Performance comparison of deep learning models for MRI-based brain tumor detection

Abdulmajeed Alsufyani; Abdulmajeed Alsufyani

doi:10.3934/bioeng.2025001

AIMS Bioengineering

2025, Volume 12, Issue 1: 1-21. doi: 10.3934/bioeng.2025001

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Research article Special Issues

Performance comparison of deep learning models for MRI-based brain tumor detection

Abdulmajeed Alsufyani ^,

Department of Computer Science, College of Computers and Information Technology, Taif University, Taif, Saudi Arabia

Received: 02 November 2024 Revised: 27 December 2024 Accepted: 02 January 2025 Published: 15 January 2025

Brain tumors pose a significant threat to human health, as they can severely affect both physical well-being and quality of life. These tumors often lead to increased intracranial pressure and neurological complications. Traditionally, brain tumors are diagnosed through manual interpretation via medical imaging techniques such as Computed Tomography (CT) and Magnetic Resonance Imaging (MRI). While these methods are effective, they are time-consuming and subject to errors in consistency and accuracy. This study explored the use of several deep-learning models, including YOLOv8, YOLOv9, Faster R-CNN, and ResNet18, for the detection of brain tumors from MR images. The results demonstrate that YOLOv9 outperforms the other models in terms of accuracy, precision, and recall, highlighting its potential as the most effective deep-learning approach for brain tumor detection.

Keywords:

Citation: Abdulmajeed Alsufyani. Performance comparison of deep learning models for MRI-based brain tumor detection[J]. AIMS Bioengineering, 2025, 12(1): 1-21. doi: 10.3934/bioeng.2025001

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Abstract

1. Introduction and preliminaries

Let $\mathcal{A}$ indicate an analytic functions family, which is normalized under the condition $f$ $(0) =$ $f^{\prime}(0)-1 = 0$ in $\mathbb{U} = \{z:z\in \mathbb{C}$ and $\left \vert z \ \right \vert < 1\}$ and given by the following Taylor-Maclaurin series:

$\begin{equation} f \ (z) = z+\sum \limits_{n = 2}^{\infty}a_{n}z^{n}\ .\ \ \ \ \ \end{equation}$

(1.1)

Further, by $\mathcal{S}$ we shall denote the class of all functions in $\mathcal{A}$ which are univalent in $\mathbb{U}$ .

With a view to recalling the principle of subordination between analytic functions, let the functions $f$ and $g$ be analytic in $\mathbb{U}$ . Then we say that the function $f$ is subordinate to $g$ if there exists a Schwarz function $w\left(z\right)$ , analytic in $\mathbb{U}$ with

$\begin{equation*} \omega \left( 0\right) = 0, \ \left\vert \omega (z)\right\vert < 1, \ \left( z\in \mathbb{U}\right) \end{equation*}$

such that

$\begin{equation*} f \ \left( z\right) = g \ (\omega \left( z\right) ). \end{equation*}$

We denote this subordination by

$\begin{equation*} f\prec g\text{ or}\ f \ (z)\prec g \ (z). \end{equation*}$

In particular, if the function $g$ is univalent in $\mathbb{U}$ , the above subordination is equivalent to

$\begin{equation*} f \ (0) = g \ (0), \ f \ (\mathbb{U})\subset g \ ( \mathbb{U}). \end{equation*}$

The Koebe-One Quarter Theorem ^[11] asserts that image of $\mathbb{U}$ under every univalent function $f\in \mathcal{A}$ contains a disc of radius $\frac{1}{4}.\$ thus every univalent function $f$ has an inverse $\ f^{-1}\$ satisfying $\ f^{-1}(f(z)) = z$ and $f$ $(\ f^{-1}$ $(w)) = w$ $(\left\vert w\right\vert < r$ $_{0}(f$ $), r$ $_{0}(f$ $)$ $> \frac{ 1}{4}$ $)$ , where

$\begin{equation} \ f^{-1}(w) = w-a_{2}w^{2}+(2a_{2}^{2}-a_{3})w^{3}-(5a_{2}^{3}-5a_{2}a_{3}+a_{4})w^{4}+\cdots . \end{equation}$

(1.2)

