Sharp Coefficient Bounds for Starlike Functions Associated with Cosine Function

Preprint

Article

Sharp Coefficient Bounds for Starlike Functions Associated with Cosine Function

Altmetrics

Downloads

136

Views

Comments

A peer-reviewed article of this preprint also exists.

This version is not peer-reviewed

Submitted:

03 June 2024

Posted:

04 June 2024

You are already at the latest version

Alerts

Abstract

Let $\mathcal{\mathcal{S}}_{\cos}^{\ast}$ denote the class of normalized analytic functions $f$ in the open unit disk $\mathbb{D}$ satisfying the subordination $\dfrac{zf^{\prime}(z)}{f(z)}\prec\cos z$. In the second section of this article we find the sharp upper bounds for the initial coefficients $a_{3}$, $a_{4}$ and $a_{5}$ and the sharp upper bound for module of the Hankel determinant $|H_{2,3}(f)|$ for the functions from the class $\mathcal{S}_{\cos}^{\ast}$. The first result of the next section deals with the sharp upper bounds of the logarithmic coefficients $\gamma_{3}$ and $\gamma_{4}$ and we found in addition the sharp upper bound for $\left|H_{2,2}\left(F_{f}/2\right)\right|$. For obtaining these results we used the very useful and appropriate Lemma 2.4 of N.E. Cho et al. [Filomat 34(6) (2020), 2061--2072], and the technique for finding the maximum value of a three variable function on a closed cuboid. All the maximum found values were checked by using MAPLE\texttrademark{} computare software, and we also found the extremal functions in each cases. All of our present results are the best ones and give sharp versions of those recently published in [Hacet. J. Math. Stat. 52, 596--618, 2023].

Keywords:

Subject: Computer Science and Mathematics - Analysis

MSC: 30C45; 30C50; 30C80

1. Introduction and Preliminaries

Let

A

denote the class of all analytic and normalized functions f having the Maclaurin series expansion as follows

f (z) = z + \sum_{m = 2}^{\infty} a_{m} z^{m}, z \in D,

(1)

where

D : = {z \in C : | z | < 1}

represents the open unit disk in the complex plane

C

, and let consider a subclass of

S \subset A

containing all the univalent functions of

A

The subclass of

A

defined by

S^{*} : = \{f \in A : Re \frac{z f^{'} (z)}{f (z)} > 0, z \in D\}

is called the class of starlike functions (with respect to the origin) in

D

, and it’s well-known that

S^{*} \subset S

. Let

B

denote the family of functions w which are analytic in

D

, such that

w (0) = 0

and

| w (z) | < 1

for

z \in D

. Such functions are referred to as Schwarz functions.

Consider two analytic functions f and g in

D

. Then we say that f is subordinated to g, written as

f (z) ≺ g (z)

if there exists a Schwarz function w such that

f (z) = g (w (z))

for

z \in D

. If

f (z) ≺ g (z)

then

f (0) = g (0)

and

f (D) \subseteq g (D)

, and if g is univalent in

D

then

f (z) ≺ g (z)

if and only if

f (0) = g (0)

and

f (D) \subseteq g (D)

Based on the geometric properties of the image of

D

by some analytic functions, the functions can be categorized into different families. Thus, in 1992 Ma and Minda [19] introduced a generalized subclass of

S^{*}

denoted by

S^{*} (ϕ)

which is defined in terms of the subordination as follows:

S^{*} (ϕ) : = \{f \in A : \frac{z f^{'} (z)}{f (z)} ≺ ϕ (z)\},

where

ϕ

satisfies the conditions

ϕ (0) = 1

and

Re ϕ^{'} (z) > 0

D

and

ϕ

maps the unit disk

D

onto a star-shaped domain. Several subclasses of

S^{*}

can be obtained by varying the function

ϕ

. For example, if we choose

ϕ (z) = \frac{1 + L z}{1 + M z}

- 1 \leq M < L \leq 1

, then we get the class

S^{*} [L, M] : = S^{*} (\frac{1 + L z}{1 + M z})

which is called the Janowski starlike functions class and was investigated in [12]. Sokól and Stankiewicz [30] defined and studied the class

S_{L}^{*} : = S^{*} (\sqrt{1 + z})

, where the function

ϕ (z) = \sqrt{1 + z}

maps

D

onto the image domain bounded by

|w^{2} - 1| < 1

. The class

S_{C}^{*} : = S^{*} (1 + \frac{4}{3} z + \frac{2}{3} z^{2})

was studied by Sharma [29] and it’s related to the cardioid function.

In case of

ϕ (z) = cosh z

the class

S_{cosh}^{*}

was defined and examined in [1], whereas the class

S_{e}^{*} : = S^{*} (e^{z})

is presented and studied in [21] by Mendiratta et al. For

ϕ (z) = 1 + sin z

the class

S_{sin}^{*}

was introduced and investigated in [2,10], while if

ϕ (z) = 1 + tanh z

the class

S_{t a n h}^{*}

was defined and studied in [31]. Recently, the class

S_{cos}^{*}

defined by

S_{cos}^{*} : = \{f \in A : \frac{z f^{'} (z)}{f (z)} ≺ cos z\}

was introduced and investigated in [5], and to determine sharp results connected with those of [20] represent the main goal of the present paper.

The following part of this section is necessary for the motivation of this work. Thus, in 1916 Bieberbach presented the well-known Bieberbach conjecture which was proved by de Branges [7] in 1985. Prior to de Branges proof of this conjecture numerous mathematicians exerted considerable effort to prove it, leading to the establishment of coefficient bounds for some remarkable subclasses of the class

S

. They also developed new inequalities related to coefficient bounds for some subclasses of univalent functions, and those related to the Fekete-Szego functional, that is

|a_{3} - λ a_{2}^{2}|

, is one of those inequalities. Another coefficient problem closely related to Fekete–Szego functional is the Hankel determinant as we will see below.

Like we could see in Pommerenke’s [24], for a function

f \in A

the Hankel determinant

H_{m, n} (f)

is as follows:

H_{m, n} (f) : = |\begin{matrix} a_{n} & a_{n + 1} & \dots & a_{n + m - 1} \\ a_{n + 1} & a_{n + 2} & \dots & a_{n + m} \\ ⋮ & ⋮ & \dots & ⋮ \\ a_{n + m - 1} & a_{n + m} & \dots & a_{n + 2 m - 2} \end{matrix}|,

where

m, n \geq 1

. We remark that

\begin{matrix} H_{2, 1} (f) = a_{3} - a_{2}^{2}, H_{2, 2} (f) = a_{2} a_{4} - a_{3}^{2}, H_{2, 3} (f) = a_{3} a_{5} - a_{4}^{2}, \end{matrix}

(2)

\begin{matrix} H_{3, 1} (f) = 2 a_{2} a_{3} a_{4} - a_{4}^{2} - a_{3}^{3} - a_{2}^{2} a_{5} + a_{3} a_{5}, \end{matrix}

(3)

where

| H_{2, 1} (f) |

is the classical Fekete-Szego functional obtained for

λ = 1

. Several authors have determined the maximum value of

| H_{2, 1} (f) |

and the upper bound for

| H_{2, 2} (f) |

for various subclasses of

A

, see for example [13,14,22]. Results regarding the second Hankel determinant

H_{2, 3} (f)

could be found also in the recent papers like [23,28,32]. The determinant

H_{2, 3} (f)

is not much studied in the literature. Babalola [3] studied first time a non-sharp bounds for the determinant

H_{3, 1} (f)

for various subclasses of

S

, while in 2017 Zaprawa [33] improved the results of Babalola by using a new technique. We mention that the sharp bounds of the modulus of

H_{3, 1} (f)

for the class

S^{*}

are recently obtained by Kowalczyk et al. [17], whereas for the class

C

and bounded turning functions sharp bounds were obtained in [15,16], respectively. For sharp inequalities results of the determinant

H_{3, 1} (f)

for some subclasses of

S^{*}

we refer to [4,18,26,27,28].

Very recently Marimuthu et al. [20] have determined the coefficient bounds, the upper bounds for the second, third and fourth order Hankel determinants for the functions of the class

S_{cos}^{*}

, but most of the results presented in this paper are not sharp.

Motivated from the aforementioned study, in this paper we have established the sharp results for the upper bounds of the coefficients, the inverse coefficients and the logarithmic coefficients of the functions of the class

S_{cos}^{*}

. Also we have developed the sharp upper bounds for the modulus of the second and third order Hankel determinants for the functions of this class.

The well-known Carathéodory class

P

is the family of holomorphic functions h in

D

which satisfy the condition

Re h (z) > 0

z \in D

, and having the power series expansion of the form

h (z) = 1 + \sum_{n = 1}^{\infty} c_{n} z^{n}, z \in D .

(4)

The study of some coefficient problems in different classes of analytic functions revolves around the idea of expressing function coefficients in a given class by function coefficients that have a positive real part. Thus, inequalities known for the class

P

can be used to study coefficient functionals. We require the following results about the class

P

for our discussions.

We will recall the well-known Carathéodory lemma [8] (see also [11,25]):

Lemma 1.

h \in P

has the form (4), then

| c_{n} | \leq 2, n \geq 1 .

(5)

The inequality holds for all

n \geq 1

if and only if

h (z) = \frac{1 + λ z}{1 - λ z}

| λ | = 1

The next result represents the relations (2.7), (2.8) and (2.9) of Lemma 2.4. from [6]:

Lemma 2.

