Is a c 1 nujxjt jyj ou2 nujxjt jyj1 ou c nujxjt

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: jyj { Golub and van Loan (1996): C = 1:01 if nu < 0:01 We have jfl(xT y) ; xT yj jfl(xT y) ; xT yj jxT yj jxjT jyj C nu jxT yj (5) { The relative error may not be small since jxT yj can be much less than jxjT jyj 6 Backward Error Analysis Prior analysis is called forward error analysis { Relate rounding errors to the solution Backward error analysis: { Relate rounding errors to the original data Consider the dot-product computation { Write (4a) as ^^ fl(xT y) = xT y (6a) ^^ x, y are perturbed values of x, y The rounded dot product fl(xT y) is the exact dot ^^ product xT y of perturbed vectors { Using (4a) 2p x1 6 x2p1 + 1+ ^6 x=6 4 p .. xn 1 + 3 1 7 27 7 5 2p y1 1 + 6 y2p1 + ^6 y=6 4 p .. yn 1 + n { Let x, y be perturbations, i.e., ^ x=x+ x 7 ^ y=y+ y 3 1 7 27 7 (6b) 5 n Backward Error Analysis Observe that Thus, p 1+ k 2 = 1 + k =2 + O( k ) 2 3 2 x ( =2 + O( )) 6 x1( 1=2 + O( 12)) 7 6 7 2 x=6 2 2 . 7 . 4 5 2 xn( n=2 + O( n)) But j kj nu + O(u2), so j xj nu jxj + O(u2) (7a) j yj nu jyj + O(u2) (7b) and 2 2 These bounds are very conservative 8...
View Full Document

Ask a homework question - tutors are online