1 ( g ) 1000 | c 2 | | c 1 c 3 c 2 2 | + 1250 | c 1 | | c 2 | | c 3 | + 2025 | c 2 c 4 c 3 2 | + 135 | c 2 | | c 4 | + 1350 | c 1 | 2 | c 4 | 1000 | c 2 | ( 1 | c 1 | 2 ) + 1250 | c 1 | | c 2 | ( 1 | c 1 | 2 | c 2 | 2 1 + | c 1 | ) + 2025 ( 1 | c 1 | 2 ) 2 + 135 | c 2 | ( 1 | c 1 | 2 | c 2 | 2 ) + 1350 | c 1 | 2 ( 1 | c 1 | 2 | c 2 | 2 ) = 2025 2700 | c 1 | 2 + 675 | c 1 | 4 + ( 1135 + 1250 | c 1 | 1135 | c 1 | 2 1250 | c 1 | 3 ) | c 2 | 1350 | c 1 | 2 | c 2 | 2 135 + 1385 | c 1 | 1 + | c 1 | | c 2 | 3 .
Setting | c 1 | = c and | c 2 | = d , we have
46656 H 3 , 1 g 2025 2700 c 2 + 675 c 4 + ( 1135 + 1250 c 1135 c 2 1250 c 3 ) d 1350 c 2 d 2 135 + 1385 c 1 + c d 3 = ϵ 3 ( c , d )
where c [ 0 , 1 ] and d [ 0 , 1 c 2 ] .
Taking the partial derivative with respect to c and y respectively, and we have
ϵ 3 c = 5400 c + 2700 c 3 + ( 1250 2270 c 3405 c 2 ) d 2700 c d 2 1250 ( 1 + c ) 2 d 3
and
ϵ 3 d = 1135 + 1250 c 1135 c 2 1250 c 3 2700 c 2 d 405 + 4155 c 1 + c d 2 .
Setting ϵ 3 c = ϵ 3 d = 0 and simplifying, we yield
5400 c 10800 c 2 2700 c 3 + 5400 c 4 + 2700 c 5 + ( 1250 + 230 c 6695 c 2 9080 c 3 3405 c 4 ) d 2700 c ( 1 + c ) 2 d 2 1250 d 3 = 0 , 1135 + 2358 c + 115 c 2 2385 c 3 1250 c 4 2700 c 2 ( 1 + c ) d ( 405 + 4155 c ) d 2 = 0 .
Applying Newton’s methods, we yield
c 1 = 1 , d 1 = 0 , c 2 = 0.1018 , d 2 = 1.3080 , c 3 = 1.0726 , d 3 = 1.1909 , c 4 = 1.7878 d 4 = 0.3706 , c 5 = 3.3060 , d 5 = 6.5565 , c 6 = 1.3977 , d 6 = 0.0899 .
Thus, there is no critical point in ( 0 , 1 ) × ( 0 , 1 c 2 ) .
For c = 0 ,
ϵ 3 ( 0 , d ) = 2025 + 1135 d 135 d 3 ϵ 3 ( 0 , 1 ) = 3025 .
For d = 0 ,
ϵ 3 ( c , 0 ) = 2025 2700 c 2 + 675 c 4 2025 .
For d = 1 c 2 ,
ϵ 3 ( c , 0 ) = 3025 4665 c 2 + 1605 c 4 + 35 c 6 ϵ 3 ( 0 , 0 ) = 3025 .
Thus, we obtain
H 3 , 1 ( g ) 3025 46656 = 0.064836 .
The proof of Theorem 3 is completed.  □
Theorem 4.
If g B T 4 l , then
| b 2 b 5 b 3 b 4 | 15.503407 432 = 0.035888 .
Proof. 
Let g B T 4 l . From (8), we obtain
| b 2 b 5 b 3 b 4 | = 1 432 | 30 c 1 c 4 25 c 2 c 3 | .
Applying Lemma 1 and the triangle inequality in (14), we get
| b 2 b 5 b 3 b 4 | 1 432 [ 30 | c 1 | ( 1 | c 1 | 2 | c 2 | 2 ) + 25 | c 2 | ( 1 | c 1 | 2 | c 2 | 2 1 + | c 1 | ) ] = 1 432 [ 30 | c 1 | 30 | c 1 | 3 + ( 25 25 | c 1 | 2 ) | c 2 | 30 | c 1 | | c 2 | 2 25 1 + | c 1 | | c 2 | 3 ] .
Setting | c 1 | = c and | c 2 | = d , we yield
| b 2 b 5 b 3 b 4 | 1 432 ϵ 4 ( c , d ) .
Taking the partial derivative with respect to c and d, respectively, we obtain
ϵ 4 c = 30 90 c 2 50 c d 30 d 2 + 25 ( 1 + c ) 2 d 3
and
ϵ 4 d = 25 25 c 2 60 c d 75 1 + c d 2 .
Setting ϵ 4 c = ϵ 4 d = 0 , and simplifying, we receive
90 c 4 180 c 3 60 c 2 + 60 c + 30 50 c ( 1 + c ) 2 d 30 ( 1 + c ) 2 d 2 + 25 d 3 = 0 , 25 c 3 25 c 2 + 25 c + 25 60 ( c 2 + c ) d 75 d 2 = 0 .
Applying Newton’s methods, we have
c 1 = 1 , d 1 = 0 , c 2 = 0.4098 , d 2 = 0.8978 , c 3 = 0.4271 , d 3 = 0.4258 , c 4 = 0.4550 d 4 = 0.2931 , c 5 = 0.7258 , d 5 = 0.3023 .
Therefore, we obtain ε 4 ( c , d ) ε 4 ( 0.4271 , 0.4258 ) = 15.503407 .
(a)
For c = 0 ,
ε 4 ( 0 , d ) = 25 d 25 d 3 ε 4 ( 0 , 3 3 ) = 50 3 9 .
(b)
For d = 0 ,
ε 4 ( c , 0 ) = 30 c 30 c 3 ε 4 ( 3 3 , 0 ) = 20 3 3 = 11.546666 .
(c)
For d = 1 c 2 ,
ε 4 ( c , 1 c 2 ) = 25 c 20 c 3 5 c 5 = η 2 ( c ) = η 2 ( 0.6017 ) = 10.2913 .
Thus, we obtain
| b 2 b 5 b 3 b 4 | 15.503407 432 = 0.035888 .
Theorem 5.
If g B T 4 l , then
| H 2 , 2 ( g ) | = | b 2 b 4 b 3 2 | 200.27 2592 = 0.077265 .
Proof. 
Let g B T 4 l . From (8), we obtain
| b 2 b 4 b 3 2 | = 1 2592 | 225 c 1 c 3 200 c 2 2 | = 1 2592 | 200 ( c 1 c 3 c 2 2 ) + 25 c 1 c 3 | .
Applying Lemma 1, Lemma 2 and the triangle inequality, we receive
| b 2 b 4 b 3 2 | 1 2592 [ 200 ( 1 | c 1 | 2 ) + 25 | c 1 | ( 1 | c 1 | 2 | c 2 | 2 1 + | c 1 | ) ] = 1 2592 [ 200 + 25 | c 1 | 200 | c 1 | 2 25 | c 1 | 3 | c 1 | 1 + | c 1 | | c 2 | 2 ] 1 2592 200 + 25 | c 1 | 200 | c 1 | 2 25 | c 1 | 3 = 1 2592 η 3 ( | c 1 | ) 1 2592 η 3 ( 0.0618 ) = 200.2700 2592 = 0.077265 .
Theorem 6.
If g B T 4 l , then
| H 3 , 2 ( g ) | 0.025986 .
Proof. 
Let g B T 4 l . From Lemma 3, Theorems 1, 2, 4 and 5, we yield
| H 3 , 2 ( g ) | | b 4 | | b 3 b 5 b 4 2 | + | b 5 | | b 2 b 5 b 3 b 4 | + | b 6 | | b 2 b 4 b 3 2 | = 5 24 × 0.044516 + 1 6 × 0.035888 + 5 36 × 0.077265 = 0.025986 .
Theorem 7.
If g B T 4 l , then
| H 2 , 4 ( g ) | = | b 4 b 6 b 5 2 | 24.385252 864 = 0.028224 .
