数值分析答案

习题二2-1已知y=f(x)的数值如下：(1)x0123y2312147(2)x-2-101y154524求Lagrange插值多项式并写出截断误差。解：(1))())()(())()(()())()(())()(()(131210132003020103213xfxxxxxxxxxxxxxfxxxxxxxxxxxxxL)())()(())()(()())()(())()((32313032102321202310xfxxxxxxxxxxxxxfxxxxxxxxxxxx147)25)(15(5)2)(1(12)52)(12(2)5)(1(3)51)(21()5)(2(2)5)(2)(1()5)(2)(1(xxxxxxxxxxxx223xxx50),()5)(2)(1(241)()4(3fxxxxxR(2))())()(())()(()())()(())()(()(131210132003020103213xfxxxxxxxxxxxxxfxxxxxxxxxxxxxL)())()(())()(()())()(())()((32313032102321202310xfxxxxxxxxxxxxxfxxxxxxxxxxxx24)11)(21()1)(2(5)1(2)1)(1)(2(4)11)(1)(21()1()2(15)12)(2)(12()1()1(xxxxxxxxxxxx59923xxx12),()1)(2(241)()4(23xfxxxR2-2已知函数lnx的如下数据x8101214y2.079442.302592.484912.63906试分别用Lagrange线性插值和二次插值计算ln(11.85)的近似值，并估计它的截断误差。解：线性插值公式：)()()(101001011xfxxxxxfxxxxxL当x=11.85时，47124.248491.210121030259.2121012)85.11(1xxL321103875.1)1285.11)(1085.11(21)(xxR二次插值：)())(())(()())(())(()())(())(()(2120210121012002010212xfxxxxxxxxxfxxxxxxxxxfxxxxxxxxxL47221.263906.224)15.0(85.148491.2)2(2)15.2(85.130259.24215.215.0误差估计：4321098875.1)1485.11)(1285.11)(1085.11(31)(xxR。2-3设nxxx,,,10为任意给定的n+1个互不相同的节点，证明：(1)若f(x)为不高于n次的多项式，则f(x)关于这组节点的n次插值多项式就是它自己。(2)若),,1,0()(nixli是关于这组节点的Lagrange基函数，则有恒等式nkxxlxnikiki,,1,0,)(1nkxxxlnikii,,1,0,0))((1证明：(1))()!1()()()()(1)1(xwnfxPxfxRnnn因为f(x)是n次多项式，所以它的n+1阶导数为零。故f(x)关于这组节点的n次插值多项式就是它自己。(2)取nkxxfk,,1,0,)(，在nxxx,,,10处进行n次拉格朗日插值，则有)()!1()()()()(1)1(0xwnfxxlxRxPxnnnikiinnk由于0)()1(nf，故有nkxxlxnjkjkj,,1,0,)(0。(3)将kjxx)(按二项式展开，得kiikijikikkjxxCxx0)1()(，则njjikjkiiikinjjkiiikjikinjjkjxlxxCxlxxCxlxx00000)()1()()1()()(由上题的结论得：0)1()1(00kiikikkiikiikiCxxxC。2-4已知函数表x0.10.20.40.60.9y0.99500.98010.92110.82530.6216试构造四次Newton插值多项式，计算cos0.47的近似值并估计截断误差。解：自变量函数值一阶差商二阶差商三阶差商四阶差商0.10.99500.20.9801-0.1490.40.9211-0.295-0.48670.60.8253-0.479-0.460.05340.90.6216-0.679-0.40.08570.040375P4(x)=0.9950-0.149(x-0.1)-0.4867(x-0.1)(x-0.2)+0.0534(x-0.1)(x-0.2)(x-0.4)+0.040375(x-0.1)(x-0.2)(x-0.4)(x-0.6)当x=0.47时，P4(x)=0.891662105517.2)9.047.0)(6.047.0)(4.047.0)(2.047.0)(1.047.0(!5)9.0sin()(xR。62102576.3)9.047.0)(6.047.0)(4.047.0)(2.047.0)(1.047.0(!51)(xR2-5在区间[-4,4]上给出f(x)=ex在等距节点下的函数表，若用二次插值求ex的近似值，要使截断误差不超过10-6，问所用函数表的步长应怎样选取？