第一章緒論1．設x0,x的相對誤差為，求lnx的誤差。e*x*x解：近似值x=的相對誤差為e**x*elnx*lnx*lnx1e*x*r而lnx的誤差為x*進而有(lnx*)2．設x的相對誤差為2%，求。x的相對誤差n解：設f(x)xn，則函數(shù)的xf'(x)|C|p條件數(shù)為f(x),C|xnxn1|nf'(x)nxn1又又np((x*)n)C(x*)rpr且e(x*)為2r((x*)n)0.02nr3．下列各數(shù)都是經(jīng)過四舍五入得到的近似數(shù)，即誤差限不超過最后一位的半個單位，試指x*1.1021,x0.031,x385.6,**23出它們是幾位有效數(shù)字：1x*56.430,x71.0.*54解：x*1.1021是五位有效數(shù)字；1x*0.031是二位有效數(shù)字；2x*385.6是四位有效數(shù)字；3x*56.430是五位有效數(shù)字；4x*71.0.是二位有效數(shù)字。54．利用公式求下列各近似值的誤差限：(1)x*x*x*4,(2)x*x*x,(3)x*/x*23*4.1212其中x*,x*,x*,x均為*4第3題所給的數(shù)。123解：(x*)110421(x*)110322(x*)110123(x*)110324(x*)110125(1)(x*x*x*)124(x*)(x*)(x*)1241110411031032221.05103(2)(x*x*x*)123**13xx(x*)xx(x*)xx(x*)****23123121.10210.03111010.031385.611041.1021385.611032220.215(3)(x*/x*)24x(x*)x(x*)*2*4422x*40.031110356.43011032256.43056.4301055計算球體積要使相對誤差限為1，問度量半徑R時允許的相對誤差限是多少4R3解：球體體積為V3則何種函數(shù)的條件數(shù)為CRV'R4R23p4VR33(V*)C(R*)3(R*)rprr又(V*)1r(R*)110.33R時允許的相對誤差限為故度量半徑3r1Y28，按遞推公式Y(jié)Y783（n=1,2,…）1006．設0n1n計算到Y(jié)。若取10078327.982（5位有效數(shù)字），試問計算將有多大誤差Y1001YY783解：100n1n1YY783100100991YY78310019998YY7831009897……1YY783100101YY100有100783依次代入后，1000即YY783，100078327.982,YY27.982若取1000(Y*)(Y)(27.982)1103210001的誤差限為Y100103。27．求方程x256x10的兩個根，使它至少具有4位有效數(shù)字（78327.982）。解：x256x10，故方程的x28783根應為1,2x287832827.98255.982故1x具有5位有效數(shù)字1111x287832287832827.98255.9820.017863x具有5位有效數(shù)字21N18．當N充分大時，怎樣求dx1x2N11x2N1dxarctan(N1)arctanN解N設arctan(N1),arctanN。則tanN1,tanN.N1N11x2dxarctan(tan())1tantanarctantantanN1Narctan1(N1)N1N2N1arctan1cm29．正方形的邊長大約為了100cm，應怎樣測量才能使其面積誤差不超過解：正方形的面積函數(shù)為A(x)x2(A*)2A*(x*).若(A*)1,x*100時，當(x*)1102則2故測量中邊長誤差限不超過0.005cm時，才能使其面積誤差不超過1cm2110．設Sgt，假定g是準確的，而對t的測量有0.1秒的誤差，證明當t增加時S的22絕對誤差增加，而相對誤差卻減少。解：S12gt2,t0(S*)gt2(t*)t*增加時，S*的絕對誤差增加當(S*)S*(S*)rgt2(t*)12g(t*)22(t*)t*t*增加時，(t*)保持不變，則S*的相對誤差減少。當11．序列y滿足遞推關(guān)系ny10y1(n=1,2,…),n1n若y21.41（三位有效數(shù)字），計算到y(tǒng)時誤差有多大這個計算過程穩(wěn)定嗎010y21.410解：(y*)110220y10y1n又n1y10y110(y*)10(y*)10y10y1又21(y*)10(y*)21(y*)102(y*)20......(y*)1010(y*)100101011022110821計算到y(tǒng)時誤差為108，這個計算過程不穩(wěn)定。