[6] Let

\bar{D} : = {z \in C : | z | \leq 1}

be the closed unit disk, and

h \in P

be given by (4). Then,

\begin{matrix} c_{1} = 2 t_{1}, \\ c_{2} = 2 t_{1}^{2} + 2 (1 - | t_{1} |^{2}) t_{2}, \\ c_{3} = 2 t_{1}^{3} + 4 (1 - | t_{1} |^{2}) t_{1} t_{2} - 2 (1 - | t_{1} |^{2}) \bar{t_{1}} t_{2}^{2} + 2 (1 - | t_{1} |^{2}) (1 - | t_{2} |^{2}) t_{3}, \end{matrix}

for some

t_{k} \in \bar{D}

and

k \in {1, 2, 3}

Note that the extension of this lemma for the coefficients

c_{4}

and

c_{5}

may also be found in Lemma 2.1. of [9].

The next lemma represents the first part of the result from [19]:

Lemma 3.

h \in P

is given by (4), then

|c_{2} - ν c_{1}^{2}| \leq 2 - ν {|c_{1}|}^{2}, 0 < ν \leq \frac{1}{2} .

2. Initial Coefficients Sharp Upper Bounds

The next main results gives us the sharp upper bounds for the initial coefficients of the functions from the class

S_{cos}^{*}

Theorem 1.

Let

f \in S_{cos}^{*}

be given by (1). Then,

| a_{3} | \leq \frac{1}{4}, | a_{4} | \leq \frac{2}{27} \sqrt{3}, | a_{5} | \leq \frac{1}{8},

and these bounds are sharp.

Proof.

f \in S_{cos}^{*}

, then by the definition of subordination there exists a Schwarz function w that is analytic in

D

and satisfies the condition

w (0) = 0

and

| w (z) | < 1

for all

z \in D

such that

\frac{z f^{'} (z)}{f (z)} = cos w (z), z \in D .

(6)

Therefore, the function h defined by

h (z) = \frac{1 + w (z)}{1 - w (z)} = 1 + c_{1} z + c_{2} z^{2} + \dots, z \in D,

(7)

has the property

h \in P

Using the relations (6) and (7), by equating the first four coefficients we get

\begin{matrix} a_{2} & = 0, a_{3} = - \frac{1}{16} c_{1}^{2}, a_{4} = \frac{1}{24} c_{1}^{3} - \frac{1}{12} c_{1} c_{2}, \end{matrix}

(8)

\begin{matrix} a_{5} & = - \frac{1}{48} c_{1}^{4} - \frac{1}{32} c_{2}^{2} + \frac{3}{32} c_{1}^{2} c_{2} - \frac{1}{16} c_{1} c_{3} . \end{matrix}

(9)

(i) From the second relation of (8), according to the inequality (5) if follows that

| a_{3} | \leq \frac{1}{4},

and this inequality is attained for the function

f_{*}

from the Remark 2.2. of [20]. Thus, the above upper bound is sharp, that is the best possible.

(ii) To find the upper bound of

| a_{4} |

, we see that the third equality of (8) could be written in the form

a_{4} = - \frac{c_{1}}{12} (c_{2} - \frac{c_{1}^{2}}{2}),

and using Lemma 3 for

ν = \frac{1}{2}

we obtain

| a_{4} | \leq \frac{|c_{1}|}{12} (2 - \frac{{|c_{1}|}^{2}}{2}) .

Denoting

c : = |c_{1}|

, from Lemma 1 we have

0 \leq c \leq 2

, hence

| a_{4} | \leq \frac{c}{12} (2 - \frac{c^{2}}{2}) = : F (c), c \in [0, 2] .

(10)

It’s easy to check that the function F attained the maximum value at

c = \frac{2}{\sqrt{3}}

, and according to the above inequality we get

| a_{4} | \leq F (\frac{2}{\sqrt{3}}) = \frac{2}{27} \sqrt{3} .

For the prove the sharpness of this upper bound, let

\hat{t} = \frac{2}{\sqrt{3}}

h_{1} (z) = \frac{1 - z^{2}}{1 - \hat{t} z + z^{2}}

and

w_{1} (z) = \frac{h_{1} (z) - 1}{h_{1} (z) + 1} = \frac{z (3 z - \sqrt{3})}{\sqrt{3} z - 3}, z \in D,

thus

w_{1} (0) = 0

. To prove that

|w_{1} (z)| < 1

D

, we remarks that the function

w_{1}

could be written in the form

w_{1} (z) = z φ (z)

, where

φ (z) = \frac{3 z - \sqrt{3}}{\sqrt{3} z - 3}, z \in D .

Since

| φ (z) | < 1

z \in D

, if and only if

Re H (z) > 0

z \in D

, where

H (z) : = \frac{1 - φ (z)}{1 + φ (z)} = \frac{(\sqrt{3} - 3) (z + 1)}{(\sqrt{3} + 3) (z - 1)}, z \in D,

is a circular transform. It’s easy to check that

H (0) = \frac{3 - \sqrt{3}}{\sqrt{3} + 3} > 0, H (- 1) = 0, H (i) = \frac{(3 - \sqrt{3}) i}{\sqrt{3} + 3}, H (- i) = \frac{(\sqrt{3} - 3) i}{\sqrt{3} + 3} .

Using the fact that every circular transform maps the circles (in the large sense, that are circles or lines) of

C_{\infty} : = C \cup {\infty}

into circles of

C_{\infty}

, from the above values of H it follows that

Re H (z) > 0

z \in D

, which implies

| φ (z) | < 1

D

, hence

| w_{1} (z) | = | z | | φ (z) | < 1, z \in D .

Therefore, the function

f_{1} (z) : = z exp (\int_{0}^{z} \frac{cos w_{1} (t) - 1}{t} d t) = z - \frac{1}{12} z^{3} + \frac{2}{27} \sqrt{3} z^{4} + \dots, z \in D,

(11)

belongs to the class

S_{cos}^{*}

, hence

| a_{4} | = \frac{2}{27} \sqrt{3}

for the above function

f_{1}

that proves the sharpness of the second inequality of this theorem.

(iii) To determine the upper bound of

| a_{5} |

, using the relation (9) combined with Lemma 2 we obtain

a_{5} = \frac{1}{96} [4 t_{1}^{4} + 24 {| t_{1} |}^{2} t_{2}^{2} (1 - | t_{1} |^{2}) - 12 t_{2}^{2} {(1 - | t_{1} |^{2})}^{2} - 24 t_{1} (1 - | t_{1} |^{2}) (1 - | t_{2} |^{2}) t_{3}] .

Setting

x : = | t_{1} | \in [0, 1]

y : = | t_{2} | \in [0, 1]

and

u : = | t_{3} | \in [0, 1]

, from the triangle’s inequality we obtain

\begin{matrix} | a_{5} | & \leq \frac{1}{24} [x^{4} + 6 x^{2} (1 - x^{2}) y^{2} + 3 {(1 - x^{2})}^{2} y^{2} + 6 x (1 - x^{2}) (1 - y^{2}) u] \\ = \frac{1}{24} F (x, y, u), x, y, u \in [0, 1], \end{matrix}

(12)

where

F (x, y, u) : = x^{4} + 6 x^{2} (1 - x^{2}) y^{2} + 3 {(1 - x^{2})}^{2} y^{2} + 6 x (1 - x^{2}) (1 - y^{2}) u .

(13)

Denoting by

Ω : = [0, 1] \times [0, 1] \times [0, 1]

the closed unit cuboid, we have to find the maximum value of F in

Ω

I. First, let’s consider that

(x, y, u)

belongs to the interior of

Ω

, denoted by

int Ω

. Differentiating (13) with respect to u we obtain

\frac{\partial F (x, y, u)}{\partial u} = 6 x (1 - x^{2}) (1 - y^{2}) \neq 0, (x, y, u) \in int Ω,

therefore the function F have no extremal values in

int Ω

II. Next we will discuss the existence of the maximum value for F on the open six faces of

Ω

, as follows.

(i) On the face

x = 0

the next inequality holds:

F (0, y, u) = 3 y^{2} < 3, y, u \in (0, 1) .

(14)

(ii) On the face

x = 1

we have the equality

F (1, y, u) = 1, y, u \in (0, 1) .

(15)

(iii) On

y = 0

, let denote

κ_{1} (x, u) : = F (x, 0, u) = x^{4} + 6 (1 - x^{2}) x u, x, u \in (0, 1),

(16)

and because

\frac{\partial κ_{1} (x, u)}{\partial u} = 6 x (1 - x^{2}) \neq 0, x \in (0, 1),

it follows that the function

κ_{1}

has no extremal values on

(0, 1) \times (0, 1)

(iv) On the open face

y = 1

the function F reduces to

s_{1} (x) : = F (x, 1, u) = x^{4} + 6 x^{2} (1 - x^{2}) + 3 {(1 - x^{2})}^{2}, x \in (0, 1) .

(17)

Since

s_{1}^{'} (x) = - 8 x^{3} < 0

x \in (0, 1)

, therefore

s_{1}

is a strictly decreasing function on

(0, 1)

, hence

F (x, 1, u) < 3, x, u \in (0, 1) .

(18)

(v) For the open face

u = 0

we get

κ_{2} (x, y) : = F (x, y, 0) = x^{4} + 6 x^{2} (1 - x^{2}) y^{2} + 3 {(1 - x^{2})}^{2} y^{2}, x, y \in (0, 1),

therefore

\frac{\partial κ_{2} (x, y)}{\partial x} = - 4 x^{3} (3 y^{2} - 1), \frac{\partial κ_{2} (x, y)}{\partial y} = - 6 (x - 1) (x + 1) y (x^{2} + 1),

hence the system of equations

\frac{\partial κ_{2} (x, y)}{\partial x} = 0

and

\frac{\partial κ_{2} (x, y)}{\partial y} = 0

has no solutions in

(0, 1) \times (0, 1)

(vi) On the open face

u = 1

the function F becomes

κ_{3} (x, y) : = x^{4} + 6 x^{2} (1 - x^{2}) y^{2} + 3 y^{2} {(1 - x^{2})}^{2} + 6 x (1 - x^{2}) (1 - y^{2}), x, y \in (0, 1),

and it’s easy to check that the system of equations

\frac{\partial κ_{3} (x, y)}{\partial x} = 0

and

\frac{\partial κ_{3} (x, y)}{\partial y} = 0

has no solutions in

(0, 1) \times (0, 1)

III. Now we will investigate the existence of the maximum of F on the edges of

Ω

(i) From (13) we obtain

F (x, 0, 0) = x^{4} \leq 1, x \in [0, 1] .

