数值计算方法复习题

数值计算方法复习一、用牛顿法求解方程42()230f x x x x =+--=的解，0 1.0x =，收敛精度0.001ε=()34212420001020010 '441 23,0,1,441231213=1 1.14294414410.14280.001n n n n n n n f x x x x x x x x n x x x x x x x x x x x +=+-+--=-=+-+--+---=-=+-+--=>L 解：迭代格式42421112123112142422223222223 1.14292 1.1429 1.14293=1 1.12454414 1.14294 1.14291 0.01840.00123 1.12452 1.1245 1.1 =1441x x x x x x x x x x x x x x x x +--+⨯---=-=+-⨯+⨯--=>+--+⨯--=-+-33224531.12414 1.12454 1.124510.0003580.001x x -=⨯+⨯--=>二、应用列主元素消元法求解1230.001 2.000 3.000 1.0001.000 3.712 4.623 2.0002.000 1.072 5.643 3.000x x x ⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥-=⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥-⎣⎦⎣⎦⎣⎦的解，保留4位有效数字0.0012.0003.000 1.000 2.000 1.072 5.643 3.000 = 1.000 3.7124.623 2.000 1.000 3.712 4.623 2.0002.000 1.0725.643 3.0000.001 2.000 3.000 1.000 A -⎡⎤⎡⎤⎢⎥⎢⎥-→-⎢⎥⎢⎥⎢⎥⎢⎥-⎣⎦⎣⎦%解：扩展矩阵如下：1.0000.536 2.822 1.500 1.0000.536 2.822 1.50003.176 1.8010.500010.5670.15702.0013.003 1.00200 1.8680.6871.0000.536 2.822 1.500 010.5670.1570010.368------⎡⎤⎡⎤⎢⎥⎢⎥→→⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦---⎡⎤⎢⎥→⎢⎥⎢⎥⎣⎦原方程变换[]3211.0000.536 2.822 1.500 010.5670.1570010.3680.3680.05110.499x x x x ---⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥=⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦=--为：求解得到：三、应用雅可比迭代法求解下列方程组1231231231027.21028.35 4.2x x x x x x x x x --=⎧⎪-+-=⎨⎪--+=⎩，收敛精度0.01ε< (1)()()123(1)()()213(1)()()312 0.10.20.720.10.20.830.20.20.84k k k k k k k k k x x x x x x x x x +++⎧=++⎪=++⎨⎪=++⎩解：迭代格式求解过程如表：四、设246391541636A ⎡⎤⎢⎥=⎢⎥⎢⎥⎣⎦，应用幂法求解其最大特征值，以及对应的特征向量[]0101(1)(0)1112461120.2143 3915127 0.4821 41636156 1.0000 max 0.78570.Tj j jx x Ax x x x =⎡⎤⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥====⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦⎣⎦-=>解：取212(2)(1)0012460.21438.35710.1875 39150.482119.9821 0.4483 41636 1.000044.5714 1.0000 max 0.03380.001j j jx Ax x x x ⎡⎤⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥====⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦⎣⎦-=>323(3)(2)2460.18758.16830.1860 39150.448319.5974 0.4462 41636 1.000043.9231 1.0000 max 0.00210.001j j jx Ax x x x ⎡⎤⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥====⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦⎣⎦-=>434(4)(3)2460.18608.15660.1859 39150.446219.5735 0.4460 41636 1.000043.8827 1.0000 max 0.000130.001j j jx Ax x x x ⎡⎤⎡⎤⎡⎤⎡⎤⎢⎥⎢⎥⎢⎥⎢⎥====⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎢⎥⎣⎦⎣⎦⎣⎦⎣⎦-=<计算得到：[]118.156619.573543.882743.879430.18600.446218.156619.573543.8827Tλω===（++）对应的特征值为：五、用拉格朗日差值法构造三次多项式，求解472.0=x 处的函数值要求小数点后4位23413412121314212324123124331321441 ()()()()()()()()()()()()()()()()()()()()()()()(x x x x x x x x x x x x y f x y y x x x x x x x x x x x x x x x x x x x x x x x x y x x x x x x x x x ------==+------------++----解：构造插值函数如下：442430.472)()(0.4720.47)(0.4720.48)(0.4720.49)0.484655(0.460.47)(0.460.48)(0.460.49)(0.4720.46)(0.4720.48)(0.4720.49)0.493745(0.470.46)(0.470.48)(0.470.49)(0. y x x x y -----=⨯------+⨯---+4720.46)(0.4720.47)(0.4720.49)0.50275(0.480.46)(0.480.47)(0.480.49)(0.4720.46)(0.4720.47)(0.4720.48)0.511668(0.490.46)(0.490.47)(0.490.48)0.4956---⨯------+⨯---=20122111023111112234111 nnn i ii i i i nnn ni i i i i i i i i n n n i ii i i i y a a x a x nx x y a x x x a x y a x x x ===========++⎛⎫ ⎪⎪⎛⎫ ⎪ ⎪= ⎪ ⎪ ⎪ ⎪⎝⎭ ⎪ ⎪⎝⎭∑∑∑∑∑∑∑∑∑解：构造拟合函数如下：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21012290 3.7518.1183 0 3.7508.44373.750 2.76567.5870 2.0001 2.25160.0313n i i i x y a a a y x x =⎛⎫ ⎪ ⎪ ⎪ ⎪⎪⎪ ⎪⎝⎭⎛⎫⎛⎫⎛⎫ ⎪⎪ ⎪= ⎪⎪ ⎪⎪⎪ ⎪⎝⎭⎝⎭⎝⎭=++∑∑代入具体数据后，得计算得到各个系数如下：七、用龙贝格求积公式计算积分12041I dx x =+⎰的近似值，要求收敛精度0.0001ε=21211214224 ()11) (0)4,(1)21((0)(1)) 322) (0.5) 3.211(0.5) 3.12241= =3.13333333) (0.25) 3.76471,(0.75) 2.5611((0.25)(0.75)) 3.13118244 = 3f x x f f T f f f T T f S T T f f T T f f S T =+===+===+=-===++=解：设421218448421=3.141573161= =3.1421215154) (0.125) 3.93846,(0.375) 3.50685,(0.625) 2.8764,(0.875) 2.2654911((0.125)(0.375)(0.625)(0.875)) 3.138992841= =3.14159331 = T C S S f f f f T T f f f f S T T C --=====++++=-421211684261=3.141591515641==3.1415863635) 3.14094,=3.14159,=3.14159,=3.14159S S R C C T S C R --=八、用预测-校正的改进欧拉法求解如下常微分方程2(0)0dy x x ydxy ⎧=+-⎪⎨⎪=⎩，取步长为0.1，计算到1.0 ()()(){}21221- 2i i i i i i i i i i i i i y y h x x y h y y x x y x x y ++⎧=++-⎪⎨=++-++-⎪⎩%%解：预测校正的欧拉法迭代计算格式如下：各阶段计算结果如下：。