21012．計算f(21)6，取2，利用下列等式計算，哪一個得到的結(jié)果最好11(322)399702，。,,(21)6(322)3解：設y(x1)6，1101x2，x1.4，則x*。2若*1若通過計算y值，則(21)61y(x*1)7x**6(x1)yx***7yx**若通過(322)3計算y值，則y*(32x*)2x*632xyx***yx**1若通過(322)3計算y值，則1y(32x*)4x**1(32x)yx***7yx**1通過計算后得到的結(jié)果最好。(322)313．f(x)ln(xx21),求f(30)的值。若開平方用6位函數(shù)表，改用另一等價公式。ln(xx21)ln(xx21)大若問求對數(shù)時誤差有多計算，求對數(shù)時誤差有多大解f(x)ln(xx21),f(30)ln(30899)設u899,yf(30)則u*u*142故y*u*u*1u*0.01673若改用等價公式ln(xx21)ln(xx21)則f(30)ln(30899)此時，y*u*u*1u*59.98337第二章插值法1．當x1,1,2時，f(x)0,3,4,求f(x)的二次插值多項式。解：x1,x1,x2,012f(x)0,f(x)3,f(x)4;012(xx)(xx)12(x1)(x2)2l(x)(xx)(xx)120010l(x)(xx)(xx)1(x1)(x2)(xx)(xx)60211012(xx)(xx)1l(x)(xx)(xx)3(x1)(x1)0122021則二次拉格朗日插值多項式為2L(x)yl(x)kk2k03l(x)4l(x)0212(x1)(x2)43(x1)(x1)56x2x37232．給出f(x)lnx的數(shù)值表Xlnxln0.54用線性插值及二次插值計算的近似值。解：由表格知，x0.4,x0.5,x0.6,x0.7,x0.8;01234f(x)0.916291,f(x)0.69314701f(x)0.510826,f(x)0.35667523f(x)0.2231444若采用線性插值法計算即ln0.54f(0.54)，則0.50.540.6l(x)xx210(x0.6)xx112l(x)xx10(x0.5)1xx221L(x)f(x)l(x)f(x)l(x)111226.93147(x0.6)5.10826(x0.5)L(0.54)0.62021860.6202191ln0.54若采用二次插值法計算時，l(x)(xx)(xx)(xx)(xx)50(x0.5)(x0.6)1200102l(x)(xx)(xx)100(x0.4)(x0.6)02(xx)(xx)11012l(x)(xx)(xx)(xx)(xx)50(x0.4)(x0.5)0122021L(x)f(x)l(x)f(x)l(x)f(x)l(x)2001122500.916291(x0.5)(x0.6)69.3147(x0.4)(x0.6)0.51082650(x0.4)(x0.5)L(0.54)0.615319840.6153202cosx,0x903．給全h1(1/60),若函數(shù)表具有的函數(shù)表，步長5位有效數(shù)字，研cosx究用線性插值求近似值時的總誤差界。cosx近似值時，誤差解：求解可以分為兩個部分，一方面，x是近似值，具有5位有效數(shù)cosx字，在此后的計算過程中產(chǎn)生一定的誤差傳播；另一方面，利用插值法求函數(shù)的近似值時，采用的線性插值法插值余項不為0，也會有一定的誤差。因此，總誤差界的計算應綜合以上兩方面的因素。當0x90時，令f(x)cosx取x0,h()116060180108000令xxih,i0,1,...,5400i0則x9025400當xx,x時，線性插值多項式為kk1L(x)f(x)xxk1f(x)xxkxxxx1kk1kk1k1k插值余項為1f()(xx)(xx)R(x)cosxL(x)12kk1cosx0,1，故計算中有誤差傳播又在建立函數(shù)表時，表中數(shù)據(jù)具有5位有效數(shù)字，且過程。(f*(x))11052kxxxxxxk1R(x)(f*(x))k(f*(x))k1x2k1xkk1k1kxxkxxxxk(f*(x))(k)xx11kk1k1k(f*(x))1(xxxx)hk1kk(f*(x))k總誤差界為RR(x)R(x)121(cos)(xx)(xx)(f*(x))2kk1k1(xx)(xx)(f*(x))2kk1k11(h)2(f*(x))k221.06108110520.501061054．