(ii) Also, from (16) we get

s_{2} (x) : = F (x, 0, 1) = x^{4} + 6 x (1 - x^{2})

. We may easily see that the zero of

s_{2}^{'}

[0, 1]

given by

x_{0} = 0.621924 \dots

satisfies

s_{2}^{''} (x_{0}) < 0

, thus

F (x, 0, 1) \leq F (x_{0}, 0, 1) = 2.437828 \dots, x \in [0, 1] .

(iii) Putting

x = 0

in (16) we have

F (0, 0, u) = 0, u \in [0, 1] .

(iv) Since (17) is independent of u, similarly as we obtained the inequality (18) we deduce

F (x, 1, 1) = F (x, 1, 0) \leq 3, x \in [0, 1],

and putting

x = 0

in (17) we obtain

F (0, 1, u) = 3, u \in [0, 1] .

(v) The function given by (15) is independent on the variables y and u, thus

F (1, y, 0) = F (1, y, 1) = F (1, 0, u) = F (1, 1, u) = 1, y, u \in [0, 1] .

(vi) Since the function defined by (14) is independent of the variable u, we have

F (0, y, 0) = F (0, y, 1) \leq 3, y \in [0, 1] .

From all the above reasons we conclude that

max \{F (x, y, u) : (x, y, u) \in Ω\} = F (0, 1, u) = 3,

and according to (12) we finally obtain that

| a_{5} | \leq \frac{1}{8}

. This upper bound for

| a_{5} |

is sharp for the function

f_{2} (z) : = z exp (\int_{0}^{z} \frac{cos (t^{2}) - 1}{t} d t) = z - \frac{1}{8} z^{5} + \dots, z \in D,

(19)

that completes our proof. □

Remark 1.

1. In [20] it is proved that

| a_{4} | \leq \frac{1}{3}

and

| a_{5} | \leq \frac{11}{24}

and the authors gave in the Remark 2.2. a function of the class

S_{cos}^{*}

for which

| a_{4} | = 0

and

| a_{5} | = \frac{1}{24}

. In the above theorem the results are sharp by giving the best upper bounds.

2. The maximum value of the function F defined by (13) could be easily found using the MAPLE™ software with the code

[> maximize(F,x=0..1,y=0..1,u=0..1,location=true);

that gives the same result as above.

In the following two theorems we determined the sharp upper bounds for the Hankel determinants

| H_{3, 1} |

and

| H_{2, 3} |

, respectively, over the class

S_{cos}^{*}

Theorem 2.

f \in S_{cos}^{*}

is given by (1), then

| H_{3, 1} (f) | \leq - \frac{3115}{164268} + \frac{671}{657072} \sqrt{1342} = 0.018446 \dots,

and this result is sharp.

Proof.

Using the relations (8)–(9) in (3) we obtain

H_{3, 1} (f) = - \frac{7}{36864} c_{1}^{6} + \frac{5}{4608} c_{1}^{4} c_{2} + \frac{1}{256} c_{1}^{3} c_{3} - \frac{23}{4608} c_{1}^{2} c_{2}^{2},

and replacing all the variables of the above relation with those of Lemma 2 it follows that

\begin{matrix} H_{3, 1} (f) = \frac{1}{576} [3 t_{1}^{6} - 36 t_{1}^{2} {| t_{1} |}^{2} t_{2}^{2} (1 - | t_{1} |^{2}) - 46 t_{1}^{2} t_{2}^{2} {(1 - | t_{1} |^{2})}^{2} \\ + 36 t_{1}^{3} (1 - | t_{1} |^{2}) (1 - | t_{2} |^{2}) t_{3}] . \end{matrix}

If we set

x : = | t_{1} | \in [0, 1]

y : = | t_{2} | \in [0, 1]

and

u : = | t_{3} | \in [0, 1]

, using the above relation and the triangle’s inequality we get

\begin{matrix} | H_{3, 1} (f) | & \leq \frac{1}{576} [3 x^{6} + 36 x^{4} (1 - x^{2}) y^{2} + 46 x^{2} {(1 - x^{2})}^{2} y^{2} + 36 x^{3} (1 - x^{2}) (1 - y^{2}) u] \\ = \frac{1}{576} F (x, y, u), x, y, u \in [0, 1], \end{matrix}

(20)

where

F (x, y, u) : = 3 x^{6} + 36 x^{4} (1 - x^{2}) y^{2} + 46 x^{2} {(1 - x^{2})}^{2} y^{2} + 36 x^{3} (1 - x^{2}) (1 - y^{2}) u .

(21)

With the same notations and method like in the proof of Theorem 1, next we will find the maximum value of F in

Ω

I. If we consider that

(x, y, u) \in int Ω

, differentiating (21) with respect to u we obtain

\frac{\partial F (x, y, u)}{\partial u} = 36 x^{3} (1 - x^{2}) (1 - y^{2}) \neq 0, (x, y, u) \in int Ω,

it follows that the function F have no maximum value in

int Ω

II. In the sequel we will study the existence of the maximum value of F in the interior of six faces of

Ω

(i) On the face

x = 0

we have

F (0, y, u) = 0, y, u \in (0, 1) .

(22)

(ii) On the face

x = 1

we get

F (1, y, u) = 3, y, u \in (0, 1) .

(23)

(iii) On

y = 0

, the function F can be written as

κ_{1} (x, u) : = F (x, 0, u) = 3 x^{6} + 36 x^{3} (1 - x^{2}) u, x, u \in (0, 1) .

(24)

Since

\frac{\partial κ_{1} (x, u)}{\partial u} = 36 x^{3} (1 - x^{2}) \neq 0, x \in (0, 1),

it implies that the function

κ_{1}

has no maximum point in this face of

Ω

(iv) On

y = 1

, the function F reduces to

s_{1} (x) : = F (x, 1, u) = 3 x^{6} + 36 x^{4} (1 - x^{2}) + 46 x^{2} {(1 - x^{2})}^{2}, x \in (0, 1) .

(25)

Since

s_{1}^{'} (x) = 78 x^{5} - 224 x^{3} + 92 x

x \in (0, 1)

, it follows that

s_{1}^{'} (x) = 0

has the zero

x_{0} = \frac{\sqrt{2184 - 39 \sqrt{1342}}}{39} = 0.704685 \dots \in (0, 1)

, that satisfy the inequality

s_{1}^{''} (x_{0}) < 0

, we obtain

F (x, 1, u) \leq - \frac{49840}{4563} + \frac{2684}{4563} \sqrt{1342} = 10.625427 \dots, x, u \in (0, 1) .

(26)

(v) On the face

u = 0

, the function F becomes

κ_{2} (x, y) : = F (x, y, 0) = 3 x^{6} + 36 x^{4} (1 - x^{2}) y^{2} + 46 x^{2} {(1 - x^{2})}^{2} y^{2}, x, y \in (0, 1),

thus

\frac{\partial κ_{2} (x, y)}{\partial x} = (60 y^{2} + 18) x^{5} - 224 x^{3} y^{2} + 92 x^{2}, \frac{\partial κ_{2} (x, y)}{\partial y} = 4 x^{2} y (5 x^{4} - 28 x^{2} + 23) .

It’s easy to see that the system of equations

\frac{\partial κ_{2} (x, y)}{\partial x} = 0

and

\frac{\partial κ_{2} (x, y)}{\partial y} = 0

have no solutions in

(0, 1) \times (0, 1)

(vi) On

u = 1

, the function F takes the form

κ_{3} (x, y) = 3 x^{6} + 36 x^{4} (1 - x^{2}) y^{2} + 46 x^{2} {(1 - x^{2})}^{2} y^{2} + 36 x^{3} (1 - x^{2}) (1 - y^{2}), x, y \in (0, 1) .

Similarly, the system of equations

\frac{\partial κ_{3} (x, y)}{\partial x} = 0

and

\frac{\partial κ_{3} (x, y)}{\partial y} = 0

have no solutions in

(0, 1) \times (0, 1)

III. Now we will investigate the existence of the maximum of F on the edges of

Ω

(i) From (24) we obtain

F (x, 0, 0) = 3 x^{6} \leq 3, x \in [0, 1] .

(ii) The relation (24) at

u = 1

becomes

s_{2} (x) : = F (x, 0, 1) = 3 x^{6} + 36 x^{3} (1 - x^{2})

x \in [0, 1]

. The solution of

s_{2}^{'} (x) = 18 x^{2} (x^{3} - 10 x^{2} + 6) = 0

x_{0} = 0.807920 \dots \in [0, 1]

and satisfy

s_{2}^{''} (x_{0}) < 0

, hence

F (x, 0, 1) \leq F (x_{0}, 0, 1) = 7.427102 \dots, x \in [0, 1] .

(iii) Putting

x = 0

in (24) we have

F (0, 0, u) = 0, u \in [0, 1]

(iv) Since (25) is independent of u, similarly as above we obtain

F (x, 1, 1) = F (x, 1, 0) \leq - \frac{49840}{4563} + \frac{2684}{4563} \sqrt{1342}, x \in [0, 1],

while if we take

x = 0

in (25) it follows

F (0, 1, u) = 0, u \in [0, 1] .

