4 习题参考答案
形式逻辑学 第四版(华东师大版)课后习题参考答案
练习答案第一章形式逻辑的对象和意义(P13-14)一、1、逻辑学;客观规律。
2、思维规律。
3、客观规律。
4、某种理论、观点、看法。
二、1、(b)。
2、(b)第二章概念(P43-49)二.(1)单独、集合;(2)普遍、非集合;(3)普遍、集合;(4)普遍、非集合;(5)普遍、非集合;(6)普遍、集合。
三.字母ABCD分别表示先后出现的概念(见下页)六.全部错误。
理由:1、使用了否定;2、循环定义;3、定义过窄;4、循环定义;5、隐喻;6、定义过宽;7、定义过窄;8、定义过宽。
1、2、3、4、5、6、7、8、orA BBDDCABCDAABCCABBCDACBAB CAA BC七、全部错误。
理由:1、是分解;2、混淆根据、子项相容;3、不是划分;4、子项相容、划分不全、混淆根据;5、混淆根据、子项相容;6、是分解;7、多出子项;8、划分不全。
九、1、内涵、外延。
2、交叉、反对。
3、不相容(全异)、同一。
4、(略)。
5、定义过窄。
6、真包含(同一)、不相容(全异)。
7、限制、概括。
8、多出子项、划分不全。
十、a c d d(c) c d a c第三章简单命题及其推理(上)(P77-81)一、(3)、(5)直接表达判断。
二、A A A E O I A(a) E三、1、不能,能。
2、能,能。
3、(略)六、(3)正确。
七、1、SOP。
2、真包含于。
3、全同、真包含于。
4、真假不定。
5、特称、肯定。
6、SI P 真。
八、c d d d c d九、de de bc bc十、SIP、SOP取值为真,SIP可换位:SIP PIS。
十一、推导一:ABC三句话分别是性质命题SAP、SaP、SEP,a与E是反对关系,必有一假,所以根据题意SAP必真,所有学生懂计算机,班长必然懂计算机。
推导二:A句与C句是反对关系,不可同真,必有一假,所以B句真,B句真则C句假,所以A句亦真,所有学生懂计算机,班长必然懂计算机。
十二、推导:SIP与SOP是下反对关系,不能同假,必有一真,所以POS必假,P真包含于S或与S全同,即S真包含P或与P全同,而前者使AB两句话均真,不合题意,所以S 与P全同。
计量经济学精要习题参考答案(第四版)
计量经济学(第四版)习题参考答案第一章 绪论1.1 一般说来,计量经济分析按照以下步骤进行:(1)陈述理论(或假说) (2)建立计量经济模型 (3)收集数据 (4)估计参数 (5)假设检验 (6)预测和政策分析1.2 我们在计量经济模型中列出了影响因变量的解释变量,但它(它们)仅是影响因变量的主要因素,还有很多对因变量有影响的因素,它们相对而言不那么重要,因而未被包括在模型中。
为了使模型更现实,我们有必要在模型中引进扰动项u 来代表所有影响因变量的其它因素,这些因素包括相对而言不重要因而未被引入模型的变量,以及纯粹的随机因素。
1.3时间序列数据是按时间周期(即按固定的时间间隔)收集的数据,如年度或季度的国民生产总值、就业、货币供给、财政赤字或某人一生中每年的收入都是时间序列的例子。
横截面数据是在同一时点收集的不同个体(如个人、公司、国家等)的数据。
如人口普查数据、世界各国2000年国民生产总值、全班学生计量经济学成绩等都是横截面数据的例子。
1.4 估计量是指一个公式或方法,它告诉人们怎样用手中样本所提供的信息去估计总体参数。
在一项应用中,依据估计量算出的一个具体的数值,称为估计值。
如Y 就是一个估计量,1nii YY n==∑。
现有一样本,共4个数,100,104,96,130,则根据这个样本的数据运用均值估计量得出的均值估计值为5.107413096104100=+++。
第二章 计量经济分析的统计学基础2.1 略,参考教材。
2.2 NS S x ==45=1.25 用α=0.05,N-1=15个自由度查表得005.0t =2.947,故99%置信限为 x S t X 005.0± =174±2.947×1.25=174±3.684也就是说,根据样本,我们有99%的把握说,北京男高中生的平均身高在170.316至177.684厘米之间。
2.3 原假设 120:0=μH备择假设 120:1≠μH 检验统计量()10/25XX μσ-Z ====查表96.1025.0=Z 因为Z= 5 >96.1025.0=Z ,故拒绝原假设, 即此样本不是取自一个均值为120元、标准差为10元的正态总体。
数字电子技术基础第四章习题及参考答案
数字电子技术基础第四章习题及参考答案第四章习题1.分析图4-1中所示的同步时序逻辑电路,要求:(1)写出驱动方程、输出方程、状态方程;(2)画出状态转换图,并说出电路功能。
CPY图4-12.由D触发器组成的时序逻辑电路如图4-2所示,在图中所示的CP脉冲及D作用下,画出Q0、Q1的波形。
设触发器的初始状态为Q0=0,Q1=0。
D图4-23.试分析图4-3所示同步时序逻辑电路,要求:写出驱动方程、状态方程,列出状态真值表,画出状态图。
CP图4-34.一同步时序逻辑电路如图4-4所示,设各触发器的起始状态均为0态。
(1)作出电路的状态转换表;(2)画出电路的状态图;(3)画出CP作用下Q0、Q1、Q2的波形图;(4)说明电路的逻辑功能。
图4-45.试画出如图4-5所示电路在CP波形作用下的输出波形Q1及Q0,并说明它的功能(假设初态Q0Q1=00)。
CPQ1Q0CP图4-56.分析如图4-6所示同步时序逻辑电路的功能,写出分析过程。
Y图4-67.分析图4-7所示电路的逻辑功能。
(1)写出驱动方程、状态方程;(2)作出状态转移表、状态转移图;(3)指出电路的逻辑功能,并说明能否自启动;(4)画出在时钟作用下的各触发器输出波形。
CP图4-78.时序逻辑电路分析。
电路如图4-8所示:(1)列出方程式、状态表;(2)画出状态图、时序图。
并说明电路的功能。
1C图4-89.试分析图4-9下面时序逻辑电路:(1)写出该电路的驱动方程,状态方程和输出方程;(2)画出Q1Q0的状态转换图;(3)根据状态图分析其功能;1B图4-910.分析如图4-10所示同步时序逻辑电路,具体要求:写出它的激励方程组、状态方程组和输出方程,画出状态图并描述功能。
1Z图4-1011.已知某同步时序逻辑电路如图4-11所示,试:(1)分析电路的状态转移图,并要求给出详细分析过程。
(2)电路逻辑功能是什么,能否自启动?(3)若计数脉冲f CP频率等于700Hz,从Q2端输出时的脉冲频率是多少?CP图4-1112.分析图4-12所示同步时序逻辑电路,写出它的激励方程组、状态方程组,并画出状态转换图。
人教版九年级数学上册课后习题参考答案
第21章第4页练习第1题答案解:(1)5x2-4x-1=0,二次相系数为5,一次项系数为-4,常数项为-1 (2)4x2-81=0,二次项系数为4,一次项系数为0,常数项为-81(3)4x2+8x-25=0,二次项系数为4,一次项系数为8,常数项为-25 (4)3x2-7x+1=0,二次项系数为3,一次项系数为-7,常数项为1【规律方法:化为一般形式即把所有的项都移到方程的左边,右边化为0的行驶,在确定二次项系数,一次项系数和常数项时,要特别注意各项系数及常数项均包含前面的符号。
】第4页练习第2题答案解:(1)4x2=25, 4x2-25=0(2)x(x-2)=100,x2-2x-100=0(3)x∙1=(1-x)2-3x+1=0习题21.1第1题答案(1)3x2-6x+1=0,二次项系数为3,一次项系数-6,常数项为1(2)4x2+5x-81=0,二次项系数为4,一次项系数为5,常数项为-81(3)x2+5x=0,二次项系数为1,一次项系数为5,常数项为0(4)x2-2x+1=0,二次项系数为1,一次项系数为-2,常数项为1(5)x2+10=0,二次项系数为1,一次项系数为0,常数项为10(6)x2+2x-2=0,二次项系数为1,一次项系数为2,常数项为-2习题21.1第2题答案(1)设这个圆的半径为Rm,由圆的面积公式得πR2=6.28,∴πR2-6.28=0(2)设这个直角三角形较长的直角边长为x cm,由直角三角形的面积公式,得1/2x(x-3)=9,∴x2-3x-18=0习题21.1第3题答案方程x2+x-12=0的根是-4,3习题21.1第4题答案设矩形的宽为x cm,则矩形的长为(x+1)cm,由矩形的面积公式,得x∙(x+1)=132,∴x2+x-132=0习题21.1第5题答案解:设矩形的长为x m,则矩形的宽为(0.5-x)m,由矩形的面积公式得:(0.5-x)=0.06∴x2-0.5x+0.06=0习题21.1第6题答案解:设有n人参加聚会,根据题意可知:(n-1)+(n-2)+(n-3)+…+3+2+1=10,即(n(n-1))/2=10,n2-n-20=0习题21.2第1题答案(1)36x2-1=0,移项,得36x2=1,直接开平方,得6x=±1,,6x=1或6x=-1,∴原方程的解是x1=1/6,x2=-1/6(2)4x2=81,直接开平方,得2=±9,,2x=9或2x=-9,∴原方程的解是x1=9/2,x2=-9/2(3)(x+5)2=25,直接开平方,得x+5=±5,∴+5=5或x+5=-5,∴原方程的解是x1=0,x2=-10(4)x2+2x+1=4,原方程化为(x+1)2=4,直接开平方,得x+1=±2,∴x+1=2或x+1=-2,∴原方程的解是x1=1,x2=-3习题21.2第2题答案(1)9;3(2)1/4;1/2(3)1;1(4)1/25;1/5习题21.2第3题答案(1)x2+10x+16=0,移项,得x2+10x=-16,配方,得x2+10x+52=-16+52,即(x+5)2=9,开平方,得x+5=±3,∴+5=3或x+5=-3,∴原方程的解为x1=-2,x2=-8(2)x2-x-3/4=0,移项,得x2-x=3/4,配方,得x2-x=3/4,配方,得x2-x+1/4=3/4+1/4,即(x-1/2)2=1,开平方,得x- 1/2=±1,∴原方程的解为x1=3/2,x2=-1/2(3)3x2+6x-5=0,二次项系数化为1,得x2+2x-5/3=0,移项,得x2+2x=5/3,配方,得x2+2x+1=5/3+1,即(x+1)2=8/3,(4)4x2-x-9=0,二次项系数化为1,得x2-1/4x-9/4=0,移项,得x2-1/4 x= 9/4,配方,得x2-1/4x+1/64=9/4+1/64,即(x-1/8)2=145/64,习题21.2第4题答案(1)因为△=(-3)2-4×2×(-3/2)=21>0,所以原方程有两个不相等的实数根(2)因为△=(-24)2-4×16×9=0,所以与原方程有两个相等的实数根(3)因为△=-4×1×9=-4<0,因为△=(-8)2-4×10=24>0,所以原方程有两个不相等的实数根习题21.2第5题答案(1)x2+x-12=0,∵a=1,b=1,c=-12,∴b2-4ac=1-4×1×(-12)=49>0,∴原方程的根为x1=-4,x2=3.∴b2-4ac=2-4×1×(-1/4)=3>0,(3)x2+4x+8=2x+11,原方程化为x2+2x-3=0,∵a=1,b=2,c=-3,∴b2-4ac=22-4×1×(-3)=16>0,∴原方程的根为x1=-3,x2=1.(4)x(x-4)=2-8x,原方程化为x2+4x-2=0,∵a=1,b=4,c=-2,∴b2-4ac=42-4×1×(-2)=24>0,(5)x2+2x=0,∵a=1,b=2,c=0,∴b2-4ac=22-4×1×0=4>0,∴原方程的根为x1=0,x2=-2.(6) x2+2x+10=0,∵a=1,b=2,c=10,∴b2-4ac=(2)2-4×1×10=-20<0,∴原方程无实数根习题21.2第6题答案(1)3x2-12x=-12,原方程可化为x2-4x+4=0,即(x-2)2=0,∴原方程的根为x1=x2=2(2)4x2-144=0,原方程可化为4(x+6)(x-6),∴x+6=0或x-6=0,∴原方程的根为x1=-6,x2=6.(3)3x(x-1)=2(x-1),原方程可化为(x-1)∙(3x-2)=0∴x-1=0或3x-2=0∴原方程的根为x1=1,x2=2/3(4)(2x-1)2=(3-x)2,原方程可化为[(2x-1)+(3-x)][(2x-1)-(3-x)]=0,即(x+2)(3x-4)=0,∴x+2=0或3x-4=0∴原方程的根为x1=-2,x2=4/3习题21.2第7题答案设原方程的两根分别为x1,x2(1)原方程可化为x2-3x-8=0,所以x1+x2=3,x1·x2=-8(2)x1+x2=-1/5,x1·x2=-1(3)原方程可化为x2-4x-6=0,所以x1+x2=4,x1·x2=-6(4)原方程可化为7x2-x-13=0,所以x1+x2=1/7,x1·x2=-13/7习题21.2第8题答案解:设这个直角三角形的较短直角边长为 x cm,则较长直角边长为(x+5)cm,根据题意得:1/2 x(x+5)=7,所以x2+5x-14=0,解得x1=-7,x2=2,因为直角三角形的边长为:答:这个直角三角形斜边的长为cm习题21.2第9题答案解:设共有x家公司参加商品交易会,由题意可知:(x-1)+(x-2)+(x-3)+…+3+2+1=45,即x(x-1)/2=45,∴x2-x-90=0,即(x-10)(x+9)=0,∴x-10=0或x+9=0,∴x1=10,x2=-9,∵x必须是正整数,∴x=-9不符合题意,舍去∴x=10答:共有10家公司参加商品交易会习题21.2第10题答案解法1:(公式法)原方程可化为3x2-14x+16=0,∵a=3,b=-14,c=16,∴b2-4ac=(-14)2-4×3×16=4>0,∴x=[-(-14)±]/(2×3)=(14±2)/6,∴原方程的根为x1=2,x2=8/3解法2:(因式分解法)原方程可化为[(x-3)+(5-2x)][(x-3)-(5-2x)]=0,即(2-x)(3x-8)=0,∴2-x=0或3x-8=0,∴原方程的根为x1=2,x2=8/3习题21.2第11题答案解:设这个矩形的一边长为x m,则与其相邻的一边长为(20/2-x)m,根据题意得:x(20/2-x)=24,整理,得x2-10x+24=0,解得x1=4,x2=6.当x=4时,20/2-x=10-4=6当x=6时, 20/2-x=10-6=4.故这个矩形相邻两边的长分别为4m和6m,即可围城一个面积为24m2的矩形习题21.2第12题答案解设:这个凸多边形的边数为n,由题意可知:1/2n(n-3)=20解得n=8或n=-5因为凸多边形的变数不能为负数所以n=-5不合题意,舍去所以n=8所以这个凸多边形是八边形假设存在有18条对角线的多边形,设其边数为x,由题意得:1/2 x(x-3)=18解得x=(3±)/2因为x的值必须是正整数所以这个方程不存在符合题意的解故不存在有18条对角线的凸多边形习题21.2第13题答案解:无论p取何值,方程(x-3)(x-2)-p2=0总有两个不相等的实数根,理由如下:原方程可以化为:x2-5x+6-p2=0△=b2-4ac=(-5)2-4×1×(6-p2)=25-24+4p2=1+4p2∵p2≥0,,1+4p2>0∴△=1+4p2>0∴无论P取何值,原方程总有两个不相等的实数根习题21.3第1题答案(1)x2+10x+21=0,原方程化为(x+3)(x+7)=0,或x+7=0,∴x1=-3,x2=-7.(2) x2-x-1=0∵a=1,b=-1,c=-1,b2-4ac=(-1)2-4×1×(-1)=5>0,(3)3x2+6x-4=0,∵a=3,b=6,c=-4,b2-4ac=62-4×4×3×(-4)=84>0,(4)3x(x+1)=3x+3,原方程化为x2=1,直接开平方,得x=±1,∴x1=1,x2=-1(5)4x2-4x+1=x2+6x+9,原方程化为(2x-1)2=(x+3)2,∴[(2x-1)+(x+3)][(2x-1)-(x+3)]=0,即(3x+2)(x-4)=0,,3x+2=0或x-4=0,∴x1=-2/3,x2=4∴a=7,b=-,c=-5,b2-4ac=(-)2-4×7×(-5)=146>0∴x= [-(-)±]/(2×7)=(±)/14,∴x1=(+)/14,x2=(-)/14习题21.3第2题答案解:设相邻两个偶数中较小的一个是x,则另一个是(x+2).根据题意,得x(x+2)=168∴x2+2x-168=0∴x1=-14,x2=12.当x=-14时,x+2=-12当x=12时,x+2=14答:这两个偶数是-14,-12或12,14习题21.3第3题答案解:设直角三角形的一条直角边长为 xcm,由题意可知1/2x(14-x)=24,∴x2-14x+48=0∴x1=6,x2=8当x=6时,14-x=8当x=8时,14-x=6∴这个直角三角形的两条直角边的长分别为6cm,8cm习题21.3第4题答案解:设每个支干长出x个小分支,则1+x+x2=91整理得x2+x-90=0,(x-9)∙(x+10)=0解得x1=9,x2=-10(舍)答:每个支干长出来9个小分支习题21.3第5题答案解:设菱形的一条对角线长为 x cm,则另一条对角线长为(10-x)cm,由菱形的性质可知:1/2 x∙(10-x)=12,整理,的x2-10x+24=0,解得x1=4,x2=6.当x=4时,10-x=6当x=6时,10-x=4所以这个菱形的两条对角线长分别为6cm和4cm.由菱形的性质和勾股定理,得棱长的边长为:所以菱形的周长是4cm习题21.3第6题答案解:设共有x个队参加比赛,由题意可知(x-1)+(x-2)+(x-3)+…+3+2+1=90/2,即1/2x(x-1)=45整理,得x2-x-90=0解得x1=10,x2=-9因为x=-9不符合题意,舍去所以x=10答:共有10个队参加比赛习题21.3第7题答案解:设水稻每公顷产量的年平均增长率为x,则7200(1+x)2=8450解得x1=1/12,x2=-25/12因为x=- 25/12 不符合题意,舍去所以x= 1/12≈0.083=8.3%答:水稻每公顷产量的年平均增长率约为8.3%习题21.3第8题答案解:设镜框边的宽度应是x cm,根据题意得:(29+2x)(22+2x)-22×29=1/4×29×22整理,得8x2+204x-319=0解得x= [-204±]/16所以x1=[-204+)]/16,x2=[-204-)]/16因为x= [-204-)]/16<0不合题意,舍去所以x= [-204+)]/16≈1.5答:镜框边的宽度约 1.5cm习题21.3第9题答案解:设横彩条的宽度为3x cm,则竖彩条的宽为2x cm.根据题意得:30×20×1/4=30×20-(30-4x)(20-6x),整理,得12x2-130x+75=0解得x1=[65+5)]/12,x2=(65-5)/12因为30-4x>0,且20-6x>0所以x<10/3所以x= (65+5)/12不符合题意,舍去所以x=(65-5)/12≈0.6所以3x≈1.8,2x≈1.2答:设计横彩条的宽度约为1.8cm,竖彩条的宽度约为1.2cm习题21.3第10题答案(1)设线段AC的长度为x,则x2=(1-x)×1,解得x1=(-1+)/2,x2=(-1-)/2(舍),∴AC=(-1+)/2(2)设线段AD的长度为x,则x2=((-1+)/2-x)∙(1+)/2,解得x1=(3-)/2,x2=-1(舍),∴ AD=(3-)/2(3)设线段AE的长度为x,则x2=((3-)/2-x)∙(3-)/2,解得x1=-2+,x2=(1-)/2 (舍)∴AE=-2+【规律方法:若C为线段AB上一点,且满足AC2=BC∙AB,则 AC/AB=(-1)/2∙(-1)/2也叫作黄金比,C点为黄金分割点,一条线段上有两个黄金分割点.】第6页练习答案练习题答案复习题21第1题答案(1)196x2-1=0,移项,得196x2=1,直接开平方,得14x=±1,x=± 1/14,∴原方程的解为x1=1/14,x2=-1/14(2)4x2+12x+9=81,原方程化为x2+3x-18=0∵a=1,b=3,c=-18,b2-4ac=32-4×1×(-18)=81>0∴x1=-6,x2=3(3)x2-7x-1=0∵a=1,b=-7,c=-1,b2-4ac=(-7)2-4×1×(-1)=53>0,(4)2x2+3x=3,原方程化为2x2+3x-3=0,∵a=2,b=3,b=-3,b2-4ac=32-4×2×(-3)=33>0,∴x= (-3± )/(2×2)=(-3±)/4,∴x1=(-3+)/4,x2=(-3-)/4(5)x2-2x+1=25,原方程化为x2-2x-24=0,因式分解,得(x-6)(x+4)=0,∴x-6=0或x+4=0,∴x1=6,x2=-4(6)x(2x-5)=4x-10,原方程化为(2x-5)(x-2)=0,,2x-5=0或x-2=0,∴x1=5/2,x2=2(7)x2+5x+7=3x+11,原方程化为x2+2x-4=0,∵a=1,b=2,c=-4,b2-4ac=22-4×1×(-4)=20>0∴x= (-2±)/(2×1)=(-2±2)/2=-1±∴x1=-1+,x2=-1-(8)1-8x+16x2=2-8x,原方程化为(1-4x)(-1-4x)=0,1-4x=0或-1-4x=0,∴x1=1/4,x2=-1/4复习题21第2题答案解:设其中一个数为(8-x),根据题意,得x(8-x)=9.75,整理,得x2-8x+9.75=0,解得x1=6.5,x2=1.5当x=6.5时,8-x=1.5当x=1.5时,8-x=6.5答:这两个数是6.5和1.5复习题21第3题答案解:设矩形的宽为x cm,则长为(x+3)cm由矩形面积公式可得x(x+3)=4整理,得x2+3x-4=0解得x1=-4整理,得x2+3x-4=0解得x1=-4,x2=1因为矩形的边长是正数,所以x=-4不符合题意,舍去所以x=1所以x+3=1+3=4答:矩形的长是4cm,宽是1cm复习题21第4题答案解:设方程的两根分别为x1,x2(1)x1+x2=5,x1∙x2=-10(2) x1+x2=-7/2,x1∙x2=1/2(3)原方程化为3x2-2x-6=0,∴x1+x2=2/3,x1∙x2=-2(4)原方程化为x2-4x-7=0,∴x1+x2=4,x1∙x2=-7复习题21第5题答案解:设梯形的伤低长为x cm ,则下底长为(x+2)cm,高为(x-1)cm,根据题意,得1/2 [x+(x+2)]∙(x-1)=8,整理,得x2=9,解得x1=3,x2=-3.因为梯形的低边长不能为负数,所以x=-3不符合题意,舍去,所以x=3,所以x+2=5,x-1=2.