數(shù)值分析答案

1、習(xí)題二2-1己知y=f(x)的數(shù)值如下：(1)TOC o 1-5 h zx0123y2312147x-2-101y154524求Lagrange插值多項(xiàng)式并寫解：心占簽簽1）+壽君塔心（X-X。）（兀一“）（兀一兀3）/（丫）十（X_Xo）（X_Xi）（X_D）/（t）（七-Xo）（X2一Xj（x2一勺）（“3一人0）（勺一勺）（兀3一兀2）(x_1)(x_2)(x_5)2十xt_2)(x_5)*3十x(x_l)(x_5)乂】？十x(x_l)(x_2)(-1)(-2)(-5)(1-2)(!-5)2(2-1)(2-5)5(5-1)(52)=X3+X2-x+2R(x)=丄A(x-l)(x一2)(x

2、-5)/(g),0vv5厶=(中-專-)+弩一晉“(X。一勺)(-X2)(XO一兀3)U1一Xo)(一X2)(X1一勺)十(X-Xo)(X-“)(X-X3)幾丫)|(X_Xo)(X_)(X_X2)幾丫)(也一勺)(2一X1)(X2一勺)(“3一人0)(勺一習(xí))(兀3一七)2(-1)=(x+l)x(x-l)X15+(x+2)x(xT)x4+(x+2)(x+l)(x-1)x5+(x+2)(x+l)x%24(-2+1)(-2)(-2-1)(-1+2)(-1)(-1-1)21)(1+2)(1+1)2(-1)=x3+9.V2+9.V+5&(x)誌(x+2)(+-DV(),-212己知函數(shù)lnx的如下數(shù)據(jù)

3、X8101214y2.079442.302592.484912.63906試分別用Lagrange線性插值和二次插值計(jì)算ln（ll85）的近似值，并估計(jì)它的截?cái)嗾`差。解：線性插值公式:TOC o 1-5 h z、X-X.一、_解：線性插值公式:厶(X)=/(0)+/(V1)兀0一M一當(dāng)x=11.85時(shí),區(qū)(x)|=亠(11.85一10)(11.85一12)1.3875X103厶(11.85)=x-1210-12x2.30259+x-10厶(11.85)=x-1210-12x2.30259+x-1012-10 x2.48491=2.47124二次插值：5苯迸為5+鋁舲心需禺心=ggpx2.302

4、59+18*一21x2.48491+吩。x2.63906=2.472212x42x(-2)4x2誤差估計(jì)：|/?.(x)|x(11.85-10)(11.85-12)(11.85-14)九3y=oy=oy=o/r=oy=o由上題的結(jié)論得：=(一1)%宀旳-1心0。1=01=02-4已知函數(shù)表X0.10.20.40.60.9y0.99500.98010.92110.82530.6216試構(gòu)造四次Nelon插值多項(xiàng)式，計(jì)算cosO.47的近似值并估計(jì)截?cái)嗾`差。解:自變量函數(shù)值一階差商二階差商三階差商四階差商0.10.99500.20.9801-0.1490.40.9211-0.295-0.48670

5、.60.8253-0.479-0.460.05340.90.6216-0.679-0.40.08570.040375P4(x)=0.9950-0.149(x-0.1)-0.4867(x-0.1)(x-0.2)+0.0534(x-0.1)(x-0.2)(x-0.4)+0.040375(x-0.1)(x-0.2)(x-0.4)(x-0.6)當(dāng)x=0.47時(shí),P4(x)=0.8916|/?2(x)|Smx(0.47-0.1)(0.47-0.2)(0.47-0.4)(0.47-0.6)(0.47-0.9)2.5517x106。R2(x)(047-01)(047-02)(047-04)(047-06)(

6、047-09)S32576xl()Y5在區(qū)間44上給出f(x)=e-在等距節(jié)點(diǎn)下的函數(shù)表，若用二次插值求i的近似值，要使截?cái)嗾`差不超過10問所用函數(shù)表的步長應(yīng)怎樣選?。拷猓涸趨^(qū)間x1.bx1,記x=xx+-S2誤差陶(X)|=尋-x-s(5-l)(5+l)-e4A3/z/0.6585xl0-26對區(qū)間仏b作步長為h的剖分，且|/(x)|M,xwd,b,證明：在任意相鄰兩節(jié)點(diǎn)間做線性插值，其誤差限為R(x)|Zm/F。8證明：區(qū)間上的誤差限：RU(x)|x-|x-x/+1h2,xexi9xi+l28誤差限:Rk(x)|Rd)=U-Ao)x-“)。9用下列函數(shù)值表構(gòu)造不超過3次的插值多項(xiàng)式，并建立