Proof. 
Let f B T 4 l . From (8) and (9), we have
| b 4 b 6 b 5 2 | = 1 864 | 25 c 3 c 5 24 c 4 2 + 5 c 3 c 1 5 | = 1 864 | 24 ( c 3 c 5 c 4 2 ) + c 3 c 5 + 5 c 3 c 1 5 | .
Using Lemmas 1 and 2, we have
| b 4 b 6 b 5 2 | 1 864 ( 24 | c 3 c 5 c 4 2 | + | c 3 | | c 5 | + 5 | c 3 | | c 1 5 | ) 1 864 [ 24 ( 1 | c 1 | 2 ) + | c 3 | ( 1 | c 1 | 2 | c 2 | 2 | c 3 | 2 1 + | | c 1 ) + 5 | c 1 | 5 | c 3 | ] = 1 864 [ 24 24 | c 1 | 2 + ( 1 | c 1 | 2 + 5 | c 1 | 5 | c 2 | 2 ) | c 3 | 1 1 + | c 1 | | c 3 | 3 ] .
Setting c = | c 1 | , d = | c 2 | and e = | c 3 | , we yield
| b 4 b 6 b 5 2 | 1 864 ϵ 5 ( c , d , e )
where ϵ 5 ( c , d , e ) = 24 24 c 2 + ( 1 c 2 + 5 c 5 d 2 ) e 1 1 + c e 3 and ( c , d , e ) Ω = { ( c , d , e ) : 0 c 1 , 0 d 1 c 2 , 0 e 1 c 2 d 2 1 + c } . Consider
ϵ 5 d = 2 d e 0 ,
thus, there are no points in Ω .
(1)
For c = 0 ,
ϵ 5 ( 0 , d , e ) = 24 + ( 1 d 2 ) e e 3 = η 4 ( d , e ) .
It is evident that there is on point in ( 0 , 1 ) × ( 0 , 1 d 2 ) .
(2)
For d = 0 ,
ϵ 5 ( c , 0 , e ) = 24 24 c 2 + ( 1 c 2 + 5 c 5 ) e 1 1 + c e 3 = η 5 ( c , e ) .
Partial derivative of η 5 with respect to c, and then with respect to e, we achieve
η 5 c = 48 c + ( 2 c + 25 c 4 ) e + 1 ( 1 + c ) 2 e 3 ,
and
η 5 e = 1 c 2 + 5 c 5 3 1 + c e 2 .
Setting η 5 c = η 5 e = 0 and simplifying, we yield
48 c ( 1 + c ) 2 + ( 25 c 6 + 50 c 5 + 25 c 4 2 c 3 4 c 2 2 c ) e + e 3 = 0 , 5 c 6 + 5 c 5 c 3 c 2 + c + 1 3 e 2 = 0 .
We obtain a critical point ( 0.0039 , 0.5785 ) ; thus, we have η 5 ( c , e ) η 5 ( 0.0039 , 0.5785 ) = 24.385252 .
(3)
For e = 0 ,
ϵ 5 ( c , d , 0 ) = 24 24 c 2 24 .
(4)
For e = 1 c 2 d 2 1 + c ,
ϵ 5 ( c , d , 1 c 2 d 2 1 + c ) = 24 + c 24 c 2 2 c 3 + 6 c 5 5 c 7 + 5 c 5 + 4 c 3 c 2 4 c + 1 1 + c d 2 + 2 + 4 c ( 1 + c ) 2 d 4 + 1 ( 1 + c ) 4 d 6 = η 6 ( c , d ) η 6 ( 0.0016 , 0.5767 ) = 24.148169 .
(5)
For c = d = 0
ϵ 5 ( 0 , 0 , e ) = 24 + e e 3 = η 7 ( e ) η 7 ( 3 3 ) = 24 + 2 3 9 = 24.384889 .
(6)
For c = e = 0 ,
ϵ 5 ( 0 , d , 0 ) = 24 .
(7)
For d = e = 0 ,
ϵ 5 ( c , 0 , 0 ) = 24 + e e 3 24.384889 .
(8)
For e = 0 , d = 1 c 2 ,
ϵ 5 ( c , 1 c 2 , 0 ) = 24 24 c 2 24 .
(9)
For d = 0 and e = 1 c 2 ,