画出这个直角梯形如下图所示:复习题21第6题答案解:设这个长方体的长为5x cm,则宽为2 x cm,根据题意,得2x2+7-4=0,解得x1=1/2,x2=-4.因为长方体的棱长不能为负数,所以x=-4不合题意,舍去,所以x= 1/2.所以这个长方体的长为5x=1/2×5=2.5(cm),宽为2x=1(cm).画这个长方体的一个展开图如下图所示.(注意:长方体的展开图不唯一)复习题21第7题答案解:设应邀请x个球队参加比赛,由题意可知:(x-1)+(x-2)+…+3+2+1=15,即1/2 x(x-1)=15解得x1=6,x2=-5因为球队的个数不能为负数所以x=-5不符合题意,应舍去所以x=6答:应邀请6个球队参加比赛复习题21第8题答案解:设与墙垂直的篱笆长为x m,则与墙平行的篱笆为(20-2x)m根据题意,得x(20-2x)=50整理,得x2-10x+25=0解得x1=x2=5所以20-2x=10(m)答:用20m长的篱笆围城一个长为10m,宽为5m的矩形场地.(其中一边长为10m,另两边均为5m)复习题21第9题答案解:设平均每次降息的百分率变为x,根据题意得:2.25%(1-x)2=1.98%整理,得(1-x)2=0.88解得x1=1 -x2=1+因为降息的百分率不能大于1所以x=1+不合题意,舍去所以x=1-≈0.0619=6.19%答:平均每次降息的百分率约是6.19%复习题21第10题答案解:设人均收入的年平均增长率为x,由题意可知:12000(x+1)2=14520,解这个方程,得x+1=±x=-1或x=--1又∵x=--1不合题意,舍去∴x=(-1)×100%=10%答:人均收入的年平均增长率是10%复习题21第11题答案解:设矩形的一边长为x cm,则与其相邻的一边长为(20-x)cm,由题意得:x(20-x)=75整理,得x2-20x+75=0解得x1=5,x2=15,从而可知矩形的一边长15cm,与其相邻的一边长为5cm当面积为101cm2时,可列方程x(20-x)=101,即x2-20x+101=0∵△=-4<0∴次方程无解∴不能围成面积为101cm2的矩形复习题21第12题答案解:设花坛中甬道的宽为x m.梯形的中位线长为1/2 (100+180)=140(m),根据题意得:1/2(100+180)×80×1/6=80∙x∙2+140x-2x2整理,得3x2-450x+2800=0解得x1=(450+)/6=75+5/3,x2=(450-)/6=75-5/3因为x=75+5/3不符合题意,舍去所以x=75-5/3≈6.50(m)故甬道的宽度约为6.50m复习题21第13题答案(1)5/4=1.25(m/s),所以平均每秒小球的滚动速度减少1.25m/s (2)设小球滚动5m用了x s·(5+(5-1.25x))/2x=5,即x2-8x+8=0解得x1=4+2(舍),x2=4-2≈1.2答:小球滚动5 m 约用了1.2s第9页练习答案练习第1题答案练习第2题答案第14页练习答案练习第1题答案练习第2题答案第16页练习答案练习题答案第22章习题22.1第1题答案解:设宽为x,面积为y,则y=2x2习题22.1第2题答案y=2(1-x)2习题22.1第3题答案列表:x ... -2 -1 0 1 2 ...y=4x2... 16 4 0 4 16 ...y=-4x2... -16 -4 0 -4 -16 ...y=(1/4)x2... 1 1/4 0 1/4 1 ... 描点、连线,如下图所示:习题22.1第4题答案解:抛物线y=5x2的开口向上,对称轴是y轴,顶点坐标是(0,0)抛物线y= -1/5x2的开口向下,对称轴是y轴,顶点坐标是(0,0)习题22.1第5题答案提示:图像略(1)对称轴都是y轴,顶点依次是(0,3)(0, -2)(2)对称轴依次是x=-2,x=1,顶点依次是(-2,-2)(1,2)习题22.1第6题答案(1)∵a=-3,b=12,c=-3∴-b/2a=-12/(2×(-3))=2,(4ac-b2)/4a=(4×(-3)×(-3)-122)/(4×(-3))=9∴抛物线y=-3x2+12x-3的开口向下,对称轴为直线x=2,顶点坐标是(2,9)(2)∵a=4,b=-24,c=26∴- b/2a=-(-24)/(2×4)=3, (4ac-b2)/4a=(4×4×26-(-24)2)/(4×4)=-10∴抛物线y=4x2 - 24x+26的开口向上,对称轴为直线x=3,顶点坐标是(3, -10)(3)∵a=2,b=8,c=-6∴- b/2a=-8/(2×2)=-2, (4ac-b2)/4a= (4×2×(-6)-82)/(4×2)= -14∴抛物线y=2x2 +8x-6的开口向上,对称轴是x=-2,顶点坐标为(-2,-14)(4)∵a=1/2,b =-2,c=-1∴- b/2a=-(-2)/(2×1/2)=2, (4ac-b2)/4a=(4×1/2×(-1)- (-2)2)/(4×1/2)=-3 ∴抛物线y=1/2x2-2x-1的开口向上,对称轴是x=2,顶点坐标是(2, -3).图略习题22.1第7题答案(1)-1;-1(2)1/4;1/4习题22.1第8题答案解:由题意,可知S=1/2×(12-2t)×4t=4t(6-t)∴S=-4t2+24t,即△PBQ的面积S与出发时间t之间的关系式是S=-4t2+24t 又∵线段的长度只能为正数∴∴0<t<6,即自变量t的取值范围是0<t<6习题22.1第9题答案解:∵s=9t+1/2t2∴当t=12时,s=9×12+1/2×122=180,即经过12s汽车行驶了180m当s=380时,380=9t+1/2t2∴t1=20,t2=-38(不合题意,舍去),即行驶380m需要20s习题22.1第10题答案(1)抛物线的对称轴为(-1+1)/2=0,设该抛物线的解析式为y=ax2+k(a≠0)将点(1,3)(2,6)代入得∴函数解析式为y=x2+2(2)设函数解析式为y=a x2+bx+c(a≠0),将点(-1,-1)(0,-2)(1,1)代入得∴函数解析式为y=2x2+x-2(3)设函数解析式为y=a(x+1)(x-3) (a≠0),将点(1,-5)代入,得-5=a(1+1)(1-3)解得a=5/4∴函数解析式为y=5/4(x+1)(x-3),即y=5/4x2-5/2x-15/4(4)设函数解析式为y=a x2+ bx+c(a≠0),将点(1,2)(3,0)(-2,20)代入得∴函数解析式为y=x2-5x+6习题22.1第11题答案解:把(-1,-22)(0,-8)(2,8)分别代入y=a x2+bx+c,得a=-2,b=12, c=-8所以抛物线的解析式为y=-2x2+12x-8将解析式配方,得y=-2(x-3)2+10又a=-2<0所以抛物线的开口向下,对称轴为直线x=3,顶点坐标为(3,10)习题22.1第12题答案(1)由已知vt=v0+at=0+1.5t=1.5t,s=vt=(v0+vt)/2t=1.5t/2t=3/4t2,即s=3/4t2(2)把s=3代入s=3/4t2中,得t=2(t=-2舍去),即钢球从斜面顶端滚到底端用2s第29页练习答案练习第1题答案练习第2题答案习题22.2第1题答案(1)图像如下图所示:(2)有图像可知,当x=1或x=3时,函数值为0 习题22.2第2题答案(1)如下图(1)所示:方程x2-3x+2=0的解是x1=1,x2=2(2)如下图所示:方程-x2-6x-9=0的解是x1=x2=-3习题22.2第3题答案(1)如下图所示:(2)由图像可知,铅球推出的距离是10m习题22.2第4题答案解法1:由抛物线的轴对称性可知抛物线的对称轴是直线x=(-1+3)/2=1 解法2:设抛物线的解析式为y=a(x+1)(x-3),即y=ax2-2ax-3a,∴x=-(-2a)/2a=1,即这条抛物线的对称轴是直线x=1习题22.2第5题答案提示:图像略(1)x1=3,x2=-1(2)x<-1或x>3(3)-1<x<3习题22.2第6题答案提示:(1)第三或第四象限或y轴负半轴上(2)x轴上(3)第一或第二象限或y轴正半轴上,当a<0时(1)第一或第二象限或y轴正半轴上(2)x轴上(3)第三或第四象限或y轴负半轴上第32页练习答案练习题答案习题22.3第1题答案(1)∵a=-4<0∴抛物线有最高点∵x=-3/[2×(-4)]=3/8,y=[4×(-4)×0-32]/[2×(-4)]=9/16∴抛物线最高点的坐标为(3/8,9/16)(2)∵a=3>0∴抛物线有最低点∵x=-1/(2×3)=-1/6,y=(4×3×6-12)/(4×3)=71/12∴抛物线最低点的坐标为(-1/6,71/12)习题22.3第2题答案解:设所获总利润为y元.由题意,可知y=(x-30)(100-x),即y=-x2+130x-3000 =-(x-65)2+1225∴当x=65时,y有最大值,最大值是1225,即以每件65元定价才能使所获利润最大习题22.3第3题答案解:s=60t-1.5t2=-1.5(t2-40t+400)+1.5×400=-1.5(t-20)2+600∴当t=20时,s取最大值,且最大值是600,即飞行着陆后滑行600m才能停下来习题22.3第4题答案解:设一条直角边长是x,那么另一条直角边长是8-x设面积为y,则y=1/2x•(8-x),即y=-(1/2)x2+4x对称轴为直线x=-b/2a=-4/(2×(-1/2))=4当x=4时,8-x=4,ymax=8∴当两条直角边长都为4时,面积有最大值8习题22.3第5题答案解:设AC的长为x,四边形ABCD 的面积为y.由题意,可知y=1/2AC•BD ∴y= 1/2 x(10-x), 即y=-1/2x2+5x=-1/2(x-5)2+25/2∴当x=5时,y有最大值,y最大值=25/2此时,10-x=10-5=5,故当AC=BD=5时,四边形ABCD的面积最大,最大面积为25/2习题22.3第6题答案解:∵∠A=30°,∠C=90°,且四边形CDEF是矩形∴FE//BC,ED//AC∴∠DEB=30°在Rt△AFE中,FE=1/2AE在Rt△EDB中,BD=1/2EB,设AE=x,则FE=1/2x令矩形CDEF的面积为S,则S=FE•ED= 1/2 x •/2(12-x)=/4(12x-x2)∴当x=6时,S最大值=9,此时AE=6,EB=12-x=6∴AE=EB,即点E是AB的中点时,剪出的矩形CDEF面积最大习题22.3第7题答案解:设AE=x,AB=a,正方形EFGH的面积为S,由正方形的性质可知AE=DH,即AH=a-x在Rt△AEH中:HE2=AH2+AE2=(a-x)2+x2=2x2-2ax+a2=2(x-1/2 a) 2+1/2a2∴当x=1/2a时,S有最小值,且S最小值=1/2a2,此时AE=1/2a,EB=1/2a,即点E是AB边的中点∴当点E是AB边的中点时,正方形EFGH的面积最小习题22.3第8题答案解:设房价定为每间每天增加x元,宾馆利润为y元由题意可知,y=(180+x-20)(50-x/10)=-1/10x2+34x+8000=-1/10(x-170)2+10890∴当x=170时,y取最大值,且y最大值=10890,此时180+x=350(元)∴房间每天每间定价为350元时,宾馆利润最大习题22.3第9题答案解:用定长为L的线段围成矩形时,设矩形的一边长为x则S矩形=x•(1/2L-x)=-x2+1/2 Lx=-(x-1/4L)2+1/16L2,当x=1/4 L时,S最大值=1/16L2用定长为L的线段围成圆时,设圆的半径为R,则2R=L,S圆=R2=(L/2)2=L2/4ᅲ∵1/16L2=/16L2,L2/4=4/16L2,且π<4∴1/16L2<L2/4∴S矩形<S圆∴用定长为L的线段围成圆的面积大第33页练习答案练习题答案复习题第1题答案解:由题意可知,y=(4+x)(4-x)= -x2+16,即y与x之间的关系式是y=-x2+16 复习题第2题答案解:由题意可知,y=5000(1+x)2=5000x2+10000x+5000,即y与x之间的函数关系式为:y=5000x2+10000x+5000复习题第3题答案D复习题第4题答案(1)∵a=1>0∴抛物线开口向上又∵x=-2/(2×1)=-1,y=(4×1×(-3)-22)/(4×1)=-4∴抛物线的对称轴是直线x=-1,顶点坐标是(-1,-4).图略(2)∵a=-1<0∴抛物线开口向下又∵x=-6/(2×(-1))=3,y=(4×(-1)×1-62)/(4×(-1))=10∴抛物线的对称轴是直线x=3,顶点坐标是(3,10).图略(3)∵a=1/2>0∴抛物线开口向上又∵x=-2/(2×1/2)=-2, y= (4×1/2×1-22)/(4×1/2)=-1∴抛物线的对称轴是直线x=-2,顶点坐标是(-2,-1).图略(4)∵a=-1/4<0∴抛物线开口向下又∵x=-1/(2×(-1/4))=2,y=(4×(-1/4)×(-4)-12)/(4×(-1/4))=-3∴抛物线的对称轴是直线x=2,顶点坐标是(2, -3).图略复习题第5题答案解:∵s=15t-6t2∴当t=-15/(2×(-6))=5/4时,s最大值=(4×(-6)×0-152)/(4×(-6))=75/8,即汽车刹车后到停下来前进了75/8m复习题第6题答案(1)分别把(-3,2),(-1,-1),(1,3)代入y=ax2+bx+c得a=7/8,b=2,c=1/8所以二次函数的解析式为y=7/8x2+2x+1/8(2)设二次函数的解析式为y=a(x+1/2)(x-3/2)把(0, -5)代入,得a=20/3所以二次函数的解析式为y=20/3x2-20/3 x-5复习题第7题答案解:设垂直于墙的矩形一边长为xm,则平行于墙的矩形的另一边长为(30-2x)m设矩形的面积为ym2,则y=x(30-2x)=-2x2+30x=-2(x-15/2)2+112.5∴当x=15/2时,y有最大值,最大值为112.5,此时30-2x=15∴当菜园垂直于墙的一边长为15/2m,平行于墙的另一边长为15m时,面积最大,最大面积为112.5m2复习题第8题答案解:设矩形的长为x cm,则宽为(18-x)cm,S侧=2x•(18-x)=-2x2+36x=-2(x-9)2+162当x=9时,圆柱的侧面积最大,此时18-x=18-9=9当矩形的长与宽都为9cm时旋转形成的圆柱的侧面积最大复习题第9题答案(1)证明:∵四边形ABCD是菱形∴AB=BC=CD=AD又∵BE=BF=DG=DH∴AH=AE=CG=CF∴∠AHE∠AEH,∠A+∠AEH+∠AHE=180,∠A+2∠AHE=180〬又∵∠A+∠D=180〬∴∠D=2∠AHE,同理可得∠A=2∠DHG∴2∠AHE+2∠DHG=180〬∴∠AHE+∠DHG=90〬∴∠EHG=90〬,同理可得∠HGF=∠GFE=90〬∴四边形EFGH是矩形(2)解:连接BD交EF于点K,如图7所示,设BE的长为x,BD=AB=a∴四边形ABCD为菱形,∠A=60〬∴∠EBK=60〬,∠KEB=30〬在Rt△BKE中,BE=x,则BK=1/2x,EK=/2xS矩形EFGH=EF•FG=2EK•(BD-2BK)=2×/2 x(a-2×1/2x)=x(a-x)=-(x2-ax)=-(x2-ax+a2/4-a2/4)=-(x-a/2)2+/4a2当x=a/2时,即BE=a/2时,矩形EFGH的面积最大第35页练习答案第37页练习答案第39页练习答案第40页练习答案练习第1题答案练习第2题答案第23章习题23.1第1题答案(1)如下图所示:(2)如下图所示:(3)如下图所示:(4)如下图所示:习题23.1第2题答案解:如下图所示,旋转中心为O点,旋转角为OA所转的角度习题23.1第3题答案解:如下图所示:习题23.1第4题答案解:旋转图形分别为△A₁B₁C₁,△A₂B₂C₂,如下图所示:习题23.1第5题答案(1)旋转中心为O₁点,旋转角为60〬,如下图所示:(2)旋转中心为O₂点,旋转角为90〬,如下图所示:习题23.1第6题答案提示:旋转角就是以旋转中心为顶点的周角被均匀地等分问题(360〬÷5=72〬 ,360〬÷3=120〬)解:(1)旋转角为72°,114°,216°,288°,360°时,旋转后的五角星与自身重合(2)等边三角形绕中心点O旋转120〬,240〬,360〬时与自身重合习题23.1第7题答案风车图案由四个全等的基本图形构成,可由其中一个基本图形绕中心旋转90〬,180〬,270〬得到习题23.1第8题答案提示:旋转中心在等腰三角形的外部解:五角星中间的点为旋转中心,旋转角为72〬,114〬,216〬,288〬习题23.1第9题答案(1)如下图所示:(2)∵BC=3,AC=4,∠C=90〬习题23.1第10题答案提示:线段BE与DC在形状完全相同的两个三角形中,可考虑旋转变换,点A是两个三角形的公共点,因此点A是旋转中心解:BE=DC,理由如下:因为△ABD与△ACE都是等边三角形所以AE=AC, AB=AD,∠DAB=∠CAE=60〬所以∠DAB+∠BAC=∠CAE+∠BAC,即∠DAC=∠BAE所以△BAE绕点A顺时针旋转60〬时,BA与DA重合,AE与AC重合,则△BAE与△DAC完全重合所以BE=DC第59页练习答案练习第1题答案练习第2题答案练习第3题答案习题23.2第1题答案如下图所示:习题23.2第2题答案解:依题可知,是中心对称图形的有:禁止标志、风轮叶片、正方形、正六边形它们的对称中心分别是圆心,叶片的轴心,正方形对角线的交点,正六边形任意两条最长的对角线的交点习题23.2第3题答案如下图所示,四边形ABCD关于原点O对称的四边形为A\\\\\\\'B\\\\\\\'C\\\\\\\'D\\\\\\\'习题23.2第4题答案解:∵A(a,1)与A\\\\\\\'(5,b)关于原点O对称习题23.2第5题答案解:依题意可知此图形时中心对称图形,对称中心是O₁O₂的中点习题23.2第6题答案解:如下图所示,做出△ABC以BC的中点O为旋转中心旋转180〬°后的图形△DCB,则四边形ABCD即为以AC,AB为一组邻边的平行四边形习题23.2第7题答案解:如下图(1)中的△DCE是由△ACB以C为旋转中心,顺时针旋转90〬得到的.在下图(2)中,先以AC为对称轴作△ABC的轴对称图形△AFC,再把△AFC以C为旋转中心,逆时针旋转90〬,即可得到△DCE习题23.2第8题答案解:依题意知这两个梯形是全等的因为菱形是以它的对角线的交点为对称中心的中心对称图形根据中心对称的性质过对称中心的任意一条直线都将图形分成两个全等的图形所以它们全等习题23.2第9题答案不一定当两个全等的梯形的上底与下底之和等于它的一条腰长的时候,这两个全等的梯形可以拼成一个菱形,其他情况不行习题23.2第10题答案解:如下图所示:连接BE,DF,EF,BD,AC,BD与EF交于点O∵四边形ABCD是平行四边形∴AD//BC,AD=BC∴∠1=∠2∵△ADE是等边三角形∴DE=AD,∠3=60〬∵△BCF为等边三角形∴BC=BF,∠4=60〬∴DE=BF∴∠1+∠3=∠2+∠4,即∠BDE=∠DBF∴DE//BF∴四边形BEDF为平行四边形∴BD与EF互相平分于点O又∵四边形BEDF为平行四边形∴BD与AC互相平分于点O,即OD=OB,OE=OF,OA=OC ∴△ADE和△BCF成中心对称第61页练习答案练习第1题答案练习第2题答案练习第3题答案。
自动控制理论第4版全套参考答案
第一章习题参考答案1-1多速电风扇的转速控制为开环控制。
家用空调器的温度控制为闭环控制。
1-2 设定温度为参考输入,室内温度为输出。
1-3 室温闭环控制系统由温度控制器、电加热装置、温度传感器等组成,其中温度控制器可设定希望达到的室温,作为闭环控制系统的参考输入,温度传感器测得的室温为反馈信号。
温度控制器比较参考输入和反馈信号,根据两者的偏差产生控制信号,作用于电加热装置。
1-4 当实际液面高度下降而低于给定液面高度h r ,产生一个正的偏差信号,控制器的控制作用使调节阀增加开度,使液面高度逼近给定液面高度。
第二章 习题参考答案2-1 (1)()()1453223++++=s s s s s R s C ; (2)()()1223+++=s s s ss R s C ; (3)()()1223+++=-s s s e s R s C s2-2 (1)单位脉冲响应t t e e t g 32121)(--+=;单位阶跃响应t t e e t h 3612132)(----=; (2)单位脉冲响应t e t g t 27s i n 72)(2-=;单位阶跃响应)21.127sin(7221)(2+-=-t e t h t 。
2-3 (1)极点3,1--,零点2-;(2) 极点11j ±-.2-4)2)(1()32(3)()(+++=s s s s R s C . 2-5 (a)()()1121211212212122112+++⋅+=+++=CS R R R R CS R R R R R R CS R R R CS R R s U s U ; (b)()()1)(12221112212121++++=s C R C R C R s C C R R s U s U 2-6 (a)()()RCsRCs s U s U 112+=;(b)()()141112+⋅-=Cs RR R s U s U ; (c)()()⎪⎭⎫⎝⎛+-=141112Cs R R R s U s U . 2-7 设激磁磁通f f i K =φ恒定()()()⎥⎦⎤⎢⎣⎡++++=Θφφπφm e a a a a m a C C f R s J R f L Js L s C s U s 2602. 2-8()()()φφφπφm A m e a a a a m A C K s C C f R i s J R f L i Js iL C K s R s C +⎪⎭⎫⎝⎛++++=26023.2-9 ()2.0084.01019.23-=⨯--d d u i .2-10 (2-6) 2-11(2-7)2-12 前向传递函数)(s G 改变、反馈通道传递函数)(s H 改变可引起闭环传递函数)()(s R s C 改变。
大学精品课件:哈工大_结构力学(王焕定第二版)第四章习题参考答案解密版
4-2(b) 答:本题三次超静定,但在横向荷载下轴力为零,因此可如下求解:
FP
X1 FP
X2
A l/2
B l/2
A B
基本体系
FP A
B FPl/4 MP 图
1
1
A X1 =1
B
M 1图
FPl/8
FPl/8
A
X2 =1 B
M 2图
A M图
B FPl/4
在上述荷载及单位弯矩图下,可图乘求系数、建立力法方程并求解,最后叠加出最终
F
G
3 C
3 3D
E 3
X1 =1
A
B
48 120
