Toeplitz Determinants for a Certain Family of Analytic Functions Endowed with Borel Distribution

Abstract

In this work, we derive coefficient bounds for the symmetric Toeplitz matrices $T_2\left(2\right)$, $T_2\left(3\right)$, $T_3\left(1\right)$, and $T_3\left(2\right)$, which are the known first four determinants for a new family of analytic functions with Borel distribution series in the open unit disk U. Further, some special cases of results obtained are also pointed.

1. Introduction

Let $\mathcal{A}$ stand for the class of functions f given by:

$$f\left(z\right)=z+\sum _{n=2}^{\infty }{a}_n{z}^n,$$

(1)

which are analytic in the open unit disk $U=\left\{z\in C:\left|z\right|<1\right\}$. Let S indicate the family of functions in the analytic function class, which are univalent in U.

The primary distributions such as the Binomial, Poisson, Logarithmic, Pascal, beta and negative binomial have been partly researched in Geometric Function Theory on a theorical basis (see, for example, [1,2,3,4,5,6,7]).

Very recently, Wanas and Khuttar [8] introduced the following power series whose coefficients are predictions of the Borel distribution:

$$\mathcal{M}(\delta ,z)=z+\sum _{n=2}^{\infty }\frac{{\left(\delta (n-1)\right)}^{n-2}{e}^{-\delta (n-1)}}{(n-1)!}{z}^n(z\in U;0<\delta \le 1).$$

It can be noted that the common ratio test expresses infinity for the radius convergence if the series given above is infinity.

The linear operator ${\mathcal{B}}_{\delta }:\mathcal{A}⟶\mathcal{A}$ is given by (see [8]):

$${\mathcal{B}}_{\delta }f\left(z\right)=\mathcal{M}(\delta ,z)\ast f\left(z\right)=z+\sum _{n=2}^{\infty }\frac{{\left(\delta (n-1)\right)}^{n-2}{e}^{-\delta (n-1)}}{(n-1)!}{a}_n{z}^{n}z\in U,$$

where the symbol “*” stands the Hadamard product (or convolution) of two series.

In the univalent function theory, an extensive focus has been given to estimate the bounds of Hankel matrices. Determinants and Hankel matrices have a big role in many areas of mathematics and have many practices [9]. Hankel determinants and Toeplitz determinants are closely related to each other. Toeplitz matrices have constant entries along the diagonal, while Hankel matrices have constant entries along the reverse diagonal. Recently, Thomas and Halim [10] introduced the symmetric Toeplitz determinant $T_q\left(n\right)$ for $f\in \mathcal{A}$, given by

$$T_q\left(n\right)=\left|\begin{array}{cccc}{a}_n& a_{n+1}& \dots & a_{n+q-1}\\ a_{n+1}& a_n& \dots & a_{n+q-2}\\ ⋮& ⋮& ⋮& ⋮\\ a_{n+q-1}& a_{n+q-2}& \dots & a_n\end{array}\right|,$$

where $n\ge 1$, $q\ge 1$ and $a_1=1$. Particulary,

$$T_2\left(2\right)=\left|\begin{array}{cc}{a}_2& a_3\\ a_3& a_2\end{array}\right|,T_2\left(3\right)=\left|\begin{array}{cc}{a}_3& a_4\\ a_4& a_3\end{array}\right|,$$

and

$$T_3\left(1\right)=\left|\begin{array}{ccc}1& a_2& a_3\\ a_2& 1& a_2\\ a_3& a_2& 1\end{array}\right|,T_3\left(2\right)=\left|\begin{array}{ccc}{a}_2& a_3& a_4\\ a_3& a_2& a_3\\ a_4& a_3& a_2\end{array}\right|.$$

The notion of Toeplitz matrices acts a significant part in applied mathematics, functional analysis as well as in physics and technical sciences (for more details, see [9]). Very recently, several authors established estimates of the Toeplitz determinant $\left|T_q\left(n\right)\right|$ for functions belonging to various classes of univalent functions (see, for example, [10,11,12,13,14,15,16,17,18,19,20,21]). In order to obtain wanted bounds in our study, we need the lemmas given below.

