RD-A172 772 THE CHINESE REMAINDER PROBLEM AND POLYNOMIAL 1/1 INTERPOLATION(U) WISICONSIN UNIV-MADISON MATHEMATICS RESEARCH CENTER I J SCHOENBERG AUG 86 MRC-TSR-2954 I EhiLA NLSSIFIED DAAG29-8 -C-8041 F,'G 1211i U 1 8 1 i
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MRC Technical Summary Report #2954 THE CHINESE REMAINDER PROBLEM AND POLYNOMIAL INTERPOLATION Isaac J. Schoenberg Mathematics Research Center University of Wisconsin-Madison * 610 Walnut Street Madison, Wisconsin 53705 August 1986 (Received August 13, 1986) DTIC OCTS 0 8 1986 B C31 LAU A- Approved for public release Distribution unlimited Sponsored by U. S. Army Research Office P. 0. Box 12211 Research Triangle Park North Carolina 27709 86a 10 7 16
UNIVERSITY OF WISCONSIN-MADISON MATEMATICS RESEARCH CENTER THE CHINESE REMAINDER PROBLEM AND POLYNOMIAL INTERPOLATION Isaac J. Schoenberg Technical Summary Report #2954 August 1986 ABSTRACT The Chinese Remainder Theorem is as follows: Given integers a i (i - 1,2,...,n) and corresponding moduli m i, which are pairwise relatively prime, than the n congruences (1) x ai mod m (i - have a unique solution x mod m, where m - mlm 2 ***' n. Sometimes in the 1950s the late Hungarian-Swedish mathematician Marcel Riesz visited the University of Pennsylvania and told us informally that the above theorem is an analogue of the unique interpolation at n distinct data by a polynomial of degree n - I. It follows that (1) can be solved in two different ways: 1. By an analogue of Lagrange's interpolation formula. 2. By an analogue of Newton's solution by divided differences. This analogy gives sufficient insight to furnish a proof of the theorem that V(mim 2... mn) - V(Ml )... (sn), where 1(m) is Euler's function. ANS (MOS) Subject Classifications: loa10, 41A10 Key Words: Chinese Remainder Theorem, Polynomial Interpolation Work Unit Number 3 (Numerical Analysis and Scientific Computing) Sponsored by the United States Army under Contract No. DAAG29-80-C-0041... uy,
SIGNIFICANCE AND EXPLANATION The Chinese Remainder Theorem is one of the most important results of elementary Number Theory as it was used by Kurt G~del in one of his most fundamental papers in Logic. The paper uses the analogy with the theorem of polynomial interpolation to solve it in two different ways. Nl' Dist 1 The responsibility for the wording and views expressed in this descriptive summary Iles with NRC, and not with the author of this report. P 1 1
THU CHINESE REMAINDER PROBLEM AND POLYNOMIAL INTERPOLATION Isaac J. Schoenberg For given Integers ai (1 a I 5 n) and positive integers mj1 (1 S ± S n) that are pairvise relatively prime, the Chxinese Remainder Problem (abbreviated to C.R.P.) my be stated as follows: 2be Problin. To find an integer x satisfying the congruences x S a I (mod m I), (i - 1,2,...,n). (1) If we have found one solution x then clearly all solutions of (1) belong to a residue class modulo N - m Im 2 **mn Sometimes in the 19501s the late Hungarian-Swedish mathematician Marcel Ries: visited the University of Pennsylvania and told us informally that the C.R.P. (1) can be thought of as an analogue of the interpolation by polynomials: Given real values y I (1 5i n) and distinct real values i *to find a polynomial P(x) of degree n I such that we can solve (2) by Lagrange's formula P(x) - Yi yil(x) (3) where the fundamental functions Li (x) -H n1 x-x* Sponsored by the United States Army under Contract No. D&AG29-0-C-0041.