M 1图
F
G
72
E
CD
72
120 48
A
B
M (kN ⋅ m)
4-4 (c)本题求解过程如下面图形所示
A
MB
D
C l
基本体系
X1 l
M
M
M
M
图
P
l
M 1图
X1 =1
0.75M 0.25M
M图
δ11 X1 + ∆1P = 0
∑ ∫ δ11 =
2
M1 dx =
A
M图
本体也可取变 B 结点为铰以一对力矩为未知力求解,工作量和上述解法相同。
4-3 (b) 本题也一次超静定,基本体系、荷载与单位弯矩图如下所示
10 kN/m
10 kN/m
B
C
20 kN
B
C
20 kN
X1
3m 3m
A 6m
A
基本体系
180 B
180
结构力学 第四章 作业参考答案
结构力学 第四章习题 参考答案2005级4-1 图示抛物线拱的轴线方程24(fy x l l=−)x ,试求截面K 的内力。
解:(1) 求支座反力801155 kN 16AV AV F F ×=== 0805(5580)0.351500.93625 kN 16BV BV F F ×==−×+×== 0Mc 55880350 kN 4H F f ×−×===(2) 把及代入拱轴方程有:16m l =4m f =(16)16xy =−x (1)由此可得:(8)tan '8x y θ−==(2) 把截面K 的横坐标 ,代入(1),(2)两式可求得: 5m x ==>, 3.44m y =tan 0.375θ= 由此可得:20.56θ= 则有sin 0.351θ=,cos 0.936θ=最后得出截面k 处的内力为: (上标L 表示截面K 在作用力左边,R 则表示截面在作用力右边)055550 3.44103 kN m K H M M F y =−=×−×=i0cos sin 550.936500.35133.93 kN L sK s H F F F θθ=−=×−×= (5580)0.936500.35140.95 kN R sK F =−×−×==40.95 KN 0sin cos 550.351500.93666.1 kN L NK s H F F F θθ=+=×+×= (5580)0.351500.93638.03 kN R NK F =−×+×=4-2 试求拉杆的半圆三铰拱截面K 的内力。
解:(1)以水平方向为X 轴,竖直方向为Y 轴取直角坐标系,可得K 点的坐标为:2m6mK K x y =⎧⎪⎨==⎪⎩ (2)三铰拱整体分别对A ,B 两点取矩,由平衡方程可解得支座反力:0 20210500 20210500 2100A By B Ay x Ax M F M F F F ⎧=×−××⎪⎪=×+××⎨⎪=−×=⎪⎩∑∑∑=== => 5 kN ()20 kN () 5 kN ()Ay Ax By F F F =−⎧⎪=−⎨⎪=⎩向下向上向左(3)把拱的右半部分隔离,对中间铰取矩,列平衡方程可求得横拉杆轴力为:CN 0 105100MF =×−×∑=>N 5 kN F =(4)去如图所示的α角,则有:=>cos 0.6sin 0.8θθ=⎧⎨=⎩于是可得出K 截面的内力,其中:22(6)206525644 kN m 2K M ×=−+×−×−×=isK F (20265)sin 5cos 0.6 kN θθ=−×−×−×=− NK F (20265)cos 5sin 5.8 kN θθ=−−×−×−×=−13K M F r Fr ==(内侧受拉) K 截面作用有力,剪力有突变 且有01sin3032LSK 2F F F F =−=−×=− (2) 22R SK F FF F =−=(3)011sin30(326NKF F F F ==×=拉力)(4)4-4 试求图示三铰拱在均布荷载作用下的合理拱轴线方程。
概率论与数理统计(茆诗松)第二版课后第四章习题参考答案
第四章 大数定律与中心极限定理习题4.11. 如果X X Pn →,且Y X Pn →.试证:P {X = Y } = 1.证:因 | X − Y | = | −(X n − X ) + (X n − Y )| ≤ | X n − X | + | X n − Y |,对任意的ε > 0,有⎭⎬⎫⎩⎨⎧≥−+⎭⎬⎫⎩⎨⎧≥−≤≥−≤2||2||}|{|0εεεY X P X X P Y X P n n ,又因X X Pn →,且Y X Pn →,有02||lim =⎭⎬⎫⎩⎨⎧≥−+∞→εX X P n n ,02||lim =⎭⎫⎩⎨⎧≥−+∞→εY X P n n ,则P {| X − Y | ≥ ε} = 0,取k 1=ε,有01||=⎭⎬⎫⎩⎨⎧≥−k Y X P ,即11||=⎭⎬⎫⎩⎨⎧<−k Y X P , 故11||lim1||}{1=⎭⎬⎫⎩⎨⎧<−=⎭⎬⎫⎩⎨⎧⎭⎬⎫⎩⎨⎧<−==+∞→+∞=k Y X P k Y X P Y X P k k I . 2. 如果X X Pn →,Y Y Pn →.试证:(1)Y X Y X Pn n +→+; (2)XY Y X Pn n →.证:(1)因 | (X n + Y n ) − (X + Y ) | = | (X n − X ) + (Y n − Y )| ≤ | X n − X | + | Y n − Y |,对任意的ε > 0,有⎭⎫⎩⎨⎧≥−+⎭⎬⎫⎩⎨⎧≥−≤≥+−+≤2||2||}|)()({|0εεεY Y P X X P Y X Y X P n n n n ,又因X X P n →,Y Y P n →,有02||lim =⎭⎫⎩⎨⎧≥−+∞→εX X P n n ,02||lim =⎭⎬⎫⎩⎨⎧≥−+∞→εY Y P n n ,故0}|)()({|lim =≥+−++∞→εY X Y X P n n n ,即Y X Y X Pn n +→+;(2)因 | X n Y n − XY | = | (X n − X )Y n + X (Y n − Y ) | ≤ | X n − X | ⋅ | Y n | + | X | ⋅ | Y n − Y |,对任意的ε > 0,有⎭⎬⎫⎩⎨⎧≥−⋅+⎭⎬⎫⎩⎨⎧≥⋅−≤≥−≤2||||2||||}|{|0εεεY Y X P Y X X P XY Y X P n n n n n ,对任意的h > 0,存在M 1 > 0,使得4}|{|1h M X P <≥,存在M 2 > 0,使得8}|{|2hM Y P <≥, 存在N 1 > 0,当n > N 1时,8}1|{|h Y Y P n <≥−, 因| Y n | = | (Y n − Y ) + Y | ≤ | Y n − Y | + | Y |,有4}|{|}1|{|}1|{|22h M Y Y Y P M Y P n n <≥+≥−≤+≥, 存在N 2 > 0,当n > N 2时,4)1(2||2h M X X P n <⎭⎬⎫⎩⎨⎧+≥−ε,当n > max{N 1, N 2}时,有244}1|{|)1(2||2||||22h h h M Y P M X X P Y X X P n n n n =+<+≥+⎭⎬⎫⎩⎨⎧+≥−≤⎭⎬⎫⎩⎨⎧≥⋅−εε,存在N 3 > 0,当n > N 3时,42||1hM Y Y P n <⎭⎬⎫⎩⎨⎧≥−ε,有244}|{|2||2||||11h h h M X P M Y Y P X Y Y P n n =+<≥+⎭⎬⎫⎩⎨⎧≥−≤⎭⎬⎫⎩⎨⎧≥⋅−εε,则对任意的h > 0,当n > max{N 1, N 2, N 3} 时,有h h h Y Y X P Y X X P XY Y X P n n n n n =+<⎭⎬⎫⎩⎨⎧≥−⋅+⎭⎬⎫⎩⎨⎧≥⋅−≤≥−≤222||||2||||}|{|0εεε,故0}|{|lim =≥−+∞→εXY Y X P n n n ,即XY Y X Pn n →.3. 如果X X Pn →,g (x )是直线上的连续函数,试证:)()(X g X g Pn →. 证:对任意的h > 0,存在M > 0,使得4}|{|h M X P <≥, 存在N 1 > 0,当n > N 1时,4}1|{|h X X P n <≥−, 因| X n | = | (X n − X ) + X | ≤ | X n − X | + | X |,则244}|{|}1|{|}1|{|h h h M X P X X P M X P n n =+<≥+≥−≤+≥, 因g (x ) 是直线上的连续函数,有g (x ) 在闭区间 [− (M + 1), M + 1] 上连续,必一致连续, 对任意的ε > 0,存在δ > 0,当 | x − y | < δ 时,有 | g (x ) − g ( y ) | < ε ,存在N 2 > 0,当n > N 2时,4}|{|hX X P n <≥−δ,则对任意的h > 0,当n > max{N 1, N 2} 时,有{}}|{|}1|{|}|{|}|)()({|0M X M X X X P X g X g P n n n ≥+≥≥−≤≥−≤U U δεh hh h M X P M X P X X P n n =++<≥++≥+≥−≤424}|{|}1|{|}|{|δ, 故0}|)()({|lim =≥−+∞→εX g X g P n n ,即)()(X g X g Pn →.4. 如果a X P n →,则对任意常数c ,有ca cX Pn →. 证:当c = 0时,有c X n = 0,ca = 0,显然ca cX Pn →;当c ≠ 0时,对任意的ε > 0,有0||||lim =⎭⎬⎫⎩⎨⎧≥−+∞→c a X P n n ε, 故0}|{|lim =≥−+∞→εca cX P n n ,即ca cX Pn →.5. 试证:X X P n →的充要条件为:n → +∞ 时,有0||1||→⎟⎟⎠⎞⎜⎜⎝⎛−+−XX X X E n n .证:以连续随机变量为例进行证明,设X n − X 的密度函数为p ( y ),必要性:设X X Pn →,对任意的ε > 0,都有0}|{|lim =≥−+∞→εX X P n n ,对012>+εε,存在N > 0,当n > N 时,εεε+<≥−1}|{|2X X P n , 则∫∫∫≥<∞+∞−+++=+=⎟⎟⎠⎞⎜⎜⎝⎛−+−εε||||)(||1||)(||1||)(||1||||1||y y n n dy y p y y dy y p y y dy y p y y XX X X E εεεεεεεεεεεεε=+++<≥−+<−+=++≤∫∫≥<11}|{|}|{|1)()(12||||X X P X X P dy y p dy y p n n y y ,故n → +∞ 时,有0||1||→⎟⎟⎠⎞⎜⎜⎝⎛−+−XX X X E n n ; 充分性:设n → +∞ 时,有0||1||→⎟⎟⎠⎞⎜⎜⎝⎛−+−XX X X E n n , 因∫∫∫≥≥≥++≤++==≥−εεεεεεεεεε||||||)(||1||1)(11)(}|{|y y y n dy y p y y dy y p dy y p X X P ⎟⎟⎠⎞⎜⎜⎝⎛−+−+=++≤∫∞+∞−||1||1)(||1||1X X X X E dy y p y y n n εεεε, 故0}|{|lim =≥−+∞→εX X P n n ,即X X Pn →.6. 设D (x )为退化分布:⎩⎨⎧≥<=.0,1;0,0)(x x x D试问下列分布函数列的极限函数是否仍是分布函数?(其中n = 1, 2, ….)(1){D (x + n )}; (2){D (x + 1/n )}; (3){D (x − 1/n )}.解:(1)对任意实数x ,当n > −x 时,有x + n > 0,D (x + n ) = 1,即1)(lim =++∞→n x D n ,则 {D (x + n )} 的极限函数是常量函数f (x ) = 1,有f (−∞) = 1 ≠ 0,故 {D (x + n )} 的极限函数不是分布函数; (2)若x ≥ 0,有01>+n x ,11=⎟⎠⎞⎜⎝⎛+n x D ,即11lim =⎟⎠⎞⎜⎝⎛++∞→n x D n ,若x < 0,当x n 1−>时,有01<+n x ,01=⎟⎠⎞⎜⎝⎛+n x D ,即01lim =⎟⎠⎞⎜⎝⎛++∞→n x D n ,则⎩⎨⎧≥<=⎟⎠⎞⎜⎝⎛++∞→.0,1;0,01lim x x n x D n 这是在0点处单点分布的分布函数,满足分布函数的基本性质,故⎭⎬⎫⎩⎨⎧⎟⎠⎞⎜⎝⎛+n x D 1的极限函数是分布函数;(3)若x ≤ 0,有01<−n x ,01=⎟⎠⎞⎜⎝⎛−n x D ,即01lim =⎟⎠⎞⎜⎝⎛−+∞→n x D n ,若x > 0,当x n 1>时,有01>−n x ,11=⎟⎠⎞⎜⎝⎛−n x D ,即11lim =⎟⎠⎞⎜⎝⎛−+∞→n x D n ,则⎩⎨⎧>≤=⎟⎠⎞⎜⎝⎛−+∞→.0,1;0,01lim x x n x D n 在x = 0处不是右连续,故⎭⎬⎫⎩⎨⎧⎟⎠⎞⎜⎝⎛−n x D 1的极限函数不是分布函数.7. 设分布函数列 {F n (x )} 弱收敛于连续的分布函数F (x ),试证:{F n (x )} 在 (−∞, +∞) 上一致收敛于分布函数F (x ). 证:因F (x ) 为连续的分布函数,有F (−∞) = 0,F (+∞) = 1,对任意的ε > 0,取正整数ε2>k ,则存在分点x 1 < x 2 < … < x k −1,使得1,,2,1,)(−==k i kix F i L ,并取x 0 = −∞,x k = +∞, 可得k k i k x F x F i i ,1,,2,1,21)()(1−=<=−−L ε, 因 {F n (x )} 弱收敛于F (x ),且F (x ) 连续,有 {F n (x )} 在每一点处都收敛于F (x ),则存在N > 0,当n > N 时,1,,2,1,2|)()(|−=<−k i x F x F i i n L ε,且显然有20|)()(|00ε<=−x F x F n ,20|)()(|ε<=−k k n x F x F ,对任意实数x ,必存在j ,1 ≤ j ≤ k ,有x j −1 ≤ x < x j ,因2)()()()(2)(11εε+<≤≤<−−−j j n n j n j x F x F x F x F x F ,则εεεε−=−−>−−>−−222)()()()(1x F x F x F x F j n ,且εεεε=+<+−<−222)()()()(x F x F x F x F j n ,即对任意的ε > 0和任意实数x ,总存在N > 0,当n > N 时,都有 | F n (x ) − F (x ) | < ε , 故 {F n (x )} 在 (−∞, +∞) 上一致收敛于分布函数F (x ).8. 如果X X Ln →,且数列a n → a ,b n → b .试证:b aX b X a Ln n n +→+. 证:设y 0是F aX + b ( y ) 的任一连续点,则对任意的ε > 0,存在h > 0,当 | y − y 0 | < h 时,4|)()(|0ε<−++y F y F b aX b aX ,又设y 是满足 | y − y 0 | < h 的F aX + b ( y ) 的任一连续点,因⎟⎠⎞⎜⎝⎛−=⎭⎬⎫⎩⎨⎧−≤=≤+=+a b y F a b y X P y b aX P y F X b aX }{)(,有a b y x −=是F X (x )的连续点,且X X L n→, 有)()(lim x F x F X X n n =+∞→,存在N 1,当n > N 1时,4|)()(|ε<−x F x F X X n ,即4|)()(|ε<−++y F y F b aX b aX n ,则当n > N 1且 | y − y 0 | < h 时,2|)()(||)()(||)()(|00ε<−+−≤−++++++y F y F y F y F y F y F b aX b aX b aX b aX b aX b aX n n , 因X 的分布函数F X (x ) 满足F X (−∞) = 0,F X (+∞) = 1,F X (x ) 单调不减且几乎处处连续, 存在M ,使得F X (x ) 在x = ± M 处连续,且41)(ε−>M F X ,4)(ε<−M F X ,因X X Ln →,有41)()(lim ε−>=+∞→M F M F X X n n ,4)()(lim ε<−=−+∞→M F M F X X n n ,则存在N 2,当n > N 2时,41)(ε−>M F n X ,4)(ε<−M F n X ,可得2)(1)(}|{|ε<−+−=>M F M F M X P n n X X n ,因数列a n → a ,b n → b ,存在N 3,当n > N 3时,M h a a n 4||<−,4||h b b n <−, 可得当n > max{N 2, N 3}时,⎭⎫⎩⎨⎧>−+−=⎭⎬⎫⎩⎨⎧>+−+2|)()(|2|)()(|h b b X a a P h b aX b X a P n n n n n n n2}|{|24||42||||||ε<>=⎭⎬⎫⎩⎨⎧>+⋅≤⎭⎬⎫⎩⎨⎧>−+⋅−≤M X P h h X M hP h b b X a a P nn n n n , 则⎭⎬⎫⎩⎨⎧⎭⎬⎫⎩⎨⎧>+−+⎭⎬⎫⎩⎨⎧+≤+≤≤+=+2|)()(|2}{)(000h b aX b X a h y b aX P y b X a P y F n n n n n n n n b X a n n n U222|)()(|200ε+⎟⎠⎞⎜⎝⎛+<⎭⎬⎫⎩⎨⎧>+−++⎭⎬⎫⎩⎨⎧+≤+≤+h y F h b aX b X a P h y b aX P b aX n n n n n n , 且⎭⎬⎫⎩⎨⎧⎭⎬⎫⎩⎨⎧>+−+≤+≤⎭⎬⎫⎩⎨⎧−≤+=⎟⎠⎞⎜⎝⎛−+2|)()(|}{22000h b aX b X a y b X a P h y b aX P h y F n n n n n n n n b aX n U2)(2|)()(|}{00ε+<⎭⎬⎫⎩⎨⎧>+−++≤+≤+y F h b aX b X a P y b X a P n n n b X a n n n n n n n , 即22)(22000εε+⎟⎠⎞⎜⎝⎛+<<−⎟⎠⎞⎜⎝⎛−+++h y F y F h y F b aX b X a b aX n n n n n ,因当n > N 1且 | y − y 0 | < h 时,2)()(2)(00εε+<<−+++y F y F y F b aX b aX b aX n ,在区间⎟⎠⎞⎜⎝⎛++h y h y 00,2取F aX + b ( y ) 的任一连续点y 1,满足 | y 1 − y 0 | < h ,当n > max{N 1, N 2, N 3}时,εεε+<+≤+⎟⎠⎞⎜⎝⎛+<++++)(2)(22)(0100y F y F h y F y F b aX b aX b aX b X a n n n n n ,在区间⎟⎠⎞⎜⎝⎛−−2,00h y h y 取F aX + b ( y ) 的任一连续点y 2,满足 | y 2 − y 0 | < h ,当n > max{N 1, N 2, N 3}时,εεε−>−≥−⎟⎠⎞⎜⎝⎛−>++++)(2)(22)(0200y F y F h y F y F b aX b aX b aX b X a n n n n n ,即对于F aX + b ( y ) 的任一连续点y 0,当n > max{N 1, N 2, N 3}时,ε<−++|)()(|00y F y F b aX b X a n n n , 故)()(y F y F b aX Wb X a n n n ++→,b aX b X a Ln n n +→+. 9. 如果X X Ln →,a Y Pn →,试证:a X Y X Ln n +→+. 证:设y 0是F X + a ( y ) 的任一连续点,则对任意的ε > 0,存在h > 0,当 | y − y 0 | < h 时,4|)()(|0ε<−++y F y F a X a X ,又设y 是满足 | y − y 0 | < h 的F X + a ( y )的任一连续点,因F X + a ( y ) = P {X + a ≤ y } = P {X ≤ y − a } = F X ( y − a ),有x = y − a 是F X (x )的连续点,且X X Ln →, 有)()(lim x F x F X X n n =+∞→,存在N 1,当n > N 1时,4|)()(|ε<−x F x F X X n ,即4|)()(|ε<−++y F y F a X a X n , 则当n > N 1且 | y − y 0 | < h 时,2|)()(||)()(||)()(|00ε<−+−≤−++++++y F y F y F y F y F y F a X a X a X a X a X a X n n ,因a Y Pn →,有02||lim =⎭⎫⎩⎨⎧>−+∞→h a Y P n n ,存在N 2,当n > N 2时,22||ε<⎭⎬⎫⎩⎨⎧>−h a Y P n , 则⎭⎬⎫⎩⎨⎧⎭⎫⎩⎨⎧>−⎭⎬⎫⎩⎨⎧+≤+≤≤+=+2||2}{)(000h a Y h y a X P y Y X P y F n n n n Y X n n U222||200ε+⎟⎠⎞⎜⎝⎛+<⎭⎬⎫⎩⎨⎧>−+⎭⎬⎫⎩⎨⎧+≤+≤+h y F h a Y P h y a X P a X n n n , 且⎭⎬⎫⎩⎨⎧⎭⎫⎩⎨⎧>−≤+≤⎭⎬⎫⎩⎨⎧−≤+=⎟⎠⎞⎜⎝⎛−+2||}{22000h a Y y Y X P h y a X P h y F n n n n a X n U2)(2||}{00ε+<⎭⎬⎫⎩⎨⎧>−+≤+≤+y F h a Y P y Y X P n n Y X n n n , 即22)(22000εε+⎟⎠⎞⎜⎝⎛+<<−⎟⎠⎞⎜⎝⎛−+++h y F y F h y F a X Y X a X n n n n ,因当n > N 1且 | y − y 0 | < h 时,2)()(2)(00εε+<<−+++y F y F y F a X a X a X n ,在区间⎟⎠⎞⎜⎝⎛++h y h y 00,2取F X + a ( y ) 的任一连续点y 1,满足 | y 1 − y 0 | < h ,当n > max{N 1, N 2}时,εεε+<+≤+⎟⎠⎞⎜⎝⎛+<++++)(2)(22)(0100y F y F h y F y F a X a X a X Y X n n n n ,在区间⎟⎠⎞⎜⎝⎛−−2,00h y h y 取F X + a ( y ) 的任一连续点y 2,满足 | y 2 − y 0 | < h ,当n > max{N 1, N 2}时,εεε−>−≥−⎟⎠⎞⎜⎝⎛−>++++)(2)(22)(0200y F y F h y F y F a X a X a X Y X n n n n ,即对于F X + a ( y ) 的任一连续点y 0,当n > max{N 1, N 2}时,ε<−++|)()(|00y F y F a X Y X n n , 故)()(y F y F a X WY X n n ++→,a X Y X Ln n +→+. 