數(shù)值分析第四章數(shù)值積分與數(shù)值微分知識(shí)題目解析

1、_第四章數(shù)值積分與數(shù)值微分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：(1)hA 1 f (h)A0 f (0)A1 f ( h);f ( x)dxh(2)2hA 1 f (h)A0 f (0)A1 f (h);f (x)dx2h(3)1 f ( 1)2 f (x1)3 f ( x2 )/ 3;f ( x)dx1(4)hh f (0)f (h)/ 2ah2 f (0) f (h);f ( x)dx0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m 的多項(xiàng)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)行驗(yàn)證

2、性求解。h（1 ）若 (1)f ( x)dxA 1 f ( h)A0 f (0)A1 f ( h)h令 f (x) 1 ，則2hA 1A0A1令 f(x)x ，則0A 1hA1h令 f(x)x2 ，則2 h3h2 A1h2 A31從而解得A04 h3A11h3A 11h3_令 f (x)x3，則hh0f (x)dxx3dxhhA 1 f (h)A0 f (0)A1 f ( h) 0h故f ( x)dx A 1 f (h) A0 f (0) A1 f (h) 成立。h令 f (x)x4，則hf (x)dxhA 1 f ( h)h x4 dxhA0 f (0)2 h55A1 f (h)2 h53故

3、此時(shí)，hf (x)dxA 1 f ( h)A0 f (0)A1 f (h)hh故f ( x)dx A 1 f (h)A0 f (0) A1 f (h)h具有 3次代數(shù)精度。（2 ）若2 hf ( x)dxA 1 f ( h)A0 f (0) A1 f ( h)2 h令 f (x)1 ，則4hA 1A0 A1令f (x)x ，則0A 1hA1h令 f (x)x2，則16 h3h2 A 1 h2 A13從而解得_A04 h3A18h3A 18 h3令 f (x)x3，則2 hf ( x)dx2 h02 hx3dx2 hA 1 f (h)A0 f (0)A1 f ( h)02 h故f ( x)dx

4、A 1 f ( h)A0f (0) A1 f (h) 成立。2h令 f (x)x4，則2 hf ( x)dxx4dx64h52 h2 h2 h516 h5A1f (h)A f (0)A f (h)013故此時(shí)，2 hf ( x)dxA 1 f (h)A0 f (0)A1 f (h)2 h因此，2 hf ( x)dxA 1 f (h)A0 f (0)A1 f (h)2 h具有 3 次代數(shù)精度。12 f ( x1 ) 3 f (x2 )/ 3（3 ）若f (x)dx f (1)1令 f (x)1 ，則1f (x)dx2 f (1)2 f ( x1 )3 f (x2 )/ 31令 f (x)x ，則

5、012 x13x2令 f (x)x2，則2221 2x13x2_從而解得x10.2899x10.6899x2或x20.12660.5266令 f (x)x3 ，則110f ( x)dxx3dx11 f (1)2 f (x1)3 f ( x2 )/ 3012 f ( x1 )3 f (x2 )/ 3 不成立。故f (x)dx f (1)1因此，原求積公式具有2 次代數(shù)精度。hh f (0) f ( h)/ 2ah2 f (0) f ( h)（4 ）若f (x)dx0令 f (x)1 ，則hh,0f (x)dxh f (0)f (h)/2ah2 f(0)f (h)h令 f (x)x ，則hh1 h

6、20f (x)dxxdx02h f (0)f (h)/2ah2 f(0)f (h)1 h22令 f (x)x2，則hh2dx1 h30f ( x)dxx03h f (0)f (h)/2ah2 f(0)f (h)1 h32ah22故有1h31h32ah232a112_令 f (x)x3 ，則hh3dx1 h4f (x)dxx004h f (0)f (h)/21 h2 f (0)f(h)1 h41 h41 h412244令 f (x)x4 ，則hh4dx1 h5f (x)dxx005h f (0)f (h)/21 h2 f (0)f(h)1 h51 h51 h512236故此時(shí)，hh f (0)f