ϵ 5 ( c , 0 , 1 c 2 ) = 24 + c 24 c 2 2 c 3 + 6 c 5 5 c 7 = η 8 ( c ) η 8 ( 0.0206 ) = 24.0104 .
(10)
For c = 0 , e = 1 d 2
ϵ 5 ( 0 , d , 1 d 2 ) = 24 + d 2 2 d 4 + d 6 = η 9 ( d ) η 9 ( 3 3 ) = 24.148148 .
Thus, we get
| b 4 b 6 b 5 2 | 24.385252 864 = 0.035888 .
Theorem 8.
If g B T 4 l , then
| b 3 b 6 b 4 b 5 | 5 144 .
Proof. 
Let g B T 4 l . From (8) and (9), we receive
| b 3 b 6 b 4 b 5 | = 1 1296 | 50 c 2 c 5 45 c 3 c 4 + 50 c 2 c 1 5 | .
Using Lemma 1, we obtain
| b 3 b 6 b 4 b 5 | 1 1296 ( 50 | c 2 | | c 5 | + 45 | c 3 | | c 4 | + 50 | c 2 | | c 1 5 | ) 1 1296 [ 50 | c 2 | ( 1 | c 1 | 2 | c 2 | 2 | c 3 | 2 1 + | c 1 | ) + 45 | c 3 | ( 1 | c 1 | 2 | c 2 | 2 ) + 50 | c 1 | 5 | c 2 | ] = 1 1296 [ ( 50 50 | c 1 | 2 + 50 | c 1 | 5 ) | c 2 | 50 | c 2 | 3 + ( 45 45 | c 1 | 2 45 | c 2 | 2 ) | c 3 | 50 | c 2 | 1 + | c 1 | | c 3 | 2 ] .
By setting | c 1 | = c , | c 2 | = d and | c 3 | = e , we have
| b 3 b 6 b 4 b 5 | 1 1296 Ψ ( c , d , e )
where Ψ ( c , d , e ) = ( 50 50 c 2 + 50 c 5 ) d 50 d 3 + ( 45 45 c 2 45 d 2 ) e 50 d 1 + c e 2 , ( c , d , e ) Ω .
Differentiating partially with respect to c, d and e, respectively, we obtain
Ψ c = ( 100 c + 250 c 4 ) d 90 c e + 50 d ( 1 + c ) 2 e 2 ,
Ψ d = 50 50 c 2 + 50 c 5 150 d 2 90 d e 50 1 + c e 2 ,
and
Ψ e = 45 45 c 2 45 d 2 100 d 1 + c e .
By putting Ψ c = Ψ d = Ψ e = 0 , and simplifying, we obtain
( 250 c 6 + 500 c 5 + 250 c 4 100 c 3 200 c 2 100 c ) d 900 c ( 1 + c ) 2 e + 50 d e 2 = 0 , 50 c 6 + 50 c 5 50 c 3 50 c 2 + 50 c + 50 150 ( 1 + c ) d 2 90 ( 1 + c ) d e 50 e 2 = 0 , 45 c 3 45 c 2 + 45 c + 45 45 ( 1 + c ) d 2 100 d e = 0 .
By a numberical calculation, we obtain
c 1 = 1 , d 1 = d , e 1 = 0 , c 2 = 0.8441 , d 2 = 0.4127 , e 2 = 0.2354 , c 3 = 0.8441 , d 3 = 0.4127 , e 3 = 0.2354 .
Thus, there is no critical point that satisfies 0 c 1 and 0 d 1 c 2 .
(1)
For c = 0 ,
Ψ ( 0 , d , e ) = 50 d 50 d 3 + ( 45 45 d 2 ) e 50 d e 2 = Λ 1 ( d , e ) .
Consider
Λ 1 d = 50 150 d 2 90 d e 50 e 2 = 0 , Λ 1 e = 45 45 d 2 100 d e = 0 .
A numerical calculation is that there is no critical point in ( 0 , 1 ) × ( 0 , 1 d 2 ) .
(2)
For d = 0 ,
Ψ ( c , 0 , e ) = ( 45 45 c 2 ) e 45 ( 1 c 2 ) 2 45 .
(3)
For e = 0 ,
