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

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

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Submitted:

15 April 2024

Posted:

24 April 2024

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Abstract

Let BT4l denotes a subclass of bounded turning functions connected with a four-leaf-type domain. The goal of the study is to probe into coefﬁcients of |b6|, |b7|, |b8|, the bounds of the logarithmic coefﬁcients and the third-order determinants of H3,1, H3,2, H3,3 for functions in this class.

Keywords:

Subject: Computer Science and Mathematics - Computational Mathematics

1. Introduction and Definitions

Let

H

be a representation for a family of mappings of the following kind, denoted by g

g (ξ) = ξ + \sum_{n = 2}^{\infty} b_{n} ξ^{n}

(1)

in the unit disc

U = {ξ \in C : | z | < 1} .

We refer to

S

as a subfamily of

H

, which consists of univalent functions in

U .

For functions

g \in H

of the form

g (ξ) = ξ + b_{2} ξ^{2} + b_{3} ξ^{3} + \cdot \cdot \cdot

and positive integers i and n, the Hankel determinant

H_{i, n} (g)

is defined by

H_{i, n} (g) : = |\begin{matrix} b_{n} & a_{n + 1} & \dots & b_{n + i - 1} \\ b_{n + 1} & a_{n + 2} & \dots & b_{n + i} \\ ⋮ & ⋮ & \dots & ⋮ \\ b_{n + i - 1} & b_{n + i} & \dots & b_{n + 2 i - 2} \end{matrix}| (b_{1} = 1) .

The Hankel determinant

H_{i, n} (g)

was introduced by Pommerrenke [1,2]. The third Hankel determinants are 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},

(2)

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}),

(3)

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}) .

(4)

For the first time, the bounds of the third-order Hankel determinant

H_{3, 1} (g)

for the families of

S^{*}

K

and

R

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 \in S

, Let

F_{g} (ξ)

be defined as the logarithmic coefficients of

g (ξ)

F_{g} (ξ) = l o g \frac{g (ξ)}{ξ} = 2 \sum_{n = 1}^{\infty} δ_{n} ξ^{n} .

(5)

The

δ_{n} = δ_{n} (g)

are referred to as the 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} | \leq \frac{1}{n} (n \geq 1)

and that this is not true in general [11].

Using (1) and differentiating (5), we have

\{\begin{matrix} δ_{1} = \frac{1}{2} b_{2}, \\ δ_{2} = \frac{1}{2} (b_{3} - \frac{1}{2} b_{2}^{2}), \\ δ_{3} = \frac{1}{2} (b_{4} - b_{2} b_{3} + \frac{1}{3} b_{2}^{3}), \\ δ_{4} = \frac{1}{2} (b_{5} - b_{4} b_{2} + b_{3} b_{2}^{2} - \frac{1}{2} b_{3}^{2} - \frac{1}{4} b_{2}^{4}), \\ δ_{5} = \frac{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} + \frac{1}{5} b_{2}^{5}) . \end{matrix}

(6)

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 \in S : g^{'} (ξ) ≺ 1 + \frac{5}{6} ξ + \frac{1}{6} ξ^{5}, (ξ \in U)\} .

Sunthrayuth et al. [12] 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 of the logarithmic coefficients for

g \in B T_{4 l}

Let

ω_{0}

be the family of Schwarz functions. Thus, the function

w \in ω_{0}

may be expressed as a power series

w (ξ) = \sum_{n = 1}^{\infty} c_{n} ξ^{n} .

(7)

Lemma 1

(see [13]). Suppose a function w is member of

ω_{0}

. Then

|c_{2}| \leq 1 - {|c_{1}|}^{2}, | c_{3} | \leq 1 - | c_{1} |^{2} - \frac{| c_{2} |^{2}}{1 + | c_{1} |}, | c_{4} | \leq 1 - | c_{1} |^{2} - {| c_{2} |}^{2},

| c_{5} | \leq 1 - | c_{1} |^{2} - | c_{2} |^{2} - \frac{| c_{3} |^{2}}{1 + | c_{1} |}, | c_{6} | \leq 1 - | c_{1} |^{2} - | c_{2} |^{2} - {| c_{3} |}^{2},

| c_{7} | \leq 1 - | c_{1} |^{2} - | c_{2} |^{2} - {| c_{3} |}^{2} - \frac{| c_{4} |^{2}}{1 + | c_{1} |} .

Lemma 2

(see [14,15]). Suppose a function w is member of

ω_{0}

. then

| c_{1} c_{3} - c_{2}^{2} | \leq 1 - | c_{1} |^{2}, | c_{2} c_{4} - c_{3}^{2} | \leq {(1 - c_{1}^{2})}^{2}, | c_{3} c_{5} - c_{4}^{2} | \leq 1 - c_{1}^{2} .

Lemma 3

(see [12]). If

g \in B T_{4 l}

, then

| b_{2} | \leq \frac{5}{12}, | b_{3} | \leq \frac{5}{18}, | b_{4} | \leq \frac{5}{24}, | b_{5} | \leq \frac{1}{6} .

Lemma 4

(see [11]). If

w \in ω_{0}

, then

|c_{n}| \leq 1 (n \geq 1) .

Lemma 5

(see [16]). Let

w \in ω_{0}

, then for

γ \in C

, we have

|c_{2} + γ c_{1}^{2}| \leq m a x \{1, |γ|\}

Lemma 6

(see [17]). If

w \in ω_{0}

is in the form (7). Then, we get

| c_{3} + γ c_{1} c_{2} + ς c_{1}^{3} | \leq 1,

where

(γ, ς) \in D_{0} \cup D_{1},

with

D_{1} = {(γ, ς) \in R^{2} : | γ | \leq \frac{1}{2}, - 1 \leq ς \leq 1},

D_{2} = {(γ, ς) \in R^{2} : \frac{1}{2} \leq | γ | \leq 2, \frac{4}{27} {(| γ | + 1)}^{3} - (| γ | + 1) \leq ς \leq 1},

Lemma 7

(see [18]). Let

w \in ω_{0}

and

γ \in 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}| \leq {m a x {1, | γ |}^{3}} .

Lemma 8

(see [18]). If

w \in ω_{0}

, then for all

γ \in 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} | \leq 1 .

2. The bounds of the third Hankel determinant for $g \in B T_{4 l}$

Theorem 1.

g \in B T_{4 l}

, then

\begin{matrix} | b_{6} | \leq \frac{5}{36}, | b_{7} | \leq \frac{5}{42}, | b_{8} | \leq \frac{5}{48} . \end{matrix}

The bounds are sharp.

