武漢理工大學(xué)數(shù)值分析考試試題

1、武漢理工大學(xué)考試試題紙考試試題(A卷)課程名稱(chēng)數(shù)值分析專(zhuān)業(yè)班級(jí)信息專(zhuān)業(yè)題號(hào)一二三四五六七八九十總分題分10101010101010101010100備注：學(xué)生不得在試題紙上答題(含填空題、選擇題等客觀題)1、已知 f (T) = 2, f (1) = 1, f (2) =1,求 f (x)的 Lagrange 插值多項(xiàng)式。2、已知列表函數(shù)y = f (x):x1234y0-5-63試求滿(mǎn)足上述插值條件的3次Newton插值多項(xiàng)式px)3、已知函數(shù)y = f (x)的數(shù)值表x0123y121764試分別求出f (x)的三次Newton向前和向后插值公式；并分別計(jì)算x = 0.5和x = 2.5時(shí)

2、，f (x)的近似值。.(1 1)4、設(shè)A =計(jì)算A的各種范數(shù)。(-3 3 )_. (23 ) 一 5、計(jì)算矩陣A = 1 2 3000J的條件數(shù)Cond(A)1.10 x - x - 2x = 7.26、 分別寫(xiě)出方程組-x1 +10 x2 -2x3 = 8.3的Jacobi迭代格式和Gauss-Seidel迭代格式。x x + 5 x = 4.2 TOC o 1-5 h z 1237、 用 Newton 迭代法求方程 f (x) = x3 + 2x 5 = 0 的根，要求 I x 1 x 1 10-6.8、試確定求積公式F f (x)dx電0 f (0) + A1 f (h) + A2f(

3、0)使其具有盡可能高的代數(shù)精度。9、設(shè)I =1Sinxdx，利用復(fù)化梯形公式T計(jì)算I的近似值，要使I - T 1x 10-3 , n應(yīng)取多少?0 xnn 210、確定x , x , A , A，使下面公式成為Gauss型求積公式L)dx r A f (x ) + A f (x ) 12120 x1122武漢理工大學(xué)教務(wù)處試題標(biāo)準(zhǔn)答案及評(píng)分標(biāo)準(zhǔn)用紙 課程名稱(chēng) 高等數(shù)學(xué)（下）（A卷）(x - x )(x - x )(x - 1)(x - 2) TOC o 1-5 h z 1、設(shè)x =-L x = L x = 2, y = 2, y = L y = L 則012012l0(x) = (x -x)(x

4、 -x ) = (-1-1)(-1-2) = 6(x2 -3x + 2)， 0102l (x) = (x-x0)(x-x2)= (x + 1)(x - 2) =- 1(x2 -x + 2),1(x1 - x0)( x1 - x2)(1+1)(1- 2)2(x) = (x-x0)(x-x) = (x + 1)(x-1) = 1(-1)尤一(x2 -x0)(x2 -x1) (2 +1)(2-1) f -.故所求插值多項(xiàng)式為p (x) = y l (x) + yl (x) + y l (x) = 1(x2 - 3x + 8).20 01 12 26則所求3次Newton插值多項(xiàng)式為p (x) = f

5、 (x ) + f x , x (x - x ) + f x , x , x (x - x )(x - x )3001001201+ f x ,x ,x ,x (x - x )(x - x )(x - x )0123012=0 - 5( x -1) + 2( x - 1)(x - 2) + 1x (x - 1)(x - 2)( x - 3)t At(t-1)人t(t 1)(t 2)p (x) = y + Ay + 險(xiǎn) y +A3 y30 1! 02!03!0=1 + - X1 + t X14 + t(t -1)(t - 2) X181!2!3!=1 - 2t 2 + 3t 3x = 0.5 =

6、 x0 + th, h = 1 n t = 0.5， f (0.5)注 p3(0.5) = 0.875.所求三次Newton向后插值公式為tt (t + 1)t (t + 1)(t + 2) TOC o 1-5 h z p (x) = y + W + c, v2y +v3y331!32!33!3=64 + - x 47 +x 32 + *(t +1)(t + 2) x 181!2!3!=64 - 69t + 25t 2 + 3t 3x = 2.5 = x3 + th, h = 1 n t = -0.5，f (2.5) p3(2.5) = 35.375.|A| = max1 + |-3|, 1

7、+ 3 = 4;4、=max1 +1, |-3| + 3 = 6 ；8A =12 +12 + -32 + 3212 =何二25;At AE(1(10 -8At A 人I =10 人-8-3 3)-810 人I-8 10 J=(X-10)2 -82 =X2 -20X + 36 = (X- 18)(X-2)得ATA的兩個(gè)特征值% =18,=：x = 18 = 3* .2m=maxX , X = max18,2 = 18 ,5、(23 、A = I2 3.0001)，IAII=max|2| +12|, |3| +13.0001 = 6.0001 ；A-1A-1 =|A|1 (3.0001 -2_ 1