Ψ ( c , d , 0 ) = ( 50 50 c 2 + 50 c 5 ) d 50 d 3 = Λ 2 ( c , d ) Λ 2 ( 0.7368 , 0.2828 ) = 8.403278 .
(4)
For e = 1 c 2 d 2 1 + c ,
Ψ ( c , d , 1 c 2 d 2 1 + c ) = 45 90 c 2 + 45 c 4 + ( 50 c 50 c 3 + 50 c 5 ) d + ( 90 + 45 c + 45 c 2 ) d 2 + 50 150 c 1 + c d 3 + 45 1 + c d 4 50 ( 1 + c ) 3 d 5 = Λ 3 ( c , d ) .
Differentiating Λ 3 partially with respect to c and d, we yield
Λ 3 c = 180 c + 180 c 3 + ( 50 150 c 2 + 250 c 4 ) d + ( 45 + 90 c ) d 2 100 d 3 ( 1 + c ) 2 45 d 4 ( 1 + c ) 2 + 150 d 5 ( 1 + c ) 4
and
Λ 3 d = 50 c 50 c 3 + 50 c 5 + ( 180 + 90 c + 90 c 2 ) d + 150 450 c 1 + c d 2 + 180 d 3 1 + c 250 d 4 ( 1 + c ) 3 .
Setting Λ 3 c = Λ 3 d = 0 and simplifying, we receive
180 c 7 + 720 c 6 + 900 c 5 900 c 3 720 c 2 180 c + ( 250 c 8 + 1000 c 7 + 1350 c 6 + 400 c 5 600 c 4 400 c 3 + 150 c 2 + 200 c + 50 ) d + ( 90 c 5 + 405 c 4 + 720 c 3 + 630 c 2 + 270 c + 45 ) d 2 100 ( 1 + c ) 2 d 3 45 ( 1 + c ) 2 d 4 + 150 d 5 = 0 , 50 c 8 + 150 c 7 + 100 c 6 100 c 5 100 c 4 + 100 c 3 + 150 c 2 + 50 c + ( 90 c 5 + 360 c 4 + 360 c 3 180 c 2 450 c 180 ) d + ( 450 c 3 750 c 2 150 c + 150 ) d 2 + 180 ( 1 + c ) 2 d 3 250 d 4 = 0
Applying Newton’s methods, we recieve
c 1 = 1 d 1 = 0 , c 2 = 0 , d 2 = 0 , c 3 = 0.8336 , d 3 = 0.4141 , c 4 = 0.9194 , d 4 = 0.0957 .
Thus, there is no critical point satisfing 0 c 1 and 0 d 1 c 2 .
(5)
For d = c = 0 ,
Ψ ( 0 , 0 , e ) = 45 e 45 .
(6)
For e = d = 0 ,
Ψ ( c , 0 , 0 ) = 0 .
(7)
For c = e = 0 ,
Ψ ( 0 , d , 0 ) = 50 d 50 d 3 = Λ 4 ( d ) Λ 4 ( 3 3 ) = 100 3 9 = 19.2444 .
(8)
For d = 1 c 2 and e = 0 ,
Ψ ( c , 1 c 2 , 0 ) = 50 c 2 100 c 4 + 50 c 5 + 50 c 6 50 c 7 = Λ 5 ( c ) Λ 5 ( 0.7145 ) = 10.6734 .
(9)
For e = 1 c 2 and d = 0 ,
Ψ ( c , 0 , 1 c 2 ) = 45 ( 1 c 2 ) 2 45 .
(10)
For e = 1 d 2 and c = 0 ,
Ψ ( 0 , d , 1 c 2 ) = 45 90 d 2 + 50 d 3 + 45 d 4 50 d 5 = Λ 6 ( d ) Λ 6 ( 0 ) = 45 .
Hence, we obtain
| b 3 b 6 b 4 b 5 | 45 1296 = 5 144 .
Theorem 9.
If g B T 4 l , then
| H 3 , 3 ( g ) | 0.016103 .
Proof. 
Let g B T 4 l . From Lemma 3 and Theorems 1–4, we receive
| H 3 , 3 ( g ) | | b 5 | | b 4 b 6 b 5 2 | + | b 6 | | b 3 b 6 b 4 b 5 | + | b 7 | | b 3 b 5 b 4 2 | 1 6 × 0.035888 + 5 36 × 5 144 + 5 42 × 0.044516 = 0.016103 .

