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 numbers 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 interval 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. Then, for all unit length x in Ck, the This has input: N, O(aO), N~, (~l), (S., Ej), U = (~1,. . .,9s), length of Mx is bounded by II M Il. where U is a list of polynomials. The output is either FAIL If M is a square matrix of polynomials in or a verification that (=a E N.l (al)) ((Sr, Ej) = OA U = O), xl, ezl ,yl, . . ..xn. ezn, y., we can get an upper bound for together with a new definition of a as a solution of (A, Ej ) = II Lf(x) II for x in N. (a) by applying the crude interval O in N., (al), where A is an ascending set, and every poly- arithmetic estimate above to the lengths of the columns of nomial in either S, or U has Wu Pseudoremainder zero with M(x). We will say respect to ascending set .4. The procedure uses the assumption that N,, (al ) C N,, (CYO) +,m (VZ e N.(~))(11 M(x) 1/

105 the larger neighbourhood, we may conclude that it is also If t + j < 2n, there are less equations than variables in the smaller neighborhood. in At = O, E~ = O. There must be a curve through a on This triangulate procedure is an extension of the basic which these are zero. On this curve, we must also have LNTest procedure. (We may have r + s + j > 2n here.) ST = O,E~ = O, U = O. But we supposed the solution a was The procedure depends strongly on the fact that we have unique. So this can not happen. an approximation to a. This approximation can be used to Suppose that t + j = 2n. In this case, if a is a non sin- simplify and speed up construction of the Wu decomposi- gular solution of (At, E3 ) = O, the Newton test will succeed tion. We only need to consider branches of the decomposi- eventually (i.e. for sufficiently small < and sufficiently large tion which could be true at a. pm) and will return N6 (al) C N,, (CIO). The procedure works as follows. Suppose on the other hand that a is a singular solution of (At, E~ ) = O. If the Wu Pseudo remainder of Jg with Definition 5 Procedure respect to At is non zero, triangulate is called recursively, ~iangulate(N,O (CYO),N., (al), (s,, E~ ), U) producing a lower order ascending set, giving the correct If some polynomial in U is provable non zero, using the result. crude interval arithmetic estimate, in N,l (al), return FAIL. The possibility which seems to remain is that Jy has zero Else, Wu pseudo remainder with respect to A,. It is claimed, how- Apply Wu’s method to (S,, U) to get a characteristic set ever, that this can not happen. For the sake of a contradic- A = (al, ,a~). Let II, . . ,It be the initials of A, and let tion, suppose that it does. We have Jy is identically zero on D1, . . . . Dt be the derivatives, with Di equal to the partial the solution set of At = O in N,, (al). On this solution set, derivative of a~ with respect to the leading variable of ai. the gradients of At are linearly independent, since At is as- Let B be the subset of{Il, . . . ,It, Dl, . ,Dt} which are cending and none of the first partial derivatives with respect not provably non zero in N,l (al). If B is non empty, add to the leading variables are zero. Pick a generic point on the B to U, and return the result of recursively applying trian- solution set of At = O in the neighborhood. Consider the gulate. vector space spanned by the rows of .JV. Extend the set of Else, gradients of At by adding other rows of .lV until a maxi- Apply LNTest(al, (A, E~ )). If this succeeds and returns mal independent set is obtained. Partition E3 into El, Ez, NJ (al) C N,O (so), then return SUCCESS and the definition where El is the subset of exponential terms corresponding Ofcl: to rows of Jv which are in the maximal independent set, and (A, I?j) = O in N.2 (aI) with ez = minimum(el, 6) E2 are the others. Each term in Ez has a gradient which Else, is orthogonal to the tangent space of At = O, El = O in the if the number of equations in (A, E~) = O is 2n, let J(z) neighborhood. Of course At = O, El = O, EZ = O at cr. be the Jacobian of (A, El ), and let Jv be obtained from J(x) The number of equations in At = O, El = O is less than the by replacing each e’i by y~. If JU is not provably non zero in number of variables. It follows that there must be a curve N,l (al) and if the Wu Pseudoremainder of Jy with respect through a on which .4i = O,El = O. On this curve we would to ascending set A is non zero, add Ju to U and return the also have (S,, E2 ) and U identically zero. But this can not result of recessively calling triangulate happen, since we supposed that the solution a was unique. Else return FAIL. ❑

