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1、第二章2.当x1,1,2时，f(x)数值分析0,3,4，求f(x)的二次插值多项式。解:x1,x11,x22,f(x)0,f(x，3,f(x2)4;l0(x)(xx1)(xx2)1(x1)(x2)(x0X)(x。x2)2八)，/、(xx0)(xx2)1,l1(x)-2(x1)(x2)1(X%)3x2)6x)J-x0)(xx1)、1(x1)(x1)(x2x0)(x2X)3则二次拉格朗日插值多项式为2L2(x)yJk(x)k0) o31)2411-2(X 1(X4 一 3)1 (X16.设Xj,j0,1,L,n为互异节点，求证：n(1)x：(x)xk(k0,1,L,n);j0n(Xjx)klj(x
2、)0(k0,1,L,n);j0证明(1)令f(x)xkn若插值节点为xj,j0,1,L,n,则函数f(x)的n次插值多项式为Ln(x)x：(x)。j0插值余项为Rn(x)f(x) Ln(x)f(n1)()(n 1)!n 1(x)又Qkn,f(n1)()0R(x)0nxklj(x)xk(k0,1,L,n);j0n(xjx)klj(x)j0nn(C：xj(x)ki)lj(x)j0i0nn_ikiiCk(x)(xjlj(x)i0j0又Q0in由上题结论可知nxklj(x)xij0n原式Ck(x)kixii0(xx)k0得证。7设f(x)C2a,b且f(a)f(b)0,求证:maxf(x)axb8(b
3、a)2maxf(x).axb解：令x0a,x1b,以此为插值节点，则线性插值多项式为x为xx0Li(x)f(x)一2f(x1)0x0x1xx0xbxa=f(a)f(b)abxa又Qf(a)f(b)0L1(x)01-插值余项为R(x)f(x)L1(x)3f(x)(xx0)(xx1)1.f(x)2f(x)(x%)(xx1)又Q(x%)(xXi)2(X 4(X1 4(bXo)Xo)2a)2(XiX)max a x bf(x)8(ba)2 max f (x)7 a x b 8.在 4 x4上给出f(x) ex的等距节点函数表，若用二次插值求ex的近似值，要使截断误差不超过10 6,问使用函数表的步长
4、h应取多少?4yn (E 1)4yn解：若插值节点为Xi1,Xi和Xi1,则分段二次插值多项式的插值余项为-,、1R(x)f()(xXi1)(xx)(xXi1)3!R2(X)1/、(-(X Xi i)(X6x)(x Xi i) max f (x)设步长为h,即 Xi 1Xih,X 1 X hR2(x)1e4 -2-h363,3 e4h3. 27若截断误差不超过106,R2(x)10634,36eh1027h0.0065.9.若 yn2n，求 4yn及4yn.，解：根据向前差分算子和中心差分算子的定义进行求解。nyn24(j04(j04(j01)j1)j1)j(21)4VnE4jyny，24yn
5、yn2n1E 2)4yn14yn(E21(E2)4(E1)4%4VnVn2n16.f(x)3x1,求F20,21,L,27及F20,21,L,28解:Qf(x)x43x1右xi2i,i0,1,L,8Xo,Xi,Lf(n)()n!19xo,xi,L,x7x0,x1，L,%f()7!f(8)()8!次数不7!一17!高于4次的多项式P(x),使它满足P(0)P(0)0,P(1)P(1)0,P(2)解法一：利用埃米尔特插值可得到次数不高于4的多项式x0,xi1Vo0,y11m00m111%(x)yjj(x)mjj(x)j0j00(x)(12)(土)2X0X1X0X12(12x)(x1)21(x)(1
6、2)(上)2X1X0X1X02(32x)x22o(x)X(X1)21(x)(X1)x3 c 2x 2x_22H3(x)(32x)x(x1)x设 P(x) H3(x) A(x x0)2(xX1)2其中，A为待定常数Q P(2) 1P(x)X3 2x2 Ax2(x 1)2从而P(x)1221X2(X3)24解法二：采用牛顿插值，作均差表:Xif(Xi)一阶均差二阶均差00111210-1/2p(x)p(Xo)(xXo)fXo,Xi(xXo)(XXi)fXo,Xi,X2(ABx)(xx(x1)(又由p(0)p(x)所以0,p(1)2X1，得x0)(xx1)(xX2)1/2)(ABx)x(x1)(x2