3. The Bounds of the Logarithmic Coefficients for g B T 4 l

Theorem 10.
If g B T 4 l , then
| δ 1 | 5 24 , | δ 2 | 5 36 , | δ 3 | 5 48 , | δ 4 | 1 12 , | δ 5 | 173400 2488320 = 0.069686 .
The first four bounds are the best possible.
Proof. 
Let g B T 4 l . From (6), (8), and (9), we have
δ 1 = 5 c 1 24 ,
δ 2 = 5 36 ( c 2 5 8 c 1 2 ) ,
δ 3 = 5 48 ( c 3 5 9 c 1 c 2 + 25 216 c 1 3 ) ,
δ 4 = 1 165888 ( 13824 c 4 7200 c 1 c 3 + 4000 c 1 2 c 2 3200 c 2 2 625 c 1 4 ) .
δ 5 = 1 2488320 ( 172800 c 5 86400 c 1 c 4 72000 c 2 c 3 + 45000 c 1 2 c 3 + 40000 c 1 c 2 2 5000 c 1 3 c 2 + 37685 c 1 5 ) .
Applying Lemma 4 to (15), we have
| δ 1 | 5 24
Applying Lemma 5 to (16), we obtain
| δ 2 | 5 36 .
Utilizing the triangle inequality and Lemma 6 with γ = 5 9 and ζ = 25 216 , we yield
| δ 3 | 5 48 .
Rearranging (18), we obtain
| δ 4 | = 1 165888 | 7200 ( c 4 c 1 c 3 1 2 c 2 2 + 3 4 c 1 2 c 2 1 8 c 1 4 ) + 6624 c 4 + 400 c 2 2 1400 c 1 2 c 2 + 275 c 1 4 | 1 165888 | 7200 ( c 4 c 1 c 3 1 2 c 2 2 + 3 4 c 1 2 c 2 1 8 c 1 4 ) | + 1 165888 | 6624 c 4 + 400 c 2 2 1400 c 1 2 c 2 + 275 c 1 4 | = 1 165888 1 + 1 165888 2 ,
where 1 = | 7200 ( c 4 c 1 c 3 1 2 c 2 2 + 3 4 c 1 2 c 2 1 8 c 1 4 ) |
2 = | 6624 c 4 + 400 c 2 2 1400 c 1 2 c 2 + 275 c 1 4 | .
Using Lemma 7 with γ = 1 2 , we obtain 1 7200 .
Using Lemma 6 and the triangle inequality in (20), we receive
2 6624 ( 1 | c 1 | 2 | c 2 | 2 ) + 400 | c 2 | 2 + 1400 | c 1 | 2 | c 2 | + 275 | c 1 | 4 = 6624 6624 | c 1 | 2 + 275 | c 1 | 4 + 1400 | c 1 | 2 | c 2 | 6224 | c 2 | 2 = Υ 1 ( c , d )
where | c 1 | = c , | c 2 | = d .
Consider
Υ 1 c = 13248 c + 1100 c 3 + 2800 c d = 0 , Υ 1 d = 1400 c 2 12448 d = 0
Applying Newton’s methods, we have
c 1 = 0 d 1 = 0 , c 2 = 3.0599 , d 2 = 1.0531 , c 3 = 3.0599 , d 3 = 1.0531 .
Thus, in ( 0 , 1 ) × ( 0 , 1 c 2 ) , there is no critical point.
(1)
For c = 0 ,
Υ 1 ( 0 , d ) = 6624 6224 d 2 6624 .
(2)
For d = 0 ,
Υ 1 ( c , 0 ) = 6624 6624 c 2 + 275 c 4 = γ 1 ( c ) γ 1 ( 0 ) = 6624 .
(3)
For d = 1 c 2 ,
Υ 1 ( c , 1 c 2 ) = 400 + 7224 c 2 7349 c 4 = γ 2 ( c ) γ 2 ( 0.7011 ) = 2175.3 .
Therefore, we have
| γ 4 | 13824 165888 = 1 12 .
Rearranging (19), we obtain
| δ 5 | = 1 2488320 | 86400 ( c 5 c 1 c 4 c 2 c 3 + 3 4 c 1 c 2 2 + 3 4 c 1 2 c 3 1 2 c 1 3 c 2 + 1 16 c 1 5 ) + 86400 c 5 + 14400 c 2 c 3 19800 c 1 2 c 3 24800 c 1 c 2 2 + 38200 c 1 3 c 2 + 32285 c 1 5 | 1 2488320 | 86400 ( c 5 c 1 c 4 c 2 c 3 + 3 4 c 1 c 2 2 + 3 4 c 1 2 c 3 1 2 c 1 3 c 2 + 1 16 c 1 5 ) | + 1 2488320 | 86400 c 5 + 14400 c 2 c 3 19800 c 1 2 c 3 24800 c 1 c 2 2 + 38200 c 1 3 c 2 + 32285 c 1 5 | = 1 2488320 3 + 1 2488320 4 ,
