一、单项选择题 1 A 2 B 3 C 4 B 5 C 6 D 7 B 8 A 9 D 10 A 11 B 12 D 13 B 14 C 15 C

Let A and B be sets. A function f from A to B is an assignment of exactly one element of B to each element of A. We write f(a) = b if b is the unique element of B assigned by the function f to the element a of A. If f is a function from A to B, we write f : A→B

一、选择题 (1)B)(2)D)(3)C)(4)A)()D)(6)B)(7)A)(8)B) (9)C)(10)D)(11)D)(12)A)(13)C)(14)A)(15)D)(16)B) (17)B)(18)D)(19)B)(20)C)(21A)(22)A)(23)B)(24)C) (25)A)(26)D)(27)A)(28)C)(29)B)(30)C)(31)B)(32)D) (33)A)(34)A)(35)D)

1. 根据算符∇ 的微分性与矢量性 推导下列公式 A B B A B A A B A B r r r r r r r r r r ∇( ⋅ ) = × (∇ × ) + ( ⋅∇) + × (∇ × ) + ( ⋅∇) A A A A A r r r r r ( ) 2

One parse tree only a The role of the grammar a distinguish between syntactically legal and illegal programs a But that's not enough: it must also define a parse tree a the parse tree conveys the meaning of the program What if a string can be parsed with multiple parse trees? a we say the grammar is ambiguous a must fix the grammar(the problem is not in the parser) a Note: often a string can be derived in more than one way D ie, with more than one derivation sequence o this does not mean the grammar is ambiguous

8-2同余式 8.2.1有理整数环中的同余的定义 定义8.5设m是一个正整数,若a,b∈Z,且ba∈(m),亦即m(b-a),则 称b与a模m同余,记作b=a(modm)。不难得到,b与a模m同余就是它们用m做带 余除法所得的余数相同。整数模m同余为一等价关系,验证如下: 1、反身性:a=a(modm) 2、对称性:若b=a(modm),则a=b(modm) 3、转递性:若a=b(modm),b=c(modm),则

习题1 2.设A,B,C为三个事件,用A,B,C的运算表示下列事件: (1)A,B,C都发生; 解A,B,C都发生表示为ABC. (2)A,B发生,C不发生; 解A,B发生,C不发生表示为ABC=AB-C. (3)A,B,C都不发生; 解A,B,C都不发生表示为ABC. (4)A,B中至少有一个发生而C不发生; 解A,B中至少有一个发生而C不发生表示为(AUB)C=AB-C

一.填空题(本题共5小题,每小题3分,满分15分) (1)设常数a≠,则 Inn-2na+1=1-2a nn(1-2a) (2)交换积分序:f,y)axy)x=ay)dy 12-2)

1.证明:若微分方程xy\+y+xy=0有多项式解y=a+ax+a2x2+…+anx,则必有a1=0(i=1,2,…,n)

第六章二次型 变量x1,x2,…,xn的二次齐次多项式 f(x1,x2,,xn)=a1x2+2a12x1x2+2a13x1x3+…+2anx1xn +a22x2+2a23x2x3+…+2a2nx2xn +amx 称为n元二次型,简称为二次型 a∈R:称f(x1,x2,…,xn)为实二次型(本章只讨论实二次型) a∈C:称f(x1,x2,…,xn)为复二次型 6.1二次型的矩阵表示 1.矩阵表示:令an=a(>i),则有