(v) The relation (23) is independent on the variables

y, u \in [0, 1]

, hence

F (1, y, 0) = F (1, y, 1) = F (1, 0, u) = F (1, 1, u) = 3, y, u \in [0, 1] .

(vi) Finally, since (22) is independent on the variable

u \in [0, 1]

, we have

F (0, y, 0) = F (0, y, 1) = 0, y \in [0, 1] .

All the inequalities we obtained above show that

\begin{matrix} max \{F (x, y, u) : (x, y, u) \in Ω\} = F (\frac{\sqrt{2184 - 39 \sqrt{1342}}}{39}, 1, u) \\ = - \frac{49840}{4563} + \frac{2684}{4563} \sqrt{1342} = 10.625427 \dots, \end{matrix}

and using (20) our inequality is proved.

For proving the sharpness of this inequality, let consider

\tilde{t} = \frac{\sqrt{2184 - 39 \sqrt{1342}}}{39}

h_{3} (z) = \frac{1 + (1 + i) \tilde{t} z + i z^{2}}{1 - (1 - i) \tilde{t} z - i z^{2}}

and

w_{3} (z) = \frac{h_{2} (z) - 1}{h_{2} (z) + 1} = \frac{z (\sqrt{2184 - 39 \sqrt{1342}} + 39 i z)}{39 + \sqrt{2184 - 39 \sqrt{1342}} i z}, z \in D .

We can see that

w_{3} (0) = 0

, and let’s write the function

w_{3}

w_{3} (z) = z ψ (z)

, where

ψ (z) = \frac{\sqrt{2184 - 39 \sqrt{1342}} + 39 i z}{39 + \sqrt{2184 - 39 \sqrt{1342}} i z}, z \in D .

From the same reasons regarding the circular transforms like in the proof of the sharpness of Theorem 1(ii), we will show that

| ψ (z) | < 1

D

by proving that

Re H (z) > 0

z \in D

, where

H (z) : = \frac{1 - ψ (z)}{1 + ψ (z)} = \frac{(\sqrt{2184 - 39 \sqrt{1342}} - 39) (z + i)}{(\sqrt{2184 - 39 \sqrt{1342}} + 39) (- z + i)}, z \in D,

is a circular transform. Since

\begin{matrix} H (0) = \frac{- \sqrt{2184 - 39 \sqrt{1342}} + 39}{\sqrt{2184 - 39 \sqrt{1342}} + 39} = 0.173236 \dots > 0, H (- i) = 0, \\ H (- 1) = \frac{- (\sqrt{2184 - 39 \sqrt{1342}} - 39) i}{\sqrt{2184 - 39 \sqrt{1342}} + 39}, H (1) = \frac{(\sqrt{2184 - 39 \sqrt{1342}} - 39) i}{\sqrt{2184 - 39 \sqrt{1342}} + 39}, \end{matrix}

like in the above mentioned proof, these values of H lead us to

Re H (z) > 0

z \in D

, which implies

| ψ (z) | < 1

D

, therefore

| w_{3} (z) | = | z | | ψ (z) | < 1, z \in D .

Thus, the function

\begin{matrix} f_{3} (z) = z exp (\int_{0}^{z} \frac{cos w_{3} (t) - 1}{t} d t) = z + (- \frac{14}{39} + \frac{\sqrt{1342}}{156}) z^{3} \\ - \frac{1}{4563} (- 17 + \sqrt{1342}) \sqrt{2184 - 39 \sqrt{1342}} z^{4} + (\frac{23135}{36504} - \frac{163 \sqrt{1342}}{9126}) z^{5} + \dots, z \in D, \end{matrix}

belongs to the class

S_{cos}^{*}

, with the initial coefficients

\begin{matrix} a_{2} = 0, a_{3} = - \frac{14}{39} + \frac{\sqrt{1342}}{156}, \\ a_{4} = - \frac{1}{4563} ((- 17 + \sqrt{1342}) \sqrt{2184 - 39 \sqrt{1342}}) i, a_{5} = \frac{23135}{36504} - \frac{163 \sqrt{1342}}{9126} . \end{matrix}

From the relation (3) we get

|H_{3, 1} (f_{3})| = - \frac{3115}{164268} + \frac{671}{657072} \sqrt{1342} = 0.018446 \dots

and the proof is complete. □

Remark 2.

In [20] it is proved that

| H_{3, 1} (f) | \leq \frac{139}{576} = 0.241319 \dots

for all

f \in S_{cos}^{*}

, but that result wasn’t the best possible. If we compare the upper bounds of

| H_{3, 1} (f) |

for

f \in S_{cos}^{*}

, obtained here with those of [20], the result of Theorem 2 is a significant improvement of the previous one. Moreover, the inequality obtained in above theorem is sharp, thus the found upper bound for

| H_{3, 1} (f) |

f \in S_{cos}^{*}

cannot be improved.

Theorem 3.

f \in S_{cos}^{*}

has the form (1), then

| H_{2, 3} (f) | \leq \frac{77}{15552} + \frac{29}{15552} \sqrt{58} = 0.019152 \dots,

and the result is sharp.

Proof.

Replacing in (2) the values of

a_{2}

a_{3}

a_{2}

, and

a_{5}

given by (8)–(9) we obtain

H_{2, 3} (f) = - \frac{1}{2304} c_{1}^{6} + \frac{5}{4608} c_{1}^{4} c_{2} + \frac{1}{256} c_{1}^{3} c_{3} - \frac{23}{4608} c_{1}^{2} c_{2}^{2} .

According to the Lemma 2, from the above relation we deduce that

\begin{matrix} H_{2, 3} (f) = \frac{1}{288} [- 3 t_{1}^{6} - 18 t_{1}^{2} {| t_{1} |}^{2} t_{2}^{2} (1 - | t_{1} |^{2}) - 23 t_{1}^{2} t_{2}^{2} {(1 - | t_{1} |^{2})}^{2} \\ + 18 t_{1}^{3} (1 - | t_{1} |^{2}) (1 - | t_{2} |^{2}) t_{3}] . \end{matrix}

Setting

x : = | t_{1} | \in [0, 1]

y : = | t_{2} | \in [0, 1]

and

u : = | t_{3} | \in [0, 1]

, then using the triangle’s inequality, the above relation lead us to

\begin{matrix} | H_{2, 3} (f) | & \leq \frac{1}{288} [3 x^{6} + 18 x^{4} (1 - x^{2}) y^{2} + 23 x^{2} {(1 - x^{2})}^{2} y^{2} + 18 x^{3} (1 - x^{2}) (1 - y^{2}) u] \\ = \frac{1}{288} F (x, y, u), x, y, u \in [0, 1], \end{matrix}

(27)

where

F (x, y, u) : = 3 x^{6} + 18 x^{4} (1 - x^{2}) y^{2} + 23 x^{2} {(1 - x^{2})}^{2} y^{2} + 18 x^{3} (1 - x^{2}) (1 - y^{2}) u .

(28)

With the same notations like in the proofs of the two previous theorems, we have to find the maximum value of F on

int Ω

, on the six faces, and on the twelve edges of

Ω

I. First we consider the arbitrary interior point

(x, y, u) \in int Ω

. Differentiating (28) with respect to u we obtain

\frac{\partial F (x, y, u)}{\partial u} = 18 x^{3} (1 - x^{2}) (1 - y^{2}) \neq 0, (x, y, u) \in int Ω,

therefore we have no maximum value of F in

int Ω

II. Next we will study the existence of the maximum value of the function F in the interior of six faces of

Ω

(i) On the face

x = 0

the function F reduces to

F (0, y, u) = 0, y, u \in (0, 1) .

(29)

(ii) On the face

x = 1

it takes the form

F (1, y, u) = 3, y, u \in (0, 1) .

(30)

(iii) On

y = 0

, the function F can be written as

κ_{1} (x, u) : = F (x, 0, u) = 3 x^{6} + 18 x^{3} (1 - x^{2}) u, x, u \in (0, 1) .

(31)

Since

\frac{\partial κ_{1} (x, u)}{\partial u} = 18 x^{3} (1 - x^{2}) \neq 0, x \in (0, 1),

it follows that

κ_{1}

has no maximum point in this face of

Ω

(iv) On

y = 1

, the function F reduces to

s_{1} (x) : = F (x, 1, u) = 3 x^{6} + 18 x^{4} (1 - x^{2}) + 23 x^{2} {(1 - x^{2})}^{2}, x \in (0, 1),

(32)

thus

s_{1}^{'} (x) = 48 x^{5} - 112 x^{3} + 46 x, x \in (0, 1) .

Hence

s_{1}^{'} (x) = 0

has the zero

x_{0} = \frac{\sqrt{42 - 3 \sqrt{58}}}{6} = 0.729396 \dots \in (0, 1)

and

s_{1}^{''} (x_{0}) < 0

. Therefore,

F (x, 1, u) \leq F (x_{0}, 1, u) = \frac{77}{54} + \frac{29}{54} \sqrt{58} = 5.515878 \dots, x, u \in (0, 1) .

(33)

(v) On the face

u = 0

the function F becomes

κ_{2} (x, y) : = F (x, y, 0) = 3 x^{6} + 18 x^{4} (1 - x^{2}) y^{2} + 23 x^{2} {(1 - x^{2})}^{2} y^{2}, x, y \in (0, 1) .