A function $f\in \mathcal{A}$ is said to be bi-univalent functions in $\mathbb{U}$ if both $f$ and $\ f^{-1}$ are univalent in $\mathbb{U}$ . A function $f\in \mathcal{S}$ is said to be bi-univalent in $\mathbb{U}$ if there exists a function $g\in \mathcal{S}$ such that $g(z)$ is an univalent extension of $f^{-1}$ to $\mathbb{U}$ . Let $\Lambda$ denote the class of bi-univalent functions in $\mathbb{U}$ . The functions $\frac{z}{ 1-z}$ , $-\log (1-z)$ , $\frac{1}{2}\log \left(\frac{1+z}{1-z}\right)$ are in the class $\Lambda$ (see details in ^[20]). However, the familiar Koebe function is not bi-univalent. Lewin ^[17] investigated the class of bi-univalent functions $\Lambda$ and obtained a bound $|a_{2}|\leq 1.51$ . Motivated by the work of Lewin ^[17], Brannan and Clunie ^[9] conjectured that $|a_{2}|\leq \sqrt{2}.$ The coefficient estimate problem for $|a_{n}|\, \, \, \, (n\in \mathbb{N}, \, \, \, \, n\geq 3)$ is still open (^[20]). Brannan and Taha ^[10] also worked on certain subclasses of the bi-univalent function class $\Lambda$ and obtained estimates for their initial coefficients. Various classes of bi-univalent functions were introduced and studied in recent times, the study of bi-univalent functions gained momentum mainly due to the work of Srivastava et al. ^[20]. Motivated by this, many researchers ^[1], ^[4,5,6,7,8], ^[13,14,15], ^[20], ^[21], and ^[27,28,29], also the references cited there in) recently investigated several interesting subclasses of the class $\Lambda$ and found non-sharp estimates on the first two Taylor-Maclaurin coefficients. Recently, many researchers have been exploring bi-univalent functions, few to mention Fibonacci polynomials, Lucas polynomials, Chebyshev polynomials, Pell polynomials, Lucas–Lehmer polynomials, orthogonal polynomials and the other special polynomials and their generalizations are of great importance in a variety of branches such as physics, engineering, architecture, nature, art, number theory, combinatorics and numerical analysis. These polynomials have been studied in several papers from a theoretical point of view (see, for example, ^{[23,24,25,26,27,28,29,30]} also see references therein).

We recall the following results relevant for our study as stated in ^[3].

Let $p(x)$ and $q(x)$ be polynomials with real coefficients. The $(p, q)-$ Lucas polynomials $\mathcal{L}_{p, q, n}(x)$ are defined by the recurrence relation

$\begin{equation*} \mathcal{L}_{p, q, n}(x) = p(x)\mathcal{L}_{p, q, n-1}(x)+q(x)\mathcal{L} _{p, q, n-2}(x)\quad (n\geq 2), \end{equation*}$

from which the first few Lucas polynomials can be found as

$\begin{eqnarray} \mathcal{L}_{p, q, 0}(x) & = &2, \\ \mathcal{L}_{p, q, 1}(x) & = &p(x), \\ \mathcal{L}_{p, q, 2}(x) & = &p^{2}(x)+2q(x), \\ \mathcal{L}_{p, q, 3}(x) & = &p^{3}(x)+3p(x)q(x), ...\text{.} \end{eqnarray}$

(1.3)

For the special cases of $p(x)$ and $q(x),$ we can get the polynomials given $\mathcal{L}_{x, 1, n}(x)\equiv \mathcal{L}_{n}(x)$ Lucas polynomials, $\mathcal{L}_{2x, 1, n}(x)\equiv \mathcal{D}_{n}(x)$ Pell–Lucas polynomials, $\mathcal{L}_{1, 2x, n}(x)\equiv j_{n}(x)$ Jacobsthal–Lucas polynomials, $\mathcal{L}_{3x, -2, n}(x)\equiv F_{n}(x)$ Fermat–Lucas polynomials, $\mathcal{L}_{2x, -1, n}(x)\equiv T_{n}(x)$ Chebyshev polynomials first kind.

