鸡内金活性炭对酸碱性染料的吸附

李敏; 洪孝挺; 吕向红; 钟昌成

doi:10.6054/j.jscnun.2014.12.010

摘要: 本文以鸡内金（ECGG）为原料，先用450 ℃高温炭化，后添加KOH作为活化剂，分别选取800 ℃ 、900 ℃和1000 ℃作为活化温度，制备出三种类型活性炭（ECGG-800，EGGG-900和EGGG-1000）. 通过SEM、BET等对三种类型活性炭进行分析表征，并选取ECGG-900做酸性品红和亚甲基蓝的吸附饱和，进行两种最常用的吸附模型Freundlich和Langmuir的拟合. 结果表明用Freundlich吸附等温线模型能更好地解释鸡内金活性炭对酸性染料的吸附；而对于碱性染料的吸附，则Langmuir吸附等温线模型更有说服力；且该活性炭对酸性品红的吸附量可达1.682 g/g，对亚甲基蓝的吸附量更达2.045 g/g. 随着吸附时间的延长，三种活性炭对染料的去除率也随即增大. 由此可知，鸡内金活性炭对处理酸性和碱性染料效果均佳，是一种具有发展潜力的吸附剂.

关键词:

吸附平衡

Abstract: Activated carbon was prepared from endothelium corneum gigeriae galli ( ECGG) by preliminary carbonization at 450 ℃ and final KOH activation at high temperatures( ECGG-800 ℃, ECGG-900 ℃ and ECGG-1000 ℃ ). The surface and structural properties are analyzed with SEM and BET. And then selected ECGG-900 to draw its adsorption isotherms about acidic and basic dyes, the adsorption isotherm data were fitted by the two most used models, the Freundlich and Langmuir models. Results show that adsorption isotherm to acid fuchsin is best fitted with Freundlich model , while methylene blue is more accord with Langmuir model. The adsorption capacity of acid fuchsin and methylene blue reached 1.682 g/g, 2.045 g/g. As the adsorption time sessions, three kinds of activated carbons removal rate of dyes also immediately increased. It can be know from the foregoing that the Endothelium corneum gigeriae galli activated carbon can achieve a best effectiveness for removing the acid fuchsin and methylene blue in the water solution and it is a potential adsorbent.

Keywords:

Adsorption equilibrium

设 $\mathcal{A}$ 表示单位圆盘D={z∈ ${\mathbb{C}}$ : |z| < 1}内解析且具有如下形式

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

(1)

的函数族.

设 $\mathcal{P}$ 表示单位圆盘D内满足条件p(0)=0及Re p(z)>0的函数p组成的解析函数族.

1992年, MA和MINDA^[1]引入某类星像函数类S^*(ϕ):

$f \in S^{*}(\phi) \Leftrightarrow \frac{z f^{\prime}(z)}{f(z)}\prec\phi(z),$

其中ϕ ∈ $\mathcal{P}$ . 该函数类将单位圆盘映射到关于实轴对称、关于ϕ(0)=1星像且满足ϕ′(0)>0的星像区域.

1996年, SOKÓȽ和STANKIEWICZ^[2]引入函数类S^*( $\sqrt{1+z}$ ), 该函数类将单位圆盘映射为伯努利双纽线{w∈ ${\mathbb{C}}$ : |w²-1| < 1}的右半部分星像区域.

2014年, MENDIRATTA等^[3]引入函数类S^* $(\sqrt{2}- (\sqrt{2}-1) \sqrt{(1-z) /(1+2(\sqrt{2}-1)}))$ , 该函数类将单位圆盘映射为伯努利双纽线{w ∈ ${\mathbb{C}}$ : |w²-1| < 1}的左半部分星像区域.

2015年, MENDIRATTA等^[4]引入了与指数函数有关的函数类S_e^*, 该函数类将单位圆盘映射到关于实轴对称、关于1星像的右半平面区域.

另一方面, 对于不同解析函数类的Hankel行列式研究一直是热点问题之一. 1966年, POMMERENKE^[5]定义了解析函数f的q阶Hankel行列式H_q(n). 很明显, 当q=2, n=1时, |H₂(1)|即是Fekete-Szegö泛函^[6-11]. 近年来, 许多学者对各类解析函数的二、三阶Hankel行列式做了大量研究^[12-24]. 如: 研究了与指数函数有关的星像函数类的二、三阶Hangkel行列式^[12-14]; 研究了有界转动、星像和凸像函数类的三阶Hankel行列式^[15]; 研究了近于凸函数类的三阶Hankel行列式^[17].

但是, 目前对于与指数函数有关的函数类的Hankel行列式的研究都仅基于二阶和三阶的情形, 而对四阶Hankel行列式的研究还不多见. 基于以上启发, 本文主要研究了与指数函数有关的星像函数类的四阶Hankel行列式H₄(1), 得到其上界估计.

1. 预备知识

定义1^[25] 设函数f和g在单位圆盘D内解析. 如果存在D内的Schwarz函数ω, 满足: ω(0)=0, |ω(z)| < 1且f(z)=g(ω(z)), 则称f从属于g, 记为f $\prec$ g. 特别地, 如果g在D上是单叶的, 则

$f\prec g \quad(z \in D) \Leftrightarrow f(0)=g(0), f(D) \subset g(D) .$

设函数f(z)∈ $\mathcal{A}$ , 若满足条件: Re[zf′(z)/f(z)]>0, 则称f属于星像函数类, 记为f∈S^*. 显然, f∈S^*将单位圆盘映射到右半平面且星像的区域^[1].