Lemma 1

([22]). If the function $p\in \mathcal{P}$ is presented by the series $p\left(z\right)=1+p_{1}z+p_2{z}^2+p_3{z}^3+\cdots$, then the sharp estimate $\left|p_k\right|\le 2$$\left(k=1,2,3,\cdots \right)$ holds.

Lemma 2

([23]). If the function $p\in \mathcal{P}$, then

$$2p_2=p_1^2+\left(4-p_1^2\right)x$$

$$4p_3=p_1^3+2p_1\left(4-p_1^2\right)x-p_1\left(4-p_1^2\right)x^2+2\left(4-p_1^2\right)\left(1-{\left|x\right|}^2\right)z,$$

for some $x,z$ with $\left|x\right|\le 1$ and $\left|z\right|\le 1$.

2. Main Results

Firstly, in this section, we introduce the family $\mathcal{K}(\gamma ,\alpha ,\delta )$ as follows:

Definition 1.

A function $f\in \mathcal{A}$ is said to be in the class $\mathcal{K}(\gamma ,\alpha ,\delta )$ if it satisfies the condition:

$$Re\left\{(1-\gamma ){\left(\frac{{\mathcal{B}}_{\delta }f\left(z\right)}{z}\right)}^{\alpha }+\gamma {\left({\mathcal{B}}_{\delta }f\left(z\right)\right)}^{\prime }{\left(\frac{{\mathcal{B}}_{\delta }f\left(z\right)}{z}\right)}^{\alpha -1}\right\}>0,$$

where $\gamma \ge 1$, $\alpha \ge 0$, $0<\delta \le 1$ and $z\in U$.

Theorem 1.

Let $f\in \mathcal{K}(\gamma ,\alpha ,\delta )$ be given by (1). Then

$$\left|a_2\right|\le \frac{2}{W},$$

$$\left|a_3\right|\le \frac{2}{M}+\frac{4L}{M{W}^2}$$

and

$$\left|a_4\right|\le \frac{2}{V}+\frac{4H}{WMV}+\frac{8E}{3W^{3}MV},$$

where

$$\begin{array}{cc}\hfill V& =\frac{3}{2}{\delta }^2{e}^{-3\delta }\left(\alpha +3\gamma \right),\hfill \\ \hfill W& =e^{-\delta }\left(\alpha +\gamma )\right),\hfill \\ \hfill M& =\delta e^{-2\delta }\left(\alpha +2\gamma \right),\hfill \\ \hfill H& =\delta e^{-3\delta }\left(\alpha -2\right)\left(\alpha +3\gamma \right),\hfill \\ \hfill L& =\frac{1}{2}{e}^{-2\delta }(\alpha -2)\left(\alpha +2\gamma \right),\hfill \\ \hfill E& =\delta e^{-2\delta }\left(\alpha +2\gamma \right)\left[1+\frac{1}{2}{e}^{-3\delta }{\left(\alpha -2\right)}^2\left(\alpha +3\gamma \right)\right].\hfill \end{array}$$

(2)

Proof. 

Let $f\in \mathcal{K}(\gamma ,\alpha ,\delta )$, then there exists $p\in \mathcal{P}$ such that

$$(1-\gamma ){\left(\frac{{\mathcal{B}}_{\delta }f\left(z\right)}{z}\right)}^{\alpha }+\gamma {\left({\mathcal{B}}_{\delta }f\left(z\right)\right)}^{\prime }{\left(\frac{{\mathcal{B}}_{\delta }f\left(z\right)}{z}\right)}^{\alpha -1}=p\left(z\right),$$

