數(shù)值分析第四版答案

1、第一章 緒論1設(shè)x 0, x的相對(duì)誤差為，求In x的誤差解：近似值x*的相對(duì)誤差為e*x* xx*x*而In x的誤差為e In x*In x* Inx1x*e*進(jìn)而有 (In x*)2設(shè)x的相對(duì)誤差為2%求xn的相對(duì)誤差。xf ( x)解：設(shè)f(x) xn，則函數(shù)的條件數(shù)為 Cp |f(x)n 1n 1X nx |又 f (x) nx , Cp I| nn又;r(x*) n) Cp r(x*)且 er (x*)為 2r(x*)n)0.02 n3 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(gè)單位，試指385.6出它們是幾位有效數(shù)字：X； 1.1021 , X；0.031

2、 ,X；x456.430, x 7 1.0.解：X；1.1021是五位有效數(shù)字；x；0.031是二位有效數(shù)字；X3385.6是四位有效數(shù)字；X456.430是五位有效數(shù)字；x57 1.0.是二位有效數(shù)字。, * *x2/ x4.4利用公式(2.3)求下列各近似值的誤差限：(1) X1 X2 X4 ,(2) X1 X2X3 ,(3) 其中x*,x2,x3,x4均為第3題所給的數(shù)。解:*14（X1）210*1,亠3（X2）210*11（X3）210*1,亠3210*11（X5）210* *(1) (X1X2X4)*(X1 )（X2）1 41 “-10-10221.051033(X4)1 1032X

3、2X3* * *X1X3(X2)(2)(X1X2X3)* * *X1X2 (X3)1.1021 0.03110 1 *10.031 385.6 -10411.1021 385.6 ? 10 30.215(3) (X2/X4)(X4) X4*(X2)2103R2C3則何種函數(shù)的條件數(shù)為R3 R3* 2X41 30.03110 56.4302r(V*)Cp(R*)3 r(R*)又（V*）11故度量半徑R時(shí)允許的相對(duì)誤差限為r（R*）-36設(shè)丫0 28，按遞推公式 Yn 丫 7831001 0.33(n=1,2,)計(jì)算到丫00。若取.78327.982 （ 5位有效數(shù)字），試問計(jì)算Yoo將有多大誤差

4、?解：Y, Yn 11.7831001 論0 丫99 “亦1001 丫987831001 ,丫97V 783100丫99丫981丫0 100茨依次代入后，有00綣100即丫。Y）砲，若取,78327.982，丫100（丫0。）（丫0）(27.982)14 3Y00的誤差限為一10 。丫1Yo 27.9827 求方程X211031 78310056x 10的兩個(gè)根,使它至少具有4位有效數(shù)字（78327.982 ）。故方程的根應(yīng)為X1,228、783故 X128783 2827.98255.982x1具有5位有效數(shù)字x228、78328、78328 27.9820.01786355.982X2具有

5、5位有效數(shù)字、 1&當(dāng)N充分大時(shí)，怎樣求2 dx ?N 1 x2pdx arctan(N 1) arctanN xarcta n(N 1),arctan N。則 tan N 1,tanN.1 11 x2dxarcta n(ta n( )丄 tantanarctan1 tan tan丄 N 1 N arctan1 (N 1)N9.正方形的邊長(zhǎng)大約為了100cm,應(yīng)怎樣測(cè)量才能使其面積誤差不超過1cm2 ?解：正方形的面積函數(shù)為 A(x) x2(Agt2| (t*)1 * 2g(t)(t*) )2A*| (x*).當(dāng) x* 100時(shí)，若(A*)1,1 2則(x*)1022故測(cè)量中邊長(zhǎng)誤差限不超過0