where 3 = | 86400 ( c 5 c 1 c 4 c 2 c 3 + 3 4 c 1 c 2 2 + 3 4 c 1 2 c 3 1 2 c 1 3 c 2 + 1 16 c 1 5 ) | and
4 = | 86400 c 5 + 14400 c 2 c 3 19800 c 1 2 c 3 24800 c 1 c 2 2 + 38200 c 1 3 c 2 + 32285 c 1 5 | .
Using Lemma 8 with γ = 1 2 , we obtain 3 86400 . Rearranging (21), we obtain
4 = | 86400 c 5 + 14400 c 2 ( c 3 2 c 1 c 2 + c 1 3 ) 19800 c 1 2 ( c 3 c 1 c 2 c 1 3 ) + 4000 c 1 c 2 2 + 4000 c 1 3 c 2 + 12485 c 1 5 | .
Utilizing the triangle inequality, Lemma 1, Lemma 6 and 5, we obtain
4 86400 | c 5 | + 14400 | c 2 | | c 3 2 c 1 c 2 + c 1 3 | + 19800 | c 1 | 2 | c 3 c 1 c 2 c 1 3 | + 4000 | c 1 | | c 2 | 2 + 4000 | c 1 | 3 | c 2 | + 12485 | c 1 | 5 86400 ( 1 | c 1 | 2 | c 2 | 2 | c 3 | 2 1 + | c 1 | ) + 14400 | c 2 | + 19800 | c 1 | 2 + 4000 | c 1 | | c 2 | 2 + 4000 | c 1 | 3 | c 2 | + 12485 | c 1 | 5 = 86400 666000 | c 1 | 2 + 12485 | c 1 | 5 + ( 14400 + 4000 | c 1 | 3 ) | c 2 | + ( 86400 + 4000 | c 1 | ) | c 2 | 2 86400 | c 3 | 2 1 + | c 1 | 86400 666000 | c 1 | 2 + 12485 | c 1 | 5 + ( 14400 + 4000 | c 1 | 3 ) | c 2 | + ( 86400 + 4000 | c 1 | ) | c 2 | 2 = Υ 2 ( c , d )
where | c 1 | = c , | c 2 | = d . Consider
Υ 2 c = 133200 c + 62425 c 4 + 12000 c 2 d + 4000 d 2 = 0 , Υ 2 d = 14400 + 4000 c 3 + ( 172800 + 8000 c ) d = 0 .
We have a critical point ( 0.000209 , 0.083334 ) . Thus, we obtain Υ 2 ( c , d ) Υ 2 ( 0.000209 , 0.083334 ) = 86999.968 .
(1)
For c = 0 ,
Υ 2 ( 0 , d ) = 86400 + 14400 d 86400 d 2 = γ 3 ( d ) γ 3 ( 1 12 ) = 87000 .
(2)
For d = 0 ,
Υ 2 ( c , 0 ) = 86400 66600 c 2 + 12485 c 5 864000 .
(3)
For d = 1 c 2 ,
Υ 2 ( c , 1 c 2 ) = 14400 + 4000 c + 91800 c 2 4000 c 3 86400 c 4 + 12485 c 5 = γ 4 ( c ) γ 4 ( 0.7771 ) = 43098 .
Therefore, we receive
| δ 5 | 173400 2488320 = 0.069686 .
These bounds can be determined by the following extremal functions:
g 4 ( ξ ) = 0 ξ ( 1 + 5 6 t + 1 6 t 5 ) d t = ξ + 5 12 ξ 2 + 1 36 ξ 6 + · · · ,
g 5 ( ξ ) = 0 ξ ( 1 + 5 6 t 2 + 1 6 t 10 ) d t = ξ + 5 18 ξ 3 + 1 66 ξ 11 + · · · ,
g 6 ( ξ ) = 0 ξ ( 1 + 5 6 t 3 + 1 6 t 15 ) d t = ξ + 5 24 ξ 4 + 1 96 ξ 16 + · · · ,
g 7 ( ξ ) = 0 ξ ( 1 + 5 6 t 4 + 1 6 t 20 ) d t = ξ + 1 6 ξ 5 + 1 126 ξ 21 + · · · .
The proof of Theorem 10 is completed.  □