Theorem 2 Suppose that there is a unique solution a of 2.6 Resolving Matrix (( S,, E,) = O,U = O) in N, O(aO). There exists a number < so that if c1 < ( and a is in N,l (m) C N,, (~o), Let a be an elementary point in C2’, defined by (S., Ek) = then triangulate succeeds ijprn is large enough, unless Ej = 0. Suppose the exponential Ek are WI – e:’, . . . . w~ – e’k, {wl–e”,. . . . w~–e’~}, andzl,. . . . .zj are linearly dependent with x variables Z1, . . . . .Zk. Define a Resolving Matrix for over Z at Q, or {z1, . . . ,Zj, f3z1, . .,e’i} is a counterexample this situation to be a d by k matrix of integers, of rank d, to Schanuel’s conjecture at CY. with d maximal, so that RX = O, where X is the column vector zl, . . ..z~. Proof Ascending sets are well ordered by the Wu ordering. The idea is that a resolving matrix is supposed to show Suppose Wu’s method finds characteristic set At for {S,, U}, all the integral linear dependencies among the x variables and that the proposition is true for triangulate applied to which occur in exponent ials. arguments which give ascending sets below At in the well We can find a candidate resolving matrix with the fol- ordering. lowing technique. For prn sufficiently large, the subset B which is not provably non zero is the same as the subset which is zero at a. Definition 6 FindResolvingMatrix procedure. If B is non empty, the recursive calls to triangulate pro- Given current approximation N6 (IY* ), and exponential duce ascending sets below A, in the well ordering. We can system (S,, Ek). inductively suppose that this works. Suppose the x variables which occur in Ek are, in or- Assume that B is empty. So all the initials and partial der, .z1,. . . . Zk, and the current approximations to these are derivatives of At are provably non zero. We have * * ‘1,...,~k. S,= O>Ej=O, U=O Set Threshold = 10 f100’(prn12k). Set M = 10prn. Let I be the k by k identity matrix. u If z is complex, let Re(z) be the real part of z and let Af=O, Ej=O Irn(z) be the imaginary part of z. If z is real, let [z] be the nearest integer to x, or the integer part of x if x is just in in N,l (al). between two integers. If t+ j > 2n the transcendence rank at a is less than j. Let v~,k+z = [~, c,, ~~], where So either Z1, . , zj are linearly dependent over the integers, C, is the Transpose of or we have a counterexample to Schanual’s conjecture.

106 ~$:(Af*z~)], [Re(AI*z~)],. . . . [Re(fil*z~)]) Proof For the first part, see Mignotte [1991 ], proposition 7.1 corollary 2. C, is the Transpose of For the second part, suppose R is a resolving matrix. Let ([Im(Af*z~)], [Im(M*z;)], . . . . [~?7t(A!f*Z;)]) M = 10p”n. For vector z = (cl, . . ., c~), define ht(x) to be Apply LLL to the lattice generated by the rows of V~,~+2. the maximum of lc~]. Let H be the maximum of ht (v) where Get back reduced basis w~,k+z. Suppose that d of the rows v is a row of R. in W have all entries below Threshold in absolute value. RC~ and RCT have components bounded by kH. The If d = O return FAIL. rows of [R, RCi, RC,] are in the lattice generated by the Otherwise, if the Tows below the threshold are rl, . . . . rd, rows of vk ,k+z, are linearly independent, and have Iengt h let wd,k~~ be the matrix with these rows, and obtain R by bounded by 3kH. removing the last two columns of Wd, h+z; and return R, Let bl, . . . . bd be the first d rows in a reduced basis. By which is d by k. part 1), the lengths of these are bounded, independent of pm, by 3kH2k We can use the triangulate procedure to verify that a However, if v is any vector which is not orthogonal to candidate resolving mat rix R is correct. (2’1,..., ~k) at a, then (v, wo C~, v o C,) grows to unbounded This works as follows. The z vaxiables which occur in length as prn tends to infinity. Eh arezl, ..., ~k, and the corresponding g variables are There are only finitely many integral vectors v with M(v) WI, ..., wk. Let X be the