《计算方法》课后作业参考答案（1-6章）打印版.docx

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1、第1章绪论1.3 设准确值为 x= 3.78695,y =10 ,它们的近似值 /= 3.7869/*= 3.7870 ,y；=9.9999,y*亍10.1,旷亍10.0001分别具有几位有效数字？解：(1)工；=3.7869 = 0.37869X101, x = 0.378695x10周=k-X, | = p.378695xio- 0.37869 xlO11 = 0.000005x10 LxlO1-5所以，彳*具有5位有效数字。芯=3.7870 = 0.37870X101, x = 0.378695X101E= |x-x*| = p.378695xl0- 0.37870 xlO11 = 0.

2、000005x10 Lx IO1-5所以，/具有5位有效数字。(2)y = O.lxlOy = 0.99999 xlO1, = O.lOlxlO2, O.lOOOOlxlO2E1= y y； = 1x101 0.99999x1()1E1= y y； = 1x101 0.99999x1()1=O.OOOOlxlOLxlO142所以，x*具有4位有效数字。E2= |y-| = ).lxio2-0.101 XlO2 | = 0.001xl02 LxlO2-2所以，液具有2位有效数字。E3= |y-| = ),lxl02-0.100001xl02 | = 0.000001 xlO2 LxlO2 5所以

3、，才具有5位有效数字。L4设9=0.0056731是龙的具有五位有效数字的近似值，试计算其绝对误差限和相对误差限。解：根据定理1.1, %* =0.0056731具有五位有效数字，则*的相对误差尸满足石下 1xl0-n+1=1xl0-5+1 = 0.5xl0-4= 即为相对误差限。rl 22(2) A二2 1-3-13 10 7-1 2 4 -2 10-152 1 3 110 0 013 1 0-12 41 0 -17 0 1-2 0 05 0 0 0 1J03 出事&出少出血1?0433180 13 311 I ID01100331 3 12 -4 -2 0-1520 0F1 031-0 1

4、312-37170721 3307001 +2x_0 _J_ -180 -J_ 0 03333J牛郎0 10 00 03 19250522510010013 0000I1312r0 1025400 0-17707- 70 0-3719 0 -21T 7 71785ofof78 -517o o O 1O 3 一7 1 一至25 n -311 - 7 11 - - 5O 。-7J25 7 _31 7 8 -51750127 1 O- -3 1 o O14+I3xma23 _ 161-器 tr4113 万38410tr 336打19585T7 8517517o o i -A _L A8517850

5、0 01 0 00 1 03.7对以下矩阵进行LU分解。2011(1) I -3 4-2 I；Il 17-5 ij(1)=2,3 = 173；1212312331二 4 一 (二)x0 = 4 w=-2-(-l)xl=-l22317=(7 x 0) / 4 =3232ixl-4)=-二11237T（注意:最后一个与3的求法）1_3 22三角分解的紧凑格式:112 -3/21/20 14 2 I7 W201-3/241/2 -1/2 7/4-52-3/21/2014-1/2I7/4 -37/8J1L= -3/21ll/27/4-1 2 3 4| 2 u = |IL0 14 -1/2-37/8 1

6、(2)13 4 52 14 423 2511=1 必 2=2,3= 3 921 -【，A31 . Ju22 =3 2 = 1u23 =43=1, “23 = 54 = 1，32= 1 2x3 = 3,，42= 3 2x 2 = 1“33 = 4 _ 2x3 1 = 3，34 = 4 2x 4 + 3 = 1/43=2-2x3-1x(-1)x1 = 344= 5-2x4 -(-1)x1- (-3)x(-l) = -5三角分解的紧凑格式:三角分解的紧凑格式:1 2 3 41 2 3 414 5 一 14 4l 122 523 4 5 一 11 4 41|23 2 51221-3-131424 12

7、341114|2-31-15 12-1-3-52-10 0-3510(3)0 2 4 -1 0 0 7 10122131 二 ,，41 = u22= 5-12义1，12U233=1 x 0 = 1, w = 024I =0I =(2-0x1)323.547 24u33，八4= 4-0- xl =I =74377x2_ = 4924 245u = 1 x34xO = -lU4449= 10-0-0-x(-l) =24289241_3-240-1712242 124三角分解的紧凑格式:0 14 72-2OOO 14 7n Im J oo-llo289 I24-17 -2 4 - 7 o-7-24-