解：在区间[xi-1,xi]上，记shxxi221误差634342112109243)]1)(1([8!3)])())([(!3)(hessshexxxxxxexRiiii210317.1h则用二次插值的步长应：2106585.0h2-6对区间[a,b]作步长为h的剖分，且],[,)(baxMxf，证明：在任意相邻两节点间做线性插值，其误差限为2181)(MhxR。证明：区间上的误差限：],[,82)()(1211iiiiixxxhMxxxxfxR误差限：nihMxRxRi,,1,0,8)(max)(2112-7设999237125)(357xxxxf,计算差商]2,2[10f,]2,,2,2[710f及]2,,2,2[810f.解：自变量函数值一阶差商1-8862-2975-2089]2,2[10f=-2089，1!7)()2,,2,2,2()7(7210ff，0!8)()2,,2,2,2()8(8210ff。2-8设)(xf在],[ba有三阶导数，],[,10baxx，证明：当],[bax)()()()('))(()()()2)(()(120120010100201101xfxxxxxfxxxxxxxfxxxxxxxxf),(),(''')()(61120bafxxxx证明：根据已知条件可得到如下表所示的插值条件：xx0x1yf(x0)f(x1)y’f’(x0)建立差商表：自变量函数值一阶差商二阶差商x0f(x0)x0f(x0)f’(x0)x1f(x1)0101)()(xxxfxf0100101)(')()(xxxfxxxfxf则由newton插值公式可得：)()()(')()())((')()(200100101000xRxxxxxfxxxfxfxxxfxfxf整理得：)()()()()('))(()()()2)(()(120120010100201101xRxfxxxxxfxxxxxxxfxxxxxxxxf其中R(x)由以下计算得到：构造辅助函数：))()(()()()()()()()(21201202xNxfxxxxxtxttNtft)(x有0x,x,1x三个零点，)('x有0x,1,2三个零点，则)('''x至少有一个零点，记作。则)()(!3)(''')(0))()(()()(!3)(''')('''1202120xxxxfxRxNxfxxxxf。2-9用下列函数值表构造不超过3次的插值多项式，并建立误差估计式。x012f(x)129f’(x)3解：建立差商表：自变量函数值一阶差商二阶差商三阶差商01121123229741则由newton插值公式可得：321)1()1(21)(xxxxxxxP。误差估计式：)2()1(!4)()(2)4(xxxfxR。2-10求满足下列条件的Hermite插值多项式xi12yi23y’i1-1解：)]2()1()2)(1()1)(1221(3)2)(1121(2[1122222xxxxxxxxH538223xxx2-11求一个不高于4次的插值多项式P4(x)，使得1)2(,1)1(')1(,0)0(')0(ppppp。解：根据已知条件可得到如下表所示的插值条件：x012P011P’01建立差商表：自变量函数值一阶差商二阶差商三阶差商四阶差商0000011111110-1210-1-0.50.25则由newton插值公式可得：234222225.25.125.0)1()0(25.0)1()0(1)0(1)(xxxxxxxxxP。2-12根据下表建立三次样条插值函数x123f(x)242f’1(x)1-1解：211,211,0]),[],[(32111011xxfxxfg列方程：00212211210mmmm则三次样条插值函数为：11200210120102101)()())(()()(21)()(21)(mxxxxmxxxxyxxxxyxxxxxS=8-16x+13x2-3x3，]2,1[x。22211221221212212)()())(()()(21)()(21)(mxxxxmxxxxyxxxxyxxxxxS=-40+56x-23x2+3x3，]3,2[x。2-13已知y=f(x)的如下数值x01234y-8-701956求三次样条插值函数S(x)，满足边界条件(1)S’(0)=0，S’(4)=48(2)S”(0)=0，S”(4)=24解：用三转角算法计算：(1)212121hhh,211,12]),[],[(32111011xxfxxfg213232hhh,212,39]),[],[(33222122xxfxxfg214343hhh,213,84]),[],[(34333233xxfxxfg列方程组：271236039122210212210212321321mmmmmm则三次样条插值函数为：11200210120102101)()())(()()(21)()(21)(mxxxxmxxxxyxxxxyxxxxxS=x3-8，]1,0[x。22211221221212212)()())(()()(21)()(21)(mxxxxmxxxxyxxxxyxxxxxS=x3-8，]2,1[x。33222232322322323)()())(()()(21)()(21)(mxxx