10.如果X X Ln →,0Pn Y →,试证:0Pn n Y X →.证:因X 的分布函数F X (x ) 满足F X (−∞) = 0,F X (+∞) = 1,F X (x ) 单调不减且几乎处处连续,则对任意的h > 0,存在M ,使得F X (x ) 在x = ± M 处连续,且41)(h M F X −>,4)(hM F X <−, 因X X L n →,有41)()(lim h M F M F X X n n −>=+∞→,4)()(lim h M F M F X X n n <−=−+∞→,则存在N 1,当n > N 1时,41)(h M F n X −>,4)(hM F n X <−,可得2)(1)(}|{|hM F M F M X P n n X X n <−+−=>,因0Pn Y →,对任意的ε > 0,有0||lim =⎭⎬⎫⎩⎨⎧>+∞→M Y P n n ε,存在N 2,当n > N 2时,2||h M Y P n <⎭⎬⎫⎩⎨⎧>ε, 则当n > max{N 1, N 2}时,有h M Y P M X P M Y M X P Y X P n n n n n n <⎭⎬⎫⎩⎨⎧>+>≤⎭⎬⎫⎩⎨⎧⎭⎬⎫⎩⎨⎧>>≤>εεε||}|{|||}|{|}|{|U ,故0}|{|lim =>+∞→εn n n Y X P ,即0Pn n Y X →.11.如果X X Ln →,a Y Pn →,且Y n ≠ 0,常数a ≠ 0,试证:aXY X L n n →. 证:设y 0是F X / a ( y ) 的任一连续点,则对任意的ε > 0,存在h > 0,当 | y − y 0 | < h 时,4|)()(|0//ε<−y F y F a X a X ,又设y 是满足 | y − y 0 | < h 的F X / a ( y ) 的任一连续点,因)(}{)(/ay F ay X P y a X P y F X a X =≤=⎭⎬⎫⎩⎨⎧≤=,有x = ay 是F X (x )的连续点,且X X Ln →,有)()(lim x F x F X X n n =+∞→,存在N 1,当n > N 1时,4|)()(|ε<−x F x F X X n ,即4|)()(|//ε<−y F y F a X a X n ,则当n > N 1且 | y − y 0 | < h 时,2|)()(||)()(||)()(|0////0//ε<−+−≤−y F y F y F y F y F y F a X a X a X a X a X a X n n ,因X 的分布函数F X (x )满足F X (−∞) = 0,F X (+∞) = 1,F X (x )单调不减且几乎处处连续,存在M ,使得F X (x ) 在x = ± M 处连续,且121)(ε−>M F X ,12)(ε<−M F X ,因X X Ln →,有121)()(lim ε−>=+∞→M F M F X X n n ,12)()(lim ε<−=−+∞→M F M F X X n n ,则存在N 2,当n > N 2时,121)(ε−>M F n X ,12)(ε<−M F n X ,可得6)(1)(}|{|ε<−+−=>M F M F M X P n n X X n ,因0≠→a Y Pn ,有02||lim =⎭⎬⎫⎩⎨⎧>−+∞→h a Y P n n ,存在N 3 > 0,当n > N 3时,62||||ε<⎭⎬⎫⎩⎨⎧>−a a Y P n ,有62||||ε<⎭⎬⎫⎩⎨⎧<a Y P n ,且64||2ε<⎭⎬⎫⎩⎨⎧>−M h a a Y P n , 可得当n > max{N 1, N 2, N 3}时,⎭⎬⎫⎩⎨⎧>⋅−⋅=⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧>−=⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧>−2||||||||2)(2h Y a a Y X P h aY Y a X P h a X Y X P n n n n n n n n n ⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎭⎬⎫⎩⎨⎧<⎭⎬⎫⎩⎨⎧>−>≤2||||4||}|{|2a Y M h a a Y M X P n n n U U22||||4||}|{|2ε<⎭⎬⎫⎩⎨⎧<+⎭⎬⎫⎩⎨⎧>−+>≤a Y P M h a a Y P M X P n n n ,则⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎪⎭⎪⎫⎪⎩⎪⎨⎧>−⎭⎬⎫⎩⎨⎧+≤≤⎭⎬⎫⎩⎨⎧≤=22)(000/h a X Y X h y a XP y Y X P y F n n n n n n Y X n n U22220/0ε+⎟⎠⎞⎜⎝⎛+<⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧>−+⎭⎬⎫⎩⎨⎧+≤≤h y F h a X Y X P h y a X P a X n n n n n ,且⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧>−⎭⎬⎫⎩⎨⎧≤≤⎭⎬⎫⎩⎨⎧−≤=⎟⎠⎞⎜⎝⎛−222000/h a X Y X y Y X P h y a X P h y F n n n nn n a X n U2)(20/0ε+<⎪⎭⎪⎬⎫⎪⎩⎪⎨⎧>−+⎭⎬⎫⎩⎨⎧≤≤y F h a X Y X P y Y X P n n Y X n n n n n ,即22)(220/0/0/εε+⎟⎠⎞⎜⎝⎛+<<−⎟⎠⎞⎜⎝⎛−h y F y F h y F a X Y X a X n n n n ,因当n > N 1且 | y − y 0 | < h 时,2)()(2)(0//0/εε+<<−y F y F y F a X a X a X n ,在区间⎟⎠⎞⎜⎝⎛++h y h y 00,2取F X / a ( y ) 的任一连续点y 1,满足 | y 1 − y 0 | < h ,当n > max{N 1, N 2, N 3}时,εεε+<+≤+⎟⎠⎞⎜⎝⎛+<)(2)(22)(0/1/0/0/y F y F h y F y F a X a X a X Y X n n n n ,在区间⎟⎠⎞⎜⎝⎛−−2,00h y h y 取F X / a ( y ) 的任一连续点y 2,满足 | y 2 − y 0 | < h ,当n > max{N 1, N 2, N 3}时,εεε−>−≥−⎟⎠⎞⎜⎝⎛−>)(2)(22)(0/2/0/0/y F y F h y F y F a X a X a X Y X n n n n ,即对于F X / a ( y ) 的任一连续点y 0,当n > max{N 1, N 2, N 3}时,ε<−|)()(|0/0/y F y F a X Y X n n ,故)()(//y F y F a X WY X n n →,aX Y X L n n →. 12.设随机变量X n 服从柯西分布,其密度函数为+∞<<∞−+=x x n nx p n ,)1π()(22.试证:0Pn X →.证:对任意的ε > 0,)arctan(π2)arctan(π1)1π(}|{|22εεεεεεn nx dx x n n X P n ==+=<−−∫, 则12ππ2)arctan(lim π2}|0{|lim =⋅==<−+∞→+∞→εεn X P n n n , 故0Pn X →.13.设随机变量序列{X n }独立同分布,其密度函数为⎪⎩⎪⎨⎧<<=.,0;0,1)(其他ββx x p其中常数β > 0,令Y n = max{X 1, X 2, …, X n },试证:βPn Y →.证:对任意的ε > 0,P {| Y n − β | < ε} = P {β − ε < Y n < β + ε} = P {max{X 1, X 2, …, X n } > β − ε}= 1 − P {max{X 1, X 2, …, X n } ≤ β − ε} = 1 − P {X 1 ≤ β − ε} P {X 2 ≤ β − ε} … P {X n ≤ β − ε}n⎟⎟⎠⎞⎜⎜⎝⎛−−=βεβ1, 则11lim }|{|lim =⎥⎥⎦⎤⎢⎢⎣⎡⎟⎟⎠⎞⎜⎜⎝⎛−−=<−+∞→+∞→nn n n Y P βεβεβ, 故βPn Y →.14.设随机变量序列{X n }独立同分布,其密度函数为⎩⎨⎧<≥=−−.,0;,e )()(a x a x x p a x 其中Y n = min{X 1, X 2, …, X n },试证:a Y Pn →.证:对任意的ε > 0,P {| Y n − a | < ε} = P {a − ε < Y n < a + ε} = P {min{X 1, X 2, …, X n } < a + ε}= 1 − P {min{X 1, X 2, …, X n } ≥ a + ε} = 1 − P {X 1 ≥ a + ε} P {X 2 ≥ a + ε} … P {X n ≥ a + ε}εεεn na a x n a a x dx −∞++−−∞++−−−=⎟⎠⎞⎜⎝⎛−−=⎟⎠⎞⎜⎝⎛−=∫e 1e 1e 1)()(, 则1)e 1(lim }|{|lim =−=<−−+∞→+∞→εεn n n n a Y P ,故a Y Pn →.15.设随机变量序列{X n }独立同分布,且X i ~ U(0, 1).令nni i n X Y 11⎟⎟⎠⎞⎜⎜⎝⎛=∏=,试证明:c Y P n →,其中c 为常数,并求出c .证:设∑∏===⎟⎟⎠⎞⎜⎜⎝⎛==n i i n i i n n X n X n Y Z 11ln 1ln 1ln ,因X i ~ U (0, 1), 则1)ln (ln )(ln 101−=−==∫x x x xdx X E i ,2)2ln 2ln (ln )(ln 12122=+−==∫x x x x x xdx X E i ,1)](ln [)(ln )Var(ln 22=−=i i i X E X E X , 可得1)(ln 1)(1−==∑=n i i n X E n Z E ,n X nZ ni in 1)Var(ln 1)Var(12==∑=,由切比雪夫不等式,可得对任意的ε > 0,221)Var(}|)({|εεεn Z Z E Z P n n n =≤≥−,则01lim }|)({|lim 02=≤≥−≤+∞→+∞→εεn Z E Z P n n n n ,即0}|)({|lim =≥−+∞→εn n n Z E Z P ,1)(−=→n P n Z E Z ,因n Z n Y e =,且函数e x 是直线上的连续函数,根据本节第3题的结论,可得1e e −→=PZ n n Y , 故c Y Pn →,其中1e −=c 为常数.16.设分布函数列{F n (x )}弱收敛于分布函数F (x ),且F n (x ) 和F (x ) 都是连续、严格单调函数,又设 ξ 服从(0, 1)上的均匀分布,试证:)()(11ξξ−−→F F Pn. 证:因F (x ) 为连续的分布函数,有F (−∞) = 0,F (+∞) = 1,则对任意的h > 0,存在M > 0,使得21)(h M F −>,2)(h M F <−, 因F (x ) 是连续、严格单调函数,有F −1( y ) 也是连续、严格单调函数, 可得F −1( y ) 在区间 [F (− M − 1), F (M + 1)] 上一致连续, 对任意的ε > 0,存在δ > 0,当y , y * ∈ [F (− M − 1), F (M + 1)] 且 | y − y * | < δ 时,| F −1( y ) − F −1( y *) | < ε, 设y * 是 [F (−M ), F (M )] 中任一点,记x * = F −1( y *),有x * ∈ [−M , M ],不妨设0 < ε < 1, 则对任意的x 若满足 ε≥−|*|x x ,就有 δ≥−|*)(|y x F ,根据本节第7题的结论知,{F n (x )} 在 (−∞, +∞) 上一致收敛于分布函数F (x ), 则对δ > 0和任意实数x ,总存在N > 0,当n > N 时,都有 | F n (x ) − F (x ) | < δ, 因当n > N 时,δ<−|)()(|x F x F n 且δ≥−|*(|y x F ,有*)(y x F n ≠,即*)(1y F x n −≠, 则对任意的0 < ε < 1,当n > N 时,*)(1y F n −满足ε<−=−−−−|*)(*)(||**)(|111y F y F x y F n n , 可得对任意的0 < ε < 1,当n > N 时,h M F M F P F F P n −>−∈≥<−−−1)]}(),([{}|)()({|11ξεξξ由h 的任意性可知1}|)()({|lim 11=<−−−+∞→εξξF F P n n ,故)()(11ξξ−−→F F Pn.17.设随机变量序列{X n }独立同分布,数学期望、方差均存在,且E (X n ) = µ,试证:µP n k k X k n n →⋅+∑=1)1(2.证:令∑=⋅+=nk k n X k n n Y 1)1(2,并设Var (X n ) = σ 2, 因µµµ=+⋅+=+=∑=)1(21)1(2)1(2)(1n n n n k n n Y E nk n , 且222212222)1(324)12)(1(61)1(4)1(4)Var(σσσ++=++⋅+=+=∑=n n n n n n n n k n n Y nk n , 则由切比雪夫不等式可得,对任意的ε > 0,222)1(3241)Var(1}|{|1σεεεµ++−=−≥<−≥n n n Y Y P n n , 因1)1(3241lim 22=⎥⎦⎤⎢⎣⎡++−+∞→σεn n n n ,由夹逼准则可得1}|{|lim =<−+∞→εµn n Y P , 故µP n k kn X k n n Y →⋅+=∑=1)1(2. 18.设随机变量序列{X n }独立同分布,数学期望、方差均存在,且E (X n ) = 0,Var (X n ) = σ 2.试证:E (X n ) = 0,Var (X n ) = σ 2.试证:2121σP n k k X n →∑=. 注:此题与第19题应放在习题4.3中,需用到4.3节介绍的辛钦大数定律.证:因随机变量序列}{2n X 独立同分布,且222)]([)Var()(σ=+=n n n X E X X E 存在,故}{2nX 满足辛钦大数定律条件,}{2nX 服从大数定律,即2121σP n k k X n →∑=.19.设随机变量序列{X n }独立同分布,且Var (X n ) = σ 2存在,令∑==n i i X n X 11,∑=−=n i i n X X n S 122)(1.试证:22σPnS →.证:2122112122122121)2(1)(1X X n X n X X X n X X X X n X X n S n i i ni i n i i n i i i n i i n−=⎟⎟⎠⎞⎜⎜⎝⎛+−=+−=−=∑∑∑∑∑=====,设E(X n ) = µ,{X n }满足辛钦大数定律条件,{X n }服从大数定律,即µP nk k X n X →=∑=11,则根据本节第2题第(2)小问的结论知,22µPX →,因随机变量序列}{2n X 独立同分布,且2222)]([)Var()(µσ+=+=n n n X E X X E 存在,则}{2nX 满足辛钦大数定律条件,}{2nX 服从大数定律,即22121µσ+→∑=P n k k X n ,故根据本节第2题第(1)小问的结论知,22222122)(1σµµσ=−+→−=∑=P n i i nX X n S .20.将n 个编号为1至n 的球放入n 个编号为1至n 的盒子中,每个盒子只能放一个球,记⎩⎨⎧=.,0;,1反之的盒子的球放入编号为编号为i i X i 且∑==ni i n X S 1,试证明:0)(Pn n n S E S →−. 证:因n X P i 1}1{==,nX P i 11}0{−==,且i ≠ j 时,)1(1}1{−==n n X X P j i ,)1(11}0{−−==n n X X P j i , 则n X E i 1)(=,⎟⎠⎞⎜⎝⎛−=n n X i 111)Var(, 且i ≠ j 时,)1(1)(−=n n X X E j i ,)1(11)1(1)()()(),Cov(22−=−−=−=n n n n n X E X E X X E X X j i j i j i , 有1)()(1==∑=ni i n X E S E ,1)1(1)1(11),Cov(2)Var()Var(211=−⋅−+−=+=∑∑≤<≤=n n n n n X X X S nj i j i ni i n , 可得0)]()([1)(=−=⎥⎦⎤⎢⎣⎡−n n n n S E S E n n S E S E ,221)Var(1)(Var n S n n S E S n n n ==⎥⎦⎤⎢⎣⎡−, 由切比雪夫不等式,可得对任意的ε > 0,2221)(Var 1)()(εεεn n S E S n S E S E n S E S P n n n n n n =⎥⎦⎤⎢⎣⎡−≤⎭⎬⎫⎩⎨⎧≥⎥⎦⎤⎢⎣⎡−−−, 则01lim )()(lim 022=≤⎭⎬⎫⎩⎨⎧≥⎥⎦⎤⎢⎣⎡−−−≤+∞→+∞→εεn n S E S E n S E S P n n n n n n , 故0)(Pn n nS E S →−.习题4.21. 设离散随机变量X 的分布列如下,试求X 的特征函数.1.02.03.04.03210PX解:特征函数ϕ (t ) = e it ⋅ 0 × 0.4 + e it ⋅ 1 × 0.3 + e it ⋅ 2 × 0.2 + e it ⋅ 3 × 0.1 = 0.4 + 0.3 e it + 0.2 e 2it + 0.1 e 3it .2. 设离散随机变量X 服从几何分布P {X = k } = (1 − p ) k − 1 p , k = 1, 2, … .试求X 的特征函数.并以此求E (X ) 和Var (X ). 解:特征函数ititk k ititk k itk p p p p p p t e)1(1e )]1([ee)1(e )(1111−−=−=−⋅=∑∑+∞=−+∞=−ϕ; 因22]e )1(1[e ]e )1(1[]e )1([e ]e )1(1[e )(it it it it it it it p ip p i p p p i p t −−=−−⋅−−⋅−−−⋅⋅=′ϕ,有)()0(2X iE pip ip ===′ϕ,故pX E 1)(=; 因332]e )1(1[]e )1(1[e ]e )1([]e )1(1[e 2]e )1(1[e )(it it it itit itit itp p p i p p ip p i ip t −−−+−=⋅−−⋅−−−−−⋅⋅=′′−−ϕ, 有)(2)2()0(2223X E i pp p p p =−−=−−=′′ϕ,可得222)(p p X E −=, 故222112)Var(p pp p p X −=⎟⎟⎠⎞⎜⎜⎝⎛−−=. 3. 设离散随机变量X 服从巴斯卡分布rk r p p r k k X P −−⎟⎟⎠⎞⎜⎜⎝⎛−−==)1(11}{,k = r , r + 1, …试求X 的特征函数.解:特征函数∑∑+∞=−−+∞=−−+−−−=−⎟⎟⎠⎞⎜⎜⎝⎛−−⋅=r k r k it r k itr r r k r k r itkp r k k r p p p r k t )(e)1)(1()1()!1(e )1(11e )(L ϕ ∑∑+∞=−=−−−+∞=−=−−=+−−−=r k p x r k r r it rk p x r k r it ititdx x d r p x r k k r p e )1(111e )1()()!1()e ()1()1()!1()e (L itit it p x r r it p x r r r it p x k k r r r it x r r p x dx d r p x dx d r p e )1(e )1(11e )1(1111)1()!1()!1()e (11)!1()e ()!1()e (−=−=−−−=+∞=−−−−−⋅−=⎟⎠⎞⎜⎝⎛−⋅−=⎟⎟⎠⎞⎜⎜⎝⎛⋅−=∑rit itr it r it p p p p ⎥⎦⎤⎢⎣⎡−−=−−=e )1(1e ]e )1(1[)e (. 4. 求下列分布函数的特征函数,并由特征函数求其数学期望和方差.(1))0(,e 2)(||1>=∫∞−−a dt a x F x t a ; (2))0(,1π)(222>+=∫∞−a dt at a x F x . 