Proof.

For a function

g \in B T_{4 l}

, there exists a Schwarz function

w (ξ)

, such that

g^{'} (ξ) = 1 + \frac{5}{6} w (ξ) + \frac{1}{6} w^{5} (ξ),

Comparing the coefficients, we yield

b_{2} = \frac{5}{12} c_{1}, b_{3} = \frac{5}{18} c_{2}, b_{4} = \frac{5}{24} c_{3}, b_{5} = \frac{1}{6} c_{4},

(8)

b_{6} = \frac{5 c_{5} + c_{1}^{5}}{36}

(9)

b_{7} = \frac{5 c_{6} + 5 c_{1}^{4} c_{2}}{42},

(10)

b_{8} = \frac{5 c_{7} + 5 c_{1}^{4} c_{3} + 10 c_{1}^{3} c_{2}^{2}}{48} .

(11)

From (9) and Lemma 1, we have

\begin{matrix} |b_{6}| & \leq \frac{1}{36} [5 | c_{5} | + | c_{1} |^{5}] \\ \leq \frac{1}{36} [5 (1 - | c_{1} |^{2} - | c_{2} |^{2} - \frac{| c_{3} |^{2}}{1 + | c_{1} |}) + | c_{1} |^{5}] \\ = \frac{1}{36} (5 - 5 | c_{1} |^{2} + | c_{1} |^{5} - 5 | c_{2} |^{2} - \frac{5 | c_{3} |^{2}}{1 + | c_{1} |}) \\ \leq \frac{1}{36} (5 - 5 | c_{1} |^{2} + | c_{1} |^{5}) \\ \leq \frac{5}{36} . \end{matrix}

From (10) and Lemma 1, we achieve

\begin{matrix} |b_{7}| & \leq \frac{1}{42} [5 | c_{6} | + 5 | c_{1} |^{4} | c_{2} |] \\ \leq \frac{1}{42} [5 (1 - | c_{1} |^{2} - | c_{2} |^{2} - | c_{3} |^{2}) + 5 | c_{1} |^{4} | c_{2} |] \\ = \frac{1}{42} (5 - 5 | c_{1} |^{2} + 5 | c_{1} |^{4} | c_{2} | - 5 | c_{2} |^{2} - 5 | c_{3} |^{2}) \\ \leq \frac{1}{42} [5 - 5 | c_{1} |^{2} + 5 | c_{1} |^{4} (1 - | c_{1} |^{2})] \\ = \frac{1}{42} (5 - 5 | c_{1} |^{2} + 5 | c_{1} |^{4} - 5 {| c_{1} |}^{6}) \\ \leq \frac{5}{42} . \end{matrix}

From (11) and Lemma 1, we get

\begin{matrix} |b_{8}| & \leq \frac{1}{48} [5 | c_{7} | + 5 | c_{1} |^{4} | c_{3} | + 10 | c_{1} |^{3} {| c_{2} |}^{2}] \\ \leq \frac{1}{48} [5 (1 - | c_{1} |^{2} - | c_{2} |^{2} - | c_{3} |^{2} - \frac{| c_{4} |^{2}}{1 + | c_{1} |}) + 5 | c_{1} |^{4} | c_{3} | + 10 | c_{1} |^{3} {| c_{2} |}^{2}] \\ = \frac{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 \frac{| c_{4} |^{2}}{1 + | c_{1} |}] \\ \leq \frac{1}{48} [5 - 5 | c_{1} |^{2} + (- 5 + 10 | c_{1} |^{3}) | c_{2} |^{2} + 5 | c_{1} |^{4} (1 - | c_{1} |^{2} - \frac{| c_{2} |^{2}}{1 + | c_{1} |})] \\ = \frac{1}{48} [5 - 5 | c_{1} |^{2} + 5 | c_{1} |^{4} - 5 | c_{1} |^{6} + \frac{- 5 - 5 | c_{1} | + 10 | c_{1} |^{3} + 5 {| c_{1} |}^{4}}{1 + | c_{1} |} | c_{2} |^{2}] . \end{matrix}

Setting

c = | c_{1} |

and

d = | c_{2} |

, we get

\begin{matrix} |b_{8}| \leq \frac{1}{48} ϵ_{1} (c, d) \end{matrix}

where

\begin{matrix} ϵ_{1} (c, d) = 5 - 5 c^{2} + 5 c^{4} - 5 c^{6} + \frac{- 5 - 5 c + 10 c^{3} + 5 c^{4}}{1 + c} d^{2} . \end{matrix}

The critical points of

ϵ_{1}

satisfy

\{\begin{matrix} \frac{\partial ϵ_{1}}{\partial c} = - 10 c + 20 c^{3} - 30 c^{5} + \frac{15 c^{4} + 40 c^{3} + 30 c^{2}}{{(1 + c)}^{2}} d^{2}, \\ \frac{\partial ϵ_{1}}{\partial d} = \frac{- 10 - 10 c + 20 c^{3} + 10 c^{4}}{1 + c} d = 0 . \end{matrix}

Applying numerical computations, we have

\{\begin{matrix} c_{0} = 0, \\ d_{0} = 0, \end{matrix}

\{\begin{matrix} c_{1} = 0.8668, \\ d_{1} = 0.7940, \end{matrix}

\{\begin{matrix} c_{2} = 0.8668, \\ d_{2} = - 0.7940 . \end{matrix}

Thus, in

(0, 1) \times (0, 1 - c^{2})

, there is no critical points which satisfies

0 \leq d \leq 1 - c^{2} .

For

c = 0,

ϵ_{1} (0, d) = 5 - 5 d^{2} \leq 5 .

For

d = 0,

ϵ_{1} (c, 0) = 5 - 5 c^{2} + 5 c^{4} - 5 c^{6} \leq 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) \leq η_{1} (0.7202) = 2.5799 .

Thus, we have

\begin{matrix} | b_{8} | & \leq \frac{5}{48} . \end{matrix}

The bounds hold for

c = 0 .

The proof of Theorem 1 is completed. □

Theorem 2.

g \in B T_{4 l}

, then

\begin{matrix} |H_{2, 3} (g)| & \leq & 0.044516 . \end{matrix}

Proof.

