武汉理工大学数值分析(高秉建)的作业

1、①工程中数值方法的主要思想答：工程总，把理论与实际情况相结合，用数值方法直接求解较少简化的模型，及忽略一些无关的因素求出近似值，又使得到的景近似解满足程变得要求 ②数值方法中误差产生的原因答：当数值模型不能得到精确解释，通常要用数值方法求接触他的近似解，七近似解与精确解之间的误差称为截断误差。

当用计算机做数值计算时，由于计算机的字长有限，原始数据在计算机上表示会产生误差，计算过程总中有产生误差，这种误差称为舍入误差。

③数值方法应用对象由数学模型给出的数值计算方法，以及根据计算方法编制的法程序2、取x=1、2、2时f （x ）=2、0、1，计算f(x)在x=21处得近似解x i 1 2 3f （x i ）2 0 1解：二次拉格朗日插值多项式为L （x ）=∑=2k k k )x (l yl 0（x ）=)x x )(x x ()x x )(x x (201021----=)31)(21()3x )(2x (----=21（x-2）（x-3）l 1（x ）=)x x )(x x ()x x )(x x (210120----=)32)(12()3x )(1x (----=-（x-1）（x-3）l 2（x ）=)x x )(x x ()x x )(x x (120210----=)23)(13()2x )(1x (----=21（x-1）（x-2）则L （x ）=∑=2k k k )x (l y =l 0（x ）+l 1（x ）+l 2（x ）=21（x-2）（x-3）+（x-1）（x-3）+21（x-1）（x-2）=23x 2-213x+7 所以L （21）=23×（21）2_213×（21）+7=833即f(x)在x=21处得景近似解为8333、f （x ）=（x-1）4，在[]1,1-上计算范数1,ff∞与2f解f （x ）=(x-1)4，x ∈[]1,1-,则 f ’(x)=4（x-1）3≦0所以f （x ）=（x-1）4在[]1,1-上单调递减 ∞f =}{)1(f ),1(f max )x (f max1x 1-=≤≤-=}{160,16max = ⎰-⎰-==114ba dxdx )x (f 1)1x (f=115)5x (51--=5322111x 42d )1x (f⎥⎦⎤⎢⎣⎡-=⎰-=2111x 8d )1x (⎥⎦⎤⎢⎣⎡-⎰-=21119|)1x (91⎥⎦⎤⎢⎣⎡--=3216929=4、对权函数2()1x x ρ=-，区间[1,1]-，试求首项系数为1的正交多项式(),0,1,2,3.n x n ϕ= 解：若2()1x x ρ=-，则区间[1,1]-上内积为 11(,)()()()f g f x g x x dx ρ-=⎰定义0()1x ϕ=，则11()()()()n n n n n x x x x ϕαϕβϕ+-=--其中1101211211211321122111221121((),())/((),())((),())/((),())(,1)/(1,1)(1)(1)0()(,)/(,)(1)(1)0(,)/(1,1)(1)(1)n n n n n n n n n n x x x x x x x x x x x x dx x dxx xx x x x x x dx x x dxx x x x dx x αϕϕϕϕβϕϕϕϕαϕαβ--------==∴=+=+=∴==+=+==+=+⎰⎰⎰⎰⎰22162158532()5dxx x ϕ==∴=-⎰32222132211222122212221122132332222(,)/(,)555522()()(1)5522()()(1)5522(,)/(,)5522()()(1)55(1)136175251670152179()57014x x x x x x x x x dxx x x dx x x x x x x x dxx x dxx x x x x xαβϕ----=------+=--+==----+=+==∴=--=-⎰⎰⎰⎰5、求[]()0,1xf x e =在[]0,1上的最佳一次逼近多项式。