8、7 o 2 3 -2 o o IIyIlJI o o T 叫 o 124一 74924? 32 j )2 - o O - -I O1O-11OJO1- O 11一 o o T10 O 14 7 7 0?- 7 OT28924I 24一 749243.13判断用雅克比迭代法和高斯赛德尔迭代法解下列线性方程组是否收敛。(1)解：雅克比迭代矩阵:13O- I -U10-23 Ojoj0.2C3.+U) =,0.25j0 -2 -110 -0,4 -0.2- 1 -2 = 0.25 0 -0.50.1JL-2 30 J 0.2 0.302AE-Cj = -0.250.20.4 0.220.5 = 23

9、+0.212+ 0.055 = 0-0.3 2求得G的特征根4 = -0.21,4= 0.11+ 0.49z , /L3= 0.11- 0.49i/XG) = O51，故该线性方程组收敛。高斯-赛德尔迭代法的迭代矩阵:5C =(D-LTU = -1 GI1|_2-2 110 0.4- 0.2 10 -2 j =-0.1 -0.550 I |_0 0.05 -0.125J2|4_Cg卜()00.40.22 + 0.10.55= 2 2 + 0.2252 + 0.04) = 0-0.05 2+0.125% = 0, %= -0.1125 + 01654i,-0.1125- 0.1654/XCg)=

10、 021,故该线性方程组发散。-0.5-0.5-0.52 -0.5 AE-Cg 02 + 0.5000.50.5 =2(2 + 0.5)2=02 + 0.54 = 4=-0.5,4=0 AQ)1故该方程组收敛。15第4章插值与拟合4.1已知一组数据，见表4-15o试用线性插值与二次插值计算sin (0.629)的近 似值。表 4-15X00.5235990.7853981.0471981.570796sinx00.50.7071070.8660251解：由 题意设 y=sinx,取 xo=O yo=O ； xi=0.523599yi=0.5 ； X2=0.785398y2=0.707107；

11、x3= 1.047198 y3=0.866025 ； x4= 1.570796 y4= 1L线性插值法：lxi=0.523599, X2=0.785398为节点线性插值多项式：(通式)：PAx)= X-X. y.+ XXy = X-0.785398x0.5+ x-0.523599 x 0.7071071221=0.5833820.022459+0.207107%0.261799或者sin(0.629)=% + 为一必.x) = 0583382x -x212.二次插值计算：以xi=0.523599, x2=0.785398, X3=l.047198 为节点(通式)= (x 0.785398)(x

12、 1.047198)x 0.5 + (0.523599 - 0.785398)(0.523599 -1.047198)23599)(1 .。47198)+(0.785398 0.523599)(0.785398 -1.047198)(x - 0.523599)(x - 0.785398)八环小”x 0.866025(1.047198 - 0.523599)(1.047198 0.785398)_ -0.3303x2+1.1756x - 0.0552-0.9395= 0.589177 4.5给定一组数据，见表牛19,用牛顿基本插值公式计算/(Q1581) 和/(0.6367)的值。表 4-19解

13、:1012345X0.1250.250.3750.50.6250.75/(X)0.796180.773340.743710.704130.656320.6022816，5（%）=/（/）+%0，P （1-%）+/K，%1，%2 1（% - A）（% 一%1）+f x0, x19 x2, x3 (% - x0)(x - X1 )(x-x2) +f x0,X2, x3, x4(x - x0)(x - %! )(x - x2)(x - x3) +fxQ, %1，x2, x3, x4, x5 (x - x0 )(x - -)(% 一 x2 )(x - x3 )(x - x4)Xiyi一阶均差二阶均差三

14、阶均差四阶均差五阶均差0.1250.7961810.250.77334-0.18272x-0.1250.3750.74371-0.23704-0.21728(x-0.125)(x-0.25)0.50.70413-0.31664-0.3184-0.26965(x-0.125)(x-0.25) (x-0.375)0.6250.65632-0.38248-0.263360.146770.83284(x-O.I25)(x-O,25) (x-0.375)(x-0.5)0.750.60228-0.43232-0.199360.170670.0478-1.25603(x-0.125)(x-0.25) (x-