Let

g \in B T_{4 l}

. From (8), we receive

\begin{matrix} |H_{2, 3} (g)| = | b_{3} b_{5} - b_{4}^{2} | = \frac{5}{1728} |16 c_{2} c_{4} - 15 c_{3}^{2}| = \frac{5}{1728} | 15 (c_{2} c_{4} - c_{3}^{2}) + c_{2} c_{4} | . \end{matrix}

(12)

Utilizing inequality, Lemma 1 and Lemma 2 in (12), we get

\begin{matrix} |H_{2, 3} (g)| & \leq \frac{5}{1728} [15 | c_{2} c_{4} - c_{3}^{2} | + | c_{2} | | c_{4} |] \\ \leq \frac{5}{1728} [15 (1 - | c_{1} |^{2})^{2} + | c_{2} | (1 - | c_{1} |^{2} - | c_{2} |^{2})] \\ = \frac{5}{1728} [15 - 30 | c_{1} |^{2} + 15 | c_{1} |^{4} + (1 - | c_{1} |^{2}) | c_{2} | - | c_{2} |^{3})] \\ \leq \frac{5}{1728} [15 - 30 | c_{1} |^{2} + 15 | c_{1} |^{4} + (1 - | c_{1} |^{2}) | c_{2} | - | c_{2} |^{3}] . \end{matrix}

Let

| c_{1} | = c

and

| c_{2} | = d

, we obtain

\begin{matrix} |H_{2, 3} (g)| & \leq \frac{5}{1728} ϵ_{2} (c, d) . \end{matrix}

Consider

\{\begin{matrix} \frac{\partial ϵ_{2}}{\partial c} = - 60 c + 60 c^{3} - 2 c d = 0, \\ \frac{\partial ϵ_{2}}{\partial d} = 1 - c^{2} - 3 d^{2} = 0 . \end{matrix}

Applying numerical computations, we get

\{\begin{matrix} c_{1} = 0, \\ d_{1} = - 0.5774, \end{matrix}

\{\begin{matrix} c_{2} = 0, \\ d_{2} = 0.5774, \end{matrix}

\{\begin{matrix} c_{3} = - 1, \\ d_{3} = 0, \end{matrix}

\{\begin{matrix} c_{4} = 1, \\ d_{4} = 0, \end{matrix}

\{\begin{matrix} c_{5} = 0.9998, \\ d_{5} = - 0.0111, \end{matrix}

\{\begin{matrix} c_{6} = - 0.9998, \\ d_{6} = - 0.0111 . \end{matrix}

Thus, there is no critical point in

(0, 1) \times (0, 1 - c^{2}) .

(1)For

d = 0,

\begin{matrix} ϵ_{2} (c, 0) = 15 - 30 c^{2} + 15 c^{4} \leq 15 . \end{matrix}

(2)For

c = 0

\begin{matrix} ϵ_{2} (0, d) = 15 + d - d^{3} \leq ϵ_{2} (0, \frac{\sqrt{3}}{3}) = \frac{2 \sqrt{3} + 135}{9} = 15.384888 . \end{matrix}

(3)For

d = 1 - c^{2},

\begin{matrix} ϵ_{2} (c, 1 - c^{2}) = 15 - 29 c^{2} + 13 c^{4} + c^{6} \leq 15 . \end{matrix}

Therefore, we yield

\begin{matrix} |H_{2, 3} (g)| \leq \frac{10 \sqrt{3} + 675}{15552} = 0.044516 . \end{matrix}

The proof of Theorem 2 is completed. ☐

Theorem 3.

g \in B T_{4 l}

, then

\begin{matrix} |H_{3, 1} (g)| & \leq & \frac{3025}{46656} = 0.064836 . \end{matrix}

Proof.

Assume that

g \in B T_{4 l}

. From (8) and (2), we achieve

\begin{matrix} |H_{3, 1} (g)| & = \frac{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}| \\ = \frac{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}| \end{matrix}

(13)

By applying the triangle inequality, Lemma 1 and Lemma 2 in (13), we receive

\begin{matrix} 46656 |H_{3, 1} (g)| & \leq 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} | \\ \leq 1000 | c_{2} | (1 - | c_{1} |^{2}) + 1250 | c_{1} | | c_{2} | (1 - | c_{1} |^{2} - \frac{| 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} - \frac{135 + 1385 | c_{1} |}{1 + | c_{1} |} {| c_{2} |}^{3} . \end{matrix}

Setting

| c_{1} | = c

and

| c_{2} | = d

, we have

\begin{matrix} 46656 |H_{3, 1} (g)| & \leq & 2025 - 2700 c^{2} + 675 c^{4} + (1135 + 1250 c - 1135 c^{2} - 1250 c^{3}) d \\ - 1350 c^{2} d^{2} - \frac{135 + 1385 c}{1 + c} d^{3} = ϵ_{3} (c, d) \end{matrix}

where

c \in [0, 1]

and

d \in [0, 1 - c^{2}] .

Taking the partial derivative with respect to c and y respectively, and we have

\begin{matrix} \frac{\partial ϵ_{3}}{\partial c} & = - 5400 c + 2700 c^{3} + (1250 - 2270 c - 3405 c^{2}) d - 2700 c d^{2} - \frac{1250}{{(1 + c)}^{2}} d^{3} \end{matrix}

and

\begin{matrix} \frac{\partial ϵ_{3}}{\partial d} & = 1135 + 1250 c - 1135 c^{2} - 1250 c^{3} - 2700 c^{2} d - \frac{405 + 4155 c}{1 + c} d^{2} . \end{matrix}

Setting

\frac{\partial ϵ_{3}}{\partial c} = \frac{\partial ϵ_{3}}{\partial d} = 0

and simplifying, we yield

\{\begin{matrix} - 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 . \end{matrix}

Applying Newton’s methods, we yield

\{\begin{matrix} c_{1} = - 1, \\ d_{1} = 0, \end{matrix}

\{\begin{matrix} c_{2} = 0.1018, \\ d_{2} = - 1.3080, \end{matrix}

\{\begin{matrix} c_{3} = 1.0726, \\ d_{3} = - 1.1909, \end{matrix}

\{\begin{matrix} c_{4} = 1.7878 \\ d_{4} = - 0.3706, \end{matrix}

\{\begin{matrix} c_{5} = 3.3060, \\ d_{5} = - 6.5565, \end{matrix}

\{\begin{matrix} c_{6} = - 1.3977, \\ d_{6} = 0.0899 . \end{matrix}

Thus, there is no critical point in

(0, 1) \times (0, 1 - c^{2}) .

