The solution of the Brannan conjecture

TL;DR

This paper completes the proof of the longstanding Brannan conjecture using Maclaurin series and integral estimates, advancing mathematical understanding in complex analysis.

Contribution

It provides the final proof of the Brannan conjecture, a significant unresolved problem in mathematical analysis.

Findings

Proof of the Brannan conjecture established

Use of Maclaurin series and integral estimation techniques

Advancement in complex analysis theory

Abstract

We make the final step to give a proof for the Brannan's conjecture. The basic tool of the study is a Mac-Laurin development and an adequately estimation of an integral.

Equations

$$\frac{(1+xz{)}^{\alpha }}{(1-z{)}^{\beta }}=\sum _{n=0}^{\infty }{A}_n(\alpha ,\beta ,x)z^n,$$

$$|A_{2n-1}(\alpha ,\beta ,x)|\le A_{2n-1}(\alpha ,\beta ,1),$$

$$|A_{2n-1}(\alpha ,x)|\le |A_{2n-1}(\alpha ,1)|,\text{for all}|x|=1.$$

$$|A_{2n-1}(\alpha ,x)|\le |A_{2n-1}(\alpha ,1)|,\text{for all}|x|=1\text{with}\arg (x)\in [-\frac{2\pi }{3},\frac{2\pi }{3}].$$

$$\displaystyle 0.61>\frac{\pi}{3\sqrt{3}}\geq\sqrt{\frac{1-y}{1+y}}\Big{(}\arctan\sqrt{\frac{1-y}{1+y}}+\arctan{\frac{y}{\sqrt{1-y^{2}}}}\Big{)}\geq 0.$$

$$\sqrt{\frac{1-y}{1+y}}\Big{(}\arctan\sqrt{\frac{1-y}{1+y}}+\arctan{\frac{y}{\sqrt{1-y^{2}}}}\Big{)}=t\left(\arctan{t}+\arctan\frac{1-t^{2}}{2t}\right).$$

$$0.60459\dots =\frac{\pi }{3\sqrt{3}}\ge \phi (t)>0,t\in (0,\frac{1}{\sqrt{3}}).$$

$$(1+e^{i\theta }{)}^{\alpha }=1+\frac{\alpha }{1!}{e}^{i\theta }+\frac{\alpha (\alpha -1)}{2!}{e}^{2i\theta }+\dots +\frac{\alpha (\alpha -1)\dots (\alpha -2n+2)}{(2n-1)!}{e}^{(2n-1)i\theta }$$

$$+e^{2ni\theta }\frac{\alpha (\alpha -1)\dots (\alpha -2n+1)}{(2n-1)!}{\int }_0^1(1-t{)}^{2n-1}(1+e^{i\theta }t{)}^{\alpha -2n}dt,\theta \in (-\pi ,\pi ).$$

$$\displaystyle|A_{2n-1}(\alpha,e^{i\theta})|=\big{|}1+\frac{\alpha}{1!}e^{i\theta}$$

$$\displaystyle+\frac{\alpha(\alpha-1)}{2!}e^{2i\theta}+\ldots+\frac{\alpha(\alpha-1)\ldots(\alpha-2n+2)}{(2n-1)!}e^{(2n-1)i\theta}\big{|}=\big{|}(1+e^{i\theta})^{\alpha}\ \ \$$

$$\displaystyle-e^{2ni\theta}{\alpha(1-\alpha)(1-\frac{\alpha}{2})\ldots(1-\frac{\alpha}{2n-1})}\int_{0}^{1}(1-t)^{2n-1}(1+e^{i\theta}t)^{\alpha-2n}dt\big{|},$$

$$\theta \in (-\pi ,\pi ).$$

$$A_{2n-1}(\alpha ,1)\ge 2^{\alpha }-\frac{\alpha {\beta }_n}{2n},\alpha \in (0,1).$$

$$\displaystyle A_{2n-1}(\alpha,1)=2^{\alpha}-{\alpha\beta_{n}}\int^{1}_{0}\Big{(}\frac{1-t}{1+t}\Big{)}^{2n-1}(1+t)^{\alpha-1}dt.$$

$$\displaystyle\int^{1}_{0}\Big{(}\frac{1-t}{1+t}\Big{)}^{2n-1}(1+t)^{\alpha-1}dt\leq\int^{1}_{0}\Big{(}\frac{1-t}{1+t}\Big{)}^{2n-1}dt.$$