6、.005cm時(shí)，才能使其面積誤差不超過1cm210.設(shè)S 1gt2，假定g是準(zhǔn)確的，而對(duì)t的測(cè)量有 0.1秒的誤差，證明當(dāng)t增加時(shí)S的 絕對(duì)誤差增加，而相對(duì)誤差卻減少。1 2解: , S 尹,t 0(S*) gt2| (t*)當(dāng)t *增加時(shí)，S*的絕對(duì)誤差增加r(S*)(S*) S*當(dāng)t*增加時(shí)，(t*)保持不變，則 S*的相對(duì)誤差減少。11序列 y 滿足遞推關(guān)系yn I0yn 1 1 (n=1,2,)，若y0,2 1.41 (三位有效數(shù)字)，計(jì)算到y(tǒng)10時(shí)誤差有多大？這個(gè)計(jì)算過程穩(wěn)定嗎?解：:y0邁 1.411 2(y。*) 2 102又；yn 10yn1 1y1 10 y0 1S) 10

7、(y。*)又：y2 10y1 1(y2*) 10 (y1*)(y2*)102 (y*)10(Y10*)10 (y*)1010 1 10221 10821計(jì)算到y(tǒng)e時(shí)誤差為? 108，這個(gè)計(jì)算過程不穩(wěn)定。12計(jì)算f C-.2 1)6，取 .2，利用下列等式計(jì)算，哪一個(gè)得到的結(jié)果最好?1(/2 1)6(32)3,1,99 70、2 。(3 2*2)3解：設(shè) y (x 1)6, *I若 x .2 , x 1.4，貝Ux102若通過(,，21)6計(jì)算y值，則1(X 1)7*(X若通過(32.2)3計(jì)算y值，則* 21 i *y(3 2X ) I x6y x3 2x* *y x若通過計(jì)算y值，則(3 2

8、.2)3通過 3計(jì)算后得到的結(jié)果最好。(3 2、2)31(3 2X*)1 *7 y x (3 2X )7 y* *y x13. f (x) In(x . x2 1),求f (30)的值。若開平方用 6位函數(shù)表，問求對(duì)數(shù)時(shí)誤差有多大？若改用另一等價(jià)公式。ln(xx2 1) ln(x . x2 1)計(jì)算，求對(duì)數(shù)時(shí)誤差有多大？解:f(x) ln(x Jx2 1), f (30) In(30 7899)設(shè) u .899, y f (30)則u*(xXj(xX2)(X。X1)(XX2)(xX0)(XX2)(X1X0)(XX2)(xX)(XX1)(X2X)(X2X1)l(x)h(x)l2(x)則二次拉格朗

9、日插值多項(xiàng)式為2L2(x)yk(x)k 0X2X/V1 - 22)(X4 - 3*y uu1 *u0.01673若改用等價(jià)公式ln(x 、x 1) ln(x -X 1)則 f (30)In(30. 899)此時(shí)，* *y uu1 *u59.98337第二章插值法1當(dāng)x 1, 1,2時(shí)，f(x) 0, 3,4 ,求f (x)的二次插值多項(xiàng)式。解：x0 x1h x2尹 1)(x 1)f(x。) 0,f(X1)3,f(X2) 4;1(x 1)(x 2)21(x 1)(x 2)62.給出f(x) lnx的數(shù)值表X0.40.50.60.70.8lnx-0.916291-0.693147-0.510826

10、-0.356675-0.223144用線性插值及二次插值計(jì)算 In0.54的近似值。 解：由表格知，x00.4, x10.5, x20.6, x3 0.7, x4f(x。)0.916291, f (x1)0.693147f(X2)0.510826, f (x3)0.356675f(X4)0.223144若采用線性插值法計(jì)算In0.54即f (0.54),則 0.5 0.54 0.6h(x)XXIX2X210(x0.6)J(x)XX2X1X110(x0.5)Li(x)f(xjli(x) f(X2)2(x)6.93147(x 0.6) 5.10826( x 0.5)L1 (0.54)0.62021

11、860.620219l(x)(xX1)(xX2)(X。X1)(XX2)h(x)(XX0)(XX2)(X1X0)(X1X2)若采用二次插值法計(jì)算In0.54時(shí),50( x 0.5)(x 0.6)100(x 0.4)( x 0.6)l2(x) (x x0)(x x1)50(x 0.4)(x 0.5)(X2 X0)(X2 X1)L2(x)f(x)l(x) f(X1)h(X) f(X2)2(X)50 0.916291(x 0.5)(x0.6) 69.3147( x 0.4)(x 0.6) 0.510826 50(x 0.4)(x0.5)L2(0.54)0.615319840.615320(1/60)，