7、誤差估計(jì)式。X012f(x)129f(x)3解：建立差商表:自變量函數(shù)值一階差商二階差商三階差商01121123229741則由newton插值公式可得：P(x)=l+x+2xx(x-l)+x(xI)2=1+x3o誤差估計(jì)式：R(x)=x(x-1)2(x-2)o4!10求滿足下列條件的Hemiite插值多項(xiàng)式X】12V123yi1-1解：/2=i2(l+2)(x-2)2+3(l-2)(x-l)2+(x-l)(x-2)2-(A-l)2U-2)=2.1+8.v+3x+52-11求一個(gè)不高于4次的插值多項(xiàng)式P(x),使得P(0)=p(0)=0,p(l)=pl)=1,p(2)=1o解：根據(jù)已知條件可得

8、到如下表所示的插值條件：X012P011P，01建立差商表:自變量函數(shù)值一階差商二階差商三階差商四階差商0000011111110-1210-1-0.50.25則由newton插值公式可得:P(x)=x(x_0)2_lx(x_0)2(X_l)+0.25x(x_0)2(x_1)2=025J_l5p+2.25/o2-12根據(jù)卜表建立三次樣條插值函數(shù)X123f(x)fl(x)2421-1解：Z7!=2gl=%兀0,H】+“，兀2】)=列方程:-i/H0+2叫+/?2=0=叫列方程:則三次樣條插值函數(shù)為:s!(X)=1+2(x一x0)(x-x1)2y0+l-2(x一xj(x-x0)?”+(x-x0)(

9、x-mQ+(x-x0)2(x-x1z1=8-16x+13x2=8-16x+13x2-3x3,-veL2os2(Aj=1+2(x一也)b7)2Ji+1-2(x-x2)(x一羽)2y2+(x七)(兀一七)+(x-xx)2(x-x2)m2=-40+56x-23x2+3x3,xe2,3。2-13已知y=f(x)的如下數(shù)值X01234y-8-701956求三次樣條插值函數(shù)S(x),滿足邊界條件SXO)=O,S5(4)=48Sn(0)=0,S”(4=24解：用三轉(zhuǎn)角算法計(jì)算：(1)人=-=,“I=t，g=3(幾勺，“+yx2)=12%+幾22兄2=了=“2=廳，=%幾三/I,X2+“2于“2，“3)=39

10、h2+%22心,務(wù)=3陽匹宀+“丿心百)=84列方程組:=3=12列方程組:=3=12=27則三次樣條插值函數(shù)為:5i(x)=14-2(x-xo)(x-x1)2yo+l-2(x-x1)(x-xo)2yi+(x-xo)(x-x1)2/no+(x-xo)2(x-xi)w1=x3-8,xe0,1oS2(x)=14-2(x-)(x-x2)2y!+1-2(x-x2)(x-x1)2,24-(x-x1)(x-x2)2/?1+(x-)2(x-x2)m2=x3-8,xe1,2oS3(x)=1+2(x一x2)(x-七)2y2+1一2(x-屯)b一X2)23+(兀一x2)(x一和IU2+(牙一(兀一xz)nh=x3

11、-8,xe2,3oS4(x)=1+2(X_+1_2(x_x4)x一耳)2兒+(A-.3)(X_心尸竹+(X_(X_些)心=x3-8,xel4o(2)go=3/1心，勺-牛W=3g4=3/n+織：=123210001宀210001宀12-002200丄2丄2200012叫=o12加i=3m2=39m2=1284叫=27“4123ni4=48列方程組：0*2*0Si(-v)=|i+2(A：-x0)(x-x1)2y0+l-2(x-x1)(x-x0)2y1+(x-x0)(x-x1)2m0+(x-x0)2(x-x1)w1=x3-8,xe0,1oS2(x)=14-2(x-)(x-x2)2y!+|1-2(x

12、-x2)(x-x1)2y24-(x-x1)(x-x2)2/?14-(x-)2(x-x2)m2=x3-8,xel,2oS3(x)=1+2(x一x2)(x-.v3)2y2+1-2(x-v3)(x一x2)2y3+(x一x2)x-.v3)2叫+(x一x2)2(x一x3)w3=x3-8,xe2,3oSO=1+2(x-x3)(x-x4)2”+1-2(x-)(x-羽尸y4+(入一-心)訕3+(兀一x3)2(x一x4)m4=x3-8,xe3,4o用三彎矩算法計(jì)算：(1)幾=,“1=t，gi=6/“，勺勺=18TOC o 1-5 h z/?1+h2222=r=斤，“三=牙，S2=6/1“、2，“】=36/?2+

13、/?322113=Tr=09皿g3=6/七川3，心=54/?3+力422丿心心-必fhh4列方程組:6864661356-一一-01234MMMMMooo1-22oO1-221O1-221-20121-2oO21-2ooO01234MMMMMXV=fhh4列方程組:6864661356-一一-01234MMMMMooo1-22oO1-221O1-221-20121-2oO21-2ooO01234MMMMMXV=A/】(2)列方程組:18=36=M=6=122嚴(yán)42M3=18第三章最佳逼近3求下列函數(shù)在指定區(qū)間上得一次最佳平方逼近多項(xiàng)式并估計(jì)平方誤差（1）/（a）=Jx,xe0,1解：設(shè)PiU）

