浅谈求函数y=a1x2+b1x+c1/a2x2+b2x+c2值域的判别式法

函数y=a1x2+b1x+c1/a2x2+b2x+c2值域的求法,很多资料上给出方法是判别式(即△)法,而一旦自变量的范围给以限定,当△法失效时,还有其他方法吗?一般资料上就避而不谈了.要全面系统解决函数y=a1x2+b1x+c1/a2x2+b2x+c2值域的问题,本文以为需解决以下三个事情:①判别式法的过程和依据,②自变量有限制时还能用判别式法吗?③自变量有范围限制,问题可以归结为三类常见函数:反比例函数;y=t+c/t(c＞0);y=t+c/t(c＜0)的值域求法.     

关于不定方程y(y+1)(y+2)(y+3)=4n~2x(x+1)(x+2)(x+3)

设1n∈N*,运用Pell方程的一些结果以及代数数论和p-adic分析方法证明了不定方程y(y+1)(y+2)(y+3)=4n~2x(x+1)(+2)(x+3)(x,y∈N*)除开n=1189时仅有一组解(x,y)=(33,1680)外,无其他解.     

关于不定方程5x(x+1)(x+2)(x+3)=9y(y+1)(y+2)(y+3)

运用pell方程、递归序列、二次平方剩余的方法,并使用数学软件Mathematica的计算,求出了丢番图方程5x(x+1)(x+2)(x+3)=9y(y+1)(y+2)(y+3)的所有20组整数解,其中只有一组正整数解为(x,y)=(6,5).     

关于不定方程3x(x+1)(x+2)(x+3)=10y(y+1)(y+2)(y+3)

通过利用pell方程、递归序列、平方剩余、Legendre符号、同余关系等初等证明方法,并利用Mathematica软件对Legendre符号等进行计算,证明了方程3x(x+1)(x+2)(x+3)=10y(y+1)(y+2)(y+3)共有16组整数解,并且无正整数解.     

椭圆曲线y~2=(x+2)(x~2-2x+43)的整数点

利用初等方法证明了椭圆曲线y~2=(x+2)(x~2-2x+43)仅有整数点(x,y)=(-2,0).     

关于丟番图方程x(x+1)(x+2)(x+3)=27y(y+1)(y+2)(y+3)

先后运用了pell方程、勒让德符号,同余关系,递归序列、二次平方剩余,分类讨论的有关方法,并通过使用数学软件Mathematica进行计算,证明了以下结论:不定方程x(x+1)(x+2)(x+3)=27y(y+1)(y+2)(y+3)没有正整数解,并找出了该方程的全部16组整数解.     

椭圆曲线y2=(x+p)(x2+p2)的本原整数点

设p是奇素数.本文给出了椭圆曲线y2=(x+p)(x2+p2)存在可使y为偶数的本原整数点(x,y)的充要条件.     

椭圆曲线y~2=(x+6)(x~2-6x+23)的正整数点

利用初等方法证明了椭圆曲线y~2=(x+6)(x~2-6x+23)无正整数点,研究结果对于a,p∈Z时,椭圆曲线y~2=(x+a)(x~2-ax+p)的求解有一定的借鉴作用,同时此结果推进了该类椭圆曲线的研究.     

不定方程x1+x2+…+xn=m的解的个数及在计数问题中的作用

不少计数问题归结为不定方程 x1+ x2+… + xn =m在特定条件下的解的个数问题便迎刃而解 .本文研究不定方程 x1+ x2 +… + xn =m在有关条件下的解的个数问题 ,并举例说明其在计数问题中的应用 .(注 :文中约定 :当 m 相似文献   

从(x±1)(x~2±x+1)=x~3±1谈起

乘法公式中有 (x+1)(x~2-x+1)=x~3+1,(x-1)(x~2+x+1)=x~3-1。等式两边互换,就得到因式分解 x~3+1=(x+1)(x~2-x+1),x~3-1=(x-1)(x~2+x+1)。进而有 x~4+1=(x+1)(x~3-x~2+x-1),x~4-1=(x-1)(x~3+x~2+x+1)。推广这些公式,可以得到定理1 (1)对任意正整数n,有 x~n-1=(x-1)(x~(n-1)+x~(n-2)+…+x+1)     

Calculation of coupling factors for the equation x2ϕ″−(x3+a2x2+a1x+a0)ϕ

A numerical realization of the method of finding coupling factors, described in the author's previous papers, is given in the present paper and its efficiency is studied. It is shown that this method is a generalization of the traditional method of finding coupling factors for Bessel functions. Several relations containing coupling factors are also derived. In a number of cases it is possible to judge the calculation error from the accuracy with which they are satisfied.Translated from Zapiski Nauchnykh Seminarov Leningradskogo Otdeleniya Matematicheskogo Instituta im. V. A. Steklova AN SSSR, Vol. 156, pp. 109–124, 1986.     

Über die Gleichung (x + y)(x 2 +y 2)

Ohne Zusammenfassung     

Weitere Untersuchungen über die ternäre biquadratische Formf =x 1 3 x 2+x 2 3 x 3+x 3 3 x 1

Ohne Zusammenfassung     

Le equazioni diofanteeax 4-bx 2+1=□, 4x 4+x 2 y 2+4y 4=□.

Summary Si determinano tre classi di quartiche u²=ax⁴+1, a e b interi, b²−4a≠□ per le quali l'appartenenza di un punto razionale (x₀,y₀),x₀≠0 porta l'esistenza nella stessa quartica di infiniti punti razionali. Si dimostra pure che l'equazione 4x²+x²y²+4y⁴=□ non ammette soluzioni intere con |xy| > 1. A BeniaminoSegre nel suo 70^mo compleanno. Entrata in Redazione il 26 febbraio 1973.     

Reducibility of polynomials a0(x) + a1 (x)y + a2 (x) y2 modulo p

Let f=a₀(x)+a₁(x)y+a₂(x)y² ? \Bbb Z[x,y]f=a_0(x)+a_1(x)y+a_2(x)y^2\in {\Bbb Z}[x,y] be an absolutely irreducible polynomial of degree m in x. We show that the reduction f mod p will also be absolutely irreducible if p 3 c_m·H(f)^e_mp\ge c_m\cdot H(f)^{e_m} where H (f) is the height of f and e₁ = 4,e₂ = 6, e₃ = 6 [2/3]{2}\over{3} and e_m = 2 m for m S 4. We also show that the exponents e_m are best possible for m 1 3m\ne 3 if a plausible number theoretic conjecture is true.