清华大学杨顶辉数值分析第6次作业

9.令*()(21),[0,1]nnTxTxx，试证*{()}nTx是在[0,1]上带权21()xxx的正交多项式，并求****0123(),(),(),()TxTxTxTx.证明：11**20011**20112121**201()()()(21)(21)2111()()()()()211()221()()11()11()()()()()1nmnmnmnmnmnnmnmxTxTxdxTxTxdxxxtxxTxTxdxTtTtdtttTtTtdttTxxxTxTxdxTtTtt令，则由切比雪夫多项式关于权函数的正交性，可得1101=020mndtmnmn所以*{()}nTx是在[0,1]上带权21()xxx的正交多项式*00*11*2222*33233()(21)1()(21)21()(21)2(21)188()(21)4(21)3(21)3248181TxTxTxTxxTxTxxxxTxTxxxxxx14.已知实验数据如下：ix1925313844iy19.032.349.073.397.8用最小二乘法求形如2yabx的经验公式，并求均方误差解：法方程为22222(1,)(1,1)(1,)(,)(,1)(,)ayxbxyxxx即55327271.453277277699369321.5ab解得0.9725790.050035ab拟合公式为20.9725790.050035yx均方误差24220[]0.015023iiiyabx21.给出()lnfxx的函数表如下：x0.40.50.60.7lnx-0.916291-0.693147-0.510826-0.356675用拉格朗日插值求ln0.54的近似值并估计误差（计算取1n及2n）解：1n时，取010.5,0.6xx由拉格朗日插值定理有1100.60.50.6931470.5108260.50.(60.60.51.82321)01.()6047()52jjjxxxLxfxlx所以1ln0.54(0.54)0.620219L误差为ln0.54(0.620219)=0.0040322n时，取0120.4,0.5,0.6xxx由拉格朗日插值定理有2022(0.5)(0.6)(0.4)(0.6)(0.4)(0.5)0.9162910.6931470.510826(0.40.5)(0.40.6)(0.(50.4)(0.50.6)(0.60.4)(0.60.4)2.0411504.0684752)().217097()jjjLxxxxxxxxxfxlx所以2ln0.54(0.54)0.615320L误差为4ln0.54(0.615320)8.6629941023.建立三次样条插值函数()sx，并求(0)f的近似值(0)s，这里已给函数表。ix-0.3-0.10.10.3()ifx-0.20431-0.089930.110070.39569边界条件''(0.3)''(0.3)0ss解：由剖分节点可知0121212110.2,,22hhh101221236[,,]6.4215,6[,,]6.4215dfxxxdfxxx得到方程组12126.4215216.421522MM解得122.5686MM注意到030MM得到三次样条插值函数()sx322322.14051.926451.0642150.000632,[0.3,0.1]1.28430.002773,[0.1,0.1]2.14051.926450.9357850.000632,[0.1,0.()3]xxxxxxxxxxxsx29.确定下列求积公式的待定参数，使其代数精度尽量高，并指出所构造出的求积公式所具有的代数精度：（1）()()(0)()hhfxdxAfhBfCfh解：取2()1,,fxxx，有21hhhdxABC0hhxdxAhCh322223hhhxdxAhCh解以上方程，得14,33AChBh求积公式为141()()(0)()333hhfxdxhfhhfhfh取3()fxx，30,()(0)()0hhxdxAfhBfCfh取4()fxx，4544542112,()05333()(0)()hhhhxdxhhhBhhhxdxAfhBfCfh所以因此构造的求积公式代数精度为330.证明求积公式11133()[5()8(0)5()]955fxdxfff的代数精度为5证明：取()ifxx则对0i任意的，有1111+11-1110,11(1)()|=2111iiiiifxdxxdxxiiii为奇数，为偶数101121()133133133[5()8(0)5()]=[5()5()]=[-5()5()]=095595595513310[5()8(0)5()]=[585]=2955913313322[5()8(0)5()]=[55]=9559553134[5()895ifxfffffffifffxdxifffxdxif为奇数时，为奇函数时，时，时，14116131992(0)5()]=[55]=59252551331272766[5()8(0)5()]=[55]=955912512525ffxdxifffxdx时，从而有6268()0,05,()0725175iExiEx所以求积公式的代数精度为533.求1212,,,xxAA,使公式1112201()()()fxAfxAfxx为高斯型求积公式解：构造区间[0,1]上关于权函数1x的二次正交多项式22()Pxxaxb由2()Px与1x和正交，可得00220011220()()253Pxdxxaxbdxabxx002200112220()()753xPxdxxxaxbdxabxx联立以上方程，可得63,735ab所以2263()735Pxxx求2()Px的零点即为高斯求积公式的节点121(326/5)0.115587110071(326/5)0.74155574717xx1112100120001111211210001115()()()11.304290310361115()()()10.695709690336xxAxlxdxlxdxdxxxxxxxAxlxdxlxdxdxxxxx