For

c = 0,

ϵ_{3} (0, d) = 2025 + 1135 d - 135 d^{3} \leq ϵ_{3} (0, 1) = 3025 .

For

d = 0,

ϵ_{3} (c, 0) = 2025 - 2700 c^{2} + 675 c^{4} \leq 2025 .

For

d = 1 - c^{2}

ϵ_{3} (c, 0) = 3025 - 4665 c^{2} + 1605 c^{4} + 35 c^{6} \leq ϵ_{3} (0, 0) = 3025 .

Thus, we obtain

\begin{matrix} |H_{3, 1} (g)| & \leq \frac{3025}{46656} = 0.064836 . \end{matrix}

The proof of Theorem 3 is completed. ☐

Theorem 4.

g \in B T_{4 l}

, then

\begin{matrix} | b_{2} b_{5} - b_{3} b_{4} | \leq \frac{15.503407}{432} = 0.035888 . \end{matrix}

Proof.

Let

g \in B T_{4 l}

. From (8), we obtain

| b_{2} b_{5} - b_{3} b_{4} | = \frac{1}{432} | 30 c_{1} c_{4} - 25 c_{2} c_{3} | .

(14)

Applying Lemma 1 and the triangle inequality in (14), we get

\begin{matrix} | b_{2} b_{5} - b_{3} b_{4} | & \leq \frac{1}{432} [30 | c_{1} | (1 - | c_{1} |^{2} - | c_{2} |^{2}) + 25 | c_{2} | (1 - | c_{1} |^{2} - \frac{| c_{2} |^{2}}{1 + | c_{1} |})] \\ = \frac{1}{432} [30 | c_{1} | - 30 | c_{1} |^{3} + (25 - 25 | c_{1} |^{2}) | c_{2} | - 30 | c_{1} | | c_{2} |^{2} - \frac{25}{1 + | c_{1} |} {| c_{2} |}^{3}] . \end{matrix}

Setting

| c_{1} | = c

and

| c_{2} | = d

, we yield

\begin{matrix} | b_{2} b_{5} - b_{3} b_{4} | \leq \frac{1}{432} ϵ_{4} (c, d) . \end{matrix}

Taking the partial derivative with respect to c, and d respectively, we get

\begin{matrix} \frac{\partial ϵ_{4}}{\partial c} & = 30 - 90 c^{2} - 50 c d - 30 d^{2} + \frac{25}{{(1 + c)}^{2}} d^{3} \end{matrix}

and

\begin{matrix} \frac{\partial ϵ_{4}}{\partial d} & = 25 - 25 c^{2} - 60 c d - \frac{75}{1 + c} d^{2} . \end{matrix}

Setting

\frac{\partial ϵ_{4}}{\partial c} = \frac{\partial ϵ_{4}}{\partial d} = 0

,and simplifying, we receive

\{\begin{matrix} - 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 . \end{matrix}

Applying Newton’s methods, we have

\{\begin{matrix} c_{1} = - 1, \\ d_{1} = 0, \end{matrix}

\{\begin{matrix} c_{2} = 0.4098, \\ d_{2} = - 0.8978, \end{matrix}

\{\begin{matrix} c_{3} = 0.4271, \\ d_{3} = 0.4258, \end{matrix}

\{\begin{matrix} c_{4} = - 0.4550 \\ d_{4} = - 0.2931, \end{matrix}

\{\begin{matrix} c_{5} = - 0.7258, \\ d_{5} = 0.3023 . \end{matrix}

Therefore, we get

ε_{4} (c, d) \leq ε_{4} (0.4271, 0.4258) = 15.503407 .

(a)For

c = 0,

ε_{4} (0, d) = 25 d - 25 d^{3} \leq ε_{4} (0, \frac{\sqrt{3}}{3}) = \frac{50 \sqrt{3}}{9} .

(b)For

d = 0

ε_{4} (c, 0) = 30 c - 30 c^{3} \leq ε_{4} (\frac{\sqrt{3}}{3}, 0) = \frac{20 \sqrt{3}}{3} = 11.546666 .

(b)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 get

| b_{2} b_{5} - b_{3} b_{4} | \leq \frac{15.503407}{432} = 0.035888 .

☐

Theorem 5.

g \in B T_{4 l}

, then

\begin{matrix} | b_{2} b_{4} - b_{3}^{2} | \leq \frac{200.27}{2592} = 0.077265 . \end{matrix}

Proof.

Let

g \in B T_{4 l}

. From (8), we get

\begin{matrix} | b_{2} b_{4} - b_{3}^{2} | & = \frac{1}{2592} | 225 c_{1} c_{3} - 200 c_{2}^{2} | \\ = \frac{1}{2592} | 200 (c_{1} c_{3} - c_{2}^{2}) + 25 c_{1} c_{3} | . \end{matrix}

Applying Lemma 1, Lemma 2 and the triangle inequality, we receive

\begin{matrix} | b_{2} b_{4} - b_{3}^{2} | & \leq \frac{1}{2592} [200 (1 - | c_{1} |^{2}) + 25 | c_{1} | (1 - | c_{1} |^{2} - \frac{| c_{2} |^{2}}{1 + | c_{1} |})] \\ = \frac{1}{2592} [200 + 25 | c_{1} | - 200 | c_{1} |^{2} - 25 | c_{1} |^{3} - \frac{| c_{1} |}{1 + | c_{1} |} {| c_{2} |}^{2}] \\ \leq \frac{1}{2592} (200 + 25 | c_{1} | - 200 | c_{1} |^{2} - 25 {| c_{1} |}^{3}) \\ = \frac{1}{2592} η_{3} (| c_{1} |) \leq \frac{1}{2592} η_{3} (0.0618) = \frac{200.2700}{2592} = 0.077265 . \end{matrix}

☐

Theorem 6.

g \in B T_{4 l}

, then

\begin{matrix} | H_{3, 2} (g) | \leq 0.025986 . \end{matrix}

Proof.

Let

g \in B T_{4 l}

. From Lemma 3, Theorem 1, Theorem 2, Theorem 4 and Theorem 5, we yield

\begin{matrix} | H_{3, 2} (g) | & \leq | 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} | \\ = \frac{5}{24} \times 0.044516 + \frac{1}{6} \times 0.035888 + \frac{5}{36} \times 0.077265 \\ = 0.025986 . \end{matrix}

☐

Theorem 7.

g \in B T_{4 l}

, then

\begin{matrix} | b_{4} b_{6} - b_{5}^{2} | \leq \frac{24.385252}{864} = 0.028224 . \end{matrix}

Proof.