解:(1)因密度函数||11e 2)()(x a ax F x p −=′=,故⎥⎥⎦⎤⎢⎢⎣⎡−++=⎥⎦⎤⎢⎣⎡+=⋅=+∞−∞−+∞+−∞−+∞+∞−−∫∫∫0)(0)(0)(0)(||1e e 2e e 2ee 2)(ait a it a dx dx a dx a t x a it x a it x a it x a it x a itx ϕ 222112at a a it a it a +=⎟⎠⎞⎜⎝⎛−−+=; 因222222221)(22)()(a t ta t a t a t +−=⋅+−=′ϕ,有)(0)0(1X iE ==′ϕ, 故E (X ) = 0;因32242242222222221)(26)(2)(22)(2)(a t a t a a t t a t t a a t a t +−=+⋅+⋅−+⋅−=′′ϕ, 有)(22)0(222641X E i a a a =−=−=′′ϕ,可得222)(a X E =, 故222202)Var(aa X =−=;(2)因密度函数22221π)()(ax a x F x p +⋅=′=, 则∫+∞∞−+⋅=dx a x a t itx 2221e π)(ϕ, 由第(1)小题的结论知∫∞+∞−=+=dx x p a t a t itx )(e )(12221ϕ,根据逆转公式,可得∫∫∞+∞−−∞+∞−−−+⋅===dt at a dt t a x p itx itx x a 2221||1e π21)(e π21e 2)(ϕ, 可得||||222e πe 2π21e y a y a itya a a dt a t −−−+∞∞−=⋅=+⋅∫, 故||||222e e ππ1e π)(t a t a itx a a dx ax a t −−+∞∞−=⋅=+⋅=∫ϕ; 因⎩⎨⎧>−<=′−,0,e ,0,e )(2t a t a t atat ϕ 有a a −=+′≠=−′)00()00(22ϕϕ,即)0(2ϕ′不存在, 故E (X ) 不存在,Var (X ) 也不存在.5. 设X ~ N (µ, σ 2),试用特征函数的方法求X 的3阶及4阶中心矩. 解:因X ~ N (µ, σ 2),有X 的特征函数是222e)(t t i t σµϕ−=,则)(e)(2222t i t t t i σµϕσµ−⋅=′−,)(e)(e )(222222222σσµϕσµσµ−⋅+−⋅=′′−−t t i t t i t i t ,因)()(3e)(e)(2223222222σσµσµϕσµσµ−⋅−⋅+−⋅=′′′−−t i t i t t t i t t i ,有ϕ″′(0) = e 0 ⋅ (i µ )3 + e 0 ⋅ 3i µ ⋅ (−σ 2) = − i µ 3 − 3i µσ 2 = i 3E (X 3) = − i E (X 3), 故E (X 3) = µ 3 + 3µσ 2; 又因2222222422)4()(3e)()(6e)(e)(222222σσσµσµϕσµσµσµ−⋅+−⋅−⋅+−⋅=−−−t t i t t i t t i t i t i t ,有ϕ (4)(0) = e 0 ⋅ (i µ )4 + e 0 ⋅ 6(i µ)2 ⋅ (−σ 2) + e 0 ⋅ 3σ 4 = µ 4 + 6µ 2σ 2 + 3σ 4 = i 4E (X 4) = E (X 4), 故E (X 4) = µ 4 + 6µ 2σ 2 + 3σ 4.6. 试用特征函数的方法证明二项分布的可加性:若X ~ b (n , p ),Y ~ b (m , p ),且X 与Y 独立,则X + Y ~ b (n + m , p ).证:因X ~ b (n , p ),Y ~ b (m , p ),且X 与Y 独立,有X 与Y 的特征函数分别为ϕ X (t ) = ( p e it + 1 − p ) n ,ϕ Y (t ) = ( p e it + 1 − p ) m , 则X + Y 的特征函数为ϕ X + Y (t ) = ϕ X (t ) ⋅ϕ Y (t ) = ( p e it + 1 − p ) n + m ,这是二项分布b (n + m , p )的特征函数, 故根据特征函数的唯一性定理知X + Y ~ b (n + m , p ).7. 试用特征函数的方法证明泊松分布的可加性:若X ~ P (λ1),Y ~ P (λ2),且X 与Y 独立,则X + Y ~ P (λ1 + λ2).证:因X ~ P (λ1),Y ~ P (λ2),且X 与Y 独立,有X 与Y 的特征函数分别为)1(e1e )(−=itt X λϕ,)1(e2e )(−=itt Y λϕ,则X + Y 的特征函数为)1)(e(21e )()()(−++==itt t t Y X Y X λλϕϕϕ,这是泊松分布P (λ1 + λ2)的特征函数,故根据特征函数的唯一性定理知X + Y ~ P (λ1 + λ2).8. 试用特征函数的方法证明伽马分布的可加性:若X ~ Ga (α1, λ),Y ~ Ga (α2, λ),且X 与Y 独立,则X + Y ~ Ga (α1 + α2 , λ).证:因X ~ Ga (α1, λ),Y ~ Ga (α2, λ),且X 与Y 独立,有X 与Y 的特征函数分别为11)(αλϕ−⎟⎠⎞⎜⎝⎛−=it t X ,21)(αλϕ−⎟⎠⎞⎜⎝⎛−=it t Y ,则X + Y 的特征函数为)(211)()()(ααλϕϕϕ+−+⎟⎠⎞⎜⎝⎛−==it t t t Y X Y X ,这是伽马分布Ga (α1 + α2 , λ)的特征函数,故根据特征函数的唯一性定理知X + Y ~ Ga (α1 + α2 , λ).9. 试用特征函数的方法证明χ 2分布的可加性:若X ~ χ 2 (n ),Y ~ χ 2 (m ),且X 与Y 独立,则X + Y ~ χ 2 (n + m ).证:因X ~ χ 2 (n ),Y ~ χ 2 (m ),且X 与Y 独立,有X 与Y 的特征函数分别为2)21()(n X it t −−=ϕ,2)21()(m Y it t −−=ϕ,则X + Y 的特征函数为2)21()()()(m n Y X Y X it t t t +−+−==ϕϕϕ,这是χ 2分布χ 2 (n + m )的特征函数,故根据特征函数的唯一性定理知X + Y ~ χ 2 (n + m ).10.设X i 独立同分布,且X i ~ Exp(λ),i = 1, 2, …, n .试用特征函数的方法证明:),(~1λn Ga X Y ni i n ∑==.证:因X i ~ Exp (λ),i = 1, 2, …, n ,且X i 相互独立,有X i 的特征函数为11)(−⎟⎠⎞⎜⎝⎛−=−=λλλϕit it t i X ,则∑==ni i n X Y 1的特征函数为nni X Y it t t i n −=⎟⎠⎞⎜⎝⎛−==∏λϕϕ1)()(1,这是伽马分布Ga (n , λ)的特征函数,故根据特征函数的唯一性定理知Y n ~ Ga (n , λ).11.设连续随机变量X 的密度函数如下:+∞<<∞−−+⋅=x x x p ,)(π1)(22µλλ, 其中参数λ > 0, −∞ < µ < +∞,常记为X ~ Ch (λ, µ ).(1)试证X 的特征函数为exp{i µ t − λ | t |},且利用此结果证明柯西分布的可加性; (2)当µ = 0, λ = 1时,记Y = X ,试证ϕ X + Y (t ) = ϕ X (t ) ⋅ϕ Y (t ),但是X 与Y 不独立;(3)若X 1, X 2, …, X n 相互独立,且服从同一柯西分布,试证:)(121n X X X n+++L 与X 1同分布. 证:(1)根据第4题第(2)小题的结论知:若X *的密度函数为22π1)(*xx p +⋅=λλ,即X * ~ Ch (λ, 0), 则X *的特征函数为ϕ * (t ) = e −λ | t |,且X = X * + µ 的密度函数为22)(π1)(µλλ−+⋅=x x p , 故X 的特征函数为ϕ X (t ) = e i µ t ϕ * (t ) = e i µ t ⋅ e −λ | t | = e i µ t −λ | t |; 若X 1 ~ Ch (λ1, µ1),X 2 ~ Ch (λ2, µ2),且相互独立,有X 1与X 2的特征函数分别为||111e )(t t i X t λµϕ−=,||222e )(t t i X t λµϕ−=, 则X 1 + X 2的特征函数为||)()(21212121e )()()(t t i X X X X t t t λλµµϕϕϕ+−++==,这是柯西分布Ch (λ1 + λ2, µ1 + µ2)的特征函数,故根据特征函数的唯一性定理知X 1 + X 2 ~ Ch (λ1 + λ2, µ1 + µ2); (2)当µ = 0, λ = 1时,X ~ Ch (1, 0),有X 的特征函数为ϕ X (t ) = e −| t |,又因Y = X ,有Y 的特征函数为ϕ Y (t ) = e −| t |,且X + Y = 2X ,故X + Y 的特征函数为ϕ X + Y (t ) = ϕ 2X (t ) = ϕ X (2t ) = e −| 2t | = e −| t | ⋅ e −| t | =ϕ X (t ) ⋅ϕ Y (t ); 但Y = X ,显然有X 与Y 不独立;(3)因X i ~ Ch (λ, µ ),i = 1, 2, …, n ,且X i 相互独立,有X i 的特征函数为||e )(t t i X t i λµϕ−=, 则)(121n n X X X nY +++=L 的特征函数为 )(e e )()(1||111t n t t t X t t i n t n ti n ni X ni X nY i in ϕϕϕϕλµλµ===⎟⎠⎞⎜⎝⎛==−⎟⎟⎠⎞⎜⎜⎝⎛⋅−⋅==∏∏,故根据特征函数的唯一性定理知)(121n X X X n+++L 与X 1同分布. 12.设连续随机变量X 的密度函数为p (x ),试证:p (x ) 关于原点对称的充要条件是它的特征函数是实的偶函数.证:方法一:根据随机变量X 与−X 的关系充分性:设X 的特征函数ϕ X (t )是实的偶函数,有ϕ X (t ) = ϕ X (−t ),则−X 的特征函数ϕ −X (t ) = ϕ X (−t ) = ϕ X (t ),根据特征函数的唯一性定理知−X 与X 同分布,因X 的密度函数为p (x ),有−X 的密度函数为p (−x ),故由−X 与X 同分布可知p (−x ) = p (x ),即p (x ) 关于原点对称; 必要性:设X 的密度函数p (x ) 关于原点对称,有p (−x ) = p (x ), 因−X 的密度函数为p (−x ),即−X 与X 同分布,则−X 的特征函数ϕ −X (t ) = ϕ X (−t ) = ϕ X (t ),且)(][e ][e ][e )()()(t E E E t t X itX itX X it X X ϕϕϕ=====−−−, 故X 的特征函数ϕ X (t )是实的偶函数. 方法二:根据密度函数与特征函数的关系充分性:设连续随机变量X 的特征函数ϕ X (t )是实的偶函数,有ϕ X (t ) = ϕ X (−t ),因∫+∞∞−−=dt t x p itx )(e π21)(ϕ,有∫∫+∞∞−+∞∞−−−==−dt t dt t x p itxx it )(e π21)(e π21)()(ϕϕ, 令t = −u ,有dt = −du ,且当t → −∞时,u → +∞;当t → +∞时,u → −∞,则)()(e π21)(e π21))((e π21)()(x p du u du u du u x p iuxiux x u i ==−=−−=−∫∫∫+∞∞−−+∞∞−−−∞∞+−ϕϕϕ, 故p (x ) 关于原点对称;必要性:设X 的密度函数p (x ) 关于原点对称,有p (−x ) = p (x ),因∫+∞∞−−==dx x p E t itxitX)(e )(e)(ϕ,有∫∫+∞∞−−+∞∞−−==−dx x p dx x p t itx xt i )(e )(e)()(ϕ,令x = −y ,有dx = −dy ,且当x → −∞时,y → +∞;当x → +∞时,y → −∞, 则)()(e )(e ))((e )()(t dy y p dy y p dy y p t X ity ity y it X ϕϕ==−=−−=−∫∫∫+∞∞−+∞∞−−∞∞+−−,且)(][e ][e ][e )()()(t E E E t t X itX itX X t i X X ϕϕϕ====−=−−, 故X 的特征函数ϕ X (t )是实的偶函数.13.设X 1, X 2, …, X n 独立同分布,且都服从N(µ , σ 2)分布,试求∑==ni i X n X 11的分布.证:因X i ~ N (µ , σ 2),i = 1, 2, …, n ,且X i 相互独立,有X i 的特征函数为222e)(t t i X t i σµϕ−=,则∑==n i i X n X 11的特征函数为nt t i n t n t i n ni X n i X n X n t t t i i 2211112222ee)()(σµσµϕϕϕ−⎥⎥⎦⎤⎢⎢⎣⎡⎟⎠⎞⎜⎝⎛−⋅====⎟⎠⎞⎜⎝⎛==∏∏,这是正态分布⎟⎟⎠⎞⎜⎜⎝⎛n N 2,σµ的特征函数,故根据特征函数的唯一性定理知⎟⎟⎠⎞⎜⎜⎝⎛=∑=n N X n X ni i 21,~1σµ. 14.利用特征函数方法证明如下的泊松定理:设有一列二项分布{b (k , n , p n )},若λ=→∞n n np lim ,则L ,2,1,0,e !),,(lim ==−∞→k k p n k b kn n λλ.证:二项分布b (n , p n )的特征函数为ϕ n (t ) = ( p n e it + 1 − p n ) n = [1 + p n (e it − 1)] n ,且n → ∞时,p n → 0,因)1(e)1(e )1(e 1e )]1(e 1[lim )]1(e 1[lim )(lim −−⋅−→→∞→∞=−+=−+=itit n it n n np p itn p n it n n n n p p t λϕ,。
杭电测试技术第四章习题参考答案
解:(1)若假设电阻应变与钢质弹性元件不粘贴,温度变化20℃之后长度 变化为:
应变片:Ls Ls0 Ls0 s 20 3.2 104 Ls0
Ls (1 3.2 104 )Ls0
弹性元件:Lg Lg0 Lg0 g 20 2.4 104 Lg0
解:(1)
R k 2.05 800106 1.64 103
R R 1.64 103 120 0.1968
(2)
u0
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3 1.64 103 4
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u' E( R1 R1 R3 ) 1.229mv
0
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u' 0
u0
100%
解:参见教材P58
1
第4章 应变式传感器
习题参考答案
4-3 一应变片的电阻R=120Ω,K=2.05,用做最大应变为ε=800μm/m的传
感元件。当弹性体受力变形至最大应变时,
(1)求ΔR和ΔR/R; (2)若将应变片接入电桥单臂,其余桥臂电阻均为120Ω的固定电阻, 供桥电压U=3V,求传感元件最大应变时单臂电桥的输出电压U。和非 线性误差。
Lg (1 2.4 104 )Lg0
5
第4章 应变式传感器
习题参考答案
粘贴在一起后,L s0
Lg0
L0
则附加应变为:
L L0
Ls g L0
8105
附加电阻变化为:R KR0 0.0192
(2)应变片粘贴后的电阻温度系数为:
0 K (s g ) 2.8 105
单位温度变化引起的虚应变为:
0.082%
信号与系统 第4章-作业参考答案
题图 4-3-1 解:
11
第四章 傅立叶分析
第 4 章 习题参考答案
4-3-7
1)x(t)是实周期信号,且周期为 6; 3)x(t) = −x(t − 3)
1 3
设某信号x(t)满足下述条件:
2)x(t)的傅里叶系数为ak ,且当k = 0 和 k > 2时,有ak = 0;
1
4) ∫−3 |x(t)|2dt = 6 2 5)a1是正实数。
第四章 傅立叶分析
第 4 章 习题参考答案
第 4 章 习题参考答案
4-1 思考题 答案暂略 4-1 练习题 4-2-2 已知三个离散时间序列分别为 x1 ( n) = cos
2πn 2πn , x3 (n) = e , x 2 (n) = sin 25 10
π x (t ) = sin 4π t + cos 6π t + 时,试求系统输出 y (t ) 的傅立叶级数。 4
解:
3
第四章 傅立叶分析
第 4 章 习题参考答案
4因果系统: y(t) + 4y(t) = x(t)
式中x(t) 为系统输入,y(t)是系统输出。在下面两种输入条件下,求输出y(t)的傅里叶级数 展开: 1)x(t) = cos2πt ;
2
2
= 3 ) f ( t ) Sa (100t ) + Sa
解:
( 60t ) 4)
sin(4π t ) , −∞ < t < ∞ πt
9
第四章 傅立叶分析
第 4 章 习题参考答案
4)T=1/4 4-2-27 设 x(t ) 是一实值信号,在采样频率 ω s = 10000π 时, x(t ) 可用其样本值唯一确定
概率论与数理统计第四章习题参考答案
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=
0.95
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故,正态总体均值 µ 的 95%的置信区间为 (X − 2.94, X + 2.94)
代入样本值得正态总体均值 µ 的 95%的置信区间为(-2.565,3.315)。
(2)当σ 未知时,由 T = X − µ ~ t(n − 1) 即T = X − µ ~ t(3) ,所以
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计量经济学精要习题参考答案(第四版)
计量经济学(第四版)习题参考答案第一章 绪论1.1 一般说来,计量经济分析按照以下步骤进行:(1)陈述理论(或假说) (2)建立计量经济模型 (3)收集数据 (4)估计参数 (5)假设检验 (6)预测和政策分析1.2 我们在计量经济模型中列出了影响因变量的解释变量,但它(它们)仅是影响因变量的主要因素,还有很多对因变量有影响的因素,它们相对而言不那么重要,因而未被包括在模型中。
为了使模型更现实,我们有必要在模型中引进扰动项u 来代表所有影响因变量的其它因素,这些因素包括相对而言不重要因而未被引入模型的变量,以及纯粹的随机因素。
1.3时间序列数据是按时间周期(即按固定的时间间隔)收集的数据,如年度或季度的国民生产总值、就业、货币供给、财政赤字或某人一生中每年的收入都是时间序列的例子。
横截面数据是在同一时点收集的不同个体(如个人、公司、国家等)的数据。
如人口普查数据、世界各国2000年国民生产总值、全班学生计量经济学成绩等都是横截面数据的例子。
1.4 估计量是指一个公式或方法,它告诉人们怎样用手中样本所提供的信息去估计总体参数。
在一项应用中,依据估计量算出的一个具体的数值,称为估计值。
如Y 就是一个估计量,1nii YY n==∑。
现有一样本,共4个数,100,104,96,130,则根据这个样本的数据运用均值估计量得出的均值估计值为5.107413096104100=+++。
第二章 计量经济分析的统计学基础2.1 略,参考教材。
2.2 NS S x ==45=1.25 用α=0.05,N-1=15个自由度查表得005.0t =2.947,故99%置信限为 x S t X 005.0± =174±2.947×1.25=174±3.684也就是说,根据样本,我们有99%的把握说,北京男高中生的平均身高在170.316至177.684厘米之间。
2.3 原假设 120:0=μH备择假设 120:1≠μH2检验统计量()10/25XX μσ-Z ====查表96.1025.0=Z 因为Z= 5 >96.1025.0=Z ,故拒绝原假设, 即此样本不是取自一个均值为120元、标准差为10元的正态总体。
有机化学课后习题参考答案(汪小兰第四版)
绪论1.1 扼要归纳典型的以离子键形成的化合物与以共价键形成的化合物的物理性质。
1.2答案: NaCl 与KBr 各1mol 与NaBr 及KCl 各1mol 溶于水中所得溶液相同。
因为两者溶液中均为Na + , K + , Br -, Cl -离子各1mol 。
由于CH 4 与CCl 4及CHCl 3与CH 3Cl 在水中是以分子状态存在,所以是两组不同的混合物。
1.3C+624HCCH 4中C 中有4个电子与氢成键为SP 3杂化轨道,正四面体结构CH 4SP 3杂化2p y2p z2p xH1.4答案:a.C C H H H HCC HH HH或 b.H C H c.H N Hd.H S He.H O NO f.O C H Hg.OP O O H H Hh.H C C HHH H HO P O O H HH或i.H C C Hj.O S O HH OH H或1.5 答案:b.ClClc.H Brd.He.H3COH H3COCH3f.1.6根据S与O的电负性差别,H2O与H2S相比,哪个有较强的偶极-偶极作用力或氢键?答案:电负性O > S , H2O与H2S相比,H2O有较强的偶极作用及氢键。
1.7下列分子中那些可以形成氢键?答案:d. CH3NH2e. CH3CH2OH1.8醋酸分子式为CH3COOH,它是否能溶于水?为什么?答案:能溶于水,因为含有C=O和OH两种极性基团,根据相似相容原理,可以溶于极性水。
第二章饱和烃2.1卷心菜叶表面的蜡质中含有29个碳的直链烷烃,写出其分子式。
C29H602.2用系统命名法(如果可能的话,同时用普通命名法)命名下列化合物,并指出(c)和(d)中各碳原子的级数。
答案:a. 2,4,4-三甲基-5-正丁基壬烷b. 正己烷hexanec. 3,3-二乙基戊烷d.3-甲基-5-异丙基辛烷 e. 2-甲基丙烷(异丁烷) f. 2,2-二甲基丙烷(新戊烷)g. 3-甲基戊烷h. 2-甲基-5-乙基庚烷2.3下列各结构式共代表几种化合物?用系统命名法命名。
《语言学概论》课后习题参考答案
《语言学概论》练习(一)参考答案一、填空1、中国、印度、希腊—罗马具有悠久的历史文化传统,是语言学的三大发源地。
2、历史比较语言学是在19世纪逐渐发展和完善的,它是语言学走上独立发展道路的标志。
3、人的大脑分左右两半球,大脑的左半球控制语言活动,右半球掌管不需要语言的感性直观思维。
4、一个符号,如果没有意义,就失去了存在的必要,如果没有形式,我们就无法感知,符号也就失去了存在的物质基础。
5、用什么样的语音形式代表什么样的意义,完全是由使用这种语言的社会成员约定俗成。
6、语言符号具有任意性和线条性特点。
7、语言的底层是一套音位,上层是符号和符号的序列,可以分为若干级,第一级是语素,第二级是词,第三级是句子。
