数值计算方法复习

一、用牛顿法求解方程f(x)x42x2x30的解，x01.0，收敛精度0.001

解： f'x4x4x13 迭代格式 xn1xn4xn2xnxn34x4xn1222n42,n0,1,1213441 x1=x0x02x0x034x04x0111.1429 x1x00.14280.001 x2=x1x12x1x134x4x112142

1.142921.14291.1429341.142941.1429134211.1245 x2x10.01840.001 x3=x2x22x2x234x24x2124211.124521.12451.1245341.124541.124513421.1241 x3x20.0003580.001

0.001二、应用列主元素消元法求解1.0002.0002.0003.7121.0723.000x11.0004.623x22.000的解，保留4位有

5.3x33.000效数字

解： 扩展矩阵如下：0.001=1.000 A2.0001.000 001.000 00原方程变换为：1.000 000.536102.822x31.5000.567x20.1571x10.3680.4992.0003.7121.0720.5363.1762.0010.536103.0004.6235.31.0002.0002.0001.0003.0000.0011.0723.7122.0005.34.6233.0003.0002.0001.0001.5000.1570.6872.8221.8013.0032.8220.56711.5001.0000.50001.00201.5000.1570.3680.536102.8220.5671.868求解得到：x0.3680.0511

三、应用雅可比迭代法求解下列方程组

10x1x22x37.2x110x22x38.3，收敛精度0.01 xx5x4.2123解： 迭代格式(k)(k)x1(k1)0.1x20.2x30.72 (k1)(k)(k) x20.1x10.2x30.83x(k1)0.2x(k)0.2x(k)0.84123求解过程如表：

k x1 x2 x3 kkk0 0 0 0 (k1)1 0.72 0.83 0.84 0.84 2 0.971 1.070 1.150 0.31 3 1.057 1.1571 1.2482 0.0982 4 1.0853 1.1853 1.2828 0.0346 5 1.0951 1.1951 1.2941 0.0113 6 1.0983 1.1983 1.2980 0.0039 Maxxixi(k)

2四、设A34解： 4916615，应用幂法求解其最大特征值，以及对应的特征向量 36 取x01112 x1Ax034(1)jT491661120.214315127 x10.4821 361561.0000(0) maxxjxj2 x2Ax134 maxxjj(2)0.78570.0014916xj4916xj4916xj60.21438.35710.1875150.482119.9821 x20.4483 361.000044.57141.0000(1)0.03380.0012 x3Ax234 maxxjj(3)60.18758.16830.1860150.448319.5974 x30.4462 361.000043.92311.0000(2)0.00210.0012 x4Ax334 maxxjj(4)60.18608.15660.1859150.446219.5735 x40.4460 361.000043.88271.0000(3)0.000130.001

计算得到：18.156619.5733.8827 1（++）43.879430.18600.44621对应的特征值为：8.156619.5733.8827T

五、用拉格朗日差值法构造三次多项式，求解x0.472处的函数值 x 0.46 0.47 0.493745 0.48 0.50275 0.49 0.511668 y 0.484655 要求小数点后4位

解： 构造插值函数如下： yf(x) (xx2)(xx3)(xx4)(x1x2)(x1x3)(x1x4)(xx1)(xx2)(xx4)(x3x1)(x3x2)(x1x4)y1y3(xx1)(xx3)(xx4)(x2x1)(x2x3)(x2x4)(xx1)(xx2)(xx3)(x4x1)(x4x2)(x4x3)y2y4y0.472(0.4720.47)(0.4720.48)(0.4720.49)(0.460.47)(0.460.48)(0.460.49)(0.4720.46)(0.4720.48)(0.4720.49)(0.470.46)(0.470.48)(0.470.49)(0.4720.46)(0.4720.47)(0.4720.49)(0.480.46)(0.480.47)(0.480.49)(0.4720.46)(0.4720.47)(0.4720.48)(0.490.46)(0.490.47)(0.490.48)0.4846550.4937450.502750.511668    0.4956

六、已知如下表的函数，试用最小二乘法求二次多项式来拟合这组数据 x y -1.00 -0.75 -0.5 -0.25 0 0.25 0.5 0.75 1 -0.2209 0.3295 0.8826 1.4392 2.0003 2.55 3.1334 3.7061 4.2836 2解： 构造拟合函数如下： ya0a1xa2xn应用最小二乘拟合的原理，可以得到上式中各系数的计算方程组nn xii1n2xii19 03.75ni1ni1nxi2ii1nxxi1i1n3ii1xa03xia1a24xi2inyii1nxyiii1n2xiyii1代入具体数据后，得03.7503.75a00a12.7656a218.11838.44377.58702计算得到各个系数如下： y2.00012.2516x0.0313x

七、用龙贝格求积公式计算积分I解： 设f(x)41x21041x2dx的近似值，要求收敛精度0.0001

1) f(0)4,f(1)2 T11212431243(f(0)f(1))3 2) f(0.5)3.2 T2 S1= T1T212131413f(0.5)3.1T1=3.133333) f(0.25)3.771,f(0.75)2.56 T4 S2= C1= T2T4(f(0.25)f(0.75))3.13118T2=3.14157115S1=3.1421216151243S24) f(0.125)3.93846,f(0.375)3.50685,f(0.625)2.87,f(0.875)2.269 T8 S4= C2= R1=T4T81813(f(0.125)f(0.375)f(0.625)f(0.875))3.139T4=3.14159115163S2=3.14159C1=3.14158

1615S463C25) T163.14094,S8=3.14159,C4=3.14159,R2=3.14159 八、用预测-校正的改进欧拉法求解如下常微分方程

dy2xxy，取步长为0.1，计算到1.0 dxy(0)0解：预测-校正的欧拉法迭代计算格式如下：yi1yihxi2xiyi  h22ixixiyixixiyyi1yi2各阶段计算结果如下： x  yy 0 0 0 0.6 0.2055 0.2129 0.1 0 0.0055 0.7 0.2877 0.29 0.2 0.0160 0.0219 0.8 0.38 0.3929 0.3 0.0437 0.0501 0.9 0.4976 0.5059 0.4 0.0841 0.0909 1.0 0.6263 0.6348 0.5 0.1378 0.1450 x  yy