12、若函數(shù)表具有5位有效數(shù)字，研3.給全cosx,0； x 90的函數(shù)表，步長(zhǎng)h 1究用線性插值求cosx近似值時(shí)的總誤差界。解：求解cosx近似值時(shí)，誤差可以分為兩個(gè)部分，一方面，x是近似值，具有 5位有效數(shù)字，在此后的計(jì)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)cosx的近似值時(shí)，采用的線性插值法插值余項(xiàng)不為0,也會(huì)有一定的誤差。因此，總誤差界的計(jì)算應(yīng)綜合以上兩方面的因素。當(dāng) 0： x 901 時(shí),令 f (x) cosxXo0，h(6060 180 10800令 xx0 ih,i 0,1,.,5400則滋290當(dāng)x Xk,xki時(shí)，線性插值多項(xiàng)式為L(zhǎng)i(x) f(Xk)x xk1

13、f(xki) x xkxk xk 1xk 1 xk插值余項(xiàng)為R(x) cosx L1(x)12 f ( )(x xk)(x xk1)又；在建立函數(shù)表時(shí)，表中數(shù)據(jù)具有5位有效數(shù)字，且cosx 0,1 ，故計(jì)算中有誤差傳播過程。*15(f (xk)2 105R2(x)(f(f*g)(Xk)(XkxXkXk 1Xk 1Xk 1xk 1(f*(Xk1)X Xk 1Xk 1Xkx Xk 1Xk 1XkR R(x)甩(x)I*2( cos )(xxk)(x Xk1) (f(xk)1 *(x Xk)(Xki x) (f (Xk)2(222108 i1.06 10 -250.50106 104.設(shè)為互異節(jié)點(diǎn)，

14、求證:n(1)x：lj(x) xk (k 0,1,川,n);j 01n(2)(Xjx)k| j (x)0 (k 0，1，|，n)；j 0證明(1 )令 f (x) xk若插值節(jié)點(diǎn)為xj, j0,1川，n，則函數(shù)f (x)的n次插值多項(xiàng)式為 Ln(x)x：lj(x)j 0插值余項(xiàng)為R,(x)f(x) Ln(x)(n1)()(n 1)!n 1(X)I I又k n,f(n1)( )0Rn(x)0nkl /kXjlj(x) Xj 0(k 0,1,川,n);n k(Xj X) lj(x)j 0n n(Cxj( x)k i)lj(x)j 0 i 0nnik iiCk( X) ( Xjlj(x)i 0j 0

15、又0 i n 由上題結(jié)論可知nj O原式iCk(okX)i iX(X X)kO得證。5 設(shè) f (x)C2 a,b 且 f (a)f(b)O,求證：maxf(x)i(ba)2maxa x bf (x).解：令xoa, Xib,以此為插值節(jié)點(diǎn)，則線性插值多項(xiàng)式為L(zhǎng)i(x)f(Xo) XXf (Xi)X XoXoXiX Xof(b)x aX b= f (a)- a b,11又、f(a) f(b) OJ(x) O插值余項(xiàng)為R(x) f(x) J(x)1-f (x)(X Xo)(X Xi)f(x)i又，(X8(x8 (Xi8(b12 f (x)(x xo)(x Xi)xo)(x Xi)X。)Xo)2a

16、)2maxa x bf(x)8(b6 在 4 x截?cái)嗾`差不超過2X)a)2 maxf(X)4上給出f(x) ex的等距節(jié)點(diǎn)函數(shù)表，若用二次插值求ex的近似值，要使10 6，問使用函數(shù)表的步長(zhǎng) h應(yīng)取多少?解：若插值節(jié)點(diǎn)為 x i,xi和xi i，則分段二次插值多項(xiàng)式的插值余項(xiàng)為R(x)石 f ( )(x Xi 1)(x x)(x Xi 1) 3!1 (x Xi 1)(X6R2(x)x)(x x j maxf (x)設(shè)步長(zhǎng)為h,即 xi 1xih,Xi 1XihR2(X)6 3y324h3.27若截?cái)嗾`差不超過10 6 ,R2(x)10 6e4h310 627h 0.0065.7若 yn 2n,

