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A Simplified Method of Recognizing Zero Among Elementary Constants

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A Simplified Method of Recognizing Zero Among Elementary Constants A simplified method of recognizing zero among elementary constants Daniel Richardson, Department of Mathematics, University of Bath, Bath BA2, England. [email protected]. maths Abstract unique non singular solution in N, (r). The question of how such a proof can be given will be discussed later. In ISSAC ’94, a method was given for deciding whether or A basic problem about the elementary numbers is how not an elementary constant, given as a polynomial image of to decide, given a description, as above, of an elementary a solution of a system of exponential polynomial equations, number, whether or not the number is zero. This is called represents the famous object zero. the element ary constant problem [see Richardson, 1992]. In this article the technique is considerably simplified The solution given below is an improved version of the and speeded up. The main improvement has been to in- solution given in the 1994 ISSAC [Richardson and Fitch]. tegrate the numerical and symbolic computations in such a Both solutions rely upon not stumbling over a counterex- way that unnecessary branches of the symbolic computation ample to the following conjecture. are avoided. Schanuel’s Conjecture If ZI, ..., Zk are ~ comPlex num- bers which are linearly independent over the rationals, then 1 The elementary numbers the transcendence rank of An exponential system is a system of equations (S = O,E = zk,l,l, ezk}ezk} O), where S is a finite set of polynomials in QIx1, Y1, ..., xn> Yn], {.2,,..., and E is a subset of {ul —ezl, . ,Y~ —e“ } is at least k We will write (S., E~ ) for an exponential system with r polynomials and k exponential terms. 2 Description of the algorithm Definition 1 An elementary point is a point, a, in C2n which is a non-singular isolated solution of an exponential Suppose we are given an elementary point a as the non sin- system (S,, Ek) = O. gular solution of an exponential system (s., Ek) = O in a neighbourhood N~ (r) in C2n. We are also given a polyno- To say that the point a is a non singular solution of mial q, and we wish to decide whether or not q(a) = O. (S,, Ek) = O means that the number of equations, r + k is It must be the case that (S,, E~) = O is 2n equations in the same as the number of variables, 2n, and the Jacobian 2n unknowns. We will assume also that our approximation mat rix is non singular at the point. to a is good enough so that the Newton sequence, or some variant of the Newton sequence converges to a. This implies Definition 2 An elementary number is a complex number that if q(a) # 0, we will eventually discover this fact. of the form p(z I,yI, . ,zn, yn), w~e~e P is a polynomial Suppose q(a) = O. If this is true algebraically, as a con- with rational coefficients, and (XI, YI, . ., xn, yn ) is an ele- sequence of S, = O and membership of N6 (r), we can verify mentary point. this by using Wu’s method on (S,, q) = O to get ascending set A~ = O, and then verifying that (A,, Ek ) = O has a so- If a is a point in C2n and 6 is a positive real number, lution, /3, in NJ (r). Since /5’then also satisfies the defining IVJ(a) will be the set of points in C2n which have distance conditions for a, we must have ~ = a, and q(a) = q(/3) = O. no more than d from a. It might happen that q(a) = O, but that this identity A rational point in C ‘n is a point all of whose coordi- depends in some way on Ek = O. The Schanuel conjecture nates have real and imaginary parts which are rational. An then implies that there is some linear relationship with inte- elementary point can be specified by giving a rational point gral coefficients among the z variables which are arguments r, and a neighbourhood, N,(r) in C2n, and an exponential to the exponential function. This linear relationship can be system (S = O,E = O), and a proof that this system has a discovered by use of the LLL algorithm. Once the linear relationship has been discovered, it can be used to eliminate Permission to copy without fee all or part of this material is granted some of the exponential terms from the definition of a, thus provided that the