数值分析第一章思考题第一章思考题(2012级本科学生作品)1、什么样的算法被称为不稳定算法？试列举一个例子进行说明。

在算法执行过程中，舍入算法对计算结果影响大的一类算法被称为数值不稳定的一种算法。

例如，假设初始数据有一点微小误差，就会对一个算法的数据结构产生很大的影响，造成误差扩散。

用计算公式ln 1ln n n =-，构造出的递推算法是一个数值不稳定的算法；而另一公式ln 1(1ln)/n -=-则可以构造出一个数值稳定的算法。

2、我们都知道秦九韶算法能够减少运算次数，高中也学过他的具体过程，请举出一个例子并用秦九韶算法计算。

答；一般的，一元n 次多项式的求值需要经过(1)/2n n +次乘法和n 次加法，而秦九韶算法只需要n 次乘法和n 次加法。

具体的不太会了。

3、为什么要设立相对误差的概念？答：相对误差是近似值误差与精确值的比值，用来衡量近似值的近似程度。

x=10±1，y=1000±5。

虽然x 的误差比y 的误差小，但y 的近似程度比x 更好。

这单用误差无法表现出来，而相对误差可以解决这个问题。

4、误差在生活中有什么作用？答：误差的作用不仅仅体现在数学课题研究中，在生活中误差的作用也非常大，比如在建筑行业中，设计图纸时必须要达到一定的精确度才行。

5、有效数字以及计算规则答：有效数字是指实际上能测量到的数值，在该数值中只有最后一位是可疑数字，其余的均为可靠数字。

它的实际意义在于有效数字能反映出测量时的准确程度。

例如，用最小刻度为0.1cm 的直尺量出某物体的长度为11.23cm ，显然这个数值的前3位数是准确的，而最后一位数字就不是那么可靠，医|学教育网搜集整理因为它是测试者估计出来的，这个物体的长度可能是11.24cm ，亦可能是11.22cm ，测量的结果有±0.01cm 的误差。

我们把这个数值的前面3位可靠数字和最后一位可疑数字称为有效数字。

这个数值就是四位有效数字。

武汉理工大学研究生课程考试试题纸（A）课程名称：数值分析专业年级：2013级专业研究生一简述题，给出简述步骤（每题5分，共25分）1、f（x）=x*x+2在[0，2]上的复化梯形积分（n=2）值为（）2、已知y’=x+2y，y（0）=1；h=0.1；一步欧拉法计算的y（0.1）=（）3、给定f（0）=1；f（1）=2；f（2）=7；则中心倒数f’（1）=（）4、若X=X*X+aX+b；根为3，当迭代收敛阶p=2时，（a，b）=（）5、f（x）在[0，b]上的积分公式J=0.5b[f（0.2b）+f（0.8b）]的代数精度为（）二计算（每题10分，共40分）1、若P（0）=0；P（1）=2；P’（0）=f’（0）；P’（1）=f’（1）；且4f’（0）+f’（1）=2；f’（0）-4f’（1）=3；求3次多项式P（x=0.5)的值？2、若y’=xy-x+1；y（0）=0；步长h为方程0.11=x*exp（x）的根，试用隐式欧拉法计算y（h）？误差限取0.05。

3、用收敛的高赛迭代法公式求（x，y，z）的一次迭代值；x-6y+z=2；x+2y+7z=2；5x+y-z=1；初值（0,0,0）。

4、设a，b分别为f=x和f=x3在[0,1]上的二点复化高斯积分值，步长h=0.5，试用收敛的迭代法公式计算函数f（x）=ax-bx2的零点，迭代2次，x0=1.0。

三分析（10分）设f（x）=P+R，R为误差，P为3次多项式，且满足条件：P（a）=f（a），P’（a）=f’（a）；P’’（a）=f’’（a）；P（b）=f（b）。

RI为R在[a，b]上的积分值，试估计RI。

四程序（25分）1、设a，b分别为f=x和f=x3在[0,1]上的复化辛普生积分值，步长h=0.25，试用收敛的迭代法公式编辑计算f（x）=ax-bx2的零点。

（10分）2、我机考平均用时（）分钟，作业一共交（）个程序，11月11日交（）个，最难的程序为（）请写出该程序。

武汉理工大学计算机学‎院数值分析实验报告‎武汉理工大学计算机学‎院数值分析实验报告‎‎篇一：数‎值分析实验报告学‎生实验报告‎书实验课程名称开‎课学院指导教‎师姓名学生姓‎名学生专业班级数‎值分析计算机科学与‎技术学院熊盛武 2‎01X—— 201X‎学年第二学期‎实验课程名称：‎数值分析‎篇‎二：数值分‎析实验报告武汉理工‎大学学生实验‎报告书实验课‎程名称：数‎值分析开课‎名学生姓名‎：201X‎1—— 201X学年‎第二学期第一‎次试验（1）‎二分法计算流程图：‎简单迭代法算‎法流程图：（‎2）（3）牛‎顿迭代法流程图：‎（4）弦截法算法‎程序流程图：‎‎篇三：‎数值分析实验报告湖‎北民族学院理学院《数‎值分析》课程实验报告‎（一）湖北民‎族学院理学院《数值分‎析》课程实验报告‎（二） xn?)篇‎四：数值分析‎实验报告数值分析实‎验报告姓名：‎学号：学院‎：老师：‎ XXX XXX‎X实验一一‎、实验内容用雅克比‎迭代法和高斯塞德尔迭‎代法求解课本例‎3.1，设置精度为1‎0-6。