17、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)行求解。nyn24yn(E1)4yn4(j 04(j 04(j 0(21)j1)j1)j4j4j4jE4 jyny4 n j24j Yn1)4ynyn2n1E 2)4yn4yn11(E 2)4(E 1)4yn24Eyn yn 22“ 28 如果f (x)是 m次多 項(xiàng)式，記f(x) f(x h) f (x)，證明f (x)的k階差分f(x)(Ok m)是m k次多項(xiàng)式，并且m 1f (x)0 (l為正整數(shù))。解：函數(shù)f (x)的Taylor展式為f(x h) f (x)f (x)h jf(x)h2 卅1m!(m)(x)hm1(m

18、1)!(m 1)()h其中 (x,x h)又:f (x)是次數(shù)為m的多項(xiàng)式f(m1)()of(x) f(x h) f(x)f (x)h 1f (x)h2卻叭x)hmf (x)為m 1階多項(xiàng)式f (x)(f(x)f (x)為 m2階多項(xiàng)式依此過程遞推，得 k f (x)是m k次多項(xiàng)式mf (x)是常數(shù)當(dāng)I為正整數(shù)時(shí),m1f(x)09證明(fkgk)fk gkgk 1 fk證明(fkgk)fk 1gk 1fkgkfk 1gk 1fkgk 1fkgk 1fkgkgk 1( fk1fk )fk(gk 1gk)gk 1 fkfk gkfk gkgk 1 fk得證n 1n110.證明fk gkfngnf

19、0g0gk 1 fkk 0k0證明：由上題結(jié)論可知fk gk (fkgk) gk 1 fkn 1fk gkk 0n 1gk1 fk)gk1k 0(fkgk)k 0n 1(fkgk)k 0(fkgk) fk 1gk1 fkgkn 1(fkgk)k 0n 1fk gkk 0fngnf0g0n 1gk 1k 0得證。n 111 .證明2yjynyj 0n 1n 1證明2yj(yj 1yj)j 0j 0(y1y0)(y2yny0(f1g1 fog。)(f2g2 仏1)卅(fngn fn 1gn 1) fngn fog。ydl|l ( y % 1)得證。12.若 f (x) a ajx 川 an 1xn

20、 1anXn有n個(gè)不同實(shí)根X1,X2，川，Xn,證明：nX：0,0 kn 2；j 1 f (Xj)n01,kn 1證明：-f (x)有個(gè)不同實(shí)根X1.X2.IIXna0 a1X 卅an 1Xn1且 f (x)nanXf(X) an(X Xj(X X2) |(x Xn) 令 n(x) (X Xj(X X2)|(X Xn)XjXj而 n(X) (X X2)(X X3)川(X Xn) (X XJ(X X3)|(X Xn)n(Xj)川(X Xi)(X X2)川(xXn l)(XjXi)(Xj X2) (Xj Xj i)(Xj Xj i) (Xj Xn)令 g(x)kXj1 n(Xj)則 g x1,x2

21、,|,XnkXj1 n(Xj)nkn X,又L-j 1 f (Xj)1g Xi,X2 an，|幾nknXjj i f (Xj)0,0 k n 2;1no ,k n 1得證。13.證明n階均差有下列性質(zhì):(J )若 F(x) Cf(x)，則 F Xg,XiJ|,Xncf X0,XiJ|,Xn ；(2)若 F(x) f(x) g(x)，則 F Xo,Xi,卅,Xnf 心,川,人 g Xo,Xi,|”，Xn 證明:)1 f Xi,X2?,Xnn f(xj)j o(Xj Xo)|(Xj Xj J(Xj XjJ 卅(XjXn)F Xi,X2,川,Xnj 0F(Xj)(Xj Xo 川 |(Xj XjJ(X

