數(shù)值計(jì)算方法習(xí)題答案(習(xí)題3習(xí)題6)

習(xí)題三解：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習(xí)題四2.證明：迭代函數(shù)(x)1cosx2'(x)1sinx'(x)11,x(,)22上，迭代過(guò)程1對(duì)全部[,]均收斂。因此在cosxk1xkx0ab(,)2解：記I02,I12則有6.2,.In2In1,n1,2由上述迭代格式之迭代函數(shù)為(n)2x，則11'(x)(2x)2故關(guān)于隨意的x>0，均有11'(x)122x迭代是收斂的。不如假定limInI,則有I2I即I22I解之得I=2,及I=-1,負(fù)根不合題意舍去，故limnIn2,即lim22222證明：（1）(x)11,(x)2x2x3x1.3,1.6時(shí)，(x)1112,111.3,1.6x211.61.3220.911且(x)21.3因此迭代過(guò)程xk11在區(qū)間[1.3,1.6]上收斂。1x2231x2,x22（2）(x)(x)x133當(dāng)x1.3,1.6時(shí)，(x)311.32,311.621.3,1.621x225(x)3,(x)8x21x2339令(x)0得3x3(x)在x3,3上單一遞加。在x3,單一遞減。又(1.6)0.461,(1.3)0.451x1.3,1.6時(shí)，(x)1因此迭代過(guò)程xk131xk2在區(qū)間[1.3,1.6]上收斂。18.解：方程x3-a=0的根x*=3a.用Newton迭代法xk1xkxk3a2xka,k0,12323xk3xk此公式的迭代函數(shù)(x)2xa33x2'(x)22a.因?yàn)?'(x)0.'(x)2則3330x3a故迭代法二階收斂。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下證此格式是線性收斂的因迭代函數(shù)(x)5xa,而'(x)5ax3,又x3a,則6263'(3a)51(3a)3511063632故此迭代格式是線性收斂的。21.解：因?yàn)橐蟛缓_方，又無(wú)除法運(yùn)算，故將計(jì)算1(a0)等價(jià)化為求10a2ax的正根。而此時(shí)有f(x)a1,f'(x)2x2x3故計(jì)算1a

的Newton迭代公式為a1xn23a33a2xn1xn22xn2xn(22xn)xnxn324.解：牛頓法迭代公式為:xk1xkf(xk)xk32xk210xk20f'(xk)xk3xk24xk10迅速弦截法的迭代公式為:xk1xkf(xk)(xkxk1)xkxk32xk210xk20f(xk)f(xk1)xk2xkxk1xk22xk2xk1101第六章2（2）解235534763476347623550113222133513350529347634760529052901130036222x14,x21,x32解：ALDLT24A21710109100400L110,D016021310044L(DLTx)bx1165y11032Lyb61y28x2LTxD1y16y323x3234增補(bǔ)一些計(jì)算題11分別運(yùn)用梯形公式、Simpson公式、Cotes公式計(jì)算積分exdx（要求小數(shù)點(diǎn)后起碼保存0五位）解：運(yùn)用梯形公式1xbafafb1e0e11.8591409edx022運(yùn)用Simpson公式bxdxbafa4fabfbf62a11e04e2e11.71886126運(yùn)用Cotes公式117e01137e11.718282688exdx32e412e232e4902利用復(fù)化11dx（將積分區(qū)間5平分）Simpson公式計(jì)算積分I01x解：區(qū)間長(zhǎng)度b-a=1,n=5,故1，在每個(gè)小區(qū)間[h0.2，節(jié)點(diǎn)xi、xi,xi1]5ihi01...5中還需計(jì)算xi1xih，i=0，1.4。22nn1ns6fa4fx12fxifb6i0ii12121111111010.210.410.610.865411111110.110.310.510.710.9110.69315用歐拉法求初值問題代4步)解：歐拉格式為

dy1x3y3dx(取步長(zhǎng)h=0.1，結(jié)果起碼保存六位有效數(shù)字，迭y00由計(jì)算得：4.用改良的歐拉法求解初值問題y'12ty2,0t21ty(0)0要求步長(zhǎng)h=0.5，計(jì)算結(jié)果保存6位小數(shù)。解：改良?xì)W拉法的計(jì)算公式為~ynhf(tn,yn)yn1yn1ynhf(tn,yn)2

~f(tn1,yn1),n0,1,22ty代入得將h=0.5，f(t,y)121t~ynf(0.5tnyn)yn121tnyn1yntnyntn1~0.5122yn11tn1tn1由y00計(jì)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.對(duì)初值問題y'=-y+x+1,y(0)=1取步長(zhǎng)h=0.1，用經(jīng)典四階龍格—庫(kù)塔法求y(0.2)的近似值，并求解函數(shù)yxex在x=0.2處的值比較。解：經(jīng)典四階龍格—庫(kù)塔格式為yn1ynhk16k1f(xn,yn)k2f(xnh,yn2k3f(xnh,yn2k4f(xnh,yn

2k22k3k4hk1),n1,2,32k2)2hk3)將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，則n=0時(shí)k10.000000000,k20.050000000k30.047500000,k40.0952500000.12k22k3k4)1.004837500y(0.1)y1y0(k16n=1時(shí)k10.095162500,k20.140404375k30.138142281,k40.1813482710.1y(0.2)y2y1(k12k22k3k4)1.018730901精準(zhǔn)解y(0.2)0.2e0.2偏差為y(0.2)y21.47107給定線性方程組10.40.4x110.410.8x220.40.81x33試成立求解該方程組的矩陣形式的高斯-塞德爾迭代公式,并判斷該公式能否收斂,假如收斂,以x0=[0.50.50.5]T為初始向量迭代一步.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