?8-32‎??x1??20??‎?411?‎1??x233‎??6312??x?‎?36? ??3‎??二、实验‎公式 ?? 雅克‎比迭代法的基本思想：‎设方程组Ax‎?b的系数矩阵的对角‎线元素 ??aii?‎0(i?1,2,..‎.,n)，根据方程组‎A x?b推导出一个迭‎代公式，然后将任意选‎取的?(0)?‎(1)?(1)‎?(2) xx‎x x一初始向量代入迭‎代公式，求出，再以代‎入同一迭代公式，求出‎，1、雅克比‎迭代法 ?(k)?(‎k) {x}{x}收‎敛时，如此反复进行，‎得到向量序列。

当其极‎限即为原方程组的解。

‎2、高斯塞德‎尔迭代法：‎在雅可比（Jacbi‎）迭代法中，如果当新‎的分量求出后，马上用‎它来代替旧的分量，‎则可能会更快地接近方‎程组的准确解。

基于这‎种设想构造的迭代公式‎称为高斯-塞德尔(G‎a uss-Seide‎l)迭代法。

《数值分析》大作业四一、算法设计方案：复化梯形积分法，选取步长为1/500=0.002，迭代误差控制在E ≤1.0e-10①复化梯形积分法：11()[()()2()]2n bak hf x dx f a f b f a kh -=⎰≈+++∑，截断误差为：322()''()''(),[,]1212T b a b a R f h f a b n ηηη--=-=-∈其中。

复化Simpson 积分法，选取步长为1/50=0.02，迭代误差控制在E ≤1.0e-10②Simpson 积分法：121211()[()()4()2()]3m m bi i a i i hf x dx f a f b f x f x --==≈+++∑∑⎰， 截断误差为：4(4)(),[,]180s b a R h f a b ηη-=-∈。

③Guass积分法选用Gauss-Legendre 求积公式：111()()ni i i f x dx A f x -=≈∑⎰截断误差为：R= ()()n 2n 422n!2×(2[2!]2n 1f n n ⨯（2）η（））＋ η∈(1,1)。

选择9个节点：-0.9681602395,-0.8360311073,-0.6133714327,-0.3242534234,0,0.3242534234,0.6133714327,0.8360311073,0.9681602395， 对应的求积系数依次为：0.0812743884,0.1806481607,0.2606106964,0.3123470770,0.3302393550,0.3123470770,0.2606106964,0.1806481607,0.0812743884。