8、=0.00013.0001 -3、( 30001-20000=max|30001| + |-20000|, |-30000| + |20000 = 50001 ；1Cond(A)1 = |A| A-11 = 6.0001 x 50001 = 300011.00016、從方程組0.5）中分離出氣,，% :據(jù)此建立Jacobi迭代公式x = 0.1 x + 0.2 x + 0.72x = 0.1x + 0.2 x + 0.83x = 0.2 x + 0.2 x + 0.84 312x( k+1) = 0.1x( k) + 0.2 x( k) + 0.72x; k+1) = 0.1x( k) + 0

9、.2 x；k)+ 0.83己 k+1) = 0.2 x( k) + 0知)+ 0.84r x(k+1) = 0.1 x(k)+ 0.2 x(k)+ 0.72及 Gauss-Seidel 迭代公式 x(k+1) = 0.1 x(k+1)+ 0.2x(k) + 0.832 1 3x (k+1) = 0.2 x (k+1) + 0.2 x (k+1) + 0.84317、f (x) = x3 + 2x 5 = 0 f(x) = 3x3 + 2，據(jù)此建立 Newton 迭代公式f (x )x . = x -f-k=xk-3000020000 )x3+2x 5, k=0,1, 3x2 +2k取金=1.5

10、迭代結(jié)果列于下表中。kxx 一 xkkk-101.511.342857140.15714321.328384140.01447331.328268860.00011541.328268867.261620 x 10-8由表結(jié)果知君=1.32826886是x *的滿(mǎn)足條件的近似值8、這里有三個(gè)待定常數(shù)A0, A1, A2，將f (x) = 1, x, x2代入，得 TOC o 1-5 h z h = A + A ,=A h + A , = A h 2,0121231解得A = -h, A =如,A = h2 .于是jh f (x)dx a h4f (0) + 2f (h) + hf (0).03

11、132606直接驗(yàn)證，當(dāng)f (x) = x3時(shí)，(2.6)的左邊=1 h4，右邊=1 h4.故求積公式的最高代數(shù)精度d = 2 .43一、sin x9、因?yàn)?f (x) = j 1cos(xt)dt，所以x0f (k) (x)=(cos(xt)dt = j1 tk cos(xt + !)dt,0dxk02f (k) (x) j1 tk0cos( xt +k兀、i， )dt 2j1 tkdt =01k+1(1) a = 0，b = 1,要使J滿(mǎn)足誤差要求，由式(4.2)，只需|氏(f, J)| = | f - T =- 1021 f (門(mén))112n 211 55.55556，亦即n 7.4535

12、6，故應(yīng)取n = 8.貝步長(zhǎng)h = ba = 1，相應(yīng)地取9個(gè)節(jié)點(diǎn)，見(jiàn)表 n 8xf (x)xf (x)01.00000005/80.93615561/80.99739786/80.90885162/80.98961587/80.87719253/80.976726710.84147094/80.9588510用復(fù)化梯形公式得T8 =我1 + 0.8414709 + 2 x (0.9973978 + 0.9896158+ 0.9767267 + 0.9588510 + 0.9361556+ 0.9088516 + 0.8771925) = 0.9456909 .10、因?yàn)閮牲c(diǎn)Gauss型求積公

13、式具有2n-1 = 2x2-1 = 3次代數(shù)精度，所以當(dāng)f=1, x, x2, x3時(shí)，上述兩點(diǎn)Gauss型求積公式應(yīng)準(zhǔn)確成立，由此得:2yx 10=2=2 f1 2 x3 233021 _ 2-x5 25=502 .1 2-x 72770A + A ,12=A X2 + A X2,=Ax3 + A x3,1122=Ax + A x ,1122_3 2x17 73 2x+27 7A-1 + -13A=1-二2315566解得 解法二因?yàn)樯鲜鰞牲c(diǎn)Gauss型求積公式的Gauss點(diǎn)氣,氣是0, 1上以P=E為權(quán)函數(shù)的某2次正交多項(xiàng)式七的零點(diǎn)，不妨設(shè)七=（x-氣心-x ）.于是IL 1 、/ cJx x 1x (x 一 x )(x 一 x ) d x = 0J1 1 Xq(*r )(尤一尤)d* = 0 j x x X (x x )(x x )U x u1,、1(x + x ) 一 x x =,3121 251，、11(x+x ) 一一xx