column of the z variables ZI, . . . . zk, bounded by 3kH2k. For prn sufficiently large, all of these and let Y be the corresponding column of y variables. If R are excluded from the reduced basis, except those which are really is a resolving matrix, we will have RX = O. Corre- orthogonal to (.zl, . . . . XL.) at a. ❑ sponding to the set of d linear relations among the x variables, is a set of d binomial equations among the y variables. Write this set of y equations as YR – 1 = O. The two equa- 2.7 The elementary zero recognize tions RX = O and YR – 1 = O define d of the exponential Given definition of elementary point a as solution of (S,, Ek ) == in terms of the others. Remove d defined exponential terms O, in N, (a”), with r + k = 2n, the number of variables. Also from Ek to get Ek_d. Now apply Triangulate procedure to given an equation to test q(a) = O. (S, = O,R-Y = O, Y~ – 1 = O,E~-d = O). If it fails, return 1) Set pm = 4k + 20 FAIL. 2) Apply LNTest(a*, (S,, Ek)). If it fails, stop and com- Otherwise we have a solution, say ~ to (S, = O,RX = plain about the definition. Else, we get Na (a”). Let co = 6, O,}-R – 1 = O,Ek...d = O) in N,l (al). We want to be sure and CEO= a“. Set prn = prn + 5 + ceiling(lloglo( d)l) that this solution is the same as a which was defined as 3) Iterate lazy Newton procedure until we get approxi- the unique solution to (S7, Ek ) = O in the same neighborh- mation N,l (al) with el < 10–p’n. If ~P,~ al @ N, O(aO), ood. It is sufficient to show that E~ (f?) = O. We know writ e “Warning: original neighbourhood in correct.” Else that Ek..-d(~) = O. E~–d is obtained by removing d ex- 4) If +P.n q(al) # O, stop with g(a) # O. Else, ponential terms, say & from Ek. We have to check that 5) Triangulate (N,, (ao), N,l (al), (S,, ~k), {q}). If SUC- Ed(~) = O. Suppose, for simplicity of notation, that Ed is CESS, stop and announce q(a) = O. Else, {wl–ez’,..., Wd – ezd }. By solving the system RX = O, 6) FindResolvingMatrix R. If this fails, double pm, and we get ZI, . . . . Zd expressed in terms of the other variables. go back to step 3). Else, We have a candidate resolving So we have matrix with d rows, corresponding to an LLL reduced basis. UzZi = b~,d+lZ’~+1 + . . . + b,,k~k 7) While (d > O), try to verify R as a resolving matrix. If this succeeds, bi, d+l . . . Wbi, k w:< = wd+l we have a new definition of a with less exponential terms, k (.4, ~&d) = Oin N$(al). Go back to step 3) using this new fori=l,..., d. Since wj=e’~at ~forj=d+l,..., k, definition. Otherwise, delete last row of R, and set d to d – 1 we can be sure that w, = eziy~, for i = 1, . . . . d at ~, where 8) double pm and go back to step 3). the -y, me ai roots of unity. We hope that all these ~~ are actually 1. Any of these which are not actually 1 will be called ghosts. 3 Implementation in Reduce Note that a is in the neighborhood and at this point we do have w~ = ez’, for all i. We know the imaginary part of An implementation of the above has been written in Reduce at Bath, as a group MSC project, by Michael Bahr, Mark zi at a to within 10–P’”, and so we know the argument of wi at a to the same precision. We need to check therefore that Hindess and Andrew Smith, using an LLL procedure written the neighbourhood N,l (al) determines the argument of w~ by John Abbott, and a Wu procedure written by Russell with error bound below ~/ai, in order to exclude ghosts. Bradford. The LLL and the Wu procedure are in symbolic This almost always is the case. Of course if it does not mode, and all the other procedures are in algebraic mode. happen, the precision is too small and the verification fails. The times that are given below are therefore quite hard to Otherwise, return SUCCESS and the new definition of a interpret. It seems that most of the computation time is produced by triangulate. spent in algebraic mode writing unnecessary lists.

Theorem 3 1)If bl, . . . . b~ is a reduced basis of a lattice , 4 Some examples and al,..., ad are linearly independent vectors in the lattice, then 1) Define a in C2 by lbj\2 ~ 2n-’maZ{lall’,la212,...,lad12} for all indices j = 1,2,... ,d X2 — 2=0 2) If there is a resolving matrix of rank d, than for prn sufficiently large, the first d vectors in a reduced basis are (Z-l) y-l=o the rows of a resolving marriz.

