《数值分析》课程PPT教学课件（Numerical Analysis）Chapter 6 Iterative Methods for Solving Linear Equations

3-81-41-0(0)2.54003x(1) = B x(0) +f +011-310124r(0)(1r= 4.924=[2.5,3,3]]31018-412.5)(2.5)4330+x(2) =B x(1) +f-11-233104x(2) - x(1)= [2.875, 2.3636,1]= 2.1320

                − − − − = 0 4 1 2 1 11 1 0 11 4 4 1 8 3 0 (1) (0) x B x f = +           + 3 3 2.5            0 0 0 T = [2.5, 3, 3] 4.924 (1) (0) x − x =                 − − − − = 0 4 1 2 1 11 1 0 11 4 4 1 8 3 0 (2) (1) x B x f = +           + 3 3 2.5            3 3 2.5 T = [2.875, 2.3636,1] 2.1320 (2) (1) x − x =

3-81-4102.5)2.875403x(3) =B x(2) + f =2.3636+11-11-21-4310x(3) - x(2)1=0.4127=[3.1364,2.0455, 0.9716]7And so on, the solution of the eguations tomeet the precision is x12.Jacobi. mx4 =3.02411.94780.9205d=0.1573The numberofx53.00031.98401.0010d=0.0914二iterations is 12.x6I2.99382.00001.0038d=0.01753.0000X1x70.00592.99902.00261.0031d=2.0000x8X23.00020.9998d=0.0040=2.0006x93.00031.99990.9997d= 7.3612e-004三1.0000X3)x10 =3.00001.99990.9999d= 2.8918e-004x11 =3.00002.00001.0000d = 1.7669e-004x12 =3.00002.00001.0000d= 3.0647e-005

                − − − − = 0 4 1 2 1 11 1 0 11 4 4 1 8 3 0 (3) (2) x B x f = +           + 3 3 2.5            1 2.3636 2.875 T = [3.1364,2.0455, 0.9716] 0.4127 (3) (2) x − x = And so on, the solution of the equations to meet the precision is x12. The number of iterations is 12. x4 = 3.0241 1.9478 0.9205 d = 0.1573 x5 = 3.0003 1.9840 1.0010 d = 0.0914 x6 = 2.9938 2.0000 1.0038 d = 0.0175 x7 = 2.9990 2.0026 1.0031 d = 0.0059 x8 = 3.0002 2.0006 0.9998 d = 0.0040 x9 = 3.0003 1.9999 0.9997 d = 7.3612e-004 x10 = 3.0000 1.9999 0.9999 d = 2.8918e-004 x11 = 3.0000 2.0000 1.0000 d = 1.7669e-004 x12 = 3.0000 2.0000 1.0000 d = 3.0647e-005 Jacobi. m           =           1.0000 2.0000 3.0000 3 2 1 x x x

Analyzing the Jacobi iteration process,x(+1)=x()+=(b-之ay()a11j=11(k+1)(k)(b, -Za2,x()X2X2a22Jj=1we can find that before solving x(k+), x(+),++),.-(k+1)拉.hasbeensolved.e x(+1) , we also use x(*), x*,.., x()But when we solve3to iterate.Can we use x(++, +),., + to iterate when wesolve x(k+1) ?

Analyzing the Jacobi iteration process, ( ) 1 1 ( ) 1 1 11 ( ) 1 ( 1) 1 = + = + − n j k j j k k b a x a x x ( ) 1 1 ( ) 2 2 22 ( ) 2 ( 1) 2 = + = + − n j k j j k k b a x a x x we can find that before solving , ( 1) k i x + has been solved. ( 1) ( 1) ( 1) 1 2 1 , , , k k k i x x x + + + − But when we solve , we also use ( 1) k i x + ( ) ( ) ( ) 1 2 1 , , , k k k i x x x − to iterate. Can we use to iterate when we ( 1) ( 1) ( 1) 1 2 1 , , , k k k i x x x + + + − ( 1) k i x + solve ?