数值计算方法习题答案(习题3习题6)

习题三解：yn1yn0.2(ynxnyn2)y110.210.8y20.80.20.80.20.820.6144同理，y30.4613解：ypynhfxn,ynyn0.1(12yn2)1xn2ycynhfxn1,ynyn0.1(12yp2)1xn21yn11ypyc2yp10.1,yc10.097,y10.0985同理，y20.1913,y30.273711.解：yn1yn0.2k12k22k3k46k183ynk283yn30.2k21k383yn30.2k22k483yn30.2k3k11,k21.4,k31.58,k41.05,y(0.2)2.3004同理，k11.0986,k20.7692,k30.8681,k40.5780,y(0.4)2.4654解：hyn1yn3ynyn12y1y,y00,y(0.2)0.181y(0.4)y(0.2)02.23y(0.2)10.1810.1310.18110.3267y(0.6)y(0.4)02.23y(0.4)y(0.2)0.32670.1310.3267(10.181)0.4468同理，y(0.8)0.5454,y(1)0.6265习题四2.证明：迭代函数(x)1cosx2'(x)1sinx'(x)11,x(,)22上，迭代过程1对全部[,]均收敛。因此在cosxk1xkx0ab(,)2解：记I02,I12则有6.2,.In2In1,n1,2由上述迭代格式之迭代函数为(n)2x，则11'(x)(2x)2故关于随意的x>0，均有11'(x)122x迭代是收敛的。不如假定limInI,则有I2I即I22I解之得I=2,及I=-1,负根不合题意舍去，故limnIn2,即lim22222证明：（1）(x)11,(x)2x2x3x1.3,1.6时，(x)1112,111.3,1.6x211.61.3220.911且(x)21.3因此迭代过程xk11在区间[1.3,1.6]上收敛。1x2231x2,x22（2）(x)(x)x133当x1.3,1.6时，(x)311.32,311.621.3,1.621x225(x)3,(x)8x21x2339令(x)0得3x3(x)在x3,3上单一递加。在x3,单一递减。又(1.6)0.461,(1.3)0.451x1.3,1.6时，(x)1因此迭代过程xk131xk2在区间[1.3,1.6]上收敛。18.解：方程x3-a=0的根x*=3a.用Newton迭代法xk1xkxk3a2xka,k0,12323xk3xk此公式的迭代函数(x)2xa33x2'(x)22a.因为''(x)0.'(x)2则3330x3a故迭代法二阶收敛。19.解：因f(x)=(x3-a)2,故f'(x)=6x2(x3-a)由Newton迭代公式：xn1xnf(xn),n0,1,f'(xn)得xn1xn(xn3a)25xna,n0,1,23a)626xn(xn6xn下证此格式是线性收敛的因迭代函数(x)5xa,而'(x)5ax3,又x3a,则6263'(3a)51(3a)3511063632故此迭代格式是线性收敛的。21.解：因为要求不含开方，又无除法运算，故将计算1(a0)等价化为求10a2ax的正根。而此时有f(x)a1,f'(x)2x2x3故计算1a

的Newton迭代公式为a1xn23a33a2xn1xn22xn2xn(22xn)xnxn324.解：牛顿法迭代公式为:xk1xkf(xk)xk32xk210xk20f'(xk)xk3xk24xk10迅速弦截法的迭代公式为:xk1xkf(xk)(xkxk1)xkxk32xk210xk20f(xk)f(xk1)xk2xkxk1xk22xk2xk1101第六章2（2）解235534763476347623550113222133513350529347634760529052901130036222x14,x21,x32解：ALDLT24A21710109100400L110,D016021310044L(DLTx)bx1165y11032Lyb61y28x2LTxD1y16y323x3234增补一些计算题11分别运用梯形公式、Simpson公式、Cotes公式计算积分exdx（要求小数点后起码保存0五位）解：运用梯形公式1xbafafb1e0e11.8591409edx022运用Simpson公式bxdxbafa4fabfbf62a11e04e2e11.71886126运用Cotes公式117e01137e11.718282688exdx32e412e232e4902利用复化11dx（将积分区间5平分）Simpson公式计算积分I01x解：区间长度b-a=1,n=5,故1，在每个小区间[h0.2，节点xi、xi,xi1]5ihi01...5中还需计算xi1xih，i=0，1.4。22nn1ns6fa4fx12fxifb6i0ii12121111111010.210.410.610.865411111110.110.310.510.710.9110.69315用欧拉法求初值问题代4步)解：欧拉格式为

dy1x3y3dx(取步长h=0.1，结果起码保存六位有效数字，迭y00由计算得：4.用改良的欧拉法求解初值问题y'12ty2,0t21ty(0)0要求步长h=0.5，计算结果保存6位小数。解：改良欧拉法的计算公式为~ynhf(tn,yn)yn1yn1ynhf(tn,yn)2

~f(tn1,yn1),n0,1,22ty代入得将h=0.5，f(t,y)121t~ynf(0.5tnyn)yn121tnyn1yntnyntn1~0.5122yn11tn1tn1由y00计算得~0.500000,y(0.5)y10.400000y1~0.740000,y(1.0)y20.635000y2~0.817500,y(1.5)y30.787596y3~0.924049,y(2.0)y40.921025y45.对初值问题y'=-y+x+1,y(0)=1取步长h=0.1，用经典四阶龙格—库塔法求y(0.2)的近似值，并求解函数yxex在x=0.2处的值比较。解：经典四阶龙格—库塔格式为yn1ynhk16k1f(xn,yn)k2f(xnh,yn2k3f(xnh,yn2k4f(xnh,yn

2k22k3k4hk1),n1,2,32k2)2hk3)将f(x,y)=1+x-y及h=0.1代入得yn1yn0.1k12k22k3k46k11xnynk21(xn0.05)(yn0.05k1),n0.1,2,3k31(xn0.05)(yn0.05k2)k41(xn0.1)(yn0.1k3)因初值y01，则n=0时k10.000000000,k20.050000000k30.047500000,k40.0952500000.12k22k3k4)1.004837500y(0.1)y1y0(k16n=1时k10.095162500,k20.140404375k30.138142281,k40.1813482710.1y(0.2)y2y1(k12k22k3k4)1.018730901精准解y(0.2)0.2e0.2偏差为y(0.2)y21.47107给定线性方程组10.40.4x110.410.8x220.40.81x33试成立求解该方程组的矩阵形式的高斯-塞德尔迭代公式,并判断该公式能否收敛,假如收敛,以x0=[0.50.50.5]T为初始向量迭代一步.解：10.40.4000A0.410.8,L0.400,0.40.810.40.8000.40.4100U000.8,D010000001有x(k1)x1(k1)x2(k1)x3(k1)令G

1(k)(DL)1bDLUx00.4