15、0.375)(x-0.5) (x-0.625)f(x) = 0.79618 + (-0.18272) X(x- 0.125) + (-0.21728)x (x - 0.125) x(x- 0.25) +(-0.26965) x(x- 0.125) x(x - 0.125) x(x - 0.375)+(0.83284) x(x - 0.125) x(x - 0.25) x(x - 0.375) x(x - 0.5)+(-1.25603) x(x - 0.125) x(x - 0.25) x(x - 0.375) x(x - 0.5) x(x - 0.625)=-1.25603x5+ 3.1878

16、96/-2.978865x3+ 0.958676x2- 0.296086x + 0.8232895 所以：f (0.1581) = -1.25603x(0.1581)5+ 3.187896 x(0.1581)4-2.978865x(0.1581)3 +0.958676x(0.1581)2-0.296086 x(0.1581) + 0.8232895 = 0.79029同理可得：/(0.6367) = 0.65152 4.11设实验数据见表4-24,求其二次拟合多项式。表 4-24解：设拟合曲线为一+k + Xi0.10.20.30.40.50.60.7y5.12345.30535.56845.

17、93786.42707.07987.9493建立正规方程组为:1.4 0.784 0.4676。29.765317由此解得:5.1623-0.7601lL 6.7024 lj即可求得二次拟合多项式为：y=5.1623-0.7601 x+6.7024x2*1%X21JC31X41yMX， /10.10.010.0010.00015.12340.512340.05123420.20.040.0080.00165.30531.061060.21221230.30.090.0270.00815.56841.670520.50115640.40.160.0640.02565.93782.375120.9

18、5004850.50.250.1250.06256.42703.21351.6067560.60.360.2160.12967.07984.247882.54872870.70.490.3430.24017.94935.564513.895157和2.81.40.7840.467643.39118.64499.765318第5章数值微积分5.1分别利用梯形公式和辛普森公式计算下列积分。(1) J】x In xdx ；解：梯形公式：lx(Z?-Q)(Q)+ /(/?) 2原式=x(2-l)x(2xln2 + lxln 1) = 0.693147 2辛普森公式：S-)己/()+2/(c) + l/

19、(。) , c= 6662a + b 3 c=22i4331原式=(2 -1) x_xlxlnl+ _x _xln _+ _x 2xln 2 = 0.63651466 22 6解：梯形公式：原式= 1 x 703-0) = 0.082158 03x(2辛普森公式：ci + b 0 + 0.3 八 1 .c= 0.151z+ -xV0.3) = 0.1048466一i 4原式= (0.3-0)x(_x0 + _x J。566,0.5 2e cos 3xdx；o解：梯形公式：原式=-x(0.5 一 0)x(e cos(1.5)+ eQcos(O) = 0.298071辛普森公式：a + bc=0.

20、252xe() xcos0 + xe0,5 xcos0.75 + xexcosl.5) = 0.501474 66xe() xcos0 + xe0,5 xcos0.75 + xexcosl.5) = 0.501474 66原式= (0.5-0)x(：(4) f Jl+ddx o Jo解：梯形公式：19根据屋,rX得： = ；%* =0.5x 10-4x 0.0056731 =0.283655x10-6,即为绝对误差限。或者E(x )+3)02k-2kk=k=1 )/ 4 4 4 4、/ 2 1 2、 Lf 厂=_xl+ 4x(_ 上上 +_) +2x +_ 上)上 =1.0987251257

21、9 113 2 53yexdx,取N = 8。Jo解：复合梯形公式：20T =-(a) + 2Z/(x) + /S)N 9kNk=7 b - a 1n =一N 81 /78=；7(0)+ 今 y(xj+/(i)16f=i11：,277=_x0 + 2 xQ2x +2 xQ28+ +2x(_)2g8+e16888=0.728890复合辛普森公式：h也S =+ 4Z/(x ) + 2N-，/(x ) +f(b)N G2k-2k0Z=1k=l.b - a 1n =N 8k k ZJ 八 ix = a + it K 0,1, 6)2161848482k-2kk=k=-x /(0) + 4f(x ) +