Lemma 1.1. ^[16] Let $G\{ \mathcal{L}(x)\}(z)$ be the generating function of the $(p, q)-$ Lucas polynomial sequence $\mathcal{L}_{p, q, n}(x).$ Then,

$\begin{equation*} G\{ \mathcal{L}(x)\}(z) = \sum\limits_{n = 0}^{\infty }\mathcal{L}_{p, q, n}(x)z^{n} = \frac{2-p(x)z}{1-p(x)z-q(x)z^{2}} \end{equation*}$

and

$\begin{equation*} \mathcal{G}_{\{ \mathcal{L}(x)\}}(z) = G\{ \mathcal{L}(x)\}(z)-1 = 1+ \sum\limits_{n = 1}^{\infty }\mathcal{L}_{p, q, n}(x)z^{n} = \frac{1+q(x)z^{2}}{ 1-p(x)z-q(x)z^{2}}. \end{equation*}$

Definition 1.2. ^[22] For $\vartheta \geq 0,$ $\delta \in \mathbb{R},$ $\vartheta +i\delta \neq 0$ and $f\in \mathcal{A}$ , let $\mathcal{B} (\vartheta, \delta)$ denote the class of Bazilevič function if and only if

$\begin{equation*} Re\left[ {\left( {\frac{{z{f^\prime }(z)}}{{f(z)}}} \right){{\left( {\frac{{f(z)}}{z}} \right)}^{\vartheta + i\delta }}} \right] > 0\;. \end{equation*}$

Several authors have researched different subfamilies of the well-known Bazilevič functions of type $\vartheta$ from various viewpoints (see ^[3] and ^[19]). For Bazilevič functions of order $\vartheta +i\delta$ , there is no much work associated with Lucas polynomials in the literature. Initiating an exploration of properties of Lucas polynomials associated with Bazilevič functions of order $\vartheta +i\delta$ is the main goal of this paper. To do so, we take into account the following definitions. In this paper motivated by the very recent work of Altinkaya and Yalcin ^[3] (also see ^[18]) we define a new class $\mathcal{B}(\vartheta, \delta)$ , bi-Bazilevič function of $\Lambda$ based on $(p, q)-$ Lucas polynomials as below:

Definition 1.3. For $f\in \Lambda$ , $\vartheta \geq0,$ $\delta \in \mathbb{R},$ $\vartheta+i\delta \neq0$ and let $\mathcal{B}(\vartheta, \delta)$ denote the class of Bi-Bazilevič functions of order $t\ \$ and type $\vartheta+i\delta$ if only if

$\begin{equation} \left[ \left( \frac{zf^{\prime}(z)}{f(z)}\right) \left( \frac{f(z)}{z} \right) ^{\vartheta+i\delta}\right] \prec \mathcal{G}_{\{ \mathcal{L} (x)\}}(z)\quad (z\in \mathbb{U}) \end{equation}$

(1.4)

and

$\begin{equation} \left[ \left( \frac{zg^{\prime}(w)}{g(w)}\right) \left( \frac{g(w)}{w}% \right) ^{\vartheta+i\delta}\right] \prec \mathcal{G}_{\{ \mathcal{L}% (x)\}}(w)\quad (w\in \mathbb{U}), \label{ieq4} \end{equation}$

(1.5)

where $\mathcal{G}_{\mathcal{L}_{p, q, n}} (z) \in \Phi$ and the function $g$ is described as $g(w) = f^{-1}(w).$

Remark 1.4. We note that for $\delta = 0$ the class $R(\vartheta, 0) = R(\vartheta)$ is defined by Altinkaya and Yalcin ^[2].

The class $\mathcal{B}(0, 0) = \mathcal{S}_\Lambda^*$ is defined as follows:

Definition 1.5. A function $f\in \Lambda$ is said to be in the class $\mathcal{S }_\Lambda^*$ , if the following subordinations hold

$\begin{equation*} \frac{zf^{\prime}(z)}{f(z)}\prec \mathcal{G}_{\{ \mathcal{L}(x)\}}(z)(z\in \mathbb{U}) \end{equation*}$

and

$\begin{equation*} \frac{wg^{\prime }(w)}{g(w)}\prec \mathcal{G}_{\{ \mathcal{L}(x)\}}(w)(w\in \mathbb{U}) \end{equation*}$

where $g(w) = f^{-1}(w).$

We begin this section by finding the estimates of the coefficients $|a_{2}|$ and $|a_{3}|$ for functions in the class $\mathcal{B}(\vartheta, \delta)$ .