Therefore,

\frac{\partial κ_{2} (x, y)}{\partial x} = 2 x (15 x^{4} y^{2} + 9 x^{4} - 56 x^{2} y^{2} + 23 y^{2}), \frac{\partial κ_{2} (x, y)}{\partial y} = 2 x^{2} y (5 x^{4} - 28 x^{2} + 23),

and the system of equations

\frac{\partial κ_{2} (x, y)}{\partial x} = 0

and

\frac{\partial κ_{2} (x, y)}{\partial y} = 0

have no solutions in

(0, 1) \times (0, 1)

(vi) On

u = 1

, the function F takes the form

κ_{3} (x, y) = 3 x^{6} + 18 x^{4} (1 - x^{2}) y^{2} + 23 x^{2} {(1 - x^{2})}^{2} y^{2} + 18 x^{3} (1 - x^{2}) (1 - y^{2}), x, y \in (0, 1),

hence

\begin{matrix} \frac{\partial κ_{3} (x, y)}{\partial x} = 6 (5 y^{2} + 3) x^{5} - 90 (1 - y^{2}) x^{4} - 112 x^{3} y^{2} + 54 (1 - y^{2}) x^{2} + 46 x y^{2}, \\ \frac{\partial κ_{3} (x, y)}{\partial y} = 10 {(x - 1)}^{2} (x + 1) y (x + \frac{23}{5}) x^{2} . \end{matrix}

Therefore, the system of equations

\frac{\partial κ_{3} (x, y)}{\partial x} = 0

and

\frac{\partial κ_{3} (x, y)}{\partial y} = 0

have no solutions in

(0, 1) \times (0, 1)

III. Now we investigate the maximum of F on the edges of

Ω

(i) From (31) we obtain

F (x, 0, 0) = 3 x^{6} \leq 3, x \in [0, 1] .

(ii) Also, from (31) at

u = 1

we get

s_{2} (x) : = F (x, 0, 1) = 3 x^{6} + 18 x^{3} (1 - x^{2})

. The solution in

[0, 1]

of the equation

s_{2}^{'} (x) = 18 x^{5} - 90 x^{4} + 54 x^{2} = 0

for which

s_{2}^{''} (x) < 0

x_{0} = 0.850256 \dots \in [0, 1]

, thus

F (x, 0, 1) \leq F (x_{0}, 0, 1) = 4.199005 \dots, x \in [0, 1] .

(iii) Putting

x = 0

in (31) we have

F (0, 0, u) = 0, u \in [0, 1] .

(iv) Since (32) is independent on u, according to (33) we obtain

F (x, 1, 1) = F (x, 1, 0) \leq \frac{77}{54} + \frac{29}{54} \sqrt{58}, x \in [0, 1] .

(v) If we take

x = 0

in (32) we get

F (0, 1, u) = 0, u \in [0, 1] .

Using the fact that the relation (30) is independent on the variables y and u, we deduce

F (1, y, 0) = F (1, y, 1) = F (1, 0, u) = F (1, 1, u) = 3, y, u \in [0, 1] .

(vi) Since the relation (29) is independent of the variable u, we have

F (0, y, 0) = F (0, y, 1) = 0, y \in [0, 1] .

Consequently, from the above reasons we conclude that

\begin{matrix} max \{F (x, y, u) : (x, y, u) \in Ω\} = F (\frac{\sqrt{42 - 3 \sqrt{58}}}{6}, 1, u) \\ = \frac{77}{54} + \frac{29}{54} \sqrt{58} = 5.515878 \dots, \end{matrix}

and combining with (27) it follows

|H_{2, 3} (f)| \leq \frac{77}{15552} + \frac{29}{15552} \sqrt{58} = 0.019152 \dots .

To prove the sharpness of the above result, let us consider

t^{*} = \frac{1}{3} \sqrt{42 - 3 \sqrt{58}}

h_{4} (z) = \frac{1 + t^{*} z + z^{2}}{1 - z^{2}}

and

w_{4} (z) = \frac{h_{4} (z) - 1}{h_{4} (z) + 1} = \frac{z (\sqrt{42 - 3 \sqrt{58}} + 6 z)}{6 + \sqrt{42 - 3 \sqrt{58}} z}, z \in D .

First,

w_{4} (0) = 0

, and we will write

w_{4} (z) = z χ (z)

, where

χ (z) = \frac{\sqrt{42 - 3 \sqrt{58}} + 6 z}{6 + \sqrt{42 - 3 \sqrt{58}} z}, z \in D .

Using the same property of the circular transforms like in the proof of the sharpness of Theorem 1(ii) and Theorem 2, we will show that

| χ (z) | < 1

D

by proving that

Re H (z) > 0

z \in D

, where

H (z) : = \frac{1 - χ (z)}{1 + χ (z)} = \frac{(\sqrt{42 - 3 \sqrt{58}} - 6) (z - 1)}{(\sqrt{42 - 3 \sqrt{58}} + 6) (z + 1)},

is a circular transform. Computing the below values

\begin{matrix} H (0) = - \frac{\sqrt{42 - 3 \sqrt{58}} - 6}{\sqrt{42 - 3 \sqrt{58}} + 6} = 0.156472 \dots > 0, H (1) = 0, \\ H (i) = \frac{(\sqrt{42 - 3 \sqrt{58}} - 6) i}{\sqrt{42 - 3 \sqrt{58}} + 6}, H (- i) = - \frac{(\sqrt{42 - 3 \sqrt{58}} - 6) i}{\sqrt{42 - 3 \sqrt{58}} + 6}, \end{matrix}

from the similar reasons as in the above mentioned proofs, these values of H implies

Re H (z) > 0

z \in D

, that yields

| χ (z) | < 1

D

, therefore

| w_{4} (z) | = | z | | χ (z) | < 1, z \in D .

Consequently, the function

\begin{matrix} f_{4} (z) & = z exp (\int_{0}^{z} \frac{cos w_{4} (t) - 1}{t} d t) = z + (- \frac{7}{24} + \frac{1}{48} \sqrt{58}) z^{3} \\ - \frac{1}{216} (- 2 + \sqrt{58}) \sqrt{42 - 3 \sqrt{58}} z^{4} + (- \frac{7}{54} + \frac{5}{216} \sqrt{58}) z^{5} + \dots, z \in D, \end{matrix}

belongs to the class

S_{cos}^{*}

. In the above power series expansion we have

a_{3} = - \frac{7}{24} + \frac{1}{48} \sqrt{58}

a_{4} = - \frac{1}{216} (- 2 + \sqrt{58}) \sqrt{42 - 3 \sqrt{58}}

and

a_{5} = - \frac{7}{54} + \frac{5}{216} \sqrt{58}

, hence

|H_{2, 3} (f_{4})| = |a_{3} a_{5} - a_{4}^{2}| = \frac{77}{15552} + \frac{29}{15552} \sqrt{58},

that completes our proof. □

Remark 3.

The maximum values of the functions F defined by (21) and (28) could be also found by using the MAPLE™ computer software codes like in the Remark 1 2., and we obtain the same values like in both of the above two theorems.

3. Logarithmic Coefficients Sharp Upper Bounds

The logarithmic coefficients

γ_{n} : = γ_{n} (f)

n \in N

, for the function

f \in S

are defined by

F_{f} (z) : = log \frac{f (z)}{z} = 2 \sum_{n = 1}^{\infty} γ_{n} z^{n}, z \in D .

(34)

Since the function

ϕ (z) : = cos z

has positive real part in

D

, and moreover

Re (cos z) > \frac{1}{2}, z \in D,

it follows that

S_{cos}^{*} \subset S^{*} \subset S

(see [20, p. 610]). Therefore, it is possible to define the logarithmic coefficients for the functions

f \in S_{cos}^{*}

In this section we give the sharp upper bounds estimates for the third and fourth logarithmic coefficients of the functions that belong to the class

S_{cos}^{*}

Theorem 4.

f \in S_{cos}^{*}

is given by (1), then

| γ_{3} | \leq \frac{\sqrt{3}}{27}, | γ_{4} | \leq \frac{1}{16} .

These bounds are sharp.

Proof.

f \in S_{cos}^{*}

has the form (1), then

log \frac{f (z)}{z} = a_{2} z + (- \frac{a_{2}^{2}}{2} + a_{3}) z^{2} + (- a_{2} a_{3} + a_{4} + \frac{a_{2}^{3}}{3}) z^{3} + \dots, z \in D .

(35)

and equating the first four coefficients of (34) with those of (35) we get

\begin{matrix} γ_{1} = \frac{a_{2}}{2}, γ_{2} = \frac{1}{4} (2 a_{3} - a_{2}^{2}), γ_{3} = \frac{1}{6} (a_{2}^{3} - 3 a_{2} a_{3} + 3 a_{4}), \end{matrix}

(36)

\begin{matrix} γ_{4} = \frac{1}{8} (- a_{2}^{4} + 4 a_{2}^{2} a_{3} - 4 a_{2} a_{4} - 2 a_{3}^{2} + 4 a_{5}) . \end{matrix}

(37)

With the same notation like in the proof of Theorem 1, replacing in (36) and (37) the values of

a_{2}

a_{3}

a_{4}

and

a_{5}

from the relations (8) and (9), we obtain

\begin{matrix} γ_{1} = 0, γ_{2} = - \frac{1}{32} c_{1}^{2}, γ_{3} = \frac{1}{48} c_{1}^{3} - \frac{1}{24} c_{1} c_{2}, \end{matrix}

(38)

\begin{matrix} γ_{4} = - \frac{35}{3072} c_{1}^{4} - \frac{1}{64} c_{2}^{2} + \frac{3}{64} c_{1}^{2} c_{2} - \frac{1}{32} c_{1} c_{3} \end{matrix}

(39)

For the upper bound of

| γ_{3} |

, using (38) we write

γ_{3} = - \frac{c_{1}}{24} (c_{2} - \frac{c_{1}^{2}}{2}),

and according to Lemma 3 for

ν = \frac{1}{2}

we obtain

| γ_{3} | \leq \frac{|c_{1}|}{24} (2 - \frac{|c_{1}^{2}|}{2}) .

