《數(shù)值分析》第二章答案

習(xí)題21.分析下列方程各存在幾個(gè)根，并找出每個(gè)根的含根區(qū)間：(1)；xcosx0(2)；3xcosx0(3)0；sinxex(4)x0。2xe解：(1)xcosx0(A)，f(x)1sinx0,x(,)f(x)xcosx,f(0)0cos01f(1)1cos(1)1cos10方程(A)有唯一根x*[1,0](2)3xcosx0(B),,x(,)時(shí)f(x)3xcosxf(x)3sinx0，f(1)31cos13cos10f(0)30cos010方程(B)有唯一根*[0,1]x(3)(C)sinxex0sinxex,f1(x)sinxf2x()ex方程(C)有無(wú)窮個(gè)正根，無(wú)負(fù)根在[2k,2k]內(nèi)有一根，且()1klim[()2]0xkkx21k在[]內(nèi)有一根，且2k,2kxk1)]0()2klim[()(2kx22k(示圖如下)k0,1,2,3-1-f2(x)1234x(4)(D)x2ex0x2exy2f1(x)f2(x)1f1(x)x2,f2(x)ex方程(D)有唯一根x*[0,1]121x當(dāng)x0時(shí)(D)與方程xxe2x(E)e2y同解21x當(dāng)x0時(shí)(E)無(wú)根2.給定方程；xx10221x(1)試用二分法求其正根，使誤差不超過(guò)0.05；(2)若在[0,2]上用二分法求根，要使精確度達(dá)到6位有效數(shù)，需二分幾次？解：x210x1),,f(x)x2x10f(1)1f(1.5)0.250f(2)1x*1521.618034*[1.5,2],x-2-1.75(+)2(+)1.5()1.625(+)1.75(+)1.5625(+)1.625(+)1.5()1.5()1.625(+)1.5625()1.5937(5)(1.6251.5625)1101220.03125x*1.593751.62位有效近似值為1.6，2)aa00bb20c1(ab)k2kkba2k11x*ck2k112，210105k15ln10ln216.60k152k只要2等分18次3.為求的正根，試構(gòu)造x5x303種簡(jiǎn)單迭代格式，判斷它們是否3收斂，且選擇一種較快的迭代格式求出具有3位有效數(shù)的近似根。解：f(x)x35x3x(x25)3f(x)3x253(x25)355時(shí)3當(dāng)x時(shí)，;當(dāng)f(x)0xf(x)035(5)31053055f()33333-3-53)1053033，f(0)30f(，f(3)3(95)39f(2)2(45)35y23553x33由草圖可知唯一正根x*(2,3)11(1)5xx33，，,x(x33)(x)(x33)155xk11(x33)3構(gòu)造迭代格式(I)x25k15(x)3225125當(dāng),x[2,3]迭代格式(I)發(fā)散112)x,x35x3,構(gòu)造迭代格式35x3，(II)xk135x3k215531(5x3)，x()2(x)35x33233(5x3)2當(dāng)x[2,3]時(shí)(x)5315133169511123312533(523)2當(dāng)x[2,3]時(shí)2(x)[(2),(3)][3523,3533][313,318][2,3]22迭代格式(II)對(duì)任意均收斂x0[2,3]-4-x25x33，533)x5xxx3構(gòu)造迭代格式x(III)5k1xk135x1331，(x)xx2()(5)2(3)332x235x2x當(dāng)x[2,3]時(shí)3131x253213(x)13235222585x2x當(dāng)時(shí)x[2,3](x)[(3),(2)][6,6.5][2,3]333迭代格式(III)對(duì)任意均收斂x[2,3]0(x)(2)510.3014534)max22331692x33131max(x)2x332minx23523233min{225,325}x2x3310.06802min{46.5,96}3取格式(III)xk15xk，1，，x2.49095x2.490863x02.5x2.489982x*2.494.用簡(jiǎn)單迭代格式求方程0.20的所有實(shí)根，精確至有3位有效數(shù)。