are such that they satisfy the equations Lj(xj) ' Sij (ij -,.,).(4) Here the Sij, called the Kronecker deltas, are defined by I if i - j 8i 0 if i 0j(5 To solve the C.R.P. suppose that we proceed similarly, letting the integers a i be the analogues of the yi, and defining integers bi such that bi~ =45i (mod i.) (i,j -1,...,n),(6) as the analogues of the functions L i(x). This leads to Theores 1. A solution of the system (1) is given by n x a ib 1(7 Indeed, as the bi satisfy (6), we find from (7) that n n x aib i ai6ij-, aj (mod mji) for all j =1,.n 3xaxple I. To find x satisfying x -=2(mod 5), x E6 (mod 7), x 5 (mod 11).(8) we are to solve (6) which in our case is bi 1 (mod 5), b, 0 (mod 7), b 1 0 (mod 11), b 0 (mod 5), bs 1 (mod 7), b 0O (mod 11), b 0 (mod 5), b 0 (mod 7), b 3 1 (mod 11), from which we obtain that bi=231, b - 330, b 3 1 By (7) we find that all solutions of (8) are given by x 527 (mod 385), where 385 567 11. -2-
The solution (7) of the C.R.P. (1) is essentially the solution as given by G. E. Andrews in [1], and by E. Grosswald in (2), without mentioning the analogy with Lagrange's formula. My colleague Richard Askey tells me that Riesz' remark is well known to computer scientists, but apparently not to mathematicians. Besides recording Riesz' remark, the author's contribution is the following remark: Newton solves the interpolation problem (2) using successive divided differences ci to obtain P(x) = c 1 + c 2 (x - x 1 ) + c 3 (x - x 1 )(x - x 2 ) + *.. + Cn(x - xl)(x- x 2 )... (x - Xn.i), (9) where the coefficients ci are obtained by solving Yl Cl Y2 cl + c 2 (x 2 - x 1 ) Yn C 1 + c2(xn xl) + c 3 (xn " Xl)(Xn " x 2 ) + 000 + Cn (xn - xl)(x n - x 2 )... (xn - Xni) * (10) Applying Newton's idea to the solution of the C.R.P. (1), we consider the mi to be the analogues of the x - xi and seek to determine the integer di (1 1 i j n) from the system of congruences ad I a I (mod m 1 ) 1 + d 2 m 1 _ a 2 (mod m 2 ) d 1 + d 2 m 1 + d 3 m 1 m 2 a 3 (mod m 3 ) (11) d 1 + d 2 m 1 + d 3 m 1 m 2 +.. + dnmlm 2... n- a n (mod m n ) In this way we obtain!.~ -3- -AI
2heoren 2. A solution of the C.R.P. (1) is obtained as follows: We first determine the inteqrs di as solutions of the congruences (11), and then a solution of (1) is given by x - d 1 + d 2 m I + d 3 mim 2 + *'" + dnmlm 2... _ (12) Indeed, notice that by (11), the x given by (12), satisfies all congruences (1): For any k, 1 I k I n, from (12) we get that x = di + d 2 ml + oo, + dkmlm2... ik. I (mod m x ) and therefore, by the k-th congruence (11), 4e have that x - ak (mod mk). 3xmple 2. Let us solve the C.R.P. (8) by the Newton approach. For (8) we have n - 3, a 1 2, a 2 - = 6, a 3 5, m I - 5, m 2-7, m 3-11. As we can always choose d 1 a 1 = 2, the remaining n - 1-2 congruence (11) are 2 + 5<2 6 (mod 7), 2 + 5d 2 + 35d 3 5 (mod 11) The first has the solution d 2-5 and the second now becomes 2 + 25 + 35d 3-5 (mod 11) whose solhv ion is d 3-0 (mod 11). From (12), for n = 3 we obtain that x - 27 is a solution of (8). A consequence of Theorem 1, or of Theorem 2, is the following Corollary I. The Chinese Remainder Problem (1) has always a unique solution x, mod M, where M = mlm 2...mn - Moreover, either of the theorems gives a method of finding this unique solution. Let us keep fixed the n pairwise relatively prime moduli mi'm 2,... n. How many Chinese Residue Problems (1) correspond to them? Evidently their number is M for we may restrict the ai to assume the values of a residue system mod mi, for instance For every choice of the n-tuple (a,a 2,...,an) there corresponds a unique -4-.1ji
