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附录A磁路和铁心线圈 本章重点 A-1磁场和磁路 A-2铁磁物质的磁化曲线 A-3磁路的基本定律 A-4恒定磁通磁路的计算 A-5交变磁通磁路简介 A6铁心线圈
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where a subscribed element of a matrix is always read as arou, column. Here we confine the element to be real number a vector is a matrix with one row or one column. Therefore a row vector is Alxk and a column vector is AixI and commonly denoted as ak and ai,respec- tively. In the followings of this course, we follow conventional custom to say that a vector is a columnvector except for
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设V是复线性空间.VV上的一个函数(,·),如果满足: (i)(,)对第一个变量是线性的; (ii)(a,)=(B,a); (iii)a∈v,(a,a)≥0,且(a,a)=0a=0
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WHATIS DESIGN??? A DESIGN IS PLAN FOR: Manufacturing an artifact refrigerator or a rug Building the mean of manufacture A factory or a chemical plant Building structure A bridge or a baseball stadium Implementing an organization A hospital emergency room A air transportation system
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7-1幂零线性变换的 Jordan标准型 A是数域K上n维线性空间V上的线性变换,如果存在正整数m,使A=0,则称A是一个 幂零线性变换. 对数域K上n阶方阵A,如果存在正整数m,使Am=0,则称A为幂零矩阵 命题幂零线性变换的特征值等于0 证明设是V上幂零线性变换A的特征值,则存在V中非零向量a,使得 Aa= 假设A=0
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设V是复线性空间.V×V上的一个函数,如果满足 (i)(·,·)对第一个变量是线性的 (i)(a,B)=(B (ii1)ya∈V,(a,a)≥0,且(a,a)=0分a=0 则称(a,B)为向量a,B的内积,具有内积的复线性空间称为酉空间(欧氏空间在复线性 空间上的推广)
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2007年4月笔试试卷参考答案 一、选择题 (1)B)(2)D)(3)A)(4)C)(5)D(6)C)(7)A)(8)B) (9)C)(10)A)(11)A)(12)C)(13)B)(14)C)(15)D)(16)D) (17)B)(18)D)(19)A)(20)C)(21)C)(22)B)(23)B)(24)C) (25)C)(26)D)(27)C)(28)A)(29)C)(30)A)(31)D)(32)D) (33)B)(34)A)(35)B)
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一、选择题 (1)D) (2)B) (3)C) (4)A) (5)A) (6)D) (7)C) (8)A) (9)B) (10)A) (11)B) (12)D) (13)C) (14)C) (15)A) (16)B) (17)A) (18)A) (19)C) (20)A)
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1. Unit 1 Life and logic Text A Love and logic: The story of a fallacy 2. Unit 2 Secrets to beauty Text A The confusing pursuit of beauty 3. Unit 3 Being entrepreneurial Text A Fred Smith and FedEx: The vision that changed the world 4. Unit 4 Nature: To worship or to conquer Text A Achieving sustainable environmentalism 5. Unit 5 Why culture counts Text A Speaking Chinese in America 6. Unit 6 Gender equality Text A The weight men carry 7. Unit 7 Energy and food crises Text A The coming energy crisis
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Guessing a particular solution. Recall that a general linear recurrence has the form: f(n)=a1f(n-1)+a2f(n-2)+…+aaf(n-d)+g(n) As explained in lecture, one step in solving this recurrence is finding a particular solu- tion; i.e., a function f(n)that satisfies the recurrence, but may not be consistent with the boundary conditions. Here's a recipe to help you guess a particular solution:
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