7、 (h)/ 21 h2 f(0) f ( h),f (x)dx0121 h2 fhf ( x)dx h f (0) f (h)/ 2(0)f (h)因此，012具有 3次代數(shù)精度。2.分別用梯形公式和辛普森公式計(jì)算下列積分：1xdx, n8;(1)0 4x211 (1 ex ) 210;(2)dx, n0 x9(3) xdx, n 4;1(4)6 4sin2d , n 6;0解：(1)n8,a0,b1x1,h, f ( x)284 x復(fù)化梯形公式為h f ( a) 27T8f ( xk )f (b) 0.111402k 1復(fù)化辛普森公式為_h f ( a)77S84f ( x1 )2f ( x

8、k )f (b) 0.111576k 0k2k11(2) n10, a0,b1,h1(1e x ) 210, f ( x)x復(fù)化梯形公式為h f (a)9T102f ( xk )f (b)1.391482k1復(fù)化辛普森公式為h f (a)99S104f ( xk1 )2f (xk )f (b)1.454716k02k 1(3) n4, a 1,b9, h2, f (x)x,復(fù)化梯形公式為h f (a)3T42f ( xk )f (b)17.227742k 1復(fù)化辛普森公式為h f (a)33S44f ( x1 )2f (xk )f (b)17.322226k0k2k 1(4) n6,a 0,b

9、, h, f (x)4sin2636復(fù)化梯形公式為h5T62f ( xk )f (b)1.03562 f (a)2k1復(fù)化辛普森公式為h55f ( xk 1 )S6 f (a)4k02f (xk ) f (b) 1.0357762k13 。直接驗(yàn)證柯特斯教材公式（2 。 4）具有 5 交代數(shù)精度。證明：_柯特斯公式為bbaf (x)dx7 f (x0 )32 f ( x1) 12 f ( x2 ) 32 f ( x3 )7 f ( x4 )a90令 f (x)1 ，則bbaf (x)dx90ab a7 f ( x0 ) 32 f (x1) 12 f (x2 ) 32 f ( x3 ) 7 f

10、( x4 ) b a90令 f (x)x ，則bb1 (b2a2 )f (x)dxxdxaa2b a 7 f ( x0 ) 32 f ( x1 ) 12 f (x2 ) 32 f ( x3 ) 7 f ( x4 )1 (b2a2 )902令 f (x)x2 ，則bb2dx1 (b3a3)f (x)dxxaa3b a 7 f ( x0 ) 32 f ( x1) 12 f (x2 ) 32 f ( x3 ) 7 f ( x4 )1 (b3a3 )903令 f (x)x3 ，則bb1 (b4a4 )f (x)dxx3dxaa4b a 7 f ( x0 ) 32 f ( x1 ) 12 f (x2 )

11、 32 f ( x3 ) 7 f ( x4 )1 (b4a4 )904令 f (x)x4 ，則bb4dx1 (b5a5 )f (x)dxxaa5b a 7 f ( x0 ) 32 f ( x1) 12 f (x2 ) 32 f ( x3 ) 7 f ( x4 )1 (b5a5 )905令 f (x)x5 ，則_bb1 (b6a6 )f (x)dxx5dxaa6b a 7 f ( x0 ) 32 f ( x1) 12 f (x2 ) 32 f ( x3 ) 7 f ( x4 )1 (b6a6 )906令 f (x)x6 ，則hb af ( x)dx7 f (x0) 32 f ( x1) 12 f

12、 ( x2 ) 32 f (x3) 7 f ( x4 )090因此，該柯特斯公式具有5 次代數(shù)精度。4 。用辛普森公式求積分1e xdx 并估計(jì)誤差。0解：辛普森公式為Sb a f (a)ab64 f () f (b)2此時(shí)，a0,b1, f ( x)e x ,從而有1 (11e 1 ) 0.63233S4e 26誤差為R( f )ba ( b a )4 f (4) ( )180 21 1 e0 0.00035,(0,1)180 245 。推導(dǎo)下列三種矩形求積公式：bf (x)dx (b a) f ( a)f ( ) (b a)2 ;a2bf (x)dx (b a) f (b)f () (b