定义2^[4] 设S_e^*表示单位圆盘D={z: |z| < 1} 内满足

$\frac{z f^{\prime}(z)}{f(z)}\prec\mathrm{e}^{z} \quad(z \in D, f \in \mathcal{A})$

(2)

的解析函数类的全体. 实际上, f∈S_e^*当且仅当

$\left|\log \frac{z f^{\prime}(z)}{f(z)}\right|<1 \quad(z \in D).$

定义3^[5] 设函数f∈ $\mathcal{S}$ ,

$H_{q}(n)=\left|\begin{array}{cccc} a_{n} & a_{n+1} & \cdots & a_{n+q-1} \\ a_{n+1} & a_{n+2} & \cdots & a_{n+q} \\ \vdots & \vdots & & \vdots \\ a_{n+q-1} & a_{n+q} & \cdots & a_{n+2 q-2} \end{array}\right| \quad\left(a_{1}=1\right),$

其中, a₁=1, n≥1, q≥1. 特别地, 有

$H_{4}(1)=\left|\begin{array}{cccc} a_{1} & a_{2} & a_{3} & a_{4} \\ a_{2} & a_{3} & a_{4} & a_{5} \\ a_{3} & a_{4} & a_{5} & a_{6} \\ a_{4} & a_{5} & a_{6} & a_{7} \end{array}\right| \quad(n=1, q=4) .$

下面给出本文所需用的引理.

引理1^[26] 如果 $p(z)=1+\sum\limits_{k=1}^{\infty} c_{k} z^{k} \in \mathcal{P}$ , 则存在复数x和z，且|x|≤1, |z|≤1, 使得

$2 c_{2}=c_{1}^{2}+x\left(4-c_{1}^{2}\right)$

及

$4 c_{3}=c_{1}^{3}+2 c_{1} x\left(4-c_{1}^{2}\right)-\left(4-c_{1}^{2}\right) c_{1} x^{2}+2\left(4-c_{1}^{2}\right)\left(1-|x|^{2}\right) z.$

引理2^[27] 如果 $p(z)=1+\sum\limits_{k=1}^{\infty} c_{k} z^{k} \in \mathcal{P}$ , 则

$\begin{array}{c} \left|c_{1}^{4}+c_{2}^{2}+2 c_{1} c_{3}-3 c_{1}^{2} c_{2}-c_{4}\right| \leqslant 2,\\\left|c_{1}^{5}+3 c_{1} c_{2}^{2}+3 c_{1}^{2} c_{3}-4 c_{1}^{3} c_{2}-2 c_{1} c_{4}-2 c_{2} c_{3}+c_{5}\right| \leqslant 2, \end{array} \\ \mid c_{1}^{6}+6 c_{1}^{2} c_{2}^{2}+4 c_{1}^{3} c_{3}+2 c_{1} c_{5}+2 c_{2} c_{4}+c_{3}^{2}-c_{2}^{3}-5 c_{1}^{4} c_{2}- \\ \ \ \ \ \ \ \ \ 3 c_{1}^{2} c_{4}-6 c_{1} c_{2} c_{3}-c_{6} \mid \leqslant 2$

及|c_n|≤2 (n=1, 2, …).

引理3^[28] 如果 $p(z)=1+\sum\limits_{k=1}^{\infty} c_{k} z^{k} \in \mathcal{P}$ , 那么对0≤μ≤1, 有

$\begin{array}{c} \left|c_{2}-\frac{c_{1}^{2}}{2}\right| \leqslant 2-\frac{\left|c_{1}\right|^{2}}{2},\\\left|c_{n+k}-\mu c_{n} c_{k}\right|<2,\\\left|c_{n+2 k}-\mu c_{n} c_{k}^{2}\right| \leqslant 2(1+2 \mu) . \end{array}$

2. 主要结果

下面给出本文的主要定理.

定理1 若f(z)∈ $\mathcal{S}$ _e^*且具有式(1)的形式, 则

$\begin{gathered} \left|a_{2}\right| \leqslant 1,\left|a_{3}\right| \leqslant \frac{3}{4},\left|a_{4}\right| \leqslant \frac{31}{72},\left|a_{5}\right| \leqslant \frac{1}{3}, \\ \left|a_{6}\right| \leqslant \frac{17}{60},\left|a_{7}\right| \leqslant \frac{59}{80}. \end{gathered}$

(3)

证明设f∈S_e^*, 则由定义1和式(2), 易知

$\frac{z f^{\prime}(z)}{f(z)}=\mathrm{e}^{\omega(z)},$

其中, ω(z)是Schwarz函数, 满足ω(0)=0, |ω(z)| < 1, z∈D. 又

$\begin{aligned} &\frac{z f^{\prime}(z)}{f(z)}=\frac{z+\sum\limits_{n=2}^{\infty} n a_{n} z^{n}}{z+\sum\limits_{n=2}^{\infty} a_{n} z^{n}}= \\ &\ \ \ \ \left(1+\sum\limits_{n=2}^{\infty} n a_{n} z^{n-1}\right)\left[1-a_{2} z+\left(a_{2}^{2}-a_{3}\right) z^{2}-\left(a_{2}^{3}-2 a_{2} a_{3}+\right.\right. \\ &\ \ \ \ \left.\left.a_{4}\right) z^{3}+\left(a_{2}^{4}-3 a_{2}^{2} a_{3}+2 a_{2} a_{4}-a_{5}\right) z^{4}+\cdots\right]=1+a_{2} z+ \\ &\ \ \ \ \left(2 a_{3}-a_{2}^{2}\right) z^{2}+\left(a_{2}^{3}-3 a_{2} a_{3}+3 a_{4}\right) z^{3}+\left(4 a_{5}-a_{2}^{4}+4 a_{2}^{2} a_{3}-\right. \\ &\ \ \ \ \left.4 a_{2} a_{4}-3 a_{3}^{2}\right) z^{4}+\left(5 a_{6}-5 a_{2} a_{5}+a_{2}^{5}-5 a_{3} a_{4}-5 a_{2}^{3} a_{3}+\right. \\ &\ \ \ \ \left.5 a_{2}^{2} a_{4}+5 a_{2} a_{3}^{2}\right) z^{5}+\left(6 a_{7}-6 a_{2} a_{6}+6 a_{2}^{2} a_{5}-6 a_{3} a_{5}+\right. \\ &\ \ \ \ 18 a_{2} a_{3} a_{4}-a_{2}^{6}-6 a_{2}^{3} a_{4}-3 a_{4}^{2}+2 a_{3}^{3}-9 a_{2}^{2} a_{3}^{2}+6 a_{2}^{4} a_{3}- \\ &\ \ \ \ \left.3 a_{2} a_{4}\right) z^{6}+\cdots. \end{aligned}$