2. Coefficient bounds for the function class $\mathcal{B}(\vartheta, \delta)$

Theorem 2.1. Let the function $f(z)$ given by 1.1 be in the class $\mathcal{B}(\vartheta, \delta)$ . Then

$\begin{equation*} \left \vert a_{2}\right \vert \leq \frac{p(x)\sqrt {2p(x)}}{ \sqrt{\vert \{ \left( \left( \vartheta +i\delta \right) ^{2}+3\left( \vartheta +i\delta \right) +2\right)-2\left( \vartheta +i\delta +1\right) ^{2}\}p^2(x)-4q(x)\left( \vartheta +i\delta +1\right) ^{2} \vert }}. \end{equation*}$

and

$\begin{equation*} \left \vert a_{3}\right \vert \leq \frac{p^{2}(x)}{(\vartheta +1)^2+ \delta^2 } +\frac{p(x)}{\sqrt {(\vartheta +2)^2+ \delta^2} } . \end{equation*}$

Proof. Let $f\in \mathcal{B}(\vartheta, \delta, x)$ there exist two analytic functions $u, v:\mathbb{U}\rightarrow \mathbb{U}$ with $u(0) = 0 = v(0)$ , such that $|u(z)| < 1$ , $|v(w)| < 1,$ we can write from (1.4) and (1.5), we have

$\begin{equation} \left[ \left( \frac{zf^{\prime }(z)}{f(z)}\right) \left( \frac{f(z)}{z} \right) ^{\vartheta +i\delta }\right] = \mathcal{G}_{\{ \mathcal{L} (x)\}}(z)\quad (z\in \mathbb{U}) \quad \end{equation}$

(2.1)

and

$\begin{equation} \left[ \left( \frac{zg^{\prime}(w)}{g(w)}\right) \left( \frac{g(w)}{w} \right) ^{\vartheta+i\delta}\right] = \mathcal{G}_{\{ \mathcal{L} (x)\}}(w)\quad (w\in \mathbb{U}), \end{equation}$

(2.2)

It is fairly well known that if

$\begin{equation*} |u(z)| = |u_{1}z+u_{2}z^{2}+\cdots | < 1 \end{equation*}$

and

$\begin{equation*} |v(w)| = |v_{1}w+v_{2}w^{2}+\cdots | < 1. \end{equation*}$

then

$\begin{equation*} |u_{k}|\leq 1\qquad and\qquad |v_{k}|\leq 1\quad (k\in \mathbb{N}) \end{equation*}$

It follows that, so we have

$\begin{eqnarray} \mathcal{G}_{\{ \mathcal{L}(x)\}}(u(z)) & = &1+\mathcal{L}_{p, q, 1}(x)u(z)+ \mathcal{L}_{p, q, 2}(x)u^{2}(z)+\ldots \\ & = &1+\mathcal{L}_{p, q, 1}(x)u_{1}z+[\mathcal{L}_{p, q, 1}(x)u_{2}+\mathcal{L} _{p, q, 2}(x)u_{1}^{2}]z^{2}+\ldots \end{eqnarray}$

(2.3)

and

$\begin{eqnarray} \mathcal{G}_{\{ \mathcal{L}(x)\}}(v(w)) & = &1+\mathcal{L}_{p, q, 1}(x)v(w)+ \mathcal{L}_{p, q, 2}(x)v^{2}(w)+\ldots \\ & = &1+\mathcal{L}_{p, q, 1}(x)v_{1}w+[\mathcal{L}_{p, q, 1}(x)v_{2}+\mathcal{L} _{p, q, 2}(x)v_{1}^{2}]w^{2}+\ldots \end{eqnarray}$

(2.4)

From the equalities (2.1) and (2.2), we obtain that

$\begin{equation} \left[ \left( \frac{zf^{\prime }(z)}{f(z)}\right) \left( \frac{f(z)}{z} \right) ^{\vartheta +i\delta }\right] = 1+\mathcal{L}_{p, q, 1}(x)u_{1}z+[ \mathcal{L}_{p, q, 1}(x)u_{2}+\mathcal{L}_{p, q, 2}(x)u_{1}^{2}]z^{2}+\ldots, \end{equation}$