22、j Xji)|(Xj X)ncf(XjjO(Xj X。) (Xj Xji)(Xj XjJ (Xj Xn)c( nf(xj)j 0 (Xj Xo)(XjXj i)(XjXj i)| (Xj Xn)Cf X),Xi, |,Xn得證。；F(x) f(x) g(x)F(xj)j 0(XjX0)|(Xj Xj J(Xj Xj1)M(Xj XjXj 1 f(xj) g(xj) j 0 (j X0)卅(Xj Xj 1)(Xj xj 1 川 l(Xj Xn) f(xj)j 0(Xj X。川 |(Xj Xj1)(Xj Xj1)川(Xj Xn)得證。j 0 (Xj X0川 |(Xjg(xj)Xj 1)(Xj Xj

23、 1)川(XjXn)f X0,|,XngX0 J|,Xn14. f(x)X7X4 3x 1,求 F20,21J|,27 及 F 202,卅,28。解：一 f(x) X7 X4 3x 1若 X 2i,i0,1，川,8-H-則 f Xo,X1，川,Xn(n)()n!f Xo,X1, |,X7(7)()7!7!7!f X0,X1,|,X8A 08!15證明兩點(diǎn)三次埃爾米特插值余項(xiàng)是(Xk,Xk 1)R3(x) f ()(x Xk)2(x Xk1)2/4!,解: 若X xk, xk 1，且插值多項(xiàng)式滿足條件H3(Xk)f(Xk),H3(Xk)f (Xk)H3(Xk1) f (Xk 1), H3(Xk

24、1) f (Xk 1)插值余項(xiàng)為R(x) f (x) H3(x)且 R(xJ R(XkJ 022R(x)可寫成 R(x) g(x)(x Xk) (x Xk i)其中g(shù)(x)是關(guān)于x的待定函數(shù)，現(xiàn)把x看成Xk,Xk1上的一個(gè)固定點(diǎn)，作函數(shù)2 2(t) f(t) H3(t) g(x)(t Xk) (t Xki)根據(jù)余項(xiàng)性質(zhì)，有(Xk) 0, (Xki)0(X) f (X) H3(x) g(x)(x Xk) (X Xk i)f(x) H3(x) R(X)02 2(t) f (t) H3(t) g(x)2 (t Xk)(t Xki)2(t Xk 1 )(t Xk)(Xk)0(Xk i)0由羅爾定理可知

25、，存在(xk,x)和(x,xk 1)，使(i)0，( 2)0即(x)在xk,i上有四個(gè)互異零點(diǎn)。根據(jù)羅爾定理，(t)在(t)的兩個(gè)零點(diǎn)間至少有一個(gè)零點(diǎn)，故 (t)在(xk,xki)至少有三個(gè)互異零點(diǎn)，依此類推，(t)在(xk ,xk J至少有一個(gè)零點(diǎn)。記為(Xk,Xk i)使(4)( ) f(4)( ) H3(4)( ) 4!g(x)0又1 H3(4)(t)0f(4)()g(x),(Xk,Xk i)4!其中依賴于xR(x) (x x$(x xki)24!分段三次埃爾米特插值時(shí)，若節(jié)點(diǎn)為xk（k0,1，川，n），設(shè)步長(zhǎng)為h，即XkX。kh,k0,1，川，n在小區(qū)間兀入上R(x)()(x4!R(x

26、)Xk)2(x Xki)2)(x Xk)2(x Xki)2(X Xk )2 (Xk 1 4!1 x Xk Xk i 4!(Tx)2 max f (x)a x bv -)22max f(4)(x)a x b16 .P(0)丄4!h4-h4 max f1 * * (4) (x)24a x bv max f (x)384 a x b)求一個(gè)次數(shù)不高于 4 次的多項(xiàng)式P(0) 0,P(1) P(1) 0,P(2) 0P ( x ),使它滿足X 0,X11y。0,y1 1m00,mi 1(x)1yjj 00（X）(1 2(12x)( x1(x)(1 2(32x)x20（X）x(xj(x)X X1)(X