(4)

令

$p(z)=\frac{1+\omega(z)}{1-\omega(z)}=1+c_{1} z+c_{2} z^{2}+\cdots \text {. }$

显然有p(z)∈ $\mathcal{P}$ 且

$\omega(z)=\frac{p(z)-1}{1+p(z)}=\frac{c_{1} z+c_{2} z^{2}+c_{3} z^{3}+\cdots}{2+c_{1} z+c_{2} z^{2}+c_{3} z^{3}+\cdots}.$

(5)

由式(5)知,

$\begin{aligned} \mathrm{e}^{\omega(z)} &=1+\frac{1}{2} c_{1} z+\left(\frac{c_{2}}{2}-\frac{c_{1}^{2}}{8}\right) z^{2}+\left(\frac{c_{1}^{3}}{48}+\frac{2 c_{3}-c_{1} c_{2}}{4}\right) z^{3}+\\ &\left(\frac{2 c_{4}-c_{1} c_{3}}{4}+\frac{c_{1}^{2} c_{2}}{16}-\frac{c_{2}^{2}}{8}+\frac{c_{1}^{4}}{384}\right) z^{4}+\\ &\left(\frac{2 c_{5}-c_{1} c_{4}-c_{2} c_{3}}{4}+\frac{c_{1}^{2} c_{3}+c_{1} c_{2}^{2}}{16}+\frac{c_{1}^{3} c_{2}}{96}-\frac{19 c_{1}^{5}}{3\,840}\right) z^{5}+\\ &\left(\frac{2 c_{6}-c_{1} c_{5}-c_{2} c_{4}}{4}+\frac{c_{1} c_{2} c_{3}}{8}+\frac{c_{2}^{3}}{48}-\frac{c_{3}^{2}}{8}+\frac{151 c_{1}^{6}}{46\,080}-\frac{19 c_{1}^{4} c_{2}}{768}+\right.\\ &\left.\frac{c_{1}^{2} c_{2}^{2}}{64}+\frac{c_{1}^{2} c_{4}}{16}+\frac{c_{1}^{3} c_{3}}{96}\right) z^{6}+\cdots. \end{aligned}$

(6)

分别比较式(4)、(6)两边关于z、z²、z³、z⁴、z⁵、z⁶的系数, 可得

$\begin{gathered} a_{2}=\frac{c_{1}}{2}, a_{3}=\frac{c_{2}}{4}+\frac{c_{1}^{2}}{16}, a_{4}=\frac{c_{3}}{6}+\frac{c_{1} c_{2}}{24}-\frac{c_{1}^{3}}{288},\\ a_{5}=\frac{c_{4}}{8}+\frac{c_{1} c_{3}}{48}+\frac{c_{1}^{4}}{1\,152}-\frac{c_{1}^{2} c_{2}}{96}, \\ a_{6}=\frac{c_{1} c_{4}}{80}-\frac{c_{2} c_{3}}{120}-\frac{17 c_{1}^{5}}{57\,600}+\frac{11 c_{1}^{3} c_{2}}{2\,880}+\frac{c_{1}^{2} c_{3}}{480}-\frac{c_{1} c_{2}^{2}}{120}+\frac{c_{5}}{10}, \\ \begin{aligned} a_{7}=&\frac{-13 c_{1}^{2} c_{4}}{1\ 920}-\frac{97 c_{1} c_{2} c_{3}}{2\ 880}+\frac{1\,781 c_{1}^{6}}{8\ 294\ 400}+\frac{7 c_{1}^{2} c_{2}^{2}}{11\ 520}-\frac{341 c_{1}^{4} c_{2}}{138\ 240}- \\ &\frac{c_{1} c_{5}}{120}-\frac{c_{2} c_{4}}{96}-\frac{c_{2}^{3}}{576}-\frac{c_{3}^{2}}{144}+\frac{c_{1} c_{3}}{24}+\frac{c_{6}}{12}+\frac{c_{1}^{2} c_{2}}{96}-\frac{c_{1}^{4}}{1\ 152}+ \\ &\frac{211 c_{1}^{3} c_{3}}{34\ 560} . \end{aligned} \end{gathered}$

(7)

由引理2, 易证|a₂|=|c₁/2|≤1. 而由引理1, 可得

$\left|a_{3}\right|=\left|\frac{c_{2}}{4}+\frac{c_{1}^{2}}{16}\right|=\left|\frac{3 c_{1}^{2}}{16}+\frac{x\left(4-c_{1}^{2}\right)}{8}\right|$

设|x|=t (t∈[0, 1]), c₁=c (c ∈[0, 2]), 则利用三角不等式, 得

$\left|a_{3}\right| \leqslant \frac{t\left(4-c^{2}\right)}{8}+\frac{3 c^{2}}{16}.$

令 $F(c, t)=\frac{t\left(4-c^{2}\right)}{8}+\frac{3 c^{2}}{16}$ , 对任意的t ∈(0, 1), c ∈(0, 2), 有

$\frac{\partial F}{\partial t}=\frac{4-c^{2}}{8}>0,$

故F(c, t)在[0, 1]关于t是单调递增函数. 因此, F(c, t)在t=1取得最大值, 即

$\max F(c, t)=F(c, 1)=\frac{4-c^{2}}{8}+\frac{3 c^{2}}{16}=\frac{1}{2}+\frac{c^{2}}{16}=G(c).$

易证G(c)在c=2处取得最大值, 即|a₃|≤G(2)=3/4.