(2.5)

and

$\begin{equation} \left[ \left( \frac{zg^{\prime}(w)}{g(w)}\right) \left( \frac{g(w)}{w} \right) ^{\vartheta+i\delta}\right] = 1+\mathcal{L}_{p, q, 1}(x)v_{1}w+[ \mathcal{L}_{p, q, 1}(x)v_{2}+\mathcal{L}_{p, q, 2}(x)v_{1}^{2}]w^{2}+\ldots, \end{equation}$

(2.6)

It follows from (2.5) and (2.6) that

$\begin{equation} \left( \vartheta +i\delta +1\right) a_{2} = \mathcal{L}_{p, q, 1}(x)u_1, , \end{equation}$

(2.7)

$\begin{equation} \frac{\left( \vartheta +i\delta -1\right) \left( \vartheta +i\delta +2\right) }{2}a_{2}^{2}-\left( \vartheta +i\delta +2\right) a_3 = \mathcal{L} _{p, q, 1}(x)u_2+\mathcal{L}_{p, q, 2}(x)u_1^2, \end{equation}$

(2.8)

and

$\begin{equation} -\left( \vartheta+i\delta+1\right) a_{2} = \mathcal{L}_{p, q, 1}(x)v_1, \end{equation}$

(2.9)

$\begin{equation} \frac{\left( \vartheta+i\delta+2\right) \left( \vartheta+i\delta+3\right) }{2 }a_{2}^{2}+\left( \vartheta+i\delta+2\right) a_3 = \mathcal{L}_{p, q, 1}(x)v_2+ \mathcal{L}_{p, q, 2}(x)v_1^2, \end{equation}$

(2.10)

From (2.7) and (2.9)

$\begin{equation} u_{1} = -v_{1} \end{equation}$

(2.11)

and

$\begin{equation} 2\left( \vartheta+i\delta+1\right) ^{2}a_{2}^{2} = \mathcal{L} _{p, q, 1}^{2}(x)(u_{1}^{2}+v_{1}^{2})., \end{equation}$

(2.12)

by adding (2.8) to (2.10), we get

$\begin{equation} \left( \left( \vartheta +i\delta \right) ^{2}+3\left( \vartheta +i\delta \right) +2\right) a_{2}^{2} = \mathcal{L}_{p, q, 1}(x)(u_{2}+v_{2})+\mathcal{L} _{p, q, 2}(x)(u_{1}^{2}+v_{1}^{2}), \end{equation}$

(2.13)

by using (2.12) in equality (2.13), we have

$\begin{eqnarray} \left[ \left( \left( \vartheta +i\delta \right) ^{2}+3\left( \vartheta +i\delta \right) +2\right) -\frac{2\mathcal{L}_{p, q, 2}(x)\left( \vartheta +i\delta +1\right) ^{2}}{\mathcal{L}^{2}_{p, q, 1}(x)}\right] a_{2}^{2}& = & \mathcal{L}_{p, q, 1}(x)(u_{2}+v_{2}), \end{eqnarray}$

$\begin{eqnarray} a_{2}^{2}& = &\frac{\mathcal{L}^3_{p, q, 1}(x)(u_{2}+v_{2})} {\left[ \left( \left( \vartheta +i\delta \right) ^{2}+3\left( \vartheta +i\delta \right) +2\right)\mathcal{L}^{2}_{p, q, 1}(x)-2\mathcal{L}_{p, q, 2}(x)\left( \vartheta +i\delta +1\right) ^{2}\right]}. \end{eqnarray}$

(2.14)

Thus, from (1.3) and (2.14) we get

$\begin{equation*} \left \vert a_{2}\right \vert \leq \frac{p(x)\sqrt {2p(x)}}{\sqrt{\vert \{ \left( \left( \vartheta +i\delta \right) ^{2}+3\left( \vartheta +i\delta \right) +2\right)-2\left( \vartheta +i\delta +1\right) ^{2}\}p^2(x)-4q(x)\left( \vartheta +i\delta +1\right) ^{2} \vert }}. \end{equation*}$

Next, in order to find the bound on $|a_{3}|$ , by subtracting (2.10) from (2.8), we obtain