27、X0)2XI X0 X1X01)21(x) (X 1)x2解：利用埃米爾特插值可得到次數(shù)不高于4的多項(xiàng)式3H3(x)(3 2x)x (x 1)x x 2x2 2設(shè) P(x) H3(x) A(x X0)(x X1)其中，A為待定常數(shù)11x3 2x2 Ax2(x 1)2(P(2) 1P(x)a*從而P(x)1 24x(x 3)17-設(shè) f (x)1/(1 X2)，在 5 x 5上取n 10，按等距節(jié)點(diǎn)求分段線性插值函數(shù)lh(x)，計(jì)算各節(jié)點(diǎn)間中點(diǎn)處的| h(x)與f (x)值，并估計(jì)誤差。解:r右 X05, X10則步長(zhǎng)h 1,XiX0 ih,i0,1,川,10f(X)七在小區(qū)間xi, x1上，分

28、段線性插值函數(shù)為lh(x)f(Xi)Xi Xi 1X XiX 1 Xf (Xi 1 )(X 1X)1 x2(xx)11各節(jié)點(diǎn)間中點(diǎn)處的lh(X)與f (x)的值為當(dāng)X4.5 時(shí)，f (x)0.0471, I h(x)0.0486當(dāng)X3.5 時(shí)，f (x)0.0755, Ih(x)0.0794當(dāng)X2.5 時(shí)，f (x)0.1379, h(x)0.1500當(dāng)X1.5 時(shí)，f (x)0.3077, Ih(x)0.3500當(dāng)X0.5 時(shí)，f (x)0.8000, Ih(x)0.75002 Xi 14早圭 誤差maxxi x x. 1f(x)Ih(x)h2max f ()85x5又、f(x)f (x)f

29、 (x)f (x)11 x22xdF6x2 23(1 x ) 24x 24x3(1 x2)4令 f (x)0得f (x)的駐點(diǎn)為Xi,21 和 x301f (X1,2)2, f (X3)max f(x) Ih (x)5x518求f(x) x2在a, b上分段線性插值函數(shù)lh(x)，并估計(jì)誤差解：在區(qū)間ab上，X。 a,Xn b,h. x. 1 x.,i 0,1，川,n 1,h max h.0 i n 1 .2t f(x) X函數(shù)f (x)在小區(qū)間X ,x.上分段線性插值函數(shù)為lh(x)f(Xi)X. X. 1XX.X 1 Xf (X. 1)122X. (X. 1 X)X. 1 (XX.)h.誤

30、差為maxX x Xi 1f (X)lh(x)i 12.f(x) xf (X)2x, f (x)max f (x)a x blh(x)1max f8 a b24()|hi219求f(x) x4在a,b上分段埃爾米特插值，并估計(jì)誤差。解：在a,b區(qū)間上，Xo a, Xn b,hiXi,i 0,1|,n 1,令 h max hi0 i n 1f (x) x4, f (x) 4x3函數(shù)f (x)在區(qū)間xi, xi 1上的分段埃爾米特插值函數(shù)為lh(x) ()2(1 24 f(x)x x 1x 1 人(42(12)f(XiJN 1 Nx xi 1(2)2(x xi)f (x)X x 1(X Xi )2

31、 (x Xi1)f(X1)x 1 X4討x x 1)2(h 2X 2xi)4X 1 /r(xh3xj2(h4x3帚(XXi 1)2(xx 3(XG2(X2x 2xi 1)Xi)Xi 1)誤差為f(4) ( ) (x X)2(x X 1)2f(x) lh(x) 1 4!一max f(4)()124 a x b又：f(x) X4f (x)4! 24max f (x) lh (x a x b44h h max 0 i n 1 1616X0.250.300.390.450.53Y0.50000.54770.62450.67080.728020.給定數(shù)據(jù)表如下:試求三次樣條插值，并滿足條件:(1) S

32、(0.25)(2) S (0.25)1.0000, S (0.53)0.6868;S (0.53)0.hxX00.05h1xX10.09h2X3x0.06h3x4x30.08i *hj1jhj 1hj 解:Jhj 1hj hj5114,914,225, 3f(xj47, 0f(X)x, X1X1X0X1, X0.8533X2,X30.771723110.9540Jx3, x40.7150S(Xo) 1.0000,S(X4) 0.6868d。6(f x1,x2f0)5.5200hofX1,X2fX0,X1hh1fX2,X3fX1,X2h1hzfX3,X4fX2,X3h2h364.315763.2