$${\int }_0^1\phi (s)\psi (s)ds\le {\int }_0^1\phi (s)ds{\int }_0^1\psi (s)ds.$$

$$\displaystyle\int^{1}_{0}\Big{(}\frac{1-t}{1+t}\Big{)}^{2n-1}dt=\int^{1}_{0}s^{2n-1}\frac{2}{(1+s)^{2}}ds\leq\int^{1}_{0}s^{2n-1}ds\int^{1}_{0}\frac{2}{(1+s)^{2}}ds=\frac{1}{2n}.$$

$$\displaystyle\int^{1}_{0}\Big{(}\frac{1-t}{1+t}\Big{)}^{2n-\alpha}(1-t)^{\alpha-1}dt\leq\frac{1}{2n}.$$

$$\displaystyle I_{n}(y)=\int_{0}^{1}\Big{(}\frac{1-t}{\sqrt{1+t^{2}-2ty}}\Big{)}^{2n-1}\big{(}\sqrt{1+t^{2}-2ty}\big{)}^{\alpha-1}dt$$

$$\displaystyle\leq\sqrt{\frac{1-y}{1+y}}\Big{(}\arctan\sqrt{\frac{1-y}{1+y}}+\arctan{\frac{y}{\sqrt{1-y^{2}}}}\Big{)}-\frac{|2-2y|^{\frac{\alpha}{2}}}{\alpha}+\frac{1}{\alpha}.$$

$$I_n(y)\le I_1(y)={\int }_0^1\frac{1-t}{(1+t^2-2ty{)}^{1-\frac{\alpha }{2}}}dt={\int }_0^1\frac{1-y}{(1+t^2-2ty{)}^{1-\frac{\alpha }{2}}}dt$$

$$-\frac{1}{2}{\int }_0^1\frac{2(t-y)}{(1+t^2-2ty{)}^{1-\frac{\alpha }{2}}}dt\ge (1-y){\int }_0^1\frac{1}{1-y^2+(t-y{)}^2}dt$$

$$\displaystyle-\frac{1}{2}\int_{0}^{1}\frac{2(t-y)}{(1+t^{2}-2ty)^{1-\frac{\alpha}{2}}}dt=\sqrt{\frac{1-y}{1+y}}\arctan\frac{t-y}{\sqrt{1-y^{2}}}\Big{|}_{0}^{1}$$

$$\displaystyle-\frac{(1+t^{2}-2ty)^{\frac{\alpha}{2}}}{\alpha}\Big{|}_{0}^{1}=\sqrt{\frac{1-y}{1+y}}\Big{(}\arctan\sqrt{\frac{1-y}{1+y}}+\arctan{\frac{y}{\sqrt{1-y^{2}}}}\Big{)}$$

$$-\frac{|2-2y{|}^{\frac{\alpha }{2}}}{\alpha }+\frac{1}{\alpha },$$

$$A_{2n-1}(\alpha ,1)\ge |A_{2n-1}(\alpha ,e^{i\theta })|,\text{for all}\theta \in [-\pi ,-\frac{2\pi }{3}]\cup [\frac{2\pi }{3},\pi ].$$

$$|A_{2n-1}(\alpha ,e^{i\theta })|$$

$$\displaystyle=\big{|}1+\frac{\alpha}{1!}e^{i\theta}+\frac{\alpha(\alpha-1)}{2!}e^{2i\theta}+\ldots+\frac{\alpha(\alpha-1)\ldots(\alpha-2n+2)}{(2n-1)!}e^{(2n-1)i\theta}\big{|}$$

$$\displaystyle\leq\frac{\alpha(1-\alpha)(2-\alpha)\ldots(2n-1-\alpha)}{(2n-1)!}\int_{0}^{1}(1-t)^{2n-1}\big{|}1+e^{i\theta}t\big{|}^{\alpha-2n}dt$$

$$\displaystyle+\big{|}1+e^{i\theta}\big{|}^{\alpha}=\big{|}1+e^{i\theta}\big{|}^{\alpha}+\alpha(1-\alpha)(1-\frac{\alpha}{2})(1-\frac{\alpha}{3})\ldots(1-\ \$$

$$\displaystyle\frac{\alpha}{2n-1})\int_{0}^{1}(1-t)^{2n-1}\big{|}1+e^{i\theta}t\big{|}^{\alpha-2n}dt,\ \theta\in(-\pi,\pi).$$

