一元二次方程的根系关系

一、启发提问一元二次方程ａｘ２＋ｂｘ＋ｃ＝０（ａ≠０）的求根公式的推导过程中知道实数根的个数是由方程的系数ａ、ｂ、ｃ（△＝ｂ２－４ａｃ）决定时,当△≥０,方程有两个实数根：ｘ１＝－ｂ＋ｂ２－４ａｃ２ａ,ｘ２＝－ｂ－ｂ２－４ａｃ２ａ,比较ｘ１和ｘ２式中的结构,你发现了什么？１．分母相同,为２ａ２．分子－ｂ－ｂ２＋４ａｃ与－ｂ＋ｂ２－４ａｃ是互为共轭根式,３．计算：ｘ１＋ｘ２＝－ｂ＋ｂ２－４ａｃ２ａ＋－ｂ－ｂ２－４ａｃ２ａ＝,ｘ１·ｘ２＝－ｂ＋ｂ２－４ａｃ２ａ·－ｂ－ｂ２－４ａｃ２ａ＝．二、读书自学…     

关于一个题目的再修正

文［１］曾指出一个流行题目：“已知ａ,ｂ,α∈Ｒ,ａ,ｂ同号且ａ＞ｂ,求证：ａ－ｂａ＋ｂ≤ａ＋ｂｓｉｎαａ－ｂｓｉｎα≤ａ＋ｂａ－ｂ．”是一道错题．然后将题目改为“已知：α∈Ｒ,ａ＞ｂ≥０,求证：ａ－ｂａ＋ｂ≤ａ＋ｂｓｉｎαａ－ｂｓｉｎα≤ａ＋ｂａ－ｂ．　（※）其实,题目改后仍有失全面性和完整性．因为,当ａ＜ｂ≤０时,有－ａ＞－ｂ≥０,此时应用改后题目的结论（※）有：－ａ＋ｂ－ａ－ｂ≤－ａ－ｂｓｉｎα－ａ＋ｂｓｉｎα≤－ａ－ｂ－ａ＋ｂ．进一步化简得：－（ａ－ｂ）－（ａ＋ｂ）≤－（ａ＋ｂｓｉｎα）…     

一道IMO题的对偶式与面积解释

第２４届ＩＭＯ第６题是：在△ＡＢＣ中，ａ、ｂ、ｃ是三边长，求证：ａ２ｂ（ａ－ｂ）＋ｂ２ｃ（ｂ－ｃ）＋ｃ２ａ（ｃ－ａ）≥０．（１）文［１］指出了它的下述对偶形式：ａｂ２（ａ－ｂ）＋ｂｃ２（ｂ－ｃ）＋ｃａ２（ｃ－ａ）≤０，（２）并给出了统一的距离解释．即不等式（１）、（２）的几何解释为：三角形内Ｂｒｏｃａｒｄ点到内心的距离非负．受此启发，笔者研究了第６届ＩＭＯ第２题：在△ＡＢＣ中，ａ、ｂ、ｃ是三边长，求证：　ａ２（ｂ＋ｃ－ａ）＋ｂ２（ａ＋ｃ－ｂ）＋ｃ２（ａ＋ｂ－ｃ）≤３ａｂｃ，（３）发现它也有如下的…     

一类无理函数值域问题的统一解法

不少文献研究了无理函数ｙ＝ｔｘ＋ｖ＋ｋａｘ２＋ｂｘ＋ｃ（ａｋ≠０）（）的值域问题（设ｂ２－４ａｃ≠０）．本文利用三角变换结合直线斜率数形结合给出一种统一解法．原函数式配方,得ｙ＝ｔｘ＋ｖ＋ｋａ（ｘ＋ｂ２ａ）２＋４ａｃ－ｂ２４ａ．作替换ｚ＝ｘ＋ｂ２ａ,则ｙ＝ｔｚ＋（ｖ－ｂｔ２ａ）＋ｋａｚ２＋４ａｃ－ｂ２４ａ．若ａ＜０,则有ｙ＝ｔｚ＋（ｖ－ｂｔ２ａ）＋ｋ－ａ　·ｂ２－４ａｃ４ａ２－ｚ２．若ａ＞０,则有ｙ＝ｔｚ＋（ｖ－ｂｔ２ａ）＋ｋａ　·ｚ２＋４ａｃ－ｂ２４ａ２．因此,函数式的根号内可化为ｒ２－ｚ２…     

