2.5 Iterative Improvement of a Solution to Linear Equations 55 x+δx b+ob 83 旦 granted for 11800 (including this one) /Cambridge from NUMERICAL RECIPES IN 19881992 Figure 2.5.1.Iterative improvement of the solution to A.x b.The first guess x +6x is multiplied by A to produce b+5b.The known vector b is subtracted,giving b.The linear set with this right-hand (Nort server side is inverted,giving &x.This is subtracted from the first guess giving an improved solution x. America computer, users to make one paper UnN电.t THE ART 2.5 Iterative Improvement of a Solution to Linear Equations 9 ictly proh Progra Obviously it is not easy to obtain greater precision for the solution of a linear set than the precision of your computer's floating-point word.Unfortunately,for to dir large sets of linear equations,it is not always easy to obtain precision equal to,or even comparable to,the computer's limit.In direct methods of solution,roundoff OF SCIENTIFIC COMPUTING(ISBN errors accumulate,and they are magnified to the extent that your matrix is close 1988-1992 to singular.You can easily lose two or three significant figures for matrices which (you thought)were far from singular. 10-621 If this happens to you,there is a neat trick to restore the full machine precision, 43106 called iterative improvement of the solution.The theory is very straightforward(see Numerical Recipes Figure 2.5.1):Suppose that a vector x is the exact solution of the linear set (outside A·x=b (2.5.1) Software. You don't,however,know x.You only know some slightly wrong solution x +ox, where &x is the unknownerror.When multiplied by the matrix A,your slightly wrong visit website solution gives a product slightly discrepant from the desired right-hand side b,namely machine A·(x+6x)=b+b (2.5.2) Subtracting (2.5.1)from (2.5.2)gives A·6x=6b (2.5.3)2.5 Iterative Improvement of a Solution to Linear Equations 55 Permission is granted for internet users to make one paper copy for their own personal use. Further reproduction, or any copyin Copyright (C) 1988-1992 by Cambridge University Press. Programs Copyright (C) 1988-1992 by Numerical Recipes Software. Sample page from NUMERICAL RECIPES IN C: THE ART OF SCIENTIFIC COMPUTING (ISBN 0-521-43108-5) g of machine￾readable files (including this one) to any server computer, is strictly prohibited. To order Numerical Recipes books or CDROMs, visit website http://www.nr.com or call 1-800-872-7423 (North America only), or send email to directcustserv@cambridge.org (outside North America). A A−1 δx x + δx x b b + δb δb Figure 2.5.1. Iterative improvement of the solution to A · x = b. The first guess x + δx is multiplied by A to produce b + δb. The known vector b is subtracted, giving δb. The linear set with this right-hand side is inverted, giving δx. This is subtracted from the first guess giving an improved solution x. 2.5 Iterative Improvement of a Solution to Linear Equations Obviously it is not easy to obtain greater precision for the solution of a linear set than the precision of your computer’s floating-point word. Unfortunately, for large sets of linear equations, it is not always easy to obtain precision equal to, or even comparable to, the computer’s limit. In direct methods of solution, roundoff errors accumulate, and they are magnified to the extent that your matrix is close to singular. You can easily lose two or three significant figures for matrices which (you thought) were far from singular. If this happens to you, there is a neat trick to restore the full machine precision, called iterative improvement of the solution. The theory is very straightforward (see Figure 2.5.1): Suppose that a vector x is the exact solution of the linear set A · x = b (2.5.1) You don’t, however, know x. You only know some slightly wrong solution x + δx, where δx is the unknown error. When multiplied by the matrix A, your slightly wrong solution gives a product slightly discrepant from the desired right-hand side b, namely A · (x + δx) = b + δb (2.5.2) Subtracting (2.5.1) from (2.5.2) gives A · δx = δb (2.5.3)