$$\displaystyle 2^{\alpha}-\frac{\alpha\beta_{n}}{2n}\geq\big{|}1+e^{i\theta}\big{|}^{\alpha}+\alpha(1-\alpha)(1-\frac{\alpha}{2})(1-\frac{\alpha}{3})\ldots(1-\ \$$

$$\displaystyle\frac{\alpha}{2n-1})\int_{0}^{1}(1-t)^{2n-1}\big{|}1+e^{i\theta}t\big{|}^{\alpha-2n}dt,\ \theta\in[-\pi,-\frac{2\pi}{3}]\cup[\frac{2\pi}{3},\pi].$$

$$\displaystyle|2-2y|^{\frac{\alpha}{2}}+\alpha{\beta_{n}}I_{n}(y)=\big{|}1+e^{i\theta}\big{|}^{\alpha}+\alpha(1-\alpha)(1-\frac{\alpha}{2})(1-\frac{\alpha}{3})\ldots(1-\ \$$

$$\displaystyle\frac{\alpha}{2n-1})\int_{0}^{1}(1-t)^{2n-1}\big{|}1+e^{i\theta}t\big{|}^{\alpha-2n}dt,\ \theta\in[-\pi,-\frac{2\pi}{3}]\cup[\frac{2\pi}{3},\pi].$$

$$\displaystyle 2^{\alpha}\geq\frac{\alpha\beta_{n}}{2n}+\big{(}2-2y\big{)}^{\frac{\alpha}{2}}+\alpha\beta_{n}I_{n}(y).\ y\in[\frac{1}{2},1].$$

$$\displaystyle\big{(}2-2y\big{)}^{\frac{\alpha}{2}}(1-\beta_{n})+\beta_{n}+\alpha\beta_{n}\Big{(}\frac{1}{2n}+\frac{\pi}{3\sqrt{3}}\Big{)}=\frac{\alpha\beta_{n}}{2n}+\big{(}2-2y\big{)}^{\frac{\alpha}{2}}$$

Full text

111Keywords: Mac-Laurin development, integral inequality

1 Introduction

We consider the following Mac-Laurin development:

[TABLE]

where $\alpha>0,\ \beta>0,\ x=e^{i\theta},\ \theta\in[-\pi,\pi],$ and $z\in{U.}$ The radius of convergence of the series (1) is equal to $1.$ In [5] the author conjectured, that if $\alpha>0,\ \beta>0$ and $|x|=1,$ then

[TABLE]

where $n$ is a natural number. Partial results regarding this question were already proved in [1], [2], [5], [8]. Concerning the case $\beta=1$, and $\alpha\in(0,1)$ partial results have been proved in [3], [4], [6], [7], [9].

Using the results from [6] and [9] we will give the solution of the case $\beta=1,$ $\alpha\in(0,1).$ We introduce the notation $A_{2n-1}(\alpha,1,x)=A_{2n-1}(\alpha,x).$ The following theorems have been proved in [6] and [9].

Theorem 1**.**

[6]** If $n$ is a natural number, $n\leq 26$ and $\alpha\in(0,1),$ then the following inequality holds:

[TABLE]

Theorem 2**.**

[9]**  If $n$ is a natural number, $n\geq 27$ and $\alpha\in(0,1),$ then the following inequality holds:

[TABLE]

The aim of this paper is to extend this theorem to the case $\arg(x)\in[-\pi,\pi].$

In order to prove our main result, we need the lemmas from the next section.

2 Preliminaries

Lemma 1**.**

If $y\in[\frac{1}{2},1],$ then the following inequalities hold

[TABLE]

Proof.

We denote $t=\sqrt{\frac{1-y}{1+y}}.$ It is easily seen that $y\in[\frac{1}{2},1]$ is equivalent to $t\in[0,\frac{1}{\sqrt{3}}],$ and we have

[TABLE]

Let $\varphi:[0,\frac{1}{\sqrt{3}}]\rightarrow\mathbb{R}$ be the function defined by \varphi(t)=t\Big{(}\arctan{t}+\arctan\frac{1-t^{2}}{2t}\Big{)}. We have $\varphi^{\prime}(t)=\arctan{t}+\arctan\frac{1-t^{2}}{2t}-\frac{t}{1+t^{2}},$ and $\varphi^{\prime\prime}(t)=\frac{-2}{(1+t^{2})^{2}}.$