設為互異節(jié)點，求證：nxkl(x)xk(k0,1,,n);jj（1）j0n(xx)kl(x)0(k0,1,,n);（2）證明jjj0（1）令f(x)xk若插值節(jié)點為x,j0,1,,n，則函數(shù)f(x)的n次插值多項式為L(x)xkl(x)。njnjjj0()(n1)插值余項為R(x)f(x)L(x)f(n1)!(x)n1nn又kn,f(n1)()0R(x)0nnxkl(x)xkjj(k0,1,,n);j0(2)n(xx)kl(x)jjj0n(Cjxi(x)ki)l(x)nkjjj0i0nnCi(x)(xil(x))kikjji0j0又0in由上題結(jié)論可知nxkl(x)xijjj0原式nCi(x)xikiki0(xx)k0得證。5設f(x)C2a,b且f(a)f(b)0,求證：maxf(x)1(ba)2maxf(x).8axbaxbxa,xb，以此為插值節(jié)點，則線性插值多項式為解：令01L(x)f(x)xx1f(x)xx0xxxx101010=f(a)axbbf(b)xxaa又f(a)f(b)0L(x)011f(x)(xx)(xx)插值余項為R(x)f(x)L(x)12011f(x)f(x)(xx)(xx)012又(xx)(xx)0112(xx)(xx)01214(xx)21014(ba)2maxf(x)1axb(ba)max().fx28axb6．在4x4exf(x)ex的等距節(jié)點函數(shù)表，若用二次插值求的近似值，要使上給出10截斷誤差不超過，問使用函數(shù)表的步長h應取多少6x,x和x，則分段二次插值多項式的插值余項為解：若插值節(jié)點為ii1i1R(x)1f()(xx)(xx)(xx)3!2i1ii1R(x)1(xx)(xx)(xx)maxf(x)62i1ii14x4設步長為xxh,xxhh，即i1ii1iR(x)1e2h3e4h3.3463327210若截斷誤差不超過，則6R(x)10623e4h310627h0.0065.y2,求4y及4y.，7．若nnnn解：根據(jù)向前差分算子和中心差分算子的定義進行求解。y2nn4y(E1)4ynn44(1)E4jynjjjjj044(1)y4njjj044(1)2y4jjnj0(21)4ynyn2n114y(E2E2)4ynn1(E2)4(E1)4ynE24ynyn22n2f(x)是m次多項式，記f(x)f(xh)f(x)，證明f(x)的k階差分f(x)0（l為正整數(shù)）8．如果kf(x)(0km)是mk次多項式，并且1。m解：函數(shù)f(x)的Taylor展式為f(x)h21f(m)(x)hm11(m1)!f(m1)f(xh)f(x)f(x)h其中(x,xh)()hm12m!f(x)是次數(shù)為m的多項式又f(m1)()0f(x)f(xh)f(x)f(x)hf(x)h21f(x)hm1m()2m!f(x)為階多項式m12f(x)(f(x))2f(x)m2階多項式為f(x)是mk次多項式依此過程遞推，得kmf(x)是常數(shù)當為正整數(shù)時，lm1f(x)09．證明(fg)fggfkkkkkk1證明(fg)fgfgkkk1k1kkfgfgfgfgk1k1kk1kk1kkg(ff)f(gg)k1k1kkk1kgffgk1kkkkfggfkkk1得證n1k1n110．證明fgfgfggfkknn00kk0k0證明：由上題結(jié)論可知fg(fg)gfkkkkk1kn1fgkkk0n1((fg)gf)kkk1kk0n1n1k0(fg)gfk1kkkk0(fg)fgfgkkk1k1kkn1(fg)kkk0(fgfg)(fgfg)(fgfg)11002211nnn1n1fgfg00nnn1k1n1k0fgfgfggfkknn00kk0得證。

11．證明n1yyy02jnj0n1n1y)證明2y(yj1jjj0j0(yy)(yy)(yy)1021nn1yyn0得證。f(x)aaxaxn1axn有n個不同實根x,x,,x，若12．01n1n12n0,0kn2;f(x)n01,kn1n：xkj證明j1jf(x)有個不同實根x,x,,xn證明：12且f(x)aaxaxn1axn01n1nf(x)a(xx)(xx)(xx)n12n(x)(xx)(xx)(xx)n令則n12xxkknnjf(x)ja(x)j1j1jnnj而(x)(xx)(xx)(xx)(xx)(xx)(xx)n23n13n(xx)(xx)(xx)12n1(x)(xx)(xx)(xx)(xx)(xx)njj1j2jj1jj1jn令g(x)xk,xkngx,x,,xnj(x)12j1njxkn則gx,x,,xnj(x)12j1nj1xkj又nf(x)agx,x,,xn12j1jn0,0kn2;f(x)n01,kn1xkjnj1j得證。