數(shù)值分析課后題答案

1、第二章2.當(dāng)x1,1,2時(shí)，f(x)數(shù)值分析0,3,4，求f(x)的二次插值多項(xiàng)式。解:x1,x11,x22,f(x°)0,f(x，3,f(x2)4;l0(x)(xx1)(xx2)1(x1)(x2)(x0X)(x。x2)2'八)，/、(xx0)(xx2)1,l1(x)-2(x1)(x2)1(X%)3x2)6"x)J-x0)(xx1)、1(x1)(x1)(x2x0)(x2X)3則二次拉格朗日插值多項(xiàng)式為2L2(x)yJk(x)k0) o31)2411-2(X 1(X4 一 3)1 (X16.設(shè)Xj,j0,1,L,n為互異節(jié)點(diǎn)，求證：n(1)x："(x)xk(

2、k0,1,L,n);j0n(Xjx)klj(x)0(k0,1,L,n);j0證明(1)令f(x)xkn若插值節(jié)點(diǎn)為xj,j0,1,L,n,則函數(shù)f(x)的n次插值多項(xiàng)式為L(zhǎng)n(x)x："(x)。j0插值余項(xiàng)為Rn(x)f(x) Ln(x)f(n1)()(n 1)!n 1(x)又Qkn,f(n1)()0R(x)0nxklj(x)xk(k0,1,L,n);j0n(xjx)klj(x)j0nn(C：xj(x)ki)lj(x)j0i0nn_ikiiCk(x)(xjlj(x)i0j0又Q0in由上題結(jié)論可知nxklj(x)xij0n原式Ck(x)kixii0(xx)k0得證。7設(shè)f(x)C2a

3、,b且f(a)f(b)0,求證:maxf(x)axb8(ba)2maxf(x).axb解：令x0a,x1b,以此為插值節(jié)點(diǎn)，則線(xiàn)性插值多項(xiàng)式為x為xx0Li(x)f(x°)一2f(x1)0x0x1xx0xbxa=f(a)f(b)abxa又Qf(a)f(b)0L1(x)01-插值余項(xiàng)為R(x)f(x)L1(x)3f(x)(xx0)(xx1)1.f(x)2f(x)(x%)(xx1)又Q(x%)(xXi)2(X 4(X1 4(bXo)Xo)2a)2(XiX)max a x bf(x)8(ba)2 max f (x)7 a x b ' '8.在 4 x4上給出f(x) ex的

4、等距節(jié)點(diǎn)函數(shù)表，若用二次插值求ex的近似值，要使截?cái)嗾`差不超過(guò)10 6,問(wèn)使用函數(shù)表的步長(zhǎng) h應(yīng)取多少?4yn (E 1)4yn解：若插值節(jié)點(diǎn)為Xi1,Xi和Xi1,則分段二次插值多項(xiàng)式的插值余項(xiàng)為-,、1R(x)f()(xXi1)(xx)(xXi1)3!R2(X)1/、(-(X Xi i)(X6x)(x Xi i) max f (x)設(shè)步長(zhǎng)為h,即 Xi 1Xih,X 1 X hR2(x)1e4 -2-h363,3 e4h3. 27若截?cái)嗾`差不超過(guò)106,R2(x)10634,36eh1027h0.0065.9.若 yn2n，求 4yn及4yn.，解：根據(jù)向前差分算子和中心差分算子的定義進(jìn)行

5、求解。nyn24(j04(j04(j01)j1)j1)j(21)4VnE4jyny，24ynyn2n1E 2)4yn14yn(E21(E2)4(E1)4%4VnVn2n16.f(x)3x1,求F20,21,L,27及F20,21,L,28解:Qf(x)x43x1右xi2i,i0,1,L,8Xo,Xi,Lf(n)()n!19xo,xi,L,x7x0,x1，L,%f()7!f(8)()8!次數(shù)不7!一17!高于4次的多項(xiàng)式P(x),使它滿(mǎn)足P(0)P(0)0,P(1)P(1)0,P(2)解法一：利用埃米爾特插值可得到次數(shù)不高于4的多項(xiàng)式x0,xi1Vo0,y11m00m111%(x)yjj(x)m

