数值分析第3章习题答案

本章习题中有几道题不会做，待再复习31fC[ab]，写出三种常用范数||f||1,|f||2,|f||b||f||1|f(x)a||f||2 f(x)2a||f||max|f(x)a2、fgC[ab]，它们的内积是什么？如何判断函数族{01,…n}C[ab]在[ab]上线解：fgC[ab]b(f,g)fa, (1,2 (1,n)(, (, (,G ndetG0 (, (, (, n3、什么是函数fC[a,b]在区[a,b]上的n次最佳一致近多项式设pn(x)为最佳近函数，则fC[a,b]在区[a,b]上的n次最佳一致近多项||f(x)p*(x)||min||f(x)p(x)n||f(x)p*(x)||min{max|f(x)p(x) 4f在[a,b]n次最佳平方近多项式？什么是数据fmi设pn(x)为最佳近函数，则fC[a,b]在区[a,b]上的n次最佳平方近多项||f(x)p*(x)||2min||f(x)p(x)nb||f(x)p*(x)||2min{f(x)p(x)}2na设n(x是ab]an0n(x为[a,b]序列n 0 0j(j,k)(x)j(x)k(x)dxAjk 则称多项式序列 为在[a,b]上带权(x)正交，称(x)为在[a,b]的带权 当区间为-1,1](x1，由1,xx2xn正交化得到的多项式称为勒让德多项 d Pn(x) (x (2n)! mP(x)P(x)dx m P(x)(1)mP (n1)Pn1(x)(2n1)xPn(x)nPn1(x),n1,4）Pn(x在区间[-1,1]上具有n当区间为[-1,1]，权函数(x) ，由1,x,x2,...xn正交化得到的多项式称为切1Tn(xcos(narccosx，若零xcos()，则Tn(x)cos(nTn1(x)2xTn(x)Tn1(x),n1,T0(x)1,T1(x)1T(x)T n dx nm 1 nmT2n(x)xT2n1xxTn(x在区间[-1,1]上具有nxcos2j1,j1,2, T(x)的首项xn的系数为2n1,n1, nn(k(x),j(x))aj(f(x),k(x)),k0,1,...,j 1/ 1/(n1) 1/ 1/ 1/(n2)H 1/(n1)1/(n ...1/(2nf(xC[ab]nPn(xHn，使|f(xPn(x|为任P*(x)H是f(x)在[a,b]上的最佳一致近多项式，则limP*(x)f(x) nx[ab（3）f(x)C[a,b]在[a,b]上的最佳平方近多项式Pn(x)Hn则limPn(x)f(x)~Pn(x1的勒让德多项式，Qn(xHn1的多项 则[P(x)]2dxQ2x)dx 1~Tn(x)是[-1,1]上首项系数为1的切比多项式。Qn(x)Hn是任一首项系数 T(x)maxQ(x) N2pDFTFFTFFTfxP* min||fxPx minmax|fxPnxPHnamfP*2min[fxPx |min[fxiPnxi]||min[fxiPnxi]p pm|min[fxPx] [P(x)]2dxQ2 1正确。书P62正确，书1、f(x) x，给出0,1上的伯恩斯坦多项式B( ,及B(,。解：B(f,x)f()P(x)，P(x) xk(1n kk 1B1(f,x)f(k)1kk 所以f(0)1x01x)10f(1)1x1(10 03B3(f,x)f(k)33kk f(0)3x0(1x)30f(13x1(1x)2f(23x2(1x)1f(1)3x3(1 0 3 32 3x1x2 注意：伯恩斯坦多项式在0,1上有效，如果区间超出，则应进行区间不变。那么常系 2、当f(xx时，求证Bn(fx) kn B(f,x)f()P(x)f( x(1 nkk k f(0)0n (n k (n1)(kn x(1 (1k!(nk)!xk1(n k0k!(n13、证明函数1,xx2xn证明：假设1,x,x2,....,xn线性相关。则存在一组系数a,a,a,..., x，均a0a1xa2x2....axnnaxj有方程 jj xn x0G n xn n有detG(xixj0a0a1a2an0.i,i所以1,x,x2 ,xn线性无。