华中科技大学出版社―数值分析第四版―课后习题及答案

第四版数值分析习题答案第一章绪论习题参考答案1．ε（lnx）≈***()()rxxx。2．1******()()()()0.02nnnrnnnxxxnxxnxxx。3．*1x有5位有效数字，*2x有2位有效数字，*3x有4位有效数字，*4x有5位有效数字，*5x有2位有效数字。4．******4333124124()()()()0.5100.5100.5101.0510xxxxxx************123231132123()()()()0.214790825xxxxxxxxxxxx****62224***24441()()()8.85566810xxxxxxx。5．33323131()1()()()/()0.003333436433rrrVVVRVVVV。6．33100111()100101010022Y。7．12878355.982x，211287830.0178655.98228783x。8．21arc12NdxtgNx9．121()()()0.0052xSSS。10．()()0.1Sgttgt，2()2()0.2()12rgtttSttgt，故t增加时S的绝对误差增加，相对误差减小。11．1081001()10()102yy，计算过程不稳定。12．6(21)0.005051f，如果令21.4，则61(21)0.004096f，2610.005233(21)f，33(322)0.008f，4310.005125(322)f，5997021f，4f的结果最好。13．(30)4.094622f，开平方时用六位函数表计算所得的误差为41102，分别代入等价公式)1xx(ln)x(f),1xx(ln)x(f2221中计算可得2431221()ln(1)(1)6010310211fxxxxxx，4722211()ln(1)108.331060211fxxxx。14．方程组的真解为1210000000009999999981.000000,1.000000999999999999999999xx，而无论用方程一还是方程二代入消元均解得121.00,1.00xx，结果十分可靠。15．sinsincostansinsbcaacbabccabcccsabcabc第二章插值法习题参考答案1.1010)()()(nijjiniinxxxxxV；101101)(),,,(nijjinnxxxxxV.2.)12)(12()1)(1(4)21)(11()2)(1()3()21)(11()2)(1(0)(2xxxxxxxL3723652xx.3.线性插值：取510826.0,693147.0,6.0,5.01010yyxx，则620219.0)54.0()54.0(54.0ln0010101xxxyyyL；二次插值：取510826.0,693147.0,916291.0,6.0,5.0,4.0210210yyyxxx，则)54.0(54.0ln2L))(()54.0)(54.0())(()54.0)(54.0())(()54.0)(54.0(120210221012012010210xxxxxxyxxxxxxyxxxxxxy＝－0.616707.4.))()((21)()()(1011xxxxfxLxfxR，其中],[10xx.所以总误差界|))((|max|)(sco|max21|)(|1011010xxxxxxRxxxxxx822011006.1180601814)(121xx.5.))()(())()(()(3212023102xxxxxxxxxxxxxl当hxx3740时，取得最大值277710|)(|max230xlxxx.6.i)对),,1,0(,)(nkxxfk在nxxx,,,10处进行n次拉格朗日插值，则有)()(xRxPxnnk)())(()!1(1)(0)1(0nnnikjjxxxxfnxxl由于0)()1(nf，故有knikjjxxxl0)(.ii)构造函数,)()(ktxxg在nxxx,,,10处进行n次拉格朗日插值，有nijkjnxltxxL0)()()(.插值余项为njjnnkxxngxLtx0)1()()!1()()()(，由于).,,2,1(,0)()1(nkgn故有.)()()()(0nijkjnkxltxxLtx令,xt即得nijkjxltx00)()(.7.以a,b两点为插值节点作)(xf的一次插值多项式)()()()()(1axabafbfafxL，据余项定理，],[),)()((21)()(1babxaxfxLxf，由于,0)()(bfaf故.|)(|max)(81|))((|max|)(|max21|)(||)()(|21xfabbxaxxfxfxLxfbxabxabxa8.