6、jj(x)j0j00(x)(12±)(土)2X0X1X0X12(12x)(x1)21(x)(12±)(上)2X1X0X1X02(32x)x22o(x)X(X1)21(x)(X1)x3 c 2x 2x_22H3(x)(32x)x(x1)x設(shè) P(x) H3(x) A(x x0)2(xX1)2其中，A為待定常數(shù)Q P(2) 1P(x)X3 2x2 Ax2(x 1)2從而P(x)1221X2(X3)24解法二：采用牛頓插值，作均差表:Xif(Xi)一階均差二階均差00111210-1/2p(x)p(Xo)(xXo)fXo,Xi(xXo)(XXi)fXo,Xi,X2(ABx)(xx

7、(x1)(又由p(0)p(x)所以0,p(1)2X1，得x0)(xx1)(xX2)1/2)(ABx)x(x1)(x2)301一，B,44(X3)2.第四章1.確定下列求積公式中的特定參數(shù)，使其代數(shù)精度盡量高，并指明所構(gòu)造出的求積公式所具有的代數(shù)精度：hhf(x)dxAif(h)AofOAf(h)；2h2hf(x)dxAif(h)Af(0)Af(h)；11f(x)dxf(1)2f(xi)3f(x2)/3;h20f(x)dxhf(0)f(h)/2ah2f(0)f(h);解：求解求積公式的代數(shù)精度時(shí)，應(yīng)根據(jù)代數(shù)精度的定義，即求積公式對(duì)于次數(shù)不超過(guò)式均能準(zhǔn)確地成立，但對(duì)于m+1次多項(xiàng)式就不準(zhǔn)確成立，進(jìn)

8、行驗(yàn)證性求解。h(1)若hf(x)dxA1f(h)Aof(0)Af(h)令f(x)1,則2hA1A0Am的多項(xiàng)令f(x)x,則0A1hAh“o2.oo令f(x)x,則一hhA1hA3A03h1從而解得A-h3A11h3hh令f(x)x,則hf(x)dxhxdx0h故hf(x)dxA1f(h)A0f(0)Af(h)hh42Gf(x)dxxdx-hhh5,25A1f(h)A3f(0)A1f(h)-h3A1f(h)A0f(0)A/(h)成立。令f(x)x4故此時(shí),f(x)dxAf(h)A0f(0)Af(h)f(x)dxA1f(h)A0f(0)A1f(h)h具有3次代數(shù)精度。2h(2)若f(x)dxA

9、1f(h)A0f(0)A1f(h)2h令f(x)1,貝U4hA1A0A令f(x)x,則0A1hAh令f(x)x2,則Th3h2Alh2A飛3h8從而解得A-h3A18h332h2h3令f(x)x,則f(x)dxxdx02h2h2h故2hf(x)dxA1f(h)A0f(0)Af(h)成立。/2h2h/64;令f(x)x,則f(x)dxxdxh2h2h5A#(h)A0f(0)Af(h)16h532h故此時(shí)，2h"x)dxA1f(h)A0f(0)Af(h)A1f(h)A0f(0)Af(h)02h因此，f(x)dxA1f(h)A0f(0)Af(h)2h具有3次代數(shù)精度。11f(x)dxf(1

10、)2f(x1)3f%)/31令f(x)1,則f(x)dx2f(1)2f(x1)3f(x2)/3令f(x)x,則012x13x2人222令f(x)x,則212x13x2從而解得x1x20.2899或x10.68990.5266x20.1266o11o令f(x)x3,則f(x)dxx3dx0f(1)2f(x1)3f(x2H/30111故1f(x)dxf(1)2f(x1)3f(x2)/3不成立。因此，原求積公式具有2次代數(shù)精度。(4)若f(x)dxhf(0)f(h)/2ah2f(0)f(h)h令f(x)1,則0f(x)dxh,令f(x)x,則2hf(0)f(h)/2ahf(0)f(h)hhh12f(