copies are not made or distributed for direct com- simplifying the problem. mercial advantages, the ACM copyright notice and the title of the publication and its date appear, and notice is given that copying is The central claim is therefore that any true identity must by permission of the Association for Computing Machinery. To copy either be a consequence of the algebraic part of the defini- otherwise, or to republish, requires a fee and/or specific permission. tion or must result from a linear relationship among the ISSAC’95 7/95 Montreal, Canada arguments to the exponential function. @1995 ACM 0-89791-699-9/95/0007 $3.50 104 The parts of the algorithm are discussed in more detail for k in Q, if k >0 and it4 = (ci,j (x)) and C$i,j(e, a) is calcu- below. lated, as above, by the crude interval arithmetic procedure applied to c~,j(z), and if, for all j, 2.1 Approximation In the following we are dealing with various computable real and complex numbers. Each of these numbers will have an associated definition, which allows us to approximate it. At Using this, we define a lazy Newton test. each stage it will be supposed that we have a current working precision, pm, and that we have calculated all the numbers Definition 4 Procedure LNTest(cro, (f%, Ek )) at issue to within tolerance 10-p”n. If z and y are such This returns FAIL if r+k # 2n or ifJ(x) = Jacobian(S,, Eh ) approximate reals, we will write and J(x) is not provably non zero at QO, using the crude in- terval arithmetic estimate. Otherwise, let J = 2 II J-l(cro) [1[(Sr, Eh)(~o)[ to mean that Z* + 10–P”” < V* – 10-P”” where z* and y* are the current approximations to x and y. ~. ~P~~ (VX ● Na(~o))(l/ J(z) J-l(ao) – I It< 1/2, then If >P?n z < y, then z < y is a consequence of the current return SUCCESS and N6 (~o). approximation. Of course it may happen that the current Else return FAIL. precision does not distinguish between x and y although they are not equal; it also may happen that x and y are Theorem 1 If the LNTest succeeds, then the lazy Newton really equal although we do not know this. sequence converges to a solution of (s., Ek ) = o an NJ (~o) Also, there is only one such solution. Also the solution is non singular. 2.2 Interval arithmetic procedure Supposepis in Q[zl, eX1,yl, x2, eX2,y2, . .,xn, ez”, yn]. Let Proof Let F(z) = (S., Eh ) (z), and let J(s) be the Jacobian DP be the set of expressions obtained by formal first par- matrix of F(x). Define tial derivatives of p, and let D; be obtained from DP by replacing every negative coefficient by its absolute value ai+l = ai – J–l(ao)F(cti) in every expression in D=. Suppose we are given a point Cr=(al, bl, ..., an, b~) in C2m and a positive real e. Define Suppose, for the moment, that a~ and a~+l are in Na (ao), a“=([al [+e, lbll+ e,..., la~l + c, lb~l + e). Let J(e, cr) = Let S1be the straight line from a, to ai+l. .41azirnum(q(a” ) : g G D;). Then we have F(ai+l) = F(cri) + J(x)dx JSt Using this notation, Let x = (1 – t)ai + tcri+l. Then dx = –J-l(ao)F(ai)dt Definition 3 p(z) is provably non zero in N,(a), using the 1 crude interval arithmetic estimate, ‘if ~p~~ l~(e, a) ] < [P(a) I F(ai+l)= (1- J(z) J-l(ao))F(cw)dt / o Otherwise we will say that p(z) is not provably non zero, (using the crude interval arithmetic estimate). so p(cri+l)l s lF(@)l/2. We got this result with the assumption that ai and cu+l were both in NA(ao). Then, if ao, al,... ,~h are all in 2.3 Lazy Newton sequence NJ (ao), we have Let (ST, .Eh) be an exponential system with r + k = 2n. Let J(z) be the Jacobian of (S,, ~h ). The lazy Newton lF(~k)l < lF(cro)l/2k sequence, starting at aO in C2” is defined recursively by It follows that lCYh+l – Ck’kl < 6/2k+l. By induction, the a,+, = a, – J-l(ao)(Sr, E~)(ai) sequence remains in the neighbourhood and converges to a solution of F(x) = O. A similar argument shows that there can be only one 2.4 Lazy Newton Test solution in the neighbourhood, and that J(z) is non singular If A4 is a square k by k matrix of complex numbers, let in the neighbourhood. R [1 M II be the maximum of the lengths of the columns of M, using the usual notion of length in complex k space, i.e. 2.5 Triangulate procedure the square root of the dot product of the column with its complex conjugate.
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