(3)

where p have the following series representations:

$$p\left(z\right)=1+p_{1}z+p_2{z}^2+p_3{z}^3+\cdots .$$

By equating the coefficients in (3), we have the relations

$$e^{-\delta }\left(\alpha +\gamma )\right)a_2=p_1,$$

(4)

$$\delta e^{-2\delta }\left(\alpha +2\gamma \right)a_3+\frac{1}{2}{e}^{-2\delta }(\alpha -2)\left(\alpha +2\gamma \right)a_2^2=p_2$$

(5)

and

$$\begin{array}{ll}\frac{3}{2}{\delta }^2{e}^{-3\delta }(\alpha +3\gamma )a_4-\delta e^{-3\delta }\left(\alpha -2\right)(\alpha +& 3\gamma )a_2{a}_3\\ & +\frac{1}{6}{e}^{-3\delta }\left[(\alpha -1)(\alpha -2)+2\right]\left(\alpha +3\gamma \right)a_2^3=p_3.\end{array}$$

(6)

In light of (4), (5) and (6), after simplifying, we find that

$$a_2=\frac{1}{W}{p}_2,$$

(7)

$$a_3=\frac{1}{M}{p}_2-\frac{L}{M{W}^2}{p}_1^2$$

(8)

and

$$a_4=\frac{1}{V}{p}_3+\frac{H}{WMV}{p}_1{p}_2-\frac{E}{3W^{3}MV}{p}_1^3,$$

(9)

where the values of $V,W,M,H,L$ and E are given by (2). By applying Lemma 1, we get

$$\left|a_2\right|\le \frac{2}{W},$$

$$\left|a_3\right|\le \frac{2}{M}+\frac{4L}{M{W}^2}$$

and

$$\left|a_4\right|\le \frac{2}{V}+\frac{4H}{WMV}+\frac{8E}{3W^{3}MV}.$$

□

Theorem 2.

Let $f\in \mathcal{K}(\gamma ,\alpha ,\delta )$ be given by (1). Then

$$\left|T_2\left(2\right)\right|\le \frac{4(W^4-4W^{2}L+4L^2)}{W^4{M}^2}-\frac{4}{W^2},$$

(10)

where W and L are given by (2).

Proof. 

In view of (7), (8) and (2), it is easy to see that

$$\begin{array}{cc}\hfill \left|T_2\left(2\right)\right|& =\left|a_3^2-a_2^2\right|\hfill \\ \hfill & =\left|\frac{p_2^2}{M^2}-\frac{2L{p}_1^2{p}_2}{M^2{W}^2}+\frac{L^2{p}_1^4}{M^2{W}^4}-\frac{p_1^2}{W^2}\right|.\hfill \end{array}$$

By applying Lemma 2 to express $p_2$ in terms $p_1$, it follows that

$$\begin{array}{cc}\hfill & \left|a_3^2-a_2^2\right|\hfill \\ \hfill & =\left|\frac{(W^4-4W^{2}L+4L^2)p_1^4}{4W^4{M}^2}-\frac{p_1^2}{W^2}+\frac{(W^2-2L)p_1^{2}x(4-p_1^2)}{2W^2{M}^2}+\frac{x^2{(4-p_1^2)}^2}{4M^2}\right|.\hfill \end{array}$$

For convenience of notation, we choose $p_1=p$ and since the function p is in the family $\mathcal{P}$ simultaneously, we can suppose without loss of generality that $p\in [0,2]$. Thus, by applying the triangle inequality with $P=4-p^2$, we deduce that

$$\begin{array}{cc}\hfill \left|a_3^2-a_2^2\right|& \le \left|\frac{(W^4-4W^{2}L+4L^2)p^4}{4W^4{M}^2}-\frac{p^2}{W^2}\right|+\frac{(W^2-2L)p^2\left|x\right|P}{2W^2{M}^2}+\frac{{\left|x\right|}^2{P}^2}{4M^2}=:F(p,\left|x\right|).\hfill \end{array}$$

It is obvious that $F^{\prime }(p,\left|x\right|)>0$ on $[0,1]$ and thus $F(p,\left|x\right|)\le F(p,\left|1\right|)$.