Let

f \in B T_{4 l}

. From (8) and (9), we have

\begin{matrix} | b_{4} b_{6} - b_{5}^{2} | = \frac{1}{864} | 25 c_{3} c_{5} - 24 c_{4}^{2} + 5 c_{3} c_{1}^{5} | = \frac{1}{864} | 24 (c_{3} c_{5} - c_{4}^{2}) + c_{3} c_{5} + 5 c_{3} c_{1}^{5} | . \end{matrix}

Using Lemma 1 and 2, we have

\begin{matrix} | b_{4} b_{6} - b_{5}^{2} | & \leq \frac{1}{864} (24 | c_{3} c_{5} - c_{4}^{2} | + | c_{3} | | c_{5} | + 5 | c_{3} | | c_{1}^{5} |) \\ \leq \frac{1}{864} [24 (1 - | c_{1} |^{2}) + | c_{3} | (1 - | c_{1} |^{2} - | c_{2} |^{2} - \frac{| c_{3} |^{2}}{1 + | | c_{1}}) + 5 | c_{1} |^{5} | c_{3} |] \\ = \frac{1}{864} [24 - 24 | c_{1} |^{2} + (1 - | c_{1} |^{2} + 5 | c_{1} |^{5} - | c_{2} |^{2}) | c_{3} | - \frac{1}{1 + | c_{1} |} | c_{3} |^{3}] . \end{matrix}

Setting

c = | c_{1} |, d = | c_{2} |

and

e = | c_{3} |

, we yield

\begin{matrix} | b_{4} b_{6} - b_{5}^{2} | & \leq \frac{1}{864} ϵ_{5} (c, d, e) \end{matrix}

where

ϵ_{5} (c, d, e) = 24 - 24 c^{2} + (1 - c^{2} + 5 c^{5} - d^{2}) e - \frac{1}{1 + c} e^{3}

and

(c, d, e) \in Ω = {(c, d, e) : 0 \leq 1, 0 \leq d \leq 1 - c^{2}, 0 \leq e \leq 1 - c^{2} - \frac{d^{2}}{1 + c}}

. Consider

\frac{\partial ϵ_{5}}{\partial d} = - 2 d e \leq 0,

thus there are no points in

Ω

(1)For

c = 0,

\begin{matrix} ϵ_{5} (0, d, e) = 24 + (1 - d^{2}) e - e^{3} = η_{4} (d, e) . \end{matrix}

It is evident that there is on point in

(0, 1) \times (0, 1 - d^{2}) .

(2)For

d = 0,

\begin{matrix} ϵ_{5} (c, 0, e) = 24 - 24 c^{2} + (1 - c^{2} + 5 c^{5}) e - \frac{1}{1 + c} e^{3} = η_{5} (c, e) . \end{matrix}

Partial derivative of

η_{5}

with respect to c, and then with respect to e, we achieve

\begin{matrix} \frac{\partial η_{5}}{\partial c} = - 48 c + (- 2 c + 25 c^{4}) e + \frac{1}{{(1 + c)}^{2}} e^{3}, \end{matrix}

and

\begin{matrix} \frac{\partial η_{5}}{\partial e} = 1 - c^{2} + 5 c^{5} - \frac{3}{1 + c} e^{2} . \end{matrix}

Setting

\frac{\partial η_{5}}{\partial c} = \frac{\partial η_{5}}{\partial e} = 0

and simplifying, we yield

\{\begin{matrix} - 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 . \end{matrix}

We obtain a critical point

(0.0039, 0.5785)

, thus, we have

η_{5} (c, e) \leq η_{5} (0.0039, 0.5785) = 24.385252 .

(3)For

e = 0,

\begin{matrix} ϵ_{5} (c, d, 0) = 24 - 24 c^{2} \leq 24 . \end{matrix}

(4)For

e = 1 - c^{2} - \frac{d^{2}}{1 + c},

\begin{matrix} ϵ_{5} (c, d, 1 - c^{2} - \frac{d^{2}}{1 + c}) & = 24 + c - 24 c^{2} - 2 c^{3} + 6 c^{5} - 5 c^{7} + \frac{- 5 c^{5} + 4 c^{3} - c^{2} - 4 c + 1}{1 + c} d^{2} \\ + \frac{- 2 + 4 c}{{(1 + c)}^{2}} d^{4} + \frac{1}{{(1 + c)}^{4}} d^{6} = η_{6} (c, d) \leq η_{6} (0.0016, 0.5767) = 24.148169 . \end{matrix}

(5)For

c = d = 0

\begin{matrix} ϵ_{5} (0, 0, e) = 24 + e - e^{3} = η_{7} (e) \leq η_{7} (\frac{\sqrt{3}}{3}) = 24 + \frac{2 \sqrt{3}}{9} = 24.384889 . \end{matrix}

(6)For

c = e = 0,

\begin{matrix} ϵ_{5} (0, d, 0) = 24 . \end{matrix}

(7)For

d = e = 0,

\begin{matrix} ϵ_{5} (c, 0, 0) = 24 + e - e^{3} \leq 24.384889 . \end{matrix}

(8)For

e = 0, d = 1 - c^{2}

\begin{matrix} ϵ_{5} (c, 1 - c^{2}, 0) = 24 - 24 c^{2} \leq 24 . \end{matrix}

(9)For

d = 0

and

e = 1 - c^{2}

\begin{matrix} ϵ_{5} (c, 0, 1 - c^{2}) = 24 + c - 24 c^{2} - 2 c^{3} + 6 c^{5} - 5 c^{7} = η_{8} (c) \leq η_{8} (0.0206) = 24.0104 . \end{matrix}

(10)For

c = 0, e = 1 - d^{2}

\begin{matrix} ϵ_{5} (0, d, 1 - d^{2}) = 24 + d^{2} - 2 d^{4} + d^{6} = η_{9} (d) \leq η_{9} (\frac{\sqrt{3}}{3}) = 24.148148 . \end{matrix}

Thus, we get

\begin{matrix} | b_{4} b_{6} - b_{5}^{2} | \leq \frac{24.385252}{864} = 0.035888 . \end{matrix}

☐

Theorem 8.

g \in B T_{4 l}

, then

\begin{matrix} | b_{3} b_{6} - b_{4} b_{5} | \leq \frac{5}{144} . \end{matrix}

The bound is sharp.