Denoting

c : = |c_{1}|

, from Lemma 1 we have

| γ_{3} | \leq \frac{c}{24} (2 - \frac{c^{2}}{2}) = F (c), c \in [0, 2] .

Using the result we got for the computation of the maximum F given by (10) we get

| γ_{3} | \leq \frac{\sqrt{3}}{27}

To prove the sharpness of this bound let consider the function

f_{1}

given by (11), were

a_{1} = 1

a_{2} = 0

a_{3} = - \frac{1}{12}

and

a_{4} = \frac{2}{27} \sqrt{3}

. Therefore, for this function, by using the last of the relations from (36) we obtain

γ_{3} = \frac{\sqrt{3}}{27}

To find the upper bound of

| γ_{4} |

, from (37) combined with Lemma 2 we can write

γ_{4} = \frac{1}{192} [t_{1}^{4} + 24 {| t_{1} |}^{2} t_{2}^{2} (1 - | t_{1} |^{2}) - 12 t_{2}^{2} {(1 - | t_{1} |^{2})}^{2} - 24 t_{1} (1 - | t_{1} |^{2}) (1 - | t_{2} |^{2}) t_{3}],

and setting

x : = | t_{1} | \in [0, 1]

y : = | t_{2} | \in [0, 1]

and

u : = | t_{3} | \in [0, 1]

, using the triangle’s inequality we obtain

\begin{matrix} | γ_{4} | & \leq \frac{1}{192} [x^{4} + 24 x^{2} (1 - x^{2}) y^{2} + 12 {(1 - x^{2})}^{2} y^{2} + 24 x (1 - x^{2}) (1 - y^{2}) u] \\ = \frac{1}{192} F (x, y, u), x, y, u \in [0, 1], \end{matrix}

(40)

where

F (x, y, u) : = x^{4} + 24 x^{2} (1 - x^{2}) y^{2} + 12 {(1 - x^{2})}^{2} y^{2} + 24 x (1 - x^{2}) (1 - y^{2}) u .

(41)

Using the notations and the technique from the proofs of the previous theorems, we will determine the maximum of F on

Ω

as follows.

I. In the points

(x, y, u) \in int Ω

, differentiating (41) with respect to u we obtain

\frac{\partial F (x, y, u)}{\partial u} = 24 x (1 - x^{2}) (1 - y^{2}) \neq 0, (x, y, u) \in int Ω,

therefore the function F doesn’t attained its maximum value in

int Ω

II. In the next items we will discuss the existence of the maximum value of F in the interior of six faces of

Ω

(i) On the face

x = 0

we get

F (0, y, u) = 12 y^{2} \leq 12, y, u \in (0, 1) .

(42)

(ii) On the face

x = 1

the function F takes the form

F (1, y, u) = 1, y, u \in (0, 1) .

(43)

(iii) On

y = 0

, the function can be written as

κ_{1} (x, u) : = F (x, 0, u) = x^{4} + 24 x (1 - x^{2}) u, x, u \in (0, 1),

(44)

and because

\frac{\partial κ_{1} (x, u)}{\partial u} = 24 x (1 - x^{2}) \neq 0, x \in (0, 1),

it implies that the function

κ_{1}

has no maximum in

(0, 1) \times (0, 1)

(iv) On

y = 1

, the function F reduces to

s_{1} (x) : = F (x, 1, u) = - 11 x^{4} + 12, x \in (0, 1) .

(45)

Since

s_{1}^{'} (x) = - 44 x^{3} < 0

x \in (0, 1)

, the function

s_{1}

is strictly decreasing on

(0, 1)

, hence

F (x, 1, u) \leq F (0, 1, u) = 12, x, u \in (0, 1) .

(46)

(v) On the face

u = 0

, F will have the form

κ_{2} (x, y) : = F (x, y, 0) = x^{4} + 24 x^{2} (1 - x^{2}) y^{2} + 12 {(1 - x^{2})}^{2} y^{2}, x, y \in (0, 1) .

Therefore

\frac{\partial κ_{2} (x, y)}{\partial x} = - 4 x^{3} (12 y^{2} - 1), \frac{\partial κ_{2} (x, y)}{\partial y} = - 24 y (x - 1) (x + 1) (x^{2} + 1),

thus the system of equations

\frac{\partial κ_{2} (x, y)}{\partial x} = 0

and

\frac{\partial κ_{2} (x, y)}{\partial y} = 0

have no solutions in

(0, 1) \times (0, 1)

(vi) On

u = 1

, the function F will be

κ_{3} (x, y) : = - 12 x^{4} y^{2} + 24 x^{3} y^{2} + x^{4} - 24 x^{3} - 24 x y^{2} + 12 y^{2} + 24 x, x, y \in (0, 1) .

Similarly, it’s easy to check that the system of equations

\frac{\partial κ_{3} (x, y)}{\partial x} = 0

and

\frac{\partial κ_{3} (x, y)}{\partial y} = 0

have no solutions in

(0, 1) \times (0, 1)

III. Now we will investigate the existence of the maximum of F on the edges of

Ω

(i) From (44) at

u = 0

we have

F (x, 0, 0) = x^{4} \leq 1, x \in [0, 1] .

(ii) Using (44) at

u = 1

, we get

s_{2} (x) : = F (x, 0, 1) = x (x^{3} - 24 x^{2} + 24)

, hence

s_{2}^{'} (x) = 4 x^{3} - 72 x^{2} + 24

. The solution of

s_{2}^{'} (x) = 0

[0, 1]

x_{0} = 0.587000 \dots \in [0, 1]

and

s_{2}^{''} (x_{0}) < 0

it follows that

F (x, 0, 1) \leq F (x_{0}, 0, 1) = 9.352439 \dots, x \in [0, 1] .

(iii) Putting

x = 0

in (44) we get

F (0, 0, u) = 0, u \in [0, 1],

(iv) Since (45) is independent on u we obtain

F (x, 1, 1) = F (x, 1, 0) \leq 12, x \in [0, 1],

while for

x = 0

in (45) we get

F (0, 1, u) = 12 .

(v) The relation (43) is independent with respect to the variables y and u, thus

F (1, y, 0) = F (1, y, 1) = F (1, 0, u) = F (1, 1, u) = 1, y, u \in [0, 1] .

(vi) Finally, since (42) is independent on the variable u, consequently

F (0, y, 0) = F (0, y, 1) \leq 1, z \in [0, 1] .

The above computations lead to

max \{F (x, y, u) : (x, y, u) \in Ω\} = F (0, 1, u) = 12,

and from (40) we conclude that

| γ_{4} | \leq \frac{1}{16}

For proving the sharpness of the above inequality, we consider the function

f_{2}

given by (19). In this case

a_{1} = 1

a_{2} = a_{3} = a_{4} = 0

and

a_{5} = - \frac{1}{8}

, and from (37) we get

| γ_{4} | = \frac{1}{16}

, that completes our proof. □

Remark 4.

In [20] it is proved that

| γ_{3} | \leq \frac{1}{6}

and

| γ_{4} | \leq \frac{85}{384}

, while our above results are sharp and give the best upper bounds for these coefficients.

Theorem 5.

Let

f \in S_{cos}^{*}

be given by (1). Then, for the function

F_{f}

given by (34) the next inequality holds

|H_{2, 2} (F_{f} / 2)| \leq - \frac{10577}{1028376} + \frac{1549}{4113504} \sqrt{1549} = 0.015678 \dots,

and this result is sharp.

Proof.

Replacing the values of (38) and (39) in the relation 2 we obtain

H_{2, 2} (F_{f} / 2) = γ_{2} γ_{4} - γ_{3}^{2} = - \frac{23}{294912} c_{1}^{6} + \frac{5}{18432} c_{1}^{4} c_{2} + \frac{1}{1024} c_{1}^{3} c_{3} - \frac{23}{18432} c_{1}^{2} c_{2}^{2} .

Using the Lemma 2 we obtain

\begin{matrix} H_{2, 2} (F_{f} / 2) = \frac{1}{4608} [- 3 t_{1}^{6} - 72 t_{1}^{2} {| t_{1} |}^{2} t_{2}^{2} (1 - | t_{1} |^{2}) - 92 t_{1}^{2} t_{2}^{2} {(1 - | t_{1} |^{2})}^{2} \\ + 72 t_{1}^{3} (1 - | t_{1} |^{2}) (1 - | t_{2} |^{2}) t_{3}], \end{matrix}

and denoting

x : = | t_{1} | \in [0, 1]

y : = | t_{2} | \in [0, 1]

u : = | t_{3} | \in [0, 1]

, using the triangle’s inequality the above relation yields

\begin{matrix} |H_{2, 2} (F_{f} / 2)| \leq & \frac{1}{4608} [3 x^{6} + 72 x^{4} (1 - x^{2}) y^{2} + 92 x^{2} {(1 - x^{2})}^{2} y^{2} \\ + 72 x^{3} (1 - x^{2}) (1 - y^{2}) u] = \frac{1}{4608} F (x, y, u), x, y, u \in [0, 1], \end{matrix}

(47)

where

F (x, y, u) : = 3 x^{6} + 72 x^{4} (1 - x^{2}) y^{2} + 92 x^{2} {(1 - x^{2})}^{2} y^{2} + 72 x^{3} (1 - x^{2}) (1 - y^{2}) u .