又

$\left|a_{4}\right|=\left|\frac{c_{3}}{6}+\frac{c_{1} c_{2}}{24}-\frac{c_{1}^{3}}{288}\right|=\left|\frac{-49 c_{3}}{288}-\frac{c_{1} c_{2}}{48}+\frac{c_{1}^{3}-2 c_{1} c_{2}+c_{3}}{288}\right|.$

令c₁=c∈[0, 2], 则由引理2可得

$\left|a_{4}\right|=\left|\frac{-49 c_{3}}{288}-\frac{c_{1} c_{2}}{48}+\frac{c_{1}^{3}-2 c_{1} c_{2}+c_{3}}{288}\right| \leqslant \frac{25}{72}+\frac{c}{24} \text {. }$

令 $F(c)=\frac{25}{72}+\frac{c}{24}$ , 易证F(c)在c=2处取得最大值, 即|a₄|≤F(2)=31/72.

类似地, 因为

$\begin{aligned} \left|a_{5}\right|=&\left|\frac{c_{4}}{8}+\frac{c_{1} c_{3}}{48}+\frac{c_{1}^{4}}{1\,152}-\frac{c_{1}^{2} c_{2}}{96}\right|=\\ &\left|\frac{23 c_{1}}{1\,152}\left[c_{3}-\frac{10 c_{1} c_{2}}{23}\right]+\frac{c_{4}}{8}+\frac{c_{1}\left(c_{1}^{3}-2 c_{1} c_{2}+c_{3}\right)}{1\ 152}\right|, \end{aligned}$

故对c₁=c∈[0, 2], 由引理2和引理3可得

$\left|a_{5}\right| \leqslant \frac{1}{4}+\frac{c}{576}+\frac{23 c}{576}.$

令 $F(c)=\frac{1}{4}+\frac{c}{24}$ , 进而可得F(c)在c=2处取得最大值, 即|a₅|≤F(2)=1/3.

又

$\begin{aligned} \left|a_{6}\right|&=\left|\frac{c_{1} c_{4}}{80}-\frac{c_{2} c_{3}}{120}-\frac{17 c_{1}^{5}}{57\ 600}+\frac{11 c_{1}^{3} c_{2}}{2\ 880}+\frac{c_{1}^{2} c_{3}}{480}-\frac{c_{1} c_{2}^{2}}{120}+\frac{c_{5}}{10}\right|= \\ &\left| \frac{1}{20}\left[c_{5}-\frac{c_{1} c_{4}}{4}\right]+\frac{1}{20}\left[c_{5}-\frac{c_{3} c_{2}}{12}\right]-\frac{c_{3}}{240}\left[c_{2}-\frac{c_{1}^{2}}{2}\right]+ \right. \\ &\left.\frac{17 c_{1}^{3}}{28\ 800}\left[c_{2}-\frac{c_{1}^{2}}{2}\right]-\frac{c_{1} c_{2}}{120}\left(c_{2}-\frac{c_{1}^{2}}{2}\right)-\frac{17 c_{1}^{3} c_{2}}{28\ 800}\right|. \end{aligned}$

令c₁=c∈[0, 2], 由引理3可得

$\left|a_{6}\right| \leqslant \frac{1}{5}+\frac{17 c^{3}\left(2-c^{2} / 2\right)}{28\ 800}+\frac{5}{120}\left(2-\frac{c^{2}}{2}\right)+\frac{17 c^{3}}{14\ 400}.$

令

$F(c)=\frac{1}{5}+\frac{17 c^{3}\left(2-c^{2} / 2\right)}{28\ 800}+\frac{5}{120}\left(2-\frac{c^{2}}{2}\right)+\frac{17 c^{3}}{14\ 400},$

则

$F^{\prime}(c)=\frac{17 c^{2}}{3\ 600}-\frac{17 c^{4}}{11\ 520}-\frac{c}{24},$

因此, c=0是方程F′(c)=0的根. 又因为F″(0) < 0, 所以, F(c)在c=0处取得最大值, 即|a₆|≤F(0)=17/60.