4. Conclusions

In this paper, we considered a subclass of bounded turning functions linked with a four-leaf-type domain. Utilizing the estimates of the coefficients of the Schwartz function, we obtained the estimates of coefficients of | b 6 | , | b 7 | , | b 8 | , and the third-order determinants of | H 3 , 1 | , | H 3 , 2 | , | H 3 , 3 | for the class B T 4 l for the first time. Also, one can easily use this new methodology to obtain the bounds the coefficients of | b 6 | , | b 7 | , | b 8 | , and the third-order Hankel determinant of | H 3 , 1 | , | H 3 , 2 | , | H 3 , 3 | for other subclasses of univalent functions. The bounds provided in Theorems 2–9 lack sharpness and improve these bounds in further research.

Author Contributions

C.W.; Writing—original draft, Z.L.; Methodology; D.G.; Validation. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

No new data were created in this study.

Conflicts of Interest

The authors state that they have no conflicts of interest.

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Wen, C.; Li, Z.; Guo, D. Some Results on Coefficient Estimate Problems for Four-Leaf-Type Bounded Turning Functions. Mathematics 2024, 12, 1875. https://doi.org/10.3390/math12121875

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Wen C, Li Z, Guo D. Some Results on Coefficient Estimate Problems for Four-Leaf-Type Bounded Turning Functions. Mathematics. 2024; 12(12):1875. https://doi.org/10.3390/math12121875

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Wen, Chuanjun, Zongtao Li, and Dong Guo. 2024. "Some Results on Coefficient Estimate Problems for Four-Leaf-Type Bounded Turning Functions" Mathematics 12, no. 12: 1875. https://doi.org/10.3390/math12121875

APA Style

Wen, C., Li, Z., & Guo, D. (2024). Some Results on Coefficient Estimate Problems for Four-Leaf-Type Bounded Turning Functions. Mathematics, 12(12), 1875. https://doi.org/10.3390/math12121875

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