11、x)dxxdxh002_2_,12hf(0)f(h)/2ah2f(0)f(h)h22令f(x)x2,則hh213f(x)dxxdxh0032_132hf(0)f(h)/2ah2f(0)f(h)h32ah22故有13132h-h2ah321a一12令f(x)x3,則f(x)dxh314xdxh0412hf(0)f(h)/2/"0)141414f(h)hhh244令f(x)x4,則hh415f(x)dxxdxh00512-hf(0)f(h)/2hf(0)12f(h)1h521h531h56故此時(shí),h0f(x)dxhf(0)f(h)/h因此，of(x)dxhf(0)具有3次代數(shù)精度。12,

12、2h2f(0)f(h),12,12,f(h)/2h2f(0)f(h)127。若用復(fù)化梯形公式計(jì)算積分1Iexdx,問(wèn)區(qū)間0,1應(yīng)多少等分才能使截?cái)嗾`差不超過(guò)0106?解：ban米用復(fù)化梯形公式時(shí)，余項(xiàng)為R(f)ah2f(),(a,b)121又QIoexdx故f(x)ex,f(x)ex,a0,b1.Rn(f)h2f()h21212若|Rnf|106,則當(dāng)對(duì)區(qū)間0,1進(jìn)行等分時(shí)，h1,n故有nJe°-因此，將區(qū)間476等分時(shí)可以滿(mǎn)足誤差要求12第五章2.用改進(jìn)的歐拉方法解初值問(wèn)題yxy,0x1;y(0)1,x取步長(zhǎng)h=0.1計(jì)算，并與準(zhǔn)確解yx12e相比較。近似解準(zhǔn)確解近似解準(zhǔn)確解0.1

13、1.111.110340.62.040862.044240.21.242051.242810.72.323152.327510.31.398471.399720.82.645582.651080.41.581811.583650.93.012373.019210.51.794901.797441.03.428173.436563、解：改進(jìn)的歐拉法為1、一.、yn1yn二hf(xn,yn)"%1,丫卜"4,丫口)2將f(x,y)x2xy代入上式，得2h_h,%i1hyn-1hxnlx1%加22同理，梯形法公式為yn1Uhyn2hhxn(1xn)xn1(1xn1)將y00,h0

14、.1代入上二式，計(jì)算結(jié)果見(jiàn)表95表95xn改進(jìn)歐拉yn|y(Xn)yn1梯形法yn|y(xn)yn|0.10.00550030.00523809540.20.0219275000.337418036100.0214058960.755132781100.30.05014438830.04936723930.40.0909306710.6582530781030.0899036920.1366487781030.5014372238830.962608182100.185459653100.1250716721020.22373844310320/p>

15、230.253048087103可見(jiàn)梯形方法比改進(jìn)的歐拉法精確。4、用梯形方法解初值問(wèn)題yy0;y(0)1,證明其近似解為yn并證明當(dāng)h0時(shí)，它原初值問(wèn)題的準(zhǔn)確解證明：梯形公式為h_.Vn1Vn2f(Xn，yn)f(xn1,Yn1)代f(x,y)y入上式，得hyn1yn2ynyn1解得Vn 1)yn (2)Vn 1n 1)V。因?yàn)閂。Vn)n以h為步長(zhǎng)經(jīng)n步運(yùn)算可求得y(x)的近似值Vn ,故nh, nyni(2h)Vn2 h - lim()h h 0 2 hlhmo(12h2 h)lhm0(12 h 2h x2h C2h T'h h2 h，10.證明解V f (x, y)的下列差分公

