吉林大学工程数学计算方法(第三章习题答案)

1、1Gdx0.5907f0.532f512f32f7f114.9497525.2982218010.3923029.933260.430961-Txdx,并估计证差.0.5解：1)1)用梯形公式有：1-1051.2,xdx-f1f0,510.426780.5242事实上,Qfx.x,I1,xdx0.43096440.5105I-f0.5f10.42677672T1-10.5Ef.xdxf0.5f10.522)Simpson2)Simpson 公式1212.3一0.43093122第三章习题答案ETf3baf120.5312124_32.6042103o327.3657101,1,分别用梯形公式、

2、SimpsonSimpson 公式、CotesCotes 公式计算积分I0.00418774fET(f)=-(b-a)123-f(h)?248事实上,ESf0.5xdx0.51-2-1-2-15157-27-21.18377100.54f0.510.00003043 3由 CotesCotes 公式有:0.510.5故该公式的代数精度是而当f32左侧不等于右侧.所以 SimpsonSimpson 具有三次代数精度. .1x,、(1)rdx,n8,(3)04x23.3.分别用复化梯形公式和复化公式SimpsonSimpson 计算以下积分. .ECf2129459456411/2.6974106

3、EC(f)=-2(b-a)945*4f(6)(h)?.2.2V事实上,ECf0.00000032.2.证实SimpsonSimpson 公式2.8具有三次代数精度.证实:x3,b3左边bfxdxab44右边ba,fa64fb4左侧:dxbx4dxab5右侧:4fb45b55a5a4b3,22ab2,32abab4xdx,n1卜4sin2d,n解：1 1用复化梯形公式有:Tn-0162(0.0311280.0615380.0905660.117650.142350.164380.18361)0.20.1114由复化SimpsonSimpson 公式有:S8(ff4f3f1解:T4240.1115

4、70删去(3):(3):0.06153812-dx,n10、xdx,n42,由复化Simpson公式有:S4(4)解:0614sin2由复化梯形公式:0665360.117650.1643840.031128由复化梯形公式有:17.22772f,n17.3220kh,k123,4,50.0905660.412350.183510.2T6机2f(k)f(b)36f(0)f361.035621936由复化Simpson公式:c12S4-T6H6,H633k22k22k9o1123,4,55H6f36k072361.035834878S41.0357638864.给定求积节点x012,X2dx的插值

5、型求积公式,并写出它的截断误差.解：1fxdx0Af12121dx23,考虑到对称性,有AAo,xdx由于原式含有 3 3 个节点,阶精度.事实上,对此外,容易验证原式对5 5.给定积分(1)(1)(2)(2)(3)(3)113424于是有求积公式故它至少有 2 2阶精度.考虑到其对称性,可以猜测到它可能有原式左右两端相等:1x3dx04八,x不准确,102sinxdx.故所构造出的求积公式有3 3 阶精度.利用复化梯形公式计算上述积分值,使其截断误差不超过2取同样的求积节点,改用复化 SimpsonSimpson 公式计算时,截断误差是103;多少?如果要求截断误差不超过106,那么使用复化

6、 SimpsonSimpson 公式计算时,应将积分区间分成多少等分?当误差E；(f).513时,n25.6,25.6,所以取n=26=26.皿h25那么:h=52Tn韶()f(2)2kif(xk)1芯012sin(-)sin(需sin(3-).sin(失)25252525252那么ESf2(1)4(n26)2(1)47109n1822n1822n21Y解(1):.屋dx21.f(0)f(1)0.771743332,210.68439656,T2(T1H1)0.728069946,2c41S1-T2-T10.7135121533(c)0,1四等分:21131H2ff-0.705895578,T

7、4一(T2H2)0.7169827622442解：E；(f)()2f(96n,fx= =sinx, ,f(x)cosx,f(x)sinxE；(f)32sin96n23TT296n2,20.9465(2)ESfnn18(2赤)4sin(ESf那么n7.6218n=8166,用 RombergRomberg 求积方法计算以下积分,使误差不超过_51.(1)(1)T(2)(2)xsinxdx;(3)x.1x2dx;(4);(4)0114-2dxx(a)在0,1上用梯形公式：工(b)0,1厂等分:21H1f2计算可以停止2.解：xsinxdx(a)在0,2上用梯形公式得：2T1f(0)f(2)02(b

8、)将0,2二等分:,1_41H12f()0,T2-(T1H1)0,S1-T2-T10233(c)将0,2四等分:321H2f-f一29.869604401,T4-(T2H2)4.934802201222_41_4_S2T4T26.579736267,01二S233421(d)将0,2八等分:3ri1H4-f-6.9788642,T8(T42i042241-4cleS4-T8-T46.2975102C2二S4二S2334214213_41_RC2C16.266954014431431(e)将0,2十六等分i1Mf-6.447629792T163M)4i0842c41S2-T4-Ti33(d)将0

9、,1八等分:0.713287034,0142Fc1cS2S10.713272026421H4f8.3.5.7f-f-f-0.711417571888T802R1141(T4H4)0.714200166,S4-T8-T423342c1cS4S20.713271674,414143-1一一一二一02二一010.713271669,R1014314310.7132726343.52107105S17.018385352421H4)5.9568332016.2786951296.2022314976.2831323112c41-42cleG-T16-T86.28403092904FS433421421

10、(9)将0,2六十四等分：(3)解：x1x2dx0(a)在0,3上用梯形公式T13f(3)f(0)14,230249472(b)将0,3二等分:3141H13f8.11249037,T2-(T1H1)11.17136992,ST2-T110c)将0,3四等分91f9,71622377,T4T2H2)10.4437968542c4.1.S27T4T233(d)将0,3八等分10,20217249,C1f42S21421S110.20457443H4处10.08893752,4H4 10,26636719S43T83T410,20722396,C24242-S411421