3xx解：f(x)x3x0.2x(x21)0.2f(x)3x213(x21)3-5-當(dāng)1時(shí)，，xf(x)03*1*3xyx*2x121x11331當(dāng)時(shí)xf(x)03111(1)0.23332f()0.20333f(0)0.2132，f(2)820.25.8f()0.20f(1)0.2333f()(1)(11)0.230.201，f(1)0.2224811，2，*[1,2]x3*[1,x*[,0]x]1321)xx30.2迭代格式x，x30.2kk1(x)x30.2,(x)3x20當(dāng)1時(shí)，23，4x[,0](x)(x)[(),(0)][10.2,0.2][1,0]12821任取迭代格式收斂于x[,0]*2x02取0得,,x10.215625x20.210025x30.209264x0.25x40.209164x*0.2092-6-2)x,xx0.233x0.2迭代格式xk13x0.2k23(x)1(x0.2)31,(x)3x0.233(x0.2)2當(dāng)時(shí)x[1,2](x)[(1),(2)][31.2,32.2][1,2]1113(x)33(10.2)2任意迭代格式收斂于x[1,2]*x30取計(jì)算得，，,，x1.5x1.19348x1.11695012x1.09612x1.0903134x1.08867x1.0882156*1.09x3210.23)xxx10.2x10.2迭代格式xk1(III)xk(x)10.2x1)2(0.2)x2(x)1(120.20.1210.2xxx當(dāng)x[1,1]時(shí)3-7-1(x)[(1),()][10.2,10.23][0.8944,0.8084][1,31]3g(x)x210.2,x1)2(0.2)x2g(x)2x10.2xx212(10.2x(10.2)1[2x(10.2)0.1]xx112(10.2)2(2x0.40.1)(2x0.3)(10.2)xx1當(dāng)x[1,時(shí)，g(x)0]31)110.233g(3當(dāng)x[1,1]時(shí)30.10.3(x)0.3711110.23g()31迭代格式(III)對(duì)任意均收斂于，取，x*x0.80x[1,]03計(jì)算得1,,,x20.876961x30.878601*0.879x0.866025x40.878843x15.已知x(x)在區(qū)間[]內(nèi)有且只有一個(gè)根，而當(dāng)a<x<b時(shí)，,ab'(x)k1(1)試問(wèn)如何將x(x)化為適用于迭代的形式？-8-(2)將xtanx化為適用于迭代的形式，并求x4.5(弧度)附近的根。dydx1解：(1)由dxdy1(x)d(x)d1將改寫(xiě)為,則1()xx(x)xdxdx1()1d時(shí)，x1，這時(shí)迭代當(dāng)x[a,b]格式為dxk，k0,1,2,1(x)kxk1是局部收斂的。(2)由圖可知xtanx在x4.5附近有一根，但tgxtgx1(tanx)|x4.522.505(cos4.5)2234.71232將xtanx改寫(xiě)為2xarctanx(x)arctanx1(x)1x2x[,3]時(shí)當(dāng)yarctanx223且(x)[,]yarctanx222321(x)121()22迭代格式-9-，k0,1,2,xk1arctaxnk3對(duì)任意0]均收斂x[,22取得x4.5，2，x4.49342x4.4934103x4.4937201具有5位有效數(shù)的根為*4.4934x6.設(shè)(1)方程有根；(2)對(duì)一切,存在且0＜f(x)0xRf'(x)*xmf'(x)M。證明對(duì)于任意的，迭代格式(0,2M)(k0,1,2,)xxkf(x)kk1是局部收斂的。解：0mf(x)Mxk1xf(x)kk(x)xf(x)(x)1f(x)(x*)1f(x*)2當(dāng)時(shí)，1M(x*)1m(*)1x(0,)M迭代格式局部收斂。7.給定方程f(x)0，并設(shè)x*是其單根，且()足夠光滑，證明迭代格式fxf(x)f(x)f(x)2xxkkkkf(x)2f(x)f(x)k1kkk是3階局部收斂的。