solution x of (1) which assmes one of the values x C 1} ),1,...,M- (14-, ). (14) Corollary 2. There is a one-to-one correspondence between the n-tuples (a1,...,an), subject to (13), and the M possible values (14) of x. For if two distinct n-tuples (a,a 2,...,an ) JO (a' a 2..... (15) lead to equal x's: x - x' we would get from (1) that ai = a,' (mod mi), (i - 1,...,n), in contradiction to our assumption (15). Rzale 3. we choose the simplest possible example: Let n = 2, m, = 2, M 2 = 3, hence M = 6. Here, by (13) we may choose a 1 = 0,1 and ir a 2 = 0,1,2. Denoting by x r C.R.Ps to be the solutions of the 6 C.R.Ps. we find these (a) x 1 = 0 (mod 2) (b) x 2-0 (mod 2) (c) x 3 0 (mod 2) x 1 = 0 (mod 3) x 2 1 (mod 3) x 3 2 (mod 3) (d) x 4 1 (mod 2) (e) x 5 " I (mod 2) (f) x 6 1 (mod 2) x 0 (mod 3) x 5 1 (mod 3) x 6 2 (mod 3) 4-6 Their solutions are easily found to be (16) x 1 = 0, x 2 " 4, x 3 = 2, x 4 = 3, x 5 " 1, x 6 5, (17) which indeed form a residue system modulo M - 6. We wish to close our note with an elementary application of the one-toone mapping expressed by Corollary 2. For this we need Corollary 3. In the Chinese Remainder Problem (1) we have (ai,mi) - I for all i - 1,...,n (18) if and only if for the solution x of (1) we have (x,mim 2...m n ) 1. (19) -5- %~.. -. ;.:a
Indeed, by (1) we see that (18) holds iff (x,m I ) - 1 for all i, which is equivalent to (19). As usual we denote by 9(m) the Euler function giving the number of positive numbers < m which are relatively prime to m. The application we had in mind is Corollary 4. r'or the pairwise relatively prime m i we have (P(mlm 2...m) = (ml)(m 2 )... 4(m n ). (20) Because the left side is = number of solutions x of (1) satisfying (19), while the right side gives the number of C.R.Ps. (1) satisfying the conditions (18). Example 4. For the moduli m, = 1 and m 2 = 3 of Example 3 only two C.R.Ps. (e) and (f) satisfy the conditions (18). Also notice that their solutions x 5 = 1 and x 6 = 5 indeed form a reduced residue system mod 6 as they should. Remarks. 1. The second Newton approach is slightly more economical then the first approach: while the first requires to determine the n integers b i (i = 1,2,...,n), the Newton approach requires only to find the n - 1 integers di (i = 2,3,...,n). 2. I owe to Gerald Goodman the reference [3] in which Ulrich Oberst shows that appropriate abstract formulations of the Chinese Remainder Problem can be made the basis of much of Modern Algebra including the main theorems of Galois theory. 4V 3. My colleague Stephen C. Kleene informs me that Kurt G8del uses the solution of the Chinese Remainder Problem (without its name) in his fundamental paper "On formally undecidable propositions of Principia Mathematica and related systems 1" in (4], 145-195, especially Lemma 1 on page 135. See also Footnote i on page 136. I, Rk II' IM e-6-
