數(shù)值積分與數(shù)值微分習題課

1、數(shù)值積分與數(shù)值微分習題課一、已知x01,x11,x23，給出以這3個點為求積節(jié)424點在0.1上的插值型求積公式解：過這3個點的插值多項式基函數(shù)為l02xx1xx1x0l12xx0 xx0 x1l22xx0 xx0 x21x2x0 x2x2x1x2x1x2x1Ak0lk2xdx,k0,1,2A1xx1xx2dx00 xxxx20101x2011423x4342dx3x13xA11xxxxx10 x12dx0 x0 x2A1xx0 xx1dx20 xxxx02214011241x40314443412121dx32dx3故所求的插值型求積公式為fxdx2f11f12f310343234二、確立求

2、積公式115f0.68f05f0.6fxdx19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)點所有給定，直接檢驗挨次取fx1,x,x2,x3,x4,x5，有2115181511dx1901150.68050.6xdx192302501x2dx11x3dx11x4dx11x5dx15551590.62025280.60.63035380.60.64045480.60.65055580.6本題已經(jīng)達到2n-1=5。故它是Gauss公式。三、試應用復合梯形公式計算積分21Idx12x要求偏差不超出103，并把計算結果與正確值比較。解：復合梯形公式的余項為Rf,Tbf(x)dxTbah2

3、f()nan12ban1Tnf(a)f(b)2f(xk)2nk1xakh,hba,k0,1,2,L,nkn本題fx本題余項為Rf,Tn要使Rf,Tn=b-a2-1n0.1h1，fx13,M2maxfx12xxx1,221h2f()h2maxf(x)h21212x1,212h2，得h0.109545，取h0.110312于是有IT1011111.12102221.121.20.346886421.9檢驗：IT101ln2T103.121111041032四、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)，并借助此結果用Newtong插值余項證明梯形求積公式的余項為Rffxdxbafaf

4、bba3fb1a212證明：不如設一階差商函數(shù)為fx,a，x0a,b，有l(wèi)imfx0h,afx0hfalimx0hah0h0fx0fhfalimx0hah0fx0fafhfx0falimx0hax0hax0fx0,ah0a由x0的隨便性，可知一階差商函數(shù)是連續(xù)函數(shù)。由插值特色，明顯有R1bfxL1xdxbffxN1xdxaa線性插值的Newton余項公式為fxN1xfx,a,bxaxb故有R1fbfx,a,bxaxbdxa由limfxh,a,bfxh,afa,blimxhbh0h0limfxh,afa,bfx,afa,bh0 xbxbfx,a,b可知fx,a,b是變量x在a,b上的連續(xù)函數(shù)，而

5、函數(shù)axb在a,b上可積，不變號，依據(jù)積分中值定理，存在a,b，使bafxN1xdxf,a,b由差商性質，存在a,b，使fbxN1xffdxa2b3af12baxbdxxa,a,bf。所以2bxbdxxaa結論得證。五、導出中矩形公式bxdxbabfaf2a的余項。解：將fx在xa2b處進行泰勒睜開fxfabfabxab1fxab222222a,b。對上式兩邊在a,b上積分，有bbabbabab1bab2fdxfxxdxfxdx22dxf2aaa22aabbabab1bxab2baf2f2xdxf2dxa22a中矩形公式的余項RMbfxdxbabaaf2babab1bfxdxfa222aa2b

6、xdx2bQfabxabdx0;a1b2af22ab2x2dx232babf2bat32axdx0223Rfba3,M24fba24a,b六、設數(shù)值求積公式bnaf(x)dxAkf(xk),k1代數(shù)精度最少為n-1的充分必需條件是它為插值型求積公式.證:充分性.設原式是插值型求積公式,則式中的求積系數(shù)bAkalkn(x)dxnnbInAkf(xk)alkn(x)dxf(xk)k1k1bnlkn(x)f(xk)dxbaLn(x)dxk1a余項為RnfIInbf(n)()(x)dxan!n由知代數(shù)精度最少為n-1必需性.設原式代數(shù)精度最少為n-1,則對次數(shù)不超出n-1的多項式pr(x)(rn1)原

7、式成立等號,特別地取Lagrange插值基函數(shù)lkn(x),有bnlkn(x)dxAjlkn(xj),k1,2,L,na1j由于lkn(xj)1,ik,0,jk.所以bAkalkn(x)dx故原式為插值型求積公式.七、令P(x)是n次實多項式,滿足bP(x)xkdx0,k0,L,n1.a證明P(x)在開區(qū)間(a,b)中有n個實單根.bP(x)dx0，所以P(x)在a,b上最少有證明:由于a一個零點。若P(x)有k(1)個零點xi，i=1,2,k在a,b上，則有P(x)(xx1)(xx2)L(xxk)g(x)Qkxg(x)g(x)0,或g(x)0，Qk(x)(xx1)(xx2)L(xxk)kQQ

8、k(x)akxkak1xk1La1xa0aixi,(kn1)i0及bP(x)xkdx0,k0,1,L,n1，所以abbkkbP(x)aixidxP(x)Qk(x)dxaaiP(x)xidx0ai0i0a若零點個數(shù)kn1，有bb2(x)dx0P(x)Q(x)dxg(x)Qakak矛盾，所以kn，即P(x)在a,b最少有n個零點,但P(x)是n次實多項式，故k=n。八、已知點(a,f(a),f(a)和(b,f(b),f(b)，用該信息計算b定積分af(x)dx。解：記H3(x)為f(x)關于節(jié)點a,b的Hermite插值多項式：H3(x)h0(x)f(a)h1(x)f(b)g0(x)f(a)g1(

9、x)f(b)bbbbf(x)dxH3(x)dxh0(x)dxf(a)h1(x)dxf(b)aaaabf(a)bf(b)g0(x)dxg1(x)dxaabh0(x)dxbbxaxb2ba12dxabaab2bxbxa2baah1(x)dxa12dxbaba222bbxbbaag0(x)dxaxadxab1222bbxabaag1(x)dxaxbdxba12所以有bbaf(a)ba2f(b)f(a)f(b)f(x)dxa212偏差為bf4()2x2f4()b5R(f)xabdxaa4720九、驗證Gauss型求積公式exf(x)dxA0f(x0)A1f(x1)0求積系數(shù)及節(jié)點分別為2121A0A1x022x12222，22,，。解：由于