又

$\begin{aligned} &\left|a_{7}\right|=\left|\frac{-13 c_{1}^{2} c_{4}}{1\ 920}-\frac{97 c_{1} c_{2} c_{3}}{2\ 880}+\frac{1\ 781 c_{1}^{6}}{8\ 294\ 400}+\frac{7 c_{1}^{2} c_{2}^{2}}{11\ 520}-\frac{341 c_{1}^{4} c_{2}}{138\ 240}-\right.\\ &\left.\frac{c_{1} c_{5}}{120}-\frac{c_{2} c_{4}}{96}-\frac{c_{2}^{3}}{576}-\frac{c_{3}^{2}}{144}+\frac{c_{1} c_{3}}{24}+\frac{c_{6}}{12}+\frac{c_{1}^{2} c_{2}}{96}-\frac{c_{1}^{4}}{1\ 152}+\frac{211 c_{1}^{3} c_{3}}{34\ 560}\right|=\\ &\left|\frac{1\ 781 c_{1}^{4}}{4\ 147\ 200}\left[c_{2}-c_{2}^{2}\right]+\frac{c_{1}^{2}}{576}\left[c_{2}-c_{1}^{2}\right]-\frac{13 c_{1}^{2}\left[c_{4}-\frac{211 c_{1} c_{3}}{234}\right]}{1\ 920}+\right.\\ &\left.\frac{\left[c_{6}-\frac{c_{2} c_{4}}{8}\right]}{12}+\frac{c_{1}\left[c_{5}-\frac{23}{24} c_{2} c_{3}\right]}{120}-\frac{37 c_{1} c_{2}\left[c_{3}-\frac{7 c_{1} c_{2}}{296}\right]}{1440}+\frac{5 c_{1}^{2} c_{2}}{576}-\right.\\ &\left.\frac{c_{2}^{3}}{576}-\frac{c_{3}^{2}}{144}+\frac{c_{1} c_{3}}{24} \right| . \end{aligned}$

令c₁=c∈[0, 2], 由引理2和引理3可知

$\begin{aligned} &\left|a_{7}\right| \leqslant \frac{1}{6}+\frac{c}{60}+\frac{37 c}{360}+\frac{1\ 781 c^{4}\left(2-\frac{c^{2}}{2}\right)}{4\ 147\ 200}+\frac{c^{2}\left(2-\frac{c^{2}}{2}\right)}{576}+\frac{13 c^{2}}{960}+ \\ &\ \ \ \ \ \ \ \ \frac{1}{72}+\frac{1}{36}+\frac{5 c^{2}}{288}+\frac{c}{12} . \end{aligned}$

令

$\begin{aligned} F(c)=&\frac{1}{6}+\frac{c}{60}+\frac{37 c}{360}+\frac{1\ 781 c^{4}\left(2-\frac{c^{2}}{2}\right) c^{2}\left(2-\frac{c^{2}}{2}\right)}{4\ 147\ 200}+\frac{13 c^{2}}{576}+\frac{5}{960}+ \\ &\frac{1}{72}+\frac{1}{36}+\frac{5 c^{2}}{288}+\frac{c}{12}, \end{aligned}$

进而可得F′(c)≥0. 因此, F(c)在c=2处取得最大值, 即|a₇|≤F(2)=59/80. 证毕.

定理2 若f∈S_e^*且具有式(1)的形式, 则

$\left|a_{3}-a_{2}^{2}\right| \leqslant \frac{1}{2}$

(8)

证明通过式(7)可得

$\left|a_{3}-a_{2}^{2}\right|=\left|\frac{c_{2}}{4}-\frac{3 c_{1}^{2}}{16}\right|,$

由引理1可知

$\left|a_{3}-a_{2}^{2}\right|=\left|\frac{x\left(4-c_{1}^{2}\right)}{8}-\frac{c_{1}^{2}}{16}\right|.$

设|x|=t∈[0, 1], c₁=c∈[0, 2], 则由三角不等式可得

$\left|a_{3}-a_{2}^{2}\right| \leqslant \frac{t\left(4-c^{2}\right)}{8}+\frac{c^{2}}{16}.$

令 $F(c, t)=\frac{t\left(4-c^{2}\right)}{8}+\frac{c^{2}}{16}$ , 则对∀t ∈(0, 1), ∀c ∈(0, 2), 有

$\frac{\partial F}{\partial t}=\frac{4-c^{2}}{8}>0$

因此, F(c, t)在[0, 1]关于t是单调递增函数. 故F(c, t)在t=1处取得最大值, 即

$\max F(c, t)=F(c, 1)=\frac{\left(4-c^{2}\right)}{8}+\frac{c^{2}}{16}=G(c) .$

同理易证G(c)在c=0处取得最大值, 即

$\left|a_{3}-a_{2}^{2}\right| \leqslant G(0)=\frac{1}{2}.$

证毕.

定理3 若f∈S_e^*且具有式(1)的形式, 则

$\left|a_{2} a_{3}-a_{4}\right| \leqslant \frac{1}{3}$

(9)

证明由式(7)可得

$\begin{aligned} &\left|a_{2} a_{3}-a_{4}\right|=\left|\frac{c_{1} c_{2}}{8}+\frac{5 c_{1}^{3}}{144}-\frac{c_{3}}{6}+\frac{c_{1} c_{2}}{12}\right|= \\ &\ \ \ \ \ \ \ \ \left|\frac{5\left(c_{1}^{3}-2 c_{1} c_{2}+c_{3}\right)}{144}-\frac{29\left[c_{3}-\frac{22}{29} c_{1} c_{2}\right]}{144}\right| . \end{aligned}$

由引理2和引理3可得

$\left|a_{2} a_{3}-a_{4}\right| \leqslant \frac{5}{72}+\frac{29}{72}=\frac{17}{36}.$

证毕.