(48)

With the notations used in the proofs of the previous three theorems we will determine the maximum value of F in

Ω

I. For all the interior points

(x, y, u) \in int Ω

, differentiating (48) with respect to u we obtain

\frac{\partial F (x, y, u)}{\partial u} = 72 x^{3} (1 - x^{2}) (1 - y^{2}) \neq 0, (x, y, u) \in int Ω,

hence the function doesn’t get its maximum in

int Ω

II. Next we will study if it’s possible to obtain the maximum value of F in the interior of six faces of

Ω

(i) On the face

x = 0

we have

F (0, y, u) = 0, y, u \in (0, 1) .

(49)

(ii) On the face

x = 1

the function F takes the form

F (1, y, u) = 3, y, u \in (0, 1) .

(50)

(iii) On

y = 0

the function F can be written as

κ_{1} (x, u) : = F (x, 0, u) = 3 x^{6} + 72 x^{3} (1 - x^{2}) u, x, u \in (0, 1) .

(51)

Since

\frac{\partial κ_{1} (x, u)}{\partial u} = 72 x^{3} (1 - x^{2}) \neq 0, x \in (0, 1),

it follows that the function

κ_{1}

has no maximum in

(0, 1) \times (0, 1)

(iv) On

y = 1

, the function reduces to

s_{1} (x) : = F (x, 1, u) = 3 x^{6} + 72 x^{4} (1 - x^{2}) + 92 x^{2} {(1 - x^{2})}^{2}, x \in (0, 1) .

(52)

Since

s_{1}^{'} (x) = 138 x^{5} - 448 x^{3} + 184 x = 0

has on

(0, 1)

the root

x_{0} = \frac{\sqrt{7728 - 138 \sqrt{1549}}}{69} = 0.694547 \dots \in (0, 1)

and

s_{1}^{''} (x_{0}) < 0

, we deduce that

F (x, 1, u) \leq F (x_{0}, 1, u) = - \frac{676928}{14283} + \frac{24784}{14283} \sqrt{1549} = 20.899268 \dots, x \in (0, 1) .

(53)

(v) On the face

u = 0

the function F reduces to

κ_{2} (x, y) : = F (x, y, 0) = 3 x^{6} + 72 x^{4} (1 - x^{2}) y^{2} + 92 x^{2} {(1 - x^{2})}^{2} y^{2}, x, y \in (0, 1),

therefore

\frac{\partial κ_{2} (x, y)}{\partial x} = 2 x (60 x^{4} y^{2} + 9 x^{4} - 224 x^{2} y^{2} + 92 y^{2}), \frac{\partial κ_{2}}{\partial y} = 8 x^{2} (x^{2} - 1) y (5 x^{2} - 23) .

Thus, the system of equations

\frac{\partial κ_{2} (x, y)}{\partial x} = 0

and

\frac{\partial κ_{2} (x, y)}{\partial y} = 0

has no solutions in

(0, 1) \times (0, 1)

(vi) On

u = 1

the function F takes the form

\begin{matrix} κ_{3} (x, y) : = F (x, y, 1) = 3 x^{6} + 72 x^{4} (1 - x^{2}) y^{2} + 92 x^{2} {(1 - x^{2})}^{2} y^{2} \\ + 72 x^{3} (1 - x^{2}) (1 - y^{2}), x, y \in (0, 1) . \end{matrix}

We get

\frac{\partial κ_{3} (x, y)}{\partial x} = (120 y^{2} + 18) x^{5} + (360 y^{2} - 360) x^{4} - 448 x^{3} y^{2} + (- 216 y^{2} + 216) x^{2} + 184 x y^{2},

and

\frac{\partial κ_{3}}{\partial y} = 8 y x^{2} (5 x + 23) (x + 1) {(x - 1)}^{2},

hence the system of equations

\frac{\partial κ_{3} (x, y)}{\partial x} = 0

and

\frac{\partial κ_{3} (x, y)}{\partial y} = 0

has no solutions in

(0, 1) \times (0, 1)

III. Now we will investigate the maximum of F on the edges of

Ω

(i) From (51) for

u = 0

we have

F (x, 0, 0) = 3 x^{6} \leq 3, x \in [0, 1] .

(ii) From (51) for

u = 1

, let

s_{2} (x) : = F (x, 0, 1) = 3 x^{6} - 72 x^{5} + 72 x^{3}

. The solutions in

[0, 1]

of the equation

s_{2}^{'} (x) = 18 x^{5} - 360 x^{4} + 216 x^{2} = 0

are

x_{0} = 0.790371 \dots \in [0, 1]

and

x_{1} = 0

. Since

s_{2}^{''} (x_{0}) < 0

and

s_{2} (0) = 0

we deduce that

F (x, 0, 1) \leq F (x_{0}, 0, 1) = 14.073287 \dots, x \in [0, 1] .

(iii) Putting

x = 0

in (51) we have

F (0, 0, u) = 0, u \in [0, 1] .

(iv) Since the relation (52) is independent on u, from (53) we obtain

F (x, 1, 1) = F (x, 1, 0) \leq - \frac{676928}{14283} + \frac{24784}{14283} \sqrt{1549}, x \in [0, 1] .

(v) If we take

x = 0

in (52), we obtain

F (0, 1, u) = 0, u \in [0, 1]

(vi) The relation (50) is independent on the variables y and u, thus

F (1, y, 0) = F (1, y, 1) = F (1, 0, u) = F (1, 1, u) = 3, y, u \in [0, 1] .

Similarly, since the formula (49) is independent on the variable u we have

F (0, y, 0) = F (0, y, 1) = 0, y \in [0, 1] .

From the above reasons it follows

max \{F (x, y, u) : (x, y, u) \in Ω\} = F (x_{0}, 1, u) = - \frac{676928}{14283} + \frac{24784}{14283} \sqrt{1549},

and using (47) we conclude that

|H_{2, 2} (F_{f} / 2)| \leq - \frac{10577}{1028376} + \frac{1549}{4113504} \sqrt{1549} = 0.015678 \dots .

To prove the sharpness of the above inequality, we denote

t_{▵} = \frac{2}{69} \sqrt{7728 - 138 \sqrt{1549}}

and

h_{5} (z) = \frac{1 + t_{▵} z + z^{2}}{1 - z^{2}}

such that

w_{5} (z) = \frac{h_{5} (z) - 1}{h_{5} (z) + 1} = \frac{z (\sqrt{7728 - 138 \sqrt{1549}} + 69 z)}{69 + \sqrt{7728 - 138 \sqrt{1549}} z}, z \in D .

It is easy to see that

w_{7} (0) = 0

and to prove that

| w_{7} (z) | < 1

D

we will write the function

w_{5}

w_{5} (z) = z ρ (z)

, where

ρ (z) = \frac{\sqrt{7728 - 138 \sqrt{1549}} + 69 z}{69 + \sqrt{7728 - 138 \sqrt{1549}} z}, z \in D .

According to the same reasons regarding the circular transforms like in the proofs of the sharpness of Theorem 1(ii) and Theorem 2, we will show that

| ρ (z) | < 1

z \in D

, by proving that

Re H (z) > 0

z \in D

, where

H (z) : = \frac{1 - ρ (z)}{1 + ρ (z)} = \frac{(\sqrt{7728 - 138 \sqrt{1549}} - 69) (z - 1)}{(\sqrt{7728 - 138 \sqrt{1549}} + 69) (z + 1)}, z \in D,

is a circular transform. It’s easy to compute

\begin{matrix} H (0) = \frac{\sqrt{7728 - 138 \sqrt{1549}} - 69}{\sqrt{7728 - 138 \sqrt{1549}} + 69} = 0.1802561912 \dots > 0, H (1) = 0, \\ H (i) = \frac{(\sqrt{7728 - 138 \sqrt{1549}} - 69) i}{\sqrt{7728 - 138 \sqrt{1549}} + 69}, H (- i) = - \frac{(\sqrt{7728 - 138 \sqrt{1549}} - 69) i}{\sqrt{7728 - 138 \sqrt{1549}} + 69}, \end{matrix}

and these values of the circular transform H implies that

Re H (z) > 0

z \in D

, which leads us to

| ρ (z) | < 1

D

, hence

| w_{5} (z) | = | z | | ρ (z) | < 1, z \in D .

Consequently, the function

\begin{matrix} f_{5} (z) = z exp (\int_{0}^{z} \frac{cos w_{5} (t) - 1}{t} d t) = z + (- \frac{28}{69} + \frac{1}{138} \sqrt{1549}) z^{3} \\ - \frac{1}{14283} (- 43 + 2 \sqrt{1549}) \sqrt{7728 - 138 \sqrt{1549}} z^{4} \\ + (- \frac{71467}{114264} + \frac{241}{14283} \sqrt{1549}) z^{5} + \dots, z \in D, \end{matrix}

belongs to the class

S_{cos}^{*}

. The initial coefficients of

f_{5}

are

\begin{matrix} a_{3} = - \frac{28}{69} + \frac{1}{138} \sqrt{1549}, a_{4} = - \frac{1}{14283} (- 43 + 2 \sqrt{1549}) \sqrt{7728 - 138 \sqrt{1549}}, \\ a_{5} = - \frac{71467}{114264} + \frac{241}{14283} \sqrt{1549}, \end{matrix}

and from (36) and (37) we obtain

\begin{matrix} γ_{2} = - \frac{14}{69} + \frac{1}{276} \sqrt{1549}, γ_{3} = - \frac{(- 43 + 2 \sqrt{1549}) \sqrt{7728 - 138 \sqrt{1549}}}{28566}, \\ γ_{4} = - \frac{42761}{114264} + \frac{283}{28566} \sqrt{1549} . \end{matrix}

Hence,

|H_{2, 2} (F_{f_{5}} / 2)| = |γ_{2} γ_{4} - γ_{3}^{2}| = - \frac{10577}{1028376} + \frac{1549}{4113504} \sqrt{1549},

that proves the sharpness of our estimation. □

Remark 5.