二、程序源代码：#include<stdio.h>#include<math.h>#include<stdlib.h>#define E 1.0e-10/****定义函数g和K*****/double g(double a){double b;b=exp(4*a)+(exp(a+4)-exp(-a-4))/(a+4);return b;}double K(double a,double b){double c;c=exp(a*b);return c;}/******复化梯形法******/void Tixing( ){double u[1001],x[1001],h,c[1001],e;int i,j,k;FILE *fp;fp=fopen("f:/result0. xls ","w");h=1.0/1500;for(i=0;i<3001;i++){x[i]=i*h-1;u[i]=g(x[i]);}for(k=0;k<100;k++){e=0;for(i=0;i<1001;i++){for(j=1,c[i]=0;j<N-1;j++)c[i]+=K(x[i],x[j])*u[j];u[i]=g(x[i])-h*c[i]-h/2*(K(x[i],x[0])*u[0]+K(x[i],x[N-1])*u[N-1]);e+=h*(exp(4*x[i])-u[i])*(exp(4*x[i])-u[i]);}if(e<=E) break;}for(i=0;i<1001;i++)fprintf(fp,"%.12lf,%.12lf\n",x[i],u[i]);fclose(fp);}/******复化Simpson法******/void simpson( ){double u[101],x[101],h,c[101],d[101],e;int i,j,k;FILE *fp;fp=fopen("f:/result1.xls","w");h=1.0/50;for(i=0;i<101;i++){x[i]=i*h-1;u[i]=g(x[i]);}for(k=0;k<50;k++){e=0;for(i=0;i<101;i++){for(j=1,c[i]=0,d[i]=0;j<51;j++){c[i]+=K(x[i],x[2*j-1])*u[2*j-1];if(j<50)d[i]+=K(x[i],x[2*j])*u[2*j];}u[i]=g(x[i])-4*h/3*c[i]-2*h/3*d[i]-h/3*(K(x[i],x[0])*u[0]+K(x[i],x[M-1])*u[M-1]);e+=h*(exp(4*x[i])-u[i])*(exp(4*x[i])-u[i]);}if(e<=E) break;}for(i=0;i<101;i++)fprintf(fp,"%.12lf,%.12lf\n",x[i],u[i]);fclose(fp);}/******Gauss积分法******/void gauss( ){double x[9]={-0.9681602395,-0.8360311073,-0.6133714327,-0.3242534234,0,\0.3242534234,0.6133714327,0.8360311073,0.9681602395},A[9]={0.0812743884,0.1806481607,0.2606106964,0.3123470770,0.3302393550,\0.3123470770,0.2606106964,0.1806481607,0.0812743884},u[9],c[9],e;int i,j,k;FILE *fp;fp=fopen("f:/result2. xls ","w");for(i=0;i<9;i++)u[i]=g(x[i]);for(k=0;k<50;k++){e=0;for(i=0;i<9;i++){for(j=0,c[i]=0;j<9;j++)c[i]+=A[j]*K(x[i],x[j])*u[j];u[i]=g(x[i])-c[i];e+=A[i]*(exp(4*x[i])-u[i])*(exp(4*x[i])-u[i]);}if(e<=E) break;}for(i=0;i<9;i++)fprintf(fp,"%.12lf,%.12lf\n",x[i],u[i]);fclose(fp);}/******主函数******/main(){Tixing ( );Simpson( );Gauss( );return 0;}三、运算结果复化梯形数据-10.018323-0.920.02523-0.9980.018471-0.9180.025433-0.9960.018619-0.9160.025637-0.9940.018768-0.9140.025843-0.9920.018919-0.9120.026051-0.990.019071-0.910.02626-0.9880.019224-0.9080.026471-0.9860.019378-0.9060.026683-0.9840.019534-0.9040.026897-0.9820.019691-0.9020.027113-0.980.019849-0.90.027331-0.9780.020008-0.8980.02755-0.9760.020169-0.8960.027772-0.9740.020331-0.8940.027995-0.9720.020494-0.8920.028219-0.970.020658-0.890.028446-0.9680.020824-0.8880.028674-0.9660.020992-0.8860.028905-0.9640.02116-0.8840.029137-0.9620.02133-0.8820.029371-0.960.021501-0.880.029607-0.9580.021674-0.8780.029844-0.9560.021848-0.8760.030084-0.9540.022023-0.8740.030326-0.9520.0222-0.8720.030569-0.950.022378-0.870.030815-0.9480.022558-0.8680.031062-0.9460.022739-0.8660.031311-0.9440.022922-0.8640.031563-0.9420.023106-0.8620.031816-0.940.023291-0.860.032072-0.9380.023478-0.8580.032329-0.9360.023667-0.8560.032589-0.9340.023857-0.8540.032851-0.9320.024048-0.8520.033114-0.930.024241-0.850.03338-0.9280.024436-0.8480.033648-0.9260.024632-0.8460.033918-0.9240.02483-0.8440.034191-0.9220.025029-0.8420.034465-0.840.034742-0.760.047841-0.8380.035021-0.7580.048225-0.8360.035302-0.7560.048613 