4. Originally I wrote this note very briefly, even tersely. I owe to the Editor an expanded version of this note which I found very helpful in casting it in the present form. 5. In a sequel to the present paper it will be shown how to apply the Chinese Remainder theorem to obtain indices for moduli which do not admit primitive roots. These indices will be vectors...:: IP U. &-7-7- 4.. 4.'.-.-... ',I. '-".;,-4 - -""-.-'..';',-.'-.-' ':-,
REFERENCES 1. G. E. Andrews, Number Theory, W. B. Saunders Co., Philadelphia, 1971. 2. Emil Grosswald, Topics from the Theory of Numbers, The Macmillan Co., New York, 1966. 3. Ulrich Oberst, Anwendungen des chinesischen Restsatzes, Expositiones Mathematicae, vol. 3 (1985), 97-148. 4. Kurt G6del, Collected Works, volume 1, Oxford University Press, New York, 1986. 5. I. J. Schoenberg, On the theory and practice of indices mod m, to appear. IJS :scr -8-
SECURITY CLASSIFICATION OF THIS PAGE (1rhe Data Bnter E REPORT DOCUMENTATION PAGE READ INSTRUCTIONS BEFORE COMPLETING FORM 1. REPORT NUMBER 2. GOVT ACCESSION NO. 3. RECIPIENT'S CATALOG NUMBER 2954 1- It 9 I 2. 4. TITLE (and Subtitle) S. TYPE OF REPORT & PERIOD COVERED THE CHINESE REMAINDER PROBLEM AND POLYNOMIAL Summary Report - no specific INTERPOLATION reporting period 6. PERFORMING ORG. REPORT NUMBER 7. AUTHOR(e) S. CONTRACT OR GRANT NUMBER() Isaac J. Schoenberg DAAG9-80-C-0041 9. PERFORMING ORGANIZATION NAME AND AOORESS 10. PROGRAM ELEMENT. PROJECT. TASK AREA & WORK UNIT NUMBERS Mathematics Research Center, University of ;ork Unit Number 3-610 Walnut Street Wisconsin Numerical Analysis and Madison, Wisconsin 53705 Scientific Computing " 11. CONTROLLING OFFICE NAME AND ADDRESS 12. REPORT DATE U. S. Army Research Office August 1986 P.O. Box 12211 I3. NUMBER OF PAGES Research Triangle Park, North Carolina 27709 8 14. MONITORING XGENCY NAME A ADDRESS(I1 dilferent from Controlling Office) IS. SEC'URITY CLASS. (of this report) 16. DISTRIBUTION STATEMENT (of thl Report) Approved for public release; distribution unlimited. UNCLASSIFIED 15a. DECL ASSI FICATION/DOWN GRADING SCHEDULE 17. DISTRIBUTION STATEMENT (of the abetract metered In Block 20, It different from Report) IS. SUPPLEMENTARY NOTES -I. IS. KEY WORDS (Continue on reverse side if necessary and identify by block number) Chinese Remainder Theorem Polynomial Interpolation 20. ABSTRACT (Continue an reverse side if ncessar and identify by block number) The Chinese Remainder Theorem is as follows: Given integers a. 1 (i = 1,2,...,n) and corresponding moduli mi., which are pairwise relatively prime, than the n congruences (1) x a. mod m. (i = 1,...,n) 1 1 have a unique solution x mod m, where m = m 2... ran. 4 DD I 'i AN 1473 EDITION OF I NOV Of 1 OBSOLETE UNCLASSIFIED SECURITY CLASSIFICATION OF THIS PAGE (When Date Entered).:% %-'-
20. ABSTRACT - cont'd. Sometimes in the 1950s the late Hungarian-Swedish mathematician Marcel Riesz visited the University of Pennsylvania and told us informally that the above theorem is an analogue of the unique interpolation at n distinct data by a polynomial of degree n - 1. It follows that (1) can be solved in two different ways: 1. By an analogue of Lagrange's interpolation formula. 2. By an analogue of Newton's solution by divided differences. This analogy gives sufficient insight to furnish a proof of the theorem that ( 1 m 2... m = p (ml )v(m 2 )...p (m ), where 'p(m) is Euler's function. n 1 2 n "4
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