等比数列前n项和公式的七种推导方法

［题目］　在等比数列｛ａｎ｝中,已知首项ａ１和公比ｑ,求前ｎ项和Ｓｎ．［方法１］——先让学生演算Ｓ１,Ｓ２,Ｓ３,Ｓ４,然后启发学生猜想结论,让学生在探索过程中发现公式,培养学生的探索精神．当ｑ≠１时,Ｓ１＝ａ１＝ａ１（１－ｑ）１－ｑＳ２＝ａ１＋ａ１ｑ＝ａ１（１－ｑ）１－ｑ（１＋ｑ）＝ａ１（１－ｑ２）１－ｑＳ３＝ａ１＋ａ１ｑ＋ａ１ｑ２＝ａ１（１－ｑ２）１－ｑ＋ａ１ｑ２（１－ｑ）１－ｑ＝ａ１（１－ｑ３）１－ｑＳ４＝ａ１＋ａ１ｑ＋ａ１ｑ２＋ａ１ｑ３＝ａ１（１－ｑ３）１－ｑ＋ａ１ｑ３（１－ｑ）１－ｑ＝…     

图象分析引导我们深入探索——完善一道易错的题

１　一道易错的题题目　已知ｆ（ｘ）＝ａｘ２－ｃ,且－４≤ｆ（１）≤－１,－１≤ｆ（２）≤５,求ｆ（３）的范围．错解　依题意得　－４≤ａ－ｃ≤－１－１≤４ａ－ｃ≤５①②消元可得　０≤ａ≤３１≤ｃ≤７③④∵　ｆ（３）＝９ａ－ｃ,∴　－７≤ｆ（３）≤２６．正解　先用ｆ（１）、ｆ（２）表出ａ、ｃ,即有　　ｆ（１）＝ａ－ｃｆ（２）＝４ａ－ｃ　ａ＝１３［ｆ（２）－ｆ（１）］ｃ＝１３［ｆ（２）－４ｆ（１）］⑤⑥∵　ｆ（３）＝９ａ－ｃ＝８３ｆ（２）－５３ｆ（１）,∴　直接运用已知条件可得－１≤ｆ（３）≤２０．…     

两类数列变换在解题中的应用

利用数列｛ａｎ｝的如下两类变换：ａｎ＝ａ１＋（ａ２－ａ１）＋（ａ３－ａ２）＋…＋（ａｎ－ａｎ－１）及ａｎ＝ａ１ａ２ａ１ａ３ａ２…ａｎａｎ－１（ａｉ≠０,ｉ＝１,２,…,ｎ－１）不仅能简便地推导出等差数列和等比数列的通项公式,而且灵活运用它们还能简捷、...     

一个命题的再研究

命题　△ＡＢＣ中，∠Ａ、∠Ｂ、∠Ｃ所对边分别是ａ、ｂ、ｃ，求证　　ｓｉｎＡ－ｓｉｎＢｂｃ＋ｓｉｎＢ－ｓｉｎＣｃａ＋ｓｉｎＣ－ｓｉｎＡａｂ　≥０．（１）（《数学通报》１９９７年５月号问题１０７２）文［１］对上述命题给出了一种简捷证法．通过对（１）式证法的研究，笔者得到了以下几个命题．命题１　设△ＡＢＣ中，∠Ａ、∠Ｂ、∠Ｃ所对边分别是ａ、ｂ、ｃ，则有：　　ｓｉｎＡ－ｓｉｎＢｃａ＋ｓｉｎＢ－ｓｉｎＣａｂ＋ｓｉｎＣ－ｓｉｎＡｂｃ　≤０．（２）证明　由正弦定理知，不等式（２）等价于ａ－ｂｃａ＋ｂ－ｃａｂ＋…     

1999年全国高中数学联合竞赛试题及参考答案

一、选择题１．给定公比为ｑ（ｑ≠１）的等比数列｛ａｎ｝,设ｂ１＝ａ１＋ａ２＋ａ３,ｂ２＝ａ４＋ａ５＋ａ６,…,ｂｎ＝ａ３ｎ－２＋ａ３ｎ－１＋ａ３ｎ,…,则数列｛ｂｎ｝（　　）．　（Ａ）是等差数列　　（Ｂ）是公比为ｑ的等比数列　（Ｃ）是公比为ｑ３的等比数列　（Ｄ）既非等差数列又非等比数列解　由题设,ａｎ＝ａ１ｑｎ－１,则　ｂｎ＋１ｂｎ＝ａ３ｎ＋１＋ａ３ｎ＋２＋ａ３ｎ＋３ａ３ｎ－２＋ａ３ｎ－１＋ａ３ｎ＝ａ１ｑ３ｎ＋ａ１ｑ３ｎ＋１＋ａ１ｑ３ｎ＋２ａ１ｑ３ｎ－３＋ａ１ｑ３ｎ－２＋ａ１ｑ３ｎ－１＝ａ１ｑ３…     

关于Hurwitz zeta—函数的四次均值

对ｓ＝σ＋ｉｔ，ζ（ｓ，ａ）是Ｈｕｒｗｉｔｚｚｅｔａ－函数，ζ１（ｓ，ａ）＝ζ（ｓａ）－ａ＾－ｓ，。本文的主要目的是用解析的方法给出了Ｈｕｒｗｉｔｚｚｅｔａ－函数四次均值较强的渐近公式。     

What is the basic integration rule?

<正>In this paper you will learn the basic integration rule from the basic differentiation rule.Last time we have learned anti-derivative,indefinite integral,and the power rule in integration.Let us recall what the power rule in differentiation is.If n is a real number,and y=xn,then y'=nxn-1.Proof:Let y=xn,we get ln|y|=ln|xn|,Since ln|xn|=n ln|x|,so we get ln|y|=n ln|x|.Take differentiation for both side above:y'y=n x.     

On extremal properties for the polar derivative of polynomials

If p(z) is a polynomial of degree n having all its zeros on |z| = k, k ≤ 1, then it is proved[5] that max |z|=1 |p′(z)| ≤ kn1n + kn m|z|=ax1 |p(z)|. In this paper, we generalize the above inequality by extending it to the polar derivative of a polynomial of the type p(z) = cnzn + ∑n j=μ cn jzn j, 1 ≤μ≤ n. We also obtain certain new inequalities concerning the maximum modulus of a polynomial with restricted zeros.     

On the Springer''s Conjecture of the Coefficients of Univalent Meromorphic Functions

Let $\sigma$ denote the family of univalent functions $\[F(z) = z + \sum\limits_{n = 1}^\infty {\frac{{{b_n}}}{{{z^n}}}} \]$ in l< |z| <\infty if G(w) is the inverse of a function $F(z) \in \sigma ^'$, the expansion of G(w) in some neighborhood of w=\infty is $\[G(w) = w - \sum\limits_{n = 1}^\infty {\frac{{{B_n}}}{{{w^n}}}} \]$ It is well known that |B_1|\leq 1 for any F(z) \in \sigma ^'. Springer^[1] proved that | B_3| \leq 1 and conjectured that $\[|{B_{2n - 1}}| \le \frac{{(2n - 2)!}}{{n!(n - 1)!}}{\rm{ }}(n = 3,4, \cdots )\]$ (1) Kubota^[2] proved (1) for n=3, 4, 5. Schober^[3] proved (1) for n = 6, 7. Ren Fuyao[4,5] has verified (1) for n=6, 7, 8. In this article we are going to verify (1) for n=9.     

On approximation of smooth functions from null spaces of optimal linear differential operators with constant coefficients

For a real valued function f defined on a finite interval I we consider the problem of approximating f from null spaces of differential operators of the form Ln(ψ) = n ∑ k=0 akψ(k), where the constant coefficients ak ∈ R may be adapted to f . We prove that for each f ∈ C(n)(I), there is a selection of coefficients {a1, ,an} and a corresponding linear combination Sn( f ,t) = n ∑ k=1 bkeλkt of functions ψk(t) = eλkt in the nullity of L which satisfies the following Jackson’s type inequality: f (m) Sn(m )( f ,t) ∞≤ |an|2n|Im|1/1q/ep|λ|λn|n|I||nm1 Ln( f ) p, where |λn| = mka x|λk|, 0 ≤ m ≤ n 1, p,q ≥ 1, and 1p + q1 = 1. For the particular operator Mn(f) = f + 1/(2n) f(2n) the rate of approximation by the eigenvalues of Mn for non-periodic analytic functions on intervals of restricted length is established to be exponential. Applications in algorithms and numerical examples are discussed.     

关于虚二次域类数的可除性

设α＞１，ｂ＞１，（α，ｂ）＝１，ｈ（－αｂ）表虚二次域的类数。如果有正整数ｘ，ｙ，ｎ，ｋ满足（１）αｘ￣２＋ｂｙ￣２＝４ｋ￣ｎ，ｂ且；或（２）αｘ￣２＋ｂｙ￣２＝ｋ￣ｎ，ｘ｜α，ｙ｜ｂ且αｂ≡２（ｍｏｄ４），则本文证明了关于ｈ（－αｂ）的可除性的两个定理（见定理１，２），其中符号ｘ｜α表示ｘ的每一个素因子整除α。     

Tests for the convergence of continued fractions,based on the fundamental system of inequalities

We have proved that if the partial numerators of the continued fraction f(c)=1/1+c₂/l+c₃/l+... are all nonzero and for at least some number n?1 satisfy the inequalities $$p_n \left| {1 + c_n + c_{n + 1} } \right| \ge p_{n - 2} p_n \left| {c_n } \right| + \left| {c_{n + 1} } \right|(n \ge 1,p_{ - 1} = p_0 = c_1 = 0,p_n \ge 0),$$ then f(c) converges in the wide sense if and only if at least one of the series $$\begin{array}{l} \sum\nolimits_{n = 1}^\infty {\left| {c_3 c_5 \ldots c_{2n - 1} /(c_2 c_4 \ldots c_{2n} )} \right|} , \\ \sum\nolimits_{n = 1}^\infty {\left| {c_2 c_4 \ldots c_{2n} /(c_3 c_5 \ldots c_{2n + 1} )} \right|} \\ \end{array}$$      

The Krzyz Problem and Polynomials with Zeros on the Unit Circle

Let $ \Pi_{n,M} $ be the class of all polynomials $ p(z) = \sum _{0}^{n} a_{k}z^{k} $ of degree n which have all their zeros on the unit circle $ |z| = 1$ , and satisfy $ M = \max _{|z| = 1}|\,p(z)| $ . Let $ \mu _{k,n} = \sup _{p\in \Pi _{n,M}} |a_{k}| $ . Saff and Sheil-Small asked for the value of $\overline {\lim }_{n\rightarrow \infty }\mu _{k,n} $ . We find an equivalence between this problem and the Krzyz problem on the coefficients of bounded non-vanishing functions. As a result we compute $$ \overline {\lim }_{n\rightarrow \infty }\mu _{k,n} = {{M} \over {e}}\quad {\rm for}\ k = 1,2,3,4,5.$$ We also obtain some bounds for polynomials with zeros on the unit circle. These are related to a problem of Hayman.     

一类二分图的k-匹配数

设G是一个具有二分类(X_1,X_2)的简单偶图,|X_1|=|X_2|=n,如果对于给定的c>0,|M(S)|≥(1+c)|S|对任意满足|S|≤n/2的S(?)X_i(i=1,2)都成立,其中N(S)是S的邻集,则称G是(n,c)-扩张图.给出了(n,c)-扩张图的k-匹配数与完美匹配数之比的顺从界.     

一个和单位分数有关的不定方程的解

给出了一个和单位分数有关的满足α|n+1,b|n+1,c|n+1,d|n+1,a相似文献   

On the Maximum Spectral Radius of （m, n）-trees with Given Diameter

Let (V, U) be the vertex-partition of tree T as a bipartite graph. T is called an (m,n)-tree if |V|=m and |U| = n. For given positive integers m,n and d, the maximum spectral radius of all (m,n)-trees on diameter d are obtained, and all extreme graphs are determined.