22、 f(x ) +/(!)1=X481=x(0) + 4x(/3 )+/(9)+/(跖5)+ 2*(/氏)+ /(9)+ + /区4)+/0 + 4x()2x ,6 + (_)2x ,6 + .+4816= 0.71828416515ill i 7 T-)2 X 6)+ 2x( )2 x .8 +( )2 X H+ ( A X * ) + 416o4o21第6章常微分方程初值问题的数值解法6.1用欧拉方法求解下列初值问题，取步长h=0.1。a = -y + 2(%+l)(0Kl) b(o)= i解：根据欧拉公式，有：y+ = y+ O.lx-X.+ 2a+1) = 0.9y+ 0.2再+ 0.2

23、 (i = 0,1, 2,.,9)当 y(0)= 1,%0=0,%= 1i = o, y = o.9xl+ 0.2x0 + 0.2 =1.1i =1, %= 0.9x1.14- 0.2x0.1+ 0.2 =1.21i = 2, %= 0.9x1.21+ 0.2x0.2 + 0.2 =1.329i = 3, %= 0.9x1.329 + 0.2x0.3 + 0.2 =1.4561i = 4,%=0.9x 1.4581 + 0.2x0.4 + 0.2=1.59049i = 5, %= 0.9x1.59049 + 0.2x0.5 + 0.2 =1.731441i = 6,乃=0.9x 1.73144

24、1+0.2x0.6 + 0.2 =1.878297i = 7, % = 0.9x 1.878297 + 0.2x0.7 + 0.2 = 2.030467i = 8, %= 0.9x 2.030467 + 0.2x0.8 + 0.2 = 2.18742i = 9, y10= 0.9x 2.18742 + 0.2x0.9 + 0.2 = 2.348678f12=(%+i)y(o% i)(2) J2y(o)= i解：/(x,y)= Jx +I)y2 n n 2根据欧拉公式，有：%+1 =y+/(x,J)=yt05(x+J)y2 5 = 0,1,2,，9)0/ y(。)=1, x()= , y()=i

25、22 = 0,y=l+0.05=l.05=1,、= 1.05 + 0.05x(0.1+1)x1.052= 1.110638n = 2,y.= 1.110638 + 0.05x(0.2 +1) xl.H06382= LI84648 =3, % = 1.184648 + 0.05x(0.3 +1) xl. 184648?= 1.275868 =4,弘=1.275868 + 0.05x(0.4 +1) xl.2758682= 1.389817 = 5,% = 1.389817+ 0.05x(0.5+l)xl.3898172 =1.534686 =6, % = 1.534686 + 0.05x(0.6

26、 +l)xl.5346862 = 1.723107 =7,为=1.723107 + 0.05x (0.7 +1) xl.7231072= 1.975480 = 8,%= 1.975480 + 0.05x(0.8+1) xl.9754802= 2.326707n = 9,y0 = 2.326707 + 0.05x (0.9 +1) x 2.3267072= 2.8410006.2用改进欧拉方法求解下列初值问题，取步长h=0.1。f y=x + y(Qx 1)(1) ；I M。) = 1解：/(为,y)+y根据欧拉公式，有： +1 = y + 0.1x(%+ y) = 0.1%,+l.ly, (i

27、 = 0,1, 2,.,9)根据改进欧拉公式，有：( 町港次+1.切-、”。,1,2,.,9)y =y + 0.05(x +y+x +y )I M ii i Z+l z+1 . y(0) = 1, x()= 0, % = 1.，= 0,y = 0.1x0+1.1x1 = l.l,y= l+0.05x(0+l+0.1+1.1)= 1.11i=i= 1.005 +0.0 lxl.0052= 1.0151,%= 2+l.ll+ 0.2 +1.231) = 1.020354i =2= 1.041176,丁3= 1.020354 + 0.05x(0.2 +1.24205 + 0.3+1.386255)