16、式是二階的,1 , Vn 1"(Vn2 并求出截?cái)嗾`差的首項(xiàng)。yn 1)h4(4Vn 1Vn3Vn1)Vn 1VnhynVn 1Vn.(1) hynh25 Vh2TV1h3 /h3飛'o(h3)o(h3)(1)Vn(1).(2)Vnhynh2(3)2-VnO(h )2 ,hyn2), y2o(h2),代入得3 (，8hyn32o(h ) o(h )工,截?cái)嗾`差首項(xiàng)為5h38Vn12.將下列方程化為一階方程組：V3y2y0,1)V(0)1,y(0)1;(1)V'z,z'3z2y,其中y(0)1,z(0)1。2、V0.1(1V)VV0,22)y(0)1,V(0)0

17、;(2)V'z,z'0.1(1V)zy,其中y(0)1,z(0)第六章21、用二分法求方程xx10的正根，要求誤差小于0.05.2解設(shè)f(x)xx1，f10，f10,故1,2為f(x)的有根區(qū)間.又f'(x)2x1,故當(dāng)11x2時(shí)，f(x)單增，當(dāng) 2時(shí)f (x)單增.而1f(2)1一一 k 1計(jì)式(7.2)知要求誤差小于 0.05,只需20.05，解得k 1 5.322,故至少應(yīng)二分6次.5f14,，由單調(diào)性知f(x)0的惟一正根x*(1,2).根據(jù)二分法的誤差估具體計(jì)算結(jié)果見(jiàn)表7-7.表7-7kakbkxkf(x。的符號(hào)0121.5-11.521.75+21.51.

18、751.625+31.51.6251.5625-41.56251.6251.59375-51.593751.6251.609375-即x*x51.6093753、為求x3x210在x01.5附近的一個(gè)根，設(shè)將方程改寫(xiě)成下列等價(jià)形式，并建立相應(yīng)的迭代公式：/1xx 1 xk 1x2,迭代公式3.2x(2) x 1 x ,迭代公式xk 1x2，xk 1 x 1,迭代公式試分析每種迭代公式的收斂性1(1xk2義;1xk 1并選取一種公式求出具有四位有效數(shù)字的近似根解取x01.5的鄰域1.3,1.6來(lái)考察.一 (x) 1當(dāng) x 1.3,1.6時(shí),122/ (x)11 丁癡L1,故迭代公式xk 111x

19、7在1.3,1.6上整體收斂.(2)當(dāng)X1.3,1.6時(shí)(X)(1X2)1/31.3,1.62 x21.6I'(x)|3|2|32L0.52213 (1X2)33(11.32)31故Xk1(1人尸在口.3,1.6上整體收斂.1111(x),|(x)|3Z2-I1"1''G2(x1)2(1.61)故外1發(fā)散.，只需由于(2)的L叫小，故取(2)中迭代式計(jì)算.要求結(jié)果具有四位有效數(shù)字|xkL13x*|IL區(qū)xk11210I xkxk 11T 2 10 3 0.5 10 3|x6 % |取x01.5計(jì)算結(jié)果見(jiàn)表7-8.kk11.48124803441.4670479

20、7321.47270573051.46624301031.46881731461.465876820表7-8由于37、用下列方法求f(x)x3x10在x02附近的根.根的準(zhǔn)確值x*1.87938524.,要求計(jì)算結(jié)果準(zhǔn)確到四位有效數(shù)字.用牛頓法；(2)用弦截法，取x02,x11.9;(3)用拋物線(xiàn)法，取x01,x13,x22.解f(1)0,f(2)0,f(x)3x233(x21)0,f''(x)6x0,對(duì)x1,2.xkxk取x02,用牛頓迭代法Xk 1Xkxk3Xk 13xk2 32xJ 13(42 1)13x11.888888889,x21.879451567,|x2x*|10計(jì)算得2,故x*x21.879451567(2)取“2,x11.9，利用弦截法vv(xkxk1)f(xk)xk1xkf(xk)f(xk1)13x21.981093936,x31.880840630,x41.879489903,|x4x*|-10得，2，故取x*x41.8