截断误差].4,4[),)()((61)(2102xxxxxxexR其中,,1210hxxhxx则hxx331时取得最大值321044392|))()((|maxhxxxxxxx.由题意，,10)392(61|)(|6342hexR所以，.006.0h9.,221nnny,2)22()22(1122nnnnnny则可得.2)(224nnnyy2/12/122nnny，11122)22()22(nnnnnny，则可得.2)(2224nnnyy10.数学归纳法证当1k时，)()()(xfhxfxf为m－1次多项式；假设)0)((mkxfk是m-k次多项式，设为)(xg，则)()()(1xghxgxfk为m-(k+1)次多项式，得证。11.右)()(111kkkkkkffgggfkkkkgfgf11左12.,1111121001010nnnnnkkkgfgfgfgfgfgfgf.121221011101nnnnknkkgfgfgfgfgfgffg13.102njjy)()()()()()(1112230112nnnnyyyyyyyyyyyy0011)()(yyyyyynnn.14.由于nxxx,,,21是)(xf的n个互异的零点，所以)())(()(210nxxxxxxaxf,)()()(1010njiiijniixxxxaxxa对)(xf求导得njiinjiiijixxxxxxaxf010))()(()()(，则njiiijjxxaxf10)()(，.)(1)(1110njnjnjiiijkjjkjxxxaxfx记,)(kkxxg则.1,)!1(,20,0)()1(nknnkxgn由以上两式得],,,[1)()(1)(2101110nknjnjnjiiijjkjkjxxxgaxxxgaxfx.1,,20,0)!1()(110)1(0nkankngank15.i)njnjjjjjjjnxxxxxxxxxFxxxF011010)())(()()(],,,[],,,[)())(()()(100110nnjnjjjjjjjxxxfcxxxxxxxxxfc.ii)证明同上。16.;1!7!7!7)(]2,,2,2[)7(710ff.0!7)(]2,,2,2[)8(810ff17.,0)()()(3jjjxpxfxR.1,,0)()()(3kkjxpxfxRjjj即1,kkxx均为)(3xR的二重零点。因而有形式：.)())(()(2123kkxxxxxKxR作辅助函数.)())(()()()(212kkxtxtxKtptft则.0)(,0)(,0)(,0)(,0)(11kkkkxxxxx由罗尔定理，存在),,(),,(121kkxxxx使得.0)(,0)(21类似再用三次罗尔定理，存在),,(),(121kkxx使得,0)()4(又),(!4)()()4()4(xKtft可得,!4)()()4(fxK即).,(.,!4)())(()(1212)4(3kkkkxxxxxxfxR18.采用牛顿插值，作均差表：ix)(ixf一阶均差二阶均差01201110-1/2],,[))((],[)()()(210101000xxxfxxxxxxfxxxpxp))()()((210xxxxxxBxA)2)(1()()2/1)(1(0xxxBxAxxx又由,1)1(,0)0(pp得,41,43BA所以.)3(4)(22xxxp19.记.,khaxnabhk则].,[,)()()(11111iiiiiiiiiinxxxxxxxxfxxxxxfx因为],[)(baCxf，所以)(xf在],[ba上一致连续。当Nn时，nabh，此时有|)()(|maxmax|)()(|max110xxfxxfnxxxninbxaiiiiiiiiiixxxnixxxxxfxxxxxfxfii111110)()()(maxmax1iiiiiiiixxxnixxxxxfxfxxxxxfxfii111110)]()([)]()([maxmax1.maxmax111101iiiiiixxxnixxxxxxxxii由定义知当n时，)(xn在],[ba上一致收敛于)(xf。20.)(xIh在每个小区间],[1kkxx上表示为).(,)(11111kkkkkkkkkkhxxxfxxxxfxxxxxI计算各值的C程序如下：#includestdio.h#includemath.hfloatf(floatx){return(1/(1+x*x));}floatI(floatx,floata,floatb){return((x-b)/(a-b)*f(a)+(x-a)/(b-a)*f(b));}voidmain(){inti;floatx[11],xc,xx;x[0]=-5;printf(x[0]=%f\n,x[0]);for(i=1;i=10;i++){x[i]=x[i-1]+1;printf(x[%d]=%f\n,i,x[i]);}for(i=0;i10;i++){xc=(x