8、语言系统中的所有符号,既可以同别的符号组合,又可以被别的符号替换,符号之间的这种关系是组合关系和聚合关系。
9、组合关系是指符号与符号相互之间在功能上的联系,聚合关系是指符号在性质上的归类。
二、判断正误1、文字是人类最重要的交际工具。
(错)2、地主阶级和农民阶级之间没有共同语言,这说明语言是有阶级性的。
(错)3、在现代社会,文字比语言更加重要。
(错)4、现代社会,沟通的方式很多,语言的重要性日渐削弱。
(错)5、语言是思维的工具,没有语言,人类就无法思维。
(对)6、语言和思维互相依存,共同发展。
(对)7、任何一种符号,都是由内容和意义两个方面构成的。
(错)8、从本质上看,语言其实是一种符号系统。
(对)9、人类选择语音而不是色彩、手势作为语言符号的形式,是因为语音比较好听。
(错)10、语言符号的约定俗成是指语音形式和意义内容的结合是社会成员共同约定认同的。
(对)三、问答题:1、语言的作用是什么,举例说明。
为什么说语言是最重要的交际工具?答:它是人类社会的交际工具。
每个社会,无论它是经济发达的社会,还是经济十分落后的社会,都必须有属于自己的语言,都离不开语言这个交际工具,语言是组成社会必不可少的一个因素,是人类与动物区别的重要特征之一。
高等院校计算机基础教育规划教材《C++程序设计》课后习题参考答案武汉大学出版社
程序设计如下:
# include <iostream.h>
void main()
{
int a[3][3],i,j,suml=0,sum2=0;
cout<<"输入数组元素:\n";
for(i=0;i<3;i++)
for(j=0;j<3:j++)
cin a[i][j];
高等院校计算机基础教育规划教材《C++程序设计》课后习题参考答案
――武汉大学出版社
习题1参考答案
一、选择题
1.C 2.C
二、填空题
1.面向过程,面向对象
2.编辑,编译,链接,运行
}
5.编写程序,显示如下图形。
*
* * *
* * * * *
* * * * * * *
* * * * *
* * *
*
程序设计如下:
# include<iostream.h>
void main()
void main()
{
int i,j,m,a[15];
for(i=0;i<15;i++)
cina[i];
for(j=1;j<14;j++)
for(i=0;i<15-j;i++)
if(a[i]>a[i+1])
{
m=a[i];a[i]=a[i+1];a[i+1]=m;
void main()
{
char str[80],c1,c2=′′;
第四章习题参考答案
4-1题答:金属电阻应变片与半导体应变片的区别:工作原理不同:金属电阻应变片是基于金属的电阻应变效应,即金属丝在外力作用下产生机械变形时,其电阻阻值发生变化。
半导体应变片是基于晶体的压阻效应,即单晶体材料在沿某一方向受到外力作用时,电阻率发生相应变化而引起电阻阻值变化。
各自优缺点:金属电阻应变片灵敏度一般在1.7-4.0之间,但温度稳定性较好,用于测量精度要求较高场合;具有机械滞后性、蠕变等缺点。
半导体应变片的灵敏度比金属电阻应变片的高50-70倍,横向效应和机械滞后小、体积小,但温度温度性差,在较大应变下,灵敏度的非线性误差大。
如何选用:在温度变化较大,要求精度较高且应变变形相对较大的场合选用金属电阻应变片;在温度变化较小,应变变形较小的情况下优选半导体应变片。
4-2题 解:62120100010024.R K R KR R-∆=ε⇒∆=ε=⨯⨯⨯=Ω (1)1500125125120...U I A mA R ==== (2)15001247512475120024....U I A mA R R ====+∆+ (3)125124750025...I mA ∆=-=(4)示值差异太小,无法区分。
4-3题答:差动传感器的优点:测量精度高、线性范围大、稳定性好和使用方便等优点。
原理:将两个结构相同的传感器正反相连组成一个差动传感器,当活动端(如衔铁)处于中间位置时,位移为零,输出电压为零(处于平衡态)。
当活动端向一个方向偏移时,其中一个传感器感受正偏移,另一个传感器感受等值的负偏移,由于两个传感器是反向串接,则实际输出量是由一个传感器感受变化的两倍,且将由外界引起的等值偏差抵消掉,起到补偿作用,因而精度较高、线性范围大、稳定性好。
4-4题 解:312222221088510449403C A r PF --∆δ∆δ±∆=-ε=-⨯⨯π⨯⨯=-π⨯⨯=δδ... 4-5题答:压电效应:某些材料,当沿着一定方向受到作用力时,不但产生机械变形,而且内部极化,表面有电荷出现,当外力去掉后,又重新恢复到不带电状态的现象。
现代大学英语精读(4)课后习题参考答案Unit 1 to Unit 5复习课程
(1)浴巾 (2)(美)小学 (3)永恒的真理 (4)文件柜 (5)纯属无稽之谈 (6)违规行为 (7)常客 (8)新鲜空气 (9)格调很高的独自(一个人唱高调) (10)一种固定的观点 (11)时事(当前国内外大事) (12)身体障碍 (13)可怕吓人的风 (14)令人厌恶的景象 (15)言语障碍(16)使人兴奋冲动的爱国激情 (17)无情的人侵者 (18)首相(19)国际联盟(国联) (20)思维过程(思想方式) (21)条理清楚的文章 (22)一个完整的体系 (23)一位口译好手(24)一种不可阻挡的趋向 (25)烂苹果(26)根据事实(启示)写成的 (27)一位点头之交现代大学英语精读(4)课后习题参考答案Unit 1 to Unit 5Unit 1I Translate_Phrases1) Into Chinese2) Into EnglishⅡ. Translate_Vocabulary1) I knew I could expect my brother to stand by me whatever happened.2) As a general rule, young people tend to be more interested in the present and the future. 3) Both sides will stand to lose if they do not compromise.4) It is our hope to integrate all the courses and teaching materials.5) The Chinese written language has been a major factor for integrating our nation. 6) In traditional Chinese art, the bamboo stands for moral integrity and uprightness. 7) The great majority of the people stand for reform.(1) to sink one's head (2) to sink the ship (3) to contemplate the meaning of life (4) to catch the light (5) to ruin one's health (6) to ruin the country (7) to bang the desk (8) to play a prominent role (9) to hold a prominent position (10) a pious Buddhist (11) to gain a reputation (12) to satisfy one's ego(13) to give sb the third degree (14) to devise a teaching method (15) to slide a gun into sb's hand8) Queen Elizabeth the First ruled England for 45 years, and the country prospered under her rule.9) The truth is always in the hands of a small minority at first. That's the rule.10) Democracy means that the majority rules, but the minority's right to disagree is also respected. These two basic rules are of equal importance.11) A nation cannot be strong unless it is well-integrated economically, politically and culturally as well as geographically.12) The party was boring, so she slipped out of the room and went home.13) The road was muddy. He slipped and fell into the river.14) One day I was drowning my sorrows in a restaurant because I was broke when he came and slipped a roll of money into my hand.15) The Court of Florida ruled that it was necessary to recount the votes.16) The idea that the sun moves round the earth ruled ancient scholars for more than a thousand years.17) The hutongs are an integral part of old Beijing.18) Days slipped by and I still had not made much progress.19) He weighed every word carefully lest he should make a mistake.20) Her health was such that she would not go out in the sun even in winter lest she got sunstroke.Ⅲ. Translate_Grammar1) You can force a student to attend classes, but you cannot force him to think.Or: You can force a student to attend classes, but not to think.2) The study of literature can help you to understand not only other people but also yourself.3) You can improve your writing by reading good models and by practicing writing.4) In the Middle Ages, people believed that the earth was flat and that it was the center of the universe.5) I can afford neither the time nor the money to play golf with those big cheeses.6) Xiao Jin could not decide whether to apply for graduate studies right after college or to get a job first.7) Love cures people-both those who give it and those who receive it.8) Excellent firms don't believe in perfection, only in constant improvement and constant change.9) Many things cannot be learned in the classroom, such as planning one's time, working on one's own and managing one's own affairs.10) In the past ten years people, especially old people, have been concerned more about their health than about their income.Ⅳ. Paraphrase1) Nature had endow everybody except me the ability to think which is born with.2) You could hear that the fresh air had to struggle with difficulty to find its way to his chest, because he was unaccustomed to this as his lungs had been harmed by drinking. His body would lose balance and his face would become pale as a result of the unexpected visit of the wind. He would go back to his desk unsteadily and fall into the chair, unable to do anything for the rest of the morning.3) Mr. Houghton’s deeds told me that he was not ruled by thought; instead, he would feela strong urge to turn his head and look at the girls.4) Technically speaking, it is as skillful as most businessmen’s golf playing, as honest as most politicians’ purpose, and as consistent as most books’ content.5) As they are everywhere and so daunting in number that we’d better not offend them.6) Humans enjoy following the crowd as it can bring them peace, security, comfort and harmony, which is like cows eating grass on the same side of a hill.7) Our Prime Minister was a hypocrite to say that the imprisonment of the two major leaders of Free-India Movement-Nehru and Gandhi-was good for India. The American politicians were dissimulators to talk about peace but refuse to join the League of Nations. Those moments made me feel happy.8) I slid my arm around her waist and whispered that if we were talking about the number of people who believed in a certain religion, I believed the Buddhists were greater in number. My “indecent” behavior and the daunting number of the Buddhists scared her away.9) What had happened to Ruth and me now happened again. Although some close friends of mine still stuck by me, my grad-one thinking scared away many of my acquaintances.(1)夜生活 (2)吃和住(3)供吃住的寄宿舍 (4)一秒钟都不到 (5)玻璃弹子(6)抽打死马 (做徒劳无益的事) (7)阿司匹林片 (8)在此情况下 (9)提前/事先 (10)走过场(11)楼梯间平台的窗户 (12)紧张气氛 (13)毛线针 (14)梦游 (15)飞机翼展Unit 2I Translate_Phrases1) Into Chinese 2) Into EnglishⅡ. Translate_Vocabulary1) It is a miracle how our company has become a multinational in such a short span of time.2) The average life span in that country has increased from 42 years to 50 years in a matter of two decades.3) The conflict between the two countries has spanned more than half a century. 4) There are four bridges spanning the river.5) I’m much obliged t o you. Without your help, I would never have finished the book. 6) No, you are not obliged to go to the party. You don’t have to go if you don’t want to. 7) She’s always ready to oblige when people come to her for help.8) In the valley is a small lake right between a meadow and a hill. It is a perfect spot for a picnic.9) Sitting in a shady spot, he soon dozed off.10) He criticized me on the spot when he saw me throw a plastic bad down by the roadside.(1) to whip up a little interest (2) to keep the ball rolling (3) to set the ball rolling (4) an eccentric millionaire (5) to allot capital(6) to tighten one’s belt (7) to make a remark(8) to stretch out one’s hand (9) to moisten one’s lips(10) to complain of the weather(11) to plunge the stick into the sand (12) to turn on me(13) to get on one’s nerve(14) to put something out of someone’s mind (15) to come off the hook (16) to do a crossword puzzle (17) to blow one’s nose (18) to powder one’s nose (19) to give an alibi11) It was a white shirt with blue spots. It looked quite pretty.12) The detective spotted the suspect, and he walked over and arrested him.13) One of the balloons popped, and it gave me quiet a start.14) It is very impolite to keep popping in and out of the classroom when the class is still going on.15) When he saw the young man ready to pay for the BMW in cash, his eyes almost popped out of his head.16) In those countries, water is worth a lot more than oil. Friends will often bring ten dollars’ worth of water as a gift.17) When this project is completed, it will benefit about a hundred thousand people. It will be well worth the effort and investment.18) This movie is not worth seeing twice. In fact, it is not worth seeing at all.19) I think it is worthwhile to visit that place. I hear they have kept all their traditional house intact – houses that were built in Ming-Qing styles.20) Jia Baoyu was sick and tired of being his father’s worthy son. He yearned for freedom.