13、a)2 ;a2f ( ) (b a)3;bf (x)dx (b a) f ( a b )a224_證明：(1)f ( x)f (a) f ()( xa),( a, b)兩邊同時(shí)在 a, b 上積分，得bf (x)dx(ba) f ( a)f()ba)dxa( xa即bf (x)dx (b a) f ( a)f ( ) (b a)2a2(2)f ( x)f (b) f ()(bx),( a,b)兩邊同時(shí)在 a, b 上積分，得bf (x)dx(ba) f ( a)f()bx)dxa(ba即bf (x)dx(ba) f (b)f() (ba)2a2(3)f ( x)f ( a b )f ( ab

14、)( xab )f ( ) (xa b )2 ,(a, b)22222兩連邊同時(shí)在 a, b 上積分，得bf (x)dx(ba bfa b ba bf ( ) ba b2dxaa) f ()()( x)dx( x)22a22a2即bf (x)dx(ba) f ( ab )f() (ba)3;a224I16 。若用復(fù)化梯形公式計(jì)算積分exdx ，問區(qū)間 0,1 應(yīng)人多少等分才能使截?cái)嗾`差不超0過110 5 ？若改用復(fù)化辛普森公式，要達(dá)到同樣精度區(qū)間0,1 應(yīng)分多少等分？2解：采用復(fù)化梯形公式時(shí)，余項(xiàng)為Rn ( f )ba h2 f ( ),12(a, b)_I1又ex dx0故 f (x)ex

15、, f( x) ex, a0,b 1.R ( f )1 h2 f ( )e h2n1212若 Rn ( f )110 5，則2h26 105e當(dāng)對(duì)區(qū)間 0,1 進(jìn)行等分時(shí)，1h ,n故有ne 10 5212.856因此，將區(qū)間213 等分時(shí)可以滿足誤差要求采用復(fù)化辛普森公式時(shí)，余項(xiàng)為Rn ( f )ba ( h )4 f (4) ( ),(a, b)1802又f (x)ex ,f (4) (x)ex ,Rn ( f )1h4 | f (4) ( ) |eh428802880若 Rn ( f )110 5，則14402h410 5e當(dāng)對(duì)區(qū)間 0,1 進(jìn)行等分時(shí)1nh故有1n(1440105 )

16、4 3.71e_因此，將區(qū)間8 等分時(shí)可以滿足誤差要求。7 。如果 f (x) 0b，證明用梯形公式計(jì)算積分 Iaf ( x)dx 所得結(jié)果比準(zhǔn)確值I 大，并說明其幾何意義。解：采用梯形公式計(jì)算積分時(shí)，余項(xiàng)為RTf () (b a)3 , a,b12又f ( x)0 且 b aRT0又RT1TIT即計(jì)算值比準(zhǔn)確值大。其幾何意義為，f (x)0 為下凸函數(shù)，梯形面積大于曲邊梯形面積。8 。用龍貝格求積方法計(jì)算下列積分，使誤差不超過10 5 .(1) 21x dxe02(2) xsin xdx0(3)3x2 dx.x 10解：(1)I21x dxe0kT0(k)T1(k )T2( k)T3( k)

17、00.771743310.72806990.713512120.71698280.71328700.713272030.71420020.71327260.71327170.7132717_因此 I0.713727(2) I2x sin xdx0kT0(k )T1( k )03.45131310 618.62828310 7-4.44692310因此 I0(3) I3x 1 x2 dx0kT0( k )T1(k)T2( k)T3(k )T4( k)014.2302495111.1713610.151749934210.4437910.2012710.20457692544310.2663610

18、.2072210.2076210.2076672400791410.2222710.2075710.2075910.2075910.207590212433936510.2112610.2075910.2075910.2075910.20759070922222221T5( k)10.2075922因此 I10.20759229 。用 n2,3 的高斯 - 勒讓德公式計(jì)算積分_3ex sin xdx.1解：I3ex sin xdx.1x1,3, 令 t x2 ，則 t 1,1用 n2的高斯勒讓德公式計(jì)算積分I0.5555556 f ( 0.7745967)f (0.7745967)0.8888