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0.89235.445520.94844.344880.89435.730220.9544.701070.89636.017210.95245.060110.89836.306510.95445.422040.936.598120.95645.786870.90236.892080.95846.154630.90437.188410.9646.525350.90637.487110.96246.899050.90837.788210.96447.275750.9138.091730.96647.655470.91238.397680.96848.038240.91438.70610.9748.424090.91639.016990.97248.813040.91839.330380.97449.205110.9239.646280.97649.600330.92239.964720.97849.998720.92440.285720.9850.400320.92640.60930.98250.805140.92840.935480.98451.213210.9341.264280.98651.624560.93241.595720.98852.039210.93441.929820.9952.45720.93642.26660.99252.878540.93842.606090.99453.303270.9442.948310.99653.73140.94243.293270.99854.162980.94443.64101154.59802复化Simpson数据：-1 0.018319929 -0.34 0.256658088 0.32 3.596641805 -0.98 0.0198445 -0.32 0.278035042 0.34 3.896195298-0.96 0.021494322 -0.3 0.301192133 0.36 4.220697765-0.94 0.023283225 -0.28 0.326278124 0.38 4.572227037-0.92 0.025220379 -0.26 0.353453177 0.4 4.95303418-0.9 0.027320224 -0.24 0.382891765 0.42 5.365557596-0.88 0.029594431 -0.22 0.41478194 0.44 5.812438891-0.86 0.032059069 -0.16 0.527292277 0.54 8.671138204-0.84 0.034728638 -0.14 0.571209036 0.56 9.39333156-0.82 0.037621263 -0.12 0.61878367 0.58 10.17567433-0.8 0.040754615 -0.1 0.670320427 0.6 11.02317608-0.78 0.044149394 -0.08 0.726149698 0.62 11.94126383-0.76 0.047826844 -0.06 0.78662861 0.64 12.93581634-0.74 0.051810827 -0.04 0.85214479 0.66 14.01320231-0.72 0.056126648 -0.02 0.92311742 0.68 15.1803205-0.7 0.060802006 0 1.0000013 0.7 16.44464467 -0.68 0.065866854 0.02 1.083288424 0.72 17.81427057 -0.66 0.071353499 0.04 1.173512427 0.74 19.29796874 -0.64 0.077297255 0.06 1.271250748 0.76 20.90523965 -0.62 0.083735917 0.08 1.377129533 0.78 22.64637562 -0.6 0.090711017 0.1 1.491826493 0.8 24.53252554 -0.58 0.098266855 0.12 1.616076341 0.82 26.57576756 -0.56 0.106452202 0.14 1.750674449 0.84 28.78918506 -0.54 0.11531904 0.16 1.896482943 0.86 31.18695183 -0.52 0.12492459 0.18 2.054435268 0.88 33.78442141 -0.5 0.135329888 0.2 2.225543071 0.9 36.59822683 -0.48 0.14660204 0.22 2.410901825 0.92 39.64638571 -0.46 0.158812728 0.24 2.611698647 0.94 42.94841704 -0.44 0.17204064 0.26 2.829219145 0.96 46.52546475 -0.42 0.18636997 0.28 3.064856356 0.98 50.40043451 -0.4 0.201892977 0.3 3.320119013 1 54.59813904 -0.38 0.218708553 0.46 6.296539601-0.36 0.236924875 0.48 6.820959636-0.2 0.449328351 0.5 7.389057081-0.18 0.486751777 0.52 8.0044696750102030405060四、讨论①在满足相同精度要求的情况下复化梯形积分法比复化Simpson 积分法计算所需节点数多，计算量大。