Ⅲ. Translate_Grammar1) It seems the patient has no relatives in this city.2) It’s easy to make friends but difficult to keep them.3) He can always understand what his friends are thinking and worrying about.4) Perhaps it wasn’t Xiao Jiang at all who had left the tap running all night.5) The boy had no idea how he had become an old man in half a day.6) I wonder how Wang Ning has been doing in London. I haven’t heard from him for almost a year.7) It doesn’t matter what family you were born into. The real test is how far you can go from where you started.8) She simply couldn’t be lieve what she saw. It was only yesterday that the twin towers were standing there.9) Juror No. 8 pointed out that it might have been someone else who had stabbed the boy’s father to death.10) In the market economy, it is primarily by individuals and firms rather than by government agencies that decisions about what to produce, how much to produce are made.Ⅳ. Paraphrase1) Bella was young and pretty and was seen as the beauty of the boarding-house, but no one had shown any particular interest in her.2) Mr. Penbury was intelligent, but no one in the boarding-house liked him for that. He was too smart for them, and everybody felt annoyed.3) But Mrs. Mayton would not tolerate any silence for more than three minutes. So when no one broke the silence within three minutes she lost her patience and, turning to Penbury and asked.4) Mr. Calthrop was urging Mr. Penbury to give an answer immediately so that he would not have the time to make up a story.5. the weapon went right through his heart.6) We all know you are a sleep walker, so you may commit the murder in your sleep.7) Mr. Penbury advises Mr. Calthrop not to put so much emphasis on his statement when talking to the police if he does not want to arouse their suspicion about his story.8) “No,” Miss Wicks answered, “I have come to put an end to your cough.”Unit 3I Translate_Phrases1) Into Chinese2) Into EnglishⅡ. Translate_Vocabulary1) The cause of the aircraft crash is so far unknown.2) The cause of global warming is still hotly debated among scientists.(1)专业的历史工作者 (2)基于常识的反应 (3)事物的这种状况 (4)意见不一的历史学家(5)已经准备好了的现成的东西 (6)一个个人喜好不同的问题 (7)截然不同的观点 (8)民间故事 (9)书面文件(10)过去的遗留物 (11)人的动机和行为 (12)复杂和精细 (13)商船(14)一旦发生潜艇战 (15)一个粗糙的理论 (16)好战的行为;战争行为 (17)宣传机器 (18)德国外交部长 (19)实力平衡 (20)(事物的)因果 (21)海岸炮兵(22)终极关怀 (23)(事物的)近因 (24)人们常说的一句话(25)不会出错的解释 (26)绝对有效的模式 (27)永不停止的探索(28)一个难以达到但又十分诱人的目标(1) to gain new insights (2) to revise one’s ideas (3) to trace the cause(4) to begin from this premise (5) to open fire on/at(6) to give equal weight to sth. (7) to support a certain view (8) to influence the government (9) to destroy the balance of power (10) to form an alliance (11) to repay the loans(12) to contemplate war (13) to fill in the gaps (14) to conclude the quest(15) to view sth from a certain perspective(16) to benefit from the comparison (17) to eliminate from the comparison (18) to dig into the problem (19) to be immersed in a vast sea(20) to stem from a different point of view (21) to be destined to do sth. (22) to ignore the fact(23) to make an assumption (24) to defeat the enemy(25) to win back one’s lost territory (26) to sink a boat(27) to intercept the secret message (28) to piece together evidence (29) to approximate the truth (30) to master new techniques3) He devoted all his life to the cause of environmental protection.4) The river has caused us a lot of trouble in history.5) What do you think caused the upsurge in international terrorism?6) We must try and unite with those who have opposed us.7) There is always opposition to any progress and reform.8) Some people are always opposed to new things.9) A lot of those loans were never repaid. That high ratio of bad debts finally led to the financial crisis in this second economic power in the world.10) The Business Bank now offers a special loan to students who can’t pay for their education.11) The boy asked Mrs. Stow for the loan of her binoculars.12) She concluded her speech by saying that she hoped she could come again someday and see more of the country.13) As soon as they concluded the investigation, they were to report to the Security Council.14) During his visit, he will conclude a new trade agreement with India.15) Based on those reasonable doubts, the jury had to conclude that the boy was not guilty.16) She is flying to New York by way of Tokyo.17) I’d like to say a few words about the situation in the sixties of the last century by way of an introduction to the movie.18) They decided to recall their ambassador by way of protest.Ⅲ. Translate_Grammar1) Heroes and heroines are people with unusual qualities.2) Celebrities are people who become famous because of publicity through the media.3) In China's mainland, "sweetheart" often refers to a person's husband or wife.4) A fair-weather friend is one who will desert you as soon as you are in trouble.5) Broadly speaking, money refers to anything generally accepted in exchange for other goods and services.6) An armchair revolutionary is one who talks about revolution, but who doesn’t put what he says into practice.7) Professor Lu says that a good teacher is one who does all he/she can to make himself/herself unnecessary for the students.8) Economics is defined as the social science that deals with the production, distribution and consumption of goods and services.9) DVD is a disk on which large amounts of information, esp. photographs and video can be stored in a computer.10) The Oxford Advanced Learner's Dictionary defines "workaholic" as "a person who works most of the time and finds it difficult to stop working and do other things".Ⅳ. Paraphrase1) Most students usually come to have their first experience of the study of history throughthe reading of a thick history textbook and soon are overwhelmed by a large number of names, dates, events and statistics.2) People used to believe history study was just an effort of memorizing “facts. “Now history means different things to different people, because they choose the best description and interpretation according to their own preferences among those given by historians.3) They cannot help feeling that two absolutely opposite ideas about an event cannot both be correct, but they do not have the ability to judge which one is right.4) They will come across the historical interception of the “Zimmerman Note. “In that telegraph, the German foreign secretary gave order to German minister in Mexico and asked him to propose an alliance with Mexico Government in case there would be war and to promise that Mexico Government would like to help Mexico win back the land that was taken away from Mexico by the US in the Mexico war.5) We can get rid of all disagreements if our knowledge could give us a perfect model that completely explained human behavior. Unfortunately, such model does not exist.Unit 4I Translate_Phrases1) Into Chinese2) Into EnglishⅡ. Translate_Vocabulary1. Import of that country’s beef was suspend ed because of the mad cow scare.2. During the war, they had to suspend the construction of the railway.3. It was a serious offence to take drugs. Robert was suspend ed from school for two weeks.4. She was reading in a hammock suspend ed from two tree branches.5. The sales suspension has brought us heavy losses.6. This is perhaps the longest suspension bridge in Asia.7. The author is very good at creating suspense .8. He used to watch with great envy children of wealthy people go to school. 9. I rather envy their school for its beautiful campus.10. She avert ed her face so that people would not see her blush.(1)难以解决的两难困境 (2)一本难以看懂的书 (3)一个爱交际的女人 (4)黑市 (5)黑色幽默 (6)害群之马 (7)黑人权力 (8)缺少表达能力的人 (9)全国性的运动 (10)赞扬和恭维的话 (11)调皮的男孩 (12)某些大人物们 (13)种族隔离的学校 (14)他的无可争议的权威 (15)一个地位很高的人士 (16)公海 (17)上流社会 (18)机密消息(19)冷淡而缺少人情味的门 (20)冷淡的公文式的信 (21)真诚的羡慕(22)不自然的额、紧张的说话声1) to celebrate its Golden Jubilee 2) to excite admiration 3) to touch the conscience 4) to win the prize 5) to receive a reprimand 6) to omit the words7) to renounce their prizes 8) to avert a crisis9) to attend the ceremony 10) to exhibit a work of art11) to indulge in pleasures 12) to guard a child 13) to feel up to it14) to bring sth to a cloze 15) to wave sb in16) to save sb from a situation 17) to talk out one’s heart to sb 18) to knock sb down 19) to pour sb a drink20) to raise (lift) one’s glass11. He has always had an aversion to publicity.12. The government's policy succeeded in avert ing a serious economic recession.13. Michael Jordan is the envy of many black kids.14. Every summer, hundreds of thousands of people are sent to guard the riverbanks against floods.15. It was not easy to get the golden apple, for it was guard ed by a ferocious giant.16. Xicheng was practically un guard ed so Zhuge Liang narrowly escaped being captured.17. The prisoners of war killed the guard s and escaped into the woods.18. He is probably the greatest guard in the history of basketball.19. They took the Americans off guard by launching a sudden attack on a Sunday.20. There were two armed soldiers standing guard at the bridge.21. Napoleon exhibit ed (his) military talent early in life.22. These exhibit s are all insured and carefully guarded.23. When the exhibition is over, the exhibits will be given to the host country as gifts.24. She is going to exhibit some of her most recent sculptures at the National Art Gallery.