107 in No.2(Qo), with QO= (7/5, 5/2). Decide if g – z – 1 = O. The LLL algorithm is called, producing two candidate In this case the Lazy Newton test gives 6 = 0.2, and then linear relations: (just) fails. However, the Newton sequence does converge, and after one iteration, the Lazy Newton test succeeds at 16x4 – 4ZG – X2 = O the new approximation. Triangulate, given X2 – 2, (z – l)y – 1, ~ – z – 1, finds characteristic set 16x5 – 4x7 – x3 = O If this is RX = O, the corresponding YR – 1 = O equations –2, y–z–l X2 are and once again the Lazy Newton test succeeds, verifying this Yi6 = Y:Y2 equality. This takes about 2 seconds. Y:6 = Y;Y3 2) Define a in C6 by With these four equations we can eliminate two of the exponential terms. We eliminate yT = ez7, yG = e’G. A call to triangulate is now produces a set of 14 equations in 14 un- x3 — X1X2 = o knowns, with a solution in our current neighbourhood which can be verified by the Lazy Newton test. y;–l=o This takes about 180 seconds. Almost all of the time yl–e’l =() is spent doing interval arithmetic, and Newton iteration. Neither the Wu procedure nor the LLL algorithm takes more y2–e’2=o than a few seconds. y3–e’3=(j 5 Remarks in No.1 (ao), with It is not very difficult to implement the elementary zero CYo= (fCI, yI, X2, y2, X3, y3) = (i, ei, s.lA, e3”14,i3.147 ‘l). recognize, provided you have an LLL and a Wu implementation. Decide if g3 + 1 = O. In this case the Lazy Newton test The computational difficulty of the algorithm depends again fails initially although the Newton sequence and the heavily on how the elementary point is given. In particu- lazy Newton sequence both converge. After one iteration, lar, we assume here that the initial approximation is good the lazy Newton test succeeds at the new approximation. The characteristic set produced by triangulate essentially enough so that our crude test guarantees that the lazy New- ton sequence converges. There are, of course, other crite- replaces y; – 1 by ya + 1, the lazy Newton test succeeds, and the equality is verified. ria which imply Newt on convergence. For example, see the This takes about 23 seconds, discussion of the Kantoravich theorem in Rabinovitz [1970]. There are also alternative ways, using the notion of topolog- 3) Define a in C14 by ical degree, to verify the existence of solut ions of exponential systems. See Lloyd [1978], Cronin [1964]. This needs more Z;+l=o research. Z3 – Z1Z2 = o It is not expected that the triangulate procedure will often call itself recursively. g3+l=o It is expected that elementary zero recognition is about Z5 —$1X4 = o as hard, in most cases, as computation of a Wu characteristic set. The zero recognize works quickly when the initial 5(y; – 1) = Zl(yg + 1) definition of the point a is approximately triangular. The elementary zero recognize, given a definition of a X7 — Xlxfj = O point a, and an equality q(a) = O to test, either quickly 239(Y: – 1) = Zl(y; + 1) decides the truth or falsity of the equality, or it eventually not only decides but also produces an improved definition of cr. The improved definition either has an algebraic part fori= l,... ,7in No.o1(Qo), with aO = (zI, Y1,. . . ,z7, y7) = which generates a larger ideal than the algebraic part of the (i,, e’,3.14, e3’4, i3.14, –1, 0.1973955, eO’lg73955,20.1973955, original definition, or it uses less exponential terms than the e’0”1973g55,0.004184,1, iO.004184, 1). original definition. Decide if 16x4 – 4%Ij– X2 = O. We initially set p?m to 48. The problem of how to set the threshold used to find In this case, several Newton iterations are needed be- candidate resolving matrices deserves more than the ad hoc fore the lazy Newton test succeeds. Triangulate produces solution given above. I think this is related to deep questions a characteristic set with more than seven polynomials, so in transcendental number theory. the equality is not verified immediately. This is because It should be clear that there are a number of ways in there are seven exponential equations, and the Lazy New- which the elementary zero recognize may fail in practice. ton test automatically fails if given more equations than For example, if z = –101000, ez is so small that we will never unknowns. The fact that there are more than seven inde- be able to compute with enough precision to distinguish it pendent polynomial equations here