Proof.

Let

g \in B T_{4 l}

. From (8) and (9), we receive

\begin{matrix} | b_{3} b_{6} - b_{4} b_{5} | = \frac{1}{1296} | 50 c_{2} c_{5} - 45 c_{3} c_{4} + 50 c_{2} c_{1}^{5} | . \end{matrix}

Using Lemma 1, we get

\begin{matrix} | b_{3} b_{6} - b_{4} b_{5} | & \leq \frac{1}{1296} (50 | c_{2} | | c_{5} | + 45 | c_{3} | | c_{4} | + 50 | c_{2} | | c_{1}^{5} |) \\ \leq \frac{1}{1296} [50 | c_{2} | (1 - | c_{1} |^{2} - | c_{2} |^{2} - \frac{| c_{3} |^{2}}{1 + | c_{1} |}) + 45 | c_{3} | (1 - | c_{1} |^{2} - | c_{2} |^{2}) + 50 | c_{1} |^{5} | c_{2} |] \\ = \frac{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} | - \frac{50 | c_{2} |}{1 + | c_{1} |} | c_{3} |^{2}] . \end{matrix}

By setting

| c_{1} | = c, | c_{2} | = d

and

| c_{3} | = e

, we have

\begin{matrix} | b_{3} b_{6} - b_{4} b_{5} | & \leq \frac{1}{1296} Ψ (c, d, e) \end{matrix}

where

Ψ (c, d, e) = (50 - 50 c^{2} + 50 c^{5}) d - 50 d^{3} + (45 - 45 c^{2} - 45 d^{2}) e - \frac{50 d}{1 + c} e^{2}

(c, d, e) \in Ω

. Differentiating partially with respect to c, d and e, respectively, we get

\begin{matrix} \frac{\partial Ψ}{\partial c} & = (- 100 c + 250 c^{4}) d - 90 c e + \frac{50 d}{{(1 + c)}^{2}} e^{2}, \end{matrix}

\begin{matrix} \frac{\partial Ψ}{\partial d} & = 50 - 50 c^{2} + 50 c^{5} - 150 d^{2} - 90 d e - \frac{50}{1 + c} e^{2}, \end{matrix}

and

\begin{matrix} \frac{\partial Ψ}{\partial e} & = 45 - 45 c^{2} - 45 d^{2} - \frac{100 d}{1 + c} e . \end{matrix}

By putting

\frac{\partial Ψ}{\partial c} = \frac{\partial Ψ}{\partial d} = \frac{\partial Ψ}{\partial e} = 0

, and simplifying, we obtain

\{\begin{matrix} (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 . \end{matrix}

By a numberical caculation, we get

\{\begin{matrix} c_{1} = - 1, \\ d_{1} = d, \\ e_{1} = 0, \end{matrix}

\{\begin{matrix} c_{2} = 0.8441, \\ d_{2} = 0.4127, \\ e_{2} = 0.2354, \end{matrix}

\{\begin{matrix} c_{3} = 0.8441, \\ d_{3} = - 0.4127, \\ e_{3} = - 0.2354 . \end{matrix}

Thus, there’s no critical point which satisfies

0 \leq c \leq 1

and

0 \leq d \leq 1 - c^{2} .

(1)For

c = 0,

\begin{matrix} Ψ (0, d, e) = 50 d - 50 d^{3} + (45 - 45 d^{2}) e - 50 d e^{2} = Λ_{1} (d, e) . \end{matrix}

Consider

\{\begin{matrix} \frac{\partial Λ_{1}}{\partial d} = 50 - 150 d^{2} - 90 d e - 50 e^{2} = 0, \\ \frac{\partial Λ_{1}}{\partial e} = 45 - 45 d^{2} - 100 d e = 0 . \end{matrix}

A numberrical caculation that there is no critical point in

(0, 1) \times (0, 1 - d^{2}) .

(2)For

d = 0,

\begin{matrix} Ψ (c, 0, e) = (45 - 45 c^{2}) e \leq 45 {(1 - c^{2})}^{2} \leq 45 . \end{matrix}

(3)For

e = 0,

\begin{matrix} Ψ (c, d, 0) = (50 - 50 c^{2} + 50 c^{5}) d - 50 d^{3} = Λ_{2} (c, d) \leq Λ_{2} (0.7368, 0.2828) = 8.403278 . \end{matrix}

(4)For

e = 1 - c^{2} - \frac{d^{2}}{1 + c},

\begin{matrix} Ψ (c, d, 1 - c^{2} - \frac{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} \\ + \frac{50 - 150 c}{1 + c} d^{3} + \frac{45}{1 + c} d^{4} - \frac{50}{{(1 + c)}^{3}} d^{5} = Λ_{3} (c, d) . \end{matrix}

Differentiating

Λ_{3}

partially with respect to c and d, we yield

\begin{matrix} \frac{\partial Λ_{3}}{\partial c} & = - 180 c + 180 c^{3} + (50 - 150 c^{2} + 250 c^{4}) d + (45 + 90 c) d^{2} - \frac{100 d^{3}}{{(1 + c)}^{2}} \\ - \frac{45 d^{4}}{{(1 + c)}^{2}} + \frac{150 d^{5}}{{(1 + c)}^{4}} \end{matrix}

and

\begin{matrix} \frac{\partial Λ_{3}}{\partial d} & = 50 c - 50 c^{3} + 50 c^{5} + (- 180 + 90 c + 90 c^{2}) d + \frac{150 - 450 c}{1 + c} d^{2} + \frac{180 d^{3}}{1 + c} \\ - \frac{250 d^{4}}{{(1 + c)}^{3}} . \end{matrix}

Setting

\frac{\partial Λ_{3}}{\partial c} = \frac{\partial Λ_{3}}{\partial d} = 0

and simplifying, we receive

\{\begin{matrix} 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 \end{matrix}

Applying Newton’s methods, we recieve

\{\begin{matrix} c_{1} = - 1 \\ d_{1} = 0, \end{matrix}

\{\begin{matrix} c_{2} = 0, \\ d_{2} = 0, \end{matrix}

\{\begin{matrix} c_{3} = 0.8336, \\ d_{3} = 0.4141, \end{matrix}

\{\begin{matrix} c_{4} = - 0.9194, \\ d_{4} = - 0.0957 . \end{matrix}

Thus, there is no critical point satisfing

0 \leq c \leq 1

and

0 \leq d \leq 1 - c^{2} .