Like we already mentioned in the Remark 3, the same maximum values of the functions F defined by (41) and (48) could be also found by using the MAPLE™ computer software codes like in the Remark 1 2.

4. Conclusions

In this article we have obtained the sharp coefficient bounds for the starlike functions that are connected to the cosine function, and we have also obtained the sharp coefficient bounds for the logarithmic coefficients of such functions.

The main tools of our results was those of N.E. Cho et al. [6] that seems to be a very efficient for the estimation of the first coefficients of Carathéodory’s function, and in addition we have determined the sharp bound for the fifth coefficient. The technique of this paper can be used to determine the sharp upper bound of the initial coefficients for a variety of classes of analytic functions.

Moreover, we have used these results to determine the sharp upper bounds for the Hankel determinants up to order three, and we emphasize that all the results we obtained are the best possible, so there cannot be improved.

Author Contributions

Conceptualization, R.A., M.R. and T.B.; methodology, R.A., M.R. and T.B.; software, R.A., M.R. and T.B.; validation, R.A., M.R. and T.B.; formal analysis, R.A., M.R. and T.B.; investigation, R.A., M.R. and T.B.; resources, R.A., M.R. and T.B.; data curation, R.A., M.R. and T.B.; writing—original draft preparation, M.R. and T.B.; writing—review and editing, M.R. and T.B.; visualization, R.A., M.R. and T.B.; supervision, M.R. and T.B.; project administration, R.A., M.R. and T.B.. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Informed Consent Statement

Not applicable.

Conflicts of Interest

The authors declare no conflicts of interest.

References

Alotaibi, A.; Arif, M.; Alghamdi, M.A.; Hussain, S. Starlikeness associated with cosine hyperbolic function. Mathematics 2020, 8(7), 1118. [Google Scholar] [CrossRef]
Arif, M.; Raza, M.; Tang, H.; . Hussain, S.; Khan, H. Hankel determinant of order three for familiar subsets of analytic functions related with sine function. Open Math. 2019, 17(1), 1615–1630. [Google Scholar] [CrossRef]
Babalola, K.O. On H₃(1) Hankel determinant for some classes of univalent functions. In Inequality Theory and Applications; Cho, Y.J. (ed.); Nova Science Publishers, New York, 2010, vol. 6, pp. 1–7.
Banga, S.; Kumar, S.S. The sharp bounds of the second and third Hankel determinants for the class SL^*. Math. Slovaca 2020, 70(4), 849–862. [Google Scholar] [CrossRef]
Bano, K.; Raza, M. Starlike functions associated with cosine functions. Bull. Iranian Math. Soc. 2021, 47(5), 1513–1532. [Google Scholar] [CrossRef]
Cho, N.E.; Kowalczyk, B.; Lecko, A. Sharp bounds of some coefficient functionals over the class of functions convex in the direction of the imaginary axis. Bull. Aust. Math. Soc. 2019, 100(1), 86–96. [Google Scholar] [CrossRef]
De Branges, L. A proof of the Bieberbach conjecture. Acta Math. 1985, 154(1-2), 137–152. [Google Scholar]
Carathéodory, C. Über den Variabilitätsbereich der Koeffizienten von Potenzreihen, die gegebene Werte nicht annehmen. Math. Ann. 1907, 64(1), 95–115. [Google Scholar]
Cho, N.E.; Kowalczyk, B.; Lecko, A.; Śmiarowska, B. On the Fourth and Fifth coefficients in the Carathéodory class. Filomat 2020, 34(6), 2061–2072. [Google Scholar] [CrossRef]
Cho, N.E.; Kumar, V.; Kumar, S.S.; Ravichandran, V. Radius problems for starlike functions associated with the sine function. Bull. Iranian Math. Soc. 2019, 45(1), 213–232. [Google Scholar] [CrossRef]
Duren, P.L. Univalent Functions. Grundlehren der mathematischen Wissenschaften, vol. 259, Springer New York, NY, 2001.
Janowski, W. Extremal problems for a family of functions with positive real part and for some related families. Ann. Pol. Math. 1970, 23(2), 159–177. [Google Scholar] [CrossRef]
Janteng, A.; Halim, S.A.; Darus, M. Coefficient inequality for a function whose derivative has a positive real part. J. Inequal. Pure Appl. Math. 2006, 7(2), 50. [Google Scholar]
Janteng, A.; Halim, S.A.; Darus, M. Hankel determinant for starlike and convex functions. Int. J. Math. Anal. 2007, 1(13), 619–625. [Google Scholar]
Kowalczyk, B.; Lecko, A. The sharp bound of the third Hankel determinant for functions of bounded turning. Bol. Soc. Mat. Mex. 2021, 27(3), 69. [Google Scholar] [CrossRef]
Kowalczyk, B.; Lecko, A.; Sim, Y.J. The sharp bound for the Hankel determinant of the third kind for convex functions. Bull. Aust. Math. Soc. 2018, 97(3), 435–445. [Google Scholar] [CrossRef]
Kowalczyk, B.; Lecko, A.; Thomas, D.K. The sharp bound of the third Hankel determinant for starlike functions. Forum Math. 2022, 34(5), 1249–1254. [Google Scholar] [CrossRef]
Lecko, A.; Sim, Y.J.; Śmiarowska, B. The sharp bound of the Hankel determinant of the third kind for starlike functions of order 1/2. Complex Anal. Oper. Theory 2019, 13(5), 2231–2238. [Google Scholar] [CrossRef]
Ma, W.C.; Minda, D. A unified treatment of some special classes of univalent functions. In Proceedings of the Conference on Complex Analysis (Tianjin; 1992); pp. 157–169, Conf. Proc. Lecture Notes Anal., I, Int. Press, Cambridge, MA, 1994.. [Google Scholar]
Marimuthu, K.; Uma, J.; Bulboacă, T. Coefficient estimates for starlike and convex functions associated with cosine function. Hacet. J. Math. Stat. 2023, 52(3), 596–618. [Google Scholar]
Mendiratta, R.; Nagpal, S.; Ravichandran, V. On a subclass of strongly starlike functions associated with exponential function. Bull. Malays. Math. Sci. Soc. 2015, 38(1), 365–386. [Google Scholar] [CrossRef]
Obradović, M. Tuneski, N. Hankel determinants of second and third order for the class S of univalent functions. Math. Slovaca 2021, 71(3), 649–654. [Google Scholar] [CrossRef]
Obradović, M. Tuneski, N. Two types of the second Hankel determinant for the class U and the general class S. Acta Comment. Univ. Tartu. Math. 2023, 27(1), 59–67. [Google Scholar]
Pommerenke, Ch. On the coefficients and Hankel determinants of univalent functions. J. London Math. Soc. 1966, 41(1), 111–122. [Google Scholar] [CrossRef]
Pommerenke, Ch. Univalent functions. Vandenhoeck and Ruprecht, Göttingen, 1975.
Riaz, A.; Raza, M. The third Hankel determinant for starlike and convex functions associated with lune. Bull. Sci. Math. 2023, 187, 103289. [Google Scholar] [CrossRef]
Riaz, A.; Raza, M.; Thomas, D.K. Hankel determinants for starlike and convex functions associated with sigmoid functions. Forum Math. 2022, 34(1), 137–156. [Google Scholar] [CrossRef]
Riaz, A. Raza, M.; Binyamin, M.A.; Saliu, A. The second and third Hankel determinants for starlike and convex functions associated with Three-Leaf function. Heliyon, 2023; 9, 1, 12748. [Google Scholar]
Sharma, K. , Jain, N.K.; Ravichandran, V. Starlike functions associated with a cardioid. Afr. Mat. 2016, 27(5-6), 923–939. [Google Scholar]
Sokół, J.; Stankiewicz, J. Radius of convexity of some subclasses of strongly starlike functions. Zesz. Nauk. Politech. Rzesz., Mat. Fiz. 1996, 19, 101–105. [Google Scholar]
Ullah, K.; Srivastava, H.M.; Rafiq, A.; Arif, M.; Arjika, S. A study of sharp coefficient bounds for a new subfamily of starlike functions. J. Inequal. Appl. 2021, 1(2021), 194. [Google Scholar] [CrossRef]
Zaprawa, P. On Hankel determinant H₂(3) for univalent functions. Results Math. 2018, 73(3), 89. [Google Scholar] [CrossRef]
Zaprawa, P. Third Hankel determinants for subclasses of univalent functions. Mediterr. J. Math. 2017, 14(1), 19. [Google Scholar] [CrossRef]

Disclaimer/Publisher’s Note: The statements, opinions and data contained in all publications are solely those of the individual author(s) and contributor(s) and not of MDPI and/or the editor(s). MDPI and/or the editor(s) disclaim responsibility for any injury to people or property resulting from any ideas, methods, instructions or products referred to in the content.

© 2024 by the authors. Licensee MDPI, Basel, Switzerland. This article is an open access article distributed under the terms and conditions of the Creative Commons Attribution (CC BY) license (http://creativecommons.org/licenses/by/4.0/).

Copyright: This open access article is published under a Creative Commons CC BY 4.0 license, which permit the free download, distribution, and reuse, provided that the author and preprint are cited in any reuse.

MDPI Initiatives

Important Links

Choose an area of interest and we will send you notifications of new preprints at your preferred frequency.

Disclaimer

Sharp Coefficient Bounds for Starlike Functions Associated with Cosine Function

Abstract

1. Introduction and Preliminaries

2. Initial Coefficients Sharp Upper Bounds

3. Logarithmic Coefficients Sharp Upper Bounds

4. Conclusions

Author Contributions

Funding

Informed Consent Statement

Conflicts of Interest

References

MDPI Initiatives

Important Links

Subscribe