13．證明n階均差有下列性質(zhì)：（1）若F(x)cf(x)，則Fx,x,,xcfx,x,,x;01n01n（2）若F(x)f(x)g(x)，則Fx,x,,xfx,x,,xgx,x,,x.01n01n01n證明：()fxjnfx,x,,xn（1）(xx)(xx)(xx)(xx)12j0j0jj1jj1jn()FxjnFx,x,,xn(xx)(xx)(xx)(xx)j112j0j0jjj1jncf(xj)n(xx)(xx)(xx)(xx)j0j0jj1jj1jnf(xj)c(n)(xx)(xx)(xx)(xx)j0j0jj1jj1jncfx,x,,xn01得證。(2)F(x)f(x)g(x)()FxjFx,,xn(xx)(xx)(xx)(xx)0nj0j0jj1jj1jnf(xj)g(xj)(xx)(xx)(xx)(xx)nj0j0jj1jj1jnf(xj)n)(xx)(xx)(xx)(xx)j0j0jj1jj1jng(xj)n)+(xx)(xx)(xx)(xx)j0j0jj1jj1jnfx,,xgx,,x0n0n得證。14．f(x)x7x43x1,求F2,21,,2,21,,2。8070f(x)x7x43x1解：若x2i,i0,1,,8i()則fx,x,,xf(n)n!01n()7!fx,x,,xf(7)17!7!017()0fx,x,,xf(8)8!01815．證明兩點三次埃爾米特插值余項是R(x)f(4)()(xx)2(xx)2/4!,(x,x)3kk1kk1解：若x[x,x]，且插值多項式滿足條件kk1H(x)f(x),H(x)f(x)3kk3kkH(x)f(x),H(x)f(x)3k1k13k1k1插值余項為R(x)f(x)H(x)3由插值條件可知R(x)R(x)0kk1且R(x)R(x)0kk1R(x)可寫成R(x)g(x)(xx)2(xx)2kk1其中g(shù)(x)是關(guān)于x的待定函數(shù)，現(xiàn)把x看成[x,x]上的一個固定點，作函數(shù)k1k(t)f(t)H(t)g(x)(tx)2(tx)2k3k1根據(jù)余項性質(zhì)，有(x)0,(x)0kk1(x)f(x)H(x)g(x)(xx)2(xx)23kk1f(x)H(x)R(x)30(t)f(t)H(t)g(x)[2(tx)(tx)22(tx)(tx)2]3kk1k1k(x)0k(x)0k1由羅爾定理可知，存在(x,x)，使(x,x)和kk12()0,()01(x)[,]即在xx上有四個互異零點。kk1(t)(t)根據(jù)羅爾定理，在的兩個零點間至少有一個零點，故(t)(,)在xx內(nèi)至少有三個互異零點，kk1(t)(x,x)依此類推，在內(nèi)至少有一個零點。kk1(4)(x,x)使k記為k1()f(4)()H(4)()4!g(x)0(4)3H(4)(t)03又g(x)f(4)(),(x,x)k4!k1其中依賴于xR(x)f(4)()(xx)2(xx)24!kk1x(k0,1,,n)，設步長為h，即米特插值時，若節(jié)點為k分段三次埃爾xxkh,k0,1,,n在小區(qū)間[x,x]上k0kk1R(x)f(4)()(xx)2(xx)24!kk1R(x)4!1f(4)()(xx)2(xx)2kk14!1(xx)2(xx)2maxf(4)(x)kk1axb1xxxx)2]2maxf(4)(x)[(kk124!axb11h4maxf(4)(x)4!24axbh4maxf(4)(x)axb38416．求一個次數(shù)不高于4次的多項式P（x），使它滿足P(0)P(0)0,P(1)P(1)0,P(2)0解：利用埃米爾特插值可得到次數(shù)不高于4的多項式x0,x101y0,y101m0,m10111H(x)y(x)m(x)j3jjjj0j0xx)(xxxxxx010(x)(12)20110(12x)(x1)2xx)(xx(x)(12)21xxxx001101(32x)x2(x)x(x1)2(x)(x1)x201H(x)(32x)x2(x1)x2x32x23設P(x)H(x)A(xx)2(xx)2301其中，A為待定常數(shù)P(2)1P(x)x32x2Ax2(x1)2A141P(x)x2(x3)2從而417．設f(x)1/(1x2)，在5x5n10I(x)上取，按等距節(jié)點求分段線性插值函數(shù)，hI(x)與f(x)值，并估計誤差。