$\begin{eqnarray} 2\left( \vartheta +i\delta +2\right) a_{3}-2\left( \vartheta +i\delta +2\right) a_{2}^{2}& = &\mathcal{L}_{p, q, 1}(x)(u_{2}-v_{2})+\mathcal{L} _{p, q, 2}(x)(u_{1}^{2}-v_{1}^{2}) \\ 2\left( \vartheta +i\delta +2\right) a_{3}& = &\mathcal{L} _{p, q, 1}(x)(u_{2}-v_{2})+2\left( \vartheta +i\delta +2\right) a_{2}^{2} \\ a_{3}& = &\frac{\mathcal{L}_{p, q, 1}(x)(u_{2}-v_{2})}{2\left( \vartheta +i\delta +2\right)}+a_{2}^{2} \end{eqnarray}$

(2.15)

Then, in view of (2.11) and (2.12), we have from (2.15)

$\begin{equation*} a_{3} = \frac{\mathcal{L}_{p, q, 1}^{2}(x)}{2\left( \vartheta +i\delta +2\right) ^{2}}(u_{1}^{2}+v_{1}^{2})+\frac{\mathcal{L}_{p, q, 1}(x)}{2\left( \vartheta +i\delta +2\right) }(u_{2}-v_{2}). \end{equation*}$

$\begin{eqnarray*} |a_{3}|&\leq&\frac{p^{2}(x)}{\left|\vartheta +i\delta +1\right| ^{2}}+\frac{ p(x)}{|\vartheta +i\delta +2| } \notag \\ & = &\frac{p^{2}(x)}{(\vartheta +1)^2+ \delta^2 } +\frac{p(x)}{\sqrt { (\vartheta +2)^2+ \delta^2} } \end{eqnarray*}$

This completes the proof.

Taking $\delta = 0,$ in Theorem 2.1, we get the following corollary.

Corollary 2.2. Let the function $f(z)$ given by (1.1) be in the class $\mathcal{B} (\vartheta)$ . Then

$\begin{equation*} \left \vert a_{2}\right \vert \leq \frac{p(x)\sqrt {2p(x)}}{\sqrt{\vert \{ \left( \vartheta ^{2}+3\vartheta +2\right)-2\left( \vartheta+1\right) ^{2}\}p^2(x)-4q(x)\left( \vartheta +1\right) ^{2} \vert }} \end{equation*}$

and

$\begin{equation*} \left \vert a_{3}\right \vert \leq \frac{p^{2}(x)}{(\vartheta +2)^2} +\frac{ p(x)}{\vartheta +2} \end{equation*}$

Also, taking $\vartheta = 0$ and $\delta = 0,$ in Theorem 2.1, we get the results given in ^[18].

3. Fekete-Szegö inequality for the class $\mathcal{B}(\vartheta, \delta)$

Fekete-Szegö inequality is one of the famous problems related to coefficients of univalent analytic functions. It was first given by ^[12], the classical Fekete-Szegö inequality for the coefficients of $f\in \mathcal{S}$ is

$\begin{equation*} \left \vert a_{3}-\mu a_{2}^{2}\right \vert \leq1+2\exp \left( -2\mu /(1-\mu)\right) \text{ for }\mu \in \left[ 0, 1\right) . \end{equation*}$

As ${\mu} \rightarrow1^{-}$ , we have the elementary inequality $\left \vert a_{3}-a_{2}^{2}\right \vert \leq1$ . Moreover, the coefficient functional

$\begin{equation*} \varsigma_{\mu}(f) = a_{3}-\mu a_{2}^{2} \end{equation*}$

on the normalized analytic functions $f$ in the unit disk $\mathbb{U}$ plays an important role in function theory. The problem of maximizing the absolute value of the functional $\varsigma {\mu} (f)$ is called the Fekete-Szegö problem.

In this section, we are ready to find the sharp bounds of Fekete-Szegö functional $\varsigma _{\mu }(f)$ defined for $f\in \mathcal{B}(\vartheta, \delta)$ given by (1.1).