(5)For

d = c = 0,

\begin{matrix} Ψ (0, 0, e) & = 45 e \leq 45 . \end{matrix}

(6)For

e = d = 0,

\begin{matrix} Ψ (c, 0, 0) & = 0 . \end{matrix}

(7)For

c = e = 0,

\begin{matrix} Ψ (0, d, 0) & = 50 d - 50 d^{3} = Λ_{4} (d) \leq Λ_{4} (\frac{\sqrt{3}}{3}) = \frac{100 \sqrt{3}}{9} = 19.2444 . \end{matrix}

(8)For

d = 1 - c^{2}

and

e = 0,

\begin{matrix} Ψ (c, 1 - c^{2}, 0) & = 50 c^{2} - 100 c^{4} + 50 c^{5} + 50 c^{6} - 50 c^{7} = Λ_{5} (c) \leq Λ_{5} (0.7145) = 10.6734 . \end{matrix}

(9)For

e = 1 - c^{2}

and

d = 0,

\begin{matrix} Ψ (c, 0, 1 - c^{2}) & = 45 {(1 - c^{2})}^{2} \leq 45 . \end{matrix}

(10)For

e = 1 - d^{2}

and

c = 0

\begin{matrix} Ψ (0, d, 1 - c^{2}) & = 45 - 90 d^{2} + 50 d^{3} + 45 d^{4} - 50 d^{5} = Λ_{6} (d) \leq Λ_{6} (0) = 45 . \end{matrix}

Hence, we get

\begin{matrix} | b_{3} b_{6} - b_{4} b_{5} | \leq \frac{45}{1296} = \frac{5}{144} . \end{matrix}

The equality holds for

c_{1} = c_{2} = 0

and

c_{3} = c_{4} = 1 .

☐

Theorem 9.

g \in B T_{4 l}

, then

\begin{matrix} | H_{3, 3} (g) | \leq 0.016103 . \end{matrix}

Proof.

Let

g \in B T_{4 l}

. From Lemma 3, Theorem 1, Theorem 2, Theorem 3 and 4, we receive

\begin{matrix} | H_{3, 3} (g) | & \leq | 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} | \\ \leq \frac{1}{6} \times 0.035888 + \frac{5}{36} \times \frac{5}{144} + \frac{5}{42} \times 0.044516 \\ = 0.016103 . \end{matrix}

☐

3. The bounds of the logarithmic coefficients for $g \in B T_{4 l}$

Theorem 10.

g \in B T_{4 l}

, then

\begin{matrix} | δ_{1} | \leq \frac{5}{24}, | δ_{2} | \leq \frac{5}{36}, | δ_{3} | \leq \frac{5}{48}, | δ_{4} | \leq \frac{13827}{165888} = 0.083351, | δ_{5} | \leq \frac{173400}{2488320} = 0.069686 . \end{matrix}

The first three bounds are the best possible.

Proof.

Let

g \in B T_{4 l}

. From(6),(8) and (9), we have

\begin{matrix} δ_{1} = \frac{5 c_{1}}{24}, \end{matrix}

(15)

\begin{matrix} δ_{2} = \frac{5}{36} (c_{2} - \frac{5}{8} c_{1}^{2}), \end{matrix}

(16)

\begin{matrix} δ_{3} = \frac{5}{48} (c_{3} - \frac{5}{9} c_{1} c_{2} + \frac{25}{216} c_{1}^{3}), \end{matrix}

(17)

\begin{matrix} δ_{4} = \frac{1}{165888} (13824 c_{4} - 7200 c_{1} c_{3} + 4000 c_{1}^{2} c_{2} - 3200 c_{2}^{2} - 625 c_{1}^{4}) . \end{matrix}

(18)

\begin{matrix} δ_{5} = \frac{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}) . \end{matrix}

(19)

Applying Lemma 4 to (15), we have

\begin{matrix} | δ_{1} | \leq \frac{5}{24} \end{matrix}

Applying Lemma 5 to (16), we get

\begin{matrix} | δ_{2} | \leq \frac{5}{36} . \end{matrix}

Utilizing the triangle inequality and Lemma 6 with

γ = - \frac{5}{9}

and

ζ = \frac{25}{216}

, we yield

\begin{matrix} | δ_{3} | \leq \frac{5}{48} . \end{matrix}

Rearranging (18), we obtain

\begin{matrix} | δ_{4} | & = \frac{1}{165888} | 6912 (c_{4} - c_{1} c_{3} - \frac{1}{2} c_{2}^{2} + \frac{3}{4} c_{1}^{2} c_{2} - \frac{1}{8} c_{1}^{4}) + 6912 c_{4} - 288 c_{1} c_{3} + 256 c_{2}^{2} - 1184 c_{1}^{2} c_{2} + 239 c_{1}^{4} | \\ \leq \frac{1}{165888} | 6912 (c_{4} - c_{1} c_{3} - \frac{1}{2} c_{2}^{2} + \frac{3}{4} c_{1}^{2} c_{2} - \frac{1}{8} c_{1}^{4}) | + \frac{1}{165888} | 6912 c_{4} - 288 c_{1} c_{3} + 256 c_{2}^{2} - 1184 c_{1}^{2} c_{2} + 239 c_{1}^{4} | \\ = \frac{1}{165888} D_{1} + \frac{1}{165888} D_{2}, \end{matrix}

where

D_{1} = | 6912 (c_{4} - c_{1} c_{3} - \frac{1}{2} c_{2}^{2} + \frac{3}{4} c_{1}^{2} c_{2} - \frac{1}{8} c_{1}^{4}) |

\begin{matrix} D_{2} & = | 6912 c_{4} - 288 c_{1} c_{3} + 256 c_{2}^{2} - 1184 c_{1}^{2} c_{2} + 239 c_{1}^{4} | . \end{matrix}

(20)

Using Lemma 7 with

γ = - \frac{1}{2}

, we get

D_{1} \leq 6912 .