計算各節(jié)點間中點處的h解：若x5,x5010則步長h1,xxih,i0,1,,10i01f(x)1x2[x,x]在小區(qū)間上，分段線性插值函數(shù)為ii1I(x)xxf(x)xxf(x)i1i1ixxxxhiii1i1i11(xx)i11x2(xx)1xi2ii1各節(jié)點間中點處的I(x)與f(x)的值為h當x4.5時，f(x)0.0471,I(x)0.0486h當x3.5時，f(x)0.0755,I(x)0.0794h當x2.5時，f(x)0.1379,I(x)0.1500h當x1.5時，f(x)0.3077,I(x)0.3500h當x0.5時，f(x)0.8000,I(x)0.7500h誤差maxf(x)I(x)hmaxf()28hxixxi15x51f(x)1x2又2xf(x)(1x2)2,6x22f(x)(1x)2324x24x3(1x2)4f(x)f(x)0令1和x03f(x)的駐點為得x1,2f(x)1,f(x)221,23maxf(x)I(x)14h5x518．求f(x)x2在[a,b]上分段線性插值函數(shù)I(x)，并估計誤差。h解：在區(qū)間[a,b]hmaxhxa,xb,hxx,i0,1,,n1,上，0nii1ii0in1f(x)x2f(x)函數(shù)在小區(qū)間[x,x]上分段線性插值函數(shù)為ii1I(x)xxf(x)xxf(x)ixxi1i1xxhii1ii1i1[x2(xx)x2(xx)]hi1i1iii誤差為maxf(x)I(x)1maxf()h28hixxxabii1f(x)x2f(x)2x,f(x)2maxf(x)I(x)h24haxb19．求f(x)x4在[a,b]上分段埃爾米特插值，并估計誤差。解：在區(qū)間上，xa,xb,hx[a,b]x,i0,1,,n1,i0nii1令hmaxhi0in1f(x)x4,f(x)4x3函數(shù)f(x)[x,x]i在區(qū)間上的分段埃爾米特插值函數(shù)為i1I(x)(xxxxi1)2(12xx)f(x)iixxihii1i1(xx)2(12xx1)f(x)iixxixxi1i1ii1xx(xxi1)2(xx)f(x)iiii1xx()2(xx)f(x)i1i1ixxii1x4(xx)2(h2x2x)ihi1ii3ix(xx)2(h2x2x)4i1h3iii1i4x3(xx)2(xx)iih2ii14x(xx)2(xx)i3i1h2ii1誤差為f(x)I(x)h4!1f(4)()(xx)2(xx)2ii11maxf(4)()(i)4axbh242f(x)x4又f(4)(x)4!24maxf(x)I(x)maxhih416160in14haxb20．給定數(shù)據(jù)表如下：XjYj試求三次樣條插值，并滿足條件：(1)S(0.25)1.0000,S(0.53)0.6868;(2)S(0.25)S(0.53)0.解：hxx0.05010hxx0.09121hxx0.06232hxx0.08343hh,jj1jhhj1hhj1jjj5334,,,114571239241,,,11457230()()fxfxxx10fx,x0.95400101fx,x0.853312fx,x0.771723fx,x0.7150344(1)S(x)1.0000,S(x)0.686806f)5.52000d(fx,x0h120d6fx,xfx,x14.31573.26402.4300120hh011d6fx,xfx,x2231hh122d6fx,xfx,x3342hh2336(ffx,x)2.1150434d4h3由此得矩陣形式的方程組為5.520021M0592M4.315714141322M3.2640552342M2.43007732.115012M4求解此方程組得M2.0278,M1.464301M1.0313,M0.8070,M0.6539234三次樣條表達式為(xx)3M(xx)3S(x)Mj1j6h6hjj1jj(yMh2)xxMhxx2(yj1)j(j0,1,,n1)hj6jj1hj1j6jjj將M,M,M,M,M代入得012346.7593(0.30x)34.8810(x0.25)310.0169(0.30x)10.9662(x0.25)x0.25,0.302.7117(0.39x)31.9098(x0.30)36.1075(0.39x)6.9544(x0.30)x0.30,0.39S(x)2.8647(0.45x)32.2422(x0.39)310.4186(0.45x)10.9662(x0.39)x0.39,0.451.6817(0.53x)31.3623(x0.45)38.3958(0.53x)9.1087(x0.45)x0.45,0.53(2)S(x)0,S(x)0044.3157,d3.2640d2f0,d002.4300,d2f012d344