Trivially, when $p=2$, we note that the expression $F\left(\left|x\right|\right)$ has a maximum value on $[0,2]$. Consequently,

$$\left|T_2\left(2\right)\right|=\left|a_3^2-a_2^2\right|\le \frac{4(W^4-4W^{2}L+4L^2)}{W^4{M}^2}-\frac{4}{W^2}.$$

This concludes the proof. □

Theorem 3.

Let $f\in \mathcal{K}(\gamma ,\alpha ,\delta )$ be given by (1). Then

$$\left|T_2\left(3\right)\right|=|a_4^2-a_3^2|\le \frac{4{\psi }_1}{9V^2{W}^6{M}^2}-\frac{4(W^4-4W^{2}L+4L^2)}{W^4{M}^2},$$

where

$${\psi }_1=9W^6{M}^2+36W^{5}MH-24W^{3}ME+36W^4{H}^2+16E^2-48W^{2}HE.$$

Proof. 

Applying (8), (9), (2) and using Lemma 2, we have

$$\begin{array}{ll}|a_4^2-a_3^2|& =|\frac{(W^4-4W^{2}L+4L^2)p_1^4}{4W^4{M}^2}+\frac{{\psi }_1{p}_1^6}{144V^2{W}^6{M}^2}-\frac{(W^2-2L)p_1^{2}x(4-p_1^2)}{2W^2{M}^2}+\frac{{\psi }_2{p}_1^4(4-p_1^2)x}{12V^2{W}^4{M}^2}\\ & -\frac{{\psi }_3{p}_1^4(4-p_1^2)x^2}{24V^2{W}^{3}M}-\frac{x^2{(4-p_1^2)}^2}{4M^2}+\frac{{\psi }_4{p}_1^2{(4-p_1^2)}^2{x}^2}{4V^2{W}^2{M}^2}-\frac{(WM+H)p_1^2{(4-p_1^2)}^2{x}^3}{4V^{2}WM}+\frac{p_1^2{(4-p_1^2)}^2{x}^4}{16V^2}\\ & \hspace{1em}+\frac{{\psi }_3{p}_1^3(4-p_1^2){(1-|x|}^2)z}{12V^2{W}^{3}M}+\frac{(WM+H)p_1{(4-p_1^2)}^2{(1-|x|}^2)xz}{2V^{2}WM}-\frac{p_1{(4-p_1^2)}^2{(1-|x|}^2)x^{2}z}{4V^2}\\ & \hfill +\frac{{(4-p_1^2)}^2{(1-|x|}^2{)}^2{z}^2}{4V^2}|,\end{array}$$

where

$$\begin{array}{cc}\hfill {\psi }_2& =3W^4{M}^2+9W^{3}MH-4WME+6W^2{H}^2-4HE,\hfill \\ \hfill {\psi }_3& =3W^{3}M+6W^{2}H-4E,\hfill \\ \hfill {\psi }_4& =W^2{M}^2+2WMH+H^2.\hfill \end{array}$$

We select $p_1=p$ for ease of notation, and because the function p is in the family $\mathcal{P}$ at the same time, we may assume that $p\in [0,2]$ without losing generality. As a result, using the triangle inequality with $P=4-r^2$ and $Z=(1-|x{|}^2)$, we may conclude