33、64062.4300d1d2d3d42.11506(f4 f X3,X4 )h3由此得矩陣形式的方程組為M15M14141475.52004.31573.26402.43002.1150求解此方程組得M02.0278,M11.4643M21.0313,M30.8070,M40.6539(xj 1 x)3S(x) Mj3j 6hj(y Mjhj2)Xj1 (yj 丁)三次樣條表達(dá)式為(X Xj)36hjMj1hj2 x Xj（比1j-)j(j 0,1, ,n 1)6 hj將 MoMjMzMs, M4 代入得6.7593(0.30x)34.8810(x0.25)3x 0.25,0.302.7117

34、(0.39x)31.9098(x0.30)3x 0.30,0.392.8647(0.45x)32.2422(x0.39)3x 0.39,0.451.6817(0.53x)31.3623(x0.45)3x 0.45,0.53S(x) S(xo)0,S(x4)0do 2f0,d14.3157, d23.2640d32.4300,d4 2f4 0040由此得矩陣開工的方程組為10.0169(0.30 x) 10.9662(x 0.25)6.1075(0.39 x) 6.9544(x 0.30)10.4186(0.45 x) 10.9662(x 0.39)8.3958(0.53 x) 9.1087(x

35、 0.45)M。M4014M1M2M34.31573.26402.4300求解此方程組，得M00,M11.8809M20.8616,M31.0304,M4 0又；三次樣條表達(dá)式為S(x)(x 1 x)36hj(x %)36hjM jhj2 Xj 1 x6)hj(yj 1Mj 1hj2 x xj6) hj將 Mo,M1,M2,M3,M4 代入得10(0.3 x) 10.9697(x 0.25)6.2697(x 0.25)3x 0.25,0.303.4831(0.39x 0.30,0.39 S(x)x)31.5956(x 0.3)3 6.1138(0.39 x) 6.9518(x 0.30)21.

36、若2.3933(0.45x 0.39,0.452.1467(0.53x 0.45,0.53f(x)x)3x)32.8622(x 0.39)3 10.4186(0.45 x) 11.1903(x 0.39)8.3987(0.53 x)9.1(x0.45)C2 a,b ,S(x)是三次樣條函數(shù)，證明:b(1)a f (x)ba f (x)2dx2b2dx S (x) dxa2 bS (x) dx 2 S (x) f (x) S (x)a(2)若 f(xjS(xJ(i0,1，川,n)，式中Xi為插值節(jié)點(diǎn)，且axqx-i川Xnb，則baS(x) f (x) S (x)S (b) f (b) S (b)

37、dxS(a) f (a) S(a)證明：b2(1)a f (x)S (x) dxb2b2bf (x)adxS(x)adx 2f (x)S (x)dxab2b2ba f (x)dxa S(X)dx 2aSf (x) S (x)從而有b2b2a f (X)(dxaS (x) dxb2bf (x)aS(x)dx 2S(x)af (x)S (x) dx第三章函數(shù)逼近與曲線擬合1 - f(x)sinx,給出0,1上的伯恩斯坦多項(xiàng)式B(f,x)及B3(f ,x)。解:,f(x) sin 2 x 0,1伯恩斯坦多項(xiàng)式為Bn(f,X)kf ()Pk(x)k 0 n其中Pk(x)kn kx (1 x)當(dāng)n 1時(shí)

38、,1P0(x) 0(1 x)P(x) xB(f,x)10f (O)R(x)f(1)P(x)xsin 2當(dāng)n3時(shí),R(x)10(13x)P(x)1x(120x) 3x(1F2(x)32 / ,21x (1x) 3x (1B(x)33 x3 xxx)3x)2(1 叫 0)B3(f,x)kf()Pk(x)n0 3x(132-x(1 x)225 3乜3x21.5xx)2lsi n3x2(16!3 3x (1 x) 23、3 6 23x x 2 20.402x2 0.098x3x)|sin3f(x) x 時(shí)，求證 Bn( f ,x) x證明:若 f(x)x，則Bn(f,x)n kf T)Pk(x)k o