4、计算下列函数f(x)关于C[0,1]的||f||||f||1和|f||21）f(x)(x2）f(x)(x123）f(x)xm(1x)n，mn||f||max|f(x|f(xx1)3在C[0,1 af(x3(x1)20x1，即在C[0,1]f(xf(xx1)3是单调函数。因此|max|f(x)|出现在端点a或b处。ax1，||f||max|f(x)|max|(x1)3| x0，||f||max|f(x)|max|(01)3 即||f||max|f(x)|max|(x1)3 ||f||1|f(x)|dx|(x1)3| |(1x)3dx| (1 ||f||2(f(x)dx)2((x1)dx) ((x1)6dx)2 (x ||f||max|f(x)|max|x1|max(max|x1|,max|x1|) a 0 0x0 205 0 ||f||1|f(x)|dx|x0.5|dx(0.5x)dx(x 50.5(0.5x)20.50.5(x0.5)21 ||f||2(f2(x)dx)(|x1|2 ((x1)2dx)2 (x 要点1：f(x)xm(1在C[0,1]xm(1x)nf(xxm(1在C[0,1]xm与(1x)n在区间C[0,1]上的交点。要点2：||f||1|f(x)|dx|x(1x)| 1xm(1x)n01(1)n01(1)n 1|(1)nmm||f||2(f(x)b 0a1((xm(1x)n)2dx)051(x2m(1x)2ndx)05 (1)2nx2m11)0 2m (1)2n)02m (1)2n2m1(1 2m 2m5、证明||fg||||f||||g由于各范式收敛性相同，可以选择适合证明的范式来做，本题可以选择-范式||fg||max|f(x)g(x)|max||f(x)||g(x)max|f(x)|max|g(x)|||f||||g6、对f(x),g(xC1[ab]bab2），(f,g)f(x)g(x)dxfa（1）(f,g)(g,f（2）(af,g)a(f,（3）(fv,g)(f,g)(v,（4）(f,f)0 (f,g)f(x)g(x)dxg(x)f(x)dx(g,f (af,g)af(x)g(x)dxaf(x)g(x)dxa(f, b(fv,g)(f(x)ab(f(x)g(x)a f(x)g(x)dx (f,g)(v,b(f,f)f(x)2abyf(x)2abyf(x)2abf(x)2abf()2a其中(ba0,f()20f(0,a,b ff()dx0dx0, b(f,g)fab2）(f,g)f(x)g(x)dxfa (f,g)f(x)g(x)dxf(a)g(a)g(x)f(x)dxg(a)f(a)(g,f (af,g)af(x)g(x)dxaf(a)g(a)a(f(x)g(x)dxf(a)g(a))a(f, (fv,g)(f,g)(v,b(fv,g)(f(x)v(x))g(x)dx(f(a)a (f(x)g(x)dxv(x)g(x)dxf(a)g(a) (f(x)g(x)dxf(a)g(a)v(x)g(x)dx (f,g)(v,b(f,f)f(x)2dxf(a)2ab其中f(x)2dx0,f(0,a,ba ff()dx0dx0, b(f,f)f(x)2dxfa0bfgf(x)g(x)dxf(a)g(aa7、令T*(x)T(2x1),x[0,1]，试证T*(x)是在[0,1]上带权(x) 的正 x多项式，并求T*(x)T*(x)T*(x)T*(x T T T本题考查带权多项式的求法 表示切比多项式。考查 ， 的区别1由T(x)T(2x1),x0,1可知，令x t1,1， T1t)T(t),tn( 8(x1x2，区间[-1,1],1的正交多项式(x)n利用正交化求解n1((x)xn,(x)xn ((x), j 0(x)1 x(x)x x ((x)x2 ((x)x2,2(x)x (x, x2(1x2 x2(1x2x2 x23 ((x)x3, ((x)x3,x2)3(x)x (x, x (x2,x2 x3(1x2 x3(1x2 x3(1x2 x3 x (x2 (x2)2 x3651(0(x),1(x))xdx((x),(x))1x1dx1x1x122 ((x),(x))1x6xdx1x3 1 ((x),(x))1x6xdx1x3 1 2((x),(x))1x6xdx1x1 1 5 5 53((x),(x))1(x1)(x6x)dx1x1x1 1 4 9、试证明由（2.16）式给出的第二类切 多项式族un(x)是[-1,1]上的带(x 1x210、证明对每一个切比多项式Tn(x)，1T dx 1 11、用Tx)f(xex在区间[-1,1]3xcos2k1,k1,2, xcos1 xcos3 xcos5 L(x)(xx1)(xx2)ex0(xx0)(xx2)ex1(xx0)(xx1) (xx)(xx (xx)(xx (xx)(xx x(x 3 (x 3)(x 3 (x 3 e3 e 3 xe3(3x23x)3x21e3 2 x2 1.7813(3x1.732x)+0.5614( )3x2=8.6246x22.5990xf(n1)( Rn(f)(n1)!wn1(x)6(xx0)(xx1)(xx2 )(x x(x2 yx(x21在区间[-1,1])3y(x31x)3x210 x13所以max|y||x(x21|2x) Rn(f)627818181)最佳平方近多项式，若取span(1,x,x2)，那么最佳平方近多项式是什么当span(1,x 1 1 1dx1，xdx ，0xdx3,0x3x2dx3226 x(x23x2)dx 11 可知1 1 2a6 9 4 6 0 4 12a11,b所以g(x) 4x,x当span(1,xx2 1 1 1 1dx1，xdx ，xdx ,xdx ,xdx x23x2dx1 2 ，x(x23x2)dx 11 1x2(x23x013 4 3 1 23 3 6 1 9 b 4c 4 1 197 5 23 2 6 60 23 1 3 610 45 1 3 1 180 a2,b3,c13、求f(x)x3在区间[-1,1]上关于(x)1的最佳平方近二次多项span(1,x,x2 1dx2，1xdx0，1xdx3,xdx0,xdx5，xdx 有 2 3 0 2 0b 5 2c 0 5 0 2 5 0 5 3 2 50 0 45 14、求函数f(x)在指定区间上对span(1,x)的最佳平方近多项f(x)1,[1,xf(x)exf(x)cos（1 3 311dx21xdx41xdx31xdxln3 4 ln 26 b 3 2 a11ln36,b3(ln31)1 其它的类似求解，注意区间的变化，其中0xlnxdx、0xcosxdx以用分部积分 1 1 01dx1，0xdx2，0xdx3,0xdxlnx，0x(x3x2)dx41115、f(x) x，在区间[-1,1]上按勒让德多项式展开求三次最佳平方近多项2span{1,x,(3x21)/2,(5x33x)/1 1xdx(3x21)2/4dx1(9x416x2 41 x5x2x24 5(5x33x)2/4dx1(25x69x230x4 41 x73x36x5) 7 1sin2xdxcos2x 2 1xsin2xdx1xdcos212(xcos 1cos 1cos2xd(2 sin 1 (3x21)/2 xdx(3x21)d 11((3x21)cos 1cosxd(3x2 11cosxd(3x2 31xcos 21 xd 2 16xsin 1sin 1 (5x33x)/2 xdx(5x33x)d 11((5x33x)cos 1cosxd(5x3 11cosxd(5x3 11(15x23)cos 1(151x2cosxdx31cos 11(151x2cosxdx6sin 2(151x2dsinx) 130(x2sin 21xsinxdx) 60 11xdx1(3x21)/2dx1(5x33x)/2dx1x(3x21)/2dx1x(5x33x)/2dx 1(3x1)(5x3x)/4dx a b c d 72 2 2 7有a0b3c0,d p(x)abx (3x21)(5x23x),，即为所求。使 f(x)abx有span(1,x,x261(xi),1(xi)161(xi),2(xi)xi6(x),(x)(x),(x)x2 6(x),(x)x3