2f(x)f(x)f(x)證明(x)xf(x)f(x)2f(x)，*g(x)0f(x)(xx*)g(x)-10-fxg(x)(xx*)g(x)f(x)2g(x)(xx*)g(x)(xx*)g(x)(x)xg(x)(xx*)g(x)22g(x)(xx*)g(x)2[g(x)(xxgx(xx*)g(x)gxxx*)g(x)*)()]()(lim(x)x*xx*(x)(x*)(x*)limxx*xx*2g(x)(xx*)g(x)(xx*)g(x)2g(x)2lim1g(x)(xx*)g(x)2g(x)(xx*0g(x)g(x)()()*xxgxxx*1g(x*)2g(x*)0g(x*)20g(x*)2g(x*)g(x*)211g(x)2g(x)(x)x(xx*)1(xx*)g(x)g(x)(xx*)g(x)(xx*)211g(x)g(x)1(xx*)g(x)21(xx*)g(x)h(x)g(x)(xx*)g(x)-11-11g(x)(xx*)g(x)1x(xx*)(xx*)21h(x)2g(x)(1h(x))3x(xx*)[1h(x)h(x)2]1[g(x)(xx*)g(x)][13h(x)](xx*)22g(x)x*(xxhxOxx*)()(xx*)h(x)2(*)41[2g(x)(xx*)g(x)]1[3h(x)O(xx*)]x(x*)22g(x)g(x)2*g(x)(*)2(xx*)3g(x)xxxg(x)g(x)2(xx*)g(x)(xx*)(xx*)2O(xx*)412g(x)2g(x)6g(x)(xx*)3g(x)(xxOxx；*)3(*)4g(x)2x*2g(x)2g(x)x*(x)gxg(x)()21()gx2(x*x)32g(x)f(x)f(x)f(x)2方法二xk1xkkkf(x)2f(x)f(x)kkkkf(x*)fxxxfx1()(*)())2fx1()(x*xk(x*x)3f()0kkkkk2k6f(x)x*x1f(x)f()0(x*x)21(x*x)3kf(x)kk2f(x)kk6kf(x)kkk-12-f(x)f(x)(x*x)2f()x*xk1(x*x)3kkkfx()2()fxk6k()fxkkkf(x)f(x)21(x*x)3()fx*xk1(x*x)2kkkk2f(x)f(x)6k()fxkkkk2f(x)x*x1f(x)(x*x)2(x*x)kk2f(x)2f(x)kkkk1(x*x)3f()k6kf(x)kx*xk1(*)1(*)fxfx(*)fx2(*)fxfxfx(*)(x*x)3(*)6ka11.應(yīng)用Newton法分別導(dǎo)出求方程na0和f(x)10的f(x)xxn。nalim(ax)(nax)2根的迭代格式，并求nk1kk解：1)解方程f(x)0的Newton迭代格式f(x)xk1xkf(x)kkf(x)(x)x(x)1f(x)f(x)2f(x)f(x)f(x)f(x)2f(x),lim(x)0f(x)2xx*f(x)f(x)f(x*),(x)lim(x)f(x)f(x)f(x)fx(*)xx*x*xk1(x*)(*)12fx,x*nalimk(x*x)22f(x*)k-13-2),a,f(x)n(n1)x2,f(x)nxn1nf(x)xn1f(x)nxn1()fxn(n1)xn2n1x，f(x)xna()fxnxn1n11nxkNewton迭代格式xxkk1xnakna(11)xa1nxxxkkkn1knnnxknaxk11n11nlimk22nanaxnaka3),,f(x)anx(n1)f(x)an(n1)x(n2)f(x)1xna1xn1nx,anf(x)xnfx()anx(n1)fx()an(n1)x(n2)n1xf(x)anx(n1)Newton迭代格式n1n1fx()xnx(1n)xxxk1xkxkkkfx()kkkananknaxk1fxn1(*)lim)22f(x*)2na(knaxk12.試寫(xiě)出求方程1c0(其中c為已知正常數(shù))的Newton迭代格式，并證x明當(dāng)初值滿足0x2時(shí)迭代格式收斂。該迭代格式中是否含有除法運(yùn)x00c算？f(x)c11,則求等價(jià)于求方程f(x)0的根.解：記xc-14-1,x2f(x)2x3f(x)Newton迭代格式為1cf(x)xxkxk1xx(2cx)，k0,1,2,kf(x)1kkkkk2xk對(duì)任意2，存在充分小的δ(δ<1),(δ<1)使得x0(0,)ccx0[δ,2δ]現(xiàn)在考慮區(qū)間[]=[δ,2δ]ab,cc111of(a)f()c(c1)02cc2cc(1c)0f(b)f(2)cc1cc22c2c2ccf(a)f(b)02o當(dāng)時(shí)[,]xabf(x)0f(x)0當(dāng)x[a,b]時(shí)3o4of()(2c)2f()c(2c)2{}(c1)(c2)0c(2c)2ccf(2)c2c(2)2c(2)f(2)ccc1c(2c)c-15-因而當(dāng)2時(shí)，Newton迭代格式收斂。x(0,)0c直接證明xk1x(2cx)kk1cx1cx(2cx)(1cx)2kkkk1cx(1cxk1)2(1cx0)2kklimx