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1、重庆交通大学学生实验报告实验课程名称数值分析开课实验室数学实验室实验内容P238第二题,P240第二题,P238第七题学院数学与统计2013年级信息与计算科学专业班2班学生姓名 高腾 学号开课时间 2015至 2016学年第1 学期、P238 第二题为求方程X3 - x2 - 1 = 0在x二1.5附近的一个根，设将方程改写成下列等价形 0式，并建立相应的迭代公式。1)X = 1 + 1/X 2，迭代公式X= 1 + 1/X2k+1k2)X 3 = 1 + X 2，迭代公式X= 3.1 + X2 ；k+1、k3)x2 =1 /(x -1)，迭代公式X= 1/.JXk+1勺k()程序：第一小问：
2、zyoye1.mfunction y = zuoye1()%UNTITLED1 Summary of this function goes here% Detailed explanation goes herex = 1.5;while 1 = 1y = 1 + 1/xA2;if abs(y-x) 10人(-5)x = y;elsebreakendend第二小问：zuoye2.mfunction y = zuoye2()%UNTITLED1 Summary of this function goes here % Detailed explanation goes herex = 1.5;w
3、hile 1 = 1y = (1 + xA2)A(1/3);if abs(y-x) 10A(-5)x = y;elsebreakendend第三小问：zuoye3.mfunction y = zuoye2()%UNTITLED1 Summary of this function goes here% Detailed explanation goes herex = 1.5;i = 0;while i 10A(-5)x = y;elsebreakendend(二)结果分析与检验：第一小问：x = 1.4656第二小问：x = 1.4656第三小问：迭代前十次结果为 y =1.4142, y=
4、1.5538, y= 1.3438, y= 1.7055, y = 1.1906，y = 2.2907，y = 0.8802，y = 0.0000 - 2.8892i，y = 0.3317 + 0.4659i,y = 0.3321 - 1.0570i由上述实验结果可知：第三个迭代公式是发散的,无法逐渐逼近跟。二、P240 第二题多项式求根是一个病态问题,考虑多项式p(x) = (x 一 1)(x 一 2)(x 一 10) = a + a x HF a x9 + xio019求解扰动方程p(x) + x9 = 0。(1) 产生系数a , a , a。019(2) 取8 = 10-6,10-8,1
5、0-10用MATLAB求根函数计算扰动方程的根，分析e对根的影响。()程序:zuoye.mfunction output_args = zuoye( e ) %ZUOYE Summary of this function goes here % Detailed explanation goes here syms xp = 1;for i = 1:10p = p *(x-i);end expand(p) t = p + e*xA9; sort(solve(t) end(二)结果分析与检验-丽畑9 + 1320魅8 - 1815(J*k7 + L57773*k6 - 9此此吕楹飞亠 34L69
6、30*k4 - 840950a*xA3 + 12753576*x12 - 10S2S640*x 十 362S800 当 8 = 10-6 时：1.00000000000275573192248841894131. 99999398730158830421468(858509433.0000019526941774966077438208823.99993932907L27034518201302401.0006790904997270024297168898265.99651655449390050343409?56148177.OO94OL16O5732430894759588O648O77
7、.98668735313789210956204137360099.009544086284896a2462421L38699299.9972294859405487925470351321844当 = 10-8 时:1.00000000000002755;31922399487361.99339999987301587311613576635693.000000015526&727457334700008513. 99999939318624075827087264450875.00000678177601251985568675418717.000093416966817860571777
8、02821187.99386684805884561936612733325099.0000960798045712800337997546729. 9933724412157742646211485804067当 = 10-10 时：1.0000000000000002755731822398598lv999999S9999S730L53730iee75643383.00000000019526785729891588695793.99999999393185L957409458425o9o5.0000000678184947&02769380779965.99g9996500801a917
9、7043209884S8827.C00000334111S5693656317034550247.99999866847497761379207048264729.000000?608636452685292fi5535L1449.5999997244266612730328546354016a =3416930，4根的准确值由上述实验结果可知：a = 3628800 , a = -10628640 , a = 12753576 , a = -8409500, 0123a = - 902055 , a = 157773 , a = -18150 , a = 1320 , a = - 55567
10、89当越来越小时，求得的根越逼近方程的根。二、P238第七题用下列方法求f (x) = x3 - 3x -1 = 0在x = 2附近的根。 0X* = 1.87938524 ，要求计算结果准确到四位有效数字。(1)用牛顿法；(2)用弦截法，取x = 2,x = 1.9 ；01(一)程序：nd.mfunction y = nd( x )%ND Summary of this function goes here% Detailed explanation goes herey = xA3 - 3*x - 1;endndd.mfunction y = ndd( x )%NDD Summary of
11、 this function goes here% Detailed explanation goes herey = 3*xA2 - 3;end第一小问：zuoye_nd.mfunction y = zuoye_nd( )%ZUOYE3 Summary of this function goes here% Detailed explanation goes herex = 2;for i = 1:100 a(i) = x if i = 3c = log(abs(a(i)-a(i-1) / log(abs(a(i-1)-a(i-2) 实% 验收敛速度end y = x - nd(x)/ndd
12、(x); if abs(y-x) 0.0001i = i + 1;x = y;else break endendend第二小问：zuoye_xj.mfunction y = zuoye_xj()%ZUOYE_XJ Summary of this function goes here% Detailed explanation goes herex1 = 2;x2 = 1.9;for i = 1:100a(i) = x1;a(i+1) = x2;if i+1 = 3c = log(abs(a(i+1)-a(i)/ log(abs(a(i)-a(i-1)实验收敛速度endy = x2 - nd(x2)/(nd(x2)-nd(x1)*(x2-x1);if abs(y-x2) 0.00000001x1 = x2;x2 = y;elsebreakendendend（二）结果分析与检验第一小问： suDye_nd.2-1222ans =1.S794 I其中C表示实验收敛速度输出结果为根的值第二小问:C =1.63131. 8794其中c表示实验收敛速度 输出结果为根的值