39、 n3x sin2nk n k 一 、n k x (1 x) k 0 n kkn(n 1)|(k 1)!1k、n kx (1 x)3證明函數(shù)xx(1n 1x)1,x,lllxn線性無關(guān)n k 1) k n k x (1 x) k!(n 1)卅(n 1) (k 1) 1xk(1 x)nkxk1(1 X)(n1)(k1)證明:若aa1X2a2xlbnanX0, x R分別取xk(k0,1,2,|, n),對(duì)上式兩端在0,1上作帶權(quán) (x)1的積，得IIIa1h1n此方程組的系數(shù)矩陣為希爾伯特矩陣，對(duì)稱正定非奇異, 只有零解a=0。函數(shù)1,xJ|,xn線性無關(guān)。III12n 1an4。計(jì)算下列函數(shù)f

40、 (x)關(guān)于C0,1的f , f1與f 2 :(1)f(x) (x 1)3,x 0,11 f(x) x 2，(3)f(x) xm(1 x)n,m與 n 為正整數(shù), f(x) (x 1)10ex解：(1)若 f (x) (x 1)3,x 0,1，則f (x)3(x 1)20f(x) (x 1)3在(0,1)單調(diào)遞增 f| maxlf(x)max f (0), f(1)max 0,11ic 二f 2(0(4 x)6dx)2中1 x)To*丄71若 f (x) x -,x 0,1，則max f (x)0 x 110f (x) dx112 1(x )dx2214i1f 2(f2(x)dx)21 1 -

41、(x 嚴(yán)2J6(3) 若 f(x) xm(1 x)n, m與 n 為正整數(shù)當(dāng) x 0,1 時(shí)，f (x)0f (x) mxm1(1 x)n xmn(1 x)n1( 1)xm1(1 x)n1m(1當(dāng) x (0,-)時(shí),f (x)0n mf(X)在(o, m)單調(diào)遞減 n m當(dāng) x (,1)時(shí)，f (x)0n mf (x)在(m ,1)單調(diào)遞減。n mx ( m ,1)f (x) 0n mPImaxlf(x)max f(0),f()f (x) dxn m11 01 mn ,ox (1 x) dx22 m2 n2sin2m t cos2 nt costas in tdtf2 0門10 2m , 2n

42、 (sin t) (1 sin t) d sin tf （x）在0,1單調(diào)遞減。maxf(x)f(0), f(i)max2e11(x0(x10 f(x) dx1)10e xdx10i)10 xI e；10(x 1)9e xdx5 201) e12xdx2e1 f 2(X7(3自5。證明f g f g證明：(f g)gf ggf gf g6。對(duì) f (x), g(x) C1a,b，定義(1)(f,g)bf (x)g (x)dxa(f,g)ba f (x)g (x)dx f(a)g(a)問它們是否構(gòu)成積。解：(1)令 f(x)C（ C為常數(shù)，且C 0）則 f (x)(0而(f, f)bf (x)f

43、 (x)dxa這與當(dāng)且僅當(dāng)f 0時(shí)，（f,f）0矛盾不能構(gòu)成C1a,b上的積。b若(f ,g)f(x)g(x)dx f (a)g(a)，則b(g, f) a g (x) f (x)dx g(a)f (a) (f ,g),bf (X) g (x)dx af (a)g(a)a(f ,g)bf (x)g (x)dx a(f ,g)f (a)g(a)1h Ca,b,則b(f g,h) af(x)bf (x)h (x)dx f (a)h(a) a(f,h)g(x) h (x)dx f(a)g(a)h(a)bf (x)h (x)dx g(a)h(a)a(h,g)(f, f)b 2 2af(x)dx f (a)0若(f, f)0,則b 2 2f (x)2dx 0 ,且 f2(a)0f (X)0, f(a) 0f(x) 0即當(dāng)