$$\begin{array}{c}|a_4^2-a_3^2|=\left|\frac{{\psi }_1{p}^6}{144V^2{W}^6{M}^2}-\frac{(W^4-4W^{2}L+4L^2)p^4}{4W^4{M}^2}\right|+\frac{(W^2-2L)p^2\left|x\right|P}{2W^2{M}^2}+\frac{{\psi }_2{p}^{4}P\left|x\right|}{12V^2{W}^4{M}^2}\hfill \\ +\frac{{\psi }_3{p}^{4}P{\left|x\right|}^2}{24V^2{W}^{3}M}+\frac{{\left|x\right|}^2{P}^2}{4M^2}+\frac{{\psi }_4{p}^2{P}^2{\left|x\right|}^2}{4V^2{W}^2{M}^2}+\frac{(WM+H)p^2{P}^2{\left|x\right|}^3}{4V^{2}WM}+\frac{p^2{P}^2{\left|x\right|}^4}{16V^2}+\frac{{\psi }_3{p}^{3}PZ}{12V^2{W}^{3}M}\\ \hfill +\frac{(WM+H)p\left|x\right|P^{2}Z}{2V^{2}WM}+\frac{{p\left|x\right|}^2{P}^{2}Z}{4V^2}+\frac{P^2{Z}^2}{4V^2}=:F_1(p,\left|x\right|).\end{array}$$

Using elementary calculus to differentiate $F_1(p,\left|x\right|)$ w.r.t. $\left|x\right|$, we have

$$\begin{array}{c}\frac{\partial F_1(p,\left|x\right|)}{\partial \left|x\right|}=\frac{(W^2-2L)p^2(4-p^2)}{2W^2{M}^2}+\frac{{\psi }_2{p}^4(4-p^2)}{12V^2{W}^4{M}^2}-\frac{2{\psi }_3{p}^3(4-p^2)\left|x\right|}{12V^2{W}^{3}M}+\frac{{\psi }_3{p}^4(4-p^2)\left|x\right|}{12V^2{W}^{3}M}\hfill \\ +\frac{{\psi }_4{p}^2{(4-p^2)}^2\left|x\right|}{2V^2{W}^2{M}^2}-\frac{p(WM+H){(4-p^2)}^2{\left|x\right|}^2}{V^{2}WM}+\frac{3(WM+H)p^2{(4-p^2)}^2{\left|x\right|}^2}{4V^{2}WM}-\frac{p{(4-p^2)}^2{\left|x\right|}^3}{2V^2}\\ \hfill \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}+\frac{p^2{(4-p^2)}^2{\left|x\right|}^3}{4V^2}+\frac{(WM+H)p{(4-p^2)}^2{(1-|x|}^2)}{2V^{2}WM}-\frac{\left|x\right|{(4-p^2)}^2{(1-|x|}^2)}{V^2}+\frac{p\left|x\right|{(4-p^2)}^2{(1-|x|}^2)}{2V^2}.\end{array}$$

It is shown that $(\partial F_1(p,\left|x\right|)/\partial |x\left|\right)\ge 0$ for $\left|x\right|\in [0,1]$ and fixed $p\in [0,2]$. As a result, $F_1(p,\left|x\right|)$ is an increasing function of $\left|x\right|$. So, $F_1(p,\left|x\right|)\le F_1(p,\left|1\right|)$. Therefore,

$$\begin{array}{c}|a_4^2-a_3^2|\le \left|\frac{{\psi }_1{p}^6}{144V^2{W}^6{M}^2}-\frac{(W^4-4W^{2}L+4L^2)p^4}{4W^4{M}^2}\right|+\frac{(W^2-2L)p^2(4-p^2)}{2W^2{M}^2}+\frac{{(4-p^2)}^2}{4M^2}\hfill \\ \hfill +\frac{(2{\psi }_2+H_{3}WM)p^4(4-p^2)}{24V^2{W}^4{M}^2}+\frac{(4{\psi }_4+4(WM+H)WM+W^2{M}^2)p^2{(4-p^2)}^2}{16V^2{W}^2{C}^2}.\hfill \end{array}$$

Now, on $[0,2]$ at $P=2$, we have

$$|a_4^2-a_3^2|\le \frac{4{\psi }_1}{9V^2{W}^6{M}^2}-\frac{4(W^4-4W^{2}L+4L^2)}{W^4{M}^2}.$$

□

Theorem 4.

