Eco514 Game Theory Lecture 4: Games with Payoff Uncertainty(1) Marciano Siniscalchi September 28, 1999 Introduction The vast majority of games of interest in economics, finance, political economy etc. involve some form of payoff uncertainty. A simple but interesting example is provided by auctions: an object is offered for sale, and individuals are required to submit their bids in sealed

市场营销者(Marketer) 市场营销者——是指希望从别人那 里取得资源并愿意以某种有价值的东 西作为交换的人。 A marketer is someone seeking one more prospects who might engage in an exchange of values

1向量的定义 定义n个有次序的数a1,a2,…,an所组成的 数组称为n维向量这n个数称为该向量的分量 第i个数a称为第i个分量 分量全为实数的向量称为实向量 分量全为复数的向量称为复向量

TAC Three address code Three-address code (TAC) will be the intermediate representation used in our Decaf compiler. It is essentially a generic assembly language that falls in the lower-end of the mid-level IRs. Many compilers use an IR similar to TAC. It is a sequence of instructions, each of which can have at most three operands. The operands could be

TAC Handout written by Maggie Johnson and revised by me. Three address code Three-address code(TAC) will be the intermediate representation used in our Decaf compiler. It is essentially a generic assembly language that falls in the lower-end of the mid-level IRs. Some variant of 2, 3 or 4 address code is fairly commonly used as an IR, since it maps well to most

MessageBox The MessageBox function creates, displays, and operates a message box. The message box contains an application-defined message and title plus any combination of predefined icons and push buttons

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

3.4转置以及特殊矩阵 定义4.1设A=(an),把A的行写成列而得的 mxn nxm矩阵称为A的转置矩阵,记为AT,即

第三章矩阵的初等变换 3.1矩阵的秩 1.子式:在An中,选取k行与k列,位于交叉处的k2个数按照原来的 相对位置构成k阶行列式,称为A的一个k阶子式,记作D 对于给定的k,不同的k阶子式总共有C个 2.矩阵的秩:在A中,若 (1)有某个r阶子式D,≠0; (2)所有的r+1阶子式D+1=0(如果有r+1阶子式的话) 称A的秩为r,记作 rankA=r,或者r(A)r.规定:rank

12.3.2用一个多项式的根和另一个多项式计算结式的公式 命题设 f(x)=ax+a1x-+…+an(a≠0 (x) box\+b- + (bo=0) 如果f(x),g(x)在C[x]中的分解式为 g()= bo (x-B) ).(x-)(1) 那么 R(f,g)=ag(a)=(-1)f(B)(*) 证明在数域K上的n+m+1元多项式环K[x,y1yn21m]中,令 f(x,y,yn)=a(x-y)…(x-yn)(2) g(x,z1,m)=b(x-z)…(x-m)(3)