Ⅲ. Translate_Grammar1) Using “It is/was said/believed, etc.” to express general beliefs.1) It's widely rumored that Linda's being promoted.2) It is estimated that the project will cost RMB three billion.3) It is assumed that the Labor Party will remain in power.4) It was proposed a few years ago that the president be elected for one term only.5) It was announced that another bridge across the Yangtse would be built next year.6) It was believed that even them that the abnormal state of affairs wouldn’t last long. 2) Paying special attention ton subject-verb agreement.1) The jury is having trouble reaching a verdict.2) Whenever either of us is in a tight corner, we always come to each other’s help.3) Statistics are facts obtained from analyzing information given in numbers.4) Statistics is a branch of mathematics concerned with the study of information that is expressed in numbers.5) Neither his friends and nor his father was surprised when he was admitted by Tsinghua University.6) Xiao Li is one of best football players at our university who have ever participated in intercollegiate championships.Ⅳ. Paraphrase1) “My parents, my wife’s parents and our priest all thought that I’d pretend to be not feeling well enough as an excuse to be absent from the awarding ceremony. So I decided not to attend the ceremony.”2) “I’m a sculptor, and I don’t want to show any antagonistic feeling towards the whiteworld by receiving an award.”3) In Orlando you (the blacks) gradually develop a throat as strong as iron.4) So I thought I’d go and see my sculpture in the window and have some pleasant feelings of pride by enjoying my own work, which is natural to human beings.5) “What is extraordinary about the wonderful sculpture is that it is made by a black man like you. Do yo u know?”6) “She knows that her child will live a hard life in South Africa because they are black people.”7) I didn’t want to drink because if the police caught me drinking late at night I would be in great trouble.8) He wasn’t afraid of being seen walking with a black man.9) I answered “Yes”, but actually I didn’t want to tell him the truth.10) Drinking in the passage was certainly beyond my expectation. What was in my mind was not what you may be thinking…11) “Our country is beautiful. But the apartheid made me very sad.”12) as though they wanted to communicate with me emotionally but didn’t know the way to do it13) And I thought it was a sad thing, because if you don’t understand each other and don’t care for each other, they will hurt each other someday.14) Nobody knows what he was thinking. But I was thinking that he was much like a man trying to run but couldn’t because he was still not completely free from racist prejudices which were dragging his feet like iron shoes.Unit 5I Translate_Phrases1) Into Chinese2) Into EnglishⅡ. Translate_Vocabulary1. We need to increase our oil import in the coming years to meet the growing demand for energy.2. Our profits have increase d by 20% over the past two years.3. The number of privately-owned cars has increase d five times (fivefold) in five years.(1)假日别墅 (2)著名的电视明星 (3)下流语言 (4)黄色故事 (5)银行抢劫 (6)生产双层玻璃的公司 (7)联合抵押 (8)一句气话 (9)永远达不到的目标 (10)乡村音乐和西部音乐 (11)加重了的潜水腰带 (12)心碎的(13)一切以自我为中心的人 (14)光是重量 (15)光凭运气 (16)纯粹是胡言 (17)军号声 (18)人工呼吸 (19)一溜气泡(20一丁点儿的关心1) to trickle down her legs 2) to puff like a whale 3) to melt like snow 4) to sum up the scene 5) to do their interview 6) to bear resentment7) to feel sick in the stomach 8) to come in like thunder 9) to drift away gradually 10) to value one’s opinion11) t o increase one’s standing with sb 12) to have a liking for sb 13) to pull oneself together 14) to serve dinner15) to tackle an armed robber 16) to stick with sb17) to fasten the seatbelt / a belt18) to abuse one’s wife19) to screw his way around 20) to lay a finger on sb21) to catch sb trying to do sth 22) to try the kiss of life23) to heaven oneself up 24) to float to the surface25) to thrush about under the water 26) to prop oneself against 27) to break surface 28) to seize hold of sth 29) to haul oneself up 30) to pinion one’s arms 31) to fend sb off 32) to land a blow33) to stay under in the water4. The number of mobile phone owners in our city has increase d from 20 thousand to about half a million in less than 5 years.5. She prop ped her bike against a tree.6. The local economy is largely prop ped up by tourism.7. He was fast asleep with his head prop ped on a big rock.8. You need strong prop s to keep the tower from leaning any further.9. Flight 901 is due to land at the airport at 8:55.10. Many foreign observers say that the next man to land on the moon may very well be a Chinese.11. One of the stones they threw land ed on the head of a young soldier.12. If you go on like this, you will land yourself in a jail.13. I land ed a powerful punch to his chin and sent him sprawling.14. When you approach a drowning person, you must not let him grip your hands.15. Reports of the sudden appearance of these whales grip ped the interest of the whole city.16. If she lost grip on the rope, she’d fall 1000 feet to sure death.17. You must keep a grip on yourself. Don’t despair.18. Our government did everything possible to defend the value of our currency.19. The total value of our exports to that country in the first nine months this year has reached $94 billion.20. A man who dares to waste one hour of time has not discovered the value of life.21. At college, students acquire certain values. This is an important part of their education.22. If you value your life, quit smoking!23. This painting is value d at $20 million.Ⅲ. Paraphrase1) The man Nerys was engaged to leave her after she had become disfigured. But before the bank raid, he behaved like a lover. Many man, not just me, could have done the same if we had engaged with her.2) This man loved her only because she was beautiful. So he left her when she was no longer beautiful.3) We used to love this music when we were in love.4) I’m sorry abo ut what happened to Netys.5) I didn’t mean to hurt you by offering money, because I know it’s impossible for us to compensate in any way for the distress and suffering that Nerys and you have gone through.6) You don’t accept our help only because Vic was responsible for her suffering.7) You are with a hero if you are looking for a hero and that’s Vic rather than me, so don’t leave him.8) I love Vic very much. I feel guilty about this because Vic is your husband.9) Sharon, I can assure you that this experience is transient and won’t last long. We allhave the feeling when we are young.10) You never hear people speak ill of you, do you? People gossip about you.11) You know clearly that you have been hurting her.12) Beware of your manners. Stop shouting at me!13) If Sharon gets drowned, you will be held responsible.14) I will make you pay what you have done to me. You will be punished for what you have done to me.15) Sharon,you’d better not do anything. You have done enough to him.。
半导体物理第四章习题参考答案
9. 由于光的照射在半导体中产生了非平衡载流子 n p 1012 cm-3 ,分别计算
施主掺杂浓度为 ND 1016 cm-3 的 N 型硅和本征硅在这种情况下的准费米能 级的位置,并与原来的费米能级的位置做比较,画出相应的能带图。 答:有:
n
ni
exp
E fn kT
Ei
,
n
E fn
答:(1) 电离杂质散射是由电离的杂质对载流子的库仑相互作用引起的,其特点 为:掺杂浓度越高,电离杂质散射越显著;温度越高,载流子的动能越大,受库 仑相互作用力的影响相对减弱,因此,电离杂质散射在低温时起主要作用,其 、
与温度的关系为:
3
3
I T 2 , I T 2
(2) 声学波散射是晶格振动对载流子散射中作用大的一种,属于晶格自身的特
10. 设空穴浓度是线性分布,在 3μm 内浓度分布差 1015cm-3,μp=400cm2·V-1·s-1, 试计算空穴扩散电流密度。
答:由爱因斯坦关系:
Dp
kT q
p
有:
jp
qDp
p x
kT p
p x
5.52 A
cm2
11. 考虑平衡情形,证明:
en
Vthn nni
exp
Et Ei kT
i niqn piqp 4.45106 Ω cm
(2)
当掺入百万分之一的
As
时,施主浓度为:
ND
5 1022 106
cm-3
51016 cm-3
(其中 N 51022 cm-3 为 Si 的原子密度)。
由于杂质全部电离,从而: n
ND
51016 cm-3,
p
第4章 习题参考答案
习题四一、用适当内容填空1. 凡将地理位置不同且具有独立功能的【计算机及辅助设备】,通过【通信设备】和【传输线路】将其连接,由功能完善的【网络软件】实现网络【资源共享及信息通信】的系统称为计算机网络。
2. 从网络范围和计算机之间互连的距离来看,有【局域网】、城域网和【广域网】3种类型。
根据计算机网络各部分的功能,计算机网络可分成【资源子网】和【通信子网】两种类型。
3. Internet是一个基于【 TCP/IP 】协议,将各个国家、各个部门、各种机构的内部网络连接起来的计算机通信网络。
4. 将提供资源的计算机叫做【服务器】,而将使用资源的计算机叫做【客户机】。
5. HTTP的中文含义是【超文本传输协议】。
6.宽带传输是基带信号经过调制后形成【频分复用模拟信号】,有时也称【频带传输】。
7. 写出一个以机构区分域名的域【 COM、END、GOV、MIL、NET、ORG】,写出一个以国别或地区区分域名的域【 CN、US、GB、TW、JP】。
8. 在WEB上,每一信息资源都有统一的且在网上唯一的地址,该地址就叫【 URL或统一资源定位标志】。
9. 写出两个由URL地址表示的资源类型【 HTTP 】和【 FTP 】。
10. 连接Internet主要方式有:【终端方式】、【拨号方式】、【局域网方式】和【宽带网方式】。
11. ISDN的中文含义是【综合业务数字网】。
12. 通信线路中允许的最大数据传输速率是【带宽】。
13.【主页】是指用户进入网站后看到的第一个页面。
当用户在浏览器的地址栏输入网站的【 URL 】地址后,浏览器就会自动连接到主页。
14. 超文本有两个含义:【信息的表达式】、【信息间的超链接】。
15. URL的3个组成部分是:【资源类型】、【存放资源的主机域名】、【资源文件名】。
16. HTML文档也称为【Web 】文档,它由文本、图形、声音和超链接组成。
17. 网络地址格式为WWW.Z.Y.X,其中X表示【最高层域名或顶级域名】。
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高分子第四章习题参考答案
思考题:
1、无规、交替、嵌段、接枝共聚物的结构有何差异?在这些共聚物名称中,对前后单体的位置有何规定?
参考答案:
无规共聚物是聚合物中两单元M1、M2无规排列,而且M1、M2连续的单元树不多。
交替共聚物是聚合物中两单元M1、M2严格相间。
嵌段共聚物是聚合物中由较长的M1链段和另一较长的M2链段构成的大分子。
接枝共聚物是聚合物主链由单元M1组成,而支链则由另一单元M2组成。
无规共聚物名称中前一单体为主单体,后为第二单体。
嵌段共聚物名称中的前后单体则代表单体聚合的次序。
接枝共聚物中前单体为之链,后单体则为支链。
2、试用共聚合动力学来推导二元共聚物微分方程,推导时有哪些基本假设?
参考答案:
4、当r1= r2=1;r1= r2=0;r1>0,r2=0;r1·r2=1等特殊情况下,d[M1]/ d[M2]=f([M1]/ [M2]),F1=f(f1)的函数关系如何?
参考答案:
当r1= r2=1时,d[M1]/ d[M2]= [M1]/ [M2],F1= f1;
当r1= r2=0时,d[M1]/ d[M2]=1,F1= 0.5;
当r1>0,r2=0时,d[M1]/ d[M2]=1+r1· [M1]/ [M2],F1>50%;
当r1·r2=1 ,d[M1]/ d[M2]= r1· [M1]/ [M2] ,F1= r1· f1。
高分子第四章补充习题
1、为了提高聚丙睛纤维的柔软性和手感以其染色性,请你设计一个改性配方。
并说明改性原理。
参考答案:主单体:丙烯腈90~92%。
作用:耐光耐侯,保暖性好,轻软;
第二单体:丙烯酸甲酯7~10%。
作用:降低分子间吸引力,增加柔软性和手感,有利于染料分子扩散入内;
第三单体:带羧基或碱性基团,用量1%,如烯丙基磺酸钠:CH2=CHCH2SOONa,乙烯基吡啶。
作用:增加染色性,
2、简述在乳液聚合中,影响聚合反应速率和分子量的因素?
答:在乳液聚合中,影响聚合反应速率和分子量的因素有反应温度,乳化剂用量,引发剂的用量三个主要因素。
参考答案:反应温度升高聚合速率增加、分子量下降;增加引发剂浓度可提高反应速率、对分子量的影响较小;增加乳化剂浓度可提高反应速率、同时可提高分子量。
3、在丙苯乳液聚合中,增加哪种单体的用量可提高丙苯乳胶漆成膜后的弹性?参考答案:增加丙烯酸酯(丙烯酸酯丁酯)的用量可提高丙苯乳胶漆成膜后的弹性。
4、两单体的竞聚率r1=2.0,r2=0.5,如f10=0.5,转化率C=50%,试求共聚物组成。
参考答案:因为换化率大于10%,只能用积分公式计算。
式中
因为:γ=0
所以:
故而:
整理上式得:f12-3f1+1=0 解方程得:
将f1和f10及C代入:
解得:F1=0.62。