Rearranging (20), we get

\begin{matrix} D_{2} & = | 6912 c_{4} - 288 c_{1} (c_{3} + 2 c_{1}^{2} c_{2} + c_{1}^{4}) + 256 c_{2}^{2} - 608 c_{1}^{2} c_{2} + 527 c_{1}^{4} | . \end{matrix}

Using Lemma 1, Lemma 6 and the triangle inequality, we receive

\begin{matrix} D_{2} & \leq 6912 (1 - | c_{1} |^{2} - | c_{2} |^{2}) + 288 | c_{1} | + 256 | c_{2} |^{2} + 608 | c_{1} |^{2} | c_{2} | + 527 | c_{1} |^{4} \\ = 6912 + 288 | c_{1} | - 6912 | c_{1} |^{2} + 527 | c_{1} |^{4} + 608 | c_{1} |^{2} | c_{2} | - 6656 | c_{2} |^{2} = Υ_{1} (c, d) \end{matrix}

where

| c_{1} | = c, | c_{2} | = d .

Consider

\{\begin{matrix} \frac{\partial Υ_{1}}{\partial c} = 288 - 13824 c + 2108 c^{3} + 1216 c d = 0, \\ \frac{\partial Υ_{1}}{\partial d} = 608 c^{2} - 13312 d = 0 \end{matrix}

Applying Newton’s methods, we have

\{\begin{matrix} c_{1} = 2.5173 \\ d_{1} = 0.2894, \end{matrix}

\{\begin{matrix} c_{2} = - 2.5173, \\ d_{2} = 0.2942, \end{matrix}

\{\begin{matrix} c_{3} = 0.0208, \\ d_{3} = 0 . \end{matrix}

Thus, in

(0, 1) \times (0, 1 - c^{2}),

there is no critical point.

(1)For

c = 0,

\begin{matrix} Υ_{1} (0, d) = 6912 - 6912 d^{2} \leq 6912 . \end{matrix}

(2)For

d = 0,

\begin{matrix} Υ_{1} (c, 0) = 6912 + 288 c - 6912 c^{2} + 527 c^{4} = γ_{1} (c) \leq γ_{1} (0.0208) = 6915 . \end{matrix}

(3)For

d = 1 - c^{2},

\begin{matrix} Υ_{1} (c, 1 - c^{2}) = 256 + 288 c + 7008 c^{2} - 6737 c^{4} = γ_{2} (d) \leq γ_{2} (0.7313) = 2287.6 . \end{matrix}

Therefore, we have

\begin{matrix} | γ_{4} | \leq \frac{13827}{165888} = 0.083351 . \end{matrix}

Rearranging (19), we obtain

\begin{matrix} | δ_{5} | & = \frac{1}{2488320} | 86400 (c_{5} - c_{1} c_{4} - c_{2} c_{3} + \frac{3}{4} c_{1} c_{2}^{2} + \frac{3}{4} c_{1}^{2} c_{3} - \frac{1}{2} c_{1}^{3} c_{2} + \frac{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} | \\ \leq \frac{1}{2488320} | 86400 (c_{5} - c_{1} c_{4} - c_{2} c_{3} + \frac{3}{4} c_{1} c_{2}^{2} + \frac{3}{4} c_{1}^{2} c_{3} - \frac{1}{2} c_{1}^{3} c_{2} + \frac{1}{16} c_{1}^{5}) | \\ + \frac{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} | \\ = \frac{1}{2488320} D_{3} + \frac{1}{2488320} D_{4}, \end{matrix}

where

D_{3} = | 86400 (c_{5} - c_{1} c_{4} - c_{2} c_{3} + \frac{3}{4} c_{1} c_{2}^{2} + \frac{3}{4} c_{1}^{2} c_{3} - \frac{1}{2} c_{1}^{3} c_{2} + \frac{1}{16} c_{1}^{5}) |

and

\begin{matrix} D_{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} | . \end{matrix}

(21)

Using Lemma 8 with

γ = - \frac{1}{2}

, we get

D_{3} \leq 86400 .

Rearranging (21), we get

\begin{matrix} D_{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} | . \end{matrix}

Utilizing the triangle inequality, Lemma 1, Lemma 6 and 5, we obtain

\begin{matrix} D_{4} & \leq 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} \\ \leq 86400 (1 - | c_{1} |^{2} - | c_{2} |^{2} - \frac{| 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 \frac{| c_{3} |^{2}}{1 + | c_{1} |} \\ \leq 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) \end{matrix}

where

| c_{1} | = c

| c_{2} | = d

. Consider

\{\begin{matrix} \frac{\partial Υ_{2}}{\partial c} = - 133200 c + 62425 c^{4} + 12000 c^{2} d + 4000 d^{2} = 0, \\ \frac{\partial Υ_{2}}{\partial d} = 14400 + 4000 c^{3} + (- 172800 + 8000 c) d = 0 . \end{matrix}

We have a critical point

(0.000209, 0.083334)

. Thus, we get

Υ_{2} (c, d) \leq Υ_{2} (0.000209, 0.083334) = 86999.968 .

(1)For

c = 0,

\begin{matrix} Υ_{2} (0, d) = 86400 + 14400 d - 86400 d^{2} = γ_{3} (d) \leq γ_{3} (\frac{1}{12}) = 87000 . \end{matrix}

(2)For

d = 0,

\begin{matrix} Υ_{2} (c, 0) = 86400 - 66600 c^{2} + 12485 c^{5} \leq 864000 . \end{matrix}

(3)For

d = 1 - c^{2},

\begin{matrix} Υ_{2} (c, 1 - c^{2}) = 14400 + 4000 c + 91800 c^{2} - 4000 c^{3} - 86400 c^{4} + 12485 c^{5} = γ_{4} (c) \leq γ_{4} (0.7771) = 43098 . . \end{matrix}

Therefore, we receive

\begin{matrix} | δ_{5} | \leq \frac{173400}{2488320} = 0.069686 . \end{matrix}

The proof of Theorem 10 is completed. ☐

4. Conclusion

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 coefficients of

| b_{6} |, | b_{7} |, | b_{8} |,

and the third-order determinants of

H_{3, 2}, H_{3, 3}

of 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, 2}, H_{3, 3}

for other subclasses of univalent functions.

Funding

No Funding.

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Conflicts of Interest

The authors state that they have no conflicts of interest.

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

Abstract

1. Introduction and Definitions

2. The bounds of the third Hankel determinant for g ∈ B T 4 l

3. The bounds of the logarithmic coefficients for g ∈ B T 4 l

4. Conclusion

Funding

Institutional Review Board Statement

Informed Consent Statement

Conflicts of Interest

References

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2. The bounds of the third Hankel determinant for $g \in B T_{4 l}$

3. The bounds of the logarithmic coefficients for $g \in B T_{4 l}$