19、889f (0)10.9484用 n3 的高斯勒讓德公式計(jì)算積分I 0.3478548 f ( 0.8611363)f (0.8611363)0.6521452 f ( 0.3399810)f (0.3399810)10.9501410 地球衛(wèi)星軌道是一個(gè)橢圓，橢圓周長的計(jì)算公式是Sa 21( c )2 sin 2d ,0 a這是 a 是橢圓的半徑軸， c 是地球中心與軌道中心（橢圓中心） 的距離， 記 h 為近地點(diǎn)距離，H 為遠(yuǎn)地點(diǎn)距離，R=6371 （ km ）為地球半徑，則a(2 RHh)/ 2, c(Hh)/ 2.我國第一顆地球衛(wèi)星近地點(diǎn)距離h=439(km)，遠(yuǎn)地點(diǎn)距離H=2384(

20、km）。試求衛(wèi)星軌道的周長。解：R6371,h439, H2384從而有。_a(2 RHh) / 2 7782.5c( Hh) / 2972.5S4a2 1( c )2 sin 2 d0 akT0(k )T1(k)T2( k)01.56464011.5646461.56464821.5646461.5646461.564646I 1.564646 S 48708( km)即人造衛(wèi)星軌道的周長為48708km11 。證明等式35nsin3! n25! n4n試依據(jù) nsin()( n3,6,12)的值，用外推算法求的近似值。n解若 f (n)n sin,n又 sin xx1 x31 x53!5!

21、此函數(shù)的泰勒展式為f (n)n sinnn1 ()31 ()5n3!n5!n353!n25! n4Tn( k)_當(dāng) n3 時(shí) ,nsin2.598076n當(dāng) n6時(shí) ,nsin3n當(dāng) n12 時(shí),n sin3.105829n由外推法可得nT0(n )T1(n)T2( n)32.59807663.0000003.13397593.1058293.1411053.141580故3.1415812 。用下列方法計(jì)算積分3 dy ，并比較結(jié)果。1y(1) 龍貝格方法；(2) 三點(diǎn)及五點(diǎn)高斯公式；(3) 將積分區(qū)間分為四等分，用復(fù)化兩點(diǎn)高斯公式。解3 dyI1 y(1) 采用龍貝格方法可得kT0(k )

22、T1(k )T2(k)T3( k )T4(k )01.33333311.1666671.09925921.1166671.1000001.099259_31.1032111.0987261.0986411.09861341.0997681.0986201.0986131.0986131.098613故有 I1.098613(2) 采用高斯公式時(shí)3 dyI1 y此時(shí) y1,3,令 xyz, 則 x1,1,1 1Idx,1 x 21f ( x),x2利用三點(diǎn)高斯公式，則I0.5555556 f (0.7745967)f (0.7745967)0.8888889 f (0)1.098039利用五點(diǎn)高

23、斯公式，則I0.2369239 f (0.9061798)f (0.9061798)0.4786287 f ( 0.5384693)f (0.5384693)0.5688889 f (0)1.098609(3) 采用復(fù)化兩點(diǎn)高斯公式將區(qū)間 1,3 四等分，得II1 I2I 3I 41.5 dy2dy2.5 dy3dy1y1.5y2y2.5y作變換 yx5，則4_1 1I11 xdx,5f ( x)1,x5I1f ( 0.5773503) f (0.5773503) 0.4054054作變換 yx7，則41 1I 21 xdx,7f ( x)x1,7I 2f (0.5773503)f (0.57

24、73503)0.2876712作變換 yx9，則4I 311dx,1 x9f ( x)1,x9I 3f (0.5773503)f (0.5773503)0.2231405作變換 yx11，則4I 411dx,1 x11f ( x)1,x11I 4f (0.5773503)f (0.5773503)0.1823204因此，有I1.09853813. 用三點(diǎn)公式和積分公式求f (x)1在 x 1.0,1.1，和 1.2處的導(dǎo)數(shù)值，并估計(jì)誤x)2(1差。 f (x) 的值由下表給出：x1.01.11.2F(x)0.25000.22680.2066_解：f ( x)1(1x) 2由帶余項(xiàng)的三點(diǎn)求導(dǎo)公式可知