定理4 若f∈S_e^*且具有式(1)的形式, 则

$\left|a_{2} a_{4}-a_{3}^{2}\right| \leqslant \frac{1}{4}.$

(10)

证明由式(7)可得

$\left|a_{2} a_{4}-a_{3}^{2}\right|=\left|\frac{c_{1} c_{3}}{12}-\frac{c_{1}^{2} c_{2}}{48}-\frac{c_{1}^{4}}{288}-\frac{c_{2}^{2}}{16}\right|.$

由引理1可得

$\begin{aligned} &\left|a_{2} a_{4}-a_{3}^{2}\right|=\left|\frac{c_{1} c_{3}}{12}-\frac{c_{1}^{2} c_{2}}{48}-\frac{c_{1}^{4}}{288}-\frac{c_{2}^{2}}{16}\right|= \\ &\ \ \ \ \left|-\frac{5 c_{1}^{4}}{576}-\frac{x^{2} c_{1}^{2}\left(4-c_{1}^{2}\right)}{48}-\frac{x^{2}\left(4-c_{1}^{2}\right)^{2}}{64}+\frac{c_{1}\left(4-c_{1}^{2}\right)\left(1-|x|^{2}\right) z}{24}\right| . \end{aligned}$

设|x|=t∈[0, 1], c₁=c∈[0, 2], 则由三角不等式可得

$\begin{aligned} &\left|a_{2} a_{4}-a_{3}^{2}\right| \leqslant \frac{t^{2} c^{2}\left(4-c^{2}\right)}{48}+\frac{\left(1-t^{2}\right) c\left(4-c^{2}\right)}{24}+ \\ &\ \ \ \ \ \frac{t^{2}\left(4-c^{2}\right)^{2}}{64}+\frac{5 c^{4}}{576}. \end{aligned}$

令

$F(c, t)=\frac{t^{2} c^{2}\left(4-c^{2}\right)}{48}+\frac{\left(1-t^{2}\right) c\left(4-c^{2}\right)}{24}+\frac{t^{2}\left(4-c^{2}\right)^{2}}{64}+\frac{5 c^{4}}{576},$

则对∀t∈(0, 1), ∀c∈(0, 2), 有

$\frac{\partial F}{\partial t}=\frac{t\left(c^{2}-8 c+12\right)\left(4-c^{2}\right)}{96}>0.$

因此, F(c, t)在[0, 1]上关于t是单调递增函数, 从而可得F(c, t)在t=1处取得最大值, 即

$\max F(c, t)=F(c, 1)=\frac{c^{2}\left(4-c^{2}\right)}{48}+\frac{\left(4-c^{2}\right)^{2}}{64}+\frac{5 c^{4}}{576}.$

设

$G(c)=\frac{c^{2}\left(4-c^{2}\right)}{48}+\frac{\left(4-c^{2}\right)^{2}}{64}+\frac{5 c^{4}}{576} ,$

则

$G^{\prime}(c)=\frac{c\left(4-c^{2}\right)}{24}-\frac{c^{3}}{24}-\frac{c\left(4-c^{2}\right)}{16}+\frac{5 c^{3}}{144} .$

若G′(c)=0, 则c=0. 又因为G″(0)=-1/12 < 0, 故G(c)在c=0处取得最大值, 即|a₂a₄-a₃²|≤G(0)=1/4.证毕.

定理5 若f∈S_e^*且具有式(1)的形式, 则

$\left|a_{2} a_{5}-a_{3} a_{4}\right| \leqslant \frac{73}{144}$

(11)

证明若f(z)∈ $\mathcal{S}$ _e^*, 则由式(7)可得

$\begin{aligned} &\left|a_{2} a_{5}-a_{3} a_{4}\right|=\left|\frac{-c_{1}^{5}}{1\ 536}-\frac{c_{1} c_{4}}{16}+\frac{c_{1} c_{2}^{2}}{96}+\frac{c_{1}^{3} c_{2}}{144}+\frac{c_{2} c_{3}}{24}\right|= \\ &\ \ \ \ \ \ \ \ \left| \frac{c_{1}^{3}\left[c_{2}-\frac{c_{1}^{2}}{2}\right]}{144}+ \frac{c_{2}\left[c_{3}-\frac{c_{1} c_{2}}{2}\right]}{24} -\frac{c_{1}\left[c_{4}-\frac{c_{1}^{2}}{2}\right]}{16}+\frac{13 c_{1}^{5}}{4\ 608}\right| . \end{aligned}$

令c₁=c∈[0, 2], 由引理3可得

$\left|a_{2} a_{5}-a_{3} a_{4}\right| \leqslant \frac{c^{3}\left[2-\frac{c^{2}}{2}\right]}{144}+\frac{1}{6}+\frac{13 c^{5}}{4\ 608}+\frac{c}{8} \text {. }$

令

$F(c)=\frac{c^{3}\left[2-\frac{c^{2}}{2}\right]}{144}+\frac{1}{6}+\frac{13 c^{5}}{4\ 608}+\frac{c}{8},$

则对∀c∈(0, 2), 有

$F^{\prime}(c)=\frac{c^{2}}{24}-\frac{5 c^{4}}{1\ 536}+\frac{1}{8}>0 .$

因此, F(c)关于c单调递增, 故F(c)在c=2处取得最大值, 即

$\left|a_{2} a_{5}-a_{3} a_{4}\right| \leqslant F(2)=\frac{73}{144}.$

证毕.

定理6 若f∈S_e^*且具有式(1)的形式, 则

$\left|a_{5}-a_{2} a_{4}\right| \leqslant \frac{35}{96}.$

(12)

证明若f(z)∈ $\mathcal{S}$ _e^*, 则由式(7)可得

$\begin{aligned} &\left|a_{5}-a_{2} a_{4}\right|=\left|\frac{c_{1}^{4}}{384}-\frac{c_{1} c_{3}}{16}-\frac{c_{1}^{2} c_{2}}{32}+\frac{c_{4}}{8}\right|= \\ &\qquad\left|\frac{c_{1}\left[c_{1}^{3}-2 c_{1} c_{2}+c_{3}\right]}{384}+\frac{5 c_{1}\left[c_{3}-c_{1} c_{2}\right]}{192}-\frac{\left[c_{4}-\frac{35}{48} c_{1} c_{3}\right]}{8}\right|. \end{aligned}$

令c₁=c∈[0, 2], 由引理2和引理3可得

$\left|a_{5}-a_{2} a_{4}\right| \leqslant \frac{c}{192}+\frac{5 c}{96}+\frac{1}{4} .$

令 $F(c)=\frac{1}{4}+\frac{11 c}{192}$ , 则易证F(c)在c=2处取得最大值, 即

$\left|a_{5}-a_{2} a_{4}\right| \leqslant F(2)=\frac{35}{96}.$

证毕.