11、S210.207620734c1cR2C4C26,283202742,X1431431(f)将0,2三十二等分44hQR1416.28326646315.H16f8i01686.323740394,T321一(T162H16)6.262985945S16R44TI32343431453T16C841-7X21产1416.283237428,C8416.283185356,X2S1614244.44R4S86.283184528-R26.28318528811X16.28318520931H3216i0f32166.293289853,T641(T32H32)6.278137899S32R84T

12、643433T326.283188551,C16421S32丫24344-7C1615一X411C86.283185304,X4142144S166.2831852929.510六X24185106.283185304,Z144ZR86一丫211二一R46.283185304411Y16,2831853044133H22f4R43cR1-C24112C110.20766908431(e)将0,3十六等分:H8f1163i10.1781732,T1681(T8H8)10.22227022R23T164310.2075712,C442421S8-C411-3C210.20759393,X14311

13、42144S410.2075943544R211R110.2075936441(f)将0,3三十二等分H1631516i0fA323i10.20025127,T32i(T16H16)10.21126074S16R44T3T32433T1610.20759091,C8f-41S16Y143C8X2451,Q4311-X145110.20759219,X210.20759219,Y计算可以停止1解(4):-01dx(a)在0,1上用梯形公式1T-f(1)f(0)2(b)将0,1二等分:.1H1f-32T22(C)将0,1四等分：2(TIHI)3.1,SIH212f3.162352941,T4S2I

14、T23.141568627,C1f4(d)将0,1八等分H4i3.146800518,T8S4RI43T84313T43.141592502,C2C2XI4T2-S2112(T4421CI3.141585784411.43T1H2)S810.207592231441R210.207592196510103.13333333334)3.1421176473.1489884951S23.141594094421(e)0,1升六等分f(x)f(a)f()(xa),(a,x),两边在a,b上积分,得b(ba)f(a)f()(xa)dxa,1,2(ba)f(a)2f()(ba)

15、2,a,b.将f(x)在xb处 TaylorTaylor 展开,得两边在a,b上积分,得bf()(xb)dxa12(ba)f(b)1f()(ba)2,a,b.H8&RXi11i1-f3.14289473,T16(T8H8)3,1409416138i0168241_4_1_T16T83.141592652,C4S8S433421421344_1_4_C4-；C23.141592639,X1R2431431441R16.88106105,算可以停止.3.1415926621441R13.1415926667.推导以下三种矩形求积公式:b1fxdxab2fxdxab3fxdxaf2bafab

16、a;2f2bafbba;2abfbafba224证实：(1)将f(x)在xa处 TaylorTaylor 展开,得bf(x)dxabf(a)dxa)(xa)dxf(x)f(b)f()(xb),(x,b),bf(x)dxabf(b)dxa)(xb)dx(ba)f(b)ab将f(x)在x处 TaylorTaylor 展开,得两边在a,b上积分,得说明其几何意义.证实：复化梯形公式为假设f(x)在a,b上连续,那么复化梯形公式的余项为n1卜3ETfITn-f(k),k(xk,xk1)(1)k012,一一_2由于f(x)Ca,b,且1n10min1f(k(Jmnx1f(Jnk0所以(a,b)使1n1f

17、()f(k)nk0那么(1)(1)式成为：E：fWh2f()12又由于f(x)0,所以E：fbah2f()0.12b即用复化梯形公式计算积分If(x)dx所得结果比准确值大.a其几何意义：曲线f(x)在定义域内是向下凹的,即曲线在曲线上任两点连线的下方.f(x)f)f(abab)(x)22)(xT)2,a,b.bf(x)dxaabf(三)dxbaf()(x(baa)f(baa)f(-b2b2-f(af()248.如果fxb)a(Xab-)dxab.)dx2(ba)3.0,证实用复化梯形公式计算积分I()(xf()(xab.2.)dx2ab.2.)dx2Xdx所得结果比准确值大,并Tnk02f(

18、xki)2f(a)if(Xk)f(b)i9 9.对3fxdx构造一个至少具有三次代数精度的求积公式.0点 0,1,2,3,0,1,2,3,那么插值型求积公式为：30f(x)dxA0f(x0)Af(x1)AbnxxIAkdxai0 xxoX35xoo.iissgTiogggsxiAoiodagoso972Ao.74i555747i46,o.6957o969o284.因此所求的两点 GaussGauss 求积公式:i.3o429o3o972f(o.ii5587io9995)o.6957o969o284f(O.741555747146).或依下面的思想:(2)在0,1上构造权函数(ix)=一次正父多

19、项式g2(x)%xgo(x)=igi(x)=x-2g/x)=x35令2(X)(XX0)(xXi)63-x735xo1523035,xii523035代入:A0Ax.A0AiAiAxiixii,x0 xdxdxi8:30i8i8.30i8i8且fi5i82,30)35i8.30fi52.30i83535i3.i3.分别用三点和四点 Gauss-ChebyshevGauss-Chebyshev 求积公式计算积分Ho解：(i)用三点(n2)Gauss-ChebyshevGauss-Chebyshev 求积公式来计算:此时,f(x).2一(6)945印覆(2x)ii2N4dx,并估计误ix22n(k0,i,L,n),x0cos6,3y,xi3cos一60,x25cos一6、32由公式可得:2xk3(22320：)4.368939556i97(2)用四点(n3)Gauss-ChebyshevGauss-Chebyshev 求积公式来计算:15(8)945143于此时,f