數(shù)值積分與數(shù)值微分習(xí)題課

1、數(shù)值積分與數(shù)值微分習(xí)題課（共21頁(yè)）-本頁(yè)僅作為文檔封面，使用時(shí)請(qǐng)直接刪除即可-內(nèi)頁(yè)可以根據(jù)需求調(diào)整合適字體及大小-數(shù)值積分與數(shù)值微分習(xí)題課26一、已知Xo1,Xi1,X2給出以這3個(gè)點(diǎn)為求積424節(jié)點(diǎn)在0.1上的插值型求積公式解：過(guò)這3個(gè)點(diǎn)的插值多項(xiàng)式基函數(shù)為XX1XX2l02XX0X1X0X2XX0XX2li2XXiX0XiX2XX0XXil22XX2X0X2X11A01k2XdX,k0,1,2x0x1x0x2xx0xx2(x1x0x1乂2xx0xKx2x0x2x1A0101dx01xx1xx213xx241113424413xx441113242411xx42313144421dxdx0

2、1dx01dx0故所求的插值型求積公式為xdx2f:、確定求積公式1.1-fxdx5f.0.68f05f.0.619的代數(shù)精度，它是Gauss公式嗎證明：求積公式中系數(shù)與節(jié)點(diǎn)全部給定，直接檢驗(yàn)2345,依次取fx1,x,x,x,x,x,有11dx11xdx115181519x 0.615怎80591o12ox2dx5x0.680251901x3dx151靛38035例192114a一x4dx-50.68045、0.651901x5dx15J06580551 9本題已經(jīng)達(dá)到2n-1=5。故它是Gauss公式、試應(yīng)用復(fù)合梯形公式計(jì)算積分21dx12x要求誤差不超過(guò)103,并把計(jì)算結(jié)果與準(zhǔn)確值比較。

3、解：復(fù)合梯形公式的余項(xiàng)為b._ba2.Rfaf(X)dXTnb a2nn1f(a)f(b)2f(Xk)k1baXkakh,h,k0,1,2,，nn本題fx1x3,M2maxfx1,2本題余項(xiàng)為)于是有100.1要使Rb-a2-1n=一IT1021021.1檢驗(yàn):2ln2h2maxf(x)12x1,20.109545,21.221.9x1丫12取h0.10.3468863.12111104103四、證明若函數(shù)fxC1a,b,則其上的一階差商函數(shù)是連續(xù)函數(shù)，并借助此結(jié)果用Newtong插值余項(xiàng)證明梯形求積公式的余項(xiàng)為bba12x0Rffxdxafafba2證明：不妨設(shè)一階差商函數(shù)為fx,a,fx0

4、hfalimfx0h,alimh0h0x0hafx0fhfalimf x0f af x0,ax0 ah0x0hafx0fafhlimh0x0hax0ha由x的任意性，可知一階差商函數(shù)是連續(xù)函數(shù)由插值特點(diǎn)，顯然有bbRffxL1xdxfxN1xdxaa線性插值的Newton余項(xiàng)公式為fxN1xfx,a,bxaxb故有bR1ffx,a,bxaxbdxa由fxh,afa,blimfxh,a,blimlim f x h,a f a,b h 0x bh0h0xhbfx,afa,bfx,a,bxb可知fx,a,b是變量x在a,b上的連續(xù)函數(shù)，而函數(shù)xaxb定理，存在bfxa在a,b上可積，不變號(hào)，根據(jù)積分

5、中值a,b由差商性質(zhì),N1x存在Ni3fdxfdx,a,bbxaxbdxa,a,bf-o所以xbdx12結(jié)論得證。五、導(dǎo)出中矩形公式babfxdxbafa2的余項(xiàng)。解：將fx在x?處進(jìn)行泰勒展開(kāi)fxfa2bfa2bab1xfx22a,bo對(duì)上式兩邊在a，b上積分，有bfxdxabfabdxbfaabab,1xdx-bfa2xabdxdx1bf2a2abxdx2中矩形公式的余項(xiàng)bfxdxbfadxbf aabab.fxdx0;222dxfbab2_fxdx2a22ft3f3ba24f24a,b六、設(shè)數(shù)值求積公式f(x)dxAkf(xk)k1代數(shù)精度至少為n-1的充分必要條件是它為插值型求積公式.

6、證:充分性.設(shè)原式是插值型求積公式,則式中的求積系數(shù)bAkalkn(x)dxlkn(x)f(xk) dxk k 1nInAkf(xk)k1bnlkn(x)dxf(Xk)bLn(x)dxa余項(xiàng)為由知代數(shù)精度至少為bf(，)(an!n-1)n(x)dx必要性.設(shè)原式代數(shù)精度至少為n-1,則對(duì)次數(shù)不超過(guò)n-1的多項(xiàng)式Pr(x)(rn1)原式成立等號(hào)，特別地取Lagrange插值基函數(shù)lkn(x)，有nlkn(X)dX aAjlkn(Xj),k1,2，nji因?yàn)?,ik,lkn(Xi)所以j，0,jk.bAlkn(X)dXa故原式為插值型求積公式七、令P(x慮n次實(shí)多項(xiàng)式，滿足bk.aP(x)xdx0

7、,k0,n1.證明P(x*E開(kāi)區(qū)間(a,b)中有n個(gè)實(shí)單根.b證明：因?yàn)镻(x)dx0所以P(x底a,b上至少有一a個(gè)零點(diǎn)。若P(xXk(1)個(gè)零點(diǎn)x-i=1,2；k在a,b上，則有P(x)(xxi)(xx2)(x4)g(x)Qkxg(x)g(x)0,或g(x)0Qk(x)(xxi)(xx2)(xxk)k二Qk(x)akxkak1xk1a1xa0aix,(kn1)i0及bP(x)xkdx0,k0,1，,n1所以a,bP(x)Qk(x)dx akP(x)a, xi dxi 0k ba, & P(x)xdx 00若零點(diǎn)個(gè)數(shù)kn1,有b2g(x)Qk(x)dx 0abP(x)Qk(x)dxa矛盾，因

8、此kn,即P(x)在a,b至少有n個(gè)零點(diǎn)，但P(x)是n次實(shí)多項(xiàng)式，故k=n。八、已知點(diǎn)(a,f(a),f(a)和(b,f(b),f(b)，用該信息計(jì)b算定積分f(x)dx。a解：記山為f(x)關(guān)于節(jié)點(diǎn)a,b的Hermite插值多項(xiàng)H3(x)h0(x)f(a)h1(x)f(b)g0(x)f(a)g1(x)f(b)bh1(x)dx f (b) abbbf(x)dxH3(x)dxh0(x)dxf(a)bbg0(x)dxf(a)g1(x)dxf(b)aabho(x)dx abh1(x)dx ax ax b2 b aa b22 b a dx 2dxbgo(x)dxaa2dxbgi(x)dx a所以有2x a .dxb ab a22 a122b a122bbabaaf(x)dx2f(a)f(b)-2f(a)f(b)誤差為-4-4bf4()22f4()5R(f)，xaxbdxbaa4720九、驗(yàn)證Gauss型求積公式eXf(x)dxAof(xo)Af(x)o求積系數(shù)及節(jié)點(diǎn)分別為A 2 1 A 2 1-A0好A玄心2行