Thus $\varphi^{\prime}$ is strictly decreasing and $\frac{\pi}{2}=\lim_{\searrow 0}\varphi^{\prime}(t)\geq\varphi^{\prime}(t)\geq\varphi^{\prime}(\frac{1}{\sqrt{3}})=\frac{\pi}{3}-\frac{\sqrt{3}}{4}>0,\ t\in[0,\frac{1}{\sqrt{3}}].$ Consequently, $\varphi$ is strictly increasing and

[TABLE]

∎

In [9] the author proved the equality

[TABLE]

This equality is a basic tool in our study. Taking the absolute values on both sides, we infer

[TABLE]

We introduce the notation $\beta_{n}=(1-\alpha)(1-\frac{\alpha}{2})\ldots(1-\frac{\alpha}{2n-1}).$

We will prove an estimation for $A_{2n-1}(\alpha,1)$ in the followings.

Lemma 2**.**

The following inequality holds

[TABLE]

Proof.

According to (2) we have

[TABLE]

The condition $t,\alpha\in(0,1)$ implies $(1+t)^{\alpha-1}\leq 1$ and so it follows that

[TABLE]

The change of variable $s=\frac{1-t}{1+t}$ gives \int^{1}_{0}\Big{(}\frac{1-t}{1+t}\Big{)}^{2n-1}dt=\int^{1}_{0}s^{2n-1}\frac{2}{(1+s)^{2}}ds. Chebyshev’s inequality for monotonic functions claims: if $\varphi,\psi:[0,1]\rightarrow[0,\infty)$ are two integrable functions of different monotony, then the following integral inequality holds:

[TABLE]

Putting $\varphi(s)=s^{2n-1},$ and $\psi(s)=\frac{2}{(1+s)^{2}}$ in this inequality, we infer

[TABLE]

From (6) and (7) we deduce

[TABLE]

∎

Finally, we infer (4) from (5) and (8).

Lemma 3**.**

If $\alpha\in(0,1),$ y\in\big{[}\frac{1}{2},1\big{]} and $n\in\mathbb{N}^{*},$ then the following integral inequality holds

[TABLE]

Proof.

The sequence (I_{n}(y))_{n\geq 1},\ \ I_{n}(y)=\int_{0}^{1}\Big{(}\frac{1-t}{\sqrt{1+t^{2}-2ty}}\Big{)}^{2n-1}\big{(}\sqrt{1+t^{2}-2ty}\big{)}^{\alpha-1}dt is decreasing with respect to $n,$ and consequently it is enough to prove the inequality (3) in case $n=1.$ We have

[TABLE]

and the proof is done. ∎

Lemma 1 and Lemma 3 imply the following result.

Corollary 1**.**

If $\alpha\in(0,1),$ y\in\big{[}\frac{1}{2},1\big{]} and $n\in\mathbb{N}^{*},$ then  \frac{\pi}{3\sqrt{3}}-\frac{|2-2y|^{\frac{\alpha}{2}}}{\alpha}+\frac{1}{\alpha}\geq\int_{0}^{1}\Big{(}\frac{1-t}{\sqrt{1+t^{2}-2ty}}\Big{)}^{2n-1}\big{(}\sqrt{1+t^{2}-2ty}\big{)}^{\alpha-1}dt=I_{n}(y).

3 Main Result

Theorem 3**.**

If $n$ is a natural number, $n\geq 27$ and $\alpha\in(0,1)$ then the following inequality holds

[TABLE]

Proof.

We use equality (2) again

[TABLE]

Taking into account Lemma 2 and (3), it follows that in order to prove (11) we have to show that the following inequality holds

[TABLE]

We denote $y=-\cos\theta,$ and we get

[TABLE]

Thus inequality (3) can be rewritten in the following equivalent form

[TABLE]

On the other hand, according to Corollary 1, we have

[TABLE]

The conditions $y\in[\frac{1}{2},1]$ and $n\geq 27$ imply that \big{(}2-2y\big{)}^{\frac{\alpha}{2}}\leq 1, and $\frac{1}{2n}+\frac{\pi}{3\sqrt{3}}\leq\frac{65}{100}.$ Consequently we have

[TABLE]

Since $\ln 2>0.65$ and $\beta_{n}<1,$ we get

[TABLE]

Finally, inequalities (3), (17) and (18) imply (15), and the proof is done. ∎

Summarizing Theorem 1, Theorem 2 and Theorem 3 imply the following corollary.

Corollary 2**.**

If $x\in\mathbb{C}$ with $|x|=1,$ then the inequality

[TABLE]

holds for every $\alpha\in(0,1),$ and $n\in\mathbb{N}^{*}.$