定理7 若f∈S_e^*且具有式(1)的形式, 则

$\left|a_{5} a_{3}-a_{4}^{2}\right| \leqslant \frac{121}{216}.$

(13)

证明若f(z)∈ $\mathcal{S}$ _e^*, 则由式(7)可得

$\begin{aligned} &\left|a_{5} a_{3}-a_{4}^{2}\right|=\left| \frac{-c_{1}^{4} c_{2}}{6\ 912}+\frac{c_{2} c_{4}}{32}-\frac{5 c_{1} c_{2} c_{3}}{576}+\frac{17 c_{1}^{3} c_{3}}{6\ 912}-\frac{5 c_{1}^{2} c_{2}^{2}}{1\ 152}-\frac{c_{3}^{2}}{36}+\right. \\ &\ \ \ \ \left.\frac{c_{1}^{2} c_{4}}{128}+\frac{7 c_{1}^{6}}{165\ 888}\right|=\left| \frac{c_{3}\left[c_{3}-\frac{5 c_{1} c_{2}}{16}\right] c_{4}\left[c_{2}-\frac{c_{1}^{2}}{2}\right]}{36}+\frac{c_{3}^{2}}{32}- \right.\\ &\ \ \ \ \left.\frac{7 c_{1}^{4}\left[c_{2}-\frac{c_{1}^{2}}{2}\right]}{82\ 944}+\frac{17 c_{1}^{3}\left[c_{3}-\frac{5}{204} c_{1} c_{2}\right]}{6\ 912}+\frac{3 c_{1}^{2}\left[c_{4}-\frac{5 c_{2}^{2}}{27}\right]}{128}-\frac{c_{3}^{2}}{18}\right|. \end{aligned}$

令c₁=c∈[0, 2], 由引理3可得

$\left|a_{5} a_{3}-a_{4}^{2}\right| \leqslant \frac{1}{9}+\frac{2}{9}+\frac{\left[2-\frac{c^{2}}{2}\right]}{16}+\frac{7 c^{4}\left[2-\frac{c^{2}}{2}\right]}{82\ 944}+\frac{17 c^{3}}{3\ 456}+\frac{3 c^{2}}{64} \text {. }$

令

$F(c)=\frac{1}{9}+\frac{2}{9}+\frac{\left[2-\frac{c^{2}}{2}\right]}{16}+\frac{7 c^{4}\left[2-\frac{c^{2}}{2}\right]}{82\ 944}+\frac{17 c^{3}}{3\ 456}+\frac{3 c^{2}}{64},$

则对∀c∈(0, 2), 有

$F^{\prime}(c)=\frac{c}{32}+\frac{17 c^{2}}{1\ 152}+\frac{7 c^{3}}{10\ 368}-\frac{7 c^{5}}{27\ 648}>0 .$

因此, F(c)关于c单调递增, 故F(c)在c=2处取得最大值, 即

$\left|a_{5} a_{3}-a_{4}^{2}\right| \leqslant F(2)=\frac{121}{216}.$

证毕.

定理8 若f∈S_e^*且具有式(1)的形式, 则

$\left|H_{4}(1)\right| \leqslant \frac{52\ 253\ 339}{55\ 987\ 200} \approx 0.933\ 3.$

(14)

证明因为

$\begin{aligned} H_{4}&(1)=a_{7}\left\{a_{3}\left(a_{2} a_{4}-a_{3}^{2}\right)-a_{4}\left(a_{4}-a_{2} a_{3}\right)+a_{5}\left(a_{3}-a_{2}^{2}\right)\right\}- \\ &a_{6}\left\{a_{3}\left(a_{2} a_{5}-a_{3} a_{4}\right)-a_{4}\left(a_{5}-a_{2} a_{4}\right)+a_{6}\left(a_{3}-a_{2}^{2}\right)\right\}+ \\ &a_{5}\left\{a_{3}\left(a_{3} a_{5}-a_{4}^{2}\right)-a_{5}\left(a_{5}-a_{2} a_{4}\right)+a_{6}\left(a_{4}-a_{2} a_{3}\right)\right\}- \\ &a_{4}\left\{a_{4}\left(a_{3} a_{5}-a_{4}^{2}\right)-a_{5}\left(a_{2} a_{5}-a_{3} a_{4}\right)+a_{6}\left(a_{4}-a_{2} a_{3}\right)\right\}, \end{aligned}$