Theorem 3.1. Let f given by (1.1) be in the class $\mathcal{B} (\vartheta, \delta)$ and ${\mu} \in \mathbb{R}$ . Then

$\begin{equation*} \left \vert a_{3}- {\mu} a_{2}^{2}\right \vert \leq \left \{ \begin{array}{c} \frac{p(x)}{\sqrt{(\vartheta+2)^{2}+\delta ^{2}}}, \ \ \ \ \ \ \ \ 0\leq \left \vert h(\mu )\right \vert \leq \frac{1}{2\sqrt{ (\vartheta+2)^{2}+\delta ^{2}}} \\ 2p(x)\left \vert h(\mu )\right \vert , \ \ \ \ \ \ \ \ \ \ \ \ \ \left \vert h(\mu )\right \vert \geq \frac{1}{2\sqrt{(\vartheta+2)^{2}+ \delta ^{2}}} \end{array} \right. \end{equation*}$

where

$\begin{equation*} h(\mu) = \frac{\mathcal{L}^2_{p, q, 1}(x)(1-\mu)}{\left( \left( \vartheta +i\delta \right) ^{2}+3\left( \vartheta +i\delta \right) +2\right)\mathcal{L} ^{2}_{p, q, 1}(x)-2\mathcal{L}_{p, q, 2}(x)\left( \vartheta +i\delta +1\right) ^{2}}. \end{equation*}$

Proof. From (2.14) and (2.15), we conclude that

$\begin{eqnarray*} a_{3}- {\mu} a_{2}^{2}& = &(1-\mu )\frac{\mathcal{L}^3_{p, q, 1}(x)(u_{2}+v_{2})} {\left[ \left( \left( \vartheta +i\delta \right) ^{2}+3\left( \vartheta +i\delta \right) +2\right)\mathcal{L}^{2}_{p, q, 1}(x)-2\mathcal{L}_{p, q, 2}(x)\left( \vartheta +i\delta +1\right) ^{2}\right]} \\ &+&\frac{\mathcal{L}_{p, q, 1}(x)}{2\left( \vartheta +i\delta +2\right) } (u_{2}-v_{2}) \end{eqnarray*}$

$\begin{eqnarray*} = \mathcal{L}_{p, q, 1}(x)\left[ \left( h(\mu )+\frac{1}{2(\vartheta +i\delta +2)}\right) u_{2}+\left( h(\mu )-\frac{1}{2(\vartheta +i\delta +2)}\right) v_{2}\right] \end{eqnarray*}$

where

Then, in view of (1.3), we obtain

We end this section with some corollaries.

Taking ${\mu} = 1$ in Theorem 3.1, we get the following corollary.

Corollary 3.2. If $f\in \mathcal{B}(\vartheta, \delta)$ , then

$\begin{equation*} \left \vert a_{3}-a_{2}^{2}\right \vert \leq \frac{p(x)}{\sqrt{ (\vartheta+2)^{2}+\delta ^{2}}}. \end{equation*}$

Taking $\delta = 0$ in Theorem 3.1, we get the following corollary.

Corollary 3.3. Let $f$ given by (1.1) be in the class $\mathcal{B}(\vartheta, 0)$ . Then

$\begin{equation*} \left \vert a_{3}- {\mu} a_{2}^{2}\right \vert \leq \left \{ \begin{array}{c} \frac{p(x)}{\vartheta+2}, \ \ \ \ \ \ \ \ 0\leq \left \vert h(\mu )\right \vert \leq \frac{1}{2(\vartheta+2)} \\ 2p(x)\left \vert h(\mu )\right \vert , \ \ \ \ \ \ \ \ \ \ \ \ \ \left \vert h(\mu )\right \vert \geq \frac{1}{2(\vartheta+2)} \end{array} \right. \end{equation*}$

Also, taking $\vartheta = 0,$ $\delta = 0$ and ${\mu} = 1$ in Theorem 3.1, we get the following corollary.

Corollary 3.4. Let $f$ given by (1.1) be in the class $\mathcal{B}$ . Then

$\begin{equation*} \left| {{a_3} - a_2^2} \right| \le \frac{{p(x)}}{2}. \end{equation*}$

Conflict of interest

All authors declare no conflicts of interest in this paper.

https://www.kaggle.com/datasets/pkdarabi/medical-image-dataset-brain-tumor-detection/data

Conflict of interest

The author declares no conflict of interest.

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