7、)301一，B,44(X3)2.第四章1.确定下列求积公式中的特定参数，使其代数精度尽量高，并指明所构造出的求积公式所具有的代数精度：hhf(x)dxAif(h)AofOAf(h)；2h2hf(x)dxAif(h)Af(0)Af(h)；11f(x)dxf(1)2f(xi)3f(x2)/3;h20f(x)dxhf(0)f(h)/2ah2f(0)f(h);解：求解求积公式的代数精度时，应根据代数精度的定义，即求积公式对于次数不超过式均能准确地成立，但对于m+1次多项式就不准确成立，进行验证性求解。h(1)若hf(x)dxA1f(h)Aof(0)Af(h)令f(x)1,则2hA1A0Am的多项令f(
8、x)x,则0A1hAh“o2.oo令f(x)x,则一hhA1hA3A03h1从而解得A-h3A11h3hh令f(x)x,则hf(x)dxhxdx0h故hf(x)dxA1f(h)A0f(0)Af(h)hh42Gf(x)dxxdx-hhh5,25A1f(h)A3f(0)A1f(h)-h3A1f(h)A0f(0)A/(h)成立。令f(x)x4故此时,f(x)dxAf(h)A0f(0)Af(h)f(x)dxA1f(h)A0f(0)A1f(h)h具有3次代数精度。2h(2)若f(x)dxA1f(h)A0f(0)A1f(h)2h令f(x)1,贝U4hA1A0A令f(x)x,则0A1hAh令f(x)x2,则
9、Th3h2Alh2A飞3h8从而解得A-h3A18h332h2h3令f(x)x,则f(x)dxxdx02h2h2h故2hf(x)dxA1f(h)A0f(0)Af(h)成立。/2h2h/64;令f(x)x,则f(x)dxxdxh2h2h5A#(h)A0f(0)Af(h)16h532h故此时，2hx)dxA1f(h)A0f(0)Af(h)A1f(h)A0f(0)Af(h)02h因此，f(x)dxA1f(h)A0f(0)Af(h)2h具有3次代数精度。11f(x)dxf(1)2f(x1)3f%)/31令f(x)1,则f(x)dx2f(1)2f(x1)3f(x2)/3令f(x)x,则012x13x2人
10、222令f(x)x,则212x13x2从而解得x1x20.2899或x10.68990.5266x20.1266o11o令f(x)x3,则f(x)dxx3dx0f(1)2f(x1)3f(x2H/30111故1f(x)dxf(1)2f(x1)3f(x2)/3不成立。因此，原求积公式具有2次代数精度。(4)若f(x)dxhf(0)f(h)/2ah2f(0)f(h)h令f(x)1,则0f(x)dxh,令f(x)x,则2hf(0)f(h)/2ahf(0)f(h)hhh12f(x)dxxdxh002_2_,12hf(0)f(h)/2ah2f(0)f(h)h22令f(x)x2,则hh213f(x)dxxd
11、xh0032_132hf(0)f(h)/2ah2f(0)f(h)h32ah22故有13132h-h2ah321a一12令f(x)x3,则f(x)dxh314xdxh0412hf(0)f(h)/2/0)141414f(h)hhh244令f(x)x4,则hh415f(x)dxxdxh00512-hf(0)f(h)/2hf(0)12f(h)1h521h531h56故此时,h0f(x)dxhf(0)f(h)/h因此，of(x)dxhf(0)具有3次代数精度。12,2h2f(0)f(h),12,12,f(h)/2h2f(0)f(h)127。若用复化梯形公式计算积分1Iexdx,问区间0,1应多少等分才能使截断误差不超过0106?解：ban米用复化梯形公式时，余项为R(f)ah2f(),(a,b)121又QIoexdx故f(x)ex,f(x)ex,a0,b1.Rn(f)h2f()h21212若|Rnf|106,则当对区间0,1进行等分时，h1,n故有nJe-因此，将区间476等分时可以满足误差要求12第五章2.用改进的欧拉方法解初值问题yxy,0x1;y(0)1
