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1、 Numerical AnalysisrSchool of Mathematics and Statistics, Xian Jiaotong UniversityrNumerical AnalysisSK11-1 2 = 1.414213562373OO(35 ?Cq . .9k?i .)2 x 1= 1.414,2 x 2= 1.4142,|2 1.414| 1 2 103,|2 1.4142| 1 2 104,1=|2 1.414|21 2 1032 0.35355339 1032=|2 1.4142|21 2 1042 0.35355339 104rNumerical Analysis
2、SK11-6 |x| ? 1XOe(JO(.(1)1 1 + 2x1 x 1 + x,(2)1 + x21 x2 x,(3)1 cos2x x,(4)ln1 1 x2|x|.)(1)1 1 + 2x1 x 1 + x=1 + x (1 x)(1 + 2x) (1 + 2x)(1 + x)=2x2 (1 + 2x)(1 + x).(2)1 + x21 x2 x=2x2x(1 + x2+1 x2).(3)1 cos2x x=2sin2x x.(4)ln1 1 x2|x|= ln|x| 1 +1 x2.rNumerical AnalysisSK11-6 |x| ? 1XOe(JO(.(1)1 1 +
3、 2x1 x 1 + x,(2)1 + x21 x2 x,(3)1 cos2x x,(4)ln1 1 x2|x|.)(1)1 1 + 2x1 x 1 + x=1 + x (1 x)(1 + 2x) (1 + 2x)(1 + x)=2x2 (1 + 2x)(1 + x).(2)1 + x21 x2 x=2x2x(1 + x2+1 x2).(3)1 cos2x x=2sin2x x.(4)ln1 1 x2|x|= ln|x| 1 +1 x2.rNumerical AnalysisSK11-6 |x| ? 1XOe(JO(.(1)1 1 + 2x1 x 1 + x,(2)1 + x21 x2 x,(
4、3)1 cos2x x,(4)ln1 1 x2|x|.)(1)1 1 + 2x1 x 1 + x=1 + x (1 x)(1 + 2x) (1 + 2x)(1 + x)=2x2 (1 + 2x)(1 + x).(2)1 + x21 x2 x=2x2x(1 + x2+1 x2).(3)1 cos2x x=2sin2x x.(4)ln1 1 x2|x|= ln|x| 1 +1 x2.rNumerical AnalysisSK11-6 |x| ? 1XOe(JO(.(1)1 1 + 2x1 x 1 + x,(2)1 + x21 x2 x,(3)1 cos2x x,(4)ln1 1 x2|x|.)(1
5、)1 1 + 2x1 x 1 + x=1 + x (1 x)(1 + 2x) (1 + 2x)(1 + x)=2x2 (1 + 2x)(1 + x).(2)1 + x21 x2 x=2x2x(1 + x2+1 x2).(3)1 cos2x x=2sin2x x.(4)ln1 1 x2|x|= ln|x| 1 +1 x2.rNumerical AnalysisSK11-8 zUe?$g.(1)(x 5)4+ 9(x 5)3+ 7(x 5)2+ 6(x 5) + 4;(2)1 + x +x2 2!+x3 3!+ +xn n!)(1)(x 5)4+ 9(x 5)3+ 7(x 5)2+ 6(x 5)
6、+ 4= (x 5) + 9)(x 5) + 7)(x 5) + 6)(x 5) + 4(2) */, ?n = 4 1 + x +x2 2!+x3 3!+x4 4!=h?x4+ 1?x3+ 1ix2+ 1x + 1.rNumerical AnalysisSK11-9 3ONOy = 2971/71!U;.)y =2971 71!=29 7129 7029 69.29 229 1= 79.54162637.rNumerical AnalysisSK22-3 e A = LU)1)147 258 3610= LU =100 210 321147 036 0012-6 )e|424 21710 4