400由此得矩陣開工的方程組為MM004920144.3157M352M3.264012522.4300M33027求解此方程組，得M0,M1.880901M0.8616,M1.0304,M0234又三次樣條表達式為(xx)3M(xx)3S(x)Mj1j6h6hjj1jj(yMh2)xxMhxx2(yj1)j6jj1hj1jj6hjjj將M,M,M,M,M代入得012346.2697(x0.25)310(0.3x)10.9697(x0.25)x0.25,0.303.4831(0.39x)31.5956(x0.3)36.1138(0.39x)6.9518(x0.30)x0.30,0.39S(x)2.3933(0.45x)32.8622(x0.39)310.4186(0.45x)11.1903(x0.39)x0.39,0.452.1467(0.53x)38.3987(0.53x)9.1(x0.45)x0.45,0.5321．若f(x)C2a,b,S(x)是三次樣條函數(shù)，證明：S(x)2dx(1)f(x)dxb2baaf(x)S(x)dx2S(x)f(x)S(x)2dxb2baa若f(x)S(x)(i0,1,,n)，式中為插值節(jié)點，axxxb，則x(2)且iii01nS(x)f(x)S(x)dxbS(b)f(b)S(b)S(a)f(a)S(a)a證明：(1)f(x)S(x)2dxbaS(x)dx2bf(x)S(x)dxf(x)dxb2b2aaaS(x)dx2S(x)f(x)S(x)dxf(x)dxb2b2baaa從而有S(x)2dxf(x)dxb2baaf(x)S(x)dx2S(x)f(x)S(x)dxb2baa第三章函數(shù)逼近與曲線擬合1．f(x)sin2x，給出[0,1]上的伯恩斯坦多項式B(f,x)及B(f,x)。13解：f(x)sin,x[0,1]2伯恩斯坦多項式為B(f,x)nnf(k)P(x)nkk0n其中P(x)xk(1x)nkkk當n1時，1P(x)(1x)00P(x)x1B(f,x)f(0)P(x)f(1)P(x)1011(1x)sin(0)xsin220x當n3時，1P(x)(1x)3001P(x)x(1x)23x(1x)2013P(x)x2(1x)3x2(1x)123P(x)x3x333B(f,x)f(k)P(x)33nkk003x(1x)2sin3x2(1x)sinx3sin63232x(1x)253333x2(1x)x32x3336x232x221.5x0.402x20.098x32．當f(x)x時，求證B(f,x)xn證明：若f(x)x，則B(f,x)nnf(k)P(x)nkk0nknnknkk0kn(n1)(nk1)nxk(1x)nknk!k0(n1)[(n1)(k1)1]nxk(1x)nk(k1)!k1n1xk(1x)nkk1nk1nn1xk1x(1x)k1(n1)(k1)k1x[x(1x)]n1x3．證明函數(shù)1,x,,x線性無關(guān)n證明：aaxax2axn0,xR若012nx(k0,1,2,,n)，對上式兩端在[0,1]上作帶權(quán)(x)1的內(nèi)積，得分別取k1n101a00a1110n1a2n1n此方程組的系數(shù)矩陣為希爾伯特矩陣，對稱正定非奇異，只有零解a=0。函數(shù)1,x,,xn線性無關(guān)。4。計算下列函數(shù)f(x)關(guān)于C[0,1]的f,f與f：12(1)f(x)(x1)3,x[0,1](2)f(x)x1,2(3)f(x)xm(1x)n,m與n為正整數(shù)，(4)f(x)(x1)10ex解：(1)若f(x)(x1)3,x[0,1]，則f(x)3(x1)02(0,1)f(x)(x1)3在內(nèi)單調(diào)遞增fmaxf(x)maxf(0),f(1)0x1max0,11fmaxf(x)maxf(0),f(1)0x1max0,11f(1(1x)6dx)12201[1(1x)7]1270771(2)若f(x)x,x0,1，則2fmaxf(x)120x1f1f(x)dx102(x12)dx11214f(1f2(x)dx)1220[1(x1)2dx]212036(3)若f(x)xm(1x)n,m與n為正整數(shù)x0,1時,f(x)0當f(x)mx(1x)xmn(1x)(1)m1nn1m(1nmx)mx(1x)m1n1mnmx(0,當)時,f(x)0mnmf(x)(0,)內(nèi)單調(diào)遞減在mnmx(當,1)時,f(x)0mnmf(x)(,1)在內(nèi)單調(diào)遞減。