14、=c0+qx(Po4o)(io)pof2解得：co=58/35=1.6571,6=0,c2=-3/7=-0.4286求得擬合線性多項(xiàng)式函數(shù)p：(x)=1.6571-0.4286X26誤差為：滬-兒F2=0先計(jì)算出擬合函塑豈TOC o 1-5 h zXI-2-1012P2-0.05731.22851.65711.2285-0.0573得到：尸=0.22866給出下列數(shù)XI-3-224yi14.38.34.78.322.7用最小二乘法求解：根據(jù)基函數(shù)給出法方程組Ar=11111,y7=14.38.34.78.322.7941416L求得534aty求得534aty583=563法方程組為:534解

15、得：a=3.5b=1.2o3-7確定經(jīng)驗(yàn)公式尸一中的參數(shù)，使之與下列數(shù)據(jù)擬合:l+ax+bxxi0.10.20.30.40.50.6VI0.1720.3230.4840.6901.0001.579解：將經(jīng)驗(yàn)公式轉(zhuǎn)化為：L=+ax+hA=Lx+l+LLyexcccx令乙=丄，6ZO=-,=-9則上式轉(zhuǎn)化為：z=aQx+at+a丄。ycccx上表的的數(shù)據(jù)變?yōu)椋簒i0.10.20.30.40.50.6zi5.8143.0962.0661.4491.0000.633取Po(X)=1、0U)=YX，(p2(x)=X111111這時(shí)111111這時(shí)D1=10510/32.525/30.10.2030.40

16、.50.66DtD=24.5149L3624.5149362.1660.91zT=5.8143.0962.0661.4491.0000.633,14.058Dtz=87.18423.2798解得：a0=-1.9674,山=09761,心=0.5034。則a=l939,b=-3.908,c=1.987o8在某化學(xué)反應(yīng)里，生成物的質(zhì)量濃度訊10屯/皿3)與時(shí)間t(nun)的關(guān)系式為y=a+pt現(xiàn)測得一組數(shù)據(jù)如下:XI12346810121416yi4.006.418.018.799.539.8610.3310.4210.5310.61試確定出參數(shù)Q、Po解：將經(jīng)驗(yàn)公式轉(zhuǎn)化為：卜畔上表的的數(shù)據(jù)變?yōu)?

17、XI12346810121416zi0.250.1560.1250.1140.1050.1010.09680.0960.0950.094取o(x)=l,這時(shí)Dt=ill111111110.50.3330.250.166670.1250.10.083330.071430.0625g1092262.692261.492967zT=0.250.1560.1250.1140.1050.1010.09680.0960.0950.094,7.r1.23281D1z=0.45863解得：a=0.1650,B=0.0789。Xj+2x2=5Xj+2x2=5(1)2+AS=621Tx2=6“+x2=411_4簡

18、化為：Ax=Y兩邊同乘以系數(shù)矩陣的轉(zhuǎn)置矩陣，就得到所需要的法方程組：具體計(jì)算如下:具體計(jì)算如下:5216_20_65解得最小二乘解：Xi=26/lbx2=15/112A+4y=ll241f3x-5y=3s3-5門-3x+2y=612LJ-64x+2y=1442.14簡化為：Ax=Y兩邊同乘以系數(shù)矩陣的轉(zhuǎn)置矩陣，就得到所需要的法方程組：具體計(jì)算如下:2314-54-5223-5111413具體計(jì)算如下:2314-54-5223-511141314解得最小二乘解：x=1450/487=2.9774,y=597/487=1.2259第四章數(shù)值積分與數(shù)值微分1用四節(jié)點(diǎn)復(fù)化梯形公式計(jì)算積分(1)+x2d

19、x,(2)sin(cosx2)dx(1)解：(1)點(diǎn)解：(1)點(diǎn)WW+2x孰護(hù)護(hù)后爭+1.1543(2)7-3/(!+/+2x/(l+f)=0.3517(2)2用四節(jié)點(diǎn)復(fù)化Simpson公式計(jì)算積分(1)a/2-shi2xdx,(2)e2dx解：炸皿皿勒氐-l)+2xt/()=0.7241(2)53=/(1)+/(2)+4x/(1+辛)+2x/(l+i)=0.3407f=lf=l3分別用復(fù)化梯形和復(fù)化Smipson公式計(jì)算積分并使絕對誤差限|xl0-6不超過，問需要將區(qū)間0,1多少等分?解：復(fù)化梯形：所以區(qū)間應(yīng)該409等分。復(fù)化Smipson公式：_(i)：嚴(yán)(帀)7f()28852880f一!一/5.12880/42所以區(qū)間應(yīng)該6等分。44利用枳分fg/x=21n