means that the system from zero. In general, it seems to be difficult to compute could only be satisfied if there is some linear relationship with elementary points whose definitions involve applying among the x variables, or if this example contains a coun- the exponential function to numbers far away from zero. terexample to Schanuel’s conjecture. Our approximation to For some potential applications of the elementary zero the solution point implies that we do not need to consider recognition technique described above, see Shackell [1989, the branches associated with conditions such as ys(2z1 – 10) 1993], Salvy [1991], Richardson [1991, 1992 ]. For a slightly or yT(2xl —478) since these are provably non zero. different approach to proving decidability of the elementary

108 constant problem see Macintyre and Wilkie. For a different J. Shackell, Zero–Equivalence in function fields defined approach to the problem of finding a resolving matrix, see by algebraic differential equations, Trans . oft he AMS, vol Ferguson [1982], or Hastad [1986]. 336, Number 1, pp 151-171, 1993 J. Shackell, A Differential equations approach to func- tional equivalence, in G. Gonnet, editor, ISSAC ’89 Pro- 6 References ceedings, pp 7–10, ACM press, Portalnd Oregon, 1989 N. N. Vorobjov, Deciding consistency of systems of poly- B. F. Caviness, On canonical forms and simplification, nomial in exponent inequalities in subexponential time, Pro- Journal of the ACM, vol 17, no 2, April 1970, pp 385-396 ceedings of MEGA 90, 1990. D. Richardson, Some undecidable problems involving el- S.C. Chou, W. F. Schelter, and J. G. Yangj Characteris- ementary functions of a real variable, Journal of Symbolic tic Sets and Grobner Bases in Geometry Theorem Proving, Logic, 1968, pp 514-520 Draft, Institute for Computing Science, The University of J. Ax, Schanuel’s Conjecture, Ann Math 93 (1971), 252- Texas, Austin, TX 78712 68 W. T. Wu, Basic Principles of Mechanical Theorem Prov- A. Baker, Transcendental Number Theory, Cambridge ing in Elementary Geometries, J. Sys. Sci. and Math. Scis, University Press, 1975 f(3), 1984, 207-235 B. F. Caviness and M. J. Prelle, A note on algebraic inde- pendence of logarithmic and exponential constants, SIGSAM Bulletin, vol 12, no 2, pp 18–20, 1978 Cronin, J., Fixed Points and Topological Degree in Non- linear Analysis, American Mathematical Society, 1964 Ferguson, H. R, P., and R. W. Forcade, Multidimensional Euclidean Algorithms, J. Reine Ange. Math. 33, (1982), pp 171-181 N. G. Lloyd, Degree Theory, Cambridge Universit y Press, 1978 J. Hastad, B. Just, J.C. Lagarias, C. P. Schnorr, Poly- nomial time algorithms for finding integer relations among real numbers, Proceedings of STACS ’86, Lecture Notes in Computer Science A. G. Khovanskii, Fewnomials, AMS Translation Math- ematical monographs; v 88, AMS, Providence, RI, 1991. (Russian Original: Malochleny, Moscow 1987). A.K. Lenstra, H. W. Lenstra, L. Lovasz, Factoring Poly- nomials with rational coefficients. Math Ann 261, pp 513- 534 T. Leinekugel, Etre ou ne pas etre zero, report on work done at Bath University in summer 1994 A.J. Macintyre and A.J. Wilkie, On the decidability of the real exponential field, preprint, Mathematical Institute, Oxford. M. Mignotte, Mathematics for Computer Algebra, Springer Verlag, 1991 M. Pohst, A Modification of the LLL Reduction Algo- rithm, J. Symbolic Computation (1987) 4, pp 123-127 P. Rabinowitz (Ed), Numerical Methods for Nonlinear Algebraic Equations, Gordon and Breach, 1970 D. Richardson, Finding roots of equations involving solutions of first order algebraic differential equations, pp 427- 440 in Effective Methods in Algebraic Geometry, (Teo Mora and Carlo Traverso Eds), Birkhauser 1991 D. Richardson, Computing the topology of a bounded non-algebraic curve in the plane, Journal of Symbolic Com- putation, 14, 1992, pp 619–643 D. Richardson, The Elementary Constant Problem, IS- SAC 92. D. Richardson, A zero structure theorem for exponential systems, ISSAC ’93 D. Richardson and J. P. Fitch, Simplification of Elemen- tary Constants and Functions, ISSAC ’94 M. Rosenlicht, On Liouville’s Theory of Elementary Func- tions, Pacific Journal of Mathematics, VOI65, no 2, 1976, pp 485–492 B. Salvy, Asymptotique Automatique et Fonctions G6n& ratrices, Thesis, Ecole Polytechnique, 1991

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