mx(nm,1)f(x)0fmaxf(x)0x1maxf(0),f(mnm)mmnn(mn)mnf1f(x)dx101xm(1x)ndx02(sin2t)m(1sin2t)ndsin2t02sin2mtcos2ntcost2sintdt0n!m!(nm1)!f[1x2m(1x)2ndx]122012[2sin4mtcos4ntd(sin2t)]01[22sin4m1tcos4n1tdt]20(2n)!(2m)![2(nm)1]!(4)若f(x)(x1)10exx0,1時，f(x)0當f(x)10(x1)9e(x1)9ex(9x)0x)(x1)(ex10f(x)[0,1]在內(nèi)單調(diào)遞減。fmaxf(x)maxf(0),f(1)0x1210ef1f(x)dx101(x1)10exdx0(x1)10ex110(x1)9exdx100105ef[1(x1)20e2xdx]1220347()4e2fg5。證明fg證明：f(fg)gfggfgfg6。對f(x),g(x)C1[a,b]，定義(1)(f,g)f(x)g(x)dxbaf(x)g(x)dxf(a)g(a)(2)(f,g)ba問它們是否構(gòu)成內(nèi)積。解：(1)令f(x)C（C為常數(shù)，且C0）則f(x)0而(f,f)bf(x)f(x)dxa這與當且僅當f0時，(f,f)0矛盾C[a,b]上的內(nèi)積1不能構(gòu)成。fxgxdxf(a)g(a)，則(2)(,)()()b若fga(g,f)bg(x)f(x)dxg(a)f(a)(f,g),Ka(f,g)[f(x)]g(x)dxaf(a)g(a)ba[bf(x)g(x)dxf(a)g(a)]a(f,g)hC1[a,b],則(fg,h)b[f(x)g(x)]h(x)dx[f(a)g(a)]h(a)abf(x)h(x)dxf(a)h(a)f(x)h(x)dxg(a)h(a)baa(f,h)(h,g)(f,f)b[f(x)]2dxf2(a)0a若(f,f)0，則b[f(x)]2dx0,且f2(a)0af(x)0,f(a)0f(x)0即當且僅當f0時，(f,f)0.C[a,b]上的內(nèi)積1故可以構(gòu)成。17。令T*(x)T(2x1),x[0,1]，試證T(x)是在[0,1]上帶權(quán)(x)*的正交xx2nnn多項式，并求T*(x),T*(x),T*(x),T*(x)。0123解：若T*(x)T(2x1),x[0,1]，則nn1T*xTxPxdx()()()*nm01(21)(21)1TxTx0dxxx2nmxt1t[1,1]，且令t(2x1)，則，故2()()()1T*xTxxdx*nm01d(t1)1T(t)T(t)mt1(t1)22n1221T(t)T(t)1dt1t2nm11T(x)在區(qū)間[0,1]上帶權(quán)(x)又切比雪夫多項式*正交，且1x2k0,nm,nm1T(xT)(xd)0x1t22nm1,nm01的正交多項式。xx2T(x)[0,1](x)是在上帶權(quán)*nT(x)1,x[1,1]又0T*(x)T(2x1)1,x[0,1]00T(x)x,x[1,1]1T*(x)T(2x1)2x1,x[0,1]11T(x)2x21,x[1,1]2T*(x)T(2x1)222(2x1)218x28x1,x[0,1]T(x)4x33x,x[1,1]3T*(x)T(2x1)334(2x1)33(2x1)32x348x218x1,x[0,1]8。對權(quán)函數(shù)(x)1x2試求首項系數(shù)為1的正交多項式n(x),n0,1,2,3.[1,1]，區(qū)間，解：(x)1x2[1,1]，則區(qū)間上內(nèi)積為若(f,g)1f(x)g(x)(x)dx1定義，則(x)10(x)(x)(x)(x)n1n1nnn其中(x(x),(x))/((x),(x))nnnnn((x),(x))/((x),(x))nnnn1n1(x,1)/(1,1)011xxdx(1)2(1)xdx1210(x)x1(x2,x)/(x,x)11x3xdx(1)211x2(1)xdx210(x,x)/(1,1)11x2xdx(1)211(1)dxx21162515832(x)x252(x32x,x22)/(x22,x22)52555521x)(x22)(1x2)dx(x3512)(x22)(1x2)dx1(x25510(x22,x22)/(x,x)525521)(x22)(1x2)dx(x2511x2(1x2)dx1136(x)x32x2xx39x173570149。試證明由教材式(2.14)給出的第二類切比雪夫多項式族ux是[0,1]上帶權(quán)()n(x)1x2的正交多項式。