数值分析实验一、95页第17题已知实验数据如下：用最小二乘法求一个形如的经验公式，使它拟合下列数据，并计算均方误差. 解：本题给出拟合曲线，即，故法方程系数法方程为解得最小二乘拟合曲线为均方程为二、238页第2题第二问求方程在=1.5附近的一个根解：设将方程改写为建立迭代公式，k=0,1,2,3…取，则原方程的根为=1.4665572三、书238页第7题第一问1．用牛顿法求在=2附近的根解：Newton 迭代法取，则，取2.用牛顿法求 的实根解：令，则，取四、49页第20 题 X j 0.25 0.30 0.39 0.45 0.53 Y j0.50000.54770.62450.67080.7280试求三次样条插值，并满足条件：(1)(0.25) 1.0000,(0.53)0.6868;(2)(0.25)(0.53)0.S S S S ''==''''==解：0101212323430.050.090.060.08h x x h x x h x x h x x =-==-==-==-=1111234,533,,,11457j j j j j jj jh h h h h h μλμμμμ---==--∴====[][][][]1230100110122334924,,,11457()(),0.9540,0.8533,0.7717,0.7150f x f x f x x x x f x x f x x f x x λλλλ====-==-===[][][][][][][][]040120012011012312212342332344343(1)() 1.0000,()0.68686(,) 5.5200,,6 4.3157,,6 3.2640,,6 2.43006(,) 2.1150S x S x d f x x f h f x x f x x d h h f x x f x x d h h f x x f x x d h h d f f x x h ''=='=-=--==-+-==-+-==-+'=-=- 由此得矩阵形式的方程组为2 1 M 0 5.5200-514 2 914M 1 4.3157- 35 2 25M 2 = 3.2640-37 2 47M 3 2.4300-1 2 M 4 2.1150-求解此方程组得012342.0278, 1.46431.0313,0.8070,0.6539M M M M M =-=-=-=-=-三次样条表达式为331122111()()()66()()(0,1,,1)66j j j j jj j jj j jj j j jjx x x x S x M M h h M h x x M h x x y y j n h h +++++--=+--+-+-=-∴将01234,,,,M M M M M 代入得[][]3333336.7593(0.30) 4.8810(0.25)10.0169(0.30)10.9662(0.25)0.25,0.302.7117(0.39) 1.9098(0.30) 6.1075(0.39) 6.9544(0.30)0.30,0.39() 2.8647(0.45) 2.2422(0.39)10.4186(0.45x x x x x x x x x x S x x x x ----+-+-∈----+-+-∈=----+-[][]33)10.9662(0.39)0.39,0.451.6817(0.53) 1.3623(0.45)8.3958(0.53)9.1087(0.45)0.45,0.53x x x x x x x ⎧⎪⎪⎪⎪⎪⎪⎨+-⎪⎪∈⎪⎪----+-+-⎪∈⎪⎩04001234404(2)()0,()020, 4.3157, 3.26402.4300,20S x S x d f d d d d f λμ''''==''===-=-''=-====由此得矩阵开工的方程组为0412******* 4.3157322 3.264055 2.43003027M M M M M ==⎛⎫⎪-⎛⎫⎛⎫⎪ ⎪ ⎪ ⎪=- ⎪ ⎪ ⎪ ⎪ ⎪-⎪⎝⎭⎝⎭ ⎪⎪⎝⎭求解此方程组，得012340, 1.88090.8616, 1.0304,0M M M M M ==-=-=-=又三次样条表达式为331122111()()()66()()66j j j j jj j jj j jj j j jjx x x x S x M M h h M h x x M h x x y y h h +++++--=+--+-+-将01234,,,,M M M M M 代入得[][]333336.2697(0.25)10(0.3)10.9697(0.25)0.25,0.303.4831(0.39) 1.5956(0.3) 6.1138(0.39) 6.9518(0.30)0.30,0.39() 2.3933(0.45) 2.8622(0.39)10.4186(0.45)11.1903(0.39)0.3x x x x x x x x x S x x x x x x --+-+-∈----+-+-∈∴=----+-+-∈[][]39,0.452.1467(0.53)8.3987(0.53)9.1(0.45)0.45,0.53x x x x ⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪--+-+-⎪∈⎪⎩五、50页计算实习题第一题①/*拉格朗日差值*/ #include<stdio.h> #include<conio.h> #define N 4 void main(){int checkvalid(double x[],int n);double Largrange(double x[],double y[],double varx,int n); double x[N+1]={0.4,0.55,0.8,0.9,1};double y[N+1]={0.41075,0.57815,0.88811,1.02652,1.17520}; double varx=0.5;if(checkvalid(x,N)==1)printf("\n\n 插值结果：P(%f)=%f\n",varx,Largrange(x,y,varx,N)); elseprintf("输入的插值节点的x 值必须互异！"); getch(); }int checkvalid(double x[],int n) {int i,j;for(i=0;i<n+1;i++) for(j=i+1;j<n+1;j++) if(x[i]==x[j])return -1;elsereturn 1;}double Largrange(double x[],double y[],double varx,int n) {int k,j;double A,B,C=1,D=0;for(k=0;k<=n;k++){C=1;for(j=0;j<=n;j++){if(j!=k){A=(varx-x[j]);B=(x[k]-x[j]);C=C*A/B;}}D=D+C*y[k];}return D;}②/*牛顿插值*/#include<stdio.h>#include<conio.h>#define N 4int checkvalid(double x[],int n){int i,j;for(i=0;i<N;i++)for(j=i+1;j<=N;j++){if(x[i]==x[j])return(-1);}return(1);}void chashang(double x[N],double y[N],double f[N][N]){int i,j,h;for(j=0;j<=N;j++){f[j][j]=y[j];}for(h=1;h<=N;h++){for(i=0;i<=N-h;i++){f[i][i+h]=(f[i+1][i+h]-f[i][i+h-1])/(x[i+h]-x[i]);}}}double compvalue(double f[N][N],double x[N],double y[N],double varx) {int i;double t=1.000000,n=y[0];chashang(x,y,f);for(i=1;i<=N;i++){t=t*(varx-x[i-1]);n=n+f[0][i]*t;}return n;printf("the result is %f.",n);}void main(){int i,j;double varx,x[N],y[N],f[N][N];printf("input the value of x:");for(i=0;i<N;i++)scanf("%f",&x[i]);if(checkvalid(x,N)==1){printf("input the value of y:"); for(j=0;j<N;j++) scanf("%f",&y[j]);printf("input the value of varx:"); scanf("%f",&varx); compvalue(f,x,y,varx); } elseprintf("the value of x must be different!\n"); }六、238页第8题用牛顿迭代法求tan 0x x -=的最小正根.解 显然*0x =满足tan 0x x -=.另外当||x 较小时,2131tan ......321k x x x x k +=+++++,故当(0,)2x π∈时,tan x x >,因此,方程tan 0x x -=的最小正根应在3(,)22ππ内.记3()tan ,(,)22f x x x x ππ=-∈,容易算得(4) 2.842...0,(4.6) 4.26...0f f =>=-<,因此[4,4.6]是()0f x =的有限区间. 对于二分法,计算结果见下表此时391011|*|1021024x x --<=<.若用牛顿迭代法求解,由于221'()(tan )0,''()2tan 0cos f x x f x xx =-<=-<,故取0 4.6x =,迭代计算结果如表7-13所示.所以tan 0x x -=的最小正根为* 4.493409458x ≈.。