Let $f\in \mathcal{K}(\gamma ,\alpha ,\delta )$ be given by (1). Then

$$|T_3\left(1\right)|=|1+2a_2^2(a_3-1)-a_3^2|\le 1+\frac{4{\psi }_5}{W^4{M}^2}-\frac{8}{W^2},$$

(11)

where

$${\psi }_5=4W^{3}M-8ML-W^4+4W^{2}L-4L^2.$$

(12)

Proof. 

From (7), (8), (2), applying Lemma 2 and some calculations, we have

$$\begin{array}{c}\hfill |T_3\left(1\right)|=|1+\frac{p_1^4}{WM}+\frac{p_1^{2}x(4-p_1^2)}{WM}-\frac{2L{p}_1^4}{W^{4}M}-\frac{2p_1^2}{W^2}-\frac{(W^4-4W^{2}L+4L^2)p_1^4}{4W^4{M}^2}\\ \hfill -\frac{(W^2-2L)p_1^{2}x(4-p_1^2)}{2W^2{M}^2}-\frac{x^2{(4-p_1^2)}^2}{4M^2}|.\end{array}$$

We select $p_1=p$ for ease of notation, and because the function p is in the family $\mathcal{P}$ at the same time, we may assume that $p\in [0,2]$ without losing generality. As a result, using the triangle inequality with $P=4-r^2$, we have

$$|T_3\left(1\right)|\le \left|1+\frac{{\psi }_5{p}^4}{4W^4{M}^2}-\frac{2p^2}{W^2}\right|+\frac{(W^2-2L)p^2(4-p^2)}{2W^2{M}^2}+\frac{{(4-p^2)}^2}{4M^2}.$$

Hence, at $p=2$, we have

$$|T_3\left(1\right)|\le 1+\frac{4{\psi }_5}{W^4{M}^2}-\frac{8}{W^2}.$$

□

Theorem 5.

Let $f\in \mathcal{K}(\gamma ,\alpha ,\delta )$ be given by (1). Then

$$|T_3\left(2\right)|=|(a_2-a_4)(a_2^2-2a_3^2+a_2{a}_4)|\le \left[\frac{2}{W}-\frac{8(3W^{2}M+6WH-4E)p^3}{12V{W}^{2}M}\right]\left[\frac{4}{W^2}-\frac{4{\psi }_6}{3W^4{M}^{2}V}\right],$$

(13)

where

$${\psi }_6=6V(W^4-4W^{2}L+4L^2)-3W^{2}M(WM+2H)+4ME.$$

Proof. 

From (7), (9), (2) and applying Lemma 2, we have

$$\begin{array}{cc}|a_2-a_4|=|\frac{p_1}{W}-\frac{p_1^3}{4V}-\frac{p_1(4-p_1^2)x}{2V}+\frac{p_1(4-p_1^2)x^2}{4V}-\frac{(4-p_1^2){(1-|x|}^2)z}{2V}-\frac{H{p}_1^3}{2WMV}-\frac{H{p}_1(4-p_1^2)x}{2WMV}\hfill & \\ & \hfill +\frac{E{p}_1^3}{3W^{2}MV}|.\end{array}$$

Applying triangle inequality and $p_1=p$, we have

$$|a_2-a_4|\le \left|\frac{p}{W}-\frac{(3W^{2}M+6WH-4E)p^3}{12V{W}^{2}M}\right|+\frac{p(WM+H)\left|x\right|P}{WM}+\frac{{p\left|x\right|}^{2}P}{4V}+\frac{PZ}{2V}+\frac{Hp\left|x\right|P}{2WMV}.$$

Using the same methods as Theorems 2 and 3, we have

$$|a_2-a_4|\le \frac{2}{W}-\frac{8(3W^{2}M+6WH-4E)p^3}{12V{W}^{2}M}.$$

(14)