7、109x1 x2 x3=10 3 7,A = GGT=200 140 221212 042 001?Ax = b A = GGT= GGTx = b. =?Gy = b y = (5,2,1)T GTx = y x = (2,1,1)T.rNumerical AnalysisSK22-3 e A = LU)1)147 258 3610= LU =100 210 321147 036 0012-6 )e|424 21710 4109x1 x2 x3=10 3 7,A = GGT=200 140 221212 042 001?Ax = b A = GGT= GGTx = b. =?Gy = b y
8、 = (5,2,1)T GTx = y x = (2,1,1)T.rNumerical AnalysisSK22-8 OA =547 424 7451!2!A.) kAk1= kAk= max16,10,16 = 16.ATA =90054 0360 54090|I ATA| = (36)2(144) = 0.1= 2= 36,3= 144.kAk2=q max(ATA) =144 = 12.rNumerical AnalysisSK2A1=1 72185 848 581kAk11= kA1k= maxn1472,20 72,14 72o =5 18Cond1(A) = kAk1kA1k1=
9、16 +5 18=40 9Cond(A) = kAkkA1k=40 9kA1k2=q max(A1)TA1=q max(ATA)1=r 1 36=1 6.Cond2(A) = kAk2kA1k2=12 6= 2.rNumerical AnalysisSK22-15 A,B n R n ? c y1Cond(cA) = Cond(A)2Cond(AB) Cond(A)Cond(B)3Cond2(RA) = Cond2(A) = Cond2(AR)4Cond2(A) = max(ATA) min(ATA)5?A Cond2(A) =|max |min.yyy(1)Cond(cA) = kcAkk(
10、cA)1k = |c|kAk|c1|kAk = kAkkA1k = Cond(A).rNumerical AnalysisSK2(2)Cond(AB) = kABkk(AB)1k kAkkBkkA1kkB1k = Cond(A)Cond(B).(3)du(RA)T(RA) = ATRTRA = ATA.1o.kRAk2=q max(RA)T(RA) =q max(ATA) = kAk2.(RA)1T(RA)1= (A1R1)T(RA)1= R(A1)TA1R1.=(RA)1T(RA)1qu(A1)TA1.dq kA, ,rNumerical AnalysisSK2kA1k2=q max(A1)
11、TA1 =rmaxn (RA)1T(RA)1o = kRA1k2.Cond2(RA) = kRAk2k(RA)1k2= kAk2kA1k2= Cond(A).2o.kARk2=q max(AR)T(AR) =q max(RTATAR)=q max(R1ATAR) =q max(ATA) = kAk2.(AR)1T(AR)1= (R1A1)T(AR)1= (A1)TRR1A1= (A1)TA1.,k(AR)1k2=rmaxn (AR)1T(AR)1o =q max(A1)TA1 = kA1k2.rNumerical AnalysisSK2Cond2(AR) = kARk2k(AR)1k2= kA
12、k2kA1k2= Cond(A).(4)kyAB BA A. AB A, =k(AB)x = x.B(AB)x = Bx,=(BA)(Bx) = (Bx),kA1k2=q max(A1)TA1 =q max(AT)1A1=q max(AAT)1=s 1 min(AAT)=s 1 min(ATA)Cond2(A) = kAk2kA1k2=s max(ATA) min(ATA).rNumerical AnalysisSK2(5)Cond2(A)=kAk2kA1k2=q max(ATA) q max(A1)TA1=q max(A2) q max(A2)1=s max(A2) min(A2)=|max
13、 |min.rNumerical AnalysisSK22-16 O3dCdC A = QR)(1) A =111 211 245)(1)3dC11, A 11?a21= 2 “,c =15,s =25?E3dCP12=15250 25150 001rNumerical AnalysisSK2KP12A =515150353524512, P12A 11?a31= 2 “,c =53,s =2 3 ?E3dCP13=5302 3010 23053KA(1)= P13P12A =333 0353506595rNumerical AnalysisSK213, A(1)12?a32= 65“,c = 15,s = 25?EP23=100 0152502515KA(2)= P2