数值分析大作业数值分析大作业姓名：黄晨晨学号：S1*******学院：储运与建筑工程学院学院班级：储建研17-2实验3.1 Gauss消去法的数值稳定性实验实验目的：理解高斯消元过程中出现小主元即很小时引起方程组解数值不定性实验内容：求解方程组Ax=b，其中（1）A1=0.3×10?1559.14315.291?6.130?1211.29521211，b1=59.1746.7812;（2）A2=10?7013 2.099999999999625?15?10102，b2=85.90000000000151;实验要求：(1)计算矩阵的条件数，判断系数矩阵是良态的还是病态的(2)用Gauss列主元消去法求得L和U及解向量x1,x2∈R4(3)用不选主元的高斯消去法求得L和U及解向量x1,x2∈R4(4)观察小主元并分析对计算结果的影响（1）计算矩阵的条件数，判断系数矩阵是良态的还是病态的代码：format longeformat compactA1=[0.3*10^-15,59.14,3,1;5.291,-6.130,-1,2;11.2,9,5,2;1,2,1,1] b1=[59.17;46.78;1;2]n=4C1=cond(A1,1) %C1为A1矩阵1范数下的条件数C2=cond(A1,2) %C2为A1矩阵2范数下的条件数C3=cond(A1,inf) %C3为1矩阵谱范数下的条件数结果：C1 =1.362944708720448e+02C2 =6.842955771253409e+01C3 =8.431146*********e+01显然A1矩阵为病态矩阵将矩阵A2，b2输入上述代码中求得A2矩阵的条件数为：C1 =1.928316831682894e+01C2 =8.993938090170119e+00C3 =1.835643564356072e+01显然A2矩阵也为病态矩阵(2)用Gauss列主元消去法求得L和U及解向量x1,x2∈R4代码：for k=1:n-1a=max(abs(A1(k:n,k)))if a==0returnendfor i=k:nif abs(A1(i,k))==ay=A1(i,:)A1(i,:)=A1(k,:)A1(k,:)=yx=b1(i,:)b1(i,:)=b1(k,:)b1(k,:)=xbreakendendif A1(k,k)~=0A1(k+1:n,k)=A1(k+1:n,k)/A1(k,k)A1(k+1:n,k+1:n)=A1(k+1:n,k+1:n)-A1(k+1:n,k)*A1(k,k+1:n) elsebreakendendL=tril(A1,0);for i=1:nL(i,i)=1;endLU=triu(A1,0)y1=L\b1x1=U\y1得到如下结果：x1 =3.845714853511634e+001.609517394778522e+00-1.547605454206655e+011.041130489899787e+01将A2=[10,-7,0,1;-3,2.0999********,6,2;5,-1,5,-1;0,1,0,2]b2=[8;5.900000000001;5;1]代入上述代码求得结果如下：X2 =4.440892098500626e-16-9.999999999999993e-019.999999999999997e-011.000000000000000e+00(3)用不选主元的高斯消去法求得L和U及解向量x1,x2∈R4代码：format longeformat compactA1=[0.3*10^-15,59.14,3,1;5.291,-6.130,-1,2;11.2,9,5,2;1,2,1,1] b1=[59.17;46.78;1;2][L,U]=lu(A1)y1=L\b1x1=U\y1求得如下结果：x1=3.845714853511634e+001.609517394778522e+00-1.547605454206655e+011.041130489899787e+01将A2=[10,-7,0,1;-3,2.0999********,6,2;5,-1,5,-1;0,1,0,2] b2=[8;5.900000000001;5;1]代入上述代码，求得结果如下：x 2 =4.440892098500626e-16 -9.999999999999993e-01 9.999999999999997e-01 9.999999999999999e-01（2）（3）求得结果相同，可验证结果正确。

