电场作用下金属串配合物[Ru3(dpa)4]L2(L=Cl,C≡N,C≡CPh)结构的理论研究
Theoretical Study on the Structures of Metal String Complexes [Ru_3(dpa)_4]L_2(L=Cl, C≡N, C≡CPh) under the Electric Field
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关键词:
- 金属串配合物 /
- 密度泛函理论 /
- 电场 /
- Ru^6+_3离域多重键 /
- 分子导线
Abstract: As a potential molecular wire, metal string complexes [Ru_3(dpa)_4]L_2(1: L=Cl, 2: CN, 3: CCPh) under an external electric field along the Ru^6+_3 chain were investigated with the density functional theory at BP86 level and the Natural Bond Orbital Theo -
设A表示单位圆盘D={z∈C: |z| < 1}内解析且具有如下形式
f(z)=z+∞∑n=2anzn (1) 的函数族.
设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∈S∗(ϕ)⇔zf′(z)f(z)≺ϕ(z), 其中ϕ ∈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*(√1+z), 该函数类将单位圆盘映射为伯努利双纽线{w∈C: |w2-1| < 1}的右半部分星像区域.
2014年, MENDIRATTA等[3]引入函数类S*(√2−(√2−1)√(1−z)/(1+2(√2−1))), 该函数类将单位圆盘映射为伯努利双纽线{w ∈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Se*, 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Hq(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≺g. 特别地, 如果g在D上是单叶的, 则
f≺g(z∈D)⇔f(0)=g(0),f(D)⊂g(D). 设函数f(z)∈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Se*表示单位圆盘D={z: |z| < 1} 内满足
zf′(z)f(z)≺ez(z∈D,f∈A) (2) 的解析函数类的全体. 实际上, f∈Se*当且仅当
|logzf′(z)f(z)|<1(z∈D). 定义3[5] 设函数f∈S,
Hq(n)=|anan+1⋯an+q−1an+1an+2⋯an+q⋮⋮⋮an+q−1an+q⋯an+2q−2|(a1=1), 其中, a1=1, n≥1, q≥1. 特别地, 有
H4(1)=|a1a2a3a4a2a3a4a5a3a4a5a6a4a5a6a7|(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+∞∑k=1ckzk∈P, 则存在复数x和z,且|x|≤1, |z|≤1, 使得
2c2=c21+x(4−c21) 及
4c3=c31+2c1x(4−c21)−(4−c21)c1x2+2(4−c21)(1−|x|2)z. 引理2[27] 如果p(z)=1+∞∑k=1ckzk∈P, 则
|c41+c22+2c1c3−3c21c2−c4|⩽2,|c51+3c1c22+3c21c3−4c31c2−2c1c4−2c2c3+c5|⩽2,∣c61+6c21c22+4c31c3+2c1c5+2c2c4+c23−c32−5c41c2− 3c21c4−6c1c2c3−c6∣⩽2 及|cn|≤2 (n=1, 2, …).
引理3[28] 如果p(z)=1+∞∑k=1ckzk∈P, 那么对0≤μ≤1, 有
|c2−c212|⩽2−|c1|22,|cn+k−μcnck|<2,|cn+2k−μcnc2k|⩽2(1+2μ). 2. 主要结果
下面给出本文的主要定理.
定理1 若f(z)∈Se*且具有式(1)的形式, 则
|a2|⩽1,|a3|⩽34,|a4|⩽3172,|a5|⩽13,|a6|⩽1760,|a7|⩽5980. (3) 证明 设f∈Se*, 则由定义1和式(2), 易知
zf′(z)f(z)=eω(z), 其中, ω(z)是Schwarz函数, 满足ω(0)=0, |ω(z)| < 1, z∈D. 又
zf′(z)f(z)=z+∞∑n=2nanznz+∞∑n=2anzn= (1+∞∑n=2nanzn−1)[1−a2z+(a22−a3)z2−(a32−2a2a3+ a4)z3+(a42−3a22a3+2a2a4−a5)z4+⋯]=1+a2z+ (2a3−a22)z2+(a32−3a2a3+3a4)z3+(4a5−a42+4a22a3− 4a2a4−3a23)z4+(5a6−5a2a5+a52−5a3a4−5a32a3+ 5a22a4+5a2a23)z5+(6a7−6a2a6+6a22a5−6a3a5+ 18a2a3a4−a62−6a32a4−3a24+2a33−9a22a23+6a42a3− 3a2a4)z6+⋯. (4) 令
p(z)=1+ω(z)1−ω(z)=1+c1z+c2z2+⋯. 显然有p(z)∈P且
ω(z)=p(z)−11+p(z)=c1z+c2z2+c3z3+⋯2+c1z+c2z2+c3z3+⋯. (5) 由式(5)知,
eω(z)=1+12c1z+(c22−c218)z2+(c3148+2c3−c1c24)z3+(2c4−c1c34+c21c216−c228+c41384)z4+(2c5−c1c4−c2c34+c21c3+c1c2216+c31c296−19c513840)z5+(2c6−c1c5−c2c44+c1c2c38+c3248−c238+151c6146080−19c41c2768+c21c2264+c21c416+c31c396)z6+⋯. (6) 分别比较式(4)、(6)两边关于z、z2、z3、z4、z5、z6的系数, 可得
a2=c12,a3=c24+c2116,a4=c36+c1c224−c31288,a5=c48+c1c348+c411152−c21c296,a6=c1c480−c2c3120−17c5157600+11c31c22880+c21c3480−c1c22120+c510,a7=−13c21c41 920−97c1c2c32 880+1781c618 294 400+7c21c2211 520−341c41c2138 240−c1c5120−c2c496−c32576−c23144+c1c324+c612+c21c296−c411 152+211c31c334 560. (7) 由引理2, 易证|a2|=|c1/2|≤1. 而由引理1, 可得
|a3|=|c24+c2116|=|3c2116+x(4−c21)8| 设|x|=t (t∈[0, 1]), c1=c (c ∈[0, 2]), 则利用三角不等式, 得
|a3|⩽t(4−c2)8+3c216. 令F(c,t)=t(4−c2)8+3c216, 对任意的t ∈(0, 1), c ∈(0, 2), 有
∂F∂t=4−c28>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 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∈Se*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∈Se*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∈Se*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∈Se*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∈Se*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∈Se*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∈Se*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.
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