28、= 1.047026i = 3,亍4= O.lx 0.3 +1.1x1.398465 = 1.079914,%= 1.398465 + 0.05x(0.3 +1.398465 + 0.4 +1.568312) = 1.086794Z = 49y5=0.1x 0.4+1.1x1.581804= 1.134039%=1.581804 + 0.05x(0.4 +1.581804 + 0.5 +1.779984) = 1.14256823i = 5,为=0lx 0.5 +1.1x1.794893 = 2.024382, 乂= 1.794893 + 0.05x(0.5 +1.794893 + 0.6 +

29、 2.024382) = 2.040857i = 6,y=O.lx 0.6+1.lx 2.040857 = 2.304943, y产 2.040857 + 0.05x(0.6 + 2.040857 + 0.7 + 2.304943) = 2.323147i = 7, % = 0.1 x 0.7 +1 . 1 x 2.323147 = 2.625462, %= 2.323147 + 0.05x(0.7 + 2.323147 + 0.8 + 2.625462) = 2.645577i = 8,%= O.lx 0.8 +l.lx 2.645577 = 2.990135, %= 2.645577 +

30、0.05x(0.8 + 2.645577 + 0.9 + 2.990135) = 3.012363i = 9,y= O.lx 0.9 +l.lx 3.012363 = 3.403599, Mo= 3.012363 + 0.05x(0.9 + 3.012363 +1.0 + 3.403599) = 3.42816f y 1 = xy1 (0 x log() / log 2 = 9.9658 ,取左=10。ICT?kW及/(%)的符号Xk及f(xk)的符rbk及/(4)的符号13 (-)3.5 (-)4 ( + )23.5 (-)3.75 ( + )4 ( + )33.5 (-)3.625 (-)

31、3.75 ( + )43.625 (-)3.6875 ( + )3.75 ( + )53.625 (-)3.65625 ( + )3.6875 ( + )63.625 (-)3.640625 ( + )3.65625 ( + )73.625 (-)3.632813 ( + )3.640625 ( + )83.625 (-)3.628906 (-)3.632813 ( + )93.628906 (-)3.630859 (-)3.632813 ( + )103.630859 (-)3.6318363.632813 ( + )2- 222_xl.5x(1.52 * + l)3 322_xl.5x(1

32、.52 * + l)3 3(1 )令在X)= 3T+17 (pX)= _X(X +1)3=0.4558 1,收敛。（这种方法欠妥）k12345xk1.4812481.4727061.4688171.4670481.466243所以，（px）x=.5由表数据可知，迭代公式收敛。（2） xk+ =k12345xk1.5411041.6309891.8271932.2583913.243230由表数据可知，迭代公式发散。2 玉f4+ik1234541.3333332.2500030.355554-4.3642330.0427154由表数据可知，迭代公式发散。(4) 4+尸 /由表数据可知，迭代公式发

33、散。k1234541.4142141.5537731.3437981.7054871.190572用牛顿法求下列方程的根，精度要求IO。(1)x - e x - 0，取初值 1;(2)(3)(1)令/(x)=x贝|J, ,(x)=l+e7/WX - ”1六 Xk +1 一入 k _入 k _f(x)l+e-xl+ex所以，/= 0.567143 ok01234Xk10.5378830.5669870.5671430.567143七一工人 一 2x 4 3_ x +3七一工人 一 2x 4 3_ x +3_/(x)Xk+ -Xk 、(2)令/(x) = d 2x 3,贝I，f(x) = 3x x

34、3-2x-5 = 0?取初值 x = 2,x =3。-2x -23% 2 2元2 3x * 2x 2k kk kk01234522.52.3829792.3744682.3744242.374424所以，7=2.374424。(3)令/(x) = x sinx=1 ,贝！J, fx) = 1-cosx 21 . 1x- smx- smx-x cos x +f (x)k k 2 k k k 2Xk+ Xk / Xk -1-COS X - 1-COS Xkkk012345Xk11.7428161.5228871.4976441.4973001.497300所以，/= 1.497300 o25用弦割法求解下列方程，精度要求ICT4。(1) x -3x2 2x+8 = 0?取初值 x=-2,x =-1.5 ； 015(1)令 /(x)=x3-3x2-2x+8kXk+七一飞1-2-1.5-1.549300.52-1.5-1.54930-1.561960.04933-1.54930-1.56196-1.5615500.01266