證明：若U(x)sin[(n1)arccosx]1x2n令xcos，可得1U(xU)(x)1x2dxmn11sin[(m1)arccosx]sin[(n1)arccosx]dx1x210sin[(m1)sin[(n1)]d1cos2sin[(m1)sin[(n1)]d0當mn時，sin2[(m1)d01cos[2(m1)]d202當mn時，sin[(m1)sin[(n1)]d0sin[(m1)d{1cos(n1)}n100n11cos(n1)d{sin[(m1)]}m1n1cos(n1)cos(m1)d0mn11cos[(m1)]d{1sin[(n1)]}n10m1sin[(n1)]d{cos[(m1)]}0(n1)2(mn11)2sin[(n1)]sin[(m1)]d00[1(mn11)2]sin[(n1)]sin[(m1)]d00m1n1又mn，故()21sin[(n1)]sin[(m1)]d00得證。明切比雪夫多項式T(x)滿足微分方程10。證n(1x2)T(x)xT(x)n2T(x)0nnn證明：切比雪夫多項式為T(x)cos(narccosx),x1n從而有1T(x)sin(narccosx)n(n)1x2nsin(narccosx)1x2nn2T(x)nsin(narccosx)3cos(narccosx)1x2(1x2)2(1x2)T(x)xT(x)n2T(x)nnnnxsin(narccosx)n2cos(narccosx)1x2nxsin(narccosx)n2cos(narccosx)1x20得證。11。假設f(x)在[a,b]上連續(xù)，求f(x)的零次最佳一致逼近多項式解：f(x)在閉區(qū)間[a,b]上連續(xù)2x,x[a,b]，使存在1f(x)minf(x),1axbf(x)maxf(x),2axb1取P[f(x)f(x)]212[a,b]則x和x是上的2個輪流為“正”、“負”的偏差點。12由切比雪夫定理知P為f(x)的零次最佳一致逼近多項式。maxx3ax達到極小，又問這個解是否唯一12。選取常數(shù)a，使0x1解：令f(x)x3ax則f(x)在[1,1]上為奇函數(shù)maxx3ax0x1maxx3ax1x1f又f(x)的最高次項系數(shù)為1，且為3次多項式。(x)1()Tx與0的偏差最小。2333(x)14T(x)x3x3343從而有a3413。求f(x)sinx在[0,]上的最佳一次逼近多項式，并估計誤差。2解：f(x)sinx,x[0,]2f(x)cosx,f(x)sinx0af(b)f(a)2,ba12cosx,2xarccos20.880692f(x)0.771182af(a)f(x)f(b)f(a)ax2222ba00.10526于是得f(x)的最佳一次逼近多項式為2P(x)0.10526x1即2sinx0.10526x,0x2誤差限為sinxP(x)1sin0P(0)10.1052614。求f(x)ex0,1在0,1上的最佳一次逼近多項式。解：f(x)ex,x0,1f(x)ex,f(x)e0xaf(b)f(a)e1ba1ex2e1xln(e1)2f(x)ex2e12af(a)f(x)f(b)f(a)ax2222ba01(e1)(e1)ln(e1)221ln(e1)2于是得f(x)的最佳一次逼近多項式為P(x)e(e1)[x1ln(e1)]221(e1)x12[e(e1)ln(e1)]15。求f(x)x43x31在區(qū)間[0,1]上的三次最佳一致逼近多項式。解：f(x)x43x31,x[0,1]令t2(x1)，則t[1,1]21122且xt1111f(t)(t)43(t)312222161(t410t324t222t9)令g(t)16f(t)，則g(t)t10t24t22t9324g(t)為區(qū)間[1,1]上的最佳三次逼近多項式P(t)應滿足*3若maxg(t)P*(t)min31t1當g(t)P*(t)T(t)1(8t48t21)128334時，多項式g(t)P*(t)與零偏差最小，故31(t)g(t)T(t)*323410t325t222t7381f(x)的三次最佳一致逼近多項式為P(t)，則f(x)的三次最佳一致逼近多項式*3進而，16為P*(t)1[10(2x1)325(2x1)222(2x1)]7316835x354x2x11294128f(x)x，在1,1上求關(guān)于span1,x,x16。24的最佳平方逼近多項式。解：f(x)x,x1,1若(f,g)1f(x)g(x)dx121,x2,x4，則且0122,522,0222,29122(f,)1,(f,)1,(f,)1,23012(,)1,(,),(,)2,257010212則法方程組為2221352221357222a0a