Furthermore, using (7), (8), (9), and (2), applying Lemma 2 and taking $p_1=p\in [0,2]$, we have

$$\begin{array}{c}|a_2^2-2a_3^2+a_2{a}_4|\le \left|\frac{p^2}{W^2}-\frac{{\psi }_6{p}^4}{12W^4{M}^{2}V}\right|+\frac{{\psi }_7{p}^2(4-p^2)\left|x\right|}{2W^2{M}^{2}V}+\frac{p^2(4-p^2){\left|x\right|}^2}{4WV}+\frac{{(4-p^2)}^2{\left|x\right|}^2}{2M^2}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hfill \\ \hfill +\frac{p(4-p^2){(1-|x|}^2)}{2WV}:=F_2(p,\left|x\right|),\end{array}$$

where

$${\psi }_7=2V(W^2-2L)-W{M}^2-MH.$$

On the closed area $[0,2]\times [0,1]$, we need to find the maximum value of $F_2(p,\left|x\right|)$. Assume that a maximum of $[0,2]\times [0,1]$ exists at an interior point $(p_0,|x\left|\right)$. After that, by differentiating $F_2(p,\left|x\right|)$ w.r.t $\left|x\right|$, we have

$$\frac{\partial F_2(p,\left|x\right|)}{\partial \left|x\right|}=\frac{{\psi }_7{p}^2(4-p^2)}{2W^2{M}^{2}V}+\frac{p^2(4-p^2)\left|x\right|}{2WV}+\frac{{(4-p^2)}^2\left|x\right|}{M^2}-\frac{p(4-p^2)\left|x\right|)}{WV}.$$

If $p=0$,

$$F_2(0,\left|x\right|)=\frac{8}{M^2}{\left|x\right|}^2\le \frac{8}{M^2}$$

If $p=2$,

$$F_2(2,\left|x\right|)=\frac{4}{W^2}-\frac{4{\psi }_6}{3W^4{M}^{2}V}$$

If $\left|x\right|=0$,

$$F_2(p,0)=\left|\frac{p^2}{W^2}-\frac{{\psi }_6{p}^4}{12W^4{M}^{2}V}\right|+\frac{p(4-p^2)}{2WV}$$

which has the highest possible value

$$\frac{4}{W^2}-\frac{4{\psi }_6}{3W^4{M}^{2}V}$$

on $[0,2]$. Furthermore, if $\left|x\right|=1$, we have

$$F_2(p,1)=\left|\frac{p^2}{W^2}-\frac{{\psi }_6{p}^4}{12W^4{M}^{2}V}\right|+\frac{(2{\psi }_7+W{M}^2)p^2(4-p^2)}{4W^2{M}^{2}V}+\frac{{(4-p^2)}^2}{2M^2}$$

which has the highest possible value

$$\frac{4}{W^2}-\frac{4{\psi }_6}{3W^4{M}^{2}V}$$

on $[0,2]$. So,

$$|T_3\left(2\right)|=|(a_2-a_4)(a_2^2-2a_3^2+a_2{a}_4)|\le \left[\frac{2}{W}-\frac{8(3W^{2}M+6WH-4E)p^3}{12V{W}^{2}M}\right]\left[\frac{4}{W^2}-\frac{4{\psi }_6}{3W^4{M}^{2}V}\right].$$

□

3. Conclusions

Our present investigation is motived by a number of recent developments on the determinants of the Toeplitz matrices. In this paper, we have introduced a new family of analytic functions defined by the Borel distribution series in the open unit disk U. We also generated a Taylor–Maclaurin coefficient estimates for the first four determinants of the Toeplitz matrices $T_2\left(2\right)$, $T_2\left(3\right)$, $T_3\left(1\right)$, and $T_3\left(2\right)$ for the functions belonging to this newly introduced family.