所以, 由三角不等式可得

$\begin{aligned} &\left|H_{4}(1)\right| \leqslant\left|a_{7}\right|\left|a_{3}\right|\left|a_{2} a_{4}-a_{3}^{2}\right|+\left|a_{7}\right|\left|a_{4}\right|\left|a_{4}-a_{2} a_{3}\right|+\\ &\ \ \ \ \left|a_{7}\right|\left|a_{5}\right|\left|a_{3}-a_{2}^{2}\right|+\left|a_{6}\right|\left|a_{3}\right|\left|a_{2} a_{5}-a_{3} a_{4}\right|+ \\ &\ \ \ \ \left|a_{6}\right|\left|a_{4}\right|\left|a_{5}-a_{2} a_{4}\right|+\left|a_{6}\right|^{2}\left|a_{3}-a_{2}^{2}\right|+ \\ &\ \ \ \ \left|a_{5}\right|\left|a_{3}\right|\left|a_{3} a_{5}-a_{4}^{2}\right|+\left|a_{5}\right|^{2}\left|a_{5}-a_{2} a_{4}\right|+ \\ &\ \ \ \ \left|a_{5}\right|\left|a_{6}\right|\left|a_{4}-a_{2} a_{3}\right|+\left|a_{4}\right|^{2}\left|a_{3} a_{5}-a_{4}^{2}\right|+ \\ &\ \ \ \ \left|a_{4}\right|\left|a_{5}\right|\left|a_{2} a_{5}-a_{3} a_{4}\right|+\left|a_{4}\right|\left|a_{6}\right|\left|a_{4}-a_{2} a_{3}\right|. \end{aligned}$

(15)

将式(3)、(8)~(13)代入到式(15), 即得式(14). 证毕.

[1]Zhang L, Zhou X, Guo X, Song X, Liu X.Investigation on the degradation of acid fuchsin induced oxidation by MgFe 2lixiaoyansubend O 4lixiaoyansubend under microwave irradiation[J].Journal of Molecular Catalysis A: Chemical, 2011, 335(1):31-37
[2]Arulkumar M, Sathishkumar P, Palvannan T.Optimization of Orange G dye adsorption by activated carbon of Thespesia populnea pods using response surface methodology[J].Journal of Hazardous Materials, 2011, 186(1):827-834
[3] Stadler L B, Su L, Moline C J, Ernstoff A S, Aga D S, Love N G.Effect of redox conditions on pharmaceutical loss during biological wastewater treatment using sequencing batch reactors[J]. Journal of Hazardous Materials, 2014.
[4]Belaid C, Khadraoui M, Mseddi S, Kallel M, Elleuch B, Fauvarque J F.Electrochemical treatment of olive mill wastewater: Treatment extent and effluent phenolic compounds monitoring using some uncommon analytical tools[J].Journal of Environmental Sciences, 2013, 25(1):220-230
[5]Aber S, Sheydaei M.Removal of COD from industrial effluent containing indigo dye using adsorption method by activated carbon cloth: optimization,kinetic and isotherm studies[J].Clean–Soil, Air, Water, 2012, 40(1):87-94
[6]Li Y, Du Q, Liu T, Peng X, Wang J, Sun J, Wang Y, Wu S, Wang Z, Xia Y.Comparative study of methylene blue dye adsorption onto activated carbon,graphene oxide,and carbon nanotubes[J].Chemical Engineering Research and Design, 2013, 91(2):361-368
[7]Summers R S, Kim S M, Shimabuku K, Chae S-H, Corwin C J.Granular activated carbon adsorption of MIB in the presence of dissolved organic matter[J].Water research, 2013, 47(10):3507-3513
[8]Park K-H, Balathanigaimani M, Shim W-G, Lee J-W, Moon H.Adsorption characteristics of phenol on novel corn grain-based activated carbons[J].Microporous and Mesoporous Materials, 2010, 127(1):1-8
[9]Mohamad Nor N, Lau L C, Lee K T, Mohamed A R.Synthesis of activated carbon from lignocellulosic biomass and its applications in air pollution control—a review[J].Journal of Environmental Chemical Engineering, 2013, 1(4):658-666
[10]Rathinam A, Rao J R, Nair B U.Adsorption of phenol onto activated carbon from seaweed: determination of the optimal experimental parameters using factorial design[J].Journal of the Taiwan Institute of Chemical Engineers, 2011, 42(6):952-956
[11] Akar E, Altini?ik A, Seki Y.Using of activated carbon produced from spent tea leaves for the removal of malachite green from aqueous solution[J]. Ecological Engineering, 2013, 52:19-27.
[12] Wang Z.Efficient adsorption of dibutyl phthalate from aqueous solution by activated carbon developed from phoenix leaves[J]. International Journal of Environmental Science and Technology, 2014, 1-10.
[13] Njoku V, Islam M A, Asif M, Hameed B.Preparation of mesoporous activated carbon from coconut frond for the adsorption of carbofuran insecticide[J]. Journal of Analytical and Applied Pyrolysis, 2014.
[14]Li W-H, Yue Q-Y, Gao B-Y, Ma Z-H, Li Y-J, Zhao H-X.Preparation and utilization of sludge-based activated carbon for the adsorption of dyes from aqueous solutions[J].Chemical Engineering Journal, 2011, 171(1):320-327
[15]Sze M, McKay G.An adsorption diffusion model for removal of para-chlorophenol by activated carbon derived from bituminous coal[J].Environmental Pollution, 2010, 158(5):1669-1674
[16]Chen W, Zhang H, Huang Y, Wang W.A fish scale based hierarchical lamellar porous carbon material obtained using a natural template for high performance electrochemical capacitors[J].Journal of Materials Chemistry, 2010, 20(23):4773-4775
[17]Lim P N, Chang L, Tay B Y, Guneta V, Choong C, Ho B, Thian E S.Proposed Mechanism of Antibacterial Action of Chemically Modified Apatite for Reduced Bone Infection[J].ACS Applied Materials & Interfaces, 2014, 6(19):17082-17092
[18] Hong X, Hui K, Zeng Z, Hui K, Zhang L, Mo M, Li M.Hierarchical nitrogen-doped porous carbon with high surface area derived from endothelium corneum gigeriae galli for high-performance supercapacitor[J]. Electrochimica Acta, 2014, 130:464-469.