第二章1.试证明nn R⨯中的子集“上三角阵”对矩阵乘法是封闭的。

证明：设n n R B A ⨯∈,为上三角阵，则)( 0,0j i b a ij ij >== C=AB ，则∑==nk kjik ij b ac 1)( 0j i c ij >=∴，即上三角阵对矩阵乘法封闭。

2.已知矩阵⎥⎥⎥⎦⎤⎢⎢⎢⎣⎡-=512103421121A ，求A 的行空间)(T A R 及零空间N(A)的基。

解：对T A 进行行变换，⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡⇒⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡--⇒⎥⎥⎥⎥⎦⎤⎢⎢⎢⎢⎣⎡-=00100010121420050000121501131242121TA 3)(=∴T A r ，)(T A R 的基为[][][]T T T 5121,03421121321=-==ααα，由Ax=0可得[]Tx 0012-=∴N(A)的基为[]T0012-3.已知矩阵321230103A ⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦，试计算A 的谱半径()A ρ。

解：2321()det()230(3)(64)013A f I A λλλλλλλλ---=-=--=--+=--max 35()3 5.A λρ=+=+4、试证明22112212211221,,,R E E E E E E ⨯+-是中的一组基。

，其中11121001,0000E E ⎛⎫⎛⎫== ⎪ ⎪⎝⎭⎝⎭22210000,1001E E ⎛⎫⎛⎫== ⎪ ⎪⎝⎭⎝⎭。

1222112112211221134112212211221234134411221221122123410010000,,,00001001010110100000E E E E E E E E k k k k k k k E E E E E E k k k k k k E E E E E ⎛⎫⎛⎫⎛⎫⎛⎫==== ⎪ ⎪ ⎪ ⎪⎝⎭⎝⎭⎝⎭⎝⎭⎛⎫⎛⎫+=-= ⎪ ⎪-⎝⎭⎝⎭+⎛⎫⎛⎫++++-== ⎪ ⎪-⎝⎭⎝⎭++++-解：，（）（）令因此（）（0000O E ⎛⎫== ⎪⎝⎭）12331112212212211221111221122122112222112212211221 0 ,22,,,k k k k a a A V a a a a a aA a a E E E E E E R E E E E E E ⨯⇔====⎛⎫=∈ ⎪⎝⎭+-=+++-+∴+-对于任意二阶实矩阵有（）（）是中的一组基。

数值分析上机作业第 1 章1.1计算积分，n=9。

（要求计算结果具有6位有效数字）程序：n=1:19;I=zeros(1,19);I(19)=1/2*((exp(-1)/20)+(1/20));I(18)=1/2*((exp(-1)/19)+(1/19));for i=2:10I(19-i)=1/(20-i)*(1-I(20-i));endformat longdisp(I(1:19))结果截图及分析：在MATLAB中运行以上代码，得到结果如下图所示：当计算到数列的第10项时，所得的结果即为n=9时的准确积分值。

取6位有效数字可得.1.2分别将区间[-10.10]分为100,200,400等份，利用mesh或surf命令画出二元函数z=的三维图形。

程序：>> x = -10:0.1:10;y = -10:0.1:10;[X,Y] = meshgrid(x,y);Z = exp(-abs(X))+cos(X+Y)+1./(X.^2+Y.^2+1);subplot(2,2,1);mesh(X,Y,Z);title('步长0.1')>> x = -10:0.2:10;y = -10:0.2:10;[X,Y] = meshgrid(x,y);Z = exp(-abs(X))+cos(X+Y)+1./(X.^2+Y.^2+1);subplot(2,2,1);mesh(X,Y,Z);title('步长 0.2')>>x = -10:0.05:10;y = -10:0.05:10;[X,Y] = meshgrid(x,y);Z = exp(-abs(X))+cos(X+Y)+1./(X.^2+Y.^2+1);subplot(2,2,1);mesh(X,Y,Z);title('步长0.05')结果截图及分析：由图可知，步长越小时，绘得的图形越精确。

4.用Jacobi 迭代法求解方程组123123123202324812231530x x x x x x x x x ++=⎧⎪++=⎨⎪-+=⎩，取初值()()00,0,0T x =，计算迭代二次的),,(z y x 值；（2分）问Jacobi 迭代法是否收敛？为什么？（2分）若收敛，需要迭代多少次，才能保证各分量的误差绝对值小于610-？（提示：6ln(310)13.57ln 3⨯≈）（5分） 问Gauss-Seidel 迭代法是否收敛？为什么？（1分）5.用欧拉法求解初值问题()202y x yy '⎧=-+⎪⎨=⎪⎩在[]0,1.5上的数值解，取0.5h =，计算过程保留5位小数。

（要求写出迭代公式，不写公式扣4分） 三.分析题，请写出主要分析与认证过程（每小题5分，共10分） 1.设Ax b =，其中n n A R ⨯∈为非奇异矩阵，证明()()222T cond A A cond A ⎡⎤=⎣⎦2.证明向量X的范数满足不等式2X X∞∞≤≤四.证明（10分）对于给定的正数a ,应用牛顿法于方程1()0f x a x=-=，写出牛顿迭代格式； 证明当初值满足020x a<<时，该迭代法收敛。

武汉理工大学研究生课程考试标准答案用纸课程名称：数值计算（A ） 任课教师 ：一. 简答题，请简要写出答题过程（每小题5分，共30分） 1.将227和355113作为 3.14159265358979π=的近似值，它们各有几位有效数字，绝对误差和相对误差分别是多少？3分）2分）2.已知()8532f x x x =+-，求0183,3,,3f ⎡⎤⎣⎦，0193,3,,3f ⎡⎤⎣⎦.(5分）3.确定求积公式10120()(0)(1)(0)f x dx A f A f A f '≈++⎰中的待定系数，使其代数精度尽量高，并指明该求积公式所具有的代数精度。

解：要使其代数精度尽可能的高，只需令()1,,,m f x x x =使积分公式对尽可能大的正整数m 准确成立。