Assessment 2 knuth book

# In the multivariate case the linear coefficient

• Notes
• 27

This preview shows pages 7–9. Sign up to view the full content.

worth the effort if the Euclidean PRS algorithm were the only alternative. In the multivariate case, the linear coefficient growth is, of course, compounded through each level of the recursion, but even so, the primitive PRS algorithm has demonstrated considerable practical utility. 3.4 THE REDUCED PRS ALGORITHM. Since the bulk of the work in the primitive PRS algorithm is in primitive part computations, we would like to find a way to avoid most of them and still reduce the coefficient growth sharply from that which occurs in the Euclidean PRS algorithm. Surprisingly, we can accomplish this to a significant extent by choosing ~3 = 1, ~i = ai-1, i = 4, ... , k. (13) This is Collins' reduced PRS algorithm, and is justified in [7] and [8] by a proof that is sketched in Section 3.5. In a normal reduced PRS, the coefficient growth is essentially linear (see Section 3.5). In an abnormal reduced PRS, the growth can be exponential, but of course not as badly so as in the corresponding Euclidean PRS. In the example (4), which is distinctly abnormal, the reduced PRS is 1, 0, 1, 0, -3, -3, 8, 2, -5 3, 0, 5, 0, -4, -9, 21 -15, 0, 3, 0, -9 585, 1125, -2205 - 18885150, 24907500 527933700. (14) Journal of the Association for Computing Machinery, Vol. 18, No. 4, October 1971

This preview has intentionally blurred sections. Sign up to view the full version.

Euclid's Algorithm and Computation of Polynomial GCD's 485 Notice that the coefficient sizes appear to be doubling at each step until the last. It is easy to show that the small size of F6 results from the fact that ~4 < ~3 • If all of the ~ were the same and greater than 1, then the growth would be uniformly ex- ponential. 3.5 SUBRESULTANTS. Let F~, F2, • • • , Fk be any PRS in 9Ix]. The major result of Collins [7] is that F, ~ Sd~ (Fi, F2) ~ Sd~_~-l(F~, F2), i = 3, ..., k, (15) where ~ denotes similarity (Section 2.3) and where, for 0 _< j < d:, Si(F~, F2) is a polynomial of degree at most j, each of whose coefficients is a determinant of order d~ + d2 - 2j with coefficients of F~ and F~ as its elements. In particular, So(F~, F2) is the classical resultant [5, Ch. 12] of F1 and F2 ; in general, following Collins, we shall call S~ (F~, F~) the jth subresultant of F~ and F2. The constants of similarity for (15) may be expressed as products of powers of a3, - • • , ai, ~3, • • • , ~i,f2, • • • ,f~, and (-1). Brown and Traub [8] rederive these results in somewhat greater generality; their treatment is brief and simple, and shows clearly how the subresultants arise. We shall now establish bounds on the coefficients of the subresultants Let T~ = Sji_,_i(F1 , F:), i = 3, "" , k. (16) 1 ml = ~(dl + d2 -+- 2) -- di-1 = [dl+ d~ -- 2(4i-1 - 1)], i = 3, ..., k. (17) This is an approximate measure of degree loss. As we proceed through the PRS, it increases monotonically from m3 > 1 to mk _< ½ (dl + d2). In a normal PRS, mi in- creases by 1 when i increases by 1; in general, the increment is ~i-~. Since each coefficient of Ti is a determinant of order 2mi with coefficients of F1 and F2 as its elements, our bounds depend simply on m~. If the coefficients of F1 and F2 are integers bounded in magnitude by c, then by Hadamard's theorem [1, p. 375] the coefficients of T~ are bounded in magnitude by (2m~J) m'. (18) Taking the logarithm (to the same base as the base of the number system), we see that the coefficients of Ti are bounded in length by mi[2l -4- log (2m~)], (19) where 1 = log c bounds the lengths of the coefficients of Fa and F~. Although the
This is the end of the preview. Sign up to access the rest of the document.
• Spring '13
• MRR
• Math, Coefficient, F~

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern